TheoremProved

Maximum principle for harmonic functions

Statement

If $u$ is harmonic ( $\Delta u = 0$ ) on a bounded connected open set $\Omega \subset \mathbb{R}^n$ and continuous on $\overline{\Omega}$ , then $u$ attains its maximum and minimum on the boundary $\partial\Omega$ ; if $u$ attains an interior maximum or minimum, $u$ is constant.

Why is it true?

A harmonic function has no local bumps or dips of its own — its value at any point is exactly the average of its values on any surrounding sphere. A function that is always equal to a local average cannot have a genuine peak or valley in the interior; any extreme value must be forced by the boundary.

Proof sketch

Use the mean value property: for a harmonic $u$ and any ball $B(x_0,r) \subset \Omega$ , $u(x_0)$ equals the average of $u$ over the sphere $\partial B(x_0,r)$ . If $u$ attained an interior maximum $M$ at $x_0$ , the average over any such sphere equals $M$ while every value on the sphere is $\le M$ , forcing $u \equiv M$ on the sphere; propagating this argument through a chain of overlapping balls covering the connected domain shows $u \equiv M$ everywhere, so the maximum is also attained on the boundary, and it cannot be a strict interior maximum unless $u$ is constant. The minimum principle follows by applying the same argument to $-u$ .

Topics that use this theorem

Laplace's equation

Related theorems

Divergence theorem (Gauss–Ostrogradsky theorem)

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

David Gilbarg, Neil S. Trudinger (1983). Elliptic Partial Differential Equations of Second Order · DOI:10.1007/978-3-642-61798-0