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Fourier COSINE Transform (solving PDE - Laplace Equation)

  1. Feb 8, 2015 #1
    I'm trying to solve Laplace equation using Fourier COSINE Transform (I have to use that), but I don't know if I'm doing everything OK (if I'm doing everything OK, the exercise is wrong and I don't think so).

    NOTE: U(..) is the Fourier Transform of u(..)

    This are the equations (Laplace, boundary, etc.):

    [tex]u_{xx}+u_{yy} = 0, with: y>0, 0<x<a[/tex]
    [tex]u_y(x,0) = u(0, y) =0[/tex]
    [tex]u(a,y) = g(y)[/tex]
    [tex]|u(x,y)|<M[/tex]


    I used "Transform Methods for Solving PDE", from G. Duffy and this is what I'm doing (maybe you have a better way):

    Now, since x is between from 0 to a and y is between 0 and infinity, I use the definition of Fourier COSINE Transform and:
    [tex]\int_{0}^{\infty} u_{xx}.cos(w.y) dy + \int_{0}^{\infty} u_{yy}.cos(w.y) dy = 0[/tex]
    where:
    [tex]\int_{0}^{\infty} u_{xx}.cos(w.y) dy = U_{yy}(x,w)[/tex]
    [tex]\int_{0}^{\infty} u_{yy}.cos(w.y) dy = [u_y(x,y).cos(w.y)] - w.[u(x,y).sin(w.y)] - w.\int_{0}^{\infty}u(x,y).cos(w.y) dy[/tex]

    Note: I don't know how to write the FTC in LaTeX. Where it says [...] it's FTC from 0 to infinity

    Now, I know that:
    [tex][u_y(x,y).cos(w.y)] = 0[/tex] because of the conditions: [tex]u_y(x,0) =0 and |u(x,y)|<M[/tex] (is that ok?)

    But now, I want to solve this: [u(x,y).sen(w.y)] and I don't know why, because I don't have any condition for u(x,0)(but I have a condition for u(0,y).

    What's wrong? I searched everywhere but I couldn't find anything that helps me. Thanks!!!


    Note 2: using Fourier Cosine Transform definition, I know that:
    [tex]\int_{0}^{\infty}u(x,y).cos(w.y) dy = U(x,w)[/tex]
     
    Last edited: Feb 8, 2015
  2. jcsd
  3. Feb 9, 2015 #2

    Svein

    User Avatar
    Science Advisor

    That is the definition of a harmonic function.
     
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