# 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.

Then

• If $\beta >1$, it holds

$\varphi (k_0+d)=0$

where

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

• If $\beta=1$, one has

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

where

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

• If $\beta<1$ and $k_0>0$, then

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

where

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

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

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

that

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

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.