Finding a solution of a PDE, which is growing in time I've been dealing with a problem about partial differential equations for a while and hope you can help me.
Let be $\omega, c>0$ and $h: [0,a]\rightarrow \mathbb{R}$ regular ($a>0 \in \mathbb{R}$). Concern the inhomogenious, damped wave-equation
$u_{tt}(x,t) - c^2 u_{xx}(x,t)+ p u_t(x,t) =h(x) \sin(\omega t)~~~$ for $t>0, x \in (0,a)$
$u(x,0) = 0 = u_t(x,0)$
$u(0,t) = 0 = u(a,t)$.
(a) Let be $p=0$. For which $\omega$ do we get a solution, which is growing in time?
(b) What about $p>0$?
Well, I guess (a) requires solving the equation, so - due to $u(x,0) = 0 = u_t(x,0)$ and $u(0,t) = 0 = u(a,t)$ - my approach was as follows:
There is only one term in the universal solution which contributes. For Eigenvectors $\phi_n(x)=\sqrt{2} \sin(\frac{n\pi x}{l})$ and Eigenvalues $\lambda_n = \left(\frac{c n \pi}{l}\right)^2$ we get:
$u(x,t) = \int_{0}^{t} \sum_{n=0}^{\infty}\frac{\sin(\sqrt{\lambda_n}(t-s))}{\sqrt{\lambda_n}} f_n(s) \phi_n(s) ds $
I tried to evalueate the integral above and hoped to get some term, where I see some kind of condition for $\omega$ but this integral gets a real crap (which even doesn't result into something nice when I use some trigonometric identities before summing up).
Does anyone have an idea how I can find out some condition for $\omega$?
Furthermore, due to the $r$-Term doesn't vanish any more (and nothing I've found in my book doesn't apply), I don't have an idea for b). Can anyone help me?
Thanks in advance!
 A: I'm going to write this out in hopes that it puts you in the write direction, but this is not a complete answer.
Assuming that the solution takes the form(taking on a eigenfunction expansion technique)
$$u(x, t) = \sum_{n=0}^{\infty} c_n(t) \sin{\left( \frac{n \pi x}{a} \right)}$$
We get
$$\sum_{n=0}^{\infty}\left[c_n''(t)+rc_n'(t)+\left(\frac{n \pi c}{a} \right)^2c_n(t) \right]\sin{\left( \frac{n \pi x}{a} \right)} = h(x)\sin{(\omega t)}$$
Since $h(x)\in L^2[0,a]$, we can write that 
$$h(x) = \sum_{n=0}^{\infty}b_n \sin{\left(\frac{n \pi x}{a}\right)}$$
where $b_n$ are the fourier coefficients for $h(x)$, $b_n = \frac{2}{a}\int_0^a h(x)\sin{\left( \frac{n \pi x}{a} \right)}\,dx$. This means we get the following ODE which we only really care about to analyze, since we are concerned with long term behavior.
$$\begin{cases}
c_n''(t) + rc_n'(t) + \left( \frac{n \pi c}{a} \right)^2c_n(t) = b_n \sin{(\omega t)} \\
c_n(0) = c_n'(0) = 0
\end{cases}$$
The solution to this system completely depends on the eigenvalues of this ODE, we can actually write out a solution using variation of parameters.
Letting $$\gamma =\left( \frac{n \pi c}{a} \right)^2, \lambda_{\pm} = \frac{-r \pm \sqrt{r^2 - 4\gamma^2}}{2}$$
We can solve for $c_n(t)$ with variation of parameters, this gives
$$c_n(t) = Ae^{\lambda_{+}t}+Be^{\lambda_{-}t} + y_p(t)$$
Where
$$y_p(t) = -b_ne^{\lambda_{+}t}\int_0^t \frac{e^{\lambda_{-}s}\sin{\omega s}}{-e^{-rs}\sqrt{r^2 - 4\gamma^2}}\, ds + b_ne^{\lambda_{-}t}\int_0^t \frac{e^{\lambda_{+}s}\sin{\omega s}}{-e^{-rs}\sqrt{r^2 - 4\gamma^2}}\, ds$$
It can be shown that $y_p(0)=y_p'(0)=0$, so that $A = B = 0$, and our solution is just 
$$c_n(t) = \frac{b_n}{\sqrt{r^2 - 4\gamma^2}}\int_0^t e^{rs}\sin{(\omega s)}\left(e^{\lambda_{-}s + \lambda_{+}t} - e^{\lambda_{+}s + \lambda_{-}t} \right)\, ds$$
We can now use this expression to divulge on some cases:
When $r = 0$, 
$$c_n(t) = b_n\frac{\gamma \sin{\omega t} - \omega \sin{\gamma t}}{\gamma^3 - \gamma \omega^2}$$
When $r > 0$, it may be worth looking at some specific cases, such as $r = 2\gamma$ which gives a repeated eigenvalue (which you should be able to conclude blow up). It looks like $\omega = \gamma$ might be of interest as well.
