## Boundary condition and uniqueness

$\mathbf{Problem:}$ Prove that if $\Delta u=0$ in $\Omega\subset \mathbb{R}^n$ and $u=\partial u/\partial\nu=0$ on an open smooth portion of $\partial \Omega$, then $u$ 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.

$\mathbf{Proof:}$ Extend $u$ to be zero near the smooth part outside the domain(suppose it is $\Omega\cup B_r(x_0)$). Then we get a $C^1$ function which vanishes on an open set. If we can show $u$ is a harmonic function, then by the analyticity, $u\equiv 0$.

To do that we only need to show that $u$ satisfies $\int_{B_{\varepsilon}(z)}\frac{\partial u}{\partial \nu}ds=0$ for any $\varepsilon$ small enough(more or less like the local mean value property) where $\nu$ is the outer unit normal of $B_\varepsilon(z)$. For any point $z\in \Omega$ and $z\in B_r(x_0)\backslash \Omega$, mean value property holds locally. For any point $a\in \partial \Omega\cap B_r(x_0)$, we need to show for any $\varepsilon$ small enough

$\int_{B_\varepsilon(z)}\frac{\partial u}{\partial \nu}ds=0$

However, $B_{\varepsilon}(z)\cap \Omega$ is a domain with piecewise smooth boundary when $\varepsilon$ is small enough, therefore

$\int_{B_\varepsilon(z)}\frac{\partial u}{\partial \nu}ds=\int_{B_\varepsilon(z)\cap \Omega}\frac{\partial u}{\partial \nu}ds=\int_{B_\varepsilon(z)\cap \Omega}\Delta u(y)dy=0.$

where we have used $u=\partial u/\partial\nu=0$ on $B_\varepsilon(z)\cap \partial\Omega$.

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

$\mathbf{Proof:}$ Suppose $\Gamma\subset \partial\Omega$ is this portion. Choose $x_0\in \Gamma$ and a ball $B$ centered at $x_0$ small enough.

Define $\displaystyle w=u(x)$ when $x\in B\cap \Omega$ and $w=0$ when $x\in \overline{B}\backslash\Omega$. Then $w$ is harmonic on $B\cap \Omega$ and $B\backslash\overline{\Omega}$. If we can prove $w$ has continuous second partial derivatives, then $w$ will be harmonic on whole $B$, thus $w$ must be identically zero since harmonic functions are analytic ones.

For any point $x\in B\cap\Gamma$. Since Laplace equation is invariant with respect to rotation and translation, we can assume $x$ is the origin and the outer normal vector is negative $x_n$ axis. And $\displaystyle \phi(x_1,x_2,\cdots,x_{n-1})$ is the boundary $B\cap \Gamma$

Since $\Gamma$ is smooth and $\displaystyle u|_{\Gamma}$ is smooth along $\Gamma$, we can assume $u\in C^2(\overline{B\cap \Omega})$. Then we have

$u_{x_i}(0)=\left(u|_{\Gamma}\right)_{x_i}(0)=0$, for $i=1,2,\cdots, n-1$

$u_{x_ix_i}(0)=\left(u|_{\Gamma}\right)_{x_ix_i}(0)=0$, for $i=1,2,\cdots, n-1$.

$\displaystyle u_{x_n}(0)=-\frac{\partial u}{\partial \nu}(0)=0$ and $\displaystyle u_{x_nx_n}(0)=-\sum\limits_{i=1}^{n-1}u_{x_ix_i}=0$

Noticing $u(x_1,x_2,\cdots,x_{n-1}, \phi(x_1,x_2,\cdots,x_{n-1}))=0\quad (1)$

Differentiate (1) twice

$\displaystyle u_{x_ix_i}+u_{x_ix_n}\phi_{x_i}+u_{x_nx_n}\phi^2_{x_i}+u_{x_n}\phi_{x_ix_i}=0$ for $i=1,2,\cdots,n-1$.

Using the previous equalities, at origin we have $u_{x_ix_n}=0$ for $i=1,2,\cdots,n-1$. Thus we have proved all the second partial derivatives of $u$ are zeros.

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

$\mathbf{Remark:}$ 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 $u_{x_i}(0)=\left(u|_{\Gamma}\right)_{x_i}(0)$. It is such a trivial fact that I never notice. I would like to thank the insightful help of Jingang Xiong.

$\mathbf{Remark:}$ See a new proof

Suppose ${\Gamma\subset \partial\Omega}$ is this portion. Choose ${x_0\in \Gamma}$ and a ball ${B}$ centered at ${x_0}$ small enough. Define ${\displaystyle w=u(x)}$ when ${x\in B\cap \Omega }$ and ${w=0}$ when ${x\in \overline{B}\backslash\Omega}$. Then ${w}$ is harmonic on ${B\cap \Omega }$ and ${B\backslash\overline{\Omega}}$. If we can prove ${w}$ has continuous second partial derivatives, then ${w}$ will be harmonic on whole ${B}$, thus ${w}$ must be identically zero since harmonic functions are analytic ones.

