We consider the null space (corresponding to the possible eigenvalue zero) of the linear indefinite elliptic PDE $\Delta u+k^2(x)u=0$ in $\Omega$ with $u=\partial_{\nu}u=0$ on the boundary. If the solution $u$ is analytic, then from the uniqueness of continuation $u\equiv 0$ in $\Omega.$ In other words, Neumann and Dirichlet eigenfunctions can not coincide. But what happens if the domain $\Omega$ and the coefficients $k(x)$ are not analytic so that $u$ is not likely analytic? Do we still have $u\equiv0$?

Can we approximate the domain and the coefficients by analytic ones? Then, we get also the solution approximated by analytic sequence of zero functions?

My motivation is to prove the uniqueness of the solution with the boundary condition $\partial_{\nu}u+\mathrm{i}\,k u=0.$ This is typical for the Helmholtz equation. The energy identity gives $$ \int_{\Omega}|\nabla u|^2-k^2(x)|u|^2+\int_{\partial\Omega}\mathrm{i}k|u|^2=0 $$ so $u=0$ on $\partial\Omega$ because $k(x)\in\mathbb{R}$ (we assume). And further $$ \int_{\Omega}\nabla u\cdot\nabla \bar{v}-k^2(x)u\cdot\bar{v}=\int_{\partial\Omega}\partial_{\nu}u\cdot\bar{v}=0, $$ for arbitrary test function $v$ so that $\partial_{\nu}u=0$ on $\Gamma.$ My goal is to prove $u\equiv0$ in $\Omega.$


1 Answer 1


Yes. Even if $k$ is not smooth, we still have $u\equiv 0$ in $\Omega$. The reason is, we can extend the function $u$ to be zero outside $\Omega$. Since $u=\partial_\nu u=0$ on $\partial\Omega$, after this extension $u$ becomes a solution of the PDE in the whole space ${\mathbb R}^d$. The unique continuation theorem says that: if the solution $u$ vanishes in a neighborhood, then it vanishes everywhere.

See the paper of Thomas H. Wolff:

Recent Work on Sharp Estimates in Second-Order Elliptic Unique Continuation Problems

In Page 622: "The basic result is (1.1) has the SUCP if (aij) are locally Lipschitz and A and B are locally bounded. "

  • $\begingroup$ Thanks, Buyang! A good answer. The website told me I can not add the bounty to you until 16 hours later. So I will do it later. $\endgroup$
    – Hui Zhang
    Jul 21, 2014 at 16:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .