Fourier Transforms, Green's function, Helmholtz

Tags:
1. Oct 13, 2015

ognik

1. The problem statement, all variables and given/known data
I've gotten myself mixed up here , appreciate some insights ...

Using Fourier Transforms, shows that Greens function satisfying the nonhomogeneous Helmholtz eqtn
$$\left(\nabla ^2 +k_0^2 \right) G(\vec{r_1},\vec{r_2})= -\delta (\vec{r_1} -\vec{r_2}) \:is\: G(\vec{r_1},\vec{r_2})= \frac{1}{{2\pi}^{3}} \int \frac{e^{i\vec{k}.(\vec{r_1} -\vec{r_2})}}{k^2 - k_0^2} d^3k$$
2. Relevant equations
$$\left(\nabla ^2 +k_0^2 \right) G(\vec{r_1},\vec{r_2})= -\delta (\vec{r_1} -\vec{r_2})$$
3. The attempt at a solution
Taking the Fourier Transforms:
LHS: $$F\left[ \nabla^2G+k_0^2 G \right] = (- k^2 +k_0^2) \hat{u}$$
RHS: $$F\left[ -\delta (\vec{r_1} -\vec{r_2}) \right] = - \int 1 e^{-i\vec{k}.(\vec{r_1} -\vec{r_2})} d^3r$$

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

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

2. Oct 13, 2015

Orodruin

Staff Emeritus
First of all, you must decide which variable you are keeping fixed in your Fourier transformation. In this case, the PDE is in terms of $\vec r_1$ and $\vec r_2$ is a constant. You may want to call them $\vec r$ and $\vec r_0$ instead to emphasise this.

Second, the Fourier transform of the delta function is not an integral - or rather, it is an integral which is trivial to perform. Because of your notation, it is unclear what you mean by the integral you give. The Fourier transform you should be interested in is
$$F[-\delta(\vec r - \vec r')] = -\int e^{-i \vec k\cdot \vec r} \delta(\vec r - \vec r') dV = - e^{-i\vec k \cdot \vec r'}.$$
You should be able to take it from there.

3. Oct 13, 2015

ognik

Awesome, thanks, I clearly need to better understand about the variables in play - may I please get some confirmation of what I have absorbed ...using r1 and r0 as you suggested - and I think r, r1 & r0 are all vectors?

The formula I have for a 3D FT is $$F[f(r)] ≡ f ̃(ω)= ∫_{-∞}^∞ f(r) e^{-iω.r} d^3 r$$

In the above problem (I'm a bit mixed up, sorry) should I be using r or r1?

Then you use r and r' for the FT of \delta, shouldn't that (for this problem) be $$F[-\delta (r_1 - r_0)]=- \int e^{-ik.r} \delta (r_{1} - r_{0})] d^3r = e^{-ik.r}$$
(and I think I want the r in the exponent to be r0?)

For the inverse transform, I have $$F^{-1} [\tilde{f} (ω)] ≡ f (r)= \frac{1}{(2\pi)^3} ∫_{-∞}^∞ \tilde{f}(\omega) e^{+iω.r} d^3 \omega$$

What lead to this doubt was I could see that the (r1 - r0) emerged from the product of the 2 exponents, but I keep mixing up r, r1 & r0 - so would prefer to be certain and not guess. Thanks again, important for me this.

4. Nov 7, 2015

ognik

Hi would really appreciate if someone could briefly check my thoughts above - let me know if they're right, let me know what's wrong ....thanks

5. Nov 8, 2015

vela

Staff Emeritus
Actually, he suggested you use $\vec{r}$ and $\vec{r}_0$ instead of $\vec{r}_1$ and $\vec{r}_2$, so the original equation becomes
$$(\nabla^2 + k_0^2)G(\vec{r},\vec{r_0}) = -\delta(\vec{r}-\vec{r}_0).$$ Clearly, $\vec{r}$ is the variable, so multiply both sides by $e^{-i \vec{k}\cdot \vec{r}}$ and integrate to compute the Fourier transforms.

6. Nov 8, 2015

ognik

Thanks Vela, indeed Orodruin did, and I am not seeing my r's clearly, but for the delta function part he used r and r'... should I instead use r and $r_0$?

7. Nov 8, 2015

vela

Staff Emeritus
Just be consistent. What's the argument of the delta function in
$$(\nabla^2 + k_0^2)G(\vec{r},\vec{r_0}) = -\delta(\vec{r}-\vec{r}_0)?$$