Ngô Quốc Anh

April 4, 2011

An iteration by Stampacchia

Filed under: Nghiên Cứu Khoa Học — Tags: — Ngô Quốc Anh @ 3:09

Recently, my friend, CR, has shown me an iteration by Stampacchia. Stampacchia proposed his iteration in a preprint [here] entitled Équations elliptiques du second ordre à coefficients discontinus published in Séminaire Jean Leray in 1963-1964.

Suppose \varphi : [k_0, \infty) \to \mathbb R is a non-negative non-decreasing function satisfying

\displaystyle \varphi (h) \leqslant \frac{c}{(h-k)^\alpha}\big(\varphi(k)\big)^\beta

for any h>k \geqslant k_0 where c, \alpha, \beta are positive given constants.


  • If \beta >1, it holds

    \varphi (k_0+d)=0


    \displaystyle d^\alpha=c\big(\varphi(k_0)\big)^{\beta-1}2^\frac{\alpha\beta}{\beta-1}.

  • If \beta=1, one has

    \displaystyle \varphi (h)\leqslant e e^{-\eta (h-k_0)}\varphi (k_0)


    \displaystyle \eta=(ec)^{-\frac{1}{\alpha}}.

  • If \beta<1 and k_0>0, then

    \displaystyle \varphi (h) \leqslant {2^{\frac{\mu }{{1 - \beta }}}}\left\{ {1 + {c^{\frac{1}{{\beta - 1}}}}{{(2{k_0})}^\mu }\varphi ({k_0})} \right\} = \frac{{{2^{\frac{\mu }{{1 - \beta }}}}}}{{{c^{\frac{1}{{1 - \beta }}}}}}\left\{ {{c^{\frac{1}{{1 - \beta }}}} + {{(2{k_0})}^\mu }\varphi ({k_0})} \right\}


    \displaystyle \mu=\frac{\alpha}{1-\beta}.

We shall only prove the case of \beta>1.

Proof. Since d is given, we can consider the following quantity

\displaystyle k_s=k_0+d-\frac{d}{2^s}.

Then using the assumption for h=k_s and k=k_{s+1} it holds by

\displaystyle k_{s+1}-k_s=d2^{-(s+1)}


\displaystyle\varphi ({k_{s + 1}}) \leqslant c\frac{{{2^{(s + 1)\alpha }}}}{{{d^\alpha }}}{[\varphi ({k_s})]^\beta }.

We now prove by induction the following

\displaystyle\varphi ({k_{s + 1}}) \leqslant \frac{{\varphi ({k_0})}}{{{2^{ - s\mu }}}}, \quad \text{ where }\mu = \frac{\alpha }{{1 - \beta }}.

Indeed, for s=0, the inequality is trivial. Suppose it holds for s, we are going to prove for s+1. We now have from the recursive equation that

\displaystyle\begin{gathered} \varphi ({k_{s + 1}}) \leqslant c\frac{{{2^{(s + 1)\alpha }}}}{{{d^\alpha }}}{[\varphi ({k_s})]^\beta } \\\qquad\quad\quad\;\,\leqslant c\frac{{{2^{(s + 1)\alpha }}}}{{{d^\alpha }}}\frac{{{{[\varphi ({k_0})]}^\beta }}}{{{2^{ - s\beta \mu }}}} \hfill \\\qquad\quad\quad\;\, = c\frac{{{2^{(s + 1)\alpha }}}}{{c{{(\varphi ({k_0}))}^{\beta - 1}}{2^{\frac{{{\alpha\beta }}}{{\beta - 1}}}}}}\frac{{{{[\varphi ({k_0})]}^\beta }}}{{{2^{ - s\beta \mu }}}} \\\qquad\quad\quad\;\,= \frac{{\varphi ({k_0})}}{{{2^{ - (s + 1)\mu }}}}. \hfill \\ \end{gathered}

Now sending s \to +\infty with the fact that \mu<0 we can complete our proof.

For the case of \beta=1, we just consider

\displaystyle k_s=k_0+s(ec)^\frac{1}{\alpha}

then it holds

\displaystyle\varphi (k_s) \leqslant \frac{1}{e}\varphi (k_{s-1}) \leqslant \cdots \leqslant e^{-s} \varphi (k_0).

For the latter case, we shall consider the following auxiliary function

\displaystyle\varphi (h) = \psi (h)\frac{{{c^{\frac{1}{{1 - \beta }}}}}}{{{h^\mu }}}.

Then it holds

\displaystyle\psi (h) \leqslant {\left( {\frac{h}{{{{(h - k)}^{1 - \beta }}{k^\beta }}}} \right)^\mu }{[\psi (k)]^\beta }.

For the complete proof, we refer the reader to Stampacchia’s paper for details.

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 )

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

%d bloggers like this: