Suppose $(-\Delta)^s u=f \geq 0$ in a ball $B_2$ and $u=0$ in $ R^N \setminus B_2.$ Also suppose $u$ is $C^{s}$ non-negative and $(-\Delta)^s u=0$ in $B_2 \setminus B_1$ and $u\leq a$ on $\partial B_1$ where $B_1, B_2$ is a ball of radius $1$ and $2$ and $a$ is a positive constant. Can one claim that $u\leq a$ in $B_2 \setminus B_1$ or can any estimate be obtained on the upper bound of $u$. In the case $s=1$, this is maximum principle.
0
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
Yes.
We have $$ u(x) = \int_{B_1} G_{B_2}(x,y) f(y) dy , $$ where $$ G_{B_2}(x,y) = C_{N,s} \frac{1}{|x - y|^{N - 2s}} \int_0^{T(x,y)} \frac{t^{s-1}}{(t+1)^{N/2}} dt , \\ T(x, y) = \frac{(4-|x|^2)(4-|y|^2)}{4|x-y|^2} $$ is the corresponding Green function. Fortunately, $1/|x - y|^{N-2s}$ and $T(x,y)$ are radially decreasing on $B_2 \setminus B_1$, so indeed the maximal value of $u$ over $B_2 \setminus B_1$ is taken somewhere over $\partial B_1$.
answered May 21, 2020 at 7:00
-
$\begingroup$ Can anything be deduced if $B_2$ is replaced by a smooth bounded domain $\Omega.$ Thank you. $\endgroup$– SpalCommented May 22, 2020 at 6:31
-
$\begingroup$ I guess it is not: if $B_2$ is replaced by $\Omega = B_2 \setminus (\overline{B}_{1+\varepsilon} \setminus B_1)$ for a sufficiently small $\varepsilon > 0$, then the result is no longer true. I bet it could possibly be true if, say, $\Omega$ is convex, but these kind of problems are usually very difficult. $\endgroup$ Commented May 22, 2020 at 9:24
-
$\begingroup$ I think I meant something like $\Omega = B_3 \setminus (\overline{B}_2 \setminus B_{1+\varepsilon})$, sorry. $\endgroup$ Commented May 22, 2020 at 9:31
-
$\begingroup$ If $\Omega$ is smooth bounded domain and $(-\Delta)^s$ denotes the spectral fractional laplacian, do the result hold. That is $(-\Delta)^s u=0$ in $\Omega \setminus B$ and $u\leq a$ on $\partial B$ and $u=0$ on $\partial \Omega$ where $B$ is a ball of radius $1$ and $a$ is a positive constant. Can one claim that $u\leq a$ in $\Omega \setminus B.$ $\endgroup$– SpalCommented Jun 25, 2020 at 9:54
-
$\begingroup$ @Spal: No idea! Sounds much more plausible, but on the other hand I would not be surprised by a smart counter-example. I do not think such questions have been studied much, and my feeling is that they can easily get very difficult if true. $\endgroup$ Commented Jun 25, 2020 at 11:23