Let $\Omega = [0,L] \times [0,2\pi]$ and split its boundary into $\Gamma_d = \{0,L\} \times [0,2\pi]$, $\Gamma^1_p = [0,L] \times \{0\}$, $\Gamma^2_p = [0,L] \times\{2\pi\}$. Consider the following subspace of $H^2(\Omega)$ $$ V = \left\{ v \in H^2(\Omega) \mid v|_{\Gamma_d} = \partial_\nu v|_{\Gamma_d} = 0 \text{ and } v|_{\Gamma^1_p} = v|_{\Gamma^2_p}, \partial_\nu v|_{\Gamma^1_p} = -\partial_\nu v|_{\Gamma^2_p} \right\}\text{,} $$ where $\nu$ is the outer unit normal vector. I.e., we have Dirichlet boundary conditions on the left and right boundary and periodic boundary conditions on the upper and lower boundary. On $V$ define the bilinear form $$ a(u,v) = \int_{\Omega} (\Delta u + u)(\Delta v + v) \, \mathrm{d}x\text{.} $$
My question: Is it true that $a(u,u) = \int_{\Omega} (\Delta u)^2 + 2u\Delta u + u^2 \,\mathrm{d}x$ is coercive on $V$? Or, more specifically, can I find $c>0$ such that $$ \Vert \Delta u\Vert_{L^2(\Omega)}^2 + \Vert u\Vert_{L^2(\Omega)}^2 \leq c\, a(u,u) $$ holds true for all $u \in V$?
I see how by using integration by parts, one may equivalently consider the energy $$ \int_{\Omega} (\Delta u)^2 - 2\vert \nabla u\vert^2 + u^2 \,\mathrm{d}x\text{,} $$ but it is unclear to me how to go from there because I fail at controlling the $\nabla u$-term against the other two, at least for general lengths $L$.
