## Tag Archives: uniqueness continuation

### Unique continuation property on the boundary

I am writing a theorem proved by Jin Zhiren in his thesis.

Suppose $\Omega$ is a smooth domain in $\mathbb{R}^n$, $x_0\in \partial \Omega$ and $u$ is a harmonic function in $\Omega$. If there exists $A, b>0$ such that

$\displaystyle |u(x)|\leq Ae^{-\frac{b}{|x-x_0|}}\quad x\in \Omega$

for $|x-x_0|$ small, then $u=0$. If $n=2$, the same conclusion holds for the solutions of a general second order linear elliptic equation.

A borderline example for this theorem is $u$ be the real part of $e^{-1/z^\alpha}$, $\alpha\in (0,1)$. $u$ is harmonic in the right half plane and $u\leq Ae^{-1/|x|^\alpha}$ and consequently $D^\beta u(0)=0$.

### 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 ${u_{x_ix_j}=0}$, ${\forall\, i,j=1,2\cdots, n-1}$.