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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:

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

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

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  • $\begingroup$ For $a=0$, $f=0$ (this is just the Laplace equation). $\endgroup$
    – Florian
    Nov 11, 2012 at 9:47
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    $\begingroup$ Your system of PDEs is overdetermined. If you know $L$ explicitly, you should be able to explicitly write out the equations you have, and differentiate both sides until you encounter some obstructions, or reach involution (library.msri.org/books/Book18/MSRI-v18-Bryant-Chern-et-al.pdf), proving the existence of local solutions. However, since your operator $L$ seems to be mysterious, it would help if you could say more about it. Do you know the order of $L$, or whether $L$ is elliptic or hyperbolic etc., or if $L$ is linear? $\endgroup$
    – Ben McKay
    Mar 18, 2013 at 17:20

1 Answer 1

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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.

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  • $\begingroup$ you missed a A before the $\Phi(x,y)$. But yes, thats an idea to directly evaluate the equation. It doesnt give me the actual answere what restrictions there are, but it seems like to be a non-trivial problem. Thanks in any way. I will try some FEMs to solve it, if there is no simple analytical way. $\endgroup$
    – MarkusWave
    Nov 12, 2012 at 0:48

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