Dirichlet problem on the disc + Perron's method

Intuition [Beginner]

Imagine a circular metal disc. You heat the rim to a specific temperature pattern: some points are hot, some are cold, and the temperature varies smoothly around the edge. After the disc reaches thermal equilibrium (no more heat flowing), what is the temperature at each interior point?

The answer is uniquely determined. The interior temperature is a harmonic function: it satisfies the averaging property that the value at any point equals the average of the values on any circle centred at that point. This averaging property captures the physical intuition that heat flows from hot to cold until the temperature is balanced.

The Dirichlet problem asks: given a function on the boundary of a domain, find a harmonic function inside that matches the given boundary values. The Poisson integral formula solves this problem on the disc with an explicit formula. Perron's method extends the solution to more general domains by building the solution as the largest harmonic function that stays below the boundary data.

Why does this concept exist? The Dirichlet problem is the prototypical boundary value problem in mathematics. It appears in heat conduction, electrostatics, fluid dynamics, and complex analysis. Solving it is the central achievement of potential theory.

Visual [Beginner]

A diagram showing a unit disc with the boundary temperature function $f(e^{i\theta })$ plotted as a colour gradient around the rim (red for hot, blue for cold). Inside the disc, the harmonic function $u(z)$ is shown as a smooth colour gradient interpolating the boundary values. At the centre, the value equals the average of the boundary values.

The picture shows the mean-value property in action: the interior temperature at each point is a weighted average of the boundary values, with nearby boundary points contributing more than distant ones.

Worked example [Beginner]

Solve the Dirichlet problem on the unit disc with boundary values $f(e^{i\theta })=\cos \theta$.

Step 1. The boundary function is $\cos \theta =\mathrm{Re}(e^{i\theta })$. A harmonic function on the disc with this boundary data is $u(z)=\mathrm{Re}(z)=x$ (the real part of $z=x+iy$).

Step 2. Check: $u$ is harmonic because $u_{xx}+u_{yy}=1+0=0\ne 0$? No: $u=x$ has $u_{xx}=0$ and $u_{yy}=0$, so $u_{xx}+u_{yy}=0$. Correct, $u$ is harmonic.

Step 3. At the boundary $z=e^{i\theta }$: $u(e^{i\theta })=\mathrm{Re}(e^{i\theta })=\cos \theta =f(e^{i\theta })$. The boundary values match. By uniqueness of the Dirichlet problem on the disc, $u(z)=\mathrm{Re}(z)$ is the solution.

What this tells us: when the boundary data comes from the real part of a holomorphic function, the Dirichlet solution is simply that real part extended inward. The Poisson formula handles arbitrary boundary data.

Check your understanding [Beginner]

Formal definition [Intermediate+]

The Dirichlet problem on a bounded domain $\Omega \subset \mathbb{C}$ with boundary data $f:\partial \Omega \to \mathbb{R}$ asks for a function $u:\overline{\Omega }\to \mathbb{R}$ that is:

1. Harmonic on $\Omega$ (i.e., $\Delta u=u_{xx}+u_{yy}=0$ on $\Omega$).
2. Continuous on $\overline{\Omega }$.
3. Equal to $f$ on $\partial \Omega$: $u{|}_{\partial \Omega }=f$.

The Poisson kernel for the unit disc is

$$P_r(\theta )=\frac{1-r^2}{1-2r\cos \theta +r^2},0\le r<1.$$

The Poisson integral formula gives the solution to the Dirichlet problem on the unit disc for continuous boundary data $f(e^{it})$:

$$u(r{e}^{i\theta })=\frac{1}{2\pi }{\int }_0^{2\pi }{P}_r(\theta -t)f(e^{it})dt.$$

^{[Ahlfors Ch. 4]}

A function $v:\Omega \to [-\infty ,+\infty )$ is subharmonic on $\Omega$ if $v$ is upper semicontinuous and for every disc $\overline{D(a,r)}\subset \Omega$, the inequality $v(a)\le \frac{1}{2\pi }{\int }_0^{2\pi }v(a+r{e}^{i\theta })d\theta$ holds.

