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