Ngô Quốc Anh

July 4, 2010

Nash-Moser’s iteration technique

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 23:05

We now continue this topic. Today we discuss a very powerful technique known as the Nash-Moser iteration technique. Let us consider a sub-solution u of the following variational inequality

\displaystyle\int_\Omega {{A^{\alpha \beta }}(x){\nabla _\alpha }u{\nabla _\beta }\varphi dx} \leqslant 0, \quad \varphi \in H_0^1(\Omega )

where \varphi \geqslant 0. We assume A is of class L^\infty(\Omega) and do satisfy an ellipticity condition. For the sake of simplicity we assume u\geqslant 0. Then by using the test function

\varphi=u^p\eta^2, \quad p>1

with \eta a cut-off function defined as in this topic we arrive at

\displaystyle\int_{B_R} {{A^{\alpha \beta }}(x){\nabla _\alpha }u{\nabla _\beta }({u^p}{\eta ^2})dx} \leqslant 0

which yields

\displaystyle p\int_{B_R} {{A^{\alpha \beta }}(x){u^{p - 1}}{\eta ^2}{\nabla _\alpha }u{\nabla _\beta }udx} + 2\int_{B_R} {{A^{\alpha \beta }}(x){u^p}\eta {\nabla _\alpha }u{\nabla _\beta }\eta dx} \leqslant 0.

Thus by the ellipticity condition

\displaystyle\int_{{B_R}} {{u^{p - 1}}{\eta ^2}{{\left| {\nabla u} \right|}^2}dx} \leqslant \frac{c}{p}\int_{{B_R}} {\left| {\nabla u} \right|{u^{\frac{{p - 1}}{2}}}{u^{\frac{{p + 1}}{2}}}\left| {\nabla \eta } \right|\eta dx}

and thus

\displaystyle\int_{{B_R}} {{u^{p - 1}}{\eta ^2}{{\left| {\nabla u} \right|}^2}dx} \leqslant \frac{c}{{{p^2}}}\int_{{B_R}} {{u^{p + 1}}{{\left| {\nabla \eta } \right|}^2}dx} .


\displaystyle\left| {\nabla {u^{\frac{{p + 1}}{2}}}} \right| = {\left( {\frac{{p + 1}}{2}} \right)^2}{u^{p - 1}}{\left| {\nabla u} \right|^2}

we get

\displaystyle\int_{{B_R}} {\left| {\nabla {u^{\frac{{p + 1}}{2}}}} \right|{\eta ^2}dx} \leqslant c{\left( {\frac{{p + 1}}{p}} \right)^2}\int_{{B_R}} {{u^{p + 1}}{{\left| {\nabla \eta } \right|}^2}dx}

and together with

\displaystyle\left| {\nabla (\eta {u^{\frac{{p + 1}}{2}}})} \right| \leqslant c\left[ {{{\left| {\nabla {u^{\frac{{p + 1}}{2}}}} \right|}^2}{\eta ^2} + {u^{p + 1}}{{\left| {\nabla \eta } \right|}^2}} \right]


\displaystyle\int_{{B_R}} {\left| {\nabla {u^{\frac{{p + 1}}{2}}}} \right|dx} \leqslant c\left[ {1 + {{\left( {\frac{{p + 1}}{p}} \right)}^2}} \right]\int_{{B_R}} {{u^{p + 1}}{{\left| {\nabla \eta } \right|}^2}dx} \leqslant c{(p + 1)^2}\left( {1 + \frac{1}{{{p^2}}}} \right)\int_{{B_R}} {{u^{p + 1}}{{\left| {\nabla \eta } \right|}^2}dx} .

By the Sobolev inequality we finally have

\displaystyle {\left[ {\int_{{B_R}} {{{({u^{\frac{{p + 1}}{2}}}\eta )}^{{2^ \star }}}dx} } \right]^{\frac{2}{{{2^ \star }}}}} \leqslant c{(p + 1)^2}\left( {1 + \frac{1}{{{p^2}}}} \right)\frac{1}{{{{(R - \rho )}^2}}}\int_{{B_R}} {{u^{p + 1}}dx} .