Since Laplace equation is invariant with respect to rotation and translation, we can assume ${\Gamma\cap B}$ can be represented locally ${x_n=\phi(x_1,x_2,\cdots,x_{n-1})}$, ${x'=(x_1,\cdots,x_{n-1})\in V\subset\mathbb{R}^{n-1}}$. Since ${\Gamma}$ is smooth, and ${u=0}$ is ${C^\infty}$ on ${\Gamma}$, we can assume ${u\in C^2(\overline{B\cap \Omega})}$. We will prove ${u_{x_i}|_{\Gamma\cap B}=0}$ and ${u_{x_ix_j}|_{\Gamma\cap B}=0}$.

Since ${u(x',\phi(x'))=0}$, ${x'\in V}$, differentiating with respect to ${x_i}$, we get

$\displaystyle u_{x_i}+u_{x_n}f_{x_i}=0\text{ for } i=1,2,\cdots, n-1\quad (1)$

Also ${\displaystyle \frac{\partial u}{\partial \nu}=0}$ on ${\Gamma\cap B}$ implies

$\displaystyle \sum\limits_{i=1}^{n-1} u_{x_i}f_{x_i}-u_{x_n}=0\quad (2)$

Multiply ${(1)}$ by ${f_{x_i}}$, sum ${i}$ from ${1}$ to ${n-1}$ and then subtract ${(2)}$ we get

$\displaystyle u_{x_n}=0\text{ hence } u_{x_i}=0, i=1,\cdots,n-1.$

Differentiating ${(1)}$ furthermore with respect to ${x_j}$, use ${u_{x_n}=0}$ on ${B\cap \Gamma}$ we get

$\displaystyle u_{x_ix_j}+u_{x_ix_n}f_{x_j}+u_{x_ix_j}f_{x_i}+u_{x_nx_n}f_{x_j}f_{x_i}=0, \forall\, i,j=1,2\cdots, n-1\quad (3)$

Differentiating ${(2)}$ furmore with respect to ${x_j}$, use ${u_{x_i}=0}$ on ${B\cap \Gamma}$ we get

$\displaystyle \sum_{i=1}^{n-1}u_{x_ix_j}f_{x_i}+u_{x_ix_n}f_{x_i}f_{x_j}-u_{x_nx_j}-u_{x_nx_n}f_{x_j}=0, \forall\, j=1,2,\cdots,n-1\quad (4)$

Multiplying ${(3)}$ by ${f_{x_i}}$ and sum ${i}$, then subtract ${(4)}$ we get

$\displaystyle u_{x_nx_j}+u_{x_nx_n}f_{x_j}=0\quad (5)$

From this ${(3)}$ is equivalent to

$\displaystyle u_{x_ix_j}+u_{x_ix_n}f_{x_j}=0\quad (6)$

Let ${j=i}$ in ${(6)}$ and sum ${i}$, we get

$\displaystyle \sum\limits_{i}^{n-1}u_{x_ix_i}=-\sum\limits_{i=1}^{n-1}u_{x_ix_n}f_{x_i}=\sum\limits_{i=1}^{n-1}u_{x_nx_n}f^2_{x_i}$

the last equality is obtained from ${(5)}$. Note that ${\Delta u=0}$, the above equality means that

$\displaystyle \sum\limits_{i=1}^{n-1}u_{x_nx_n}f^2_{x_i}=-u_{x_nx_n}$

this implies that ${u_{x_nx_n}=0}$. From ${(5)}$, ${u_{x_ix_n}=0}$, ${i=1,2,\cdots,n-1}$. Then ${(6)}$ implies ${u_{x_ix_j}=0}$, ${\forall\, i,j=1,2\cdots, n-1}$.

• xia kn  On December 20, 2013 at 9:55 am

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

• Sun's World  On December 20, 2013 at 4:47 pm

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}).

• Sun's World  On December 20, 2013 at 4:50 pm

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.

• foop  On November 24, 2014 at 7:24 am

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.

• Sun's World  On November 24, 2014 at 2:18 pm

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?

• foop  On November 24, 2014 at 4:47 pm

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

• Sun's World  On November 25, 2014 at 1:38 am

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.

• foop  On November 25, 2014 at 5:00 pm

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.

• Sun's World  On November 25, 2014 at 11:36 pm

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?

• foop  On November 26, 2014 at 12:57 am

$\Delta w=0$, 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.

• Sun's World  On November 27, 2014 at 2:52 pm

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

• Dingqian Gao  On June 24, 2018 at 4:35 pm

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