1. ## Nonhomogeneous wave equation

Hi, how should I start to solve the following PDE?

$u_{tt} - u_{xx} = sin(x) cos(5x) sin(w t)$ 0<x<pi, t>0

u(0,t)=u(pi, t)=0
u(x,0)= $u_{t}(x,0)$=0.

My aim is actually to find the values for w, for which the solution does not diverge. Is there a way to find it without completeley solving the problem.
Thanks for the help.

2. Originally Posted by Gaudium
Hi, how should I start to solve the following PDE?

$u_{tt} - u_{xx} = sin(x) cos(5x) sin(w t)$ 0<x<pi, t>0

u(0,t)=u(pi, t)=0
u(x,0)= $u_{t}(x,0)$=0.

My aim is actually to find the values for w, for which the solution does not diverge. Is there a way to find it without completeley solving the problem.
Thanks for the help.
Yes. If you look for solutions in the form

$
u = \sum_{n=1}^{\infty} T_n(t) \sin n \pi x
$

and sub into you PDE noting that $\sin x \cos 5x = \frac{1}{2} \sin 6x - \frac{1}{2} \sin 4x$, then comparing terms of $\sin n \pi x$ you get

$
T_n'' + n^2 \pi^2 T_n = 0\;\; \text{for}\; n \ne 4 \; \text{and}\;6
$
and

$
T_4'' + 16 \pi^2 T_4 = - \frac{1}{2} \sin \omega t
$

and

$
T_6'' + 36 \pi^2 T_6 = \frac{1}{2} \sin \omega t
$

Now the solution of the first is bounded and then you ask, for what values of $\omega$ are the solutions of the second and third unbounded?