4
$\begingroup$

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.

Thanks

Originally posted https://math.stackexchange.com/posts/357170/

$\endgroup$

3 Answers 3

4
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ But how do we know that the solution satisfies the BC without assuming the solution is smooth enough? $\endgroup$
    – maximumtag
    Apr 11, 2013 at 20:56
  • $\begingroup$ @maximumtag: You have smoothness because of elliptic regularity. $\endgroup$
    – timur
    Apr 15, 2013 at 23:08
  • $\begingroup$ @timur what if $L$ is not a linear elliptic operator? Or what if the domain boundary is very irregular? $\endgroup$
    – Numa
    Apr 8 at 3:16
3
$\begingroup$

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 http://en.wikipedia.org/wiki/Trace_operator) 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?

$\endgroup$
0
1
$\begingroup$

As @Julian said, you will do well with the space $H^1$ and test functions are the keyword. But I would like to emphasize this point, which to my opinion might close the gap here.

You need to pay attention to your test functions: If you say that different BC give rise to the same weak formulation then this is "true" and false, but more precisely it's false. The weak formulations might look the same but they are not. This is best explained by reminding of two essential properties of a weak formulation:

  1. A regular enough weak solution satisfies the original problem classically
  2. A classical solution to the original problem satisfies the weak formulation.

So if you say two different BC give the same weak formulation then this would mean that at regular points of their solutions these solutions obey different PDE classically......

The point is that in the weak formulations for different BC you employ different test functions. Otherwise how do you show that a regular enough weak solution obeys the classical problem? If your test functions all vanish at the boundary, then you could not conclude that at points where they are regular they obey a Neumann boundary condition classically. So you need test functions in $H^1 \cap C^{\infty}$ for example. (Maybe try to prove that assuming you have a regular weak solution at hand it also satisfies the classical problem. In this process you will see why the choice of the test functions is essential)

Shortly: To ask for $u$ satisfying equation xyz for all test functions $\phi \in H^1_0$ is a different problem (with different solutions) then searching for $u$ satisfying equation xyz for all test functions $\phi \in H^1$.

Addendum: I would like to emphasize that, even if you know nothing about the regularity of your weak solution(except the weak regularity provided by your weak solution functional setting, like $L^2_tH^1_x \cap L^{\infty}_tL^2_x$ for instance, depending on your operator $L$) you have perfectly different weak formulations for different BC. The only question is whether your weak formulation is really a weak formulation for your classical problem, and for checking that you only hypothetically need to say, okay what if my weak solution is regular, does it then obey the classical problem or not. So you don't need to study additional regularity questions for justifying and solving your weak problem.

$\endgroup$

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.