How can show that the prerequisites for the Lax-Milgram Lemma holds if I have different test and trial spaces (which I think is the natural thing to have if at least part of the boundary is non-homogenously Dirichlet)?To take a simple example (in my actual case I have a non-homogenous Neumann part as well, but that is unproblematic), let's say that I have the Poisson equation,\begin{align}-\nabla^2 u &= f &u&\in\Omega \\u &= u_D &u&\in\partial\Omega\end{align}and write the weak form as: find $u\in H_D^1(\Omega)$ such that for all $v\in H_0^1(\Omega)$ it holds that $a(u,v) = l(v)$ where\begin{align}a(u,v) &= (\nabla u, \nabla v)_{L^2(\Omega)}\\l(v) &= (f,v)_{L^2(\Omega)}.\end{align}Now, how can I show that $a(u,v)$ is coercive and bounded? If $u_D$ had been $0$ everywhere, then I would have had $a(u,v):H_0^1\times H_0^1\rightarrow \mathbf{R}$ so it would have been possible to show coercivity and continuity and then LM would guarantee a unique solution. But I have $a(u,v):H_D^1\times H_0^1\rightarrow \mathbf{R}$. Can I still show continuity and coercivity in the exact same way as usual and then Lax-Milgram (is it still called Lax-Milgram if it isn't from $X\times X$?) will hold? Or what should I do?
The Lax-Milgram Lemma in FEM with non-homogenous Dirichlet BC
finite element;boundary conditions