Book 6B

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Course 6Book 6B: Linear PDE: Characteristics, Waves and RegularityChapter 6

Perron’s Method and Barriers

Subharmonic functions, Perron’s solution of the Dirichlet problem, barriers, regular points and Lebesgue’s spine.

39 min read · Updated Oct 20, 2026

Read with Jost, Partial Differential Equations, section 2.2 (subharmonic functions and the maximum principle) and chapter 4 (existence techniques based on the maximum principle: difference methods, the Perron method, the alternating method of Schwarz, boundary regularity); Taylor I, §5.5 ("The Dirichlet problem on a domain with rough boundary"). Evans does not cover Perron's method; Gilbarg and Trudinger, Elliptic Partial Differential Equations of Second Order, section 2.8, is the classic reference. This chapter proves every result it uses except Wiener's criterion.

In this chapter · 8 sections
  1. 6.1Random walks and electrical networks
  2. 6.2Subharmonic functions without derivatives
  3. 6.3Perron's method
  4. 6.4Barriers and regular points
  5. 6.5Where the Dirichlet problem fails
  6. 6.6The discrete Dirichlet problem
  7. 6.7History
  8. 6.8Exercises
  9. ↗Free resources

Book 6A solved the Dirichlet problem for the Laplacian in two ways: by Poisson's formula on a ball (6A.2 Harmonic Functions, 6B.4 Green’s Functions, Separation of Variables and Transforms) and by Hilbert-space methods on any bounded domain, in the weak sense (6A.6 Weak Solutions and Elliptic Regularity). Neither answers the most classical form of the question: given a bounded domain Ω\Omega and a continuous function φ\varphi on its boundary, is there a function harmonic in Ω\Omega, continuous on Ω‾\overline\Omega, equal to φ\varphi on ∂Ω\partial\Omega? The weak solution lies in H1H^1 and has boundary values only in the trace sense; and a continuous φ\varphi need not be the trace of any H1H^1 function at all.

Oskar Perron's method (1923) answers the question with nothing but the maximum principle. It builds the solution as the supremum of all subharmonic functions lying below the data, shows that this supremum is harmonic, and then asks separately whether it takes the right boundary values. The answer to the second question depends on the geometry of the boundary, through barriers, and it can fail: at an isolated boundary point, or at the tip of a sufficiently sharp inward spike. The method separates the two halves of the problem as cleanly as possible: interior existence is automatic; boundary behaviour is local and geometric.

The same chapter treats the discrete Dirichlet problem, where the maximum principle again gives existence, and proves that finite-difference solutions converge at the rate h2h^2. Barriers and comparison functions are the everyday tools of elliptic and parabolic theory, and the barrier built from the distance to the boundary (Exercise 6.19) is the same computation as the Laplacian comparison theorem of Riemannian geometry (9B.1 Laplacian Comparison).

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

  • define subharmonic functions without derivatives, and prove their maximum and comparison principles and the harmonic lifting construction;
  • prove Harnack's convergence theorem, and use it to prove that Perron's solution is harmonic;
  • define barriers and regular boundary points, prove that a barrier gives the right boundary values, and construct barriers from exterior spheres;
  • explain why the punctured disc and Lebesgue's spine have irregular boundary points, and state Wiener's criterion;
  • set up the discrete Dirichlet problem, prove its maximum principle, and prove O(h2)O(h^2) convergence of the five-point scheme.

Random walks and electrical networks

In the world Model Voltages, currents and wandering particles

Take an electrical network: nodes joined by unit resistors. Hold some boundary nodes at fixed voltages and let the rest float. By Kirchhoff's current law, the net current into each interior node is zero, and by Ohm's law the current along a resistor is the voltage difference across it. Together they say that the voltage at each interior node is the average of the voltages at its neighbours: the voltage is a discrete harmonic function, and finding it is a discrete Dirichlet problem.

The same numbers have a probabilistic meaning. Start a random walker at an interior node; at each step it moves to a uniformly chosen neighbour; it stops at the first boundary node it reaches. Then the voltage at the start equals the expected boundary voltage at the stopping point, because both satisfy the same averaging equation with the same boundary values, and that equation has only one solution (the discrete maximum principle below). Peter Doyle and Laurie Snell's book Random Walks and Electric Networks (1984) develops this dictionary: effective resistance measures how likely a walk is to escape, and the classical fact that random walk on the integer lattice returns to its start in one and two dimensions but not in three becomes a statement about resistances to infinity.

In the continuum the walker becomes a Brownian motion, the network a domain, and the voltage the solution of the Dirichlet problem: u(x)u(x) is the expected boundary value where Brownian motion from xx first leaves the domain (Shizuo Kakutani, 1944; 6D.1 Brownian Motion). Whether uu takes the right value at a boundary point ξ\xi then has a vivid answer: it does exactly when a Brownian path started at ξ\xi leaves the domain immediately. At an isolated boundary point, a Brownian path in the plane misses the point almost surely and wanders off to the rest of the boundary, so the boundary value there is ignored.

Subharmonic functions without derivatives

In 6A.2 Harmonic Functions a C2C^2 function with Δu≥0\Delta u \geq 0 was called subharmonic, and shown to lie below its averages. Perron's method needs a notion that survives taking suprema and maxima, so it is defined by comparison with harmonic functions instead.

Definition 6.1 Subharmonic and superharmonic functions

Let Ω\Omega be open. A continuous function uu on Ω\Omega is subharmonic if for every closed ball B‾⊂Ω\overline B \subset \Omega and every function hh harmonic in BB and continuous on B‾\overline B,

u≤h  on ∂B  ⟹  u≤h  in B.u \leq h \ \text{ on } \partial B \implies u \leq h \ \text{ in } B.

It is superharmonic if −u-u is subharmonic.

In words: a subharmonic function lies below the harmonic function with the same boundary values on any ball. On a ball the harmonic function with given continuous boundary values exists and is given by Poisson's integral (6B.4 Green’s Functions, Separation of Variables and Transforms), so the definition can be checked.

Proposition 6.2 Properties of subharmonic functions

Let uu be continuous on Ω\Omega.

  1. If u∈C2u \in C^2 and Δu≥0\Delta u \geq 0, then uu is subharmonic.
  2. If uu is subharmonic, then u(x)≤  -∫∂B(x,r)u dSu(x) \leq \mathop{\rlap{\,\,-}\int}\nolimits_{\partial B(x,r)}u\,dS whenever B(x,r)‾⊂Ω\overline{B(x, r)} \subset \Omega.
  3. (Strong maximum principle.) If Ω\Omega is connected, uu is subharmonic, and uu attains its maximum over Ω\Omega, then uu is constant.
  4. (Comparison.) If Ω\Omega is bounded, uu is subharmonic, vv is superharmonic, both are continuous on Ω‾\overline\Omega, and u≤vu \leq v on ∂Ω\partial\Omega, then u≤vu \leq v in Ω\Omega (and either u<vu < v in each component or u≡vu \equiv v there).
  5. The maximum of finitely many subharmonic functions is subharmonic.

Proof. (1) If u≤hu \leq h on ∂B\partial B, then u−hu - h is C2C^2 in BB (continuous on B‾\overline B) with Δ(u−h)≥0\Delta(u - h) \geq 0; by the weak maximum principle for subharmonic C2C^2 functions (6A.2 Harmonic Functions, 6A.5 Maximum Principles), u−h≤max⁡∂B(u−h)≤0u - h \leq \max_{\partial B}(u - h) \leq 0.

(2) Let hh be the Poisson integral of uu on B=B(x,r)B = B(x, r): harmonic in BB, continuous on B‾\overline B, equal to uu on ∂B\partial B (6B.4 Green’s Functions, Separation of Variables and Transforms). By the definition, u≤hu \leq h in BB; at the centre, h(x)=  -∫∂Bu dSh(x) = \mathop{\rlap{\,\,-}\int}\nolimits_{\partial B}u\,dS by the mean value property.

(3) Let M=max⁡uM = \max u and A={u=M}A = \{u = M\}, non-empty and closed in Ω\Omega. If x∈Ax \in A and B(x,r)‾⊂Ω\overline{B(x, r)} \subset \Omega, then by (2) M=u(x)≤  -∫∂B(x,ρ)u≤MM = u(x) \leq \mathop{\rlap{\,\,-}\int}\nolimits_{\partial B(x,\rho)}u \leq M for every ρ≤r\rho \leq r, so u=Mu = M on every such sphere (a continuous function ≤M\leq M with average MM is identically MM), hence on B(x,r)B(x, r). So AA is open, and by connectedness A=ΩA = \Omega.

(4) u−vu - v is subharmonic: if hh is harmonic on BB with u−v≤hu - v \leq h on ∂B\partial B, let h~\tilde h be the harmonic function on BB with h~=v\tilde h = v on ∂B\partial B; then u≤h+h~u \leq h + \tilde h on ∂B\partial B, so u≤h+h~u \leq h + \tilde h in BB, and v≥h~v \geq \tilde h in BB (superharmonic), so u−v≤hu - v \leq h. Now apply (3) on each component: if u−vu - v attained a positive maximum inside, it would be constant and positive on a component, contradicting u−v≤0u - v \leq 0 on its boundary.

(5) If u1,u2u_1, u_2 are subharmonic and max⁡(u1,u2)≤h\max(u_1, u_2) \leq h on ∂B\partial B, then each ui≤hu_i \leq h on ∂B\partial B, hence in BB.

The definition is local in the following sense: a continuous function satisfying the sub-mean-value inequality (2) on all small spheres around each point is subharmonic (Exercise 6.14). So ∣x∣|x|, max⁡(x1,0)\max(x_1, 0) and the maximum of two harmonic functions are subharmonic, although none is C2C^2.

Harmonic lifting. The key construction raises a subharmonic function on a ball to the harmonic function with the same boundary values there.

Lemma 6.3 Harmonic lifting

Let uu be subharmonic on Ω\Omega and B‾⊂Ω\overline B \subset \Omega a closed ball. Define UU to be equal to uu outside BB, and inside BB to be the Poisson integral of the boundary values of uu on ∂B\partial B. Then UU is subharmonic on Ω\Omega, and U≥uU \geq u.

Proof. UU is continuous (the Poisson integral takes the boundary values continuously, 6B.4 Green’s Functions, Separation of Variables and Transforms), and U≥uU \geq u in BB by the definition of subharmonic. Let B′‾⊂Ω\overline{B'} \subset \Omega be another ball and hh harmonic on B′B', continuous on B′‾\overline{B'}, with U≤hU \leq h on ∂B′\partial B'. Since u≤Uu \leq U, also u≤hu \leq h on ∂B′\partial B', so u≤hu \leq h in B′B' because uu is subharmonic. In particular U=u≤hU = u \leq h on B′∖BB'\setminus B. On B′∩BB'\cap B, the function U−hU - h is harmonic, and its boundary consists of points of ∂B′\partial B' (where U≤hU \leq h by assumption) and points of ∂B\partial B lying in B′B' (where U=u≤hU = u \leq h, as just shown). By the maximum principle, U≤hU \leq h on B′∩BB'\cap B too.

Harnack's convergence theorem. The other ingredient is a compactness statement for monotone sequences of harmonic functions, which follows from the Harnack inequality of 6A.2 Harmonic Functions.

Theorem 6.4 Harnack's convergence theorem

Let Ω\Omega be connected and u1≤u2≤⋯u_1 \leq u_2 \leq \cdots harmonic in Ω\Omega. If uk(y)u_k(y) is bounded at one point yy, then uku_k converges uniformly on compact subsets of Ω\Omega to a harmonic function. Otherwise uk→+∞u_k \to +\infty uniformly on compact subsets.

Proof. For k>lk > l, uk−ul≥0u_k - u_l \geq 0 is harmonic. Let K⊂ΩK \subset \Omega be compact, and enlarge it to a compact connected set containing yy (possible since Ω\Omega is connected and open). The Harnack inequality gives a constant CC, depending on KK and Ω\Omega only, with

sup⁡K(uk−ul)≤C(uk(y)−ul(y)),inf⁡K(uk−u1)≥C−1(uk(y)−u1(y)).\sup_K(u_k - u_l) \leq C(u_k(y) - u_l(y)), \qquad \inf_K(u_k - u_1) \geq C^{-1}(u_k(y) - u_1(y)).

If uk(y)u_k(y) is bounded, it converges (it is increasing), so the first inequality shows (uk)(u_k) is uniformly Cauchy on KK; the limit is continuous, and it has the mean value property (pass to the limit in uk(x)=  -∫uku_k(x) = \mathop{\rlap{\,\,-}\int}u_k on small spheres), so it is harmonic (6A.2 Harmonic Functions). If uk(y)→∞u_k(y) \to \infty, the second inequality gives inf⁡Kuk→∞\inf_Ku_k \to \infty.

Perron's method

Let Ω\Omega be bounded and φ\varphi continuous on ∂Ω\partial\Omega. A continuous function vv on Ω‾\overline\Omega is a subfunction for φ\varphi if it is subharmonic in Ω\Omega and v≤φv \leq \varphi on ∂Ω\partial\Omega. Let SφS_\varphi be the set of subfunctions. It is not empty (the constant min⁡φ\min\varphi belongs to it), and every v∈Sφv \in S_\varphi satisfies v≤max⁡φv \leq \max\varphi, by the maximum principle. So

u(x)=sup⁡v∈Sφv(x)u(x) = \sup_{v\in S_\varphi}v(x)

is a well-defined bounded function on Ω\Omega, the Perron solution. If the Dirichlet problem has a solution ww, then w∈Sφw \in S_\varphi, and v≤wv \leq w for every v∈Sφv \in S_\varphi by comparison, so u=wu = w: the Perron solution is the only possible candidate.

Theorem 6.5 Perron's theorem

The Perron solution uu is harmonic in Ω\Omega.

Proof. Fix y∈Ωy \in \Omega and a closed ball B‾⊂Ω\overline B \subset \Omega centred at yy. We show uu is harmonic in BB.

An increasing sequence approaching u(y)u(y). By definition of the supremum there are vk∈Sφv_k \in S_\varphi with vk(y)→u(y)v_k(y) \to u(y). Replacing vkv_k by max⁡(v1,…,vk,min⁡φ)\max(v_1, \ldots, v_k, \min\varphi), which is still a subfunction by Proposition 6.2(5), we may assume vkv_k increases with kk. Let VkV_k be the harmonic lifting of vkv_k in BB (Lemma 6.3): Vk∈SφV_k \in S_\varphi (it is subharmonic and equals vkv_k near ∂Ω\partial\Omega), Vk≥vkV_k \geq v_k, and VkV_k is increasing in kk (on BB, the Poisson integrals of increasing boundary values increase). So vk(y)≤Vk(y)≤u(y)v_k(y) \leq V_k(y) \leq u(y), and Vk(y)→u(y)V_k(y) \to u(y).

The limit is harmonic. On BB the VkV_k are harmonic, increasing and bounded by max⁡φ\max\varphi, so by Theorem 6.4 they converge locally uniformly in BB to a harmonic function VV, with V≤uV \leq u in BB and V(y)=u(y)V(y) = u(y).

The limit is uu. Suppose V(z)<u(z)V(z) < u(z) at some z∈Bz \in B. Choose v~∈Sφ\tilde v \in S_\varphi with V(z)<v~(z)V(z) < \tilde v(z). Let WkW_k be the harmonic lifting in BB of max⁡(Vk,v~)\max(V_k, \tilde v); as before, the WkW_k increase to a harmonic function WW on BB with V≤W≤uV \leq W \leq u in BB. Then V−W≤0V - W \leq 0 is harmonic in BB, and at the centre V(y)−W(y)≥u(y)−u(y)=0V(y) - W(y) \geq u(y) - u(y) = 0, an interior maximum; by the strong maximum principle V≡WV \equiv W in BB. But W(z)≥v~(z)>V(z)W(z) \geq \tilde v(z) > V(z), a contradiction. So u=Vu = V in BB, and uu is harmonic there.

The proof never asked how uu behaves at the boundary, and it can't: that depends on the boundary.

Barriers and regular points

Definition 6.6 Barrier and regular point

Let ξ∈∂Ω\xi \in \partial\Omega. A barrier at ξ\xi is a function ww, continuous on Ω‾\overline\Omega and superharmonic in Ω\Omega, with w(ξ)=0w(\xi) = 0 and w>0w > 0 on Ω‾∖{ξ}\overline\Omega\setminus\{\xi\}. A boundary point is regular if the Perron solution satisfies u(x)→φ(ξ)u(x) \to \varphi(\xi) as x→ξx \to \xi, for every continuous φ\varphi.

Theorem 6.7 A barrier makes a point regular

If there is a barrier at ξ∈∂Ω\xi \in \partial\Omega, then ξ\xi is regular.

Proof. Fix ε>0\varepsilon > 0, let M=max⁡∂Ω∣φ∣M = \max_{\partial\Omega}|\varphi|, and choose δ>0\delta > 0 with ∣φ(x)−φ(ξ)∣<ε|\varphi(x) - \varphi(\xi)| < \varepsilon for x∈∂Ωx \in \partial\Omega, ∣x−ξ∣<δ|x - \xi| < \delta. On the compact set Ω‾∖B(ξ,δ)\overline\Omega\setminus B(\xi, \delta) the barrier has a positive minimum mm; let k=2M/mk = 2M/m, so that kw≥2Mkw \geq 2M there. Consider

w−=φ(ξ)−ε−kw,w+=φ(ξ)+ε+kw.w^- = \varphi(\xi) - \varepsilon - kw, \qquad w^+ = \varphi(\xi) + \varepsilon + kw.

w−w^- is subharmonic (−w-w is) and w−≤φw^- \leq \varphi on ∂Ω\partial\Omega: near ξ\xi because φ>φ(ξ)−ε\varphi > \varphi(\xi) - \varepsilon there and kw≥0kw \geq 0, and away from ξ\xi because w−≤φ(ξ)−2M≤−M≤φw^- \leq \varphi(\xi) - 2M \leq -M \leq \varphi. So w−∈Sφw^- \in S_\varphi and w−≤uw^- \leq u. Similarly w+w^+ is superharmonic with w+≥φw^+ \geq \varphi on ∂Ω\partial\Omega, so every v∈Sφv \in S_\varphi satisfies v≤w+v \leq w^+ by comparison, and u≤w+u \leq w^+. Hence

∣u(x)−φ(ξ)∣≤ε+kw(x),|u(x) - \varphi(\xi)| \leq \varepsilon + kw(x),

and since w(x)→w(ξ)=0w(x) \to w(\xi) = 0 as x→ξx \to \xi, lim sup⁡x→ξ∣u(x)−φ(ξ)∣≤ε\limsup_{x\to\xi}|u(x) - \varphi(\xi)| \leq \varepsilon. As ε\varepsilon was arbitrary, u(x)→φ(ξ)u(x) \to \varphi(\xi).

The proof uses only the barrier's sign and its superharmonicity: the barrier is a scaffold that pins uu near φ(ξ)\varphi(\xi) from above and below. It is a local property: a barrier defined only on Ω‾∩B(ξ,r)\overline\Omega\cap B(\xi, r) can be extended to a global one (Gilbarg–Trudinger, section 2.8), so regularity is decided by the shape of the boundary near ξ\xi.

Corollary 6.8 Solvability of the Dirichlet problem

If every point of ∂Ω\partial\Omega has a barrier, then for every continuous φ\varphi on ∂Ω\partial\Omega the Dirichlet problem has a unique solution u∈C2(Ω)∩C(Ω‾)u \in C^2(\Omega)\cap C(\overline\Omega).

Exterior spheres. The simplest source of barriers is a ball outside Ω\Omega touching ξ\xi: suppose there is a closed ball B(y,R)‾\overline{B(y, R)} with B(y,R)‾∩Ω‾={ξ}\overline{B(y, R)}\cap\overline\Omega = \{\xi\} (the exterior sphere condition). Then

w(x)={R2−n−∣x−y∣2−n,n≥3,log⁡∣x−y∣R,n=2,w(x) = \begin{cases}R^{2-n} - |x - y|^{2-n}, & n \geq 3,\\ \log\dfrac{|x - y|}{R}, & n = 2,\end{cases}

is harmonic away from yy, vanishes at ξ\xi (where ∣x−y∣=R|x - y| = R) and is positive at all other points of Ω‾\overline\Omega (where ∣x−y∣>R|x - y| > R): a barrier. Every bounded domain with C2C^2 boundary, and every convex domain, satisfies the exterior sphere condition at each boundary point (for convex domains use a ball beyond a supporting plane, Exercise 6.15). So the Dirichlet problem is solvable for continuous data on every bounded convex domain and every bounded domain with C2C^2 boundary. A weaker condition suffices: an exterior cone with vertex at ξ\xi (Poincaré's and Zaremba's condition) gives a barrier too, so Lipschitz domains, including polygons and polyhedra, are fine (Gilbarg–Trudinger, problem 2.12; Jost, section 4.4).

Where the Dirichlet problem fails

Example 6.9 The punctured disc (Zaremba, 1911)

Let Ω={0<∣x∣<1}\Omega = \{0 < |x| < 1\} in the plane, with φ=0\varphi = 0 on the circle ∣x∣=1|x| = 1 and φ(0)=1\varphi(0) = 1 at the puncture. Every subfunction v∈Sφv \in S_\varphi is at most 11. For ε>0\varepsilon > 0 compare vv with the harmonic function εlog⁡1∣x∣\varepsilon\log\frac{1}{|x|} on the annulus δ<∣x∣<1\delta < |x| < 1: on ∣x∣=1|x| = 1, v≤0=εlog⁡1v \leq 0 = \varepsilon\log1; on ∣x∣=δ|x| = \delta, v≤1≤εlog⁡1δv \leq 1 \leq \varepsilon\log\frac1\delta once δ\delta is small enough. By comparison v≤εlog⁡1∣x∣v \leq \varepsilon\log\frac{1}{|x|} on the annulus, and since δ\delta can be taken as small as we like, on all of Ω\Omega. Letting ε→0\varepsilon \to 0, v≤0v \leq 0. So the Perron solution is u≡0u \equiv 0, which does not tend to φ(0)=1\varphi(0) = 1 at the origin. No solution exists: a harmonic function on the punctured disc that is bounded near 00 extends harmonically across 00 (removable singularities, Exercise 6.16), and then the maximum principle forces it to be 00. The origin is an irregular boundary point. The physical picture: a single point at a different potential, inside a grounded circle, carries no charge in the limit and has no effect.

