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?

Let $u(x,y,t)=\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C(m,n,t)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}$ so that it automatically satisfies $u(0,y,t)=u(a,y,t)=u(x,0,t)=u(x,b,t)=0$ ,
Then $\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty\dfrac{m^2\pi^2}{a^2}C(m,n,t)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}+\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty\dfrac{n^2\pi^2}{b^2}C(m,n,t)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}+\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty\dfrac{1}{c^2}\dfrac{\partial^2C(m,n,t)}{\partial t^2}\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}=0$
$\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty\left(\dfrac{\partial^2C(m,n,t)}{\partial t^2}+\left(\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}\right)c^2\pi^2C(m,n,t)\right)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}=0$
$\therefore\dfrac{\partial^2C(m,n,t)}{\partial t^2}+\left(\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}\right)c^2\pi^2C(m,n,t)=0$
$C(m,n,t)=C_1(m,n)\sin\left(\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi t\right)+C_2(m,n)\cos\left(\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi t\right)$
$\therefore u(x,y,t)=\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C_1(m,n)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}\sin\left(\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi t\right)$
$+\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C_2(m,n)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}\cos\left(\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi t\right)$
$u(x,y,0)=0$ :
$\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C_2(m,n)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}=0$
$C_2(m,n)=0$
$\therefore u(x,y,t)=\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C_1(m,n)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}\sin\left(\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi t\right)$
$u_t(x,y,t)=\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C_1(m,n)\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}\cos\left(\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi t\right)$
$u_t(x,y,0)=x(x-a)(y-b)$ :
$\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C_1(m,n)\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}=x(x-a)(y-b)$
You need to handle extremely complicated double kernel inversion.
