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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?

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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.

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  • $\begingroup$ I too thought about averaging, also using Jensen's inequality. However, I had been unable to convince myself that the Laplacian of the average is the average of the Laplacian. Can you detail this? $\endgroup$
    – Iosif Pinelis
    Commented Jun 19, 2023 at 19:21
  • $\begingroup$ @IosifPinelis Yes. Let $v(r)=\int_S u(r\omega) d\omega$, $S$ being the unit sphere. Then $$v'(r)=\int_S \nabla u(r\omega) \cdot \omega=r^{1-n} \int_{S_r} \nabla u (y)\cdot \nu(y) dy$$ where $S_r$ is the sphere of radius $r$ and $\nu$ the outward normal. By the divergence theorem this last equals $r^{1-n}\int_{B_r} \Delta u$. Then the derivative of $r^{n-1} v'$ equals the integral of $\Delta u$ over $S_r$ and the rest follows by using the expression of $\Delta v$ for radial $v$. $\endgroup$
    – Giorgio Metafune
    Commented Jun 19, 2023 at 20:04
  • $\begingroup$ Thank you for your response; +1 from me. This looks to me, though, as some lucky coincidence that, after some manipulations (involving especially the use of the expression of $\Delta v$ for radial $v$), one gets that the averaging commutes with the Laplacian. I suspect there should be some more direct way to see that, involving little to no calculations. What do you think about this? $\endgroup$
    – Iosif Pinelis
    Commented Jun 19, 2023 at 22:14
  • $\begingroup$ @IosifPinelis I agree with you, it is tricky. There is another (but close) way to obtain the result. Write $v$ as above and compute $$\Delta v=v''+(n-1)/r v'=\int_S (u_{rr}+ \frac{n-1}{r}v')=\int_S (u_{rr}+ \frac{n-1}{r}v'+\frac{\Delta_0 u}{r^2})=\int_S \Delta u.$$ Here $\Delta_0$ is the Laplace Beltrami on the sphere and I used $\int_S \Delta_0 u=0$. $\endgroup$
    – Giorgio Metafune
    Commented Jun 20, 2023 at 8:11
  • $\begingroup$ Another way to obtain radial solutions is to define $w(x)=\int_{SO} u_T(x)d\nu(T)$. Here $SO$ is the orthogonal group, $d\nu$ is its Harr measure and $u_T(x)=u(Tx)$. In this case, using $\Delta u_T=(\Delta u)_T$ one sees that $\Delta w$ is the average (with respect to $SO$) of $\Delta u$. The two formuals look similar but I am not sure they are really so. If $u$ is harmonic, each $u_T$ is harmonic but this is not true for $u(r\omega)$ for a fixed $\omega$ and harmonicity for $v$ follows from the zero mean property of the Laplace Beltrami. If you see some other thing, please let me know. $\endgroup$
    – Giorgio Metafune
    Commented Jun 20, 2023 at 8:24

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