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Need help with solving PDE

  1. Sep 24, 2013 #1
    I'm new here, hope it is the right place to ask the question.

    The PDE question is ∂2u/∂x2+∂2u/∂y2=0
    and u(x,0)=f(x), u(x,1)=0, u(0,y)=0, u(1,y)=0.

    I use the method of separate the variables, with is let u(x,y)=X(x)Y(y) and get X''/X+Y''/Y=0.

    Then let X''/X=-Y''/Y=-λ, i.e. X''+λX=0, and Y''-λY=0.

    Since u(x,0)=f(x)=X(x)Y(0), u(x,1)=0=X(x)Y(1), u(0,y)=0=X(0)Y(y), u(1,y)=0=X(1)Y(y), we can conclude that X(0)=0, X(1)=0, Y(0)=f(x), Y(1)=0.

    For X''+λX=0, we have 3 cases

    1) λ=0, i.e. X''=0, i.e. X=ax+b, since X(0)=0, X(1)=0, a=b=0, which is X(x) always 0. not wavelike.

    2) λ<0, Let λ=-μ2, i.e. X''-μ2X=0, General solution is X=a1eμx+a2e-μx. Since X(0)=0, X(1)=0, a1=a2=0, which is X(x) always 0. not wavelike.

    3) λ>0, Let λ=μ2, i.e. X''+μ2X=0, General solution is X=a1cos(μx)+a2sin(μx). Since X(0)=0, X(1)=0, a2sinμ=0.
    In order to get a nontrivial solution, let μ=n∏, then X(x)=sin(n∏x).
    Then we consider Y''-λY=0, i.e. Y''-n22Y=0.
    The general solution is Y=b1en∏y+b2e-n∏y.
    Then since Y(1)=0, b1en∏+b2e-n∏=0.

    Now, here is my question.
    Why b1+b2e-2n∏=0?

    and How it change in the black font? Yh(y)=-b2e-2n∏en∏y+b2e-n∏y
  2. jcsd
  3. Sep 24, 2013 #2


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    Homework Helper

    Divide both sides of the previous equation by [itex]e^{n\pi}[/itex].

    I think that's a misprint for [itex]2b_2'\sinh(n\pi(1-y))[/itex].

    Remember that
    [tex]\sinh(x) \equiv \frac{e^x - e^{-x}}{2}.[/tex]
  4. Sep 24, 2013 #3
    Got it! Thank you very much~~!!!
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