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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.

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    $\begingroup$ 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 $\lvert u/\delta\rvert_2 < C\lvert f\rvert_2$. $\endgroup$ Jan 6, 2021 at 12:57
  • $\begingroup$ @BertramArnold Thank you so much. It seems to be an interesting paper. $\endgroup$
    – XIE
    Jan 6, 2021 at 14:12
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    $\begingroup$ (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 $z$. $\endgroup$ Jan 7, 2021 at 16:13
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    $\begingroup$ (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. $\endgroup$ Jan 7, 2021 at 16:15
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    $\begingroup$ @XIE: Yes, that is what I meant: $u(x) = \int G_\Omega(x,y) f(y) dy$. $\endgroup$ Jan 10, 2021 at 9:35

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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.

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

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

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