Differential equations and dynamical systems

Laplace's equation

The equation for harmonic functions, describing steady-state temperature, potential and equilibrium.

IntuitionSteady states: what is left once nothing changes anymore

Leave a metal plate with its edges held at fixed temperatures long enough, and the interior temperature stops changing: it settles into a steady-state pattern. The same kind of settled-down equation appears for the electric potential in a region with no charge, for the shape of a soap film stretched across a wire frame, and for the velocity potential of a smooth, swirl-free fluid flow. Laplace's equation is the common rule these steady states obey: at every interior point, the value is exactly the average of its surroundings, so there can be no lonely peaks or dips away from the boundary.

A harmonic surface has no interior bumps or dips of its own: rotate and rescale it to see that any local high point on one side is balanced by a local low point on another, exactly the saddle-like behaviour the maximum principle predicts for a non-constant solution of

\nabla^2 u = 0

UndergraduateHarmonic functions and the Dirichlet problem

Definition: Harmonic function and the Laplace operator

A twice continuously differentiable function $u(x,y)$ on an open set $\Omega \subseteq \mathbb{R}^2$ is called harmonic if it satisfies Laplace's equation $\nabla^2 u = u_{xx} + u_{yy} = 0$ everywhere in $\Omega$ , where $\nabla^2$ (also written $\Delta$ ) is the Laplace operator, the sum of unmixed second partial derivatives. The Dirichlet problem asks: given a bounded domain $\Omega$ and a prescribed boundary function $g$ on $\partial\Omega$ , find the harmonic function $u$ inside $\Omega$ with $u = g$ on $\partial\Omega$ .

\nabla^2 u = u_{xx} + u_{yy} = 0, \qquad (x,y) \in \Omega

On the unit disk, the Dirichlet problem has an explicit closed-form solution: the Poisson kernel integral formula reconstructs the harmonic function inside the disk entirely from its boundary values, without ever solving a differential equation directly. Writing points inside the disk in polar form $re^{i\theta}$ with $0 \le r < 1$ , and the boundary value at angle $\phi$ as $g(\phi)$ , the formula is a weighted average of $g$ over the whole boundary circle, with weights concentrated near $\phi \approx \theta$ when $r$ is close to $1$ .

u(re^{i\theta}) = \frac{1}{2\pi}\int_0^{2\pi} \frac{1-r^2}{1-2r\cos(\theta-\phi)+r^2}\, g(\phi)\, d\phi

Laplace's equation among the family of elliptic PDEs
Equation	Formula	Physical meaning
Laplace's equation	$\nabla^2 u = 0$	Steady state, no interior sources or sinks
Poisson's equation	$\nabla^2 u = f$	Steady state with a prescribed source density $f$
Helmholtz equation	$\nabla^2 u + k^2 u = 0$	Time-harmonic waves (frequency $k$ ); eigenvalue problem

UndergraduateCore theorems: the mean value property and the maximum principle

Theorem: Mean value property of harmonic functions

Suppose $u$ is harmonic on an open set containing the closed disk $\overline{B_r(x_0)}$ of radius $r$ centred at $x_0$ . Then $u(x_0)$ equals the average of $u$ over the boundary circle: $u(x_0) = \frac{1}{2\pi}\int_0^{2\pi} u(x_0 + r\cos\theta, y_0 + r\sin\theta)\, d\theta$ . The same value also equals the average of $u$ over the whole solid disk.

Why is it true?

A harmonic function cannot favour any direction: since it has zero net curvature at every point (the sum of its curvatures along the $x$ and $y$ axes cancels exactly), it cannot bulge upward on average as you walk around any circle centred at a point, nor dip downward — the only consistent value for the centre is the average of the circle.

Proof

Without loss of generality centre the disk at the origin ( $x_0 = 0$ ) and define $\phi(r) = \frac{1}{2\pi}\int_0^{2\pi} u(r\cos\theta, r\sin\theta)\, d\theta$ for $0 \le r \le R$ , the average of $u$ over the circle of radius $r$ . We show $\phi$ is constant by proving $\phi'(r) = 0$ .

Differentiating under the integral sign, $\phi'(r) = \frac{1}{2\pi}\int_0^{2\pi} \nabla u(r\cos\theta, r\sin\theta) \cdot (\cos\theta, \sin\theta)\, d\theta = \frac{1}{2\pi}\int_0^{2\pi} \frac{\partial u}{\partial n}\, d\theta$ , since $(\cos\theta,\sin\theta)$ is the outward unit normal on the circle of radius $r$ . Multiplying and dividing by $r$ turns this into a normalised boundary integral of the normal derivative over the circle $\partial B_r$ .

By the divergence theorem, $\int_{\partial B_r} \frac{\partial u}{\partial n}\, ds = \int_{B_r} \nabla^2 u \, dA = 0$ because $u$ is harmonic. Hence $\phi'(r) = 0$ for every $r$ in the domain, so $\phi(r) = \phi(0^+) = u(x_0)$ for all such $r$ : the average over any circle centred at $x_0$ equals $u(x_0)$ itself, and averaging this constant value over $0 \le r \le R$ shows the solid-disk average equals $u(x_0)$ as well.

Theorem: Maximum principle for harmonic functions

Let $u$ be harmonic on a bounded, connected open set $\Omega$ and continuous on its closure $\overline{\Omega}$ . If $u$ is non-constant, then both its maximum and its minimum over $\overline{\Omega}$ are attained only on the boundary $\partial\Omega$ , never at an interior point.

Why is it true?

The mean value property says every interior value equals the average of a whole circle of neighbouring values. An average can only equal the overall maximum if every value being averaged already equals that maximum — so a genuine, isolated interior peak is impossible: if the centre is as high as it can possibly be, the entire surrounding disk must be exactly that high too, and this spreads outward until it reaches the boundary.

