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I am wanting to show existence of solutions to $$u_t +L(u) = f \;\;\text{on}\;\; \Omega$$ with initial condition $u|_{t=0} = u_0$ and Neumann boundary condition $\nabla u\cdot \nu = 0$ on ${\partial\Omega}$.

How do I solve this problem via the Galerkin approximation without putting the BC in the Hilbert space? It's easy to derive the weak form (which of course uses the BC) but surely this is not enough to guarantee that the solution at the end satisfies the BC, since different BCs can give rise to the same weak formulation. And in books they say the Neumann condition is only satisfied if we assume $u$ is smooth enough to do the IBP and check, but this seems like cheating.

gerw also mentioned to me that specifying this BC may not make sense since $\partial\Omega$ is measure zero. I don't know what to say about that.


Originally posted

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The Neumann boundary condition is a "natural" BC. You don't need to impose it. You have to change the space $H^1_0$ for $H^1$ and you end up with the same weak formulation that you found in the Dirichlet BC case, only that you required that the test functions lie in $H^1$.

That will do it.

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But how do we know that the solution satisfies the BC without assuming the solution is smooth enough? – maximumtag Apr 11 '13 at 20:56
@maximumtag: You have smoothness because of elliptic regularity. – timur Apr 15 '13 at 23:08

I'll try my best:

As $\partial \Omega$ is measure zero and Lp functions are defined a.e., you need that the trace operator (see is well defined. For instance, if your solution, $u$, belongs to $H^s(\partial \Omega)=W^{2,s}(\partial \Omega)$ with $s\in(3/2,2]$ then you have $\nabla u\in H^{s-1}(\Omega)$ and its trace $T(\nabla u)\in H^{s-1-1/2}(\partial \Omega)$.

Notice that your approximate solutions $u_n$ are smooth and verify the boundary condition. If you can obtain that the approximate solutions converge to the real solution strongly in $H^{s-\epsilon}$ then you are done, I think, because then the trace operator verifies $\lim T(u_n)=T(u)$. To show this strong convergence you can try to prove that the approximate solutions are Cauchy in lower norms ($L^2$ or something) and a uniform bound in higher norms. With these properties the approximate solutions will be Cauchy in between.

Has this answer been useful for you?

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Yes,thank you, good answer. – maximumtag Apr 16 '13 at 21:06

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