# Weakly lower semi-continuous of a certain functional

Suppose $\Omega \subset \mathbb{R}^3$ is a bounded domain with smooth boundary. Let $u_0 : \partial \Omega \rightarrow \mathbb{S}^2$ be Lipschitz and $B : = \{u\in W^{1,2}(\Omega,\mathbb{S}^2)|\,u|_{\partial \Omega}=u_0\}$. Why the functional $$I=\int_{\Omega}A_1(\nabla \cdot u)^2+A_2|\nabla \times u|^2 +A_3(u \cdot\nabla \times u)^2+A_4|u\times \nabla \times u|^2+A_5|\nabla u|^2$$ is weakly lower semi-continuous on $B$? All $A_i$ are non-negative. The author just said it follows from the strong convergence in $L^2$ of any weakly convergent sequence in $W^{1,2}$ and all $A_i$ are non-negative. My idea is that strong convergence in $L^2$ implies pointwise convergent subsequence and then by Fatou's Lemma (the integrand is non-negaitve). Is it correct? And I am still confused how the limits can be taken into the derivatives (taken into the curl,divergence sign) And is it true for a general functional $$J=\int_{\Omega} G(\nabla u,u)$$ with $G$ being non-negative and continuous?

• What is $W^{1,2}(\Omega,\mathbb S_2)$? $\mathbb S_2$ is not a vector space. – daw Jan 22 '18 at 10:00

If all the $A_i$ are non-negative, the first (if corrected to $(\nabla \cdot u)^2$), the second, and the fifth are convex and continuous in $Du$ from $L^2$ to $\mathbb R$ hence sequentially weakly lower semicontinuous.
The third and fourth term are certainly non-convex. However, these are integrals of squares of something. This something is a sum of expressions of the type $u_j \frac{ \partial u_k }{\partial x_l}$.
Let me show that the simple argument provided by your reference does not work. Let me take a weakly converging sequence $(u_n)$ of elements in $B$. Then $(u_n)$ is a bounded sequence in $L^\infty$ (as the vector fields are pointwise in the unit sphere $\mathbb S_2$). After extracting subsequences, $u_n$ converges strongly in $L^p$ ($p<\infty$) and weakly in $H^1$ to $u$. This implies $u_n\times \nabla \times u_n$ converges weakly in $L^q$ to $u\times \nabla \times u$, but for $q<2$. We cannot conclude weak convergence in $L^2$, unfortunately. This is not enough to conclude weak lower semicontinuity in $L^2$. Maybe, one has to use more structure.
• But why $|\nabla \times u|^2$ is convex in $Du$? – mnmn1993 Jan 22 '18 at 14:16
• $Du \mapsto \nabla\times u$ is linear, $|\cdot|^2$ is convex – daw Jan 22 '18 at 15:11