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}.