Partial Differential Equations: Graduate Level Problems and Solutions. Igor Yanovsky

Transcription

1 Partial Differential Equations: Graduate Level Problems and Solutions Igor Yanovsky

2 Partial Differential Equations Igor Yanovsky, 5 Disclaimer: This handbook is intended to assist graduate students with qualifying examination preparation. Please be aware, however, that the handbook might contain, and almost certainly contains, typos as well as incorrect or inaccurate solutions. I can not be made responsible for any inaccuracies contained in this handbook.

3 Partial Differential Equations Igor Yanovsky, 5 3 Contents Trigonometric Identities 6 Simple Eigenvalue Problem 8 3 Separation of Variables: Quick Guide 9 4 Eigenvalues of the Laplacian: Quick Guide 9 5 First-Order Equations 5. Quasilinear Equations Weak Solutions for Quasilinear Equations Conservation Laws and Jump Conditions FansandRarefactionWaves GeneralNonlinearEquations TwoSpatialDimensions ThreeSpatialDimensions Second-Order Equations 4 6. ClassificationbyCharacteristics CanonicalFormsandGeneralSolutions Well-Posedness Wave Equation 3 7. TheInitialValueProblem WeakSolutions Initial/Boundary Value Problem Duhamel sprinciple TheNonhomogeneousEquation HigherDimensions Spherical Means ApplicationtotheCauchyProblem Three-DimensionalWaveEquation Two-DimensionalWaveEquation Huygen sprinciple EnergyMethods ContractionMappingPrinciple Laplace Equation 3 8. Green sformulas PolarCoordinates Polar Laplacian in R forradialfunctions Spherical Laplacian in R 3 and R n forradialfunctions Cylindrical Laplacian in R 3 forradialfunctions MeanValueTheorem MaximumPrinciple The Fundamental Solution RepresentationTheorem Green sfunctionandthepoissonkernel... 4

4 Partial Differential Equations Igor Yanovsky, PropertiesofHarmonicFunctions EigenvaluesoftheLaplacian Heat Equation ThePureInitialValueProblem FourierTransform Multi-IndexNotation Solution of the Pure Initial Value Problem NonhomogeneousEquation Nonhomogeneous Equation with Nonhomogeneous Initial Conditions The Fundamental Solution Schrödinger Equation 5 Problems: Quasilinear Equations 54 Problems: Shocks 75 3 Problems: General Nonlinear Equations 86 3.TwoSpatialDimensions ThreeSpatialDimensions Problems: First-Order Systems 5 Problems: Gas Dynamics Systems 7 5.Perturbation StationarySolutions PeriodicSolutions EnergyEstimates Problems: Wave Equation 39 6.TheInitialValueProblem Initial/Boundary Value Problem SimilaritySolutions TravelingWaveSolutions Dispersion EnergyMethods WaveEquationinDand3D Problems: Laplace Equation 96 7.Green sfunctionandthepoissonkernel The Fundamental Solution RadialVariables WeakSolutions Uniqueness Self-AdjointOperators Spherical Means Harmonic Extensions, Subharmonic Functions

5 Partial Differential Equations Igor Yanovsky, Problems: Heat Equation 55 8.HeatEquationwithLowerOrderTerms HeatEquationEnergyEstimates Contraction Mapping and Uniqueness - Wave 7 Contraction Mapping and Uniqueness - Heat 73 Problems: Maximum Principle - Laplace and Heat 79.HeatEquation-MaximumPrincipleandUniqueness LaplaceEquation-MaximumPrinciple... 8 Problems: Separation of Variables - Laplace Equation 8 3 Problems: Separation of Variables - Poisson Equation 3 4 Problems: Separation of Variables - Wave Equation 35 5 Problems: Separation of Variables - Heat Equation 39 6 Problems: Eigenvalues of the Laplacian - Laplace 33 7 Problems: Eigenvalues of the Laplacian - Poisson Problems: Eigenvalues of the Laplacian - Wave Problems: Eigenvalues of the Laplacian - Heat Heat Equation with Periodic Boundary Conditions in D (withextraterms Problems: Fourier Transform Laplace Transform Linear Functional Analysis Norms BanachandHilbertSpaces Cauchy-SchwarzInequality HölderInequality MinkowskiInequality SobolevSpaces

6 Partial Differential Equations Igor Yanovsky, 5 6 Trigonometric Identities cos(a + b = cosa cos b sin a sin b cos(a b = cosa cos b +sina sin b sin(a + b = sina cos b +cosa sin b sin(a b = sina cos b cos a sin b cos a cos b = sin a cos b = sin a sin b = cos(a + b+cos(a b sin(a + b+sin(a b cos(a b cos(a + b cos t = cos t sin t sin t = sint cos t cos t = +cost sin t = cos t +tan t = sec t cot t + = csc t cos x = eix + e ix sin x = eix e ix i cosh x = ex + e x sinh x = ex e x d cosh x dx = sinh(x d sinh x dx = cosh(x cosh x sinh x = du a + u = u a tan a + C du = sin u a u a + C L L L L L L L L cos nπx L sin nπx L sin nπx L cos nπx L sin nπx L L mπx cos L dx = mπx sin L dx = mπx cos L dx = mπx cos L dx = mπx sin L dx = e inx e imx dx = L e inx dx = sin xdx= x cos xdx= x { n m L n = m { n m L n = m { n m L n = m { n m L n = m { n m L n = m { n L n = sin x cos x sin x cos x + tan xdx=tanx x sin x cos xdx= cos x ln(xy=ln(x+ln(y ln x =ln(x ln(y y ln x r = r lnx R R ln xdx = x ln x x x ln xdx = x x ln x 4 e z dz = π e z dz = π

7 Partial Differential Equations Igor Yanovsky, 5 7 ( a b A = c d, A = det(a ( d b c a

8 Partial Differential Equations Igor Yanovsky, 5 8 Simple Eigenvalue Problem X + λx = Boundary conditions Eigenvalues λ n Eigenfunctions X n ( X( = X(L = nπ L sin nπ L x n =,,... [ X( = X (n ] (L = π L sin (n π L x n =,,... [ X (n ] ( = X(L= π L cos (n π L x n =,,... X ( = X ( (L = nπ L cos nπ L x n =,,,... X( = X(L, X ( = X ( (L nπ L sin nπ L x n =,,... cos nπ L x n =,,,... ( X( L=X(L, X ( L =X (L nπ L sin nπ L x n =,,... x n =,,,... cos nπ L X λx = Boundary conditions Eigenvalues λ n Eigenfunctions X n X( = X(L=,X ( = X ( (L = nπl 4 sin nπ L x n =,,... X ( = X (L =,X ( = X ( (L = nπl 4 cos nπ L x n =,,,...

9 Partial Differential Equations Igor Yanovsky, Separation of Variables: Quick Guide Laplace Equation: u =. X (x X(x = Y (y Y (y X + λx =. X (t X(t = Y (θ Y (θ Y (θ+λy (θ =. = λ. = λ. Wave Equation: u tt u xx =. X (x X(x = T (t T (t X + λx =. = λ. u tt +3u t + u = u xx. T T +3T X + = T X X + λx =. u tt u xx + u =. T X + = T X X + λx =. = λ. = λ. u tt + μu t = c u xx + βu xxt, (β > X X = λ, T c T + μ T ( c T = + β T X c T X. 4th Order: u tt = ku xxxx. X X = T k T = λ. X λx =. 4 Eigenvalues of the Laplacian: Quick Guide Laplace Equation: u xx + u yy + λu =. X X + Y Y + λ =. (λ = μ + ν X + μ X =, Y + ν Y =. u xx + u yy + k u =. X X = Y Y + k = c. X + c X =, Y +(k c Y =. u xx + u yy + k u =. Y = X Y X + k = c. Y + c Y =, X +(k c X =. Heat Equation: u t = ku xx. T T = k X X = λ. X + λ k X =. 4th Order: u t = u xxxx. T T X = λ. X λx =. = X

10 Partial Differential Equations Igor Yanovsky, 5 5 First-Order Equations 5. Quasilinear Equations Consider the Cauchy problem for the quasilinear equation in two variables a(x, y, uu x + b(x, y, uu y = c(x, y, u, with Γ parameterized by (f(s,g(s,h(s. The characteristic equations are dx dy = a(x, y, z, dt with initial conditions dt = b(x, y, z, dz dt = c(x, y, z, x(s, = f(s, y(s, = g(s, z(s, = h(s. In a quasilinear case, the characteristic equations for dx dt need not decouple from the dz equation; this means that we must take the z values into account even to find dt and dy dt the projected characteristic curves in the xy-plane. In particular, this allows for the possibility that the projected characteristics may cross each other. The condition for solving for s and t in terms of x and y requires that the Jacobian matrix be nonsingular: ( xs y J s = x s y t y s x t. x t y t In particular, at t = we obtain the condition f (s b(f(s,g(s,h(s g (s a(f(s,g(s,h(s. Burger s Equation. Solve the Cauchy problem { u t + uu x =, u(x, = h(x. (5. The characteristic equations are dx dt = z, dy dt =, dz dt =, and Γ may be parametrized by (s,,h(s. x = h(st + s, y = t, z = h(s. u(x, y=h(x uy (5. The characteristic projection in the xt-plane passing through the point (s, is the line x = h(st + s along which u has the constant value u = h(s. Two characteristics x = h(s t + s and x = h(s t + s intersect at a point (x, t with t = s s h(s h(s. y and t are interchanged here

11 Partial Differential Equations Igor Yanovsky, 5 From (5., we have u x = h (s( u x t u x = h (s +h (st Hence for h (s <, u x becomes infinite at the positive time t = h (s. The smallest t for which this happens corresponds to the value s = s at which h (s has a minimum (i.e. h (s has a maximum. At time T = /h (s thesolutionu experiences a gradient catastrophe.

12 Partial Differential Equations Igor Yanovsky, 5 5. Weak Solutions for Quasilinear Equations 5.. Conservation Laws and Jump Conditions Consider shocks for an equation u t + f(u x =, (5.3 where f is a smooth function of u. If we integrate (5.3 with respect to x for a x b, we obtain d dt b a u(x, t dx + f(u(b, t f(u(a, t =. (5.4 This is an example of a conservation law. Notice that (5.4 implies (5.3 if u is C, but (5.4 makes sense for more general u. Consider a solution of (5.4 that, for fixed t, has a jump discontinuity at x = ξ(t. We assume that u, u x,andu t are continuous up to ξ. Also, we assume that ξ(t isc in t. Taking a<ξ(t <bin (5.4, we obtain d ( ξ udx+ dt a b ξ udx + f(u(b, t f(u(a, t = ξ (tu l (ξ(t,t ξ (tu r (ξ(t,t+ + f(u(b, t f(u(a, t =, ξ a u t (x, t dx + b ξ u t (x, t dx where u l and u r denote the limiting values of u from the left and right sides of the shock. Letting a ξ(t andb ξ(t, we get the Rankine-Hugoniot jump condition: ξ (t(u l u r +f(u r f(u l =, ξ (t = f(u r f(u l u r u l. 5.. Fans and Rarefaction Waves For Burgers equation u t + ( u x =, we have f (u =u, f ( ( x ũ = x t t ( x ũ = x t t. For a rarefaction fan emanating from (s, on xt-plane, we have: x s u l, t f (u l =u l, u(x, t= x s t, u l x s t u r, x s u r, t f (u r =u r.

13 Partial Differential Equations Igor Yanovsky, General Nonlinear Equations 5.3. Two Spatial Dimensions Write a general nonlinear equation F (x, y, u, u x,u y =as F (x, y, z, p, q=. Γ is parameterized by ( Γ: f(s,g(s,h(s,φ(s,ψ(s }{{} x(s, }{{} y(s, }{{} z(s, }{{} p(s, }{{} q(s, We need to complete Γ to a strip. Find φ(s andψ(s, the initial conditions for p(s, t and q(s, t, respectively: F (f(s,g(s,h(s,φ(s,ψ(s = h (s = φ(sf (s+ψ(sg (s The characteristic equations are dx dt = F p dz dt = pf p + qf q dp dt = F x F z p dy dt = F q dq dt = F y F z q We need to have the Jacobian condition. That is, in order to solve the Cauchy problem in a neighborhood of Γ, the following condition must be satisfied: f (s F q [f, g, h, φ, ψ](s g (s F p [f, g, h, φ, ψ](s Three Spatial Dimensions Write a general nonlinear equation F (x,x,x 3,u,u x,u x,u x3 =as F (x,x,x 3,z,p,p,p 3 =. Γ is parameterized by ( Γ: f (s,s,f } {{ } (s,s,f } {{ } 3 (s,s,h(s } {{ },s,φ } {{ } (s,s,φ } {{ } (s,s,φ } {{ } 3 (s,s } {{ } x (s,s, x (s,s, x 3 (s,s, z(s,s, p (s,s, p (s,s, p 3 (s,s, We need to complete Γ to a strip. Find φ (s,s, φ (s,s, and φ 3 (s,s, the initial conditions for p (s,s,t, p (s,s,t, and p 3 (s,s,t, respectively: F ( f (s,s,f (s,s,f 3 (s,s,h(s,s,φ,φ,φ 3 = h f f f 3 = φ + φ + φ 3 s s s s h f f f 3 = φ + φ + φ 3 s s s s The characteristic equations are dx dt = F p dx dt = F p dx 3 dt = F p 3 dz dt = p F p + p F p + p 3 F p3 dp dt = F x p F z dp dt = F x p F z dp 3 dt = F x 3 p 3 F z

14 Partial Differential Equations Igor Yanovsky, Second-Order Equations 6. Classification by Characteristics Consider the second-order equation in which the derivatives of second-order all occur linearly, with coefficients only depending on the independent variables: a(x, yu xx + b(x, yu xy + c(x, yu yy = d(x, y, u, u x,u y. (6. The characteristic equation is dy dx = b ± b 4ac. a b 4ac > two characteristics, and (6. is called hyperbolic; b 4ac = one characteristic, and (6. is called parabolic; b 4ac < no characteristics, and (6. is called elliptic. These definitions are all taken at a point x R ; unless a, b, andc are all constant, the type may change with the point x. 6. Canonical Forms and General Solutions ➀ u xx u yy = is hyperbolic (one-dimensional wave equation. ➁ u xx u y = is parabolic (one-dimensional heat equation. ➂ u xx + u yy = is elliptic (two-dimensional Laplace equation. By the introduction of new coordinates μ and η in place of x and y, the equation (6. may be transformed so that its principal part takes the form ➀, ➁, or➂. If (6. is hyperbolic, parabolic, or elliptic, there exists a change of variables μ(x, y and η(x, y under which (6. becomes, respectively, u μη = d(μ, η, u, u μ,u η u x x uȳȳ = d( x, ȳ, u, u x,uȳ, u μμ = d(μ, η, u, u μ,u η, u μμ + u ηη = d(μ, η, u, u μ,u η. Example. Reduce to canonical form and find the general solution: u xx +5u xy +6u yy =. (6. Proof. a =,b =5,c =6 b 4ac => hyperbolic two characteristics. The characteristics are found by solving dy dx = 5 ± = { 3 to find y =3x + c and y =x + c.

15 Partial Differential Equations Igor Yanovsky, 5 5 Let μ(x, y =3x y, η(x, y=x y. μ x =3, η x =, μ y =, η y =. u = u(μ(x, y,η(x, y; u x = u μ μ x + u η η x =3u μ +u η, u y = u μ μ y + u η η y = u μ u η, u xx = (3u μ +u η x =3(u μμ μ x + u μη η x +(u ημ μ x + u ηη η x =9u μμ +u μη +4u ηη, u xy = (3u μ +u η y =3(u μμ μ y + u μη η y +(u ημ μ y + u ηη η y = 3u μμ 5u μη u ηη, u yy = (u μ + u η y = (u μμ μ y + u μη η y + u ημ μ y + u ηη η y =u μμ +u μη + u ηη. Inserting these expressions into (6. and simplifying, we obtain u μη =, which is the Canonical form, u μ = f(μ, u = F (μ+g(η, u(x, y = F (3x y+g(x y, General solution. Example. Reduce to canonical form and find the general solution: y u xx yu xy + u yy = u x +6y. (6.3 Proof. a = y, b = y, c = b 4ac = parabolic one characteristic. The characteristics are found by solving dy dx = y y = y to find y + c = x. Let μ = y + x. We must choose a second constant function η(x, y sothatη is not parallel to μ. Chooseη(x, y=y. μ x =, η x =, μ y = y, η y =. u = u(μ(x, y,η(x, y; u x = u μ μ x + u η η x = u μ, u y = u μ μ y + u η η y = yu μ + u η, u xx = (u μ x = u μμ μ x + u μη η x = u μμ, u xy = (u μ y = u μμ μ y + u μη η y = yu μμ + u μη, u yy = (yu μ + u η y = u μ + y(u μμ μ y + u μη η y +(u ημ μ y + u ηη η y = u μ + y u μμ +yu μη + u ηη.

16 Partial Differential Equations Igor Yanovsky, 5 6 Inserting these expressions into (6.3 and simplifying, we obtain u ηη = 6y, u ηη = 6η, which is the Canonical form, u η = 3η + f(μ, u = η 3 + ηf(μ+g(μ, u(x, y = ( y y 3 ( y + y f + x + g + x, General solution.

17 Partial Differential Equations Igor Yanovsky, 5 7 Problem (F 3, #4. Find the characteristics of the partial differential equation xu xx +(x yu xy yu yy =, x >, y >, (6.4 and then show that it can be transformed into the canonical form (ξ +4ηu ξη + ξu η = whence ξ and η are suitably chosen canonical coordinates. Use this to obtain the general solution in the form η g(η dη u(ξ, η=f(ξ+ (ξ +4η where f and g are arbitrary functions of ξ and η. Proof. a = x, b = x y, c = y b 4ac =(x y +4xy > for x>, y> hyperbolic two characteristics. ➀ The characteristics are found by solving { x x = y x = y x dy dx = b ± b 4ac = x y ± (x y +4xy x y ± (x + y = = a x x y = x + c, dy y = dx x, ln y = ln x + c, ➁ Let μ = x y and η = xy y = c x. μ x =, η x = y, μ y =, η y = x. u = u(μ(x, y,η(x, y; u x = u μ μ x + u η η x = u μ + yu η, u y = u μ μ y + u η η y = u μ + xu η, u xx = (u μ + yu η x = u μμ μ x + u μη η x + y(u ημ μ x + u ηη η x =u μμ +yu μη + y u ηη, u xy = (u μ + yu η y = u μμ μ y + u μη η y + u η + y(u ημ μ y + u ηη η y = u μμ + xu μη + u η yu ημ + xyu ηη, u yy = ( u μ + xu η y = u μμ μ y u μη η y + x(u ημ μ y + u ηη η y =u μμ xu μη + x u ηη, Inserting these expressions into (6.4, we obtain x(u μμ +yu μη + y u ηη +(x y( u μμ + xu μη + u η yu ημ + xyu ηη y(u μμ xu μη + x u ηη =, (x +xy + y u μη +(x yu η =, ( (x y +4xy u μη +(x yu η =, (μ +4ηu μη + μu η =, which is the Canonical form.

18 Partial Differential Equations Igor Yanovsky, 5 8 ➂ We need to integrate twice to get the general solution: (μ +4η(u η μ + μu η =, (uη μ μ dμ = μ +4η dμ, u η ln u η = ln (μ +4η+ g(η, ln u η = ln (μ +4η + g(η, g(η u η =, (μ +4η g(η dη u(μ, η=f(μ+, General solution. (μ +4η

19 Partial Differential Equations Igor Yanovsky, Well-Posedness Problem (S 99, #. In R consider the unit square defined by x, y. Consider a u x + u yy =; b u xx + u yy =; c u xx u yy =. Prescribe data for each problem separately on the boundary of so that each of these problems is well-posed. Justify your answers. Proof. The initial / boundary value problem for the HEAT EQUATION is wellposed: u t = u x, t>, u(x, = g(x x, u(x, t = x, t>. Existence - by eigenfunction expansion. Uniqueness and continuous dependence on the data - by maximum principle. The method of eigenfunction expansion and maximum principle give well-posedness for more general problems: u t = u + f(x, t x, t>, u(x, = g(x x, u(x, t =h(x, t x, t>. It is also possible to replace the Dirichlet boundary condition u(x, t = h(x, t by a Neumann or Robin condition, provided we replace λ n, φ n by the eigenvalues and eigenfunctions for the appropriate boundary value problem. a Relabel the variables (x t, y x. We have the BACKWARDS HEAT EQUATION: u t + u xx =. Need to define initial conditions u(x, = g(x, and either Dirichlet, Neumann, or Robin boundary conditions. b The solution to the LAPLACE EQUATION { u = in, u = g on exists if g is continuous on, by Perron s method. Maximum principle gives uniqueness. To show the continuous dependence on the data, assume { { u = in, u = in, u = g on ; u = g on.

20 Partial Differential Equations Igor Yanovsky, 5 Then (u u = in. max(u u = max (g g. max u u = max g g. Maximum principle gives Thus, Thus, u u is bounded by g g, i.e. continuous dependence on data. Perron s method gives existence of the solution to the POISSON EQUATION { u = f in, u n = h on for f C ( and h C (, satisfying the compatibility condition hds = fdx.itisunique up to an additive constant. c Relabel the variables (y t. The solution to the WAVE EQUATION u tt u xx =, is of the form u(x, y =F (x + t+g(x t. The existence of the solution to the initial/boundary value problem u tt u xx = <x<, t> u(x, = g(x, u t (x, = h(x <x< u(,t=α(t, u(,t=β(t t. is given by the method of separation of variables (expansion in eigenfunctions and by the parallelogram rule. Uniqueness is given by the energy method. Need initial conditions u(x,, u t (x,. Prescribe u or u x for each of the two boundaries.

21 Partial Differential Equations Igor Yanovsky, 5 Problem (F 95, #7. Let a, b be real numbers. The PDE u y + au xx + bu yy = is to be solved in the box =[, ]. Find data, given on an appropriate part of, that will make this a well-posed problem. Cover all cases according to the possible values of a and b. Justify your statements. Proof. ➀ ab < two sets of characteristics hyperbolic. Relabeling the variables (y t, we have u tt + a b u xx = b u t. The solution of the equation is of the form u(x, t =F (x + a b t+g(x a b t. Existence of the solution to the initial/boundary value problem is given by the method of separation of variables (expansion in eigenfunctions and by the parallelogram rule. Uniqueness is given by the energy method. Need initial conditions u(x,, u t (x,. Prescribe u or u x for each of the two boundaries. ➁ ab > no characteristics elliptic. The solution to the Laplace equation with boundary conditions u = g on exists if g is continuous on, by Perron s method. To show uniqueness, we use maximum principle. Assume there are two solutions u and u with with u = g(x, u = g(x on. By maximum principle max(u u = max (g(x g(x =. Thus, u = u. ➂ ab = one set of characteristics parabolic. a = b =. Wehave u y =, afirst-orderode. u must be specified on y =, i.e. x -axis. a =,b. Wehaveu y + bu yy =, a second-order ODE. u and u y must be specified on y =, i.e. x -axis. a>, b =. u t = au xx. WehaveaBackwardsHeatEquation. Need to define initial conditions u(x, = g(x, and either Dirichlet, Neumann, or Robin boundary conditions.

22 Partial Differential Equations Igor Yanovsky, 5 a<, b =. u t = au xx. WehaveaHeatEquation. The initial / boundary value problem for the heat equation is well-posed: u t = u x, t>, u(x, = g(x x, u(x, t = x, t>. Existence - by eigenfunction expansion. Uniqueness and continuous dependence on the data - by maximum principle.

23 Partial Differential Equations Igor Yanovsky, Wave Equation The one-dimensional wave equation is u tt c u xx =. (7. The characteristic equation with a = c, b =,c =wouldbe dt dx = b ± b 4ac a and thus 4c = ± c = ± c, t = c x + c and t = c x + c, μ = x + ct η = x ct, which transforms (7. to u μη =. (7. The general solution of (7. is u(μ, η=f (μ+g(η, where F and G are C functions. Returningtothevariablesx, t we find that u(x, t=f (x + ct+g(x ct (7.3 solves (7.. Moreover, u is C provided that F and G are C. If F, then u has constant values along the lines x ct = const, so may be described as a wave moving in the positive x-direction with speed dx/dt = c; ifg, then u is a wave moving in the negative x-direction with speed c. 7. The Initial Value Problem For an initial value problem, consider the Cauchy problem { u tt c u xx =, u(x, = g(x, u t (x, = h(x. (7.4 Using (7.3 and (7.4, we find that F and G satisfy F (x+g(x =g(x, cf (x cg (x =h(x. (7.5 If we integrate the second equation in (7.5, we get cf(x cg(x = x h(ξ dξ + C. Combining this with the first equation in (7.5, we can solve for F and G to find { F (x = g(x+ x c h(ξ dξ + C G(x = g(x c x h(ξ dξ C, Using these expressions in (7.3, we obtain d Alembert s Formula for the solution of the initial value problem (7.4: u(x, t= (g(x + ct+g(x ct + c x+ct x ct h(ξ dξ. If g C and h C, then d Alembert s Formula defines a C solution of (7.4.

24 Partial Differential Equations Igor Yanovsky, Weak Solutions Equation (7.3 defines a weak solution of (7. when F and G are not C functions. Consider the parallelogram with sides that are segments of characteristics. Since u(x, t =F (x + ct+g(x ct, we have u(a+u(c = = F (k +G(k 3 +F (k +G(k 4 = u(b+u(d, which is the parallelogram rule. 7.3 Initial/Boundary Value Problem u tt c u xx = <x<l, t> u(x, = g(x, u t (x, = h(x <x<l (7.6 u(,t=α(t, u(l, t =β(t t. Use separation of variables to obtain an expansion in eigenfunctions. Find u(x, t in the form u(x, t= a (t + a n (tcos nπx L + b n(tsin nπx L. n= 7.4 Duhamel s Principle u tt c u xx = f(x, t U tt c U xx = u(x, = U(x,,s= u t (x, =. U t (x,,s=f(x, s a n + λ na n = f n (t ã n + λ nã n = a n ( = ã n (,s= a n( = ã n(,s=f n (s u(x, t= a n (t = t t U(x, t s, s ds. ã n (t s, s ds. 7.5 The Nonhomogeneous Equation Consider the nonhomogeneous wave equation with homogeneous initial conditions: { u tt c u xx = f(x, t, (7.7 u(x, =, u t (x, =. Duhamel s Principle provides the solution of (7.7: u(x, t = t ( x+c(t s f(ξ, s dξ ds. c x c(t s If f(x, t isc in x and C in t, then Duhamel s Principle provides a C solution of (7.7.

25 Partial Differential Equations Igor Yanovsky, 5 5 We can solve (7.7 with nonhomogeneous initial conditions, { u tt c u xx = f(x, t, u(x, = g(x, u t (x, = h(x, (7.8 by adding together d Alembert s formula and Duhamel s principle gives the solution: u(x, t = (g(x + ct+g(x ct + x+ct h(ξ dξ + t ( x+c(t s f(ξ, s dξ ds. c c x ct x c(t s

26 Partial Differential Equations Igor Yanovsky, Higher Dimensions 7.6. Spherical Means For a continuous function u(x onr n,itsspherical mean or average on a sphere of radius r and center x is M u (x, r= u(x + rξds ξ, ω n ξ = where ω n is the area of the unit sphere S n = {ξ R n : ξ =} and ds ξ is surface measure. Since u is continuous in x, M u (x, r is continuous in x and r, so M u (x, = u(x. Using the chain rule, we find r M u(x, r= ω n ξ = i= n u xi (x + rξ ξ i ds ξ = To compute the RHS, we apply the divergence theorem in = {ξ R n : ξ < }, which has boundary =S n and exterior unit normal n(ξ =ξ. The integrand is V n where V (ξ =r ξ u(x + rξ = x u(x + rξ. Computing the divergence of V, we obtain n div V (ξ = r u xi x i (x + rξ = r x u(x + rξ, so, = ω n ξ < = r ω n r n x = ω n r n x i= r x u(x + rξ dξ = ξ <r r r x u(x + rξ dξ ω n ξ < u(x + ξ dξ (spherical coordinates ρ n u(x + ρξ ds ξ dρ r ξ = (ξ = rξ = ω n r n ω n x ρ n M u (x, ρ dρ. If we multiply by r n, differentiate with respect to r, and then divide by r n, we obtain the Darboux equation: ρ n M u (x, ρ dρ = r n x ( r + n M u (x, r = x M u (x, r. r r Note that for a radial function u = u(r, we have M u = u, so the equation provides the Laplacian of u in spherical coordinates Application to the Cauchy Problem We want to solve the equation u tt = c u x R n,t>, (7.9 u(x, = g(x, u t (x, = h(x x R n. We use Poisson s method of spherical means to reduce this problem to a partial differential equation in the two variables r and t. r

27 Partial Differential Equations Igor Yanovsky, 5 7 Suppose that u(x, t solves (7.9. We can view t as a parameter and take the spherical mean to obtain M u (x, r, t, which satisfies t M u(x, r, t = u tt (x + rξ, tds ξ = c u(x + rξ, tds ξ = c M u (x, r, t. ω n ξ = ω n ξ = Invoking the Darboux equation, we obtain the Euler-Poisson-Darboux equation: ( t M u(x, r, t = c r + n r r M u (x, r, t. The initial conditions are obtained by taking the spherical means: M u M u (x, r, = M g (x, r, t (x, r, = M h(x, r. If we find M u (x, r, t, we can then recover u(x, t by: u(x, t = lim r M u (x, r, t Three-Dimensional Wave Equation When n = 3, we can write the Euler-Poisson-Darboux equation as ( t rm u (x, r, t = c ( r rm u (x, r, t. For each fixed x, consider V x (r, t=rm u (x, r, t as a solution of the one-dimensional wave equation in r, t>: t V x (r, t = c r V x (r, t, V x (r, = rm g (x, r G x (r, (IC Vt x (r, = rm h (x, r H x (r, (IC V x (,t = lim rm u (x, r, t= u(x, t =. r G x ( = H x ( =. (BC We may extend G x and H x as odd functions of r and use d Alembert s formula for V x (r, t: V x (r, t = ( G x (r + ct+g x (r ct + r+ct H x (ρ dρ. c Since G x and H x are odd functions, we have for r<ct: G x (r ct = G x (ct r and r+ct r ct r ct H x (ρ dρ = ct+r ct r H x (ρ dρ. After some more manipulations, we find that the solution of (7.9 is given by the Kirchhoff s formula: u(x, t = ( t g(x + ctξds ξ + t h(x + ctξds ξ. 4π t ξ = 4π ξ = If g C 3 (R 3 andh C (R 3, then Kirchhoff s formula defines a C -solution of (7.9. It is seen by expanding the equation below.

28 Partial Differential Equations Igor Yanovsky, Two-Dimensional Wave Equation This problem is solved by Hadamard s method of descent, namely, view (7.9 as a special case of a three-dimensional problem with initial conditions independent of x 3. We need to convert surface integrals in R 3 to domain integrals in R. u(x,x,t= ( g(x + ctξ,x + ctξ dξ dξ t + t ( h(x + ctξ,x + ctξ dξ dξ 4π t ξ +ξ < ξ ξ 4π ξ +ξ < ξ ξ If g C 3 (R andh C (R, then this equation defines a C -solution of ( Huygen s Principle Notice that u(x, t depends only on the Cauchy data g, h on the surface of the hypersphere {x + ctξ : ξ =} in R n, n =k +;inotherwordswehavesharp signals. If we use the method of descent to obtain the solution for n =k, the hypersurface integrals become domain integrals. This means that there are no sharp signals. The fact that sharp signals exist only for odd dimensions n 3isknownasHuygen s principle. 3 3 For x R n : ( f(x + tξds ξ = f(x + ydy t ξ = t n y t ( ( f(x + ydy = t n f(x + tξds ξ t y t ξ =

29 Partial Differential Equations Igor Yanovsky, Energy Methods Suppose u C (R n (, solves { u tt = c u x R n, t >, u(x, = g(x, u t (x, = h(x x R n, (7. where g and h have compact support. Define energy for a function u(x, t attimet by E(t= (u t + c u dx. R n If we differentiate this energy function, we obtain de = d [ ( n ] u dt dt t + c u ( n x i dx = ut u tt + c u xi u xi t dx R n i= R n i= [ n n = u t u tt dx + c u xi u t R ] R c u xi n i= n x i u t dx R n i= = u t (u tt c u dx =, R n or de = d [ ( n ] u dt dt t + c u ( n x i dx = ut u tt + c u xi u xi t dx R n i= R n i= ( = ut u tt + c u u t dx R n [ ] = u t u tt dx + c u u t R n R n n ds u t udx R n = u t (u tt c u dx =. R n Hence, E(t is constant, or E(t E(. In particular, if u and u are two solutions of (7., then w = u u has zero Cauchy data and hence E w ( =. By discussion above, E w (t, which implies w(x, t const. But w(x, = then implies w(x, t, so the solution is unique.

30 Partial Differential Equations Igor Yanovsky, Contraction Mapping Principle Suppose X is a complete metric space with distance function represented by d(,. A mapping T : X X is a strict contraction if there exists <α< such that d(tx,ty αd(x, y x, y X. An obvious example on X = R n is Tx = αx, which shrinks all of R n, leaving fixed. The Contraction Mapping Principle. If X is a complete metric space and T : X X is a strict contraction, then T has a unique fixed point. The process of replacing a differential equation by an integral equation occurs in time-evolution partial differential equations. The Contraction Mapping Principle is used to establish the local existence and uniqueness of solutions to various nonlinear equations.

31 Partial Differential Equations Igor Yanovsky, Laplace Equation Consider the Laplace equation u = in R n (8. and the Poisson equation u = f in R n. (8. Solutions of (8. are called harmonic functions in. Cauchy problems for (8. and (8. are not well posed. We use separation of variables for some special domains to find boundary conditions that are appropriate for (8., (8.. Dirichlet problem: u(x =g(x, x Neumann problem: u(x = h(x, n x Robin problem: u + αu = β, n x 8. Green s Formulas u vdx = v u n ds v udx (8.3 ( u v n u v ds = (v u u v dx n u n ds = udx (v = in (8.3 u dx = u u n ds u udx (u = v in (8.3 u x v x dxdy = vu x n ds vu xx dxdy n =(n,n R u xk vdx = uvn k ds uv xk dx n =(n,...,n n R n. u vdx = u v n ds v u n ds + u vdx. ( u v v u dx = ( v u n v u ( v ds + u n n u v ds. n

32 Partial Differential Equations Igor Yanovsky, Polar Coordinates Polar Coordinates. Let f : R n R be continuous. Then ( fdx = fds R n B r(x dr for each x R n. In particular d ( fdx = fds dr B r(x B r(x for each r>. u = u(x(r, θ,y(r, θ x(r, θ=r cos θ y(r, θ=r sin θ u r = u x x r + u y y r = u x cos θ + u y sin θ, u θ = u x x θ + u y y θ = u x r sin θ + u y r cos θ, u rr = (u x cos θ + u y sin θ r =(u xx x r + u xy y r cosθ +(u yx x r + u yy y r sinθ = u xx cos θ +u xy cos θ sin θ + u yy sin θ, u θθ = ( u x r sin θ + u y r cos θ θ = ( u xx x θ u xy y θ r sin θ u x r cos θ +(u yx x θ + u yy y θ r cos θ u y r sin θ = (u xx r sin θ u xy r cos θr sin θ u x r cos θ +( u yx r sin θ + u yy r cos θr cos θ u y r sin θ = r (u xx sin θ u xy cos θ sin θ + u yy cos θ r(u x cos θ + u y sin θ. u rr + u r θθ = u xx cos θ +u xy cos θ sin θ + u yy sin θ + u xx sin θ u xy cos θ sin θ + u yy cos θ r (u x cos θ + u y sin θ = u xx + u yy r u r. u xx + u yy = u rr + r u r + r u θθ. x + y = r + r r + r θ. 8.3 Polar Laplacian in R for Radial Functions u = r ( rur r = ( r + r u. r 8.4 Spherical Laplacian in R 3 and R n for Radial Functions In R 3 : 4 u = ( r + n u. r r u = r ( r u r r = ( ( ru r rr = r + u. r r 4 These formulas are taken from S. Farlow, p. 4.

33 Partial Differential Equations Igor Yanovsky, Cylindrical Laplacian in R 3 for Radial Functions u = r ( rur r = ( r + r u. r 8.6 Mean Value Theorem Gauss Mean Value Theorem. If u C ( is harmonic in, letξ and pick r> so that B r (ξ ={x : x ξ r}. Then u(ξ =M u (ξ, r u(ξ + rx ds x, ω n x = where ω n is the measure of the (n -dimensional sphere in R n. 8.7 Maximum Principle Maximum Principle. If u C ( satisfies u in, then either u is a constant, or u(ξ < sup u(x x for all ξ. Proof. We may assume A =sup x u(x,sobycontinuityofu we know that {x :u(x =A} is relatively closed in. But since u(ξ n u(ξ + rx dx, ω n x if u(ξ = A at an interior point ξ, then u(x = A for all x in a ball about ξ, so {x :u(x =A} is open. The connectedness of implies u(ξ <Aor u(ξ A for all ξ. The maximum principle shows that u C ( with u can attain an interior maximum only if u is constant. In particular, if is compact, and u C ( C( satisfies u in,wehavetheweak maximum principle: max u(x =max u(x. x x

34 Partial Differential Equations Igor Yanovsky, The Fundamental Solution A fundamental solution K(x for the Laplace operator is a distribution satisfying K(x =δ(x (8.4 where δ is the delta distribution supported at x =. In order to solve (8.4, we should first observe that is symmetric in the variables x,...,x n,andδ(x isalsoradially symmetric (i.e., its value only depends on r = x. Thus, we try to solve (8.4 with a radially symmetric function K(x. Since δ(x =forx, we see that (8.4 requires K to be harmonic for r >. For the radially symmetric function K, Laplace equation becomes (K = K(r: K r + n K =. (8.5 r r The general solution to (8.5 is { c + c log r if n = K(r= c + c r n (8.6 if n 3. After we determine c, we find the fundamental solution for the Laplace operator: { K(x = π log r if n = ( nω n r n if n 3. We can derive, (8.6 for any given n. For intance, when n =3,wehave: Let K + r K =. K = r w(r, K = r w r w, K = r w r w + r 3 w. Plugging these into, we obtain: r w =, or w =. Thus, w = c r + c, K = r w(r = c + c r. See the similar problem, F 99, #, where the fundamental solution for ( I is found in the process.

35 Partial Differential Equations Igor Yanovsky, 5 35 Find the Fundamental Solution of the Laplace Operator for n =3 We found that starting with the Laplacian in R 3 for a radially symmetric function K, K + r K =, and letting K = r w(r, we obtained the equation: w = c r + c, which implied: K = c + c r. We now find the constant c that ensures that for v C (R3, we have K( x v(x dx = v(. R 3 Suppose v(x for x R and let = B R (; for small ɛ>let ɛ = B ɛ (. K( x isharmonic( K( x =in ɛ. Consider Green s identity ( ɛ = B ɛ (: ( K( x vdx = K( x v ɛ n v K( x ( ds + K( x v n B } {{ } ɛ( n v K( x ds. n =, sincev for x R [ ] lim K( x vdx ɛ ɛ = K( x vdx. On B ɛ (, K( x =K(ɛ. Thus, 5 K( x v B ɛ( n ds = K(ɛ v(x K( x ds = B ɛ( n = B ɛ( B ɛ( B ɛ( v ds n c v(x ds ɛ c v( ds + ɛ ( Since K(r=c + c r is integrable at x =. c + c ɛ B ɛ( 4πɛ max v, as ɛ. c [v(x v(] ds ɛ v(x v( = c ɛ v( 4πɛ +4πc max x B ɛ( } {{ }, (v is continuous = 4πc v( v(. Thus, taking 4πc =, i.e. c = 4π, we obtain K( x vdx = lim K( x vdx = v(, ɛ ɛ that is K(r = 4πr is the fundamental solution of. 5 In R 3,for x = ɛ, K( x = K(ɛ = c + c ɛ. K( x n = K(ɛ r = c ɛ, n points toward on the sphere x = ɛ (since n points inwards. (i.e., n = x/ x.

36 Partial Differential Equations Igor Yanovsky, 5 36 Show that the Fundamental Solution of the Laplace Operator is given by. K(x = { π log r if n = ( nω n r n if n 3. (8.7 Proof. For v C (Rn, we want to show R n K( x v(x dx = v(. Suppose v(x for x R and let = B R (; for small ɛ>let ɛ = B ɛ (. K( x isharmonic( K( x =in ɛ. Consider Green s identity ( ɛ = B ɛ (: ( K( x vdx = K( x v ɛ n v K( x ( ds + K( x v n B } {{ } ɛ( n v K( x ds. n =, sincev for x R [ ] lim K( x vdx ɛ ɛ B ɛ( = K( x vdx. On B ɛ (, K( x =K(ɛ. Thus, 6 K( x v B ɛ( n ds = K(ɛ v(x K( x ds = n Thus, B ɛ( B ɛ( ( Since K(r isintegrableatx =. v ds K(ɛ ω n ɛ n max v, as ɛ. n v(x ds ω n ɛn = v( ds + B ɛ( ω n ɛn = ω n ɛ n v( ω nɛ n max = v(. K( x vdx = lim K( x vdx = v(. ɛ ɛ B ɛ( x B ɛ( [v(x v(] ds ω n ɛn v(x v( } {{ }, (v is continuous 6 Note that for x = ɛ, K( x = K(ɛ = K( x n = K(ɛ r n points toward on the sphere x = ɛ { log ɛ if n = π ( nω n ɛ n if n 3. { if n = πɛ = ω nɛ if n 3, n (i.e., n = x/ x. = ω nɛ n, (since n points inwards.

37 Partial Differential Equations Igor Yanovsky, Representation Theorem Representation Theorem, n = 3. Let be bounded domain in R 3 and let n be the unit exterior normal to. Let u C (. Then the value of u at any point x is given by the formula u(x = [ u(y 4π x y n u(y ] ds u(y dy. (8.8 n x y 4π x y Proof. Consider the Green s identity: ( w (u w w u dy = u n w u ds, n where w is the harmonic function w(y= x y, which is singular at x. In order to be able to apply Green s identity, we consider anewdomain ɛ : ɛ = B ɛ (x. Since u, w C ( ɛ, Green s identity can be applied. Since w is harmonic ( w = in ɛ and since ɛ = B ɛ (x, we have [ u(y ɛ x y dy = u(y n x y ] u(y ds (8.9 x y n [ + u(y B ɛ(x n x y ] u(y ds. (8. x y n We will show that formula (8.8 is obtained by letting ɛ. [ ] u(y lim ɛ x y dy u(y ( = x y dy. Since is integrable at x = y. x y ɛ The first integral on the right of (8. does not depend on ɛ. Hence, the limit as ɛ of the second integral on the right of (8. exists, and in order to obtain (8.8, need [ lim u(y ɛ B ɛ(x n x y ] u(y ds = 4πu(x. x y n [ u(y B ɛ(x n x y ] [ u(y ds = x y n B ɛ(x ɛ u(y ] u(y ds ɛ n [ = u(x ds + B ɛ(x ɛ B ɛ(x ɛ [u(y u(x] ] u(y ds ɛ n [ = 4πu(x + ɛ [u(y u(x] ] u(y ds. ɛ n B ɛ(x

38 Partial Differential Equations Igor Yanovsky, The last integral tends to as ɛ : [ B ɛ(x ɛ [u(y u(x] ] u(y ds ɛ n ɛ u(y u(x + u(y ds B ɛ(x ɛ B ɛ(x n 4π max u(y u(x +4πɛmax u(y. y B ɛ(x y } {{ } } {{ }, (u continuous in, ( u is finite 7 Note that for points y on B ɛ(x, x y = ɛ and n x y = ɛ.

39 Partial Differential Equations Igor Yanovsky, 5 39 Representation Theorem, n =. Let be bounded domain in R and let n be the unit exterior normal to. Let u C (. Then the value of u at any point x is given by the formula u(x = u(ylog x y dy + [ u(y ] log x y log x y u(y ds. (8. π π n n Proof. Consider the Green s identity: ( w (u w w u dy = u n w u ds, n where w is the harmonic function w(y=log x y, which is singular at x. In order to be able to apply Green s identity, we consider anewdomain ɛ : ɛ = B ɛ (x. Since u, w C ( ɛ, Green s identity can be applied. Since w is harmonic ( w = in ɛ and since ɛ = B ɛ (x, we have u(ylog x y dy (8. ɛ [ = u(y ] log x y log x y u(y ds n n [ + u(y ] log x y log x y u(y ds. B ɛ(x n n We will show that formula (8. is obtained by letting ɛ. [ ] ( lim u(ylog x y dy = u(ylog x y dy. since log x y is integrable at x = y. ɛ ɛ The first integral on the right of (8. does not depend on ɛ. Hence, the limit as ɛ of the second integral on the right of (8. exists, and in order to obtain (8., need [ lim u(y ] log x y log x y u(y ds = πu(x. ɛ n n B ɛ(x B ɛ(x [ u(y ] log x y log x y u(y ds = n n = B ɛ(x u(x ds + ɛ = πu(x + B ɛ(x B ɛ(x [ B ɛ(x [ u(y log ɛ u(y ɛ n [u(y u(x] log ɛ u(y ɛ n [ [u(y u(x] log ɛ u(y ɛ n ] ds ] ds ] ds.

40 Partial Differential Equations Igor Yanovsky, The last integral tends to as ɛ : [ ] [u(y u(x] log ɛ u(y ds B ɛ(x ɛ n u(y u(x +logɛ u(y ds ɛ B ɛ(x B ɛ(x n π max u(y u(x +πɛlog ɛ max u(y. y B ɛ(x y } {{ } } {{ }, (u continuous in, ( u is finite 8 Note that for points y on B ɛ(x, log x y =logɛ and n log x y = ɛ.

41 Partial Differential Equations Igor Yanovsky, 5 4 Representation Theorems, n > 3 can be obtained in the same way. We use the Green s identity with w(y= x y n, which is a harmonic function in R n with a singularity at x. The fundamental solution for the Laplace operator is (r = x : { K(x = π log r if n = ( nω n r n if n 3. Representation Theorem. If R n is bounded, u C (, andx, then [ K(x y u(x = K(x y u(y dy + u(y K(x y u(y ] ds. (8.3 n n Proof. Consider the Green s identity: ( w (u w w u dy = u n w u ds, n where w is the harmonic function w(y=k(x y, which is singular at y = x. In order to be able to apply Green s identity, we consider a new domain ɛ : ɛ = B ɛ (x. Since u, K(x y C ( ɛ, Green s identity can be applied. Since K(x y is harmonic ( K(x y =in ɛ and since ɛ = B ɛ (x, we have [ K(x y K(x y u(y dy = u(y K(x y u(y ] ds (8.4 ɛ n n [ K(x y + u(y K(x y u(y ] ds. (8.5 B ɛ(x n n We will show that formula (8.3 is obtained by letting ɛ. [ ] ( lim K(x y u(y dy = K(x y u(y dy. since K(x y is integrable at x = y. ɛ ɛ The first integral on the right of (8.5 does not depend on ɛ. Hence, the limit as ɛ of the second integral on the right of (8.5 exists, and in order to obtain (8.3, need [ K(x y lim u(y K(x y u(y ] ds = u(x. ɛ n n B ɛ(x

42 Partial Differential Equations Igor Yanovsky, 5 4 B ɛ(x [ K(x y u(y K(x y u(y ] ds = n n = B ɛ(x u(x K(ɛ n ds + = ω n ɛ n u(x ds B ɛ(x ω n ɛ n = ω n ɛ n u(xω nɛ n ω } {{ } n ɛ n u(x B ɛ(x B ɛ(x B ɛ(x B ɛ(x [ u(y K(ɛ n K(ɛ u(y n [ K(ɛ n [u(y u(x] ds [u(y u(x] K(ɛ u(y n [u(y u(x] ds B ɛ(x B ɛ(x K(ɛ u(y n K(ɛ u(y n 9 The last two integrals tend to as ɛ : ω n ɛ n [u(y u(x] ds K(ɛ u(y B ɛ(x B ɛ(x n ds ω n ɛ n max u(y u(x ωn ɛ n + K(ɛ max u(y ωn ɛ n. y B ɛ(x y } {{ } } {{ }, (u continuous in, ( u is finite ] ds ] ds ds ds. 8. Green s Function and the Poisson Kernel With a slight change in notation, the Representation Theorem has the following special case. Theorem. If R n is bounded, u C ( C ( is harmonic, andξ, then [ K(x ξ u(ξ = u(x K(x ξ u(x ] ds. (8.6 n n Let ω(x beanyharmonic function in, and for x, ξ consider G(x, ξ=k(x ξ+ω(x. If we use the Green s identity (with u =and ω =, we get: ( ω = u n ω u ds. (8.7 n Adding (8.6 and (8.7, we obtain: [ G(x, ξ u(ξ = u(x G(x, ξ u(x ] ds. (8.8 n n Suppose that for each ξ we can find a function ω ξ (x that is harmonic in and satisfies ω ξ (x = K(x ξ for all x. Then G(x, ξ =K(x ξ +ω ξ (x isa fundamental solution such that G(x, ξ= x. 9 Note that for points y on B ɛ(x, K(x y = K(ɛ = { log ɛ π if n = ( nω n ɛ n if n 3. K(x y n = K(ɛ r = { πɛ if n = ω nɛ n if n 3, = ω nɛ n, (since n points inwards.

43 Partial Differential Equations Igor Yanovsky, 5 43 G is called the Green s function and is useful in satisfying Dirichlet boundary conditions. The Green s function is difficult to construct for a general domain since it requires solving the Dirichlet problem ω ξ =in, ω ξ (x = K(x ξ forx, for each ξ. From (8.8 we find u(ξ = u(x G(x, ξ ds. n Thus if we know that the Dirichlet problem has a solution u C (, then we can calculate u from the Poisson integral formula (provided of course that we can compute G(x, ξ. If we did not assume u = in our derivation, we would have (8.3 instead of (8.6, and an extra term in (8.7, which would give us a more general expression: u(ξ = G(x, ξ udx + G(x, ξ u(x ds. n

44 Partial Differential Equations Igor Yanovsky, Properties of Harmonic Functions Liouville s Theorem. A bounded harmonic function defined on all of R n must be a constant. 8. Eigenvalues of the Laplacian Consider the equation { u + λu = u = in on, (8.9 where is a bounded domain and λ is a (complex number. The values of λ for which (8.9 admits a nontrivial solution u are called the eigenvalues of in, and the solution u is an eigenfunction associated to the eigenvalue λ. (The convention u + λu = is chosen so that all eigenvalues λ will be positive. Properties of the Eigenvalues and Eigenfunctions for (8.9:. The eigenvalues of (8.9 form a countable set {λ n } n= of positive numbers with λ n as n.. For each eigenvalue λ n there is a finite number (called the multiplicity of λ n of linearly independent eigenfunctions u n. 3. The first (or principal eigenvalue, λ, is simple and u does not change sign in. 4. Eigenfunctions corresponding to distinct eigenvalues are orthogonal. 5. The eigenfunctions may be used to expand certain functions on in an infinite series.

45 Partial Differential Equations Igor Yanovsky, Heat Equation The heat equation is u t = k u for x, t>, (9. with initial and boundary conditions. 9. The Pure Initial Value Problem 9.. Fourier Transform If u C (Rn, define its Fourier transform û by û(ξ = e ix ξ u(x dx for ξ R n. (π n R n We can differentiate û: ξ j û(ξ = (π n R n e ix ξ ( ix j u(x dx = [ ( ixj u ] (ξ. Iterating this computation, we obtain ( k û(ξ = [ ( ixj ξ k u ] (ξ. (9. j Similarly, integrating by parts shows ( u ix ξ u (ξ = e (x dx = (e ix ξ u(x dx x j (π n R n x j (π n R n x j = (iξ (π n j e ix ξ u(x dx R n = (iξ j û(ξ. Iterating this computation, we obtain ( k u (ξ =(iξ j k û(ξ. (9.3 x k j Formulas (9. and (9.3 express the fact that Fourier transform interchanges differentiation and multiplication by the coordinate function. 9.. Multi-Index Notation A multi-index is a vector α =(α,...,α n whereeachα i is a nonnegative integer. The order of the multi-index is α = α α n. Given a multi-index α, define D α u = α u x α xαn n = α x x αn n u. We can generalize (9.3 in multi-index notation: D α u(ξ = e ix ξ D α u(x dx = ( α D (π n (π n x α (e ix ξ u(x dx R n = (iξ α e ix ξ u(x dx (π n R n = (iξ α û(ξ. (iξ α = (iξ α (iξ n αn.

46 Partial Differential Equations Igor Yanovsky, 5 46 Parseval s theorem (Plancherel s theorem. Assume u L (R n L (R n.thenû, u L (R n and û L (R n = u L (R n = u L (R n, or Also, u(x dx = u(x v(x dx = û(ξ dξ. û(ξ v(ξ dξ. The properties (9. and (9.3 make it very natural to consider the fourier transform on a subspace of L (R n called the Schwartz class of functions, S, which consists of the smooth functions whose derivatives of all orders decay faster than any polynomial, i.e. S = {u C (R n : forevery k N and α N n, x k D α u(x is bounded on R n }. For u S, the Fourier transform û exists since u decays rapidly at. Lemma. (i If u L (R n,thenû is bounded. (ii If u S, thenû S. Define the inverse Fourier transform for u L (R n : u (ξ = e ix ξ u(x dx for ξ R n, or (π n R n u(x = e ix ξ û(ξ dξ for x R n. (π n R n Fourier Inversion Theorem (McOwen. If u S, then(û = u; that is, u(x = (π n R n e ix ξ û(ξ dξ = (π n R n e i(x y ξ u(y dy dξ =(û (x. Fourier Inversion Theorem (Evans. Assume u L (R n.then,u =(û.

