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

Size: px
Start display at page:

Download "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

College of the Holy Cross, Spring 2009 Math 373, Partial Differential Equations Midterm 1 Practice Questions

College of the Holy Cross, Spring 2009 Math 373, Partial Differential Equations Midterm 1 Practice Questions College of the Holy Cross, Spring 29 Math 373, Partial Differential Equations Midterm 1 Practice Questions 1. (a) Find a solution of u x + u y + u = xy. Hint: Try a polynomial of degree 2. Solution. Use

More information

MATH 425, PRACTICE FINAL EXAM SOLUTIONS.

MATH 425, PRACTICE FINAL EXAM SOLUTIONS. MATH 45, PRACTICE FINAL EXAM SOLUTIONS. Exercise. a Is the operator L defined on smooth functions of x, y by L u := u xx + cosu linear? b Does the answer change if we replace the operator L by the operator

More information

This makes sense. t 2 1 + 1/t 2 dt = 1. t t 2 + 1dt = 2 du = 1 3 u3/2 u=5

This makes sense. t 2 1 + 1/t 2 dt = 1. t t 2 + 1dt = 2 du = 1 3 u3/2 u=5 1. (Line integrals Using parametrization. Two types and the flux integral) Formulas: ds = x (t) dt, d x = x (t)dt and d x = T ds since T = x (t)/ x (t). Another one is Nds = T ds ẑ = (dx, dy) ẑ = (dy,

More information

tegrals as General & Particular Solutions

tegrals as General & Particular Solutions tegrals as General & Particular Solutions dy dx = f(x) General Solution: y(x) = f(x) dx + C Particular Solution: dy dx = f(x), y(x 0) = y 0 Examples: 1) dy dx = (x 2)2 ;y(2) = 1; 2) dy ;y(0) = 0; 3) dx

More information

The Heat Equation. Lectures INF2320 p. 1/88

The Heat Equation. Lectures INF2320 p. 1/88 The Heat Equation Lectures INF232 p. 1/88 Lectures INF232 p. 2/88 The Heat Equation We study the heat equation: u t = u xx for x (,1), t >, (1) u(,t) = u(1,t) = for t >, (2) u(x,) = f(x) for x (,1), (3)

More information

Reference: Introduction to Partial Differential Equations by G. Folland, 1995, Chap. 3.

Reference: Introduction to Partial Differential Equations by G. Folland, 1995, Chap. 3. 5 Potential Theory Reference: Introduction to Partial Differential Equations by G. Folland, 995, Chap. 3. 5. Problems of Interest. In what follows, we consider Ω an open, bounded subset of R n with C 2

More information

1. First-order Ordinary Differential Equations

1. First-order Ordinary Differential Equations Advanced Engineering Mathematics 1. First-order ODEs 1 1. First-order Ordinary Differential Equations 1.1 Basic concept and ideas 1.2 Geometrical meaning of direction fields 1.3 Separable differential

More information

RAJALAKSHMI ENGINEERING COLLEGE MA 2161 UNIT I - ORDINARY DIFFERENTIAL EQUATIONS PART A

RAJALAKSHMI ENGINEERING COLLEGE MA 2161 UNIT I - ORDINARY DIFFERENTIAL EQUATIONS PART A RAJALAKSHMI ENGINEERING COLLEGE MA 26 UNIT I - ORDINARY DIFFERENTIAL EQUATIONS. Solve (D 2 + D 2)y = 0. 2. Solve (D 2 + 6D + 9)y = 0. PART A 3. Solve (D 4 + 4)x = 0 where D = d dt 4. Find Particular Integral:

More information

An Introduction to Partial Differential Equations

An Introduction to Partial Differential Equations An Introduction to Partial Differential Equations Andrew J. Bernoff LECTURE 2 Cooling of a Hot Bar: The Diffusion Equation 2.1. Outline of Lecture An Introduction to Heat Flow Derivation of the Diffusion

More information

Solutions for Review Problems

Solutions for Review Problems olutions for Review Problems 1. Let be the triangle with vertices A (,, ), B (4,, 1) and C (,, 1). (a) Find the cosine of the angle BAC at vertex A. (b) Find the area of the triangle ABC. (c) Find a vector

More information

Math 241, Exam 1 Information.

