# Spectral Properties of the Differential Operators of the Fourth-Order with Eigenvalue Parameter Dependent Boundary Condition

Authors: Ziyatkhan S. Aliyev; Nazim B. Kerimov
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

We consider the fourth-order spectral problemy(4)(x)-(q(x)y′(x))′=λy(x),x∈(0,l) with spectral parameter in the boundary condition. We associate this problem with a selfadjoint operator in Hilbert or Pontryagin space. Using this operator-theoretic formulation and analytic methods, we investigate locations (in complex plane) and multiplicities of the eigenvalues, the oscillation properties of the eigenfunctions, the basis properties in Lp(0,l), p∈(1,∞), of the system of root functions of this problem.


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## Body

## 1. Introduction

The following boundary value problem is considered:(1.1)y(4)(x)-(q(x)y′(x))′=λy(x),x∈(0,l),′:=ddx, (1.2a)y′(0)=0,(1.2b)y(0)cos⁡β+Ty(0)sin⁡β=0,(1.2c)y′(l)cos⁡⁡γ+y′′(l)sin⁡⁡γ=0,(1.2d)(aλ+b)y(l)-(cλ+d)Ty(l)=0,where λ is a spectral parameter, Ty≡y′′′-qy′, q is absolutely continuous function on [0,l], β, γ, a, b, c, and d are real constants such that 0≤β, γ≤π/2 and σ=bc-ad≠0. Moreover, we assume that the equation(1.3)y′′-qy=0,
is disfocal in [0,l], that is, there is no solution of (1.3) such that y(a)=y′(b)=0 for any a,b∈[0,l]. Note that the sign of q which satisfies the disfocal condition may change in [0,l].Problems of this type occur in mechanics. Ifβ=0, γ=π/2, b=c=0, and d=1 in the boundary conditions, then the problem (1.1), (1.2a)–(1.2d) arises when variables are separated in the dynamical boundary value problem describing small oscillations of a homogeneous rod whose left end is fixed rigidly and on whose right end a servocontrol force in acting. In particular, the case when a<0 corresponds to the situation where this is a particle of mass a at the right end of the rod. For more complete information about the physical meaning of this type of problem see [1–3].Boundary value problems for ordinary differential operators with spectral parameter in the boundary conditions have been considered in various formulations by many authors (see, e.g., [1, 4–25]). In [14–16, 20, 22] the authors studied the basis property in various function spaces of the eigen- and associated function system of the Sturm-Liouville spectral problem with spectral parameter in the boundary conditions. The existence of eigenvalues, estimates of eigenvalues and eigenfunctions, oscillation properties of eigenfunctions, and expansion theorems were considered in [4, 7, 9, 12, 17, 18, 21, 24] for fourth-order ordinary differential operators with a spectral parameter in a boundary condition. The locations, multiplicities of the eigenvalues, the oscillation properties of eigenfunctions, the basis properties in Lp(0,l),p∈(1,∞), of the system of root functions of the boundary value problem (1.1), (1.2a)–(1.2d) with q≥0, σ>0, are considered in [18] and, with q≥0, σ<0, c=0, are considered in [4, 5].The subject of the present paper is the study of the general characteristics of eigenvalue locations on a complex plane, the structure of root subspaces, the oscillation properties of eigenfunctions, the asymptotic behaviour of the eigenvalues and eigenfunctions, and the basis properties inLp(0,l), p∈(1,∞), of the system of root functions of the problem (1.1), (1.2a)–(1.2d).Note that the sign ofσ plays an essential role. In the case σ>0 we associate with problem (1.1), (1.2a)–(1.2d) a selfadjoint operator in the Hilbert space H=L2(0,l)⊕ℂ with an appropriate inner product. Using this fact and extending analytic methods to fourth-order problems, we show that all the eigenvalues are real and simple and the system of eigenfunctions, with arbitrary function removed, forms a basis in the space Lp(0,l), p∈(1,∞). For σ<0 problem (1.1), (1.2a)–(1.2d) can be interpreted as a spectral problem for a selfadjoint operator in a Pontryagin space Π1. It is proved below that nonreal and nonsimple (multiple) eigenvalues are possible and the system of root functions, with arbitrary function removed, forms a basis in the space Lp(0,l), p∈(1,∞), except some cases where the system is neither completed nor minimal.


## 2. The Operator Interpretation of the Problem (1.1), (1.2a)–(1.2d)

LetH=L2(0,l)⊕ℂ be a Hilbert space equipped with the inner product(2.1)(ŷ,û)H=({y,m},{u,s})H=(y,u)L2+|σ-1|ms¯,
where (y,u)L2=∫0lyu¯dx.We define in theH operator(2.2)Lŷ=L{y,m}={(Ty(x))′,dTy(l)-by(l)}
with domain(2.3)D(L)={y̂={y,m}∈H/y(x)∈W24(0,l),(Ty(x))′∈L2(0,l),y∈(B.C.),m=ay(l)-cTy(l)},
that is dense in H [23, 25], where (B.C.) denotes the set of separated boundary conditions (1.2a)–(1.2c).Obviously, the operatorL is well defined. By immediate verification we conclude that problem (1.1), (1.2a)–(1.2d) is equivalent to the following spectral problem:(2.4)Lŷ=λŷ,ŷ∈D(L),
that is, the eigenvalue λn of problem (1.1), (1.2a)–(1.2d) and those of problem (2.4) coincide; moreover, there exists a correspondence between the eigenfunctions and the adjoint functions of the two problems:(2.5)ŷn={yn(x),mn}⟷yn(x),mn=ayn(l)-cTyn(l).
Problem (1.1), (1.2a)–(1.2d) has regular boundary conditions in the sense of [23, 25]; in particular, it has a discrete spectrum.Ifσ>0, then L is a selfadjoint discrete lower-semibounded operator in H and hence has a system of eigenvectors {{yn(x),mn}}n=1∞, that forms an orthogonal basis in H.In the caseσ<0 the operator L is closed and non-selfadjoint and has compact resolvent in H. In H we now introduce the operator J by J{y,m}={y,-m}. J is a unitary, symmetric operator in H. Its spectrum consists of two eigenvalues: −1 with multiplicity 1, and +1 with infinite multiplicity. Hence, this operator generates the Pontryagin space Π1=L2(0,l)⊕ℂ by means of the inner products (J-metric) [26]:(2.6)(ŷ,û)Π1=({y,m},{u,s})Π1=(y,u)L2+σ-1ms¯.Lemma 2.1.
L is a J-selfadjoint operator in Π1.Proof.
JL is selfadjoint in H by virtue of Theorem 2.2 [11]. Then, J-selfadjointness of L on Π1 follows from [27, Section 3, Proposition 30].Lemma 2.2 (see [27, Section 3, Proposition 50]).
LetL* be an operator adjoined to the operator L in H.  Then, L*=JLJ.Letλ be an eigenvalue of operator L of algebraic multiplicity ν. Let us suppose that ρ(λ) is equal to ν if Imλ≠0 and equal to whole part ν/2 if Imλ=0.Theorem 2.3 (see [28]).
The eigenvalues of operatorL arrange symmetrically with regard to the real axis. ∑k=1nρ(λk)≤1 for any system {λk}k=1n(n≤+∞) of eigenvalues with nonnegative parts.From Theorem2.3 it follows that either all the eigenvalues of boundary value problem (1.1), (1.2a)–(1.2d) are simple (all the eigenvalues are real or all, except a conjugate pair of nonreal, are real) or all the eigenvalues are real and all, except one double or triple, are simple.


