From an old qualifier:
Let $$J_1 = \{e^{i\theta}: 0 < \theta < \frac\pi2\}, \,\,J_2 = \{e^{i\theta}:\frac\pi2<\theta< \pi\},\,\, J_3 = \{e^{i\theta}:\pi<\theta < 2\pi\}$$ be three arcs and $D=\{z\in \mathbb{C}: |z|<1\}.$ Find a harmonic function $u$ on $D$ with $u|_{J_1}=1, u|_{J_2}=2, u|_{J_3}=4$. Is $u$ unique? If not, give some conditions such that it is unique. Make your assumption as sharp as possible.
Ideas: First, I believe that $$u(r,\theta) = \int_0^{\pi/2} \frac{1-r^2}{1-2r\cos\theta + r^2}+ 2\int_{\pi/2}^{\pi}\frac{1-r^2}{1-2r\cos\theta + r^2} + 4\int_{\pi}^{2\pi}\frac{1-r^2}{1-2r\cos\theta + r^2}$$
is such a solution. Perhaps there is a way to write it more explicitly? (I'm not sure if the question wants this).
Regarding uniqueness, there is a very helpful Stack response here which makes the claims
- $u$ is not unique.
- $u$ is the unique such function such that $\limsup_{z\to\zeta}|u(z)|\leq 4$ for all $\zeta \in \partial D$
Before I read the answer I had thought $u$ was unique, and even Wikipedia's article would seem to suggest that the function is unique. In discussing boundary data which is $L^1$, they say:
By the maximum principle, u is the only such harmonic function on D.
This seems like a glaring error; perhaps they mean if the boundary data is continuous. (Unless someone can correct me.)
Would $\limsup_{z\to\zeta}u(z)\leq 4$ for all $\zeta \in \partial D$ be the assumption you would give to make $u$ unique? Are there any others? Can anyone sketch a proof of claim 2?
I will describe two different approaches to proving the uniqueness of harmonic functions with given boundary values.
Functional Analysis approach
Theorem 1. Suppose that $u$ is a harmonic function on $D$ and there exists $p>1$ such that $$\sup_{0<r<1} \int_0^{2\pi} |u(r,\theta)|^p\,d\theta<\infty\tag{1}$$ Then there exists an $L^p$ function $f$ on $\partial D$ such that $u$ is the Poisson integral of $f$. Also, the radial limits of $u$ are equal to $f$ a.e.
As a consequence of Theorem 1, two harmonic functions that satisfy (1) and share boundary values a.e. must be equal.
Proof of Theorem 1. Let $u_r$ be the function $\theta\mapsto u(r,\theta)$ defined on the interval $ (0,2\pi)$. By the Banach-Alaoglu theorem, closed balls in $L^p$ are weakly compact. Therefore, there is a sequence $r_k\to 1$ such that $u_{r_k}$ converge weakly to some $f\in L^p(0,2\pi)$. Fix a point with polar coordinates $(\rho,\phi)$ in $D$. For all sufficiently large $k$ we have $r_k>\rho$, hence by the Poisson integral formula (used on the disk of radius $r_k$),
$$u(\rho,\phi) = \frac{1}{2\pi} \int_0^{2\pi} \frac{r_k^2-\rho^2}{r_k^2-2r_k\rho \cos(\theta-\phi)+\rho^2}u_{r_k}(\theta) \,d\theta \tag{2}$$ We can pass to the limit $k\to \infty$ under the integral sign in (2), because $u_{r_k}$ converge weakly in $L^p$, and the other factor converges strongly in the dual space $L^q$ (indeed, it converges uniformly). Thus, $$u(\rho,\phi) = \frac{1}{2\pi} \int_0^{2\pi} \frac{1-\rho^2}{1-2\rho \cos(\theta-\phi)+\rho^2}f(\theta) \,d\theta \tag{3}$$ proving the first claim of the theorem.
Once the representation (3) is available, the claim about radial limits follows in a standard way: as $\rho\to 1$, most of the integral comes from a small interval around $\phi$, which implies convergence whenever $\phi$ is a Lebesgue point of $f$. $\quad \Box$
Potential Theory approach
Let's first state the maximum principle in a convenient form.
Maximum principle: Suppose $u$ is harmonic in $D$. If for all $\zeta\in\partial D$ we have $\limsup_{z\to\zeta } u(z)\le M$, then $u\le M$ in $D$.
Proof. Let $u_\epsilon(z)=u(z)+\epsilon |z|^2 $. The function $u_\epsilon$ does not have local maxima by the 2nd derivative test. Let $(z_k)$ be a sequence such that $u_\epsilon(z_k)\to \sup_D u_\epsilon$. Then $z_k$ approach the boundary of $D$. It follows that $\sup_D u_\epsilon \le M+ \epsilon$. Let $\epsilon\to 0$. $\quad \Box$
We can generalize the maximum principle with one weird trick.
Theorem 2. (Generalized maximum principle) Suppose that $u$ is harmonic in $D$ and bounded above. If $\limsup_{z\to \zeta} u(z)\le M$ for all $\zeta\in \partial D$, with finitely many exceptions, then $u\le M$ in $D$.
As a consequence of Theorem 2, two harmonic functions that are bounded in $D$ and share boundary values on a cofinite subset of $\partial D$ must be equal.
Proof of Theorem 2. Let $\zeta_1,\dots,\zeta_n$ be the exceptional points. Define $$u_\epsilon(z)=u(z)+\epsilon\sum_{j=1}^n \log|z-\zeta_j|$$ Observe that $u_\epsilon\le u+\epsilon\, n \log \operatorname{diam} D$. The function $u_\epsilon$ is harmonic in $D$ and tends to $-\infty$ as $z\to \zeta_j$. Hence, for all $\zeta\in\partial D$ we have $$\limsup_{z\to\zeta } u_\epsilon(z)\le M+ \epsilon\, n \log \operatorname{diam} D\tag{4}$$ By the maximum principle $u_\epsilon\le M+ \epsilon\, n \log \operatorname{diam} D$ in $D$. Let $\epsilon\to 0$. $\quad \Box$
Remarks
The potential-theoretic approach required stronger assumptions on $u$, but with it we did not actually need $D$ to be a disk; it could be any bounded domain.
The sum $\sum_{j=1}^n \log|z-\zeta_j|$ is the logarithmic potential of a finite sum of point masses. Using more general potentials for the same purpose, one can prove a stronger form of Theorem 2, in which the exceptional set is allowed to be any polar set. See Ransford's book Potential theory in the complex plane.