# Ngô Quốc Anh

## March 5, 2014

### Uniformly upper Bound for Positive Smooth Solutions To The Lichnerowicz Equation In R^N

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 16:16

In this note, we are interested in the following Lichnerowicz type equation in $\mathbb R^n$

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

As we have already seen from the previous note that solutions for the above equation are always bounded from below for certain $q$.

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

Remarkably, we are able to prove that solutions for the Lichnerowicz type equation are also bounded from above if we require $q>1$ instead of $q>0$. This result is basically due to L. Ma and X. Xu, see this paper.

Theorem (Ma-Xu). Any solution of the Lichnerowicz type equation with $q>1$ is uniformly bounded from above in $\mathbb R^n$.

Combining the two theorem above, we conclude that any solution of the Lichnerowicz equation, i.e. $q=(n+2)/(n-2)$, is uniformly bounded in $\mathbb R^n$.

The idea of the proof for Ma-Xu’s theorem is as follows: Denote

$\displaystyle f(u)=u^{-q-2} - u^q.$

Fix $x_0 \in \mathbb R^n$ but arbitrary, we then look for a positive radial super-solution $v(x)=v(|x|)>0$ of the Lichnerowicz type equation on the ball $B_R(x_0)$ with positive infinity boundary condition for some $R$ to be specified. This is equivalent to finding $v$ in such a way that

$\displaystyle\begin{array}{rcl}\displaystyle-\Delta v &\geqslant& f(v), \qquad\text{ in } B_R(x_0),\\ v&\equiv& +\infty, \qquad\text{ on }\partial B_R(x_0).\end{array}$

Note that

$\displaystyle f'(u)=-(q+2)u^{-q-3}-qu^{-q-1}<0$

for any $u>0$ and any $q>0$. Then the comparison lemma holds in the ball $B_R(x_0)$ which tells us that

$\displaystyle u(x) \leqslant v(x)$

in $B_R(x_0)$. To construct such a super-solution $v$, we initially choose

$\displaystyle v(x)=\frac{1}{(R^2-|x-x_0|^2)^\alpha}$

for some large $\alpha>1$ to be determined later. Since $v$ is radially symmetric, we calculate to find

$\displaystyle\Delta v(x) = \frac{1}{r^{n - 1}}({r^{n - 1}}w(r))' \Big|_{r = |x - {x_0}|}$

where the function $w$ is given as follows

$\displaystyle w(r)=\frac{1}{(R^2-r^2)^\alpha}.$

A simple calculation shows that

$\begin{array}{lcl} \displaystyle\Delta v(x) &=& \displaystyle \Big(w''(r) + (n - 1)\frac{{w'(r)}}{{{r^{n - 1}}}} \Big)\Big|_{r = |x - {x_0}|} \hfill \\ &=&\displaystyle \Big[ \frac{{2\alpha }}{{{{({R^2} - {r^2})}^{\alpha + 2}}}}\left( {(2\alpha + 2 - n){r^2} + n{R^2}} \right) \Big] \Big|_{r = |x - {x_0}|}. \hfill \\ \end{array}$

Hence, to obtain $-\Delta v \geqslant f(v)$, it is necessary to obtain the following

$\displaystyle {({R^2} - {r^2})^{\alpha (q + 2)}} - \frac{1}{{{{({R^2} - {r^2})}^{\alpha q}}}} \leqslant - \frac{{2\alpha }}{{{{({R^2} - {r^2})}^{\alpha + 2}}}}\left( {(2\alpha + 2 - n){r^2} + n{R^2}} \right),$

i.e.

$\displaystyle {({R^2} - {r^2})^{\alpha (q + 3) + 2}} + 2\alpha \left( {(2\alpha + 2 - n){r^2} + n{R^2}} \right) \leqslant \frac{1}{{{{({R^2} - {r^2})}^{\alpha (q - 1) - 2}}}}.$

However, provided $R \ll 1$ and $\alpha$ is such that $\alpha (q - 1) - 2 > 0$ and $2\alpha + 2 - n>0$, we obtain

$\displaystyle {({R^2} - {r^2})^{\alpha (q + 3) + 2}} + 2\alpha \left( {(2\alpha + 2 - n){r^2} + n{R^2}} \right) < 1 \leqslant \frac{1}{{{{({R^2} - {r^2})}^{\alpha (q - 1) - 2}}}}$

for any $r \in (0,R)$. Hence, in $B_{R/2}(x_0)$, we obtain the following key estimate

$\displaystyle u(x) \leqslant v(x) = \frac{1}{{{{({R^2} - {r^2})}^\alpha }}} \leqslant \frac{1}{{{{({R^2} - {R^2}/4)}^\alpha }}} < \frac{{{2^\alpha }}}{{{R^{2\alpha }}}}.$

Since $x_0$ is arbitrary and $R$ is fixed, the preceding estimate guarantees that $u$ is bounded from above.

I thank Mr. Vu Van Khu from NTU for useful discussion concerning this note.