Really could use some advice with this PDE

Question

so I'm faced with a partial differential equation that I have been able to solve only in one limiting case. I have asked the mathematics stack exchange twice for literally any help whatsoever to no avail, so I thought I might finally appeal to the experts on PDEs ;). The equation arises out of quantum mechanics. Specifically, it is a differential form of a condition for a specific system I'm studying to minimize the Robertson-Schrödinger relation (sometimes called the "generalized uncertainty principle") for angular momentum. As a result, the solutions are wavefunctions and the usual boundary conditions apply (i.e. fading out at the boundaries (in this case infinity), normalized, etc.) I can give you a semi-coordinate-free version of the equation, and also a version in parabolic coordinates that I have found to be the most fruitful (I have tried it also in elliptic cylindrical coordinates, cylindrical, cartesian and of course spherical, among other odd coordinate systems I defined in the process, and by an enormous margin parabolic coordinates produces the most compact version of the equation). I say "semi-coordinate-free" because in order to make it coordinate-free I have to define a vector that can be written in Cartesian coordinates as:

$$\mathbf{u}_\pm = \hat{\mathbf{x}} \pm i\hat{\mathbf{y}} = (1,\pm i,0)$$ But, in doing so, I can give you guys the following form, on the loosest leash I've made an effort to obtain:

$$\frac{1}{2}(\mathbf{u}_-\cdot\mathbf{r})(\rho^2\psi + \nabla^2\psi) - (1+\mathbf{r}\cdot\nabla)(\mathbf{u}_-\cdot\nabla\psi) = (\gamma/\hbar^2)\psi$$ where $\mathbf{r}$ is the position vector, $\rho \in \mathbb{R}^+$ and $\gamma \in \mathbb{C}$ are constants.

In parabolic coordinates, by which I mean

$$x = \sigma\tau\cos\varphi\\ y = \sigma\tau\sin\varphi\\ z = \frac{1}{2}(\sigma^2 - \tau^2)$$ it becomes $$\frac{1}{2}\hbar^2\left[(\partial_\sigma - \frac{i}{\sigma}\partial_\varphi)(\partial_\tau - \frac{i}{\tau}\partial_\varphi) - \rho^2\tau\sigma\right]\psi = \gamma e^{i\varphi}\psi$$

I was able to solve the case $\gamma = 0$ analytically, and the solutions are of the form

$$\psi(\sigma,\tau,\varphi) = Ae^{-\frac{\rho}{2}(\sigma^2 + \tau^2)}(\sigma\tau)^{\lambda}e^{-i\lambda\varphi} \hspace{6mm} \lambda = 0,1,2,3,...$$ Beyond this, however, I'm sort of at a loss. I've been reading up on numerical solutions to PDEs, something that, as an aspiring physicist, I assume I will require a solid familiarity with. My question about this is pretty straightforward - almost any information that might lead to a general solution or a visual depiction of a solution to the 'inhomogeneous' (not sure if that word is technically applicable outside of ODEs) case $\gamma \neq 0$, would be greatly appreciated. If anyone can classify or describe the form of the equation, that would be extremely helpful. Pointers for good places to look for a method of obtaining solutions that might be particularly helpful for this specific equation would be excellent. If you can provide a solution, well then, of course that would also be wonderful, but even then I would really love to know how you got it and what you looked for in doing so. My apologies for a vague-ish request, but I it's not so much that I am not looking for an answer, I am just accepting a very wide array of them equally. I can only hope that you all will be forgiving about that and want to attempt to answer my question anyway. That said, I admit I am rather new to the whole stack exchange thing (which seems pretty sweet in general, by the way), and understand if this is a little too broad or narrow...Thanks for taking the time to read through!

This post imported from StackExchange Physics at 2014-03-06 21:22 (UCT), posted by SE-user Nicola

Answer 1

It is a linear PDE, hence the superposition principle holds. You can eliminate the φ dependence by making the ansatz $ψ(σ,τ,φ)=\Psi(\sigma,\tau)e^{−iλφ}$. Then you are left with a linear hyperbolic equation in σ,τ that can probably be solved by integrating one variable at a time.