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

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

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

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

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.