My question is related to this one: Minima & maxima of Laplace's equation

I know that Laplace equation (like all elliptic 2nd order PDE) prohibits local extrema, and that maximal values may be found only on the edges of the domain.

My question is: Does Poisson equation (adding sources) change this? Intuitively, I would assume that it should, as if I solve the equation over a disk with a point charge in the middle, I would expect the potential to be maximal (and even explode) in the middle.

Does it mean that Poisson equation isn't elliptic, or does the original general claim regarding local extrema in elliptic equations hold only for homogeneous cases?

  • 2
    $\begingroup$ Might Mathematics be better suited for this math question? $\endgroup$
    – Kyle Kanos
    Mar 1, 2017 at 10:58
  • $\begingroup$ The context is physical and so is the source of intuition $\endgroup$
    – Yair M
    Mar 1, 2017 at 11:44
  • 1
    $\begingroup$ The statement "all elliptic 2nd-order PDEs prohibit local extrema" is false, as you have found. I believe it is true for homogeneous elliptic 2nd-order PDEs, but Poisson's equation is not homogeneous. $\endgroup$ Mar 1, 2017 at 15:07

1 Answer 1


These types of theorems assume smoothness, or at least some amount of differentiability, so when you add sources, you need to be careful what types of sources they are.

Delta-function sources are not smooth. So, if you are solving Poisson's equation in an open region $\Omega$ and you put a delta-function source at $P \in \Omega$, then you've violated the assumptions of the theorem. To restore the assumptions of the theorem, you can instead consider the open region $\Omega' = \Omega \setminus P$; that is, remove the point $P$. On $\Omega'$, your sources are zero, so you have Laplace's equation on an open region, and it should be simple to convince yourself that the solution does not achieve a maximum in $\Omega'$.

What about other types of sources, which are more smooth? E.g., consider an electrostatics problem with some nicely-behaved, smooth charge distribution $\rho$. In this case, it is possible for the potential $\Phi$ to achieve a maximum somewhere where $\rho \neq 0$. But there's nothing wrong with this, because if $\rho \neq 0$ then we are not talking about Laplace's equation.


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