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Green function Helmholtz differential equation, monodimensional limit 
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#1
Feb613, 06:32 AM

P: 8

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(yy_1)\delta(zz_1)[/itex] in the second equation I find [itex]G(\mathbf{r},\mathbf{r}_1)=\frac{e^{ik\vert xx_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 xx_1\vert}[/itex] ?Can you help me please? 


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