A Sharp Upper Bound on the Spectral Gap for Graphene Quantum Dots

Abstract

The main result of this paper is a sharp upper bound on the first positive eigenvalue of Dirac operators in two dimensional simply connected C³-domains with infinite mass boundary conditions. This bound is given in terms of a conformal variation, explicit geometric quantities and of the first eigenvalue for the disk. Its proof relies on the min-max principle applied to the squares of these Dirac operators. A suitable test function is constructed by means of a conformal map. This general upper bound involves the norm of the derivative of the underlying conformal map in the Hardy space $\mathcal {H}^{2}(\mathbb {D})$. Then, we apply known estimates of this norm for convex and for nearly circular, star-shaped domains in order to get explicit geometric upper bounds on the eigenvalue. These bounds can be re-interpreted as reverse Faber-Krahn-type inequalities under adequate geometric constraints.

Appendices

Appendix: The Massless Dirac Operator with Infinite Mass Boundary Conditions on a Disk

The goal of this Appendix is to prove Proposition 8. Namely, we are aiming to characterize the principal eigenvalue $\mu _{\mathbb {D}}$ and the associated eigenfunctions for the self-adjoint operator $\mathsf {D}_{\mathbb {D}}$ on the unit disk

$$ \mathbb{D} = \{x\in\mathbb{R}^{2}\colon |x| < 1\}. $$

The material of this appendix is essentially known (see for instance [36, App. D]). However, we recall it here for the sake of completeness.

1.1 A.1 The Representation of the Operator $\mathsf {D}_{\mathbb {D}}$ in Polar Coordinates

First, we introduce the polar coordinates (r,𝜃) on the disk $\mathbb {D}$. They are related to the Cartesian coordinates x = (x₁,x₂) via the identities

$$ x(r,\theta) = \left( \begin{array}{ccc} x_{1}(r,\theta)\\ x_{2}(r,\theta) \end{array}\right)\!,\!\quad\text{where}\!\quad x_{1} = x_{1}(r,\theta) = r\cos\theta,\!\quad x_{2} = x_{2}(r,\theta) = r\sin\theta, $$

for all $r \in \mathbb {I} := (0,1)$ and $\theta \in \mathbb {T}$. Further, we consider the moving frame (e_rad,e_ang) associated with the polar coordinates

$$ \mathbf{e}_{\text{rad}}(\theta) = \frac{\mathsf{d} x}{\mathsf{d} r} = \left( \begin{array}{ccc} \cos\theta\\ \sin\theta \end{array}\right) \quad\text{and}\quad \mathbf{e}_{\text{ang}}(\theta) = \frac{\mathsf{d} \mathbf{e}_{\text{rad}}}{\mathsf{d}\theta} = \left( \begin{array}{ccc} -\sin\theta\\ \cos\theta \end{array}\right). $$

The Hilbert space $L^{2}_{\text {cyl}}(\mathbb {D},\mathbb {C}^{2}) := L^{2}(\mathbb {I}\times \mathbb {T},\mathbb {C}^{2};r\mathsf {d} r \mathsf {d} \theta )$ can be viewed as the tensor product ${L^{2}_{r}}(\mathbb {I})\otimes L^{2}(\mathbb {T},\mathbb {C}^{2})$, where ${L^{2}_{r}}(\mathbb {I}) = L^{2}(\mathbb {I};r\mathsf {d} r)$. Let us consider the unitary transform

$$ V \colon L^{2}(\mathbb{D},\mathbb{C}^{2}) \rightarrow L^{2}_{\text{cyl}}(\mathbb{D},\mathbb{C}^{2}),\qquad (Vv)(r,\theta) = u\left( r\cos\theta,r\sin\theta\right), $$

and introduce the cylindrical Sobolev space by

$$ H^{1}_{\text{cyl}}(\mathbb{D},\mathbb{C}^{2}) \!:=\! V\!\left( \!H^{1}(\mathbb{D},\mathbb{C}^{2})\!\right) = \left\{v \!\in\! L^{2}_{\text{cyl}}(\mathbb{D},\mathbb{C}^{2}) \colon \partial_{r} v, r^{-1}(\partial_{\theta} v) \!\in\! L^{2}_{\text{cyl}}(\mathbb{D},\mathbb{C}^{2})\right\} $$

We consider the operator acting in the Hilbert space $L^{2}_{\text {cyl}}(\mathbb {D},\mathbb {C}^{2})$ defined as

$$ \widetilde{\mathsf{D}_\mathbb{D}} := V \mathsf{D}_\mathbb{D} V^{-1},\quad \text{dom}\left( \widetilde{\mathsf{D}_\mathbb{D}}\right) = V\left( \text{dom}\left( {\mathsf{D}_\mathbb{D}}\right)\right). $$

(1)

Now, let us compute the action of $\widetilde {\mathsf {D}_{\mathbb {D}}}$ on a function $v\in \text {dom}\left (\widetilde {\mathsf {D}_{\mathbb {D}}}\right )$. Notice that there exists $u\in \text {dom}\left ({\mathsf {D}_{\mathbb {D}}}\right )$ such that v = V u and the partial derivatives of v with respect to the polar variables (r,𝜃) can be expressed through those of u with respect to the Cartesian variables (x₁,x₂) via the standard relations (for x = x(r,𝜃))

$$ \begin{array}{lll} (\partial_{r} v)(r,\theta) & = \sin\theta(\partial_{2} u)(x) + \cos\theta(\partial_{1} u)(x), \\ r^{-1}(\partial_{\theta} v)(r,\theta) & = \cos\theta(\partial_{2} u)(x) - \sin\theta(\partial_{1} u)(x), \end{array} $$

and the other way round

$$ \begin{array}{lll} (\partial_{1} u)(x) & = \cos\theta (\partial_{r} v)(r,\theta) - \sin\theta\frac{(\partial_{\theta} v)(r,\theta)}{r},\\ (\partial_{2} u)(x) &= \sin\theta (\partial_{r} v)(r,\theta) + \cos\theta\frac{(\partial_{\theta} v)(r,\theta)}{r}. \end{array} $$

Using the latter formulæ we can express the action of the differential expression −i(σ ⋅∇) in polar coordinates as follows (for x = x(r,𝜃))

$$ \begin{array}{lll} (-\mathsf{i}(\sigma\cdot\nabla) u)(x) & = -\mathsf{i}\left( \begin{array}{ccc} \partial_{1} u_{2}(x) - \mathsf{i} \partial_{2} u_{2}(x)\\ \partial_{1} u_{1}(x) + \mathsf{i} \partial_{2} u_{1}(x) \end{array}\right)\\ & = -\mathsf{i}\left( \begin{array}{ccc} \text{e}^{-\mathsf{i}\theta} (\partial_{r} v_{2})(r,\theta) - \mathsf{i} \text{e}^{-\mathsf{i}\theta}r^{-1}(\partial_{\theta} v_{2})(r,\theta) \\ \text{e}^{\mathsf{i}\theta} (\partial_{r} v_{1})(r,\theta) + \mathsf{i} \text{e}^{\mathsf{i}\theta}r^{-1}(\partial_{\theta} v_{1})(r,\theta) \end{array}\right). \end{array} $$

Note that a basic computation yields

$$ \sigma\cdot \mathbf{e}_{\text{rad}} = \cos\theta\sigma_{1} + \sin\theta\sigma_{2} = \left( \begin{array}{ccc} 0 & \text{e}^{-\mathsf{i}\theta}\\ \text{e}^{\mathsf{i}\theta} &0 \end{array}\right). $$

(2)

Hence, the operator $\widetilde {\mathsf {D}_{\mathbb {D}}}$ acts as

$$ \begin{array}{@{}rcl@{}} \widetilde{\mathsf{D}_\mathbb{D}} v &=& -\mathsf{i}(\sigma\cdot \mathbf{e}_{\text{rad}}) \left( \partial_{r} v + \frac{v - \sigma_{3} {\mathsf K} v}{2r}\right),\\ \text{dom}\left( \widetilde{\mathsf{D}_\mathbb{D}}\right) & =& \left\{ v\in H_{\text{cyl}}^{1}(\mathbb{D},\mathbb{C}^{2})\colon v_{2}(1,\theta) = \mathsf{i} \text{e}^{\mathsf{i}\theta} v_{1}(1,\theta)\right\}, \end{array} $$

(3)

where K is the spin-orbit operator in the Hilbert space $L^{2}(\mathbb {T};\mathbb {C}^{2})$ defined as

$$ {\mathsf K} = -2\mathsf{i}\partial_{\theta} + \sigma_{3}, \qquad \text{dom}\left( {\mathsf K}\right) = H^{1}(\mathbb{T},\mathbb{C}^{2}). $$

(4)

Let us investigate the spectral properties of the spin-orbit operator K.

