3
$\begingroup$

$\newcommand{\avint}{⨍}$ Let $B_r$ be a call of radius $r$ and centre origin and $k<1$.$w$ satisfy the following PDE: $$ \begin{cases} -\Delta w = 0 \qquad \mbox{in $B_r\setminus B_{kr}$}\\ w=0 \qquad \mbox{on $\partial B_{kr}$}\\ w=\varphi\qquad \mbox{on $\partial B_r$} \end{cases} $$

Show that, for all $x\in \partial B_{kr}$, $$ |\nabla w(x)| \,\le\, \frac{C}{r} \avint_{\partial B_r} \varphi\,d\sigma $$

$d\sigma$ is the surface measure.

(one can take $\varphi\geq 0$ is it helps in the proof. The issue has occured in the Theorem 3.1 of [1], page 436, the estimate mentioned between equation 3.2 and 3.3)

[1] Alt, Hans Wilhelm; Caffarelli, Luis A.; Friedman, Avner, Variational problems with two phases and their free boundary, Trans. Am. Math. Soc. 282, No. 2, 431-461 (1984). ZBL0844.35137.

$\endgroup$
10
  • 2
    $\begingroup$ You should try a scaling argument to remove the $r$ . Can the $C$ depend on $k$ ? (I assume it can otherwise its probably false) $\endgroup$
    – Math604
    Feb 13, 2021 at 16:43
  • $\begingroup$ @Math604 yes, constant $C$ can depend on $k$. $\endgroup$
    – Harish
    Feb 13, 2021 at 17:30
  • 1
    $\begingroup$ If $G(x,y)$ is the Green function for $B_r \setminus B_{kr}$, then $w(x) = \int_{\partial B_r} P(x, y) \varphi(y) \sigma(dy)$ and $|\nabla w(x)| = |\partial_n w(x)| = |\int_{\partial B_r} Q(x, y) \varphi(y) \sigma(dy)|$, where $P(x,y) = \partial_{n(y)} G(x, y)$ is the Poisson kernel and $Q(x, y) = \partial_{n(x)} \partial_{n(y)} G(x,y)$ is the normal derivative of the Green function with respect to both variables. It remains to note that $Q(x,y)$ is bounded (by known estimates of the Green function, for example). $\endgroup$ Feb 13, 2021 at 17:35
  • $\begingroup$ @MateuszKwaśnicki Thank you, but these are internal estimates, do such estimate also hold true on the boundary of interior sphere? (that is on $\partial B_{kr}$)? $\endgroup$
    – Harish
    Feb 13, 2021 at 17:42
  • 1
    $\begingroup$ I may have not been clear enough, sorry: in the expression for $w(x)$, $x$ is in the interior, but in the expression for $\nabla w(x)$, $x$ is on $\partial B_{kr}$. And for smooth enough domains it is known that $G(x,y) \approx \delta(x) \delta(y)$ when $x$ and $y$ are not too close (and $\delta(x)$ is the distance to the boundary). $\endgroup$ Feb 13, 2021 at 17:50

2 Answers 2

3
$\begingroup$

OK, here are some additional details to what I wrote in my comments.

Due to scaling, we can choose $r = 1$. Let $D = B_1 \setminus B_k$, and let $P_D(x, y)$ be the Poisson kernel of $D$. Thus, $$ w(x) = \int_{\partial B_1} P_D(x, y) \varphi(y) \sigma(dy) $$ for $x \in D$. Let $n$ denote the unit normal vector at a boundary point of $D$. Note that $\nabla w(x) = (\partial_n w(x)) n$ for $x \in \partial B_k$. It follows that $$ |\nabla w(x)| = \biggl| \int_{\partial B_1} \partial_n P_D(x, y) \varphi(y) \sigma(dy) \biggr| $$ for $x \in \partial B_k$; here and below the derivative acts on the $x$ variable. Finally, by the boundary Harnack inequality, $$ \partial_n P_D(x, y) $$ is a bounded function of $x \in \partial B_k$ and $y \in \partial B_1$ (I can give more details on this if needed). We conclude that $$ |\nabla w(x)| \leqslant C \int_{\partial B_1} |\varphi(y)| \sigma(dy) ,$$ as desired.

$\endgroup$
4
  • $\begingroup$ my objection is the following: assuming that the poissin's kernel formulae(first equation) holds for the points on the inner boundaey. (Since it is taken to be true for gradient, hence must be true for values of $w$). We have $w=0$ for every $x\in \partial B_{kr}$ then $\int_{\partial B_1} P_D(x,y) \varphi(y) d\sigma(y)=0$ for every $\varphi$. Thus $P_D=0$ on $\partial B_1$. Which I suspect is not true. $\endgroup$
    – Harish
    Feb 15, 2021 at 13:11
  • $\begingroup$ It is perfectly true, $P_D(x, y)$ does vanish on the boundary — or, if one prefers this formulation, $P_D(x,y) = \delta_x(dy)$ when $x$ is on the boundary. (OK: this is not strictly true for irregular boundary points, where definitions vary. But here all boundary points are regular for the Dirichlet problem.) $\endgroup$ Feb 15, 2021 at 21:14
  • $\begingroup$ I am still not convinced, because we can have arbitrary boundary data on inner and outer bounary. Now say we do not have zero boundary data in inner boundary, then we will have non zero integration $\int_{\partial B_1}P_D(x,y) \varphi (y)d\sigma(y) \neq 0$. but as per your claim $P_D$ is zero, which is a contradiction. $\endgroup$
    – Harish
    Feb 21, 2021 at 17:16
  • $\begingroup$ Wait a minute, you wrote that $w = 0$ on the inner boundary, did you not? Otherwise of course no such gradient estimate holds. $\endgroup$ Feb 21, 2021 at 20:01
2
$\begingroup$

Below is a maximum principle-based alternative to the proof of Mateusz.

We may assume that $r = 1$ by scaling. Let $w$ be the harmonic function on $B_1 \subset \mathbb{R}^n$ with boundary data $|\varphi|$. By the usual representation formula we have $$w|_{\partial B_{\frac{k+1}{2}}} \leq \frac{C(n,k)}{|\partial B_1|}\int_{\partial B_1}|\varphi| := A.$$ By the maximum principle, $|u| \leq w$ in $B_1 \backslash B_k$ (in particular, on $\partial B_{\frac{k+1}{2}}$). We conclude using the maximum principle again that $$|u| \leq A\frac{k^{2-n}-|x|^{2-n}}{k^{2-n}-\left(\frac{k+1}{2}\right)^{2-n}} := v$$ on $B_{\frac{k+1}{2}} \backslash B_k$. (Here we assumed $n \geq 3$; when $n = 2$, the function $v$ is obtained in similar way using $\log$). Since $u = v = 0$ on $\partial B_k$ it follows that $$|\nabla u| \leq |\nabla v| = \frac{(n-2)k^{1-n}}{k^{2-n}-\left(\frac{k+1}{2}\right)^{2-n}}A$$ on $\partial B_k$, which is an estimate of the desired form.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.