I want to construct special solutions to wave equations, specifically to Helmholtz-equation and paraxial wave equation (PWE).

Let us consider the PWE $$(\partial_x^2 + \partial_y^2 - 2ik\partial_z)\psi(x,y,z)=0$$ with the constraint $$\psi(x,y,z=0)=A(x,y)\cdot \exp(i\cdot g(x,y))$$ with a specific predefined $g: (x,y) \to \mathbb{R}$, and arbitrary $A: (x,y) \to \mathbb{R}$, with natural boundary conditions $\lim_{x||y \to \infty} \psi(x,y,z)=0$. How can I find $\psi(x,y,z)$ for general $g(x,y)$?


Suppose we are in cylindrical coordinates, and I set g(r,$\phi$)=$m\cdot \phi$. In that case I will get $A(r,\phi)=\exp(-r^2)\cdot r^m$, which are Laguerre-Gaussian solutions.


1 Answer 1


I'm not sure that your original question is well posed. If only $g$ is specified, then the boundary conditions are not fully-specified and the problem of solving for $\psi$ is ill-posed. I will continue, assuming that both $A$ and $g$ are specified and that you want to solve the problem for $\psi$ given the fully-specified boundary conditions.

We have the PWE and boundary conditions $$ \left| \begin{array}{l} (\partial_x^2 + \partial_y^2 - 2ik\partial_z)\psi(x,y,z)=0, \\ \psi(x,y,0) = A(x,y) \exp(ig(x,y)), \\ \psi(x,y,z)\text{ goes to zero fast enough for large }x\text{ and }y, \end{array} \right. $$ where $A$ and $g$ are given functions, regular and smooth enough for our purposes. You'll notice I handwave the degree of regularity and the rate of decay as $x\rightarrow\infty$ and $y\rightarrow\infty$ because, well, I'm not mathematician.

Suppose $\psi$ has the following Fourier integral expansion $$ \psi(x,y,z) = \int_{-\infty}^\infty\int_{-\infty}^\infty \tilde\psi(k_x,k_y,z) e^{ik_x x+ik_y y}\;dk_x\;dk_y. $$

Inserting into the original equation, we have

$$ \int_{-\infty}^\infty\int_{-\infty}^\infty(-k_x^2 - k_y^2 - 2ik\partial_z)\tilde\psi(k_x,k_y,z) e^{ik_x x+ik_y y}\;dk_x\;dk_y=0. $$

For that to integrate to zero, it is sufficient (and I'm certain it can be proved necessary) for the integrand to identically be zero for all values of $k_x$ and $k_y$, $$ (-k_x^2 - k_y^2 - 2ik\partial_z)\tilde\psi = 0, $$ where I've dropped the explicit $k_x$ and $k_y$ notation and cancelled the non-zero term $e^{ik_x x+ik_y y}$. Now noting the $z$ derivative with a prime ($'$), we have $$ 2ik\tilde\psi' = -(k_x^2 + k_y^2)\tilde\psi. $$ Separating $\tilde\psi$ terms all on one side gives $$ \frac{\tilde\psi'}{\tilde\psi} = -\frac{k_x^2 + k_y^2}{2ik}. $$ Now integrating $dz$, $$ \int \frac{1}{\tilde\psi}\;d\tilde\psi = -\int\frac{k_x^2 + k_y^2}{2ik}\;dz, $$ $$ \log\tilde\psi = -\frac{k_x^2 + k_y^2}{2ik}z + c, $$ for some constant $c$ (a function of $k_x$ and $k_y$). Now exponentiating gives $$ \tilde\psi = C(k_x,k_y)\exp\left(-\frac{k_x^2 + k_y^2}{2ik}z\right), $$ for some other constant (in $z$, but still a function of $k_x$ and $k_y$) $C=\exp(c)$. Now inserting back into the original integral representation, we have $$ \psi(x,y,z) = \int_{-\infty}^\infty\int_{-\infty}^\infty C(k_x,k_y)\exp\left(ik_x x+ik_y y-\frac{k_x^2 + k_y^2}{2ik}z\right) \;dk_x\;dk_y. $$ This is our general solution, and you can verify that is solves the original PWE. To meet the boundary condition you require, we must find the specific $C(k_x,k_y)$ that satisfies the condition. So we plug in the value $z=0$ into the general solution, giving $$ \psi(x,y,0) = \int_{-\infty}^\infty\int_{-\infty}^\infty C(k_x,k_y)\exp\left(ik_x x+ik_y y\right) \;dk_x\;dk_y = A(x,y)\exp(i g(x,y)) $$ It's pretty clear then that $C(k_x,k_y)$ should be the Fourier transform of $A(x,y)\exp(i g(x,y))$. In the Fourier convention I am using (a pretty non-standard one but I like it), the transform from $(x,y)$ to $(k_x,k_y)$ works out to $$ C(k_x,k_y) = \frac{1}{(2\pi)^2}\int_{-\infty}^\infty\int_{-\infty}^\infty A(x,y)\exp(i g(x,y)) \exp(-ik_xx-ik_yy)\;dx\;dy $$ So, for some particular amplitude/phase screen $A\exp(ig)$, you could in principle calculate the Fourier transform above, and insert it into the general solution. You would then have an integral expression for $\psi$ that solves the PWE and the boundary condition. For some particular choices of $A$ and $g$, closed form expressions would result. However, for arbitrary boundary conditions, I expect that no closed form integrals could be calculated.

  • $\begingroup$ This is very helpful, through in a bit different way than I expected. Thanks for pointing out that the problem is ill-defined (i somehow "felt" that), and now i understand how my original problem can become well-defined. Great, thanks! $\endgroup$ Aug 19, 2014 at 10:18
  • $\begingroup$ I just reformulated my original question, adding the additional restriction I have. Maybe you have an idea for that one aswell: Form invariant solutions $\endgroup$ Aug 21, 2014 at 9:23

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