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<p<\infty 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


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


\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 '')


\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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