2
$\begingroup$

Given some bounded domain $\Omega\subset \mathbb{R}^n$ with sufficiently regular boundary (e.g. smooth boundary). Then I saw two slightly different definitions for the Dirichlet-Laplacian. Some books consider the Laplacian on the initial domain $\lbrace f\in C^{\infty}(\Omega) \vert f_{\vert\partial\Omega}=0 \rbrace$, i.e. smooth functions which vanish on the boundary. Then the Laplacian is positive and symmetric, therefore there exists a self-adjoint extension (Friedrichs extension). On the other hand, the domain is sometimes $C^{\infty}_0(\Omega)$, i.e. the set of smooth functions which have compact support in $\Omega$. The Laplacian is also symmetric and positive on the domain $C^{\infty}_0(\Omega)$ and therefore for the same reason has a self-adjoint extension. But both initial domains are different. Therefore I'm a little bit worried about the self-adjoint extensions I end up with. Are both self-adjoint extensions the same? Are they different? Is there a reason, why the people like to define it the one or the other way around?

Thanks a lot!

$\endgroup$

2 Answers 2

3
$\begingroup$

If you take $C_0^\infty$ as the initial domain, then there are many self-adjoint extensions and the Dirichlet Laplacian is only one of them. The Neumann Laplacian is another one.

$\endgroup$
2
  • $\begingroup$ Ok thank you. So I assume that the Friedrichs extension of $-\Delta:C^{\infty}_0 (\Omega)\subset L^2(\Omega)\rightarrow L^2(\Omega)$ is equal to the other one. Is that right? $\endgroup$
    – supersnail
    Commented Jan 4, 2014 at 19:20
  • $\begingroup$ The Friedrichs extension is the Dirichlet Laplacian as already pointed out by Andras Batkai. $\endgroup$ Commented Jan 4, 2014 at 23:59
2
$\begingroup$

Expanding the answer of Michael, have a look at the discussion of Example 1 in Chapter X.3. of Reed-Simon (II: FOURIER ANALYSIS, SELF-ADJOINTNESS). There is a detailed discussion of the various selfadjoint extensions of the (one-dimensional) Laplacian defined on $C_0(0,1)$. The Dirichlet-Laplacian (the closure of your first) is the Friedrichs extension.

$\endgroup$
4
  • $\begingroup$ Do you know where I can find a proof of the statement "The Dirichlet-Laplacian (the closure of your first) is the Friedrichs extension." I would really appreciate that! $\endgroup$
    – N.U.
    Commented Mar 31, 2014 at 8:03
  • $\begingroup$ @N.U.: I gave the reference in my answer. In Reed-Simon, Example 1 in Chapter X.3. $\endgroup$ Commented Mar 31, 2014 at 11:02
  • $\begingroup$ @N.U.: If you are aon an arbitrary domain, then you actually (usually) define the Dirichlet Laplacian by the Friedrichs extension procedure: you take the quadratic form and close it. $\endgroup$ Commented Mar 31, 2014 at 11:10
  • $\begingroup$ But is it obvious that the closure of the Laplacian on smooth functions which vanish on the boundary will have the same domain as the Friedrich extension which is described in Reed and Simon? It would be a nice result if the Laplacian is essentially self adjoint on the set of smooth functions which vanish on the boundary. By the way, my definition of the Dirichlet Laplacian is by the Friedrich extension. $\endgroup$
    – N.U.
    Commented Mar 31, 2014 at 13:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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