Earnshaw's theorem

Remarks

Proof: suppose that $\varphi(x) > \varphi(a)$ for all $x \ne a$ in a ball $B(a, \rho) \subseteq U$. Take $0 < r < \rho$. By the mean value property, $\varphi(a)$ is the average of $\varphi$ over the sphere of radius $r$ about $a$. But every value on that sphere is larger than $\varphi(a)$, a contradiction. A strict local maximum is excluded in the same way, or by applying the argument to $-\varphi$. In physics, $\varphi$ is the potential of charges that lie outside $U$; it is harmonic in the charge-free region $U$. Example: in $\R^3$, with $e_1 = (1, 0, 0)$ and $e_2 = (0, 1, 0)$, let $\varphi(x) = 1/\abs{x - e_1} + 1/\abs{x + e_1}$ be the potential of two unit charges at $\pm e_1$. The origin is an equilibrium point by symmetry. Along the $x_1$-axis, $\varphi(t e_1) = 2/(1 - t^2)$ has a minimum at $t = 0$, but along the $x_2$-axis, $\varphi(t e_2) = 2/\sqrt{1 + t^2}$ has a maximum, so the origin is a saddle point. The theorem is about potentials that are harmonic; diamagnetic materials, feedback control and time-varying fields are classical ways around it in physics.

Going deeper

For a student

The theorem is a direct consequence of the mean value property, and it is a weak form of the strong maximum and minimum principles. It applies to every inverse-square force law, so it also rules out stable static configurations of point masses under Newtonian gravity. At an equilibrium of a harmonic potential, the Hessian has trace $0$, so it cannot be positive definite. This is another way to see that an equilibrium cannot be a nondegenerate minimum.