Ngô Quốc Anh

May 8, 2010

Achieving regularity results via bootstrap argument

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 2:35

Let us now consider a very important technique in PDEs called the  bootstrap method or bootstrap argument. Actually, this method is a part of the proof of the Schauder estimates, etc. I will follow the recent paper due to Li, Strohmer and Wang published in Proc. Amer. Math. Soc. in 2009.

The equation considered here is the following $\displaystyle u(x) = A\int_\Omega {\frac{{|u|^p (y)}}{{|x - y{|^{n - \alpha }}}}dy} + B$

together with $u\big|_{\partial\Omega}=\beta$

where $\Omega$ is a $C^1$ bounded domain. We also assume $p,A >0, \quad \beta, B \geqslant 0, \quad 1<\alpha .

Theorem. If $u \in L^q(\Omega)$ is a solution to the PDE for some $q>1$ then $u \in C(\overline \Omega)$.

In order to run the bootstrap argument, we need the following auxiliary result

Lemma. Suppose $w\in L^r(\Omega)$ with $1\leqslant r<\infty$ and $\displaystyle v(x) = A\int_\Omega {\frac{{|w| (y)}}{{|x - y{|^{n - \alpha }}}}dy} + B$

then $v \in W^{1,s}(\Omega)$ where $\begin{cases}\frac{1}{s}+\frac{\alpha-1}{n}=\frac{1}{r}, & {\rm if }\; 1 \leqslant r <\frac{n}{\alpha-1},\\{\rm any} \; s \geqslant 1, & {\rm if } \; r \geqslant \frac{n}{\alpha-1}.\end{cases}$

Proof of theorem. Keep in mind from the Rellich-Kondrachov theorem, the following embedding $W^{1,k}(\Omega) \hookrightarrow C(\overline \Omega)$

holds for any $k>n$. So if there is some $q'$ satisfying $\displaystyle \frac{q'}{p} \geqslant \frac{n}{\alpha-1}$

such that $u \in L^{q'}(\Omega)$

we then see that $\displaystyle u^p \in L^\frac{q'}{p}(\Omega)$.

From the PDE and our lemma above we get $u \in W^{1,k}(\Omega)$

for any $k \geqslant 1$. Choosing $k=n+1$ gives us the desired result. Therefore we only need to consider the case that $\displaystyle \frac{q}{p} < \frac{n}{\alpha-1}$.

The idea of the bootstrap argument is to try to increase the regularity of $u$, in this situation, our aim is to obtain such $q'$ from the fact that $u \in L^q$. Indeed, firstly using our lemma we actually have $u \in W^{1,s}(\Omega)$

where $\displaystyle \frac{1}{s}+\frac{\alpha-1}{n}=\frac{1}{q}$.

If $s > n$ then we are done. If $s=n$ by the Rellich-Kondrachov theorem, we can find such $q'$. Otherwise, $W^{1,s}(\Omega) \hookrightarrow L^{s^\star}(\Omega)$

where $\displaystyle \frac{1}{s^\star}=\frac{1}{s}-\frac{1}{n}$.

We need to compare $s^\star$ and $q$. Actually, we need to show that $s^\star - q > {\rm const.}>0$.

Indeed, $\displaystyle\frac{1}{{{s^ \star }}} - \frac{1}{q} = \frac{1}{s} - \frac{1}{n} - \frac{1}{q} = \frac{1}{q} - \frac{{\alpha - 1}}{n} - \frac{1}{n} - \frac{1}{q} = - \frac{\alpha }{n}<0$

which implies $\displaystyle {s^ \star } - q = \frac{\alpha }{n}q{s^ \star } > \frac{{\alpha q}}{n} > \frac{\alpha }{n}$.

In other words, we reach to the case $u \in L^{q'}$ where $q'-q> {\rm const.}>0$. If $q'$ does not satisfy $\displaystyle \frac{q'}{p} \geqslant \frac{n}{\alpha-1}$

we then replace $q$ by $q'$ and repeat the above argument again until $q'$ does satisfy the above inequality. The proof then follows.