Volume 2012, Article ID 456517,28pages doi:10.1155/2012/456517
Research Article
Spectral Properties of the Differential
Operators of the Fourth-Order with Eigenvalue
Parameter Dependent Boundary Condition
Ziyatkhan S. Aliyev
1and Nazim B. Kerimov
21Department of Mathematics, Baku State University, Baku AZ 1148, Azerbaijan
2Department of Mathematics, Mersin University, 33343 Mersin, Turkey
Correspondence should be addressed to Ziyatkhan S. Aliyev,z [email protected]
Received 15 August 2011; Accepted 12 November 2011
Academic Editor: Amin Boumenir
Copyrightq2012 Z. S. Aliyev and N. B. Kerimov. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We consider the fourth-order spectral problemy4x−qxyxλyx, x∈0, lwith 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 locationsin complex planeand multiplicities of the eigenvalues, the oscillation properties of the eigenfunctions, the basis properties inLp0, l,p∈1,∞, of the system of root
functions of this problem.
1. Introduction
The following boundary value problem is considered:
y4x−qxyxλyx, x∈0, l, : d
dx, 1.1
y0 0, 1.2a
y0cosβTy0sinβ0, 1.2b
ylcosγylsinγ0, 1.2c
aλbyl−cλdTyl 0, 1.2d
whereλis a spectral parameter,Ty≡y−qy,qis absolutely continuous function on0, l,
we assume that the equation
y−qy0, 1.3
is disfocal in0, l, that is, there is no solution of1.3such thatya yb 0 for any a, b∈0, l. Note that the sign ofqwhich satisfies the disfocal condition may change in0, l.
Problems of this type occur in mechanics. If β 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 whena <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 see1–3.
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 in4,7,9,12,17,18,21,24for 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 Lp0, l, p ∈ 1,∞, of the system of root functions of the boundary value problem1.1,1.2a–1.2dwithq≥0,σ >0, are considered in18and, withq≥0,σ <0, c0, are considered in4,5.
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 inLp0, l,p ∈ 1,∞, of the system of root functions of the problem1.1,
1.2a–1.2d.
Note that the sign ofσ plays an essential role. In the caseσ > 0 we associate with problem1.1,1.2a–1.2da selfadjoint operator in the Hilbert spaceHL20, l⊕Cwith 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 Lp0, l, p ∈
1,∞. Forσ < 0 problem1.1,1.2a–1.2dcan be interpreted as a spectral problem for a selfadjoint operator in a Pontryagin spaceΠ1. It is proved below that nonreal and nonsimple
multipleeigenvalues are possible and the system of root functions, with arbitrary function removed, forms a basis in the spaceLp0, 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
LetHL20, l⊕Cbe a Hilbert space equipped with the inner product
y,uHy, m,{u, s}Hy, uL2σ−1ms, 2.1
We define in theHoperator
LyLy, m Tyx, dTyl−byl 2.2
with domain
DL yy, m∈H/yx∈W240, l,
Tyx∈L20, l, y∈B.C., mayl−cTyl,
2.3
that is dense inH 23,25, whereB.C.denotes the set of separated boundary conditions
1.2a–1.2c.
Obviously, the operatorLis well defined. By immediate verification we conclude that problem1.1,1.2a–1.2dis equivalent to the following spectral problem:
Lyλy, y∈DL, 2.4
that is, the eigenvalueλnof problem1.1,1.2a–1.2dand those of problem2.4coincide;
moreover, there exists a correspondence between the eigenfunctions and the adjoint functions of the two problems:
yn
ynx, mn
←→ynx, mnaynl−cTynl. 2.5
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, thenLis a selfadjoint discrete lower-semibounded operator inHand hence
has a system of eigenvectors{{ynx, mn}}∞n1, that forms an orthogonal basis inH.
In the caseσ <0 the operatorLis closed and non-selfadjoint and has compact resolvent
inH. InHwe now introduce the operatorJbyJ{y, m}{y,−m}.Jis a unitary, symmetric
operator inH. Its spectrum consists of two eigenvalues:−1 with multiplicity 1, and1 with infinite multiplicity. Hence, this operator generates the Pontryagin spaceΠ1L20, l⊕Cby means of the inner productsJ-metric 26:
y,uΠ1 y, m,{u, s}Π1y, uL2σ−1ms. 2.6
Lemma 2.1. Lis aJ-selfadjoint operator inΠ1.
Proof. JLis selfadjoint inHby virtue of Theorem 2.211. Then, J-selfadjointness ofLonΠ1
follows from27, Section 3, Proposition 30.
Lemma 2.2see27, Section 3, Proposition 50. LetL∗be an operator adjoined to the operatorL
inH. Then,L∗JLJ.
Theorem 2.3see28. The eigenvalues of operatorLarrange symmetrically with regard to the real
axis.nk1ρλk≤1 for any system{λk}nk1n≤∞of eigenvalues with nonnegative parts.
FromTheorem 2.3it follows that either all the eigenvalues of boundary value problem
1.1,1.2a–1.2dare simpleall the eigenvalues are real or all, except a conjugate pair of nonreal, are realor all the eigenvalues are real and all, except one double or triple, are simple.
3. Some Auxiliary Results
As in17,19, 29,30for the analysis of the oscillation properties of eigenfunctions of the problem1.1,1.2a–1.2dwe will use a Pr ¨ufer-type transformation of the following form:
yx rxsinψxcosθx,
yx rxcosψxsinϕx,
yx rxcosψxcosϕx,
Tyx rxsinψxsinθx.
3.1
Consider the boundary conditionssee29,30
y0cosα−y0sinα0, 1.2a∗
ylcosδ−Tylsinδ0, 1.2d∗
whereα∈0, π/2,δ∈0, π.
Alongside the spectral problem 1.1, 1.2a–1.2d we will consider the spectral
problem1.1,1.2a–1.2c, and1.2d∗. In30, Banks and Kurowski developed an extension of the Pr ¨ufer transformation 3.1 to study the oscillation of the eigenfunctions and their derivatives of problem1.1,1.2a∗,1.2b,1.2c, and1.2d∗withq ≥0,δ ∈ 0, π/2and in some cases when1.3is disfocal andαγ 0,δ∈0, π/2. In19, the authors used the Pr ¨ufer transformation3.1to study the oscillations of the eigenfunctions of the problem1.1,
1.2a∗,1.2b,1.2c, and1.2d∗withq≥0 andδ ∈π/2, π. In this work it is proved that problem1.1,1.2a∗,1.2b,1.2c, and1.2d∗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 wayit is equal to the serial number of an eigenvalue increasing by 1; the function associated with the lowest eigenvalue has no zeros in0, l however in reality, this eigenfunction has no zeros in0, lif the least eigenvalue is positive; the number of zeros can by arbitrary if the least eigenvalue is negative. In31, Ben Amara developed an extension of the classical Sturm theory32to study the oscillation properties for the eigenfunctions of the problem
1.1,1.2a–1.2c, and1.2d∗withβ0, in particular, given an asymptotic estimate of the number of zeros in0, lof the first eigenfunction in terms of the variation of parameters in the boundary conditions.
sufficiently small constant, then we have alsohx>0 on0, l. Thus,hx>0 in0, l, and hence the following substitutions33, Theorem 12.1:
ttx lω−1
x
0
hsds, ω
l
0
hsds, 3.2
transform0, linto the interval0, land1.1into
py¨··λry, 3.3
where p lω−1h3, r l−1ωh−1; hx, yx are taken as functions of t and · : d/dt. Furthermore, the following relations are useful in the sequel:
˙
yl−1ωh−1y, l2ω−2h3y¨ hy−hy, Ty ≡lω−1h3y¨
·
Ty. 3.4
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 when1.3is disfosal on 0, l, they partially proved it30, Lemma 7.1, and therefore they were able to study problem1.1,1.2a–1.2c, and1.2d∗withγ0,δ∈0, π/2. In31, Ben Amara shows howLemma 3.1together with the transformation3.2can be applicable to investigate boundary conditions1.2a–1.2c, and1.2d∗withβ0.
Lemma 3.1see33, Lemma 2.1. Letλ >0, and letybe a nontrivial solution of 3.3. Ify,y,˙ y¨,
andTy are nonnegative atta(but not all zero), they are positive for allt > a. Ify,−y˙, ¨y, and−Ty
are nonnegative atta(but not all zero), they are positive for allt < a.
We also need the following results which are basic in the sequel.
Lemma 3.2. All the eigenvalues of problem1.1,1.2a–1.2c, and1.2d∗forδ ∈ 0, π/2or
δπ/2,β∈0, π/2are positive.
Proof. In this case, the transformed problem is determined by 3.3 and the boundary
conditions
˙
y0 0, 3.5a
y0cosβTy 0sinβ0, 3.5b
˙
ylcosγ∗ply¨lsinγ∗0, 3.5c
ylcosδ−Ty lsinδ0, 3.5d
It is known that the eigenvalues of 3.3, 3.5a–3.5d are given by the max-min principle13, Page 405using the Rayleigh quotient
Ry
l
0py¨2dtN
y
l
0y2dt
, 3.6
whereNy y20cotβy˙2lcotγ∗y2lcotδ. It follows by inspection of the numeratorR
in3.6that zero is an eigenvalue only in the caseβδ π/2. Hence, all the eigenvalues of problem3.3,3.5a–3.5dforδ∈0, π/2orδπ/2,β∈0, π/2, are positive.Lemma 3.2
is proved.
Lemma 3.3. LetEbe the space of solution of the problem1.1,1.2a–1.2c. Then, dimE1.
The proof is similar to that of19, Lemma 2using transformation3.2, Lemmas3.1
and 3.2see also31, Lemma 2.2. However, it is not true ifπ/2 < γ < π see, e.g., 31, Page 9. Therefore,Lemma 3.1together with the transformation3.2cannot be applicable to investigate more general boundary conditions, for example,1.2a∗,1.2b, and1.2cfor 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). Ifais a zero ofuand ¨uin the interval0,l, then ˙utTu t<0
in a neighborhood ofa. Ifais a zero of ˙uorTu in0,l, thenutu¨t<0 in a neighborhood ofa.
Theorem 3.5. Letube a nontrivial solution of the problem1.1,1.2aand1.2cforλ > 0. Then
the JacobianJu r3cosψsinψof the transformation3.1does not vanish in0,l.
Proof. Letube a nontrivial solution of1.1which satisfies the boundary conditions1.2a
and1.2c. Assume first that the corresponding angleψsatisfiesψx0 nπfor some integer nand for somex0 ∈ 0, l. Then, the transformation3.1implies thatux0 Tux0 0.
Using the transformation3.2, the solutionuof3.3also satisfiesut0 Tu t0 0, where t0l−1ω
x0
0 hsds∈0, l. However, it is incompatible with the conclusion ofLemma 3.4.
The proof of the inequality cosψx/0,x ∈0, l, proceeds in the same fashion as in the previous case. The proof ofTheorem 3.5is complete.
Letyx, λbe a nontrivial solution of the problem1.1,1.2a–1.2c forλ > 0 and θx, λ,ϕx, λthe corresponding functions in3.1. Without loss of generality, we can define the initial values of these functions as followssee30, Theorem 3.3:
θ0, λ β−π
2, ϕ0, λ 0. 3.7
With obvious modifications, the results stated in30, Sections 3–5 are true for the solution of the problem1.1,1.2a–1.2c, and 1.2d∗ forδ ∈ 0, π/2. In particular, we have the following results.
Theorem 3.7. Problem1.1,1.2a–1.2c, and1.2d∗forδ ∈0, π/2(except the caseβ δ π/2) has a sequence of positive and simple eigenvalues
λ1δ< λ2δ<· · ·< λnδ−→ ∞. 3.8
Moreover,θl, λnδ 2n−1π/2−δ,n∈N; the corresponding eigenfunctionsυnδxhaven−1
simple zeros in0,l.
Remark 3.8. In the caseβ δ π/2 the first eigenvalue of boundary value problem1.1,
1.2a–1.2c, and1.2d∗ is equal to zero and the corresponding eigenfunction is constant; the statement ofTheorem 3.7is true forn≥2.
Obviously, the eigenvaluesλnδ,n∈N, of the problem1.1,1.2a–1.2c, and1.2d∗
are zeros of the entire functionyl, λcosδ−Tyl, λcosδ0. Note that the functionFλ Tyl, λ/yl, λis defined forλ∈A≡C/R∪∞n1λn−10, λn0, whereλ00 −∞.
Lemma 3.9see19, Lemma 5. Letλ∈A. Then, the following relation holds:
d
dλFλ
l
0y2x, λdx
y2l, λ . 3.9
In1.1we setλρ4. As is knownsee34, Chapter II, Section 4.5, Theorem 1in each
subdomainT of the complexρ-plane equation1.1has four linearly independent solutions zkx, ρ,k1,4, regular inρfor sufficiently largeρand satisfying the relations
zksx, ρρωk
s
eρωkx1, k1,4, s0,3, 3.10
whereωk,k1,4, are the distinct fourth roots of unity,1 1O1/ρ.
