# Idea for deriving Green's functions - Helmholtz equation

Let $x,y\in\mathbb{R}$. Then the Green's function for the Helmholtz equation is given by

$$\left(\Delta+\frac{\omega^{2}}{c^{2}}\right)G(x,y,\omega)=\delta(x-y).$$

Now what is the idea for deriving the Green's function here? Intuitively, I would take Fourier transforms on both sides, which would give me a convolution on the LHS and an exponential function on the RHS, but this would be very messy and I think I am missing something here. I believe the answer should be

$$G(x,y,\omega)=\frac{ic}{2\omega}e^{i\omega|x-y|/c}.$$

Apologies if this is a duplicate, but I could not find what I was looking for on the search bar.

• If I remember correctly, in this case the Green's function should look like $\frac{\exp(\pm i |x-y|)}{|x-y|}$ up to some multiplicative normalisation factors. – TZakrevskiy Jan 18 '17 at 12:28

$$(\omega^2/c^2-|\xi|^2) G(\xi,y,\omega ) = \exp(i y \xi),$$ which should be easy to manipulate - I think a residue theorem would do the trick.
• Because if $f$ and $g$ are functions, then $\mathcal{F}\{f(\cdot) g(\cdot)\}(\xi)=\hat{f}(\xi)\ast\hat{g}(\xi)$. – Jason Born Jan 18 '17 at 14:49
• @user3482534 in the LHS I see a differential operator $\Delta$ (it gives $-|\xi|^2$) and a constant (with respect to $x$) factor $\omega^2/c^2$, which remains as it is - a constant factor. – TZakrevskiy Jan 18 '17 at 14:59
• Also, I want to determine $G(x,y,\omega)$, not $G(\xi,y,\omega)$. – Jason Born Jan 28 '17 at 20:32
• @user3482534 when you apply the inverse FT to your $G(\xi, y,\omega)$, what did you obtain? Does it correspond to your initial hypothesis? – TZakrevskiy Jan 29 '17 at 9:29
• Well, multiplying both sides by $(\omega^2/c^2-|\xi|^2)$, from where you left off, I get $$\hat{g}(\omega,\xi,y)=\exp(iy\xi)(c^2/\omega^2-|\xi|^{-2})$$ So applying the IFT gives \begin{aligned}g(\omega,x,y)&=\frac{1}{2\pi}\int_{\mathbb{R}}\exp(iy\xi)(c^2/\omega^2-|\xi|^{-2})\cdot\exp(-ix\xi)\,dx\\&=\frac{c^2}{2\pi\omega^2}\int_{\mathbb{R}}\exp(iy\xi-ix\xi)\,dx-\frac{1}{2\pi|\xi|^{2}}\int_{\mathbb{R}}\exp(iy\xi-ix\xi)\,dx\end{aligned} But this seems to be heading in the wrong direction as those integrals are equivalent to Dirac distributions (which is what I began with anyway(!) – Jason Born Feb 7 '17 at 17:38