i'm trying to solve the Poisson equation: $$ \begin{split} -\Delta u &= f \quad \text{in } \Omega\\ u &= g \quad \text{on } \partial \Omega, \end{split} $$
where $ \Omega$ is bounded a Lipschitz-continues domain and $u \in H^2(\Omega), f,g \in L^2(\Omega)$. The least squares finite element method minimizes the following energy functional:
$$ J(u;f,g) := \frac{1}{2} \int_{\Omega} (-\Delta u-f)^2 + \int_{\partial\Omega} (u -g)^2 $$
this method leads to the following variational problem: $$ a(u,v) = l(v), \quad \forall v \in V \subset H^2(\Omega) $$ where
$$ \begin{split} a(u,v) &= \int_{\Omega} \Delta u\Delta v + \int_{\partial\Omega} uv \\ l(v) &= \int_{\Omega} -f\Delta v + \int_{\partial\Omega} gv \end{split} $$
It's clear that this bilinear operator is continues. Now i'm trying to proof that $a(v,v)$ is coercive, e.q.
$$ a(v,v) = \int_{\Omega} (\Delta v)^2 + \int_{\partial\Omega} v^2 \geq C\| v\|^2_{H^2(\Omega)} $$
It seems to be hard to prove that, so I tried to show this weaker inequality: $$a(v,v) = \int_{\Omega} (\Delta v)^2 + \int_{\partial\Omega} v^2 \geq C\| v\|^2_{L^2(\Omega)} $$ but with no success.
I don't have the opportunity to choose $u \in H^2(\Omega) \cap H^1_0(\partial \Omega)$, so i had to add the boundary residual to the energy functional $J$. I guess i'm working with the wrong energy.
This is my first question here and i'm thankful for every hint.
skymath
The first coercivity inequality does not hold. Take, for example, the disk-shape domain $\Omega = \{x, y: x^2 + y^2 \leq 1\}$ and consider a sequence of boundary data $g_n$, $n \in \mathbb R$, defined by $g_n = \cos(n\theta(x, y))$, where $\theta(\cdot, \cdot)$ denotes the angle of the argument in $[0, 2\pi]$. Take $v_n$ as being the solution of \begin{align} - \Delta v_n = 0, \qquad & \text{in $\Omega$}, \\ v_n = g_n, \qquad & \text{in $\partial \Omega$}. \end{align} Clearly, $$ \int_{\Omega} (\Delta v_n)^2 + \int_{\partial\Omega} v_n^2 = \int_{\partial \Omega} g_n^2 \leq C. $$ If the first coercivity inequality you wrote was true, it would imply: $$ \|v_n\|_{H^2(\Omega)} \leq C, $$ and thus, by the trace inequality $$ \|v_n\|_{H^{m}(\partial \Omega)} = \|g_n\|_{H^m(\partial\Omega)}\leq C, \qquad 0 \leq m \leq 3/2. $$ Given the form of $g_n$, this is clearly false for $m = 1$.