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## Homework Statement

Solve the initial-value problem for the heat equation u

_{t}= K[tex]\nabla[/tex]

^{2}u in the column 0< x < L

_{1}, 0< y < L

_{2}with the boundary conditions u(0,y;t)=0, u

_{x}(L

_{1},y,t)=0, u(x,0;t)=0, u

_{y}(x,L

_{2};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.. =/

## Homework Equations

We are told that :

where m,n=1 to infinity

u(x,y;t)= [tex]\Sigma[/tex] B

_{mn}sin (m*pi*x / L

_{1}) sin (n*pi*y)/L

_{2}) * e^-lambda

_{mn}Kt

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

## The Attempt at a Solution

I really don't get how to solve for the B

_{mn}... the book really doesn't give a good explanation.

The solution is u(x,y;t) = 4/pi

^{2}[tex]\Sigma[/tex]

_{m,n=1}[ sin[(m-(1/2))([tex]\pi[/tex]x/L

_{1})] / (m-1/2) ] * [ sin[n-1/2)(pi y/L

_{2})\ / (n-1/2) ] * [ e^-lambda

_{mn}Kt ]

lambda

_{mn}= (m-1/2)

^{2}(pi/L

_{1})

^{2}+ (n-1/2)

^{2}(pi/L

_{1})

^{2}

relaxation time = (4/pi

^{2}K)[L1

^{2}L2

^{2}/L1

^{2}+L2

^{2})]