## 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.