I come across an interesting question.



> 
> Let $ B\_r=\{x\in\mathbb{R}^3:|x|\leq r\} $ be the ball in $ \mathbb{R}^3 $ with radius $ r $. Assume that $ u \in C(\mathbb{R}^3\setminus B\_1) $ satisfies
> $$
> \Delta u\leq -u^3,\,\,u\geq 0,\,\,\forall |x|\geq 1.
> $$  
> 
> where $ \Delta u $ is defined in the sense of distribution. Show that $ u=0 $ in $ \mathbb{R}^3\setminus B\_1 $.
> 
> 
> 


I guess I need to use maximal principle but cannot go on, can you give me some hints or references?



How to show that $ u $ is vanishing in $ \mathbb{R}^3\setminus B_1 $?
A proof that $u=0$ is as follows. Assume that such $u \geq 0 $ is not identically zero. Since $\Delta u \leq 0$, $u$ can never vanish, otherwise it would have an interior minimum. Taking averages on the unit sphere of $\mathbb R^3$ we may assume also that $u$ is radial (the inequality in the differential equation is preserved by Jensen inequality). Then $u''+\frac 2r u' \leq -u^3$ or $(r^2u')' \leq -r^2u^3 \leq 0$, in particular $r^2 u'$ is decreasing.


There exists $r\_0$ such that $u'(r\_0)<0$. If not, $u$ would be increasing and $u \geq c>0$ gives $\Delta u \leq -c^2 u$ which has no positive solutions in $\mathbb R^3 \setminus B\_R$ (I assume this to be known but a proof follows with the arguments below). Then $r^2 u'(r) \leq -\alpha <0$ for $r \geq r\_0$ and then $u$ is decreasing and $\lim\_{r \to \infty}u(r)=0$ (by the same argument, if the limit is $\ell>0$ we have the inequality $\Delta u \leq -\ell^2 u$).


For $r \geq r\_0$ we have $u \geq \frac c r$ ($c=r\_0 u(r\_0))$ by subharmonic comparison, since both function tend to zero at infinity.


The inequality $(r^2u')' \leq -r^2 u^3 \leq - c^3/r$ gives $u' \leq -\alpha /r^2 -c^3/r^2 \log (r/r\_0)$. Integrating this inequality between $r$ and $\rho$ and letting $\rho \to \infty$ we obtain $u(r) \geq a/r+(b/r) \log r$ for suitable $b>0$ and then $u(r) \geq k/r$ for $r \geq M$ and $k \geq 1$.


Form this we obtain $u''+\frac 2r u' + \frac {k^2}{r^2} u \leq 0$ for $r \geq M$ and, setting $u=w/r$, $w''+\frac{k^2}{r^2} w \leq 0$. However this last equation does not have positive solutions. In fact every solution of the equation $v''+\frac{k^2}{r^2} v=0$ oscillates (it has power like solutions) and if $w>0$ satisfies the above inequality, setting $p=w''/w$ it follows that $w''-pw=0$ with $-p \geq k^2/r^2$ contradicitng the Sturm oscillation theorem.


