# Wave Equation in a Rectangle

I am trying to work through this question about the wave equation and I just don't know what to do.

Solve the wave equation in the rectangle $R=\{(x,y):0<x<1, 0<y<3\}\;$, with homogenous Dirichlet conditions on the boundary and initial conditions $u(x,y,0) \equiv xy(3-y)(1-x)\;$ and
$u_t(x,y,0) \equiv 0$.

Thank you in advance for the help.

edit for more general example:

Solve the wave equation in the rectangle $R=\{(x,y):0<x<a, 0<y<b\}\;$, with homogenous Dirichlet conditions on the boundary, and the initial conditions $u(x,y,0) \equiv xy(b-y)(a-x)\;$ and $u_t(x,y,0) \equiv 0$.

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Does the PDE be $u_{tt}=u_{xx}+u_{yy}$ ? – doraemonpaul Oct 24 '12 at 2:29

Do you have a set of modes of the rectangle? These are solutions to the wave equation without regard to the initial conditions and are of the form $f(x,y)e^{i\omega t}$. If not, do you have a set for the square?

If you do, you need to expand your $u(x,y,0)$ in these modes. They tend to be numbered $u_{mn}(x,y)$ where $m$ is the number of half waves in the $x$ direction and $n$ is the number in the $y$ direction. You are looking for a set of coefficients $a_{mn}$ so the solution $u(x,y,t)=a_{mn}u_{mn}e^{i \omega_{mn}t}$There is probably a proof that the modes are orthogonal, in which case you can just do like a Fourrier expansion. $a_{mn}=\int_{rectangle}u(x,y,0)u_{mn}(x,y)dxdy$ This only works if the modes are orthogonal.

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I have a more general example of the question that I edited in the above and if you could help me work through that example it would be much appreciated. – Steve May 7 '12 at 6:02
@Steve: the new version is identical to the old except for changing 3 to a and 1 to b. What are the modes? I would think they are something like $u_{mn}(x,y)\sin \pi mx \sin \frac {n\pi y}3 e^{i \omega_{mn} t}$ with an equation to calculate $\omega_{mn}$ from $m, n$. Just like a Fourrier expansion, you can multiply the modes by the given $u$ to get the expansion because they are orthogonal. – Ross Millikan May 7 '12 at 13:51
I know it is identical, I thought it might help though. I do not know the modes, I was only given what is in the question above, no more. Thank you for the help any way. – Steve May 7 '12 at 16:26
I worked something out I think is correct, thank you for all your help. – Steve May 7 '12 at 21:05