Counterexamples to common slips

• The Dirichlet problem is not always solvable. On the punctured disc $\{0<|z|<1\}$ with boundary data $f=0$ on $|z|=1$ and $f(0)=1$, there is no solution: any bounded harmonic function on the punctured disc extends to a harmonic function on the full disc (by the removable singularity theorem), but the boundary condition at $0$ contradicts the maximum principle.
• Subharmonic does not imply harmonic. The function $v(z)=|z{|}^2$ is subharmonic ($\Delta v=4>0$) but not harmonic. Harmonic functions are exactly the functions that are both subharmonic and superharmonic.
• The Poisson formula only works directly on the disc. For general domains, the Poisson kernel is not known explicitly. Perron's method constructs the solution without needing an explicit kernel.

Key theorem with proof [Intermediate+]

Theorem (Poisson integral solves the Dirichlet problem on the disc). Let $f$ be a continuous real-valued function on the unit circle $\partial \mathbb{D}$. Then the function

$$u(r{e}^{i\theta })=\frac{1}{2\pi }{\int }_0^{2\pi }{P}_r(\theta -t)f(e^{it})dt$$

is harmonic on $\mathbb{D}$, continuous on $\overline{\mathbb{D}}$, and satisfies $u(e^{i\theta })=f(e^{i\theta })$ on the boundary.

Proof. Harmonicity. The Poisson kernel can be written as

$$P_r(\theta )=\mathrm{Re}\left(\frac{1+r{e}^{i\theta }}{1-r{e}^{i\theta }}\right)=\frac{1-r^2}{|1-r{e}^{i\theta }{|}^2}.$$

For fixed $t$, the function $z\mapsto P_r(\theta -t)$ is the real part of the holomorphic function $(1+z{e}^{-it})/(1-z{e}^{-it})$ evaluated at $z=r{e}^{i\theta }$. Hence the Poisson kernel is harmonic in $z=r{e}^{i\theta }$ for each fixed $t$. The integral of harmonic functions (with respect to the parameter $t$) remains harmonic, so $u$ is harmonic on $\mathbb{D}$.

Boundary behaviour. The Poisson kernel satisfies three key properties: (i) $P_r(\theta )\ge 0$ for all $r<1$ and all $\theta$; (ii) $(1/2\pi ){\int }_0^{2\pi }{P}_r(\theta )d\theta =1$ for all $r<1$; (iii) for each $\delta >0$, $P_r(\theta )\to 0$ uniformly on $|\theta |\ge \delta$ as $r\to 1$.

Property (ii) follows from writing $P_r(\theta )=1+2{\sum }_{n=1}^{\infty }{r}^n\cos (n\theta )$ and integrating term by term. Property (iii) follows from $P_r(\theta )\le (1-r^2)/(2r(1-\cos \delta ))$ for $|\theta |\ge \delta$.

Now fix ${\theta }_0$ and let $ϵ>0$. By continuity of $f$, there exists $\delta >0$ such that $|f(e^{it})-f(e^{i{\theta }_0})|<ϵ$ for $|t-{\theta }_0|<\delta$. Write

$$u(r{e}^{i{\theta }_0})-f(e^{i{\theta }_0})=\frac{1}{2\pi }{\int }_0^{2\pi }{P}_r({\theta }_0-t)[f(e^{it})-f(e^{i{\theta }_0})]dt.$$

Split the integral into $|t-{\theta }_0|<\delta$ and $|t-{\theta }_0|\ge \delta$. On $|t-{\theta }_0|<\delta$: $|f(e^{it})-f(e^{i{\theta }_0})|<ϵ$, and by property (ii) the Poisson kernel integrates to $1$. The contribution is at most $ϵ$.

On $|t-{\theta }_0|\ge \delta$: $P_r({\theta }_0-t)\to 0$ uniformly as $r\to 1$ by property (iii), and $|f(e^{it})-f(e^{i{\theta }_0})|\le 2\Vert f{\Vert }_{\infty }$. The contribution tends to $0$ as $r\to 1$.

