Greens Function for Hemmholtz using Fourier

[ognik]

I've gotten myself mixed up here , appreciate some insights ...

Using Fourier Transforms, shows that Greens function satisfying the nonhomogeneous Helmholtz eqtn $ \left(\nabla ^2 +k_0^2 \right) G(\vec{r_1},\vec{r_2})= -\delta (\vec{r_1} -\vec{r_2}) $ is $ G(\vec{r_1},\vec{r_2})= \frac{1}{{2\pi}^{3}} \int \frac{e^{i\vec{k}.(\vec{r_1} -\vec{r_2})}}{k^2 - k_0^2} d^3k $

My attempt is: $ \left(\nabla ^2 +k_0^2 \right) G(\vec{r_1},\vec{r_2})= -\delta (\vec{r_1} -\vec{r_2}) $

Taking the Fourier Transform of the LHS, $F\left[ \nabla^2G+k_0^2 G \right] = (- k^2 +k_0^2) \hat{u} $
RHS: $F\left[ -\delta (\vec{r_1} -\vec{r_2}) \right] = - \int 1 e^{-i\vec{k}.(\vec{r_1} -\vec{r_2})} d^3r $

$ \therefore \hat{u} = \int \frac{1}{(k^2 - k_0^2)} e^{i\vec{k}.(\vec{r_1} -\vec{r_2})} d^3r = \frac{1}{ik(k^2 - k_0^2)} e^{i\vec{k}.(\vec{r_1} -\vec{r_2})}$

$ \therefore G(\vec{r_1},\vec{r_2})= F^{-1} \left[\frac{1}{ik(k^2 - k_0^2)} e^{i\vec{k}.(\vec{r_1} -\vec{r_2})} \right] = \frac{1}{{(2\pi)}^{3}} \int_{-\infty}^{\infty}\frac{1}{ik(k^2 - k_0^2)} e^{i\vec{k}.(\vec{r_1} -\vec{r_2})} e^{ik.r}d^3k$

 

[Jhoneine]

$= \frac{1}{{2\pi}^{3}} \int \frac{e^{i\vec{k}.(\vec{r_1} -\vec{r_2})}}{k^2 - k_0^2} d^3k $ Is this correct? Yes, your answer is correct.

 

[math771]

First of all, great job on attempting to solve the problem! It's always good to see someone putting in the effort to understand a concept.

Your approach seems to be correct, but there are a few minor mistakes that I noticed. Firstly, in the first step, the Fourier Transform of the LHS should be $(-k^2-k_0^2)\hat{G}$, not just $(-k^2+k_0^2)\hat{u}$. Also, when taking the inverse Fourier Transform, the integration should be over all values of $k$ (from $-\infty$ to $\infty$), not just from 0 to $\infty$. Lastly, in the final expression, the integration should also include a factor of $e^{ik.r}$.

So, the correct expression for $G(\vec{r_1},\vec{r_2})$ should be:

$G(\vec{r_1},\vec{r_2})=\frac{1}{{(2\pi)}^{3}} \int_{-\infty}^{\infty}\frac{e^{ik.r}}{ik(k^2 - k_0^2)} e^{i\vec{k}.(\vec{r_1} -\vec{r_2})}d^3k$

Overall, your approach is correct and all you need to do is fix these minor mistakes. Keep up the good work!