I'm stuck on the following problem.
Let $u(x, y, t)$ denote a solution to the linear wave equation $k^2(u_{xx}+u_{yy}) = u_{tt}$ with $k = 2$ on a square domain with corners at (0, 0), (0, 1), (1, 0) and (1, 1) such that u ≡ 0 on the boundary of the domain. Write down $u(x, y, t)$ if
i)
$u(x,y,0) = \sin(5\pi x)\sin(5 \pi y)$
$\frac{\partial u}{\partial t}(x,y,0)= \sin(7 \pi x)\sin(\pi y)$
ii)
$u(x,y,0) = \sin(5\pi x)\sin(5 \pi y) + 3\sin(7\pi x)\sin(\pi y)$
$\frac{\partial u}{\partial t}(x,y,0)= 0$
I'll start off with i). I know the solution is of the form
$u(x,y,t) = \sum \sin(\frac{m\pi x}{a}) \sin(\frac{n\pi y}{b})(A_{mn}\cos(\lambda t) + B_{mn}\sin(\lambda t))$
with
$A_{mn} = \frac{4}{ab}\int_0^b \int_0^a u(x,y,0)\sin(\frac{m\pi x}{a})\sin(\frac{n\pi y}{b})dxdy$
and a similar integral for $B_{mn}$. So I put $a=1$ and $b=1$ in and try to find $A$ and $B$. However, when I evaluate the integrals using maple, end up with solutions like
$A_{mn} = \frac{100\sin(m\pi)\sin(n\pi)}{\pi^2(m^2 n^2-25m^2-25n^2+625)}$
$B_{mn} = \frac{28sin(m\pi)sin(n\pi)}{\pi^2(m^2n^2-m^2-49n^2+49)}$
but this seems wrong as this just tells me that $A$ and $B$ are zero for any $m,n$... When I try ii) I get a similar answer - $A$ and $B$ are zero.
have I gone wrong anywhere? Any advice would be greatly appreciated!
First: $2 = k = 2\pi/\lambda \Rightarrow \lambda = \pi$
i) Assuming $$ u(x,y,t) = \sin(5\pi x)\sin(5 \pi y)\cos(\sqrt{200}\pi t) + (1/\sqrt{200}\pi) \sin(7 \pi x)\sin(\pi y)\sin(\sqrt{200}\pi t) $$
we get $$ u_t(x,y,t) = -\sqrt{200}\pi\sin(5\pi x)\sin(5\pi y)\sin(\sqrt{200}\pi t) + \sin(7\pi x)\sin(\pi y)\cos(\sqrt{200}\pi t) $$
and \begin{align} u(x,y,0) &= \sin(5\pi x)\sin(5 \pi y) \\ u_t(x,y,0) &= \sin(7 \pi x)\sin(\pi y) \end{align}
Checking the PDE: \begin{align} u_x(x,y,t) &= 5\pi\cos(5\pi x)\sin(5\pi y)\cos(\sqrt{200}\pi t) + \frac{7}{\sqrt{200}} \cos(7\pi x)\sin(\pi y)\sin(\sqrt{200}\pi t) \\ u_y(x,y,t) &= 5\pi \sin(5\pi x)\cos(5\pi y)\cos(\sqrt{200}\pi t) + \frac{1}{\sqrt{200}} \sin(7\pi x)\cos(\pi y)\sin(\sqrt{200}\pi t) \\ u_{xx}(x,y,t) &= -25\pi^2\sin(5\pi x)\sin(5\pi y)\cos(\sqrt{200}\pi t) - \frac{49\pi}{\sqrt{200}} \sin(7\pi x)\sin(\pi y)\sin(\sqrt{200}\pi t) \\ u_{yy}(x,y,t) &= -25\pi^2 \sin(5\pi x)\sin(5\pi y)\cos(\sqrt{200}\pi t) - \frac{\pi}{\sqrt{200}} \sin(7\pi x)\sin(\pi y)\sin(\sqrt{200}\pi t) \\ u_{tt}(x,y,t) &= -200\pi^2\sin(5\pi x)\sin(5\pi y)\cos(\sqrt{200}\pi t) - \sqrt{200} \pi\sin(7\pi x)\sin(\pi y)\sin(\sqrt{200}\pi t) \end{align}
Then \begin{align} 4(u_{xx} + u_{yy}) &= -200\pi^2\sin(5\pi x)\sin(5\pi y)\cos(\sqrt{200}\pi t) - \frac{200\pi}{\sqrt{200}}\sin(7\pi x)\sin(\pi y)\sin(\sqrt{200}\pi t) \\ &= -200\pi^2\sin(5\pi x)\sin(5\pi y)\cos(\sqrt{200}\pi t)- \sqrt{200}\pi \sin(7\pi x)\sin(\pi y)\sin(\sqrt{200}\pi t) \\ &= u_{tt} \end{align}
I used the Maxima system to verify this, because I had to revise this a couple of times due to errors I made.
ii) Using the ansatz $$ u(x,y,t) = \sin(5\pi x)\sin(5\pi y)\cos(L\pi t) + 3 \sin(7\pi x)\sin(\pi y)\cos(M\pi t) $$
gives $$ u_t(x,y,t) = -L\pi \sin(5\pi x)\sin(5\pi y)\sin(L\pi t) - 3 M\pi \sin(7\pi x)\sin(\pi y)\sin(M\pi t) $$
and one gets \begin{align} u(x,y,0) &= \sin(5\pi x)\sin(5\pi y) + 3 \sin(7\pi x)\sin(\pi y) \\ u_t(x,y,0) &= 0 \end{align} Second order partial derivatives are: \begin{align} u_{xx}(x,y,t) &= -147\pi^2 \sin(7\pi x)\sin(\pi y)\cos(M\pi t)- 25\pi^2 \sin(5\pi x)\sin(5\pi y)\cos(L\pi t) \\ u_{yy}(x,y,t) &= -3\pi^2 \sin(7\pi x)\sin(\pi y)\cos(M\pi t)- 25\pi^2 \sin(5\pi x)\sin(5\pi y)\cos(L\pi t) \\ u_{tt}(x,y,t) &= -3M^2\pi^2 \sin(7\pi x)\sin(\pi y)\cos(M\pi t)- L^2\pi^2 \sin(5\pi x)\sin(5\pi y)\cos(L\pi t) \\ \end{align}
Comparing both sides of the PDE \begin{align} 4(u_{xx}+u_{yy}) &= -600\pi^2\sin(7\pi x)\sin(\pi y)\cos(M\pi t)- 200\pi^2\sin(5\pi x)\sin(5\pi y)\cos(L\pi t) \\ u_{tt}(x,y,t) &= -3M^2\pi^2 \sin(7\pi x)\sin(\pi y)\cos(M\pi t)- L^2\pi^2 \sin(5\pi x)\sin(5\pi y)\cos(L\pi t) \\ \end{align} gives the conditions $600=3 M^2$ and $200 = L^2$, which means $M = L = \sqrt{200}$ and the solution $$ u(x,y,t) = \left[\sin(5\pi x)\sin(5\pi y) + 3 \sin(7\pi x)\sin(\pi y)\right] \cos(\sqrt{200}\pi t) $$