Harmonic Functions in Lipschitz domains

 

(29 Jun 2020)

When you start studying certain free boundary problems you soon appreciate the importance of understanding the boundary behavior of harmonic functions, especially harmonic functions which vanish on a portion of the boundary. Motivated by questions in the Hele-Shaw problem and related free boundary problems, I am going to review several important theorems proved by Dahlberg in the late 1970’s.

Let $D\subset \mathbb{R}^n$ be a Lipschitz domain. For such a domain it is well known that for any continuous function $f:\partial D\to\mathbb{R}$ there is a unique function $U_f$ which is continuous in $\overline D$ and twice differentiable in $D$ solving

\[\left \{ \begin{array}{rl}\Delta U_f = 0 & \text{ in } D \\ U_f = f & \text{ on } \partial D\end{array} \right.\]

This defines an operator $f \mapsto U_f$ which is linear, it is also monotone in the sense that if $f\leq g$ in $\partial D$ then $U_f \leq U_g$ in $D$. In particular, for every $X \in D$ the linear functional

\[f \mapsto U_f(X)\]

is positive. By the representation theorem for each $X \in D$ there is a Borel measure $\omega_X$ in $\partial D$ such that

\[U_f(X) = \int_{\partial D}f(y)d\omega_X(y).\]

The measure $\omega_X$ is called the harmonic measure associated with the domain $D$ at the point $X$. When $D$ is a Lipshitz domain, the measure $\omega_X$ is absolutely continuous with respect to the surface measure of $\partial D$, so

\[d\omega_X(y) = k(X,y)\;d\sigma(y)\]

The function $K$ is called the Poisson kernel of the domain $D$. The best known instances of the Poisson kernel are those where $D$ is a ball or a half space. For the unit ball in $\mathbb{R}^n$ the kernel takes the form

\[k(X,y) = \frac{1}{|\partial B_1|}\frac{1-|X|^2}{|X-y|^n}\]

Dahlberg showed that for a general Lipschitz domain the Poisson kernel for each fixed $X$ yields a function on the boundary which satisfies a reverse Harnack inequality.

Theorem.
There is a constant $C$ such that given $X\in D$, $y_0 \in \partial D$, and $r\in(0,1)$ we have $$ \left ( \frac{1}{\sigma(\Delta)}\int_{\Delta} k(X,y)^2\;d\sigma(y) \right )^{\frac{1}{2}} \leq \frac{C}{\sigma(\Delta)}\int_\Delta k(X,y)\;d\sigma(y),\;\;\Delta := B_r(y_0) \cap \partial D.$$

There is the Boundary Harnack Principle. It says that if $U$ is a positive harmonic function in $D$ which vanishes on a portion of $\partial D$, then the value of $U$ at a point properly place away from said portion will control the values of $U$ everywhere.

Before we state the principle, let us state two preliminary facts about harmonic functions.

$\bullet$ (De Giorgi’s oscillation lemma) Fix $\delta \in(0,1)$. Let $U$ be a subharmonic function $U$ in some ball $B_{2r}(X_0)$, and suppose that $U \leq 1$ in $B_{2r}(X_0)$ and

\[|\{ U\leq 0 \} \cap B_{2r}(X_0)| \geq \delta |B_r|\]

then \(U\leq \mu\) in $B_r(X_0)$, where $\mu = \mu(\delta) \in (0,1)$.

$\bullet$ (Harnack inequality) If $P \in D$ is such that $d(P,\partial D) = \rho$, then

\[U(P) \leq C \left ( \frac{r}{\rho}\right )^\alpha U(X_1)\]

Theorem.
Let $U$ be a positive harmonic function in $B_r(X_0) \cap D$ such that $U \equiv 0$ on $B_r(X_0) \cap \partial D$ and let $X_1$ be a point in $B_{r/2}$ such that $B_{r/4}(X_1) \subset B_R(X_0) \cap D$. Then $$U(X) \leq CU(X_1) \text{ for any } X \in B_{r/2}(X_0) \cap D$$ where the constant $C$ is determined by $D$.

