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


where \Omega is a C^1 bounded domain. We also assume

p,A >0, \quad \beta, B \geqslant 0, \quad 1<\alpha <n.

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)


\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)


\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.


\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.

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