## 3. Some Auxiliary Results

As in [17, 19, 29, 30] forthe analysis of the oscillation properties of eigenfunctions of the problem (1.1), (1.2a)–(1.2d) we will use a Prüfer-type transformation of the following form:(3.1)y(x)=r(x)sin⁡⁡ψ(x)cos⁡⁡θ(x),y′(x)=r(x)cos⁡⁡ψ(x)sin⁡⁡φ(x),y′′(x)=r(x)cos⁡⁡ψ(x)cos⁡⁡φ(x),Ty(x)=r(x)sin⁡⁡ψ(x)sin⁡⁡θ(x).Consider the boundary conditions (see [29, 30])(1.2a*)y′(0)cos⁡⁡α-y′′(0)sin⁡⁡α=0,(1.2d*)y(l)cos⁡⁡δ-Ty(l)sin⁡⁡δ=0,
where α∈[0,π/2], δ∈[0,π).Alongside the spectral problem (1.1), (1.2a)–(1.2d) we will consider the spectral problem (1.1), (1.2a)–(1.2c), and (1.2d*). In [30], Banks and Kurowski developed an extension of the Prüfer transformation (3.1) to study the oscillation of the eigenfunctions and their derivatives of problem (1.1), (1.2a*), (1.2b), (1.2c), and (1.2d*) with q≥0, δ∈[0,π/2] and in some cases when (1.3) is disfocal and α=γ=0, δ∈[0,π/2]. In [19], the authors used the Prüfer transformation (3.1) to study the oscillations of the eigenfunctions of the problem (1.1), (1.2a*), (1.2b), (1.2c), and (1.2d*) with q≥0 and δ∈(π/2,π). In this work it is proved that problem (1.1), (1.2a*), (1.2b), (1.2c), and (1.2d*) may have at most one negative and simple eigenvalue and sequence of positive and simple eigenvalues tending to infinity, the number of zeros of the eigenfunctions corresponding to positive eigenvalues behaves in that usual way (it is equal to the serial number of an eigenvalue increasing by 1); the function associated with the lowest eigenvalue has no zeros in (0,l) (however in reality, this eigenfunction has no zeros in (0,l) if the least eigenvalue is positive; the number of zeros can by arbitrary if the least eigenvalue is negative). In [31], Ben Amara developed an extension of the classical Sturm theory [32] to study the oscillation properties for the eigenfunctions of the problem (1.1), (1.2a)–(1.2c), and (1.2d*) with β=0, in particular, given an asymptotic estimate of the number of zeros in (0,l) of the first eigenfunction in terms of the variation of parameters in the boundary conditions.Letu be a solution of (1.3) which satisfies the initial conditions u(0)=0, u′(0)=1. Then the disfocal condition of (1.3) implies that u′(x)>0 in [0,l]. Therefore, if h denotes the solution of (1.3) satisfying the initial conditions u(0)=c>0, u′(0)=1, where c is a sufficiently small constant, then we have also h′(x)>0 on [0,l]. Thus, h(x)>0 in [0,l], and hence the following substitutions [33, Theorem 12.1]:(3.2)t=t(x)=lω-1∫0xh(s)ds,ω=∫0lh(s)ds,
transform [0,l] into the interval [0,l] and (1.1) into(3.3)(pÿ)⋅⋅=λry,
where p=(lω-1h)3, r=l-1ωh-1; h(x), y(x) are taken as functions of t and ·:=d/dt. Furthermore, the following relations are useful in the sequel:(3.4)ẏ=l-1ωh-1y′,l2ω-2h3ÿ=hy′′-h′y′,T̃y≡((lω-1h)3ÿ)⋅=Ty.
It is clear from the second relation (3.4) that the sign of y′′ is not necessarily preserved after the transformation (3.2). For this reason this transformation cannot be used in any straightforward way. The following lemma of Leighton and Nehari  [33] will be needed throughout our discussion. In [30, Lemma 2.1], Banks and Kurowski gave a new proof of this lemma for q≥0. However, in the case when (1.3) is disfosal on (0,l], they partially proved it [30, Lemma 7.1], and therefore they were able to study problem (1.1), (1.2a)–(1.2c), and (1.2d*) with γ=0, δ∈[0,π/2]. In [31], Ben Amara shows how Lemma 3.1 together with the transformation (3.2) can be applicable to investigate boundary conditions (1.2a)–(1.2c), and (1.2d*) with β=0.Lemma 3.1 (see [33, Lemma 2.1]).
Letλ>0, and let y be a nontrivial solution of (3.3). If y, ẏ, ÿ, and T̃y are nonnegative at t=a (but not all zero), they are positive for all t>a. If y, -ẏ, ÿ, and -T̃y are nonnegative at t=a (but not all zero), they are positive for all t<a.We also need the following results which are basic in the sequel.Lemma 3.2.
All the eigenvalues of problem (1.1), (1.2a)–(1.2c), and (1.2d*) for δ∈[0,π/2) or δ=π/2, β∈[0,π/2) are positive.Proof.
In this case, the transformed problem is determined by (3.3) and the boundary conditions(3.5a)ẏ(0)=0,(3.5b)y(0)cos⁡⁡β+T̃y(0)sin⁡⁡β=0,(3.5c)ẏ(l)cos⁡⁡γ*+p(l)ÿ(l)sin⁡⁡γ*=0,(3.5d)y(l)cos⁡⁡δ-T̃y(l)sin⁡⁡δ=0,where γ*=arctg{l-2ω2h-1(l)[h(l)cosγ+h′(l)sinγ]-1}∈[0,π/2).
It is known that the eigenvalues of (3.3), (3.5a)–(3.5d) are given by the max-min principle [13, Page 405] using the Rayleigh quotient
(3.6)R[y]=(∫0lpÿ2dt+N[y])(∫0ly2dt),
where N[y]=y2(0)cotβ+ẏ2(l)cotγ*+y2(l)cotδ. It follows by inspection of the numerator R in (3.6) that zero is an eigenvalue only in the case β=δ=π/2. Hence, all the eigenvalues of problem (3.3), (3.5a)–(3.5d) for δ∈[0,π/2) or δ=π/2, β∈[0,π/2), are positive. Lemma 3.2 is proved.Lemma 3.3.
LetE be the space of solution of the problem (1.1), (1.2a)–(1.2c). Then, dimE=1.The proof is similar to that of [19, Lemma 2] using transformation (3.2), Lemmas 3.1 and 3.2 (see also [31, Lemma 2.2]). However, it is not true if π/2<γ<π (see, e.g., [31, Page 9]). Therefore, Lemma 3.1 together with the transformation (3.2) cannot be applicable to investigate more general boundary conditions, for example, (1.2a*), (1.2b), and (1.2c) for α∈(0,π/2].Lemma 3.4 (see [29, Lemma 2.2]).
Letλ>0 and u be a solution of (3.3) which satisfies the boundary conditions (3.5a)–(3.5c). If a is a zero of u and ü in the interval (0,l), then u̇(t)T̃u(t)<0 in a neighborhood of a. If a is a zero of u̇ or T̃u in (0,l), then u(t)ü(t)<0 in a neighborhood of a.Theorem 3.5.
Letu be a nontrivial solution of the problem (1.1), (1.2a) and (1.2c) for λ>0. Then the Jacobian J[u]=r3cosψsinψ of the transformation (3.1) does not vanish in (0,l).Proof.
Letu be a nontrivial solution of (1.1) which satisfies the boundary conditions (1.2a) and (1.2c). Assume first that the corresponding angle ψ satisfies ψ(x0)=nπ for some integer n and for some x0∈(0,l). Then, the transformation (3.1) implies that u(x0)=Tu(x0)=0. Using the transformation (3.2), the solution u of (3.3) also satisfies u(t0)=T̃u(t0)=0, where t0=l-1ω∫0x0h(s)ds∈(0,l). However, it is incompatible with the conclusion of Lemma 3.4.
The proof of the inequalitycosψ(x)≠0, x∈(0,l), proceeds in the same fashion as in the previous case. The proof of Theorem 3.5 is complete.Lety(x,λ) be a nontrivial solution of the problem (1.1), (1.2a)–(1.2c) for λ>0 and θ(x,λ), φ(x,λ) the corresponding functions in (3.1). Without loss of generality, we can define the initial values of these functions as follows (see [30, Theorem 3.3]):(3.7)θ(0,λ)=β-π2,φ(0,λ)=0.With obvious modifications, the results stated in [30, Sections 3–5] are true for the solution of the problem (1.1), (1.2a)–(1.2c), and (1.2d*) for δ∈[0,π/2]. In particular, we have the following results.Theorem 3.6.
θ(l,λ) is a strictly increasing continuous function on λ.Theorem 3.7.
Problem (1.1), (1.2a)–(1.2c), and (1.2d*) for δ∈[0,π/2] (except the case β=δ=π/2) has a sequence of positive and simple eigenvalues
(3.8)λ1(δ)<λ2(δ)<⋯<λn(δ)⟶∞.
Moreover, θ(l,λn(δ))=(2n-1)π/2-δ, n∈ℕ; the corresponding eigenfunctions υn(δ)(x) have n-1 simple zeros in (0,l).Remark 3.8.
In the caseβ=δ=π/2 the first eigenvalue of boundary value problem (1.1), (1.2a)–(1.2c), and (1.2d*) is equal to zero and the corresponding eigenfunction is constant; the statement of Theorem 3.7 is true for n≥2.Obviously, the eigenvaluesλn(δ), n∈ℕ, of the problem (1.1), (1.2a)–(1.2c), and (1.2d*) are zeros of the entire function y(l,λ)cosδ-Ty(l,λ)cosδ=0. Note that the function F(λ)=Ty(l,λ)/y(l,λ) is defined for λ∈A≡(ℂ/ℝ)∪(⋃n=1∞(λn-1(0),λn(0))), where λ0(0)=-∞.Lemma 3.9 (see [19, Lemma 5]).
Letλ∈A. Then, the following relation holds:
(3.9)ddλF(λ)=(∫0ly2(x,λ)dx)y2(l,λ).In (1.1) we set λ=ρ4. As is known (see [34, Chapter II, Section 4.5, Theorem 1]) in each subdomain T of the complex ρ-plane equation (1.1) has four linearly independent solutions zk(x,ρ), k=1,4¯, regular in ρ (for sufficiently large ρ) and satisfying the relations(3.10)zk(s)(x,ρ)=(ρωk)seρωkx[1],k=1,4¯,s=0,3¯,
where ωk, k=1,4¯, are the distinct fourth roots of unity, [1]=1+O(1/ρ).For brevity, we introduce the notations(δ1,δ2)=sgnδ1+sgnδ2. Using relation (3.10) and taking into account boundary conditions (1.2a)–(1.2c), we obtain(3.11)y(x,λ)={(sin⁡⁡(ρx+π2sin⁡⁡β)-cos⁡⁡(ρl+π2s(β,γ))eρ(x-l))[1]ifβ∈(0,π2],2sin⁡⁡(ρx-π4)-e-ρx+(-1)1-sgn⁡γ2sin⁡⁡(ρl+(-1)sgn⁡γπ4)eρ(x-l)[1]ifβ=0,(3.12)Ty(x,λ)={-ρ3(cos⁡⁡(ρx+π2sgn⁡β)+cos⁡⁡(ρl+π2s(β,γ))eρ(x-l))[1]ifβ∈(0,π2],-ρ3(2sin⁡⁡(ρx+π4)-e-ρx-(-1)1-sgn⁡γ2sin⁡⁡(ρl+π4(-1)sgn⁡γ)eρ(x-l))[1]ifβ=0.Remark 3.10.
As an immediate consequence of (3.11), we obtain that the number of zeros in the interval (0,l) of function y(x,λ) tends to +∞ as λ→±∞.Taking into account relations (3.11) and (3.12), we obtain the asymptotic formulas(3.13)F(λ)={(2)1-sgn⁡γρ3cos⁡⁡(ρl+(π/2)sgn⁡β+(π/4)sgn⁡γ)cos⁡⁡(ρl+(π/2)sgn⁡β+(π/4)(1+sgn⁡γ))[1]ifβ∈(0,π2],(2)1-sgn⁡γρ3cos⁡(ρl+(π/4)(sgn⁡γ-1))cos⁡(ρl+(π/4)(1+sgn⁡γ))[1]ifβ=0.
Furthermore, we have(3.14)F(λ)=-(2)1-sgn⁡γ|λ|34(1+O(|λ|)-1/4),asλ⟶-∞.We define numbersτ, ν, η, αn, βn, ηn, n∈ℕ, and a function z(x,t), x∈[0,l], t∈ℝ, as follows:(3.15)τ={3(1+s(β,δ))4-1ifγ∈(0,π2],54-38((-1)sgn⁡β+(-1)sgn⁡δ)-1ifγ=0,η={3(2+sgn⁡β)4-1ifγ∈(0,π2],54-38((-1)sgn⁡β-1)-1ifγ=0,ν={3(1+s(β,|c|))4ifγ∈(0,π2],54-38((-1)sgn⁡β+(-1)sgn⁡|c|)ifγ=0,αn=(n-τ)πl,ηn=(n-η)πl,βn=(n-ν)πl,z(x,t)={sin⁡⁡(tx+π2sgn⁡β)-cos⁡⁡(tl+π2s(β,γ))e-t(l-x)ifβ∈(0,π2],2sin⁡⁡(tx-π4)+e-tx+(-1)sgn⁡γ2sin⁡⁡(tl+(-1)sgn⁡γπ4)e-t(x-l)ifβ=0.
By virtue of [18, Theorem 3.1], one has the asymptotic formulas(3.16)λn(δ)4=αn+O(n-1),(3.17)υn(δ)(x)=z(x,αn)+O(n-1),
where relation (3.17) holds uniformly for x∈[0,l].By (3.14), we have(3.18)lim⁡λ→-∞F(λ)=-∞.
From Property 1 in [30] and formulas (3.9), one has the relations(3.19)λ1(π2)<λ1(0)<λ2(π2)<λ2(0)<⋯.Remark 3.11.
It follows by Theorem3.7, Lemma 3.9, and relations (3.18) and (3.19) that if λ>0 or λ=0, β∈[0,π/2), then F(λ)<0; besides, if λ=0 and β=π/2, then F(λ)=0.Lets(λ) be the number of zeros of the function y(x,λ) in the interval (0,l).Lemma 3.12.
Ifλ>0 and λ∈(λn-1(0),λn(0)], n∈ℕ,  then s(λ)=n-1.The proof is similar to that of [19, Lemma 10] using Theorems 3.6 and 3.7 and Remark 3.11.Theorem 3.13.
The problem (1.1), (1.2a)–(1.2c), and (1.2d*) for δ∈(π/2,π) has a sequence of real and simple eigenvalues
(3.20)λ1(δ)<λ2(δ)<⋯<λn(δ)⟶+∞,
including at most one negative eigenvalue. Moreover, (a) if β∈[0,π/2), then λ1(δ)>0 for δ∈(π/2,δ0); λ1(δ)=0 for δ=δ0;λ1(δ)<0 for δ∈(δ0,π), where δ0=arctgTy(l,0)/y(l,0); (b) if β=π/2, then λ1(δ)<0; (c) the eigenfunction υn(δ)(x), corresponding to the eigenvalue λn(δ)≥0, has exactly n-1 simple zeros in (0,l).The proof parallels the proof of [19, Theorem 4] using Theorems 3.5–3.7 and Lemmas 3.9 and 3.12.Lemma 3.14.
The following non-selfadjoint boundary value problem:(3.21)y(4)(x)-(q(x)y′(x))′=λy(x),x∈(0,l),y(0)=y′(0)=Ty(0)=y′(l)cos⁡⁡γ+y′′(l)sin⁡⁡γ=0,
has an infinite set of nonpositive eigenvalues ρn tending to -∞ and satisfying the asymptote
(3.22)λn=-(n-14(1+sgn⁡γ))4π4l4+o(n4),n⟶∞.Settingx=0 in (3.12), we obtain (3.22).Remark 3.15.
By Remark3.10 the number of zeros of the eigenfunction y1(δ)(x) corresponding to an eigenvalue λ1(δ)<0 can by arbitrary. In views of [31, Corollary 2.5], as λ1(δ)<0 varies, new zeros of the corresponding eigenfunction y1(δ)(x) enter the interval (0,l) only through the end point x=0 (since y1(δ)(l)≠0), and hence the number of its zeros, in the case β∈(0,π/2], is asymptotically equivalent to the number of eigenvalues of the problem (3.21) which are higher than λ1(δ). In the case β=0 see [31, Theorem 5.3].We consider the following boundary conditions:(1.2d′)ay(l)-cTy(l)=0,(1.2d′′)cy(l)+aTy(l)=0.Note that(a,c)≠0  since  σ<0. The boundary condition (1.2d′) coincides the boundary condition (1.2d*) for δ=π/2 (resp., δ=0) in the case a=0 (resp., c=0), and the boundary condition (1.2d′′) coincides the boundary condition (1.2d*) for δ=0 (resp., δ=π/2) in the case a=0 ( resp., c=0).Letac≠0. The eigenvalues of the problem (1.1), (1.2a)–(1.2c), and (1.2d′) (resp., (1.1) (1.2a)–(1.2c), and (1.2d′′))  are the roots of the equation F(λ)=a/c (resp., F(λ)=-c/a). By (3.9), this equation has only simple roots; hence all the eigenvalues of the problems (1.1), (1.2a)–(1.2c), and (1.2d′) and (1.1), (1.2a)–(1.2c), and (1.2d′′) are simple. On the base of (3.9), (3.18), and (3.19) in each interval An, n=1,2,…, the equation F(λ)=a/c (resp., F(λ)=-c/a) has a unique solution μn (resp., νn); moreover,(3.23)ν1<λ1(π2)<μ1<λ1(0)<ν2<λ2(π2)<μ2<λ2(0)<⋯
if a/c>0 and(3.24)μ1<λ1(π2)<ν1<λ1(0)<μ2<λ2(π2)<ν2<λ2(0)<⋯
if a/c<0. Besides, μ1=0 if a/c<0 and F(0)=a/c;ν1=0 if a/c>0 and F(0)=-c/a.Taking into account (1.2d′), (1.2d′′), (3.23), and (3.24) and using the corresponding reasoning [18, Theorem 3.1] we have(3.25)μn4=ηn+O(n-1),νn4=ηn+O(n-1),(3.26)φn(x)=z(x,ηn)+O(n-1),ψn(x)=z(x,ηn)+O(n-1),
where relation (3.26) holds uniformly for x∈[0,l] and eigenfunctions φn(x) and ψn(x), n∈ℕ, correspond to the eigenvalues μn and νn, respectively.Let us denotem(λ)=ay(l,λ)-cTy(l,λ).Remark 3.16.
Note that ifλ is the eigenvalue of problem (1.1), (1.2a)–(1.2d), then m(λ)≠0 since σ≠0.It is easy to see that the eigenvalues of problem (1.1), (1.2a)–(1.2d) are roots of the equation(3.27)(aλ+b)y(l)-(cλ+d)Ty(l)=0.By virtue of Remark3.16 and formula (3.27), a simple calculation yields that the eigenvalues of the problem (1.1), (1.2a)–(1.2d) can be realized at the solution of the equation(3.28)cy(l,λ)+aTy(l,λ)ay(l,λ)-cTy(l,λ)=a2+c2-σλ+ab+cd-σ.DenoteBn=(μn-1,μn), n∈ℕ, where μ0=-∞.We observe that the functionG(λ)=(cy(l,λ)+aTy(l,λ))/(ay(l,λ)-cTy(l,λ)) is well defined for λ∈B=(ℂ∖ℝ)∪(⋃n=1∞Bn) and is a finite-order meromorphic function and the eigenvalues νn and μn, n∈ℕ, of  boundary value problems (1.1), (1.2a)–(1.2c), and (1.2d′′) and (1.1), (1.2a)–(1.2c), and (1.2d′) are zeros and poles of this function, respectively.Letλ∈B. Using formula (3.9), we get(3.29)ddλG(λ)=(a2+c2)m-2(λ)∫0ly2(x,λ)dx.Lemma 3.17.
The expansion(3.30)G(λ)={G(0)+∑n=1∞λcnμn(λ-μn)ifμ1≠0,c0+c1λ+∑n=2∞λcnμn(λ-μn)ifμ1=0,
holds, where cn, n∈ℕ, are some negative numbers.Proof.
It is known (see [35, Chapter 6, Section 5]) that the meromorphic function G(λ) with simple poles μn allows the representation
(3.31)G(λ)=G1(λ)+∑n=1∞(λμn)scnλ-μn,
where G1(λ) is an entire function,
(3.32)cn=resλ=μnG(λ)=(cy(l,μn)+aTy(l,μn))(a∂y(l,μn)∂λ-cTy(l,μn)∂λ)-1,
and integers sn, n∈ℕ, are chosen so that series (3.31) are uniformly convergent in any finite circle (after truncation of terms having poles in this circle).
We consider the casea/c>0. By virtue of relation (3.18), we have limλ→-∞G(λ)=-a/c. Hence, G(λ)<0 for λ∈(-∞,ν1) and G(λ)>0 for λ∈(ν1,μ1). Without loss of generality, we can assume ay(l,λ)-cTy(l,λ)>0 for λ∈(-∞,μ1). Then, cy(l,λ)+aTy(l,λ)<0 for λ∈(-∞,ν1). Since the eigenvalues μn and νn, n∈ℕ, are simple zeros of functions ay(l,λ)-cTy(l,λ) and cy(l,λ)+aTy(l,λ), respectively, then by (3.29) the relations
(3.33)(-1)n+1(cy(l,μn)+aTy(l,μn))>0,(-1)n+1(a∂y(l,μn)∂λ-c∂Ty(l,μn)∂λ)<0,n∈N,
are true.
Taking into account (3.33), in (3.32) we get cn<0, n∈ℕ. The cases a/c<0, a=0, and c=0 can be treated along similar lines.
DenoteΩn(ɛ)={λ∈ℂ∣|λ4-μn4|<ɛ} where ɛ>0 is some small number. From the asymptotic formula (3.25), it follows that for ɛ<π/4l the domains Ωn(ɛ) asymptotically do not intersect and contain only one pole μn of the function G(λ).
By (3.11), (3.12), (3.23), (3.24), and (3.25), we see that outside of domains Ωn(ɛ) the asymptotic formulae are true:
(3.34)G(λ)={-ac+O(ρ-1)ifac≠0,ρ3z(ρ)(1+O(ρ-1))ifc=0,-ρ-3(z(ρ))-1(1+O(ρ-1))ifa=0,
where
(3.35)z(ρ)={cos⁡⁡((π/4)sgn⁡γ)cos⁡⁡(ρl+(π/2)sgn⁡β+(π/4)sgn⁡γ)sin⁡⁡((π/4)(1+sgn⁡γ))cos⁡⁡(ρl+(π/2)sgn⁡β+(π/4)(1+sgn⁡γ))ifβ∈(0,π2],(2)1-2sgn⁡γcos⁡(ρl-(1-sgn⁡γ)(π/4))cos⁡⁡(ρl+(π/4)sgn⁡γ)ifβ=0.
Following the corresponding reasoning (see [36, Chapter VII, Section  2, formula (27)]), we see that outside of domains Ωn(ɛ) the estimation
(3.36)|G(λ)|≤{M̃1ifac≠0,M̃2|λ|34ifc=0,M̃3|λ|-34ifa=0,
holds; using it in (3.32) we get
(3.37)|cn|=|12πi∫∂Ωn(ɛ)G(λ)dλ|=2π|∫|ν-μn4|=ɛν3G(ν4)dν|≤{M1n3ifac≠0,M2n6ifc=0,M3ifa=0,
where M̃1, M̃2, M̃3, M1, M2, M3 are some positive constants. By (3.37) and asymptotic formula (3.25) the series ∑n=1∞cn|μn|-2 converges. Then, according to Theorem 2 in [35, Chapter 6, Section 5], in formula (3.31) we can assume sn=1, n∈ℕ.
Let{Γn}n=1∞ be a sequence of the expanding circles which are not crossing domains Ωn(ɛ). Then, according to Formula (9) in [37, Chapter V, Section 13], we have
(3.38)G(λ)-∑μk∈int⁡Γnckλ-μk=12πi∫ΓnG(ξ)ξ-λdξ,G(0)+∑μk∈int⁡Γnckμk=12πi∫ΓnG(ξ)ξdξ.
By (3.38), we get
(3.39)G(λ)-G(0)=∑μk∈int⁡Γnλckμk(λ-μk)=12πi∫ΓnλG(ξ)ξ(ξ-λ)dξ.
From (3.36), the right side of (3.39) tends to zero as n→∞. Then, passing to the limit as n→∞ in (3.39), we obtain
(3.40)G(λ)=G(0)+∑n=1∞λcnμn(λ-μn),
which implies G1(λ)≡G(0).
Differentiating the right side of the least equality, we have(3.41)G(s)(λ)=(-1)ss!∑n=1∞cn(λ-μn)s+1,s=1,2,3.
Note that the function F(λ) has the following expansion:
(3.42)F(λ)=F(0)+∑n=1∞λc̃nλn(0)(λ-λn(0)),
where
(3.43)c̃n=resλ=λn(0)F(λ),n=1,2,….
Now letμ1=0, that is, F(0)=a/c. G(λ) has the following expansion:
(3.44)G(λ)=G1(λ)+c1λ+∑n=2∞λcnμn(λ-μn).
Again, according to Formula (9) in [37, Chapter V, Section 13], we have
(3.45)G(λ)-c1λ-∑μk∈int⁡Γnk≠1ckλ-μk=12πi∫ΓnG(ξ)ξ-λdξ.
By (2.6) [18] and (3.9), we get
(3.46)c1=-c-2(a2+c2)(F′(0))-1.
Using (3.42), (3.41), and (3.46), we obtain
(3.47)G(λ)-c1λ=-ac+c-2(a2+c2)∑n=1∞(c̃n/λn2(0)(λ-λn(0)))F′(0)∑n=1∞(c̃n/λn(0)(λ-λn(0))).
Passing to the limit as λ→0 in (3.47), we get
(3.48)lim⁡λ→0(G(λ)-c1λ)=-ac+c-2(a2+c2)(∑n=1∞c̃nλn2(0))-1(∑n=1∞c̃nλn3(0))-1=ac+2c-2(a2+c2)(F′(0))-2F′′(0)=c0.
Using (3.48) in (3.45), we have
(3.49)c0+∑μk∈int⁡Γnk≠1ckλ-μk=12πi∫ΓnG(ξ)ξdξ.
In view of (3.49) and (3.45), we get
(3.50)G(λ)-c0-c1λ-∑μk∈int⁡Γnk≠1λckμk(λ-μk)=12πi∫ΓnλG(ξ)ξ(ξ-λ)dξ.
Passing to the limit as n→∞ in (3.50), we obtain
(3.51)G(λ)=c0+c1λ+∑n=2∞λcnμn(λ-μn).
Lemma3.17 is proved.


