# The solution of Poisson equation and the distance function from the boundary

Let $$D$$ be a domain in $$\mathbb{R}^n$$, and let $$u$$ be the solution to the Poisson equation, that is

$$\begin{cases} \Delta u = f & \text{in} ~ D, \\ u=0 & \text {on} ~ \partial D, \end{cases}$$ for some function $$f \in C(D)$$ (the function which is identically $$1$$ on $$D$$ could be a good candidate for $$f$$). Is the values of $$u(x)$$ comparable to the distance function to the boundary, that is $$\delta (x)= \mathrm{dist} (x , \partial D)$$? I think of an inequality of (boundary) Harnack type. I have not made any regularity assumption (on $$\partial D$$, or the function $$f$$), such assumptions can be applied if necessary.

• You can bound $\int_\Omega (u/\delta)^2 \mathrm dx$ by $\int_\Omega \lvert \nabla u\rvert^2\mathrm dx$ using the Hardy inequality (compare Brezis, Haïm, and Moshe Marcus. "Hardy's inequalities revisited." Annali della Scuola Normale Superiore di Pisa-Classe di Scienze 25.1-2 (1997): 217-237.numdam.org/article/ASNSP_1997_4_25_1-2_217_0.pdf ) which equals $\int_\Omega uf\dd x$ by partial integration. The Poisson formula should then give a bound $\lvert u/\delta\rvert_2 < C\lvert f\rvert_2$. Jan 6, 2021 at 12:57
• @BertramArnold Thank you so much. It seems to be an interesting paper.
– XIE
Jan 6, 2021 at 14:12
• (a) If $f = 0$ near a boundary point $z$, then this is precisely the boundary Harnack inequality at $z$, with an explicit linear decay rate, valid in $C^{1,1}$ domains. (b) If $f$ bounded near $z$, then nothing really changes, because the solution of $\Delta u = -1$ has a linear decay rate near the boundary. The last property follows by a comparison with explicit solutions for a ball $B(p, r)$ contained in $D$ and tangent to $\partial D$ at $z$, and for $B(q, R) \setminus B(q, r)$, where $R$ is large enough and $B(q, r)$ is contained in $D^c$ and tangent to $\partial D$ at $z$. Jan 7, 2021 at 16:13
• (c) If you like more singular functions $f$ near the boundary, then you can use explicit estimates of the Green function to get an integral condition for $f$ that still asserts linear decay near the boundary. Jan 7, 2021 at 16:15
• @XIE: Yes, that is what I meant: $u(x) = \int G_\Omega(x,y) f(y) dy$. Jan 10, 2021 at 9:35

Let us assume some regularity on $$\partial D$$ (bounded and $$C^2$$ suffices). Then the problem above has a unique solution $$u \in W^{2,p}(\Omega) \cap W^{1,p}_0(\Omega)$$ for every $$p<\infty$$ and taking $$p>n$$ we get $$\|\nabla u\|_\infty \le C\|f\|_\infty$$. Therefore the upper estimate $$|u(x)| \le C\|f\|_\infty \delta (x)$$ always holds. The lower estimate does not hold for every $$f$$ (take $$u$$ with support far away from $$\partial D$$ and $$f=\Delta u$$). However it holds if $$-\Delta u=f$$ (note the minus sign) and $$f \geq c >0$$. In fact, by the regularity assumption on $$D$$, $$\Delta \delta \geq -\kappa$$ in $$D$$ and then $$-\Delta (u-\epsilon \delta)=f+\epsilon \Delta \delta \geq c-\epsilon \kappa \geq 0$$ for small $$\epsilon$$. By the maximum principle $$u-\epsilon \delta \geq 0$$, which gives the lower estimate.

EDIT: The lower estimate holds assuming only that $$-\Delta u= f \geq 0$$ and $$u\neq 0$$. In fact, $$u(x)>0$$ for every $$x \in D$$, by the strong maximum principle and then $$\frac{\partial u}{\partial \nu}(x_0)<0$$ for every $$x_0 \in \partial D$$, by Hopf Lemma ($$\nu$$ is the unit exterior normal). The minimum of $$\frac{\partial u}{\partial \nu}$$ on $$\partial D$$ is then strictly negative and form this one obtains the lower bound.

• Thank you. I learned much from it.
– XIE
Jan 10, 2021 at 7:42

You can use Feynman-Kac formula: $$u(x) = -\mathbb{E}_x\int_0^T f(X_t)dt$$where $$X_t$$ is a Brownian process starting at $$x$$ and $$T$$ the stopping time $$T=\inf\{t\geq 0:X(t)\in \partial D\}$$. So $$u(x)$$ is essentially given by the expected time for the process to get out of $$D$$. For example if $$\epsilon \leq f\leq M$$ we get $$\epsilon\mathbb{E}_x(T)\leq -u(x) \leq M \mathbb{E}_x(T).$$

Consider the case $$\partial D$$ smooth around a point $$x_0$$. As a simplification we suppose that locally $$D=\mathbb{R}_+\times\mathbb{R}^{n-1}$$, $$x_0 = (0,\cdots,0)$$ and $$x=(\delta,0,\cdots,0)$$ with $$\delta>0$$ and $$X_s = (x+B_s^{(1)},B_s^{(2)},\cdots,B_s^{(n)})$$, where $$B^{(i)}$$ are iid Brownian motion. In that case it is easy to gives the escape time of $$X$$: we have $$\mathbb{P}_x(T>t)=\mathbb{P}_x(\inf_{0\leq s\leq t} B_s^{(1)} > - \delta) = 1-2\mathbb{P}_x(B_t^{(1)}<-\delta) \approx \frac{2\delta}{\sqrt{2\pi t}}$$ (see reflection) and then $$\mathbb{E}_x(T)= \mathbb{E}_x(\int_0^\infty 1_{t< T}dt)=\mathbb{E}_x(\int_0^1 1_{t< T}dt)+\mathbb{E}_x(1_{T>1}\int_1^\infty 1_{t\leq T}dt) \\ = \int_0^1 \mathbb{P}_x(t< T)dt+\mathbb{P}_x(T>1) \mathbb{E}_x\left( \int_1^\infty 1_{t\leq T}dt|T>1\right) \approx \delta C$$ for some $$C>0$$.

If the boundary is not regular at $$x_0$$, it can be possible for the process to avoid $$\partial D$$ such that $$\mathbb{P}(T>1)$$ is bounded away from $$0$$ (or decay very slowly) as $$x\rightarrow x_0$$ and then $$u(x)$$ is no more comparable with $$d(x,\partial D)$$. This lead to a nice physical phenomena call " Effet de pointe" that occures in Lightning rod.

• Thank you so much! It is interesting for me to see the problem from an stochastic analysis viewpoint.
– XIE
Jan 10, 2021 at 7:41