Math 241, Exam 1 Information. Math 241, Exam 1 Information. 9/24/12, LC 310, 11:15-12:05. Exam 1 will be based on: Sections 12.1-12.5, 14.1-14.3. The corresponding assigned homework problems (see http://www.math.sc.edu/ boylan/sccourses/241fa12/241.html)

More information

Class Meeting # 1: Introduction to PDEs

Class Meeting # 1: Introduction to PDEs MATH 18.152 COURSE NOTES - CLASS MEETING # 1 18.152 Introduction to PDEs, Fall 2011 Professor: Jared Speck Class Meeting # 1: Introduction to PDEs 1. What is a PDE? We will be studying functions u = u(x

More information

SOLUTIONS. f x = 6x 2 6xy 24x, f y = 3x 2 6y. To find the critical points, we solve

SOLUTIONS. f x = 6x 2 6xy 24x, f y = 3x 2 6y. To find the critical points, we solve SOLUTIONS Problem. Find the critical points of the function f(x, y = 2x 3 3x 2 y 2x 2 3y 2 and determine their type i.e. local min/local max/saddle point. Are there any global min/max? Partial derivatives

More information

FINAL EXAM SOLUTIONS Math 21a, Spring 03

FINAL EXAM SOLUTIONS Math 21a, Spring 03 INAL EXAM SOLUIONS Math 21a, Spring 3 Name: Start by printing your name in the above box and check your section in the box to the left. MW1 Ken Chung MW1 Weiyang Qiu MW11 Oliver Knill h1 Mark Lucianovic

More information

A QUICK GUIDE TO THE FORMULAS OF MULTIVARIABLE CALCULUS

A QUICK GUIDE TO THE FORMULAS OF MULTIVARIABLE CALCULUS A QUIK GUIDE TO THE FOMULAS OF MULTIVAIABLE ALULUS ontents 1. Analytic Geometry 2 1.1. Definition of a Vector 2 1.2. Scalar Product 2 1.3. Properties of the Scalar Product 2 1.4. Length and Unit Vectors

More information

2 Integrating Both Sides

2 Integrating Both Sides 2 Integrating Both Sides So far, the only general method we have for solving differential equations involves equations of the form y = f(x), where f(x) is any function of x. The solution to such an equation

More information

PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.

PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5. PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include

More information

Practice Final Math 122 Spring 12 Instructor: Jeff Lang

Practice Final Math 122 Spring 12 Instructor: Jeff Lang Practice Final Math Spring Instructor: Jeff Lang. Find the limit of the sequence a n = ln (n 5) ln (3n + 8). A) ln ( ) 3 B) ln C) ln ( ) 3 D) does not exist. Find the limit of the sequence a n = (ln n)6

More information

Calculus. Contents. Paul Sutcliffe. Office: CM212a.

Calculus. Contents. Paul Sutcliffe. Office: CM212a. Calculus Paul Sutcliffe Office: CM212a. www.maths.dur.ac.uk/~dma0pms/calc/calc.html Books One and several variables calculus, Salas, Hille & Etgen. Calculus, Spivak. Mathematical methods in the physical

More information

Practice Problems for Midterm 2

Practice Problems for Midterm 2 Practice Problems for Midterm () For each of the following, find and sketch the domain, find the range (unless otherwise indicated), and evaluate the function at the given point P : (a) f(x, y) = + 4 y,

More information

EXISTENCE AND NON-EXISTENCE RESULTS FOR A NONLINEAR HEAT EQUATION

EXISTENCE AND NON-EXISTENCE RESULTS FOR A NONLINEAR HEAT EQUATION Sixth Mississippi State Conference on Differential Equations and Computational Simulations, Electronic Journal of Differential Equations, Conference 5 (7), pp. 5 65. ISSN: 7-669. UL: http://ejde.math.txstate.edu

More information

1 if 1 x 0 1 if 0 x 1

1 if 1 x 0 1 if 0 x 1 Chapter 3 Continuity In this chapter we begin by defining the fundamental notion of continuity for real valued functions of a single real variable. When trying to decide whether a given function is or

More information

Calculus C/Multivariate Calculus Advanced Placement G/T Essential Curriculum

Calculus C/Multivariate Calculus Advanced Placement G/T Essential Curriculum Calculus C/Multivariate Calculus Advanced Placement G/T Essential Curriculum UNIT I: The Hyperbolic Functions basic calculus concepts, including techniques for curve sketching, exponential and logarithmic

More information

4. Complex integration: Cauchy integral theorem and Cauchy integral formulas. Definite integral of a complex-valued function of a real variable

4. Complex integration: Cauchy integral theorem and Cauchy integral formulas. Definite integral of a complex-valued function of a real variable 4. Complex integration: Cauchy integral theorem and Cauchy integral formulas Definite integral of a complex-valued function of a real variable Consider a complex valued function f(t) of a real variable