Proof

Suppose $u$ attains its maximum value $M$ over $\overline{\Omega}$ at an interior point $x_0 \in \Omega$ . Let $S = \{x \in \Omega : u(x) = M\}$ ; by continuity $S$ is closed in $\Omega$ , and it is nonempty since $x_0 \in S$ .

$S$ is also open: for any $x_1 \in S$ , choose a small disk $B_r(x_1) \subset \Omega$ . The mean value property gives $M = u(x_1) = \frac{1}{2\pi}\int_0^{2\pi} u(x_1 + r\cos\theta, x_1 + r\sin\theta)\, d\theta$ , an average of values all $\le M$ . An average of quantities bounded above by $M$ can equal $M$ only if every one of those quantities equals $M$ , so $u \equiv M$ on the whole circle $\partial B_r(x_1)$ , and by applying the same argument to every radius up to $r$ , on the whole disk $B_r(x_1)$ . Hence a neighbourhood of $x_1$ lies in $S$ , so $S$ is open.

Since $\Omega$ is connected and $S$ is a nonempty subset that is both open and closed in $\Omega$ , we must have $S = \Omega$ , i.e. $u \equiv M$ throughout $\Omega$ . But $u$ was assumed non-constant, a contradiction. Therefore no interior maximum exists; the same argument applied to $-u$ (which is also harmonic) rules out an interior minimum, so both extremes occur only on $\partial\Omega$ .

UndergraduateReal-World Applications and Worked Examples

Laplace's equation is everywhere steady-state physics is. In electrostatics, the potential in any charge-free region satisfies it; in fluid dynamics, the velocity potential of an incompressible, irrotational flow around an aircraft wing satisfies it; in gravitation, the potential outside any mass distribution satisfies it; and in computer graphics and computer vision, harmonic extension (solving the Dirichlet problem numerically) is used for image inpainting and mesh smoothing, filling in missing pixels or vertices with the smoothest possible values consistent with the known surrounding data.

Example: A saddle-shaped harmonic function on the unit disk

Verify that $u(x,y) = x^2 - y^2$ is harmonic, then find its maximum and minimum values on the closed unit disk $x^2+y^2 \le 1$ .

Solution

Check harmonicity by computing the two unmixed second partial derivatives: $u_x = 2x$ so $u_{xx} = 2$ , and $u_y = -2y$ so $u_{yy} = -2$ . Adding them, $u_{xx} + u_{yy} = 2 + (-2) = 0$ everywhere, so $u$ is indeed harmonic on all of $\mathbb{R}^2$ , including the closed unit disk.

By the maximum principle, since $u$ is non-constant and harmonic, its extreme values over the closed disk must occur on the boundary circle, not in the interior. Parametrise the boundary as $x=\cos\theta$ , $y=\sin\theta$ , giving $u = \cos^2\theta - \sin^2\theta = \cos(2\theta)$ .

As $\theta$ ranges over $[0,2\pi)$ , $\cos(2\theta)$ attains its maximum value $1$ at $\theta = 0$ (the point $(1,0)$ ) and its minimum value $-1$ at $\theta = \pi/2$ (the point $(0,1)$ ). Notice the origin, which looks like a natural candidate for an extremum by ordinary calculus (it is a critical point of $u$ ), is actually a saddle point of the surface, exactly as the maximum principle forbids any genuine interior extremum for a non-constant harmonic function.

Example: Electrostatic potential inside a circular capacitor cross-section

The boundary of a unit-radius circular region is held at potential $g(\phi) = V_0\cos\phi$ with $V_0 = 10$ volts. Using the harmonic extension of a single Fourier mode, find the potential $u(r,\theta)$ at the interior point $r = 0.5$ , $\theta = 60^\circ$ .

Solution

The boundary data $g(\phi) = V_0\cos\phi$ is already a single Fourier mode. The harmonic extension of $\cos\phi$ into the disk is simply $r\cos\phi$ (one can check directly: $r\cos\phi = x$ , and $x$ satisfies Laplace's equation trivially since $x_{xx}=0$ , $x_{yy}=0$ ). So instead of evaluating the Poisson kernel integral directly, we can read off the answer: $u(r,\theta) = V_0\, r\cos\theta$ .

Substitute the given values $V_0 = 10$ , $r = 0.5$ , $\theta = 60^\circ$ : first compute $\cos(60^\circ) = \tfrac{1}{2}$ .

Then $u(0.5, 60^\circ) = 10 \times 0.5 \times \tfrac{1}{2} = 2.5$ : the potential at that interior point is $2.5$ volts, a value strictly between the boundary extremes $-10$ and $10$ volts, consistent with the maximum principle.

Which of the following functions satisfies Laplace's equation $\nabla^2 u = 0$ on $\mathbb{R}^2$ ?

According to the mean value property, if $u$ is harmonic and $u(x_0) = 7$ , what is the average of $u$ over any circle of radius $2$ centred at $x_0$ (as long as the circle and its interior lie in the domain of harmonicity)?

A thin metal plate shaped like a disk has its boundary held at temperatures ranging between $10^\circ\text{C}$ and $30^\circ\text{C}$ , and has reached steady state (so the interior temperature is harmonic). What can you conclude about the interior temperature?

In image inpainting (filling in a missing or damaged region of a photo), a common technique solves the Dirichlet problem for Laplace's equation, treating the known pixels around the hole as boundary data. Why is this a sensible approach?

References

Lawrence C. Evans (2010). Partial Differential Equations
Walter A. Strauss (2007). Partial Differential Equations: An Introduction