Let $B\subset \mathbb R^2$ be a unit ball. Let $B^+:=B\cap \{x_2\geq 0\}$ where we set $x=(x_1,x_2)\in \mathbb R^2$. Let $\omega\in C^1(\partial B)$ be given such that $|\nabla \omega|>0$ for all $x\in\partial B$. We consider the following two equations: $$ \begin{cases} -\Delta u_1=0& \text{ if }x\in B\\ u_1=\omega&\text{ if }x\in\partial B \end{cases} $$ and $$ \begin{cases} -\Delta u_2=0& \text{ if }x\in B^+\\ u_2=\omega&\text{ if }x\in(\partial B)\cap \{x_2\geq 0\}\\ \frac{\partial }{\partial \nu}u_2=0&\text{ if }x\in B\cap \{x_2=0\} \end{cases} $$ where $\nu$ is the outer normal vector at $B\cap \{x_2=0\}$.
My question: would it be possible to prove that $$ \int_{B^+}|\nabla u_2|^2dx<\int_{B^+}|\nabla u_1|^2dx $$ with "$<$" but not "$=$"?
I think the condition $|\nabla \omega|>0$ for all $x\in\partial B$ is the key, but I don't know how to use it...
Of course, if $\omega\equiv c$ for some constant $C$, then the above strict inequality does not hold.
Thanks for your help!
Yes, this is true. First of all, the nonstrict inequality $$\int_{B^+}|\nabla u_2|^2dx \le \int_{B^+}|\nabla u_1|^2dx \tag{1}$$ holds because $u_2$ minimizes the Dirichlet integral (the LHS of (1)) on the linear space of all functions with boundary value $\omega$ on $\partial B\cap \{x_2\ge 0\}$, and no condition on $B\cap \{x_2=0\}$. (I.e., when a part of the boundary is free of condition, the minimizer satisfies the Neumann condition there: this can be proved by perturbing its boundary values, and in any case is standard.)
If equality holds in (1), then $u_1$ also has zero normal derivative on the boundary (so, $u_2=u_1$). Then the function $\tilde u(x_1,x_2)=u_1(x_1,|x_2|)$ is also harmonic in $B$, which implies $\tilde u=u_1$ in $B$. Therefore, the boundary values $\omega$ are symmetric with respect to the plane $x_2=0$. But this implies that the derivative of $\omega$ vanishes at the points $(\pm 1,0)$, a contradiction.