More information

Recall that the gradient of a differentiable scalar field ϕ on an open set D in R n is given by the formula:

Recall that the gradient of a differentiable scalar field ϕ on an open set D in R n is given by the formula: Chapter 7 Div, grad, and curl 7.1 The operator and the gradient: Recall that the gradient of a differentiable scalar field ϕ on an open set D in R n is given by the formula: ( ϕ ϕ =, ϕ,..., ϕ. (7.1 x 1

More information

The Math Circle, Spring 2004

The Math Circle, Spring 2004 The Math Circle, Spring 2004 (Talks by Gordon Ritter) What is Non-Euclidean Geometry? Most geometries on the plane R 2 are non-euclidean. Let s denote arc length. Then Euclidean geometry arises from the

More information

AB2.5: Surfaces and Surface Integrals. Divergence Theorem of Gauss

AB2.5: Surfaces and Surface Integrals. Divergence Theorem of Gauss AB2.5: urfaces and urface Integrals. Divergence heorem of Gauss epresentations of surfaces or epresentation of a surface as projections on the xy- and xz-planes, etc. are For example, z = f(x, y), x =

More information

Nonhomogeneous Linear Equations

Nonhomogeneous Linear Equations Nonhomogeneous Linear Equations In this section we learn how to solve second-order nonhomogeneous linear differential equations with constant coefficients, that is, equations of the form ay by cy G x where

More information

Critical Thresholds in Euler-Poisson Equations. Shlomo Engelberg Jerusalem College of Technology Machon Lev

Critical Thresholds in Euler-Poisson Equations. Shlomo Engelberg Jerusalem College of Technology Machon Lev Critical Thresholds in Euler-Poisson Equations Shlomo Engelberg Jerusalem College of Technology Machon Lev 1 Publication Information This work was performed with Hailiang Liu & Eitan Tadmor. These results

More information

MATH 381 HOMEWORK 2 SOLUTIONS

MATH 381 HOMEWORK 2 SOLUTIONS MATH 38 HOMEWORK SOLUTIONS Question (p.86 #8). If g(x)[e y e y ] is harmonic, g() =,g () =, find g(x). Let f(x, y) = g(x)[e y e y ].Then Since f(x, y) is harmonic, f + f = and we require x y f x = g (x)[e

More information

1 Completeness of a Set of Eigenfunctions. Lecturer: Naoki Saito Scribe: Alexander Sheynis/Allen Xue. May 3, 2007. 1.1 The Neumann Boundary Condition

1 Completeness of a Set of Eigenfunctions. Lecturer: Naoki Saito Scribe: Alexander Sheynis/Allen Xue. May 3, 2007. 1.1 The Neumann Boundary Condition MAT 280: Laplacian Eigenfunctions: Theory, Applications, and Computations Lecture 11: Laplacian Eigenvalue Problems for General Domains III. Completeness of a Set of Eigenfunctions and the Justification

More information

Second Order Linear Partial Differential Equations. Part I

Second Order Linear Partial Differential Equations. Part I Second Order Linear Partial Differential Equations Part I Second linear partial differential equations; Separation of Variables; - point boundary value problems; Eigenvalues and Eigenfunctions Introduction

More information

Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: f (x) =

Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: f (x) = Vertical Asymptotes Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: lim f (x) = x a lim f (x) = lim x a lim f (x) = x a

More information

MATH 425, HOMEWORK 7, SOLUTIONS

MATH 425, HOMEWORK 7, SOLUTIONS MATH 425, HOMEWORK 7, SOLUTIONS Each problem is worth 10 points. Exercise 1. (An alternative derivation of the mean value property in 3D) Suppose that u is a harmonic function on a domain Ω R 3 and suppose

More information

ORDINARY DIFFERENTIAL EQUATIONS

ORDINARY DIFFERENTIAL EQUATIONS ORDINARY DIFFERENTIAL EQUATIONS GABRIEL NAGY Mathematics Department, Michigan State University, East Lansing, MI, 48824. SEPTEMBER 4, 25 Summary. This is an introduction to ordinary differential equations.

