MHB 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.