Scaling property in generalized linear equation.

{ \mathbf{Problem:}} Suppose

\displaystyle \displaystyle Lu=D_i(a^{ij}(x)D_ju+b^i(x)u)+c^i(x)D_iu+d(x)u=g+D_if^i

satisfies

\displaystyle \displaystyle a^{ij}(x)\xi_i\xi_j\geq \lambda|\xi|^2

\displaystyle \displaystyle \sum |a^{ij}(x)|^2\leq \Lambda

\displaystyle \lambda^{-2}\sum (|b^i(x)|^2+|c^i(x)|^2)+\lambda^{-1}|d(x)|\leq \theta^2

\displaystyle f_i\in L^q(\Omega), i=1,2,\cdots, n, \quad g\in L^{q/2}(\Omega)\text{for some } q>n

Then if { u\in W^{1,2}(\Omega)} is a weak subsolution of { Lu=f}, we have for any { R>0} such that { B_{2R}(y)\subset \Omega} and { p>1}

\displaystyle \displaystyle \sup\limits_{B_R(y)}u\leq C(R^{-n/p}||u^+||_{L^p(B_{2R}(y))}+k(R))

where { C=C(n,\Lambda/\lambda,\theta R,q,p)} , { k(R)=\lambda^{-1}(R^{1-n/q}||f||_q+R^{2-2n/q}||g||_{q/2})}.

{ \mathbf{Proof:}} The technique of proof this theorem is assuming that { R=1} and then use dilation to show the general case. We will omit the proof the { R=1}. Let us just assume this.

If { u} is a function on { B_{2R}(x_0)}, denote { \tilde{u}(x)=u(xR)}, {\tilde{u}} is a function on { B_2(x_0)}, { (Lu,v)=(g+D_if^i,v)} means

\displaystyle \int_{B_R}a^{ij}(x)D_juD_iv+b^i(x)uD_iv+c^i(x)D_iuv+d(x)uv dx=\int_{B_R}g(x)v+f^iD_iv dx

Changing variables { x=yR}, then

\displaystyle \int_{B_R}\frac{a^{ij}(yR)}{R^2}D_j\tilde{u}D_i\tilde{v}+\frac{b^i(yR)}{R}\tilde{u}D_i\tilde{v}+\frac{c^i(yR)}{R}D_i\tilde{u}\tilde{v}+d(yR)\tilde{u}\tilde{v} dy

\displaystyle =\int_{B_R}g(yR)\tilde{v}+\frac{f^i(yR)}{R}D_i\tilde{v} dy\quad (1)

If we define

\displaystyle \tilde{L}\tilde{u}=D_i(\tilde{a}^{ij}(y)D_j\tilde{u}+\tilde{b}^i(y)\tilde{u})+\tilde{c}^i(y)D_iu+\tilde{d}(y)u

and

\displaystyle \begin{cases}\tilde{a}^{ij}(y)=\frac{a^{ij}(yR)}{R^2},\tilde{\lambda}=\frac{\lambda}{R^2},\quad \Lambda=\frac{\Lambda}{R^2}\quad\\ \tilde{b}^i(y)=\frac{b^i(yR)}{R},\quad \tilde{c}^i(y)=\frac{c^i(yR)}{R},\quad \tilde{d}(y)=d(yR)\\ \tilde{g}(y)=g(yR),\quad \tilde{f^i}(y)=\frac{f^i(yR)}{R},\quad \tilde{\theta}=\theta R\end{cases}

Then (1) is equivalent to { \tilde{L}\tilde{u}=\tilde{g}+D_i\tilde{f^i}}.

Since { \tilde{u}} is a function on { B_2(x_0)}, then by assumption

\displaystyle \sup\limits_{B_1(y)}\tilde{u}\leq C(||\tilde{u}^+||_{L^p(B_{2}(y))}+\tilde{k}(1))

It is easy to verify that this means

\displaystyle \sup\limits_{B_R(y)}u\leq C(R^{-n/p}||u^+||_{L^p(B_{2R}(y))}+k(R))

\text{Q.E.D}\hfill \square

{ \mathbf{Remark:}} Gilbarg, Trudinder’s book, p195.

 

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