More information

LINEAR ALGEBRA W W L CHEN

LINEAR ALGEBRA W W L CHEN LINEAR ALGEBRA W W L CHEN c W W L Chen, 1997, 2008 This chapter is available free to all individuals, on understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied,

More information

Introduction to Algebraic Geometry. Bézout s Theorem and Inflection Points

Introduction to Algebraic Geometry. Bézout s Theorem and Inflection Points Introduction to Algebraic Geometry Bézout s Theorem and Inflection Points 1. The resultant. Let K be a field. Then the polynomial ring K[x] is a unique factorisation domain (UFD). Another example of a

More information

Solutions to old Exam 1 problems

Solutions to old Exam 1 problems Solutions to old Exam 1 problems Hi students! I am putting this old version of my review for the first midterm review, place and time to be announced. Check for updates on the web site as to which sections

More information

1 The 1-D Heat Equation

1 The 1-D Heat Equation The 1-D Heat Equation 18.303 Linear Partial Differential Equations Matthew J. Hancock Fall 006 1 The 1-D Heat Equation 1.1 Physical derivation Reference: Guenther & Lee 1.3-1.4, Myint-U & Debnath.1 and.5

More information

Høgskolen i Narvik Sivilingeniørutdanningen STE6237 ELEMENTMETODER. Oppgaver

Høgskolen i Narvik Sivilingeniørutdanningen STE6237 ELEMENTMETODER. Oppgaver Høgskolen i Narvik Sivilingeniørutdanningen STE637 ELEMENTMETODER Oppgaver Klasse: 4.ID, 4.IT Ekstern Professor: Gregory A. Chechkin e-mail: chechkin@mech.math.msu.su Narvik 6 PART I Task. Consider two-point

More information

L 2 : x = s + 1, y = s, z = 4s + 4. 3. Suppose that C has coordinates (x, y, z). Then from the vector equality AC = BD, one has

L 2 : x = s + 1, y = s, z = 4s + 4. 3. Suppose that C has coordinates (x, y, z). Then from the vector equality AC = BD, one has The line L through the points A and B is parallel to the vector AB = 3, 2, and has parametric equations x = 3t + 2, y = 2t +, z = t Therefore, the intersection point of the line with the plane should satisfy:

More information

Section 12.6: Directional Derivatives and the Gradient Vector

Section 12.6: Directional Derivatives and the Gradient Vector Section 26: Directional Derivatives and the Gradient Vector Recall that if f is a differentiable function of x and y and z = f(x, y), then the partial derivatives f x (x, y) and f y (x, y) give the rate

More information

Parabolic Equations. Chapter 5. Contents. 5.1.2 Well-Posed Initial-Boundary Value Problem. 5.1.3 Time Irreversibility of the Heat Equation

Parabolic Equations. Chapter 5. Contents. 5.1.2 Well-Posed Initial-Boundary Value Problem. 5.1.3 Time Irreversibility of the Heat Equation 7 5.1 Definitions Properties Chapter 5 Parabolic Equations Note that we require the solution u(, t bounded in R n for all t. In particular we assume that the boundedness of the smooth function u at infinity

More information

Exam 1 Sample Question SOLUTIONS. y = 2x

Exam 1 Sample Question SOLUTIONS. y = 2x Exam Sample Question SOLUTIONS. Eliminate the parameter to find a Cartesian equation for the curve: x e t, y e t. SOLUTION: You might look at the coordinates and notice that If you don t see it, we can

More information

General Theory of Differential Equations Sections 2.8, 3.1-3.2, 4.1

General Theory of Differential Equations Sections 2.8, 3.1-3.2, 4.1 A B I L E N E C H R I S T I A N U N I V E R S I T Y Department of Mathematics General Theory of Differential Equations Sections 2.8, 3.1-3.2, 4.1 Dr. John Ehrke Department of Mathematics Fall 2012 Questions

More information

8 Hyperbolic Systems of First-Order Equations

8 Hyperbolic Systems of First-Order Equations 8 Hyperbolic Systems of First-Order Equations Ref: Evans, Sec 73 8 Definitions and Examples Let U : R n (, ) R m Let A i (x, t) beanm m matrix for i,,n Let F : R n (, ) R m Consider the system U t + A

More information

An Introduction to Partial Differential Equations in the Undergraduate Curriculum

An Introduction to Partial Differential Equations in the Undergraduate Curriculum An Introduction to Partial Differential Equations in the Undergraduate Curriculum J. Tolosa & M. Vajiac LECTURE 11 Laplace s Equation in a Disk 11.1. Outline of Lecture The Laplacian in Polar Coordinates

More information

Multiplicity. Chapter 6

Multiplicity. Chapter 6 Chapter 6 Multiplicity The fundamental theorem of algebra says that any polynomial of degree n 0 has exactly n roots in the complex numbers if we count with multiplicity. The zeros of a polynomial are

