## Strong barrier function and generalized linear elliptic equation

Suppose $\Omega$ is a domain in $\mathbb{R}^n$. $x_0\in \partial \Omega$ satisfies the exterior sphere condition. That is there exists a ball $B=B_R(y)$ such that $\bar{B}\cap \overline{\Omega}=x_0$. Then the function
$\displaystyle w(x)=\begin{cases}R^{2-n}-|x-y|^{2-n} \quad n\geq 3\\log\frac{|x-y|}{R} \qquad n=2\end{cases}$
is a barrier function for $\Delta$ at $x_0$.
For a strong barrier function $w$ at $x_0$, we mean $\Delta u\leq -1$ or $Lu\leq -1$, at such $x_0$ Define

$w=K(R^{-\alpha}-|x-y|^{-\alpha})$

Then $\displaystyle Lw=K\alpha|x-y|^{-\alpha-4}\left[-(\alpha+2)a^{ij}(x_i-y_i)(x_j-y_j)+|x-y|^2( a^{ii}+b^i(x_i-y_i))\right]+cw$
Assume $c\leq 0$ and let $|x-y|=r$
$Lw\leq K\alpha r^{-\alpha-4}\left[-(\alpha+2)a^{ij}(x_i-y_i)(x_j-y_j)+r^2(a^{ii}+b^i(x_i-y_i))\right]\leq -1$
when $\alpha=\alpha(\lambda, \Lambda,diam(\Omega),R)$ and $K$ large enough.
$\mathbf{Problem: }$ Let $u\in C^2(\Omega)\cap C^0(\overline{\Omega})$ be a solution of $Lu=f$ in a bounded $C^1$ domain $\Omega$ $n\geq 3$. Suppose $x_0\in\partial \Omega$ satisfy the exterior ball condition with $B_R(y)$. Moreover, assume

$a^{ij}\xi_i\xi_j\geq \lambda |\xi|^2$
$|a^{ij}|, |b^i|, |c|\leq \Lambda$

Suppose $u|_{\partial \Omega}=\phi$, $\phi\in C^2(\overline{\Omega})$. Show that $u$ satisfies a Lipschitz condition at $x_0$

$|u(x)-u(x_0)|\leq K(x-x_0)$, $x\in \Omega$.

where $K=K(\lambda,\Lambda,R,diam(\Omega), \sup|f|)$. If the sign $c$ is unrestricted, show that the same result holds provided $K$ also depends on $\sup|u|$.

$\mathbf{Proof:}$ When $c\leq 0$ suppose $w_0$ is the strong barrier function of $L$ at $x_0$, then there exists $M=M(\sup\limits_{\Omega}|f|, |\phi|_{2;\Omega})$ such that $\displaystyle w=Mw_0$ satisfies
$w+u(x_0)-u\geq 0$ on $\partial \Omega$ and $L(w+u(x_0)- u)\leq 0$

$-w+u(x_0)-u\leq 0$ on $\partial \Omega$ and $L(-w+u(x_0)-u)\geq 0$
By the maximum principle, we know that

$-w(x)\leq u(x)-u(x_0)\leq w(x)$ $|u(x)-u(x_0)|\leq w(x)=w(x)-w(x_0)\leq K|x-x_0|$.

When the sign of $c$ is undefined, let $\tilde{L}u=Lu-c^+u=f-c^+u$, apply the above procedure to $\tilde{L}$, only need to adjust that $M=M(\sup\limits_{\Omega}|f|, |\phi|_{2;\Omega},\Lambda,\sup|u|)$.

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

$\mathbf{Remark:}$Gilbarg Trudinger’s book. Chapter, exercise 3.6.