MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Our team of undergraduate physicists are familiar with finding numerical approximations to the following Poisson-like PDE central to our plasma research in a torus:

$\nabla^2 V = \frac{f(V)}{R^2}$

where $f(V)$ is an arbitrary function of the electric potential V, and $R$ is the radial cylindrical coordinate (assume $\phi$ symmetry, so we have two variables: $(R,z)$). The boundary condition is that $V=0$ on the torus boundary (the variable $R$ resides in the torus).

We know the following about $f$ and the physical solutions $V$:

  • $V$ is a smooth and always-negative function, which goes to zero at the toroidal boundary, and has an extremum near the geometric center of the torus' innards.

  • $f(V)$ is a specified smooth function of the potential such that $f(0)=0$, is always-positive, and goes to zero past some maximum potential $V_0$.

We've noticed that if $f(V)$ is expanded in a power series in $V$, then we'd might as well solve only:

$ \nabla^2 V = \frac{V^n}{R^2}$

where n is any positive (or negative, if that's convenient to the maths) integer.

Could you provide a name for either PDE, a method of approximating it algebraically, or some guidance in simplifying it?

Thanks for reading this!

share|cite|improve this question
this is a nonlinear Poisson equation --- there's no algebraic solution, you'll have to resort to numerics. – Carlo Beenakker Jun 14 '13 at 20:29
Thanks - we'll keep going with numerical methods. But one more question: does anyone notice a clever choice of f(V) to yield a specific solution? – Alex Patterson Jun 15 '13 at 5:55
For $n=1$ this is a linear PDE and it appears a Schroedinger-like equation with a $\frac{1}{R^2}$ potential. One must verify that the zero eigenvalue does occur. – Jon Jun 18 '13 at 6:38
Thanks Jon, I did check & as it turns out, the f(V)=V solution doesn't converge in the torus. The source f(V)/R^2 (interpreting the equation as Poisson's equation) needs to go to zero eventually. So any finite series for f(V) is out of the question. I guess that's it. – user35122 Jun 19 '13 at 22:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.