For brevity, we introduce the notationsδ1, δ2 sgnδ1sgnδ2. Using relation3.10
and taking into account boundary conditions1.2a–1.2c, we obtain
yx, λ
⎧ ⎪ ⎨ ⎪ ⎩
sinρxπ 2 sinβ
−cosρlπ
2s
β, γeρx−l1 ifβ∈0,π
2
,
√
2 sinρx−π 4
−e−ρx −11−sgnγ√2 sinρl−1sgnγπ
4
eρx−l1 ifβ0,
3.11
Tyx, λ
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
−ρ3cosρxπ
2 sgnβ
cosρl π
2s
β, γeρx−l1
if β∈0,π 2
,
−ρ3√2 sinρxπ
4
−e−ρx−−11−sgnγ√2 sinρlπ
4−1
sgnγ
eρx−l1
ifβ0.
Remark 3.10. As an immediate consequence of3.11, we obtain that the number of zeros in the interval0, lof functionyx, λtends to∞asλ → ±∞.
Taking into account relations3.11and3.12, we obtain the asymptotic formulas
Fλ
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ √
21−sgnγρ3 cos
ρl π/2sgnβ π/4sgnγ
cosρl π/2sgnβ π/41sgnγ1 ifβ∈
0,π 2 , √
21−sgnγρ3cos
ρl π/4sgnγ−1
cosρl π/41sgnγ1 ifβ0.
3.13
Furthermore, we have
Fλ −√21−sgnγ 4
|λ|31O|λ|−1/4, asλ−→ −∞. 3.14
We define numbersτ,ν,η,αn,βn,ηn,n∈N, and a functionzx, t,x∈0, l,t∈R, as
follows: τ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
31sβ, δ
4 −1 ifγ∈
0,π 2 , 5 4 − 3 8
−1sgnβ −1sgnδ−1 ifγ0,
η ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
32sgnβ
4 −1 ifγ ∈
0,π 2 , 5 4 − 3 8
−1sgnβ−1−1 ifγ 0,
ν ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
31sβ,|c|
4 if γ∈
0,π 2 , 5 4 − 3 8
−1sgnβ −1sgn|c| if γ0,
αn n−τπ
l , ηn
n−ηπ
l , βn
n−νπ
l ,
zx, t
⎧ ⎪ ⎨ ⎪ ⎩
sintxπ 2 sgnβ
−costlπ
2s
β, γe−tl−x if β∈0,π
2
,
√
2 sintx−π 4
e−tx −1sgnγ√2 sin
tl−1
sgnγ
π 4
e−tx−l if β0.
3.15
By virtue of18, Theorem 3.1, one has the asymptotic formulas
4
λnδ αnO
n−1, 3.16
υnδx zx, αn O
n−1, 3.17
By3.14, we have
lim
λ→ −∞Fλ −∞. 3.18
From Property 1 in30and formulas3.9, one has the relations
λ1
π
2
< λ10< λ2
π
2
< λ20<· · ·. 3.19
Remark 3.11. It follows byTheorem 3.7,Lemma 3.9, and relations3.18and3.19that ifλ >0
orλ0,β∈0, π/2, thenFλ<0; besides, ifλ0 andβπ/2, thenFλ 0.
Letsλbe the number of zeros of the functionyx, λin the interval0, l.
Lemma 3.12. Ifλ >0 andλ∈λn−10, λn0,n∈N, thensλ 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 problem1.1,1.2a–1.2c, and1.2d∗forδ ∈ π/2, πhas a sequence of real and simple eigenvalues
λ1δ< λ2δ<· · ·< λnδ−→∞, 3.20
including at most one negative eigenvalue. Moreover, (a) if β ∈ 0, π/2, then λ1δ > 0 for δ ∈
π/2, δ0;λ1δ 0 forδδ0; λ1δ<0 forδ∈δ0, π, whereδ0arctgTyl,0/yl,0; (b) if
β π/2, thenλ1δ< 0; (c) the eigenfunctionυnδx, corresponding to the eigenvalueλnδ≥0,
has exactlyn−1 simple zeros in0, l.
The proof parallels the proof of19, Theorem 4using Theorems3.5–3.7and Lemmas
3.9and3.12.
Lemma 3.14. The following non-selfadjoint boundary value problem:
y4x−qxyxλyx, x∈0, l,
y0 y0 Ty0 ylcosγylsinγ0,
3.21
has an infinite set of nonpositive eigenvaluesρntending to−∞and satisfying the asymptote
λn−
n−1
4
1sgnγ
4π4
l4 o
n4, n−→ ∞. 3.22
Settingx0 in3.12, we obtain3.22.