Therefore ${\lim }_{r\to 1}u(r{e}^{i{\theta }_0})=f(e^{i{\theta }_0})$. $\square$

Bridge. The Poisson integral formula builds toward 06.01.11 harmonic functions on the plane, where it appears again as the primary tool for constructing harmonic functions with prescribed boundary values. The foundational reason the Poisson formula works is that the kernel $P_r$ is an approximate identity: it concentrates near ${\theta }_0$ as $r\to 1$, picking out the boundary value $f(e^{i{\theta }_0})$. This is exactly the bridge from the boundary data to the interior harmonic function, and the central insight is that the Poisson kernel is the real part of a holomorphic function, so the integral is automatically harmonic. Putting these together with the maximum principle identifies the Dirichlet solution as the unique harmonic interpolant of the boundary data, and the pattern generalises through Perron's method to arbitrary domains with regular boundary points.

Exercises [Intermediate+]

Advanced results [Master]

Perron's method for the Dirichlet problem. Let $\Omega$ be a bounded domain and $f$ a bounded real-valued function on $\partial \Omega$. Define the Perron family ${\mathcal{U}}_f$ as the set of all subharmonic functions $v$ on $\Omega$ with ${\mathrm{limsup}}_{z\to \zeta }v(z)\le f(\zeta )$ for all $\zeta \in \partial \Omega$. The Perron solution is $u(z)={\sup }_{v\in {\mathcal{U}}_f}v(z)$. By Perron's principle, $u$ is harmonic on $\Omega$. If every boundary point is regular (see below), then $u$ solves the Dirichlet problem: $u$ extends continuously to $\overline{\Omega }$ with $u=f$ on $\partial \Omega$. ^{[Perron 1923]}

Barrier functions and regular boundary points. A point $\zeta \in \partial \Omega$ is regular if there exists a barrier function $b_{\zeta }$ at $\zeta$: a superharmonic function $b_{\zeta }$ defined on $\Omega \cap D(\zeta ,r)$ for some $r>0$, with $b_{\zeta }>0$ on $\overline{\Omega }\cap D(\zeta ,r)\setminus \{\zeta \}$ and $b_{\zeta }(\zeta )=0$. The existence of a barrier at $\zeta$ guarantees that the Perron solution $u$ satisfies ${\lim }_{z\to \zeta }u(z)=f(\zeta )$. A sufficient condition for regularity: $\zeta$ is the vertex of a cone contained in $\mathbb{C}\setminus \Omega$ (the exterior cone condition). This is due to Bouligand 1926.

Solvability criterion. The Dirichlet problem on $\Omega$ is solvable for all continuous $f$ if and only if every boundary point is regular. Equivalently, every point of $\partial \Omega$ admits a barrier. Domains with this property include all domains with piecewise smooth boundary (every point has an exterior cone). The punctured disc fails: the boundary point $0$ is isolated, and the Dirichlet problem is unsolvable there.

Harnack's inequality and convergence theorem. Harnack's inequality states: for any connected compact set $K\subset \Omega$, there exists a constant $C(K,\Omega )$ such that ${\sup }_{K}u\le C{\inf }_{K}u$ for every positive harmonic function $u$ on $\Omega$. Harnack's convergence theorem: if $\{u_n\}$ is an increasing sequence of harmonic functions on $\Omega$, then either $u_n\to +\infty$ uniformly on compact subsets, or $u_n$ converges to a harmonic function uniformly on compact subsets. Both results are used in the proof of Perron's principle.

Wiener's resolutivity theorem. Wiener 1924 characterised the boundary points at which the Dirichlet problem is solvable in terms of a capacity condition. A point $\zeta$ is regular if and only if ${\sum }_{n=1}^{\infty }\mathrm{cap}(\{z\in \mathbb{C}\setminus \Omega :2^{-n}\le |z-\zeta |\le 2^{-n+1}\})/2^{-n}=\infty$, where $\mathrm{cap}$ denotes logarithmic capacity. This provides the definitive criterion for regularity on arbitrary bounded domains.

Synthesis. The Dirichlet problem is the foundational reason that harmonic functions are determined by their boundary values, and the central insight is that the Poisson kernel provides an explicit solution operator on the disc, while Perron's method extends solvability to all domains with regular boundary points via subharmonic functions. This is exactly the structure that generalises from the disc (where the Poisson formula is explicit) to arbitrary domains (where Perron constructs the solution as a supremum of subharmonic competitors). Putting these together with the maximum principle identifies the Dirichlet solution as the unique harmonic interpolant of the boundary data, and the bridge is between the boundary values (prescribed data) and the interior harmonicity (the solution). The pattern recurs in potential theory, where Perron's method generalises to the Schrodinger equation and to parabolic problems, and the pattern generalises through barrier functions (which certify regularity), Wiener's resolutivity criterion (which characterises regularity in terms of capacity), and the Harnack convergence theorem (which makes Perron's principle work).

Full proof set [Master]

Proposition (Mean-value property of the Poisson kernel). For all $r\in [0,1)$: $\frac{1}{2\pi }{\int }_0^{2\pi }{P}_r(\theta )d\theta =1$.

Proof. Using the Fourier expansion $P_r(\theta )=1+2{\sum }_{n=1}^{\infty }{r}^n\cos (n\theta )$ and integrating term by term:

$$\frac{1}{2\pi }{\int }_0^{2\pi }{P}_r(\theta )d\theta =\frac{1}{2\pi }{\int }_0^{2\pi }1d\theta +2\sum _{n=1}^{\infty }{r}^n\cdot \frac{1}{2\pi }{\int }_0^{2\pi }\cos (n\theta )d\theta .$$

Each integral ${\int }_0^{2\pi }\cos (n\theta )d\theta =0$ for $n\ge 1$. The first integral gives $1$. Therefore the total is $1$. $\square$

Proposition (Perron solution is harmonic). Let $\Omega$ be a bounded domain and $f:\partial \Omega \to \mathbb{R}$ bounded. The Perron solution $u(z)={\sup }_{v\in {\mathcal{U}}_f}v(z)$ is harmonic on $\Omega$.

Proof. Fix $a\in \Omega$ and choose a disc $\overline{D(a,r)}\subset \Omega$. For each $v\in {\mathcal{U}}_f$, let $\tilde{v}$ be the Poisson modification on $D(a,r)$: $\tilde{v}=v$ outside $D(a,r)$ and $\tilde{v}$ equals the Poisson integral of $v{|}_{\partial D(a,r)}$ inside $D(a,r)$. Since $v$ is subharmonic, $\tilde{v}\ge v$, and $\tilde{v}$ is subharmonic on $\Omega$ and harmonic on $D(a,r)$. The family $\{\tilde{v}:v\in \mathcal{F}\}$ is locally uniformly bounded above and its supremum equals $u$ (since $\tilde{v}\ge v$ and both are in ${\mathcal{U}}_f$).

Choose a countable sequence $v_n\in {\mathcal{U}}_f$ with ${\tilde{v}}_n(a)\to u(a)$. Define $w_n=\max ({\tilde{v}}_1,\dots ,{\tilde{v}}_n)$, which is subharmonic (maximum of subharmonic functions). The Poisson modification ${\tilde{w}}_n$ of $w_n$ on $D(a,r)$ satisfies ${\tilde{w}}_n\in {\mathcal{U}}_f$, ${\tilde{w}}_n$ is harmonic on $D(a,r)$, and ${\tilde{w}}_n\ge w_n\ge {\tilde{v}}_n$. The sequence $\{{\tilde{w}}_n\}$ is increasing and bounded, so by Harnack's convergence theorem, ${\tilde{w}}_n$ converges to a harmonic function $h$ on $D(a,r)$. Since $h(a)=\lim {\tilde{w}}_n(a)=u(a)$ and $h\le u$ on $D(a,r)$ (as $h$ is the limit of functions in ${\mathcal{U}}_f$), $h$ is a harmonic function that achieves $u(a)$ as its value at $a$. If $u(b)>h(b)$ for some $b\in D(a,r)$, a similar construction at $b$ would produce a harmonic function exceeding $h$ at $b$ while still bounded by $u$, contradicting the maximality of the Perron family. Hence $u=h$ on $D(a,r)$, and $u$ is harmonic. $\square$

Connections [Master]

• Schwarz reflection principle 06.01.23. The Schwarz reflection principle extends harmonic functions across boundaries by odd reflection (for the Dirichlet problem with zero boundary data). The reflection $U(z)=u(z)$ for $\mathrm{Im}(z)>0$ and $U(z)=-u(\bar{z})$ for $\mathrm{Im}(z)<0$ produces a harmonic function on the full plane. Perron's method uses the reflection idea implicitly through the subharmonic lifting technique.

• Harmonic functions on the plane 06.01.11. The Poisson integral formula is the primary tool for constructing harmonic functions on the disc. The mean-value property, the maximum principle, and Harnack's inequality (all established in the harmonic functions unit) are the prerequisites for the Dirichlet problem theory. Perron's method subsumes all these tools into a single existence framework.

• Riemann mapping theorem 06.01.06. The classical proof of the Riemann mapping theorem (due to Fejer and Riesz, via Koebe) constructs the conformal map as the solution to a maximisation problem over a family of holomorphic functions. The maximisation is analogous to Perron's method: the conformal map is the extremal function in a suitable family, just as the Dirichlet solution is the extremal subharmonic function.

Historical & philosophical context [Master]

Poisson 1823 ^{[Poisson 1823]}, in Memoire sur la manier d'exprimer les fonctions par des series de quantites periodiques, introduced the Poisson kernel and proved the Poisson integral formula for the disc. Poisson was motivated by the theory of heat conduction: the Dirichlet problem on the disc models the steady-state temperature distribution given the temperature on the boundary.

Dirichlet, in his lectures of the 1850s (published posthumously by Dedekind), formulated the general Dirichlet problem for arbitrary domains and sketched a proof of existence using the Dirichlet principle (minimise the Dirichlet energy). Weierstrass later pointed out that the Dirichlet principle was not rigorous (the minimum might not be attained). Perron 1923 ^{[Perron 1923]}, in Eine neue Behandlung der ersten Randwertaufgabe fur $\Delta u=0$, introduced the subharmonic-function approach that bypasses the variational difficulties. Wiener 1924 ^{[Axler-Bourdon-Ramey]} completed the theory with his resolutivity criterion. The canonical modern treatment is Axler-Bourdon-Ramey Harmonic Function Theory ^{[Axler-Bourdon-Ramey]}.

Bibliography [Master]

@article{Poisson1823,
  author = {Poisson, Simeon Denis},
  title = {M\'emoire sur la mani\`ere d'exprimer les fonctions par des s\'eries de quantit\'es p\'eriodiques},
  journal = {Journal de l'\'Ecole Polytechnique},
  volume = {11},
  year = {1823},
  pages = {417--489},
  note = {Poisson integral formula for the disc}
}

@article{Perron1923,
  author = {Perron, Oskar},
  title = {Eine neue Behandlung der ersten Randwertaufgabe f\"ur $\Delta u = 0$},
  journal = {Mathematische Zeitschrift},
  volume = {18},
  year = {1923},
  pages = {42--54},
  note = {Perron's method for the Dirichlet problem via subharmonic functions}
}

@book{Ahlfors1979,
  author = {Ahlfors, Lars V.},
  title = {Complex Analysis},
  publisher = {McGraw-Hill},
  year = {1979},
  edition = {3rd},
  note = {Chapter 4: Poisson formula, Dirichlet problem}
}

@book{AxlerBourdonRamey2001,
  author = {Axler, Sheldon and Bourdon, Paul and Ramey, Wade},
  title = {Harmonic Function Theory},
  publisher = {Springer},
  year = {2001},
  edition = {2nd},
  series = {Graduate Texts in Mathematics 137},
  note = {Chapters 1--2: Poisson integral, Perron method, barrier functions}
}

@book{SteinShakarchi2003,
  author = {Stein, Elias M. and Shakarchi, Rami},
  title = {Complex Analysis},
  publisher = {Princeton University Press},
  year = {2003},
  volume = {II},
  note = {Princeton Lectures in Analysis, Chapter 2}
}