# Ngô Quốc Anh

## June 13, 2010

### Achieving regularity results via bootstrap argument, 2

Filed under: PDEs — Tags: , — Ngô Quốc Anh @ 1:37

Today, we shall discuss a very strong tool in the theory of elliptic PDEs in order to achieve the smoothness of solution. The tool we just mentioned is known as the Calderón-Zygmund $L^p$ estimates or the Calderón-Zygmund inequality. Precisely,

Theorem (Calderón-Zygmund). Let $1 and $f \in L^p(\Omega)$ ( $\Omega$ is open and bounded). Let $u$ be the weak solution of the following PDE $\displaystyle \Delta u = f$.

Then $u\in W^{2,p}(\Omega')$ for any $\Omega' \Subset \Omega$.

Let us consider the regularity of solution of $\displaystyle \Delta u +\Gamma(u)|\nabla u|^2=0$

with a smooth $\Gamma$. We also require that $\Gamma$ is bounded.

Motivation. The above PDE occurs as the Euler-Lagrange equation of the variational problem $\displaystyle I(u)=\int_\Omega g(u(x))|\nabla u(x)|^2dx \to {\rm min}$

with a smooth $g$ with is bounded and bounded away from zero. Moreover, $g'$ is bounded.

In fact, to derive the Euler-Lagrange equation, we consider $\displaystyle I(u + t\varphi ) = \int_\Omega {g(u + t\varphi ){{\left| {\nabla (u + t\varphi )} \right|}^2}dx}$

where $\varphi \in H_0^{1,2}(\Omega)$. In that case $\displaystyle \frac{d}{{dt}}I(u + t\varphi ) = \int_\Omega {\left[ { - 2g(u)\Delta u - g'(u){{\left| {\nabla u} \right|}^2}} \right]\varphi dx}$

after integrating by parts and assuming for the moment $u \in C^2$. Thus, the minimizer will verify $\displaystyle - 2g(u)\Delta u - g'(u){\left| {\nabla u} \right|^2} = 0$

which turns out to be $\displaystyle\Delta u + \underbrace {\frac{{g'(u)}}{{2g(u)}}}_{\Gamma (u)}{\left| {\nabla u} \right|^2} = 0.$

In order to apply the $L^p$-theory, we assume that $u$ is a weak solution of the PDE with $\displaystyle u \in W^{1,p_1}(\Omega) \quad p_1 >n$.

Step 1. Since $\Gamma$ is bounded, we have $\displaystyle -\Gamma (u){\left| {\nabla u} \right|^2} = f = :{f_1} \in {L^{\frac{{{p_1}}}{2}}}(\Omega )$.

Since $u$ verifies the PDE, we get $\displaystyle u \in W^{2,p_1}(\Omega') \quad \Omega' \Subset \Omega$.

By the Sobolev embedding theorem $\displaystyle u \in W^{1,p_2}(\Omega') \quad \Omega' \Subset \Omega$

with $\displaystyle {p_2} = \frac{{n\frac{{{p_1}}}{2}}}{{n - \frac{{{p_1}}}{2}}} > {p_1}$.

Thus $\displaystyle f_1\in {L^{\frac{{{p_2}}}{2}}}(\Omega ')$.

Applying the Sobolev embedding theorem again gives us $\displaystyle u \in {W^{2,\frac{{{p_2}}}{2}}} \cap {W^{1,{p_3}}}(\Omega '')$

with $\displaystyle {p_3} = \frac{{n\frac{{{p_2}}}{2}}}{{n - \frac{{{p_2}}}{2}}} > {p_2}$

and $\Omega'' \Subset \Omega' \Subset \Omega$. Iterating this procedure, we eventually obtain $\displaystyle u \in W^{2,q}(\Omega') \quad \Omega' \Subset \Omega$

for any $q$.

Step 2. We now differentiate the PDE in order to obtain an equation for $D_i u$ for all $i=\overline{1,n}$. In fact, we get $\displaystyle\Delta ({D_i}u) + \Gamma '(u){D_i}u{\left| {\nabla u} \right|^2} + 2\Gamma (u)\sum\limits_j {{D_j}u{D_{ji}}u} = 0$.

This time, we put $\displaystyle {f_2}: = - \Gamma '(u){D_i}u{\left| {\nabla u} \right|^2} - 2\Gamma (u)\sum\limits_j {{D_j}u{D_{ji}}u}$.

Since $u \in W^{2,q}(\Omega)$ for any $q$, it is clear to see that $f_3 \in L^q(\Omega')$ for any $q$. Thus $\displaystyle D_iu \in W^{2,q}(\Omega') \quad \Omega' \Subset \Omega$

for any $q$ which is equivalent to $\displaystyle u \in W^{3,q}(\Omega') \quad \Omega' \Subset \Omega$

for any $q$.

Step 3. By repeating step 2 we finally get $\displaystyle u \in W^{k,q}(\Omega') \quad \Omega' \Subset \Omega$

for any $k$ and $q$. As a consequence of the Sobolev embedding theorem, $u$ is smooth.

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