47 Partial Differential Equations Igor Yanovsky, 5 47 Shift: Let u(x } {{ a } = v(x, and determinte v(ξ: y π u(x a(ξ = v(ξ = = u(x a(ξ = e iaξ û(ξ. π R R e ixξ v(x dx = π e iyξ e iaξ u(y dy = e iaξ û(ξ. R e i(y+aξ u(y dy Delta function: δ(x(ξ = π R e ixξ δ(x dx = π, δ(x a(ξ = e iaξ δ(ξ = e iaξ. π ( since u(x = (using result from Shift R δ(x y u(y dy. Convolution: (f g(x = f(x yg(y dy, R n (f g(ξ = e ix ξ f(x y g(y dy dx = e ix ξ f(x y g(y dy dx (π n R n R n (π n R n R n [ ][ ] = e i(x y ξ f(x y dx e iy ξ g(y dy (π n R n R n = e iz ξ f(z dz e iy ξ g(y dy = (π n (π n f(ξĝ(ξ. R n R n (f g(ξ = (π n f(ξ ĝ(ξ. Gaussian: (completing the square ( e x (ξ = e ixξ e x dx = e x +ixξ dx = e x +ixξ ξ dx e ξ π R π R π R = π R ( e x (ξ = e ξ. e (x+iξ dx e ξ = π R e y dy e ξ = π πe ξ = e ξ. Multiplication by x: ixu(ξ = e ixξ( ixu(x dx = π R d dξ û(ξ. xu(x(ξ = i d dξ û(ξ.

48 Partial Differential Equations Igor Yanovsky, 5 48 Multiplication of u x by x: (using the above result xu x (x(ξ = e ixξ( xu x (x dx = [e ixξ xu π π R ] } {{ } = iξ e ixξ xudx e ixξ udx π R π R = = iξ xu(x(ξ û(ξ = iξ [i d ] dξ û(ξ xu x (x(ξ = ξ d û(ξ û(ξ. dξ π û(ξ = ξ d û(ξ û(ξ. dξ R ( ( iξe ixξ x + e ixξ udx Table of Fourier Transforms: ( e ax (ξ = a e ξ a, e ibx f(ax(ξ = a f ( ξ b, { a, x L f(x =, x >L, ê a x (ξ = a + x (ξ = H(a x (ξ = Ĥ(x(ξ = (Gaussian a π a, (a > + ξ π a e a ξ, (a > sin aξ, π ξ ( πδ(ξ+ π iξ f(x(ξ = sin(ξl, π ξ, ( H(x H( x (ξ = π iξ, (sign (ξ = πδ(ξ. Results with marked with were taken from W. Strauss, where the definition of Fourier Transform is different. An extra multiple of was added to each of these results. π

49 Partial Differential Equations Igor Yanovsky, Solution of the Pure Initial Value Problem Consider the pure initial value problem { u t = u for t>, x R n u(x, = g(x for x R n. (9.4 We take the Fourier transform of the heat equation in the x-variables. (u t (ξ, t = e ix ξ u (π n t (x, t dx = t R n tû(ξ, u(ξ, t = n (iξ j û(ξ, t= ξ û(ξ, t. j= The heat equation therefore becomes tû(ξ, t= ξ û(ξ, t, which is an ordinary differential equation in t, with the solution û(ξ, t =Ce ξ t. The initial condition û(ξ, = ĝ(ξ gives û(ξ, t = ĝ(ξ e ξ t, ( u(x, t = ĝ(ξ e ξ t = [g ( e (π n ξ t ] [ ] = g e ξ t e ix ξ dξ (π n (π n R n [ ] = g e ix ξ ξ t dξ = (4π n R n (4π n ] = g [e x (4πt n 4t = (4πt n R n e Thus, solution of the initial value problem (9.4 is u(x, t = K(x, y, t g(y dy = R n (4πt n R n e x y 4t x y 4t ( g [e x π n ] 4t t g(y dy. g(y dy. Uniqueness of solutions for the pure initial value problem fails: there are nontrivial solutions of (9.4 with g =. 3 Thus, the pure initial value problem for the heat equation is not well-posed, as it was for the wave equation. However, the nontrivial solutions are unbounded as functions of x when t > is fixed; uniqueness can be regained by adding a boundedness condition on the solution. Identity (Evans, p. 87. : R n e ix ξ ξ t dξ = e x 4t ( π t n. 3 The following function u satisfies u t = u xx for t>withu(x, = : u(x, t = k= d k (k! xk dt k e /t.

50 Partial Differential Equations Igor Yanovsky, Nonhomogeneous Equation Consider the pure initial value problem with homogeneous initial condition: { u t = u + f(x, t for t>, x R n u(x, = for x R n. (9.5 Duhamel s principle gives the solution: u(x, t = t R n K(x y, t s f(y, s dy ds Nonhomogeneous Equation with Nonhomogeneous Initial Conditions Combining two solutions above, we find that the solution of the initial value problem { u t = u + f(x, t for t>, x R n u(x, = g(x for x R n (9.6. is given by u(x, t = K(x y, t g(y dy + R n t R n K(x y, t s f(y, s dy ds The Fundamental Solution Suppose we want to solve the Cauchy problem { u t = Lu x R n,t> u(x, = g(x x R n. (9.7 where L is a differential operator in R n with constant coefficients. Suppose K(x, t is a distribution in R n for each value of t, K is C in t and satisfies { K t LK =, (9.8 K(x, = δ(x. We call K a fundamental solution for the initial value problem. The solution of (9.7 is then given by convolution in the space variables: u(x, t = K(x y, t g(y dy. R n

51 Partial Differential Equations Igor Yanovsky, 5 5 For operators of the form t L, the fundamental solution of the initial value problem, K(x, t as defined in (9.8, coincides with the free space fundamental solution, which satisfies ( t L K(x, t = δ(x, t, provided we extend K(x, t by zero to t<. e K(x, x t = (4πt n/ 4t t> t. Notice that K is smooth for (x, t (,. For the heat equation, consider (9.9 K defined as in (9.9, is the fundamental solution of the free space heat equation. Proof. We need to show: ( t K(x, t = δ(x, t. (9. To verify (9. as distributions, we must show that for any v C (Rn+ : 4 ( K(x, t t vdxdt = δ(x, t v(x, t dx dt v(,. R n+ R n+ To do this, let us take ɛ> and define e K x ɛ (x, t = (4πt n/ 4t t>ɛ t ɛ. Then K ɛ K as distributions, so it suffices to show that ( t K ɛ δ as distributions. Now ( ( ( K ɛ t vdxdt = K(x, t t v(x, t dx dt ɛ R n ( ( = K(x, t t v(x, t dx dt K(x, t v(x, t dx dt ɛ R n ɛ R [ ] n t= ( ( = K(x, t v(x, t dx + t K(x, t v(x, t dx dt K(x, t v(x, t dx dt R n t=ɛ ɛ R n ɛ R n ( ( = t K(x, t v(x, t dx dt + K(x, ɛ v(x, ɛ dx. ɛ R n But for t>ɛ, ( t K(x, t = ; moreover, since lim t + K(x, t =δ (x =δ(x, we have K(x, ɛ δ (x as ɛ, so the last integral tends to v(,. R n 4 Note, for the operator L = / t, theadjoint operator is L = / t.

52 Partial Differential Equations Igor Yanovsky, 5 5 Schrödinger Equation Problem (F 96, #5. The Gauss kernel G(t, x, y= e (x y (4πt 4t is the fundamental solution of the heat equation, solving G t = G xx, G(,x,y=δ(x y. By analogy with the heat equation, find the fundamental solution H(t, x, y of the Schrödinger equation H t = ih xx, H(,x,y=δ(x y. Show that your expression H(x is indeed the fundamental solution for the Schrödinger equation. You may use the following special integral e ix 4 dx = i4π. Proof. Remark: Consider the initial value problem for the Schrödinger equation { u t = i u x R n, t >, u(x, = g(x x R n. If we formally replace t by it in the heat kernel, we obtain the Fundamental Solution of the Schrödinger Equation: 5 H(x, t = u(x, t = (4πit n (4πit n e x 4it (x R n, t R n e x y 4it g(y dy. In particular, the Schrödinger equation is reversible in time, whereas the heat equation is not. Solution: We have already found the fundamental solution for the heat equation using the Fourier transform. For the Schrödinger equation is one dimension, we have tû(ξ, t= iξ û(ξ, t, which is an ordinary differential equation in t, with the solution û(ξ, t =Ce iξt. The initial condition û(ξ, = ĝ(ξ gives û(ξ, t = ĝ(ξ e iξt, ( u(x, t = ĝ(ξ e iξ t [ = g ( e iξ t ] π [ ] = g e iξt e ix ξ dξ π π R = [ ] π R g e ix ξ iξt dξ = (need some work = ] = g [e x 4it = e x y 4it g(y dy. 4πit 4πit 5 Evans, p. 88, Example 3. R

53 Partial Differential Equations Igor Yanovsky, 5 53 For the Schrödinger equation, consider e Ψ(x, x t = (4πit n/ 4it t> t. (. Notice that Ψ is smooth for (x, t (,. Ψ defined as in (., is the fundamental solution of the Schrödinger equation. We need to show: ( t i Ψ(x, t = δ(x, t. (. To verify (. as distributions, we must show that for any v C (Rn+ : 6 ( Ψ(x, t t i vdxdt = δ(x, t v(x, t dx dt v(,. R n+ R n+ To do this, let us take ɛ> and define e Ψ x ɛ (x, t = (4πit n/ 4it t>ɛ t ɛ. Then Ψ ɛ Ψ as distributions, so it suffices to show that ( t i Ψ ɛ δ as distributions. Now ( ( ( Ψ ɛ t i vdxdt = Ψ(x, t t i v(x, t dx dt ɛ R n ( ( = t i Ψ(x, t v(x, t dx dt + Ψ(x, ɛ v(x, ɛ dx. ɛ R n R n But for t>ɛ, ( t i Ψ(x, t = ; moreover, since lim Ψ(x, t + t=δ (x =δ(x, we have Ψ(x, ɛ δ (x as ɛ, so the last integral tends to v(,. 6 Note, for the operator L = / t, theadjoint operator is L = / t.

54 Partial Differential Equations Igor Yanovsky, 5 54 Problems: Quasilinear Equations Problem (F 9, #7. Use the method of characteristics to find the solution of the first order partial differential equation x u x + xyu y = u which passes through the curve u =, x = y. Determine where this solution becomes singular. Proof. We have a condition u(x = y =. Γ is parametrized by Γ : (s,s,. dx = x x = x(,s= dt t c (s c (s = s x = t + = s ts s, dy = xy dy dt dt = s y ts y = c (s s ts y(s, = c (s =s y = ts, dz = z z = z(,s= dt t c 3 (s c 3 (s = z = t. Thus, x x y = s y = y t = y t x x x. y u(x, y = y x + x = x x + x y. The solution becomes singular when y = x + x. It can be checked that the solution satisfies the PDE and u(x = y = Problem (S 9, #7. Solve the first order PDE f x + x yf y + f = f(x =,y=y using the method of characteristics. Proof. Rewrite the equation u x + x yu y = u, u(,y = y. Γ is parameterized by Γ : (,s,s. dx = x = t, dt dy = x y dy dt dt = t y y = se t 3 dz = z z = s e t. dt Thus, x = t and s = ye t3 3 = ye x3 3, and u(x, y=(ye x3 3 e x = y e 3 x3 x. The solution satisfies both the PDE and initial conditions. 3, y 4 y 4 +y y =.

55 Partial Differential Equations Igor Yanovsky, 5 55 Problem (S 9, #. Consider the Cauchy problem u t = xu x u + <x<, t u(x, = sin x <x< and solve it by the method of characteristics. Discuss the properties of the solution; in particular investigate the behavior of u x (,t for t. Proof. Γ is parametrized by Γ : (s,, sins. We have dx dt = x x = se t, dy dt = y = t, dz = z z = sin s dt e t. Thus, t = y, s = xe y,and u(x, y= e y + sin(xey e y. It can be checked that the solution satisfies the PDE and the initial condition. As t, u(x, t. Also, u x (x, y = cos(xe y =. u x (x, y oscillate between and.ifx =,u x =. Problem (W, #6. Solve the Cauchy problem u t + u u x =, t >, u(,x=+x. Proof. Solved

56 Partial Differential Equations Igor Yanovsky, 5 56 Problem (S 97, #. Find the solution of the Burgers equation u t + uu x = x, t u(x, = f(x, <x<. Proof. Γ is parameterized by Γ : (s,,f(s. dx dt = z, dy dt = y = t, dz dt = x. Note that we have a coupled system: { ẋ = z, ż = x, which can be written as a second order ODE: ẍ + x =, x(s, = s, ẋ(s, = z( = f(s. Solving the equation, we get x(s, t = s cos t + f(ssint, and thus, z(s, t = ẋ(t = s sin t + f(scost. { { x = s cos y + f(ssiny, x cos y = s cos y + f(ssiny cos y, u = s sin y + f(scosy. u sin y = s sin y + f(scosy sin y. x cos y u sin y = s(cos y +sin y=s. u(x, y =f(x cos y u sin ycosy (x cos y u sin ysiny. Problem (F 98, #. Solve the partial differential equation u y u u x =3u, u(x, = f(x using method of characteristics. (Hint: find a parametric representation of the solution. Proof. Γ is parameterized by Γ : (s,,f(s. dx = z dx dt dt = f (se 6t x = 6 f (se 6t + 6 f (s+s, dy dt = y = t, dz dt = 3z z = f(se 3t.

57 Partial Differential Equations Igor Yanovsky, 5 57 Thus, { x = 6 f (se 6y + 6 f (s+s, f(s = z x = 6 e 3y s = x z 6e z = f (x z u(x, y=f 6. 6y + z z + 6e6y 6 e 3y. (x u u + 6e6y 6 e 3y. z z e 6y e6y + 6 e 6y + s = z z 6e6y 6 + s,

58 Partial Differential Equations Igor Yanovsky, 5 58 Problem (S 99, # Modified Problem. a Solve ( u 3 u t + 3 = (. x for t>, <x< with initial data { a( e x, x < u(x, = h(x = a( e x, x > where a> is constant. Solve until the first appearance of discontinuous derivative and determine that critical time. b Consider the equation ( u 3 u t + = cu. (. 3 x How large does the constant c> has to be, so that a smooth solution (with no discontinuities exists for all t>? Explain. Proof. a Characteristic form: u t + u u x =. Γ: (s,,h(s. dx dt = dy z, dt =, dz dt =. x = h(s t + s, y = t, z = h(s. u(x, y=h(x u y (.3 The characteristic projection in the xt-plane 7 passing through the point (s, is the line x = h(s t + s along which u has the constant value u = h(s. The derivative of the initial data is discontinuous, and that leads to a rarefaction-like behavior at t =. However, if the question meant to ask to determine the first time when a shock forms, we proceed as follows. Two characteristics x = h(s t + s and x = h(s t + s intersect at a point (x, t with s s t = h(s h(s. From (.3, we have u x = h h (s (s( uu x t u x = +h(sh (st Hence for h(sh (s <, u x becomes infinite at the positive time t = h(sh (s. The smallest t for which this happens corresponds to the value s = s at which h(sh (s has a minimum (i.e. h(sh (s has a maximum. At time T = /(h(s h (s the 7 y and t are interchanged here

59 Partial Differential Equations Igor Yanovsky, 5 59 solution u experiences a gradient catastrophe. Therefore, need to find a minimum of { f(x =h(xh a( e x ae x (x = a( e x ( ae x = { a e x ( e x, x < a e x ( e x, x > f (x = { a e x ( e x, x < a e x ( e x, x > = { x =ln( = ln(, x < x =ln(, x > {f(ln( = a e ln( ( e ln( = a ( ( a =, x < f(ln( = a ( ( a =, x > t = min{h(sh (s} = a Proof. b Characteristic form: u t + u u x = cu. Γ : (s,,h(s. dx dt = z = h(s e ct x = s + c h(s ( e ct, dy dt = y = t, dz dt = cz z = h(se ct ( h(s =ue cy. Solving for s and t in terms of x and y, weget: Thus, t = y, s = x c h(s ( e cy. u(x, y = h ( x c u e cy ( e cy e cy. u x = h (se cy ( c uu xe cy ( e cy, h (se cy u x = + c h (se cy u ( e cy = h (se cy + c h (sh(s( e cy. Thus, c> that would allow a smooth solution to exist for all t> should satisfy + c h (sh(s( e cy. We can perform further calculations taking into account the result from part (a: min{h(sh (s} = a.

60 Partial Differential Equations Igor Yanovsky, 5 6 Problem (S 99, #. Original Problem. a. Solve u t + u3 x 3 = (.4 for t>, <x< with initial data { a( e x, x < u(x, = h(x = a( e x, x > where a> is constant. Proof. Rewrite the equation as F (x, y, u, u x,u y = u3 x 3 + u y =, F (x, y, z, p, q = p3 3 + q =. Γ is parameterized by Γ : (s,, h(s, φ(s,ψ(s. We need to complete Γ to a strip. Find φ(s andψ(s, the initial conditions for p(s, t and q(s, t, respectively: F (f(s,g(s,h(s,φ(s,ψ(s =, φ(s 3 + ψ(s =, 3 ψ(s = φ(s3 3. h (s = φ(sf (s+ψ(sg (s { ae s = φ(s, x < ae s = φ(s, x > Therefore, now Γ is parametrized by {Γ : (s,, a( e s, ae s, a3 e 3s 3, x < Γ: (s,, a( e s, ae s, a 3 e 3s 3, x > dx dt = F p = p = { a e s a e s x(s, t = {ψ(s = a3 e 3s 3, x < ψ(s = a3 e 3s 3, x > { a e s t + c 4 (s a e s t + c 5 (s x = { a e s t + s a e s t + s dy dt dz dt = F q = y(s, t =t + c (s y = t { = pf p + qf q = p 3 + q = { a 3 e 3s a3 e 3s 3 = 3 a3 e 3s, x < a 3 e 3s + a3 e 3s 3 = 3 a3 e 3s, x > { 3 a 3 e 3s t + c 6 (s, x < 3 a 3 e 3s t a( e s, x < z(s, t = 3 a3 e 3s z = t + c 7 (s, x > 3 a3 e 3s t a( e s, x > dp dt { ae s, x < = F x F z p = p(s, t=c (s p = ae s, x >

61 Partial Differential Equations Igor Yanovsky, 5 6 Thus, dq dt = F y F z q = q(s, t =c 3 (s q = { a3 e 3s 3, x < a 3 e 3s 3, x > { u(x, y= 3 a3 e 3s y a( e s, x < 3 a3 e 3s y a( e s, x > where s is defined as { a e s y + s, x < x = a e s y + s, x >. b. Solve the equation u t + u3 x = cu. (.5 3 Proof. Rewrite the equation as F (x, y, u, u x,u y = u3 x 3 + u y + cu =, F (x, y, z, p, q = p3 + q + cz =. 3 Γ is parameterized by Γ : (s,, h(s, φ(s,ψ(s. We need to complete Γ to a strip. Find φ(s andψ(s, the initial conditions for p(s, t and q(s, t, respectively: F (f(s,g(s,h(s,φ(s,ψ(s =, φ(s 3 + ψ(s+ch(s =, 3 ψ(s = φ(s3 3 ch(s = h (s = φ(sf (s+ψ(sg (s { ae s = φ(s, x < ae s = φ(s, x > { φ(s3 3 + ca( e x, x < φ(s3 3 + ca( e x, x > {ψ(s= a3 e 3s 3 + ca( e x, x < ψ(s= a3 e 3s 3 + ca( e x, x > Therefore, now Γ is parametrized by {Γ : (s,, a( e s, ae s, a3 e 3s 3 + ca( e x, x < Γ: (s,, a( e s, ae s, a 3 e 3s 3 + ca( e x, x >

62 Partial Differential Equations Igor Yanovsky, 5 6 dx dt = F p = p dy dt = F q = dz dt = pf p + qf q = p 3 + q dp dt = F x F z p = cp dq dt = F y F z q = cq We can proceed solving the characteristic equations with initial conditions above.

63 Partial Differential Equations Igor Yanovsky, 5 63 Problem (S 95, #7. a Solve the following equation, using characteristics, u t + u 3 u x =, { a( e x, for x< u(x, = a( e x, for x> where a> is a constant. Determine the first time when a shock forms. Proof. a Γ is parameterized by Γ : (s,,h(s. dx dt = z3, dy dt =, dz dt =. x = h(s 3 t + s, y = t, z = h(s. u(x, y=h(x u 3 y (.6 The characteristic projection in the xt-plane 8 passing through the point (s, is the line x = h(s 3 t + s along which u has a constant value u = h(s. Characteristics x = h(s 3 t + s and x = h(s 3 t + s intersect at a point (x, t with s s t = h(s 3 h(s 3. From (.6, we have u x = h (s( 3u h (s u x t u x = +3h(s h (st Hence for 3h(s h (s <, u x becomes infinite at the positive time t = 3h(s h (s. The smallest t for which this happens corresponds to the value s = s at which h(s h (s has a minimum (i.e. h(s h (s has a maximum. At time T = /(3h(s h (s the solution u experiences a gradient catastrophe. Therefore, need to find a minimum of { f(x =3h(x h 3a ( e x ae x = 3a 3 e x ( e x, x < (x = 3a ( e x ae x = 3a 3 e x ( e x, x > { f 3a 3[ e x ( e x e x ( e x e x] = 3a 3 e x ( e x ( 3e x, x < (x = 3a 3[ e x ( e x + e x ( e x e x] = 3a 3 e x ( e x ( +3e x =, x > { The zeros of f x =, x = ln 3, x <, (x are We check which ones give the minimum of f(x : x =, x = ln3, x >. {f( = 3a 3, f( ln 3 = 3a 3 3 ( 3 = 4a3 9, x < f( = 3a 3, f(ln 3 = 3a 3 3 ( 3 = 4a3 9, x > 8 y and t are interchanged here

64 Partial Differential Equations Igor Yanovsky, 5 64 t = min{3h(s h (s} = min f(s = 3a 3.

65 Partial Differential Equations Igor Yanovsky, 5 65 b Now consider u t + u 3 u x + cu = with the same initial data and a positive constant c. How large does c need to be in order to prevent shock formation? b Characteristic form: u t + u 3 u x = cu. Γ : (s,,h(s. dx dt dy dt dz dt = z 3 = h(s 3 e 3ct x = s + 3c h(s3 ( e 3ct, = y = t, = cz z = h(se ct ( h(s =ue cy. ( z(s, t =h x 3c h(s3 ( e 3ct e ct, ( u(x, y=h x 3c u3 e 3cy ( e 3cy e cy. ( u x = h (s e cy c u u x e 3cy ( e 3cy, h (se cy u x = + c h (su e cy ( e 3cy = h (se cy + c h (sh(s ( e 3cy. Thus, we need + c h (sh(s ( e 3cy. We can perform further calculations taking into account the result from part (a: min{3h(s h (s} = 3a 3.

66 Partial Differential Equations Igor Yanovsky, 5 66 Problem (F 99, #4. Consider the Cauchy problem u y + a(xu x =, u(x, = h(x. Give an example of an (unbounded smooth a(x for which the solution of the Cauchy problem is not unique. Proof. Γ is parameterized by Γ : (s,,h(s. dx t = a(x x(t x( = a(xdt dt x = dy dt = y(s, t =t + c (s y = t, dz dt = z(s, t =c (s z = h(s. Thus, ( y u(x, t=h x a(xdy t a(xdt + s, Problem (F 97, #7. a Solve the Cauchy problem u t xuu x = <x<, t, u(x, = f(x <x<. b Find a class of initial data such that this problem has a global solution for all t. Compute the critical time for the existence of a smooth solution for initial data, f, which is not in the above class. Proof. a Γ is parameterized by Γ : (s,,f(s. dx = xz dx dt dt = xf(s x = se f(st, dy = y = t, dt dz = z = f(s. dt z = f ( xe f(st, u(x, y=f ( xe uy. Check: { u x = f (s (e uy + xe uy u x y u y = f (s xe uy (u y y + u { u x f (sxe uy u x y = f (se uy u y f (sxe uy u y y = f (sxe uy u { ux = f (se uy f (sxye uy u y = f (se uy xu f (sxye uy u y xuu x = f (seuy xu f (sxye uy xu f (se uy f =. (sxyeuy u(x, = f(x.

67 Partial Differential Equations Igor Yanovsky, 5 67 b The characteristics would intersect when f (sxye uy =. Thus, t c = f (sxe utc.

68 Partial Differential Equations Igor Yanovsky, 5 68 Problem (F 96, #6. Find an implicit formula for the solution u of the initial-value problem u t =(x tu x +sin(πx t, u(x, t ==. Evaluate u explicitly at the point (x =.5, t=. Proof. Rewrite the equation as u y +( xyu x =sin(πx y. Γ is parameterized by Γ : (s,,. dx = ( xy =( xt x = dt (s e t + (, s =(x et +, dy = y = t, dt dz ( π = sin(πx y =sin dt (s e t + π t. t [ ( π z(s, t = sin (s e t + π ] t dt + z(s,, t [ ( π z(s, t = sin (s e t + π ] t dt. y [ ( π u(x, y = sin (s e y + π ] y dy = = y y [ ( π sin (x ey e y + π ] y [ ( π sin (x + π ] y dy = u(x, y = y sin(πx y. Note: This solution does not satisfy the PDE. Problem (S 9, #8. Consider the Cauchy problem u t = xu x u, <x<, t, u(x, = f(x, f(x C. dy y [ ] sin(πx y dy, Assume that f for x. Solve the equation by the method of characteristics and discuss the behavior of the solution. Proof. Rewrite the equation as u y xu x = u, Γ is parameterized by Γ : (s,,f(s. dx dt = x x = se t, dz dt = z z = f(se t. u(x, y=f(xe y e y. dy dt = y = t,

69 Partial Differential Equations Igor Yanovsky, 5 69 The solution satisfies the PDE and initial conditions. As y +, u. u =for xe y u =for x e y.

70 Partial Differential Equations Igor Yanovsky, 5 7 Problem (F, #4. Consider the nonlinear hyperbolic equation u y + uu x = <x<. a Find a smooth solution to this equation for initial condition u(x, = x. b Describe the breakdown of smoothness for the solution if u(x, = x. Proof. a Γ is parameterized by Γ : (s,,s. dx = z = s x = st + s s = x dt t + = x y +. dy dt = y = t, dz dt = z = s. u(x, y= x ; solution is smooth for all positive time y. y + b Γ is parameterized by Γ : (s,, s. dx dt dy dt dz dt = z = s x = st + s s = x t = x y. = y = t, = z = s. u(x, y= x ; solution blows up at time y =. y

71 Partial Differential Equations Igor Yanovsky, 5 7 Problem (F 97, #4. Solve the initial-boundary value problem u t +(x + u x = x for x>, t> u(x, = f(x <x<+ u(,t = g(t <t<+. Proof. Rewrite the equation as u y +(x + u x = x for x>, y> u(x, = f(x <x<+ u(,y = g(y <y<+. For region I, we solve the following characteristic equations with Γ is parameterized 9 by Γ : (s,,f(s. dx dt dy dt dz dt = (x + x = s + (s +t, = y = t, = x = s + (s +t, z = ln (s +t t + c (s, z = ln (s +t t + f(s. In region I, characteristics are of the form s + x = (s +y. Thus, region I is bounded above by the line x = x, or y = y x +. Since t = y, s = x xy y xy+y+,wehave ( x xy y ( u(x, y = ln xy + y + + x xy y y y + f, xy + y + ( x xy y u(x, y = ln y + f. xy + y + xy + y + For region II, Γ is parameterized by Γ : (,s,g(s. dx = (x + x = dt t, dy = y = t + s, dt dz = x = dt t, z = ln t t + c (s, z = ln t t + g(s. 9 Variable t as a third coordinate of u and variable t used to parametrize characteristic equations are two different entities.

72 Partial Differential Equations Igor Yanovsky, 5 7 Since t = x+ x,s= y x+ x,wehave x u(x, y= ln x + Note that on y = x ( x + + g y x. x + x x+, both solutions are equal if f( = g(.

73 Partial Differential Equations Igor Yanovsky, 5 73 Problem (S 93, #3. Solve the following equation u t + u x + yu y =sint for t, x, <y< and with u = x + y for t =, x and u = t + y for x =, t. Proof. Rewrite the equation as (x x, y x,t x 3 : u x3 + u x + x u x =sinx 3 for x 3, x, <x <, u(x,x, = x + x, u(,x,x 3 =x 3 + x. For region I, we solve the following characteristic equations with Γ is parameterized by Γ : (s,s,,s + s. dx dt dx dt dx 3 dt dz dt = x = t + s, = x x = s e t, = x 3 = t, = sinx 3 =sint z = cos t + s + s +. Since in region I, in x x 3 -plane, characteristics are of the form x = x 3 + s,region I is bounded above by the line x = x 3. Since t = x 3, s = x x 3, s = x e x 3,we have u(x,x,x 3 = cos x 3 + x x 3 + x e x 3 +, or u(x, y, t = cos t + x t + ye t +, x t. For region II, we solve the following characteristic equations with Γ is parameterized by Γ : (,s,s 3,s + s 3. dx dt = x = t, dx dt = x x = s e t, dx 3 dt = x 3 = t + s 3, dz dt = sinx 3 =sin(t + s 3 z = cos(t + s 3 +coss 3 + s + s 3. Since t = x, s 3 = x 3 x, s = x e x 3,wehave u(x,x,x 3 = cos x 3 +cos(x 3 x +x e x 3 +(x 3 x, or u(x, y, t = cos t +cos(t x+ye t +(t x, x t. Note that on x = t, both solutions are u(x = t, y = cos x + ye x +. Variable t as a third coordinate of u and variable t used to parametrize characteristic equations are two different entities.

74 Partial Differential Equations Igor Yanovsky, 5 74 Problem (W 3, #5. Find a solution to xu x +(x + yu y = which satisfies u(,y=y for y. Find a region in {x, y } where u is uniquely determined by these conditions. Proof. Γ is parameterized by Γ : (,s,s. dx = x x = e t. dt dy = x + y y y = e t. dt dz = z = t + s. dt The homogeneous solution for the second equation is y h (s, t = c (se t. Since the right hand side and y h are linearly dependent, our guess for the particular solution is y p (s, t =c (ste t. Plugging in y p into the differential equation, we get c (ste t + c (se t c (ste t = e t c (s =. Thus, y p (s, t =te t and y(s, t =y h + y p = c (se t + te t. Since y(s, = s = c (s, we get y = se t + te t. With and, we can solve for s and t in terms of x and y to get t = ln x, y = sx + xlnx s = y xlnx. x u(x, y = t + s = ln x + y xlnx. x u(x, y= y x. We have found that the characteristics in the xy-plane are of the form y = sx + xlnx, where s is such that s. Also, the characteristics originate from Γ. Thus, u is uniquely determined in the region between the graphs: y = xlnx, y = x + xlnx.

75 Partial Differential Equations Igor Yanovsky, 5 75 Problems: Shocks Example. Determine the exact solution to Burgers equation u t + ( u x =, t > with initial data if x<, u(x, = h(x = if <x<, if x>. Proof. Characteristic form: u t + uu x =. The characteristic projection in xt-plane passing through the point (s, is the line x = h(st + s. Rankine-Hugoniot shock condition at s = : shock speed: ξ (t = F (u r F (u l u r u l = The /slope of the shock curve = /. Thus, x = ξ(t = t + s, u r u l u r u l = =. and since the jump occurs at (,, ξ( = =s. Therefore, x = t. Rankine-Hugoniot shock condition at s = : shock speed: ξ (t = F (u r F (u l u r u l = The /slope of the shock curve = /. Thus, x = ξ(t = t + s, u r u l u r u l = and since the jump occurs at (,, ξ( = = s. Therefore, x = t +. At t =, Rankine-Hugoniot shock condition at s = : shock speed: ξ (t = F (u r F (u l u r u l = The /slope of the shock curve =. Thus, x = ξ(t =s, u r u l u r u l = =. =. and since the jump occurs at (x, t=(,, ξ( = = s. Therefore, x =.

76 Partial Differential Equations Igor Yanovsky, 5 76 For t<, if x< t, u(x, t= if t <x< t +, if x> t +. { and for t>, u(x, t = if x<, if x>.

77 Partial Differential Equations Igor Yanovsky, 5 77 Example. Determine the exact solution to Burgers equation ( u t + u =, t > x with initial data if x<, u(x, = h(x = if <x<, if x>. Proof. Characteristic form: u t + uu x =. The characteristic projection in xt-plane passing through the point (s, is the line x = h(st + s. For Burgers equation, for a rarefaction fan emanating from (s, on xt-plane, we have: x s u l, t u l, u(x, t= x s t, u l x s t u r, x s u r, t u r., x < t, x+ t, t <x<, i.e. < x+ t < u(x, t=, <x<, x t, <x<t+, i.e. < x t <, x > t+.

78 Partial Differential Equations Igor Yanovsky, 5 78

79 Partial Differential Equations Igor Yanovsky, 5 79 Example 3. Determine the exact solution to Burgers equation u t + ( u x =, t > with initial data { if <x<, u(x, = h(x = if otherwise. Proof. Characteristic form: u t + uu x =. The characteristic projection in xt-plane passing through the point (s, is the line x = h(st + s. Shock: Rankine-Hugoniot shock condition at s = : shock speed: ξ (t = F (u r F (u l u r u l = The /slope of the shock curve =. Thus, x = ξ(t =t + s, u r u l u r u l = =. and since the jump occurs at (,, ξ( = = s. Therefore, x = t +. Rarefaction: A rarefaction emanates from (, on xt-plane. if x<, x For <t<, u(x, t = t if <x<t, if t<x<t+. if x>t+. Rarefaction catches up to shock at t =. Shock: At (x, t=(,, u l = x/t, u r =. Rankine-Hugoniot shock condition: ξ (t = F (u r F (u l u r u l = dx s = x dt t, x = c t, u r u l u r u l = ( x t x t = x t, and since the jump occurs at (x, t=(,, x( = = c. Therefore, x = t. if x<, For t>, u(x, t= x t if <x< t, if x> t.

80 Partial Differential Equations Igor Yanovsky, 5 8

81 Partial Differential Equations Igor Yanovsky, 5 8 Example 4. Determine the exact solution to Burgers equation u t + ( u x =, t > with initial data u(x, = h(x = { +x if x<, if x>. Proof. Characteristic form: u t + uu x =. The characteristic projection in xt-plane passing through the point (s, is the line x = h(st + s. ➀ For s>, the characteristics are x = s. ➁ For s<, the characteristics are x =(+st + s. There are two ways to look for the solution on the left half-plane. One is to notice that the characteristic at s = is x = t and characteristic at s = isx = and that characteristics between s = and s = are intersecting at (x, t =(,. Also, for a fixed t, u is a linear function of x, i.e. for t =,u =+x, allowing a continuous change of u with x. Thus, the solution may be viewed as an implicit rarefaction, originating at (,, thus giving rise to the solution u(x, t= x + t +. Another way to find a solution on the left half-plane is to solve ➁ for s to find s = x t t. Thus, u(x, t=h(s =+s =+x +t +t = x + t +. Shock: At (x, t=(,, u l = x+ t+, u r =. Rankine-Hugoniot shock condition: ξ (t = F (u r F (u l u r u l = dx s = x + dt t +, x = c t +, u r u l t+ = ( x+ u r u l x+ t+ = x + t +, and since the jump occurs at (x, t =(,, x( = = c, or c =. Therefore, theshockcurveis x = t +. { x+ u(x, t= t+ if x< t +, if x> t +.

82 Partial Differential Equations Igor Yanovsky, 5 8

83 Partial Differential Equations Igor Yanovsky, 5 83 Example 5. Determine the exact solution to Burgers equation u t + ( u x =, t > with initial data u if x<, u(x, = h(x = u ( x if <x<, if x, where u >. Proof. Characteristic form: u t + uu x =. The characteristic projection in xt-plane passing through the point (s, is the line x = h(st + s. ➀ For s>, the characteristics are x = s. ➁ For <s<, the characteristics are x = u ( st + s. ➂ For s<, the characteristics are x = u t + s. The characteristics emanating from (s,, <s<onxt-plane intersect at (, u. Also, we can check that the characteristics do not intersect before t = u for this problem: ( t c =min h =. (s u To find solution in a triangular domain between x = u t and x =,wenotethat characteristics there are x = u ( st + s. Solving for s we get s = x u t u t. Thus, u(x, t=h(s =u ( s =u ( x u t u t = u ( x. u t We can also find a solution in the triangular domain as follows. Note, that the characteristics are the straight lines dx dt = u =const. Integrating the equation above, we obtain x = ut + c Since all characteristics in the triangular domain meet at (, u, we have c = u u, and x = ut + ( u or u = u ( x. u u t For <t< u if x<u t, u, u(x, t = ( x u u t if u t<x<, if x>.

84 Partial Differential Equations Igor Yanovsky, 5 84 Shock: At (x, t=(, u, Rankine-Hugoniot shock condition: ξ (t = F (u r F (u l u r u l = ξ(t = u t + c, u r u l u r u l = u u = u, and since the jump occurs at (x, t=(, u, theshockcurveis x = u t+. x ( u == + c, orc =. Therefore,

85 Partial Differential Equations Igor Yanovsky, 5 85 For t> u, u(x, t= { u if x< u t+, if x> u t+. Problem. Show that for u = f(x/t to be a nonconstant solution of u t + a(uu x =, f must be the inverse of the function a. Proof. If u = f(x/t, u t = f ( x x t t and u x = f ( x t t. Hence, u t + a(uu x = implies that f ( x x ( ( x t t + a f f ( x t t t = or, assuming f is not identically to rule out the constant solution, that ( ( x a f = x t t. This shows the functions a and f to be inverses of each other.

86 Partial Differential Equations Igor Yanovsky, Problems: General Nonlinear Equations 3. Two Spatial Dimensions Problem (S, #3. Solve the initial value problem u x u y = x, u(x, = x. You will find that the solution blows up in finite time. Explain this in terms of the characteristics for this equation. Proof. Rewrite the equation as F (x, y, z, p, q= p q + x =. Γ is parameterized by Γ : (s,,s,φ(s,ψ(s. We need to complete Γ to a strip. Find φ(s andψ(s, the initial conditions for p(s, t and q(s, t, respectively: F (f(s,g(s,h(s,φ(s,ψ(s =, F (s,,s,φ(s,ψ(s =, φ(s ψ(s+ s =, ψ(s = φ(s + s. h (s = φ(sf (s+ψ(sg (s, =φ(s. ψ(s = s +. Therefore, now Γ is parametrized by Γ : (s,,s,, s +. dx dt = F p = p, dy dt = F q = y(s, t = t + c (s y = t, dz dt = pf p + qf q = p q, dp dt = F x F z p = x, dq = F y F z q = q(s, t=c (s q = s +. dt Thus, we found y and q in terms of s and t. Note that we have a coupled system: { x = p, p = x, which can be written as two second order ODEs: x + x =, x(s, = s, x (s, = p(s, =, p + p =, p(s, =, p (s, = x(s, = s.

87 Partial Differential Equations Igor Yanovsky, 5 87 Solving the two equations separately, we get x(s, t = s cos t +sint, p(s, t = cost s sin t. From this, we get dz dt = p q = ( cos t s sin t s + t [ z(s, t = cos t s cos t sin t + s sin t s + [ t sin t cos t z(s, t = + [ sin t cos t = + s cos t s sin t cos t ] t + s, sin t cos t = + s cos t s sin t cos t s + s = sin t cos t = + s cos t s sin t cos t. Plugging in x and y found earlier for s and t, weget u(x, y = sin( ycos( y sin y cos y = = cos t s cos t sin t + s sin t s +. ] dt + z(s,, + s cos t + s t s sin t cos t + x sin( y cos( y + x +siny cos y cos y + cos ( y (x +siny cos y sin y cos y = +(x +sinycosy + (x +siny sin y cosy sin y cos y = x cos y + + (x +siny sin y. cosy t(s + ] t + s, (x sin( y cos ( y sin y cos y sin( ycos( y

88 Partial Differential Equations Igor Yanovsky, 5 88 Problem (S 98, #3. Find the solution of u t + u x = x, t, <x< u(x, = h(x, <x<, where h(x is smooth function which vanishes for x large enough. Proof. Rewrite the equation as F (x, y, z, p, q= p + q + x =. Γ is parameterized by Γ : (s,, h(s, φ(s,ψ(s. We need to complete Γ to a strip. Find φ(s andψ(s, the initial conditions for p(s, t and q(s, t, respectively: F (f(s,g(s,h(s,φ(s,ψ(s =, F (s,,h(s,φ(s,ψ(s =, φ(s + ψ(s+ s =, ψ(s = φ(s + s. h (s = φ(sf (s+ψ(sg (s, h (s =φ(s. ψ(s = h (s + s. Therefore, now Γ is parametrized by Γ : (s,,s,h (s, h (s +s. dx dt = F p = p, dy dt = F q = y(s, t =t + c (s y = t, dz dt = pf p + qf q = p + q, dp dt = F x F z p = x, dq = F y F z q = q(s, t=c (s q = h (s + s. dt Thus, we found y and q in terms of s and t. Note that we have a coupled system: { x = p, p = x, which can be written as a second order ODE: x + x =, x(s, = s, x (s, = p(s, = h (s. Solving the equation, we get x(s, t = s cos t + h (s sint, p(s, t = x (s, t =h (s cost s sin t.

89 Partial Differential Equations Igor Yanovsky, 5 89 From this, we get dz dt = p + q = ( h (scost s sin t h (s + s = h (s cos t sh (scost sin t + s sin t h (s + s. t z(s, t = [h (s cos t sh (scost sin t + s sin t h (s + s t = [h (s cos t sh (scost sin t + s sin t h (s + s ] dt + z(s, ] dt + h(s. We integrate the above expression similar to S #3 to get an expression for z(s, t. Plugging in x and y found earlier for s and t, wegetu(x, y.

90 Partial Differential Equations Igor Yanovsky, 5 9 Problem (S 97, #4. Describe the method of the bicharacteristics for solving the initial value problem ( ( x u(x, y + y u(x, y = +y, u(x, = u (x = x. Assume that x u (x < and consider the solution such that u y >. Apply all general computations for the particular case u (x=x. Proof. We have u x + u y = +y u(x, = u (x =x. Rewrite the equation as F (x, y, z, p, q=p + q y =. Γ is parameterized by Γ : (s,,s,φ(s,ψ(s. We need to complete Γ to a strip. Find φ(s andψ(s, the initial conditions for p(s, t and q(s, t, respectively: F (f(s,g(s,h(s,φ(s,ψ(s =, F (s,,s,φ(s,ψ(s =, φ(s + ψ(s =, φ(s + ψ(s =. h (s = φ(sf (s+ψ(sg (s, =φ(s. ψ(s =±. Since we have a condition that q(s, t >, we choose q(s, = ψ(s=. Therefore, now Γ is parametrized by Γ : (s,,s,,. dx dt dy dt dz dt = F p =p dx = x =t + s, dt = F q =q dy dt =t + y = t +t, = pf p + qf q =p +q =y +4 dz dt =t +4t +4, z = 3 t3 +t +4t + s = 3 t3 +t +4t + x t = 3 t3 +t +t + x, dp dt = F x F z p = p =, dq dt = F y F z q = q = t +. We solve y = t +t, a quadratic equation in t, t +t y =,fort in terms of y to get: t = ± +y. u(x, y= 3 ( ± +y 3 +( ± +y +( ± +y+x. Both u ± satisfy the PDE. u x =,u y = ± y + u x + u y = y + u + satisfies u + (x, = x. However, u does not satisfy IC, i.e. u (x, = x 4 3.

91 Partial Differential Equations Igor Yanovsky, 5 9 Problem (S, #6. Consider the equation u x + u x u y =, u(x, = f(x. Assuming that f is differentiable, what conditions on f insure that the problem is noncharacteristic? If f satisfies those conditions, show that the solution is u(x, y=f(r y + y f (r, where r must satisfy y =(f (r (x r. Finally, show that one can solve the equation for (x, y in a sufficiently small neighborhood of (x, with r(x, = x. Proof. Solved. In order to solve the Cauchy problem in a neighborhood of Γ, need: f (s F q [f, g, h, φ, ψ](s g (s F p [f, g, h, φ, ψ](s, ( h (s + h (s h, (s h (s. Thus, h (s ensures that the problem is noncharacteristic. To show that one can solve y = (f (s (x s for (x, y in a sufficiently small neighborhood of (x, with s(x, = x,let G(x, y, s = (f (s (x s y =, G(x,,x =, G r (x,,x = (f (s. Hence, if f (s, s, theng s (x,,x and we can use the implicit function theorem in a neighborhood of (x,,x toget G(x, y, h(x, y = and solve the equation in terms of x and y.

92 Partial Differential Equations Igor Yanovsky, 5 9 Problem (S, #. Find the solutions of (u x +(u y = in a neighborhood of the curve y = x satisfying the conditions u (x, x = and u y (x, x >. Leave your answer in parametric form. Proof. Rewrite the equation as F (x, y, z, p, q=p + q =. Γ is parameterized by Γ : (s, s,,φ(s,ψ(s. We need to complete Γ to a strip. Find φ(s andψ(s, the initial conditions for p(s, t and q(s, t, respectively: F (f(s,g(s,h(s,φ(s,ψ(s =, F ( s, s,,φ(s,ψ(s =, φ(s + ψ(s =. h (s = φ(sf (s+ψ(sg (s, =φ(s+sψ(s, φ(s = sψ(s. Thus, s ψ(s + ψ(s = ψ(s = s +. Since, by assumption, ψ(s >, we have ψ(s = (. s + Therefore, now Γ is parametrized by Γ : s, s,, s, s + s +. dx dt dy dt dz dt dp dt dq dt = F p =p = s s + = F q =q = s + x = st s + + s, y = t s + + s, = pf p + qf q =p +q = z =t, = F x F z p = p = s s +, = F y F z q = q = Thus, in parametric form, z(s, t = t, x(s, t = st s + + s, s +. y(s, t = t s + + s.

93 Partial Differential Equations Igor Yanovsky, Three Spatial Dimensions Problem (S 96, #. Solve the following Cauchy problem : u x + u y + u z =, u(,y,z=y z. Proof. Rewrite the equation as u x + u x + u x 3 =, u(,x,x 3 =x x 3. Write a general nonlinear equation F (x,x,x 3,z,p,p,p 3 =p + p + p 3 =. Γ is parameterized by Γ: ( }{{} x (s,s,, s }{{} x (s,s,, s }{{} x 3 (s,s,, s s }{{} z(s,s,,φ (s,s,φ } {{ } (s,s,φ } {{ } 3 (s,s } {{ } p (s,s, p (s,s, p 3 (s,s, We need to complete Γ to a strip. Find φ (s,s, φ (s,s, and φ 3 (s,s, the initial conditions for p (s,s,t, p (s,s,t, and p 3 (s,s,t, respectively: F ( f (s,s,f (s,s,f 3 (s,s,h(s,s,φ,φ,φ 3 =, F (,s,s,s s,φ,φ,φ 3 = φ + φ + φ 3 =, φ + φ + φ 3 =. h f f f 3 = φ + φ + φ 3, s s s s s = φ. h f f f 3 = φ + φ + φ 3, s s s s s = φ 3. Thus, we have: φ = s, φ 3 = s, φ = s s ( +. Γ: }{{}, s, s }{{}, s }{{} s, s }{{} s +, s } {{ } x (s,s, x (s,s, x 3 (s,s, z(s,s, p (s,s, }{{} p (s,s, This problem is very similar to an already hand-written solved problem F 95 #., s }{{} p 3 (s,s,

94 Partial Differential Equations Igor Yanovsky, 5 94 The characteristic equations are dx dt = F p = x = t, dx = F p =p dx dt dt =s x =s t + s, dx 3 = F p3 =p 3 dx 3 dt dt =s x 3 =s t + s, dz dt = p F p + p F p + p 3 F p3 = p +p +p 3 = s s ++s +s = s + s + z =(s + s +t + s s, dp dt = F x p F z = p = s s +, dp dt = F x p F z = p = s, dp 3 dt = F x3 p 3 F z = p 3 = s. Thus, we have x = t t = x t = x x =s t + s s = x s t s = x x x 3 4x x 3 =s t + s s = x 3 s t s = x 3 x x z =(s + s +t + s s z =(s 4x + s +t + s s z =(s + s +t + s s [ (x x x 3 ( x3 x x ] ( x x x ( 3 x3 x x u(x,x,x 3 = + + 4x 4x x + 4x 4x. Problem (F 95, #. Solve the following Cauchy problem u x + u y + u 3 z = x + y + z, u(x, y, = xy. Proof. Solved

95 Partial Differential Equations Igor Yanovsky, 5 95 Problem (S 94, #. Solve the following PDE for f(x, y, t: f t + xf x +3t f y = f(x, y, = x + y. Proof. Rewrite the equation as (x x, y x,t x 3,f u: x u x +3x 3 u x + u x3 =, u(x,x, = x + x. F (x,x,x 3,z,p,p,p 3 =x p +3x 3p + p 3 =. Γ is parameterized by Γ: ( s }{{} x (s,s,, s }{{} x (s,s,, }{{} x 3 (s,s,,s + s,φ } {{ } (s,s,φ (s,s,φ 3 (s,s } {{ } } {{ } } {{ } z(s,s, p (s,s, p (s,s, p 3 (s,s, We need to complete Γ to a strip. Find φ (s,s, φ (s,s, and φ 3 (s,s, the initial conditions for p (s,s,t, p (s,s,t, and p 3 (s,s,t, respectively: F ( f (s,s,f (s,s,f 3 (s,s,h(s,s,φ,φ,φ 3 =, F ( s,s,,s + s,φ,φ,φ 3 = s φ + φ 3 =, φ 3 = s φ. h s f f f 3 = φ + φ + φ 3, s s s s = φ. h s f f f 3 = φ + φ + φ 3, s s s s = φ. Thus, we have: φ =s, φ =s, φ 3 =s (. Γ: s }{{}, s, }{{} }{{},s + s, s } {{ } x (s,s, x (s,s, x 3 (s,s, z(s,s, }{{} p (s,s,, s }{{} p (s,s,, s }{{} p 3 (s,s, The characteristic equations are dx = F p = x x = s e t, dt dx = F p =3x 3 dx dt dt =3t x = t 3 + s, dx 3 = F p3 = x 3 = t, dt dz = p F p + p F p + p 3 F p3 = p x + p 3x 3 + p 3 = z = s + s dt, dp = F x p F z = p p =s e t, dt dp = F x p F z = p =s, dt dp 3 = F x3 p 3 F z = 6x 3 p dp 3 dt dt = ts p 3 = 6t s +s. With t = x 3, s = x e x3,s = x x 3 3, we have ( u(x,x,x 3 =x e x3 +(x x 3 3. f(x, y, t=x e t +(y t 3. The solution satisfies the PDE and initial condition.

96 Partial Differential Equations Igor Yanovsky, 5 96 Problem (F 93, #3. Find the solution of the following equation f t + xf x +(x + tf y = t 3 f(x, y, = xy. Proof. Rewrite the equation as (x x, y x,t x 3,f u: x u x +(x + x 3 u x + u x3 = x 3, u(x,x, = x x. Method I: Treat the equation as a QUASILINEAR equation. Γ is parameterized by Γ : (s,s,,s s. dx = x x = s e t, dt dx = x + x 3 dx dt dt = s e t + t x = s e t + t + s s, dx 3 = x 3 = t, dt dz = x 3 3 dz dt dt = t3 z = t4 4 + s s. Since t = x 3,s = x e x 3,s = x s e t t + s = x x x 3 + x e x 3,wehave u(x,x,x 3 = x x e x 3 (x x x 3 + x e x 3, or f(x, y, t = t4 4 + xe t (y x t + xe t. The solution satisfies the PDE and initial condition. Method II: Treat the equation as a fully NONLINEAR equation. F (x,x,x 3,z,p,p,p 3 =x p +(x + x 3 p + p 3 x 3 3 =. Γ is parameterized by Γ: ( s }{{} x (s,s,, s }{{} x (s,s,, }{{} x 3 (s,s,, s s }{{},φ (s,s,φ } {{ } (s,s,φ } {{ } 3 (s,s } {{ } z(s,s, p (s,s, p (s,s, p 3 (s,s, We need to complete Γ to a strip. Find φ (s,s, φ (s,s, and φ 3 (s,s, the initial conditions for p (s,s,t, p (s,s,t, and p 3 (s,s,t, respectively: F ( f (s,s,f (s,s,f 3 (s,s,h(s,s,φ,φ,φ 3 =, F ( s,s,,s s,φ,φ,φ 3 = s φ + s φ + φ 3 =, φ 3 = s (φ + φ. h s f f f 3 = φ + φ + φ 3, s s s s = φ. h s f f f 3 = φ + φ + φ 3, s s s s = φ. Thus, we have: φ = s, φ = s, φ 3 = s s s. ( Γ: s }{{}, s, }{{} }{{}, s s, s }{{} x (s,s, x (s,s, x 3 (s,s, z(s,s, }{{} p (s,s,, s }{{} p (s,s,, s s s } {{ } p 3 (s,s,

97 Partial Differential Equations Igor Yanovsky, 5 97 The characteristic equations are dx = F p = x x = s e t, dt dx = F p = x + x 3 dx dt dt = s e t + t x = s e t + t + s s, dx 3 = F p3 = x 3 = t, dt dz = p F p + p F p + p 3 F p3 = p x + p (x + x 3 +p 3 = x 3 3 = t 3 z = t4 dt 4 + s s, dp = F x p F z = p p = p s p =s e t s, dt dp = F x p F z = p = s, dt dp 3 = F x3 p 3 F z =3x 3 p =3t s p 3 = t 3 s t s s s. dt With t = x 3,s = x e x3,s = x s e t t + s = x x x 3 + x e x3, we have u(x,x,x 3 = x x e x3 (x x x 3 + x e x3, or f(x, y, t = t4 4 + xe t (y x t + xe t. The solution satisfies the PDE and initial condition. Variable t in the derivatives of characteristics equations and t in the solution f(x, y, t are different entities.

98 Partial Differential Equations Igor Yanovsky, 5 98 Problem (F 9, #. Solve the initial value problem u t + αu x + βu y + γu = for t> u(x, y, = ϕ(x, y, in which α, β and γ are real constants and ϕ is a smooth function. Proof. Rewrite the equation as (x x, y x,t x 3 3 : αu x + βu x + u x3 = γu, u(x,x, = ϕ(x,x. Γ is parameterized by Γ : (s,s,,ϕ(s,s. dx dt = α x = αt + s, dx dt = β x = βt + s, dx 3 dt = x 3 = t, dz = γz dz dt z = γdt z = ϕ(s,s e γt. J det ( (x,x,x 3 = (s,s,t α β Since t = x 3,s = x αx 3,s = x βx 3,wehave u(x,x,x 3 = ϕ(x αx 3,x βx 3 e γx 3, or u(x, y, t = ϕ(x αt, y βte γt. The solution satisfies the PDE and initial condition. 4 = J is invertible. 3 Variable t as a third coordinate of u and variable t used to parametrize characteristic equations are two different entities. 4 Chain Rule: u(x,x,x 3=ϕ(f(x,x,x 3,g(x,x,x 3, then u x = ϕ f + ϕ g f g x. x

99 Partial Differential Equations Igor Yanovsky, 5 99 Problem (F 94, #. Find the solution of the Cauchy problem u t (x, y, t+au x (x, y, t+bu y (x, y, t+c(x, y, tu(x, y, t= u(x, y, = u (x, y, where <t<+, <x<+, <y<+, a, b are constants, c(x, y, t is a continuous function of (x, y, t, andu (x, y is a continuous function of (x, y. Proof. Rewrite the equation as (x x, y x,t x 3 : au x + bu x + u x3 = c(x,x,x 3 u, u(x,x, = u (x,x. Γ is parameterized by Γ : (s,s,,u (s,s. dx dt = a x = at + s, dx dt = b x = bt + s, dx 3 dt = x 3 = t, dz = c(x,x,x 3 z dz dt dt = c(at + s,bt+ s,tz dz z = c(at + s,bt+ s,tdt t ln z = c(aξ + s,bξ+ s,ξdξ + c (s,s, J det z(s,s,t=c (s,s e t c(aξ+s,bξ+s,ξdξ z(s,s, = c (s,s =u (s,s, z(s,s,t=u (s,s e t c(aξ+s,bξ+s,ξdξ. ( (x,x,x 3 = (s,s,t a b Since t = x 3,s = x ax 3,s = x bx 3,wehave = J is invertible. u(x,x,x 3 = u (x ax 3,x bx 3 e x 3 c(aξ+x ax 3,bξ+x bx 3,ξdξ = u (x ax 3,x bx 3 e x 3 c(x +a(ξ x 3,x +b(ξ x 3,ξdξ, or u(x, y, t = u (x at, y bte t c(x+a(ξ t,y+b(ξ t,ξdξ.

100 Partial Differential Equations Igor Yanovsky, 5 Problem (F 89, #4. Consider the first order partial differential equation u t +(α + βtu x + γe t u y = (3. in which α, β and γ are constants. a For this equation, solve the initial value problem with initial data u(x, y, t ==sin(xy (3. for all x and y and for t. b Suppose that this initial data is prescribed only for x (and all y and consider (3. in the region x, t and all y. For which values of α, β and γ is it possible to solve the initial-boundary value problem (3., (3. with u(x =,y,t given for t? For non-permissible values of α, β and γ, where can boundary values be prescribed in order to determine a solution of (3. in the region (x, t, ally. Proof. a Rewrite the equation as (x x,y x,t x 3 : (α + βx 3 u x + γe x 3 u x + u x3 =, u(x,x, = sin(x x. Γ is parameterized by Γ : (s,s,, sin(s s. dx = α + βx 3 dx dt dt = α + βt x = βt + αt + s, dx = γe x 3 dx dt dt = γet x = γe t γ + s, dx 3 dt = x 3 = t, dz dt = z =sin(s s. ( (x,x,x 3 J det = (s,s,t = βt + α γe t J is invertible. Since t = x 3,s = x βx 3 αx 3,s = x γe x 3 + γ, wehave u(x,x,x 3 = sin((x βx 3 αx 3(x γe x 3 + γ, or u(x, y, t = sin((x βt αt(y γet + γ. The solution satisfies the PDE and initial condition. b We need a compatibility condition between the initial and boundary values to hold on y-axis (x =,t=: u(x =,y, = u(,y,t=, =.

101 Partial Differential Equations Igor Yanovsky, 5

102 Partial Differential Equations Igor Yanovsky, 5 4 Problems: First-Order Systems Problem (S, #a. Find the solution u = to the (strictly hyperbolic equation satisfying ( u t 5 3 ( u (x, u (x, u x =, ( e ixa = Proof. Rewrite the equation as ( U t + U 5 3 x =, ( u U(x, = ( (x, u ( = (x,, a R. ( e ixa ( u (x, t u (x, t, (x, t R R,. The eigenvalues of the( matrix A are( λ =, λ = 3 and the corresponding eigenvectors are e =, e 5 =. Thus, Λ= ( 3 Let U =ΓV. Then,, Γ= U t + AU x =, ΓV t + AΓV x =, V t +Γ AΓV x =, V t +ΛV x =. ( 5, Γ = ( det Γ Γ= 5 Thus, the transformed problem is ( V t + V 3 x =, ( ( V (x, = Γ U(x, = e ixa 5 = ( eixa 5 We have two initial value problems { { v ( t v x ( =, v ( t 3v x ( =, v ( (x, = eixa ; v ( (x, = 5 eixa, which we solve by characteristics to get v ( (x, t= eia(x+t, v ( (x, t = 5 eia(x+3t. ( v ( We solve for U: U =ΓV =Γ v ( Thus, U = ( u ( (x, t u ( (x, t ( = = ( 5 e ia(x+t 5 eia(x+t + 5 eia(x+3t. ( eia(x+t 5 eia(x+3t Can check that this is the correct solution by plugging it into the original equation....

103 Partial Differential Equations Igor Yanovsky, 5 3 Part (b of the problem is solved in the Fourier Transform section.

104 Partial Differential Equations Igor Yanovsky, 5 4 Problem (S 96, #7. Solve the following initial-boundary ( value problem in the domain x>, t>, for the unknown vector U = u ( : ( 3 U t + U(x, = ( sin x u ( U x =. (4. and u ( (,t=t. Proof. The eigenvalues( of the matrix( A are λ =, λ = and the corresponding eigenvectors are e =, e =. Thus, ( ( Λ=, Γ=, Γ = ( det Γ Γ=. Let U =ΓV. Then, U t + AU x =, ΓV t + AΓV x =, V t +Γ AΓV x =, V t +ΛV x =. Thus, the transformed problem is ( V t + V (x, = Γ U(x, = V x =, (4. ( ( sin x = ( sin x Equation (4. gives traveling wave solutions of the form v ( (x, t=f (x +t, We can write U in terms of V : ( ( v ( U =ΓV = = v ( v ( (x, t=g(x t. ( ( F (x +t G(x t =. (4.3 ( F (x +t+g(x t G(x t (4.4.

105 Partial Differential Equations Igor Yanovsky, 5 5 For region I, (4. and (4.3 give two initial value problems (since any point in region I can be traced back along both characteristics to initial conditions: { { v ( t v x ( =, v ( t + v x ( =, v ( (x, = sin x; v ( (x, =. which we solve by characteristics to get traveling wave solutions: v ( (x, t=sin(x +t, v ( (x, t =. Thus, for region I, U =ΓV = ( ( ( sin(x +t sin(x +t = For region II, solutions of the form F (x+t can be traced back to initial conditions. Thus, v ( is the same as in region I. Solutions of the form G(x t can be traced back to the boundary. Since from (4.4, u ( = v (, we use boundary conditions to get u ( (,t=t = G( t. Hence, G(x t = (x t. Thus, for region II, U =ΓV = ( ( sin(x +t (x t = Solutions for regions I and II satisfy (4.. Solution for region I satisfies both initial conditions. Solution for region II satisfies given boundary condition.. ( sin(x +t (x t (x t.

106 Partial Differential Equations Igor Yanovsky, 5 6 Problem (S, #7. Consider the system ( ( ( u u =. (4.5 t v x v Find an explicit solution for the following mixed problem for the system (4.5: ( ( u(x, f(x = for x>, v(x, u(,t = for t>. You may assume that the function f is smooth and vanishes on a neighborhood of x =. Proof. Rewrite the equation as ( U t + U x =, ( u U(x, = ( (x, u ( = (x, ( f(x The eigenvalues of( the matrix A( are λ = 3, λ = and the corresponding eigenvectors are e =, e =. Thus, ( ( 3 Λ=, Γ=, Γ = ( det Γ Γ=. 5 Let U =ΓV. Then, U t + AU x =, ΓV t + AΓV x =, V t +Γ AΓV x =, V t +ΛV x =. Thus, the transformed problem is ( 3 V t + V x =, (4.6 V (x, = Γ U(x, = ( ( f(x = f(x ( 5 5 Equation (4.6 gives traveling wave solutions of the form:.. (4.7 v ( (x, t=f (x +3t, v ( (x, t=g(x t. (4.8 We can write U in terms of V : U =ΓV = ( ( v ( v ( = ( ( F (x +3t G(x t = ( F (x +3t G(x t F (x +3t+G(x t (4.9.

107 Partial Differential Equations Igor Yanovsky, 5 7 For region I, (4.6 and (4.7 give two initial value problems (since value at any point in region I can be traced back along both characteristics to initial conditions: { { v ( t 3v x ( =, v ( t +v x ( =, v ( (x, = 5 f(x; v ( (x, = 5 f(x. which we solve by characteristics to get traveling wave solutions: v ( (x, t= f(x +3t, 5 v( (x, t = f(x t. 5 ( ( ( Thus, for region I, U =ΓV = 5f(x +3t = 5 f(x +3t+ 4 5f(x t 5f(x t 5 f(x +3t 5f(x t For region II, solutions of the form F (x+3t can be traced back to initial conditions. Thus, v ( is the same as in region I. Solutions of the form G(x t can be traced back to the boundary. Since from (4.9, u ( = v ( v (, we have u ( (x, t=f (x +3t G(x t = f(x +3t G(x t. 5. The boundary condition gives u ( (,t== f(3t G( t, 5 G( t = 5 f(3t, G(t = ( f 3, t G(x t = ( f 3. (x t ( Thus, for region II, U =ΓV = Solutions for regions I and II satisfy (4.5. Solution for region I satisfies both initial conditions. Solution for region II satisfies given boundary condition. ( ( 5 f(x +3t 5 f(x +3t f( 3 = 5 f( 3 (x t (x t 5 f(x +3t+ f( 3 (x t.

108 Partial Differential Equations Igor Yanovsky, 5 8 Problem (F 94, #; S 97, #7. Solve the initial-boundary value problem u t +3v x =, v t + u x +v x = in the quarter plane x, t <, with initial conditions 5 u(x, = ϕ (x, v(x, = ϕ (x, <x<+ and boundary condition u(,t=ψ(t, t >. Proof. Rewrite the equation as U t + AU x =: ( 3 U t + U x =, (4. U(x, = ( u ( (x, u ( (x, = ( ϕ (x ϕ (x. The eigenvalues of( the matrix A( are λ =, λ = 3 and the corresponding eigenvectors are e =, e 3 =. Thus, ( ( 3 Λ=, Γ=, Γ = ( 3 det Γ Γ=. 4 3 Let U =ΓV. Then, U t + AU x =, ΓV t + AΓV x =, V t +Γ AΓV x =, V t +ΛV x =. Thus, the transformed problem is ( V t + V 3 x =, (4. V (x, = Γ U(x, = ( ( ϕ (x = ( ϕ (x+ϕ (x 4 3 ϕ (x 4 ϕ (x+3ϕ (x Equation (4. gives traveling wave solutions of the form:. (4. v ( (x, t=f (x + t, v ( (x, t =G(x 3t. (4.3 We can write U in terms of V : U =ΓV = ( 3 ( v ( v ( = ( 3 ( F (x + t G(x 3t = ( 3F (x + t+g(x 3t F (x + t+g(x 3t ( In S 97, #7, the zero initial conditions are considered.

109 Partial Differential Equations Igor Yanovsky, 5 9 For region I, (4. and (4. give two initial value problems (since value at any point in region I can be traced back along characteristics to initial conditions: { { v ( t v x ( =, v ( t +3v x ( =, v ( (x, = 4 ϕ (x+ 4 ϕ (x; v ( (x, = 4 ϕ (x+ 3 4 ϕ (x, which we solve by characteristics to get traveling wave solutions: v ( (x, t= 4 ϕ (x + t+ 4 ϕ (x + t, v ( (x, t = 4 ϕ (x 3t+ 3 4 ϕ (x 3t. Thus, for region I, ( ( 3 U = ΓV = 4 ϕ (x + t+ 4 ϕ (x + t 4 ϕ (x 3t+ 3 4 ϕ (x 3t = ( 3ϕ (x + t 3ϕ (x + t+ϕ (x 3t+3ϕ (x 3t 4 ϕ (x + t+ϕ (x + t+ϕ (x 3t+3ϕ (x 3t For region II, solutions of the form F (x + t can be traced back to initial conditions. Thus, v ( is the same as in region I. Solutions of the form G(x 3t can be traced back to the boundary. Since from (4.4, u ( = 3v ( + v (, we have u ( (x, t= 3 4 ϕ (x + t 3 4 ϕ (x + t+g(x 3t. The boundary condition gives u ( (,t=ψ(t = 3 4 ϕ (t 3 4 ϕ (t+g( 3t,. G( 3t =ψ(t 3 4 ϕ (t+ 3 4 ϕ (t, ( G(t =ψ t 3 ( 3 4 ϕ t ϕ ( G(x 3t =ψ x 3t 3 ( 3 4 ϕ ( t, 3 x 3t ϕ ( x 3t. 3 Thus, for region II, ( ( 3 U = ΓV = 4 ϕ (x + t+ 4 ϕ (x + t ψ( x 3t ϕ ( x 3t ϕ ( x 3t 3 ( 3 = 4 ϕ (x + t 3 4 ϕ (x + t+ψ( x 3t ϕ ( x 3t ϕ ( x 3t 3 4 ϕ (x + t+ 4 ϕ (x + t+ψ( x 3t ϕ ( x 3t ϕ ( x 3t 3.

110 Partial Differential Equations Igor Yanovsky, 5 Solutions for regions I and II satisfy (4.. Solution for region I satisfies both initial conditions. Solution for region II satisfies given boundary condition.

111 Partial Differential Equations Igor Yanovsky, 5 Problem (F 9, #. Solve explicitly the following initial-boundary value problem for linear hyperbolic system u t = u x + v x v t =3u x v x, where <t<+, <x<+ with initial conditions u(x, = u (x, v(x, = v (x, <x<+, and the boundary condition u(,t+bv(,t=ϕ(t, <t<+, where b 3 is a constant. What happens when b = 3? Proof. Let us change the notation (u u (, v u (. Rewrite the equation as ( U t + 3 U(x, = ( u ( (x, u ( (x, U x =, (4.5 ( = u ( (x u ( (x The eigenvalues of( the matrix A( are λ =, λ = and the corresponding eigenvectors are e =, e =. Thus, 3 ( ( Λ=, Γ=, Γ = ( Let U =ΓV. Then, U t + AU x =, ΓV t + AΓV x =, V t +Γ AΓV x =, V t +ΛV x =. Thus, the transformed problem is ( V t + V x =, (4.6. V (x, = Γ U(x, = 4 ( 3 ( u ( (x, u ( (x, ( = 4 3u ( (x+u( (x u ( (x u( (x. (4.7 Equation (4.6 gives traveling wave solutions of the form: v ( (x, t=f (x +t, v ( (x, t=g(x t. (4.8

112 Partial Differential Equations Igor Yanovsky, 5 We can write U in terms of V : U =ΓV = ( 3 ( v ( v ( = ( 3 ( F (x +t G(x t = ( F (x +t+g(x t F (x +t 3G(x t (4.9 For region I, (4.6 and (4.7 give two initial value problems (since value at any point in region I can be traced back along characteristics to initial conditions: { { v ( t v x ( =, v ( t +v x ( =, v ( (x, = 3 4 u( (x+ 4 u( (x; v ( (x, = 4 u( (x 4 u( (x, which we solve by characteristics to get traveling wave solutions: v ( (x, t= 3 4 u( (x +t+ 4 u( (x +t; v( (x, t= 4 u( (x t 4 u( (x t. Thus, for region I, ( ( 3 U = ΓV = 4 u( (x +t+ 4 u( (x +t 3 4 u( (x t 4 u( (x t ( 3 = 4 u( (x +t+ 4 u( (x +t+ 4 u( (x t 4 u( (x t 3 4 u( (x +t+ 4 u( (x +t 3 4 u( (x t u( (x t For region II, solutions of the form F (x+t can be traced back to initial conditions. Thus, v ( is the same as in region I. Solutions of the form G(x t can be traced back to the boundary. The boundary condition gives u ( (,t+bu ( (,t=ϕ(t. Using (4.9, v ( (,t+g( t+bv ( (,t 3bG( t=ϕ(t, ( + bv ( (,t+( 3bG( t=ϕ(t,. ( 3 ( + b 4 u( (t+ 4 u( (t +( 3bG( t=ϕ(t, ( ϕ(t ( + b 3 4 u( (t+ 4 u( (t G( t =, ( 3b ϕ( t ( + b 34 u ( ( t+ 4 u( ( t G(t =, 3b G(x t = Thus, for region II, U = ΓV = = ( u( 3 4 u( ϕ( x t ( + b ( 34 u ( ( (x t + 4 u( (x +t+ 4 u( (x +t+ 4 u( 3b ( (x t 3 4 u( (x ( +t+ 4 u( (x +t ϕ( x t (+b 3 4 u( ( (x t+ 4 u( ( (x t 3b ( x t ϕ( (+b 3 4 (x +t+ u( ( (x t+ 4 u( ( (x t ( 3b x t 3ϕ( 3(+b 3 4 (x +t u( ( (x t+ 4 u( ( (x t 3b The following were performed, but are arithmetically complicated: Solutions for regions I and II satisfy (4.5...

113 Partial Differential Equations Igor Yanovsky, 5 3 Solution for region I satisfies both initial conditions. Solution for region II satisfies given boundary condition. If b = 3, u( (,t+ 3 u( (,t=f (t+g( t+ 3 F (t G( t = 4 3 F (t =ϕ(t. Thus, the solutions of the form v ( = G(x t are not defined at x =,whichleads to ill-posedness.

114 Partial Differential Equations Igor Yanovsky, 5 4 Problem (F 96, #8. Consider the system u t = 3u x +v x v t = v x v in the region x, t. Which of the following sets of initial and boundary data make this a well-posed problem? a u(x, =, x v(x, = x, x v(,t=t, t. b u(x, =, x v(x, = x, x u(,t=t, t. c u(x, =, x v(x, = x, x u(,t=t, t v(,t=t, t. Proof. Rewrite the equation as U t + AU x = BU. Initial conditions are same for (a,(b,(c: ( ( 3 U t + U x = U, ( u U(x, = ( (x, u ( = (x, x. The eigenvalues of( the matrix ( A are λ = 3, λ =, and the corresponding eigenvectors are e =, e =. Thus, ( ( 3 Λ=, Γ=, Γ = (. Let U =ΓV. Then, U t + AU x = BU, ΓV t + AΓV x = BΓV, V t +Γ AΓV x =Γ BΓV, V t +ΛV x =Γ BΓV. Thus, the transformed problem is ( ( 3 V t + V x = V, (4. V (x, = Γ U(x, = ( ( ( x = x Equation (4. gives traveling wave solutions of the form. (4. v ( (x, t=f (x +3t, v ( (x, t =G(x t. (4.

115 Partial Differential Equations Igor Yanovsky, 5 5 We can write U in terms of V : ( ( v ( U =ΓV = = v ( ( ( F (x +3t G(x t = ( F (x +3t+G(x t G(x t (4.3. For region I, (4. and (4. give two initial value problems (since a value at any point in region I can be traced back along both characteristics to initial conditions: { { v ( t 3v x ( = v (, v ( t + v x ( = v (, v ( (x, = x ; v ( (x, = x, which we do not solve here. Thus, initial conditions for v ( and v ( have to be defined. Since (4.3 defines u ( and u ( in terms of v ( and v (, we need to define two initial conditions for U. For region II, solutions of the form F (x+3t can be traced back to initial conditions. Thus, v ( is the same as in region I. Solutions of the form G(x t are traced back to the boundary at x =. Since from (4.3, u ( (x, t = v ( (x, t = G(x t, i.e. u ( is written in term of v ( only, u ( requires a boundary condition to be defined on x =. Thus, a u ( (,t=t, t. Well-posed. b u ( (,t=t, t. Not well-posed. c u ( (,t=t, u ( (,t=t, t. Not well-posed.

116 Partial Differential Equations Igor Yanovsky, 5 6 Problem (F, #3. Consider the first order system u t + u x + v x = v t + u x v x = on the domain <t< and <x<. Which of the following sets of initialboundary data are well posed for this system? Explain your answers. a u(x, = f(x, v(x, = g(x; b u(x, = f(x, v(x, = g(x, u(,t = h(x, v(,t = k(x; c u(x, = f(x, v(x, = g(x, u(,t = h(x, v(,t = k(x. Proof. Rewrite the equation as U t +AU x =. Initial conditions are same for (a,(b,(c: ( U t + U x =, ( u U(x, = ( ( (x, f(x u ( =. (x, g(x The eigenvalues of the matrix A are λ =, λ = ( ( and the corresponding eigenvectors are e = +, e =. Thus, ( ( Λ=, Γ= +, Γ = ( + + Let U =ΓV. Then, U t + AU x =, ΓV t + AΓV x =, V t +Γ AΓV x =, V t +ΛV x =. Thus, the transformed problem is ( V t + V x =, (4.4 V (x, = Γ U(x, = ( + + ( f(x g(x Equation (4.4 gives traveling wave solutions of the form:. = ( ( + f(x+g(x ( + f(x g(x (4.5 v ( (x, t=f (x t, v ( (x, t =G(x + t. (4.6 However, we can continue and obtain the solutions. We have two initial value problems { ( v t + v x ( { ( =, v t v x ( =, v ( (x, = (+ f(x+ g(x; v ( (x, = ( + f(x g(x, which we solve by characteristics to get traveling wave solutions: v ( (x, t = ( + f(x t+ g(x t, v ( (x, t = ( + f(x + t g(x + t..

117 Partial Differential Equations Igor Yanovsky, 5 7 We can obtain general solution U by writing U in terms of V : ( v ( U =ΓV =Γ v ( ( = + ( ( + f(x t+g(x t ( + f(x + t g(x + t (4.7. In region I, the solution is obtained by solving two initial value problems(since a value at any point in region I can be traced back along both characteristics to initial conditions. In region II, the solutions of the form v ( = G(x + t can be traced back to initial conditions and those of the form v ( = F (x t, to left boundary. Since by (4.7, u ( and u ( are written in terms of both v ( and v (, one initial condition and one boundary condition at x = need to be prescribed. In region III, the solutions of the form v ( = G(x + t can be traced back to right boundary and those of the form v ( = F (x t, to initial condition. Since by (4.7, u ( and u ( are written in terms of both v ( and v (, one initial condition and one boundary condition at x = need to be prescribed. To obtain the solution for region IV, two boundary conditions, one for each boundary, should be given. Thus, a No boundary conditions. Not well-posed. b u ( (,t=h(x, u ( (,t=k(x. Not well-posed. c u ( (,t=h(x, u ( (,t=k(x. Well-posed.

118 Partial Differential Equations Igor Yanovsky, 5 8 Problem (S 94, #3. Consider the system of equations f t + g x = g t + f x = h t +h x = on the set x, t, with the following initial-boundary values: a f, g, h prescribed on t =, x ; f, h prescribed on x =, t. b f, g, h prescribed on t =, x ; f g, h prescribed on x =, t. c f + g, h prescribed on t =, x ; f, g, h prescribed on x =, t. For each of these 3 sets of data, determine whether or not the system is well-posed. Justify your conclusions. Proof. The third equation is decoupled from the first two and can be considered separately. Its solution can be written in the form h(x, t=h(x t, and therefore, h must be prescribed on t =andonx =, since the characteristics propagate from both the x and t axis. We rewrite the first two equations as (f u, g u : ( U t + U x =, U(x, = ( u ( (x, u ( (x,. The eigenvalues of( the matrix A( are λ =, λ = and the corresponding eigenvectors are e =, e =. Thus, ( ( Λ=, Γ=, Γ = (. Let U =ΓV. Then, U t + AU x =, ΓV t + AΓV x =, V t +Γ AΓV x =, V t +ΛV x =. Thus, the transformed problem is ( V t + V x =, (4.8 V (x, = Γ U(x, = ( ( u ( (x, u (. (4.9 (x, Equation (4.8 gives traveling wave solutions of the form: v ( (x, t=f (x + t, v ( (x, t =G(x t. (4.3

119 Partial Differential Equations Igor Yanovsky, 5 9 We can write U in terms of V : ( ( ( ( ( v ( F (x + t F (x + t+g(x t U =ΓV = v ( = = G(x t F (x + t+g(x t (4.3 For region I, (4.8 and (4.9 give two initial value problems (since a value at any point in region I can be traced back along both characteristics to initial conditions. Thus, initial conditions for v ( and v ( have to be defined. Since (4.3 defines u ( and u ( in terms of v ( and v (, we need to define two initial conditions for U. For region II, solutions of the form F (x + t can be traced back to initial conditions. Thus, v ( is the same as in region I. Solutions of the form G(x t are traced back to the boundary at x =. Since from (4.3, u ( (x, t =v ( (x, t +v ( (x, t = F (x + t +G(x t, i.e. u ( is written in terms of v ( = G(x t, u ( requires a boundary condition to be defined on x =.. a u (, u ( prescribed on t =; u ( prescribed on x =. Since u ( (x, t= F (x + t+g(x t, u ( (x, t =F (x + t+g(x t, i.e. both u ( and u ( are written in terms of F (x + t and G(x t, we need to define two initial conditions for U (on t =. A boundary condition also needs to be prescribed on x = to be able to trace back v ( = G(x t. Well-posed. b u (, u ( prescribed on t =; u ( u ( prescribed on x =. As in part (a, we need to define two initial conditions for U. Since u ( u ( = F (x + t, its definition on x = leads to ill-posedness. On the contrary, u ( + u ( =G(x t should be defined on x = in order to be able to trace back the values through characteristics. Ill-posed. c u ( + u ( prescribed on t =; u (, u ( prescribed on x =. Since u ( + u ( =G(x t, another initial condition should be prescribed to be able to trace back solutions of the form v ( = F (x + t, without which the problem is ill-posed. Also, two boundary conditions for both u ( and u ( define solutions of both v ( = G(x t and v ( = F (x + t on the boundary. The former boundary condition leads to ill-posedness. Ill-posed.

120 Partial Differential Equations Igor Yanovsky, 5 Problem (F 9, #8. Consider the system u t + u x + av x = v t + bu x + v x = for <x< with boundary and initial conditions u = v = for x = u = u, v = v for t =. a For which values of a and b is this a well-posed problem? b For this class of a, b, state conditions on u and v so that the solution u, v will be continuous and continuously differentiable. Proof. a Let us change the notation (u u (,v u (. Rewrite the equation as ( a U t + U b x =, (4.3 U(x, = U(,t = ( u ( ( (x, u ( = (x, ( u ( (,t u ( =. (,t u ( (x u ( (x, The eigenvalues of the matrix A are λ = ab, λ =+ ab. ( ab Λ= +. ab Let U =ΓV, where Γ is a matrix of eigenvectors. Then, U t + AU x =, ΓV t + AΓV x =, V t +Γ AΓV x =, V t +ΛV x =. Thus, the transformed problem is ( ab V t + + V x =, (4.33 ab V (x, = Γ U(x,. The equation (4.33 gives traveling wave solutions of the form: v ( (x, t=f (x ( abt, v ( (x, t =G(x ( + abt. (4.34 We also have U =ΓV, i.e. both u ( and u ( (and their initial and boundary conditions are combinations of v ( and v (. In order for this problem to be well-posed, both sets of characteristics should emanate from the boundary at x =. Thus, the eigenvalues of the system are real (ab > and λ, > (ab <. Thus, <ab<. b For U to be C, we require the compatibility condition, u ( ( =, u( ( =.

121 Partial Differential Equations Igor Yanovsky, 5 Problem (F 93, #. Consider the initial-boundary value problem u t + u x = v t ( cx v x + u x = on x and t, with the following prescribed data: u(x,, v(x,, u(,t, v(,t. For which values of c is this a well-posed problem? Proof. Let us change the notation (u u (, v u (. The first equation can be solved with u ( (x, = F (x togetasolutionintheform u ( (x, t =F (x t, which requires u ( (x, and u ( (,t to be defined. With u ( known, we can solve the second equation u ( t ( cx u ( x + F (x t =. Solving the equation by characteristics, we obtain the characteristics in the xt-plane are of the form dx dt = cx. We need to determine c such that the prescribed data u ( (x, and u ( (,t makes the problem to be well-posed. The boundary condition for u ( (,t requires the characteristics to propagate to the left with t increasing. Thus, x(t is a decreasing function, i.e. dx dt < cx < for <x< c<. We could also do similar analysis we have done in other problems on first order systems involving finding eigenvalues/eigenvectors of the system and using the fact that u ( (x, t is known at both boundaries (i.e. values of u ( (,t can be traced back either to initial conditions or to boundary conditions on x =.

122 Partial Differential Equations Igor Yanovsky, 5 Problem (S 9, #4. Consider the first order system u t + au x + bv x = v t + cu x + dv x = for <x<, with prescribed initial data: u(x, = u (x v(x, = v (x. a Find conditions on a, b, c, d such that there is a full set of characteristics and, in this case, find the characteristic speeds. b For which values of a, b, c, d can boundary data be prescribed on x =and for which values can it be prescribed on x =? How many pieces of data can be prescribed on each boundary? Proof. a Let us change the notation (u u (,v u (. Rewrite the equation as ( a b U t + U c d x =, (4.35 ( u U(x, = ( ( (x, u ( u ( = (x (x, u ( (x. The system is hyperbolic if for each value of u ( and u ( the eigenvalues are real and the matrix is diagonalizable, i.e. there is a complete set of linearly independent eigenvectors. The eigenvalues of the matrix A are λ, = a + d ± (a + d 4(ad bc = a + d ± (a d +4bc. We need (a d +4bc >. This also makes the problem to be diagonalizable. Let U =ΓV, where Γ is a matrix of eigenvectors. Then, U t + AU x =, ΓV t + AΓV x =, V t +Γ AΓV x =, V t +ΛV x =. Thus, the transformed problem is ( λ V t + V λ x =, (4.36 Equation (4.36 gives traveling wave solutions of the form: v ( (x, t=f (x λ t, v ( (x, t =G(x λ t. (4.37 The characteristic speeds are dx dt = λ, dx dt = λ. b We assume (a + d 4(ad bc >. a + d>, ad bc > λ,λ > B.C.onx =. a + d>, ad bc < λ <,λ > B.C.onx =,B.C.onx =. a + d<, ad bc > λ,λ < B.C.onx =.

123 Partial Differential Equations Igor Yanovsky, 5 3 a + d<, ad bc < λ <,λ > B.C.onx =,B.C.onx =. a + d>, ad bc = λ =,λ > B.C.onx =. a + d<, ad bc = λ =,λ < B.C.onx =. a + d =, ad bc < λ <,λ > B.C.onx =,B.C.on x =.

124 Partial Differential Equations Igor Yanovsky, 5 4 Problem (S 94, #. Consider the differential operator ( ( u ut +9v L = x u xx v v t u x v xx ( u(x, t on x π, t, in which the vector consists of two functions that v(x, t are periodic in x. a Find the eigenfunctions and eigenvalues of the operator L. b Use the results of (a to solve the initial value problem ( u L = for t, v ( ( u e ix = for t =. v Proof. a We find the space eigenvalues and eigenfunctions. We rewrite the system as ( ( 9 U t + U x + U xx =, and find eigenvalues ( ( 9 U x + ( ( u(x, t n= Set U = = n= u n(te inx v(x, t n= n= v n(te inx ( 9 ( inun (te inx invn (te inx + U xx = λu. (4.38 ( ( ( ( ( 9 inun (t n + u n (t inv n (t n v n (t ( ( ( ( 9in un (t n un (t + in v n (t n v n (t ( n ( λ 9in un (t in n λ v n (t (n λ 9n =, which gives λ = n +3n, λ = n 3n, are eigenvalues, and v = are corresponding eigenvectors.. Plugging this into (4.38, we get ( n u n (te inx ( un (te n v n (te inx = λ inx vn (te inx, ( un (t = λ v n (t ( un (t = λ v n (t =, ( 3i,, ( 3i, v =,

125 Partial Differential Equations Igor Yanovsky, 5 5 ( u b We want to solve v ( ( ( t u u u = L = λ v v v 6 7 t U(x, t = = U(x, = ( u + L v, i.e. ( u =, L v ( u v ( un (te inx vn (te inx = n= n= n= ( a n e (n +3nt 3i a n ( 3i a n = b n =,n ; e inx + b n ( 3i ( 9vx u = xx u x v xx. We have = e λt. We can write the solution as a n e λ t v e inx + b n e λ t v e inx e inx + b n e (n 3nt ( e inx e ix = ( 3i, a + b = and a = b a = b = 3i 6i. U(x, t = ( 3i 6i e 4t e ix + ( 3i 6i et e ix ( (e = 4t + e t 6i (e 4t e t e ix. e inx. 6 ChiuYen s and Sung-Ha s solutions give similar answers. 7 Questions about this problem:. Needed to find eigenfunctions, not eigenvectors.. The notation of L was changed. The problem statement incorporates the derivatives wrt. t into L. 3. Why can we write the solution in this form above?

126 Partial Differential Equations Igor Yanovsky, 5 6 Problem (W 4, #6. Consider the first order system u t u x = v t + v x = in the diamond shaped region <x+ t<, <x t<. For each of the following boundary value problems state whether this problem is well-posed. If it is well-posed, find the solution. a u(x + t =u (x + t on x t =, v(x t =v (x t on x + t =. b v(x + t =v (x + t on x t =, u(x t =u (x t on x + t =. Proof. We have u t u x =, v t + v x =. u is constant along the characteristics: x + t = c (s. Thus, its solution is u(x, t=u (x + t. It the initial condition is prescribed at x t =, the solution can be determined in the entire region by tracing back through the characteristics. v is constant along the characteristics: x t = c (s. Thus, its solution is v(x, t=v (x t. It the initial condition is prescribed at x + t =, the solution can be determined in the entire region by tracing forward through the characteristics.

127 Partial Differential Equations Igor Yanovsky, Problems: Gas Dynamics Systems 5. Perturbation Problem (S 9, # Consider the gas dynamic equations u t + uu x +(F (ρ x =, ρ t +(uρ x =. Here F (ρ is a given C -smooth function of ρ. Att =, π-periodic initial data u(x, = f(x, ρ(x, = g(x. a Assume that f(x =U + εf (x, g(x=r + εg (x where U,R > are constants and εf (x, εg (x are small perturbations. Linearize the equations and given conditions for F such that the linearized problem is well-posed. b Assume that U > and consider the above linearized equations for x, t. Construct boundary conditions such that the initial-boundary value problem is well-posed. Proof. a We write the equations in characteristic form: u t + uu x + F (ρρ x =, ρ t + u x ρ + uρ x =. Consider the special case of nearly constant initial data u(x, = u + εu (x,, ρ(x, = ρ + ερ (x,. Then we can approximate nonlinear equations by linear equations. Assuming u(x, t=u + εu (x, t, ρ(x, t=ρ + ερ (x, t remain valid with u = O(, ρ = O(, we find that u t = εu t, ρ t = ερ t, u x = εu x, ρ x = ερ x, F (ρ=f (ρ + ερ (x, t = F (ρ +ερ F (ρ +O(ε. Plugging these into, gives εu t +(u + εu εu x + ( F (ρ +ερ F (ρ +O(ε ερ x =, ερ t + εu x (ρ + ερ +(u + εu ερ x =. Dividing by ε gives u t + u u x + F (ρ ρ x = εu u x ερ ρ x F (ρ +O(ε, ρ t + u x ρ + u ρ x = εu x ρ εu ρ x. 8 See LeVeque, Second Edition, Birkhäuser Verlag, 99, p This problem has similar notation with S 9, #4.

128 Partial Differential Equations Igor Yanovsky, 5 8 For small ε, wehave { u t + u u x + F (ρ ρ x =, ρ t + u x ρ + u ρ x =. This can be written as ( ( u u F + ( ( (ρ u =. ρ ρ t u ρ x u λ F (ρ ρ u λ =(u λ(u λ ρ F (ρ =, λ u λ + u ρ F (ρ =, λ, = u ± ρ F (ρ, u >, ρ >. For well-posedness, need λ, R or F (ρ. b We have u >, and λ = u + ρ F (ρ, λ = u ρ F (ρ. If u > ρ F (ρ λ >,λ > BCatx =. If u = ρ F (ρ λ >,λ = BCatx =. If <u < ρ F (ρ λ >,λ < BCatx =,BCatx =. 5. Stationary Solutions Problem (S 9, #4. 3 Consider u t + uu x + ρ x = νu xx, ρ t +(uρ x = for t, <x<. Give conditions for the states U +,U,R +,R, such that the system has stationary solutions (i.e. u t = ρ t = satisfying lim x + ( u ρ = ( U+ R +, lim x ( u ρ = ( U R. Proof. For stationary solutions, we need ( u u t = ρ x + νu xx =, x ρ t = (uρ x =. Integrating the above equations, we obtain u ρ + νu x = C, uρ = C. 3 This problem has similar notation with S 9, #3.

129 Partial Differential Equations Igor Yanovsky, 5 9 Conditions give u x =atx = ±. Thus U + + R + = U + R, U + R + = U R.

130 Partial Differential Equations Igor Yanovsky, Periodic Solutions Problem (F 94, #4. Let u(x, t be a solution of the Cauchy problem u t = u xxxx u xx, u(x, = ϕ(x, <x<+, <t<+, where u(x, t and ϕ(x are C functions periodic in x with period π; i.e. u(x +π, t=u(x, t, x, t. Prove that u(,t Ce at ϕ where u(,t = π u(x, t dx, ϕ = π ϕ(x dx, C, a are some constants. Proof. METHOD I: Since u is π-periodic, let u(x, t= a n (te inx. n= Plugging this into the equation, we get a n (teinx = n 4 a n (te inx + n= n= a n(t = ( n 4 +n a n (t, a n (t = a n (e ( n4 +n t. Also, initial condition gives u(x, = a n (e inx = ϕ(x, u(x, t = = n= π n= n= π a n (e inx = ϕ(x. u (x, t dx = π ( n= π a n (t e inx e inx dx =π n= = C e t ϕ. a n ( n= e ( n4 +n t n= a n (te inx ( n= n a n (te inx, a n =π m= (t =π n= a n (te imx dx n= a n ( } {{ } ϕ e t a n (e( n4 +n t n= e (n t } {{ } = C,(convergent u(x, t Ce t ϕ.

131 Partial Differential Equations Igor Yanovsky, 5 3 METHOD II: Multiply this equation by u and integrate uu t = uu xxxx uu xx, d dt (u = uu xxxx uu xx, d π π π u dx = uu xxxx dx uu xx dx, dt d π π π π dt u = uu xxx + u x u xx u xx dx uu xx dx, } {{ } } {{ } d dt u = = π π u xx dx u xx dx + = π π d dt u u, u u( e t, u u( e t. uu xx dx ( ab a + b (u + u xx dx = π u dx = u, METHOD III: Can use Fourier transform. See ChiuYen s solutions, that have both Method II and III.

132 Partial Differential Equations Igor Yanovsky, 5 3 Problem (S 9, #4. Let f(x C be a π-periodic function, i.e., f(x =f(x +π and denote by f = π f(x dx the L -norm of f. a Express d p f/dx p in terms of the Fourier coefficients of f. b Let q>p> be integers. Prove that ɛ>, K = N(ε, p, q, constant, such that dp f d q f ε + K f dx p dx q. c Discuss how K depends on ε. Proof. a Let 3 f(x = d p f dx p = f dx p = = f n e inx, f n (in p e inx, π π f n (in p e inx dx = π f n n p e inx dx = π fn np. n= b We have dp f dx p ε dq f + K f dx q, π fn np ε π fn nq + K π fn, n= n= n p εn q K, n p ( εn q } {{ } K, some q >. <, fornlarge n= i p f n n p e inx dx Thus, the above inequality is true for n large enough. The statement follows. 3 Note: L e inx e imx dx = { n m L n = m

133 Partial Differential Equations Igor Yanovsky, 5 33 Problem (S 9, #5. 3 Consider the flame front equation u t + uu x + u xx + u xxxx = with π-periodic initial data u(x, = f(x, f(x =f(x +π C. a Determine the solution, if f(x f = const. b Assume that f(x =+εg(x, <ε, g =, g(x=g(x +π. Linearize the equation. Is the Cauchy problem well-posed for the linearized equation, i.e., do its solutions v satisfy an estimate v(,t Ke α(t t v(,t? c Determine the best possible constants K, α. Proof. a The solution to u t + uu x + u xx + u xxxx =, u(x, = f =const, is u(x, t=f =const. b We consider the special case of nearly constant initial data u(x, = + εu (x,. Then we can approximate the nonlinear equation by a linear equation. Assuming u(x, t=+εu (x, t, remain valid with u = O(, from, we find that εu t +(+εu εu x + εu xx + εu xxxx =. Dividing by ε gives u t + u x + εu u x + u xx + u xxxx =. For small ε, wehave u t + u x + u xx + u xxxx =. Multiply this equation by u and integrate d d dt dt u + u π u x } {{ } = π π 3 S 9 #5, #6, #7 all have similar formulations. u u t + u u x + u u xx + u u xxxx =, d ( u ( u + dt + u u xx + u u xxxx =, x u dx + u π π π + u u xx dx + u u xxxx dx =, } {{ } = u x dx + u π π u xxx u x u xx + } {{ } } {{ } = = d π dt u = π u x dx u xx dx =, π u xx dx.

134 Partial Differential Equations Igor Yanovsky, 5 34 Since u is π-periodic, let u = a n (te inx. Then, u x = i u xx = n= n= Thus, d dt u = ( na n (te inx u x = n= ( n a n (te inx u xx = n= n= = π π π u x dx u xx dx na n (te inx, n a n (te inx. ( nan (te inx π ( dx n a n (te inx dx = π n a n (t π n 4 a n (t = π a n (t (n + n 4. u (,t u (,, where K =,α =. Problem (W 3, #4. Consider the PDE u t = u x + u 4 for t> u = u for t = for <x<π. Define the set A = {u = u(x : û(k = if k < }, inwhich {û(k, t} is the Fourier series of u in x on [, π]. a If u A, show that u(t A. b Find differential equations for û(,t, û(,t, andû(,t. Proof. a Solving u t = u x + u 4 u(x, = u (x by the method of characteristics, we get u (x + t u(x, t=. ( 3t(u (x + t 3 3 Since u A, û k =ifk<. Thus, u (x = û k e i kx. Since û k = π k= π kx i u(x, t e dx,

135 Partial Differential Equations Igor Yanovsky, 5 35 we have u(x, t = û k e i kx, k= that is, u(t A.

136 Partial Differential Equations Igor Yanovsky, Energy Estimates Problem (S 9, #6. Let U(x, t C be π-periodic in x. Consider the linear equation u t + Uu x + u xx + u xxxx =, u(x, = f(x, f(x =f(x +π C. a Derive an energy estimate for u. b Prove that one can estimate all derivatives p u/ x p. c Indicate how to prove existence of solutions. 33 Proof. a Multiply the equation by u and integrate uu t + Uuu x + uu xx + uu xxxx =, d dt (u + U(u x + uu xx + uu xxxx =, d π u dx + π π π U(u x dx + uu xx dx + uu xxxx dx =, dt d dt u + π Uu π U x u π π dx + uu x } {{ } u x dx = π +uu xxx u π π xu xx + u xx dx =, d dt u π π π U x u dx u x dx + u xx dx =, d dt u = π π π U x u dx + u x dx u xx dx (from S 9, #5 π U x u dx π max U x u dx. x d dt u max U x u, x u(x, t u(x, e (maxx Uxt. This can also been done using Fourier Transform. See ChiuYen s solutions where the above method and the Fourier Transform methods are used. 33 S 9 #5, #6, #7 all have similar formulations.

137 Partial Differential Equations Igor Yanovsky, 5 37 Problem (S 9, #7. 34 Consider the nonlinear equation u t + uu x + u xx + u xxxx =, u(x, = f(x, f(x =f(x +π C. a Derive an energy estimate for u. b Show that there is an interval t T, T depending on f, such that also u(,t/ x can be bounded. Proof. a Multiply the above equation by u and integrate uu t + u u x + uu xx + uu xxxx =, d dt (u + 3 (u3 x + uu xx + uu xxxx =, d π u dx + π π (u 3 x dx + uu xx dx + dt 3 d dt u + 3 u3 π π π u x dx + u xx dx =, } {{ } = π d dt u = u x dx u(,t u(,. π π u xx dx, (from S 9, #5 uu xxxx dx =, b In order to find a bound for u x (,t, differentiate with respect to x: u tx +(uu x x + u xxx + u xxxxx =, Multiply the above equation by u x and integrate: u x u tx + u x (uu x x + u x u xxx + u x u xxxxx =, d dt π (u x dx + π u x (uu x x dx + π u x u xxx dx + π u x u xxxxx dx =. We evaluate one of the integrals in the above expression using the periodicity: π u x (uu x x dx = We have d dt u x + π π π π u xx uu x = u xx uu x = u x (uu x x = u 3 x dx + π 34 S 9 #5, #6, #7 all have similar formulations. π π π u x (u x + uu xx = u 3 x, u 3 x. u x u xxx dx + π π u x u xxxxx dx =. π u 3 x + uu x u xx,

138 Partial Differential Equations Igor Yanovsky, 5 38 Let w = u x,then d dt w = = π π w 3 dx + d dt u x = π w 3 dx π π w x dx u 3 x dx. π ww xx dx π w xx dx ww xxxx dx π w 3 dx,

139 Partial Differential Equations Igor Yanovsky, Problems: Wave Equation 6. The Initial Value Problem Example (McOwen 3. #. Solve the initial value problem: u tt c u xx =, u(x, = x 3, u t (x, = sin x. }{{} g(x }{{} h(x Proof. D Alembert s formula gives the solution: u(x, t = (g(x + ct+g(x ct + c = (x + ct3 + (x ct3 + c x+ct x ct x+ct x ct h(ξ dξ sin ξdξ = x 3 +xc t c cos(x + ct+ cos(x ct = c = x 3 +xc t + sin x sin ct. c Problem (S 99, #6. Solve the Cauchy problem { u tt = a u xx +cosx, u(x, = sin x, u t (x, = + x. (6. Proof. We have a nonhomogeneous PDE with nonhomogeneous initial conditions: u tt c u xx =cosx } {{ }, f(x,t u(x, = sin x }{{}, u t (x, = } {{ + x }. g(x h(x The solution is given by d Alembert s formula and Duhamel s principle. 35 u A (x, t = (g(x + ct+g(x ct + c u D (x, t = c = c t x c(t s x+ct x ct x+ct h(ξ dξ = (sin(x + ct+sin(x ct + ( + ξ dξ c x ct = sinx cos ct + [ξ + ξ ] ξ=x+ct =sinx cos ct + xt + t. c ξ=x ct t ( x+c(t s f(ξ, s dξ ds = t ( x+c(t s c ( sin[x + c(t s] sin[x c(t s] x c(t s cos ξdξ ds ds = (cos x cos x cos ct. c u(x, t = u A (x, t+u D (x, t =sinx cos ct + xt + t + (cos x cos x cos ct. c 35 Note the relationship: x ξ, t s.

140 Partial Differential Equations Igor Yanovsky, 5 4 We can check that the solution satisfies equation (6.. Can also check that u A, u D satisfy { { u A tt c u A xx =, u D tt c u D xx =cosx, u A (x, = sin x, u A t (x, = + x; u D (x, =, u D t (x, =.

141 Partial Differential Equations Igor Yanovsky, Initial/Boundary Value Problem Problem. Consider the initial/boundary value problem u tt c u xx = <x<l, t> u(x, = g(x, u t (x, = h(x <x<l u(,t=, u(l, t= t. (6. Proof. Find u(x, t intheform u(x, t= a (t + a n (tcos nπx L n= + b n(tsin nπx L. Functions a n (t andb n (t are determined by the boundary conditions: =u(,t= a (t + a n (t a n (t =. Thus, u(x, t= n= n= b n (tsin nπx L. (6.3 If we substitute (6.3 into the equation u tt c u xx =, weget ( nπ bn b n (tsinnπx L + c (tsin nπx =, or L L n= n= ( nπc bn b n(t+ (t =, L whose general solution is b n (t =c n sin nπct L Also, b n(t =c n ( nπc + d n cos nπct L. (6.4 L cosnπct L d n( nπc L sinnπct L. The constants c n and d n are determined by the initial conditions: g(x =u(x, = b n ( sin nπx L = d n sin nπx L, h(x =u t (x, = L n= n= b nπx n ( sin L n= = n= c n nπc L sin nπx L. By orthogonality, we may multiply by sin(mπx/l and integrate: L g(xsin mπx L L dx = d n sin nπx L Thus, d n = L h(xsin mπx L dx = L n= L n= c n nπc L g(xsin nπx L dx, c n = nπc mπx sin L dx = d L m, nπx sin L L The formulas (6.3, (6.4, and (6.5 define the solution. sin mπx L dx = c m h(xsin nπx L mπc L L. dx. (6.5

142 Partial Differential Equations Igor Yanovsky, 5 4 Example (McOwen 3. #. Consider the initial/boundary value problem u tt u xx = <x<π, t> u(x, =, u t (x, = <x<π u(,t=, u(π, t= t. (6.6 Proof. Find u(x, t intheform u(x, t= a (t + a n (tcosnx + b n (tsinnx. n= Functions a n (t andb n (t are determined by the boundary conditions: =u(,t= a (t + a n (t a n (t =. Thus, u(x, t= n= b n (tsinnx. (6.7 n= If we substitute this into u tt u xx =, weget b n(tsinnx + b n (tn sin nx =, n= n= b n(t+n b n (t =, whose general solution is b n (t =c n sin nt + d n cos nt. (6.8 Also, b n(t =nc n cos nt nd n sin nt. The constants c n and d n are determined by the initial conditions: =u(x, = b n ( sinnx = d n sin nx, =u t (x, = n= n= or b n ( sinnx = nc n sin nx. n= By orthogonality, we may multiply both equations by sin mx and integrate: π π sin mx dx = d m, π π dx = nc n. Thus, n= d n = ( cos nπ = nπ { 4 nπ, n odd,, n even, and c n =. (6.9 Using this in (6.8 and (6.7, we get { 4 b n (t = nπ cos nt, n odd,, n even,

143 Partial Differential Equations Igor Yanovsky, 5 43 u(x, t = 4 π n= cos(n +t sin(n +x. (n +

144 Partial Differential Equations Igor Yanovsky, 5 44 We can sum the series in regions bouded by characteristics. Wehave u(x, t= 4 cos(n +t sin(n +x, or π (n + n= u(x, t= sin[(n +(x + t] + sin[(n +(x t]. π (n + π (n + n= n= (6. The initial condition may be written as =u(x, = 4 sin(n +x π (n + for <x<π. (6. n= We can use (6. to sum the series in (6.. In R, u(x, t = + =. Since sin[(n +(x t] = sin[(n +( (x t], and < (x t <πin R, in R, u(x, t = =. Since sin[(n +(x + t] = sin[(n +(x + t π] = sin[(n + (π (x + t], and < π (x + t <πin R 3, in R 3, u(x, t = + =. Since < (x t <π and < π (x + t <π in R 4, in R 4, u(x, t = =.

145 Partial Differential Equations Igor Yanovsky, 5 45 Problem. Consider the initial/boundary value problem u tt c u xx = <x<l, t> u(x, = g(x, u t (x, = h(x <x<l u x (,t=, u x (L, t = t. Proof. Find u(x, t intheform u(x, t= a (t + a n (tcos nπx L n= + b n(tsin nπx L. Functions a n (t andb n (t are determined by the boundary conditions: ( nπ u x (x, t= a n (t sin nπx nπ n(t( L L + b cos nπx L L, n= ( nπ =u x (,t= b n (t b n (t =. Thus, L u(x, t= a (t n= + n= (6. a n (tcos nπx L. (6.3 If we substitute (6.3 into the equation u tt c u xx =, weget a (t + a n(tcos nπx ( nπ L + nπx c a n (t cos L L =, n= n= ( nπc an a (t = and a n (t+ (t =, L whose general solutions are a (t =c t + d and a n (t =c n sin nπct L Also, a (t =c and a n(t =c n ( nπc d n( nπc L cosnπct L + d n cos nπct L. (6.4 L sinnπct L. The constants c n and d n are determined by the initial conditions: g(x =u(x, = a ( + a n ( cos nπx L = d + n= n= h(x =u t (x, = a ( + a nπx n ( cos L = c + nπc c n L n= n= d n cos nπx L, cos nπx L. By orthogonality, we may multiply both equations by cos(mπx/l, including m =, and integrate: Thus, L L d n = L g(x dx = d L, h(x dx = c L, L L L g(xcos mπx L h(xcos mπx L g(xcos nπx L dx, c n = nπc L dx = d L m, dx = c m mπc L L. h(xcos nπx L dx, c = L The formulas (6.3, (6.4, and (6.5 define the solution. L h(x dx. (6.5

146 Partial Differential Equations Igor Yanovsky, 5 46 Example (McOwen 3. #3. Consider the initial/boundary value problem u tt u xx = <x<π, t> u(x, = x, u t (x, = <x<π u x (,t=, u x (π, t= t. (6.6 Proof. Find u(x, t intheform u(x, t= a (t + a n (tcosnx + b n (tsinnx. n= Functions a n (t andb n (t are determined by the boundary conditions: u x (x, t= a n (tn sin nx + b n (tn cos nx, n= =u x (,t= u(x, t= a (t + b n (tn b n (t =. Thus, n= a n (tcosnx. (6.7 n= If we substitute (6.7 into the equation u tt u xx =, weget a (t + a n(tcosnx + a n (tn cos nx =, n= n= a (t = and a n (t+n a n (t =, whose general solutions are a (t =c t + d and a n (t =c n sin nt + d n cos nt. (6.8 Also, a (t =c and a n(t =c n n cos nt d n n sin nt. The constants c n and d n are determined by the initial conditions: x = u(x, = a ( + a n ( cos nx = d + d n cos nx, n= n= =u t (x, = a ( + a n( cos nx = c + c n n cos nx. n= By orthogonality, we may multiply both equations by cos mx, including m =,and integrate: Thus, π π xdx= d π, dx = c π, π π x cos mx dx = d m π, cosmx dx = c m m π. d = π, d n = πn (cos nπ, c n =. (6.9 Using this in (6.8 and (6.7, we get n= a (t =d = π, a n (t = (cos nπ cos nt, πn

147 Partial Differential Equations Igor Yanovsky, 5 47 u(x, t = π + π n= (cos nπ cos nt cos nx n.

148 Partial Differential Equations Igor Yanovsky, 5 48 We can sum the series in regions bouded by characteristics. Wehave u(x, t= π + (cos nπ cos nt cos nx π n, n= or u(x, t= π + (cos nπ cos[n(x t] π n + (cos nπ cos[n(x + t] π n. (6. n= n= The initial condition may be written as u(x, = x = π + (cos nπ cos nx π n n= for <x<π, which implies x π 4 = (cos nπ cos nx π n n= for <x<π, (6. We can use (6. to sum the series in (6.. In R, u(x, t = π + x t π 4 + x + t π 4 = x. Since cos[n(x t] = cos[n( (x t], and < (x t <πin R, in R, u(x, t = π (x t + π 4 + x + t π 4 = t. Since cos[n(x+t] = cos[n(x+t π] = cos[n(π (x+t], and < π (x+t < π in R 3, in R 3, u(x, t = π + x t π π (x + t + π 4 4 = π t. Since < (x t <πand < π (x + t <πin R 4 in R 4, u(x, t = π + (x t π 4 + π (x + t π 4 = π x.

149 Partial Differential Equations Igor Yanovsky, 5 49 Example (McOwen 3. #4. Consider the initial boundary value problem u tt c u xx = for x>, t> u(x, = g(x, u t (x, = h(x for x> u(,t= for t, (6. where g( = = h(. Ifweextendg and h as odd functions on <x<, show that d Alembert s formula gives the solution. Proof. Extend g and h as odd functions on <x< : { { g(x, x h(x, x g(x = h(x = g( x, x < h( x, x <. Then, we need to solve { ũ tt c ũ xx = for <x<, t> ũ(x, = g(x, ũ t (x, = h(x for <x<. (6.3 To show that d Alembert s formula gives the solution to (6.3, we need to show that the solution given by d Alembert s formula satisfies the boundary condition ũ(,t=. ũ(x, t = ( g(x + ct+ g(x ct + c ũ(,t = ( g(ct+ g( ct + c ct ct x+ct x ct h(ξ dξ h(ξ dξ, = ( g(ct g(ct + (H(ct H( ct c = + (H(ct H(ct =, c whereweusedh(x = x h(ξ dξ; and since h is odd, thenh is even. Example (McOwen 3. #5. Find in closed form (similar to d Alembet s formula the solution u(x, t of u tt c u xx = for x, t > u(x, = g(x, u t (x, = h(x for x> (6.4 u(,t=α(t for t, where g, h, α C satisfy α( = g(, α ( = h(, andα ( = c g (. Verify that u C, even on the characteristic x = ct. Proof. As in (McOwen 3. #4, we can extend g and h to be odd functions. We want to transform the problem to have zero boundary conditions. Consider the function: U(x, t=u(x, t α(t. (6.5

150 Partial Differential Equations Igor Yanovsky, 5 5 Then (6.4 transforms to: U tt c U xx = α (t } {{ } f U (x,t U(x, = g(x α(, U } {{ } t (x, = h(x α ( } {{ } g U (x h U (x U(,t=. }{{} α u(t We use d Alembert s formula and Duhamel s principle on U. After getting U, we can get u from u(x, t =U(x, t+α(t.

151 Partial Differential Equations Igor Yanovsky, 5 5 Example (Zachmanoglou, Chapter 8, Example 7.. Find the solution of u tt c u xx = for x>,t> u(x, = g(x, u t (x, = h(x for x> (6.6 u x (,t= for t>. Proof. Extend g and h as even functions on <x< : { { g(x, x h(x, x g(x = h(x = g( x, x < h( x, x <. Then, we need to solve { ũ tt c ũ xx = for <x<, t> ũ(x, = g(x, ũ t (x, = h(x for <x<. (6.7 To show that d Alembert s formula gives the solution to (6.7, we need to show that the solution given by d Alembert s formula satisfies the boundary condition ũ x (,t=. ũ(x, t = ( g(x + ct+ g(x ct + c x+ct x ct h(ξ dξ. ũ x (x, t = ( g (x + ct+ g (x ct + c [ h(x + ct h(x ct], ũ x (,t = ( g (ct+ g ( ct + c [ h(ct h( ct] =. Since g is even, theng is odd. Problem (F 89, #3. 36 Let α c, constant. Find the solution of u tt c u xx = for x>,t> u(x, = g(x, u t (x, = h(x for x> u t (,t=αu x (,t for t>, (6.8 where g, h C for x> and vanish near x =. Hint: Use the fact that a general solution of (6.8 can be written as the sum of two traveling wave solutions. Proof. D Alembert s formula is derived by plugging in the following into the above equation and initial conditions: u(x, t=f (x + ct+g(x ct. As in (Zachmanoglou 7., we can extend g and h to be even functions. 36 Similar to McOwen 3. #5. The notation in this problem is changed to be consistent with McOwen.

152 Partial Differential Equations Igor Yanovsky, 5 5 Example (McOwen 3. #6. Solve the initial/boundary value problem u tt u xx = for <x<π and t> u(x, =, u t (x, = for <x<π u(,t=, u(π, t= π / for t. (6.9 Proof. If we first find a particular solution of the nonhomogeneous equation, this reduces the problem to a boundary value problem for the homogeneous equation ( as in (McOwen 3. # and (McOwen 3. #3. Hint: You should use a particular solution depending on x! ❶ First, find a particular solution. This is similar to the method of separation of variables. Assume which gives u p (x, t=x(x, X (x =, X (x =. The solution to the above ODE is X(x = x + ax + b. The boundary conditions give u p (,t = b =, u p (π, t = π + aπ + b = π, a = b =. Thus, the particular solution is u p (x, t= x. This solution satisfies the following: u p tt u pxx = u p (x, = x, u pt (x, = u p (,t=, u p (π, t = π. ❷ Second, we find a solution to a boundary value problem for the homogeneous equation: u tt u xx = u(x, = x, u t(x, = u(,t=, u(π, t=. This is solved by the method of Separation of Variables. See Separation of Variables subsection of Problems: Separation of Variables: Wave Equation McOwen 3. #. The only difference there is that u(x, =. We would find u h (x, t. Then, u(x, t =u h (x, t+u p (x, t.

153 Partial Differential Equations Igor Yanovsky, 5 53 Problem (S, #. a Given a continuous function f on R which vanishes for x >R, solve the initial value problem { u tt u xx = f(xcost, u(x, =, u t (x, =, <x<, t< by first finding a particular solution by separation of variables and then adding the appropriate solution of the homogeneous PDE. b Since the particular solution is not unique, it will not be obvious that the solution to the initial value problem that you have found in part (a is unique. Prove that it is unique. Proof. a ❶ First, find a particular solution by separation of variables. Assume which gives u p (x, t=x(xcost, X(xcost X (xcost = f(xcost, X + X = f(x. The solution to the above ODE is written as X = X h +X p. The homogeneous solution is X h (x = a cos x + b sin x. To find a particular solution, note that since f is continuous, G C (R, such that Thus, G + G = f(x. X p (x =G(x. X(x=X h (x+x p (x =a cos x + b sin x + G(x. u p (x, t= [ a cos x + b sinx + G(x ] cos t. It can be verified that this solution satisfies the following: { u p tt u pxx = f(xcost, u p (x, = a cos x + b sinx + G(x, u p t (x, =. ❷ Second, we find a solution of the homogeneous PDE: u tt u xx =, u(x, = a cos x b sin x G(x, } {{ } u t (x, = g(x }{{} h(x The solution is given by d Alembert s formula (with c =: x+t u h (x, t = u A (x, t = (g(x + t+g(x t + h(ξ dξ x t = ( ( ( a cos(x + t b sin(x + t G(x + t + a cos(x t b sin(x t G(x t = ( ( a cos(x + t+b sin(x + t+g(x + t a cos(x t+b sin(x t+g(x t..

154 Partial Differential Equations Igor Yanovsky, 5 54 It can be verified that the solution satisfies the above homogeneous PDE with the boundary conditions. Thus, the complete solution is: u(x, t =u h (x, t+u p (x, t. Alternatively, we could use Duhamel s principle to find the solution: 37 u(x, t = t ( x+(t s f(ξ cossdξ ds. x (t s However, this is not how it was suggested to do this problem. b The particular solution is not unique, since any constants a, b give the solution. However, we show that the solution to the initial value problem is unique. Suppose u and u are two solutions. Then w = u u satisfies: { w tt w xx =, w(x, =, w t (x, =. D Alembert s formula gives x+t w(x, t = (g(x + t+g(x t + h(ξ dξ =. x t Thus, the solution to the initial value problem is unique. 37 Note the relationship: x ξ, t s.

155 Partial Differential Equations Igor Yanovsky, Similarity Solutions Problem (F 98, #7. Look for a similarity solution of the form v(x, t=t α w(y = x/t β for the differential equation v t = v xx +(v x. (6.3 a Find the parameters α and β. b Find a differential equation for w(y and show that this ODE can be reduced to first order. c Find a solution for the resulting first order ODE. Proof. We can rewrite (6.3 as v t = v xx +vv x. (6.3 We look for a similarity solution of the form v(x, t=t α w(y, (y = x t β. v t = αt α w + t α w y t = αt α w + t α( βx t β+ w = αt α w t α βyw, v x = t α w y x = t α w t β = t α β w, v xx = (t α β w x = t α β w y x = t α β w t β = t α β w. Plugging in the derivatives we calculated into (6.3, we obtain αt α w t α βyw = t α β w +(t α w(t α β w, αw βyw = t β w +t α β+ ww. The parameters that would eliminate t from equation above are β =, α =. With these parameters, we obtain the differential equation for w(y: w yw = w +ww, w +ww + yw + w =. We can write the ODE as w +ww + (yw =. Integrating it with respect to y, we obtain the first order ODE: w + w + yw = c.

156 Partial Differential Equations Igor Yanovsky, Traveling Wave Solutions Consider the Korteweg-de Vries (KdVequationintheform 38 u t +6uu x + u xxx =, <x<, t >. (6.3 We look for a traveling wave solution u(x, t=f(x ct. (6.33 We get the ODE cf +6ff + f =. (6.34 We integrate (6.34 to get cf +3f + f = a, (6.35 where a is a constant. Multiplying this equality by f,weobtain cff +3f f + f f = af. Integrating again, we get c f + f 3 + (f = af + b. (6.36 We are looking for solutions f which satisfy f(x, f (x, f (x asx ±. (In which case the function u having the form (6.33 is called a solitary wave. Then (6.35 and (6.36 imply a = b =,sothat c f + f 3 + (f =, or f = ±f c f. The solution of this ODE is f(x = c c sech [ (x x ], where x is the constant of integration. A solution of this form is called a soliton. 38 Evans, p. 74; Strauss, p. 367.

157 Partial Differential Equations Igor Yanovsky, 5 57 Problem (S 93, #6. The generalized KdV equation is u t = u (n +(n +un x 3 u x 3, where n is a positive integer. Solitary wave solutions are sought in which u = f(η, where η = x ct and f, f,f, as η ; c, the wave speed, is constant. Show that f = f n+ + cf. Hence show that solitary waves do not exist if n is even. Show also that, when n =, all conditions of the problem are satisfied provided c> and u = csech [ ] c(x ct. Proof. We look for a traveling wave solution u(x, t=f(x ct. We get the ODE cf = (n +(n +f n f f, Integrating this equation, we get cf = (n +f n+ f + a, (6.37 where a is a constant. Multiplying this equality by f,weobtain cff = (n +f n+ f f f + af. Integrating again, we get cf = f n+ (f + af + b. (6.38 We are looking for solutions f which satisfy f, f,f asx ±. Then (6.37 and (6.38 imply a = b =,sothat cf = f n+ (f, (f = f n+ + cf. We show that solitary waves do not exist if n is even. f = ± f n+ + cf = ± f f n + c, f dη = ± f f n + cdη, f = ± f f n + cdη, = ± f f n + cdη. We have

158 Partial Differential Equations Igor Yanovsky, 5 58 Thus, either ➀ f f =,or ➁ f n + c =. Sincef asx ±,wehavec = f =. Thus, solitary waves do not exist if n is even.

159 Partial Differential Equations Igor Yanovsky, 5 59 When n =,wehave (f = f 3 + cf. (6.39 We show that all conditions of the problem are satisfied provided c>, including u = csech [ ] c(x ct, or ] c ]. f = csech [ η c = cosh [ η c ] = c cosh We have [ η c ] 3 [ η c ] c f = c cosh sinh [ c 3 sinh η ] c (f = [ cosh 6 η ], c f 3 = cf = c 3 cosh 6 [ η c c 3 cosh 4 [ η c ]. ], Plugging these into (6.39, we obtain: 39 [ c 3 sinh η ] c [ c cosh 6 η ] = [ 3 c c cosh 6 η ] + [ 3 c cosh 4 η ], c [ c 3 sinh η ] [ c [ c 3 + c 3 cosh η ] c cosh 6 η ] = [ c cosh 6 η ], c [ c 3 sinh η ] [ c [ c 3 sinh η ] c cosh 6 η ] = [ c cosh 6 η ]. c [ η c = c [ η c ] 3 [ η c ] c cosh sinh, Also, f, f,f, as η, since f(η = csech [ η c ] ( c = cosh [ η c ] = c e [ η c, as η. ] + e [ η c ] Similarly, f,f, as η. 39 cosh x sinh x =. cosh x = ex + e x, sinh x = ex e x

160 Partial Differential Equations Igor Yanovsky, 5 6 Problem (S, #5. Look for a traveling wave solution of the PDE u tt +(u xx = u xxxx of the form u(x, t =v(x ct. In particular, you should find an ODE for v. Under the assumption that v goes to a constant as x, describe the form of the solution. Proof. Since (u x =uu x, and (u xx =u x +uu xx, we have u tt +u x +uu xx = u xxxx. We look for a traveling wave solution u(x, t=v(x ct. We get the ODE c v +(v +vv = v, c v +((v + vv = v, c v +(vv = v, (exact differentials c v +vv = v + a, s = x ct c v + v = v + as + b, v + c v + v = a(x ct+b. Since v C = const as x,wehavev, v, as x.thus, implies c v + v = as + b. Since x, but v C, wehavea =: v + c v b =. v = c ± c 4 +4b.

161 Partial Differential Equations Igor Yanovsky, 5 6 Problem (S 95, #. Consider the KdV-Burgers equation u t + uu x = ɛu xx + δu xxx in which ɛ>, δ>. a Find an ODE for traveling wave solutions of the form u(x, t=ϕ(x st with s> and lim ϕ(y = y and analyze the stationary points from this ODE. b Find the possible (finite values of ϕ + = lim y ϕ(y. Proof. a We look for a traveling wave solution u(x, t=ϕ(x st, y = x st. We get the ODE sϕ + ϕϕ = ɛϕ + δϕ, sϕ + ϕ = ɛϕ + δϕ + a. Since ϕ asy,thenϕ, ϕ asy. Therefore, at y =, a =. We found the following ODE, ϕ + ɛ δ ϕ + s δ ϕ δ ϕ =. In order to find and analyze the stationary points of an ODE above, we write it as a first-order system. φ = ϕ, φ = ϕ. φ = ϕ = φ, φ = ϕ = ɛ δ ϕ s δ ϕ + δ ϕ = ɛ δ φ s δ φ + δ φ. { φ = φ =, φ = δ ɛ φ s δ φ + δ φ =; { φ = φ =, φ = s δ φ + δ φ =; { φ = φ =, φ = δ φ (s φ =. Stationary points: (,, (s,, s >. φ = φ = f(φ,φ, φ = ɛ δ φ s δ φ + δ φ = g(φ,φ.

162 Partial Differential Equations Igor Yanovsky, 5 6 In order to classify a stationary point, need to find eigenvalues of a linearized system at that point. [ ] f f [ ] φ J(f(φ,φ,g(φ,φ = φ = s δ + δ φ. g φ g φ For (φ,φ =(, : det(j (, λi = λ s δ ɛ δ λ = λ + δ ɛ λ + s δ =. λ ± = ɛ δ ± ɛ s 4δ δ. If 4δ ɛ >s λ ± R, λ ± <. (, is Stable Improper Node. If ɛ 4δ <s λ ± C, Re(λ ± <. (, is Stable Spiral Point. For (φ,φ =(s, : det(j (s, λi= λ s δ ɛ δ λ = λ + ɛ δ λ s δ =. λ ± = ɛ δ ± ɛ + s 4δ δ. λ + >, λ <. (s, is Untable Saddle Point. b Since lim ϕ(y = = lim ϕ(x st, y t we may have lim ϕ(y = lim ϕ(x st =s. y + t That is, a particle may start off at an unstable node (s, and as t increases, approach the stable node (,. A phase diagram with (, being a stable spiral point, is shown below. ɛ δ

163 Partial Differential Equations Igor Yanovsky, 5 63

164 Partial Differential Equations Igor Yanovsky, 5 64 Problem (F 95, #8. Consider the equation u t + f(u x = ɛu xx where f is smooth and ɛ>. We seek traveling wave solutions to this equation, i.e., solutions of the form u = φ(x st, under the boundary conditions u u L u u R and u x as x, and u x as x +. Find a necessary and sufficient condition on f, u L, u R and s for such traveling waves to exist; in case this condition holds, write an equation which defines φ implicitly. Proof. We look for traveling wave solutions u(x, t=φ(x st, y = x st. The boundary conditions become φ u L and φ as x, φ u R and φ as x +. Since f(φ(x st x = f (φφ,wegettheode sφ + f (φφ = ɛφ, sφ +(f(φ = ɛφ, sφ + f(φ =ɛφ + a, φ sφ + f(φ = + b. ɛ We use boundary conditions to determine constant b: 4 At x =, =φ = su L + f(u L ɛ At x =+, =φ = su R + f(u R ɛ s = f(u L f(u R u L u R. + b b = su L f(u L. ɛ + b b = su R f(u R. ɛ 4 For the solution for the second part of the problem, refer to Chiu-Yen s solutions.

165 Partial Differential Equations Igor Yanovsky, 5 65 Problem (S, #5; F 9, #. Fisher s Equation. Consider u t = u( u+u xx, <x<, t >. The solutions of physical interest satisfy u, and lim u(x, t =, lim x u(x, t=. x + One class of solutions is the set of wavefront solutions. These have the form u(x, t= φ(x + ct, c. Determine the ordinary differential equation and boundary conditions which φ must satisfy (to be of physical interest. Carry out a phase plane analysis of this equation, and show that physically interesting wavefront solutions are possible if c, but not if c<. Proof. We look for a traveling wave solution u(x, t=φ(x + ct, s = x + ct. We get the ODE cφ = φ( φ+φ, φ cφ + φ φ =, φ(s, as s, φ(s, as s +, φ. In order to find and analyze the stationary points of an ODE above, we write it as a first-order system. y = φ, y = φ. y = φ = y, y = φ = cφ φ + φ = cy y + y. { y = y =, y = cy y + y =; Stationary points: (,, (,. y = y = f(y,y, y = cy y + y = g(y,y. g y { y =, y (y =. In order to classify a stationary point, need to find eigenvalues of a linearized system at that point. [ ] f f [ ] y J(f(y,y,g(y,y = y =. y c g y

166 Partial Differential Equations Igor Yanovsky, 5 66 For (y,y =(, : det(j (, λi = λ c λ = λ cλ +=. λ ± = c ± c 4. If c λ ± R, λ ± >. (, is Unstable Improper (c > / Proper(c =Node. If c< λ ± C, Re(λ ±. (, is Unstable Spiral Node. For (y,y =(, : det(j (, λi = λ c λ = λ cλ =. λ ± = c ± c +4. If c λ + >, λ <. (, is Unstable Saddle Point. By looking at the phase plot, a particle may start off at an unstable node (, and as t increases, approach the unstable node (,.

167 Partial Differential Equations Igor Yanovsky, 5 67

168 Partial Differential Equations Igor Yanovsky, 5 68 Problem (F 99, #6. For the system t ρ + x u = t u + x (ρu = x u look for traveling wave solutions of the form ρ(x, t=ρ(y = x st, u(x, t =u(y = x st. Inparticular a Find a first order ODE for u. b Show that this equation has solutions of the form u(y =u + u tanh(αy + y, for some constants u, u, α, y. Proof. a We rewrite the system: ρ t + u x = u t + ρ x u + ρu x = u xx We look for traveling wave solutions ρ(x, t=ρ(x st, u(x, t =u(x st, y = x st. We get the system of ODEs { sρ + u =, su + ρ u + ρu = u. The first ODE gives ρ = s u, ρ = s u + a, where a is a constant, and integration was done with respect to y. The second ODE gives su + ( s u u + s u + a u = u, su + s uu + au = u. Integrating, we get su + s u + au = u + b. u = s u +(a su b. b Note that the ODE above may be written in the following form: u + Au + Bu = C, which is a nonlinear first order equation.

169 Partial Differential Equations Igor Yanovsky, 5 69 Problem (S, #7. Consider the following system of PDEs: f t + f x = g f g t g x = f g a Find a system of ODEs that describes traveling wave solutions of the PDE system; i.e. for solutions of the form f(x, t =f(x st and g(x, t=g(x st. b Analyze the stationary points and draw the phase plane for this ODE system in the standing wave case s =. Proof. a We look for traveling wave solutions f(x, t=f(x st, We get the system of ODEs Thus, sf + f = g f, sg g = f g. f = g f s, g = f g s. b If s =, the system becomes { f = g f, g = g f. g(x, t=g(x st. Relabel the variables f y, g y. { y = y y =, y = y y =. Stationary points: (,, (,, (,. { y = y y = φ(y,y, y = y y = ψ(y,y. In order to classify a stationary point, need to find eigenvalues of a linearized system at that point. [ ] φ φ [ ] y J(φ(y,y,ψ(y,y = y y y =. y ψ y ψ y For (y,y =(, : det(j (, λi = λ λ = λ( λ =. λ =,λ = ; eigenvectors: v = (, is Unstable Node. ( (, v =.

170 Partial Differential Equations Igor Yanovsky, 5 7 For (y,y =(, : det(j (, λi= λ λ = λ 3λ =. λ ± = 3 7 ±. λ <, λ + >. (-, is Unstable Saddle Point. For (y,y =(, : det(j (, λi = λ λ = λ + λ +=. λ ± = ± i 7. Re(λ ± <. (, is Stable Spiral Point.

171 Partial Differential Equations Igor Yanovsky, Dispersion Problem (S 97, #8. Consider the following equation u t =(f(u x x αu xxxx, f(v=v v, (6.4 with constant α. a Linearize this equation around u =and find the principal mode solution of the form e ωt+ikx. For which values of α are there unstable modes, i.e., modes with ω = for real k? For these values, find the maximally unstable mode, i.e., the value of k with the largest positive value of ω. b Consider the steady solution of the (fully nonlinear problem. Show that the resulting equation can be written as a second order autonomous ODE for v = u x and draw the corresponding phase plane. Proof. a We have u t = (f(u x x αu xxxx, u t = (u x u x x αu xxxx, u t = u x u xx u xx αu xxxx. However, we need to linearize (6.4 around u =. To do this, we need to linearize f. f(u =f( + uf ( + u f ( + =+u( + = u +. Thus, we have u t = u xx αu xxxx. Consider u(x, t=e ωt+ikx. ωe ωt+ikx =(k αk 4 e ωt+ikx, ω = k αk 4. To find unstable nodes, we set ω =,toget α = k. To find the maximally unstable mode, i.e., the value of k with the largest positive value of ω, consider ω(k = k αk 4, ω (k = k 4αk 3. To find the extremas of ω, wesetω =. Thus,the extremas are at k =, k,3 = ± α. To find if the extremas are maximums or minimums, we set ω =: ω (k = αk =, ω ( = > k = is the minimum. ( ω ± = 4 < k = ± is the maximum unstable mode. α α ( ω ± = is the largest positive value of ω. α 4α

172 Partial Differential Equations Igor Yanovsky, 5 7 b Integrating, weget u x u x αu xxx =. Let v = u x. Then, v v αv xx =, or v = v v α. In order to find and analyze the stationary points of an ODE above, we write it as a first-order system. y = v, y = v. y = v = y, y = v = v v = y y. α α { { y = y =, y =, y = y y α =; y (y =. Stationary points: (,, (,. y = y = f(y,y, y = y y = g(y,y. α In order to classify a stationary point, need to find eigenvalues of a linearized system at that point. [ ] f f [ ] y J(f(y,y,g(y,y = y = y. α g y g y For (y,y =(,, λ ± = ± α. If α<, λ ± R, λ + >, λ <. (, is Unstable Saddle Point. If α>, λ ± = ±i α C, Re(λ ±=. (, is Spiral Point. For (y,y =(,, λ ± = ± α. If α<, λ ± = ±i α C, Re(λ ±=. (, is Spiral Point. If α>, λ ± R, λ + >, λ <. (, is Unstable Saddle Point.

173 Partial Differential Equations Igor Yanovsky, 5 73

174 Partial Differential Equations Igor Yanovsky, Energy Methods Problem (S 98, #9; S 96, #5. Consider the following initial-boundary value problem for the multi-dimensional wave equation: u(x, = f(x, u tt = u in (,, u (x, = g(x for x, t u n + a(x u t = on. Here, is a bounded domain in R n and a(x. Define the Energy integral for this problem and use it in order to prove the uniqueness of the classical solution of the problem. Proof. dẽ dt = = (u tt uu t dx = u tt u t dx = t (u t dx + t u dx + Thus, a(xu t dx = u t t + u dx. } {{ } Let Energy integral be E(t = u t + u dx. u n u t ds + u u t dx a(xu t ds. In order to prove that the given E(t from scratch, take its derivative with respect to t: de dt (t = ( ut u tt + u u t dx u = u t u tt dx + u t n ds u t udx = u t (u tt u dx a(xu t dx. } {{ } = Thus, E(t E(. To prove the uniqueness of the classical solution, suppose u and u are two solutions of the initial boundary value problem. Let w = u u. Then, w satisfies w tt = w in (,, w(x, =, w t (x, = for x, w n + a(x w t = on. We have E w ( = (w t (x, + w(x, dx =.

175 Partial Differential Equations Igor Yanovsky, 5 75 E w (t E w ( = E w (t =. Thus,w t =,w xi = w(x, t =const=. Hence, u = u. Problem (S 94, #7. Consider the wave equation c (x u tt = u x u u α(x = on R. t n Assume that α(x is of one sign for all x (i.e. α always positive or α always negative. For the energy E(t= c (x u t + u dx, show that the sign of de dt is determined by the sign of α. Proof. We have de ( dt (t = c (x u tu tt + u u t dx = c (x u u tu tt dx + u t n ds u t udx ( = u t c (x u tt u dx + α(x u t dx } {{ } = { >, if α(x >, x, = α(x u t dx = <, if α(x <, x.

176 Partial Differential Equations Igor Yanovsky, 5 76 Problem (F 9, #. Let R n. Let u(x, t be a smooth solution of the following initial boundary value problem: u tt u + u 3 = u(x, t= for (x, t [,T] for (x, t [,T]. a Derive an energy equality for u. (Hint: Multiply by u t and integrate over [,T]. b Show that if u t= = u t t= = for x, then u. Proof. a Multiply by u t and integrate: = (u tt u + u 3 u t dx = u tt u t dx u n u t ds } {{ } = = t (u t dx + t u dx + 4 t (u4 dx = d dt Thus, the Energy integral is ( E(t = u t + u + u4 dx =const = E(. b Since u(x, =, u t (x, =, we have ( E( = ut (x, + u(x, + u(x, 4 dx =. Since E(t =E( =, we have ( E(t= ut (x, t + u(x, t + u(x, t4 dx =. Thus, u. + u u t dx + u 3 u t dx ( u t + u + u4 dx.

177 Partial Differential Equations Igor Yanovsky, 5 77 Problem (F 4, #3. Consider a damped wave equation { u tt u + a(xu t =, (x, t R 3 R, u t= = u, u t t= = u. Here the damping coefficient a C (R3 is a non-negative function and u, u C (R3. Show that the energy of the solution u(x, t at time t, E(t= ( x u + u t dx R 3 is a decreasing function of t. Proof. Take the derivative of E(t with respect to t. Note that the boundary integral is by Huygen s principle. de dt (t = ( u ut + u t u tt dx R 3 u = u t R 3 n ds u t udx+ u t u tt dx R } {{ } 3 R 3 = = u t ( u + u tt dx = u t ( a(xu t dx = a(xu t dx. R 3 R 3 R 3 Thus, E(t E(, i.e. E(t is a decreasing function of t. de dt

178 Partial Differential Equations Igor Yanovsky, 5 78 Problem (W 3, #8. a Consider the damped wave equation for high-speed waves ( <ɛ<< in a bounded region D ɛ u tt + u t = u with the boundary condition u(x, t=on D. Show that the energy functional E(t= ɛ u t + u dx D is nonincreasing on solutions of the boundary value problem. b Consider the solution to the boundary value problem in part (a with initial data u ɛ (x, =, u ɛ t (x, = ɛ α f(x, wheref does not depend on ɛ and α<. Use part (a to show that u ɛ (x, t dx D uniformly on t T for any T as ɛ. c Show that the result in part (b does not hold for α =. To do this consider thecasewheref is an eigenfunction of the Laplacian, i.e. f + λf =in D and f =on D, andsolveforu ɛ explicitly. Proof. a de dt = = D D ɛ u t u tt dx + u u t dx ɛ u t u tt dx + D u n u t ds D } {{ } =, (u= on D = (ɛ u tt uu t dx = = D Thus, E(t E(, i.e. E(t is nonincreasing. D uu t dx D u t dx. b From (a, we know de dt. We also have E ɛ ( = ɛ (u ɛ t(x, + u ɛ (x, dx D = ɛ (ɛ α f(x +dx = ɛ ( α f(x dx as ɛ. D Since E ɛ ( E ɛ (t = D ɛ (u ɛ t + u ɛ dx, thene ɛ (t asɛ. Thus, D uɛ dx asɛ. c If α =, E ɛ ( = D ɛ ( α f(x dx = D D f(x dx. Since f is independent of ɛ, E ɛ ( does not approach as ɛ. We can not conclude that D uɛ (x, t dx.

179 Partial Differential Equations Igor Yanovsky, 5 79 Problem (F 98, #6. Let f solve the nonlinear wave equation f tt f xx = f( + f for x [, ], withf(x =,t=f(x =,t=and with smooth initial data f(x, t = f (x. a Find an energy integral E(t which is constant in time. b Show that f(x, t <cfor all x and t, inwhichc is a constant. Hint: Note that f +f = d df log( + f. Proof. a Since f(,t=f(,t=, t, wehavef t (,t=f t (,t=. Let de ( = = ftt f xx + f( + f f t dx dt ff t = f tt f t dx f xx f t dx + +f dx = f tt f t dx [f x f }{{} t ] ff t + f x f tx dx + +f dx = = t (f t dx + t (f x dx + t (ln( + f dx = d ( f dt t + fx + ln( + f dx. Thus, E(t= ( f t + fx + ln( + f dx. b We want to show that f is bounded. For smooth f(x, = f (x, we have E( = ( ft (x, + f x (x, + ln( + f(x, dx <. Since E(t is constant in time, E(t = E( <. Thus, ln( + f dx ( f t + fx + ln( + f dx = E(t <. Hence, f is bounded.

180 Partial Differential Equations Igor Yanovsky, 5 8 Problem (F 97, #. Consider initial-boundary value problem u tt + a (x, tu t u(x, t = x R n, <t<+ u(x = x u(x, = f(x, u t (x, = g(x x. Prove that L -norm of the solution is bounded in t on (, +. Here is a bounded domain, and a(x, t, f(x, g(x are smooth functions. Proof. Multiply the equation by u t and integrate over : u t u tt + a u t u t u =, u t u tt dx + a u t dx u t udx=, d u t dx + a u u t dx u t dt n ds + u u t dx =, } {{ } =, (u=, x d u t dx + a u t dx + d u dx =, dt dt d ( u dt t + u dx = a u t dx. Let Energy integral be ( E(t = u t + u dx. de dt We have, i.e. E(t E(. ( E(t E( = ut (x, + u(x, dx = ( g(x + f(x dx <, since f and g are smooth functions. Thus, ( E(t= u t + u dx <, u dx <, u dx C u dx <, by Poincare inequality. Thus, u is bounded t.

181 Partial Differential Equations Igor Yanovsky, 5 8 Problem (S 98, #4. a Let u(x, y, z, t, equation u tt + u t = u xx + u yy + u zz u(x, y, z, = f(x, y, z, u t (x, y, z, = g(x, y, z. < x,y,z < be a solution of the (6.4 Here f, g are smooth functions which vanish if x + y + z is large enough. Prove that it is the unique solution for t. b Suppose we want to solve the same equation (6.4 in the region z, < x, y <, with the additional conditions u(x, y,,t = f(x, y, t u z (x, y,,t = g(x, y, t with the same f, g as before in (6.4. What goes wrong? Proof. a Suppose u and u are two solutions. Let w = u u. Then, w tt + w t = w, w(x, y, z, =, w t (x, y, z, =. Multiply the equation by w t and integrate: w t w tt + wt = w t w, w t w tt dx + wt dx = w t wdx, R 3 R 3 R 3 d wt dt dx + wt dx = w w t R 3 R 3 R 3 n dx w w t dx, R } {{ } 3 = d wt dt dx + wt dx = d w dx, R 3 R 3 dt R 3 d ( w dt t + w dx = wt dx, R } 3 R {{ } 3 E(t de, dt ( E(t E( = wt (x, + w(x, dx =, R 3 ( E(t = w t + w dx =. R 3 Thus, w t =, w =,andw = constant. Since w(x, y, z, =, we have w. b

182 Partial Differential Equations Igor Yanovsky, 5 8 Problem (F 94, #8. The one-dimensional, isothermal fluid equations with viscosity and capillarity in Lagrangian variables are v t u x = u t + p(v x = εu xx δv xxx in which v(= /ρ is specific volume, u is velocity, and p(v is pressure. The coefficients ε and δ are non-negative. Find an energy integral which is non-increasing (as t increases if ε > and conserved if ε =. Hint: if δ =, E = u / P (v dx where P (v=p(v. Proof. Multiply the second equation by u and integrate over R. We use u x = v t. Note that the boundary integrals are due to finite speed of propagation. uu t + up(v x = εuu xx δuv xxx, uu t dx + up(v x dx = ε uu xx dx δ uv xxx dx, R R R R R t (u dx + up(v ds+ u x p(v dx R R } {{ } = = ε uu x dx ε u x dx δ uv xx dx +δ u x v xx dx, R R R R } {{ } } {{ } = = R t (u dx + v t p(v dx = ε u x dx + δ v t v xx dx, R R R R t (u dx + P (v dx = ε u x dx + δ v t v x dx δ v xt v x dx, R t R R R } {{ } = R t (u dx + R t P (v dx + δ R t (v x dx = ε u x dx, R d ( u dt R + P (v+ δ v x dx = ε u x dx. R ( u E(t = + P (v+ δ v x dx R is nonincreasing if ε>, and conserved if ε =.

183 Partial Differential Equations Igor Yanovsky, 5 83 Problem (S 99, #5. Consider the equation u tt = x σ(u x (6.4 with σ(z a smooth function. This is to be solved for t>, x, with periodic boundary conditions and initial data u(x, = u (x and u t (x, = v (x. a Multiply (6.4 by u t and get an expression of the form d F (u t,u x = dt that is satisfied for an appropriate function F (y, z with y = u t, z = u x, where u is any smooth, periodic in space solution of (6.4. b Under what conditions on σ(z is this function, F (y, z, convex in its variables? c What à priori inequality is satisfied for smooth solutions when F is convex? d Discuss the special case σ(z =a z 3 /3, witha> and constant. Proof. a Multiply by u t and integrate: u t u tt = u t σ(u x x, u t u tt dx = d dt u t σ(u x x dx, u t dx = u tσ(u x } {{ } =, (π-periodic Let Q (z =σ(z, then d dt Q(u x=σ(u x u xt. Thus, b We have = u tx σ(u x dx = d dt d ( u t dt + Q(u x dx =. F (u t,u x = u t + Q(u x. u tx σ(u x dx = Q(u x dx. 4 For F to be convex, the Hessian matrix of partial derivatives must be positive definite. 4 A function f is convex on a convex set S if it satisfies f(αx +( αy αf(x+( αf(y for all α andforallx, y S. If a one-dimensional function f has two continuous derivatives, then f is convex if and only if f (x. In the multi-dimensional case the Hessian matrix of second derivatives must be positive semi-definite, that is, at every point x S y T f(x y, for all y. The Hessian matrix is the matrix with entries [ f(x] ij f(x. x i x j For functions with continuous second derivatives, it will always be symmetric matrix: f xi x j = f xj x i.

184 Partial Differential Equations Igor Yanovsky, 5 84 The Hessian matrix is Futu F (u t,u x =( t F utux F uxut F uxux y T F (x y = ( y y ( σ (u x = ( σ (u x ( y y. = y + σ (u x y }{{} need Thus, for a Hessian matrix to be positive definite, need σ (u x, so that the above inequality holds for all y. c We have d dt F (u t,u x dx =, F (u t,u x dx = const, F (u t,u x dx = ( u t + Q(u x dx = d If σ(z =a z 3 /3, we have F (u t (x,,u x (x, dx, ( v + Q(u x dx. F (u t,u x = u t + Q(u x = u t + a u 4 x, d ( u t dt + a u 4 x dx =, ( u t + a u 4 x ( u t + a u 4 x dx = const, dx = ( v + a u 4 x dx..

185 Partial Differential Equations Igor Yanovsky, 5 85 Problem (S 96, #8. 4 Let u(x, t be the solution of the Korteweg-de Vries equation u t + uu x = u xxx, x π, with π-periodic boundary conditions and prescribed initial data u(x, t ==f(x. a Prove that the energy integral I (u = π u (x, t dx is independent of the time t. b Prove that the second energy integral, π ( I (u = u x(x, t+ 6 u3 (x, t dx is also independent of the time t. c Assume the initial data are such that I (f+i (f <. Use (a + (b to prove that the maximum norm of the solution, u =sup x u(x, t, isboundedintime. Hint: Use the following inequalities (here, u p is the L p -norm of u(x, t at fixed time t: u π 6 ( u + u x u 3 3 u u (one of Sobolev s inequalities; (straightforward. Proof. a Multiply the equation by u and integrate. Note that all boundary terms are due to π-periodicity. d dt π π uu t dx + u dx + 3 d dt π uu t + u u x = uu xxx, π π u u x dx = π uu xxx dx, (u 3 x dx = uu xx π π π u dx + 3 u3 π = (u x x dx, d π u dx = dt u π x =. I (u = π u dx = C. u x u xx dx, Thus, I (u = π u (x, t dx is independent of the time t. Alternatively, wemaydifferentiatei (u: di dt (u = d π π π u dx = uu t dx = u( uu x + u xxx dx dt π π π = u u x dx + uu xxx dx = π 3 (u3 x dx +uu xx π u x u xx dx = π 3 u3 π (u x x dx = u π x =. 4 Also, see S 9, #7.

186 Partial Differential Equations Igor Yanovsky, 5 86 b Note that all boundary terms are due to π-periodicity. di dt (u = d π ( dt u x + π 6 u3 dx = (u x u xt + u u t dx = We differentiate the original equation with respect to x: u t = uu x + u xxx u tx = (uu x x + u xxxx. = = π π u x ( (uu x x + u xxxx dx + u x (uu x x dx + π = u x uu x π + π ( u 4 4 = = π π π π u ( uu x + u xxx dx π u x u xxxx dx π u xx uu x dx + u x u xxx π π u 3 u x dx + u xx u xxx dx π u u xxx dx dx + x u u xx π uu x u xx dx π u xx uu x dx u xx u xxx dx u 4 π π 4 uu x u xx dx u xx u xxx dx = u π π π xx + u xxx u xx dx = u xxx u xx dx =, since π u xx u xxx dx =+ π u xx u xxx dx. Thus, π ( I (u = u x(x, t+ 6 u3 (x, t dx = C, and I (u is independent of the time t. c From (a and (b, we have I (u = I (u = π π Using given inequalities, we have u π 6 ( u + u x π 6 u dx = u, ( u x + 6 u3 dx = u x + 6 u 3 3. ( I (u+i (u 3 u 3 3 π 6 I (u+ π 3 I (u+ π 8 u u π 6 I (u+ π 3 I (u+ π 8 I (u u = C + C u. u C + C u, u C. Thus, u is bounded in time. Also see Energy Methods problems for higher order equations (3rd and 4th in the section on Gas Dynamics.

187 Partial Differential Equations Igor Yanovsky, Wave Equation in D and 3D Problem (F 97, #8; (McOwen 3. #9. Solve u tt = u xx + u yy + u zz with initial conditions u(x, y, z, = x + y, } {{ } u t (x, y, z, = g(x }{{} h(x Proof. ➀ We may use the Kirchhoff s formula: u(x, t = ( t g(x + ctξ ds ξ + t h(x + ctξ ds ξ 4π t ξ = 4π ξ = = ( ( t (x + ctξ +(x + ctξ ds ξ + = 4π t ξ =. ➁ We may solve the problem by Hadamard s method of descent, since initial conditions are independent of x 3. We need to convert surface integrals in R 3 to domain integrals in R. Specifically, we need to express the surface measure on the upper half of the unit sphere S+ in terms of the two variables ξ and ξ.todothis,consider f(ξ,ξ = ξ ξ over the unit disk ξ + ξ <. ds ξ = +(f ξ +(f ξ dξ dξ dξ dξ =. ξ ξ

188 Partial Differential Equations Igor Yanovsky, 5 88 u(x,x,t = ( g(x + ctξ,x + ctξ dξ dξ t 4π t ξ +ξ < ξ ξ + t ( h(x + ctξ,x + ctξ dξ dξ 4π ξ +ξ < ξ ξ = ( ξ+ξ< (x + tξ +(x + tξ t dξ 4π t ξ ξ dξ +, = ( x t +x tξ + t ξ + x +x tξ + t ξ dξ π t ξ +ξ < ξ ξ dξ = ( tx +x t ξ + t 3 ξ + tx +x t ξ + t 3 ξ dξ π t ξ +ξ < ξ ξ dξ = ( x +4x tξ +3t ξ + x +4x tξ +3t ξ dξ π ξ +ξ < ξ ξ dξ = ( (x + x +4t(x ξ + x ξ +3t (ξ + ξ dξ π ξ +ξ < ξ ξ dξ = π (x + x dξ dξ + 4t x ξ + x ξ dξ ξ +ξ < ξ ξ π ξ } {{ } +ξ < ξ ξ dξ } {{ } ❶ ❷ + 3t ξ + ξ dξ π ξ +ξ < ξ ξ dξ = } {{ } ❸ ❶ = π (x + x dξ dξ ξ +ξ < ξ ξ = π (x + x π π u du dθ = π (x + x dθ = x + x. ❷ = 4t x ξ + x ξ π ξ +ξ < ξ ξ =. ❸ = 3t ξ + ξ dξ π ξ +ξ < ξ ξ dξ = 3t π = 3t π π π = π π (x + x rdrdθ r ( u = r, du = rdr dξ dξ = 4t ξ π ξ = 3t π π x ξ + x ξ dξ ξ ξ dξ (r cos θ +(r sin θ r r 3 drdθ ( u = r, du = rdr r 3 dθ = t π π dθ = t. u(x,x,t = ❶ + ❷ + ❸ = x + x +t. rdrdθ ➂ We may guess what the solution is: u(x, y, z, t = [ (x + t +(y + t +(x t +(y t ] = x + y +t.

189 Partial Differential Equations Igor Yanovsky, 5 89 Check: u(x, y, z, = x + y. u t (x, y, z, t = (x + t+(y + t (x t (y t, u t (x, y, z, =. u tt (x, y, z, t = 4, u x (x, y, z, t = (x + t+(x t, u xx (x, y, z, t =, u y (x, y, z, t = (y + t+(y t, u yy (x, y, z, t =, u zz (x, y, z, t =, u tt = u xx + u yy + u zz.

190 Partial Differential Equations Igor Yanovsky, 5 9 Problem (S 98, #6. Consider the two-dimensional wave equation w tt = a w, with initial data which vanish for x +y large enough. Prove that w(x, y, t satisfies the decay w(x, y, t C t. (Note that the estimate is not uniform with respect to x, y since C may depend on x, y. Proof. Suppose we have the following problem with initial data: u tt = a u x R,t>, u(x, = g(x, u t (x, = h(x x R. The result is the consequence of the Huygens principle and may be proved by Hadamard s method of descent: 43 u(x, t = ( g(x + ctξ,x + ctξ dξ dξ t 4π t ξ +ξ < ξ ξ + t ( h(x + ctξ,x + ctξ dξ dξ 4π ξ +ξ < ξ ξ = th(x + ξ+g(x + ξ dξ dξ π ξ <c t ξ c t c t + t g(x + ξ (ct, ct dξ dξ π ξ <c t ξ c t. c t For a given x, lett (x be so large that T > and supp(h + g B T (x. Then for t>t we have: u(x, t = tm + M +Mct dξ dξ π ξ <c T c T c t c T 4 = πc T [( M ( M 3/4 π c t + 3/4 c Tt + Mc ] c. t u(x, t C /t for t>t. For t T : u(x, t = π TM + M +4McT ξ <c t ξ = (TM + M +4Mctπ π c t ct dξ dξ c t rdr/c t r c t M(T ++4cT du M(T ++4cT M(T ++4cT T = =. u/ t Letting C =max(c,m(t ++4cT T, we have u(x, t C(x/t. For n = 3, suppose g, h C (R3. The solution is given by the Kircchoff s formula. There is a constant C so that u(x, t C/t for all x R 3 and t>. As McOwen suggensts in Hints for Exercises, to prove the result, we need to estimate the 43 Nick s solution follows.

191 Partial Differential Equations Igor Yanovsky, 5 9 area of intersection of the sphere of radius ct with the support of the functions g and h.

192 Partial Differential Equations Igor Yanovsky, 5 9 Problem (S 95, #6. Spherical waves in 3-d are waves symmetric about the origin; i.e. u = u(r, t where r is the distance from the origin. The wave equation u tt = c u then reduces to c u tt = u rr + r u r. (6.43 a Find the general solutions u(r, t by solving (6.43. Include both the incoming waves and outgoing waves in your solutions. b Consider only the outgoing waves and assume the finite out-flux condition < lim r u r < r for all t. The wavefront is defined as r = ct. How is the amplitude of the wavefront decaying in time? Proof. a We want to reduce (6.43 to the D wave equation. Let v = ru. v tt = ru tt, v r = ru r + u, v rr = ru rr +u r. Thus, (6.43 becomes c r v tt = r v rr, c v tt = v rr, v tt = c v rr, which has the solution Thus, v(r, t=f(r + ct+g(r ct. Then u(r, t= v(r, t= r f(r + ct } r {{ } incoming, (c> + g(r ct. } r {{ } outgoing, (c> b We consider u(r, t= r g(r ct: < lim r u r <, r < lim r ( r r g (r ct r g(r ct <, ( < lim rg (r ct g(r ct <, r < g( ct <, < g( ct = G(t <, ( t g(t = G. c

193 Partial Differential Equations Igor Yanovsky, 5 93 The wavefront is defined as r = ct. We have u(r, t = ( r ct r g(r ct = r G c u(r, t =. t c G( = ct G(. The amplitude of the wavefront decays like t.

194 Partial Differential Equations Igor Yanovsky, 5 94 Problem (S, #8. a Show that for a smooth function F on the line, while u(x, t =F (ct + x / x may look like a solution of the wave equation u tt = c u in R 3, it actually is not. Do this by showing that for any smooth function φ(x, t with compact support u(x, t(φ tt φ dxdt =4π φ(,tf (ct dt. R 3 R R Note that, setting r = x, for any function w which only depends on r one has w = r (r w r r = w rr + r w r. b If F ( = F ( =, what is the true solution to u tt = u with initial conditions u(x, = F ( x / x and u t (x, = F ( x / x? c (Ralston Hw Suppose u(x, t is a solution to the wave equation u tt = c u in R 3 R with u(x, t =w( x,t and u(x, =. Show that F ( x + ct F ( x ct u(x, t = x for a function F of one variable. Proof. a We have u (φ tt φ dxdt = lim R 3 R ɛ = lim ɛ R R dt x >ɛ u (φ tt φ dx [ dt φ (u tt u dx + x >ɛ x =ɛ ] u n φ u φ n ds. The final equality is derived by integrating by parts twice in t, and using Green s theorem: ( u (v u u v dx = v n u v ds. n Since ds = ɛ sin φ dφ dθ and n = r, substituting u(x, t =F ( x + ct/ x gives: u (φ tt φ dxdt = 4πφF(ct dt. R 3 R Thus, u is not a weak solution to the wave equation. R b c We want to show that v( x,t= x w( x,t is a solution to the wave equation in one space dimension and hence must have the from v = F ( x + ct+g( x ct. Then we can argue that w will be undefined at x =forsomet unless F (ct+g( ct = for all t. We work in spherical coordinates. Note that w and v are independent of φ and θ. We have: v tt (r, t = c w = c r (r w r r = c r (rw r + r w rr, rw tt = c rw rr +w r. Thus we see that v tt = c v rr, and we can conclude that v(r, t = F (r + ct+g(r ct and w(r, t = F (r + ct+g(r ct. r

195 Partial Differential Equations Igor Yanovsky, 5 95 lim r w(r, t does not exist unless F (ct+g( ct = for all t. Hence w(r, t = u(x, t = F (ct + r+g(ct r, and r F (ct + x +G(ct x. x

196 Partial Differential Equations Igor Yanovsky, Problems: Laplace Equation A fundamental solution K(x for the Laplace operator is a distribution satisfying 44 K(x =δ(x The fundamental solution for the Laplace operator is { K(x = π log x if n = ( nω n x n if n Green s Function and the Poisson Kernel Green s function is a special fundamental solution satisfying 45 { G(x, ξ =δ(x for x G(x, ξ= for x, (7. To construct the Green s function, ➀ consider w ξ (x with w ξ (x =inandw ξ (x = K(x ξ on ; ➁ consider G(x, ξ =K(x ξ +w ξ (x, which is a fundamental solution satisfying (7.. Problem. Given a particular distribution solution to the set of Dirichlet problems { u ξ (x =δ ξ (x for x u ξ (x = for x, how would you use this to solve { u = for x u(x =g(x for x. Proof. u ξ (x =G(x, ξ, a Green s function. G is a fundamental solution to the Laplace operator, G(x, ξ=,x. In this problem, it is assumed that G(x, ξ isknownfor. Then u(ξ = G(x, ξ udx+ u(x G(x, ξ ds x n x for every u C (. In particular, if u =inandu = g on, then we obtain the Poisson integral formula u(ξ = G(x, ξ n x g(x ds x, 44 We know that u(x = K(x yf(ydy is a distribution solution of u = f when f is integrable R n and has compact support. In particular, we have u(x = K(x y u(y dy whenever u C (R n. R n The above result is a consequence of: u(x = δ(x yu(y dy = ( K u = K ( u = K(x y u(y dy. 45 Green s function is useful in satisfying Dirichlet boundary conditions.

197 Partial Differential Equations Igor Yanovsky, 5 97 where H(x, ξ= G(x,ξ n x is the Poisson kernel. Thus if we know that the Dirichlet problem has a solution u C (, then we can calculate u from the Poisson integral formula (provided of course that we can compute G(x, ξ.

198 Partial Differential Equations Igor Yanovsky, 5 98 Dirichlet Problem on a Half-Space. Solve the n-dimensional Laplace/Poisson equation on the half-space with Dirichlet boundary conditions. Proof. Use the method of reflection to construct Green s function. upper half-space in R n.ifx =(x,x n, where x R n,wecansee x ξ = x ξ, and hence K(x ξ =K(x ξ. Thus Let be an G(x, ξ=k(x ξ K(x ξ is the Green s function on. G(x, ξ isharmonicin, and G(x, ξ=on. To compute the Poisson kernel, we must differentiate G(x, ξ in the negative x n direction. For n, K(x ξ = x n ξ n x ξ n, x n ω n so that the Poisson kernel is given by x n G(x, ξ xn= Thus, the solution is u(ξ = = ξ n ω n x ξ n, for x R n. G(x, ξ g(x ds x = ξ n g(x n x ω n R n x ξ n dx. If g(x is bounded and continuous for x R n,thenu(ξ isc and harmonic in R n + and extends continuously to R n + such that u(ξ =g(ξ.

199 Partial Differential Equations Igor Yanovsky, 5 99 Problem (F 95, #3: Neumann Problem on a Half-Space. a Consider the Neumann problem in the upper half plane, ={x =(x,x : <x <, x > }: u = u x x + u x x = x, u x (x, = f(x <x <. Find the corresponding Green s function and conclude that u(ξ =u(ξ,ξ = ln [(x ξ + ξ π ] f(x dx is a solution of the problem. b Show that this solution is bounded in if and only if f(x dx =. Proof. a Notation: x =(x, y, ξ =(x,y. Since K(x ξ= π log x ξ, n =. ➀ First, we find the Green s function. We have K(x ξ = π log (x x +(y y. Let G(x, ξ=k(x ξ+ω(x. Since the problem is Neumann, we need: { G(x, ξ =δ(x ξ, G y ((x,,ξ =. G((x, y,ξ = π log (x x +(y y + ω((x, y,ξ, G y y ((x, y,ξ = y π (x x +(y y + ω y((x, y,ξ, G y ((x,,ξ = y π (x x + y + ω y ((x,,ξ =. Let ω((x, y,ξ = a π log (x x +(y + y. Then, G y ((x,,ξ = y π (x x + y + a y π (x x + y =. Thus, a =. 46 G((x, y,ξ = π log (x x +(y y + π log (x x +(y + y. ➁ Consider Green s identity (after cutting out B ɛ (ξ and having ɛ : ( (u G G u dx = u G G u ds }{{} }{{} n n = = 46 Note that for the Dirichlet problem, we would have gotten the - sign instead of + in front of ω.

200 Partial Differential Equations Igor Yanovsky, 5 Since u n = u ( y = f(x, we have uδ(x ξ dx = G((x, y,ξ f(x dx, For y =,wehave u(ξ = G((x, y,ξ f(x dx. G((x, y,ξ = π log (x x + y + (x π log x + y = (x π log x + y = π log [ (x x + y ]. Thus, u(ξ = log [ (x x + y π ] f(x dx. b Show that this solution is bounded in if and only if f(x dx =. Consider the Green s identity: u udxdy = n ds = u y dx = f(x dx =. Note that the Green s identity applies to bounded domains. R π u fdx + Rdθ =. R r???

201 Partial Differential Equations Igor Yanovsky, 5 McOwen 4. # 6. For n =, use the method of reflections to find the Green s function for the first quadrant ={(x, y:x, y > }. Proof. For x, x ξ ( x ξ ( = x ξ ( x ξ (3, x ξ ( = x ξ( x ξ (3 x ξ (. But ξ ( = ξ, soforn =, G(x, ξ = log x ξ π π log x ξ( x ξ (3 x ξ (. G(x, ξ=,x. Problem. Use the method of images to solve G = δ(x ξ in the first quadrant with G =on the boundary. Proof. To solve the problem in the first quadrant we take a reflection to the fourth quadrant and the two are reflected to the left half. G = δ(x ξ ( δ(x ξ ( δ(x ξ ( +δ(x ξ (3. G = π log x ξ( x ξ (3 x ξ ( x ξ ( = (x π log x +(y y (x + x +(y + y (x x +(y + y (x + x +(y y. Note that on the axes G =.

202 Partial Differential Equations Igor Yanovsky, 5 Problem (S 96, #3. Construct a Green s function for the following mixed Dirichlet- Neumann problem in ={x =(x,x R : x >, x > }: u = u + u x x = f, x, u x (x, =, x >, u(,x =, x >. Proof. Notation: x =(x, y, ξ =(x,y. Since K(x ξ = π log x ξ, n =. K(x ξ = π log (x x +(y y. Let G(x, ξ=k(x ξ+ω(x. At (,y, y>, G ( (,y,ξ = π log x +(y y + ω(,y =. Also, G y ( (x, y,ξ = π = π (y y (x x +(y y + w y(x, y y y (x x +(y y + w y(x, y. At (x,, x>, ( y G y (x,,ξ = π (x x + y + w y (x, =. We have ω((x, y,ξ = a π log (x + x +(y y + b π log (x x +(y + y + c π log (x + x +(y + y. Using boundary conditions, we have = G((,y,ξ= π log x +(y y + ω(,y = π log x +(y y + a π log x +(y y + b π log x +(y + y + c π log x +(y + y. Thus, a =, c = b. Also, = G y ((x,,ξ= y π (x x + y + w y (x, = y π (x x + y ( y π (x + x + y + b y π (x x + y + ( b y π (x + x + y. Thus, b =,and G((x, y,ξ = π log (x x +(y y + ω(x = [ log (x x π +(y y

203 Partial Differential Equations Igor Yanovsky, 5 3 log (x + x +(y y +log (x x +(y + y log (x + x +(y + y ]. It can be seen that G((x, y,ξ= onx =, for example.

204 Partial Differential Equations Igor Yanovsky, 5 4 Dirichlet Problem on a Ball. Solve the n-dimensional Laplace/Poisson equation on the ball with Dirichlet boundary conditions. Proof. Use the method of reflection to construct Green s function. Let = {x R n : x <a}. For ξ, define ξ = a ξ as its reflection in ; note ξ ξ /. x ξ x ξ = a ξ for x = a. x ξ = ξ a x ξ. (7. From (7. we conclude that for x (i.e. x = a, ( π K(x ξ = log ξ a x ξ if n = ( a n K(x ξ ξ if n 3. Define for x, ξ : G(x, ξ= a x ξ ( K(x ξ π log ξ if n = K(x ξ ( a n K(x ξ ξ if n 3. Since ξ is not in, the second terms on the RHS are harmonic in x. Moreover, by (7.3 we have G(x, ξ=ifx. Thus, G is the Green s function for. u(ξ = G(x, ξ g(x ds x = a ξ g(x n x aω n x =a x ξ n ds x. (7.3

205 Partial Differential Equations Igor Yanovsky, The Fundamental Solution Problem (F 99, #. ➀ Given that K a (x y and K b (x y are the kernels for the operators ( ai and ( bi on L (R n,where <a<b, show that ( ai( bi has a fundamental solution of the form c K a + c K b. ➁ Use the preceding to find a fundamental solution for,whenn =3. Proof. METHOD ❶: ➀ ( aiu = f ( biu = f u = K }{{} a u = K b }{{} f fundamental solution kernel û = K a f û = K b f if u L, ( aiu =( ξ aû = f ( biu =( ξ bû = f û = (ξ + a f(ξ û = (ξ + b f(ξ K a = K ξ b = + a ξ + b ( ai( biu = f, ( (a + b + abi u = f, û = (ξ + a(ξ + b f(ξ = K new f(ξ, K new = (ξ + a(ξ = + b b a K new = b a (K b K a, ( ξ + b + ξ + a c = b a, c = b a. ➁ n = 3 is not relevant (may be used to assume K a,k b L. For, a =,b = above, or more explicitly ( u = f, (ξ 4 + ξ û = f, û = (ξ 4 + ξ f, K = = b a ( K b K a, (ξ 4 + ξ = ξ (ξ + = ξ + + ξ = K K.

206 Partial Differential Equations Igor Yanovsky, 5 6 METHOD ❷: For u C (Rn wehave: u(x = K a (x y( ai u(y dy, R n u(x = K b (x y( bi u(y dy. R n Let u(x = c ( bi φ(x, for ➀ u(x = c ( ai φ(x, for ➁ ➀ ➁ for φ(x C (Rn. Then, c ( biφ(x = K a (x y( ai c ( biφ(y dy, R n c ( aiφ(x = K b (x y( bi c ( aiφ(y dy. R n We add two equations: (c + c φ(x (c b + c aφ(x = (c K a + c K b ( ai( bi φ(y dy. R n If c = c and (c b + c a =, that is, c = a b, we have: φ(x = R n a b (K a K b ( ai( bi φ(y dy, which means that a b (K a K b is a fundamental solution of ( ai( bi. = ( = ( I( I. ( I has fundamental solution K = 4πr in R 3. To find K, afundamental solution for ( I, we need to solve for a radially symmetric solution of ( IK = δ. In spherical coordinates, in R 3, the above expression may be written as: K + r K K =. Let K = r w(r, K = r w r w, K = r w r w + r 3 w. Plugging these into, we obtain: r w w r =, or w w =.

207 Partial Differential Equations Igor Yanovsky, 5 7 Thus, w = c e r + c e r, K = r w(r = c e r r + c e r r. Suppose v(x for x R and let = B R (; for small ɛ>let ɛ = B ɛ (. Note: ( IK( x = in ɛ. Consider Green s identity ( ɛ = B ɛ (: ( ( K( x v v K( x dx = K( x v ɛ n v K( x ds + n B } {{ } ɛ( =, sincev for x R We add ɛ K( x vdx + ɛ vk( x dx to LHS to get: ( ( K( x ( Iv v ( IK( x dx = K( x v ɛ } {{ } B ɛ( n v K( x ds. n =,in ɛ [ ] lim K( x ( Ivdx ɛ ɛ = K( x ( Ivdx. (Since K(r=c e r r + c ( K( x v n v K( x n e r r ds is integrable at x =. On B ɛ (, K( x =K(ɛ. Thus, 47 K( x v B ɛ( n ds = K(ɛ v e ɛ ds c B ɛ( n ɛ + c e ɛ 4πɛ max v, as ɛ. ɛ v(x K( x [ ( ds = c e ɛ + c e ɛ + ( c B ɛ( n B ɛ( ɛ ɛ e ɛ + c e ɛ] v(x ds = = [ ɛ [ ( c e ɛ + c e ɛ + ( c ɛ ɛ e ɛ + c e ɛ] v( ds Bɛ( ( c e ɛ + c e ɛ + ( c ɛ e ɛ + c e ɛ] v(x ds Bɛ( [ ( + c e ɛ + c e ɛ + ( c ɛ ɛ e ɛ + c e ɛ] [v(x v(] ds Bɛ( ɛ ( c e ɛ + c e ɛ v( 4πɛ 4π(c + c v( = v(. Thus, taking c = c, we have c = c = 8π,whichgives K( x ( Ivdx = lim K( x ( Ivdx = v(, ɛ ɛ 47 In R 3,for x = ɛ, e ɛ K( x = K(ɛ = c K( x n = K(ɛ r ɛ + c e ɛ ɛ. = c ( e ɛ ɛ eɛ ɛ ( c e ɛ e ɛ ɛ ɛ since n points inwards. n points toward on the sphere x = ɛ (i.e., n = x/ x. = ɛ ( ce ɛ + c e ɛ + ɛ ( ce ɛ + c e ɛ,

208 Partial Differential Equations Igor Yanovsky, 5 8 ( that is K(r = 8π e r r ( I. By part (a, + e r r = 4πr cosh(r is the fundamental solution of a b (K a K b is a fundamental solution of ( ai( bi. Here, the fundamental solution of ( I( I is (K K = 4πr cosh(r = 4πr( cosh(r. ( 4πr +

209 Partial Differential Equations Igor Yanovsky, 5 9 Problem (F 9, #3. Prove that cos k x 4π x is a fundamental solution for ( + k in R 3 where x = x + x + x 3, i.e. prove that for any smooth function f(x with compact support u(x = cos k x y f(y dy 4π R 3 x y is a solution to ( + k u = f. Proof. For v C (Rn, we want to show that for K( x = cos k x 4π x, we have ( + k K = δ, i.e. K( x ( + k v(x dx = v(. R n Suppose v(x for x R and let = B R (; for small ɛ>let ɛ = B ɛ (. ( + k K( x =in ɛ. Consider Green s identity ( ɛ = B ɛ (: ( ( K( x v v K( x dx = K( x v ɛ n v K( x ds + n } {{ } =, sincev for x R B ɛ( We add ɛ k K( x vdx ɛ vk K( x dx to LHS to get: ( ( K( x ( + k v v ( + k K( x dx = K( x v ɛ } {{ } B ɛ( n v K( x ds. n =,in ɛ [ ] lim K( x ( + k vdx ɛ ɛ On B ɛ (, K( x =K(ɛ. Thus, 48 K( x v n ds = K(ɛ B ɛ( = 48 In R 3,for x = ɛ, cos kɛ K( x = K(ɛ = 4πɛ. K( x n = K(ɛ r K( x ( + k vdx. B ɛ( v ds n = ( k sin kɛ cos kɛ 4π ɛ ɛ ( cos kr Since K(r = 4πr cos kɛ 4πɛ = ( k sin kɛ + 4πɛ since n points inwards. n points toward on the sphere x = ɛ (i.e., n = x/ x. ( K( x v n v K( x ds n is integrable at x =. 4πɛ max v, as ɛ. cos kɛ, ɛ

210 Partial Differential Equations Igor Yanovsky, 5 v(x K( x ds = B ɛ( n = ( cos kɛ k sin kɛ + 4πɛ ɛ = ( cos kɛ k sin kɛ + 4πɛ ɛ ( k sin kɛ + = 4πɛ cos kɛ ɛ cos kɛ v( v(. B ɛ( B ɛ( B ɛ( ( k sin kɛ + 4πɛ v(x ds v( 4πɛ cos kɛ v(x ds ɛ v( ds ( cos kɛ k sin kɛ + 4πɛ ɛ B ɛ( ( cos kɛ k sin kɛ + [v(x v(] 4πɛ } 4πɛ ɛ {{ }, (v is continuous Thus, K( x ( + k vdx = lim K( x ( + k vdx = v(, ɛ ɛ that is, K(r = 4π cos kr r is the fundamental solution of + k. [v(x v(] ds Problem (F 97, #. Let u(x be a solution of the Helmholtz equation u + k u = x R 3 satisfying the radiation conditions ( u = O, r u ( r iku = O r, x = r. Prove that u. Hint: A fundamental solution to the Helmholtz equation is 4πr eikr. Use the Green formula. Proof. Denote K( x = 4πr eikr, a fundamental solution. Thus, ( + k K = δ. Let x be any point and = B R (x ; for small ɛ>let ɛ = B ɛ (x. ( + k K( x =in ɛ. Consider Green s identity ( ɛ = B ɛ (x : ( ( u ( + k K K( + k u dx = u K } ɛ n K u ( ds + u K n B {{ } ɛ(x n K u ds. n } {{ } = u(x,asɛ (It can be shown by the method previously used that the integral over B ɛ (x approaches u(x asɛ. Taking the limit when ɛ, we obtain ( u(x = u K n K u ( ds = u e ik x x n r 4π x x eik x x u ds 4π x x r ( [ e ik x x ] eik x x = u ik eik x x [ u ] iku ds r 4π x x 4π x x 4π x x r } {{ } ( = O O R = O( x ; (can be shown ( ( ( R 4πR O O R R 4πR =. Taking the limit when R,wegetu(x =.

211 Partial Differential Equations Igor Yanovsky, 5 Problem (S, #. a Find a radially symmetric solution, u, to the equation in R, u = log x, π and show that u is a fundamental solution for, i.e. show φ( = u φdx R for any smooth φ which vanishes for x large. b Explain how to construct the Green s function for the following boundary value in a bounded domain D R with smooth boundary D w w = and n = w = f in D. on D, Proof. a Rewriting the equation in polar coordinates, we have u = ( rur r r + r u θθ = log r. π For a radially symmetric solution u(r, we have u θθ =. Thus, ( rur = r r log r, π ( rur = r log r, r π ru r = r log rdr = r log r r π 4π 8π, u r = r log r 4π r 8π, u = 4π u(r = 8π r( log r. r log rdr 8π rdr = 8π r( log r. We want to show that u defined above is a fundamental solution of for n =. That is u vdx = v(, v C (R n. R See the next page that shows that u defined as u(r = 8π r log r is the Fundamental Solution of. (The 8π r term does not play any role. In particular, the solution of if given by ω = f(x, ω(x = R u(x y ω(y dy = 8π R x y ( log x y f(y dy.

212 Partial Differential Equations Igor Yanovsky, 5 b Let K(x ξ = 8π x ( ξ log x ξ. We use the method of images to construct the Green s function. Let G(x, ξ=k(x ξ+ω(x. We need G(x, ξ=and G n (x, ξ=forx. Consider w ξ (x with w ξ (x =in,w ξ (x = K(x ξ and w ξ K n (x = n (x ξ on. Note, we can find the Greens function for the upper-half plane, and then make a conformal map onto the domain.

213 Partial Differential Equations Igor Yanovsky, 5 3 Problem (S 97, #6. Show that the fundamental solution of in R is given by V (x,x = 8π r ln(r, r = x ξ, and write the solution of w = F (x,x. Hint: In polar coordinates, = r r (r r + ; for example, V = r θ π ( +ln(r. Proof. Notation: x =(x,x. We have V (x = 8π r log(r, In polar coordinates: (here, V θθ = V = ( rvr r r = ( ( r r 8π r log(r = ( ( r r log(r+r r r 8π r r = (r log(r+r = ( 4r +4r log r 8π r r 8π r = ( + log r. π The fundamental solution V (x for is the distribution satisfying: V (r =δ(r. ( V = ( V = π ( + log r = ( ( + log r = r( + log rr π π r r = ( r π r r = r π r ( r = for r. Thus, V (r=δ(r V is the fundamental solution. The approach above is not rigorous. See the next page that shows that V defined above is the Fundamental Solution of. The solution of if given by ω = F (x, ω(x = R V (x y ω(y dy = 8π R x y log x y F (y dy.

214 Partial Differential Equations Igor Yanovsky, 5 4 Show that the Fundamental Solution of in R is given by: K(x = 8π r ln(r, r = x ξ, (7.4 Proof. For v C (Rn, we want to show K( x v(x dx = v(. R n Suppose v(x for x R and let = B R (; for small ɛ>let ɛ = B ɛ (. K( x isbiharmonic( K( x =in ɛ. Consider Green s identity ( ɛ = B ɛ (: K( x ( v vdx = K( x ɛ n v K( x ( v K( x ds + K( x v ds n n n } {{ } + [ ] lim K( x vdx ɛ ɛ B ɛ( = =, sincev for x R ( v K( x n v K( x ds + n K( x v dx. On B ɛ (, K( x =K(ɛ. Thus, 49 K( x v n ds = K(ɛ B ɛ( B ɛ( B ɛ( v(x K( x n B ɛ( ds = B ɛ( = 8π ɛ log(ɛ ω n ɛ max = B ɛ( B ɛ( B ɛ( ( K( x v n ( Since K(r isintegrableatx =. v ds K(ɛ ω n ɛ max ( v n x ( v, as ɛ. x v(x ds πɛ v( ds + πɛ = v( πɛ πɛ max K( x v n ds = K(ɛ v K( x ds = n B ɛ( B ɛ( x B ɛ( [v(x v(] ds B ɛ( πɛ v(x v( = v(. } {{ }, (v is continuous K( x v ds. n v ds n π ( + log ɛ πɛmax v, as ɛ. ( 4π ɛ log ɛ ɛ log ɛ + πɛ 4π 8π ɛ v(x ds max x B ɛ( x v, as ɛ. 49 Note that for x = ɛ, K( x = K(ɛ = 8π ɛ log ɛ, K = ( + log ɛ, π K( x = K(ɛ = n r 4π ɛ log ɛ 8π ɛ, K n = K = r πɛ.

215 Partial Differential Equations Igor Yanovsky, 5 5 K( x vdx = lim ɛ ɛ K( x vdx = v(.

216 Partial Differential Equations Igor Yanovsky, Radial Variables Problem (F 99, #8. Let u = u(x, t solve the following PDE in two spatial dimensions u = for r<r(t, inwhichr = x is the radial variable, with boundary condition u = on r = R(t. In addition assume that R(t satisfies dr dt = u (r = R r with initial condition R( = R. a Find the solution u(x, t. b Find an ODE for the outer radius R(t, andsolveforr(t. Proof. a Rewrite the equation in polar coordinates: u = ( r (ru r r + r u θθ =. For a radially symmetric solution u(r, we have u θθ =. Thus, r (ru r r =, (ru r r = r, ru r = r + c, u r = r + c r, u(r, t = r 4 + c log r + c. Since we want u to be defined for r =,wehavec =. Thus, u(r, t = r 4 + c. Using boundary conditions, we have u(r(t,t = R(t 4 b We have u(r, t= r 4 + R(t. 4 + c = c = R(t. Thus, 4 u(r, t = r 4 + R(t, 4 u r = r, dr dt = u r (r = R = R, (from dr R = dt, log R = t, R(t = c e t, R( = c = R. Thus,

217 Partial Differential Equations Igor Yanovsky, 5 7 R(t =R e t.

218 Partial Differential Equations Igor Yanovsky, 5 8 Problem (F, #3. Let u = u(x, t solve the following PDE in three spatial dimensions u = for R <r<r(t, inwhichr = x is the radial variable, with boundary conditions u(r = R(t,t=, and u(r = R,t=. In addition assume that R(t satisfies dr dt = u (r = R r with initial condition R( = R in which R >R. a Find the solution u(x, t. b Find an ODE for the outer radius R(t. Proof. a Rewrite the equation in spherical coordinates (n =3,radialfunctions: ( u = r + u = r r r (r u r r =. (r u r r =, r u r = c, u r = c r, u(r, t = c r + c. Using boundary conditions, we have u(r(t,t = c R(t + c = c = c R(t, u(r,t = c + c R =. This gives c = R R(t R R(t, c = R R R(t. u(r, t = R R(t R R(t r + R R R(t. b We have u(r, t = R R(t R R(t r + R R R(t, u R R(t = r R R(t r, dr = u dt r (r = R = R R(t R R(t Thus, an ODE for the outer radius R(t is R(t = R (R R(t R(t (from { dr dt = R (R(t R R(t, R( = R, R >R.

219 Partial Differential Equations Igor Yanovsky, 5 9 Problem (S, #3. Steady viscous flow in a cylindrical pipe is described by the equation ( u u + ρ p η u = ρ on the domain <x <, x +x 3 R,where u =(u,u,u 3 =(U(x,x 3,, is the velocity vector, p(x,x,x 3 is the pressure, and η and ρ are constants. p a Show that x is a constant c, and that U = c/η. b Assuming further that U is radially symmetric and U =on the surface of the pipe, determine the mass Q of fluid passing through a cross-section of pipe per unit time in terms of c, ρ, η, andr. Note that Q = ρ Udx dx 3. {x +x 3 R } Proof. a Since u =(u,u,u 3 =(U(x,x 3,,, we have ( u ( u u = (u,u,u 3, u, u 3 = (U(x,x 3,, (,, =. x x x 3 Thus, ρ p η u =, ρ p = η u, ( p, p, p = η( u, u, u 3, x x x 3 ( p, p, p = η(u x x x x x + U x3 x 3,,. 3 We can make the following observations: p = η (U x x x + U x3 x 3 } {{ } indep. of x, p x = p = f(x,x 3, p x 3 = p = g(x,x. Thus, p = h(x. But p x = η U, U = η p x p x = c η. is independent of x. Therefore, p x = c.

220 Partial Differential Equations Igor Yanovsky, 5 b Cylindrical Laplacian in R 3 for radial functions is U = ( rur r r, ( rur = c r r η, ( rur = cr r η, ru r = cr η + c, U r = cr η + c r. For U r to stay bounded for r =, we need c =. Thus, U r = cr η, U = cr 4η + c, = U(R = cr 4η + c, U = cr 4η cr 4η = c 4η (r R. Q = ρ Udx dx 3 = cρ π {x +x 3 R } 4η = cρr4 π. 8η It is not clear why Q is negative? R (r R rdrdθ = cρ π R 4 4η 4 dθ

221 Partial Differential Equations Igor Yanovsky, Weak Solutions Problem (S 98, #. Afunctionu H ( is a weak solution of the biharmonic equation u = f in u = on on provided u n = u vdx= fvdx for all test functions v H (. Prove that for each f L (, there exists a unique weak solution for this problem. Here, H ( is the closure of all smooth functions in which vanish on the boundary and with finite H norm: u = (u xx + u xy + u yy dxdy <. Hint: use Lax-Milgram lemma. Proof. Multiply the equation by v H ( and integrate over : u = f, uvdx = fvdx, u n vds u v n ds + u vdx = fvdx, } {{ } = u vdx = fvdx. Denote: V = H (. } {{ } a(u,v Check the following conditions: } {{ } L(v ❶ a(, is continuous: γ >, s.t. a(u, v γ u V v V, u, v V ; ❷ a(, is V-elliptic: α >, s.t. a(v, v α v V, v V ; ❸ L( is continuous: Λ >, s.t. L(v Λ v V, v V.

222 Partial Differential Equations Igor Yanovsky, 5 We have 5 ❶ a(u, v = ❷ a(v, v = ❸ L(v = u vdx ( ( v dx v H (. fvdx ( ( u dx f v dx ( = f L ( v L ( f L ( v H } {{ } (. Λ ( v dx ( f dx v dx Thus, by Lax-Milgram theorem, there exists a weak solution u H (. u H ( v H (. Also, we can prove the stability result. α u H ( a(u, u = L(u Λ u H (, u H ( Λ α. Let u, u be two solutions so that a(u,v = L(v, a(u,v = L(v for all v V. Subtracting these two equations, we see that: a(u u,v = v V. Applying the stability estimate (with L, i.e. Λ =, we conclude that u u H ( =, i.e. u = u. 5 Cauchy-Schwarz Inequality: (u, v u v in any norm, for example uv dx ( u dx ( v dx ; a(u, v a(u, u a(v, v ; v dx = v dx = ( v dx ( dx. Poincare Inequality: v H ( C ( v dx. Green s formula: ( u dx = (u xx + u yy +u xxu yy dxdy = (u xx + u yy u xxyu y dxdy = (u xx + u yy + u xy dxdy u H (.

223 Partial Differential Equations Igor Yanovsky, Uniqueness Problem. The solution of the Robin problem u + αu = β, n x for the Laplace equation is unique when α> is a constant. Proof. Let u and u be two solutions of the Robin problem. Let w = u u.then w = in, w + αw = n on. Consider Green s formula: u vdx = v u n ds v udx. Setting w = u = v gives w dx = w w n ds w wdx. } {{ } = Boundary condition gives w dx = αw ds. } {{ } } {{ } Thus, w, and u u. Hence, the solution to the Robin problem is unique. Problem. Suppose q(x for x and consider solutions u C ( C ( of u q(xu = in. Establish uniqueness theorems for a the Dirichlet problem: u(x =g(x, x ; b the Neumann problem: u/ n = h(x, x. Proof. Let u and u be two solutions of the Dirichlet or Neumann problem. Let w = u u.then w q(xw = in, w w = or n = on. Consider Green s formula: u vdx = v u n ds v udx. Setting w = u = v gives w dx = w w n ds } {{ } =, Dirichlet or Neumann w wdx.

224 Partial Differential Equations Igor Yanovsky, 5 4 w dx } {{ } = q(xw dx } {{ }. Thus, w, and u u. Hence, the solution to the Dirichlet and Neumann problems are unique. Problem (F, #8; S 93, #5. Let D be a bounded domain in R 3. Show that a solution of the boundary value problem is unique. u = f in D, u = u = on D Proof. Method I: Maximum Principle. Let u, u be two solutions of the boundary value problem. Define w = u u.thenw satisfies w = ind, w = w = on D. So w is harmonic and thus achieves min and max on D w. So w is harmonic, but w on D w. Hence, u = u. Method II: Green s Identities. w w =, w wdx =, w ( w n ds w ( w dx } {{ } =, = w n wds + ( w dx } {{ } =. = Thus, w. Now, multiply w =byw. Weget w dx =. Multiply the equation by w and integrate: Thus, w =andw is a constant. Since w =on, we have w. Problem (F 97, #6. a Let u(x be continuous in closed bounded domain R n, u is continuous in, u = u and u =. Prove that u. b What can you say about u(x when the condition u(x in is dropped?

225 Partial Differential Equations Igor Yanovsky, 5 5 Proof. a Multiply the equation by u and integrate: u u n ds } {{ } = u u = u 3, u u dx = u 3 dx, u dx = u 3 dx, ( u 3 + u dx =. Since u(x, we have u. b If we don t know that u(x, then u can not be nonnegative on the entire domain. That is, u(x <, on some (or all parts of. If u is nonnegative on, then u.

226 Partial Differential Equations Igor Yanovsky, 5 6 Problem (W, #5. Consider the boundary value problem u + n k= α k u x k u 3 = in, u = on, where is a bounded domain in R n with smooth boundary. If the α k s are constants, and u(x has continuous derivatives up to second order, prove that u must vanish identically. Proof. Multiply the equation by u and integrate: n u u u + α k u u 4 x k =, u u n ds } {{ } = u udx + u dx + n k= n k= k= α k u x k udx α k u x k udx } {{ } ➀ We will show that ➀ =. u α k udx = α k u ds α k u u dx, x k x } {{ } k = u α k udx =, x k n u α k udx =. x k k= Thus, we have u dx u 4 dx =, ( u + u 4 dx =. Hence, u =andu 4 =. Thus,u. u 4 dx =, u 4 dx =. Note that n k= α k u x k udx = and thus, α uudx =. α uudx = α nu ds α uudx, } {{ } =

227 Partial Differential Equations Igor Yanovsky, 5 7 Problem (W, #9. Let D = {x R : x,x }, and assume that f is continuous on D and vanishes for x >R. a Show that the boundary value problem u = f in D, u(x, = u (,x = x can have only one bounded solution. b Find an explicit Green s function for this boundary value problem. Proof. a Let u, u be two solutions of the boundary value problem. Define w = u u.thenw satisfies w = ind, w(x, = w (,x =. x Consider Green s formula: u vdx = v u D D n ds v udx. D Setting w = u = v gives w dx = w w D D n ds w wdx, D w dx = w w D R x n ds + w w R x n ds + x >R w = w(x, ds + w(,x R x } {{ } w x R x x = D w dx = w = w =const. Since w(x, = w. Thus, u = u. }{{} = w w n ds ds + x >R D w wdx w }{{} = b The similar problem is solved in the appropriate section (S 96, #3. Notice whenever you are on the boundary with variable x, x ξ ( = x ξ( x ξ (3 x ξ (. So, G(x, ξ= ( log x ξ log x ξ( x ξ (3 π x ξ ( is the Green s function. w n ds w w D }{{} = dx,

228 Partial Differential Equations Igor Yanovsky, 5 8 Problem (F 98, #4. In two dimensions x =(x, y, define the set a as in which a = + + = { x x a} {x } = { x + x a} {x } = + and x =(,. Note that a consists of two components when <a< and a single component when a>. Consider the Neumann problem in which u = f, x a u/ n =, x a + f(x dx = f(x dx = a Show that this problem has a solution for <a,butnotfor <a<. (You do not need to construct the solution, only demonstrate solveability. b Show that max a u as a from above. (Hint: Denote L to be the line segment L = +, and note that its length L goes to as a. Proof. a We use the Green s identity. For <a, u = a n ds = udx = f(x dx a a = f(x dx + f(x dx = =. + Thus, the problem has a solution for <a. For <a<, + and are disjoint. Consider + : u = + n ds = udx = f(x dx =, + + u = n ds = udx = f(x dx =. We get contradictions. Thus, the solution does not exist for <a<.

229 Partial Differential Equations Igor Yanovsky, 5 9 b Using the Green s identity, we have: (n + is the unit normal to + u udx = + + n + ds = u L n + ds, u udx = n ds = u L n ds = u L n + ds. u udx udx = + L n + ds, u f(x dx f(x dx = + L n + ds. u = n + ds, = Thus, L L u n + ds L u ds n + max u a L. As a (L max a u. L ( u ( u + n + τ L max L u L max u. a

230 Partial Differential Equations Igor Yanovsky, 5 3 Problem (F, #. Consider the Dirichlet problem in a bounded domain D R n with smooth boundary D, u + a(xu = f(x in D, u = on D. a Assuming that a(x is small enough, prove the uniqueness of the classical solution. b Prove the existence of the solution in the Sobolev space H (D assuming that f L (D. Note: Use Poincare inequality. Proof. a By Poincare Inequality, for any u C (D, we have u C u. Consider two solutions of the Dirichlet problem above. Let w = u u. Then, w satisfies { w + a(xw = ind, w = on D. w w + a(xw =, w wdx+ a(xw dx =, w dx + a(xw dx =, a(xw dx = w dx C a(xw dx w dx, C a(x w dx w dx, C ( a(x w dx. C w dx, (by Poincare inequality If a(x < C w. b Consider F (v, u= (v u + a(xvu dx = vf(x dx = F (v. F (v is a bounded linear functional on v H, (D, D =. F (v f v f C v H, (D So by Riesz representation, there exists a solution u H, (D of <u,v>= v u + a(xvu dx = vf(x dx = F (v v H, (D.

231 Partial Differential Equations Igor Yanovsky, 5 3 Problem (S 9, #8. Define the operator Lu = u xx + u yy 4(r +u in which r = x + y. a Show that ϕ = e r satisfies Lϕ =. b Use this to show that the equation Lu = f in u n = γ on has a solution only if ϕf dx = ϕγ ds(x. Proof. a Expressing Laplacian in polar coordinates, we obtain: Lu = r (ru r r 4(r +u, Lϕ = r (rϕ r r 4(r +ϕ = r (r e r r 4(r +e r = r (4rer +r re r 4r e r 4e r =. b We have ϕ = e r = e x +y = e x e y. From part (a, Lϕ =, ϕ n = ϕ n =(ϕ x,ϕ y n =(xe x e y, ye x e y n =e r (x, y ( y, x=. 5 Consider two equations: Lu = u 4(r +u, Lϕ = ϕ 4(r +ϕ. Multiply the first equation by ϕ and the second by u and subtract the two equations: ϕlu = ϕ u 4(r +uϕ, ulϕ = u ϕ 4(r +uϕ, ϕlu ulϕ = ϕ u u ϕ. Then,westartfromtheLHSoftheequalityweneedtoproveandendupwithRHS: ϕf dx = ϕlu dx = (ϕlu ulϕ dx = (ϕ u u ϕ dx = (ϕ u n u ϕ n ds = ϕ u n ds = ϕγ ds. 5 The only shortcoming in the above proof is that we assume n =( y, x, without giving an explanation why it is so.

232 Partial Differential Equations Igor Yanovsky, Self-Adjoint Operators Consider an mth-order differential operator Lu = a α (xd α u. α m The integration by parts formula u xk vdx = uvn k ds with u or v vanishing near is: u xk vdx = uv xk dx. uv xk dx n =(n,...,n n R n, We can repeat the integration by parts with any combination of derivatives D α =( / x α ( / x n αn : (D α uvdx =( m ud α vdx, (m = α. We have (Lu vdx = ( α m = ( α = α m L (v udx, for all u C m ( and v C. The operator L (v = a α (xd α u vdx = ( α D α (a α (x v α m is called the adjoint of L. The operator is self-adjoint if L = L. Also, L is self-adjoint if 5 vl(u dx = ul(v dx. α m D α (a α (x v udx = a α (x vd α udx ( α D α (a α (x v udx α m } {{ } L (v 5 L = L (Lu v =(u L v=(u Lv.

233 Partial Differential Equations Igor Yanovsky, 5 33 Problem (F 9, #6. Consider the Laplace operator in the wedge x y with boundary conditions f = on x = x f x α f = on x = y. y a For which values of α is this operator self-adjoint? b For such a value of α, suppose that f = e r / cos θ with these boundary conditions. Evaluate r fds C R in which C R is the circular arc of radius R connecting the boundaries x =and x = y. Proof. a We have Lu = u = u = on x = x u x α u = on x = y. y The operator L is self-adjoint if: (ulv vlu dx =. (ulv vlu dx = = = = = = (u v v u dx = x= x= x= + ( u v ds + n v u n ( u ( v n v ( u n x=y ( u v n v u n ( u v ds n v u n ds + x=y ds ( u ( v n v ( u n ( u ( (v x,v y (, v ( (u x,u y (, ds x=y ds ( u ( (v x,v y (/, / v ( (u x,u y (/, / ds ( u ( (,v y (, v ( (,u y (, ds x= } {{ } + x=y x=y = ( u ( (αv y,v y (/, / v ( (αu y,u y (/, / ds ( uvy (α vu y (α ds = }{{} need.

234 Partial Differential Equations Igor Yanovsky, 5 34 Thus, we need α = sothatl is self-adjoint. b We have α =. Using Green s identity and results from part (a, ( f n =on x =andx = y: f fdx = n ds = f C R n ds + f f ds + x= n x=y C R r ds. Thus, C R f r ds = = fdx = R π π 4 }{{} = ( R e r / rdr = e r / cos θ r drdθ f ds = }{{} n = ( ( e R /.

235 Partial Differential Equations Igor Yanovsky, 5 35 Problem (F 99, #. Suppose that u =intheweaksenseinr n and that there is a constant C such that u(y dy < C, x R n. { x y <} Show that u is constant. Proof. Consider Green s formula: u vdx = v u n ds v udx For v =,wehave u n ds = udx. Let B r (x be a ball in R n.wehave u u = udx = ds = rn B r(x B r(x n x = r (x + rx ds = r n ω n u(x + rx ds. r ω n x = Thus, ω n x = u(x + rx ds is independent of r. Hence, it is constant. By continuity, as r, we obtain the Mean Value property: u(x = ω n x = u(x + rx ds. If x y < u(y dy < C x Rn, we have u(x <C in R n. Since u is harmonic and bounded in R n, u is constant by Liouville s theorem Liouville s Theorem: A bounded harmonic function defined on R n is a constant.

236 Partial Differential Equations Igor Yanovsky, 5 36 Problem (S, #. For bodies (bounded regions B in R 3 which are not perfectly conducting one considers the boundary value problem = γ(x u = u = f on B. 3 j= ( γ(x u x j x j The function γ(x is the local conductivity of B and u is the voltage. We define operator Λ(f mapping the boundary data f to the current density at the boundary by Λ(f=γ(x u n, and / n is the inward normal derivative (this formula defines the current density. a Show that Λ is a symmetric operator, i.e. prove gλ(f ds = fλ(g ds. B B b Use the positivity of γ(x > to show that Λ is negative as an operator, i.e., prove fλ(f ds. B Proof. a Let { γ(x u = on, u = f on. Λ(f =γ(x u n, { γ(x v = on, v = g on. v Λ(g=γ(x n. Since / n is inward normal derivative, Green s formula is: We have }{{} v =g gλ(f ds = γ(x u n ds = = b We have γ(x >. fλ(f ds = = v γ(x udx= v γ(x udx. gγ(x u n ds = v γ(x udx v γ(x u } {{ } = uγ(x v n ds + fγ(x v n ds = u γ(x v dx } {{ } = fλ(g ds. uγ(x u n ds = u γ(x u dx γ(x u udx } {{ } = γ(x u } {{ }. dx

237 Partial Differential Equations Igor Yanovsky, 5 37 Problem (S, #4. The Poincare Inequality states that for any bounded domain in R n there is a constant C such that u dx C u dx for all smooth functions u which vanish on the boundary of. a Find a formula for the best (smallest constant for the domain in terms of the eigenvalues of the Laplacian on, and b give the best constant for the rectangular domain in R ={(x, y: x a, y b}. Proof. a Consider Green s formula: u vdx = v u n ds v udx. Setting u = v and with u vanishing on, Green s formula becomes: u dx = u udx. Expanding u in the eigenfunctions of the Laplacian, u(x = a n φ n (x, the formula above gives u dx = a n φ n (x λ m a m φ m (x dx = λ m a n a m φ n φ m dx Also, = u dx = n= λ n a n. n= a n φ n (x n= m= a m φ m (x = m= m,n= a n. Comparing and, and considering that λ n increases as n,weobtain λ u dx = λ a n λ n a n = u dx. with C =/λ. u dx λ n= u dx, n= b For the rectangular domain = {(x, y : x a, y b} R,the eigenvalues of the Laplacian are λ mn = π ( m a + n b, m,n =,,... λ = λ = π ( a + b, C = λ = π ( a + b. n=

238 Partial Differential Equations Igor Yanovsky, 5 38

239 Partial Differential Equations Igor Yanovsky, 5 39 Problem (S, #6. a Let B be a bounded region in R 3 with smooth boundary B. The conductor potential for the body B is the solution of Laplace s equation outside B V = in R 3 /B subject to the boundary conditions, V =on B and V (x tends to zero as x. Assuming that the conductor potential exists, show that it is unique. b The capacity C(B of B is defined to be the limit of x V (x as x. Show that C(B = V 4π B n ds, where B is the boundary of B and n is the outer unit normal to it (i.e. the normal pointing toward infinity. c Suppose that B B. Show that C(B C(B. Proof. a Let V, V be two solutions of the boundary value problem. Define W = V V.ThenW satisfies W = in R 3 /B W = on B W as x. Consider Green s formula: u vdx = B Setting W = u = v gives W dx = B B B v u n ds v udx. B }{{} W = W n ds W W B }{{} = dx =. Thus, W = W =const. SinceW =on B, W, and V = V. b & c For (b&(c, see the solutions from Ralston s homework (a few pages down.

240 Partial Differential Equations Igor Yanovsky, 5 4 Problem (W 3, #. Let L be the second order differential operator L = a(x in which x =(x,x,x 3 is in the three-dimensional cube C = { <x i <, i=,, 3}. Suppose that a> in C. Consider the eigenvalue problem Lu = λu for x C u = for x C. a Show that all eigenvalues are negative. b If u and v are eigenfunctions for distinct eigenvalues λ and μ, show that u and v are orthogonal in the appropriate product. c If a(x =a (x +a (x +a 3 (x 3 find an expression for the eigenvalues and eigenvectors of L in terms of the eigenvalues and eigenvectors of a set of one-dimensional problems. Proof. a We have u a(xu = λu. Multiply the equation by u and integrate: u u n ds } {{ } = u u a(xu = λu, u udx a(xu dx = λ u dx, a(xu dx = λ u dx, u dx λ = ( u + a(xu dx u dx <. b Let λ, μ, be the eigenvalues and u, v be the corresponding eigenfunctions. We have u a(xu = λu. (7.5 v a(xv = μv. (7.6 Multiply (7.5 by v and (7.6 by u and subtract equations from each other v u a(xuv = λuv, u v a(xuv = μuv. v u u v = (λ μuv. Integrating over gives ( v u u v dx = (λ μ ( v u n u v dx = (λ μ } {{ n} = Since λ μ, u and v are orthogonal on. uv dx, uv dx.

241 Partial Differential Equations Igor Yanovsky, 5 4 c The three one-dimensional eigenvalue problems are: u x x (x a(x u (x =λ u (x, u x x (x a(x u (x =λ u (x, u 3x3 x 3 (x 3 a(x 3 u 3 (x 3 =λ 3 u 3 (x 3. We need to derive how u, u, u 3 and λ, λ, λ 3 are related to u and λ.

242 Partial Differential Equations Igor Yanovsky, Spherical Means Problem (S 95, #4. Consider the biharmonic operator in R 3 u ( x + y + z u. a Show that is self-adjoint on x < with the following boundary conditions on x =: u =, u =. Proof. a We have Lu = u = u = on x = u = on x =. The operator L is self-adjoint if: (ulv vlu dx =. (ulv vlu dx = (u v v u dx = = u v n ds u ( v dx } {{ } = v u n ds + u vdx + } {{ } = v u n ds } {{ } = u v n ds } {{ } = + v ( u dx v udx =.

243 Partial Differential Equations Igor Yanovsky, 5 43 b Denote x = r and define the averages S(r = (4πr u(x ds, x =r V (r = ( 4 3 πr3 u(x dx. x r Show that d dr S(r= r 3 V (r. Hint: Rewrite S(r as an integral over the unit sphere before differentiating; i.e., S(r = (4π u(rx dx. x = c Use the result of (b to show that if u is biharmonic, i.e. u =,then S(r =u( + r 6 u(. Hint: Use the mean value theorem for u. b Let x = x/ x. We have 54 u(rx r ds = 4π S(r = 4πr u(x ds r = x =r 4πr x = ds = u dr 4π x = r (rx ds = 4π x = = 4πr udx. x r where we have used Green s identity in the last equality. Also r 3 V (r = 4πr udx. x r u n (rx ds = 4πr x = x =r u(rx ds. u n (x ds r c Since u is biharmonic (i.e. u is harmonic, u has a mean value property. We have d dr S(r = r 3 V (r = r ( πr3 u(x dx = r x r 3 u(, S(r = r 6 54 Change of variables: Surface integrals: x = rx in R 3 : u(x ds = u(rx r ds. x =r u( + S( = u( + r 6 u(. x = Volume integrals: ξ = rξ in R n : h(x + ξ dξ = h(x + rξ r n dξ. ξ <r ξ <

244 Partial Differential Equations Igor Yanovsky, 5 44

245 Partial Differential Equations Igor Yanovsky, 5 45 Problem (S, #7. Suppose that u = u(x for x R 3 is biharmonic; i.e. that u ( u =. Show that (4πr u(x ds(x =u( + (r /6 u( x =r through the following steps: a Show that for any smooth f, d f(x dx = f(x ds(x. dr x r x =r b Show that for any smooth f, d dr (4πr f(x ds(x =(4πr x =r x =r in which n is the outward normal to the circle x = r. c Use step (b to show that d dr (4πr f(x ds(x =(4πr x =r x r d Combine steps (a and (c to obtain the final result. n f(x, y ds f(x dx. Proof. a We can express the integral in Spherical Coordinates: 55 R π π f(x dx = f(φ, θ, r r sin φdφdθdr. d dr x R x R f(x dx = d dr = = R π π π π x =R f(φ, θ, r r sin φdφdθdr =??? f(φ, θ, r R sin φdφdθ f(x ds. 55 Differential Volume in spherical coordinates: d 3 ω = ω sin φdφdθdω. Differential Surface Area on sphere: ds = ω sin φdφdθ.

246 Partial Differential Equations Igor Yanovsky, 5 46 b&c We have ( d dr 4πr x =r f(x ds = d ( dr 4πr f(rx r ds x = = f 4π x = r (rx ds = 4π f = 4πr (x ds = n 4πr x =r = 4πr f dx. x r Green s formula was used in the last equality. Alternatively, ( d dr 4πr x =r f(x ds = d ( dr 4πr = d ( dr 4π = 4π = 4π = = π π π π 4πr 4πr π π π π π π x =r = 4π x = x =r ( d f(rx ds dr x = f n (rx ds f nds f(φ, θ, r r sin φdφdθ f(φ, θ, r sinφdφdθ f (φ, θ, r sinφdφdθ r f n sin φdφdθ f nr sin φdφdθ f nds = 4πr f dx. x =r d Since f is biharmonic (i.e. f is harmonic, f has a mean value property. From (c, we have 56 ( d dr 4πr x =r f(x ds(x = 4πr f(x dx = r x r πr3 = r 3 f(. 4πr f(x ds(x = r f( + f(. x =r 6 x r f(x dx 56 Note that part (a was not used. We use exactly the same derivation as we did in S 95 #4.

247 Partial Differential Equations Igor Yanovsky, 5 47 Problem (F 96, #4. Consider smooth solutions of u = k u in dimension d =with k>. a Show that u satisfies the following mean value property : M x (r+ r M x(r k M x (r =, in which M x (r is defined by π M x (r= u(x + r cos θ, y + r sin θ dθ π and the derivatives (denoted by areinr with x fixed. b For k =, this equation is the modified Bessel equation (of order f + r f f =, for which one solution (denoted as I is π I (r = e r sin θ dθ. π Find an expression for M x (r in terms of I. Proof. a Laplacian in polar coordinates written as: u = u rr + r u r + r u θθ. Thus, the equation may be written as u rr + r u r + r u θθ = k u. M x (r = π M x(r = π M x (r = π π π π udθ, u r dθ, u rr dθ. M x (r+ r M x (r k M x (r = π π = πr b Note that w = e r sin θ satisfies w = w, i.e. w = w rr + r w r + r w θθ Thus, ( urr + r u r k u dθ π u θθ dθ = πr [ uθ ] π =. = sin θe r sin θ + r sin θer sin θ + r ( r sin θer sin θ + r cos θe r sin θ =e r sin θ = w. M x (r = e y π π e r sin θ dθ = e y I.

248 Partial Differential Equations Igor Yanovsky, Check with someone about the last result.

249 Partial Differential Equations Igor Yanovsky, Harmonic Extensions, Subharmonic Functions Problem (S 94, #8. Suppose that is a bounded region in R 3 and that u =on. If u =in the exterior region R 3 / and u(x as x,provethe following: a u> in R 3 /; b if ρ(x is a smooth function such that ρ(x =for x >Rand ρ(x=near, then for x >R, u(x = 4π R 3 / ( (ρu(y x y dy. c lim x x u(x exists and is non-negative. Proof. a Let B r ( denote the closed ball {x : x r}. Given ε>, we can find r large enough that B R ( and max x BR ( u(x <ε, since u(x as x. Since u is harmonic in B R, it takes its maximum and minimum on the boundary. Assume min u(x = a < (where a <ε. x B R ( We can find an R such that max x BR ( u(x < a ; hence u takes a minimum inside B R (, which is impossible; hence u. Now let V = {x : u(x } and let α =min x V x. Since u cannot take a minimum inside B R ( (where R>α, it follows that u C and C =, but this contradicts u =on. Hence u>inr 3. b For n =3, K( x y = x y n = ( nω n 4π x y. Since ρ(x=for x >R, then for x/ B R,wehave (ρu = u =. Thus, ( (ρu(y dy 4π R 3 / x y = ( (ρu(y dy 4π B R / x y = ( y y (ρu dy 4π B R / x y 4π = ( ρu dy + 4π B R / x y 4π (B R / =??? = u(x 4πR uds y B 4πR } {{ }, asr = u(x. (B R / n B ( ( ρu n x y u n ds y } {{ }, asr x y ds y ρu ds y 4π (B R / ( ρu n x y ds y c See the next problem.

250 Partial Differential Equations Igor Yanovsky, 5 5 Ralston Hw. a Suppose that u is a smooth function on R 3 and u =for x >R. If lim x u(x =, show that you can write u as a convolution of u with the 4π x and prove that lim x x u(x=exists. b The conductor potential for R 3 is the solution to the Dirichlet problem v =. The limit in part (a is called the capacity of. Show that if, then the capacity of is greater or equal the capacity of. Proof. a If we define v(x= u(y 4π R 3 x y dy, then (u v =inallr 3, and, since v(x as x, we have lim x (u(x v(x =. Thus, u v must be bounded, and Liouville s theorem implies that it is identically zero. Since we now have x u(x= 4π x u(y R 3 x y dy, and x / x y converges uniformly to on { y R}, it follows that lim x u(x = u(y dy. x 4π R 3 b Note that part (a implies that the limit lim x x v(x exists, because we can apply (a to u(x =φ(xv(x, where φ is smooth and vanishes on, but φ(x =for x >R. Let v be the conductor potential for and v for.sincev i as x and v i =on i, the max principle says that >v i (x > forx R 3 i. Consider v v.since, this is defined in R 3, positive on, and has limit as x. Thus, it must be positive in R 3. Thus, lim x x (v v. Problem (F 95, #4. 58 Let be a simply connected open domain in R and u = u(x, y be subharmonic there, i.e. u in. Prove that if D R = {(x, y :(x x +(y y R } then u(x,y π u(x + R cos θ, y + R sin θ dθ. π Proof. Let π M(x,R = u(x + R cos θ, y + R sin θ dθ, π w(r, θ = u(x + R cos θ, y + R sin θ. Differentiate M(x,R with respect to R: d dr M(x,R = πr π 58 See McOwen, Sec.4.3, p.3, #. w r (R, θrdθ,

251 Partial Differential Equations Igor Yanovsky, See ChiuYen s solutions and Sung Ha s solutions (in two places. Nick s solutions, as started above, have a very simplistic approach.