Letting q=p+1>2 and \lambda=\frac{n}{n-2} we get

\displaystyle {\left[ {\int_{{B_R}} {{u^{\lambda q}}dx} } \right]^{\frac{1}{\lambda }}} \leqslant c\frac{{{{(q + 1)}^2}}}{{{{(R - \rho )}^2}}}\int_{{B_R}} {{u^q}dx} .

Now we choose

\displaystyle\begin{gathered} {q_0} = 2, \quad {q_i} = 2{\lambda ^i} = \lambda {q_{i - 1}}, \hfill \\ {R_0} = R, \quad {R_i} = \frac{R}{2} + \frac{R}{{{2^{i + 1}}}}, \hfill \\ \end{gathered}

and we find that

\displaystyle {\left[ {\int_{{B_{{R_{i + 1}}}}} {{u^{{q_{i + 1}}}}dx} } \right]^{\frac{1}{{{\lambda ^{i + 1}}}}}} \leqslant \prod\limits_{k = 0}^i {{{\left[ {{4^{k + 1}}\frac{{c{{(1 + {q_k})}^2}}}{{{R^2}}}} \right]}^{\frac{1}{{{\lambda ^k}}}}}} {\left[ {\int_{{B_R}} {{u^2}dx} } \right]^{\frac{1}{2}}}.

We have to estimate the above product

\displaystyle\begin{gathered} \prod\limits_{k = 0}^\infty {{{\left[ {{4^{k + 1}}\frac{{c{{(1 + {q_k})}^2}}}{{{R^2}}}} \right]}^{\frac{1}{{{\lambda ^k}}}}}} = \exp \left( {\log \prod\limits_{k = 0}^i {{{\left[ {{4^{k + 1}}\frac{{c{{(1 + {q_k})}^2}}}{{{R^2}}}} \right]}^{\frac{1}{{{\lambda ^k}}}}}} } \right) \hfill \\ \qquad= \exp \sum\limits_{k = 0}^\infty {\frac{2}{{{\lambda ^k}}}\log \widehat c\frac{{1 + {q_k}}}{{R{2^{ - k - 1}}}}} = \exp (\overline c - n\log R) \hfill \\ \end{gathered}

thus for all k

\displaystyle {\left[ {\int_{{B_{{R_k}}}} {{u^{{q_k}}}dx} } \right]^{\frac{1}{{{q_k}}}}} \leqslant \widetilde c{\left[ {\int_{{B_R}} {{u^2}dx} } \right]^{\frac{1}{2}}}

from which immediately follows that

\displaystyle\mathop {\sup }\limits_{x \in {B_{\frac{R}{2}}}} u(x) \leqslant \widetilde c{\left[ {\int_{{B_R}} {|u{|^2}dx} } \right]^{\frac{1}{2}}}.

Remark. Instead of exponent 2, one can take any exponent p>0 in the above formulas, i.e.

\displaystyle\mathop {\sup }\limits_{x \in {B_{\frac{R}{2}}}}  u(x) \leqslant \widetilde c{\left[ {\int_{{B_R}} {|u{|^p}dx} }  \right]^{\frac{1}{p}}}, \quad p>0.

If we start with a super-solution u>0

\displaystyle\int_\Omega {{A^{\alpha \beta }}(x){\nabla _\alpha  }u{\nabla _\beta }\varphi dx} \geqslant 0, \quad \varphi \in  H_0^1(\Omega )

where \varphi \geqslant 0 then we can take p<-1. Follow exactly as before we then get

\displaystyle\mathop {\inf }\limits_{x \in {B_{\frac{R}{2}}}}  u(x) \leqslant \widetilde c{\left[ {\int_{{B_R}} {|u{|^q}dx} }  \right]^{\frac{1}{q}}}, \quad q<0.

We recommend the reader to read a joint paper by L. Ma and J.C. Wei published in J. Funct. Anal. in 2008 [here]. In that paper, the authors studied some properties of solutions to an elliptic equation with negative exponent

\Delta u = u^\tau, \quad \tau <0

in either the whole space or bounded domain.

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: Logo

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

Google photo

You are commenting using your Google 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 )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Blog at

%d bloggers like this: