Two dimensional Helmholtz equation and complex function theory

Question

I am trying to find new particular solution for the Helmholtz equation and I was wondering if it is possible to use the complex function theory in this context.

The two dimensional Helmholtz equation is given by

$$\frac{\partial^2 u(x,y)}{\partial x^2}+\frac{\partial^2 u(x,y)}{\partial y^2}+\lambda^2\,u(x,y)=p(x,y)$$ with $x\in\Bbb R$ and $y\in\Bbb R$.

If I use the variable transformation $\eta=x+\mathrm{i}\,y$ and $\zeta=x-\mathrm{i}\,y$ we get $$4\,\frac{\partial^2 u(\eta,\zeta)}{\partial \eta\,\partial\zeta}+\lambda^2\,u(\eta,\zeta)=p(\eta,\zeta)$$ with $\eta\in\Bbb C$ and $\zeta\in\Bbb C$.

I am wondering now, if it is possible to derive the solution for $p(x,y)=\delta(x)\,\delta(y)$ ($\delta()$ the dirac delta function) also for the transformed equation.

Solving the problem in the cartesian coordinate system $(x,y)$ is straight forward using double Fourier Transform. Is this also possible for the transformed equation in $(\eta,\zeta)$? Is the Fourier Transform also valid for functions of complex variables?

Thank you for any hints.

Answer 1

Yes, both have the same form : $$u \ast h = p$$ where $h$ is a certain distribution and $\ast$ is the convolution.

All you have to do is finding $g$ (aka the Green function) the convolutive inverse of $h$, such that $$h \ast g= \delta$$ ($\delta$ the Dirac delta, here in $2D$ indicating the point $(0,0)$)

So that $$u = u \ast h \ast g = p \ast g$$ In the 1st case $$h(x,y) = \partial_x^2 \delta(x,y)+\partial_y^2 \delta(x,y)+\delta(x,y)$$ hence $FT[h](\xi_x,\xi_y) = (2i \pi \xi_x)^2 + (2i \pi \xi_y)^2+\lambda^2$ and $$g(x,y) = FT^{-1}[\frac{1}{(2i \pi \xi_x)^2 + (2i \pi \xi_y)^2+\lambda^2}]$$ In the second case $$h_2(\eta,\zeta) = 4\partial_\eta \partial_\zeta \delta+\lambda^2 \delta$$ $FT[h_2](\xi_\eta,\xi_\zeta) = 4(2i \pi \xi_\eta)(2i \pi \xi_\eta)+\lambda^2$ and $$g_2(\eta,\zeta) = FT^{-1}[\frac{1}{4(2i \pi \xi_\eta)(2i \pi \xi_\eta)+\lambda^2}]$$ I let you do the Fourier transforms calculus yourself (but you should get $g_2(\eta,\zeta) = g(x+iy,x-iy)$)