252 Partial Differential Equations Igor Yanovsky, 5 5 Ralston Hw (Maximum Principle. Suppose that u C( satisfies the mean value property in the connected open set. a Show that u satisfies the maximum principle in, i.e. either u is constant or u(x < sup u for all x. b Show that, if v is a continuous function on a closed ball B r (ξ and has the mean value property in B r (ξ, thenu = v on B r (ξ implies u = v in B r (ξ. Doesthis imply that u is harmonic in? Proof. a If u(x is not less than sup u for all x, then the set K = {x : u(x =supu} is nonempty. This set is closed because u is continuous. We will show it is also open. This implies that K = because is connected. Thus u is constant on. Let x K. Sinceisopen, δ >, s.t. B δ (x ={x R n : x x δ}. Let sup u = M. By the mean value property, for r δ M = u(x = A(S n u(x + rξds ξ, and = ξ = A(S n (M u(x + rξds ξ. ξ = Sinse M u(x +rξ is a continuous nonnegative function on ξ, this implies M u(x + rξ = for all ξ S n.thusu =onb δ (x. b Since u v has the mean value property in the open interior of B r (ξ, by part a it satisfies the maximum principle. Since it is continuous on B r (ξ, its supremum over the interior of B r (ξ isitsmaximumonb r (ξ, and this maximum is assumed at a point x in B r (ξ. If x in the interior of B r (ξ, then u v is constant ant the constant must be zero, since this is the value of u v on the boundary. If x is on the boundary, then u v must be nonpositive in the interior of B r (ξ. Applying the same argument to v u, one finds that it is either identically zero or nonpositive in the interior of B r (ξ. Thus, u v onb r (ξ. Yes, it does follow that u harmonic in. Take v in the preceding to be the harmonic function in the interior of B r (ξ which agrees with u on the boundary. Since u = v on B r (ξ, u is harmonic in the interior of B r (ξ. Since is open we can do this for every ξ. Thus u is harmonic in.

253 Partial Differential Equations Igor Yanovsky, 5 53 Ralston Hw. Assume is a bounded open set in R n and the Green s function, G(x, y, for exists. Use the strong maximum principle, i.e. either u(x < sup u for all x, or u is constant, to prove that G(x, y < for x, y, x y. Proof. G(x, y=k(x, y+ω(x, y. For each x, f(y =ω(x, y is continuous on, thus, bounded. So ω(x, y M x for all y. K(x y as y x. Thus, given M x,thereisδ>, such that K(x y < M x when x y = r and <r δ. So for <r δ the Green s function with x fixed satisfies, G(x, y isharmonicon B r (x, and G(x, y on the boundary of B r (x. Since we can choose r as small as we wish, we get G(x, y < fory {x}. Problem (W 3, #6. Assume that u is a harmonic function in the half ball D = {(x, y, z :x +y +z <, z } which is continuously differentiable, and satisfies u(x, y, =. Show that u can be extended to be a harmonic function in the whole ball. If you propose and explicit extension for u, explain why the extension is harmonic. Proof. We can extend u to all of n-space by defining u(x,x n = u(x, x n for x n <. Define ω(x = aω n y = a x x y n v(yds y ω(x is continuous on a closed ball B, harmonicinb. Poisson kernel is symmetric in y at x n =. ω(x =,(x n =. ω is harmonic for x B, x n,with the same boundary values ω = u. ω is harmonic u can be extended to a harmonic function on the interior of B. Ralston Hw. Show that a bounded solution to the Dirichlet problem in a half space is unique. (Note that one can show that a bounded solution exists for any given bounded continuous Dirichlet data by using the Poisson kernel for the half space. Proof. We have to show that a function, u, which is harmonic in the half-space, continuous, equal to when x n =, and bounded, must be identically. We can extend u to all of n-space by defining u(x,x n = u(x, x n for x n <. This extends u to a bounded harmonic function on all of n-space (by the problem above. Liouville s theorem says u must be constant, and since u(x, =, the constant is. So the original u must be identically. Ralston Hw. Suppose u is harmonic on the ball minus the origin, B = {x R 3 : < x <a}. Show that u(x can be extended toaharmonicfunctionontheball B = { x <a} iff lim x x u(x=. Proof. The condition lim x x u(x = is necessary, because harmonic functions are continuous. To prove the converse, let v be the function which is continuous on { x a/}, harmonic on { x <a/}, and equals u on { x = a/}. One can construct v using the Poisson kernel. Since v is continuous, it is bounded, and we can assume that v M. Since lim x x u(x =, given ɛ >, we can choose δ, <δ<a/ such that ɛ < x u(x <ɛwhen x <δ. Note that u, v ɛ/ x, andv +ɛ/ x are harmonic

254 Partial Differential Equations Igor Yanovsky, 5 54 on { < x <a/}. Choose b, <b<min(ɛ, a/, so that ɛ/b > M. Thenonboth { x = a/} and { x = b} we have v ɛ/ x <u(x <v+ɛ/ x. Thus, by max principle these inequalities hold on {b x a/}. Pick x with < x a/. u(x =v(x. v is the extension of u on { x <a/}, andu is extended on { x <a}.

255 Partial Differential Equations Igor Yanovsky, Problems: Heat Equation McOwen 5. #7(a. Consider u t = u xx for x>, t> u(x, = g(x for x> u(,t= for t>, where g is continuous and bounded for x and g( =. Find a formula for the solution u(x, t. Proof. Extend g to be an odd function on all of R: { g(x, x g(x = g( x, x <. Then, we need to solve { ũ t =ũ xx for x R, t> ũ(x, = g(x for x R. The solution is given by: 6 ũ(x, t = = = = = R K(x, y, tg(y dy = 4πt [ 4πt [ 4πt 4πt 4πt u(x, t = 4πt e (x y 4t g(y dy + e (x y 4t g(y dy e (x y 4t (e x +xy y 4t e x xy y 4t e (x +y 4t (e xy t e (x +y 4t sinh Since sinh( =, we can verify that u(,t=. e (x y 4t e (x+y 4t e xy t g(y dy. ( xy g(y dy. t g(y dy g(y dy ] g(y dy ] g(y dy 6 In calculations, we use: ey dy = e y dy, and g( y = g(y.

256 Partial Differential Equations Igor Yanovsky, 5 56 McOwen 5. #7(b. Consider u t = u xx for x>, t> u(x, = g(x for x> u x (,t= for t>, where g is continuous and bounded for x. Find a formula for the solution u(x, t. Proof. Extend g to be an even function 6 on all of R: { g(x, x g(x = g( x, x <. Then, we need to solve { ũ t =ũ xx for x R, t> ũ(x, = g(x for x R. The solution is given by: 6 ũ(x, t = = = = = R K(x, y, tg(y dy = 4πt [ 4πt [ 4πt 4πt 4πt u(x, t = 4πt e (x y 4t g(y dy + e (x y 4t g(y dy + e (x y 4t (e x +xy y 4t + e x xy y 4t e (x +y 4t (e xy t e (x +y 4t cosh e (x y 4t e (x+y 4t + e xy t g(y dy. ( xy g(y dy. t g(y dy g(y dy ] g(y dy ] g(y dy To check that the boundary condition holds, we perform the calculation: [ d u x (x, t = e (x +y ( xy ] 4t cosh g(y dy 4πt dx t [ = x (x +y ( xy e 4t cosh + e (x +y 4t y ( xy ] 4πt 4t t t sinh g(y dy, t [ ] u x (,t = e y 4t cosh+e y y 4t 4πt t sinh g(y dy =. 6 Even extensions are always continuous. Not true for odd extensions. g odd is continuous if g( =. 6 In calculations, we use: ey dy = e y dy, and g( y =g(y.

257 Partial Differential Equations Igor Yanovsky, 5 57 Problem (F 9, #5. The initial value problem for the heat equation on the whole real line is f t = f xx t f(t =,x=f (x with f smooth and bounded. a Write down the Green s function G(x, y, t for this initial value problem. b Write the solution f(x, t as an integral involving G and f. c Show that the maximum values of f(x, t and f x (x, t are non-increasing as t increases, i.e. sup x f(x, t sup x f (x sup x When are these inequalities actually equalities? f x (x, t sup f x (x. x Proof. a The fundamental solution K(x, y, t = 4πt e x y 4t. The Green s function is: 63 [ G(x, t; y, s = (π n π k(t s ] n e (x y 4k(t s. b The solution to the one-dimensional heat equation is u(x, t = K(x, y, t f (y dy = 4πt c We have R sup u(x, t = x = 4πt 4πt R R sup f (x x sup f (x x = sup f (x x e (x y 4t e y 4t R e x y 4t f (y dy 4πt f (x y dy 4πt 4πt π R R e y 4t dy e z 4tdz e z dz R } {{ } = π f (y dy. R e (x y 4t ( z = = sup f (x. x f (y dy y, dz = dy 4t 4t 63 The Green s function for the heat equation on an infinite domain; derived in R. Haberman using the Fourier transform.

258 Partial Differential Equations Igor Yanovsky, 5 58 u x (x, t = sup u(x, t x = 4πt R (x y e (x y 4t f (y dy = 4t [ ] e (x y 4t f (y 4πt } {{ } = sup f x (x 4πt x = sup x f x (x. R + 4πt e (x y 4t dy = R 4πt e (x y 4t R d ] [e (x y 4t f (y dy dy f y (y dy, sup f x (x e z 4tdz 4πt x R These inequalities are equalities when f (x andf x (x are constants, respectively.

259 Partial Differential Equations Igor Yanovsky, 5 59 Problem (S, #5. a Show that the solution of the heat equation u t = u xx, <x< with square-integrable initial data u(x, = f(x, decays in time, and there is a constant α independent of f and t such that for all t> ( max u x (x, t αt 3 4 f(x dx. x x b Consider the solution ρ of the transport equation ρ t +uρ x =with square-integrable initial data ρ(x, = ρ (x and the velocity u from part (a. Show that ρ(x, t remains square-integrable for all finite time ρ(x, t dx e Ct 4 ρ (x dx, R R where C does not depend on ρ. Proof. a The solution to the one-dimensional homogeneous heat equation is u(x, t= e (x y 4t f(y dy. 4πt R Take the derivative with respect to x, weget 64 u x (x, t = (x y e (x y 4t f(y dy = 4πt 4t π u x (x, t = 4t 3 R π R (x ye (x y 4t f(y dy 4t 3 ( 4t 3 (x y e (x y t dy f L π (R R ( z (t 3 e z dz f L π (R 4t 3 R = (t 3 ( 4 4t 3 z e z dz f L π (R R } {{ } M< = Ct 3 4 M f L (R = αt 3 4 f L (R. R (x ye (x y 4t f(y dy. (Cauchy-Schwarz ( z = x y t, dz = dy t b Note: 65 max x u = max x e (x y 4t f(y dy 4πt ( R 4πt R e z ( t dz f L (R ( 4πt = (t 4 π t R ( e z dz f L (R R } {{ } = π e (x y t = Ct 4 f L (R. dy f L (R z = x y, dz = dy t t 64 Cauchy-Schwarz: (u, v u v in any norm, for example 65 See Yana s and Alan s solutions. uv dx ( u dx ( v dx

260 Partial Differential Equations Igor Yanovsky, 5 6 Problem (F 4, #. Let u(x, t be a bounded solution to the Cauchy problem for the heat equation { u t = a u xx, t >, x R, a >, u(x, = ϕ(x. Here ϕ(x C(R satisfies lim ϕ(x =b, lim x + ϕ(x =c. x Compute the limit of u(x, t as t +, x R. Justify your argument carefully. Proof. For a =, the solution to the one-dimensional homogeneous heat equation is u(x, t= e (x y 4t ϕ(y dy. 4πt R We want to transform the equation to v t = v xx. Make a change of variables: x = ay. u(x, t =u(x(y, t=u(ay, t =v(y, t. Then, v y = u x x y = au x, v yy = au xx x y = a u xx, v(y, = u(ay, = ϕ(ay. Thus, the new problem is: { v t = v yy, t >, y R, v(y, = ϕ(ay. v(y, t= 4πt Since ϕ is continuous, R e (y z 4t ϕ(az dz. ϕ(x < M, x R. Thus, v(y, t = M 4πt M 4πt and lim x + ϕ(x =b, lim x ϕ(x=c, we have R R ( e z 4t dz s = z, ds = dz 4t 4t M e s 4tds = π e s ds = M. R } {{ } π Integral in converges uniformly lim = lim. For ψ = ϕ(a : v(y, t = = 4πt 4πt = π e (y z 4t ψ(z dz = e s ψ(y s 4t 4tds e s ψ(y s 4t ds. 4πt e z 4t ψ(y z dz

261 Partial Differential Equations Igor Yanovsky, 5 6 lim v(y, t = t + π = π = c + b. e s lim ψ(y s 4t ds + e s t + π e s bds = c π π + b π π e s cds+ π lim ψ(y s 4t ds t +

262 Partial Differential Equations Igor Yanovsky, 5 6 Problem. Consider u t = ku xx + Q, <x< u(,t =, u(,t =. What is the steady state temperature? Proof. Set u t =, and integrate with respect to x twice: ku xx + Q =, u xx = Q k, u x = Q k x + a, x u = Q + ax + b. k Boundary conditions give u(x = Q k x + ( + Q k x.

263 Partial Differential Equations Igor Yanovsky, Heat Equation with Lower Order Terms McOwen 5. #. Find a formula for the solution of { u t = u cu in R n (, u(x, = g(x on R n. (8. Show that such solutions, with initial data g L (R n, are unique, even when c is negative. Proof. McOwen. Consider v(x, t =e ct u(x, t. The transformed problem is { v t = v in R n (, v(x, = g(x on R n. (8. Since g is continuous and bounded in R n,wehave v(x, t = K(x, y, t g(y dy = R n (4πt n u(x, t = e ct v(x, t= (4πt n R n e R n e x y 4t x y 4t ct g(y dy. g(y dy, u(x, t isabounded solution since v(x, t is. To prove uniqueness, assume there is another solution v of (8.. w = v v satisfies { w t = w in R n (, w(x, = on R n (8.3. Since bounded solutions of (8.3 are unique, and since w is a nontrivial solution, w is unbounded. Thus, v is unbounded, and therefore, the bounded solution v is unique.

264 Partial Differential Equations Igor Yanovsky, Heat Equation Energy Estimates Problem (F 94, #3. Let u(x, y, t be a twice continuously differential solution of u t = u u 3 in R, t u(x, y, = in u(x, y, t= in, t. Prove that u(x, y, t in [,T]. Proof. Multiply the equation by u and integrate: uu t = u u u 4, uu t dx = u udx u 4 dx, d u dx = u u dt n ds u dx u 4 dx, } {{ } = d dt u = u dx u 4 dx. Thus, u(x, y, t u(x, y, =. Hence, u(x, y, t =, and u.

265 Partial Differential Equations Igor Yanovsky, 5 65 Problem (F 98, #5. Consider the heat equation u t u = in a two dimensional region. Define the mass M as M(t = u(x, t dx. a For a fixed domain, show M is a constant in time if the boundary conditions are u/ n =. b Suppose that =(t is evolving in time, with a boundary that moves at velocity v, which may vary along the boundary. Find a modified boundary condition (in terms of local quantities only for u, so that M is constant. Hint: You may use the fact that d dt (t f(x, t dx = (t f t (x, t dx + (t in which n is a unit normal vector to the boundary. n vf(x, t dl, Proof. a We have { u t u =, u n =, on. on We want to show that d dtm(t =. Wehave66 d dt M(t = d u(x, t dx = u t dx = udx = dt b We need d dtm(t =. = d dt M(t = d u(x, t dx = u t dx + dt (t (t = udx + n vuds= Thus, we need: = (t (t n ( u + vu ds =, (t u nds + on. (t n vuds = (t (t u ds =. n n vuds u n ds + (t (t n vuds n ( u + vu ds. 66 The last equality below is obtained from the Green s formula: u udx = n ds.

266 Partial Differential Equations Igor Yanovsky, 5 66 Problem (S 95, #3. Write down an explicit formula for a function u(x, t solving { u t + b u + cu = u in R n (, u(x, = f(x on R n (8.4. where b R n and c R are constants. Hint: First transform this to the heat equation by a linear change of the dependent and independent variables. Then solve the heat equation using the fundamental solution. Proof. Consider u(x, t = e α x+βt v(x, t. u t = βe α x+βt v + e α x+βt v t =(v t + βve α x+βt, u = αe α x+βt v + e α x+βt v =(αv + ve α x+βt, ( u = ((αv + ve α x+βt =(α v + ve α x+βt +( α v + α ve α x+βt = ( v +α v + α ve α x+βt. Plugging this into (8.4, we obtain v t + βv + b (αv + v+cv = v +α v + α v, v t + ( b α v + ( β + b α + c α v = v. In order to get homogeneous heat equation, we set α = b, β = b 4 c, which gives { vt = v in R n (, v(x, = e b x f(x on R n. The above PDE has the following solution: x y v(x, t= (4πt n 4t e b y f(y dy. Thus, R n e u(x, t =e b b x ( 4 +ct v(x, t= (4πt n e b x ( b 4 +ct R n e x y 4t e b y f(y dy.

267 Partial Differential Equations Igor Yanovsky, 5 67 Problem (F, #7. Consider the parabolic problem u t = u xx + c(xu (8.5 for <x<, inwhich c(x = for x >, c(x = for x <. Find solutions of the form u(x, t =e λt v(x in which u dx <. Hint: Look for v to have the form v(x = ae k x for x >, v(x = b cos lx for x <, for some a, b, k, l. Proof. Plug u(x, t =e λt v(x into (8.5 to get: λe λt v(x = e λt v (x+ce λt v(x, λv(x = v (x+cv(x, v (x λv(x+cv(x =. For x >, c =. We look for solutions of the form v(x =ae k x. v (x λv(x =, ak e k x aλe k x =, k λ =, k = λ, k = ± λ. Thus, v(x=c e λx + c e λx. u(x, t=ae λt e λx. For x <, c =. Since we want u dx < : We look for solutions of the form v(x =b cos lx. v (x λv(x+v(x =, bl cos lx +( λb coslx =, l +( λ =, l = λ, l = ± λ. Thus, (since cos( x = cosx u(x, t=be λt cos ( λx. We want v(x to be continuous on R, andatx = ±, in particular. Thus, ae λ = b cos ( λ, a = λ be cos ( λ. Also, v(x is symmetric: u dx = [ u dx = u dx + ] u dx <.

268 Partial Differential Equations Igor Yanovsky, 5 68 Problem (F 3, #3. ❶ The function h(x, T=(4πT e X 4T satisfies (you do not have to show this h T = h XX. Using this result, verify that for any smooth function U satisfies u(x, t=e 3 t3 xt u t + xu = u xx. U(ξ h(x t ξ, t dξ ❷ Given that U(x is bounded and continuous everywhere on x, establish that lim U(ξ h(x ξ, t dξ = U(x t ❸ and show that u(x, t U(x as t. (You may use the fact that e ξ dξ = π. Proof. We change the notation: h K, U g, ξ y. We have K(X, T= e X 4T 4πT ❶ We want to verify that satisfies We have u(x, t=e 3 t3 xt u t + xu = u xx. u t = = xu = u x = = u xx = = K(x y t,t g(y dy. d [ ] e 3 t3 xt K(x y t,t g(y dy dt [ (t x e 3 t3 xt K + e 3 t3 xt ( ] K X ( t +K T g(y dy, xe 3 t3 xt K(x y t,t g(y dy, d [ ] e 3 t3 xt K(x y t,t g(y dy dx [ ] te 3 t3 xt K + e 3 t3 xt K X g(y dy, d [ ] te 3 t3 xt K + e 3 t3 xt K X g(y dy dx ] [t e 3 t3 xt K te 3 t3 xt K X te 3 t3 xt K X + e 3 t3 xt K XX g(y dy.

269 Partial Differential Equations Igor Yanovsky, 5 69 Plugging these into, most of the terms cancel out. The remaining two terms cancel because K T = K XX. ❷ Given that g(x is bounded and continuous on x, we establish that 67 lim t K(x y, t g(y dy = g(x. Fix x R n, ε>. Choose δ> such that g(y g(x <ε if y x <δ, y R n. Then if x x < δ,wehave: ( R K(x, t dx = K(x y, t g(y dy g(x K(x y, t[g(y g(x ] dy R R K(x y, t g(y g(x dy + K(x y, t g(y g(x dy B δ (x R B } {{ } δ (x ε R K(x y,t dy = ε Furthermore, if x x δ and y x δ, then y x y x + δ y x + y x. Thus, y x y x. Consequently, = ε + g L K(x y, t dy ε + C t ε + C t = ε + C t R B δ (x R B δ (x δ R B δ (x e x y 4t e y x 6t dy dy e r 6t rdr ε + as t +. Hence, if x x < δ and t> is small enough, u(x, t g(x < ε. 67 Evans, p. 47, Theorem (c.

270 Partial Differential Equations Igor Yanovsky, 5 7 Problem (S 93, #4. The temperature T (x, t in a stationary medium, x, is governed by the heat conduction equation T t = T x. (8.6 Making the change of variable (x, t (u, t, whereu = x/ t, show that 4t T t = T u +u T u. (8.7 Solutions of (8.7 that depend on u alone are called similarity solutions. 68 Proof. We change notation: the change of variables is (x, t (u, τ, where t = τ. After the change of variables, we have T = T (u(x, t,τ(t. u = x u t = x, u t 4t 3 x = t, u xx =, τ = t τ t =, τ x =. T = T u t u t + T τ, T x = T u u x, T x = ( T x x Thus, (8.6 gives: T u u t + T τ T ( x + T u 4t 3 τ T τ 4t T τ 4t T τ = ( T x u u x = = T ( u, u x = T (, u t = T 4t u + x 4t 3 = T u + x t T u, = T T +u u u. ( T u u u x x + T u u }{{} x = T u, = T ( u. u x 68 This is only the part of the qual problem.

271 Partial Differential Equations Igor Yanovsky, Contraction Mapping and Uniqueness - Wave Recall that the solution to { u tt c u xx = f(x, t, u(x, = g(x, u t (x, = h(x, is given by adding together d Alembert s formula and Duhamel s principle: u(x, t = (g(x + ct+g(x ct + c x+ct x ct h(ξ dξ + c t ( x+c(t s x c(t s (9. f(ξ, s dξ ds. Problem (W, #8. a Find an explicit solution of the following Cauchy problem { u t = f(t, x, u u(,x=, x (,x=. (9. b Use part (a to prove the uniqueness of the solution of the Cauchy problem { u t u x u + q(t, xu =, x u(,x=, Here f(t, x and q(t, x are continuous functions. u x (,x=. (9.3 Proof. a It was probably meant to give the u t initially. We rewrite (9. as { u tt u xx = f(x, t, u(x, =, u t (x, =. Duhamel s principle, with c =, gives the solution to (9.4: u(x, t= t ( x+c(t s f(ξ, s dξ ds = t ( x+(t s c x c(t s x (t s f(ξ, s dξ ds. (9.4 b We use the Contraction Mapping Principle to prove uniqueness. Define the operator t x+(t s T (u = x (t s q(ξ, s u(ξ, s dξ ds. on the Banach space C,,. We will show Tu n Tu n+ < α u n u n+ where α <. Then {u n } n= : u n+ = T (u n converges to a unique fixed point which is the unique solution of PDE. Tu n Tu n+ = t x+(t s q(ξ, s ( u n (ξ, s u n+ (ξ, s dξ ds t x (t s q u n u n+ (t s ds t q u n u n+ α u n u n+, for small t. Thus, T is a contraction a unique fixed point. Since Tu= u, u is the solution to the PDE.

272 Partial Differential Equations Igor Yanovsky, 5 7 Problem (F, #3. Consider the Goursat problem: Find the solution of the equation u t u + a(x, tu = x in the square D, satisfying the boundary conditions u γ = ϕ, u γ = ψ, where γ, γ are two adjacent sides D. Herea(x, t, ϕ and ψ are continuous functions. Prove the uniqueness of the solution of this Goursat problem. Proof. The change of variable μ = x + t, η = x t transforms the equation to ũ μη +ã(μ, ηũ =. We integrate the equation: η η μ η μ ũ μη (u, v du dv = (ũη (μ, v ũ η (,v η dv = ũ(μ, η=ũ(μ, + ũ(,η u(, We change the notation. In the new notation: f(x, y = ϕ(x, y x y f = ϕ + Kf, f = ϕ + K(ϕ + Kf, f = ϕ + K n ϕ, n= f = Kf f =, max f δ max a max f. <x<δ μ η μ a(u, vf(u, v du dv, ã(μ, ηũdudv, ã(μ, ηũdudv, ã(μ, ηũdudv. For small enough δ, the operator K is a contraction. Thus, there exists a unique fixed point of K, andf = Kf,wheref is the unique solution.

273 Partial Differential Equations Igor Yanovsky, 5 73 Contraction Mapping and Uniqueness - Heat The solution of the initial value problem { u t = u + f(x, t for t>, x R n u(x, = g(x for x R n. (. is given by u(x, t = K(x y, t g(y dy + R n t R n K(x y, t s f(y, s dy ds where K(x, t= (4πt n x e 4t for t>, for t. Problem (F, #. Consider the Cauchy problem u t u + u (x, t = f(x, t, x R N, <t<t u(x, =. Prove the uniqueness of the classical bounded solution assuming that T is small enough. Proof. Let {u n } be a sequence of approximations to the solution, such that t S(u n = u n+ }{{} = K(x y, t s ( f(y, s u n(y, s dy ds. use Duhamel R sprinciple n We will show that S has a fixed point ( S(u n S(u n+ α u n u n+, α< {u n } converges to a uniques solution for small enough T. Since u n, u n+ C (R n C (t u n+ + u n M. t S(u n S(u n+ K(x y, t s u n+ u n dy ds R n t = K(x y, t s u n+ u n u n+ + u n dy ds R n t M K(x y, t s u n+ u n dy ds R n t MM u n+ (x, s u n (x, s ds MM T u n+ u n < u n+ u n for small T. Thus, S is a contraction a unique fixed point u C (R n C (t such that u = lim n u n. u is implicitly defined as t u(x, t = K(x y, t s ( f(y, s u (y, s dy ds. R n

274 Partial Differential Equations Igor Yanovsky, 5 74 Problem (S 97, #3. a Let Q(x such that x= Q(x dx =, and define Q ɛ = ɛ Q( x ɛ. Show that (here denotes convolution Q ɛ (x w(x L w(x L. In particular, let Q t (x denote the heat kernel (at time t, then Q t (x w (x Q t (x w (x L w (x w (x L. b Consider the parabolic equation u t = u xx + u subject to initial conditions u(x, = f(x. Show that the solution of this equation satisfies u(x, t =Q t (x f(x+ t Q t s (x u (x, s ds. (. c Fix t>. Let {u n (x, t}, n =,,... the fixed point iterations for the solution of (. u n+ (x, t =Q t (x f(x+ t Q t s (x u n(x, s ds. (.3 Let K n (t =sup m n u m (x, t L. Using (a and (b show that u n+ (x, t u n (x, t L sup K n (τ τ t t u n (x, s u n (x, s L ds. Conclude that the fixed point iterations in (.3 converge if t is sufficiently small. Proof. a We have Q ɛ (x w(x L = Q ɛ (x yw(y dy w = w Qɛ (x y dy = w ( y ɛ Q dy ɛ = w Q(z dz = w(x. Q ɛ (x yw(y dy ( x y ɛ Q dy ɛ ( z = y ɛ,dz= dy ɛ

275 Partial Differential Equations Igor Yanovsky, 5 75 Q t (x = 4πt e x 4t, the heat kernel. We have 69 Q t (x w (x Q t (x w (x L = = 4πt Q t (x yw (y dy 4πt e (x y 4t w (y dy e (x y 4t Q t (x yw (y dy e (x y 4t w (y dy w (y w (y dy w (y w (y 4πt z = x y 4t, dz = dy 4t = w (y w (y 4πt = w (y w (y π = w (y w (y. π e (x y 4t e z dz } {{ } dy e z 4tdz 69 Note: Q t(x dx = 4πt e (x y 4t dy = 4πt e z 4tdz = π e z dz =.

276 Partial Differential Equations Igor Yanovsky, 5 76 b Consider { u t = u xx + u, u(x, = f(x. We will show that the solution of this equation satisfies u(x, t =Q t (x f(x+ t t Q t s (x u (x, s ds. t Q t s (x u (x, s ds = Q t s (x y u (y, s dy ds R t ( = Q t s (x y u s (y, s u yy (y, s dy ds = R t d Qt s (x yu(y, s R ds( d ( Qt s (x y u(y, s Q t s (x yu yy (y, s dy ds ds [ ] = Q (x yu(y, t dy Q t (x yu(y, dy R R t d ( Qt s (x y u(y, s+ d R ds dy Q t s(x yu(y, s dy ds } {{ } = u(x, t R =, since Q t satisfies heat equation Q t (x yf(y dy Note: lim t + Q(x, t=δ (x =δ(x. = u(x, t Q t (x f(x. lim t + R Q(x y, tv(y dy = v(. Note that we used: D α (f g =(D α f g = f (D α g. c Let u n+ (x, t =Q t (x f(x+ u n+ (x, t u n (x, t L = }{{} (a t t t t t sup τ t Q t s (x u n (x, s ds. Q t s (x ( u n(x, s u n (x, s ds Qt s (x ( u n(x, s u n (x, s ds u n (x, s u n (x, s ds u n (x, s u n (x, s u n (x, s+u n (x, s ds u n (x, s+u n (x, s t u n (x, s u n (x, s ds sup K n (τ τ t t u n (x, s u n (x, s L ds. Also, u n+ (x, t u n (x, t L t sup K n (τ u n (x, s u n (x, s L. τ t

277 Partial Differential Equations Igor Yanovsky, 5 77 For t small enough, Tu = Q t (x f(x+ t sup τ t K n (τ α<. Thus, T defined as t Q t s (x u (x, s ds is a contraction, and has a unique fixed point u = Tu.

278 Partial Differential Equations Igor Yanovsky, 5 78 Problem (S 99, #3. Consider the system of equations u t = u xx + f(u, v v t =v xx + g(u, v to be solved for t>, <x<, and smooth initial data with compact support: u(x, = u (x, v(x, = v (x. If f and g are uniformly Lipschitz continuous, give a proof of existence and uniqueness of the solution to this problem in the space of bounded continuous functions with u(,t =sup x u(x, t. Proof. The space of continuous bounded functions forms a complete metric space so the contraction mapping principle applies. First, let v(x, t=w ( x,t,then u t = u xx + f(u, w w t = w xx + g(u, w. These initial value problems have the following solutions (K is the heat kernel: t u(x, t = K(x y, t u (y dy + K(x y, t s f(u, w dyds, R n R n t w(x, t = K(x y, t w (y dy + K(x y, t s g(u, w dyds. R n R n By the Lipshitz conditions, f(u, w M u, g(u, w M w. Now we can show the mappings, as defined below, are contractions: t T u = K(x y, t u (y dy + K(x y, t s f(u, w dy ds, R n R n t T w = K(x y, t w (y dy + K(x y, t s g(u, w dyds. R n R n t T (u n T (u n+ K(x y, t s f(u n,w f(u n+,w dy ds R n t M K(x y, t s u n u n+ dy ds R n t M sup u n u n+ K(x y, t sdy ds x R n t M sup u n u n+ ds M t sup u n u n+ x x < sup un u n+ for small t. x We used the Lipshitz condition and R K(x y, t s dy =. Thus, for small t, T is a contraction, and has a unique fixed point. Thus, the solution is defined as u = T u. Similarly, T is a contraction and has a unique fixed point. The solution is defined as w = T w.

279 Partial Differential Equations Igor Yanovsky, 5 79 Problems: Maximum Principle - Laplace and Heat. Heat Equation - Maximum Principle and Uniqueness Let us introduce the cylinder U = U T = (,T. We know that harmonic (and subharmonic functions achieve their maximum on the boundary of the domain. For the heat equation, the result is improved in that the maximum is achieved on a certain part of the boundary, parabolic boundary: Γ = {(x, t U : x or t =}. Let us also denote by C ; (U functions satisfying u t, u xi x j C(U. Weak Maximum Principle. Let u C ; (U C(U satisfy u u t in U. Then u achieves its maximum on the parabolic boundary of U: max U u(x, t = maxu(x, t. (. Γ Proof. First, assume u >u t in U. For <τ<t consider U τ = (,τ, Γ τ = {(x, t U τ : x or t =}. If the maximum of u on U τ occurs at x andt = τ, then u t (x, τ and u(x, τ, violating our assumption; similarly, u cannot attain an interior maximum on U τ. Hence (. holds for U τ : max U τ u =max Γτ u. But max Γτ u max Γ u and by continuity of u, max U u = lim τ T max U τ u. This establishes (.. Second, we consider the general case of u u t in U. Let u = v + εt for ε>. Notice that v u on U and v v t > inu. Thus we may apply (. to v: max U u = max U (v + εt max v + εt Letting ε establishes (. for u. U = max Γ v + εt max u + εt. Γ

280 Partial Differential Equations Igor Yanovsky, 5 8 Problem (S 98, #7. Prove that any smooth solution, u(x, y, t in the unit box ={(x, y x, y }, of the following equation u t = uu x + uu y + u, t, (x, y u(x, y, = f(x, y, (x, y satisfies the weak maximum principle, max u(x, y, t max{ max u(±, ±,t, max f(x, y}. [,T ] t T (x,y Proof. Suppose u satisfies given equation. Let u = v + εt for ε>. Then, v t + ε = vv x + vv y + εt(v x + v y + v. Suppose v has a maximum at (x,y,t (,T. Then v x = v y = v t = ε = v v> v has a minimum at (x,y,t, a contradiction. Thus, the maximum of v is on the boundary of (,T. Suppose v has a maximum at (x,y,t, (x,y. Then v x = v y =, v t ε v v> v has a minimum at (x,y,t, a contradiction. Thus, max v max{ max v(±, ±,t, max f(x, y}. [,T ] t T (x,y Now max u = [,T ] max (v + εt max [,T ] Letting ε establishes the result. v + εt max{ max v(±, ±,t, max f(x, y} + εt [,T ] t T (x,y max{ max u(±, ±,t, max f(x, y} + εt. t T (x,y

281 Partial Differential Equations Igor Yanovsky, 5 8. Laplace Equation - Maximum Principle Problem (S 9, #6. Suppose that u satisfies Lu = au xx + bu yy + cu x + du y eu = with a>, b>, e>, for(x, y, with a bounded open set in R. a Show that u cannot have a positive maximum or a negative minimum in the interior of. b Use this to show that the only function u satisfying Lu =in, u =on and u continuous on is u =. Proof. a For an interior (local maximum or minimum at an interior point (x, y, we have u x =, u y =. Suppose u has a positive maximum in the interior of. Then u>, u xx, u yy. With these values, we have au xx }{{ } + bu yy + cu }{{} }{{} x = + du y }{{} = eu }{{} < =, which leads to contradiction. Thus, u can not have a positive maximum in. Suppose u has a negative minimum in the interior of. Then u<, u xx, u yy. With these values, we have au xx }{{ } + bu yy + cu }{{} }{{} x = + du y }{{} = eu }{{} > =, which leads to contradiction. Thus, u can not have a negative minimum in. b Since u can not have positive maximum in the interior of, then max u =on. Since u can not have negative minimum in the interior of, then min u =on. Since u is continuous, u on.

282 Partial Differential Equations Igor Yanovsky, 5 8 Problems: Separation of Variables - Laplace Equation Problem : The D LAPLACE Equation on a Square. Let =(,π (,π, and use separation of variables to solve the boundary value problem u xx + u yy = <x,y<π u(,y==u(π, y y π u(x, =, u(x, π=g(x x π, where g is a continuous function satisfying g( = = g(π. Proof. Assume u(x, y =X(xY (y, then substitution in the PDE gives X Y +XY =. X X = Y Y = λ. From X + λx =,weget X n (x =a n cos nx + b n sin nx. Boundary conditions give { u(,y=x(y (y = u(π, y=x(πy (y = X( = = X(π. Thus, X n ( = a n =,and X n (x =b n sin nx, n =,,... n b n sin nx + λb n sin nx =, λ n = n, n =,,... With these values of λ n we solve Y n Y = to find Y n (y =c n cosh ny + d n sinh ny. Boundary conditions give u(x, = X(xY ( = Y ( = = c n. Y n (x =d n sinh ny. By superposition, we write u(x, y= ã n sin nx sinh ny, n= which satifies the equation and the three homogeneous boundary conditions. boundary condition at y = π gives u(x, π=g(x = ã n sin nx sinh nπ, π n= g(xsinmx dx = n= π ã n sinh nπ sin nx sin mx dx = π ãm sinh mπ. The

283 Partial Differential Equations Igor Yanovsky, 5 83 ã n sinh nπ = π π g(xsinnx dx.

284 Partial Differential Equations Igor Yanovsky, 5 84 Problem : The D LAPLACE Equation on a Square. Let =(,π (,π, and use separation of variables to solve the mixed boundary value problem u = in u x (,y==u x (π, y <y<π u(x, =, u(x, π=g(x <x<π. Proof. Assume u(x, y =X(xY (y, then substitution in the PDE gives X Y +XY =. X X = Y Y = λ. Consider X + λx =. If λ =, X (x =a x + b. If λ>, X n (x =a n cos nx + b n sin nx. Boundary conditions give { u x (,y=x (Y (y = u x (π, y=x (πy (y = Thus, X ( = a =,andx n( = nb n =. X ( = = X (π. X (x =b, X n (x =a n cos nx, n =,,... n a n cos nx + λa n cos nx =, λ n = n, n =,,,... With these values of λ n we solve Y n Y =. If n =, Y (y =c y + d. If n, Y n (y =c n cosh ny + d n sinh ny. Boundary conditions give u(x, = X(xY ( = Y ( =. Thus, Y ( = d =,andy n ( = c n =. We have Y (y =c y, Y n (y =d n sinh ny, n =,,... u (x, y =X (xy (y =b c y =ã y, u n (x, y=x n (xy n (y =(a n cos nx(d n sinh ny =ã n cos nx sinh ny. By superposition, we write u(x, y=ã y + ã n cos nx sinh ny, n= which satifies the equation and the three homogeneous boundary conditions. The fourth boundary condition gives u(x, π=g(x =ã π + ã n cos nx sinh nπ, n=

285 Partial Differential Equations Igor Yanovsky, 5 85 { π g(x dx = π (ã π + n= ãn cos nx sinh nπ dx =ã π, π g(xcosmx dx = n= ãn sinh nπ π cos nx cos mx dx = π ãm sinh mπ. ã = π π ã n sinh nπ = π g(x dx, π g(xcosnx dx.

286 Partial Differential Equations Igor Yanovsky, 5 86 Problem (W 4, #5 The D LAPLACE Equation in an Upper-Half Plane. Consider the Laplace equation u x + u y =, y >, <x<+ u(x, u(x, = f(x, y where f(x C (R. Find a bounded solution u(x, y and show that u(x, y when x + y. Proof. Assume u(x, y =X(xY (y, then substitution in the PDE gives X Y +XY =. X X = Y Y = λ. Consider X + λx =. If λ =, X (x =a x + b. If λ>, X n (x =a n cos λ n x + b n sin λ n x. Since we look for bounded solutions as x,wehavea =. Consider Y λ n Y =. If λ n =, Y (y =c y + d. If λ n >, Y n (y =c n e λ ny + d n e λ ny. Since we look for bounded solutions as y,wehavec =, d n =. Thus, u(x, y=ã + e λ ( ny ã n cos λ n x + b n sin λ n x. n= Initial condition gives: f(x =u y (x, u(x, = ã ( λ n + ( ã n cos λ n x + b n sin λ n x. n= f(x C (R, i.e. has compact support [ L, L], for some L>. Thus the coefficients ã n, b n are given by L L L L f(xcos λ n xdx = ( λ n +ã n L. f(xsin λ n xdx = ( λ n + b n L. Thus, u(x, y when x + y. 7 7 Note that if we change the roles of X and Y in, the solution we get will be unbounded.

287 Partial Differential Equations Igor Yanovsky, 5 87 Problem 3: The D LAPLACE Equation on a Circle. Let be the unit disk in R and consider the problem { u = in u n = h on, where h is a continuous function. Proof. Use polar coordinates (r, θ { u rr + r u r + u r θθ = u r (,θ=h(θ for r<, θ<π for θ<π. r u rr + ru r + u θθ =. Let r = e t, u(r(t,θ. u t = u r r t = e t u r, u tt = ( e t u r t = e t u r + e t u rr = ru r + r u rr. Thus, we have u tt + u θθ =. Let u(t, θ =X(tY (θ, which gives X (ty (θ+x(ty (θ =. X (t X(t = Y (θ Y (θ = λ. From Y (θ+λy (θ =, we get Y n (θ =a n cos nθ + b n sin nθ. λ n = n, n =,,,... With these values of λ n we solve X (t n X(t =. If n =, X (t =c t + d. X (r= c log r + d. If n, X n (t =c n e nt + d n e nt X n (r=c n r n + d n r n. We have u (r, θ = X (ry (θ =( c log r + d a, u n (r, θ = X n (ry n (θ =(c n r n + d n r n (a n cos nθ + b n sin nθ. But u must be finite at r =,soc n =,n =,,,... u (r, θ = d a, u n (r, θ = d n r n (a n cos nθ + b n sin nθ. By superposition, we write u(r, θ=ã + r n (ã n cos nθ + b n sin nθ. n= Boundary condition gives u r (,θ= n(ã n cos nθ + b n sin nθ =h(θ. n= The coefficients a n, b n for n are determined from the Fourier series for h(θ. a is not determined by h(θ and therefore may take an arbitrary value. Moreover,

288 Partial Differential Equations Igor Yanovsky, 5 88 the constant term in the Fourier series for h(θ must be zero [i.e., π h(θdθ =]. Therefore, the problem is not solvable for an arbitrary function h(θ, and when it is solvable, the solution is not unique.

289 Partial Differential Equations Igor Yanovsky, 5 89 Problem 4: The D LAPLACE Equation on a Circle. Let ={(x, y R : x + y < } = {(r, θ: r<, θ<π}, and use separation of variables (r, θ to solve the Dirichlet problem { u = in u(,θ=g(θ for θ<π. Proof. Use polar coordinates (r, θ { u rr + r u r + u r θθ = for r<, θ<π u(,θ=g(θ for θ<π. r u rr + ru r + u θθ =. Let r = e t, u(r(t,θ. u t = u r r t = e t u r, u tt = ( e t u r t = e t u r + e t u rr = ru r + r u rr. Thus, we have u tt + u θθ =. Let u(t, θ =X(tY (θ, which gives X (ty (θ+x(ty (θ =. X (t X(t = Y (θ Y (θ = λ. From Y (θ+λy (θ =, we get Y n (θ =a n cos nθ + b n sin nθ. λ n = n, n =,,,... With these values of λ n we solve X (t n X(t =. If n =, X (t =c t + d. X (r= c log r + d. If n, X n (t =c n e nt + d n e nt X n (r=c n r n + d n r n. We have u (r, θ = X (ry (θ =( c log r + d a, u n (r, θ = X n (ry n (θ =(c n r n + d n r n (a n cos nθ + b n sin nθ. But u must be finite at r =,soc n =,n =,,,... u (r, θ = d a, u n (r, θ = d n r n (a n cos nθ + b n sin nθ. By superposition, we write u(r, θ=ã + r n (ã n cos nθ + b n sin nθ. n= Boundary condition gives u(,θ=ã + (ã n cos nθ + b n sin nθ =g(θ. n=

290 Partial Differential Equations Igor Yanovsky, 5 9 ã = π ã n = π bn = π π π π g(θ dθ, g(θcosnθ dθ, g(θsinnθ dθ.

291 Partial Differential Equations Igor Yanovsky, 5 9 Problem (F 94, #6: The D LAPLACE Equation on a Circle. Find all solutions of the homogeneous equation u xx + u yy =, x + y <, u n u =, x + y =. Hint: = r r (r r + r θ in polar coordinates. Proof. Use polar coordinates (r, θ: { u rr + r u r + u r θθ = for r<, θ<π u r (,θ u(,θ = for θ<π. Sincewesolvetheequationonacircle,wehaveperiodicconditions: u(r, = u(r, π X(rY ( = X(rY (π Y ( = Y (π, u θ (r, = u θ (r, π X(rY ( = X(rY (π Y ( = Y (π. Also, we want the solution to be bounded. In particular, u is bounded for r =. r u rr + ru r + u θθ =. Let r = e t, u(r(t,θ, we have u tt + u θθ =. Let u(t, θ =X(tY (θ, which gives X (ty (θ+x(ty (θ =. X (t X(t = Y (θ Y (θ = λ. From Y (θ+λy (θ =, we get Y n (θ =a n cos λθ + b n sin λθ. Using periodic condition: Y n ( = a n, Y n (π=a n cos( λ n π+b n sin( λ n π =a n λ n = n λ n = n. Thus, Y n (θ =a n cos nθ + b n sin nθ. With these values of λ n we solve X (t n X(t =. If n =, X (t =c t + d. X (r= c log r + d. If n, X n (t =c n e nt + d n e nt X n (r=c n r n + d n r n. u must be finite at r = c n =,n =,,,... u(r, θ=ã + r n (ã n cos nθ + b n sin nθ. n= Boundary condition gives = u r (,θ u(,θ = ã + (n (ã n cos nθ + b n sin nθ. n= Calculating Fourier coefficients gives πã = ã =. π(n a n = ã n =, n =, 3,... a, b are constants. Thus, u(r, θ=r(ã cos θ + b sin θ.

292 Partial Differential Equations Igor Yanovsky, 5 9 Problem (S, #4. a Let (r, θ be polar coordinates on the plane, i.e. x + ix = re iθ. Solve the boudary value problem u = in r< u/ r = f(θ on r =, beginning with the Fourier series for f (you may assume that f is continuously differentiable. Give your answer as a power series in x + ix plus a power series in x ix. There is a necessary condition on f for this boundary value problem to be solvable that you will find in the course of doing this. b Sum the series in part (a to get a representation of u in the form u(r, θ= π N(r, θ θ f(θ dθ. Proof. a Green s identity gives the necessary compatibility condition on f: π u f(θ dθ = r= r dθ = u n ds = udx =. Use polar coordinates (r, θ: { u rr + r u r + u r θθ = for r<, θ<π u r (,θ = f(θ for θ<π. Sincewesolvetheequationonacircle,wehaveperiodicconditions: u(r, = u(r, π X(rY ( = X(rY (π Y ( = Y (π, u θ (r, = u θ (r, π X(rY ( = X(rY (π Y ( = Y (π. Also, we want the solution to be bounded. In particular, u is bounded for r =. r u rr + ru r + u θθ =. Let r = e t, u(r(t,θ, we have u tt + u θθ =. Let u(t, θ =X(tY (θ, which gives X (ty (θ+x(ty (θ =. X (t X(t = Y (θ Y (θ = λ. From Y (θ+λy (θ =, we get Y n (θ =a n cos λθ + b n sin λθ. Using periodic condition: Y n ( = a n, Y n (π=a n cos( λ n π+b n sin( λ n π =a n λ n = n λ n = n. Thus, Y n (θ =a n cos nθ + b n sin nθ. With these values of λ n we solve X (t n X(t =. If n =, X (t =c t + d. X (r= c log r + d.

293 Partial Differential Equations Igor Yanovsky, 5 93 If n, X n (t =c n e nt + d n e nt X n (r=c n r n + d n r n. u must be finite at r = c n =,n =,,,... Since u(r, θ=ã + u r (r, θ = r n (ã n cos nθ + b n sin nθ. n= nr n (ã n cos nθ + b n sin nθ, n= the boundary condition gives u r (,θ = n (ã n cos nθ + b n sin nθ = f(θ. n= ã n = π f(θ cosnθ dθ, nπ bn = π f(θ sinnθ dθ. nπ π ã is not determined by f(θ (since f(θ dθ =. Therefore, it may take an arbitrary value. Moreover, the constant term in the Fourier series for f(θ mustbezero [i.e., π f(θdθ = ]. Therefore, the problem is not solvable for an arbitrary function f(θ, and when it is solvable, the solution is not unique. b In part (a, we obtained the solution and the Fourier coefficients: ã n = nπ bn = nπ u(r, θ = ã + π π f(θ cosnθ dθ, f(θ sinnθ dθ. r n (ã n cos nθ + b n sin nθ n= n= π π ( [ = ã + r n f(θ cosnθ dθ ] [ cos nθ + nπ n= nπ r n π = ã + f(θ [ cos nθ cos nθ +sinnθ sin nθ ] dθ nπ n= r n π = ã + f(θ cosn(θ θ dθ nπ n= π r n = ã + nπ cos n(θ θ f(θ dθ. } {{ } N(r, θ θ f(θ sinnθ dθ ] sin nθ

294 Partial Differential Equations Igor Yanovsky, 5 94 Problem (S 9, #6. Consider the Laplace equation u xx + u yy = for x + y. Denotingbyx = r cos θ, y = r sin θ polar coordinates, let f = f(θ be a given smooth function of θ. Construct a uniformly bounded solution which satisfies boundary conditions u = f for x + y =. What conditions has f to satisfy such that lim x +y (x + y u(x, y=? Proof. Use polar coordinates (r, θ: { u rr + r u r + u r θθ = for r u(,θ=f(θ for θ<π. Since we solve the equation on outside of a circle, we have periodic conditions: u(r, = u(r, π X(rY ( = X(rY (π Y ( = Y (π, u θ (r, = u(r, π X(rY ( = X(rY (π Y ( = Y (π. Also, we want the solution to be bounded. In particular, u is bounded for r =. r u rr + ru r + u θθ =. Let r = e t, u(r(t,θ, we have u tt + u θθ =. Let u(t, θ =X(tY (θ, which gives X (ty (θ+x(ty (θ =. X (t X(t = Y (θ Y (θ = λ. From Y (θ+λy (θ =, we get Y n (θ =a n cos λθ + b n sin λθ. Using periodic condition: Y n ( = a n, Y n (π=a n cos( λ n π+b n sin( λ n π =a n λ n = n λ n = n. Thus, Y n (θ =a n cos nθ + b n sin nθ. With these values of λ n we solve X (t n X(t =. If n =, X (t =c t + d. X (r= c log r + d. If n, X n (t =c n e nt + d n e nt X n (r=c n r n + d n r n. u must be finite at r = c =,d n =,n =,,... u(r, θ=ã + r n (ã n cos nθ + b n sin nθ. n= Boundary condition gives f(θ = u(,θ = ã + (ã n cos nθ + b n sin nθ. n=

295 Partial Differential Equations Igor Yanovsky, 5 95 πã = π f(θ dθ, πã n = f =ã = π f(θ cosnθ dθ, π b n = f n =ã n = π π f(θ sinnθ dθ. f n = b n = π We need to find conditions for f such that Since we need lim x +y (x + y u(x, y=, lim r r u(r, θ }{{} =, lim r r[ f + lim r [ n> [ lim r f + r need or r n (f n cos nθ + f ] n sin nθ n= r n (f n cos nθ + f ] n sin nθ =, n= Thus, the conditions are f n, f n =, n =,,. r n (f n cos nθ + f ] n sin nθ = }{{} need = }{{} need. π π π. f(θ dθ, f(θ cosnθ dθ, π f(θ sinnθ dθ.

296 Partial Differential Equations Igor Yanovsky, 5 96 Problem (F 96, #: The D LAPLACE Equation on a Semi-Annulus. Solve the Laplace equation in the semi-annulus u =, <r<, <θ<π, u(r, = u(r, π=, <r<, u(,θ=sinθ, <θ<π, u(,θ=, <θ<π. Hint: Use the formula = r r (r r + r θ for the Laplacian in polar coordinates. Proof. Use polar coordinates (r, θ u rr + r u r + r u θθ = <r<, <θ<π, r u rr + ru r + u θθ =. With r = e t,wehave u tt + u θθ =. Let u(t, θ =X(tY (θ, which gives X (ty (θ+x(ty (θ =. X (t X(t = Y (θ Y (θ = λ. From Y (θ+λy (θ =, we get Boundary conditions give Y n (θ =a n cos λθ + b n sin λθ. u n (r, = = X n (ry n ( =, Y n ( =, u n (r, π==x n (ry n (π=, Y n (π=. Thus, = Y n ( = a n,andy n (π =b n sin λπ = λ = n λ n = n. Thus, Y n (θ =b n sin nθ, n =,,... With these values of λ n we solve X (t n X(t =. If n =, X (t =c t + d. X (r= c log r + d. If n>, X n (t =c n e nt + d n e nt X n (r=c n r n + d n r n. We have, u(r, θ= X n (ry n (θ = n= ( c n r n + d n r n sinnθ. n= Using the other two boundary conditions, we obtain sin θ = u(,θ = ( c n + d n sinnθ = u(,θ = n= { c + d =, c n + d n =, n =, 3,... ( c n n + d n n sinnθ c n n + d n n =, n =,,... n= Thus, the coefficients are given by c = 4 3, d = 3 ; c n =, d n =.

297 Partial Differential Equations Igor Yanovsky, 5 97 u(r, θ= ( 4 3r r 3 sin θ.

298 Partial Differential Equations Igor Yanovsky, 5 98 Problem (S 98, #8: The D LAPLACE Equation on a Semi-Annulus. Solve u =, <r<, <θ<π, u(r, = u(r, π=, <r<, u(,θ=u(,θ=, <θ<π. Proof. Use polar coordinates (r, θ u rr + r u r + r u θθ = for <r<, <θ<π, r u rr + ru r + u θθ =. With r = e t,wehave u tt + u θθ =. Let u(t, θ =X(tY (θ, which gives X (ty (θ+x(ty (θ =. X (t X(t = Y (θ Y (θ = λ. From Y (θ+λy (θ =, we get Boundary conditions give Y n (θ =a n cos nθ + b n sin nθ. u n (r, = = X n (ry n ( =, Y n ( =, u n (r, π==x n (ry n (π=, Y n (π=. Thus, = Y n ( = a n,andy n (θ =b n sin nθ. λ n = n, n =,,... With these values of λ n we solve X (t n X(t =. If n =, X (t =c t + d. X (r= c log r + d. If n>, X n (t =c n e nt + d n e nt X n (r=c n r n + d n r n. We have, u(r, θ= X n (ry n (θ = n= ( c n r n + d n r n sinnθ. n= Using the other two boundary conditions, we obtain u(,θ= = ( c n + d n sinnθ, u(,θ= = n= ( c n n + d n n sinnθ, n= which give the two equations for c n and d n : π π that can be solved. sin nθ dθ = π ( c n + d n, sin nθ dθ = π ( c n n + d n n,

299 Partial Differential Equations Igor Yanovsky, 5 99 Problem (F 89, #. Consider Laplace equation inside a 9 sector of a circular annulus u = a<r<b, <θ< π subject to the boundary conditions u u (r, =, θ θ (r, π =, u r (a, θ =f u (θ, r (b, θ=f (θ, where f (θ, f (θ are continuously differentiable. a Find the solution of this equation with the prescribed boundary conditions using separation of variables. Proof. a Use polar coordinates (r, θ u rr + r u r + r u θθ = for a<r<b, <θ< π, r u rr + ru r + u θθ =. With r = e t,wehave u tt + u θθ =. Let u(t, θ =X(tY (θ, which gives X (ty (θ+x(ty (θ =. X (t X(t = Y (θ Y (θ = λ. From Y (θ+λy (θ =, we get Y n (θ =a n cos λθ + b n sin λθ. Boundary conditions give u nθ (r, = X n (ry n( = Y n( =, u nθ (r, π = X n(ry n( π = Y n( π =. Y n(θ = a n λn sin λ n θ + b n λn cos λ n θ.thus,y n( = b n λn = b n =. Y n( = a n λn sin π λ n = π λ n = nπ λ n =(n. Thus, Y n (θ =a n cos(nθ, n =,,,... In particular, Y (θ =a t + b. Boundary conditions give Y (θ =b. With these values of λ n we solve X (t (n X(t =. If n =, X (t =c t + d. X (r= c log r + d. If n>, X n (t =c n e nt + d n e nt X n (r =c n r n + d n r n. u(r, θ = c log r + d + ( c n r n + d n r n cos(nθ. n= Using the other two boundary conditions, we obtain u r (r, θ = c r + ( n c n r n +n d n r n cos(nθ. n=

300 Partial Differential Equations Igor Yanovsky, 5 3 f (θ =u r (a, θ = c a + n( c n a n + d n a n cos(nθ, n= f (θ =u r (b, θ = c b + n( c n b n + d n b n cos(nθ. n= which give the two equations for c n and d n : π π f (θcos(nθ dθ = π n( c na n + d n a n, f (θsin(nθ dθ = π n( c nb n + d n b n. b Show that the solution exists if and only if π a f (θ dθ b π f (θ dθ =. Proof. Using Green s identity, we obtain: = = = = π π π udx = u (b, θ dθ + r f (θ dθ + f (θ dθ + π u n π π u (a, θ dθ + r f (θ dθ ++ f (θ dθ. b a u a (r, dr + θ b u ( r, π dr θ c Is the solution unique? Proof. No, since the boundary conditions are Neumann. The solution is unique only up to a constant.

301 Partial Differential Equations Igor Yanovsky, 5 3 Problem (S 99, #4. Let u(x, y be harmonic inside the unit disc, with boundary values along the unit circle {, y > u(x, y=, y. Compute u(, and u(,y. Proof. Since u is harmonic, u =. Use polar coordinates (r, θ u rr + r u r + u r θθ = r<, θ<π {, <θ<π u(,θ=, π θ π. r u rr + ru r + u θθ =. With r = e t,wehave u tt + u θθ =. Let u(t, θ =X(tY (θ, which gives X (ty (θ+x(ty (θ =. X (t X(t = Y (θ Y (θ = λ. From Y (θ+λy (θ =, we get Y n (θ =a n cos nθ + b n sin nθ. λ n = n, n =,,... With these values of λ n we solve X (t n X(t =. If n =, X (t =c t + d. X (r= c log r + d. If n>, X n (t =c n e nt + d n e nt X n (r=c n r n + d n r n. We have u (r, θ = X (ry (θ =( c log r + d a, u n (r, θ = X n (ry n (θ =(c n r n + d n r n (a n cos nθ + b n sin nθ. But u must be finite at r =,soc n =,n =,,,... u (r, θ = ã, u n (r, θ = r n (ã n cos nθ + b n sin nθ. By superposition, we write u(r, θ=ã + r n (ã n cos nθ + b n sin nθ. n= Boundary condition gives { u(,θ=ã + (ã n cos nθ + b, <θ<π n sin nθ =, π θ π, n= and the coefficients ã n and b n are determined from the above equation. 7 7 See Yana s solutions, where Green s function on a unit disk is constructed.

302 Partial Differential Equations Igor Yanovsky, Problems: Separation of Variables - Poisson Equation Problem (F 9, #: The D POISSON Equation on a Quarter-Circle. Solve explicitly the following boundary value problem u xx + u yy = f(x, y in the domain ={(x, y, x>, y>, x + y < } with boundary conditions u = for y =, <x<, u = x for x =, <y<, u = for x>, y >, x + y =. Function f(x, y is known and is assumed to be continuous. Proof. Use polar coordinates (r, θ: u rr + r u r + u r θθ = f(r, θ r<, θ< π u(r, = r<, u θ (r, π = r<, u(,θ = θ π. We solve r u rr + ru r + u θθ =. Let r = e t, u(r(t,θ, we have u tt + u θθ =. Let u(t, θ =X(tY (θ, which gives X (ty (θ+x(ty (θ =. X (t X(t = Y (θ Y (θ = λ. From Y (θ +λy (θ =, we get Y n (θ =a n cos λθ + b n sin λθ. Boundary conditions: { u(r, = X(rY ( = u θ (r, π =X(rY ( π = Y ( = Y ( π =. Thus, Y n ( = a n =,andy n( π = λ n b n cos π λ n = π λ n = nπ π, n =,,... λ n =(n. Thus, Y n (θ =b n sin(n θ, n =,,... Thus, we have u(r, θ= X n (r sin[(n θ]. n=

303 Partial Differential Equations Igor Yanovsky, 5 33 We now plug this equation into with inhomogeneous term and obtain ( X n(t sin[(n θ] (n X n (t sin[(n θ] = f(t, θ, n= n= The solution to this equation is ( X n(t (n X n (t sin[(n θ] = f(t, θ, π ( X 4 n(t (n X n (t = π X n(t (n X n (t = 4 π X n (t = c n e (n t + d n e (n t + U np (t, or X n (r = c n r (n + d n r (n + u np (r, where u np is the particular solution of inhomogeneous equation. u must be finite at r = c n =,n =,,... Thus, u(r, θ= ( dn r (n + u np (r sin[(n θ]. n= Using the last boundary condition, we have ( =u(,θ = dn + u np ( sin[(n θ], u(r, θ= n= = π 4 (d n + u np (, d n = u np (. ( unp (r (n + u np (r sin[(n θ]. n= π f(t, θ sin[(n θ] dθ, f(t, θ sin[(n θ] dθ. The method used to solve this problem is similar to section Problems: Eigenvalues of the Laplacian - Poisson Equation: First, we find Y n (θ eigenfunctions. Then, we plug in our guess u(t, θ =X(tY (θ into the equation u tt + u θθ = f(t, θ and solve an ODE in X(t. Note the similar problem on D Poisson equation on a square domain. The problem is used by first finding the eigenvalues and eigenfunctions of the Laplacian, and then expanding f(x, y in eigenfunctions, and comparing coefficients of f with the general solution u(x, y. Here, however, this could not be done because of the circular geometry of the domain. In particular, the boundary conditions do not give enough information to find explicit representations for μ m and ν n. Also, the condition u = for x>, y >, x +y =

304 Partial Differential Equations Igor Yanovsky, 5 34 can not be used. 7 7 ChiuYen s solutions have attempts to solve this problem using Green s function.

305 Partial Differential Equations Igor Yanovsky, Problems: Separation of Variables - Wave Equation Example (McOwen 3. #. We considered the initial/boundary value problem and solved it using Fourier Series. We now solve it using the Separation of Variables. u tt u xx = <x<π, t> u(x, =, u t (x, = <x<π (4. u(,t=, u(π, t= t. Proof. Assume u(x, t=x(xt (t, then substitution in the PDE gives XT X T =. X X = T T = λ. From X + λx =,weget X n (x =a n cos nx + b n sin nx. Boundary conditions give { u(,t=x(t (t= X( = X(π=. u(π, t=x(πt (t= Thus, X n ( = a n =, andx n (x=b n sin nx, λ n = n, n =,,... With these values of λ n,wesolve T +n T = to find T n (t =c n sin nt+d n cos nt. Thus, u(x, t = ( cn sin nt + d n cos nt sin nx, u t (x, t = n= ( n cn cos nt n d n sin nt sin nx. n= Initial conditions give =u(x, = d n sin nx, =u t (x, = n= n c n sin nx. n= By orthogonality, we may multiply both equations by sin mx and integrate: π sin mx dx = d π m, π π dx = n c n, which gives the coefficients d n = ( cos nπ = nπ { 4 nπ, n odd,, n even, and c n =. Plugging the coefficients into a formula for u(x, t, we get u(x, t = 4 π n= cos(n +t sin(n +x. (n +

306 Partial Differential Equations Igor Yanovsky, 5 36 Example. Use the method of separation of variables to find the solution to: u tt +3u t + u = u xx, <x< u(,t=, u(,t=, u(x, =, u t (x, = x sin(πx. Proof. Assume u(x, t =X(xT (t, then substitution in the PDE gives XT +3XT + XT = X T, T T +3T X + = T X = λ. From X + λx =, X n (x =a n cos λ n x + b n sin λ n x. Boundary conditions give { u(,t=x(t (t= X( = X( =. u(,t=x(t (t= Thus, X n ( = a n =, andx n (x=b n sin λ n x. X n ( = b n sin λ n =. Hence, λ n = nπ, or λ n =(nπ, n =,,... λ n =(nπ, X n (x =b n sin nπx. With these values of λ n,wesolve T +3T + T = λ n T, T +3T + T = (nπ T, T +3T +(+(nπ T =. We can solve this nd-order ODE with the following guess, T (t =ce st to obtain s = 3 ± 5 4 (nπ 5. For n, 4 (nπ <. Thus, s = 3 (nπ ± i 5 4. T n (t =e 3 t( c n cos u(x, t = X(xT (t= (nπ 5 4 t + d n sin (nπ 5 4 t. e 3 t( c n cos (nπ 5 4 t + d n sin (nπ 5 4 t sin nπx. n= Initial conditions give =u(x, = c n sin nπx. n= By orthogonality, we may multiply this equations by sin mπx and integrate: dx = c m c m =.

307 Partial Differential Equations Igor Yanovsky, 5 37 Thus, u(x, t = u t (x, t = d n e 3 t( sin (nπ 5 4 t sin nπx. n= [ 3 d ne 3 t( sin (nπ 5 4 t + d n e 3 t( (nπ 5 ( cos (nπ 4 5 ] 4 t sin nπx, n= x sin(πx = u t (x, = ( d n (nπ 5 sin nπx. 4 n= By orthogonality, we may multiply this equations by sin mπx and integrate: ( x sin(πx sin(mπx dx = d m (mπ 5, 4 d n = (nπ 5 4 u(x, t = e 3 t n= d n ( sin x sin(πx sin(nπx dx. (nπ 5 4 t sin nπx. Problem (F 4, #. Solve the following initial-boundary value problem for the wave equation with a potential term, u tt u xx + u = <x<π,t< u(,t=u(π, t= t> u(x, = f(x, u t (x, = <x<π, where { x f(x = π x if x (,π/, if x (π/,π. The answer should be given in terms of an infinite series of explicitly given functions. Proof. Assume u(x, t =X(xT (t, then substitution in the PDE gives XT X T + XT =, T X + = T X = λ. From X + λx =, X n (x =a n cos λ n x + b n sin λ n x. Boundary conditions give { u(,t=x(t (t= X( = X(π=. u(π, t=x(πt (t= Thus, X n ( = a n =, andx n (x=b n sin λ n x. X n (π =b n sin λ n π =. Hence, λ n = n, or λ n = n, n =,,... λ n = n, X n (x =b n sin nx.

308 Partial Differential Equations Igor Yanovsky, 5 38 With these values of λ n,wesolve T + T = λ n T, T + T = n T, T n +(+n T n =. The solution to this nd-order ODE is of the form: T n (t =c n cos +n t + d n sin +n t. u(x, t = X(xT (t = u t (x, t = ( cn cos +n t + d n sin +n t sin nx. n= ( cn ( +n sin +n t + d n ( +n cos +n t sin nx. n= Initial conditions give f(x = u(x, = = u t (x, = c n sin nx. n= d n ( +n sinnx. n= By orthogonality, we may multiply both equations by sin mx and integrate: π π f(x sinmx dx = c m, π π dx = d m +m, which gives the coefficients π π c n = f(x sinnx dx = x sin nx dx + π π π = [ x n π π cos nx + π ] cos nx dx + n π [ πn cos nπ + n sin nπ n ] sin π π [ π n cos nx π (π x sinnx dx π + x n cos nx π π n π π ] cos nx dx = π [ ] + π π n cos nπ + π n cos nπ + π n cos nπ π n cos nπ n sin nπ + n sin nπ = [ π n sin nπ ] + [ π n sin nπ ] = 4 πn sin nπ, n =k { = 4, n =k, n =4m + = πn 4 ( n 4, n =k +. πn, n =4m +3 πn d n =. u(x, t = ( cn cos +n t sin nx. n=

309 Partial Differential Equations Igor Yanovsky, Problems: Separation of Variables - Heat Equation Problem (F 94, #5. Solve the initial-boundary value problem u t = u xx <x<, t > u(x, = x x + x u(,t=, u(,t=3 t>. Find lim t + u(x, t. Proof. ➀ First, we need to obtain function v that satisfies v t = v xx and takes boundary conditions. Let v(x, t=u(x, t+(ax + b, (5. where a and b are constants to be determined. Then, Thus, v t = u t, v xx = u xx. v t = v xx. We need equation (5. to take boundary conditions for v(,tandv(,t: v(,t= = u(,t+b = +b b =, v(,t= = u(,t+a = a + a =. Thus, (5. becomes v(x, t=u(x, t x. (5. The new problem is v t = v xx, v(x, = (x x + x =x x, v(,t=v(,t=. ➁ We solve the problem for v using the method of separation of variables. Let v(x, t =X(xT (t, which gives XT X T =. X X = T T = λ. From X + λx =,weget X n (x =a n cos λx + b n sin λx. Using boundary conditions, we have { v(,t=x(t (t= X( = X( =. v(,t=x(t (t= Hence, X n ( = a n =,andx n (x =b n sin λx. X n ( = b n sin λ = λ = nπ λ n =( nπ. X n (x =b n sin nπx, λ n = ( nπ.

310 Partial Differential Equations Igor Yanovsky, 5 3 With these values of λ n,wesolvet + ( nπ T = to find nπ ( T n (t =c n e t. v(x, t= X n (xt n (t = n= n= c n e ( nπ t sin nπx. Coefficients c n are obtained using the initial condition: v(x, = c n sin nπx = x x. n= c n = (x x sin nπx dx = { niseven, 3 (nπ 3 n is odd. v(x, t = n=k 3 nπ e ( t sin nπx (nπ 3. We now use equation (5. to convert back to function u: u(x, t=v(x, t+x +. u(x, t = n=k lim u(x, t = x +. t + 3 nπ e ( t sin nπx (nπ 3 + x +.

311 Partial Differential Equations Igor Yanovsky, 5 3 Problem (S 96, #6. Let u(x, t be the solution of the initial-boundary value problem for the heat equation u t = u xx <x<l, t> u(x, = f(x x L u x (,t=u x (L, t=a t > (A = Const. Find v(x -thelimitofu(x, t when t. Show that v(x is one of the inifinitely many solutions of the stationary problem v xx = <x<l v x ( = v x (L =A. Proof. ➀ First, we need to obtain function v that satisfies v t = v xx and takes boundary conditions. Let v(x, t=u(x, t+(ax + b, (5.3 where a and b are constants to be determined. Then, Thus, v t = u t, v xx = u xx. v t = v xx. We need equation (5.3 to take boundary conditions for v x (,tandv x (L, t. v x = u x + a. v x (,t= = u x (,t+a = A + a a = A, v x (L, t= = u x (L, t+a = A + a a = A. We may set b = (infinitely many solutions are possible, one for each b. Thus, (5.3 becomes v(x, t=u(x, t Ax. (5.4 The new problem is v t = v xx, v(x, = f(x Ax, v x (,t=v x (L, t =. ➁ We solve the problem for v using the method of separation of variables. Let v(x, t =X(xT (t, which gives XT X T =. X X = T T = λ. From X + λx =,weget X n (x =a n cos λx + b n sin λx. Using boundary conditions, we have { v x (,t=x (T (t= v x (L, t =X X ( = X (L =. (LT (t=

312 Partial Differential Equations Igor Yanovsky, 5 3 X n(x = a n λ sin λx + bn λ cos λx. Hence, X n( = b n λn = b n =;andx n (x =a n cos λx. X n(l = a n λ sin L λ = L λ = nπ λn =( nπ L. X n (x =a n cos nπx ( nπ. L, λ n = L With these values of λ n,wesolvet + ( nπ L T = to find T (t =c, T n (t =c n e ( nπ L t, n =,,... v(x, t= X n (xt n (t = c + n= n= c n e ( nπ L t cos nπx L. Coefficients c n are obtained using the initial condition: v(x, = c + c n cos nπx = f(x Ax. L n= L c = L c n = L L (f(x Ax dx = L f(x dx AL (f(x Ax cos nπx L dx c n = L L L c = f(x dx AL L, (f(x Ax cos nπx L dx. v(x, t = L L f(x dx AL + n c n e ( nπ L t cos nπx L. We now use equation (5.4 to convert back to function u: u(x, t=v(x, t+ax. u(x, t = L L f(x dx AL + n c n e ( nπ L t cos nπx L + Ax. lim u(x, t = Ax + b, b t + arbitrary. To show that v(x isone of the inifinitely many solutions of the stationary problem v xx = <x<l v x ( = v x (L =A, we can solve the boundary value problem to obtain v(x, t=ax+b, whereb is arbitrary.

313 Partial Differential Equations Igor Yanovsky, 5 33 Heat Equation with Nonhomogeneous Time-Independent BC in N-dimensions. The solution to this problem takes somewhat different approach than in the last few problems, but is similar. Consider the following initial-boundary value problem, u t = u, x, t u(x, = f(x, x u(x, t =g(x, x, t >. Proof. Let w(x be the solution of the Dirichlet problem: { w =, x w(x =g(x, x and let v(x, t be the solution of the IBVP for the heat equation with homogeneous BC: v t = v, x, t v(x, = f(x w(x, x v(x, t=, x, t >. Then u(x, t satisfies u(x, t=v(x, t+w(x. lim u(x, t=w(x. t

314 Partial Differential Equations Igor Yanovsky, 5 34 Nonhomogeneous Heat Equation with Nonhomogeneous Time-Independent BC in N dimensions. Describe the method of solution of the problem u t = u + F (x, t, x, t u(x, = f(x, x u(x, t =g(x, x, t >. Proof. ❶ We first find u, the solution to the homogeneous heat equation (no F (x, t. Let w(x be the solution of the Dirichlet problem: { w =, x w(x =g(x, x and let v(x, t be the solution of the IBVP for the heat equation with homogeneous BC: v t = v, x, t v(x, = f(x w(x, x v(x, t=, x, t >. Then u (x, t satisfies u (x, t =v(x, t+w(x. lim u (x, t =w(x. t ❷ The solution to the homogeneous equation with boundary conditions is given by Duhamel s principle. { u t = u + F (x, t for t>, x R n u (x, = for x R n (5.5. Duhamel s principle gives the solution: t u (x, t = K(x y, t s F (y, s dy ds R n Note: u (x, t= on may not be satisfied. t u(x, t =v(x, t+w(x+ K(x y, t s F (y, s dy ds. R n

315 Partial Differential Equations Igor Yanovsky, 5 35 Problem (S 98, #5. Find the solution of u t = u xx, t, <x<, u(x, =, <x<, u(,t= e t, u x (,t=e t, t >. Prove that lim t u(x, t exists and find it. Proof. ➀ First, we need to obtain function v that satisfies v t = v xx and takes boundary conditions. Let v(x, t=u(x, t+(ax + b+(c cos x + c sin xe t, (5.6 where a, b, c, c are constants to be determined. Then, Thus, v t = u t (c cos x + c sin xe t, v xx = u xx +( c cos x c sin xe t. v t = v xx. We need equation (5.6 to take boundary conditions for v(,tandv x (,t: v(,t= = u(,t+b + c e t = e t + b + c e t. Thus, b =, c =, and (5.6 becomes v(x, t=u(x, t+(ax + (cos x + c sin xe t. (5.7 v x (x, t = u x (x, t+a +( sin x + c cos xe t, v x (,t= = u x (,t+a +( sin + c cos e t = +a +( sin + c cos e t. sin Thus, a =,c = cos, and equation (5.7 becomes v(x, t=u(x, t+(x + (cos x + sin sin xe t. (5.8 cos Initial condition tranforms to: v(x, = u(x, + (x + (cos x + sin sin x =(x + (cos x + sin sin x. cos cos The new problem is v t = v xx, v(x, = (x + (cos x + v(,t=, v x (,t=. sin cos sin x, ➁ We solve the problem for v using the method of separation of variables. Let v(x, t =X(xT (t, which gives XT X T =. X X = T T = λ.

316 Partial Differential Equations Igor Yanovsky, 5 36 From X + λx =,weget X n (x =a n cos λx + b n sin λx. Using the first boundary condition, we have v(,t=x(t (t= X( =. Hence, X n ( = a n =,andx n (x =b n sin λx. Wealsohave v x (,t = X (T (t= X ( =. X n(x = λb n cos λx, X n ( = λb n cos λ =, cos λ =, π λ = nπ +. Thus, ( X n (x =b n sin nπ + π ( x, λ n = nπ + π. ( T With these values of λ n,wesolvet + nπ + π = to find T n (t =c n e (nπ+ π t. v(x, t= X n (xt n (t = n= n= ( bn sin nπ + π xe (nπ+ π t. We now use equation (5.8 to convert back to function u: u(x, t=v(x, t (x (cos x + sin sin xe t. cos ( u(x, t = bn sin nπ + π xe (nπ+ π t (x (cos x + sin sin xe t. cos n= Coefficients b n are obtained using the initial condition: ( u(x, = bn sin nπ + π x (x (cos x + sin sin x. cos n= ➂ Finally, we can check that the differential equation and the boundary conditions are satisfied: u(,t = ( + e t = e t. u x (x, t = ( bn nπ + π ( cos nπ + π n= u x (,t = +(sin sin cos e t = +e t. cos ( u t = b n nπ + π ( sin nπ + π n= xe (nπ+ π t +(sinx sin cos xe t, cos xe (nπ+ π t +(cosx + sin sin xe t = u xx. cos

317 Partial Differential Equations Igor Yanovsky, 5 37 Problem (F, #6. The temperature of a rod insulated at the ends with an exponentially decreasing heat source in it is a solution of the following boundary value problem: u t = u xx + e t g(x for (x, t [, ] R + u x (,t=u x (,t= u(x, = f(x. Find the solution to this problem by writing u as a cosine series, u(x, t= a n (tcosnπx, n= and determine lim t u(x, t. Proof. Let g accept an expansion in eigenfunctions g(x=b + n= b n cos nπx with b n = n= g(xcosnπx dx. Plugging in the PDE gives: a (t+ a n(tcosnπx = n π a n (tcosnπx + b e t + e t n= which gives { a (t =b e t, a n(t+n π a n (t =b n e t, n =,,... n= b n cos nπx, Adding homogeneous and particular solutions of the above ODEs, we obtain the solutions { a (t =c b e t, a n (t =c n e n π t for some constants c n, u(x, t = n= Initial condition gives u(x, = n= bn e t, n =,,..., n π n =,,,... Thus, ( c n e n π t b n n π e t cos nπx. ( b n c n n π cos nπx = f(x, As, t, the only mode that survives is n =: u(x, t c + b as t.

318 Partial Differential Equations Igor Yanovsky, 5 38 Problem (F 93, #4. a Assume f, g C. Give the compatibility conditions which f and g must satisfy if the following problem is to possess a solution. u = f(x x u (s =g(s n s. Show that your condition is necessary for a solution to exist. b Give an explicit solution to u t = u xx +cosx x [, π] u x (,t=u x (π, t= t> u(x, = cos x +cosx x [, π]. c Does there exist a steady state solution to the problem in (b if u x ( = u x (π=? Explain your answer. Proof. a Integrating the equation and using Green s identity gives: u f(x dx = udx = n ds = g(s ds. b With v(x, t=u(x, t cos x the problem above transforms to v t = v xx v x (,t=v x (π, t= v(x, = cos x. We solve this problem for v using the separation of variables. Let v(x, t =X(xT (t, which gives XT = X T. X X = T T = λ. From X + λx =,weget X n (x =a n cos λx + b n sin λx. X n(x = λ n a n sin λx + λ n b n cos λx. Using boundary conditions, we have { v x (,t=x (T (t= v x (π, t=x X ( = X (π=. (πt (t= Hence, X n( = λ n b n =,andx n (x =a n cos λ n x. X n (π= λ n a n sin λ n π = λ n = n λ n =( n. Thus, X n (x =a n cos nx ( n, λ n =

319 Partial Differential Equations Igor Yanovsky, 5 39 With these values of λ n,wesolvet + ( n T = to find T n (t =c n e ( n t. v(x, t= X n (xt n (t = n= Initial condition gives v(x, = ã n cos nx n= n= = cosx. Thus, ã 4 =, ã n =,n 4. Hence, v(x, t = e 4t cos x. ã n e ( n t cos nx. u(x, t =v(x, t+cosx = e 4t cos x +cosx. c Does there exist a steady state solution to the problem in (b if u x ( = u x (π=? Explain your answer. c Set u t =. Wehave { u xx +cosx = x [, π] u x ( =, u x (π =. u xx = cos x, u x = sin x + C, u(x = cosx + Cx + D. Boundary conditions give: = u x ( = C, =u x (π = C contradiction There exists no steady state solution. We may use the result we obtained in part (a with u xx =cosx = f(x. We need u f(x dx = n ds, π cos xdx } {{ } = = u x (π u x ( = }{{}. given

320 Partial Differential Equations Igor Yanovsky, 5 3 Problem (F 96, #7. Solve the parabolic problem ( ( ( u = u, v v t xx x π, t > u(x, = sin x, u(,t=u(π, t=, v(x, = sin x, v(,t=v(π, t=. Prove the energy estimate (for general initial data π x= for come constant c. [u (x, t+v (x, t] dx c π x= [u (x, + v (x, ] dx Proof. We can solve the second equation for v and then use the value of v to solve the first equation for u. 73 ➀ We have v t =v xx, x π, t > v(x, = sin x, v(,t=v(π, t=. Assume v(x, t =X(xT (t, then substitution in the PDE gives XT =X T. T =X T X = λ. From X + λ X =, weget X n(x =a n cos λ x + b n sin λ x. Boundary conditions give { v(,t=x(t (t= v(π, t=x(πt (t= X( = X(π=. λ Thus, X n ( = a n =,andx n (x =b n sin x. λ X n (π =b n sin π =. Hence λ = n, orλ =n. λ =n, X n (x=b n sin nx. With these values of λ n,wesolve T +n T =toget T n (t =c n e nt. Thus, the solution may be written in the form v(x, t= ã n e nt sin nx. n= From initial condition, we get v(x, = ã n sin nx = sinx. n= Thus, ã =, ã n =, n =, 3,... v(x, t=e t sin x. 73 Note that if the matrix was fully inseparable, we would have to find eigenvalues and eigenvectors, just as we did for the hyperbolic systems.

321 Partial Differential Equations Igor Yanovsky, 5 3 ➁ We have u t = u xx e t sin x, x π, t > u(x, = sin x, u(,t=u(π, t=. Let u(x, t= n= u n(tsinnx. Plugging this into the equation, we get u n(tsinnx + n u n (tsinnx = e t sin x. n= For n =: n= u (t+u (t = e t. Combining homogeneous and particular solution of the above equation, we obtain: u (t = e t + c e t. For n =, 3,...: Thus, u(x, t = u n(t+n u n (t =, u n (t =c n e nt. ( e t + c e t sin x + c n e nt sin nx = e t sin x + c n e nt sin nx. n= From initial condition, we get u(x, = sin x + c n sin nx = sinx. n= Thus, c =, c n =, n =, 3,... n= u(x, t = sin x (e t + e t. To prove the energy estimate (for general initial data π x= [u (x, t+v (x, t] dx c π x= n= n= [u (x, + v (x, ] dx for come constant c, we assume that u(x, = a n sin nx, v(x, = b n sin nx.

322 Partial Differential Equations Igor Yanovsky, 5 3 The general solutions are obtained by the same method as above u(x, t = e t sin x + c n e nt sin nx, π x= v(x, t = [u (x, t+v (x, t] dx = n= b n e nt sin nx. n= π x= ( e t sin x + (b n + a n n= π x= c n e nt sin nx ( + b n e nt sin nx dx n= sin nx dx π x= n= [u (x, + v (x, ] dx.

323 Partial Differential Equations Igor Yanovsky, Problems: Eigenvalues of the Laplacian - Laplace The D LAPLACE Equation (eigenvalues/eigenfuctions of the Laplacian. Consider u xx + u yy + λu = in u(,y==u(a, y for y b, (6. u(x, = = u(x, b for x a. Proof. We can solve this problem by separation of variables. Let u(x, y=x(xy (y, then substitution in the PDE gives X Y + XY + λxy =. X X + Y Y + λ =. Letting λ = μ + ν and using boundary conditions, we find the equations for X and Y : X + μ X = Y + ν Y = X( = X(a= Y ( = Y (b =. The solutions of these one-dimensional eigenvalue problems are μ m = mπ ν n = nπ a b X m (x =sin mπx Y n (y =sin nπy a b, where m, n =,,... Thus we obtain solutions of (6. of the form λ mn = π ( m a + n b u mn (x, y=sin mπx nπy sin a b, where m, n =,,... Observe that the eigenvalues {λ mn } m,n= are positive. The smallest eigenvalue λ has only one eigenfunction u (x, y =sin(πx/asin(πy/b; notice that u is positive in. Other eigenvalues λ may correspond to more than one choice of m and n; for example, in the case a = b we have λ nm = λ nm. For this λ, there are two linearly independent eigenfunctions. However, for a particular value of λ there are at most finitely many linearly independent eigenfunctions. Moreover, b a u mn (x, y u m n (x, y dx dy = b { a b = sin nπy b sin n πy b dy = a sin mπx a nπy sin b sin m πx a { ab 4 if m = m and n = n if m m or n n. sin n πy b dx dy In particular, the {u mn } are pairwise orthogonal. We could normalize each u mn by a scalar multiple (i.e. multiply by 4/ab sothatab/4 above becomes. Let us change the notation somewhat so that each eigenvalue λ n corresponds to a particular eigenfunction φ n (x. If we choose an orthonormal basis of eigenfunctions in each eigenspace, we may arrange that {φ n } n= is pairwise orthonormal: { if m = n φ n (xφ m (x dx = if m n.

324 Partial Differential Equations Igor Yanovsky, 5 34 In this notation, the eigenfunction expansion of f(x defined on becomes f(x a n φ n (x, where a n = f(xφ n (x dx. n=

325 Partial Differential Equations Igor Yanovsky, 5 35 Problem (S 96, #4. Let D denote the rectangular D = {(x, y R :<x<a, <y<b}. Find the eigenvalues of the following Dirichlet problem: ( + λu = in D u = on D. Proof. Theproblemmayberewrittenas u xx + u yy + λu = in u(,y==u(a, y for y b, u(x, = = u(x, b for x a. We may assume that the eigenvalues λ are positive, λ = μ + ν. Then, λ mn = π ( m a + n b u mn (x, y=sin mπx a nπy sin, m,n =,,... b Problem (W 4, #. Consider the differential equation: u(x, y x + u(x, y y + λu(x, y = (6. in the strip {(x, y, <y<π, <x<+ } with boundary conditions u(x, =, u(x, π=. (6.3 Find all bounded solutions of the boundary value problem (6.4, (6.5 when a λ =, b λ>, c λ<. Proof. a λ =. u xx + u yy =. Wehave Assume u(x, y =X(xY (y, then substitution in the PDE gives X Y + XY =. Boundary conditions give { u(x, = X(xY ( = u(x, π=x(xy (π = Method I: We have X X = Y Y = c, c >. Y ( = Y (π=. From X + cx =, wehave X n (x =a n cos cx + b n sin cx. From Y cy =, wehave Y n (y =c n e cy + d n e cy. Y ( = c n + d n = c n = d n.

326 Partial Differential Equations Igor Yanovsky, 5 36 Y (π=c n e cπ c n e cπ = c n = Y n (y =. u(x, y=x(xy (y =. Method II: We have X = Y X Y = c, c >. From X cx =, wehave X n (x =a n e cx + b n e cx. Since we look for bounded solutions for <x<, a n = b n = From Y + cy =, wehave Y n (y =c n cos cy + d n sin cy. X n (x =. Y ( = c n =, Y (π=d n sin cπ = c = n c = n. Y n (y =d n sin nx =. u(x, y=x(xy (y =. b λ>. We have X X + Y Y + λ =. Letting λ = μ + ν, and using boundary conditions for Y, we find the equations: X + μ X = Y + ν Y = Y ( = Y (π=. The solutions of these one-dimensional eigenvalue problems are X m (x =a m cos μ m x + b m sin μ m x. ν n = n, Y n (y =d n sin ny, where m, n =,,... u(x, y= c λ<. We have m,n= u mn (x, y= u xx + u yy + λu =, u(x, =, u(x, π=. m,n= (a m cos μ m x + b m sin μ m xsinny. u is the solution to this equation. We will show that this solution is unique. Let u and u be two solutions, and consider w = u u. Then, w + λw =, w(x, =, w(x, π=. Multiply the equation by w and integrate: w w + λw =, w wdx+ λ w dx =, w w n ds w dx + λ w dx =, } {{ } = w dx } {{ } = λ w dx. } {{ }

327 Partial Differential Equations Igor Yanovsky, 5 37 Thus, w and the solution u(x, y is unique.

328 Partial Differential Equations Igor Yanovsky, 5 38 Problem (F 95, #5. Find all bounded solutions for the following boundary value problem in the strip <x<a, <y<, ( + k u = (k = Const >, u(,y=, u x (a, y=. In particular, show that when ak π, the only bounded solution to this problem is u. Proof. Let u(x, y=x(xy (y, then we have X Y + XY + k XY =. X X + Y Y + k =. Letting k = μ + ν and using boundary conditions, we find: X + μ X =, Y + ν Y =. X( = X (a =. The solutions of these one-dimensional eigenvalue problems are μ m = (m π, a X m (x =sin (m πx Y n (y =c n cos ν n y + d n sin ν n y, a where m, n =,,... Thus we obtain solutions of the form ( (m kmn = π +νn a, where m, n =,,... u(x, y= m,n= u mn (x, y= u mn(x, y=sin (m πx ( c n cos ν n y+d n sin ν n y, a m,n= sin (m πx ( c n cos ν n y + d n sin ν n y. a We can take an alternate approach and prove the second part of the question. We have X Y + XY + k XY =, Y = X Y X + k = c. We obtain Y n (y =c n cos cy + d n sin cy. The second equation gives X + k X = c X, X +(k c X =, X m (x =a c m e k x + b c m e k x. Thus, X m (x is bounded only if k c >, (if k c =,X =,andx m (x = a m x + b m,bc sgivex m (x =πx, unbounded, in which case X m (x =a m cos k c x + b m sin k c x.

329 Partial Differential Equations Igor Yanovsky, 5 39 Boundary conditions give X m ( = a m =. X m(x =b m k c cos k c x, X m (a =b m k c cos k c a =, k c a = mπ π, m =,,..., ( π ( k c = m, a ( π ( k = m + c a, a k >π ( m, ( ak > π m, m =,,... Thus, bounded solutions exist only when ak > π. Problem (S 9, #. Show that the boundary value problem u(x, y x + u(x, y y + k u(x, y=, (6.4 where <x<+, <y<π, k> is a constant, u(x, =, u(x, π = (6.5 has a bounded solution if and only if k. Proof. We have u xx + u yy + k u =, X Y + XY + k XY =, X X = Y Y + k = c. We obtain X m (x =a m cos cx + b m sin cx. The second equation gives Y + k Y = c Y, Y +(k c Y =, Y n (y =c c n e k y + d c n e k y. Thus, Y n (y is bounded only if k c >, (if k c =,Y =,andy n (y =c n y+d n, BC s give Y, in which case Y n (y =c n cos k c y + d n sin k c y. Boundary conditions give Y n ( = c n =. Y n (π =d n sin k c π = k c = n k c = n k = n + c, n =,,... Hence, k>n, n =,,...

330 Partial Differential Equations Igor Yanovsky, 5 33 Thus, bounded solutions exist if k. Note: Ifk =,thenc =, which gives trivial solutions for Y n (y. u(x, y= m,n= X m (xy n (y = m,n= sin ny X m (x.

331 Partial Differential Equations Igor Yanovsky, 5 33 McOwen, 4.4 #7; 66B Ralston Hw. Show that the boundary value problem { a(x u + b(xu = λu in u = on has only trivial solution with λ, whenb(x and a(x > in. Proof. Multiplying the equation by u and integrating over, we get u a udx+ bu dx = λ u dx. Since (ua u =u a u + a u, we have (ua u dx + a u dx + bu dx = λ Using divergence theorem, we obtain }{{} u a u n ds + a u dx + = }{{} a u dx + }{{} b u dx = > bu dx = λ λ }{{} u dx. (6.6 u dx, u dx, Thus, u =in,andu is constant. Since u =on, u on. Similar Problem I: Note that this argument also works with Neumann B.C.: { a(x u + b(xu = λu in u/ n = on Using divergence theorem, (6.6 becomes ua u ds + a u dx + bu dx = λ u dx, }{{} n = }{{} a u dx + }{{} b u dx = }{{} λ u dx. > Thus, u =,andu = const on. Hence, we now have }{{} b u dx = }{{} λ u dx, which implies λ =. This gives the useful information that for the eigenvalue problem 74 { a(x u + b(xu = λu u/ n =, λ = is an eigenvalue, its eigenspace is the set of constants, and all other λ s are positive. 74 In Ralston s Hw#7 solutions, there is no - sign in front of a(x u below, which is probably a typo.

332 Partial Differential Equations Igor Yanovsky, 5 33 Similar Problem II: If λ, we show that the only solution to the problem below is the trivial solution. { u + λu = in u = on u udx+ λ u dx =, u }{{} u n ds u dx + }{{} λ u dx =. = Thus, u =in,andu is constant. Since u =on, u on.

333 Partial Differential Equations Igor Yanovsky, Problems: Eigenvalues of the Laplacian - Poisson The ND POISSON Equation (eigenvalues/eigenfunctions of the Laplacian. Suppose we want to find the eigenfunction expansion of the solution of u = f u = in on, when f has the expansion in the orthonormal Dirichlet eigenfunctions φ n : f(x a n φ n (x, where a n = n= f(xφ n (x dx. Proof. Writing u = c n φ n and inserting into λu = f, weget λ n c n φ n = n= a n φ n (x. n= Thus, c n = a n /λ n,and u(x = n= a n φ n (x λ n. The D POISSON Equation (eigenvalues/eigenfunctions of the Laplacian. For the boundary value problem u = f(x u( =, u(l =, the related eigenvalue problem is φ = λφ φ( =, φ(l =. The eigenvalues are λ n =(nπ/l, and the corresponding eigenfunctions are sin(nπx/l, n =,,... Writing u = c n φ n = c n sin(nπx/l and inserting into λu = f, we get L n= c n ( nπ L ( nπ c n L n= sin nπx L sin nπx L mπx sin L dx = c n ( nπ L L = f(x, = L L c n = L f(x sin mπx L dx, f(x sin nπx L dx, L f(x sin(nπx/l dx (nπ/l.

334 Partial Differential Equations Igor Yanovsky, u(x = c n sin(nπx/l = L n= L f(ξ sin(nπx/l sin(nπξ/l dξ (nπ/l, u = L [ f(ξ sin(nπx/l sin(nπξ/l ] L (nπ/l dξ. n= } {{ } = G(x,ξ See similar, but more complicated, problem in Sturm-Liouville Problems (S 9, #(c.

335 Partial Differential Equations Igor Yanovsky, Example: Eigenfunction Expansion of the GREEN s Function. Suppose we fix x and attempt to expand the Green s function G(x, y in the orthonormal eigenfunctions φ n (y: G(x, y a n (xφ n (y, where a n (x = n= G(x, zφ n (z dz. Proof. We can rewrite u + λu =in, u =on, as an integral equation 75 u(x+λ G(x, yu(y dy =. Suppose, u(x = c n φ n (x. Plugging this into, weget c m φ m (x+λ a n (xφ n (y c m φ m (y dy =, Thus, m= m= n= c m φ m (x+λ a n (x G(x, y n= n= n= c m m= φ n(xφ n (y λ n. m= φ n (yφ m (y dy =, c n φ n (x+ λa n (xc n =, n= ( c n φn (x+λa n (x =, n= a n (x= φ n(x λ n. 75 See the section: ODE - Integral Equations.

336 Partial Differential Equations Igor Yanovsky, The D POISSON Equation (eigenvalues/eigenfunctions of the Laplacian. Solve the boundary value problem u xx + u yy = f(x, y for <x<a, <y<b u(,y==u(a, y for y b, (7. u(x, = = u(x, b for x a. f(x, y C, f(x, y=if x =, x = a, y =, y = b, f(x, y= ab m,n= c mn sin mπx a nπy sin b. Proof. ➀ First, we find eigenvalues/eigenfunctions of the Laplacian. u xx + u yy + λu = in u(,y==u(a, y for y b, u(x, = = u(x, b for x a. Let u(x, y=x(xy (y, then substitution in the PDE gives X Y + XY + λxy =. X X + Y Y + λ =. Letting λ = μ + ν and using boundary conditions, we find the equations for X and Y : X + μ X = Y + ν Y = X( = X(a= Y ( = Y (b =. The solutions of these one-dimensional eigenvalue problems are μ m = mπ ν n = nπ a b X m (x =sin mπx Y n (y =sin nπy a b, where m, n =,,... Thus we obtain eigenvalues and normalized eigenfunctions of the Laplacian: λ mn = π ( m a + n b where m, n =,,... Note that f(x, y= m,n= c mn φ mn. φ mn (x, y= sin mπx nπy sin ab a b, ➁ Second, writing u(x, y= c mn φ mn and inserting into λu = f, weget λ mn c mn φ mn (x, y= c mn φ mn (x, y. Thus, m,n= c mn = cmn λ mn. u(x, y= n= m,n= c mn λ mn φ mn (x, y,

337 Partial Differential Equations Igor Yanovsky, with λ mn,φ mn (x given above, and c mn given by b a b a f(x, yφ mn dx dy = c m n φ m n φ mn dx dy = c mn. m,n =

338 Partial Differential Equations Igor Yanovsky, Problems: Eigenvalues of the Laplacian - Wave In the section on the wave equation, we considered an initial boundary value problem for the one-dimensional wave equation on an interval, and we found that the solution could be obtained using Fourier series. If we replace the Fourier series by an expansion in eigenfunctions, we can consider an initial/boundary value problem for the n-dimensional wave equation. The ND WAVE Equation (eigenvalues/eigenfunctions of the Laplacian. Consider u tt = u for x, t > u(x, = g(x, u t (x, = h(x for x u(x, t = for x, t >. Proof. For g, h C ( with g = h =on, we have eigenfunction expansions g(x= a n φ n (x and h(x = b n φ n (x. n= n= Assume the solution u(x, t may be expanded in the eigenfunctions with coefficients depending on t: u(x, t= n= u n(tφ n (x. This implies u n(tφ n (x = λ n u n (tφ n (x, n= u n (t+λ nu n (t = n= foreachn. Since λ n >, this ordinary differential equation has general solution u n (t = A n cos λ n t + B n sin λ n t. Thus, ( u(x, t = An cos λ n t + B n sin λ n t φ n (x, u t (x, t = u(x, = u t (x, = n= ( λn A n sin λ n t + λ n B n cos λ n t φ n (x, n= A n φ n (x =g(x, n= λn B n φ n (x =h(x. n= Comparing with, weobtain A n = a n, B n = b n λn. Thus, the solution is given by u(x, t = n= ( an cos λ n t + b n λn sin λ n t φ n (x,

339 Partial Differential Equations Igor Yanovsky, with a n = g(xφ n (x dx, b n = h(xφ n (x dx.

340 Partial Differential Equations Igor Yanovsky, 5 34 The D WAVE Equation (eigenvalues/eigenfunctions of the Laplacian. Let =(,a (,b and consider u tt = u xx + u yy for x, t > u(x, = g(x, u t (x, = h(x for x (8. u(x, t = for x, t>. Proof. ➀ First, we find eigenvalues/eigenfunctions of the Laplacian. u xx + u yy + λu = in u(,y==u(a, y for y b, u(x, = = u(x, b for x a. Let u(x, y=x(xy (y, then substitution in the PDE gives X Y + XY + λxy =. X X + Y Y + λ =. Letting λ = μ + ν and using boundary conditions, we find the equations for X and Y : X + μ X = Y + ν Y = X( = X(a= Y ( = Y (b =. The solutions of these one-dimensional eigenvalue problems are μ m = mπ ν n = nπ a b X m (x =sin mπx Y n (y =sin nπy a b, where m, n =,,... Thus we obtain eigenvalues and normalized eigenfunctions of the Laplacian: λ mn = π ( m a + n b φ mn (x, y= sin mπx nπy sin ab a b, where m, n =,,... ➁ Second, we solve the Wave Equation (8. using the space eigenfunctions. For g, h C ( with g = h =on, we have eigenfunction expansions 76 g(x= a n φ n (x and h(x = b n φ n (x. n= n= Assume u(x, t = n= u n(tφ n (x. This implies u n(t+λ n u n (t = foreachn. 76 In D, φ n is really φ mn, andx is (x, y.

341 Partial Differential Equations Igor Yanovsky, 5 34 Since λ n >, this ordinary differential equation has general solution u n (t = A n cos λ n t + B n sin λ n t. Thus, ( u(x, t = An cos λ n t + B n sin λ n t φ n (x, u t (x, t = u(x, = u t (x, = n= ( λn A n sin λ n t + λ n B n cos λ n t φ n (x, n= A n φ n (x =g(x, n= λn B n φ n (x =h(x. n= Comparing with, weobtain A n = a n, B n = b n λn. Thus, the solution is given by u(x, t = m,n= with λ mn,φ mn (x given above, and a mn = g(xφ mn (x dx, b mn = h(xφ mn (x dx. ( amn cos λ mn t + b mn λmn sin λ mn t φ mn (x,

342 Partial Differential Equations Igor Yanovsky, 5 34 McOwen, 4.4 #3; 66B Ralston Hw. Consider the initial-boundary value problem u tt = u + f(x, t for x, t > u(x, t = for x, t > u(x, =, u t (x, = for x. Use Duhamel s principle and an expansion of f in eigenfunctions to obtain a (formal solution. Proof. a We expand u in terms of the Dirichlet eigenfunctions of Laplacian in. Assume a n(t = φ n + λ n φ n = in, φ n = on. u(x, t= f(x, t = = a n ( = a n( = a n (tφ n (x, a n (t = n= f n (tφ n (x, f n (t = n= φ n (xu(x, t dx. φ n (xf(x, t dx. φ n (xu tt dx = φ n ( u + f dx = φ n udx+ φ n fdx φ n udx+ φ n fdx= λ n φ n udx+ φ n fdx= λ n a n (t+f n (t. } {{ } f n φ n (xu(x, dx =. φ n (xu t (x, dx =. 77 Thus,wehaveanODEwhichisconvertedandsolvedbyDuhamel s principle: a n + λ n a n = f n (t ã n + λ n ã n = t a n ( = ã n (,s= a n (t = ã n (t s, s ds. a n ( = ã n (,s=f n(s With the anzats ã n (t, s =c cos λ n t + c sin λ n t, we get c =,c = f n (s/ λ n, or ã n (t, s =f n (s sin λ n t λn. Duhamel s principle gives a n (t = t u(x, t = ã n (t s, s ds = n= φ n (x λn t t f n (s sin( λ n (t s λn f n (ssin( λ n (t s ds. ds. 77 We used Green s formula: ( u φn n u φn n ds = (φn u φnu dx. On, u =; φ n = since eigenfunctions are Dirichlet.

343 Partial Differential Equations Igor Yanovsky, Problem (F 9, #3. Consider the initial-boundary value problem u tt = a(tu xx + f(x, t x π, t u(,t=u(π, t= t u(x, = g(x, u t (x, = h(x x π, where the coefficient a(t. a Express (formally the solution of this problem by the method of eigenfunction expansions. b Show that this problem is not well-posed if a. Hint: Take f =and prove that the solution does not depend continuously on the initial data g, h. Proof. a We expand u in terms of the Dirichlet eigenfunctions of Laplacian in. φ nxx + λ n φ n = in, φ n ( = φ n (π =. That gives us the eigenvalues and eigenfunctions of the Laplacian: λ n = n, φ n (x = sin nx. Assume u(x, t= u n (tφ n (x, u n (t = φ n (xu(x, t dx. f(x, t = g(x = h(x = n= f n (tφ n (x, f n (t = n= g n φ n (x, g n = n= h n φ n (x, h n = n= φ n (xf(x, t dx. φ n (xg(x dx. φ n (xh(x dx. u n (t = φ n (xu tt dx = φ n (a(tu xx + f dx = a(t = a(t φ nxx udx+ φ n fdx= λ n a(t = λ n a(tu n (t+f n (t. u n ( = φ n (xu(x, dx = φ n (xg(x dx = g n. u n( = φ n (xu t (x, dx = φ n (xh(x dx = h n. φ n udx+ φ n u xx dx + φ n fdx } {{ } f n Thus,wehaveanODEwhichisconvertedandsolvedbyDuhamel s principle: u n + λ n a(tu n = f n (t u n ( = g n u n( = h n. φ n fdx

344 Partial Differential Equations Igor Yanovsky, Note: The initial data is not ; therefore, the Duhamel s principle is not applicable. Also, the ODE is not linear in t, and it s solution is not obvious. Thus, u(x, t = u n (tφ n (x, n= where u n (t aresolutionsof.

345 Partial Differential Equations Igor Yanovsky, b Assume we have two solutions, u and u, to the PDE: u tt + u xx =, u tt + u xx =, u (,t=u (π, t =, u (,t=u (π, t =, u (x, = g (x, u t (x, = h (x; u (x, = g (x, u t (x, = h (x. Note that the equation is elliptic, and therefore, the maximum principle holds. In order to prove that the solution does not depend continuously on the initial data g, h, we need to show that one of the following conditions holds: max u u > max g g, max u t u t > max h h. That is, the difference of the two solutions is not bounded by the difference of initial data. By the method of separation of variables, we may obtain u(x, t = (a n cos nt + b n sin nt sinnx, u(x, = u t (x, = Not complete. n= a n sin nx = g(x, n= nb n sin nx = h(x. n= We also know that for elliptic equations, and for Laplace equation in particular, the value of the function u has to be prescribed on the entire boundary, i.e. u = g on, which is not the case here, making the problem under-determined. Also, u t is prescribed on one of the boundaries, making the problem overdetermined.

