2
$\begingroup$

Let $D_1$ be a domain with smooth boundary and assume that $D_1$ is a proper subset of $D_2$ which is itself a bounded domain in $\mathbb R^n$ with a smooth boundary. Assume also that $D_2\setminus D_1$ is connected. We write $L^2(D_2\setminus D_1)$ for the set of functions in the space $$\{f \in L^2(D_2)\,:\,\textrm{supp}(f)\subset D_2\setminus \overline{D_1}\}$$ let us define the mapping $$S: L^2(D_2\setminus \overline{D_1})\mapsto H^{\frac{3}{2}}(\partial D_1),$$ through $$ Sf:= u|_{\partial D_1},$$ where $u \in H^2(D_2)$ is the unique solution to the equation $$ \Delta u =f \quad \text{on $D_2$},$$ subject to $u|_{\partial D_2}=0$. Is it true that the image of $S$ is dense in $H^{\frac{1}{2}}(\partial D_1)$?

$\endgroup$

2 Answers 2

1
$\begingroup$

The answer is yes: take any smooth function $g_0$ on $\partial D_1$ and solve the Dirichlet problem $$ \begin{cases} \Delta g = 0 & \text{ on } D_1\\ g = g_0 & \text{ on } \partial D_1. \end{cases} $$ Now extend $g$ to a smooth function on $\mathbb{R}^n$. Multiply by a smooth cutoff function $\eta$ which is $1$ on $D_1$ and compactly supported on $D_2$.

Then $f = \Delta (\eta g)$ is smooth and supported on $D_2 \setminus \bar{D}_1$, so in particular lies in the given $L^2$ space. This shows that $g_0$ is in the image of your operator $S$, and smooth functions are dense in $H^s$.

$\endgroup$
2
  • $\begingroup$ But why is ${\rm supp}(f)\subset D_2\backslash\overline{D_1}$? $\endgroup$ Oct 5, 2021 at 2:02
  • $\begingroup$ You are correct and I misread part of the question. However, I believe essentially the same argument still applies (solve for the harmonic function first, then extend), and I have modified the answer. $\endgroup$
    – user378654
    Oct 5, 2021 at 3:03
0
$\begingroup$

The answer is yes. Suppose $g$ is orthogonal to the image of $S$, and let $v$ be the solution of the Dirichlet problem $\Delta v=g\delta(\partial D_1)$ on $D_2$, where $\delta(\partial D_1)$ is a delta function localized on $\partial D_1$. We find $$\int_{\partial D_1} gu\,dS=\int_{D_2}v\Delta u\,dx.$$ Now suppose this is true for every $u$ for which $\Delta u$ has compact support in any subregion of $D_2\backslash \overline{D_1}$. Then $v$ must vanish on that subregion. Since $\Delta v=0$ on $D_2\backslash \overline{D_1}$, it follows that $v$ also vanishes there. But $v$ is continuous across $\partial D_1$ (only the normal derivative has a jump), and $\Delta v=0$ in $D_1$, so $v$ must be zero everywhere by uniqueness of the Dirichlet problem for $D_1$. Hence $g=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.