Solve the PDE $ u_{tt}=4u_{xx}+A \sin (2 \pi t) $

Question

Solve the PDE

$$ u_{tt}=4u_{xx}+A \sin (2 \pi t) \ , \ 0<x<1 , \ t>0 \\ u(0,t)=u(1,t)=0 \\ u(x,0)=u_t(x,0)=0 , \ 0<x< 1 $$

Answer:

Can I use D'Alembert's solution in Wave equation ?

Answer 1

Hint: Look for a solution of the form

$$ u(x,t) = \sum_{n=1}^\infty T_n(t)\sin(n\pi x) $$ where the eigenfunctions in $x$ are derived from the homogeneous equation with the same boundary conditions.

Plugging this in, we find

$$ \sum_{n=0}^\infty \big[{T_n}''+4n^2T_n\big]\sin(n\pi x) = A\sin(2\pi t) $$

To compare coefficients, we'll need to find the Fourier series (in $x$) of the RHS, which is

$$ A\sin(2\pi t) = \frac{2A}{\pi}\sin(2\pi t) \sum_{n=0}^\infty \frac{1-(-1)^n}{n}\sin(n\pi x) $$

Therefore

$$ {T_n}'' + 4n^2 T_n = \begin{cases} 0, & n \text{ even} \\ \dfrac{4A}{n\pi}\sin(2\pi t), & n \text{ odd} \end{cases} $$

with initial conditions $$ T_n(0) = {T_n}'(0) = 0 $$