Proposition 1

Let the operator K be as in (30). Then the following hold.

• (i) K is self-adjoint and has a compact resolvent.

• (ii) $\mathsf {Sp}\left ({\mathsf K}\right ) = \{2k + 1\}_{k\in \mathbb {Z}}$ and $\mathcal {F}_{k} := \ker \left ({\mathsf K} - (2k + 1)\right ) = \mathsf {span} (\phi _{k}^{+}, \phi _{k}^{-})$ , where

  $$ \phi_{k}^{+} = \frac1{\sqrt{2\pi}} \left( \begin{array}{ccc} \text{e}^{\mathsf{i} k\theta}\\0 \end{array}\right) \quad\text{and}\quad \phi_{k}^{-} = \frac1{\sqrt{2\pi}} \left( \begin{array}{ccc} 0\\ \text{e}^{\mathsf{i} (k + 1)\theta} \end{array}\right). $$

• (iii) $(\sigma \cdot \mathbf {e}_{\text {rad}}) \phi _{k}^{\pm } = \phi _{k}^{\mp }$ and $\sigma _{3} \phi _{k}^{\pm } = \pm \phi _{k}^{\pm }$.

Proof

(i) The operator K is clearly self-adjoint in $L^{2}(\mathbb {T},\mathbb {C}^{2})$, because adding the matrix σ₃ can be viewed as a symmetric bounded perturbation of an unbounded self-adjoint momentum operator $H^{1}(\mathbb {T},\mathbb {C}^{2})\ni \phi \mapsto -\mathsf {i} \phi ^{\prime }$ in the Hilbert space $L^{2}(\mathbb {T},\mathbb {C}^{2})$. As $\text {dom}\left (\mathsf {K}\right ) = H^{1}(\mathbb {T},\mathbb {C}^{2})$ is compactly embedded into $L^{2}(\mathbb {T},\mathbb {C}^{2})$ the resolvent of K is compact.(ii) Let ϕ = (ϕ⁺,ϕ⁻)^⊤∈dom (K) and $\lambda \in \mathbb {R}$ be such that Kϕ = λϕ. The eigenvalue equation on ϕ reads as follows

$$ (\phi^{\pm})^{\prime} = \frac{\mathsf{i}}{2} \left( \lambda \mp 1\right)\phi^{\pm}. $$

The generic solution of the above system of differential equations is given by

$$ \phi^{\pm}(\theta) = A_{\pm} \exp\left( \mathsf{i} \tfrac{\lambda\mp 1}2 \theta\right), \qquad A_{\pm}\in\mathbb{C}. $$

Hence, the periodic boundary condition ϕ^±(0) = ϕ^±(2π) implies that the eigenvalues of K are exhausted by λ = 2k + 1 for $k\in \mathbb {Z}$ and that $\{\phi _{k}^{+},\phi _{k}^{-}\}$ is a basis of $\mathcal {F}_{k}$.(iii) These algebraic relations are obtained via basic matrix calculus using (28). □

We are now ready to introduce subspaces of $\text {dom}\left (\widetilde {\mathsf {D}_{\mathbb {D}}}\right )$ that are invariant under its action. The analysis of $\widetilde {\mathsf {D}_{\mathbb {D}}}$ reduces to the study of its restrictions to each invariant subspace.

Proposition 2

There holds

$$ L^{2}_{\text{cyl}}\left( \mathbb{D}, \mathbb{C}^{2}\right)\simeq {L^{2}_{r}}(\mathbb{I})\otimes L^{2}(\mathbb{T},\mathbb{C}^{2}) = \oplus_{k\in\mathbb{Z}} {\mathcal E}_{k}, $$

where ${\mathcal E}_{k} = {L^{2}_{r}}(\mathbb {I})\otimes \mathcal {F}_{k}$ and ${L^{2}_{r}}(\mathbb {I}):= L^{2}(\mathbb {I};r\mathsf {d} r)$. Moreover, the following hold true.

• (i) For any $k\in \mathbb {Z}$,

  $$ d_{k} u := \widetilde{\mathsf{D}_\mathbb{D}} u,\qquad \text{dom}\left( d_{k}\right) := \text{dom}\left( \widetilde{\mathsf{D}_\mathbb{D}}\right) \cap {\mathcal E}_{k} $$

  is a well-defined self-adjoint operator in the Hilbert space ${\mathcal E}_{k}$.

• (ii) For any $k \in \mathbb {Z}$, the operator d_k is unitarily equivalent to the operator d_k in the Hilbert space ${L^{2}_{r}}(\mathbb {I},\mathbb {C}^{2})$ defined as

  $$ \begin{array}{@{}rcl@{}} \mathbf{d}_{k} \!\!&=&\!\! \left( \!\begin{array}{ccc} 0 & -\mathsf{i} \frac{\mathsf{d}}{\mathsf{d} r} - \mathsf{i}\frac{k + 1}{r}\\ -\mathsf{i}\frac{\mathsf{d}}{\mathsf{d} r} + \mathsf{i} \frac{k}r &0 \end{array}\!\right)\!,\\ {} \text{dom}{}\left( \mathbf{d}_{k}\right) \!\!&=&\!\! \left\{{\kern-.5pt}u = (u_{+}{\kern-.5pt},{\kern-.5pt}u_{-}{\kern-.5pt}){\kern-.5pt} \colon{\kern-.5pt} u_{\pm}{\kern-.5pt},{\kern-.5pt} u_{\pm}^{\prime}{\kern-.5pt},{\kern-.5pt} \tfrac{ku_{+}}{r}{\kern-.5pt},{\kern-.5pt} \tfrac{(k{\kern-.5pt}+{\kern-.5pt}1)u_{-}}{r} \!\in\! {L^{2}_{r}}(\mathbb{I}){\kern-.5pt},{\kern-.5pt} u_{-}(1) = \mathsf{i} u_{+}{\kern-.5pt}({\kern-.5pt}1{\kern-.5pt}){}\right\}\!. \end{array} $$

  (5)

• (iii) $\mathsf {Sp}\left (\mathsf {D}_{\mathbb {D}}\right ) = \mathsf {Sp}\left (\widetilde {\mathsf {D}_{\mathbb {D}}}\right ) = \bigcup _{k\in \mathbb {Z}}\mathsf {Sp}\left (\mathbf {d}_{k}\right )$.

Proof

(i) Let us check that d_k is well defined. Pick a function $u\in \text {dom}\left (\widetilde {\mathsf {D}_{\mathbb {D}}}\right )\cap \mathcal {E}_{k}$. By definition, u writes as

$$ u(r,\theta) = u_{+}(r)\phi_{k}^{+}(\theta) + u_{-}(r)\phi_{k}^{-}(\theta), $$

and, since $u\in H_{\text {cyl}}^{1}(\mathbb {D},\mathbb {C}^{2})$, we have $u_{\pm },u_{\pm }^{\prime },\frac {k}ru_{+},\frac {k + 1}r u_{-}\in {L_{r}^{2}}(\mathbb {I})$. Applying the differential expression obtained in (29), we get

$$ \begin{array}{lll} (\widetilde{\mathsf{D}_\mathbb{D}} u)(r,\theta) & = -\mathsf{i}(\sigma\cdot \mathbf{e}_{\text{rad}}) \left( \partial_{r} v + \frac{v - \sigma_{3} {\mathsf K} v}{2r}\right)u(r,\theta)\\ &=\! \left[-\mathsf{i} u^{\prime}_{-}(r)\! -\! \frac{\mathsf{i}(k + 1)}{r} u_{-}(r)\right]\!\phi_{k}^{+}(\theta)\! +\! \left[-\mathsf{i} u_{+}^{\prime}(r) + \frac{\mathsf{i} k}r u_{+}(r)\right]\!\phi_{k}^{-}(\theta). \end{array} $$

(6)

It yields $\widetilde {\mathsf {D}_{\mathbb {D}}} \left (\text {dom}\left (\mathsf {D}_{\mathbb {D}}\right )\cap {\mathcal E}_{k}\right ) \subset {\mathcal E}_{k}$. It is now an easy exercise to show that d_k is self-adjoint.

(ii) Let us introduce the unitary transform

$$ W_{k} \colon {\mathcal E}_{k} \!\rightarrow\! {L_{r}^{2}}(\mathbb{I},\mathbb{C}^{2}),\quad (W_{k}u)(r) = \left( (u(r,\cdot),\phi_{k}^{+})_{L^{2}(\mathbb{T},\mathbb{C}^{2})}, (u(r,\cdot),\phi_{k}^{-})_{L^{2}(\mathbb{T},\mathbb{C}^{2})}\right)^{\top}. $$

For $u\in {\mathcal E}_{k}$ it is clear that we have $\|W_{k}u\|_{{L_{r}^{2}}(\mathbb {I},\mathbb {C}^{2})} = \|u\|_{L^{2}_{\text {cyl}}(\mathbb {D},\mathbb {C}^{2})}$ and we observe that

$$ \mathbf{d}_{k} = W_{k}d_{k} W_{k}^{-1},\qquad \text{dom}\left( \mathbf{d}_{k}\right) = W_{k}\left( \text{dom}\left( d_{k}\right)\right). $$

(iii) The first equality is a consequence of (27), while the second one is an application of [31, Theorem XIII.85]. □

1.2 A.2 Eigenstructure of the Disk

Before describing the eigenstructure of the disk recall that C denotes the charge conjugation operator introduced in (14). It is not difficult to see that C is anti-unitary and maps dom (d_k) onto dom (d_−(k+ 1)) for all $k\in \mathbb {Z}$. Furthermore, a computation yields

$$ C \mathbf{d}_{-(k + 1)}C = - \mathbf{d}_{k}. $$

(7)

In particular, C² = 1₂, which also reads C^− 1 = C. Combined with (33) and as the spectrum of d_k is discrete one immediately observes that

$$ \mathsf{Sp}\left( \mathbf{d}_{k}\right) = - \mathsf{Sp}\left( \mathbf{d}_{-(k + 1)}\right). $$

(8)

Hence, we can restrict ourselves to $k\geqslant 0$.

Lemma 3

Let $k\in \mathbb {N}_{0}$ . Let d _k be the self-adjoint operator defined in ( 31 ). Then for all $k\in \mathbb {N}$ the following hold.

• (i) dom (d_k) ⊂dom (d₀)

• (ii) $\|\mathbf {d}_{k} u\|_{{L^{2}_{r}}(\mathbb {I};\mathbb {C}^{2})}^{2} \geqslant \|\mathbf {d}_{0} u\|_{{L^{2}_{r}}(\mathbb {I};\mathbb {C}^{2})}^{2}$ forall u ∈dom (d_k).

Proof

Let $k\in \mathbb {N}$ and u = (u₊,u₋)^⊤∈dom (d_k). It is clear that u ∈dom (d₀) and that for integrability reasons u(0) = 0. Hence, we have

$$ \begin{array}{lll} \left\|u_{+}^{\prime} - \tfrac{k}r u_{+}\right\|_{{L^{2}_{r}}(\mathbb{I})}^{2} & = \left\|u_{+}^{\prime}\right\|_{{L^{2}_{r}}(\mathbb{I})}^{2} - 2k\Re\left( u_{+}^{\prime},\tfrac{1}{r}u_{+}\right)_{{L^{2}_{r}}(\mathbb{I})} + k^{2} \left\| \tfrac{1}{r} u_{+}\right\|_{{L^{2}_{r}}(\mathbb{I})}^{2}\\ & \ge \left\|u_{+}^{\prime}\right\|_{{L^{2}_{r}}(\mathbb{I})}^{2} - 2k{\Re{\int}_{0}^{1}} u_{+}^{\prime}\overline{u_{+}}\mathsf{d} r\\ & = \left\|u_{+}^{\prime}\right\|_{{L^{2}_{r}}(\mathbb{I})}^{2} - k{{\int}_{0}^{1}}(|u_{+}|^{2})^{\prime}\mathsf{d} r\\ & = \left\|u_{+}^{\prime}\right\|_{{L^{2}_{r}}(\mathbb{I})}^{2} - k|u_{+}(1)|^{2}. \end{array} $$

(9)

Analogously, we get

$$ \begin{array}{lll} \left\|u_{-}^{\prime} + \tfrac{k + 1}r u_{-}\right\|_{{L^{2}_{r}}(\mathbb{I})}^{2} & \ge \left\| u_{-}^{\prime} + \tfrac{1}r u_{-} \right\|_{{L^{2}_{r}}(\mathbb{I})}^{2} + 2k\Re\left( u_{-}^{\prime},\tfrac{1}{r}u_{-}\right)_{{L^{2}_{r}}(\mathbb{I})}\\ & = \left\| u_{-}^{\prime} + \tfrac{1}r u_{-} \right\|_{{L^{2}_{r}}(\mathbb{I})}^{2} + k{{\int}_{0}^{1}}(|u_{-}|^{2})^{\prime}\mathsf{d} r\\ & = \left\| u_{-}^{\prime} + \tfrac{1}r u_{-} \right\|_{{L^{2}_{r}}(\mathbb{I})}^{2} + k|u_{-}(1)|^{2} \end{array} $$

(10)

Combining (35) and (36) with the boundary condition u₋(1) = iu₊(1) we get

$$ \begin{array}{lll} \|\mathbf{d}_{k}u\|^{2}_{{L^{2}_{r}}(\mathbb{I},\mathbb{C}^{2})} & \ge \left\| u_{+}^{\prime}\right\|_{{L^{2}_{r}}(\mathbb{I})}^{2} + \left\| u_{-}^{\prime} + \tfrac{1}r u_{-} \right\|_{{L^{2}_{r}}(\mathbb{I})}^{2} + k\left( |u_{-}(1)|^{2} - |u_{+}(1)|^{2}\right)\\ & = \|\mathbf{d}_{0}u\|_{{L^{2}_{r}}(\mathbb{I},\mathbb{C}^{2})}^{2} + k\left( |u_{-}(1)|^{2} - |u_{+}(1)|^{2}\right) = \|\mathbf{d}_{0} u\|_{{L^{2}_{r}}(\mathbb{I},\mathbb{C}^{2})}^{2}. \end{array} $$

□

Now, we have all the tools to prove Proposition 8.

Proof of Proposition 8

As a direct consequence of Lemma 34 and the min-max principle, we obtain that

$$ \mu_{1}({\mathbf{d}_{k}^{2}}) \geqslant \mu_{1}({\mathbf{d}_{0}^{2}}) = \mu_{\mathbb{D}}^{2}. $$

Thus, by Proposition 33 (iii) and Eq. 34, in order to investigate the first eigenvalue of $\mathsf {D}_{\mathbb {D}}$, we only have to focus on the operator d₀.

Let μ > 0 be an eigenvalue of d₀ and u be an associated eigenfunction. In particular, $u = (u_{+},u_{-})^{\top } \in \text {dom}\left ({\mathbf {d}_{0}^{2}}\right )$ and we have

$$ 0 = (\mathbf{d}_{0} + \mu)(\mathbf{d}_{0} - \mu) u = \left( \begin{array}{ccc} -u_{+}^{\prime\prime} - \frac{u_{+}^{\prime}}{r} - \mu^{2} \\ -u_{-}^{\prime\prime} - \frac{u_{-}^{\prime}}{r} + \frac{u_{-}}{r^{2}} - \mu^{2} \end{array}\right) . $$

Hence, we obtain

$$ u_{+}(r) = a_{+}J_{0}(\mu r) + b_{+}Y_{0}(\mu r) \quad\text{and}\quad u_{-}(r) = a_{-} J_{1}(\mu r) + b_{-} Y_{1}(\mu r), $$

with some constants $a_{\pm },b_{\pm } \in \mathbb {C}$ and where J_ν and Y_ν (ν = 0, 1) denote the Bessel function of the first kind of order ν and the Bessel function of the second kind of order ν, respectively. Taking into account that

$$ \lim_{r\rightarrow 0^{+}} r^{2} |Y_{0}^{\prime}(r)|^{2} {=} \frac{4}{\pi^{2}},\quad \lim_{r\rightarrow 0^{+}}r^{4} |Y_{1}^{\prime}(r)|^{2} {=}\frac{4}{\pi^{2}}, $$

(see [25, §10.7(i)]), the condition u ∈dom (d₀) implies b_± = 0 or, in other words,

$$u_{+}(r) = a_{+} J_{0}(\mu r) \quad\text{and}\quad u_{-}(r) = a_{-} J_{1}(\mu r). $$

Now, as u satisfies the eigenvalue equation d₀u = μu we get $u_{+}^{\prime } = \mathsf {i}\mu u_{-}$ and the identity

$$ a_{-}\mu J_{1}(\mu r) = \mathsf{i} \mu a_{+} J_{1}(\mu r) $$

holds for all $r\in \mathbb {I}$. In particular, we obtain a₋ = ia₊ which gives

$$ u = a_{+} \left( \begin{array}{ccc} J_{0}(\mu r)\\ \mathsf{i} J_{1}(\mu r) \end{array}\right). $$

(11)

Now, the boundary condition u₋(1) = iu₊(1) reads as

$$ J_{0}(\mu) = J_{1}(\mu), $$

(12)

which gives the eigenvalue equation, whose first positive root is the principal eigenvalue of $\mathsf {D}_{\mathbb {D}}$. An eigenfunction of $\widetilde {\mathsf {D}_{\mathbb {D}}}$ corresponding to the eigenvalue $\mu _{\mathbb {D}}$ is given in polar coordinates by

$$ w(r,\theta) = u\otimes \phi_{0}^{-} = \frac1{\sqrt{2\pi}}\left( \begin{array}{ccc}J_{0}(r\mu_{\mathbb{D}} )\\\mathsf{i} \text{e}^{\mathsf{i} \theta}J_{1}(r\mu_{\mathbb{D}}) \end{array}\right) $$

where $\phi _{0}^{-}$ is as in Proposition 32 (ii), u is as in (37) (with a₊ = 1) and $\mu _{\mathbb {D}}$ is the smallest positive root of (38). □

Appendix B: Proof of Proposition 27

• Step 1. For any z ∈ ∂Ω, the map $\overline {\Omega } \ni y \to |y-z|$ is continuous. Hence, the maps defined as

  $$ r_{\mathrm{i}} := \overline{\Omega}\ni y \mapsto \inf_{z\in\partial{\Omega}}|y - z|, \qquad r_{\mathrm{o}} := \overline{\Omega}\ni y \mapsto \sup_{z\in\partial{\Omega}}|y-z|, $$

  are continuous on $\overline {\Omega }$ as well and they attain their upper and lower bounds. In particular, there exist $y_{\mathrm {i}}, y_{\mathrm {o}} \in \overline {\Omega }$ such that

  $$ r_{\mathrm{i}}(y_{\mathrm{i}})= \max_{y\in\overline{\Omega}}r_{\mathrm{i}}(y), \qquad r_{\mathrm{o}}(y_{\mathrm{o}})= \min_{y\in\overline{\Omega}}r_{\mathrm{o}}(y). $$

• Step 2. Assume that Ω has an axis of symmetry Λ. By Step 1 there exist $y_{\mathrm {i}},y_{\mathrm {o}} \in \overline {\Omega }$ such that r_i(y_i) = supy∈Ωr_i(y) and r_o(y_o) = infy∈Ωr_o(y). Our aim is to show that y_i,y_o can be both chosen in Λ. Let us suppose that y_i,y_o∉Λ and define the reflection $\mathcal {R}_{{\Lambda } }\colon \overline {\Omega }\rightarrow \overline {\Omega }$ with respect to Λ. Remark that $\mathcal {R}_{{\Lambda } } y_{\mathrm {i}}$ and $\mathcal {R}_{{\Lambda } } y_{\mathrm {o}}$ also satisfy $r_{\mathrm {i}}(\mathcal {R}_{{\Lambda } } y_{\mathrm {i}}) = \max _{y\in \overline {\Omega }}r_{\mathrm {i}}(y)$ and $r_{\mathrm {o}}(\mathcal {R}_{{\Lambda } } y_{\mathrm {o}}) = \min _{y\in \overline {\Omega }}r_{\mathrm {o}}(y)$. Set $\widetilde {y}_{\mathrm {i}} := \frac 12 y_{\mathrm {i}} + \frac 12 \mathcal {R}_{{\Lambda } } y_{\mathrm {i}}$ and $\widetilde {y}_{\mathrm {o}} := \frac 12 y_{\mathrm {o}} + \frac 12 \mathcal {R}_{{\Lambda } } y_{\mathrm {o}}$. As Ω is convex we have $\widetilde {y}_{\mathrm {i}}, \widetilde {y}_{\mathrm {o}} \in \overline {\Omega }$. Also by convexity of Ω, we get

  $$ \frac12 \mathbb{D}_{r_{\mathrm{i}}(y_{\mathrm{i}})}(y_{\mathrm{i}}) + \frac12\mathbb{D}_{r_{\mathrm{i}}(y_{\mathrm{i}})}(\mathcal{R}_{{\Lambda} } y_{\mathrm{i}}) = \mathbb{D}_{r_{\mathrm{i}}(y_{\mathrm{i}})}(\widetilde{y}_{\mathrm{i}}) \subset {\Omega}, $$

  (13)

  where $\mathbb {D}_{r}(y)$ denotes the disk of radius r > 0 centred at $y\in \mathbb {R}^{2}$. Now, (13) implies $r_{\mathrm {i}}(\widetilde {y}_{\mathrm {i}}) \geqslant r_{\mathrm {i}}(y_{\mathrm {i}})$ and we obtain $r_{\mathrm {i}}(\widetilde {y}_{\mathrm {i}}) = \max _{y\in \overline {\Omega }} r_{\mathrm {i}}(y)$.

  Similarly, by convexity of Ω, we get

  $$ \frac12 \mathbb{D}_{r_{\mathrm{o}}(y_{\mathrm{o}})}(y_{\mathrm{o}}) + \frac12\mathbb{D}_{r_{\mathrm{o}}(y_{\mathrm{o}})}(\mathcal{R}_{{\Lambda} } y_{\mathrm{o}}) = \mathbb{D}_{r_{\mathrm{o}}(y_{\mathrm{o}})}(\widetilde{y}_{\mathrm{o}}) \supset {\Omega}. $$

  In particular, $r_{\mathrm {o}}(\widetilde {y}_{\mathrm {o}}) \leqslant \min _{y\in \overline {\Omega }}r_{\mathrm {o}}(y)$ and we have equality in this inequality.

• Step 3. Suppose now that Ω has two axes of symmetry Λ₁ and Λ₂. Let y_Λ ∈Ω be the unique point of intersection of these axes. Thanks to Steps 1 and 2 for all y ∈Ω we necessarily have r_i(y) ≤ r_i(y_Λ) and $r_{\mathrm {o}}(y) \geqslant r_{\mathrm {o}}(y_{{\Lambda } })$. Next, define the function

  $$ \mathcal{G}(r_{1},r_{2}) := \left( \frac{|{\Omega}|+ \pi {r_{1}^{2}}}{2\pi}\right)^{\frac12} \exp\left( 2(r_{\mathrm{c}} -r_{2}){\Phi}(r_{1},r_{\mathrm{c}})\right),\quad r_{1} < r_{2}, r_{\mathrm{c}} < r_{2}. $$

  Remark that $\mathcal {G}$ is a non-decreasing function of r₁ whereas it is a non-increasing function of r₂. Now, we have

  $$ \mathcal{F}_{\mathrm{c}}({\Omega} -y) = \mathcal{G}\left( r_{\mathrm{i}}(y),r_{\mathrm{o}}(y)\right) \leqslant \mathcal{G}\left( r_{\mathrm{i}}(y_{\mathrm{i}}),r_{\mathrm{o}}(y_{\mathrm{o}})\right) = \mathcal{F}_{\mathrm{c}}({\Omega} - y_{{\Lambda} }). $$

  Hence, $\mathcal {F}_{\mathrm {c}}^{\star }({\Omega }) =\sup _{y\in {\Omega }} \mathcal {F}_{\mathrm {c}}({\Omega }-y)= \mathcal {F}_{\mathrm {c}}({\Omega } -y_{{\Lambda } })$, by which the proof is concluded.