More information

FIELDS-MITACS Conference. on the Mathematics of Medical Imaging. Photoacoustic and Thermoacoustic Tomography with a variable sound speed

FIELDS-MITACS Conference. on the Mathematics of Medical Imaging. Photoacoustic and Thermoacoustic Tomography with a variable sound speed FIELDS-MITACS Conference on the Mathematics of Medical Imaging Photoacoustic and Thermoacoustic Tomography with a variable sound speed Gunther Uhlmann UC Irvine & University of Washington Toronto, Canada,

More information

Surface Normals and Tangent Planes

Surface Normals and Tangent Planes Surface Normals and Tangent Planes Normal and Tangent Planes to Level Surfaces Because the equation of a plane requires a point and a normal vector to the plane, nding the equation of a tangent plane to

More information

Introduction to the Finite Element Method

Introduction to the Finite Element Method Introduction to the Finite Element Method 09.06.2009 Outline Motivation Partial Differential Equations (PDEs) Finite Difference Method (FDM) Finite Element Method (FEM) References Motivation Figure: cross

More information

1 TRIGONOMETRY. 1.0 Introduction. 1.1 Sum and product formulae. Objectives

1 TRIGONOMETRY. 1.0 Introduction. 1.1 Sum and product formulae. Objectives TRIGONOMETRY Chapter Trigonometry Objectives After studying this chapter you should be able to handle with confidence a wide range of trigonometric identities; be able to express linear combinations of

More information

Solutions to Practice Problems for Test 4

Solutions to Practice Problems for Test 4 olutions to Practice Problems for Test 4 1. Let be the line segmentfrom the point (, 1, 1) to the point (,, 3). Evaluate the line integral y ds. Answer: First, we parametrize the line segment from (, 1,

More information

3. INNER PRODUCT SPACES

3. INNER PRODUCT SPACES . INNER PRODUCT SPACES.. Definition So far we have studied abstract vector spaces. These are a generalisation of the geometric spaces R and R. But these have more structure than just that of a vector space.

More information

Solutions to Homework 10

Solutions to Homework 10 Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x

More information

Second-Order Linear Differential Equations

Second-Order Linear Differential Equations Second-Order Linear Differential Equations A second-order linear differential equation has the form 1 Px d 2 y dx 2 dy Qx dx Rxy Gx where P, Q, R, and G are continuous functions. We saw in Section 7.1

More information

(a) We have x = 3 + 2t, y = 2 t, z = 6 so solving for t we get the symmetric equations. x 3 2. = 2 y, z = 6. t 2 2t + 1 = 0,

(a) We have x = 3 + 2t, y = 2 t, z = 6 so solving for t we get the symmetric equations. x 3 2. = 2 y, z = 6. t 2 2t + 1 = 0, Name: Solutions to Practice Final. Consider the line r(t) = 3 + t, t, 6. (a) Find symmetric equations for this line. (b) Find the point where the first line r(t) intersects the surface z = x + y. (a) We

More information

Section 1.1. Introduction to R n

Section 1.1. Introduction to R n The Calculus of Functions of Several Variables Section. Introduction to R n Calculus is the study of functional relationships and how related quantities change with each other. In your first exposure to

More information

1 3 4 = 8i + 20j 13k. x + w. y + w

1 3 4 = 8i + 20j 13k. x + w. y + w ) Find the point of intersection of the lines x = t +, y = 3t + 4, z = 4t + 5, and x = 6s + 3, y = 5s +, z = 4s + 9, and then find the plane containing these two lines. Solution. Solve the system of equations

More information

The Dirichlet Unit Theorem

The Dirichlet Unit Theorem Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if

More information

Methods of Solution of Selected Differential Equations Carol A. Edwards Chandler-Gilbert Community College

Methods of Solution of Selected Differential Equations Carol A. Edwards Chandler-Gilbert Community College Methods of Solution of Selected Differential Equations Carol A. Edwards Chandler-Gilbert Community College Equations of Order One: Mdx + Ndy = 0 1. Separate variables. 2. M, N homogeneous of same degree:

More information

Chapter 9 Partial Differential Equations

Chapter 9 Partial Differential Equations 363 One must learn by doing the thing; though you think you know it, you have no certainty until you try. Sophocles (495-406)BCE Chapter 9 Partial Differential Equations A linear second order partial differential

More information

3. Let A and B be two n n orthogonal matrices. Then prove that AB and BA are both orthogonal matrices. Prove a similar result for unitary matrices.

