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).


u(x_1)\leqslant u(x_0).

Since f is increasing we deduce that

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


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.


  1. Thanks for interesting posts! I follow your blog.

    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?


    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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