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I think I misunderstand the Lax-Milgram theorem. Suppose that I have a bilinear form $a$ satisfying the conditions of the Lax-Milgram theorem on a Hilbert space H. Then it must satisfy the same conditions on every Hilbert subspace of H. Therefore, the variational problem a(u,v)=l(v), where the linear form $l$ is continuous on H, has a unique solution on every Hilbert subspace of H. But from the point of view of PDE Hilbert subspaces are just additional conditions on the function. For example, let us consider the Poisson equation $\Delta u=f$ on a bounded domain $\Omega$. This problem has a unique solution on $H^1_0(\Omega)$ by Lax-Milgram. Let's take a hyperplane $H$ in $H^1_0(\Omega)$ defined by $H=\{v\in H^1_0(\Omega)~|~\int_\Omega v=0\}$. Then once again by Lax-Milgram there's a unique solution $u'$ on $H$. But it must be the same solution since $u$ is unique. So we get that for any solution $u$ of $\Delta u=f$ on $H^1_0(\Omega)$ one has $\int_\Omega u=0$ which is not true. Where is the mistake?

Thank you!

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  • $\begingroup$ Yes, but your $u$ and $v$ are also in the new subspace, so the new $u$ and the old $u$ aren't necessarily the same. $\endgroup$
    – krc
    Feb 21 '20 at 19:50
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The mistake is in the line "But it must be the same solution since $u$ is unique". This holds in your chosen subspace, but does not hold between subspaces. The unique solution in $H_0^1(\Omega)$ is not in general the same solution as in $H$. So there is a unique solution $u_{0}\in H_0^1(\Omega)$ for the problem formulated in $H_0^1(\Omega)$, and a unique solution $u\in H$ for the problem formulated in $H$.

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  • $\begingroup$ Is that really what's happening? I think the issue is that the integral equation changes when you change the Hilbert space: the test functions get drawn from a new space as well. After restricting the space of test functions it may be that a solution to a solution to the "weak formulation" is no longer a solution to the strong formulation, even if it has enough regularity. $\endgroup$
    – Ian
    Feb 21 '20 at 20:00
  • $\begingroup$ @Ian I'm not sure I follow. The test functions will be drawn from the new space, but isn't it the case that in this new weak formulation Lax-Milgrim gives a unique solution to the weak problem in the new subspace? I think you're right that this new weak formulation does not solve the strong formulation. $\endgroup$
    – krc
    Feb 21 '20 at 20:05
  • $\begingroup$ You get a unique solution to the new problem in the new subspace, but the new problem has fewer test functions available, so you can't "un-integrate by parts" to get back to the strong solution. Thus the solution that you get in the subspace isn't really a solution to the problem you started from, it is a solution to a kind of projected problem. $\endgroup$
    – Ian
    Feb 21 '20 at 20:08
  • $\begingroup$ @Ian Okay, thats what I thought, you are solving the problem formulated in $H$ rather than in $H_0^1(\Omega)$. The projection is correct, to fully solve the PDE you also need to solve the PDE in $H^\perp$ and then superimpose the weak solutions from each since every $w\in H_0^1=H\oplus H^\perp$ can be written as $u+v$ for $u\in H$ and $v\in H^\perp$. In some sense this idea underlies the Galerkin method, i.e. the approximate solutions don't solve the original problem. $\endgroup$
    – krc
    Feb 21 '20 at 20:18
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    $\begingroup$ @Carabaev Right, the solution to the problem with domain given by a subspace of $H^1_0(\Omega)$ only solves the problem for a subspace of the test functions as well, so it generally doesn't solve the original problem...but it does solve it "modulo those test functions that you had to remove", which is basically the idea of Galerkin methods. $\endgroup$
    – Ian
    Feb 21 '20 at 21:23

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