3. Let A and B be two n n orthogonal matrices. Then prove that AB and BA are both orthogonal matrices. Prove a similar result for unitary matrices. Exercise 1 1. Let A be an n n orthogonal matrix. Then prove that (a) the rows of A form an orthonormal basis of R n. (b) the columns of A form an orthonormal basis of R n. (c) for any two vectors x,y R

More information

BANACH AND HILBERT SPACE REVIEW

BANACH AND HILBERT SPACE REVIEW BANACH AND HILBET SPACE EVIEW CHISTOPHE HEIL These notes will briefly review some basic concepts related to the theory of Banach and Hilbert spaces. We are not trying to give a complete development, but

More information

5.4 The Heat Equation and Convection-Diffusion

5.4 The Heat Equation and Convection-Diffusion 5.4. THE HEAT EQUATION AND CONVECTION-DIFFUSION c 6 Gilbert Strang 5.4 The Heat Equation and Convection-Diffusion The wave equation conserves energy. The heat equation u t = u xx dissipates energy. The

More information

Review Sheet for Test 1

Review Sheet for Test 1 Review Sheet for Test 1 Math 261-00 2 6 2004 These problems are provided to help you study. The presence of a problem on this handout does not imply that there will be a similar problem on the test. And

More information

Mathematics Course 111: Algebra I Part IV: Vector Spaces

Mathematics Course 111: Algebra I Part IV: Vector Spaces Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 1996-7 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are

More information

THREE DIMENSIONAL GEOMETRY

THREE DIMENSIONAL GEOMETRY Chapter 8 THREE DIMENSIONAL GEOMETRY 8.1 Introduction In this chapter we present a vector algebra approach to three dimensional geometry. The aim is to present standard properties of lines and planes,

More information

3.8 Finding Antiderivatives; Divergence and Curl of a Vector Field

3.8 Finding Antiderivatives; Divergence and Curl of a Vector Field 3.8 Finding Antiderivatives; Divergence and Curl of a Vector Field 77 3.8 Finding Antiderivatives; Divergence and Curl of a Vector Field Overview: The antiderivative in one variable calculus is an important

More information

TOPIC 3: CONTINUITY OF FUNCTIONS

TOPIC 3: CONTINUITY OF FUNCTIONS TOPIC 3: CONTINUITY OF FUNCTIONS. Absolute value We work in the field of real numbers, R. For the study of the properties of functions we need the concept of absolute value of a number. Definition.. Let

More information

Problem Set 5 Due: In class Thursday, Oct. 18 Late papers will be accepted until 1:00 PM Friday.

Problem Set 5 Due: In class Thursday, Oct. 18 Late papers will be accepted until 1:00 PM Friday. Math 312, Fall 2012 Jerry L. Kazdan Problem Set 5 Due: In class Thursday, Oct. 18 Late papers will be accepted until 1:00 PM Friday. In addition to the problems below, you should also know how to solve

More information

CONTROLLABILITY. Chapter 2. 2.1 Reachable Set and Controllability. Suppose we have a linear system described by the state equation

CONTROLLABILITY. Chapter 2. 2.1 Reachable Set and Controllability. Suppose we have a linear system described by the state equation Chapter 2 CONTROLLABILITY 2 Reachable Set and Controllability Suppose we have a linear system described by the state equation ẋ Ax + Bu (2) x() x Consider the following problem For a given vector x in

More information

Properties of BMO functions whose reciprocals are also BMO

Properties of BMO functions whose reciprocals are also BMO Properties of BMO functions whose reciprocals are also BMO R. L. Johnson and C. J. Neugebauer The main result says that a non-negative BMO-function w, whose reciprocal is also in BMO, belongs to p> A p,and

More information

CITY UNIVERSITY LONDON. BEng Degree in Computer Systems Engineering Part II BSc Degree in Computer Systems Engineering Part III PART 2 EXAMINATION

CITY UNIVERSITY LONDON. BEng Degree in Computer Systems Engineering Part II BSc Degree in Computer Systems Engineering Part III PART 2 EXAMINATION No: CITY UNIVERSITY LONDON BEng Degree in Computer Systems Engineering Part II BSc Degree in Computer Systems Engineering Part III PART 2 EXAMINATION ENGINEERING MATHEMATICS 2 (resit) EX2005 Date: August

More information

INTEGRATING FACTOR METHOD

INTEGRATING FACTOR METHOD Differential Equations INTEGRATING FACTOR METHOD Graham S McDonald A Tutorial Module for learning to solve 1st order linear differential equations Table of contents Begin Tutorial c 2004 g.s.mcdonald@salford.ac.uk

