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PDE semi-infinite strip

  1. Sep 17, 2006 #1
    Hi can someone please help me work through the following question. It is the two dimensional Laplace equation in a semi-infinite strip.

    [tex]
    \frac{{\partial ^2 u}}{{\partial x^2 }} + \frac{{\partial ^2 u}}{{\partial y^2 }} = 0,0 < x < a,0 < y < \infty
    [/tex]

    The boundary conditions along the edges are u(0,y) = 0, u(a,y) = 0 and u(x,0) = f(x) where f(x) is some prescribed function.

    With my limited experience in dealing with PDEs at first glance I don't think there is enough information to even get to an expression for u(x,y) in terms of f(x) but I'll try this anyway.

    Let [tex]u\left( {x,y} \right) = X\left( x \right)Y\left( y \right)[/tex] then

    [tex]
    \frac{{X''\left( x \right)}}{{X\left( x \right)}} = - \frac{{Y''\left( y \right)}}{{Y\left( y \right)}} = - \lambda
    [/tex]

    I have set the separation constant to be negative lambda because the boundary conditions suggest that I'll first need to work with the ODE in X(x).

    The ODE in X(x) I get is

    [tex]X''\left( x \right) + \lambda X\left( x \right) = 0[/tex] with X(0) = 0, X(a) = 0.

    Dirichlet BCs so lambda is positive which gives me

    [tex]
    X\left( x \right) = A\cos \left( {\sqrt \lambda x} \right) + B\sin \left( {\sqrt \lambda x} \right)
    [/tex]

    The first BC implies A = 0 and avoiding trivial solutions the other BC yields the eigenvalues:

    [tex]
    \lambda _n = \left( {\frac{{n\pi }}{a}} \right)^2
    [/tex] where n is a natural number.

    The corresponding eigenfunctions are [tex]X_n \left( x \right) = B\sin \left( {\frac{{n\pi x}}{a}} \right)[/tex]

    The ODE in Y(y) is:

    [tex]
    Y''\left( y \right) - \lambda _n Y\left( y \right) = 0
    [/tex]

    Since lambda is positive [tex]Y_n \left( y \right) = C\cosh \left( {\sqrt {\lambda _n } y} \right) + D\sinh \left( {\sqrt {\lambda _n } y} \right)[/tex].

    [tex]
    u_n \left( {x,y} \right) = X_n \left( x \right)Y_n \left( y \right)
    [/tex]

    [tex]
    u_n \left( {x,y} \right) = \left[ {a_n \cosh \left( {\sqrt {\lambda _n } y} \right) + b_n \sinh \left( {\sqrt {\lambda _n } y} \right)} \right]\sin \left( {\frac{{n\pi x}}{a}} \right)
    [/tex]

    [tex]
    u\left( {x,y} \right) = \sum\limits_{n = 1}^\infty {\left[ {a_n \cosh \left( {\sqrt {\lambda _n } y} \right) + b_n \sinh \left( {\sqrt {\lambda _n } y} \right)} \right]\sin \left( {\frac{{n\pi x}}{a}} \right)}
    [/tex]

    The only other bit of information I have left is u(x,0) = f(x) so I try that to see what happens. But that just tells me that

    [tex]
    f\left( x \right) = \sum\limits_{n = 1}^\infty {a_n \sin \left( {\frac{{n\pi x}}{a}} \right)}
    [/tex]

    [tex]
    a_n = \frac{2}{a}\int\limits_0^a {f\left( x \right)} \sin \left( {\frac{{n\pi x}}{a}} \right)dx
    [/tex]

    So that gives me a_n but what about b_n? I really don't know how to proceed. Can someone please help me out? Thanks.
     
  2. jcsd
  3. Sep 17, 2006 #2

    StatusX

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    Homework Helper

    You need to specify the B.C.'s as y->infinity. Probably they want the function to vanish, in which case you don't want to use sinh y and cosh y but instead e^y and e^-y, and only the e^-y terms will survive.
     
    Last edited: Sep 17, 2006
  4. Sep 17, 2006 #3
    Yeah I think the BC for y going to infinity is missing but I think the condition is usually as you said, u tends to zero and y gets large. In that case I can rewrite the sinh and cosh in terms of exponentials and simply set the appropriate constants to be equal to zero. Thanks for the pointers.
     
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