# A variation of Poisson's equation in cylindrical coordinates

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!

• this is a nonlinear Poisson equation --- there's no algebraic solution, you'll have to resort to numerics. math.uiowa.edu/ftp/atkinson/nonlinear_poisson.pdf – 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