346 Partial Differential Equations Igor Yanovsky, Problems: Eigenvalues of the Laplacian - Heat The ND HEAT Equation (eigenvalues/eigenfunctions of the Laplacian. Consider the initial value problem with homogeneous Dirichlet condition: u t = u for x, t > u(x, = g(x for x u(x, t = for x, t >. Proof. For g C ( with g =on, we have eigenfunction expansion g(x= a n φ n (x n= Assume the solution u(x, t may be expanded in the eigenfunctions with coefficients depending on t: u(x, t= n= u n(tφ n (x. This implies u n (tφ n(x = λ n u n (tφ n (x, n= u n(t+λ n u n (t =, n= u n (t =A n e λnt. Thus, u(x, t= A n e λnt φ n (x, u(x, = n= A n φ n (x =g(x. n= which has the general solution Comparing with, weobtaina n = a n. Thus, the solution is given by u(x, t = a n e λnt φ n (x, n= Also with a n = u(x, t = = g(xφ n (x dx. a n e λnt φ n (x = n= n= ( n= e λnt φ n (xφ n (y g(y dy } {{ } K(x,y,t, heat kernel g(yφ n (y dy e λnt φ n (x

347 Partial Differential Equations Igor Yanovsky, The D HEAT Equation (eigenvalues/eigenfunctions of the Laplacian. Let =(,a (,b and consider u t = u xx + u yy for x, t > u(x, = g(x for x (9. u(x, t = for x, t >. Proof. ➀ First, we find eigenvalues/eigenfunctions of the Laplacian. u xx + u yy + λu = in u(,y==u(a, y for y b, u(x, = = u(x, b for x a. Let u(x, y=x(xy (y, then substitution in the PDE gives X Y + XY + λxy =. X X + Y Y + λ =. Letting λ = μ + ν and using boundary conditions, we find the equations for X and Y : X + μ X = Y + ν Y = X( = X(a= Y ( = Y (b =. The solutions of these one-dimensional eigenvalue problems are μ m = mπ ν n = nπ a b X m (x =sin mπx Y n (y =sin nπy a b, where m, n =,,... Thus we obtain eigenvalues and normalized eigenfunctions of the Laplacian: λ mn = π ( m a + n b φ mn (x, y= sin mπx nπy sin ab a b, where m, n =,,... ➁ Second, we solve the Heat Equation (9. using the space eigenfunctions. For g C ( with g =on, we have eigenfunction expansion g(x= a n φ n (x. n= Assume u(x, t = n= u n(tφ n (x. This implies u n(t+λ n u n (t =, which has the general solution u n (t =A n e λnt. Thus, u(x, t= A n e λnt φ n (x, u(x, = n= A n φ n (x =g(x. n=

348 Partial Differential Equations Igor Yanovsky, Comparing with, weobtaina n = a n. Thus, the solution is given by u(x, t = a mn e λmnt φ mn (x, m,n= with λ mn,φ mn given above and a mn = g(xφ mn(x dx.

349 Partial Differential Equations Igor Yanovsky, Problem (S 9, #. Consider the heat equation u t = u xx + u yy on the square ={ x π, y π} with periodic boundary conditions and with initial data u(,x,y=f(x, y. a Find the solution using separation of variables. Proof. ➀ First, we find eigenvalues/eigenfunctions of the Laplacian. u xx + u yy + λu = in u(,y=u(π, y for y π, u(x, = u(x, π for x π. Let u(x, y=x(xy (y, then substitution in the PDE gives X Y + XY + λxy =. X X + Y Y + λ =. Letting λ = μ + ν and using periodic BC s, we find the equations for X and Y : X + μ X = Y + ν Y = X( = X(π Y ( = Y (π. The solutions of these one-dimensional eigenvalue problems are μ m = m ν n = n X m (x =e imx Y n (y =e iny, where m, n =...,,,,,,... Thus we obtain eigenvalues and normalized eigenfunctions of the Laplacian: λ mn = m + n φ mn (x, y=e imx e iny, where m, n =...,,,,,,... ➁ Second, we solve the Heat Equation using the space eigenfunctions. Assume u(x, y, t= m,n= u mn(te imx e iny. This implies u mn (t+(m + n u mn (t =, which has the general solution u n (t =c mn e (m +n t. u(x, y, t= m,n= Thus, c mn e (m +n t e imx e iny.

350 Partial Differential Equations Igor Yanovsky, 5 35 π π u(x, y, = f(x, ye imx e iny dxdy = m,n= π π π = π c mn = 4π c mn e imx e iny m,n= n= π π = f(x, y, c mn e imx e iny e im x e in y dxdy c mn e iny e in y dy = 4π c mn. f(x, ye imx e iny dxdy = f mn.

351 Partial Differential Equations Igor Yanovsky, 5 35 b Show that the integral u (x, y, t dxdy is decreasing in t, iff is not constant. Proof. We have u t = u xx + u yy Multiply the equation by u and integrate: d dt uu t = u u, d dt u = u u, u dxdy = u udxdy = u u n ds } {{ } =, (periodic BC u dxdy = u dxdy. Equality is obtained only when u = u = constant f = constant. If f is not constant, u dxdy is decreasing in t.

352 Partial Differential Equations Igor Yanovsky, 5 35 Problem (F 98, #3. Consider the eigenvalue problem d φ + λφ =, dx φ( dφ dφ ( =, φ( + ( =. dx dx a Show that all eigenvalues are positive. b Show that there exist a sequence of eigenvalues λ = λ n, each of which satisfies tan λ = λ λ. c Solve the following initial-boundary value problem on <x<, t> u t = u x, u(,t u x (,t=, u(x, = f(x. u(,t+ u x (,t=, You may call the relevant eigenfunctions φ n (x and assume that they are known. Proof. a If λ =, the ODE reduces to φ =. Tryφ(x =Ax + B. From the first boundary condition, φ( φ ( = = B A B = A. Thus, the solution takes the form φ(x = Ax+A. The second boundary condition gives φ( + φ ( = = 3A A = B =. Thus the only solution is φ, which is not an eigenfunction, and not an eigenvalue. If λ<, try φ(x =e sx,whichgivess = ± λ = ±β R. Hence, the family of solutions is φ(x =Ae βx +Be βx.also,φ (x =βae βx βbe βx. The boundary conditions give φ( φ ( = = A + B βa + βb = A( β+b( + β, (9. φ(+φ ( = = Ae β +Be β +βae β βbe β = Ae β (+β+be β ( β. (9.3 From (9. and (9.3 we get +β β = A B and +β β = B A e β, or A B = e β. From (9., β = A + B A B and thus, A B = e A+B B A, which has no solutions. b Since λ>, the anzats φ = e sx gives s = ±i λ and the family of solutions takes the form φ(x =A sin(x λ+b cos(x λ. Then, φ (x =A λ cos(x λ B λ sin(x λ. The first boundary condition gives φ( φ ( = = B A λ B = A λ.

353 Partial Differential Equations Igor Yanovsky, Hence, φ(x =A sin(x λ+a λ cos(x λ. The second boundary condition gives φ( + φ ( = = A sin( λ+a λ cos( λ+a λ cos( λ Aλ sin( λ = A [ ( λsin( λ+ λ cos( λ ] A (sincea = impliesb = andφ =, which is not an eigenfunction. Therefore, ( λsin( λ= λ cos( λ, and thus tan( λ= λ λ. c We may assume that the eigenvalues/eigenfunctins of the Laplacian, λ n and φ n (x, are known. We solve the Heat Equation using the space eigenfunctions. u t = u xx, u(,t u x (,t=, u(,t+u x (,t=, u(x, = f(x. For f, we have an eigenfunction expansion f(x = a n φ n (x. n= Assume u(x, t = n= u n(tφ n (x. This implies u n (t+λ nu n (t =, which has the general solution u n (t =A n e λnt. Thus, u(x, t= A n e λnt φ n (x, u(x, = n= A n φ n (x =f(x. n= Comparing with, wehavea n = a n. Thus, the solution is given by with u(x, t = a n = a n e λnt φ n (x, n= f(xφ n (x dx.

354 Partial Differential Equations Igor Yanovsky, Problem (W 3, #3; 66B Ralston Hw. Let be a smooth domain in three dimensions and consider the initial-boundary value problem for the heat equation u t = u + f(x for x, t > u/ n = for x, t > u(x, = g(x for x, in which f and g are known smooth functions with g/ n = for x. a Find an approximate formula for u as t. Proof. We expand u in terms of the Neumann eigenfunctions of Laplacian in. φ n + λ n φ n = in, φ n n = on. Note that here λ =andφ is the constant V /,wherev isthevolumeof. Assume u(x, t= a n (tφ n (x, a n (t = φ n (xu(x, t dx. f(x = g(x = n= f n φ n (x, f n = n= g n φ n (x, g n = n= a n (t = = a n ( = 78 Thus,wesolvetheODE: { a n + λ na n = f n a n ( = g n. φ n (xf(x dx. φ n (xg(x dx. φ n (xu t dx = φ n ( u + f dx = φ n udx+ φ n fdx φ n udx+ φ n fdx= λ n φ n udx+ φ n fdx= λ n a n + f n. } {{ } f n φ n (xu(x, dx = φ n gdx= g n. For n =,λ =,andweobtaina (t =f t + g. For n, the homogeneous solution is a nh = ce λnt. The anzats for a particular solution is a np = c t + c,whichgivesc = and c = f n /λ n. Using the initial condition, we obtain ( a n (t = g n f n e λnt + f n. λ n λ n 78 We used Green s formula: ( u φn n u φn n ds = (φn u φnu dx. On, u =; φ n = since eigenfunctions are Neumann. n n

355 Partial Differential Equations Igor Yanovsky, u(x, t =(f t + g φ (x+ n= [( g n f n λ n e λnt + f n λ n ] φ n (x. If f = If f ( ( f(x dx =, lim u(x, t=g φ + t f(x dx, lim u(x, t f φ t. t n= f n φ n λ n.

356 Partial Differential Equations Igor Yanovsky, b If g and f>, show that u> for all t>.

357 Partial Differential Equations Igor Yanovsky, Problem (S 97, #. a Consider the eigenvalue problem for the Laplace operator in R with zero Neumann boundary condition { u xx + u yy + λu = in u n = on. Prove that λ =is the lowest eigenvalue and that it is simple. b Assume that the eigenfunctions φ n (x, y of the problem in (a form a complete orthogonal system, and that f(x, y has a uniformly convergent expansion f(x, y= f n φ n (x, y. n= Solve the initial value problem u t = u + f(x, y subject to initial and boundary conditions u u(x, y, =, n u =. What is the behavior of u(x, y, t as t? c Consider the problem with Neumann boundary conditions { v xx + v yy + f(x, y= in v n v = on. When does a solution exist? Find this solution, and find its relation with the behavior of lim u(x, y, t in (b as t. Proof. a Suppose this eigenvalue problem did have a solution u with λ. Multiplying u + λu = byu and integrating over, we get u udx+ λ u dx =, u u ds u dx + λ u dx =, }{{} n = u dx = }{{} λ u dx, Thus, u =in,andu is constant in. Hence, we now have = }{{} λ u dx. For nontrivial u, wehaveλ =. For this eigenvalue problem, λ = is an eigenvalue, its eigenspace is the set of constants, and all other λ s are positive.

358 Partial Differential Equations Igor Yanovsky, b We expand u in terms of the Neumann eigenfunctions of Laplacian in. 79 φ n + λ n φ n = in, u(x, y, t= a n(t = = a n ( = φ n n = a n (tφ n (x, y, a n (t = n= 8 Thus,wesolvetheODE: { a n + λ n a n = f n a n ( =. on. φ n (x, yu(x, y, t dx. φ n (x, yu t dx = φ n ( u + f dx = φ n udx+ φ n fdx φ n udx+ φ n fdx= λ n φ n udx+ φ n fdx= λ n a n + f n. } {{ } f n φ n (x, yu(x, y, dx =. For n =,λ =,andweobtaina (t =f t. For n, the homogeneous solution is a nh = ce λnt. The anzats for a particular solution is a np = c t + c,whichgivesc = and c = f n /λ n. Using the initial condition, we obtain a n (t = f n λ n e λnt + f n λ n. u(x, t =f φ t + If f = If f ( ( ( n= f n λ n e λnt + f n λ n φ n (x. f(x dx =, lim u(x, t= t n= f(x dx, lim u(x, t f φ t. t c Integrate v + f(x, y= over: fdx = vdx = vdx = f n φ n λ n. v n ds =, where we used divergence theorem and Neumann boundary conditions. Thus, the solution exists only if fdx=. 79 We use dx dy dx. 8 We used Green s formula: ( φn u On, u n =; φ n n u φn n n ds = (φn u φnu dx. = since eigenfunctions are Neumann.

359 Partial Differential Equations Igor Yanovsky, Assume v(x, y = n= a nφ n (x, y. Since we have f(x, y = n= f nφ n (x, y, we obtain λ n a n φ n + f n φ n =, n= n= λ n a n φ n + f n φ n =, a n = f n. λ n v(x, y = n= ( fn λ n φ n (x, y.

360 Partial Differential Equations Igor Yanovsky, Heat Equation with Periodic Boundary Conditions in D (with extra terms Problem (F 99, #5. In two spatial dimensions, consider the differential equation u t = ε u u with periodic boundary conditions on the unit square [, π]. a If ε =find a solution whose amplitude increases as t increases. b Findavalueε, so that the solution of this PDE stays bounded as t,ifε<ε. Proof. a Eigenfunctions of the Laplacian. The periodic boundary conditions imply a Fourier Series solution of the form: u(x, t = m,n a mn (te i(mx+ny. u t = m,n a mn(te i(mx+ny, u = u xx + u yy = m,n(m + n a mn (te i(mx+ny, u = u xxxx +u xxyy + u yyyy = (m 4 +m n + n 4 a mn (te i(mx+ny m,n = m,n(m + n a mn (te i(mx+ny. Plugging this into the PDE, we obtain a mn(t =ε(m + n a mn (t (m + n a mn (t, a mn(t [ε(m + n (m + n ]a mn (t =, a mn (t (m + n [ε (m + n ]a mn (t =. The solution to the ODE above is a mn (t = α mn e (m +n [ε (m +n ]t. u(x, t = α mn e (m +n [ε (m +n ]t e} i(mx+ny {{ }. m,n oscillates When ε =,wehave u(x, t= m,n α mn e (m +n [ (m +n ]t e i(mx+ny. We need a solution whose amplitude increases as t increases. α mn >, with (m + n [ (m + n ] >, (m + n >, >m + n. Hence, α mn > for (m, n=(,, (m, n=(,, (m, n=(,. Else, α mn =. Thus, u(x, t = α + α e t e ix + α e t e iy =+e t e ix + e t e iy = +e t (cos x + i sin x+e t (cos y + i sin y. Thus, we need those

361 Partial Differential Equations Igor Yanovsky, 5 36 b For ε ε =,thesolution stays bounded as t.

362 Partial Differential Equations Igor Yanovsky, 5 36 Problem (F 93, #. Suppose that a and b are constants with a, and consider the equation u t = u xx + u yy au 3 + bu (9.4 in which u(x, y, t is π-periodic in x and y. a Let u be a solution of (9.4 with u(t = = π π u(x, y, t = dxdy / <ɛ. Derive an explicit bound on u(t and show that it stays finite for all t. b If a =, construct the normal modes for (9.4; i.e. find all solutions of the form u(x, y, t=e λt+ikx+ily. c Use these normal modes to construct a solution of (9.4 with a =for the initial data u(x, y, t = = + eix. e ix Proof. a Multiply the equation by u and integrate: u t = u au 3 + bu, uu t = u u au 4 + bu, uu t dx = u udx au 4 dx + bu dx, d u dx = u u dt n ds u dx } {{ } =, u periodic on [,π] d dt u b u, u u(x, e bt, u u(x, e bt εe bt. Thus, u stays finite for all t. au 4 dx } {{ } b Since a =, plugging u = e λt+ikx+ily into the equation, we obtain: Thus, u t = u xx + u yy + bu, λe λt+ikx+ily = ( k l + b e λt+ikx+ily, λ = k l + b. u kl = e ( k l +bt+ikx+ily, u(x, y, t = a kl e ( k l +bt+ikx+ily. k,l + bu dx,

363 Partial Differential Equations Igor Yanovsky, c Using the initial condition, we obtain: u(x, y, = k,l a kl e i(kx+ly = = = + eix e ix ( eix k ( + e ix k k= k= = + Thus, l =, and we have a k e ikx = + k eikx + 8 k= k= k= k eikx + k= k= k= k eikx + k eikx, k e ikx, k= a =; a k = k, k > ; a k = k, k < a =; a k =, k. k u(x, y, t=e bt + + k=,k k e( k +bt+ikx. k eikx. 8 Note a similar question formulation in F 9 #3(b.

364 Partial Differential Equations Igor Yanovsky, Problem (S, #3. Consider the initial-boundary value problem for u = u(x, y, t u t = u u for (x, y [, π],withperiodic boundary conditions and with u(x, y, = u (x, y in which u is periodic. Find an asymptotic expansion for u for t large with terms tending to zero increasingly rapidly as t. Proof. Since we have periodic boundary conditions, assume u(x, y, t = m,n u mn (t e i(mx+ny. Plug this into the equation: ( m n u mn (t e i(mx+ny, u mn(t e i(mx+ny = m,n m,n u mn(t = ( m n u mn (t, u mn (t = a mn e ( m n t, u(x, y, t = a mn e (m +n +t e i(mx+ny. m,n Since u is periodic, u (x, y= u mn e i(mx+ny, u mn = 4π m,n Initial condition gives: u(x, y, = m,n a mn e i(mx+ny = u (x, y, π π a mn e i(mx+ny = u mn e i(mx+ny, m,n m,n a mn = u mn. u(x, y, t= m,n u mn e (m +n +t e i(mx+ny. u (x, y e i(mx+ny dxdy. u mn e (m +n +t e i(mx+ny as t, since e (m +n +t as t.

365 Partial Differential Equations Igor Yanovsky, Problems: Fourier Transform Problem (S, #b. Write the solution of initial value problem ( U t U 5 3 x =, for general initial data ( u ( (x, u ( = (x, ( f(x as an inverse Fourier transform. You may assume that f is smooth and rapidly decreasing as x. Proof. Consider the original system: u ( t u ( x =, u ( t 5u ( x 3u ( x =. Take the Fourier transform in x. The transformed initial value problems are: û ( t iξû ( =, û ( (ξ, = f(ξ, û ( t 5iξû ( 3iξû ( =, û ( (ξ, =. Solving the first ODE for û ( gives: û ( (ξ, t = f(ξe iξt. With this û (, the second initial value problem becomes û ( t 3iξû ( = 5iξ f(ξe iξt, û ( (ξ, =. The homogeneous solution of the above ODE is: û ( h (ξ, t = c e 3iξt. With û ( p = c e iξt as anzats for a particular solution, we obtain: iξc e iξt 3iξc e iξt = 5iξ f(ξe iξt, iξc e iξt = 5iξ f(ξe iξt, c = 5 f(ξ. û ( p (ξ, t = 5 f(ξe iξt. û ( (ξ, t = û ( (ξ, t+û( p (ξ, t = c e 3iξt 5 f(ξe iξt. h We find c using initial conditions: û ( (ξ, = c 5 f(ξ = c = 5 f(ξ. Thus, û ( (ξ, t = 5 f (ξ ( e 3iξt e iξt.

366 Partial Differential Equations Igor Yanovsky, u ( (x, t andu ( (x, t are be obtained by taking inverse Fourier transform: u ( (x, t = ( û ( (ξ, t = e π R ixξ f(ξ e iξt dξ, n u ( (x, t = ( û ( (ξ, t = π R n e ixξ 5 f(ξ ( e 3iξt e iξt dξ.

367 Partial Differential Equations Igor Yanovsky, Problem (S, #4. Use the Fourier transform on L (R to show that du + cu(x+u(x = f (3. dx has a unique solution u L (R for each f L (R when c > - you may assume that c is a real number. Proof. u L (R. Define its Fourier transform û by û(ξ = e ixξ u(x dx for ξ R. π du (ξ = iξû(ξ. dx We can find u(x (ξ in two ways. Let u(x } {{ } = v(x, and determinte v(ξ: y R u(x (ξ = v(ξ = = π π R R e ixξ v(x dx = π e iyξ e iξ u(y dy = e iξ û(ξ. R e i(y+ξ u(y dy We can also write the definition for û(ξ and substitute x later in calculations: û(ξ = e iyξ u(y dy = e i(x ξ u(x dx π R π R = e ixξ e iξ u(x dx = e iξ u(x (ξ, π R u(x (ξ =e iξû(ξ. Substituting into (3., we obtain iξû(ξ+cû(ξ+e iξ û(ξ = f(ξ, f(ξ û(ξ = iξ + c + e iξ. ( f(ξ u(x = iξ + c + e iξ = ( f B = f B, π where B = ( B = iξ + c + e iξ, iξ + c + e iξ = π R e ixξ dξ. iξ + c + e iξ For c >, û(ξ exists for all ξ R, sothatu(x =(û(ξ and this is unique by the Fourier Inversion Theorem. Note that in R n, becomes u(x (ξ = v(ξ = = (π n (π n e ix ξ v(x dx = e i(y+ ξ u(y dy R n (π n R n e iy ξ e i ξ u(y dy = e i ξû(ξ =e ( i R n j ξ j û(ξ.

368 Partial Differential Equations Igor Yanovsky, Problem (F 96, #3. Find the fundamental solution for the equation u t = u xx xu x. (3. Hint: The Fourier transform converts this problem into a PDE which can be solved using the method of characteristics. Proof. u L (R. Define its Fourier transform û by û(ξ = e ixξ u(x dx for ξ R. π û x (ξ = iξ û(ξ, R û xx (ξ = (iξ û(ξ = ξ û(ξ. We find xu x (ξ in two steps: ➀ Multiplication by x: ixu(ξ = e ixξ( ixu(x dx = π R xu(x(ξ = i d dξ û(ξ. d dξ û(ξ. ➁ Using the previous result, we find: xu x (x(ξ = e ixξ( xu x (x dx = [e ixξ xu π π R ] } {{ } = iξ e ixξ xudx e ixξ udx π R π R = = iξ xu(x(ξ û(ξ = iξ [i d ] dξ û(ξ π û(ξ = ξ d û(ξ û(ξ. dξ R ( ( iξe ixξ x + e ixξ udx xu x (x(ξ = ξ d û(ξ û(ξ. dξ Plugging these into (3., we get: tû(ξ, t = ξ û(ξ, t û t = ξ û + ξû ξ + û, û t ξû ξ = (ξ û. ( ξ d dξ û(ξ, t û(ξ, t, We now solve the above equation by characteristics. We change the notation: û u, t y, ξ x. We have u y xu x = (x u. dx dt dy dt dz dt = x x = c e t, (c = xe t = y = t + c, = (x z = (c e t z dz z = (c e t dt log z = c e t + t + c 3 = x + t + c 3 = x + y c + c 3 z = ce x +y.

369 Partial Differential Equations Igor Yanovsky, Changing the notation back, we have û(ξ, t = ce ξ +t. Thus, we have û(ξ, t = ce ξ +t. We use Inverse Fourier Tranform to get u(x, t: 8 u(x, t = e ixξ û(ξ, t dξ = π π = = = R c e t π c e t π R R c π e t e x u(x, t = ce t e x. R e ixξ e ξ dξ = c π e t e ixξ+ξ dξ = c R π e t e ixξ ce ξ +t dξ e ixξ+ ξ R R dξ e (ξ+ix dξ e x e y dy = c π e t e x π = ce t e x. Check: Thus, u t = ce t e x, u x = ce t xe x, u xx = ce t ( e x + x e x. u t = u xx xu x, ce t e x = ce t ( e x + x e x xce t xe x. 8 We complete the square for powers of exponentials.

370 Partial Differential Equations Igor Yanovsky, 5 37 Problem (W, #4. a Solve the initial value problem u n t + a k (t u + a (tu =, x R n, x k k= u(,x=f(x where a k (t, k =,...,n,anda (t are continuous functions, and f is a continuous function. You may assume f has compact support. b Solve the initial value problem u n t + a k (t u + a (tu = f(x, t, x R n, x k k= u(,x= where f is continuous in x and t. Proof. a Use the Fourier transform to solve this problem. û(ξ, t = e ix ξ u(x, t dx for ξ R. (π n R n u = iξ k û. x k Thus, the equation becomes: { û t + i n k= a k(tξ k û + a (tû =, û(ξ, = f(ξ, or û t + i a(t ξ û + a (tû =, û t = ( i a(t ξ + a (t û. This is an ODE in û with solution: û(ξ, t = ce t (i a(s ξ+a (s ds, û(ξ, = c = f(ξ. Thus, û(ξ, t = f(ξ e t (i a(s ξ+a (s ds. Use the Inverse Fourier transform to get u(x, t: [ u(x, t = û(ξ, t = f(ξ e ] t (i a(s ξ+a (s ds (f g(x =, (π n where ĝ(ξ = e t (i a(s ξ+a (s ds. g(x = e ix ξ ĝ(ξ dξ = e ix ξ [ e t (π n R n (π n (i a(s ξ+a (s ds ] dξ. R n (f g(x u(x, t = = (π n (π n e i(x y ξ [ e t (i a(s ξ+a (s ds ] dξ f(y dy. R n R n b Use Duhamel s Principle and the result from (a. u(x, t = t U(x, t s, s ds, U n t + a k (t U + a (tu =, x k k= U(x,,s=f(x, s. where U(x, t, s solves

371 Partial Differential Equations Igor Yanovsky, 5 37 u(x, t = t U(x, t s, s ds = (π n t R n R n e i(x y ξ [ e t s (i a(s ξ+a (s ds ] dξ f(y, s dy ds.

372 Partial Differential Equations Igor Yanovsky, 5 37 Problem (S 93, #. a Define the Fourier transform 83 f(ξ = e ixξ f(x dx. State the inversion theorem. If π, ξ <a, f(ξ = π, ξ = a,, ξ >a, where a is a real constant, what f(x does the inversion theorem give? b Show that f(x b =e iξb f(x, where b is a real constant. Hence, using part (a and Parseval s theorem, show that π sin a(x + z x + z sin a(x + ξ x + ξ where z and ξ are real constants. dx = sin a(z ξ, z ξ Proof. a The inverse Fourier transform for f L (R n : f (ξ = π e ixξ f(x dx for ξ R. Fourier Inversion Theorem: Assume f L (R. Then f(x = e ixξ f(ξ dξ = e i(y xξ f(y dy dξ =( f (x. π π Parseval s theorem (Plancherel s theorem (for this definition of the Fourier transform. Assume f L (R n L (R n. Then f,f L (R n and π f L (R n = f L (R n = f L (R n, or f(x dx = π f(ξ dξ. Also, f(x g(x dx = π We can write { π, ξ <a, f(ξ =, ξ >a. f(ξ ĝ(ξ dξ. 83 Note that the Fourier transform is defined incorrectly here. There should be - sign in e ixξ. Need to be careful, since the consequences of this definition propagate throughout the solution.

373 Partial Differential Equations Igor Yanovsky, a f(x = ( f(ξ = e ixξ f(ξ dξ = dξ + π π π = a e ixξ dξ = [e ixξ] ξ=a a ix ξ= a ix b Let f(x } {{ } b= g(x, and determinte ĝ(ξ: y f(x b(ξ =ĝ(ξ = With f(x = π sin a(x + z x + z sin ax x = R R e ixξ g(x dx = R a a [ e iax e iax] = e i(y+bξ f(y dy e iyξ e ibξ f(y dy = e ibξ f(ξ. (from (a, we have sin a(x + s x + s dx = π = π = π π = π = π = π f(x + zf(x + s dx f(x + z sf(x dx a a a a a f(x + z s f(x dξ f(ξ e i(z sξ f(ξ dξ f(ξ e i(z sξ dξ π e i(z sξ dξ = e i(z sξ dξ a [ = e i(z sξ ] ξ=a i(z s ξ= a = ei(z sa e i(z sa i(z s = e ixξ πdξ+ π sin ax x. a dξ (x = x + s, dx = dx (Parseval s part (b sin a(z s. z s

374 Partial Differential Equations Igor Yanovsky, Problem (F 3, #5. ❶ State Parseval s relation for Fourier transforms. ❷ Find the Fourier transform ˆf(ξ of { e iαx / πy, x y f(x =, x >y, in which y and α are constants. ❸ Use this in Parseval s relation to show that sin (α ξy (α ξ dξ = πy. What does the transform ˆf(ξ become in the limit y? ❹ Use Parseval s relation to show that sin(α βy = sin(α ξy (α β π (α ξ sin(β ξy (β ξ dξ. Proof. f L (R. Define its Fourier transform û by f(ξ = e ixξ f(x dx for ξ R. π R ❶ Parseval s theorem (Plancherel s theorem: Assume f L (R n L (R n. Then f,f L (R n and f L (R n = f L (R n = f L (R n, or f(x dx = f(ξ dξ. Also, f(x g(x dx = f(ξ ĝ(ξ dξ. ❷ Find the Fourier transform of f: f(ξ = e ixξ f(x dx = y π π = π y i(α ξ = sin y(α ξ π y(α ξ. ❸ Parseval s theorem gives: R f(ξ dξ = sin y(α ξ π y(α ξ dξ = [e i(α ξx] x=y sin y(α ξ (α ξ dξ = π y y y e ixξ y e iαx πy dx = x= y = iπ y(α ξ f(x dx, e iαx 4πy dx, y dx, sin y(α ξ (α ξ dξ = πy. y π e i(α ξx dx y y [ e i(α ξy e i(α ξy]

375 Partial Differential Equations Igor Yanovsky, ❹ We had f(ξ = sin y(α ξ π y(α ξ. We make change of variables: α ξ = β ξ. Then, ξ = ξ + α β. We have f(ξ = f(ξ + α β = sin y(β ξ (β ξ, or f(ξ + α β = sin y(β ξ. (β ξ We will also use the following result. Let f(ξ + a =ĝ(ξ, } {{ } ξ f(ξ + a = ĝ(ξ = e ixξ ĝ(ξ dξ = π π e ix(ξ a f (ξ dξ Using these results, we have π sin(α ξy (α ξ R = e ixa f(x. sin(β ξy (β ξ dξ = π (π y f(ξ f(ξ + α β dξ = πy f(x e (α βix f(x dx = πy = πy y y y R f(x e (α βix dx e iαx 4πy e (α βix dx = e (α βix dx y = [ e (α βix ] x=y i(α β x= y = [ e (α βiy e (α βiy] i(α β = sin(α βy. α β

376 Partial Differential Equations Igor Yanovsky, Problem (S 95, #5. For the Laplace equation ( f x + y f = (3.3 in the upper half plane y, consider the Dirichlet problem f(x, = g(x; the Neumann problem yf(x, = h(x. Assume that f, g and h are π periodic in x and that f is bounded at infinity. Find the Fourier transform N of the Dirichlet-Neumann map. In other words, find an operator N taking the Fourier transform of g to the Fourier transform of h; i.e. Nĝ k = ĥk. Proof. We solve the problem by two methods. ❶ Fourier Series. Since f is π-periodic in x, wecanwrite f(x, y = a n (y e inx. n= Plugging this into (3.3, we get the ODE: ( n a n (ye inx + a n(ye inx =, n= a n (y n a n (y =. Initial conditions give: (g and h are π-periodic in x f(x, = a n (e inx = g(x = ĝ n e inx a n ( = ĝ n. f y (x, = n= n= Thus, the problems are: a n(e inx = h(x = a n (y n a n (y =, a n ( = ĝ n, (Dirichlet a n( = ĥn. (Neumann n= n= ĥ n e inx a n( = ĥn. a n (y = b n e ny + c n e ny, n =,,...; a (y=b y + c. a n (y = nb ne ny nc n e ny, n =,,...; a (y =b. Since f is bounded at y = ±, wehave: b n = for n>, c n = for n<, b =, c arbitrary.

377 Partial Differential Equations Igor Yanovsky, n>: n<: a n (y = c n e ny, a n ( = c n = ĝ n, (Dirichlet a n( = nc n = ĥn. (Neumann nĝ n = ĥn. a n (y = b n e ny, a n ( = b n = ĝ n, (Dirichlet a n( = nb n = ĥn. (Neumann nĝ n = ĥn. n ĝ n = ĥn, n. n =: a (y = c, a ( = c = ĝ, (Dirichlet a ( = = ĥ. (Neumann Note that solution f(x, y may be written as f(x, y = n= = c + = ❷ Fourier Transform. f(ξ, y = f(x, y = a n (y e inx n= = a (y+ b n e ny e inx + n= c n e ny e inx n= a n (y e inx + a n (y e inx n= { ĝ + n= ĝne ny e inx + n= ĝne ny e inx, (Dirichlet c + ĥ nn n= e ny e inx + n= ĥn n e ny e inx. (Neumann π π The Fourier transform of f(x, y inx is: e ixξ f(x, y dx, e ixξ f (ξ, y dξ. (iξ f(ξ, y+ fy y(ξ, y =, f yy ξ f =. The solution to this ODE is: f(ξ, y = c e ξy + c e ξy. For ξ>, c =; forξ<, c =. ξ>: f(ξ, y = c e ξy, f y (ξ, y = ξc e ξy, c = f(ξ, = e ixξ f(x, dx = π π ξc = f y (ξ, = π ξĝ(ξ = ĥ(ξ. e ixξ f y (x, dx = π e ixξ g(x dx = ĝ(ξ, e ixξ h(x dx = ĥ(ξ. (Dirichlet (Neumann

378 Partial Differential Equations Igor Yanovsky, ξ<: f(ξ, y = c e ξy, fy (ξ, y = ξc e ξy, c = f(ξ, = e ixξ f(x, dx = π π ξc = f y (ξ, = π ξĝ(ξ = ĥ(ξ. e ixξ f y (x, dx = π e ixξ g(x dx = ĝ(ξ, e ixξ h(x dx = ĥ(ξ. (Dirichlet (Neumann ξ ĝ(ξ = ĥ(ξ.

379 Partial Differential Equations Igor Yanovsky, Problem (F 97, #3. Consider the Dirichlet problem in the half-space x n >, n : u + a u + k u =, x n x n > u(x, = f(x, x =(x,...,x n. Here a and k are constants. Use the Fourier transform to show that for any f(x L (R n there exists a solution u(x,x n of the Dirichlet problem such that u(x,x n dx C R n for all <x n < +. Proof. 84 Denote ξ =(ξ,ξ n. Transform in the first n variables: ξ û(ξ,x n + û x (ξ,x n + a û (ξ,x n + k û(ξ,x n =. n x n Thus, the ODE and initial conditions of the transformed problem become: { û xnxn + aû xn +(k ξ û =, û(ξ, = f(ξ. With the anzats û = ce sxn, we obtain s + as +(k ξ =, and s, = a ± a 4(k ξ. Choosing only the negative root, we obtain the solution: 85 û(ξ,x n = c(ξ e a a 4(k ξ x n. û(ξ, = c = f(ξ. Thus, û(ξ,x n = f(ξ e a a 4(k ξ x n. Parseval s theorem gives: u L (R n = û L (R n = û(ξ,x n dξ R n = f(ξ e a a 4(k ξ x n dξ R n f(ξ dξ R n = f L (R n = f L (R n C, since f(x L (R n. Thus, u(x,x n L (R n. 84 Note that the last element of x =(x,x n=(x,...,x n,x n, i.e. x n, plays a role of time t. As such, the PDE may be written as u + u tt + au t + k u =. 85 Note that a> should have been provided by the statement of the problem.

380 Partial Differential Equations Igor Yanovsky, 5 38 Problem (F 89, #7. Find the following fundamental solutions a G(x, y, t = a(t G(x, y, t G(x, y, t t x + b(t x G(x, y, = δ(x y, + c(tg(x, y, t for t> where a(t, b(t, c(t are continuous functions on [, + ], a(t > for t>. G n b t (x,...,x n,y,...,y n,t= a k (t G for t>, x k k= G(x,...,x n,y,...,y n, = δ(x y δ(x y...δ(x n y n. Proof. a We use the Fourier transform to solve this problem. Transform the equation in the first variable only. That is, Ĝ(ξ, y, t = e ixξ G(x, y, t dx. π R The equation is transformed to an ODE, that can be solved: Ĝ t (ξ, y, t = a(t ξ Ĝ(ξ, y, t +ib(t ξ Ĝ(ξ, y, t +c(t Ĝ(ξ, y, t, Ĝ t (ξ, y, t = [ a(t ξ + ib(t ξ + c(t ] Ĝ(ξ, y, t, Ĝ(ξ, y, t = ce t [ a(sξ +ib(sξ+c(s] ds. We can also transform the initial condition: Ĝ(ξ, y, = δ(x y(ξ = e iyξ δ(ξ = e iyξ. π Thus, the solution of the transformed problem is: Ĝ(ξ, y, t = π e iyξ e t [ a(sξ +ib(sξ+c(s] ds. The inverse Fourier transform gives the solution to the original problem: G(x, y, t = ( Ĝ(ξ, y, t = e ixξ Ĝ(ξ, y, t dξ π R [ = e ixξ e iyξ e ] t [ a(sξ +ib(sξ+c(s] ds dξ π R π = e i(x yξ t e [ a(sξ +ib(sξ+c(s] ds dξ. π R b Denote x =(x,...,x n, y =(y,...,y n. Transform in x: Ĝ( ξ, y, t = e i x ξ G( x, y, t d x. (π n R n The equation is transformed to an ODE, that can be solved: Ĝ t ( ξ, y, n t = a k (t iξ k Ĝ( ξ, y, t, k= Ĝ( ξ, y, t = ce i t [ n k= a k(s ξ k ] ds.

381 Partial Differential Equations Igor Yanovsky, 5 38 We can also transform the initial condition: Ĝ( ξ, y, = [ δ(x y δ(x y...δ(x n y n ] (ξ = e i y ξ δ( ξ = Thus, the solution of the transformed problem is: Ĝ( ξ, y, t = e i y ξ e i t (π n [ n k= a k(s ξ k ] ds. The inverse Fourier transform gives the solution to the original problem: G( x, y, t = ( Ĝ( ξ, y, t = e i x ξ Ĝ( ξ, y, t dξ (π n R n [ = e i x ξ e i y ξ e i ] t (π n R n (π n [ n k= a k(s ξ k ] ds dξ = (π n e i( x y ξ e i t [ n k= a k(s ξ k ] ds dξ. R n (π n e i y ξ.

382 Partial Differential Equations Igor Yanovsky, 5 38 Problem (W, #7. Consider the equation ( + + x u = f in R n, (3.4 n x where f is an integrable function (i.e. f L (R n, satisfying f(x =for x R. Solve (3.4 by Fourier transform, and prove the following results. a There is a solution of (3.4 belonging to L (R n if n>4. b If R n f(x dx =, there is a solution of (3.4 belonging to L (R n if n>. Proof. u = f, ξ û(ξ = f(ξ, û(ξ = ξ f(ξ, ξ R n, u(x = ( f(ξ. ξ a Then û L (R n = ( Rn f(ξ ξ 4 dξ ( f(ξ ξ < ξ 4 dξ + } {{ } A Notice, f = f B, so B<. Use polar coordinates on A. A = f(ξ ξ < ξ 4 dξ = Sn f r 4 rn ds n dr = If n>4, A f ds n = f <. S n u L (R n = û L (R n = (A + B <. f(ξ ξ ξ 4 dξ } {{ } B. S n f r n 5 ds n dr.

383 Partial Differential Equations Igor Yanovsky, b We have ( f(ξ u(x, t = = ix ξ f(ξ ξ e (π n ξ dξ = e ix ξ ( (π n ξ e iy ξ f(y dy dξ (π n = ( e i(x y ξ (π n f(y ξ dξ dy = ( e (π n f(y Sn i(x yr r = ( (π n f(y e i(x yr r n 3 ds n dr S } n {{ } M<, if n>. u(x, t = Mf(y dy <. (π n r n ds n dr dy dy.

384 Partial Differential Equations Igor Yanovsky, Problem (F, #7. For the right choice of the constant c, the function F (x, y =c(x + iy is a fundamental solution for the equation u x + i u y = f in R. Find the right choice of c, and use your answer to compute the Fourier transform (in distribution sense of (x + iy. Proof. 86 = ( x + i ( y x i. y F (x, y= π log z is the fundamental solution of the Laplacian. z = x + iy. F (x, y = δ, ( x + i ( y x i F (x, y = δ. y h x + ih y = e i(xξ +yξ. Suppose h = h(xξ + yξ or h = ce i(xξ +yξ. c ( iξ e i(xξ +yξ i ξ e i(xξ +yξ = ic(ξ iξ e i(xξ +yξ e i(xξ +yξ, ic(ξ iξ =, c = i(ξ iξ, h(x, y = i(ξ iξ e i(xξ +yξ. Integrate by parts: ( (ξ = e i(xξ +yξ ( x + iy R i(ξ iξ x + i y (x + iy dxdy = i(ξ iξ = i(ξ + iξ. 86 Alan solved in this problem in class.

385 Partial Differential Equations Igor Yanovsky, Laplace Transform If u L (R +, we define its Laplace transform to be L[u(t] = u # (s = e st u(t dt (s >. In practice, for a PDE involving time, it may be useful to perform a Laplace transform in t, holding the space variables x fixed. The inversion formula for the Laplace transform is: u(t = L [u # (s] = πi c+i c i e st u # (s ds. Example: f(t =. L[] = Example: f(t =e at. L[e at ] = e st dt = e st e at dt = [ s e st] t= t= e (a st dt = = s for s>. [e (a st] t= a s t= Convolution: We want to find an inverse Laplace transform of s L [ s }{{} L[f] ] } s {{ + } L[g] = f g = t sin t dt = cos t. = s a s +. for s>a. Partial Derivatives: u = u(x, t L[u t ] = L[u tt ] = e st u t dt = e st u tt dt = [ ] t= e st u(x, t [e st u t ] t= = s L[u] su(x, u t (x,, L[u x ] = e st u x dt = x L[u], t= L[u xx ] = e st u xx dt = x L[u]. Heat Equation: Consider { u t u = in U (, u = f on U {t =}, t= + s + s and perform a Laplace transform with respect to time: L[u t ] = L[ u] = e st u t dt = sl[u] u(x, = sl[u] f(x, e st udt = L[u]. e st udt = sl[u] u(x,, e st u t dt = u t (x, + sl[u t ]

386 Partial Differential Equations Igor Yanovsky, Thus, the transformed problem is: have sl[u] f(x = L[u]. Writing v(x = L[u], we v + sv = f in U. Thus, the solution of this equation with RHS f is the Laplace transform of the solution of the heat equation with initial data f.

387 Partial Differential Equations Igor Yanovsky, Table of Laplace Transforms: L[f] = f # (s a L[sin at] = s + a, s > s L[cos at] = s + a, s > a L[sinh at] = s a, s > a s L[cosh at] = s a, s > a L[e at sin bt] = b (s a + b, s > a L[e at cos bt] = s a (s a + b, s > a L[t n ] = n! s n+, s > L[t n e at ] = n! (s a n+, s > a L[H(t a] = e as s, s > L[H(t a f(t a] = e as L[f], L[af(t+bg(t] = al[f]+bl[g], L[f(t g(t] = L[f] L[g], [ t L g(t t f(t dt ] = L[f] L[g], [ df ] L = sl[f] f(, dt [ d f ] ( L dt = s L[f] sf( f (, f = df dt [ d n f ] L dt n = s n L[f] s n f(... f n (, L[f(at] = a f #( s, a L[e bt f(t] = f # (s b, L[tf(t] = d ds L[f], [ f(t ] L = f # (s ds, t s [ t L f(t dt ] = s L[f], L[J (at] = (s + a, L[δ(t a] = e sa. Example: f(t =sint. After integrating by parts twice, we obtain: L[sint] = e st sin tdt = s e st sin tdt, e st sin tdt = +s.

388 Partial Differential Equations Igor Yanovsky, Example: f(t =t n. [ t L[t n ] = e st t n n e st ] dt = s + n e st t n dt = n s s L[tn ] = n ( n L[t n ] =... = n! n! L[] = s s sn s n+.

389 Partial Differential Equations Igor Yanovsky, Problem (F, #6. Consider the initial-boundary value problem u t u xx + au =, t >, x > u(x, =, x > u(,t = g(t, t >, where g(t is continuous function with a compact support, and a is constant. Find the explicit solution of this problem. Proof. We solve this problem using the Laplace transform. L[u(x, t] = u # (x, s = e st u(x, t dt (s >. [ ] t= L[u t ] = e st u t dt = e st u(x, t + s e st udt t= = su # (x, s u(x, = su # (x, s, (since u(x, = L[u xx ] = L[u(,t] = u # (,s = e st u xx dt = x u# (x, s, e st g(t dt = g # (s. Plugging these into the equation, we obtain the ODE in u # : su # (x, s x u# (x, s+au # (x, s =. { (u # xx (s + au # =, u # (,s = g # (s. This initial value problem has a solution: u # (x, s = c e s+ax + c e s+ax. Since we want u to be bounded as x,wehavec =,so u # (x, s = c e s+ax. u # (,s = c = g # (s, thus, u # (x, s = g # (se s+ax. To obtain u(x, t, we take the inverse Laplace transform of u # (x, s: u(x, t = L [u # (x, s] = L [ g # (s e ] s+ax } {{ } } {{ } = g f L[g] L[f] u(x, t = = g L [ e s+ax ] [ = g πi t [ g(t t πi c+i c i c+i c i ] e st e s+ax ds dt. ] e st e s+ax ds,

390 Partial Differential Equations Igor Yanovsky, 5 39 Problem (F 4, #8. The function y(x, t satisfies the partial differential equation x y x + y +y =, x t and the boundary conditions y(x, =, y(,t=e at, where a. FindtheLaplace transform, y(x, s, of the solution, and hence derive an expression for y(x, t in the domain x, t. Proof. We change the notation: y u. { xu x + u xt +u =, u(x, =, u(,t=e at. The Laplace transform is defined as: L[u(x, t] = u # (x, s = L[xu x ] = L[u xt ] = e st xu x dt = x e st u xt dt = We have e st u(x, t dt (s >. [ e st u x (x, t e st u x dt = x(u # x, ] t= t= + s e st u x dt = s(u # x u x (x, = s(u # x, (since u(x, = L[u(,t] = u # (,s = e st e at dt = e (s+at dt = = s + a. Plugging these into the equation, we obtain the ODE in u # : { (x + s(u # x +u # =, u # (,s = s+a, which can be solved: (u # x u # = x + s From the initial conditions: u # (,s= c s = s + a log u # = log(x + s+c u # = c e log(x+s = c = s s + a. [ s + a e (s+at] t= t= c (x + s. u # s (x, s = (s + a(x + s. To obtain u(x, t, we take the inverse Laplace transform of u # (x, s: u(x, t = L [u # (x, s] = L [ s ] (s + a(x + s = c+i [ e st s ] πi (s + a(x + s ds. u(x, t = πi c+i c i e st [ s (s + a(x + s ] ds. c i

391 Partial Differential Equations Igor Yanovsky, 5 39 Problem (F 9, #. Using the Laplace transform, or any other convenient method, solve the Volterra integral equation u(x =sinx + x Proof. Rewrite the equation: u(t = sint + t sin(x yu(y dy. sin(t t u(t dt, u(t = sint +(sint u. Taking the Laplace transform of each of the elements in : L[u(t] = u # (s = e st u(t dt, L[sin t] = +s, L[(sint u] = L[sint] L[u] = u# +s. Plugging these into the equation: u # = u # (s = s. +s + u# +s = u# + +s. To obtain u(t, we take the inverse Laplace transform of u # (s: u(t = L [u # (s] = L [ ] s = t. u(t = t.

392 Partial Differential Equations Igor Yanovsky, 5 39 Problem (F 9, #5. In what follows, the Laplace transform of x(t is denoted either by x(s or by Lx(t. ❶ Show that, for integral n, L(t n = n! s n+. ❷ Hence show that where LJ ( ut = s e u/s, J (z = n= ( n ( zn n!n! is a Bessel function. ❸ Hence show that [ L J ( ] utx(u du = ( s x. (3. s ❹ Assuming that LJ (at = a + s, prove with the help of (3. that if t J (auj ( ut du = ( t a J. a Hint: For the last part, use the uniqueness of the Laplace transform. Proof. ❶ L[t n ] = ❸ where = n s }{{} e st }{{} t n g f ( n ❷ LJ ( [ ut = L s = s [ t n e st dt = s ] } {{ } = + n s L[t n ] =... = n! n! L[] = sn s n+. n= n= [ L J ( ] ut x(u du x # (s = e us x(u du. ( n u n t n ] n!n! = e st t n dt = n s L[tn ] ( n u n L[t n ] = n!n! n= ( n ( u n = n! s s e u s. = L[J ( ut] x(u du = s = s x#(, s n= ( n u n n!s n+ e u s x(u du

393 Partial Differential Equations Igor Yanovsky, Linear Functional Analysis 3. Norms is a norm on a vector space X if i x =iffx =. ii αx = α x for all scalars α. iii x + y x + y (the triangle inequality. The norm induces the distance function d(x, y= x y so that X is a metric space, called a normed vector space. 3. Banach and Hilbert Spaces A Banach space is a normed vector space that is complete in that norm s metric. I.e. a complete normed linear space is a Banach space. A Hilbert space is an inner product space for which the corresponding normed space is complete. I.e. a complete inner product space is a Hilbert space. Examples: Let K be a compact set of R n and let C(K denote the space of continuous functions on K. Since every u C(K achieves maximum and minimum values on K, we may define u =max x K u(x. is indeed a norm on C(K and since a uniform limit of continuous functions is continuous, C(K is a Banach space. However, this norm cannot be derived from an inner product, so C(K is not a Hilbert space. C(K isnot a Banach space with norm. (Bell-shaped functions on [, ] may converge to a discontinuous δ-function. In general, the space of continuous functions on [, ], with the norm p, p<, isnot a Banach space, since it is not complete. 3 R n and C n are real and complex Banach spaces (with a Eucledian norm. 4 L p are Banach spaces (with p norm. 5 The space of bounded real-valued functions on a set S, withthesupnorm S are Banach spaces. 6 The space of bounded continuous real-valued functions on a metric space X is a Banach space. 3.3 Cauchy-Schwarz Inequality (u, v u v in any norm, for example a(u, v a(u, u a(v, v v dx = v dx = ( v dx ( dx uv dx ( u dx ( v dx 3.4 Hölder Inequality uv dx u p v q, which holds for u L p ( and v L q (, where p + q uv L (. =. In particular, this shows

394 Partial Differential Equations Igor Yanovsky, Minkowski Inequality u + v p u p + v p, which holds for u, v L p (. In particular, it shows u + v L p (. Using the Minkowski Inequality, we find that p is a norm on L p (. The Riesz-Fischer theorem asserts that L p ( is complete in this norm, so L p ( is a Banach space under the norm p. If p =,thenl ( is a Hilbert space with inner product (u, v= uv dx. Example: R n bounded domain, C ( denotes the functions that, along with their first-order derivatives, extend continuously to the compact set. Then C ( is a Banach space under the norm u, =max( u(x + u(x. x Note that C ( is not a Banach space since u, need not be finite for u C (. 3.6 Sobolev Spaces A Sobolev space is a space of functions whose distributional derivatives (up to some fixed order exist in an L p -space. Let be a domain in R n, and let us introduce <u,v> = ( u v + uv dx, (3. u, = ( <u,u> = ( u + u dx (3. when these expressions are defined and finite. For example, (3. and (3. are defined for functions in C (. However, C ( is not complete under the norm (3., and so does not form a Hilbert space. Divergence Theorem A nds= div Adx Trace Theorem u L ( C u H ( smooth or square Poincare Inequality u p C u p u(x dx C p u(x dx u C(, H, ( i.e. p = u u p u p u H,p (

395 Partial Differential Equations Igor Yanovsky, u = u(x dx Notes u n = u n = n u dx = ab a + b u u = ( u (u xy dx = (Average value of u over, is the volume of u x + n u u udx ab a + b u x u = u x + u x u u u + u u xx u yy dx u H ( square Problem (F 4, #6. Let q C (R3. Prove that the vector field u(x = q(y(x y 4π R 3 x y 3 dy enjoys the following properties: 87 a u(x is conservative; b div u(x =q(x for all x R 3 ; c u(x = O( x for large x. Furthermore, prove that the proverties (, (, and (3 above determine the vector field u(x uniquely. Proof. a To show that u(x is conservative, we need to show that curl u =. The curl of V is another vector field defined by curl V = V e e e 3 ( = det 3 V3 = V V, V 3 V, V. x V V V x 3 x 3 x x x 3 Consider V (x = x x 3 = (x,x,x 3 (x + x + x 3. 3 Then, u(x = q(y V (x y dy, 4π R 3 curl u(x = q(y curl x V (x y dy. 4π R 3 curl V (x = curl (x,x,x 3 (x + x + x 3 3 ( 3 = x x 3 (x + x + x 3 5 = (,,. 3 x 3x (x + x + x 3, 5 3 x 3x (x + x + x x x 3 (x + x + x 3, 5 3 x x (x + x + x McOwen, p

396 Partial Differential Equations Igor Yanovsky, Thus, curl u = 4π R q(y dy =, and u(x is conservative. 3 b Note that the Laplace kernel in R 3 is 4πr. u(x = q(y(x y 4π R 3 x y 3 dy = q(r r q(r 4π R 3 r 3 rdr = dr = q. R 3 4πr c Consider F (x = q(y 4π x y dy. R 3 F (x is O( x as x. Note that u = F, which is clearly O( x as x.