This is a weird problem of sorts in the text (Partial Differential Equations, Asmar), but it's one of those things where I can see sort of see where the answer came from but I am trying to follow the steps.
Here's the start: you have the wave equation: $$\frac{\partial^2 u}{\partial t^2} = c^2\left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right), 0<x<a, 0<y<b, t>0$$ and the boundary conditions, $u(0,y,t) = u(a,y,t) =0$ for $0<y<b$ and $t>0$ and $u(x,0,t) = u(x,b,t) = 0$ for $0<x<a$ and $t>0$.
and you have the initial conditions $u(x,y,0)= f(x,y)$. $\frac{\partial u}{\partial t}(x,y,0)=g(x,y)$
The problem given is $a=b=1$ and $f(x,y) = \sin \pi x \sin \pi y$ and $g(x,y) = \sin \pi x$
So far good this shoudn't be so hard. Yet: when I try to come up with the solution I get coefficients that go to zero.
The general soution is adouble sum, $$(x,y,t) = \sum^{\infty}_{n=1}\sum^{\infty}_{m=1}(A_{mn}cos(\lambda_{mn}t)+B_{mn}\sin\lambda_{mn}t)\sin\frac{m\pi}{a}x \sin\frac{n\pi}{b}y$$
where $A_{mn} = \frac{4}{ab} \int^b_0\int^a_0 f(x,y) sin\frac{m\pi}{a}x \sin\frac{n\pi}{b}y$
When I insert $f(x,y)$ I get: $\frac{4}{ab} \int^b_0\int^a_0 \sin \pi x \sin \pi y \sin(\frac{m\pi}{a}x) \sin (\frac{n\pi}{b}y) \, dx dy$
and plugging in the numbers a and b and gathering terms
$4 \int^1_0\int^1_0 [\sin \pi x \sin(m \pi x)] [\sin \pi y \sin (n\pi y)] \, dx dy$
nd i can use a trig identity here
$2 \int^1_0\int^1_0 [\cos (\pi x-m \pi x)+ \cos (\pi x+m \pi x)] [\sin \pi y \sin (n\pi y)] \, dx dy$
And this is where it gets weird. When I integrate the first part (the x's) I get:
$2 \int^1_0 [\frac{\sin (\pi x-m \pi x)}{\pi-m\pi}+ \frac{\sin (\pi x+m \pi x)}{\pi +m\pi}] [\sin \pi y \sin (n\pi y)] \, dy$
and from 0 to 1 that gos to zero and thus the whole integral is zero. If I treat m=1 as a special case and plug that in before I do the integral I STILL get zero. And of course if I plug in any other values (integers) for m the cosine terms go to zero.
The solution sheet says that I should get $A_{mn} = \delta_{m,1} \, \delta_{n,1}$. Ans that is where I am a little confused and maybe it's just me not reading the notation right. Once I move on to th eB coefficient things are a little better, I get a result of $2 \int^1_0 \int^1_0 [\cos (\pi x-m \pi x) \cos (\pi x+m \pi x)[ \sin (n\pi y)] \, dx dy$
which seems better behaved in that for n=odd I get the cosines adding up to 2, which integrates into something I can work with.
But again, I am jut trying to walk through this because I am told that the A coefficient hangs around for at least one value of m. maybe I am just missing something obvious.
thanks in advance.
The answer in the book is indeed correct. For $m=1$, we are integrating $\int_0^1\sin^2(\pi x)$, which is $\frac{1}{2}$. Doing this again with the $y$ terms cancels the 4 in front of the integral. With the other terms $0$ as you showed, we get $\delta_{m1}\delta_{n1}$ as the answer.