Have you done separation of variables yet, dubmdell? I would do something like the following. I hope you can follow LaTeX-y notation. I'm going to be sloppy and skip justifications. You can look stuff up about the details.
Assume u(t,x,y) = T(t)X(x)Y(y). (I'm assuming it's a product of three one-variable functions, hence separation of variables).
So u_{tt} = T''(t)X(x)Y(y), u_{xx} = T(t)X''(x)Y(y), and u_{yy} = T(t)X(x)Y''(y). The equation then reads
T''(t)X(x)Y(y) = c^2(T(t)X''(x)Y(y) + T(t)X(x)Y''(y))
Divide through by TXY to get
T''(t) / T(t) = c^2(X''(x) / X(x) + Y''(y)/Y(y))
The argument now goes: here you have an equality in different variables that is supposed to be true for all t,x, and y. This can only happen if each side is constant. I'm not going to go through all the details, but suffice to say that, with these boundary conditions, we're going to want to take the constant to be negative, so call it -k^2. (You can try taking it positive and going through the remaining steps, and you'll see that you come to grief).
So we have, as it's usually written,
T''(t) / T(t) = -c^2k^2, or
T''(t) = -c^2k^2T(t).
This is a second-order ODE. The two solutions are T(t) = B sin(ckt) and A cos(ckt), for constants A and B.
Since we want u_t(0) = 0, that tells us that T'(0) = 0, so B = 0, so we just have T(t) = A cos(ckt) for some A to be determined.
Now, each of the terms on the right-hand side has to be a constant, too. Let's say -a^2 and -b^2, where a^2 + b^2 = k^2. So we have
X''(t) = -a^2X(t), so X(t) = B cos(at) + C sin(at), and similarly Y(t) = D cos(bt) + E sin(bt).
We're starting to get lots of undetermined constants. Let's try to clean those up. We know u = 0 on the boundary for all times. That means X should be 0 at 0 and 1, and Y should be 0 at 0 and 1. Cosine is never going to be zero at 0, so we can scrap both cosine terms, i.e., set B = D = 0. We're left with X(x) = C sin(at) and Y(y) = D sin(at). Since we have X(1) = 0, that tells us sin(at) = 0, or a = n \pi for some nonzero integer n. Ditto for b -- say b = m \pi for some integer m.
That means k = \pi \sqrt(n^2 + m^2) = \pi \sqrt( (remember we had k^2 = a^2 + b^2). So our general solution is
u(t,x,y) = A cos(c\pi \sqrt(n^2 + m^2)t) sin(n\pi x) sin (m \pi y). Great! But wait -- we still have these undetermined numbers A, m, and n. What gives? Well, here's the amazing step, of course, and the one where we use our last boundary condition. It's pretty time-consuming, yet amazing, so I'm just going to sketch it out.
This is a linear equation, so the sum of solutions is a solution. It turns out we can write any solution as
\sum_{m,n = 1}^{\infty} A(m,n) cos(c\pi \sqrt(n^2 + m^2)t) sin(n\pi x) sin (m \pi y).
(I can leave off negative m, n, because sine is odd).
All that's left is to figure out the coefficients A(m,n)s! To do this, use the amazing fact that the double integral over R of sin(n\pi x)sin(m \pi y) sin(n' \pi x) sin (m' \pi y) is zero unless m' = m and n' = n. So multiply both sides of the above equation (the infinite sum one) by sin(n' \pi x) sin(m' \pi x), set t = 0 (which sets the cosines all to 1), and integrate over k. That kills all the terms in the sum except the (m', n') one on the right. So we might as well rename it (m,n) for convenience. You're left with
the integral over R of -16x(x - 1/2)y(y - 1/2) sin(n\pi x)sin(m \pi y) = A(m,n) times the integral over R of sin^2(n\pi x) sin^2 (m \pi y).
Perform the integrations (which could take half an hour :-)) and you'll have A(m,n)! You can then just plug that into the series above, and you have your solution written as an infinite series of sines and cosines.
I hope this was helpful at all. Probably a little confusing. I suggest you check out Chapter 6, and especially Example 6.4, in Vretblad's "Fourier Analysis and Its Applications." There's also a lot on the internet about separation of variables problems.
He also does equations with the Fourier transform in Chapter 7, if it's important to use that. (Is it, by the way? Maybe I should have asked. :-P).
)
Has Anyone Noticed Abge...
)
)
)