# Hopf引理

Theorem 1. 假设$\Omega\subset\R^n$是一个有界开区域, 且$\pt\Omega$光滑. 又设
$$L=-a^{ij}(x)D_{ij}+b^i(x)D_i+c(x),$$

1. $u$在$x_0$处可微,
2. $u(x)>u(x_0),\quad \forall x\in \Omega$,

1. $c\equiv0$, in $\Omega$,
2. $c\geq0$, $\forall x\in\Omega$且$u(x_0)\leq0$,
3. $u(x_0)=0$.

$$\frac{\pt u(x_0)}{\pt\nu}<0.$$

Proof . 分析, 我们需要计算$\frac{\pt u}{\pt\nu}$在$x_0$处的值, 于是需要考察$x_0$的一个邻域$A\subset\Omega$时$u$的值. 注意到在$B$中有$u(x)>u(x_0)$, 因此我们立即可得到(由于$\pt\Omega$光滑, 故此时$B$和$\pt\Omega$相切)
$$\frac{\pt u(x_0)}{\pt\nu}\leq0.$$

$$0\geq\frac{\pt(u+v)(x_0)}{\pt\nu}=\frac{\pt u(x_0)}{\pt\nu}+\frac{\pt v(x_0)}{\pt\nu},$$

$$v=\eps e^{\lambda r^2}-\eps e^{\lambda |x-x^0|^2}-u(x_0),$$

$$\frac{\pt v(x_0)}{\pt\nu}=-2\lambda r\eps e^{\lambda|x_0-x^0|^2}<0.$$

$$v=\eps e^{-\lambda r^2}-\eps e^{-\lambda |x-x^0|^2}-u(x_0),$$

$$\frac{\pt v(x_0)}{\pt\nu}=2\lambda r\eps e^{-\lambda r^2}>0.$$

\begin{align*}
L(v)&=-\eps\left(-a^{ij}D_{ij}+b^iD_i\right)e^{-\lambda |x-x^0|^2}+cv\\
&=\eps e^{-\lambda |x-x^0|^2}\Bigg\{
4 \lambda^2a^{ij}(x_i-x^0_i)(x_j-x_j^0)\\
\Bigg\}+cv\\
&\geq\eps e^{-\lambda |x-x^0|^2}\Bigg\{
4 \lambda^2\theta\|x-x^0\|^2\\
\Bigg\}+cv\\
&\geq\eps e^{-\lambda|x-x^0|^2}\Bigg\{
4\theta\lambda^2\|x-x^0\|^2-2\lambda\left(\sum_ia^{ii}-r\|b\|\right)-c
\Bigg\}+cc_0\\
&\geq\eps e^{-\lambda|x-x^0|^2}\Bigg\{
4\theta\lambda^2\|x-x^0\|^2-2\lambda\left(\sum_ia^{ii}-r\|b\|\right)-c
\Bigg\}-cu(x_0)\\
\end{align*}

$$L(v)\geq-cu(x_0)\geq0,\quad\forall x\in A.$$

$$L(u+v)\geq L(v)\geq0,\quad\forall x\in A.$$

$$\min_{\bar A}(u+v)=\min_{\pt A}(-(u+v)^-).$$

$$u+v\geq0.$$

\begin{align} u(x)&>u(x_0),\\ v(x)&=\eps e^{\lambda r^2}-\eps e^{\lambda |x-x^0|^2}-u(x_0)\notag\\ &\geq \eps\left(e^{-\lambda r^2}-e^{-\lambda r^2/4}\right)-u(x_0), \end{align}

$$u+v>0.$$

$$\tilde a^{ij}=a^{ij},\quad \tilde b_i=b_i,\quad \tilde c=c+c^-=c^+\geq0,$$

$$\tilde L u=Lu+c^-u\geq Lu\geq0,$$

Remark 1. thm:1的条件:$u(x)>u(x_0)$在$\Omega$中成立, 可以稍微减弱点(正如有的书上所说), 我们可以假设$u(x)>u(x_0)$在$B$中成立, 此时证明基本不变, 唯一要注意的是在$\pt A\cap B\subset\pt A\cap \bar B$上, 我们只能得到$u(x)\geq u(x_0)$(不再是严格大于), 这将导致$\eps$的存在性有问题. 但是我们只需将考察区域换成环状区域$A=B_{r}(x^0)\setminus \bar B_{r/2}(x^0)$即可.