In the plane, isolated points are the only kind of problem for domains whose complement has no tiny pieces: if each boundary point lies in a connected piece of the complement containing more than one point, it is regular. In three dimensions, a boundary can fail to be regular even at a point of a solid, connected complement.

Example 6.10 Lebesgue's spine (1913)

In R3\mathbb{R}^3, take a domain whose complement near the origin is a thin spike, a solid of revolution around the negative x3x_3-axis whose radius shrinks to zero at the origin faster than any power (like e−1/∣x3∣e^{-1/|x_3|}). Henri Lebesgue showed that the tip of such a spike is irregular: the potential of a charge distribution along the spike can be arranged so that its level surfaces near the tip do not close up around it, and the Dirichlet problem with suitable continuous data has no continuous solution. A spike whose radius shrinks only like a power of ∣x3∣|x_3| is regular (an exterior cone, flattened, still provides a barrier). Brownian motion makes the dichotomy plausible: a path started at the tip of a very sharp spike almost surely wanders into Ω\Omega without touching the spike again for a while, so it does not leave Ω\Omega immediately.

The exact criterion was found by Norbert Wiener in 1924. For n≥3n \geq 3 and λ∈(0,1)\lambda \in (0, 1), let Ek={x∉Ω:λk+1≤∣x−ξ∣≤λk}E_k = \{x \notin \Omega : \lambda^{k+1} \leq |x - \xi| \leq \lambda^k\} be the pieces of the complement in dyadic-type shells around ξ\xi. Then:

Theorem 6.11 Wiener's criterion (statement)

A boundary point ξ\xi of Ω⊂Rn\Omega \subset \mathbb{R}^n, n≥3n \geq 3, is regular if and only if

∑k=0∞cap⁡(Ek)λk(n−2)=∞,\sum_{k=0}^\infty\frac{\operatorname{cap}(E_k)}{\lambda^{k(n-2)}} = \infty,

where cap⁡\operatorname{cap} is the Newtonian capacity: cap⁡(E)=inf⁡{∫Rn∣∇v∣2:v∈Cc∞, v≥1 on E}\operatorname{cap}(E) = \inf\{\int_{\mathbb{R}^n}|\nabla v|^2 : v \in C_c^\infty,\ v \geq 1 \text{ on } E\}.

Capacity measures how much a set matters to potential theory: a ball of radius ρ\rho has capacity proportional to ρn−2\rho^{n-2}, a point has capacity zero, and a thin spike of radius ρ(t)\rho(t) has capacity per unit length about 1log⁡(1/ρ)\frac{1}{\log(1/\rho)}. The criterion says the complement must be "thick enough at all scales" near ξ\xi: the sum compares the capacity of each piece with that of the shell. For the exterior cone each term is comparable to a constant, so the sum diverges; for Lebesgue's spike with radius e−1/∣x3∣e^{-1/|x_3|} the terms decay fast enough that it converges. Taylor II, §11.6 (Newtonian capacity), develops capacity and its probabilistic meaning.

The discrete Dirichlet problem

Perron's method is an existence proof, not an algorithm. The algorithm is to discretise. On a grid of spacing hh, replace Δ\Delta by the five-point Laplacian (in two dimensions; 2n+12n + 1 points in nn)

Δhu(x)=∑i=1nu(x+hei)+u(x−hei)−2u(x)h2.\Delta_hu(x) = \sum_{i=1}^n\frac{u(x + he_i) + u(x - he_i) - 2u(x)}{h^2}.

Let Ωh\Omega_h be the grid points in Ω\Omega whose 2n2n neighbours are all in Ω‾\overline\Omega (interior points; for simplicity, assume the boundary passes through grid points, as for a rectangle), and ∂Ωh\partial\Omega_h the grid points on ∂Ω\partial\Omega that are neighbours of interior points. The discrete Dirichlet problem is: Δhu=f\Delta_hu = f on Ωh\Omega_h, u=φu = \varphi on ∂Ωh\partial\Omega_h. Note that Δhu(x)=0\Delta_hu(x) = 0 says exactly that u(x)u(x) is the average of its 2n2n neighbours: discrete harmonic, as in the electrical network.

Proposition 6.12 The discrete maximum principle

If Δhu≥0\Delta_hu \geq 0 on Ωh\Omega_h, then max⁡Ωh∪∂Ωhu=max⁡∂Ωhu\max_{\Omega_h\cup\partial\Omega_h}u = \max_{\partial\Omega_h}u. Consequently the discrete Dirichlet problem has exactly one solution for every ff and φ\varphi.

Proof. Suppose the maximum MM is attained at an interior point xx. Then Δhu(x)≥0\Delta_hu(x) \geq 0 says u(x)≤u(x) \leq the average of its neighbours, while each neighbour has u≤M=u(x)u \leq M = u(x). So all neighbours have u=Mu = M. Repeat from each neighbour that is interior; the grid is connected through interior points to the boundary (follow a coordinate direction until leaving Ωh\Omega_h), so MM is attained at a boundary point. For the consequence: the discrete problem is a square linear system, one equation and one unknown for each interior point. If uu solves it with f=0f = 0 and φ=0\varphi = 0, the maximum principle applied to uu and −u-u gives u=0u = 0. A square linear system with only the zero solution of the homogeneous problem is invertible.

Existence by uniqueness: the finite-dimensional Fredholm alternative (6A.6 Weak Solutions and Elliptic Regularity). The matrix is symmetric, negative definite and sparse, and it is solved in practice by iterative methods; the simplest, Jacobi iteration, replaces each interior value by the average of its neighbours, over and over, which is exactly the relaxation that computed 6A.2 Harmonic Functions's saddle figure. It converges, slowly, because the iteration matrix has spectral radius less than one (its largest eigenvalue is about 1−π2h221 - \frac{\pi^2h^2}{2} on a unit square), and multigrid methods accelerate it enormously.

Theorem 6.13 Convergence of the five-point scheme

Let Ω⊆B(0,R)\Omega \subseteq B(0, R) be a domain (such as a rectangle) for which the grid fits the boundary, and let u∈C4(Ω‾)u \in C^4(\overline\Omega) solve Δu=f\Delta u = f in Ω\Omega, u=φu = \varphi on ∂Ω\partial\Omega. Let uhu_h solve the discrete problem with the same ff and φ\varphi at the grid points. Then

max⁡Ωh∣uh−u∣≤R2M424h2,M4=max⁡isup⁡Ω∣∂i4u∣.\max_{\Omega_h}|u_h - u| \leq \frac{R^2M_4}{24}h^2, \qquad M_4 = \max_i\sup_\Omega|\partial_i^4u|.

Proof. Consistency. By Taylor's theorem in one variable, u(x+hei)+u(x−hei)−2u(x)h2=∂i2u(x)+h212∂i4u(ξi)\frac{u(x + he_i) + u(x - he_i) - 2u(x)}{h^2} = \partial_i^2u(x) + \frac{h^2}{12}\partial_i^4u(\xi_i) for some ξi\xi_i on the segment. Summing over ii, the truncation error τ(x)=Δhu(x)−Δu(x)\tau(x) = \Delta_hu(x) - \Delta u(x) satisfies ∣τ∣≤nh212M4|\tau| \leq \frac{nh^2}{12}M_4.

Stability. Let e=uh−ue = u_h - u on the grid. Then Δhe=f−(f+τ)=−τ\Delta_he = f - (f + \tau) = -\tau on Ωh\Omega_h, and e=0e = 0 on ∂Ωh\partial\Omega_h. Compare with w(x)=h2M424(R2−∣x∣2)w(x) = \frac{h^2M_4}{24}(R^2 - |x|^2): since Δh∣x∣2=2n\Delta_h|x|^2 = 2n exactly (the scheme is exact on quadratics), Δhw=−nh2M412≤−∣τ∣\Delta_hw = -\frac{nh^2M_4}{12} \leq -|\tau|, and w≥0w \geq 0 on ∂Ωh\partial\Omega_h. So Δh(e−w)=−τ−Δhw≥0\Delta_h(e - w) = -\tau - \Delta_hw \geq 0, and e−w≤0e - w \leq 0 on the boundary; by the discrete maximum principle e≤we \leq w. The same argument with −e-e gives −e≤w-e \leq w. Hence ∣e∣≤max⁡w≤h2M4R224|e| \leq \max w \leq \frac{h^2M_4R^2}{24}.

The proof has the structure of every convergence proof for numerical methods: consistency (the exact solution almost satisfies the scheme) plus stability (the scheme does not amplify errors) gives convergence. Here stability is the discrete maximum principle; in 6D.9 Finite Difference Methods it is the same principle for the heat equation, and the Lax equivalence theorem makes "consistency + stability = convergence" general. Jost, section 4.1, proves that solutions of the discrete problem converge to the solution of the continuous Dirichlet problem under much weaker assumptions, which gives another existence proof.

Schwarz's alternating method (1870) is a third route to existence, and an ancestor of modern domain decomposition: to solve the Dirichlet problem on the union of two overlapping domains where it is solvable separately (say a disc and a rectangle), solve on the first with the current guess as boundary data on the part of its boundary inside the second, then on the second, and alternate; the maximum principle shows the iterates converge (Jost, section 4.3).

Where this goes Barriers and comparison in the rest of the guide

Barriers control boundary behaviour throughout elliptic theory: the boundary gradient estimate for the minimal surface equation is a barrier argument, and it is why the Dirichlet problem for minimal graphs is solvable on convex domains but not on all domains (6C.6 Quasilinear Elliptic Equations and Minimal Surfaces). Sub- and supersolutions, Perron's subfunctions with a nonlinear operator, give existence for semilinear equations (6C.5 Fixed Points, Monotonicity and Sub- and Supersolutions). In Riemannian geometry the distance function to a point or a hypersurface is the universal barrier, and its Laplacian is controlled by curvature: that is the Laplacian comparison theorem (9B.1 Laplacian Comparison, and Exercise 6.19 here). Perron's method itself is used for viscosity solutions, whose existence is proved by taking the supremum of subsolutions exactly as in Theorem 6.5 (6C.9 Hamilton–Jacobi Equations and Viscosity Solutions).

History

Hermann Amandus Schwarz introduced his alternating method in 1870. Henri Poincaré used sweeping out (balayage), a precursor of Perron's method, in 1887–90, and proved solvability under an exterior cone condition. Stanisław Zaremba gave the punctured disc example in 1911, and Henri Lebesgue his spine in 1913. Oskar Perron's method was published in 1923, and Norbert Wiener's criterion in 1924. Shizuo Kakutani's representation of harmonic functions by Brownian motion appeared in 1944. Richard Courant, Kurt Friedrichs and Hans Lewy's 1928 paper on difference equations proved convergence of finite-difference schemes and introduced the CFL condition (6D.9 Finite Difference Methods). Doyle and Snell's book appeared in 1984.

Recall Where we stand

Subharmonic functions are defined by comparison with harmonic functions on balls; they satisfy the sub-mean-value inequality, the strong maximum principle and comparison, are closed under maxima, and can be lifted to harmonic functions on balls. Harnack's inequality makes increasing harmonic sequences converge locally uniformly. The Perron solution, the supremum of all subfunctions below the data, is harmonic in any bounded domain; it takes the boundary value at ξ\xi if there is a barrier there, for instance from an exterior sphere or cone, so the Dirichlet problem is solvable on convex and C2C^2 domains. Isolated points in the plane and Lebesgue's sharp spike in space are irregular, and Wiener's criterion characterises regularity by capacity. The discrete Dirichlet problem is solvable by the discrete maximum principle, and the five-point scheme converges at rate h2h^2 by consistency plus stability. 6B.7 Schauder Theory and the Continuity Method turns from existence to estimates: the Schauder theory.

Exercises

Exercise 6.14 The sub-mean-value property is enough

Let uu be continuous on Ω\Omega, and suppose every x∈Ωx \in \Omega has an rx>0r_x > 0 with u(x)≤  -∫∂B(x,r)u dSu(x) \leq \mathop{\rlap{\,\,-}\int}\nolimits_{\partial B(x,r)}u\,dS for all r<rxr < r_x. Show that uu is subharmonic. (If u≤hu \leq h on ∂B\partial B with hh harmonic, apply the argument of the strong maximum principle to u−hu - h on BB, which also satisfies the local sub-mean-value property.)

Solution

u−hu - h is continuous on B‾\overline B and satisfies the local sub-mean-value property in BB (since hh has the exact mean value property). If max⁡B‾(u−h)=M>0\max_{\overline B}(u - h) = M > 0, it is attained in BB (it is ≤0\leq 0 on ∂B\partial B); the set where u−h=Mu - h = M is closed in BB and, by the local sub-mean-value property, open, so it is all of BB, and by continuity u−h=M>0u - h = M > 0 on ∂B\partial B, a contradiction.

Exercise 6.15 Exterior spheres for convex domains

Let Ω\Omega be bounded and convex, ξ∈∂Ω\xi \in \partial\Omega, and aa a unit vector with Ω⊆{x:a⋅(x−ξ)≤0}\Omega \subseteq \{x : a\cdot(x - \xi) \leq 0\} (a supporting half-space). (a) Show that for every R>0R > 0 the closed ball B(ξ+Ra,R)‾\overline{B(\xi + Ra, R)} meets Ω‾\overline\Omega only at ξ\xi, so convex domains satisfy the exterior sphere condition everywhere. (b) Why does a bounded domain with C2C^2 boundary satisfy it too?

Solution

(a) Every point xx of the closed ball satisfies a⋅(x−ξ)≥0a\cdot(x - \xi) \geq 0, with equality only at x=ξx = \xi (the ball lies on the far side of the hyperplane and touches it only at ξ\xi). Points of Ω‾\overline\Omega satisfy a⋅(x−ξ)≤0a\cdot(x - \xi) \leq 0. So the only common point is ξ\xi. (b) Near each boundary point the boundary is the graph of a C2C^2 function with bounded second derivatives, say by KK, uniformly over the compact boundary; a ball of radius less than 1/K1/K tangent to the boundary from outside lies on the outer side of the graph and touches it only at the tangency point.

Exercise 6.16 Removable singularities

Let uu be harmonic in the punctured disc 0<∣x∣<10 < |x| < 1 in the plane and bounded near 00. Let vv be the Poisson integral of uu on the circle ∣x∣=12|x| = \frac12. Show that u=vu = v on 0<∣x∣<120 < |x| < \frac12 by comparing u−vu - v with ±εlog⁡12∣x∣\pm\varepsilon\log\frac{1}{2|x|}, as in Example 6.9. Conclude that uu extends harmonically to the disc.

Solution

u−vu - v is harmonic and bounded on the annulus δ<∣x∣<12\delta < |x| < \frac12, zero on ∣x∣=12|x| = \frac12. The function εlog⁡12∣x∣\varepsilon\log\frac{1}{2|x|} is harmonic, zero on ∣x∣=12|x| = \frac12 and at least sup⁡∣u−v∣\sup|u - v| on ∣x∣=δ|x| = \delta once δ\delta is small. By comparison ∣u−v∣≤εlog⁡12∣x∣|u - v| \leq \varepsilon\log\frac{1}{2|x|}; let δ→0\delta \to 0 and then ε→0\varepsilon \to 0. So u=vu = v away from 00, and vv is harmonic on the whole disc of radius 12\frac12.

Exercise 6.17 A small discrete problem

On the grid {0,1,2,3}2\{0, 1, 2, 3\}^2 (spacing h=1h = 1), with boundary values u=1u = 1 on the left side x=0x = 0 and u=0u = 0 on the other three sides (corners don't matter), find the four interior values of the discrete harmonic function, and check the discrete maximum principle.

Solution

Interior points (1,1),(2,1),(1,2),(2,2)(1, 1), (2, 1), (1, 2), (2, 2). By symmetry in y↔3−yy \leftrightarrow 3 - y, u(1,1)=u(1,2)=au(1, 1) = u(1, 2) = a and u(2,1)=u(2,2)=bu(2, 1) = u(2, 2) = b. Averaging: 4a=1+b+a+04a = 1 + b + a + 0 (neighbours (0,1)(0, 1), (2,1)(2, 1), (1,2)(1, 2), (1,0)(1, 0)), so 3a=1+b3a = 1 + b; and 4b=a+0+b+04b = a + 0 + b + 0, so 3b=a3b = a. Hence a=38a = \frac38, b=18b = \frac18. All values lie between 00 and 11.

