Prove that if in and on an open smooth portion of , then is identically zero.

**Remark:** I saw this nice proof from my friend and also someone anonymous on the internet notified me he/she has a similar idea.

Extend to be zero near the smooth part outside the domain(suppose it is ). Then we get a function which vanishes on an open set. If we can show is a harmonic function, then by the analyticity, .

To do that we only need to show that satisfies for any small enough(more or less like the local mean value property) where is the outer unit normal of . For any point and , mean value property holds locally. For any point , we need to show for any small enough

However, is a domain with piecewise smooth boundary when is small enough, therefore

where we have used on .

Suppose is this portion. Choose and a ball centered at small enough.

Define when and when . Then is harmonic on and . If we can prove has continuous second partial derivatives, then will be harmonic on whole , thus must be identically zero since harmonic functions are analytic ones.

For any point . Since Laplace equation is invariant with respect to rotation and translation, we can assume is the origin and the outer normal vector is negative axis. And is the boundary

Since is smooth and is smooth along , we can assume . Then we have

, for

, for .

and

Noticing

Differentiate (1) twice

for .

Using the previous equalities, at origin we have for . Thus we have proved all the second partial derivatives of are zeros.

Gilbarg Trudinger Book exercise 2.2. Up to now I am not sure this proof is right. Its proof shouldn’t be very difficult. I spent two weeks to realize that . It is such a trivial fact that I never notice. I would like to thank the insightful help of Jingang Xiong.

See a new proof

Suppose is this portion. Choose and a ball centered at small enough. Define when and when . Then is harmonic on and . If we can prove has continuous second partial derivatives, then will be harmonic on whole , thus must be identically zero since harmonic functions are analytic ones.

Since Laplace equation is invariant with respect to rotation and translation, we can assume can be represented locally , . Since is smooth, and is on , we can assume . We will prove and .

Since , , differentiating with respect to , we get

Also on implies

Multiply by , sum from to and then subtract we get

Differentiating furthermore with respect to , use on we get

Differentiating furmore with respect to , use on we get

Multiplying by and sum , then subtract we get

From this is equivalent to

Let in and sum , we get

the last equality is obtained from . Note that , the above equality means that

this implies that . From , , . Then implies , .

## Comments

How do you know that \Delta u = 0 on \Gamma \cap B?

It follows from I assume $u\in C^2(\overline{B\cap \Omega})$. I implicitly used the fact that the Dirichlet problem \Deltau u=0 and u=\phi\in C^\infty(\bar{\Omega}) on \partial\Omega has a solution u\in C^2(\bar{\Omega}).

This fact holds in general. But I don’t know how to prove it in easy way. Unless the case when \Omega is exactly a ball. Then we have the possion expression, which can be proved no very difficult.

I think both of your proofs are incorrect.

If you assume $u\in C^2(\overline{B\cap \Omega})$, then it’s all over because $\Delta u$ will be zero on the boundary of $\overline{B\cap \Omega}$ by continuity.

why $\Delta u =0 $ on the boundary? Suppose $u=x^2$ on upper half plane $(x,y),y>0$. Is $\Delta u=0$ on $x$ axis?

Because $\Delta u =0 $ in $\Omega$ and $u$ has continuous second derivative in $\overline{B\cap \Omega}$ by your assumption.

Sorry. I am a little bit sloppy on the last reply. Yes, the $\Delta u=0$ on the smooth boundary. But what do you mean the proof is over if I assume the continuous second derivative? I think you still need to verify all the second derivative vanish on the boundary.

That $\Delta u=0$ on the smooth boundary implies that $\Delta w=0$ in $B$, hence is constant, and is actually zero in $B$. Therefore, $u$ is constant.

I have to say I don’t see why $\Delta w=0$ in $B$ implies it is zero in $B$ even if on the smooth portion $w=0$. Can you say a few words about it? Did you use the fact that the normal derivative equal zero?

$\Delta w=0$, hence $w$ is real analytic. Since $w=0$ in an open ball in $B$, we know that $w=0$ in $B$ since $B$ is connected and $w$ is real analytic.

I don’t know why $w=0$ in an open ball in $B$. Can you be more accurate?

I find an easier way to prove it,and how can i send you the proof？

can you send it to sunlimingbit@gmail.com

## Trackbacks

[…] found a proof here, but I’m not sure if it’s the correct proof. Does anyone have any idea? Any hint would […]