Harmonic Functions

Leave the heat equation running long enough with fixed boundary temperatures and it settles into a steady state, where $u_t = 0$. What remains is Laplace's equation $\Delta u = 0$, and its solutions are the harmonic functions. They are the equilibria of diffusion, and their properties are the static versions of everything the heat equation does: they average, they smooth, and they cannot have interior maxima.

Book 5A met harmonic functions in the plane as real parts of holomorphic functions. Here there are no complex numbers to lean on, and every result is proved in every dimension, from one formula: the value of a harmonic function at a point is its average over any sphere around that point. From the mean value property follow the maximum principle, uniqueness for the Dirichlet problem, smoothness, derivative estimates, the Harnack inequality and Liouville's theorem. These are the prototypes of the estimates used on the Ricci flow, and this chapter is where their proofs are simplest.

By the end of this chapter you will be able to:

• write down the fundamental solution of the Laplacian and the Poisson formula for a ball;
• prove the mean value property and its converse;
• deduce the strong maximum principle, uniqueness, smoothness, derivative estimates and Liouville's theorem;
• prove a Harnack inequality from the mean value property;
• state Dirichlet's principle and show that harmonic functions minimise energy.

2.1Steady temperatures

2.2The fundamental solution

Laplace's equation is invariant under rotations, so look first for radial solutions $u = v(r)$, $r = |x|$. In $\mathbb{R}^n$, $\Delta v(r) = v'' + \frac{n - 1}{r}v'$ (Exercise 2.7), and the solutions of $v'' + \frac{n - 1}rv' = 0$ are $v = a + b\log r$ for $n = 2$ and $v = a + br^{2-n}$ for $n \geq 3$. With the right constants these give the fundamental solution

where $\alpha(n)$ is the volume of the unit ball (3A.5 Product Measures and Change of Variables). The constants are chosen so that $-\Delta\Phi = \delta_0$ in the sense of distributions (4A.8 Distributions and Weak Derivatives): $\Phi$ is the potential of a unit point charge, or the steady temperature from a unit point source. Consequently, for $f$ smooth with compact support,

solves Poisson's equation $-\Delta u = f$ (Evans, section 2.2.1). In three dimensions this is Newton's and Coulomb's inverse-square law: the potential of a mass or charge distribution is the superposition of $\frac{1}{4\pi|x - y|}$ over its points.

On a bounded region with boundary values, the fundamental solution is corrected by a harmonic function to vanish on the boundary, giving Green's function. For the ball $B_r(0)$ this can be done explicitly by reflection in the sphere, and the result is Poisson's formula: the solution of $\Delta u = 0$ in $B_r(0)$ with $u = g$ on the boundary is

(Evans, section 2.2.4). In the plane this is the Poisson kernel of 5A.4 Harmonic Functions and Conformal Mapping. The kernel is positive, has total integral $1$, and concentrates at the boundary point nearest $x$ as $x$ approaches the boundary: an approximate identity (3A.8 Convolution and Mollifiers).

2.3The mean value property

Write $\mathop{\rlap{\,\,-}\int}$ for an average: $\mathop{\rlap{\,\,-}\int}\nolimits_{\partial B_r(x)}u\,dS = \frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B_r(x)}u\,dS$ is the average over a sphere, and $\mathop{\rlap{\,\,-}\int}\nolimits_{B_r(x)}u\,dy$ the average over a ball.

The proof shows more. If $\Delta u \geq 0$ (subharmonic), then $\phi' \geq 0$ and $u(x) \leq \mathop{\rlap{\,\,-}\int}\nolimits_{\partial B_r(x)}u$: a subharmonic function lies below its averages. If $\Delta u \leq 0$ (superharmonic) the inequality reverses. This is the content of "the Laplacian measures how $u$ compares with its averages", the heuristic of 2B.7 Fourier Series and the First Heat Equation, made exact (Figure 2.1).

The converse holds too: a continuous function with the mean value property on every small sphere is harmonic, and in particular smooth (Exercise 2.8). The mean value property is therefore a complete characterisation of harmonic functions, and it needs no derivatives to state. That makes it the right definition in settings where derivatives are not available, such as graphs: a function on the vertices of a graph is called harmonic if its value at each vertex is the average of its neighbours' values. The figure in Figure 2.2 was computed by exactly that definition on a grid.

2.4Consequences of the mean value property

This is the argument of 5A.2 Cauchy’s Theorem and Its Consequences's maximum modulus principle, now in every dimension. A different proof, which doesn't use the mean value property, works for general elliptic operators: at an interior maximum the Hessian is negative semidefinite (2B.8 Calculus in Several Variables), so $\Delta u \leq 0$ there; it is the starting point of 6A.4 Maximum Principles.

Smoothness. A harmonic function is $C^\infty$, however little regularity was assumed. Let $\eta_\varepsilon$ be a radial mollifier (3A.8 Convolution and Mollifiers). Because $\eta_\varepsilon$ is radial with integral $1$, integrating the mean value property over spheres gives $u * \eta_\varepsilon = u$ wherever the convolution is defined. Convolutions with smooth kernels are smooth, so $u$ is. Harmonic functions are even real-analytic (Evans, section 2.2.3), but smoothness is what is used.