## 4. The Structure of Root Subspaces, Location of Eigenvalues on a Complex Plane, and Oscillation Properties of Eigenfunctions of the Problem (1.1), (1.2a)–(1.2d)

Forc≠0, we find a positive integer N from the inequality μN-1<-d/c≤μN.Theorem 4.1.
The problem (1.1), (1.2a)–(1.2d) for σ>0 has a sequence of real and simple eigenvalues
(4.1)λ1<λ2<⋯<λn⟶+∞,
including at most 1+sgn|c| number of negative ones. The corresponding eigenfunctions have the following oscillation properties.(a)
Ifc=0, then the eigenfunction yn(x), n≥2, has exactly n-1 zeros in (0,l), the eigenfunction y1(x) has no zeros in (0,l) in the case λ1≥0, and the number of zeros of y1(x) can be arbitrary in the case λ1<0.(b)
Ifc≠0, then the eigenfunction yn(x) corresponding to the eigenvalue λn≥0 has exactly n-1 simple zeros for n≤N and exactly n-2 simple zeros for n>N in (0,l) and the eigenfunctions associated with the negative eigenvalues may have an arbitrary number of simple zeros in (0,l).The proof of this theorem is similar to that of [18, Theorem 2.2] using Remark 3.15.Throughout the following, we assume thatσ<0.Letλ,μ(λ≠μ) be the eigenvalue of the operator L. The eigenvectors y(λ)={y(x,λ),m(λ)} and y(μ)={y(x,μ),m(μ)} corresponding to the eigenvalues λ and μ, respectively, are orthogonal in Π1, since the operator L is J-selfadjoint in Π1. Hence, by (2.4), we have(4.2)∫0ly(x,λ)y(x,μ)¯dx=-σ-1m(λ)m(μ)¯.Lemma 4.2.
Letλ*∈ℝ be an eigenvalue of boundary value problem (1.1), (1.2a)–(1.2d) and G′(λ*)≤A, where A=-(a2+c2)/σ. Then, problem (1.1), (1.2a)–(1.2d) has no nonreal eigenvalues.Proof.
Letμ∈ℂ∖ℝ be an eigenvalue of problem (1.1), (1.2a)–(1.2d). Then, from Remark 3.16 and equality (4.2), we obtain
(4.3)∫0ly(x,λ*)m(λ*)(y(x,μ)¯m(μ))dx=-σ-1,∫0ly(x,λ*)m(λ*)y(x,μ)m(μ)dx=-σ-1,(4.4)∫0l|y(x,μ)m(μ)|2dx=-σ-1.
In view of formula (3.29), the inequality
(4.5)∫0l(y(x,λ*)m(λ*))2dx≤-σ-1
is true.
By (4.3),
(4.6)∫0ly(x,λ*)m(λ*)Rey(x,μ)m(μ)dx=-σ-1.
From (4.4)–(4.6), we get
(4.7)∫0l{(y(x,λ*)m(λ*)-Rey(x,μ)m(μ))2+Im⁡2y(x,μ)m(μ)}dx<0ifG′(λ*)<A,∫0l{(y(x,λ*)m(λ*)-Rey(x,μ)m(μ))2+Im⁡2y(x,μ)m(μ)}dx=0ifG′(λ*)=A.
From the second relation it follows thatIm(y(x,μ)/m(μ))=0, which by (1.1) contradicts the condition μ∈ℂ∖ℝ. The obtained contradictions prove Lemma 4.2.Lemma 4.3.
Letλ1*,λ2*∈ℝ, λ1*≠λ2* be eigenvalues of problem (1.1), (1.2a)–(1.2d) and G′(λ1*)≤A. Then, G′(λ2*)>A.Proof.
LetG′(λ2*)≤A. By (3.29) and (4.2), we have
(4.8)∫0l(y(x,λ1*)m(λ1*))2dx≤-σ-1,∫0l(y(x,λ2*)m(λ2*))2dx≤-σ-1,∫0ly(x,λ1*)m(λ1*)y(x,λ2*)m(λ2*)dx=-σ-1.
Hence, we get(4.9)∫0l(y(x,λ1*)m(λ1*)y(x,λ2*)m(λ2*))dx<0ifG′(λ1*)<AorG′(λ2*)<A,∫0ly(x,λ1*)m(λ1*)y(x,λ2*)m(λ2*)dx=0ifG′(λ1*)=G′(λ2*)=A.
From (4.9), it follows that y(x,λ1*)/m(λ1*)=y(x,λ2*)/m(λ2*) for x∈[0,l]. Therefore, m(λ2*)y(x,λ1*)=m(λ1*)y(x,λ2*).
Sinceλ1≠λ2, then by (1.1) y(x,λ1)≡0. The obtained contradictions prove Lemma 4.3.By Lemmas4.2 and 4.3 problem (1.1), (1.2a)–(1.2d) can have only one multiple real eigenvalue. From (3.41), we get G(3)(λ)>0, λ∈B, whence it follows that the multiplicity of real eigenvalue of problem (1.1), (1.2a)–(1.2d) does not exceed three.Theorem 4.4.
The boundary value problem (1.1), (1.2a)–(1.2d) for σ<0 has only point spectrum, which is countable infinite and accumulates at +∞ and can thus be listed as λn,n≥1 with eigenvalues repeated according to algebraic multiplicity and ordered so as to have increasing real parts. Moreover, one of the following occurs.(1)
All eigenvalues are real, at thatB1 contains algebraically two (either two simple or one double) eigenvalues, and Bn, n=2,3,…, contain precisely one simple eigenvalues.(2)
All eigenvalues are real, at thatB1contains no eigenvalues but, for some  s≥2,  Bs  contains algebraically three (either three simple, or one simple and one double, or one triple) eigenvalues, and  Bn,  n=2,3,…, n≠s   contain precisely one simple eigenvalue.(3)
There are two nonreal eigenvalues appearing as a conjugate pair, at thatB1 contains no eigenvalues, and Bn, n=2,3,…, contain precisely one simple eigenvalue.Proof.
Remember that the eigenvalues of problem (1.1), (1.2a)–(1.2d) are the roots of the equation G(λ)=Aλ+B, where A=-(a2+c2)/σ, B=-(ab+cd)∖σ (see (3.28)). From (3.41), it follows that G′′(λ)>0 for λ∈B1; therefore, the function G(λ) is convex on the interval B1. By virtue of (3.18) and (3.30), we have
(4.10)lim⁡λ→-∞G(λ)={-acifc≠0,-∞ifc=0,lim⁡λ→μn-0G(λ)=+∞.
That is why for each fixed numberA there exists number BA such that the lines Aλ+BA, λ∈ℝ, touch the graph of function G(λ) at some point λ̃∈B1. Hence, in the interval B1, (3.28) has two simple roots λ1<λ2 if B>BA, one double root λ1=λ̃ if B=BA, and no roots if B<BA.
By (3.29) and (3.30) we have limλ→μn+0G(λ)=-∞, limλ→μn-0G(λ)=+∞, n∈ℕ. Therefore, (3.28) has at least one solution in the interval Bn, n=2,3,….
LetB≥BA. If B>BA, then G′(λ1)<A, G′(λ2)<A; if B=BA then G′(λ1)=A. By (3.29), (3.28) has only simple root λn+1 for B>BA,λn for B=BA in the interval Bn, n=2,3,….
LetB<BA. By Lemma 4.3 either G′(λn)>A for any λn∈ℝ, or there exists k∈ℕ such that F′(λk)≤A and F′(λn)>A, n∈ℕ∖{k}. Assume that λk∈Bs. Obviously, s≥2. Choose natural number n0 such that the inequalities
