## Hopf lemma

Hopf Lemma: Suppose ${\Omega\subset \mathbb{R}^n}$, ${n\geq 2}$ and ${u\in C^2(\Omega)\cap C^1(\overline{\Omega})}$ satisfies that

$\displaystyle Lu=a^{ij}\partial_{ij}u+b^iu\geq 0$

${L}$ is uniformly elliptic and ${|b^i/\lambda|\leq C}$, ${i=1,2,\cdots,n}$. Suppose ${\Omega}$ has an interior open ball ${B}$ such that ${B\subset \Omega}$ and there exists one point ${x_0\subset \partial \Omega}$ satisfies that ${x_0\subset \partial B}$. ${u}$ satisfies that ${u(x) if ${x\in B}$. If ${\frac{\partial u}{\partial n}}$ exists, then ${\frac{\partial u}{\partial n}>0}$

Proof: Under some translation, we assume that ${B=B_R(0)}$ has radius ${R}$. Define

$\displaystyle w(x)=e^{-\alpha |x|^2}-e^{-\alpha R^2}\text{ on } B_R\backslash B_{\frac{R}{2}}(0)$

Then

$\displaystyle Lw=\left[a^{ij}(-2\alpha\delta_{ij}+4\alpha^2 x_i x_j)-2\alpha x_ib^i\right]e^{-\alpha |x|^2}$

$\displaystyle =\left[4\alpha^2 a^{ij} x_i x_j+\alpha(-2x_ib^i-2a^{ij}\delta_{ij})\right]e^{-\alpha |x|^2}$

Since ${a^{ij}}$ is uniformly ellptic and ${|\frac{b^i}{\lambda}|, we can choose ${\alpha}$ big enough that

$\displaystyle Lw\geq 0$

Next consider ${u(x)+\epsilon w(x)}$ on ${B_R\backslash B_{\frac{R}{2}}(0)}$.

$\displaystyle L(u+\epsilon w)\geq 0$

Firstly ${u(x)+\epsilon w(x)=u(x)}$ for ${x\in\partial B_R}$
Secondly since ${\partial B_{\frac{R}{2}}}$ is compact, choose ${\epsilon}$ small enough that

$\displaystyle u(x)+\epsilon w(x)

So by the maximum principle

$\displaystyle u(x)+\epsilon w(x)

Then $\displaystyle {\frac{\partial (u+\epsilon w)}{\partial \nu}\geq 0}$, hence

$\displaystyle \frac{\partial u}{\partial \nu}\geq -\epsilon\frac{\partial w}{\partial \nu}>0$

Remark: This theorem is also true when either of the two cases
(1)${Lu=a^{ij}\partial_{ij}u+b^iu +cu}$ with ${c\leq 0}$ and ${|\frac{c}{\lambda}|\leq C}$ and ${u(x_0)\geq 0}$
(2)${u(x_0)=0}$ irrespective the sign of ${c}$.

Proof: (1) Since ${c\leq 0}$ and ${|\frac{c}{\lambda}|\leq C}$ we can also prove that

$\displaystyle Lw\geq 0$

So ${L(u+\epsilon w-u(x_0))\geq 0}$, by the maximum principle

$\displaystyle u(x)+\epsilon w(x)-u(x_0)\leq \sup\limits_{\partial B_R\cup \partial B_{R/2}}(u(x)+\epsilon w(x)-u(x_0))^+\leq 0$

the following step is the same.

(2)Since ${u(x) in ${B}$, we have

$\displaystyle (L-c^+)u\geq 0$

Replacing ${L}$ by ${L-c^+}$, the proof is unchanged.