# Ngô Quốc Anh

## June 8, 2011

### Uniformly boundedness of positive smooth solutions to the Lichnerowicz equation in R^n

Filed under: PDEs — Tags: , — Ngô Quốc Anh @ 7:26

Let us consider the following so-called Lichnerowicz equation in $\mathbb R^n$

$\displaystyle -\Delta u =-u^q+\frac{1}{u^{q+2}}, \quad q>0.$

Recently, Brezis [here] proved the following

Theorem. Any solution of the Lichnerowicz equation with $q>0$ satisfies $u\geqslant 1$ in $\mathbb R^n$.

Let us study the trick used in his paper.

Proof. Let

$f(t)=t^q-t^{q+2}, \quad t>0.$

Then $\Delta u =f(u)$. Fix any point $x_0 \in \mathbb R^n$ and consider the function

$u_\varepsilon(x)=u(x)+\varepsilon |x-x_0|^2, \quad \varepsilon>0, x \in \mathbb R^n.$

Since $u_\varepsilon(x) \to \infty$ as $|x|\to \infty$, $\min_{\mathbb R^n}u_\varepsilon(x)$ is achieved at some $x_1$. We have

$0\leqslant \Delta u_\varepsilon(x_1)=\Delta u(x_1)+2\varepsilon n=f(u(x_1))+2\varepsilon n.$

By definition,

$u(x_1)+\varepsilon |x_1-x_0|^2=u_\varepsilon(x_1) \leqslant u_\varepsilon(x_0)=u(x_0).$

Thus,

$u(x_1)\leqslant u(x_0).$

Since $f$ is increasing we deduce that

$f(u(x_1))\leqslant f(u(x_0)).$

Therefore,

$0\leqslant \Delta u_\varepsilon(x_1) \leqslant f(u(x_1))+2\varepsilon n \leqslant f(u(x_0))+2\varepsilon n.$

By sending $\varepsilon \to 0$ we deduce that $f(u(x_0)) \geqslant 0$. In other words, $u(x_0) \geqslant 1$.

As can be seen, he only uses the fact that $f$ is monotone increasing in his argument, therefore, this approach can be used for a wider class of nonlinearity.

In the second displayed equation, should there be $t^{-(q+2)}$?

Comment by timur — June 8, 2011 @ 22:53

• Thanks, you were correct. It is just equal to minus of the right hand side of the PDE.

Comment by Ngô Quốc Anh — June 8, 2011 @ 22:57

2. Hi! Good post again. 😉

I would say that when you define f(t) the second exponent should be -q-2. Am I right?

Greetings!!

Comment by Urko — June 8, 2011 @ 22:55

• Urko, thanks. As I mentioned in the foregoing reply, you were absolutely correct. It is just equal to minus of the right hand side of the PDE. So function $f$ should read

$\displaystyle f(t)=t^q-t^{-(q+2)}, \quad t>0.$

Comment by Ngô Quốc Anh — June 8, 2011 @ 22:59

• Ups sorry, I dont know why I didnt read the previous comments. My apologies!!!

Comment by Urko — June 9, 2011 @ 15:42

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