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TAMU MATH 412 - Lecture3-7web

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Math 412-501Theory of Partial Differential EquationsLecture 3-7: Poisson’s equation.Complex form of Fourier series.Fourier transforms.Poisson’s equationPoisson’s equation is the nonhomogeneous versionof Laplace’s equation: ∇2u = Q.Consider a boundary value problem:∇2u = Q in the domain D,u|∂D= α,where Q is a function on D and α is a function onthe boundary ∂D.There are two ways to solve this problem.Method 1. Suppose u0is a smooth function inthe domain D such that ∇2u0= Q.Then u = u0+ w, where w is the solution of theboundary value problem for Laplace’s equation:∇2u = 0 in the domain D,u|∂D= β,where β(x) = α(x) − u0(x) for any x ∈ ∂D.We know how to solve the latter problem if D is arectangle or a circle. Method 1 applies only tocertain functions Q as it relies on a lucky guess.Example: ∇2u = Q(x, y) = e−xcos 2y − sin 3x sin y.Lucky guess: u0(x, y) = −13e−xcos 2y +110sin 3x sin y .Method 2. u = u1+ u2,where u1is the solution of the problem∇2u1= Q in the domain D,u1|∂D= 0,while u2is the solution of the problem∇2u2= 0 in the domain D,u2|∂D= α.We know how to solve the latter problem if D is arectangle or a circle. The former problem is solvedby method of eigenfunction expansion.Boundary value problem:∇2u1= Q in the domain D,u1|∂D= 0.Let λ1< λ2≤ . . . be eigenvalues of the negativeDirichlet Laplacian in D (counting withmultiplicities), and φ1, φ2, . . . be the corresponding(orthogonal) eigenfunctions.We have thatu1(x, y) =X∞n=1cnφn(x, y),Q(x, y) =X∞n=1qnφn(x, y).The Laplacian ∇2u1can be evaluated term-by-term(it follows from the fact that u1|∂D= 0). Hence∇2u1=X∞n=1cn∇2φn= −X∞n=1λncnφn.Thus −λncn= qnfor n = 1, 2, . . . .Solution:u1= −X∞n=1λ−1nqnφn, where qn=hQ, φnihφn, φni.Heat equation on an infinite intervalInitial-boundary value problem:∂u∂t= k∂2u∂x2(0 < x < ∞),u(0, t) = 0, limx→∞u(x, t) = 0,u(x, 0) = f (x) (0 < x < ∞).The problem is supposed to describe heatconduction in a very long rod.We expect that the solution is the limit of solutionson intervals [0, L] as L → ∞.The problem on a finite interval:∂u∂t= k∂2u∂x2(0 < x < L),u(0, t) = u(L, t) = 0,u(x, 0) = f (x) (0 < x < L).Solution: Expand f into the Fourier sine series on[0, L]:f (x) =X∞n=1bnsinnπxL,where bn=2LZL0f (˜x) sinnπ˜xLd ˜x.Then u(x, t) =X∞n=1bnexp−(nπ/L)2ktsinnπxL.For any ω > 0 let B(ω) =2πZ∞0f (˜x) sin ω˜x d˜x.For simplicity, assume that f (x) = 0 for x > L0.Then B(ω) is well defined. Moreover,bn=2LZL0f (˜x) sinnπ˜xLd ˜x =πLBnπLprovided that L ≥ L0. Thereforeu(x, t) =πLX∞n=1B(ωn) exp(−ω2nkt) sin ωnx,where ωn= nπ/L. It follows thatlimL→∞u(x, t) =Z∞0B(ω)e−ω2ktsin ωx dω.The problem on the infinite interval:∂u∂t= k∂2u∂x2(0 < x < ∞),u(0, t) = 0, limx→∞u(x, t) = 0,u(x, 0) = f (x) (0 < x < ∞).Let us try and solve this problem by separation ofvariables. First we search for solutionsu(x, t) = φ(x)G(t) of the equation that satisfy theboundary conditions. The PDE holds ifd2φdx2= −λφ,dGdt= −λkt,where λ is a separation constant.Boundary conditions u(0, t) = u(∞, t) = 0 