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Homework Help: Separation of variables - Laplace's Equation

  1. May 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Use separation of variables to find the solution to Laplaces equation satisfying the boundary conditions

    u(x,0)=0 (0<x<2)
    u(x,1)=0 (0<x<2)
    u(0,y)=0 (0<y<1)
    u(2,y)= asin2πy(0<y<1)

    3. The attempt at a solution

    I am able to perform the separation of variables technique on the wave equation. The heat equation is a little harder, I struggle a bit, but eventually I get there. Laplace's equation is pretty much impossible. From my understanding, the method is very similar in all three cases, but I think there are some differences which I don't see, which is why I can't do this question.

    So I managed to separate the variables, deriving two ODE's, one in terms of x and one in terms of y, with the separation constant λ.

    F''(x) - λF(x) = 0
    G''(x) - λG(x) = 0

    For the case where λ=0, there are no solutions because nothing can satisfy the last boundary condition listed.

    For the case where λ<0, I think there are no solutions, I could sort of tell by having a look at the final answer given. I don't understand how to show this?

    For the case where λ>0
    I get F(x) = A*cosh(sigma*x) + B*sinh(sigma*y)
    G(y) = (Ccos(sigma*y) + Dsin(sigma*y))

    where sigma is the roots of the ODE's.

    so now u(x,t) = F(x)*G(y)
    I have this function with 5 unknowns, A,B,C,D, and sigma
    When I apply all the boundary conditions, I don't really get anywhere. No helpful information appears.

    What am I doing wrong? Or, what am I not doing?
  2. jcsd
  3. May 21, 2012 #2


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    You have a sign error in one of your ODEs. Also, one of them should have ys in it. It looks like what you meant to write was:
    F''(x) - \lambda F(x) &= 0 \\
    G''(y) + \lambda G(y) &= 0
    \end{align*} Depending on the sign of ##\lambda##, you get regular trig functions for one solution and hyperbolic functions for the other.

    To see why you can't have ##\lambda<0##, try to satisfy the boundary conditions when you have ##G(y) = C\sinh \sigma y + D\cosh \sigma y##. To satisfy the first two boundary conditions, you need G(0)=0 and G(1)=0. You should find the only way you can do this is if C=D=0, which isn't a useful solution.

    If ##\lambda>0##, you have ##G(y) = C\sin \sigma y + D\cos \sigma y##. What do you get from the first two boundary conditions in this case?
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