More information

LINEAR MAPS, THE TOTAL DERIVATIVE AND THE CHAIN RULE. Contents

LINEAR MAPS, THE TOTAL DERIVATIVE AND THE CHAIN RULE. Contents LINEAR MAPS, THE TOTAL DERIVATIVE AND THE CHAIN RULE ROBERT LIPSHITZ Abstract We will discuss the notion of linear maps and introduce the total derivative of a function f : R n R m as a linear map We will

More information

correct-choice plot f(x) and draw an approximate tangent line at x = a and use geometry to estimate its slope comment The choices were:

correct-choice plot f(x) and draw an approximate tangent line at x = a and use geometry to estimate its slope comment The choices were: Topic 1 2.1 mode MultipleSelection text How can we approximate the slope of the tangent line to f(x) at a point x = a? This is a Multiple selection question, so you need to check all of the answers that

More information

Recitation 4. 24xy for 0 < x < 1, 0 < y < 1, x + y < 1 0 elsewhere

Recitation 4. 24xy for 0 < x < 1, 0 < y < 1, x + y < 1 0 elsewhere Recitation. Exercise 3.5: If the joint probability density of X and Y is given by xy for < x

More information

Math 209 Solutions to Assignment 7. x + 2y. 1 x + 2y i + 2. f x = cos(y/z)), f y = x z sin(y/z), f z = xy z 2 sin(y/z).

Math 209 Solutions to Assignment 7. x + 2y. 1 x + 2y i + 2. f x = cos(y/z)), f y = x z sin(y/z), f z = xy z 2 sin(y/z). Math 29 Solutions to Assignment 7. Find the gradient vector field of the following functions: a fx, y lnx + 2y; b fx, y, z x cosy/z. Solution. a f x x + 2y, f 2 y x + 2y. Thus, the gradient vector field

More information

TOPIC 4: DERIVATIVES

TOPIC 4: DERIVATIVES TOPIC 4: DERIVATIVES 1. The derivative of a function. Differentiation rules 1.1. The slope of a curve. The slope of a curve at a point P is a measure of the steepness of the curve. If Q is a point on the

More information

The Fourth International DERIVE-TI92/89 Conference Liverpool, U.K., 12-15 July 2000. Derive 5: The Easiest... Just Got Better!

The Fourth International DERIVE-TI92/89 Conference Liverpool, U.K., 12-15 July 2000. Derive 5: The Easiest... Just Got Better! The Fourth International DERIVE-TI9/89 Conference Liverpool, U.K., -5 July 000 Derive 5: The Easiest... Just Got Better! Michel Beaudin École de technologie supérieure 00, rue Notre-Dame Ouest Montréal

More information

Numerical Methods for Differential Equations

Numerical Methods for Differential Equations Numerical Methods for Differential Equations Course objectives and preliminaries Gustaf Söderlind and Carmen Arévalo Numerical Analysis, Lund University Textbooks: A First Course in the Numerical Analysis

More information

F = 0. x ψ = y + z (1) y ψ = x + z (2) z ψ = x + y (3)

F = 0. x ψ = y + z (1) y ψ = x + z (2) z ψ = x + y (3) MATH 255 FINAL NAME: Instructions: You must include all the steps in your derivations/answers. Reduce answers as much as possible, but use exact arithmetic. Write neatly, please, and show all steps. Scientists

More information

The Convolution Operation

The Convolution Operation The Convolution Operation Convolution is a very natural mathematical operation which occurs in both discrete and continuous modes of various kinds. We often encounter it in the course of doing other operations

More information

PROBLEM SET 6: POLYNOMIALS

PROBLEM SET 6: POLYNOMIALS PROBLEM SET 6: POLYNOMIALS 1. introduction In this problem set we will consider polynomials with coefficients in K, where K is the real numbers R, the complex numbers C, the rational numbers Q or any other

More information

Calculus 1: Sample Questions, Final Exam, Solutions

Calculus 1: Sample Questions, Final Exam, Solutions Calculus : Sample Questions, Final Exam, Solutions. Short answer. Put your answer in the blank. NO PARTIAL CREDIT! (a) (b) (c) (d) (e) e 3 e Evaluate dx. Your answer should be in the x form of an integer.

