Math 412-501Theory of Partial Differential EquationsLecture 7: Eigenvalue problems.Solution of the initial-boundary value problemfor the heat equation.Boundary value problem for the heat equation∂u∂t= k∂2u∂x2, 0 ≤ x ≤ L,u(0, t) = u(L, t) = 0.We are looking for solutions u(x, t) = φ(x)G(t).PDE holds ifd2φdx2= −λφ,dGdt= −λkGfor the same constant λ.Boundary conditions hold ifφ(0) = φ(L) = 0.Boundary value problem:d2φdx2= −λφ, 0 ≤ x ≤ L,φ(0) = φ(L) = 0.There is an obvious solution: 0.When is it not unique?If for some value of λ the boundary value problemhas a nonzero solution φ, then this λ is called aneigenvalue and φ is called an eigenfunction.The eigenvalue problem is to find all eigenvalues(and corresponding eigenfunctions).Matrices vs. differential operatorsSuppose A is an n × n matrix, v ∈ Rnis a nonzerovector, and λ = const. IfAv = λv then λ is aneigenvalue of A and v is an eigenvector.d2φdx2= −λφ, 0 ≤ x ≤ L,φ(0) = φ(L) = 0.Instead of A, we have the linear operator −d2dx2.Instead of Rn, we have the linear spaceV = {φ ∈ C2[0, L] : φ(0) = φ(L) = 0}.Eigenvalue problemφ′′= −λφ, φ(0) = φ(L) = 0.We are looking only for real eigenvalues.Three cases: λ > 0, λ = 0, λ < 0.Case 1: λ > 0. φ(x) = C1cos µx + C2sin µx,where λ = µ2, µ > 0.φ(0) = φ(L) = 0 =⇒ C1= 0, C2sin µL = 0.A nonzero solution exists if µL = nπ, n ∈ Z.So λn= (nπL)2, n = 1, 2, . . . are eigenvalues andφn(x) = sinnπxLare corresponding eigenfunctions.Eigenfunctionsφn(x) = sinnπxLAre there other eigenfunctions?Case 2: λ = 0. φ(x) = C1+ C2x.φ(0) = φ(L) = 0 =⇒ C1= C1+ C2L = 0=⇒ C1= C2= 0.Case 3: λ < 0. φ(x) = C1eµx+ C2e−µx,where λ = −µ2, µ > 0.cosh z =ez+ e−z2sinh z =ez− e−z2ez= cosh z + sinh z, e−z= cosh z − sinh z.φ(x) = D1cosh µx + D2sinh µx, D1, D2= const.φ(0) = 0 =⇒ D1= 0φ(L) = 0 =⇒ D2sinh µL = 0 =⇒ D2= 0Hyperbolic functionsThere is another way to show that all eigenvaluesare positive. Given an eigenfunction φ, letI =ZL0φ′′(x)φ(x) dx.Since φ′′= −λφ, we haveI = −λZL0|φ(x)|2dx.Integrating by parts, we obtainI = φ′(L)φ(L) − φ′(0)φ(0) −ZL0φ′(x)φ′(x) dx.HenceλZL0|φ(x)|2dx =ZL0|φ′(x)|2dx.=⇒ either λ > 0, or else λ = 0 and φ = const.SummaryEigenvalue problem: φ′′= −λφ, φ(0) = φ(L) = 0.Eigenvalues: λn= (nπL)2, n = 1, 2, . . .Eigenfunctions: φn(x) = sinnπxL.Solution of the heat equation: u(x, t) = φ(x)G (t).dGdt= −λkG =⇒ G (t) = C0exp(−λkt)Theorem For n = 1, 2, . . . , the functionu(x, t) = e−λnktφn(x) = exp(−n2π2L2kt) sinnπxLis a solution of the following boundary valueproblem for the heat equation:∂u∂t= k∂2u∂x2, u(0, t) = u(L, t) = 0.Initial-boundary value problem∂u∂t= k∂2u∂x2, 0 ≤ x ≤ L,u(x, 0) = f (x), u(0, t) = u(L, t) = 0.Function u(x, t) = e−λnktφn(x) is a solution of theboundary value problem. Initial condition is satisfiedif f = φn. For any B1, B2, . . . , BN∈ R the functionu(x, t) =XNn=1Bne−λnktφn(x)is also a solution of the boundary value problem.This time the initial condition is satisfied iff (x) =XNn=1Bnφn(x) =XNn=1BnsinnπxL.From finite sums to seriesConjecture For suitably chosen coefficientsB1, B2, B3, . . . the functionu(x, t) =X∞n=1Bne−λnktφn(x)is a solution of the boundary value problem. Thissolution satisfies the initial condition withf (x) =X∞n=1Bnφn(x) =X∞n=1BnsinnπxL.Theorem IfP∞n=1|Bn| < ∞ then the conjecture istrue. Namely, u(x, t) is smooth for t > 0 and solvesthe boundary value problem. Also, u(x, t) iscontinuous for t ≥ 0 and satisfies the initialcondition.How do we solve the initial-boundary value problem?∂u∂t= k∂2u∂x2, 0 ≤ x ≤ L,u(x, 0) = f (x), u(0, t) = u(L, t) = 0.• Expand the function f into a seriesf (x) =X∞n=1BnsinnπxL.• Write the solution:u(x, t) =X∞n=1Bnexp−n2π2L2ktsinnπxL.J. Fourier, The Analytical Theory of Heat(written in 1807, published in