Derivative estimates. Each partial derivative $\partial_iu$ is harmonic too, so it has the mean value property, and by the divergence theorem

which gives $|\nabla u(x_0)| \leq \frac{2n}{r}\max_{\overline{B_{r/2}(x_0)}}|u|$ (Exercise 2.9). Iterating, the $k$-th derivatives at $x_0$ are bounded by $\frac{C_k}{r^k}\max_{B_r(x_0)}|u|$. A bound on $u$ on a ball controls all its derivatives at the centre, with the scaling $r^{-k}$: the real-variable form of the Cauchy estimates of 5A.2 Cauchy’s Theorem and Its Consequences.

2.5The Harnack inequality

The maximum principle compares a harmonic function with its boundary values. The Harnack inequality compares a positive harmonic function with itself: its values at nearby points are comparable.

The constant doesn't depend on $u$: a positive harmonic function cannot be large at one point and tiny at a nearby one. The Harnack inequality gives a second Liouville theorem: a harmonic function on $\mathbb{R}^n$ that is bounded below is constant (Exercise 2.10). In two dimensions this was the rehearsal of 5A.2 Cauchy’s Theorem and Its Consequences, there proved with complex analysis.

2.6Dirichlet's principle

Harmonic functions have a variational characterisation: they minimise energy. For $w$ on a bounded region $U$, let

the Dirichlet energy.

So Laplace's equation is the Euler–Lagrange equation of the Dirichlet energy. Riemann used this principle in 1851 to construct harmonic functions, assuming a minimiser exists; Weierstrass objected that it need not; Hilbert's direct method (4A.6 Weak Convergence and the Direct Method) and the weak solutions of 6A.5 Weak Solutions and Elliptic Regularity repair the argument. In 6A.9 Calculus of Variations and Gradient Flows the heat equation is shown to be the gradient flow of the same energy: heat flows so as to decrease $E$ as fast as possible.

2.7History

Pierre-Simon Laplace used the equation that bears his name in the 1780s in his work on gravitational attraction; Siméon Denis Poisson added the source term in 1813. George Green's 1828 Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism introduced Green's functions and the integral identities behind them. Carl Friedrich Gauss proved the mean value property for harmonic functions in three dimensions in his 1840 memoir on the theory of attraction. Riemann named the Dirichlet principle after his teacher and used it in 1851. Axel Harnack proved his inequality in 1887, for harmonic functions in the plane. Faraday's shielding experiment dates from 1836.

2.8Exercises

Solution

$\partial_iu = v'(r)\frac{x_i}{r}$, and $\partial_i\partial_iu = v''\frac{x_i^2}{r^2} + v'\big(\frac1r - \frac{x_i^2}{r^3}\big)$. Summing over $i$: $v'' + v'\frac{n - 1}{r}$. For $v = r^{2-n}$: $(2 - n)(1 - n)r^{-n} + (n - 1)(2 - n)r^{-n} = 0$. For $v = r^2$: $2 + 2(n - 1) = 2n$.

Solution

$|\partial_iu(x_0)| \leq \frac{1}{|B_{r/2}|}|\partial B_{r/2}|\max|u| = \frac{n}{r/2}\max|u| = \frac{2n}{r}\max|u|$; the same bound holds for $\nabla u\cdot e$ for any unit vector $e$, hence for $|\nabla u|$. If $|u| \leq C(1 + |x|)$, apply the estimate on $B_r(x_0)$: $|\nabla u(x_0)| \leq \frac{2n}{r}C(1 + |x_0| + r) \to 2nC$ as $r \to \infty$. So each $\partial_iu$ is bounded and harmonic, hence constant by Liouville, and $u$ is affine.

Solution

For all $x, y \in B_r(0)$, $u(x) - m \leq 3^n(u(y) - m)$, so $\sup_{B_r}(u - m) \leq 3^n\inf_{B_r}(u - m)$. As $r \to \infty$ the right side tends to $3^n\cdot0$, because $\inf_{\mathbb{R}^n}(u - m) = 0$, while the left side is non-decreasing in $r$. So $\sup_{B_r}(u - m) = 0$ for every $r$.

Solution

In polar coordinates $|\nabla w|^2 = (\partial_rw)^2 + r^{-2}(\partial_\theta w)^2 = r^{2k-2}(k^2\cos^2\theta + \sin^2\theta)$. So $2E = \int_0^{2\pi}\int_0^1r^{2k-1}(k^2\cos^2\theta + \sin^2\theta)\,dr\,d\theta = \frac{1}{2k}\cdot\pi(k^2 + 1)$. And $\frac{k^2 + 1}{k} = k + \frac1k \geq 2$ with equality only at $k = 1$.

Solution

(a) $\Delta\frac12\sum_i(\partial_iu)^2 = \sum_{i,j}\partial_j(\partial_iu\,\partial_j\partial_iu) = \sum_{i,j}(\partial_j\partial_iu)^2 + \sum_i\partial_iu\,\partial_i(\Delta u)$, using $\sum_j\partial_j\partial_j\partial_iu = \partial_i\Delta u$. (b) $\Delta|\nabla u|^2 = 2|\nabla^2u|^2 \geq 0$; the maximum principle for subharmonic functions follows from the inequality $u(x) \leq \mathop{\rlap{\,\,-}\int}\nolimits_{B_r(x)}u$ exactly as in Theorem 2.2.