PDE 83215 Solutions to exercises 5 1. (a) The general solution of utt ...

The function v must satisfyv tt − 4v xx = − 4 π + sin tv x (0, t) = v x (π, t) = 0(v(x, 0) = x2 xπ π − 3 2)v t (x, 0) = 0The boundary conditions and the initial condition for v t suggest we should lookfor a solution of this in the form∞∑v(x, t) = T 0 (t) + v n cos nx cos 2ntn=1To satisfy the differential equation T 0 must satisfyT ′′0 = − 4 π + sin tTo satisfy the initial condition for v t , T 0 must satisfyThus we must haveT ′ 0(0) = 0T 0 = − 2t2π + t + T 0(0) − sin tImposing the intial condition for v givesx 2π( xπ − 3 ∞∑= T 0 (0) + v n cos nx2)n=1Integrating givesT 0 (0) = 1 ∫ π x 2 ( xπ 0 π π − 3 dx = −2)π 4Multiplying by cos nx and integrating givesThe answer:v n = 2 π∫ π0x 2π⎧( xπ − 3 ⎨cos nx dx =2)⎩(u(x, t) = (x − 1) sin t + x 1 − x )− 2t22π π + t − π 4 +0 n even24π 3 n 4∑n=1,3,5,...n odd24cos nx cos 2ntπ 3 n4 10