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Regularising the Green's function in 2D

Physics Asked by Curiosity on December 29, 2020

The Green’s function for the 2D Helmholtz equation satisfies the following equation:

$$(nabla^2+k_0^2+mathrm{i}eta),{mathsf{G}}_{2mathrm{D}}(mathbf{r}-mathbf{r}’,k_o)=delta^{(2)}(mathbf{r}-mathbf{r}’).$$

By Fourier transforming the Green’s function and using the plane wave representation for the Dirac-delta function, it is fairly easy to show (using basic contour integration) that the 2D Green’s function is given by

$${mathsf{G}}_{2mathrm{D}}(mathbf{r}-mathbf{r}’,k_0)=displaystylelim_{etato0}intfrac{mathrm{d}^2mathbf{k}}{(2pi)^2}frac{mathrm{e}^{mathrm{i}mathbf{k}cdot(mathbf{r}-mathbf{r}’)}}{k_0^2+mathrm{i}eta-k^2}=frac{1}{4mathrm{i}}operatorname{H}_0^{(1)}left(k_0|mathbf{r}-mathbf{r}’|right)$$

where $operatorname{H}_0^{(1)}$ is the Hankel function of zeroth order and first kind.

However, this 2D Green’s function diverges (logarithmically) at $mathbf{r}=mathbf{r}’$. Therefore, if we want it to be well-defined for $mathbf{r}=mathbf{r}’$, one can introduce a Gaussian cut-off function like so

$$tilde{{mathsf{G}}}_{2mathrm{D}}(mathbf{r}-mathbf{r}’,k_0)=displaystylelim_{etato0}intfrac{mathrm{d}^2mathbf{k}}{(2pi)^2}frac{mathrm{e}^{mathrm{i}mathbf{k}cdot(mathbf{r}-mathbf{r}’)}}{k_0^2+mathrm{i}eta-k^2}mathrm{e}^{-frac{a^2k^2}{2}}$$

where $a$ is some cut-off parameter.

Question: How do you evaluate this integral?

One Answer

The poles are still in the same place so what is stopping you using residue theorem?

Answered by lux on December 29, 2020

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