wave equation with initial values and boundary condititon I have a homogenious 2-dimensional wave equation:
$$ - \frac{\partial^2 u}{\partial x^2} (x, y, t) - \frac{\partial^2 u}{\partial y^2} (x, y, t) + \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} (x, y, t) = 0$$
With:
$$ 0 < x < a, 0 < y < b, t > 0$$
$$ u(x, y, 0) = 0$$
$$ \dot u(x, y, 0) = x(x-a)(y-b)$$
And boundary condition:
$$ u(0, y, t) = u(a, y, t) = u(x, 0, t) = u(x, b, t) = 0$$
It says that I should solve it using separation of variables with $u(x, y, t) = G(x)H(y)w(t)$ as a starting point. With that approach, I got this solution:
$$  u(x, y, t)
 = \sin\left(\frac{n\pi}a x\right)
 \sin\left(\frac{n\pi}b y\right)
 \sin\left(\left(\frac{n\pi}a - \frac{n\pi}b\right) c^2 t\right)
$$
This solved everything, except the initial value for $\dot u$. A friend of mine proved that if such a seperated function would solve this initial value, the function $w(t)$ would depend on $x$ and $y$.
How do I solve this differential equation so that it satifies the $\dot u$ initial value?

In a more simple way, take $u(x,y,t) = T(t) f(x,y)$ and substitute in your equation. Then
$$
\frac{1}{c^2 T(t)} \frac{d^2 T(t)}{d t^2} = \frac{1}{f(x,y)} \Delta f(x,y)
$$
Where $\Delta$ is the Laplace operator. The only way for this equality to be satisfied for all $(x,y,t)$ is that both sides of it are a constant, lets say $-\lambda$. Then
\begin{align}
\frac{d^2 T}{d t^2} + \lambda c^2 T &= 0\\
\Delta f + \lambda f &= 0
\end{align}
Now, looking at the spatial part, take $f(x,y) = X(x)Y(y)$, then
$$
\frac{1}{X(x)}\frac{d^2 X(x)}{d x^2} + \frac{1}{Y} \frac{d^2 Y(y)}{d y^2} + \lambda = 0
$$
The same argument as above, leads to the two ode's
\begin{align}
\frac{d^2 X}{d x^2} + \lambda_x X &= 0\\
\frac{d^2 Y}{d y^2} + \lambda_y Y &= 0
\end{align}
where $\lambda_x + \lambda_y = \lambda$. The solutions for both equations are
\begin{align}
X(x) &= A \sin(\sqrt{\lambda_x} x)\\
Y(y) &= B \sin(\sqrt{\lambda_y} y)
\end{align}
where we have made use that $u(0,y,t) = u(x,0,t) = 0$. The conditions on the other side of the domain will either lead to the trivial solutions or to a restriction on the values of $\lambda_{x,y}$. Hence
$$
\lambda_x = \frac{m^2 \pi^2}{a^2}, \qquad \lambda_y = \frac{n^2 \pi^2}{b^2}
$$
where $m,\,n\, \in \mathbb{Z}$. Then
$$
f_{nm}(x,y) = A_{mn} \sin\big(\tfrac{m \pi}{a} x\big) \sin\big(\tfrac{n \pi}{b} y\big)
$$
and
$$
T_{mn}(t) = C \sin\Big(\sqrt{\tfrac{m^2 \pi^2}{a^2} + \tfrac{n^2 \pi^2}{b^2}}\, ct\Big) + D \cos\Big(\sqrt{\tfrac{m^2 \pi^2}{a^2} + \tfrac{n^2 \pi^2}{b^2}}\, ct\Big)
$$
Finally, due linearity of the PDE and homogeneous boundary conditions, the solution is
$$
u(x,y,t) = \sum_{m,n = 0}^\infty B_{mn} \sin\big(\tfrac{m \pi}{a} x\big) \sin\big(\tfrac{n \pi}{b} y\big) \sin\Big(\sqrt{\tfrac{m^2 \pi^2}{a^2} + \tfrac{n^2 \pi^2}{b^2}}\, ct\Big)
$$
given that $u(x,y,0) = 0$.
To obtain the value of $B_{nm}$, one have to use the other initial condition and the orthogonality of $\{\sin(\frac{m \pi}{a} x)\}_m$, $\{\sin(\frac{m \pi}{b} y)\}_n$ in the intervals $(0,a)$ and $(0,b)$ respectively. Multiplying by $\sin(\frac{r \pi}{a}x)$ where $r$ is an integer, and integrating from $0$ to $a$, we have
\begin{multline}
\sum_{m,n = 0}^\infty \left(B_{mn} \sqrt{\tfrac{m^2 \pi^2}{a^2} + \tfrac{n^2 \pi^2}{b^2}}\, c\right) \sin\big(\tfrac{n \pi}{b} y\big) \int_0^a \sin\big(\tfrac{m \pi}{a} x\big) \sin\big(\tfrac{r \pi}{a} x\big) dx = \\
(y-b) \int_0^a x(x-a)\sin\big(\tfrac{r \pi}{a} x\big) dx
\end{multline}
which, given orthogonality, leads to
$$
\sum_{n = 0}^\infty \frac{a}{2}\left(B_{rn} \sqrt{\tfrac{r^2 \pi^2}{a^2} + \tfrac{n^2 \pi^2}{b^2}}\, c\right) \sin\big(\tfrac{n \pi}{b} y\big) = 
\frac{2 a^3}{r^3 \pi^3}\left[(-1)^r - 1\right](y-b)
$$
Multiplying by $\sin(\frac{s \pi}{b} y)$, were $s$ is an integer, and integrating from $0$ to $b$, we have
$$
\frac{ab}{4}\left(B_{rs} \sqrt{\tfrac{r^2 \pi^2}{a^2} + \tfrac{s^2 \pi^2}{b^2}}\, c\right) = -\frac{2 a^3 b^2}{r^3 s \pi^4}\left[(-1)^r - 1\right]
$$
hence
$$
B_{mn} =  -\frac{8 a^2 b \left[(-1)^m - 1\right]}{m^3 n \pi^4 c \sqrt{\tfrac{m^2 \pi^2}{a^2} + \tfrac{n^2 \pi^2}{b^2}}}
$$
and the solution is
$$
u(x,y,t) = -\sum_{m,n = 0}^\infty \tfrac{8 a^2 b \left[(-1)^m - 1\right]}{m^3 n \pi^4 c \sqrt{\tfrac{m^2 \pi^2}{a^2} + \tfrac{n^2 \pi^2}{b^2}}} \sin\big(\tfrac{m \pi}{a} x\big) \sin\big(\tfrac{n \pi}{b} y\big) \sin\Big(\sqrt{\tfrac{m^2 \pi^2}{a^2} + \tfrac{n^2 \pi^2}{b^2}}\, ct\Big)
$$
