Solve laplace's equation on semi infinite strip

  1. [itex]\nabla[/itex]^2(Z)=0

    Z= 0 for x=0, y=0
    Z= x(1-x) for y=0
    Z=0 for y=infinity

    Range 0<x<1 and y>0 (suppose strictly speaking should be x=1 and x=0 too)

    So all I want to do is solve this

    Use separation of variables:

    X''/X = a^2 = -Y''/Y

    Gives X = Aexp(ax) + Bexp(-ax) and Y=Ccos(ay) + Dsin(ay)

    Or completely free to swap these around to give
    Y=Aexp(ay)+ Bexp(-ay) and X= Ccos(ax) +Dsin(ax)
    which I shall do as get further with boundary conditions

    Know that Z=XY

    As at y=infinity, Z=0 ==> A=0
    At x=0, Z=0 ==>C = 0
    At x=1, Z=0 ==> a=n*pi where n is an integer

    so have Z=Esin(n*pi*x)exp(-n*pi*y) where E is a new constant

    but how on earth do I make this compatible with the remaining boundary condition for y=0

    ==> Z=x(1-x) = Esin(n*pi*x) ??????

    Clearly must have gone wrong somewhere???
  2. jcsd
  3. kai_sikorski

    kai_sikorski 162
    Gold Member

    The next step is very standard in solving PDEs by separation of variables, any book/article on the topic will explain it. You have to expand [itex]x(x-1)[/itex] as a series using [itex] \sin(\pi n x)[/itex]. You should find that

    [tex] x(x-1) = \sum_{n=1}^\infty \frac{4 \left(-1+(-1)^n\right)}{n^3 \pi ^3}\sin(\pi n x)[/tex]
    Last edited: Feb 15, 2012
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