(4.11)ARn0+B>0,|G(λ)-(Aλ+BA)|>|BA-B|,λ∈SRn0,
are fulfilled; where Rn=τn+δ0,
(4.12)τn={λn(π2)ifc=0,νnifc≠0,ac>0,λn(0)-1ifc≠0,a=0,νn-1ifc≠0,ac<0,δ0 is sufficiently small positive number, and SRn={z∈C∣|z|=Rn}.
We have(4.13)ΔSRn0arg⁡(G(λ)-(Aλ+B))=ΔSRn0arg⁡(G(λ)-(Aλ+BA))+ΔSRn0arg⁡(1+BA-BG(λ)-(Aλ+BA)),
where
(4.14)ΔSRn0arg⁡f(z)=1i∫SRn0(f′(z)f(z))dz
(see [37, Chapter  IV, Section  10]). By (4.11)
(4.15)|(BA-B)(G(λ)-(Aλ+BA))|<1,λ∈SRn0,
hence, the point
(4.16)ω=(BA-B)(G(λ)-(Aλ+BA))
does not go out of circle {|ω|<1}. Therefore, vector w=1+ω cannot turn around the point w=0, and the second summand in (4.13) equals zero. Thus,
(4.17)ΔSRn0arg⁡(G(λ)-(Aλ+B))=ΔSRn0arg⁡(G(λ)-(Aλ+BA)).
By the argument principle (see [37, Chapter IV, Section 10, Theorem 1]) we have
(4.18)12πΔSRn0arg⁡(G(λ)-(Aλ+BA))=∑λn(BA)∈int⁡SRn0ϱ(λn(BA))-∑μn∈int⁡SRn0ϱ(μn),
where ρ(λn(BA)) and ϱ(μn) are multiplicity of zero λn(BA) and pole μn of the function G(λ)-(Aλ+BA), respectively (λ1(BA)=λ2(BA)). Obviously, ∑λn(BA)∈intSRn0ρ(λn(BA))=n0 and ∑μn∈intSRn0ρ(μn)=n0-1. Then, by (4.18) we obtain
(4.19)(2π)-1ΔSRn0arg⁡(G(λ)-(Aλ+BA))=1.
From (4.17) and (4.19) follows the validity of the equality
(4.20)(2π)-1ΔSRn0arg⁡(G(λ)-(Aλ+B))=1.
Using the argument principle again, by (4.20) we get
(4.21)∑λn∈int⁡SRn0ρ(λn)-∑μn∈int⁡SRn0ϱ(μn)=1,
whence it follows that
(4.22)∑λn∈int⁡SRn0ρ(λn)=n0,
where λn, n∈ℕ, are roots of the equation G(λ)=Aλ+B. From the above-mentioned reasoning, by (4.22) we have
(4.23)∑λm∈int⁡SRnρ(λm)=n,n=n0,n0+1,…,
and, therefore, problem (1.1), (1.2a)–(1.2d) in the interval Bn for n=n0, n0+1,…, has only one simple eigenvalue.
Consider the following two cases.
Case 1.
For all real eigenvaluesλn of problem (1.1), (1.2a)–(1.2d) the inequalities G′(λn)>A, λn∈⋃m=2∞Bm, are fulfilled. The problem (1.1), (1.2a)–(1.2d) in every interval Bm, m=2,3,…,n0-1, has one simple eigenvalue. Hence, problem (1.1), (1.2a)–(1.2d) in the interval (-∞,SRn), n≥n0, has n-2 simple eigenvalues, and hence, by (4.23), this problem in the circle SRn⊂ℂ has one pair of simple nonreal eigenvalues. In this case, the location of the eigenvalues will be in the following form: λ1,λ2∈ℂ∖ℝ, λ2=λ¯1, Imλ1>0, λn∈Bn-1, n=3,4,….Case 2.
LetG′(λk)≤A, G′(λn)>A, n∈N∣{k} and λk∈Bs, s≥2. By Lemma 4.2 problem (1.1), (1.2a)–(1.2d) has no nonreal eigenvalues. From the above-mentioned reasoning it follows that in each interval Bn, n≠k, n=2,3,…, problem (1.1), (1.2a)–(1.2d) has one simple eigenvalue.Subcase 1.
LetG′(λk)=A,G′′(λk)≠0, that is, the eigenvalue λk is a double one (by this λk=λk+1). Then, from (4.23) it follows that the interval Bs besides the eigenvalue λk contains one more simple eigenvalue: at that it is either λk-1 (by this k=s) or λk+2 (by this k=s-1). Hence, λn∈Bn+1, n=1,2,…,s-2, λs-1,λs,λs+1∈Bs (by this either λs-1<λs=λs+1 or λs-1=λs<λs+1), λn∈Bn-1, n=s+2,s+3,….Subcase 2.
LetG′(λk)=A, G′′(λk)=0. By (3.41), G′′′(λk)≠0. Hence, λk is a triple eigenvalue of the problem (1.1), (1.2a)–(1.2d) (by this λk=λk+1=λk+2). Then, from (4.23) it follows that in the interval Bs problem (1.1), (1.2a)–(1.2d) has unique triple eigenvalue λk, and therefore, k=s-1. At this λn∈Bn+1, n=1,2,…,s-2, λs-1=λs=λs+1∈Bs, λn∈Bn-1,n=s+2,s+3,….Subcase 3.
LetG′(λk)<A, that is, the eigenvalue λk is simple. Then, by (4.23), in the interval Bsproblem (1.1), (1.2) has an eigenvalue λk as well as two more simple eigenvalues, which, by Lemma 4.3, are λk-1  and λk+1 (and hence k=s). In this case, we have λn∈Bn+1, n=1,2,…,s-2, λs-1,λs,λs+1∈Bs (λs-1<λs<λs+1), λn∈Bn-1, n=s+2,s+3,….
Theorem4.4 is proved.By Theorem4.4 we have ϱ(λn)=2, that is, λn=λn+1 if n=s-1 or n=s; ϱ(λn)=3, that is, λn=λn+1=λn+2 if n=s-1 (If assertion (2) in Theorem 4.4 holds, then we set  s=1).Let{yn(x)}n=1∞ be a system of eigen- and associated functions corresponding to the eigenvalue system {λn}n=1∞ of problem (1.1), (1.2a)–(1.2d), where yn(x)=y(x,λn) if ρ(λn)=1; yn(x)=y(x,λn), yn+1(x)=yn+1*(x)+cnyn(x), yn+1*(x)=(∂y(x,λn))/∂λ, cn is an arbitrary constant, if ϱ(λn)=2;yn(x)=y(x,λn), yn+1(x)=yn+1*(x)+dnyn(x), yn+2(x)=yn+2*(x)+dnyn+1*(x)+hnyn(x), yn+2*(x)=∂2y(x,λn)/2∂λ2, dn, hn are arbitrary constants, if ρ(λn)=3. Here, yn(x) is an eigenfunction for λn and yn+1(x) when ρ(λn)=2; yn+1(x), yn+2(x) when ρ(λn)=3 are the associated functions (see [34, Pages 16–20] for more details).We turn now to the oscillation theorem of the eigenfunctions corresponding to the positive eigenvalues of problem (1.1), (1.2a)–(1.2d) since the eigenfunctions associated with the negative eigenvalues may have an arbitrary number of simple zeros in (0,l).Theorem 4.5.
For eachn<N (resp., n>N), yn has n-1 (resp., n) zeros in the interval (0,l). Similarly ys, ys+1 both have s-1 (resp., s) zeros if s<N (resp., s>N). Finally, if c≠0, then each of yN (and ys, ys+1 if s=N) has N-1  or  N zeros according to λN, λS,  λS+1≤ or >-d/c,  and if c=0 and s=N, then ys, ys+1 both have s zeros.The proof of this theorem is similar to that of [11, Theorem 4.4] using Lemma 3.12.


## 5. Asymptotic Formulae for Eigenvalues and Eigenfunctions of the Boundary Value Problem (1.1), (1.2a)–(1.2d)

Forc≠0, let K be an integer such that λk-1(π/2)<b/a≤λk(π/2) (interpreting λ0(π/2)=-∞).Lemma 5.1.
The following relations hold for sufficiently largen∈ℕ, n>n1=max{s,N,K}+2:
(5.1)λn-2(0)<λn<λn-1(π2)<λn-1(0)ifc≠0,ac≤0,λn-2(0)<λn-1(π2)<λn<λn-1(0)ifc≠0,ac>0orc=0.Proof.
Letac≠0. Note that the eigenvaluesλn(0) (resp., λn(π/2)), n∈ℕ, of problem (1.1), (1.2a)–(1.2c), and (1.2d′′) forδ=0 (resp., for δ=π/2) are roots of the equation G(λ)=-a/c(resp., G(λ)=c/a). The equation Aλ+B=-a/c (resp., Aλ+B=c/a) has a unique solution λ=-d/c (resp., λ=-b/a). Since n>max{N+2,K+2}, in view of (3.29), G(λn)>max{-a/c,c/a}. Hence, by (3.23), (3.24), and (3.29), the following relations hold for n>n1:
(5.2)μn-2<λn-2(π2)<νn-2<λn-2(0)<λn<μn-1ifac<0,μn-2<λn-2(0)<νn-1<λn-1(π2)<λn<μn-1ifac>0.
Leta=0. In this case μn=λn(π/2), νn=λn(0), n∈ℕ. Since n>N+2, so G(λn)>0. Then, by the equality G(λn)=Aλn+B, n∈ℕ, we obtain
(5.3)λn-2(π2)<λn-2(0)<λn<λn-1(π2)<λn-1(0),n>n1.
Now let c=0. In this case μn=λn(0), νn=λn(π/2), n∈ℕ. Since n>K+2, so G(λn)>0. Therefore, using G(λn)=Aλn+B, we have
(5.4)λn-2(0)<λn-1(π2)<λn<λn-1(0),n>n1.
Relations (5.1) are consequences of relations (5.2)–(5.4).
The proof of Lemma5.1 is complete.We define numbersχ, χn, n∈ℕ, as follows:(5.5)χ={3(1+s(β,|c|))4ifγ∈(0,π2],54-38((-1)sgn⁡β+(-1)sgn⁡|c|)ifγ=0,χn=(n-χ)πl.Using relations (5.1) and formulas (3.16), (3.17), the following corresponding reasoning [18, Theorem 3.1] can be proved.Theorem 5.2.
The following asymptotic formulae hold:(5.6)λn4=χn+O(n-1),yn(x)=z(x,χn)+O(n-1),
where relation (5.6) holds uniformly for x∈[0,l].