hold ifφ(0) = φ(∞) = 0.Eigenvalue problem on (0, ∞):φ′′= −λφ, φ(0) = φ(∞) = 0.This problem has no eigenvalues. If we drop thecondition φ(∞) = 0 then any λ ∈ C will be aneigenvalue, which is bad too.The right decision is to relax the condition:φ′′= −λφ, φ(0) = 0, |φ(∞)| < ∞.Eigenvalues: λ = ω2, where ω > 0.Eigenfunctions: φω(x) = sin ωx.Dependence on t:G′= −λkG =⇒ G (t) = c0e−λktSolutions with separated variables:uω(x, t) = e−ω2ktsin ωx, ω > 0.Now we search for the solution of theinitial-boundary value problem as a superposition ofsolutions with separated variables:u(x, t) =Z∞0B(ω)e−ω2ktsin ωx dω.The initial condition u(x, 0) = f (x) is satisfied iff (x) =Z∞0B(ω) sin ωx dω.The right-hand side is called a Fourier integral.Solution: Expand f into the Fourier integral:f (x) =Z∞0B(ω) sin ωx dω.Then u(x, t) =Z∞0B(ω)e−ω2ktsin ωx dω.How do we expand f into the Fourier integral?Approximation by finite-interval problems suggeststhatB(ω) =2πZ∞0f (x) sin ωx dx.Fourier sine transformLet f be a function on (0, ∞). The functionS[f ](ω) =2πZ∞0f (x) sin ωx dx, ω > 0is called the Fourier sine transform of f .The transform S[f ] is well defined if the integralconverges for all ω > 0.One sufficient condition isR∞0|f (x)| dx < ∞.Given a function F on (0, ∞), the functionS−1[F ](x) =Z∞0F (ω) sin ωx dω, x > 0is called the inverse Fourier sine transform of F .Theorem Suppose f is an absolutely integrablefunction on (0, ∞) and let F = S[f ] be its Fouriersine transform.(i) If f is smooth then f = S−1[F ].(ii) If f is piecewise smooth then the inverseFourier sine transform S−1[F ] is equal to f at pointsof continuity. OtherwiseS−1[F ](x) =f (x+) + f (x−)2.Fourier cosine transformGiven a function f on (0, ∞), the functionC [f ](ω) =2πZ∞0f (x) cos ωx dx, ω > 0is called the Fourier cosine transform of f .Given a function F on (0, ∞), the functionC−1[F ](x) =Z∞0F (ω) cos ωx dω, x > 0is the inverse Fourier cosine transform of F .Theorem Suppose f is an absolutely integrablefunction on (0, ∞) and let F = C[f ] be its Fouriercosine transform. If f is smooth then f = C−1[F ].A Fourier series on the interval [−L, L]:a0+X∞n=1ancosnπxL+X∞n=1bnsinnπxL.A Fourier series in the complex form:X∞n=−∞cnexpinπxL.Note that for any y ∈ R,eiy= cos y + i sin y, e−iy= cos y − i sin y,cos y =12(eiy+ e−iy), sin y =12i(eiy− e−iy).Hence both forms of the Fourier series areequivalent. Coefficients are related as follows:a0= c0, an= cn+ c−n, bn= i(cn− c−n), n ≥ 1.For any n ∈ Z, let φn(x) = einπx/L. Functions φnareorthogonal relative to the inner producthf , gi =ZL−Lf (x)g(x) dx.Indeed, if n 6= m, thenhφn, φmi =ZL−Leinπx/Leimπx/Ldx=ZL−Leinπx/Le−imπx/Ldx =ZL−Lei(n−m)πx/Ldx=Li(n − m)πei(n−m)πx/LL−L= 0.Also,hφn, φni =ZL−L|φn(x)|2dx =ZL−Ldx = 2L.Functions φnform a basis in the Hilbert spaceL2([−L, L]). Any square-integrable function f on[−L, L] is expanded into a seriesf (x) =X∞n=−∞cnφn(x) =X∞n=−∞cneinπx/Lthat converges in the mean. Coefficients areobtained as usual:cn=hf , φnihφn, φni=12LZL−Lf (x)e−inπx/Ldx.Fourier transformGiven a function f : R → C, the


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