Solution of Helmholtz-Equation where Phase is restricted by additional PDE - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T09:40:14Z http://mathoverflow.net/feeds/question/112045 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/112045/solution-of-helmholtz-equation-where-phase-is-restricted-by-additional-pde Solution of Helmholtz-Equation where Phase is restricted by additional PDE MarkusWave 2012-11-11T00:22:48Z 2013-05-13T17:22:00Z <p>Hello! Let's say I have</p> <p>$(\partial_x^2 + \partial_y^2 + a)f(x,y)=0$</p> <p>with f(x,y) $\in \mathbb{C}$, ($\lim_{x,y \to \infty} f(x,y)=0$).</p> <p>Now separate the Amplitude and Phase of the solution:</p> <p>$f(x,y)=A(x,y)\cdot \exp(i\cdot g(x,y))$</p> <p>with $A(x,y),g(x,y) \in \mathbb{R}$.</p> <p>The additional restriction to the Phase g(x,y) is a PDE in the form of:</p> <p>$\hat{L}g(x,y)=h(x,y)$</p> <p>Now my questions are: </p> <ul> <li>Are there restrictions on h(x,y) and differential operator $\hat{L}$ to get a solution f(x,y)?</li> <li>How can find such a solution f(x,y)? Are there analytical ways? Are there stochastic ways? Are there numerical ways?</li> <li>Are there ways to find solutions for the simplification a=0?</li> </ul> <p>I'm very thankful for any hint in a useful direction. Unfortunaly I'm totally stuck with this problem. Thanks alot in Advance!</p> <p>Markus</p> http://mathoverflow.net/questions/112045/solution-of-helmholtz-equation-where-phase-is-restricted-by-additional-pde/112071#112071 Answer by Jon for Solution of Helmholtz-Equation where Phase is restricted by additional PDE Jon 2012-11-11T12:08:40Z 2012-11-12T08:01:53Z <p>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.</p>