Stacked tree construction for free-fermion projected entangled pair states

Yuman He

Kangle Li

Yanbai Zhang

Hoi Chun Po
Department of Physics, Hong Kong University of Science and Technology, Clear Water Bay, Hong Kong, China
IAS Center for Quantum Technologies, Hong Kong University of Science and Technology, Clear Water Bay, Hong Kong, China
Center for Theoretical Condensed Matter Physics, Hong Kong University of Science and Technology, Clear Water Bay, Hong Kong, China

The tensor network representation of a state in higher dimensions, say a projected entangled-pair state (PEPS), is typically obtained indirectly through variational optimization or imaginary-time Hamiltonian evolution.
Here, we propose a divide-and-conquer approach to directly construct a PEPS representation for free-fermion states admitting descriptions in terms of filling exponentially localized Wannier functions.
Our approach relies on first obtaining a tree tensor network description of the state in local subregions. Next, a stacking procedure is used to combine the local trees into a PEPS. Lastly, the local tensors are compressed to obtain a more efficient description.
We demonstrate our construction for states in one and two dimensions, including the ground state of an obstructed atomic insulator on the square lattice.

§ ACKNOWLEDGEMENTS

This work is supported by the Ministry of Science and Technology, China through MOST22SC01 and the Hong Kong Research Grant Council through ECS 26308021.


# Local Spectral Optimisation for Robin Problems with Negative Boundary
Parameter on Quadrilaterals

Julie Clutterbucka and James Larsen-Scotta
aSchool of Mathematics, Monash University, 9 Rainforest Walk, Melbourne, VIC
3800, Australia email<EMAIL_ADDRESS>author email:
<EMAIL_ADDRESS>
(September 6, 2023)

###### Abstract

We investigate the Robin eigenvalue problem for the Laplacian with negative
boundary parameter on quadrilateral domains of fixed area. In this paper, we
prove that the square is a local maximiser of the first eigenvalue with
respect to the Hausdorff metric. We also provide asymptotic results relating
to the optimality of the square for extreme values of the Robin parameter.

††The work of James Larsen-Scott was supported via an Australian Government
Research Training Program Scholarship. The research of Julie Clutterbuck was
supported by grant DP220100067 of the Australian Research Council.

##  1 Introduction

Consider the Robin eigenvalue problem for the Laplacian

$\displaystyle-\Delta u=\lambda u\quad\text{in }\Omega,$ (1a)
$\displaystyle\frac{\partial u}{\partial\nu}+\alpha u=0\quad\text{on
}\partial\Omega,$ (1b)

with outward facing unit normal $\nu,$ eigenvalue $\lambda$ and Robin
parameter $\alpha,$ for a domain $\Omega$ of $\mathbb{R}^{N},N\geq 2.$ The
Robin problem can be seen as an extension of the well-studied Neumann and
Dirichlet boundary conditions, where $\alpha=0$ corresponds to the Neumann
problem, and $\alpha=+\infty$ corresponds to the Dirichlet problem. The
eigenvalues associated with the problem (1a-1b) form a non-decreasing sequence
${\lambda_{k}(\Omega,\alpha)}_{k=1}^{\infty}.$ The first eigenvalue
$\lambda_{1}$ is simple and given by the variational characterisation

$\displaystyle\lambda_{1}(\Omega,\alpha)=\min_{\begin{subarray}{c}u\in
H^{1}(\Omega)\\\ u\neq 0\end{subarray}}\frac{\int_{\Omega}|\nabla
u|^{2}dx+\alpha\int_{\partial\Omega}|u|^{2}d\sigma}{\int_{\Omega}|u|^{2}dx},$
(2)

where we understand the boundary integral in the sense of traces where
necessary.
This paper concerns optimisation problems for the first eigenvalue of the
Laplacian. In particular we are interested in the following question

######  Question 1.

Let $\Lambda$ be a collection of domains with fixed area. For each fixed
$\alpha\in\mathbb{R}$ does there exist $\Omega_{0}\in\Lambda$ such that for
each $\Omega\in\Lambda$ we have
$|\lambda_{1}(\Omega,\alpha)|\geq|\lambda_{1}(\Omega_{0},\alpha)|$?

A famous result of this type is the Faber-Krahn inequality: among domains of
fixed dimension and area, the first eigenvalue of (1a) with Dirichlet boundary
condition $u=0$ on $\partial\Omega$ is minimized by the ball. This was
initially conjectured by Lord Rayleigh [18] before later being proved
independently by Faber [8] and Krahn [13].
An analogous result holds for the Robin problem (1a-1b) with positive Robin
parameter $\alpha>0$. This was established in two dimensions by Bossel [4],
before Daners later extended this work to higher dimensions [7] and alongside
Bucur extended to the p-Laplacian [5]. For recent related results see [1, 6].
In the case of negative Robin parameter $\alpha<0,$ the first eigenvalue is
negative $(\lambda_{1}<0)$ and the relevant spectral optimisation question is
whether there is a domain $\Omega_{0}$ which maximises the first eigenvalue.
Bareket conjectured (at least in two dimensions, see [2]) that a reverse
Faber-Krahn type inequality holds, where the ball maximises the first
eigenvalue among domains of a given volume. However, this was shown to be
false in general by Freitas and Krejčiřík [9] who showed for sufficiently
large negative $\alpha$ that a spherical shell has larger first eigenvalue. In
the same paper, Freitas and Krejčiřík also showed that in 2 dimensions a
reverse Faber-Krahn equality does hold, where the ball is still the maximiser,
provided that the absolute value of $\alpha$ is sufficiently small. It remains
an open question to consider whether the disk is a maximizer once one
restricts to simply connected domains in two dimensions.

######  Question 2.

Let $\Lambda$ be the collection of $N$-gons with fixed area. Does there exist
$\Omega_{0}\in\Lambda$ such that for each $\alpha\in\mathbb{R}$ and
$\Omega\in\Lambda$ we have
$|\lambda_{1}(\Omega,\alpha)|\geq|\lambda_{1}(\Omega_{0},\alpha)|$? If so, is
$\Omega_{0}$ the regular $N$-gon?

In the case of the Dirichlet problem, for triangles and quadrilaterals, Pólya
[17] proved that indeed the equilateral triangle and square respectively
minimize the first eigenvalue. However, for $N\geq 5$ the problem is still
open in the Dirichlet case. See [3, 11] for recent developments. In the case
of rectangles, for both positive and negative $\alpha$ the square is the
optimiser, as can be seen through a separation of variables argument (see for
instance the survey paper [15].) However, for general quadrilaterals and
triangles this problem is still open for both positive and negative $\alpha$
(this problem for triangles with $\alpha>0$ is conjecture 6.4 in [10]).
Recent work by Krejčiřík et al. in [14] has made some progress on this problem
for trianglar domains when the Robin parameter $\alpha$ is negative. The first
result of this paper showed that for sufficently small $\alpha$ the
equilateral triangle is a local maximiser. The second result showed, that for
a fixed triangle of given area, if $\alpha$ is sufficiently small or large in
absolute value, the equilateral triangle of the same area has larger first
eigenvalue. In this paper we follow a similar approach to Krejčiřík et al. and
obtain similar results for the Robin problem with negative boundary parameter
on quadrilateral domains.
Our main result is the following local result.

######  Theorem 1.

The square is a local maximiser (with respect to the Hausdorff metric up to
isometry) of the first eigenvalue of the problem (1a-1b) with negative
boundary parameter $\alpha$ in the family of quadrilaterals of fixed area.

Notice that unlike the equivalent result for triangles in [14] there is no
restriction on the size of $\alpha.$ The method of proof follows that in [14]:
we parametrise an arbitrary quadrilateral of fixed area $\Omega$ in terms of 4
parameters. We replace the problem (1a-1b) on $\Omega$ with an isospectral
problem on the (rotated) square $\Omega_{0}$ whose associated quadratic form
depends analytically upon the 4 parameters. By computing the first and second
order derivatives with respect to the parameters of the eigenvalue, it can be
shown that the square is a local maximiser in this parameter space. The
argument takes advantage of the fact that the solution to (1a-1b) is known
explicitly on the square.
Our second result shows that a reverse Faber-Krahn inequality holds at least
asymptotically for large or small $\alpha.$

######  Theorem 2.

Let $\Omega_{0}$ be the square, and let $\Omega$ be any quadrilateral with the
same area. Then there exists $\alpha_{2}\leq\alpha_{1}<0$ such that for all
$\alpha\in\left(-\infty,\alpha_{2}\right]\cup\left[\alpha_{1},0\right)$ we
have $\lambda_{1}(\Omega,\alpha)\leq\lambda_{1}(\Omega_{0},\alpha).$

Note that the constants $\alpha_{1},\alpha_{2}$ depend upon $\Omega.$ This
theorem is proved for small $\alpha$ via trial function arguments and for
large $\alpha$ via an asymptotic result of Levitin and Parnovski [16].
Our third result shows that a reverse Faber-Krahn also holds in the case of
quadrilaterals far from the square in terms of Hausdorff distance.

######  Theorem 3.

Let $\Omega_{0}$ be the square, let $\Omega$ be a quadrilateral with the same
area and fix $\alpha<0.$ Then there exists a constant $C$ depending only upon
$\alpha$ such that if $d_{H}(\Omega,\Omega_{0})>C$ we have
$\lambda_{1}(\Omega,\alpha)\leq\lambda_{1}(\Omega_{0},\alpha).$

Here $d_{H}$ is the Hausdorff distance up to isometry. This theorem is proved
using a test function argument in a similar way to our proof of Theorem 2.
This paper is structured as follows; in Section 2 we provide the preliminary
setup required, this involves the parametrisation of an arbitrary
quadrilateral and construction of a transform that replaces the Robin problem
on this quadrilateral with an isopectral problem on the square. We provide the
explicit solution to the Robin problem on the square, and provide useful
properties and results that will be used later. In Section 3 we compute the
first and second order derivatives of the eigenvalues with respect to the
geometric parameters, and prove Theorem 1. In Section 4 we utilise test
function arguments to obtain sufficient conditions for
$\lambda_{1}(\Omega)\leq\lambda_{1}(\Omega_{0}),$ and as a consequence prove
Theorem 2 and 3. Section 5 is an appendix containing further details of nasty
computations required in the proof of Theorem 1.

##  2 Preliminary Setup

In this section we follow Krejčiřík et al’s approach [14]. We parametrise an
aribtrary quadrilateral of fixed area 2S, then show that the Robin spectral
problem on arbitrary quadrilaterals can be reduced to the spectral problem on
the square for an associated family of operators dependent upon the
parametrisation parameters.

###  2.1 Geometric Setup

We provide a parametrisation of a general quadrilateral as follows. Let $c>0,\
a_{1},a_{2}\in\mathbb{R},\ S_{1},S_{2}>0,$ with $S_{1}+S_{2}=2S>0.$ Let
$\Omega_{a_{1},a_{2},c,S_{1}}$ be the quadrilateral with area $2S$ formed by
connecting, in order, the vertices $(-c,0),\ (a_{1},b_{1}),\ (c,0),\
(a_{2},-b_{2}),$ where $b_{i}=S_{i}/c.$ The choice
$(a_{0},c_{0})=(0,\sqrt{S})$ leads to the (rotated) square with side length
$\sqrt{2S},$ which we shall denote as $\Omega_{0},$ when
$(a_{1},a_{2},c,S_{1})=(a_{0},a_{0},c_{0},S).$

$(a_{1},S_{1}/c$)$(c,0$)$(a_{2},-S_{2}/c$)$(-c,-0$) Figure 1: Parametrisation
of a Quadrilateral

We note that by construction the areas of the quadrilaterals are constant and
given by

$\displaystyle|\Omega_{a_{1},a_{2},c,S_{1}}|=c(b_{1}+b_{2})=S_{1}+S_{2}=2S=|\Omega_{0}|.$

We label the corners of the upper and lower triangles that make up the
quadrilateral by

$\displaystyle p_{1}^{(j)}$ $\displaystyle=(-c,0),$ $\displaystyle
p_{2}^{(j)}$ $\displaystyle=(a_{j},(-1)^{j+1}S_{j}/c),$ $\displaystyle
p_{3}^{(j)}$ $\displaystyle=(c,0).$

for $j=1,2.$
The boundary of $\Omega_{a_{1},a_{2},c,S_{1}}$ may be written as the union of
line segments in the form

$\displaystyle\partial\Omega_{a_{1},a_{2},c,S_{1}}=\Gamma_{a_{1},c,S_{1}}^{(1,1)}\cup\Gamma_{a_{1},c,S_{1}}^{(2,1)}\cup\Gamma_{a_{2},c,S_{2}}^{(1,2)}\cup\Gamma_{a_{2},c,S_{2}}^{(2,2)},$

where we define $\Gamma_{a_{j},c,S_{j}}^{(i,j)}$ for $i=1,2$ to be the line
segment connecting $p^{(j)}_{i}=(x_{i,j},y_{i,j})$ to
$p^{(j)}_{i+1}=(x_{i+1,j},y_{i+1,j}),$ that is

$\displaystyle\Gamma_{a_{j},c,S_{j}}^{(i,j)}=\\{p_{i}^{(j)}+t(p_{i+1}^{(j)}-p_{i}^{(j)}):t\in[0,1]\\}.$

Further, the lengths of the line segments are given by

$\displaystyle|\Gamma_{a_{j},c,S_{1}}|=||p_{i}^{(j)}-p_{i+1}^{(j)}||=\sqrt{\frac{S_{j}^{2}}{c^{2}}+(a_{j}+(-1)^{(i+1)}c)^{2}}.$

Note that since $|\Gamma_{0}^{(i,j)}|$ is independent of $(i,j),$ the
superscript $(i,j)$ will be omitted. The perimeters of the quadrilaterals are
given by

$\displaystyle|\partial\Omega_{a_{1},a_{2},c,S_{1}}|=$
$\displaystyle\Sigma_{i,j}|\Gamma_{a_{j},c,S_{1}}^{(i,j)}|$ $\displaystyle=$
$\displaystyle\sqrt{\frac{S_{1}^{2}}{c^{2}}+(a_{1}-c)^{2}}+\sqrt{\frac{S_{1}^{2}}{c^{2}}+(a_{1}+c)^{2}}+\sqrt{\frac{S_{2}^{2}}{c^{2}}+(a_{2}-c)^{2}}+\sqrt{\frac{S_{2}^{2}}{c^{2}}+(a_{2}+c)^{2}},$

with

$\displaystyle\quad|\partial\Omega_{0}|=4\sqrt{2S}.$

We next construct a mapping between the quadrilaterals $\Omega_{0}$ and
$\Omega_{a_{1},a_{2},c,S_{1}}.$

######  Lemma 1.

Let $a,a_{0},S\in\mathbb{R},c,c_{0}>0$ with $a\neq\pm c.$ Then, a linear
mapping between the triangle $\Omega_{0}$ defined by the vertices
$(c_{0},0),(-c_{0},0),(a_{0},S/c_{0})$ and the triangle $\Omega_{a,c}$ defined
by the vertices $(c_{1},0),(-c_{1},0),(a_{1},\tilde{S}/c_{1})$ is given by

$\displaystyle\mathcal{L}:\Omega_{0}\to\Omega_{a,c}:(x,y)\mapsto\left(\frac{c_{1}}{c_{0}}x+\frac{a_{1}c_{0}-a_{0}c_{1}}{S}y,\frac{\tilde{S}c_{0}}{{S}c_{1}}y\right).$

###### Proof.

It is easily verified that $\mathcal{L}$ is a linear mapping that maps the
points $(c_{0},0),(-c_{0},0),(a_{0},S/c_{0})$ to the images
$(c_{1},0),(-c_{1},0),(a_{1},S/c_{1}).$ Since linear maps preserve line
segments, it follows that $\mathcal{L}$ maps $\Omega_{0}$ to $\Omega_{a,c}.$ ∎

In line with Lemma 1 above, viewing $\Omega_{a_{1},a_{2},c,S_{1}}$ as the
union of the triangles $\Omega_{a_{1},c}$ and $\Omega_{a_{2},c}$ with areas
$S_{1}$ and $S_{2}$ respectively we construct the following map piecewise

$\displaystyle\mathscr{L}_{a_{1},a_{2},c,S_{1}}:\Omega^{+}_{0}\to\Omega^{+}_{a_{1},a_{2},c,S_{1}}:\left\\{(x,y)\mapsto\left(\frac{c}{c_{0}}x+\frac{a_{1}c_{0}}{S}y,\frac{c_{0}S_{1}}{cS}y\right)\right\\},$
$\displaystyle\mathscr{L}_{a_{1},a_{2},c,S_{1}}:\Omega^{-}_{0}\to\Omega^{-}_{a_{1},a_{2},c,S_{1}}:\left\\{(x,y)\mapsto\left(\frac{c}{c_{0}}x-\frac{a_{2}c_{0}}{S}y,\frac{c_{0}S_{2}}{cS}y\right)\right\\},$

with inverse

$\displaystyle\mathscr{L}^{-1}_{a_{1},a_{2},c,S_{1}}:\Omega^{+}_{a_{1},a_{2},c,S_{1}}\to\Omega^{+}_{0}:\left\\{(x,y)\mapsto\left(\frac{c_{0}}{c}x-\frac{a_{1}c_{0}}{S_{1}}y,\frac{cS}{c_{0}S_{1}}y\right)\right\\},$
$\displaystyle\mathscr{L}^{-1}_{a_{1},a_{2},c,S_{1}}:\Omega^{-}_{a_{1},a_{2},c,S_{1}}\to\Omega^{-}_{0}:\left\\{(x,y)\mapsto\left(\frac{c_{0}}{c}x+\frac{a_{2}c_{0}}{S_{2}}y,\frac{cS}{c_{0}S_{2}}y\right)\right\\}.$

Here, for a set $A\subseteq\mathbb{R}^{2},$ we set $A^{\pm}=A\cap\\{\pm y\geq
0\\}.$

We define the maps
$U_{a_{1},a_{2},c,S_{1}}:L^{2}(\Omega_{a_{1},a_{2},c,S_{1}})\to
L^{2}(\Omega_{0})$ and $U_{a_{1},a_{2},c,S_{1}}^{-1}:L^{2}(\Omega_{0})\to
L^{2}(\Omega_{a_{1},a_{2},c,S_{1}})$ via

$\displaystyle U_{a_{1},a_{2},c,S_{1}}(u)=$ $\displaystyle
u\circ\mathscr{L}_{a_{1},a_{2},c,S_{1}},$ $\displaystyle
U_{a_{1},a_{2},c,S_{1}}^{-1}(u)=$ $\displaystyle
u\circ\mathscr{L}_{a_{1},a_{2},c,S_{1}}^{-1}.$

######  Lemma 2.

The maps $U_{a_{1},a_{2},c,S_{1}},$ $U^{-1}_{a_{1},a_{2},c,S_{1}}$ satisfy the
following properties:

  1. (I)

For any $u,v\in L^{2}(\Omega_{a_{1},a_{2},c,S_{1}})$ we have

$\displaystyle\langle U_{a_{1},a_{2},c,S_{1}}\circ
u,U_{a_{1},a_{2},c,S_{1}}\circ
v\rangle_{L^{2}(\Omega_{0})}=\frac{S}{S_{1}}\langle
u,v\rangle_{L^{2}(\Omega_{a_{1},a_{2},c,S_{1}}^{+})}+\frac{S}{S_{2}}\langle
u,v\rangle_{L^{2}(\Omega_{a_{1},a_{2},c,S_{1}}^{-})}.$

Similarly, for any $\phi,\psi\in L^{2}(\Omega_{0})$ we have

$\displaystyle\langle
U_{a_{1},a_{2},c,S_{1}}^{-1}\circ\phi,U_{a_{1},a_{2},c,S_{1}}^{-1}\circ\psi\rangle_{L^{2}(\Omega_{a_{1},a_{2},c,S_{1}})}=\frac{S_{1}}{S}\langle\phi,\psi\rangle_{L^{2}(\Omega_{0}^{+})}+\frac{S_{2}}{S}\langle\phi,\psi\rangle_{L^{2}(\Omega_{0}^{-})}.$

  2. (II)

For $i,j=1,2,$

$\displaystyle\langle
u,v\rangle_{L^{2}(\Gamma^{(i,j)}_{a_{j},c,S_{1}})}=\frac{|\Gamma_{a_{j},c,S_{1}}^{(i,j)}|}{|\Gamma_{0}|}\langle
U_{a_{1},a_{2},c,S_{1}}\circ u,U_{a_{1},a_{2},c,S_{1}}\circ
v\rangle_{L^{2}(\Gamma^{(i,j)}_{0})},$

###### Proof.

  1. (I)

Make a change of variables $\tilde{x}=\mathscr{L}_{a_{1},a_{2},c,S_{1}}(x),$
where the Jacobian is

$\displaystyle
J=\det\left[\begin{matrix}\frac{c_{0}}{c}&{c_{0}}\left(-\frac{a_{1}}{S_{1}}\mbox{\Large$\chi$}_{y\geq
0}+\frac{a_{2}}{S_{2}}\mbox{\Large$\chi$}_{y<0}\right)\\\
0&\frac{cS}{c_{0}\left(S_{1}\mbox{\Large$\chi$}_{y\geq
0}+S_{2}\mbox{\Large$\chi$}_{y<0}\right)}\end{matrix}\right]=\frac{S}{\left(S_{1}\mbox{\Large$\chi$}_{y\geq
0}+S_{2}\mbox{\Large$\chi$}_{y<0}\right)},$

and the proof follows.
The proof for $U_{a_{1},a_{2},c,S_{1}}^{-1}$ is analogous.

  2. (II)

Let
$q_{i}^{j}=(\xi_{i}^{j},\eta_{i}^{j})=\mathscr{L}_{a_{1},a_{2},c,S_{1}}(p_{i}^{j}).$
Let $(x_{i}^{j},y_{i}^{j})=p_{i}^{j}\in\Omega_{a_{1},a_{2},c,S_{1}}$ such that
$\Gamma_{a_{j},c,S_{1}}^{(i,j)}$ is the line segment connecting $p_{i}^{j}$
and $p_{i+1}^{j}.$ Further, let $\psi=U_{a_{1},a_{2},c,S_{1}}\circ u$ and
$\phi=U_{a_{1},a_{2},c,S_{1}}\circ v.$ Then,

$\displaystyle\frac{\langle
u,v\rangle_{L^{2}(\Gamma^{(i,j)}_{a_{j},c,S_{1}})}}{\langle\psi,\phi\rangle_{L^{2}(\Gamma^{(i,j)}_{0})}}=$
$\displaystyle\frac{|\Gamma_{a_{j},c,S_{1}}^{(i,j)}|\int_{0}^{1}u\left(\left[\begin{matrix}x_{i}^{j}\\\
y_{i}^{j}\end{matrix}\right]+t\left[\begin{matrix}x_{i+1}^{j}-x_{i}^{j}\\\
y_{i+1}^{j}-y_{i}^{j}\end{matrix}\right]\right)v\left(\left[\begin{matrix}x_{i}^{j}\\\
y_{i}^{j}\end{matrix}\right]+t\left[\begin{matrix}x_{i+1}^{j}-x_{i}^{j}\\\
y_{i+1}^{j}-y_{i}^{j}\end{matrix}\right]\right)dt}{|\Gamma_{0}|\int_{0}^{1}\psi\left(\left[\begin{matrix}\xi_{i}^{j}\\\
\eta_{i}^{j}\end{matrix}\right]+\tau\left[\begin{matrix}\xi_{i+1}^{j}-\xi_{i}^{j}\\\
\eta_{i+1}^{j}-\eta_{i}^{j}\end{matrix}\right]\right)\phi\left(\left[\begin{matrix}\xi_{i}^{j}\\\
\eta_{i}^{j}\end{matrix}\right]+\tau\left[\begin{matrix}\xi_{i+1}^{j}-\xi_{i}^{j}\\\
\eta_{i+1}^{j}-\eta_{i}^{j}\end{matrix}\right]\right)d\tau},$ $\displaystyle=$
$\displaystyle\frac{|\Gamma_{a_{j},c,S_{1}}^{(i,j)}|}{|\Gamma_{0}|},$

where we have used the fact that
$\psi=u\circ\mathscr{L}_{a_{1},a_{2},c,S_{1}},$
$\phi=v\circ\mathscr{L}_{a_{1},a_{2},c,S_{1}},$ and
$\mathscr{L}_{a_{1},a_{2},c,S_{1}}$ is linear, so

$\displaystyle\mathscr{L}_{a_{1},a_{2},c,S_{1}}\left(\left[\begin{matrix}\xi_{i}^{j}\\\
\eta_{i}^{j}\end{matrix}\right]+\tau\left[\begin{matrix}\xi_{i+1}^{j}-\xi_{i}^{j}\\\
\eta_{i+1}^{j}-\eta_{i}^{j}\end{matrix}\right]\right)=$
$\displaystyle\mathscr{L}_{a_{1},a_{2},c,S_{1}}\left(\left[\begin{matrix}\xi_{i}^{j}\\\
\eta_{i}^{j}\end{matrix}\right]\right)+\tau\mathscr{L}_{a_{1},a_{2},c,S_{1}}\left(\left[\begin{matrix}\xi_{i+1}^{j}-\xi_{i}^{j}\\\
\eta_{i+1}^{j}-\eta_{i}^{j}\end{matrix}\right]\right),$ $\displaystyle=$
$\displaystyle\left[\begin{matrix}x_{i}^{j}\\\
y_{i}^{j}\end{matrix}\right]+\tau\left[\begin{matrix}x_{i+1}^{j}-x_{i}^{j}\\\
y_{i+1}^{j}-y_{i}^{j}\end{matrix}\right].$

∎

###  2.2 Functional Setup

We view the Robin problem (1a-1b) on the quadrilateral
$\Omega_{a_{1},a_{2},c,S_{1}}$ as the spectral problem in the Hilbert space
$L^{2}(\Omega)$ for

$\displaystyle H_{\alpha}u=-\Delta u,\quad D(H_{\alpha})=\\{u\in
H^{1}(\Omega)\ :\ \frac{\partial u}{\partial n}+\alpha u=0\ \text{ on }\
\partial\Omega\\},$ (3)

where the derivative is viewed as a distribution where necessary, and the
associated quadratic form is

$\displaystyle h_{\alpha}[u]=\int_{\Omega}|\nabla
u|^{2}+\alpha\int_{\partial\Omega}|u|^{2},\quad D(h_{\alpha})=H^{1}(\Omega).$

We define the operator $\hat{H}_{\alpha,a_{1},a_{2},c,S_{1}}$ via

$\displaystyle\hat{H}_{\alpha,a_{1},a_{2},c,S_{1}}=U_{a_{1},a_{2},c,S_{1}}H_{\alpha}U_{a_{1},a_{2},c,S_{1}}^{-1},$

where
$D(\hat{H}_{\alpha,a_{1},a_{2},c,S_{1}})=U_{a_{1},a_{2},c,S_{1}}(D(H_{\alpha})).$
Then the operators $H_{\alpha}$ and $\hat{H}_{\alpha,a_{1},a_{2},c,S_{1}}$ are
isospectral. One observes that $\hat{H}_{\alpha,a_{1},a_{2},c}$ is the
operator on $L^{2}(\Omega_{0})$ associated with the quadratic form

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\psi]=h_{\alpha}\left[\sqrt{\frac{S}{S_{1}}\mbox{\Large$\chi$}_{y>0}+\frac{S}{S_{2}}\mbox{\Large$\chi$}_{y<0}}U^{-1}_{a_{1},a_{2},c,S_{1}}\psi\right].$

This associated form can be written

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\psi,\varphi]=$
$\displaystyle\int_{\Omega_{0}}\left(\left({\partial_{1}\mathscr{L}_{a_{1},a_{2},c,S_{1}}^{-1}}\cdot\nabla\right)\psi\right)\left(\left({\partial_{1}\mathscr{L}_{a_{1},a_{2},c,S_{1}}^{-1}}\cdot\nabla\right)\varphi\right)$
$\displaystyle+$
$\displaystyle\left(\left({\partial_{2}\mathscr{L}_{a_{1},a_{2},c,S_{1}}^{-1}}\cdot\nabla\right)\psi\right)\left(\left({\partial_{2}\mathscr{L}_{a_{1},a_{2},c,S_{1}}^{-1}}\cdot\nabla\right)\varphi\right)d{x}d{y}$
$\displaystyle+$
$\displaystyle\alpha\sum_{i,j}\frac{|\Gamma_{a_{j},c,S_{1}}^{(i,j)}|S}{|\Gamma_{0}|S_{j}}\langle\psi,\varphi\rangle_{L^{2}(\Gamma^{(i,j)}_{0})},$
(4)

or more explicitly,

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\psi]=$
$\displaystyle\frac{c_{0}^{2}}{c^{2}}||\partial_{1}\psi||^{2}_{L^{2}(\Omega_{0})}$
$\displaystyle+$
$\displaystyle\left|\left|\left(\mbox{\Large$\chi$}_{y<0}\frac{a_{2}}{S_{2}}-\mbox{\Large$\chi$}_{y>0}\frac{a_{1}}{S_{1}}\right)c_{0}\partial_{1}\psi+\frac{cS}{c_{0}\left(S_{1}\mbox{\Large$\chi$}_{y>0}+S_{2}\mbox{\Large$\chi$}_{y<0}\right)}\partial_{2}\psi\right|\right|^{2}_{L^{2}(\Omega_{0})}$
$\displaystyle+\frac{\alpha
S}{\sqrt{2}cc_{0}}\sum_{i,j}\frac{\sqrt{S_{j}^{2}+c^{2}(a_{j}+(-1)^{i+1}c)^{2}}}{S_{j}}||\psi||^{2}_{L^{2}(\Gamma^{(i,j)}_{0})},$
(5)

where we write $\partial_{1}$ to mean $\frac{\partial}{\partial x}$ and
$\partial_{2}$ to mean $\frac{\partial}{\partial y}.$
In all that follows, let $\psi_{a_{1},a_{2},c,S_{1}}$ be the positive first
eigenfunction of $\hat{H}_{\alpha,a_{1},a_{2},c,S_{1}}$ corresponding to
$\lambda_{a_{1},a_{2},c,S_{1}},$ scaled so that
$||\psi_{a_{1},a_{2},c,S_{1}}||_{L^{2}(\Omega_{0})}=1.$ Then the eigenvalue
problem
$\hat{H}_{\alpha,a_{1},a_{2},c,S_{1}}\psi_{a_{1},a_{2},c}=\lambda_{a_{1},a_{2},c,S_{1}}\psi_{a_{1},a_{2},c,S_{1}}$
is equivalent to the weak problem

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\phi,\psi_{a_{1},a_{2},c,S_{1}}]=\lambda_{a_{1},a_{2},c,S_{1}}\langle\phi,\psi_{a_{1},a_{2},c,S_{1}}\rangle,\quad\forall\phi\in
H^{1}(\Omega_{0}).$ (6)

Further, both $\lambda_{a_{1},a_{2},c,S_{1}}$ and $\psi_{a_{1},a_{2},c,S_{1}}$
are real analytic functions of the parameters $a_{1},a_{2},c,S_{1}$ for
$a_{1},a_{2}\in\mathbb{R},$ $c\in(0,\infty)$ and $S_{1}\in(0,2S).$ This may be
observed by noting that $\hat{H}_{\alpha,a_{1},a_{2},c,S_{1}}$ forms a
holomorphic family of type (B) in the sense of Kato (see [12, Theorem 4.8
section VII]).
Note that for the remainder of the paper we will use $\lambda_{0}$ and
$\psi_{0}$ to denote $\lambda_{a_{1},a_{2},c,S_{1}}$ and
$\psi_{a_{1},a_{2},c,S_{1}}$ when $(a_{1},a_{2},c,S_{1})=(0,0,c_{0},S).$

###  2.3 The Solution on the Square

The solution of the problem $(\ref{opp})$ is known explicitly for rectangles
(see for instance [15]). In particular, the explicit solution on the square
and its properties will be of use in the following sections, and so for
completeness we state its solution below, as well as some of its properties
which will be of use later. We will frame the problem in terms of the rotated
square $\Omega_{0}$ of area $2S,$ for $S>0,$ with vertices $(\pm\sqrt{S},0)$
and $(0,\pm\sqrt{S}).$

$(0,\sqrt{S})$$(\sqrt{S},0)$$(0,-\sqrt{S})$$(-\sqrt{S},0)$ Figure 2: The
Rotated Square $\Omega_{0}$

######  Proposition 1.

Consider the eigenvalue problem

$\displaystyle\begin{cases}\Delta u=-\lambda u,\quad\text{in }\Omega_{0},\\\
\frac{\partial u}{\partial\nu}+\alpha u=0\quad\text{on
}\partial\Omega_{0}.\end{cases}$ (7)

Then the eigenvalue problem (7) has first eigenfunction $u_{1}$ given by

$\displaystyle\begin{cases}u_{1}=\cos\left({\frac{\sqrt{\lambda_{1}}}{2}}(x+y)\right)\cos\left({\frac{\sqrt{\lambda_{1}}}{2}}(y-x)\right),\quad\alpha>0,\\\
u_{1}=\cosh\left({\frac{\sqrt{-\lambda_{1}}}{2}}(x+y)\right)\cosh\left({\frac{\sqrt{-\lambda_{1}}}{2}}(y-x)\right),\quad\alpha<0,\end{cases}$
(8)

with corresponding first eigenvalues $\lambda_{1}$ given by

$\displaystyle\lambda_{1}=\begin{cases}2\left(\frac{f^{-1}(\alpha
L)}{L}\right)^{2},\quad\alpha>0,\\\ -2\left(\frac{g^{-1}(-\alpha
L)}{L}\right)^{2},\quad\alpha<0,\end{cases}$

where

$\displaystyle f(t)=$ $\displaystyle t\tan(t),\quad
t\in(0,\pi/2)\cup(\pi/2,\pi),$ $\displaystyle g(t)=$ $\displaystyle
t\tanh(t),\quad t\geq 0,$

and $L=\sqrt{S/2}.$

######  Corollary 1.

Let $\alpha\neq 0,$ then the function $u_{1}$ as given by Proposition 1 has
the following properties

  1. (I)

$u_{1}$ is even in $x$

  2. (II)

$u_{1}$ is even in $y$

  3. (III)

$||u_{1}||_{L^{2}(\Gamma_{0}^{i,j})}$ is independent of $i,j$

  4. (IV)

$||\partial_{1}u_{1}||_{L^{2}(\Omega_{0})}^{2}$=$||\partial_{2}u_{1}||_{L^{2}(\Omega_{0})}^{2}=\frac{1}{2}||\nabla
u_{1}||_{L^{2}(\Omega_{0})}^{2}$

  5. (V)

$\langle\partial_{1}u_{1},\partial_{2}u_{1}\rangle_{L^{2}(\Omega_{0}^{+})}=0$

  6. (VI)

$\langle\partial_{1}u_{1},\partial_{2}u_{1}\rangle_{L^{2}(\Omega_{0}^{-})}=0$

  7. (VII)

$||\partial_{1}u_{1}||^{2}_{L^{2}(\Omega_{0}^{+})}=||\partial_{1}u_{1}||^{2}_{L^{2}(\Omega_{0}^{-})}=\frac{1}{2}||\partial_{1}u_{1}||^{2}_{L^{2}(\Omega_{0})}$

  8. (VIII)

$||\partial_{2}u_{1}||^{2}_{L^{2}(\Omega_{0}^{+})}=||\partial_{2}u_{1}||^{2}_{L^{2}(\Omega_{0}^{-})}=\frac{1}{2}||\partial_{2}u_{1}||^{2}_{L^{2}(\Omega_{0})}.$

###### Proof.

Follows from checking the explicit form (8). ∎

In the following lemmas we provide a useful result that will be of use in the
following section.

######  Lemma 3.

Let

$\displaystyle\zeta(\alpha)=||\nabla\psi_{0}||_{L^{2}(\Omega_{0})}^{2}+\alpha
C||\psi_{0}||_{L^{2}(\partial\Omega_{0})}^{2},$

then $\zeta(\alpha)<0$ for all $\alpha<0,$ provided that $C\in(0,1).$

###### Proof.

Since

$\displaystyle\lambda_{0}=||\nabla\psi_{0}||_{L^{2}(\Omega_{0})}^{2}+\alpha||\psi_{0}||_{L^{2}(\partial\Omega_{0})}^{2},$

it follows that

$\displaystyle\zeta(\alpha)=\lambda_{0}+\alpha(C-1)||\psi_{0}||_{L^{2}(\partial\Omega_{0})}^{2}.$

Let $(\lambda_{0})^{\prime}_{\alpha}$ denote
$\frac{\partial\lambda_{0}}{\partial\alpha}.$ Then

$\displaystyle(\lambda_{0})^{\prime}_{\alpha}=||\psi_{0}||_{L^{2}(\partial\Omega_{0})}^{2},$

(see for instance [10, section 4.3.2] ) and so

$\displaystyle\zeta(\alpha)=\lambda_{0}+\alpha(C-1)(\lambda_{0})^{\prime}_{\alpha}.$

Thus

$\displaystyle\zeta(\alpha)<0\iff\frac{(\lambda_{0})^{\prime}_{\alpha}}{\lambda_{0}}>\frac{1}{\alpha(1-C)}.$

Recall from Proposition 1

$\displaystyle g\left(L\sqrt{\frac{-\lambda_{0}}{2}}\right)=-\alpha L,$

where $g(t)=t\tanh(t).$
Then

$\displaystyle
g^{\prime}\left(L\sqrt{\frac{-\lambda_{0}}{2}}\right)\frac{1}{\sqrt{\frac{-\lambda_{0}}{2}}}\frac{-L}{2}(\lambda_{0})^{\prime}_{\alpha}=-L\implies(\lambda_{0})^{\prime}_{\alpha}=\frac{-L\lambda_{0}}{L\sqrt{\frac{-\lambda_{0}}{2}}g^{\prime}\left(L\sqrt{\frac{-\lambda_{0}}{2}}\right)}.$

Now,

$\displaystyle\frac{(\lambda_{0})^{\prime}_{\alpha}}{\lambda_{0}}=\frac{-L}{\left(L\sqrt{\frac{-\lambda_{0}}{2}}\right)\tanh\left(L\sqrt{\frac{-\lambda_{0}}{2}}\right)+\left(L\sqrt{\frac{-\lambda_{0}}{2}}\right)^{2}\text{sech}^{2}\left(L\sqrt{\frac{-\lambda_{0}}{2}}\right)},$

where we have used

$\displaystyle g^{\prime}(t)=\tanh(t)+t\text{sech}^{2}(t).$

Thus,

$\displaystyle\zeta(\alpha)<0,$ $\displaystyle\iff$
$\displaystyle\frac{-L}{\left(L\sqrt{\frac{-\lambda_{0}}{2}}\right)\tanh\left(L\sqrt{\frac{-\lambda_{0}}{2}}\right)+\left(L\sqrt{\frac{-\lambda_{0}}{2}}\right)^{2}\text{sech}^{2}\left(L\sqrt{\frac{-\lambda_{0}}{2}}\right)}>\frac{1}{\alpha(1-C)},$
$\displaystyle\iff$
$\displaystyle\frac{\left(L\sqrt{\frac{-\lambda_{0}}{2}}\right)\tanh\left(\sqrt{L\frac{-\lambda_{0}}{2}}\right)(1-C)}{\left(L\sqrt{\frac{-\lambda_{0}}{2}}\right)\tanh\left(L\sqrt{\frac{-\lambda_{0}}{2}}\right)+\left(L\sqrt{\frac{-\lambda_{0}}{2}}\right)^{2}\text{sech}^{2}\left(L\sqrt{\frac{-\lambda_{0}}{2}}\right)}<1.$

This is always satisfied, as the lefthand side is always less than $1.$ ∎

##  3 A Local Optimisation Result

The goal of this section is to show that there is a local maxima
$\lambda_{a_{1},a_{2},c,S_{1}}$ at
$(a_{1},a_{2},c,S_{1})=(a_{0},a_{0},c_{0},S).$ We use the expression (6) for
the weak form $\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}},$ to obtain expressions
for $\lambda_{a_{1},a_{2},c,S_{1}}$ and its derivatives. These derivatives
allow us to find conditions on the Hessian matrix for
$\lambda_{a_{1},a_{2},c,S_{1}}$ corresponding to a local maximum at
$(a_{1},a_{2},c,S_{1})=(a_{0},a_{0},c_{0},S).$

###  3.1 First Derivatives

We introduce the following notation; let $v$ be any of the perturbation
parameters $a_{1},a_{2},c$ or $S_{1}$, and let $f^{v}$ denote $\frac{\partial
f}{\partial v}.$ Further define as follows

$\displaystyle\bar{h}_{\alpha,a_{1},a_{2},c,S_{1}}^{d_{v}}[f,g]=$
$\displaystyle\sum_{i=1}^{2}\left[\left\langle\left(\frac{\partial}{\partial
v}\left({\partial_{i}\mathscr{L}_{a_{1},a_{2},c,S_{1}}^{-1}}\right)\cdot\nabla\right)f,\left({\partial_{i}\mathscr{L}_{a_{1},a_{2},c,S_{1}}^{-1}}\cdot\nabla\right)g\right\rangle_{L^{2}(\Omega_{0})}\right.$
$\displaystyle+\left.\left\langle\left({\partial_{i}\mathscr{L}_{a_{1},a_{2},c,S_{1}}^{-1}}\cdot\nabla\right)f,\left(\frac{\partial}{\partial
v}\left({\partial_{i}\mathscr{L}_{a_{1},a_{2},c,S_{1}}^{-1}}\right)\cdot\nabla\right)g\right\rangle_{L^{2}(\Omega_{0})}\right],$
$\displaystyle\tilde{h}_{\alpha,a_{1},a_{2},c,S_{1}}^{d_{v}}[f,g]=$
$\displaystyle\frac{S}{|\Gamma_{0}|}\sum_{i,j}\frac{\partial}{\partial
v}\left(\frac{|\Gamma_{a_{j},c,S_{1}}^{(i,j)}}{S_{j}}|\right)\langle
f,g\rangle_{L^{2}(\Gamma^{(i,j)}_{0})},$
$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}^{d_{v}}[f,g]=$
$\displaystyle\bar{h}_{\alpha,a_{1},a_{2},c,S_{1}}^{d_{v}}[f,g]+\alpha\tilde{h}_{\alpha,a_{1},a_{2},c,S_{1}}^{d_{v}}[f,g].$
(9)

Then it follows from (4) that

$\displaystyle\frac{\partial}{\partial
v}\left(\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\phi,\psi_{a_{1},a_{2},c,S}]\right)=\hat{h}^{d_{v}}_{\alpha,a_{1},a_{2},c,S_{1}}[\phi,\psi_{a_{1},a_{2},c,S_{1}}]+\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\phi,\psi^{v}_{a_{1},a_{2},c,S_{1}}].$

Then, observing from (6) that

$\displaystyle\frac{\partial}{\partial
v}\left(\lambda_{a_{1},a_{2},c,S_{1}}\langle\phi,\psi_{a_{1},a_{2},c,S_{1}}\rangle_{L^{2}(\Omega_{0})}\right)=\frac{\partial}{\partial
v}\left(\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\phi,\psi_{a_{1},a_{2},c,S_{1}}]\right),$

it then follows that

$\displaystyle\lambda^{v}_{a_{1},a_{2},c,S_{1}}\langle\phi,\psi_{a_{1},a_{2},c,S_{1}}\rangle_{L^{2}(\Omega_{0})}+\lambda_{a_{1},a_{2},c,S_{1}}\langle\phi,\psi^{v}_{a_{1},a_{2},c,S_{1}}\rangle_{L^{2}(\Omega_{0})}$
$\displaystyle=\hat{h}^{d_{v}}_{\alpha,a_{1},a_{2},c,S_{1}}[\phi,\psi_{a_{1},a_{2},c,S_{1}}]+\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\phi,\psi^{v}_{a_{1},a_{2},c,S_{1}}],$
(10)

and so with $\phi=\psi_{a_{1},a_{2},c,S_{1}},$ we readily obtain

$\displaystyle{\lambda}^{v}_{a_{1},a_{2},c,S_{1}}=\hat{h}^{d_{v}}_{a_{1},a_{2},c,S_{1}}[\psi_{a_{1},a_{2},c,S_{1}}].$
(11)

######  Lemma 4.

$\lambda_{0}^{v}=0$ for $v=a_{1},a_{2},c,S_{1}.$

###### Proof.

Step 1. When $(a_{1},a_{2},c,S_{1})=(0,0,c_{0},S)$ the formulae (3.1) simplify
as follows (see Lemma 8 in the appendix for intermediary details)

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c}^{d_{a_{1}}}[\psi_{0}]=$
$\displaystyle-\frac{2c_{0}}{S}\langle\partial_{1}\psi_{0},\partial_{2}\psi_{0}\rangle_{L^{2}(\Omega_{0}^{+})}$
$\displaystyle+$
$\displaystyle\alpha\frac{c_{0}}{|\Gamma_{0}|^{2}}\left(||\psi_{0}||^{2}_{L^{2}(\Gamma_{0}^{(1,1)})}-||\psi_{0}||^{2}_{L^{2}(\Gamma_{0}^{(2,1)})}\right)$
$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c}^{d_{a_{2}}}[\psi_{0}]=$
$\displaystyle
2\frac{c_{0}}{S}\langle\partial_{1}\psi_{0},\partial_{2}\psi_{0}\rangle_{L^{2}(\Omega_{0}^{-})}$
$\displaystyle+$
$\displaystyle\alpha\frac{c_{0}}{|\Gamma_{0}|^{2}}\left(||\psi_{0}||^{2}_{L^{2}(\Gamma_{0}^{(1,2)})}-||\psi_{0}||^{2}_{L^{2}(\Gamma_{0}^{(2,2)})}\right)$
$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c}^{d_{c}}[\psi_{0}]=$
$\displaystyle\frac{2}{c_{0}}\left(||\partial_{2}\psi_{0}||^{2}_{L^{2}(\Omega_{0})}-||\partial_{1}\psi_{0}||_{L^{2}(\Omega_{0})}\right)$
$\displaystyle+$
$\displaystyle\alpha\frac{-\frac{S^{2}}{c_{0}^{3}}+c_{0}}{|\Gamma_{0}|^{2}}||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}$
$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c}^{d_{S_{1}}}[\psi_{0}]=$
$\displaystyle\frac{2}{S}\left(||\partial_{2}\psi_{0}||^{2}_{L^{2}(\Omega_{0}^{-})}-||\partial_{2}\psi_{0}||^{2}_{L^{2}(\Omega_{0}^{+})}\right)$
$\displaystyle+$
$\displaystyle\frac{\alpha}{|\Gamma_{0}|}\left(\frac{S}{|\Gamma_{0}|c_{0}^{2}}-\frac{|\Gamma_{0}|}{S}\right)\left(||\psi_{0}||^{2}_{L^{2}(\Gamma_{0})}+||\psi_{0}||^{2}_{L^{2}(\Gamma_{0})}\right.$
$\displaystyle-$
$\displaystyle\left.||\psi_{0}||^{2}_{L^{2}(\Gamma_{0}^{(1,2)})}-||\psi_{0}||^{2}_{L^{2}(\Gamma_{0}^{(2,2)})}\right).$

Step 2. Recall from Corollary 1 the following symmetry properties of
$\psi_{0}:$
$\langle\partial_{1}\psi_{0},\partial_{2}\psi_{0}\rangle_{L^{2}(\Omega_{0}^{+})}=0,$
$\langle\partial_{1}\psi_{0},\partial_{2}\psi_{0}\rangle_{L^{2}(\Omega_{0}^{-})}=0,$
$||\psi_{0}||_{L^{2}(\Gamma_{0}^{i,1})}=||\psi_{0}||_{L^{2}(\Gamma_{0}^{j,1})},$
$||\psi_{0}||_{L^{2}(\Gamma_{0}^{1,i})}=||\psi_{0}||_{L^{2}(\Gamma_{0}^{1,j})},$
$||\partial_{1}\psi_{0}||_{L^{2}(\Omega_{0})}=||\partial_{2}\psi_{0}||_{L^{2}(\Omega_{0})},$
$||\partial_{2}\psi_{0}||_{L^{2}(\Omega_{0}^{+})}=||\partial_{2}\psi_{0}||_{L^{2}(\Omega_{0}^{-})}.$
Step 3. Applying the symmetry properties from step 2 to the formulae in step
1, recalling that $S^{2}/c_{0}^{3}=c_{0}$ and applying equation (11) leads to
the desired result. ∎

###  3.2 Second Derivatives

We introduce the following notation. Let

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}^{d_{v_{1}},d_{v_{2}}}[\psi_{a_{1},a_{2},c,S_{1}},\psi_{a_{1},a_{2},c,S_{1}}]=$
$\displaystyle\bar{h}_{\alpha,a_{1},a_{2},c,S_{1}}^{d_{v_{1}},d_{v_{2}}}[\psi_{a_{1},a_{2},c,S_{1}},\psi_{a_{1},a_{2},c,S_{1}}]$
$\displaystyle+$
$\displaystyle\tilde{h}_{\alpha,a_{1},a_{2},c,S_{1}}^{d_{v_{1}},d_{v_{2}}}[\psi_{a_{1},a_{2},c,S_{1}},\psi_{a_{1},a_{2},c,S_{1}}],$
(12)

where

$\displaystyle\bar{h}^{d_{v_{1}}d_{v_{2}}}_{\alpha,a_{1},a_{2},c,S_{1}}[f,g]=$
$\displaystyle\sum_{i=1}^{2}\left[\left\langle\left(\frac{\partial^{2}}{\partial
v_{1}\partial
v_{2}}\left({\partial_{i}\mathscr{L}_{a_{1},a_{2},c,S_{1}}^{-1}}\right)\cdot\nabla\right)f,\left({\partial_{i}\mathscr{L}_{a_{1},a_{2},c,S_{1}}^{-1}}\cdot\nabla\right)g\right\rangle_{L^{2}(\Omega_{0})}\right.$
$\displaystyle+$
$\displaystyle\left.\left\langle\left(\frac{\partial}{\partial
v_{1}}\left({\partial_{i}\mathscr{L}_{a_{1},a_{2},c,S_{1}}^{-1}}\right)\cdot\nabla\right)f,\left(\frac{\partial}{\partial
v_{2}}\left({\partial_{i}\mathscr{L}_{a_{1},a_{2},c,S_{1}}^{-1}}\right)\cdot\nabla\right)g\right\rangle_{L^{2}(\Omega_{0})}\right.$
$\displaystyle+$
$\displaystyle\left.\left\langle\left(\frac{\partial}{\partial
v_{2}}\left({\partial_{i}\mathscr{L}_{a_{1},a_{2},c,S_{1}}^{-1}}\right)\cdot\nabla\right)f,\left(\frac{\partial}{\partial
v_{1}}\left({\partial_{i}\mathscr{L}_{a_{1},a_{2},c,S_{1}}^{-1}}\right)\cdot\nabla\right)g\right\rangle_{L^{2}(\Omega_{0})}\right.$
$\displaystyle+$
$\displaystyle\left.\left\langle\left({\partial_{i}\mathscr{L}_{a_{1},a_{2},c,S_{1}}^{-1}}\cdot\nabla\right)f,\left(\frac{\partial^{2}}{\partial
v_{1}\partial
v_{2}}\left({\partial_{i}\mathscr{L}_{a_{1},a_{2},c,S_{1}}^{-1}}\right)\cdot\nabla\right)g\right\rangle_{L^{2}(\Omega_{0})}\right],$
(13)

and

$\displaystyle\tilde{h}^{d_{v_{1}},d_{v_{2}}}_{\alpha,a_{1},a_{2},c,S_{1}}[f,g]=\frac{S}{|\Gamma_{0}|}\sum_{i,j}\frac{\partial^{2}}{\partial
v_{1}\partial
v_{2}}\left(\frac{|\Gamma_{a_{j},c,S_{1}}^{(i,j)}|}{S_{j}}\right)\langle
f,g\rangle_{L^{2}(\Gamma^{(i)}_{0})}.$ (14)

Then, differentiating (11), we have

$\displaystyle{\lambda}^{v_{1},v_{2}}_{a_{1},a_{2},c,S_{1}}=2\hat{h}_{a_{1},a_{2},c,S_{1}}^{d_{v_{1}}}[\psi_{a_{1},a_{2},c,S_{1}},\psi_{a_{1},a_{2},c,S_{1}}^{v_{2}}]+\hat{h}_{a_{1},a_{2},c,S_{1}}^{d_{v_{1}},d_{v_{2}}}[\psi_{a_{1},a_{2},c,S_{1}},\psi_{a_{1},a_{2},c,S_{1}}].$
(15)

######  Lemma 5.

The following are true:

  1. (I)

$\displaystyle{\lambda}^{v_{1},v_{2}}_{a_{1},a_{2},c,S_{1}}=$ $\displaystyle
2(\lambda_{a_{1},a_{2},c,S_{1}}^{v_{1}}\langle\psi_{a_{1},a_{2},c,S_{1}},\psi^{v_{2}}_{a_{1},a_{2},c,S_{1}}\rangle+\lambda_{a_{1},a_{2},c,S_{1}}\langle\psi_{a_{1},a_{2},c,S_{1}}^{v_{1}},\psi^{v_{2}}_{a_{1},a_{2},c,S_{1}}\rangle$
$\displaystyle-$
$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\psi^{v_{1}}_{a_{1},a_{2},c,S_{1}},\psi^{v_{2}}_{a_{1},a_{2},c,S_{1}}])+\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}^{d_{v_{1}},d_{v_{2}}}[\psi_{a_{1},a_{2},c,S_{1}},\psi_{a_{1},a_{2},c,S_{1}}],$

  2. (II)

$\displaystyle{\lambda}^{v_{1},v_{2}}_{0}=$ $\displaystyle
2(\lambda_{0}\langle\psi_{0}^{v_{1}},\psi^{v_{2}}_{0}\rangle-\hat{h}_{\alpha}[\psi^{v_{1}}_{0},\psi^{v_{2}}_{0}])+\hat{h}_{\alpha}^{d_{v_{1}},d_{v_{2}}}[\psi_{0},\psi_{0}],$

  3. (III)

$\displaystyle{\lambda}^{v,v}_{0}=$ $\displaystyle
2(\lambda_{0}||\psi_{0}^{v}||^{2}_{L^{2}(\Omega_{0})}-\hat{h}_{\alpha}[\psi^{v}_{0}])+\hat{h}_{\alpha}^{d_{v},d_{v}}[\psi_{0}],$

  4. (IV)

If $\hat{h}_{\alpha}^{d_{v},d_{v}}[\psi_{0}]<0,$ then ${\lambda}^{v,v}_{0}<0.$

###### Proof.

  1. (I)

Recall from (3.1) that

$\displaystyle\lambda_{a_{1},a_{2},c,S_{1}}^{v_{1}}\langle\psi_{a_{1},a_{2},c,S_{1}},\phi\rangle+\lambda_{a_{1},a_{2},c,S_{1}}\langle\psi_{a_{1},a_{2},c,S_{1}}^{v_{1}},\phi\rangle$
$\displaystyle=$
$\displaystyle\hat{h}^{d_{v_{1}}}_{\alpha,a_{1},a_{2},c,S_{1}}[\phi,\psi_{a_{1},a_{2},c,S_{1}}]+\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\phi,\psi^{v_{1}}_{a_{1},a_{2},c,S_{1}}],$

and so setting $\phi=\psi^{v_{2}}_{a_{1},a_{2},c_{0},S_{1}}$ we have

$\displaystyle\lambda_{a_{1},a_{2},c,S_{1}}^{v_{1}}\langle\psi_{a_{1},a_{2},c,S_{1}},\psi^{v_{2}}_{a_{1},a_{2},c,S_{1}}\rangle+\lambda_{a_{1},a_{2},c,S_{1}}\langle\psi_{a_{1},a_{2},c,S_{1}}^{v_{1}},\psi^{v_{2}}_{a_{1},a_{2},c}\rangle$
$\displaystyle=$
$\displaystyle\hat{h}^{d_{v_{1}}}_{\alpha,a_{1},a_{2},c,S_{1}}[\psi^{v_{2}}_{a_{1},a_{2},c,S_{1}},\psi_{a_{1},a_{2},c,S_{1}}]+\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\psi^{v_{1}}_{a_{1},a_{2},c,S_{1}},\psi^{v_{2}}_{a_{1},a_{2},c,S_{1}}].$

Hence, using (15) we have

$\displaystyle{\lambda}^{v_{1},v_{2}}_{a_{1},a_{2},c,S_{1}}=$ $\displaystyle
2(\lambda_{a_{1},a_{2},c,S_{1}}^{v_{1}}\langle\psi_{a_{1},a_{2},c,S_{1}},\psi^{v_{2}}_{a_{1},a_{2},c,S_{1}}\rangle+\lambda_{a_{1},a_{2},c,S_{1}}\langle\psi_{a_{1},a_{2},c,S_{1}}^{v_{1}},\psi^{v_{2}}_{a_{1},a_{2},c,S_{1}}\rangle$
$\displaystyle-$
$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\psi^{v_{1}}_{a_{1},a_{2},c,S_{1}},\psi^{v_{2}}_{a_{1},a_{2},c,S_{1}}])+\hat{h}_{\alpha,a_{1},a_{2},c}^{d_{v_{1}},d_{v_{2}}}[\psi_{a_{1},a_{2},c},\psi_{a_{1},a_{2},c}],$

  2. (II)

Follows from recalling that $\lambda_{0}^{v_{1}}=0,$ and applying (I).

  3. (III)

Apply (II) with the choice $v_{1}=v_{2}=v$.

  4. (IV)

Follows from (III) once one recalls, using the variational characterisation
(2), that
$\hat{h}_{\alpha}{[\psi_{0}^{v}]}\geq\lambda_{0}||\psi_{0}^{v}||^{2}_{L^{2}(\Omega_{0})}.$

∎

######  Lemma 6.

Explicitly, we have for $j=1,2$

  1. (I)

$\displaystyle\lambda_{0}^{a_{j},a_{j}}=$ $\displaystyle
2\left(\lambda_{0}||\psi_{0}^{a_{j}}||^{2}_{L^{2}(\Omega_{0})}-\hat{h}_{\alpha}[\psi^{a_{j}}_{0}]\right)+\frac{1}{2S}||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})}+\alpha\frac{1}{8S}||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}$

  2. (II)

$\displaystyle\lambda_{0}^{c,c}=$ $\displaystyle
2\left(\lambda_{0}||\psi_{0}^{c}||^{2}_{L^{2}(\Omega_{0})}-\hat{h}_{\alpha}[\psi^{c}_{0}]\right)+\frac{4}{S}||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})}+\alpha\frac{2}{S}||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}$

  3. (III)

$\displaystyle\lambda_{0}^{S_{1},S_{1}}=$ $\displaystyle
2\left(\lambda_{0}||\psi_{0}^{S_{1}}||^{2}_{L^{2}(\Omega_{0})}-\hat{h}_{\alpha}[\psi^{S_{1}}_{0}]\right)+\frac{3}{S^{2}}||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})}+\frac{5\alpha}{4S^{2}}||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}$

  4. (IV)

$\displaystyle\lambda_{0}^{a_{1},a_{2}}=$
$\displaystyle-2\frac{c_{0}}{S}\left(\langle\partial_{1}\psi_{0},\partial_{2}\psi_{0}^{a_{2}}\rangle_{L^{2}(\Omega_{0}^{+})}+\langle\partial_{1}\psi_{0}^{a_{2}},\partial_{2}\psi_{0}\rangle_{L^{2}(\Omega_{0}^{+})}\right)$
$\displaystyle+$ $\displaystyle 2\frac{\alpha
c_{0}}{|\Gamma_{0}|^{2}}\left(\langle\psi_{0},\psi_{0}^{a_{2}}\rangle_{L^{2}(\Gamma_{0}^{(1,1)})}-\langle\psi_{0},\psi_{0}^{a_{2}}\rangle_{L^{2}(\Gamma_{0}^{(2,1)})}\right)$

  5. (V)

$\displaystyle\lambda_{0}^{a_{j},c}=0$

  6. (VI)

$\displaystyle\lambda_{0}^{S_{1},a_{j}}=0$

  7. (VII)

$\displaystyle\lambda_{0}^{S_{1},c}=0$

###### Proof.

(I-III) Apply Lemma 5 alongside the computations found in Lemma 10 in the
appendix.
(IV) Apply equation (15) alongside the computations found in Lemma 10 in the
appendix.
(V) We have by (15) and Lemma 10

$\displaystyle{\lambda}^{a_{j},c}_{0}=$ $\displaystyle
2\hat{h}_{\alpha}^{d_{a_{j}}}[\psi_{0},\psi_{0}^{c}]+\hat{h}_{\alpha}^{d_{a_{j}},d_{c}}[\psi_{0}],$
$\displaystyle=$
$\displaystyle(-1)^{j}\frac{2c_{0}}{S}\left(\langle\partial_{1}\psi_{0},\partial_{2}\psi_{0}^{c}\rangle_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}+\langle\partial_{1}\psi_{0}^{c},\partial_{2}\psi_{0}\rangle_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}\right)$
$\displaystyle+\frac{2\alpha
c_{0}}{|\Gamma_{0}|^{2}}\left(\langle\psi_{0},\psi_{0}^{c}\rangle_{L^{2}(\Gamma_{0}^{(1,j)})}-\langle\psi_{0},\psi_{0}^{c}\rangle_{L^{2}(\Gamma_{0}^{(2,j)})}\right)$
$\displaystyle=$ $\displaystyle 0,$

where we have made use of the observation that $\psi_{0}^{c}$ is even in $x.$
(VI) We have again by (15) and Lemma 10

$\displaystyle{\lambda}^{a_{j},S_{1}}_{0}=$ $\displaystyle
2\hat{h}_{\alpha}^{d_{a_{j}}}[\psi_{0},\psi_{0}^{S_{1}}]+\hat{h}_{\alpha}^{d_{a_{j}},d_{S_{1}}}[\psi_{0}],$
$\displaystyle=$
$\displaystyle(-1)^{j}\frac{2c_{0}}{S}\left(\langle\partial_{1}\psi_{0},\partial_{2}\psi_{0}^{S_{1}}\rangle_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}+\langle\partial_{1}\psi_{0}^{S_{1}},\partial_{2}\psi_{0}\rangle_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}\right)$
$\displaystyle+\frac{2\alpha
c_{0}}{|\Gamma_{0}|^{2}}\left(\langle\psi_{0},\psi_{0}^{S_{1}}\rangle_{L^{2}(\Gamma_{0}^{(1,j)})}-\langle\psi_{0},\psi_{0}^{S_{1}}\rangle_{L^{2}(\Gamma_{0}^{(2,j)})}\right)$
$\displaystyle=$ $\displaystyle 0,$

where we have used the fact that $\psi_{0}^{S_{1}}$ is even in $x.$
(VII) Using (15) and Lemma 10 we have

$\displaystyle\lambda_{0}^{S_{1},c}=$
$\displaystyle\frac{4}{S}\left(\langle\partial_{2}\psi_{0},\partial_{2}\psi_{0}^{c}\rangle_{L^{2}(\Omega_{0}^{-})}-\langle\partial_{2}\psi_{0},\partial_{2}\psi_{0}^{c}\rangle_{L^{2}(\Omega_{0}^{+})}\right)$
$\displaystyle+2\frac{\alpha
S}{|\Gamma_{0}|}\left(\frac{1}{|\Gamma_{0}|c_{0}^{2}}-\frac{|\Gamma_{0}|}{S^{2}}\right)\left(\langle\psi_{0},\psi_{0}^{c}\rangle_{L^{2}(\Gamma_{0}^{(1,1)})}+\langle\psi_{0},\psi_{0}^{c}\rangle_{L^{2}(\Gamma_{0}^{(2,1)})}\right.$
$\displaystyle\phantom{=}\left.-\langle\psi_{0},\psi_{0}^{c}\rangle_{L^{2}(\Gamma_{0}^{(1,2)})}-\langle\psi_{0},\psi_{0}^{c}\rangle_{L^{2}(\Gamma_{0}^{(2,2)})}\right)$
$\displaystyle=$ $\displaystyle 0,$

since $\psi_{0}^{c}$ is even in $y.$ ∎

###### Proof of Theorem 1.

Recall that $\lambda_{0}$ is a local maximum if it is a critical point and the
Hessian matrix is negative definite.
The Hessian matrix for $\lambda_{0}$ is

$\displaystyle
H_{\lambda_{0}}=\left[\begin{matrix}\lambda_{0}^{a_{1},a_{1}}&\lambda_{0}^{a_{1},a_{2}}&\lambda_{0}^{a_{1},c}&\lambda_{0}^{a_{1},S_{1}}\\\
\lambda_{0}^{a_{2},a_{1}}&\lambda_{0}^{a_{2},a_{2}}&\lambda_{0}^{a_{2},c}&\lambda_{0}^{a_{2},S_{1}}\\\
\lambda_{0}^{c,a_{1}}&\lambda_{0}^{c,a_{2}}&\lambda_{0}^{c,c}&\lambda_{0}^{c,S_{1}}\\\
\lambda_{0}^{S_{1}c,a_{1}}&\lambda_{0}^{S_{1},a_{2}}&\lambda_{0}^{S_{1},c}&\lambda_{0}^{S_{1},S_{1}}\end{matrix}\right].$

From Lemma 6 we have for $j=1,2,$
$\lambda_{0}^{c,a_{j}}=\lambda_{0}^{a_{j},S_{1}}=\lambda_{0}^{c,S_{1}}=0,$ and
thus $H_{\lambda_{0}}$ takes the form of a block diagonal matrix

$\displaystyle
H_{\lambda_{0}}=\left[\begin{matrix}\lambda_{0}^{a_{1},a_{1}}&\lambda_{0}^{a_{1},a_{2}}&0&0\\\
\lambda_{0}^{a_{2},a_{1}}&\lambda_{0}^{a_{2},a_{2}}&0&0\\\
0&0&\lambda_{0}^{c,c}&0\\\
0&0&0&\lambda_{0}^{S_{1},S_{1}}\end{matrix}\right],$

and hence its eigenvalues $\mu_{i}$ are the eigenvalues of the diagonal
blocks. Hence, $\mu_{1}=\lambda_{0}^{c,c}$ and
$\mu_{2}=\lambda_{0}^{S_{1},S_{1}}.$
By Lemma 5 we see that $\lambda_{0}^{v,v}<0$ if
$\hat{h}_{\alpha}^{d_{v},d_{v}}[\psi_{0}]<0.$ But by comparing Lemmas 5 and 6
we see that for $v=a_{1},a_{2},c,S_{1},$ we have
$\hat{h}_{\alpha}^{d_{v},d_{v}}[\psi_{0}]=B(||\nabla\psi_{0}||^{2}_{L_{2}(\Omega_{0})}+\alpha
C||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}),$ for some $0<B,$ $0<C<1$
depending upon $v.$ By Lemma 3 we have
$||\nabla\psi_{0}||^{2}_{L_{2}(\Omega_{0})}+\alpha
C||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}<0,$ and so
$\lambda_{0}^{v,v}<0.$
This implies that we have $\mu_{1},\mu_{2}<0.$
The other eigenvalues are the eigenvalues of the upper left block matrix,
whose trace and determinant give the following for the sum and product of
eigenvalues

$\displaystyle\mu_{3}+\mu_{4}=$
$\displaystyle\lambda_{0}^{a_{1},a_{1}}+\lambda_{0}^{a_{2},a_{2}},$ (16)
$\displaystyle\mu_{3}\mu_{4}=$
$\displaystyle\lambda_{0}^{a_{1},a_{1}}\lambda_{0}^{a_{2},a_{2}}-\left(\lambda_{0}^{a_{1},a_{2}}\right)^{2}.$

Since $\lambda_{0}^{a_{j},a_{j}}<0,$ for $j=1,2$ (16) implies at least one of
$\mu_{3}$ or $\mu_{4}$ is strictly negative. Thus for both $\mu_{3}<0$ and
$\mu_{4}<0$ we need $\mu_{3}\mu_{4}>0.$ But

$\displaystyle\lambda_{0}^{a_{1},a_{1}}\lambda_{0}^{a_{2},a_{2}}-\left(\lambda_{0}^{a_{1},a_{2}}\right)^{2}=$
$\displaystyle\left(\frac{1}{2S}||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})}+\alpha\frac{1}{8S}||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}\right)^{2}$
$\displaystyle-$ $\displaystyle
2\left(\frac{1}{2S}||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})}+\alpha\frac{1}{8S}||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}\right)\left(\mathscr{G}[\psi_{0}^{a_{1}},\psi_{0}^{a_{1}}]+\mathscr{G}[\psi_{0}^{a_{2}},\psi_{0}^{a_{2}}]\right)$
$\displaystyle+$ $\displaystyle
4(\mathscr{G}[\psi_{0}^{a_{1}},\psi_{0}^{a_{1}}]\mathscr{G}[\psi_{0}^{a_{2}},\psi_{0}^{a_{2}}]-\mathscr{G}[\psi_{0}^{a_{1}},\psi_{0}^{a_{2}}]^{2}),$

where $\mathscr{G}$ is given by
$\mathscr{G}[f,g]=\hat{h}_{\alpha}[f,g]-\lambda_{0}\langle
f,g\rangle_{L^{2}(\Omega_{0})}.$ Clearly the first two terms are strictly
positive and the last term can be seen to be non-negative once one observes
that $\mathscr{G}$ is a positive semi-definite Hermitian form, and hence the
Cauchy-Schwarz inequality holds for $\mathscr{G}.$
Since $\lambda_{0}$ is a local maximum in the parameter space
$(a_{1},a_{2},c,S_{1})$ it is also a local maximum with respect to the
Hausdorf metric. ∎

##  4 Quadrilateral Global Results

This section collects various asymptotic results for small and large negative
$\alpha,$ provable via test function arguments. The approach is similar to
that of [14], but adapted to quadrilaterals. In particular these results will
prove Theorems 2 and 3.
Again, throughout this section we make the normalisation
$||\psi_{0}||^{2}_{L^{2}(\Omega_{0})}=1.$

######  Proposition 2.

Let $\Omega_{a_{1},a_{2},c,S_{1}}$ be a quadrilateral other than the square.
Then there exists a negative constant $\alpha_{c},$ depending only on
$a_{1},a_{2},c,S_{1}$ and $S$ such that
$\lambda_{a_{1},a_{2},c,S_{1}}<\lambda_{0}$ for all $\alpha$ satisfying
$\alpha_{c}\leq\alpha<0.$

###### Proof.

First note that

$\displaystyle\lambda_{a_{1},a_{2},c,S_{1}}\leq$
$\displaystyle{\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\psi_{0}]}$
$\displaystyle\implies\lambda_{a_{1},a_{2},c,S_{1}}\leq$
$\displaystyle\lambda_{0}+{\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\psi_{0}]-{h}_{\alpha}[\psi_{0}]},$

and so
${\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\psi_{0}]-{h}_{\alpha}[\psi_{0}]}<0$ is
a sufficient condition for $\lambda_{a_{1},a_{2},c,S_{1}}<\lambda_{0}.$

Next observe using the symmetry properties from Corollary 1 that

$\displaystyle\left|\left|\left(\mbox{\Large$\chi$}_{y<0}\frac{a_{2}}{S_{2}}-\mbox{\Large$\chi$}_{y>0}\frac{a_{1}}{S_{1}}\right)c_{0}\partial_{1}\psi_{0}+\frac{cS}{c_{0}\left(S_{1}\mbox{\Large$\chi$}_{y>0}+S_{2}\mbox{\Large$\chi$}_{y<0}\right)}\partial_{2}\psi_{0}\right|\right|^{2}_{L^{2}(\Omega_{0})}$
$\displaystyle=\frac{a_{1}^{2}c_{0}^{2}}{S_{1}^{2}}||\partial_{1}\psi_{0}||_{L^{2}(\Omega_{0}^{+})}^{2}-\frac{2a_{1}cS}{S_{1}^{2}}\langle\partial_{1}\psi_{0},\partial_{2}\psi_{0}\rangle_{L^{2}(\Omega_{0}^{+})}+\frac{c^{2}S^{2}}{c_{0}^{2}S_{1}^{2}}||\partial_{2}\psi_{0}||_{L^{2}(\Omega_{0}^{+})}^{2}$
$\displaystyle\phantom{a}+\frac{a_{2}^{2}c_{0}^{2}}{S_{2}^{2}}||\partial_{1}\psi_{0}||_{L^{2}(\Omega_{0}^{-})}^{2}+\frac{2a_{2}cS}{S_{2}^{2}}\langle\partial_{1}\psi_{0},\partial_{2}\psi_{0}\rangle_{L^{2}(\Omega_{0}^{-})}+\frac{c^{2}S^{2}}{c_{0}^{2}S_{2}^{2}}||\partial_{2}\psi_{0}||_{L^{2}(\Omega_{0}^{-})}^{2}$
$\displaystyle=\left(\frac{a_{1}^{2}c_{0}^{2}}{4S_{1}^{2}}+\frac{a_{2}^{2}c_{0}^{2}}{4S_{2}^{2}}+\frac{c^{2}S^{2}}{4c_{0}^{2}S_{1}^{2}}+\frac{c^{2}S^{2}}{4c_{0}^{2}S_{2}^{2}}\right)||\nabla\psi_{0}||_{L^{2}(\Omega_{0})}^{2},$

and so we may write equation (2.2) in the form

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\psi_{0}]=$
$\displaystyle\left(\frac{a_{1}^{2}c_{0}^{2}}{4S_{1}^{2}}+\frac{a_{2}^{2}c_{0}^{2}}{4S_{2}^{2}}+\frac{c^{2}S^{2}}{4c_{0}^{2}S_{1}^{2}}+\frac{c^{2}S^{2}}{4c_{0}^{2}S_{2}^{2}}+\frac{c_{0}^{2}}{2c^{2}}\right)||\nabla\psi_{0}||_{L^{2}(\Omega_{0})}^{2}$
$\displaystyle+$ $\displaystyle\frac{\alpha
S}{4\sqrt{2}c_{0}}l(a_{1},a_{2},c,S_{1})||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})},$

where

$\displaystyle
l(a_{1},a_{2},c,S_{1})=\sum_{i,j}\frac{|\Gamma_{a_{j},c,S_{1}}^{(i,j)}|}{S_{j}}$
$\displaystyle=$
$\displaystyle\frac{\sqrt{\frac{S_{1}^{2}}{c^{2}}+(a_{1}+c)^{2}}+\sqrt{\frac{S_{1}^{2}}{c^{2}}+(a_{1}-c)^{2}}}{S_{1}}+\frac{\sqrt{\frac{S_{2}^{2}}{c^{2}}+(a_{2}+c)^{2}}+\sqrt{\frac{S_{2}^{2}}{c^{2}}+(a_{2}-c)^{2}}}{S_{2}}.$

Via applying straightforward optimisation arguments in each variable one finds
that

$\displaystyle l(a_{1},a_{2},c,S_{1})\geq$ $\displaystyle
2\frac{\sqrt{\frac{S_{1}^{2}}{c^{2}}+c^{2}}}{S_{1}}+2\frac{\sqrt{\frac{S_{2}^{2}}{c^{2}}+c^{2}}}{S_{2}}$
(17) $\displaystyle\geq$
$\displaystyle\frac{2\sqrt{2}}{\sqrt{S}_{1}}+\frac{2\sqrt{2}}{\sqrt{S}_{2}}$
(18) $\displaystyle\geq$ $\displaystyle\frac{4\sqrt{2S}}{S},$ (19)

where we have equality only in the case $a_{1}=a_{2}=0,$ $c=c_{0},$ $S_{1}=S.$
Since we have

$\displaystyle\frac{S^{2}}{S_{1}^{2}}+\frac{S^{2}}{S_{2}^{2}}\geq 2,$

we also have

$\displaystyle\frac{c^{2}S^{2}}{4c_{0}^{2}S_{1}^{2}}+\frac{c^{2}S^{2}}{4c_{0}^{2}S_{2}^{2}}+\frac{c_{0}^{2}}{2c^{2}}\geq\frac{c^{2}}{2c_{0}^{2}}+\frac{c_{0}^{2}}{2c^{2}}=\frac{(c^{2}-c_{0}^{2})^{2}+2c^{2}c_{0}^{2}}{2c^{2}c_{0}^{2}}\geq
1.$ (20)

Let

$\displaystyle z(a_{1},a_{2},c,S_{1})=$
$\displaystyle\frac{\frac{Sl(a_{1},a_{2},c,S_{1})}{4\sqrt{2}c_{0}}-1}{\frac{a_{1}^{2}c_{0}^{2}}{4S_{1}^{2}}+\frac{a_{2}^{2}c_{0}^{2}}{4S_{2}^{2}}+\frac{c^{2}S^{2}}{4c_{0}^{2}S_{1}^{2}}+\frac{c^{2}S^{2}}{4c_{0}^{2}S_{2}^{2}}+\frac{c_{0}^{2}}{2c^{2}}-1},$
$\displaystyle g(\alpha)=$
$\displaystyle-\frac{||\nabla\psi_{0}||_{L^{2}(\Omega_{0})}^{2}}{\alpha||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}},$

observing from (19) and (20) that $z(a_{1},a_{2},c,S_{1})\geq 0$ with equality
if and only if $(a_{1},a_{2},c,S_{1})=(0,0,c_{0},S).$
Now, note that

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\psi_{0}]-\hat{h}_{\alpha}[\psi_{0}]<$
$\displaystyle 0$ $\displaystyle\iff 0>$
$\displaystyle\left(\frac{a_{1}^{2}c_{0}^{2}}{4S_{1}^{2}}+\frac{a_{2}^{2}c_{0}^{2}}{4S_{2}^{2}}+\frac{c^{2}S^{2}}{4c_{0}^{2}S_{1}^{2}}+\frac{c^{2}S^{2}}{4c_{0}^{2}S_{2}^{2}}+\frac{c_{0}^{2}}{2c^{2}}-1\right)||\nabla\psi_{0}||_{L^{2}(\Omega_{0})}^{2}$
$\displaystyle+\alpha\left(\frac{Sl(a_{1},a_{2},c,S_{1})}{4\sqrt{2}c_{0}}-1\right)||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}$
$\displaystyle\iff 0<$
$\displaystyle\frac{||\nabla\psi_{0}||_{L^{2}(\Omega_{0})}^{2}}{\alpha||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}}+\frac{\frac{Sl(a_{1},a_{2},c,S_{1})}{4\sqrt{2}c_{0}}-1}{\frac{a_{1}^{2}c_{0}^{2}}{4S_{1}^{2}}+\frac{a_{2}^{2}c_{0}^{2}}{4S_{2}^{2}}+\frac{c^{2}S^{2}}{4c_{0}^{2}S_{1}^{2}}+\frac{c^{2}S^{2}}{4c_{0}^{2}S_{2}^{2}}+\frac{c_{0}^{2}}{2c^{2}}-1}$
$\displaystyle\iff g(\alpha$ $\displaystyle)<z(a_{1},a_{2},c,S_{1}).$

Recall that

$\displaystyle\lambda_{0}=||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})}+\alpha||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})},$

with $\lambda_{0}$ converging to the first Neumann eigenvalue in the limit
$\alpha\to 0.$ Further,

$\displaystyle\frac{d\lambda_{0}}{d\alpha}=||\psi_{0}(\alpha)||^{2}_{L^{2}(\partial\Omega_{0})},$

(see for instance [10, p. 86-89].)
Recall, for the Neumann problem that the first eigenfunction is the constant
$\psi_{0}=|\Omega_{0}|^{-1/2}.$
At $\alpha=0$ we have

$\displaystyle\left.\frac{d\lambda_{0}}{d\alpha}\right|_{\alpha=0}=||\psi_{0}(0)||^{2}_{L^{2}(\partial\Omega_{0})}=\frac{|\partial\Omega_{0}|}{|\Omega_{0}|}.$

Further, by the definition of derivative, we have

$\displaystyle\frac{|\partial\Omega_{0}|}{|\Omega_{0}|}=\left.\frac{d\lambda_{0}}{d\alpha}\right|_{\alpha=0}=\lim\limits_{\alpha\to
0}\frac{\lambda_{0}(\alpha)-\lambda_{0}(0)}{\alpha}=\lim\limits_{\alpha\to
0}\frac{||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})}+\alpha||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}}{\alpha}.$
(21)

Now,

$\displaystyle\lim\limits_{\alpha\to 0}g(\alpha)=$
$\displaystyle-\lim\limits_{\alpha\to
0}\frac{||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})}}{\alpha||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}}$
$\displaystyle=$ $\displaystyle 1-\lim\limits_{\alpha\to
0}\frac{||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})}+\alpha||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}}{\alpha||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}}$
$\displaystyle=$ $\displaystyle 1-\lim\limits_{\alpha\to
0}\left(\frac{1}{||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}}\right)\lim\limits_{\alpha\to
0}\left(\frac{||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})}+\alpha||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}}{\alpha}\right),$
$\displaystyle=$ $\displaystyle
1-\frac{|\Omega_{0}|}{|\partial\Omega_{0}|}\frac{|\partial\Omega_{0}|}{|\Omega_{0}|},$
$\displaystyle=$ $\displaystyle 0,$

where we have used (21).
Now, since $g(\alpha)>0$ for $\alpha<0$ and $g(\alpha)\to 0$ as $\alpha\to 0,$
it follows that since $z(a_{1},a_{2},c,S_{1})>0$ there exists $\alpha_{c}$
such that $0>\alpha>\alpha_{c}$ implies $0<g(\alpha)<z(a_{1},a_{2},c,S_{1}).$
∎

The asymptotic behaviour of the first Robin eigenvalue with large negative
$\alpha$ is known for piecewise smooth domains, see for instance the work of
Levitin and Parnovski [16]. In particular we will make use of the following
result which follows from Theorem 3.6 in [16],

######  Lemma 7.

Let $\Omega_{a_{1},a_{2},c,S_{1}}$ be a quadrilateral with inner angles
$\theta_{1},\theta_{2},\theta_{3},\theta_{4}$ respectively. Let

$\displaystyle C_{i}=\begin{cases}1,\quad&\text{if }\quad\theta_{i}\geq\pi;\\\
\csc^{2}\left(\theta_{i}/2\right),\quad&\text{if
}\quad\theta_{i}\leq\pi.\end{cases}$

Then,

$\displaystyle\lambda_{a_{1},a_{2},c,S_{1}}=-\alpha^{2}\max\limits_{i\in\\{1,2,3,4\\}}C_{i}+o(\alpha^{2}),\quad\alpha\to-\infty.$
(22)

######  Corollary 2.

Let $\Omega_{a_{1},a_{2},c,S_{1}}$ be a quadrilateral other than a rectangle.
Then, there exists $\alpha_{c}<0$ such that for all $\alpha<\alpha_{c},$

$\displaystyle\lambda_{a_{1},a_{2},c,S_{1}}<\lambda_{0}.$

###### Proof.

Observe from Lemma (22) above that

$\displaystyle\lambda_{0}=-2\alpha^{2}+o(\alpha^{2}),\quad\alpha\to-\infty,$

and thus it is clear that the corollary follows if

$\displaystyle-\alpha^{2}\max\limits_{i\in\\{1,2,3,4\\}}C_{i}<-2\alpha^{2},$
(23)

with $\theta_{i}$ and $q_{i}$ given as in the previous lemma. However, since
$\Omega_{a_{1},a_{2},c,S_{1}}$ is not a rectangle, at least one of its inner
angles, $\theta_{1}$ say, satisfies $\theta_{1}<\pi/2,$ and so
$\max\limits_{i\in\\{1,2,3,4\\}}C_{q_{i}}>\csc^{2}(\theta_{1}/2)>2$ showing
(23). ∎

######  Remark.

Note that although the above Corollary does not hold for rectangles, it is
already known via a separation of variables argument that the square is the
maximiser of the first eigenvalue among rectangular domains, see for instance
[15].

###### Proof of Theorem 2.

Proposition 2, Corollary 2 and the above Remark together prove Theorem 2.∎∎

We can also use test function arguments to obtain asymptotic results for
extreme values of the quadrilateral parameters.

######  Proposition 3.

Let $\alpha<0,$ then there exist positive constants $A,c_{1},c_{2},\tilde{S}$
depending only upon $\alpha$ such that any of the following restrictions

  1. (I)

$|a_{1}|>A,\ a_{2}\in\mathbb{R},\ \ c>0\text{ and }0<S_{1}<2S;$

  2. (II)

$|a_{2}|>A,\ a_{1}\in\mathbb{R},\ \ c>0\text{ and }0<S_{1}<2S;$

  3. (III)

$a_{1}\in\mathbb{R},\ a_{2}\in\mathbb{R},\ \ 0<S_{1}<2S\text{ and }c>c_{1};$

  4. (IV)

$a_{1}\in\mathbb{R},\ a_{2}\in\mathbb{R},\ \ 0<S_{1}<2S\text{ and }c<c_{2};$

  5. (V)

$a_{1}\in\mathbb{R},\ a_{2}\in\mathbb{R},\ \ S_{1}<\tilde{S}\text{ and }c>0;$

  6. (VI)

$a_{1}\in\mathbb{R},\ a_{2}\in\mathbb{R},\ \ |2S-S_{1}|<\tilde{S}\text{ and
}c>0;$

imply that $\lambda_{a_{1},a_{2},c,S_{1}}<\lambda_{0}.$

###### Proof.

Consider the trial function $\mathds{1}\in L^{2}(\Omega_{0}).$
Then,

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\mathds{1}]=$
$\displaystyle\alpha\sum_{i,j}\frac{S|\Gamma_{a_{j},c,S_{1}}^{(i,j)}|}{S_{j}|\Gamma_{0}^{(i)}|}||\mathds{1}||^{2}_{L^{2}(\Gamma^{(i)}_{0})}$
$\displaystyle=$
$\displaystyle\alpha\sum_{i,j}\frac{S|\Gamma_{a_{j},c,S_{1}}^{(i,j)}|}{S_{j}},$

and hence,

$\displaystyle\lambda_{a_{1},a_{2},c,S_{1}}\leq\frac{\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}[\mathds{1}]}{||\mathds{1}||^{2}_{L^{2}(\Omega_{0})}}=\frac{\alpha}{2}\sum_{i,j}\frac{|\Gamma_{a_{j},c,S_{1}}^{(i,j)}|}{S_{j}}.$

As in the proof of Proposition (2) we set
$l(a_{1},a_{2},c,S_{1})=\sum_{i,j}\frac{|\Gamma_{a_{j},c,S_{1}}^{(i,j)}|}{S_{j}},$
and then observe that we have

$\displaystyle\lambda_{a_{1},a_{2},c,S_{1}}<\lambda_{0},$

provided that the sufficent condition

$\displaystyle l(a_{1},a_{2},c,S_{1})>\frac{2\lambda_{0}}{\alpha}$

is satisfied.
Observe that for $j=1,2,$

$\displaystyle l(a_{1},a_{2},c,S_{1})\geq$
$\displaystyle\frac{\sqrt{\frac{S_{j}^{2}}{c^{2}}+(a_{j}+c)^{2}}+\sqrt{\frac{S_{j}^{2}}{c^{2}}+(a_{j}-c)^{2}}}{S_{j}}$
$\displaystyle\geq$
$\displaystyle\frac{\sqrt{\frac{S_{j}^{2}}{c^{2}}+(|a_{j}|+c)^{2}}}{S_{j}}$
$\displaystyle\geq$ $\displaystyle\frac{|a_{j}|}{2S}\to\infty\text{ as
}|a_{j}|\to\infty.$

This proves (I) and (II).
Recall from equation (17) in the proof of Proposition 2 that

$\displaystyle l(a_{1},a_{2},c,S_{1})\geq
2\frac{\sqrt{\frac{S_{1}^{2}}{c^{2}}+c^{2}}}{S_{1}}+2\frac{\sqrt{\frac{S_{2}^{2}}{c^{2}}+c^{2}}}{S_{2}},$

and so

$\displaystyle l(a_{1},a_{2},c,S_{1})\geq$ $\displaystyle
2{\sqrt{\frac{1}{c^{2}}+\frac{c^{2}}{S_{1}^{2}}}}$ $\displaystyle\geq$
$\displaystyle 2{\sqrt{\frac{1}{c^{2}}+\frac{c^{2}}{4S^{2}}}}.$

Observing that
$\lim\limits_{c\to\infty}{\sqrt{\frac{1}{c^{2}}+\frac{c^{2}}{4S^{2}}}}=\infty$
and $\lim\limits_{c\to 0}{\sqrt{\frac{1}{c^{2}}+\frac{c^{2}}{4S^{2}}}}=\infty$
proves (III) and (IV).
Next, recall from equation (18) in the proof of Proposition 2 that

$\displaystyle
l(a_{1},a_{2},c,S_{1})\geq\frac{2\sqrt{2}}{\sqrt{S_{1}}}+\frac{2\sqrt{2}}{\sqrt{S_{2}}},$

where $\lim\limits_{S_{1}\to
0}\frac{2\sqrt{2}}{\sqrt{S_{1}}}+\frac{2\sqrt{2}}{\sqrt{S_{2}}}=\infty$ and
$\lim\limits_{S_{1}\to
2S}\frac{2\sqrt{2}}{\sqrt{S_{1}}}+\frac{2\sqrt{2}}{\sqrt{S_{2}}}=\infty$ which
proves (V) and (VI). ∎

###### Proof of Theorem 3.

Given $A,c_{1},c_{2},\tilde{S}$ from Proposition 3, we can find $C$ such that
$d_{H}(\Omega,\Omega_{0})>C$ implies that the conditions ($I$)-($VI$) hold. ∎

##  5 Appendix

This appendix consists primarily of messy computational details omitted in the
preceding sections. In particular, we provide the details needed in the
computation of the first and second derivatives of the first eigenvalue with
respect to the geometric parameters.

######  Lemma 8.

Explicitly, we have

  1. (I)

for $j=1,2,$

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}^{d_{a_{j}}}[f,\phi]=$
$\displaystyle\left\langle(-1)^{j}\frac{c_{0}}{S}\partial_{1}f,(-1)^{j}\frac{a_{j}c_{0}}{S}\partial_{1}\phi+\frac{cS}{c_{0}S_{j}}\partial_{2}\phi\right\rangle_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}$
$\displaystyle+$
$\displaystyle\left\langle(-1)^{j}\frac{c_{0}}{S}\partial_{1}\phi,(-1)^{j}\frac{a_{j}c_{0}}{S}\partial_{1}f+\frac{cS}{c_{0}S_{j}}\partial_{2}f\right\rangle_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}$
$\displaystyle+$ $\displaystyle\frac{\alpha
S}{S_{j}|\Gamma_{0}|}\sum_{i=1}^{2}\frac{a_{j}+(-1)^{i+1}c}{|\Gamma^{(i,j)}_{a_{j},c,S_{1}}|}\langle
f,\phi\rangle_{L^{2}(\Gamma_{0}^{(i,j)})},$

  2. (II)

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}^{d_{c}}[f,\phi]=$
$\displaystyle-2\frac{c_{0}^{2}}{c^{3}}\langle\partial_{1}f,\partial_{1}\phi\rangle_{L^{2}(\Omega_{0})}$
$\displaystyle+$
$\displaystyle\sum_{j=1}^{2}\left\langle\frac{S}{S_{j}c_{0}}\partial_{2}f,(-1)^{j+1}a_{j}\frac{c_{0}}{S_{j}}\partial_{1}\phi+\frac{cS}{c_{0}S_{j}}\partial_{2}\phi\right\rangle_{L^{2}(\Omega_{0}\cap{\\{(-1)^{j}y<0\\}})}$
$\displaystyle+$
$\displaystyle\sum_{j=1}^{2}\left\langle\frac{S}{S_{j}c_{0}}\partial_{2}\phi,(-1)^{j+1}a_{j}\frac{c_{0}}{S_{j}}\partial_{1}f+\frac{cS}{c_{0}S_{j}}\partial_{2}f\right\rangle_{L^{2}(\Omega_{0}\cap{\\{(-1)^{j}y<0\\}})}$
$\displaystyle+$
$\displaystyle\sum_{i,j}\alpha\frac{S\left(-\frac{S^{2}}{c^{3}}+(-1)^{i+1}a_{j}+c\right)}{S_{j}|\Gamma_{0}||\Gamma^{(i,j)}_{a_{j},c,S_{1}}|}\langle
f,\phi\rangle_{L^{2}(\Gamma_{0}^{(1,1)})},$

  3. (III)

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}^{d_{S_{1}}}[f,\phi]=\frac{\alpha
S}{|\Gamma_{0}|}\sum_{i,j}(-1)^{j+1}\left(\frac{1}{|\Gamma^{(i,j)}_{a_{j},c,S_{1}}|c^{2}}-\frac{|\Gamma^{(i,j)}_{a_{j},c,S_{1}}|}{S_{j}^{2}}\right)\langle
f,\phi\rangle_{L^{2}(\Gamma_{0}^{(i,j)})}$ $\displaystyle+$
$\displaystyle\sum_{j=1}^{2}\left\langle\frac{a_{j}c_{0}}{S_{j}^{2}}\partial_{1}f+(-1)^{j}\frac{cS}{S_{j}^{2}c_{0}}\partial_{2}f,(-1)^{j}\frac{a_{j}c_{0}}{S_{j}}\partial_{1}\phi+\frac{cS}{c_{0}S_{j}}\partial_{2}\phi\right\rangle_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})},$
$\displaystyle+$
$\displaystyle\sum_{j=1}^{2}\left\langle\frac{a_{j}c_{0}}{S_{j}^{2}}\partial_{1}\phi+(-1)^{j}\frac{cS}{S_{j}^{2}c_{0}}\partial_{2}\phi,(-1)^{j}\frac{a_{j}c_{0}}{S_{j}}\partial_{1}f+\frac{cS}{c_{0}S_{j}}\partial_{2}f\right\rangle_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}.$

###### Proof.

Via direct computation using the formulae (3.1). ∎

The following lemma contains the computations of the second order terms that
are required for the computation of
${\lambda}^{v_{1},v_{2}}_{a_{1},a_{2},c,S_{1}}$ via (15).

######  Lemma 9.

Explicitly we have

  1. (I)

for $j=1,2,$

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}^{d_{a_{j}},d_{a_{j}}}[f]=$
$\displaystyle
2\frac{c_{0}^{2}}{S_{1}^{2}}||\partial_{1}f||^{2}_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}$
$\displaystyle+\frac{\alpha
S}{S_{j}|\Gamma_{0}|}\sum_{i=1}^{2}\left(\frac{1}{|\Gamma^{(i,j)}_{a_{j},c,S_{1}}|}-\frac{(a_{1}+(-1)^{i+1}c)^{2}}{|\Gamma^{(i,j)}_{a_{j},c,S_{1}}|^{3}}\right)||f||^{2}_{L^{2}(\Gamma_{0}^{(i,j)})},$

  2. (II)

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c}^{d_{c},d_{c}}[f]=$ $\displaystyle
6\frac{c_{0}^{2}}{c^{4}}||\partial_{1}f||^{2}_{L^{2}(\Omega_{0})}+\frac{2}{c_{0}^{2}}\sum_{j=1}^{2}\frac{S^{2}}{S_{j}^{2}}\left|\left|\partial_{2}f\right|\right|^{2}_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}$
$\displaystyle+$ $\displaystyle\frac{\alpha
S}{|\Gamma_{0}|}\sum_{i,j}\frac{1}{S_{j}}\left(\frac{3\frac{S_{j}^{2}}{c^{4}}+1}{|\Gamma_{a_{j},c,S_{1}}^{(i,j)}|}-\frac{\left(-\frac{S_{j}^{2}}{c^{3}}+(-1)^{i+1}a_{j}+c\right)^{2}}{|\Gamma_{a_{j},c,S_{1}}^{(i,j)}|^{3}}\right)||f||^{2}_{L^{2}(\Gamma_{0}^{(1,1)})},$

  3. (III)

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c}^{d_{S_{1}},d_{S_{1}}}[f]=2\sum_{j=1}^{2}\left|\left|\frac{a_{j}c_{0}}{S_{j}^{2}}\partial_{1}f+(-1)^{j}\frac{cS}{c_{0}S_{j}^{2}}\partial_{2}f\right|\right|^{2}_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}$
$\displaystyle+$ $\displaystyle
2\sum_{j=1}^{2}\left\langle(-1)^{j}\frac{a_{j}c_{0}}{S_{j}}\partial_{1}f+\frac{cS}{c_{0}S_{j}}\partial_{2}f,(-1)^{j}\frac{2a_{j}c_{0}}{S_{j}^{3}}\partial_{1}f+\frac{2cS}{c_{0}S_{j}^{3}}\partial_{2}f\right\rangle_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}$
$\displaystyle+\frac{\alpha
S}{|\Gamma_{0}|}\sum_{i,j}\left(-\frac{S_{j}}{|\Gamma^{(i,j)}_{a_{j},c,S_{1}}|^{3}c^{4}}-\frac{1}{c^{2}|\Gamma^{(i,j)}_{a_{j},c,S_{1}}|S_{j}}+\frac{2|\Gamma^{(i,j)}_{a_{1},c,S_{1}}|}{S_{j}^{3}}\right)||f||^{2}_{L^{2}(\Gamma_{0}^{(i,j)})},$

  4. (IV)

for $j=1,2,$

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}^{d_{a_{j}},d_{c}}[f]=(-1)^{j}\frac{2S}{S_{j}^{2}}\langle\partial_{1}f,\partial_{2}f\rangle_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}$
$\displaystyle+$ $\displaystyle\frac{\alpha
S}{|\Gamma_{0}|S_{j}}\sum_{i=1}^{2}\left(\frac{(-1)^{i+1}}{|\Gamma^{(i,j)}_{a_{j},c,S_{1}}|}-\frac{(a_{j}+(-1)^{i+1}c)(-\frac{S_{j}^{2}}{c^{3}}+(-1)^{i+1}a_{j}+c)}{|\Gamma^{(i,j)}_{a_{j},c,S_{1}}|^{3}}\right)||f||_{\Gamma_{0}^{(i,j)}}^{2},$

  5. (V)

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c}^{d_{a_{1}},d_{a_{2}}}[f]=0,$

  6. (VI)

for $j=1,2,$

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c,S_{1}}^{d_{a_{j}},d_{S_{1}}}[f]=4\frac{c_{0}}{S_{j}^{3}}\left\langle\partial_{1}f,(-1)^{j}{a_{j}c_{0}}\partial_{1}f+\frac{cS}{c_{0}}\partial_{2}f\right\rangle_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}$
$\displaystyle+(-1)^{j}\frac{\alpha
S}{|\Gamma_{0}|}\sum_{i=1}^{2}(a_{j}+(-1)^{i+1}c)\left(\frac{1}{S_{j}^{2}|\Gamma^{(i,j)}_{a_{j},c,S_{1}}|}+\frac{1}{|\Gamma_{a_{j},c,S_{1}}^{(i,j)}|^{3}c^{2}}\right)||f||^{2}_{L^{2}(\Gamma_{0}^{(i,j)})},$

  7. (VII)

$\displaystyle\hat{h}^{d_{c},{d_{S_{1}}}}_{\alpha,a_{1},a_{2},c,S_{1}}[f]=\frac{4S}{c_{0}S_{j}^{3}}\sum_{j=1}^{2}\left\langle(-1)^{j}\partial_{2}f,(-1)^{j}{a_{j}c_{0}}\partial_{1}f+\frac{cS}{c_{0}}\partial_{2}f\right\rangle_{L^{2}(\Omega_{0}\cap\\{(-1)y^{j}<0\\})}$
$\displaystyle+$ $\displaystyle\frac{\alpha
S}{|\Gamma_{0}|}\sum_{i,j}(-1)^{j}\left(\frac{\frac{S_{j}^{2}}{c^{3}}+(-1)^{i+1}a_{j}+c}{S_{j}^{2}|\Gamma^{(i,j)}_{a_{j},c,S_{1}}|}+\frac{-\frac{S_{j}^{2}}{c^{3}}+(-1)^{i+1}a_{j}+c}{|\Gamma_{a_{j},c,S_{1}}^{(i,j)}|^{3}c^{2}}\right)||f||^{2}_{L^{2}(\Gamma_{0}^{(i,j)})},$

###### Proof.

Via direct computation, using the formulae (3.2-14). ∎

Next, the preceding lemmas applied in the case $f=\psi_{0}$ and
$(a_{1},a_{2},c,S_{1})=(0,0,c,S)$ takes the simplified form of the following
lemma.

######  Lemma 10.

Explicitly we have

  1. (I)

for $j=1,2,$

    1. (i)

$\displaystyle\hat{h}_{\alpha}^{d_{a_{j}},d_{a_{j}}}[\psi_{0}]=\frac{1}{2S}||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})}+\alpha\frac{1}{8S}||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})},$

    2. (ii)

$\displaystyle\hat{h}_{\alpha}^{d_{c},d_{c}}[\psi_{0}]=\frac{4}{S}||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})}+\alpha\frac{2}{S}||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})},$

    3. (iii)

$\displaystyle\hat{h}_{\alpha}^{{d_{S_{1}}},{d_{S_{1}}}}[\psi_{0}]=\frac{3}{S^{2}}||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})}+\frac{5\alpha}{4S^{2}}||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})},$

    4. (iv)

for $j=1,2,$

$\displaystyle\hat{h}_{\alpha}^{d_{a_{j}},d_{c}}[\psi_{0}]=0,$

    5. (v)

$\displaystyle\hat{h}_{\alpha}^{d_{a_{1}},d_{a_{2}}}[\psi_{0}]=0,$

    6. (vi)

for $j=1,2$

$\displaystyle\hat{h}_{\alpha}^{d_{a_{j}},d_{S_{1}}}[\psi_{0}]=0,$

    7. (vii)

$\displaystyle\hat{h}^{d_{c},{d_{S_{1}}}}_{\alpha}[\psi_{0}]=0,$

  2. (II)

    1. (a)

for $j=1,2,$

$\displaystyle\hat{h}_{\alpha}^{d_{a_{j}}}[\psi_{0},\phi]=$
$\displaystyle(-1)^{j}\frac{c_{0}}{S}\left(\langle\partial_{1}\psi_{0},\partial_{2}\phi\rangle_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}+\langle\partial_{1}\phi,\partial_{2}\psi_{0}\rangle_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}\right)$
$\displaystyle+$
$\displaystyle\alpha\frac{c_{0}}{|\Gamma_{0}|^{2}}\sum_{i=1}^{2}(-1)^{i+1}\langle\psi_{0},\phi\rangle_{L^{2}(\Gamma_{0}^{(i,j)})},$

    2. (b)

$\displaystyle\hat{h}_{\alpha}^{d_{c}}[\psi_{0},\phi]=\frac{2}{c_{0}}\left(\langle\partial_{2}\psi_{0},\partial_{2}\phi\rangle_{L^{2}(\Omega_{0})}-\langle\partial_{1}\psi_{0},\partial_{1}\phi\rangle_{L^{2}(\Omega_{0})}\right),$

    3. (c)

$\displaystyle\hat{h}_{\alpha}^{d_{S_{1}}}[\psi_{0},\phi]=$
$\displaystyle\frac{2}{S}\left(\langle\partial_{2}\psi_{0},\partial_{2}\phi\rangle_{L^{2}(\Omega_{0}\cap\\{y<0\\})}-\langle\partial_{2}\psi_{0},\partial_{2}\phi\rangle_{L^{2}(\Omega_{0}\cap\\{y>0\\})}\right)$
$\displaystyle+$ $\displaystyle\frac{\alpha
S}{|\Gamma_{0}|}\left(\frac{1}{|\Gamma_{0}|c_{0}^{2}}-\frac{|\Gamma_{0}|}{S^{2}}\right)\left(\sum_{i,j}(-1)^{j+1}\langle\psi_{0},\phi\rangle_{L^{2}(\Gamma_{0}^{(i,j)})}\right).$

###### Proof.

Throughout we apply Lemmas 8 and 9.

  1. (I)

    1. (i)

We have

$\displaystyle\hat{h}_{\alpha}^{d_{a_{j}},d_{a_{j}}}[\psi_{0}]=$
$\displaystyle
2\frac{c_{0}^{2}}{S^{2}}||\partial_{1}\psi_{0}||^{2}_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}$
$\displaystyle+\frac{\alpha}{2}\left(\frac{1}{|\Gamma_{0}|^{2}}-\frac{c^{2}}{|\Gamma_{0}|^{4}}\right)||\psi_{0}||^{2}_{\partial\Omega_{0}}$
$\displaystyle=$
$\displaystyle\frac{c_{0}^{2}}{2S^{2}}||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})}+\frac{\alpha}{2}\left(\frac{1}{2S}-\frac{S}{4S^{2}}\right)||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})},$
$\displaystyle=$
$\displaystyle\frac{1}{2S}||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})}+\alpha\frac{1}{8S}||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})},$

where we have made use of the symmetry properties
$||\partial_{1}\psi_{0}||^{2}_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}=\frac{1}{2}||\partial_{1}\psi_{0}||^{2}_{L^{2}(\Omega_{0})}=\frac{1}{4}||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})},$
and $||\psi_{0}||_{L^{2}(\Gamma_{0}^{i,j})}$ is independent of $i,j$ from
Corollary 1.

    2. (ii)

We have

$\displaystyle\hat{h}_{\alpha}^{d_{c},d_{c}}[\psi_{0}]=$
$\displaystyle\frac{6}{c_{0}^{2}}||\partial_{1}\psi_{0}||^{2}_{L^{2}(\Omega_{0})}+\frac{2}{c_{0}^{2}}||\partial_{2}\psi_{0}||^{2}_{L^{2}(\Omega_{0})}$
$\displaystyle+$
$\displaystyle\frac{\alpha}{|\Gamma_{0}|}\left(\frac{3\frac{S^{2}}{c_{0}^{4}}+1}{|\Gamma_{0}|}-\frac{\left(-\frac{S^{2}}{c_{0}^{3}}+c_{0}\right)^{2}}{|\Gamma_{0}|^{3}}\right)||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}$
$\displaystyle=$
$\displaystyle\frac{4}{S}||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})}+\alpha\frac{2}{S}||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})},$

where we have made use of the symmetry property
$||\partial_{1}\psi_{0}||^{2}_{L^{2}(\Omega_{0})}=||\partial_{2}\psi_{0}||^{2}_{L^{2}(\Omega_{0})}=\frac{1}{2}||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})}$
from Corollary 1.

    3. (iii)

We have

$\displaystyle\hat{h}_{\alpha}^{d_{S_{1}},d_{S_{1}}}[\psi_{0}]=$
$\displaystyle\frac{6}{S^{2}}||\partial_{2}\psi_{0}||^{2}_{L^{2}(\Omega_{0})}$
$\displaystyle+$ $\displaystyle\frac{\alpha
S}{|\Gamma_{0}|}\left(-\frac{S}{|\Gamma_{0}|^{3}c_{0}^{4}}-\frac{1}{c_{0}^{2}|\Gamma_{0}|S}+\frac{2|\Gamma_{0}|}{S^{3}}\right)||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}$
$\displaystyle=$
$\displaystyle\frac{3}{S^{2}}||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})}+\frac{5\alpha}{4S^{2}}||\psi_{0}||^{2}_{L^{2}(\partial\Omega_{0})}$

where we have made use of the symmetry properties
$||\partial_{1}\psi_{0}||^{2}_{L^{2}(\Omega_{0}\cap\\{y>0\\})}=||\partial_{1}\psi_{0}||^{2}_{L^{2}(\Omega_{0}\cap\\{y<0\\})}=\frac{1}{2}||\partial_{1}\psi_{0}||^{2}_{L^{2}(\Omega_{0})}=\frac{1}{4}||\nabla\psi_{0}||^{2}_{L^{2}(\Omega_{0})},$
from Corollary 1.

    4. (iv)

We have

$\displaystyle\hat{h}_{\alpha,a_{1},a_{2},c}^{d_{a_{j}},d_{c}}[\psi_{0}]=$
$\displaystyle(-1)^{j}\frac{2}{S}\langle\partial_{1}\psi_{0},\partial_{2}\psi_{0}\rangle_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}$
$\displaystyle+$
$\displaystyle\frac{\alpha}{|\Gamma_{0}|}\sum_{i=1}^{2}\left((-1)^{i+1}\frac{1}{|\Gamma_{0}|}-\frac{(-1)^{i+1}c_{0}(-\frac{S^{2}}{c_{0}^{3}}+c_{0})}{|\Gamma_{0}|^{3}}\right)||\psi_{0}||_{\Gamma_{0}^{(i,j)}}^{2}$
$\displaystyle=$ $\displaystyle 0,$

where we have made use of the fact that $S=c_{0}^{2}$ and so
$c_{0}-\frac{S^{2}}{c_{0}^{3}}=0,$ alongside the symmetry properties
$\langle\partial_{1}\psi_{0},\partial_{2}\psi_{0}\rangle_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}=0$
and
$||\psi_{0}||_{L^{2}(\Gamma_{0}^{(1,1)})}=||\psi_{0}||_{L^{2}(\Gamma_{0}^{(2,1)})}$
from Corollary 1.

    5. (v)

Trivial.

    6. (vi)

We have

$\displaystyle\hat{h}_{\alpha}^{d_{a_{j}},d_{S_{1}}}[\psi_{0}]=$
$\displaystyle
4\frac{c_{0}}{S^{2}}\langle\partial_{1}\psi_{0},\partial_{2}\psi_{0}\rangle_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}$
$\displaystyle+$ $\displaystyle(-1)^{j}\frac{\alpha
S}{|\Gamma_{0}^{(i)}|}c_{0}\left(\frac{1}{S^{2}|\Gamma_{0}|}+\frac{1}{|\Gamma_{0}|^{3}c_{0}^{2}}\right)\sum_{i=1}^{2}(-1)^{i+1}||\psi_{0}||^{2}_{L^{2}(\Gamma_{0}^{(i,j)})}$
$\displaystyle=$ $\displaystyle 0,$

where we have used the symmetry properties
$\langle\partial_{1}\psi_{0},\partial_{2}\psi_{0}\rangle_{L^{2}(\Omega_{0}\cap\\{(-1)^{j}y<0\\})}=0$
and
$||\psi_{0}||_{L^{2}(\Gamma_{0}^{(1,j)})}=||\psi_{0}||_{L^{2}(\Gamma_{0}^{(2,j)})}$
from Corollary 1.

    7. (vii)

We have

$\displaystyle\hat{h}^{d_{c},{d_{S_{1}}}}_{\alpha}[\psi_{0}]=$
$\displaystyle\frac{4}{c_{0}S}\left(||\partial_{2}\psi_{0}||^{2}_{L^{2}(\Omega_{0}^{-})}-||\partial_{2}\psi_{0}||^{2}_{L^{2}(\Omega_{0}^{+})}\right)$
$\displaystyle+$ $\displaystyle\frac{\alpha
S}{|\Gamma_{0}|}\left(\frac{\frac{S^{2}}{c_{0}^{3}}+c_{0}}{S^{2}|\Gamma^{(i)}_{0}|}+\frac{-\frac{S^{2}}{c_{0}^{3}}+c_{0}}{|\Gamma_{0}^{(i)}|^{3}c_{0}^{2}}\right)\left(\sum_{i,j}(-1)^{j}||\psi_{0}||^{2}_{L^{2}(\Gamma_{0}^{(i,j)})}\right)$
$\displaystyle=$ $\displaystyle 0.$

Where we have used the symmetry properties
$\langle\partial_{1}\psi_{0},\partial_{2}\psi_{0}\rangle_{L^{2}(\Omega_{0}^{+})}=\langle\partial_{1}\psi_{0},\partial_{2}\psi_{0}\rangle_{L^{2}(\Omega_{0}^{-})}=0,$
$||\psi_{0}||_{L^{2}(\Gamma_{0}^{(1,j)})}=||\psi_{0}||_{L^{2}(\Gamma_{0}^{(2,j)})}$
from Corollary 1.

  2. (II)

Follows easily from Lemma 8.

∎

We now have all the computations required to compute the
$\lambda_{0}^{v_{1},v_{2}}$ terms.

## References

  * [1] Angelo Alvino, Carlo Nitsch, and Cristina Trombetti.  A Talenti comparison result for solutions to elliptic problems with Robin boundary conditions.  Communications on Pure and Applied Mathematics, 76(3):585–603, 2023\.
  * [2] Miriam Bareket.  On an isoperimetric inequality for the first eigenvalue of a boundary value problem.  SIAM Journal on Mathematical Analysis, 8(2):280–287, 1977.
  * [3] Beniamin Bogosel and Dorin Bucur.  On the polygonal Faber-Krahn inequality.  arXiv preprint arXiv:2203.16409, 2022.
  * [4] M-H Bossel.  Membranes élastiquement liées: extension du théorème de Rayleigh-Faber-Krahn et de l’inégalité de Cheeger.  Comptes rendus de l’Académie des sciences. Série 1, Mathématique, 302(1):47–50, 1986.
  * [5] Dorin Bucur and Daniel Daners.  An alternative approach to the Faber–Krahn inequality for Robin problems.  Calculus of Variations and Partial Differential Equations, 37(1-2):75–86, 2010.
  * [6] Dorin Bucur, Vincenzo Ferone, Carlo Nitsch, and Cristina Trombetti.  The quantitative Faber–Krahn inequality for the Robin Laplacian.  Journal of Differential Equations, 264(7):4488–4503, 2018.
  * [7] Daniel Daners.  A Faber-Krahn inequality for Robin problems in any space dimension.  Mathematische Annalen, 335(4):767–785, 2006.
  * [8] Georg Faber.  Beweis, dass unter allen homogenen Membranen von gleicher Fläche und gleicher Spannung die kreisförmige den tiefsten Grundton gibt.  1923\.
  * [9] Pedro Freitas and David Krejčiřík.  The first Robin eigenvalue with negative boundary parameter.  Advances in Mathematics, 280:322–339, 2015.
  * [10] Antoine Henrot.  Shape optimization and spectral theory.  In Shape optimization and spectral theory. De Gruyter Open Poland, 2017.
  * [11] Emanuel Indrei.  On the first eigenvalue of the Laplacian for polygons.  arXiv preprint arXiv:2210.14806, 2022.
  * [12] Tosiō Kato.  Perturbation Theory for Linear Operators.  Classics in Mathematics, 132. Springer Berlin Heidelberg, 2nd ed. 1995\. edition, 1995.
  * [13] Edgar Krahn.  Über eine von Rayleigh formulierte Minimaleigenschaft des Kreises.  Mathematische Annalen, 94(1):97–100, 1925.
  * [14] David Krejčiřík, Vladimir Lotoreichik, and Tuyen Vu.  Reverse isoperimetric inequality for the lowest robin eigenvalue of a triangle.  Applied Mathematics & Optimization, 88(2):1–33, 2023.
  * [15] Richard S Laugesen.  The Robin Laplacian—spectral conjectures, rectangular theorems.  Journal of Mathematical Physics, 60(12):121507, 2019.
  * [16] Michael Levitin and Leonid Parnovski.  On the principal eigenvalue of a Robin problem with a large parameter.  Mathematische Nachrichten, 281(2):272–281, 2008.
  * [17] George Pólya and Gábor Szegö.  Isoperimetric inequalities in mathematical physics.  Princeton University Press, 1951.
  * [18] John William Strutt Rayleigh.  The theory of sound., volume 1.  Macmillan and co., England, 1878.