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.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Create a free website or blog at WordPress.com.

%d bloggers like this: