If $u_1$ and $u_2$ are solutions of (weak-form) Laplace equation on a connected domain $\Omega$, with Dirichlet boundary values $u_{\partial\Omega, 1}$ and $u_{\partial\Omega, 2}$, respectively. If $$u_{\partial\Omega,1} \neq u_{\partial\Omega, 2},$$ we know $$\|u_1-u_2\|_{L^2(\Omega)} \neq 0.$$

My question is, for any open set $\tilde{\Omega} \subset \Omega$, is $$\|u_1-u_2\|_{L^2(\tilde{\Omega})} \neq 0 $$ always true?

  • $\begingroup$ Should the last equation contain an equality instead? $\endgroup$ – knl Sep 14 '18 at 6:19
  • $\begingroup$ I changed the words, hopefully there's no ambiguity. $\endgroup$ – J.F. Yan Sep 14 '18 at 6:36
  • $\begingroup$ You will want to specify that $\Omega$ is connected because otherwise, if you change the boundary condition on one part of the domain, it doesn't affect the solution on the other part of the domain. $\endgroup$ – Wolfgang Bangerth Sep 14 '18 at 23:17
  • $\begingroup$ Also, do you assume that $\tilde\Omega$ is an open subset of $\Omega$? Or could it be just a set of individual points, or a lower-dimensional submanifold? $\endgroup$ – Wolfgang Bangerth Sep 14 '18 at 23:17
  • $\begingroup$ @Wolfgang Bangerth You are right, I need $\Omega$ to be connected, and $\tilde{\Omega}$ to be an open subset of the same dimension. Thanks. $\endgroup$ – J.F. Yan Sep 15 '18 at 0:04

Let us note $w = u_1 - u_2$. We know that $-\Delta w = 0$ in $\Omega$.

Assume that there is $\tilde{\Omega} \subset \Omega$ such that $w = 0$ on $\tilde{\Omega}$ (without any loss of generality, we can assume that $\overline{ \tilde{\Omega} } \subset \Omega$). As $w$ satisfies the Laplace equation, there is no jump for $w$ across the boundary of $\tilde{\Omega}$: $[w]_{|\partial \tilde{\Omega}} = 0$ and similarly for the normal derivative.

So, in $\Omega \backslash \tilde{\Omega}$, $w$ satisfies the Laplace equation and, on $\partial \tilde{\Omega}$, $w$ has a null trace and a null normal derivative.

By a continuation argument (see R. Leis, "Initial boundary value problems in mathematical physics", 1986), $w$ is 0 in $\Omega \backslash \tilde{\Omega}$.

  • $\begingroup$ I don't have access to that book. By the continuation argument, do you mean $u \in C^{\infty}$ at least in $\Omega\setminus\tilde{\Omega}$? $\endgroup$ – J.F. Yan Sep 18 '18 at 13:40
  • $\begingroup$ I mean that w will be zero in a neighborhood of $\partial \tilde{\Omega}$ and, as $-\Delta w = 0$, this neighborhood can grow to cover the whole region $\Omega \backslash \tilde{\Omega}$. $\endgroup$ – user7440 Sep 18 '18 at 18:24
  • $\begingroup$ The argument is sometimes called Holmgren's uniqueness theorem (that is discussed in Hormander, "Linear partial differential equations"). It is also used here: math.stackexchange.com/questions/2531504/… $\endgroup$ – user7440 Sep 18 '18 at 19:11
  • $\begingroup$ I checked Holmogren's uniqueness theorem, it says the elliptic operator is hypoelliptic, which means $u\in C^{\infty}$. It seems like all those theorems apply to strong form. What if we are talking about weak solution: $u \in H^1_{u_{\partial\Omega}}$ $$\int \nabla v \cdot \nabla u d\Omega = 0, \forall v \in H^1_0$$? $\endgroup$ – J.F. Yan Sep 18 '18 at 22:32
  • $\begingroup$ The function $w$ will be very smooth away from the boundary $\partial \Omega$, as it satisfies $\Delta w = 0$. How regular is $\Omega$? $\endgroup$ – user7440 Sep 19 '18 at 0:05

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