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)<u(x_0)} 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}|<C}, 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)<u(x_0)\text{ on }\partial B_{\frac{R}{2}}

So by the maximum principle

\displaystyle u(x)+\epsilon w(x)<u(x_0)\text{ in }B_R\backslash B_{\frac{R}{2}}(0)

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)<u(x_0)=0} in {B}, we have

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

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

 

Advertisements
Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: