Smallest eigenvalue of Laplacian of periodic lattice after removing a vertex

Question

Consider a 4-regular graph with $N^2$ vertices, which can be interpreted as a $N\times N$ lattice with periodic boundary conditions so that every vertex has degree 4.

For an unweighted and undirected graph, the Laplacian matrix can be written as $\mathbf{L}=4\mathbf{I}-\mathbf{A}$. Where $\mathbf{I}$ is the identity matrix and $\mathbf{A}$ is the adjacency matrix. In that case the eigenvalues are known and are equal to: \begin{equation} \lambda_{N(j-1)+k}=4-2 \cos \left(\frac{2 \pi j}{N}\right)-2 \cos \left(\frac{2 \pi k}{N}\right) \end{equation} for $1 \leq j,k \leq N$.

The smallest eigenvalue will be $\lambda_{N^2}=0$. And the Laplacian is a singular matrix.

In the case of adding a regulator $s$ on one of the elements of the diagonal, such that $L_{ii}=4+s\delta_{ia}-A_{ii}$, $\mathbf{L}$ becomes invertible and its smallest eigenvalue $\lambda_{N^2}>0$.

I noticed numerically that for $s\gg1$ the smallest eigenvalue seems to reach a limit that scales with some function of $N$.

How could I derive an expression for the smallest eigenvalue, or at least determine how it scales with $N$?

Edit: Using perturbation theory, for large $s$ I could show that the smallest eigenvalue of $\mathbf{L}$ tends to the smallest eigenvalue of the matrix $\tilde{\mathbf{L}}$ of size $(N-1)\times (N-1)$ where we removed the $a^\text{th}$ column and row of $\mathbf{L}$. Now the question boils down to: what is the smallest eigenvalue of $\tilde{\mathbf{L}}$?

Answer 1

(Not a true answer, but too long for a comment.)

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I doubt there is an explicit expression for that eigenvalue. However, one should be able to show that it goes to zero approximately as $1 / \log N$ when $N \to \infty$.

Indeed: the eigenvalue of the periodic $N\times N$ lattice with one vertex removed (and zero "boundary" condition a this vertex) should be approximately equal to the first eigenvalue of the usual Laplacian on the 2-D torus with a disk of radius $1/N$ removed (and Dirichlet boundary condition on the boundary).

This, in turn, should be close to the eigenvalue of the Laplace operator in the unit disk with a circular hole of radius $1/N$ in the middle, with Neumann boundary condition on the unit circle, and Dirichlet boundary condition at the boundary of the hole.

That last eigenvalue is known explicitly in terms of the Bessel functions $J_0$ and $Y_0$, and it behaves as $1 / \log N$ when $N \to \infty$.

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Another approach would be via random walk on a full 2-D lattice. The potential kernel of the latter is relatively well-studied (see, for example, DOI:10.1214/aop/1041903213) and also presents logarithmic growth. This might be enough to get the desired result, but unfortunately I have no time now to see if it really works.