## 6. Necessary and Sufficient Conditions of Basicity of Root Function System of Problem (1.1), (1.2a)–(1.2d)

Note that the elementŷn={yn(x),mn}, n∈ℕ, of the system {ŷn}n=1∞ of the root vectors of operator L satisfies the relation(6.1)Lŷn=λnŷn+θnŷn-1,
where θn equals either 0 (at that ŷn is eigenvector) or 1 (at that λn=λn-1 and ŷn is associated vector) (see, e.g., [38]).Theorem 6.1.
The system of eigen- and associated functions of operatorL is a Riesz basis in the space H.Proof.
Letμ be a regular value of operator L, that is, Rμ=(L-μI)-1 exists and a bounded in H. Then, problem (2.4) is adequate to the following problem of eigenvalues:
(6.2)Rμŷ=(λ-μ)-1ŷ,ŷ∈D(L).
By Lemma2.1,  Rμ is a completely continuous J-selfadjoint operator in Π1. Then, in view of [39] the system of the root vectors of operator Rμ (hence of operator L) forms a Riesz basis in H. Theorem 6.1 is proved.Let{υ̂n*}n=1∞, where υ̂n*={υn*(x),sn*}, be a system of the root vectors of operator L*, that is,(6.3)L*υ̂n*=λ¯nυ̂n*+θn+1υ̂n+1*.By Lemma2.2 and relations (6.1), (6.3) we have the following.Lemma 6.2.
υ̂n*=Jŷ¯n(ŷ¯n={y¯n(x),m¯n}) if  ϱ(λn)=1; υ̂n*=Jŷn+1*+c̃nJŷn,υ̂n+1*=Jŷn if ϱ(λn)=2; υ̂n*=Jŷn+2*+d̃nJŷn+1*h̃nJŷn*,υ̂n+1*=Jŷn+1*+d̃nJŷn,υ̂n+2*=Jŷn  if  ϱ(λn)=3, where ŷn+1*={yn+1*(x),mn+1*}, mn+1=m′(λn), ŷn+2*={yn+2*(x),mn+2*}, mn+2*=(1/2)m′′(λn), c̃n, d̃n, h̃n are arbitrary constants.Lemma 6.3.
Let(6.4)δ={∫0lyn2(x)dx+σ-1mn2,ifρ(λn)=1,(ŷn,ŷn+1*)Π1,ifρ(λn)=2,‖ŷn+1*‖Π12ifρ(λn)=3,
where ∥·∥Π1 is the norm in Π1. Then, δn≠0, n∈ℕ.Proof.
By Remark3.16,  mn≠0 if ŷn is the eigenvector of operator L. If ϱ(λn)=1, then G′(λn)≠A, whence by (3.29), we get δn≠0.
Letϱ(λn)=2. Then, G′(λn)=A and G′′(λn)≠0. Differentiating the right-hand side of equality (3.29) on λ, we obtain
(6.5)G′′(λ)=2(a2+c2)m2(λ){∫0ly(x,λ)∂y(x,λ)∂λdx-m′(λ)m(λ)∫0ly2(x,λ)dx}.
Assuming λ=λn in (6.5) and taking into account (3.29) and (2.6), we get
(6.6)G′′(λn)=2(a2+c2)mn-2(yn,yn+1*)Π1.
Since G′′(λn)≠0, from the last equality it follows that δn≠0.
Now letϱ(λn)=3, that is, G′(λn)=A and G′′(λn)=0, G′′′(λn)≠0. Differentiating the right-hand side of (6.5) on λ, we have
(6.7)G′′′(λ)=2(a2+c2)m-6(λ){{(∫0l(∂y(x,λ)∂λ)2dx+∫0ly(x,λ)∂2y(x,λ)∂λ2dx)m(λ)-m′(λ)∫0ly(x,λ)∂y(x,λ)∂λdx-m′′(λ)∫0ly2(x,λ)dx}m(λ)-3m′(λ){m(λ)∫0ly(x,λ)∂y(x,λ)∂λdx-m′(λ)∫0ly2(x,λ)dx}},
whence (supposing in that equality λ=λn), we obtain
(6.8)G′′′(λn)=2(a2+c2)mn-4{‖ŷn+1*‖Π12+2(ŷn,ŷn+2*)Π1}.
By (6.1) and (6.3), we have
(6.9)(ŷn,ŷn+2*)Π1=‖ŷn+1*‖Π12;
then taking into account (6.8), we get
(6.10)δn=‖ŷn+1*‖Π12=16(a2+c2)-1mn4G′′′(λn)≠0.
Lemma6.3 is proved.Lemma 6.4.
The elementsυ̂n={υn(x),sn} of the system {υ̂n}n=1∞ conjugated to the system {ŷn}n=1∞ are defined by the equality
(6.11)υ̂n=δ¯n-1υ̂n*,
where c̃n=-cn-δn-1∥ŷn+1*∥Π12 if ϱ(λn)=2; d̃n=-dn-δn-1(ŷn+1*,ŷn+2*)Π1, h̃n=-hn-δn-1∥ŷn+2*∥Π12+δn-2(ŷn+1*,ŷn+2*)Π12+dn(dn+δn-1(ŷn+1*,ŷn+2*)Π1) if ϱ(λn)=3.Proof.
On the bases of (6.1), (6.3), (2.1), (2.6), and (6.9), we have
(6.12)(ŷn,υ̂n*)=δnifϱ(λn)=1;(ŷn,υ̂m*)=0ifϱ(λn)=ϱ(λm)=1,n≠m;(ŷn,υ̂m*)=0,m=k,k+1,(ŷk,υ̂k+1*)=(ŷk+1,υ̂k*)=‖ŷk+1*‖Π12+(ck+c̃k)δk,(ŷk,υ̂k*)=(ŷk+1,υ̂k+1*)=δkifϱ(λn)=1,ϱ(λk)=2;(ŷn,υ̂m*)=0,(ŷm,υ̂m*)=δm,m=k,k+1,k+2,(ŷk,υ̂k+1*)=(ŷk+1,υ̂k*)=(yk+1*,yk+2*)Π1+(dk+d̃k)δk,(ŷk,υ̂k+2*)=(ŷk+2,υ̂k*)=‖ŷk+2*‖Π12+(dk+d̃k)(ŷk+1*,ŷk+2*)Π1+(dkd̃k+hk+h̃k)δkifϱ(λn)=1,ϱ(λk)=3.
Using relation (6.12) and taking into account (6.11), we get the validity of the equality
(6.13)(ŷn,υ̂k)=δnk,
where δnk is the Kronecker delta. The proof of Lemma 6.4 is complete.Corollary 6.5.
(i) Ifϱ(λn)=1,  then sn≠0; (ii) if ϱ(λn)=2, then sn+1≠0, sn≠0 at cn≠cn(0), sn=0at cn=cn(0), where cn(0)=mn-1mn+1*-δn-1∥ŷn+1*∥Π12; (iii) if ϱ(λn)=3, then sn+2≠0, sn+1≠0 at dn≠dn(0), sn+1=0 at dn=dn(0); sn≠0 at hn≠hn(0), sn=0 at hn=hn(0), where dn0=mn-1mn+1*-δn-1(ŷn+1*,ŷn+2*)Π1, hn(0)=mn-1mn+2*-(dn+δn-1(ŷn+1*,ŷn+2*)Π1)(dn-mn-1mn+1*)-δn-1∥ŷn+2*∥Π12+δn-2(ŷn+1*,ŷn+2*)Π12.Theorem 6.6.
Letr be an arbitrary fixed integer. If sr≠0, then the system {yn(x)}n=1,n≠r∞ forms a basis in Lp(0,l),  p∈(1,∞),  and even a Riezs basis in L2(0,l); if sr=0, the system {yn(x)}n=1,n≠r∞ is neither complete nor minimal in Lp(0,l),  p∈(1,∞).Proof.
By Theorem 7 in [40, Chapter 1, Section 4] and Theorem 6.1, the system {υ̂n}n=1∞ is a Riesz basis in H. Then, for any vector f={f,ξ}∈H, the following expansion holds:
(6.14)f̂={f,ξ}=∑n=1∞(f̂,ŷn)Hυ̂n=∑n=1∞((f,yn)L2+σ-1ξm¯n){υn,sn},
whence it follows the equalities
(6.15)f=∑n=1∞((f,yn)L2+σ-1ξm¯n)υn,ξ=∑n=1∞((f,yn)L2+σ-1ξm¯n)sn.
If ξ=0, then by (6.15), we have
(6.16)f=∑n=1∞(f,yn)L2υn,(6.17)0=∑n=1∞(f,yn)L2sn.
Letsr≠0. Then by (6.17), we obtain
(6.18)(f,yr)L2=-sr-1∑n=1n≠r∞(f,yn)L2sn,
considering which in (6.16), we get
(6.19)f=∑n=1n≠r∞(f,yn)L2(υn-sr-1snυr).
By (6.13) and (2.1), we have
(6.20)(yn,υk-sr-1skυr)L2=(yn,υk)L2-s¯r-1s¯k(yn,υr)L2=(ŷn,υ̂k)H-|σ|-1mns¯k-s¯r-1s¯k(yn,υr)L2+s¯r-1s¯k|σ|-1mns¯r=δnk,n,k≠r,
that is, the system {υn(x)-sr-1snυr(x)}n=1,n≠r∞ is conjugated to the system {yn(x)}n=1,n≠r∞. By virtue of (6.19), the system {υn(x)-sr-1snυr(x)}n=1,n≠r∞ is a Riesz basis in L2(0,l). Then, on the base of Corollary 2 [40, Chapter 1, Section 4] {yn(x)}n=1,n≠r∞ is also in a Riesz basis in L2(0,l). The basicity of the system {yn(x)}n=1,n≠r∞ in the space Lp(0,l), p∈(1,∞)∖{2}, can be proved by scheme of the proof of Theorem 5.1 in [18] using Theorem 5.2.
Now letsr=0. Then, by (2.1) and (6.13) we have
(6.21)(yn,υr)L2=(ŷn,υ̂r)H-|σ|-1mns¯r=0,n∈N,n≠r.
So, the functionυτ(x) is orthogonal to all functions of the system {yn(x)}n=1,n≠r∞, that is, the system {yn(x)}n=1,n≠r∞ is incomplete in L2(0,l).
On the basis of Corollary6.5 there exists k∈ℕ such that sk≠0 (e.g., if ϱ(λk)=1). Then, for any f(x)∈L2(0,l) the following expansion holds:
(6.22)f=∑n=1n≠k∞(f,υn-sk-1snυk)L2yn.
By (6.21) and (6.22), we get
(6.23)yk=∑n=1n≠k∞(yk,υn-sk-1snυk)L2yn=∑n=1n≠k,r∞(yk,υn-sk-1snυk)L2yn+(yk,υr-sk-1srυk)L2yr=∑n=1n≠k,r∞((yk,υn)L2-sk-1s¯n(yk,υk)L2)yn=∑n=1n≠k,r∞{(ŷk,υ̂n)H-|σ|-1mksn-s¯k-1s¯n(ŷk,υ̂k)+s¯k-1s¯n|σ|-1mks¯k}yn=-∑n=1n≠k,r∞s¯k-1s¯nyn,
whence it follows the equality ∑n=1,n≠r∞s¯nyn=0, that is, the system {yn(x)}n=1,n≠r∞ is nonminimal in L2(0,l).
Obviously, this system is neither complete nor minimal inLp(0,l), p∈(1,∞)∖{2}. The proof of Theorem 6.6 is complete.Corollary 6.7.
Ifσ>0, the system {yn(x)}n=1,n≠r∞ (for any r∈ℕ) is a basis in Lp(0,l),  p∈(1,∞),  and even a Riesz basis for p=2.



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*Source: 456517-2012-04-03.xml*