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?


1 Answer 1


Typically, you would decompose your solution into $u = u_D + u_0$, where $u_D$ satisfies inhomogeneous Dirichlet conditions. You can then solve for $u_0 \in H^1_0$ subject to $$\Delta u_0 = f-\Delta u_D.$$ You would then recover a variational formulation over $H^1_0$ again.

Note that $u_D$ is non-unique. To remedy this in proofs, $u_D$ is often taken to be the minimum-energy extension of the trace satisfying boundary conditions.

  • $\begingroup$ So as weak form I should have: find $u_0\in H_0^1(\Omega)$ such that for all $v\in H_0^1(\Omega)$ it holds that $a(u_0, v) = l(v)$ where \begin{align} a(u_0, v) &= (\nabla u_0, \nabla v)_{L^2(\Omega)} &u_0&\in\Omega \\ l(v) &= (f, v)_{L^2(\Omega)} - (\nabla u_D, \nabla v)_{L^2(\Omega)} &u&\in\partial\Omega. \end{align} This is essentially what I do when I go to the finite element form, there I take $u_{h, D}$ to be zero inside $\Omega$. Is it problematic to let $u_D$ and $u_{h, D}$ be completely different? And what is the minimum-energy extension of the trace satisfying boundary conditions? $\endgroup$
    – Christian
    Dec 3, 2013 at 16:53
  • $\begingroup$ $u_{h,D}$ shouldn't be zero inside $\Omega$; it may help to think about the 1D Laplace's equation with linear elements. The hat functions at the ends of the domain are $u_{h,D}$; the equations corresponding to them are moved to the RHS to become load/data, and we are left with hat functions that are zero on the boundary. They are zero over most of the boundary, but they do extend inside. $\endgroup$
    – Jesse Chan
    Dec 3, 2013 at 20:43
  • $\begingroup$ The minimum-energy extension of $u_D$ is defined to be $u\in H^1$ such that $u=u_D$ on the boundary and $\|u\|_H^1$ is minimized. The nice thing about this minimum energy extension is that it can be used to define norms over trace/boundary spaces in finite elements as well. $\endgroup$
    – Jesse Chan
    Dec 3, 2013 at 21:13
  • $\begingroup$ Yes, I was unclear, I meant that $u_{h,D}$ are zero except close to and on the boundary. As for which $u_D$ to take (in the weak formulation), can't I take $u_D = u_{h, D}$? It's certainly feels backwards to (in a sense) go from the finite element formulation to the weak one, but except for that: will it very difficult to show that $l(v)$ is bounded? As for the method you're suggesting... it doesn't lead to Galerkin orthogonality, does it? Because the linear forms in the weak/finite element formulations would be different. $\endgroup$
    – Christian
    Dec 3, 2013 at 21:47
  • 1
    $\begingroup$ The minimum energy extension is definitely not $u_{h,D}$ because it needs to solve a Laplace equation (with zero right hand side). But the point you seem to miss is that in order to prove existence and well-posedness, it doesn't actually matter how you choose $u_D$. You choose anything as long as it has the correct boundary values, and you can show that a solution $u_0$ exists and is unique. That's all you need. Of course, if you want, you can choose $u_D=u_{h,D}$ because that function is in $H^1$ but that's only one possible choice, not a necessary one. $\endgroup$ Dec 4, 2013 at 2:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.