Solve initial-value problem for heat equation and find relaxation time

1. Dec 17, 2008

mmo115

1. The problem statement, all variables and given/known data
Solve the initial-value problem for the heat equation ut = K$$\nabla$$2u in the column 0< x < L1, 0< y < L2 with the boundary conditions u(0,y;t)=0, ux(L1,y,t)=0, u(x,0;t)=0, uy(x,L2;t)=0 and the initial condition u(x,y;0)=1. Find the relaxation time.

Can anyone please explain how to get this solution.. I really don't understand how to arrive at the solution. I'm hoping that if i can learn specifically how to do this i can apply it to other similar problems. I have searched around the internet for some information, but it seems like most of it is using actual situations as opposed to theoretical. i mean the book doesn't even explain what relaxation time is or how to derive it.. =/

2. Relevant equations
We are told that :

where m,n=1 to infinity
u(x,y;t)= $$\Sigma$$ Bmnsin (m*pi*x / L1) sin (n*pi*y)/L2) * e^-lambdamnKt

We can use this to solve initial-value problems for the heat equation.

3. The attempt at a solution

I really don't get how to solve for the Bmn... the book really doesn't give a good explanation.
The solution is u(x,y;t) = 4/pi2 $$\Sigma$$m,n=1 [ sin[(m-(1/2))($$\pi$$x/L1)] / (m-1/2) ] * [ sin[n-1/2)(pi y/L2)\ / (n-1/2) ] * [ e^-lambdamnKt ]

lambdamn= (m-1/2)2 (pi/L1)2 + (n-1/2)2(pi/L1)2

relaxation time = (4/pi2 K)[L12L22/L12+L22)]