Exercise 6.18 Seeing the h2h^2 rate

For the harmonic function u(x,y)=sin⁡(πx)sinh⁡(πy)u(x, y) = \sin(\pi x)\sinh(\pi y), show that

Δhu=(2(cosh⁡πh−1)h2−2(1−cos⁡πh)h2)u,\Delta_hu = \Big(\frac{2(\cosh\pi h - 1)}{h^2} - \frac{2(1 - \cos\pi h)}{h^2}\Big)u,

and expand to find the truncation error τ=Δhu−Δu\tau = \Delta_hu - \Delta u to leading order in hh. Compare with the bound ∣τ∣≤nh212M4|\tau| \leq \frac{nh^2}{12}M_4 from the proof of Theorem 6.13.

Solution

sin⁡(π(x+h))+sin⁡(π(x−h))−2sin⁡πx=−2(1−cos⁡πh)sin⁡πx\sin(\pi(x + h)) + \sin(\pi(x - h)) - 2\sin\pi x = -2(1 - \cos\pi h)\sin\pi x, and similarly with sinh⁡\sinh and cosh⁡πh−1\cosh\pi h - 1. Expanding, 2(1−cos⁡πh)h2=π2−π4h212+O(h4)\frac{2(1 - \cos\pi h)}{h^2} = \pi^2 - \frac{\pi^4h^2}{12} + O(h^4) and 2(cosh⁡πh−1)h2=π2+π4h212+O(h4)\frac{2(\cosh\pi h - 1)}{h^2} = \pi^2 + \frac{\pi^4h^2}{12} + O(h^4), so τ=Δhu=π4h26u+O(h4)\tau = \Delta_hu = \frac{\pi^4h^2}{6}u + O(h^4) (since Δu=0\Delta u = 0). The bound gives 2h212π4∣u∣\frac{2h^2}{12}\pi^4|u| because ∂x4u=π4u\partial_x^4u = \pi^4u and ∂y4u=π4u\partial_y^4u = \pi^4u: it is attained to leading order. The error is genuinely of order h2h^2: ∂x4u+∂y4u=2π4u≠0\partial_x^4u + \partial_y^4u = 2\pi^4u \neq 0 although Δu=0\Delta u = 0.

Exercise 6.19 Rehearsal: the distance function as a barrier

Let Ω\Omega be a bounded domain with smooth boundary, and d(x)=dist⁡(x,∂Ω)d(x) = \operatorname{dist}(x, \partial\Omega) for x∈Ωx \in \Omega. Near ∂Ω\partial\Omega, dd is smooth, ∣∇d∣=1|\nabla d| = 1 (6B.1 First-Order Equations and Characteristics), and

Δd(x)=−∑i=1n−1κi1−κid(x),\Delta d(x) = -\sum_{i=1}^{n-1}\frac{\kappa_i}{1 - \kappa_id(x)},

where κi\kappa_i are the principal curvatures of ∂Ω\partial\Omega at the nearest boundary point, positive for a convex boundary (Gilbarg–Trudinger, Lemma 14.17). (a) Check the formula for the ball of radius RR, where d=R−∣x∣d = R - |x| and every κi=1R\kappa_i = \frac1R. (b) Deduce that if ∂Ω\partial\Omega is convex, in the sense that every κi≥0\kappa_i \geq 0, then dd is superharmonic near the boundary. (c) On a Riemannian manifold, the Laplacian of the distance function from a point satisfies Δr≤n−1r\Delta r \leq \frac{n - 1}{r} when Ric⁡≥0\operatorname{Ric} \geq 0 (9B.1 Laplacian Comparison); compare with Δ∣x∣=n−1∣x∣\Delta|x| = \frac{n - 1}{|x|} in flat space, and explain why both are statements about the mean curvature of the level sets of a distance function.

Solution

(a) d=R−rd = R - r with r=∣x∣r = |x|: Δd=−Δr=−n−1r\Delta d = -\Delta r = -\frac{n - 1}{r}. The formula gives −(n−1)1/R1−(R−r)/R=−(n−1)1/Rr/R=−n−1r-(n - 1)\frac{1/R}{1 - (R - r)/R} = -(n - 1)\frac{1/R}{r/R} = -\frac{n - 1}{r}. (b) With κi≥0\kappa_i \geq 0 and κid<1\kappa_id < 1 (which holds near the boundary), each term κi1−κid\frac{\kappa_i}{1 - \kappa_id} is ≥0\geq 0, so Δd≤0\Delta d \leq 0: dd is superharmonic there. It is positive in Ω\Omega and vanishes on the whole boundary, so it is a barrier for the entire boundary at once near ∂Ω\partial\Omega; this is how boundary gradient estimates are proved for equations on convex domains (6C.6 Quasilinear Elliptic Equations and Minimal Surfaces). (c) The Laplacian of a distance function equals the mean curvature of its level set (the sum of the principal curvatures of the parallel hypersurface, with the sign convention that spheres seen from inside have positive curvature): for ∣x∣|x| the level sets are spheres of radius rr with mean curvature n−1r\frac{n - 1}{r}; on a manifold with Ric⁡≥0\operatorname{Ric} \geq 0, distance spheres are at most as curved as Euclidean ones, which is the comparison Δr≤n−1r\Delta r \leq \frac{n - 1}{r}.

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