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.


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.


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?


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.

  • 6
    $\begingroup$ 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 $\endgroup$ Jan 2 '18 at 7:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.