More information

1 VECTOR SPACES AND SUBSPACES

1 VECTOR SPACES AND SUBSPACES 1 VECTOR SPACES AND SUBSPACES What is a vector? Many are familiar with the concept of a vector as: Something which has magnitude and direction. an ordered pair or triple. a description for quantities such

More information

MICROLOCAL ANALYSIS OF THE BOCHNER-MARTINELLI INTEGRAL

MICROLOCAL ANALYSIS OF THE BOCHNER-MARTINELLI INTEGRAL PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 00, Number 0, Pages 000 000 S 0002-9939(XX)0000-0 MICROLOCAL ANALYSIS OF THE BOCHNER-MARTINELLI INTEGRAL NIKOLAI TARKHANOV AND NIKOLAI VASILEVSKI

More information

u dx + y = 0 z x z x = x + y + 2 + 2 = 0 6) 2

u dx + y = 0 z x z x = x + y + 2 + 2 = 0 6) 2 DIFFERENTIAL EQUATIONS 6 Many physical problems, when formulated in mathematical forms, lead to differential equations. Differential equations enter naturally as models for many phenomena in economics,

More information

Some Notes on Taylor Polynomials and Taylor Series

Some Notes on Taylor Polynomials and Taylor Series Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited

More information

Systems with Persistent Memory: the Observation Inequality Problems and Solutions

Systems with Persistent Memory: the Observation Inequality Problems and Solutions Chapter 6 Systems with Persistent Memory: the Observation Inequality Problems and Solutions Facts that are recalled in the problems wt) = ut) + 1 c A 1 s ] R c t s)) hws) + Ks r)wr)dr ds. 6.1) w = w +

More information

THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS

THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS KEITH CONRAD 1. Introduction The Fundamental Theorem of Algebra says every nonconstant polynomial with complex coefficients can be factored into linear

More information

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize

More information

DIFFERENTIABILITY OF COMPLEX FUNCTIONS. Contents

DIFFERENTIABILITY OF COMPLEX FUNCTIONS. Contents DIFFERENTIABILITY OF COMPLEX FUNCTIONS Contents 1. Limit definition of a derivative 1 2. Holomorphic functions, the Cauchy-Riemann equations 3 3. Differentiability of real functions 5 4. A sufficient condition

More information

Section 3.7. Rolle s Theorem and the Mean Value Theorem. Difference Equations to Differential Equations

Section 3.7. Rolle s Theorem and the Mean Value Theorem. Difference Equations to Differential Equations Difference Equations to Differential Equations Section.7 Rolle s Theorem and the Mean Value Theorem The two theorems which are at the heart of this section draw connections between the instantaneous rate

More information

On some classes of difference equations of infinite order

On some classes of difference equations of infinite order Vasilyev and Vasilyev Advances in Difference Equations (2015) 2015:211 DOI 10.1186/s13662-015-0542-3 R E S E A R C H Open Access On some classes of difference equations of infinite order Alexander V Vasilyev

More information

A PRIORI ESTIMATES FOR SEMISTABLE SOLUTIONS OF SEMILINEAR ELLIPTIC EQUATIONS. In memory of Rou-Huai Wang

A PRIORI ESTIMATES FOR SEMISTABLE SOLUTIONS OF SEMILINEAR ELLIPTIC EQUATIONS. In memory of Rou-Huai Wang A PRIORI ESTIMATES FOR SEMISTABLE SOLUTIONS OF SEMILINEAR ELLIPTIC EQUATIONS XAVIER CABRÉ, MANEL SANCHÓN, AND JOEL SPRUCK In memory of Rou-Huai Wang 1. Introduction In this note we consider semistable

More information

Math 2280 - Assignment 6

Math 2280 - Assignment 6 Math 2280 - Assignment 6 Dylan Zwick Spring 2014 Section 3.8-1, 3, 5, 8, 13 Section 4.1-1, 2, 13, 15, 22 Section 4.2-1, 10, 19, 28 1 Section 3.8 - Endpoint Problems and Eigenvalues 3.8.1 For the eigenvalue

More information

A First Course in Elementary Differential Equations. Marcel B. Finan Arkansas Tech University c All Rights Reserved

A First Course in Elementary Differential Equations. Marcel B. Finan Arkansas Tech University c All Rights Reserved A First Course in Elementary Differential Equations Marcel B. Finan Arkansas Tech University c All Rights Reserved 1 Contents 1 Basic Terminology 4 2 Qualitative Analysis: Direction Field of y = f(t, y)

More information

RESULTANT AND DISCRIMINANT OF POLYNOMIALS

RESULTANT AND DISCRIMINANT OF POLYNOMIALS RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. All results

More information