Existence of minimum in $H^{1,2}(\Omega)$

Question

I am considering a functional

$$\mu(\Omega) = \min \{ u \in H^{1,2}(\Omega), \frac{\alpha \int_{\partial \Omega} u^2 ds + \int_{\Omega} |\nabla u|^2}{\int_{\Omega} u^2 dx} \}$$

I want to show the existence of minimum , can someone help me ?

Answer 1

The problem in proving existence is that standard techniques do not apply in an obvious way: The function to be minimized is not convex and its domain of definition is not convex either.

I assume $\alpha>0$ here. Denote $$ J(u) := (\alpha \|u\|_{L^2(\partial \Omega)}^2 + \|\nabla u \|_{L^2(\Omega)}^2 ) \|u\|_{L^2(\Omega)}^{-2}. $$ Then $J(u)$ is defined for $u\ne 0$. Moreover, it is bounded from below: Using the inequality $$ \|u\|_{L^2(\Omega)} \le c\left( \|u\|_{L^2(\partial \Omega)} + \|\nabla u \|_{L^2(\Omega)}\right) \quad \forall u\in H^1(\Omega), $$ it follows that $J(u)\ge c'>0$. Let now $(u_n)$ be a minimizing sequence, i.e. $$ \lim_{n\to \infty}J(u_n) = \inf_{u\ne 0}J(u). $$ Since $J(\lambda u)=J(u)$ for all $\lambda>0$, we can rescale the sequence, such that $\|u_n \|_{L^2(\Omega)}^2 =1$. Since $J(u_n)$ is converging, it is bounded from above, which implies that $(u_n)$ is a bounded sequence in $H^1(\Omega)$.

Hence, we can extract a weakly converging subsequence (denoted again by $(u_n)$), $u_n \rightharpoonup u$ in $H^1(\Omega)$. After extracting another subsequence (again denoted by $(u_n)$) we find $u_n\to u$ in $L^2(\Omega)$. Since $\|u_n\|_{L^2(\Omega)}=1$ and $u_n\to u$ in $L^2(\Omega)$ it follows $\|u\|_{L^2(\Omega)}=1$, in particular $u\ne 0$.

It remains to prove $J(u) \le \lim\inf J(u_n)$. Due to weak lower semicontinuity of norms, we have $$ \alpha \|u\|_{L^2(\partial \Omega)}^2 + \|\nabla u \|_{L^2(\Omega)}^2 \le \lim\inf_{n\to\infty}(\alpha \|u_n\|_{L^2(\partial \Omega)}^2 + \|\nabla u_n \|_{L^2(\Omega)}^2) . $$ Moreover, $0<\|u\|_{L^2}^2= \lim_{n\to \infty}\|u_n\|_{L^2}^2$. Hence it follows, $$ \lim\inf_{n\to \infty} J(u_n)= \lim\inf_{n\to \infty}\frac{\alpha \|u_n\|_{L^2}^2 + \|\nabla u_n \|_{L^2}^2}{\|u_n\|_{L^2}^2} = \frac{\lim\inf_{n\to \infty}(\alpha \|u_n\|_{L^2}^2 + \|\nabla u_n \|_{L^2}^2)}{\lim_{n\to \infty}\|u_n\|_{L^2}^2} \ge J(u). $$ Hence, $u$ is a solution of the problem.