# Density of traces of solutions to an elliptic equation

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 $$H^{\frac{1}{2}}(\partial D_1)$$?

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 $$H^s$$.
• But why is ${\rm supp}(f)\subset D_2\backslash\overline{D_1}$? Oct 5 '21 at 2:02
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 $$g=0$$.