We consider a Dirac operator on right triangles, subject to infinite-mass boundary conditions. We conjecture that the lowest positive eigenvalue is minimized by the isosceles right triangle under the area or perimeter constraints. We prove this conjecture under extra geometric hypotheses relying on a recent approach of Briet and Krejčiřík [J. Math. Phys. 63, 013502 (2022)].

One of the most interesting topics in spectral geometry is the determination of optimal shapes for eigenvalues of differential operators, subject to various boundary conditions and geometric constraints. Probably the most classical and well-known situation is that of the Laplace operator, subject to Dirichlet boundary conditions,
Δψ=ΛψinΩ,ψ=0onΩ,
(1)
where Ω is an open set of finite measure. The celebrated Faber–Krahn inequality states that the lowest eigenvalue Λ1 = Λ1(Ω) is minimized by a ball, among all sets of given volume. By the classical isoperimetric inequality, it follows that the ball is the minimizer under the perimeter constraint too. The optimality of the ball extends to repulsive Robin boundary conditions, but it is generally false for attractive Robin boundary conditions.1,2 The ball is generally not optimal for higher eigenvalues either. Mathematically, the optimality of the ball is closely related to the availability of symmetrization techniques. We refer to the monographs3,4 for a recent survey of this fascinating spectral-optimization subject.

By a symmetrization argument, it is also true that the Dirichlet eigenvalue Λ1 is minimized by the equilateral triangle (respectively, square), among all triangles (respectively, quadrilaterals) of a given area or perimeter. The analogous problem remains open for general polygons; see Ref. 3, Sec. 3.3.3 and Refs. 5 and 6 for a survey and the most recent progress, respectively. In general, it also remains open for Robin boundary conditions, even in the case of triangles.7 On the other hand, rectangles (or, more generally, rectangular boxes), the very special situation of quadrilaterals, can be settled by means of the availability of explicit solutions due to the separation of variables.8 

The classical physical interpretation of Λ1 in two dimensions is the square of the fundamental frequency of a vibrating membrane with fixed edges. Alternatively, Λ1 is the ground-state energy of a non-relativistic quantum particle constrained to a semiconductor nanostructure of shape Ω by hard-wall boundaries. In this paper, we are interested in the analog of the aforementioned spectral-optimization problems in the relativistic setting.

The relativistic analog of (1) (relevant for graphene materials, for instance) is the spectral problem for the Dirac operator, subject to infinite-mass (also called MIT) boundary conditions.9–14 More specifically, given an open Lipschitz set Ω in R2, the relativistic quantum Hamiltonian acts as
Tmi(1i2)i(1+i2)minL2(Ω;C2),
(2)
while the boundary conditions are encoded in the operator domain,
D(T)ψ=ψ1ψ2W1,2(Ω;C2):ψ2=i(n1+in2)ψ1 on Ω.
(3)
Here, the notations m and n=n1n2:ΩR2 stand for the non-negative mass of the relativistic (quasi-)particle and the outward unit normal of the set Ω, respectively. The operator T is self-adjoint at least if the boundary Ω is C2-smooth9 or if Ω is a polygon11 (for a general Lipschitz set, the self-adjointness can be achieved in a W1/2,2 setting15). As usual, in relativistic quantum mechanics, the spectrum of T is not bounded from below. However, it is purely discrete if Ω is bounded and the eigenvalues are symmetrically distributed on the real axis. Consequently, the lowest positive eigenvalue λ1 = λ1(Ω) of T can be characterized variationally,
λ1(Ω)2=infψD(T)ψ0Tψ2ψ2.
(4)
It is important to stress that because of the exotic boundary conditions, spinorial structure of the Hilbert space, and lack of positivity-preserving property, no symmetrization techniques are available at this moment.

In analogy with the Faber–Krahn inequality, the following conjecture is natural to expect in the relativistic setting.

Conjecture 1.
Given any m ≥ 0 and open Lipschitz set ΩR2,
λ1(Ω)λ1(Ω*),
where Ω* is the disk of the same area or perimeter as Ω.

For massless particles (i.e., m = 0), the fixed-area part of the conjecture was explicitly stated in Ref. 16. The present general statement can be found in Ref. 17. The proof of the conjecture was classified as a challenging open problem in spectral geometry during an AIM workshop in San Jose (USA) in 2019.18 Unfortunately, despite some partial attempts,16,19,20 including a numerical support, the problem remains open.

Because of the complexity of the problem in the general setting, the authors of Ref. 17 considered a rectangular version of the conjecture. More specifically, it is conjectured in Ref. 17 that λ1 is minimized by the square among all rectangles of a fixed area or perimeter. Surprisingly, even this simplified setting is not resolved and the authors of Ref. 17 managed to prove the conjecture under some additional hypotheses only (roughly, for heavy masses or eccentric rectangles). The problem is that infinite-mass boundary conditions do not allow for a separation of variables.

In this paper, we continue the study by asking whether the isosceles right triangle is the optimal geometry among all right triangles, again under the area or perimeter constraints. More specifically, let Ωa,b be the right triangle in R2 defined by the three vertexes O ≔ (0, 0), A ≔ (a, 0), and B ≔ (0, b), where a, b are any positive numbers; see Fig. 1. Note that the area and perimeter are given by 12ab and a+b+a2+b2, respectively. Define λ1(a, b) ≔ λa,b).

FIG. 1.

The right triangle OAB.

FIG. 1.

The right triangle OAB.

Close modal

Conjecture 2.

Given any m ≥ 0,

  • λ1(a, b) ≥ λ1(k, k) with any a, b, k > 0 such that ab = k2 and

  • λ1(a, b) ≥ λ1(k, k) with any a(0,(2+2)k) and b, k > 0 such that a+b+a2+b2=(2+2)k.

We have not managed to prove the conjecture in its full generality. Following Ref. 17, to get partial results, we first establish universal lower and upper bounds to λ1(a, b).

Theorem 1.
For every m ≥ 0, one has
arctan2ab+1+a2b2b2+arctan2ba+1+b2a2a2λ1(a,b)2m25π221a2+1b2.

Note that the upper bound becomes sharp in the limit m for a = b. Indeed, it is known that λ1(Ω)2m2 converges to the Dirichlet eigenvalue Λ1(Ω) as m (see, e.g., Ref. 10) and Λ1(a, a) = 5π2/a2. In contrast to the one-dimensional spectrum of the operator in Ref. 17, that in this paper is not symmetric. As in Ref. 17, the upper bound is obtained by using a suitable trial function in (4). The lower bound employs a Poincaré-type inequality for a one-dimensional Dirac problem on an interval. The latter yields an m-dependent (implicit) lower bound, while the lower bound of Theorem 1 is due to an (explicit) uniform estimate of the closest-to-zero eigenvalue of the one-dimensional problem.

As a consequence, we get the following sufficient conditions, which guarantee the validity of Conjecture 2.

Corollary 1.
Let k be defined as in Conjecture 2 and m ≥ 0.Conjecture 2. (i) Holds under the following extra hypotheses:
a9korak9.
Conjecture 2. (ii) Holds under the following extra hypotheses:
a3.5korak9.

In other words, Conjecture 2 holds true for sufficiently eccentric right triangles.

This paper is organized as follows. In Sec. II, we derive a formula for the expectation value of the square of the Dirac operator in triangles. This formula serves as the foundation for the Proof of Theorem 1. The one-dimensional Poincaré-type inequality is established in Sec. III. The main results are proved in Sec. IV. The extension of the formula for the expectation value of the square of the Dirac operator to planar polygons can be found in the Appendix.

Recall that our right triangle Ωa,b is a special planar polygon determined by the three vertices O ≔ (0, 0), A ≔ (a, 0), and B ≔ (0, b).

Let Ta,b denote operators (2) and (3) in the case of the triangle Ωa,b. The operator is self-adjoint and has a compact resolvent. The eigenvalue problem Ta,bu = λu is equivalent to the system
i(1i2)u2=(λm)u1inΩa,b,i(1+i2)u1=(λ+m)u2inΩa,b,u2=u1onOA,u2=iu1onOB,u2=aa2+b2+iba2+b2u1onAB.
(5)
The spectrum of Ta,b is symmetric with respect to zero. Indeed, u=u1u2 is an eigenfunction of Ta,b corresponding to an eigenvalue λ if, and only if, ū2ū1 is an eigenfunction of Ta,b corresponding to an eigenvalue −λ (charge conjugation symmetry). It will become evident in a moment that any solution of (5) necessarily satisfies |λ| ≥ m. Our objective is to study the smallest positive solution, λ1(a, b), of (5).
Since the analogous case of rectangles cannot be solved by separation of variables,17 there seems to be no hope to get explicit solutions of (5). As an alternative approach, we focus on the variational characterization (4). To this aim, we need a more suitable formula for the square norm Ta,bu2=(u,Ta,b2u). If Ω were a smooth bounded domain, we would have (see, e.g., Ref. 10)
Tu2=u2+m2u2+mγu212Ωκ|γu|2ds
(6)
for every uD(T), where κ is the signed curvature of the boundary Ω (with the convention that κ < 0 if Ω is convex) and γ:W1,2(Ω;C2)L2(Ω;C2) is the boundary-trace operator. Formally, this is easily seen by expanding ‖(1 + i∂2)u12 and ‖(1i∂2)u22 and integrating by parts. To justify this approach, one needs an extra regularity of u. This is certainly a non-trivial matter because, while the curvature is piece-wise zero for triangles, it is not defined at the vertices.

Our main ingredient to prove an analog of the useful formula (6) for triangles is the following density result. The idea and proof are due to Krejčiřík.21 

Lemma 1.
Let Ω be any two-dimensional polygon with the set of vertices V. Then,
DD(T)C0(R2\{V})
is a core of T.

Proof.
Clearly, it is enough to consider the massless case m = 0. Moreover, by partition of unity, it suffices to consider the sector Ω ≔ {(r cos θ, r sin θ): r ∈ (0, ), θ ∈ (0, α)} with α ∈ (0, 2π). Let us denote θΩ ≔ {(r cos θ, r sin θ): r ∈ (0, )}. Consider the Dirac operator [(2) and (3)] (with m = 0), which involves infinite-mass boundary conditions,
u2=u1on0Ω,u2=eiαu1onαΩ,
(7)
in the sense of traces in W1,2(Ω;C2)u=u1u2. More specifically, D(T)=uW1,2(Ω;C2):(7)holds.
The crucial observation is that the functions
ψ1u2u1,ψ2u2eiαu1
satisfy the Dirichlet boundary condition on 0Ω and αΩ, respectively. The inverse formulas are given by
u11eiα1(ψ1ψ2),u21eiα1(eiαψ1ψ2).
  • Step 1: Approximation by bounded functions

    For any function ϕ:ΩR, define the vertical cutoff
    ϕN(x)Nifϕ(x)>N,ϕ(x)if|ϕ(x)|N,Nifϕ(x)<N.
    By definition, ϕNL(Ω)N. If ϕW1,2(Ω), we have
    ϕϕNW1,2(Ω)2=Ω|ϕϕN|2dx+Ω|ϕϕN|2dx={|ϕ|>N}|ϕN|2dx+{|ϕ|>N}|ϕ|2dx{|ϕ|>N}4|ϕ(x)|2dx+{|ϕ|>N}|ϕ(x)|2dxN0.
    Hence, ϕNϕ in W1,2(Ω) as N.
    In our case, if uD(T), we set
    ψN(Rψ1)N+i(Iψ1)N(Rψ2)N+i(Iψ2)NanduN1eiα1ψ1Nψ2Neiαψ1Nψ2N.
    Then, uND(T)L(Ω;C2) and
    uNuW1,2(Ω;C2)2=1|eiα1|2ψ1ψ1N+ψ2ψ2NW1,2(Ω)2+1|eiα1|2eiα(ψ1ψ1N)+ψ2ψ2NW1,2(Ω)22|eiα1|2(ψ1ψ1NW1,2(Ω)2+ψ2ψ2NW1,2(Ω)2)N0.
    Therefore, uNu in W1,2(Ω;C2) as N. Consequently,
    D1D(T)L(Ω;C2) is a core of T.
  • Step 2: Approximation by compactly supported functions

    Consider the cut-off sequence ξn: [0, ) → [0, 1] defined for every n ≥ 2 by
    ξn(r)0ifr[0,n2),log(n2r)log(n)ifr[n2,n1),1ifr[n1,n),log(n2r)log(n1)ifr[n,n2),0ifr[n2,).
    For every uD1, define un(x) ≔ ξn(|x|)u(x). Clearly, unL0(Ω;C2), by which we mean that un is bounded and vanishes in a neighborhood of 0 and in a neighborhood of infinity. Moreover, unD(H).
    Since ξn → 1 pointwise as n, it is easy to see that unu in L2(Ω;C2) as n by the dominated convergence theorem. Writing
    Ω{|x|1}|(unu)|22Ω{|x|1}(ξn1)2|u|2+2Ω{|x|1}|ξn|2|u|2,
    we see that the first term on the right-hand side tends to zero as n, as above due to the dominated convergence theorem. For the second term, we estimate
    Ω{|x|1}|ξn|2|u|2uL(Ω;C2)2Ω{|x|1}|ξn|2
    and use the polar coordinates to control the last integral as follows:
    Ω{|x|1}|ξn|2=2πnn21r2log2(n)rdr=2πlog(n)n0.
    In a similar manner, we verify that
    Ω{|x|1}|(unu)|2n0.
    Consequently,
    D2D(T)L0(Ω;C2) is a core of T.
  • Step 3: Approximation by smooth functions

    Let uD2. Then, the function ψu2eu1 is well defined, where θ(x) is the unique number in [0, α] with x1 = |x|cos[θ(x)] and x2 = |x|sin[θ(x)] for every x ∈ Ω. Since ψ vanishes on Ω, there exists a sequence {ψj}C0(Ω) such that ψjψ in W1,2(Ω;C2) as j.

Since Ω satisfies the segment condition, there also exists a sequence {u1j}C0(R2) such that u1ju1 in W1,2(Ω;C2) as j. Since u1 vanishes in a neighborhood of zero, the sequence can be chosen to lie in C0(R2\{0}).

Define u2jψj+eiθu1jC0(R2\{0}). Then, u2ju2 in W1,2(Ω;C2) as j. Moreover, uj satisfies (7). Consequently,
D=D(T)L0(Ω;C2)C0(R2\{0}) is a core of T.
(8)
This concludes the proof of the lemma.□

As a special consequence, the norm of Ta,bu can be computed explicitly by using integration by parts. Formally, the result coincides with formula (6) for smooth domains with κ = 0.

Theorem 2.
For every uD(Ta,b),
Ta,bu2=uL2(Ωa,b)2+m2uL2(Ωa,b)2+mγuL2(Ωa,b)2.
(9)

Proof.
By virtue of Lemma (1), for every u=u1u2 in D(Ha,b), there exists a sequence {un=u1nu2n} in D such that unnu in H1a,b). Using integration by parts, we compute
Ta,bun2=mu1ni(1i2)u2n2+mu2n+i(1+i2)u1n2=m2un2+un22mR(u1n,i(1i2)u2n)+2mR(u2n,i(1+i2)u1n)2R(1u2n,i2u2n)+2R(1u1n,i2u1n)=m2un2+un2+2mR0biū1n(0,x2)u2n(0,x2)dx22mR0biū1nabx2+a,x2u2nabx2+a,x2dx2+2mR0aū1n(x1,0)u2n(x1,0)dx12mR0aū1nx1,bax1+bu2nx1,bax1+bdx12R0biū1n(0,x2)2u1n(0,x2)dx2+2R0biū1nabx2+a,x22u1nabx2+a,x2dx2+2R0aiū1n(x1,0)2u1n(x1,0)dx12R0aiū1nbax1+b,x12u1nx1,bax1+bdx1+2R0biū2n(0,x2)2u2n(0,x2)dx22R0biū2nabx2+a,x22u2nabx2+a,x2dx22R0aiū2n(x1,0)2u2n(x1,0)dx1+2R0aiū2nbax1+b,x12u2nx1,bax1+bdx1.
Substituting the boundary conditions, we have
0biū1n(0,x2)u2n(0,x2)dx2=0b|u1n(0,x2)|2dx2=u1nOB2=12unOB2,0aū1n(x1,0)u2n(x1,0)dx1=0b|u1n(x1,0)|2dx1=u1nOA2=12unOA2,R0biū1nabx2+a,x2u2nabx2+a,x2dx2=ba2+b20b|u1nabx2+a,x2|2dx2=b2a2+b2u1nAB2=12b2a2+b2unAB2,R0aiū1nbax1+b,x1u2nbax1+b,x1dx1=a2a2+b2u1nAB2=12a2a2+b2unAB2.
Moreover, using the boundary conditions, an integration by parts on the edge of the triangle, and the fact that the approximating sequence vanishes in a vicinity of the vertices, we have
R0biū2n(0,x2)2u2n(0,x2)dx2=R0bū1n(0,x2)2u2n(0,x2)dx2=R0b2ū1n(0,x2)u2n(0,x2)dx2+ū1n(0,0)u2n(0,0)ū1n(0,b)u2n(0,b)=R0b2ū1n(0,x2)u2n(0,x2)dx2=R0bi2ū1n(0,x2)u1n(0,x2)dx2=R0biū1n(0,x2)2u1n(0,x2)dx2.
By analogous manipulations, we have
R0biū1nabx2+a,x22u1nabx2+a,x2dx2=R0biū2nabx2+a,x22u2nabx2+a,x2dx2,R0aiū1n(x1,0)2u1n(x1,0)dx1=R0aiū2n(x1,0)2u2n(x1,0)dx1,R0aiū1nbax1+b,x12u1nx1,bax1+bdx1=R0aiū2nbax1+b,x12u2nx1,bax1+bdx1.
Putting all these identities together, we obtain the formula
Ta,bun2=unL2(Ωa,b)2+m2unL2(Ωa,b)2+mγunL2(Ωa,b)2
valid for all unD. Taking n, we obtain the desired result.□

Remark 1.

Applying the similar arguments, we can prove the validity of an analog of formula (9) for arbitrary planar polygons (see the Appendix).

For further purposes, given any m ≥ 0 and arbitrary positive numbers a, b, and L, let us consider the one-dimensional Dirac operator,
HLmimmiinL20,L;C2,D(HLm)φW1,20,L;C2,φ2(L)=aa2+b2+iba2+b2φ1(L),φ2(0)=φ1(0).
(10)

Proposition 1.

The operator HLm is self-adjoint.

Proof.
We follow (Ref. 22, Appendix A). Since the multiplication by 0mm0 generates a bounded self-adjoint operator on L2((0,L);C2), it is enough to prove the self-adjointness of HL0. To do that, we commence with the definition of the adjoint,
D(HL0*)=uL2(0,L);C2:wL2(0,L);C2 such that vD(HL0),(u,HL0v)L2((0,L);C2)=(w,v)L2((0,L);C2).
For every vC0(0,L);C2 and uD(HL0*), there holds
(HL0*u,v)L2((0,L);C2)=(u,HL0v)L2((0,L);C2)=(u1,iv1)L2((0,L);C)+(u2,iv2)L2((0,L);C)=u1u2,iv1iv2D,D=iu1iu2,v1v2D,D,
where ,D,D is the duality bracket of distributions. In particular, we know that HL0*u=iu1iu2L2(0,L);C2; thus, we get uH1(0,L);C2. Moreover, if vD(HL0), there holds
(HL0*u,v)L2((0,L);C2)=(u1,iv1)L2((0,L);C)+(u2,iv2)L2((0,L);C)=(iu1,v1)L2((0,L);C)+(iu2,v2)L2((0,L);C)+iu2̄v2|0Liu1̄v1|0L=(iu1,v1)L2((0,L);C)+(iu2,v2)L2((0,L);C)+iv1(L)u2̄(L)a+bia2+b2u1(L)̄+iv1(0)[u1(0)̄u2(0)̄].
Since it holds for any arbitrary vD(HL0), we obtain that uD(HL0), that is, D(HL0)=D(HL0*).□

By dint of the compactness embedding H1((0,L);C2) into L2((0,L);C2) and since D(HLm) is continuously embedded in H1((0,L);C2), then we deduce that the spectrum Sp(HLm) of the self-adjoint operator HLm is purely discrete. In the following, we compute the eigenvalues.

First of all, we observe that any eigenvalue λ ∈ Sp(HLm) necessarily satisfies λ2 > m2. Indeed, by computing the square norm of the operator
HLmu2=iu1+mu22+iu2mu12=u2+2Rm(iu1,u2)2Rm(u1,iu2)+m2u2=u2+m2u2+2Rm0Liu1̄u22Rm0Lu1̄iu2=u2+m2u22Rmiu1̄u2|0L=u2+m2u2+mba2+b2|u(L)|2m2u2,
(11)
we immediately obtain λ2m2. The inequality is actually strict because λ2 = m2 would imply that u is a constant, which is impossible unless u = 0.
Let λ ∈ Sp(HLm), and let u=(u1,u2)D(HLm) be an associated eigenfunction. It satisfies
iu1mu2=λu1,iu2mu1=λu2
(12)
or, equivalently,
mu2=iu1+λu1,mu1=iu2λu2.
(13)
Differentiating both sides of Eq. (12) and combining with Eq. (13), we obtain differential equations that the components of u must satisfy separately,
u1=(λ2m2)u1,u2=(λ2m2)u2.
Putting Eλ2m2 > 0, the general solutions read
u1(x)=C1cosEx+C2sinEx,u2(x)=C1̃cosEx+C2̃sinEx,
(14)
where C1,C2,C1̃,C2̃ are complex constants. The boundary condition u2(0) = u1(0) directly implies that C1=C1̃. Substituting this expression of u into (12), we have
iC1EsinEx+iC2EcosEx+mC1cosEx+mC2̃sinEx=λ(C1cosEx+C2sinEx),iC1EsinEx+iC2̃EcosExmC1cosExmC2sinEx=λ(C1cosEx+C2̃sinEx).
From these equalities, we deduce
C2=i(λ+m)EC1,C2̃=i(λm)EC1.
Putting Mλ+mE0, then we obtain
u1(x)=C1cosEx+iMC1sinEx,u2(x)=C1cosExiMC1sinEx,
(15)
with C1 being a non-zero complex constant. The boundary condition u2(L)=a+bia2+b2u1(L) requires
C1cosELiMC1sinELC1cosEL+iMC1sinEL=a+bia2+b2,
which is equivalent to
cos2ELM2sin2ELiMsin2ELcos2EL+M2sin2EL=a+bia2+b2.
Considering the real and imaginary parts separately, it is equivalent to the system
cos2ELM2sin2ELcos2EL+M2sin2EL=aa2+b2,Msin2ELcos2EL+M2sin2EL=ba2+b2.
From the second equation, we infer that Msin2EL<0, so cosEL cannot be equal zero. Putting zMtanEL<0 and dividing both the numerator and denominator of the left fractions of these equations by cos2EL, we have
1z21+z2=aa2+b2,2z1+z2=ba2+b2.
Therefore, z satisfies the equation
1z22z=ab.
Since z is the negative solution of the quadratic equation z2+2abz1=0, one has
z=z0aba2b2+1<0.
In summary, the eigenvalue λ of HLm satisfies the implicit equation,
λ+mλ2m2tanλ2m2Lz0=0.
(16)
From formula (11), we have
HLmu2m2u2u2=HL0u2
for all uD(HLm)=D(HL0). Let λ1 = λ1(m) be the closest-to-zero eigenvalue of HLm, and set E1(m) ≔ λ1(m)2m2. As a consequence,
E1(m)=λ1(m)2m2λ1(0)2=E1(0)
for every m ≥ 0. When m = 0, we have
aba2b2+1=tanE1(0)LtanE1(0)L=ab+a2b2+1>1E1(0)L=arctan(z0)>π4λ12(m)m2λ12(0)=arctan|z0|2L2>π216L2.
Applying a variational formulation analogous to (4) and a unitary equivalence, we have just established the following Poincaré-type inequality.

Lemma 2.
For every ϕD(HLm),
ϕ2+mba2+b2|ϕ(L)|2(λ12m2)ϕ2arctan2ab+a2b2+1L2ϕ2.

Simultaneously with HLm, we also consider the operator GLm, which acts as HLm but has a different domain,
D(GLm)φW1,20,L;C2,φ2(L)=aa2+b2+iba2+b2φ1(L),φ2(0)=iφ1(0).
By the similar approach, we have that GLm is self-adjoint and its spectrum is purely discrete. We compute the square norm of the operator
GLmu2=iu1+mu22+iu2mu12=u2+2Rm(iu1,u2)2Rm(u1,iu2)+m2u2=u2+m2u2+2Rm0Liu1̄u22Rm0Lu1̄iu2=u2+m2u22Rmiu1̄u2|0L=u2+m2u2+mba2+b2|u(L)|2+m|u(0)|2m2u2.
(17)
Thus, if η ∈ Sp(GLm), then |η| > m for every m ≥ 0. Let v=(v1,v2)D(GLm) be an associated eigenfunction. Then,
iv1mv2=ηv1,iv2mv1=ηv2
(18)
or, equivalently,
mv2=iv1+ηv1,mv1=iv2ηv2.
(19)
Differentiating both sides of Eq. (18) and combining with Eq. (19), we have
v1=(η2m2)v1,v2=(η2m2)v2.
Putting Fη2m2 > 0, general solutions read
v1(x)=D1cosFx+D2sinFx,v2(x)=D1̃cosFx+D2̃sinFx.
(20)
The boundary conditions u2(0) = −iu1(0) directly imply that D1̃=iD1. Substituting this expression of u into (18), we have
iD1FsinFx+iD2FcosFx+m(iD1cosFx+D2̃sinFx)=η(D1cosFx+D2sinFx),D1FsinFx+iD2̃FcosFxmD1cosFxmD2sinFx=η(iD1cosFx+D2̃sinFx).
From these equalities, we deduce that
D2=(iη+m)FD1,D2̃=(η+im)FD1.
Therefore,
v1(x)=D1cosFx+(iη+m)FD1sinFx,v2(x)=iD1cosFx+(η+im)FD1sinFx,
(21)
with D1 being a non-zero complex constant. From the boundary condition u2(L)=a+bia2+b2u1(L), we deduce that v1(L) ≠ 0 and
cosFL+(iη+m)FsinFLcosFL+(iη+m)FsinFL=ia+bia2+b2,
which is equivalent to
cosFL+mFsinFLiηFsinFLcosFL+iηFsinFL=aiba2+b2.
Putting AcosFL+mFsinFL and BηFsinFL, it is equivalent to the system
A2B2A2+B2=ba2+b2,2ABA2+B2=aa2+b2>0.
It implies that AB=ba+b2a2+1, so
FηcotFL+mη=ba+b2a2+1:=α0>0
or, equivalently,
f(η)FcotFL+mα0η=0,
(22)
where f is defined on (−, −m) ∪ (m, ). We compute the limits
limηmf(η)=1L+m+α0m>0andlimηm2+π2L2f(η)=.
By the continuity of the left-hand side of Eq. (22), we obtain that it always have solution when FL(0,π), and thus, we restrict on this range to study properties of the closest-to-zero eigenvalue η1.
If η < 0, then FcotFL+m<0, and as a result, cotFL<0 and FL(π2,π). From Eq. (22), we have
FcotFL+m=α0ηα0η=FcotFLm<FπFLmα0<ηmα0<F(πFL)m1m(α0+1)<FπFLF>π(mα0+m)1+L.
If η ≥ 0, from the implicit equation, we also obtain that
η(m)=m(sin2FL2LF)2Fsin2FLη(sin2FL2LF)2α02Fsin2FL>0,(η2m2)=4(ηmα0)Fsin2FLη(sin2EL+2FL)+2α02Fsin2FL>0.
It shows that η2m2 is a strictly increasing function with respect to m, but we cannot give an m-dependent lower bound of η12m2 due to the fact that the spectrum of GLm is not symmetric.
From the square norm of operator (17), we deduce that η12m2 is a non-decreasing function and lies in the range (π216L2,π2L2), and thus, limmm|η1|=1. We have that {η12m2}m0 is a non-decreasing sequence and is uniformly bounded; then, there exists
limmη12m2π2L2.
Evaluating the implicit equation [Eq. (22)] with limmFη1=0,limmmη1 is finite, FL(π4,π). It implies that limmcotFL must be equal minus infinity, and thus, we have FLmπ. As a consequence, η12m2π2L2, which is the first eigenvalue of the Dirichlet Laplacian. In summary, we have established the lower bound
λ12(m)m2λ2(0)=arctan2ba+1+b2a2L2>π216L2.
It yields the following Poincaré-type inequality.

Lemma 3.
For every ϕD(GLm), we have
ϕ2+mba2+b2|ϕ(L)|2+m|u(0)|2(λ12m2)ϕ2arctan2ba+1+b2a2L2ϕ2>π216L2ϕ2.

Remark 2.
When F1L=π2 with respect to m = m0, then the implicit equation [Eq. (22)] becomes
0+m0=ηα0,
so η12=m02α02 and η12m02=(1α021)m02=π24L2. It implies that m0=π2Lα021α02. For every mm0, F1Lπ2 and cotF1L0. Combining with Eq. (22), we deduce that η1α0m ≥ 0, and thus, we obtain the other lower bound
F1=η12m21α021m2.

Remark 3.
We can replace HLm by the unitarily equivalent operator H̃Lmv*G̃Lmv with vi001, where G̃Lm acts as GLm but has a different domain,
D(G̃Lm)φW1,20,L;C2,φ2(L)=aia2+b2+ba2+b2φ1(L),φ2(0)=iφ1(0).
Thus, D(H̃Lm)=D(HLm) and H̃Lm have the same spectrum as G̃Lm, which is defined by the equation
FcotFL+m=ηab+a2b2+1=η|z0|.

Now, we are in a position to prove Theorem 1.

Proof of Theorem 1.
Recall from formula (4) that the first squared positive eigenvalue of the operator Ta,b can be computed by
λ1(a,b)2=infψD(Ta,b)ψ0T̂a,bψ2ψ2
with
Ta,bψ2=ψL2(Ωa,b)2+m2ψL2(Ωa,b)2+mγψL2(Ωa,b)2
obtained from Theorem (2). We therefore have
λa,b2m2=infψD(Ta,b)ψ01ψ2+2ψ2+mγuOA2+mγuOB2+mγuAB2ψ2.
(23)
Combining Fubini’s theorem and Lemma 2, we have
1ψ2+2ψ2+mγuOA2+mγuOB2+mγuAB2=0a0bxa+b|2ψ(x,y)|2+m|u(x,0)|2+m|ux,bxa+b|2dydx+0b0ayb+a|1ψ(x,y)|2+m|u(0,y)|2+m|uayb+a,y|2dxdy0a0bxa+barctan2ba+1+b2a2bxa+b2|ψ(x,y)|2dydx+0b0ayb+aarctan2ab+1+a2b2ayb+a2|ψ(x,y)|2dxdy,arctan2ab+1+a2b2b2+arctan2ba+1+b2a2a2ψ2.
We achieve that
λ1(a,b)2m2arctan2ab+1+a2b2b2+arctan2ba+1+b2a2a2>π216a2+π216b2.
This concludes the proof of the lower bound of Theorem 1.
The upper bound follows by using the eigenfunction of the Dirichlet Laplacian on the right triangle Ω1,1 (see Ref. 23, Sec. 4.3),
ψo(x,y)sin(2πx)sin(πy)+sin(2πy)sin(πx)11,
as a trial function in (23). After computing the value of the Rayleigh quotient, we get
λ1(a,b)2m25π221a2+1b2.
This concludes the Proof of Theorem 1.□

Finally, we establish Corollary 1.

Proof of Corollary 1.

By scaling, we can assume, without loss of generality, that the double area equals 1 and the perimeter equals 2+2. These values correspond to the area and the perimeter of the isosceles right triangle in case k = 1, respectively.

For the area constraint ab = 1, we take b ≔ 1/a using Theorem 1 and we have λ1(1, 1) − m2 ≤ 5π2. If one side length a just satisfies the condition
1a2+a280,
then Conjecture 2 will be satisfied among all right triangles Ωa,b. It is easy to see that when a ≥ 9 or a19, the condition holds. This establishes Conjecture 2 (i).
For the perimeter constraint, we take b2+2a and restrict ourselves to a(0,2+2). By the similar arguments as above, we arrive at the sufficient condition
1a2+1(2+2a)280
to have the desired inequality λ1(a, b) ≥ λ1(1, 1). It is not hard to see that the inequality holds, provided that a ≥ 3.5 or a19. This establishes Conjecture 2 (ii).
Moreover, without using scaling, we deduce that when
arctan2ab+1+a2b2b2+arctan2ba+1+b2a2a25π2k2
happens, then Conjectures 2 (i) and (ii) hold.□

Remark 4.
Taking Remark (2) into account, we choose
m=m0(L=a)=π2aα021α02.
Then, λ12(a,b)m02π24a2+π216b2. By the non-decreasing property of F1(m), we obtain that
λ12(a,b)m2π24a2+π216b2
for all mπ2aα021α02. It gives a better estimate for the lower bound.

A direct consequence obtained from Remark (4) is the following corollary, which extends the range of the side long a for the validity of Conjecture 2.

Corollary 2.
Let k be defined as in Conjecture 2 and mπ2aα021α02.Conjecture 2. (i) Holds under the following extra hypotheses:
a9korak5.
Conjecture 2. (ii) Holds under the following extra hypotheses:
a3.5korak5.

In this section, we give the Proof of Remark 1, extending the validity of formula (9) to polygons. We consider a polygon OBCDA with coordinates A ≔ (a, 0), B ≔ (b, c), C ≔ (c1, c2), D ≔ (d1, d2); see Fig. 2. Without loss of generality, we can suppose 0 < b < a < c1 < d1. The proof for the general polygon is the same as in this case. Dividing the polygon into triangles is a crucial step to achieve the proof.

FIG. 2.

The pentagon OBCDA.

FIG. 2.

The pentagon OBCDA.

Close modal

Theorem 3.
Tu2=uL2(OBCDA)2+m2uL2(OBCDA)2+mγuL2((OBCDA))2.

Proof.
Using the integration by parts and the density arguments described in Theorem 2, we compute the norm ‖Tu2 in every triangle divided. First, we apply on the triangle OAB,
Tu2=mu1i(1i2)u22+i(1+i2)u1+mu22=m2u2+u22R(1u2,i2u2)+2R(1u1,i2u1)2mR(u1,i(1i2)u2)2miR((1+i2)u1,u2)=m2u2+u22R(1u2,i2u2)+2R(1u1,i2u1)2Rmi0cu1̄u2|x=bcyx=y(ba)c+ady2Rm0bu1̄u2|y=0y=cxbdx2Rmbau1̄u2|y=0y=cxacbadx=m2u2+u22R(1u2,i2u2)+2R(1u1,i2u1)+muOA2+muOB22Rmi0cu1̄u2|x=y(ba)c+ady2Rmbau1̄u2|y=cxacbadx=m2u2+u22R(1u2,i2u2)+2R(1u1,i2u1)+muOA2+muOB2+AB2mRin1ABu1̄u22mRn2ABu1̄u2.
Putting I12R(1u2,i2u2) and I22R(1u1,i2u1), we compute
I1=2R(u2,i12u2)2R0cu2̄i2u2|x=bycx=y(ba)c+ady=2R(2u2,i1u2)2R0cu2̄i2u2|x=bycx=y(ba)c+adx+2R0bu2̄i1u2|y=0y=cxbdx+2Rbau2̄i2u2|y=0y=cxacbadx=I12R0cu2̄i2u2|x=bycx=y(ba)c+adx+2R0bu2̄i1u2|y=0y=cxbdx+2Rbau2̄i2u2|y=0y=cxacbadx.
Therefore, we get
I1=Ri0cu2̄2u2|x=bycx=y(ba)c+ady+Ri0bu2̄1u2|y=0y=cxbdx+Ribau2̄2u2|y=0y=cxacbadx.
An analogous computation gives
I2=Ri0cu1̄2u1|x=bycx=y(ba)c+adxRi0bu1̄1u1|y=0y=cxbdxRibau1̄2u1|y=0y=cxacbadx.
In addition, on the side OB, we have u2=bcib2+c2u1 and
0bu2̄1u2|y=cxbdx=0bb+cib2+c2u1̄1u2|y=cxbdx=0bb+cib2+c21u1̄bcib2+c2u1|y=cxbdx=0b1u1̄u1|y=cxbdx.
As a result, we obtain
Ri0b(u2̄1u2u1̄1u1)|y=cxbdx=0.
Similarly, we also have
Ri0c(u2̄2u2u1̄2u1)|x=bycdx=0
and
Ri0a(u2̄1u2u1̄1u1)|y=0dx=0.
Hence,
I1+I2=Ri0c(u2̄2u2u1̄2u1)|x=y(ba)c+ady+Riba(u2̄1u2u1̄1u1)|y=cxacba,
which is equivalent to
I1+I2=Rin1ABAB(u2̄2u2u1̄2u1)+Rin2ABAB(u2̄1u2u1̄1u1).
In summary, the norm computed on the triangle OAB reads
TuL2(OAB)2=m2uL2(OAB)2+uL2(OAB)2+mγuOA2+mγuOB2+AB2mRin1ABu1̄u22mRn2ABu1̄u2Rin1ABAB(u2̄2u2u1̄2u1)+Rin2ABAB(u2̄1u2u1̄1u1).
We suppose that n is the outward unit vector in each triangle divided. This is the same notation, but it is different depending on each triangle. By analogous computations, we obtain the square of the operator defined on the other triangles as follows:
TuL2(ABC)2=m2uL2(ABC)2+uL2(ABC)2+mγuBC2+AB(2mRin1ABu1̄u22mRn2ABu1̄u2)Rin1ABAB(u2̄2u2u1̄2u1)+Rin2ABAB(u2̄1u2u1̄1u1)+AC(2mRin1ACu1̄u22mRn2ACu1̄u2)Rin1ACAC(u2̄2u2u1̄2u1)+Rin2ACAC(u2̄1u2u1̄1u1)
and
TuL2(ADC)2=m2uL2(ADC)2+uL2(ADC)2+mγuDC2+mγuDA2+AC2mRin1ACu1̄u22mRn2ACu1̄u2Rin1ACAC(u2̄2u2u1̄2u1)+Rin2ACAC(u2̄1u2u1̄1u1).
When dividing the polygons into triangles, we deduce that the outward normal in the inner sides in adjacent triangles is opposite; then, the integration computed in these sides will be canceled. Summarizing these computations, we obtain the following formula:
TuL2(OBCDA)2=m2uL2(OBCDA)2+uL2(OBCDA)2+mγuOA2+mγuOB2+mγuDA2+mγuCB2++mγuCD2.
Therefore, the proof for the square of the operator is completed.□

The author is grateful to David Krejčiřík for useful discussions and thanks the editor and anonymous referee for valuable comments. The author was supported by the EXPRO grant (No. 20-17749X) of the Czech Science Foundation.

The author has no conflicts to disclose.

Tuyen Vu: Writing – original draft (lead).

The data that support the findings of this study are available within the article.

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