Prove that $\|v\|_{H^1(\Omega )}\leq C(\|f\|_{L^2(\Omega )}+\|v\|_{H^{1/2}(\partial \Omega )}+\|\partial _\nu v\|_{H^{1/2}(\partial \Omega )}))$

Solution 1:

Assume by contradiction that such inequality doesn't hold for any $C$. Then for every $n \in \mathbb{N}$ one can find $v_n \in H^1(\Omega)$ such that

• $-\Delta v_n - v_n = f_n$,
• $\| v_n \|_{H^1(\Omega)} = 1$,
• $\|f_n\|_{L^2(\Omega)}, \|v_n\|_{H^{1/2}(\partial \Omega)}, \|\partial_\nu v_n\|_{H^{-1/2}(\partial \Omega)} < \frac 1n$.

Since $\Omega$ is bounded, one can choose a subsequence so that $v_n \rightharpoonup v$ in $H^1(\Omega)$ (by Banach-Alaoglu theorem) and $v_n \to v$ in $L^2(\Omega)$ (by Rellich-Kondrashov compactness theorem). It is easy to check that $-\Delta v - v = 0$ and $v = \partial_\nu u = 0$ on $\partial \Omega$. By assumption, the only solution is $v=0$. To summarize, we have that $\| v_n \|_{L^2(\Omega)} \to 0$.

What is left is to show a contradiction with $\| v_n \|_{H^1(\Omega)} = 1$. To this end, test the definition of $\partial_\nu v_n$ (given at the bottom) with $v_n$ itself: \begin{align*} \int_\Omega |\nabla v_n|^2 & = \langle \partial_\nu v_n, v_n \rangle - \int_\Omega v_n \Delta v_n \\ & = \langle \partial_\nu v_n, v_n \rangle + \int_\Omega v_n (v_n+f_n) \\ & \xrightarrow{n \to \infty} 0.\end{align*} The convergence follows from $\|\partial_\nu v_n\|_{H^{-1/2}(\partial \Omega)}, \|v_n\|_{H^{1/2}(\partial \Omega)} \to 0$ for the first term and from $\|v_n\|_{L^2(\Omega)}, \|f_n\|_{L^2(\Omega)} \to 0$ for the second. Thus we obtain $\|\nabla v_n\|_{L^2(\Omega)} \to 0$, which together with $\| v_n \|_{L^2(\Omega)} \to 0$ gives a contradiction.

Comment. For regular $v$, the divergence theorem yields the following identity: $$ \int_{\partial \Omega} \varphi \partial_\nu v = \int_\Omega \operatorname{div}(\varphi \nabla v) = \int_\Omega \nabla \varphi \nabla v + \varphi \Delta v \quad \text{for any } \varphi \in C^1(\overline{\Omega}). $$

If $v \in H^1(\Omega)$ and $\Delta v \in L^2(\Omega)$ (as it is in our case), the RHS can be used to define $\langle \partial_\nu v, \varphi \rangle$ for any $\varphi \in H^{1/2}(\partial \Omega)$, that is $$ \langle \partial_\nu v, \varphi \rangle := \int_\Omega \nabla \varphi \nabla v + \varphi \Delta v \quad \text{for } \varphi \in H^{1/2}(\partial \Omega). $$ It can be checked that this doesn't depend on the choice of the extension $\varphi \in H^{1}(\Omega)$. Look up the discussion here.