# Proving Earnshaw's theorem is subtle in three-dimensions!

The Laplace's equation for the electrostatic potential $\phi(\textbf{r})$ is given by $$\nabla^2\phi(\textbf{r})=0.\tag{1}$$

Equation (1) is said to encode the fact:

A free movable charge cannot exist in stable equilibrium under the influence of the electrostatic forces alone.

For a charge to be in stable equilibrium, it must be located at the extremum of its potential $\phi(\textbf{r})$. In textbooks, it is argued that Eq.(1) implies that the function $\phi(\textbf{r})$ does not have an extremum.

Therefore, the textbooks seems to assume that the necessary condition for a function of several variables (at least two) to not have an extremum is $\boldsymbol{\nabla}^2\phi(\textbf{r})= 0$.

However, for a function $f(x,y)$ of two variable, the actual condition for no extremum is $$f_{xx}f_{yy}-(f_{xy})^2<0\tag{2}$$ which does not look same (at least immediately) as $$\boldsymbol{\nabla}^2f(x,y)=f_{xx}+f_{yy}=0.\tag{3}$$ However, for two dimensions, it can be proved that $(3)\Rightarrow (2)$ (see the proof below). But it is not obvious in three-dimensions.

Proof

Note that for two dimensions, Eq.(1) implies $\phi_{xx}+\phi_{yy}=0.$ Squaring it we get, $$\phi_{xx}\phi_{yy}-(\phi_{xy})^2=-\frac{1}{2}(\phi^2_{xx}+\phi^2_{yy}+2\phi^2_{xy})<0$$ since the bracketed quantity on the right hand side is positive.

Question

As I understand, for more than one-dimension, $\nabla^2\phi(\textbf{r})=0$ is not the actual necessary condition for the function $\phi(\textbf{r})$ to have no extremum. The actual condition for no extremum, in case of $\phi(x,y)$ is $\phi_{xx}\phi_{yy}-(\phi_{xy})^2<0$ which however is consistent with (1).

But how is it in general obvious that $\nabla^2\phi(\textbf{r})=0$ is consistent with the condition for no extremum in three-dimensions?

• – Qmechanic Jan 2 '18 at 10:47

Geometrically, this follows from the mean value property of harmonic functions. If $\nabla^2 \phi = 0$, then the value of $\phi$ at a point $\mathbf{x}$ is the same as the average of the values of its neighbors (specifically the average of $\phi$ over a sphere centered at $\mathbf{x}$). If $\phi$ had a local extremum at $\mathbf{x}$, this would be impossible.
Analytically, consider the Hessian matrix $H$ whose entries are $$H_{ij} = \frac{\partial^2 \phi}{\partial x_i \partial x_j}.$$ This matrix is symmetric and hence has real eigenvalues. Suppose that these eigenvalues are nonzero. The condition for a local maximum or minimum is that all of the eigenvalues of $H$ have the same sign, because upon rotating to the eigenbasis, the function looks like $\sum_i \lambda_i x_i^2 /2$ to second order. But the harmonic condition $\nabla^2 \phi = 0$ is equivalent to $\text{tr}\, H = 0$, so there must be both positive and negative signs present.
• The proof of non-existence of local maxima/minima for harmonic functions is more complicated than your final argument. For instance the (non-harmonic) function $z = x^4+ y^4$ has a local minimum at the origin but the Hessian matrix has zero eigenvalues there. You should also prove that harmonic functions cannot have all zero eigenvalues... However +1 – Valter Moretti Jan 2 '18 at 7:51