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Solution HW#5 We have to solve the non-homogenous heat equation ∂u ∂2u + 4e−4t cos3 x = ∂t ∂x2 with a non-homogenous boundary condition, u( π2 , t) = 0. ∂u ∂x (0, t) = 2, and a homogenous one, We start by finding a simple function r(x) that has the same boundary values. For example, r(x) = 2x − π . Writing u = v + r and substituting it into the heat equation, we get ∂v ∂2v = + 4e−4t cos3 x ∂t ∂x2 with homogenous boundary conditions: ∂v ∂x (0, t) = 0 and v( π2 , t) = 0. We expand v(x, t) in terms of the complete set of eigenfuctions to the homogeneous problem on the given interval v(x, t) = ∞ X an (t) cos (1 + 2n)x . n=0 Because the eigenfunctions satisfy the same boundary conditions as v(x, t), we are allowed to differentiate term by term inside the sum. From the wave equation we obtain ∞ X dan n=1 dt + λn an cos (1 + 2n)x = 4e−4t cos3 x where the eigenvalues are given by λn = (1 + 2n)2 . Using the identity 4 cos3 x = 3 cos x + cos (3x), we can read off the Fourier coefficients of the source. The wave equation decomposes into an ODE for each term in the Fourier series, da0 + λ0 a0 = 3e−4t , dt da1 + λ1 a1 = e−4t , dt dan + λ n an = 0 , dt 1 n ≥ 2. Multiplying the equations by an integrating factor eλn t , we get d λ0 t a0 e = 3e(λ0 −4)t = 3e−3t , dt d λ1 t a1 e = e(λ1 −4)t = e5t , dt d an eλn t = 0 , n ≥ 2. dt The solutions to the above 1st order ODEs are −λ0 t a0 (t) = a0 (0) e +e −λ0 t Z t 3e−3τ dτ , 0 a1 (t) = a1 (0) e−λ1 t + e−λ1 t Z t e5τ dτ , 0 an (t) = an (0) e−λn t , n ≥ 2, where the terms containing integrals evaluates to (e−t − e−4t ) and spectively. 1 −4t 5 (e − e−9t ), re- The solution for v(x, t) is then ∞ X 1 v(x, t) = (e−t − e−4t ) cos x + (e−4t − e−9t ) cos (3x) + an (0) e−λn t cos (1 + 2n)x . 5 n=0 The initial condition v(x, 0) = f (x) − r(x) = f (x) − 2x + π gives the equation, f (x) − 2x + π = ∞ X an (0) cos (1 + 2n)x , n=0 which fixes the integration constants in the homogenous part of the solution 4 an (0) = π Z π/2 f (x) − 2x + π cos (1 + 2n)x dx . 0 Final answer: The solution is ∞ −t −4t u(x, t) = 2x−π+(e −e X 1 an (0) e−λn t cos (1+2n)x , ) cos x+ (e−4t −e−9t ) cos (3x)+ 5 n=0 with λn = (1 + 2n)2 , and an (0) given above. 2