Let $n\geq 3$, $K$ is a bounded function in $\mathbb{R}^n$. If $u$ is a nonegative solution but not equal to 0 of $$\Delta u=Ku^{\frac{n+2}{n2}}\quad \text{in }\,\mathbb{R}^n.$$ Can we use the strong maximum principle to deduce $u$ is positive in $\mathbb{R}^n$? I saw the strong maximum principle in the bounded domain, what about the entire space like $\mathbb{R}^n$?
$\begingroup$
$\endgroup$
9

$\begingroup$ Isn't u=0 a nonnegative solution? $\endgroup$– Michael RenardyCommented Aug 10, 2023 at 2:57

$\begingroup$ sorry, here $u$ not equal to 0. $\endgroup$– Davidi ConeCommented Aug 10, 2023 at 4:31

$\begingroup$ Yes, the usual proof for subharmonic/superharmonic functions works in any open connected set and gives $u=0$ if u vanishes somewhere. Or else you can use the strong minimun principle in any ball. $\endgroup$– Giorgio MetafuneCommented Aug 10, 2023 at 6:54

$\begingroup$ Can you give some details about the proof, thanks. $\endgroup$– Davidi ConeCommented Aug 10, 2023 at 14:32

$\begingroup$ I just use that a superharmonic function with a global miniminum is constant in a open connected set. No boundedness assumption on the set is needed. So, if your u is zero somewhere, since it is nonegative, it has a minimum and it is zero everywhere. $\endgroup$– Giorgio MetafuneCommented Aug 10, 2023 at 21:50

Show 4 more comments
1 Answer
$\begingroup$
$\endgroup$
2
Yes, $u$ is strictly positive. Assume that $F=\{x:u(x)=0\}$ is nonempty. $F$ is clearly closed and I show that is open. Let $x_0 \in F$ and $r>0$ such that $u\Delta u=u(1+Ku^{\frac{4}{n2}}) \geq 0$ in $B(x_0,r)$. Such $r$ exists since $K$ is bounded. Applying the strong minimum principle in $B(x_0,r)$ to the operator $I\Delta$, we see that $u\equiv 0$ in $B(x_0,r)$.

$\begingroup$ how to guarantee that $(1+K u^{\frac{4}{n2}}) \geq 0$ in $B\left(x_0, r\right)$? since $K$ may be always negative. $\endgroup$ Commented Aug 14, 2023 at 2:53

$\begingroup$ Just use that $u(x) \to 0$ as $x \to x_0$. $\endgroup$ Commented Aug 14, 2023 at 7:54