Solution of Helmholtz-Equation where Phase is restricted by additional PDE - MathOverflow

Solution of Helmholtz-Equation where Phase is restricted by additional PDE

2012-11-11T00:22:48Z

Hello! Let's say I have

$(\partial_x^2 + \partial_y^2 + a)f(x,y)=0$

with f(x,y) $\in \mathbb{C}$, ($\lim_{x,y \to \infty} f(x,y)=0$).

Now separate the Amplitude and Phase of the solution:

$f(x,y)=A(x,y)\cdot \exp(i\cdot g(x,y))$

with $A(x,y),g(x,y) \in \mathbb{R}$.

The additional restriction to the Phase g(x,y) is a PDE in the form of:

$\hat{L}g(x,y)=h(x,y)$

Now my questions are:

Are there restrictions on h(x,y) and differential operator $\hat{L}$ to get a solution f(x,y)?
How can find such a solution f(x,y)? Are there analytical ways? Are there stochastic ways? Are there numerical ways?
Are there ways to find solutions for the simplification a=0?

I'm very thankful for any hint in a useful direction. Unfortunaly I'm totally stuck with this problem. Thanks alot in Advance!

Markus

Answer by Jon for Solution of Helmholtz-Equation where Phase is restricted by additional PDE

2012-11-11T12:08:40Z

The condition on $g$ gives a definite pde for $A$. This can be seen in the following way. Let us insert the solution $f=A(x,y)e^{ig(x,y)}$ into the Helmholtz equation. We get $$ \Delta A+2i(\partial_xg\partial_xA+\partial_yg\partial_yA)+\Phi(x,y)A=0 $$ being $$ \Phi(x,y)=i\Delta g-(\partial_xg)^2-(\partial_yg)^2+a. $$ Now, assuiming $L$ is a linear operator with the Green function $LG=\delta$, one can write $$ g(x,y)=g_0(x,y)+\int_\Omega dx'dy'G(x,x';y,y')h(x',y') $$ being $Lg_0=0$. By substituting this into $\Phi$ and the equation for $A$ we get a partial differential equation to solve. For some operator $L$, the final equation could be simple to manage but, for the general case, maybe some approximation techniques could help.