I started reading an article and I'm having some trouble understanding a certain family operators they defined. Here the relevant parts: enter image description here enter image description here

I'm trying to understand how exactly $L_\sigma$ are defined. I noticed that if I take $\sigma=0$ then I get $B_0(u,v)=\langle Lu,v \rangle$, with respect to the standard $L^2$ inner product (and $L$ is the original Schrodinger operator). This seems like a possible direction on understanding how the operators are defined, but since they mentioned the space $H^1_0$ (which uses a different inner product), this might not be a correct idea.

I also found out that one can define self-adjoint operators from symmetric bilinear forms using the Riesz representation theorem, as described here. But this seems like a not very concrete way to define the $L_\sigma$ operators. If this is indeed the case, then I'm wondering if there's a more concrete way to interpret these operators (can we find a more concrete formula or something?), and once again, I'm not sure exactly if the writers are referring to the $L^2$ inner product or the $H^1_0$ inner product. Moreover, what's the relation to the original Schrodinger operator $L$, and why is the limit $L_\infty$ a Dirichlet boundary condition operator?

Anyway, if anyone has any ideas and can explain to me how to think of the $L_\sigma$ operators, I'd appreciate it. The original article is Nodal deficiency, spectral flow, and the Dirichlet-to-Neumann map, by Berkolaiko, Cox and Marzuola.

Thanks in advance!


1 Answer 1


The general construction is as follows: Let $H$ be a Hilbert space and $$ b\colon D(b)\times D(b)\to \mathbb{C} $$ a symmetric nonnegative bilinear form. Here $D(a)$ is a dense subspace of $H$. The form $b$ is called closed if $D(a)$ endowed with the inner product $\langle\cdot,\cdot\rangle_b$ given by $$ \langle u,v\rangle_b=b(u,v)+\langle u,v\rangle_H $$ is complete.

Then there is a positive self-adjoint operator associated with $b$ that can be described in two different ways (Kato's first and second representation theorem):

  1. The domain of $B$ is given by $$ D(B)=\{u\in D(b)\mid \exists v\in H\,\forall w\in D(b)\colon \langle v,w\rangle=b(u,w)\} $$ and $Bu=v$, where $v$ is defined as in the definition of $D(B)$ (if it exists, it is unique). This definition is similar to the one using the Riesz representation theorem, but if $D(b)$ is a proper subspace of $H$, then one has to be careful with the domain of $B$.
  2. The domain of $B^{1/2}$ is $D(b)$ and $$ \langle B^{1/2}u,B^{1/2}v\rangle_H=b(u,v) $$ for all $u,v\in D(b)$.

As noted in the OP, neither of these two representations is very explicit, but for a good reason: In many cases, the domain of $B$ does not have a nice explicit description, while the domain of $b$ can be easily written down.

If you are only interested in the action of $B$ on "nice functions", you can typically just integrate by parts. In your case that yould be $$ L_\sigma u=-\Delta u+Vu+\sigma 1_{\Gamma}u. $$ In particular, for $\sigma=0$ you get your original operator $L$ and as $\sigma\to \infty$ the last summand forces $u$ to be zero on $\Gamma$, which means that it satisfies Dirichlet boundary conditions on $\partial \Omega\cup \Gamma$ (the $\partial \Omega$ part comes from the fact that $b_\sigma$ is defined on $H^1_0(\Omega)$, which already forces Dirichlet boundary conditions on $\partial \Omega$).

If you are in Schrödinger operators and the like, I suggest you read up on these form methods as they are widely used. I can recommend Kato's Perturbation theory and Reed and Simon's Methods of Modern Mathematical Physics (you certainly don't have to read all 4 volumes, but I don't have have them at hand right now and don't know in which one form methods are treated).

  • $\begingroup$ Thank you for the detailed comment - it helps a lot. I'll definitely give the topic a read. Some follow up questions: How did you come up with this formula for $L_\sigma$ exactly via IVP? What did you integrate exactly? I couldn't quite understand. And why does the last summand force $u$ to be zero on $\Gamma$? Maybe the limit as $\sigma \rightarrow \infty$ just doesn't exist? Or am I missing something? Thanks again $\endgroup$
    – GSofer
    Jul 12, 2020 at 20:11
  • $\begingroup$ 1) Assuming all boundary terms vanish, integration by parts gives $\int_\Omega \nabla u \nabla v=-\int_\Omega (\Delta u)v$. Then you can write $b_\sigma(u,v)$ as an integral over $\Omega$ of "something" times $v$. That something is exactly $L_\sigma$ as described above. $\endgroup$
    – MaoWao
    Jul 13, 2020 at 6:04
  • $\begingroup$ 2) The correct way to take this limit is to consider $b_\sigma(u,u)$. Then the limit as $\sigma\to\infty$ always exists, it may just be $\infty$. The limiting form $b_\infty$ is defined on $\{u\in H^1_0(\Omega)\mid \lim_{\sigma\to\infty}b_\sigma(u,u)<\infty\}$ and $L_\infty$ is the positive self-adjoint operator associated with $b_\infty$. For the limit to be finite, it is clearly necessary and sufficient that $u$ vanishes on $\Gamma$. Thus $b_\infty$ acts exactly as $b_0$, just on a smaller domain. $\endgroup$
    – MaoWao
    Jul 13, 2020 at 6:07

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