Proof.

For every $X$, we shall denote by \(X^*\) a point on \(\partial D\) such that \(d(X,\partial D) = d(X^*,\partial D)\).

The proof hinges on a conflict between two effects:

  1. The positivity and harmonicity of $U$ force it to grow, meaning if $U$ reaches a certain value $h$ in a ball $B_r$ then in $B_{2r}$ it must reach at least a value above $\kappa h$, where $\kappa>1$ is a dimensional constant.
  2. The Harnack inequality prevents $U(X)$ from being much larger than $U(X_1)$ if the distance from $X$ to $X_1$ is comparable to the distance from $X$ to $\partial D$.

Let $P_0$ be the point where $U$ is the largest in $B_{r/2} \cap D$. Using the two effects above we are going to show that if the value $U(P_0)$ is much larger than $U(X_1)$, then there is a sequence of points $P_k$ all lying in $B_r \cap D$ with the property $U$ takes arbitrarily large values along this sequence.

The sequence \(\{P_k\}_k\) is built recursively, starting from $P_0$. Having determined the sequence up to some $P_k$, we determine the next element $P_{k+1}$. Let us write

\[\rho_k := d(P_k,\partial D)\]

The oscillation lemma says that

\[\sup \limits_{B_{\rho_k}(P_k)} U \leq \theta \sup \limits_{B_{2\rho_k}(P_k)} U\]

Applying the oscillation lemma in $B_{2\rho_k}(P_k^*)$ we obtain a new point $P_{k+1} \in D$ such that

\[|P_{k+1}-P_k| \leq 2\rho_k,\;\; U(P_{k+1}) \geq \theta^{-1} U(P_k)\]

This can be done as long as … so the sequence is either finite and ends at some $k=k_0$ or it goes on indefinitely.

In either case, the second inequality applied recursively means that for every $P_k$ in the sequence we have

\[U(P_k) \geq \theta^{-k} U(P_0)\]

We now use the Harnack inequality. From the H

Note there is a ball of radius $r+\rho_k$ ….containing $X_1$ and $P_k$… and which is contained in $D$, then, thanks to Harnack’s inequality

\[U(P_k) \leq C \left (\frac{r}{\rho_k} \right )^\alpha U(X_1)\]

In other words,

\[\rho_k \leq C r\left (\frac{U(X_1)}{U(P_k)}\right )^{1/\alpha}\] \[\rho_k \leq C r \left (\frac{U(X_1)}{U(P_0)}\right )^{1/\alpha}(\theta^{1/\alpha})^k\]

In summary, the sequence ${P_k}_k$ has the following properties

\[U(P_k) \geq \theta^{-k}U(P_0)\] \[|P_k-P_{k+1}| \leq Cr \left (\frac{U(X_1)}{U(P_0)} \right )^{1/\alpha} (\theta^{1/\alpha} )^{k}\]

Observe that

\[\sum \limits_{k=0}^\infty |P_k-P_{k+1}| \leq Cr\left (\frac{U(X_1)}{U(p_0)} \right )^{1/\alpha} \frac{1}{1-\theta^{1/\alpha}}\]

Let $\phi:\mathbb{R}^n \to \mathbb{R}$ be a bounded, Lipschitz function,

\[\|\phi\|_\infty \leq C,\; \|\nabla \phi\|_\infty \leq \delta\]

and consider the domain

\[D(\phi) = \{ X = (x,x_{n+1})\in \mathbb{R}^{n+1} \mid x_{n+1} > \phi(x) \}\]

Proof.

Let $V$ be superharmonic and such that $CV(y) \geq k(P,y)^q$ for $y\in \partial D$. Then

\[CV(P) \geq \int_{\partial D}k(P,y)^{1+q}\;d\sigma(y)\]