Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Solving a 2nd order ODE using Green's Function

  1. Feb 12, 2007 #1
    1. The problem statement, all variables and given/known data
    The homogeneous Helmholtz equation

    [tex]\bigtriangledown^2\psi+\lambda^2\psi=0[/tex]

    has eigenvalues [tex]\lambda^2_i[/tex] and eigenfunctions [tex]\psi_i[/tex]. Show that the corresponding Green's function that satisfies

    [tex]\bigtriangledown^2 G(\vec{r}_1, \vec{r}_2)+\lambda^2 G(\vec{r}_1, \vec{r}_2)=-\delta(\vec{r}_1-\vec{r}_2)[/tex]

    may be written as

    [tex]G(\vec{r}_1, \vec{r}_2)=\sum_{i=1}^{\infty}\frac{\psi_i(\vec{r}_1)\psi_i(\vec{r}_2)}{\lambda^2_i-\lambda^2}[/tex]


    2. Relevant equations

    [tex]\int(\psi \bigtriangledown^2 G-G\bigtriangledown^2 \psi) d\tau_2=\int(\psi \bigtriangledown G-G\bigtriangledown\psi) d\sigma[/tex]

    3. The attempt at a solution

    I'm using Arfken's Mathematical methods for physicists, and it isn't very good at explaining the examples it uses. I just need some kind of jump start to get me going. Do I need to use the equation in the relevant equations section?
     
  2. jcsd
  3. Feb 13, 2007 #2

    HallsofIvy

    User Avatar
    Science Advisor

    You might want to start by writing out the definition of "Green's function"- that's far more important than examples.

    Green's function for a linear, nonhomogeneous, differential equation, L(Y)=f(x), where L( ) is a linear differential operator, with given boundary conditions, is a function G(r, r') such that
    1) L(G(r,r'))= 0 for all r not equal to r'
    2) limit as r->r' from the right of L(G(r, r')) minus the limit as r->r' from the left of L(G(r,r'))= 1.
    3) G(r, r') satisfies the boundary conditions.
    2)
     
  4. Feb 13, 2007 #3
    How do I know the boundary conditions? The problem doesn't state any. Do I just assume there aren't any?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook