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Helmholtz Equation in a Square

  1. Jun 1, 2004 #1
    I'm solving a Helmholtz equation uxx+uyy+lambda*u=0 in a rectangle: 0<=x<=L, 0<=y<=H with the following boundary conditions:
    u(x,0)=u(x,H)=0 and ux(0,y)=ux(L,y)=0

    I found the eigenvalues to be:
    lambda(nm)=(n Pi/L)^2+(m Pi/H)^2
    and the eigenfunctions to be:
    u(nm)=Cos(n Pi x/L)*Sin(m Pi y/H)

    Now the question I'm stuck on is to show that if L=H (a square) then most eigenvalues have more than one eigenfunction
    and, Are any two eigenfunctions of this eigenvalue problem orthogonal in a two-dimensional sense?

    Any help would be greatly appreciated.
  2. jcsd
  3. Jun 2, 2004 #2

    Dr Transport

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    The eigenfunctions should be orthogonal, and if properly normalized, orthonormal. If more than one eigenfunction gives the same eigenvalue, there is degeneracy.
  4. Jun 2, 2004 #3

    matt grime

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    if L=H then lambda_(n,m)=lambda_(m,n), that is u_(n,m) and u_(m,n) have the same eigenvalue.
  5. Jun 2, 2004 #4
    Still not clear

    Thank you for the replies, but how do I show the degeneracy? Also, how do I show in the case of a square (L=H) wheather the eigenfunctions are orthogonal or not in a two-dimensional sense?
  6. Jun 2, 2004 #5

    Dr Transport

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    If the eigenvalues are the same, the system is degenerate. Now, integrate the functions, i.e.

    [tex] \int^{H} _{0}\int^{H} _{0} u_{i}(x,y) u_{j}(x,y) dxdy = \delta_{i j} [/tex]

    where i,j are the different functions. This should show that the functions are orthogonal, or if correctly normalized, orthonormal.
  7. Nov 5, 2009 #6

    The problem you are posing is quite trivial and pointless because you seem to have no driving wave or source of any type. Your boundary conditions are all homogenous and so is your equation. This means that your solution is the trivial solution u(x,y)=0. You need to do something to either your boundary condition or the RHS of your Helmholtz equation.
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