Ngô Quốc Anh

November 11, 2009

A non-existence result for positive solutions to the Lichnerowicz equation in R^N

Filed under: Nghiên Cứu Khoa Học, PDEs — Tags: , — Ngô Quốc Anh @ 22:36

In this topic, adapted from a paper due to Li Ma and Xingwang Xu published in Comptes Rendus Mathematique we shall give a non-existence result concerning the following Lichnerowicz equation in \mathbb R^N

\Delta u + R(x) u + A(x) u^{-p-1} + B(x) u^{p-1}=0, u>0 on \mathbb R^N

where R(x) \geq 0, A(x) \geq 0, and B(x) are given smooth functions of x \in \mathbb R^N. To be precise, we obtain the following

Theorem. Suppose A:=A(x) \geq 0, B := B(x) \geq 0, and R(x) \geq 0. Let \beta = \frac{p+1}{2p}. Assume that

\displaystyle \int_0^{ + \infty } {\left( {\int_{B\left( {0,r} \right)} {{A^{1 - \beta }}{B^\beta }dx} } \right){r^{1 - N}}dr} = +\infty .

Then there exists no positive solution to the above Lichnerowicz equation.

Let us denote the integral

\displaystyle\frac{1}{{{\omega _n}{r^{N - 1}}}}\int_{\partial B\left( {0,r} \right)} {f\left( x \right)dS_x}

by \overline f. We call \overline f the average of f on the sphere S(0,r) of radius r, or sphere mean of a function around the origin.

Proof. Note that a simple calculation shows us that

\displaystyle\frac{1}{{{\omega _n}{r^{N - 1}}}}\int_{\partial B\left( {0,r} \right)} {f\left( x \right)d{S_x}} = \frac{1}{{{\omega _n}}}\int_{\partial B\left( {0,1} \right)} {f\left( {rx} \right)d{S_x}} .

Therefore

\displaystyle {\overline u ^\prime }= \frac{d}{{dr}}\overline u = \frac{1}{{{\omega _n}}}\int_{\partial B\left( {0,1} \right)} {\frac{d}{{dr}}u\left( {xr} \right)d{S_x}} =\frac{1}{{{\omega _n}}}\int_{\partial B\left( {0,1} \right)} {\sum\limits_{k = 1}^N {{x_i}{u_{{x_i}}}} d{S_x}}.

Since on the sphere S(0,1), x=(x_1,...,x_N) is also the outer normal vector, therefore

\displaystyle\frac{1}{{{\omega _n}}}\int_{\partial B\left( {0,1} \right)} {\sum\limits_{k = 1}^N {{x_i}{u_{{x_i}}}\left( {xr} \right)} d{S_x}} = \frac{1}{{{\omega _n}}}\int_{\partial B\left( {0,1} \right)} {\nabla u\left( {xr} \right) \cdot {n_x}d{S_x}}

Thus by the divergence theorem, one gets

\displaystyle \frac{1}{{{\omega _n}}}\int_{\partial B\left( {0,1} \right)} {\nabla u\left( {xr} \right)\cdot {n_x}d{S_x}} = \frac{r}{{{\omega _n}}}\int_{B\left( {0,1} \right)} {\Delta u dx} = \frac{1}{{{\omega _n}{r^{N-1}}}}\int_{B\left( {0,r} \right)} {\Delta udx}.

Hence

\displaystyle{\overline u ^\prime } = \frac{1}{{{\omega _n}{r^{N - 1}}}}\int_{B\left( {0,r} \right)} {\Delta udx}.

Differentiating once more yields

\displaystyle{\overline u ^\prime }^\prime = \frac{d}{{dr}}{\overline u ^\prime } = - \underbrace {\frac{{N - 1}}{{{\omega _n}{r^N}}}\int_{B\left( {0,r} \right)} {\Delta udx} }_{\frac{{N - 1}}{r}\overline u'} + \frac{1}{{{\omega _n}{r^{N - 1}}}}\frac{d}{{dr}}\left( {\int_{B\left( {0,r} \right)} {\Delta udx} } \right).

Since

\displaystyle\frac{d}{{dr}}\left( {\int_{B\left( {0,r} \right)} {\Delta udx} } \right) =\int_{\partial B\left( {0,r} \right)} {\Delta udx}

then

\displaystyle{\overline u ^\prime }^\prime = - \frac{{N - 1}}{r}\overline u' + \frac{1}{{{\omega _n}{r^{N - 1}}}}\int_{\partial B\left( {0,r} \right)} {\Delta udx} = - \frac{{N - 1}}{r}\overline u' + \overline {\Delta u}.

Thus

\displaystyle\overline {\Delta u} = {\overline u ^\prime }^\prime + \frac{{N - 1}}{r}\overline u'

Therefore, taking this average operation we have

\displaystyle - {\overline u ^\prime }^\prime - \frac{{N - 1}}{r}\overline u' = \overline {R(x)u} + \overline {A(x){u^{ - p - 1}} + B(x){u^{p - 1}}} .

Since for each fixed x\in \mathbb R^N,

\displaystyle\begin{gathered} A{u^{ - p - 1}} + B{u^{p - 1}} = \frac{{2p}}{{p - 1}}\frac{{p - 1}}{{2p}}A{u^{ - p - 1}} + \frac{{2p}}{{p + 1}}\frac{{p + 1}}{{2p}}B{u^{p - 1}} \\\qquad\quad\geq \frac{{p - 1}}{{2p}}A{u^{ - p - 1}} + \frac{{p + 1}}{{2p}}B{u^{p - 1}}. \\ \end{gathered}

Then by using the general Cauchy inequality, one gets

\displaystyle\frac{{p - 1}}{{2p}}A{u^{ - p - 1}} + \frac{{p + 1}}{{2p}}B{u^{p - 1}} \geqslant {\left( {A{u^{ - p - 1}}} \right)^{\frac{{p - 1}}{{2p}}}}{\left( {B{u^{p - 1}}} \right)^{\frac{{p + 1}}{{2p}}}} = {A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}.

Thus,

\displaystyle\overline {A{u^{ - p - 1}} + B{u^{p - 1}}} \geq \overline {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}}.

It turns out that

\displaystyle - {\left( r^{N - 1}\overline u ' \right)^\prime } \geq {r^{N - 1}}\left( {\overline {R(x)u}+ \overline {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}} } \right),

which implies that

\displaystyle - {r^{N - 1}}\overline u ' \geq\frac{1}{\omega _n}\int_{B\left( {0,r} \right)} {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}dx} + \frac{1}{\omega _n}\int_{B\left( {0,r} \right)} {R(x)udx} \geq \frac{1}{\omega _n}\int_{B\left( {0,r} \right)} {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}dx}

after an integration. This is because, by definition of the sphere mean,

\displaystyle\begin{gathered}{r^{N - 1}}\left( {\overline {R(x)u} + \overline {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}} } \right) = {r^{N - 1}}\left( {\frac{1}{{{\omega _n}{r^{N - 1}}}}\int_{\partial B\left( {0,r} \right)} {R(x)ud{S_x}} + \frac{1}{{{\omega _n}{r^{N - 1}}}}\int_{\partial B\left( {0,r} \right)} {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}d{S_x}} } \right) \\\qquad\quad\;\;\;= \frac{1}{{{\omega _n}}}\left( {\int_{\partial B\left( {0,r} \right)} {R(x)ud{S_x}} + \int_{\partial B\left( {0,r} \right)} {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}d{S_x}} } \right) \\\qquad\qquad\qquad\qquad\;\;\;= \frac{1}{{{\omega _n}}}\frac{d}{{dr}}\left[ {\int_0^r {\left( {\int_{\partial B\left( {0,s} \right)} {R(x)ud{S_x}} + \int_{\partial B\left( {0,s} \right)} {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}d{S_x}} } \right)ds} } \right] .\\\end{gathered}

Dividing both sides by r^{N-1} and integrating this inequality over [0, r_0], we have

\displaystyle \overline u (0) \geq \overline u (0) - \overline u ({r_0}) \geqslant \int_0^{{r_0}} {\left( {{r^{1 - N}}\int_{B\left( {0,r} \right)} {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}dx} } \right)dr}.

Sending r_0 \to \infty we have

\displaystyle\overline u (0) \geq \int_0^{ + \infty } {\left( {{r^{1 - N}}\int_{B\left( {0,r} \right)} {{A^{\frac{{p - 1}}{{2p}}}}{B^{\frac{{p + 1}}{{2p}}}}dx} } \right)dr},

which is impossible by our assumption. The proof is complete.

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