# Defining a family of self-adjoint operators via a bilinear form

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

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).

• 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 Jul 12, 2020 at 20:11
• 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. Jul 13, 2020 at 6:04
• 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. Jul 13, 2020 at 6:07