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

  1. Dec 17, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve the initial-value problem for the heat equation ut = K[tex]\nabla[/tex]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)= [tex]\Sigma[/tex] 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 [tex]\Sigma[/tex]m,n=1 [ sin[(m-(1/2))([tex]\pi[/tex]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)]
  2. jcsd
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