Math 412-501Theory of Partial Differential EquationsLecture 8: Fourier’s solution of theinitial-boundary value problem (continued).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.(Fourier’s solution)How do we expand initial data into a series?f (x) =X∞n=1BnsinnπxL.Assume that such an expansion exists and the seriesconverges uniformly. ThenZL0f (x)g(x) dx =X∞n=1BnZL0sinnπxLg(x) dxfor any continuous function g : [0, L] → R.In particular, for m = 1, 2, . . . we haveZL0f (x) sinmπxLdx =X∞n=1BnZL0sinnπxLsinmπxLdx.It turns out thatZL0sinnπxLsinmπxLdx =(12L if m = n,0 if m 6= n.HenceZL0f (x) sinmπxLdx = Bm·12L.ThereforeBm=2LZL0f (x) sinmπxLdx.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,whereBn=2LZL0f (ξ) sinnπξLdξ.• Write the solution:u(x, t) =X∞n=1Bnexp−n2π2L2ktsinnπxL.Evaluation of an integralZL0sinnπxLsinmπxLdx =LπZL0sinnπxLsinmπxLd(πxL)=LπZπ0sin ny · sin my dy=L2πZπ0cos (n − m)y − cos (n + m)ydy.Let N ∈ Z. If N 6= 0 thenZπ0cos Ny dy =sin NyNπ0= 0.If N = 0 thenZπ0cos Ny dy =Zπ0dy = π.Example∂u∂t= k∂2u∂x2, 0 ≤ x ≤ L,u(x, 0) = 100, u(0, t) = u(L, t) = 0.Fourier’s expansion:100 =X∞n=1BnsinnπxL(0 < x < L),whereBn=2LZL0100 sinnπξLdξ =200πZL0sinnπξLdπξL=200πZπ0sin ny dy =200(1 − cos nπ)nπ.Bn= 0 if n is even, and Bn=400nπif n is odd.n is odd =⇒ n = 2m − 1, m a positive integer.100 =∞Xm=1400(2m−1)πsin(2m−1)πxL, 0 < x < L.Fourier’s solution:u(x, t) =∞Xm=1400(2m−1)πexp−(2m−1)2π2L2ktsin(2m−1)πxL.For a large t,u(x, t) ≈400πexp−π2L2ktsinπxL.More boundary conditions for the heat equationInitial-boundary value problem:∂u∂t= k∂2u∂x2, 0 ≤ x ≤ L,u(x, 0) = f (t),∂u∂x(0, t) =∂u∂x(L, t) = 0.(insulated ends)∂u∂t= k∂2u∂x2, −L ≤ x ≤ L,u(x, 0) = f (t),u(−L, t) = u(L, t),∂u∂x(−L, t) =∂u∂x(L, t).(periodic boundary conditions)Heat conduction in a thin circular ringSeparation of variables: u(x, t) = φ(x)G (t).PDE holds if for some λ = const,d2φdx2= −λφ,dGdt= −λkG =⇒ G (t) = C0exp(−λkt).Boundary conditions∂u∂x(0, t) =∂u∂x(L, t) = 0 hold ifφ′(0) = φ′(L) = 0.Boundary conditions u(−L, t) = u(L, t),∂u∂x(−L, t) =∂u∂x(L, t) = 0 hold ifφ(−L) = φ(L), φ′(−L) = φ′(L).Eigenvalue problem (insulated ends):φ′′= −λφ, φ′(0) = φ′(L) = 0.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 =⇒ C2= 0, −C1µ sin µL = 0.A nonzero solution exists if µL = nπ, n ∈ Z.So λn= (nπL)2, n = 1, 2, . . . are eigenvalues andφn(x) = cosnπxLare corresponding eigenfunctions.The only other eigenvalue is λ0= 0, with theeigenfunction φ0= 1.Eigenvalue problem (circular ring):φ′′= −λφ, φ(−L) = φ(L), φ′(−L) = φ′(L).Case 1: λ > 0. φ(x) = C1cos µx + C2sin µx,where λ = µ2, µ > 0.φ(−L) = φ(L) =⇒ C2sin µL = 0.φ′(−L) = φ′(L) =⇒ −C1µ sin µL = 0.A nonzero solution exists if µL = nπ, n ∈ Z.So λn= (nπL)2, n = 1, 2, . . . are multipleeigenvalues while φn(x) = cosnπxLandψn(x) = sinnπxLare corresponding eigenfunctions.The only other eigenvalue is λ0= 0, with theeigenfunction φ0= 1.Fourier’s solution (insulated ends)∂u∂t= k∂2u∂x2, 0 ≤ x ≤ L,u(x, 0) = f (x),∂u∂x(0, t) =∂u∂x(L, t) = 0.• Expand the function f into a seriesf (x) = A0+X∞n=1AncosnπxL.• Write the solution:u(x, t) = A0+X∞n=1Anexp−n2π2L2ktcosnπxL.Fourier’s solution (circular ring)∂u∂t= k∂2u∂x2, −L ≤ x ≤ L,u(x, 0) = f (t),u(−L, t) = u(L, t),∂u∂x(−L, t) =∂u∂x(L, t).• Expand the function f into a seriesf (x) = A0+∞Xn=1AncosnπxL+ BnsinnπxL.• Write the solution:u(x, t) = A0+∞Xn=1exp−n2π2L2ktAncosnπxL+