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Green function Helmholtz differential equation, monodimensional limit

  1. Feb 6, 2013 #1
    The solution of the problem
    [itex]\left(\nabla^2 + k^2 \right)\psi(\mathbf{r})=f(\mathbf{r})[/itex] is, using green function

    [itex] \psi(\mathbf{r})=-\int G(\mathbf{r},\mathbf{r}_1) f(\mathbf{r})[/itex]
    where for the tridimensional case the Green function is
    [itex]G(\mathbf{r},\mathbf{r}_1)=\frac{e^{ik\vert\mathbf{r}-\mathbf{r}_1\vert}}{4\pi \vert \mathbf{r}-\mathbf{r}_1\vert}[/itex]

    Why if I put [itex]f(\mathbf{r})=h(x)\delta(y-y_1)\delta(z-z_1)[/itex] in the second equation I find
    [itex]G(\mathbf{r},\mathbf{r}_1)=\frac{e^{ik\vert x-x_1\vert}}{4\pi \vert \mathbf{x}-\mathbf{x}_1\vert}[/itex]
    that it isn't the correct Green function for the monodimensional case, that is

    [itex]G(\mathbf{r},\mathbf{r}_1)=\frac{i}{2 k}e^{ik\vert x-x_1\vert}[/itex]

    ?Can you help me please?
     
  2. jcsd
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