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Laplace equation on a semi infinite slab

  1. Aug 2, 2010 #1
    1. The problem statement, all variables and given/known data
    Could someone check my work plaese.
    [tex]\frac{\partial^2u}{\partial x^2}(x,y)+\frac{\partial^2u}{\partial y^2}(x,y)=0[/tex]

    [tex](0<x<1, 0<y)[/tex]

    [tex]\frac{\partial u}{\partial x}(0,y)=\frac{\partial u}{\partial y}(1,y)=0[/tex]

    [tex]u(x,y)\rightarrow k[/tex] as [tex]y\rightarrow\infty[/tex]

    [tex]u(x,0)=f(x)[/tex] [tex](0\leqx\leq1)[/tex]

    2. Relevant equations
    [tex]\ddot{X}-\mu X=0[/tex] and [tex]\ddot{Y}+\mu Y=0[/tex], [tex]\mu = k^2[/tex] and [tex]k=\pi r[\tex]

    [tex]X(x)=A \cos(kx)+B \sin(kx)[/tex] so [tex]\dot{X}=-Ak\sin(kx)+Bk\cos(kx)[/tex]

    With the boundary conditions we have [tex]X_r(x)=A_r \cos(r\pi x)[/tex]

    [tex]\ddot{Y}+\mu Y=0[/tex] gives [tex]Y(y)=Ce^{r\pi y}+De^{-r\pi y}[/tex] with the boundary conditions and setting C=0, [tex]Y(y)=De^{-r\pi y}[/tex]

    3. The attempt at a solution

    So we have [tex]u(x,y)=A_0+\sum^\infty _{r=1} {A_r\cos(r\pi x)e^{-r\pi y}}[/tex]

    [tex]u(x,0)=f(x)=A_0+\sum^\infty _{r=1} {A_r\cos(r\pi x)[/tex]

    [tex]A_0 = \int^1 _0 \cos(r\pi x) dx[/tex] Here r=0 so [tex]A_0 = 1[/tex]

    [tex]A_r =2 \int^1 _0 \cos(r\pi x)\cos(r\pi x) dx[/tex] Here I get all A coefficients as 1, is this right?

    Last edited: Aug 2, 2010
  2. jcsd
  3. Aug 16, 2010 #2
    Any ideas anyone?
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