Let $\Omega\subset\mathbb{R}^N$ be a bounded, smooth domain. Assume that $\mu \in \mathcal{M}(\Omega)$ has compact support in $\Omega.$ Let $u\in W_0^{1,1}(\Omega)$ be a solution of $$ \left\{ \begin{array}{rl} -\Delta u=\mu &\mbox{ if $x\in \Omega$,} \\ u=0 &\mbox{ if $x\in \partial \Omega$,} \end{array} \right.\tag{1} $$

where by solution, it is mean that $$-\int _\Omega u\Delta \psi=\int_\Omega\nabla u\nabla \psi=\int_\Omega \psi d \mu,\ \forall \psi\in C_0^\infty(\overline{\Omega}).$$

As $\mu$ has compact support, $u$ is harmonic in a neighbourhood of $\partial\Omega$ and so the normal derivative of $u$ is well defined at the boundary. My question is, how do we prove that $$-\int _\Omega \psi d\mu=-\int _\Omega \nabla u \nabla \psi+\int _{\partial \Omega} \frac{\partial u}{\partial \eta}\psi ,\ \forall \ \psi\in C^1(\overline{\Omega}).$$

I have tried two approachs: The first consist in proving that $$\left|\int_\Omega \nabla u \nabla \psi\right |\le C\|\psi\|_\infty,\ \forall C^1(\overline{\Omega})\tag{2}.$$

If $(2)$ is true then, the linear functional $T:C^1(\overline{\Omega})\to\mathbb{R}$ defined by $$T\psi=\int_\Omega \nabla u\nabla \psi,$$

can be extended to a bounded linear functional defined in $C^0(\overline{\Omega})$. The result then follows by Riesz theorem. However, I could not prove that $(2)$ is true.

In the second approach, I was trying to use some limit arugment, for example, for small $\delta>0$, let $\Omega_\delta=\{x\in \Omega:\ \operatorname{dist}(x,\partial \Omega)<\delta\}$. As $\mu$ has compact suppport, we have that for all $C^1(\overline{\Omega})$, $$\int_\Omega \nabla u\nabla \psi=\int _{\Omega\setminus\Omega_\delta }\nabla u\nabla \psi+\int _{\Omega_\delta }\nabla u\nabla \psi=\int _{\Omega\setminus\Omega_\delta }\nabla u\nabla \psi+\int _{\partial\Omega_\delta }\psi \frac{\partial u_n}{\partial \eta}$$

Any idea is appreciated.

Remark: $C_0^\infty(\overline{\Omega})=\{u\in C^\infty (\overline{\Omega}):\ u(x)=0,\ x\in \partial \Omega\}$.


1 Answer 1


I'll use notation $\Delta u$ instead of $-\mu$, because it fits better in the formulas.

Take a smooth domain $G$ such that $\operatorname{supp} \Delta u\subset G \Subset \Omega$. Using a smooth cutoff function, write $\psi$ as $\psi_1+\psi_2$ where $\psi_1$ has compact support in $\Omega$ and $\psi_2\equiv 0$ on $\overline{G}$. Apply Green's identity to $\psi_2$ and $u$ on $\Omega\setminus \overline{G}$. Since $u$ is harmonic here, there are no smoothness issues: $$\int_{\Omega\setminus \overline{G}} \nabla \psi_2\nabla u = \int_{\partial \Omega}\psi_2\frac{\partial u}{\partial \eta} \tag1$$ (The contribution of $\partial G$ is zero.) The integral on the left of (1) can just as well be taken over $\Omega$, it stays the same. Add (1) to $$\int_{\Omega } (\psi_1 \Delta u+ \nabla \psi_1\nabla u) = 0 \tag2$$ and the result is $$\int_{\Omega } (\psi \Delta u+ \nabla \psi \nabla u) = \int_{\partial \Omega}\psi\frac{\partial u}{\partial \eta}$$

  • $\begingroup$ @Tomás It's very standard HTML, so your browser must be at fault for not rendering it. $\endgroup$
    – user147263
    Jul 9, 2014 at 21:09

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