Csdrhrt

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 $phi_n(x)=sqrt{2} sin(frac{npi x}{l})$ 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!

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 $$gamma =left( frac{n pi c}{a} right)^2, lambda_{pm} = frac{-r pm sqrt{r^2 - 4gamma^2}}{2}$$

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

$$y_p(t) = -b_ne^{lambda_{+}t}int_0^t frac{e^{lambda_{-}s}sin{omega s}}{-e^{-rs}sqrt{r^2 - 4gamma^2}}, ds + b_ne^{lambda_{-}t}int_0^t frac{e^{lambda_{+}s}sin{omega s}}{-e^{-rs}sqrt{r^2 - 4gamma^2}}, ds$$

It can be shown that $y_p(0)=y_p'(0)=0$, so that $A = B = 0$, and our solution is just

$$c_n(t) = frac{b_n}{sqrt{r^2 - 4gamma^2}}int_0^t e^{rs}sin{(omega s)}left(e^{lambda_{-}s + lambda_{+}t} - e^{lambda_{+}s + lambda_{-}t} right), ds$$

We can now use this expression to divulge on some cases:

When $r = 0$,

$$c_n(t) = b_nfrac{gamma sin{omega t} - omega sin{gamma t}}{gamma^3 - gamma omega^2}$$

When $r > 0$, it may be worth looking at some specific cases, such as $r = 2gamma$ 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.