Remark 3.15. ByRemark 3.10the number of zeros of the eigenfunctiony1δxcorresponding
new zeros of the corresponding eigenfunctiony1δxenter the interval0, lonly through the end pointx0sincey1δl/0, and hence the number of its zeros, in the caseβ∈0, π/2, is asymptotically equivalent to the number of eigenvalues of the problem3.21which are higher thanλ1δ. In the caseβ0 see31, Theorem 5.3.
We consider the following boundary conditions:
ayl−cTyl 0, 1.2d
cyl aTyl 0. 1.2d
Note thata, c/0 sinceσ <0. The boundary condition1.2dcoincides the boundary condition1.2d∗forδ π/2resp.,δ 0in the casea0resp.,c0, and the boundary condition1.2dcoincides the boundary condition1.2d∗forδ 0resp.,δ π/2in the casea0resp.,c0.
Letac /0. The eigenvalues of the problem1.1,1.2a–1.2c, and1.2d resp.,1.1
1.2a–1.2c, and1.2dare the roots of the equationFλ a/cresp.,Fλ −c/a. By
3.9, this equation has only simple roots; hence all the eigenvalues of the problems1.1,
1.2a–1.2c, and1.2dand1.1,1.2a–1.2c, and1.2dare simple. On the base of3.9,
3.18, and 3.19in each intervalAn,n 1,2, . . ., the equationFλ a/cresp.,Fλ −c/ahas a unique solutionμnresp.,νn; moreover,
ν1< λ1
π
2
< μ1< λ10< ν2< λ2
π
2
< μ2< λ20<· · · 3.23
ifa/c >0 and
μ1< λ1
π
2
< ν1< λ10< μ2< λ2
π
2
< ν2< λ20<· · · 3.24
ifa/c <0. Besides,μ10 ifa/c <0 andF0 a/c; ν10 ifa/c >0 andF0 −c/a.
Taking into account1.2d, 1.2d, 3.23, and3.24 and using the corresponding reasoning18, Theorem 3.1we have
4
μnηnO
n−1, √4ν
nηnO
n−1, 3.25
ϕnx zx, ηn
On−1, ψnx zx, ηn
On−1, 3.26
where relation3.26holds uniformly forx∈0, land eigenfunctionsϕnxandψnx,n∈N, correspond to the eigenvaluesμnandνn, respectively.
Let us denotemλ ayl, λ−cTyl, λ.
Remark 3.16. Note that if λis the eigenvalue of problem 1.1,1.2a–1.2d, thenmλ/0
It is easy to see that the eigenvalues of problem1.1,1.2a–1.2dare roots of the equation
aλbyl−cλdTyl 0. 3.27
By virtue of Remark 3.16 and formula 3.27, a simple calculation yields that the eigenvalues of the problem1.1,1.2a–1.2dcan be realized at the solution of the equation
cyl, λ aTyl, λ
ayl, λ−cTyl, λ a2c2
−σ λ
abcd
−σ . 3.28
DenoteBn μn−1, μn,n∈N, whereμ0−∞.
We observe that the functionGλ cyl, λ aTyl, λ/ayl, λ−cTyl, λis well defined forλ ∈ B C\R∪∞n1Bnand is a finite-order meromorphic function and the
eigenvaluesνn andμn,n ∈ N, of boundary value problems1.1,1.2a–1.2c, and1.2d
and1.1,1.2a–1.2c, and1.2dare zeros and poles of this function, respectively.
Letλ∈B. Using formula3.9, we get
d
dλGλ
a2c2m−2λ
l
0
y2x, λdx. 3.29
Lemma 3.17. The expansion
Gλ
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
G0
∞
n1
λcn
μn
λ−μn
ifμ1/0,
c0c1
λ
∞
n2
λcn
μn
λ−μn
ifμ10,
3.30
holds, wherecn,n∈N, are some negative numbers.
Proof. It is knownsee35, Chapter 6, Section 5that the meromorphic functionGλwith
simple polesμnallows the representation
Gλ G1λ
∞
n1
λ μn
s
cn
λ−μn, 3.31
whereG1λis an entire function,
cn res
λμn
Gλ cyl, μn
aTyl, μn
a∂y
l, μn
∂λ −c
Tyl, μn
∂λ
−1
, 3.32
and integerssn,n∈N, are chosen so that series3.31are uniformly convergent in any finite
We consider the casea/c > 0. By virtue of relation3.18, we have limλ→ −∞Gλ
−a/c. Hence,Gλ < 0 for λ ∈ −∞, ν1 and Gλ > 0 forλ ∈ ν1, μ1. Without loss of
generality, we can assumeayl, λ−cTyl, λ>0 forλ∈−∞, μ1. Then,cyl, λaTyl, λ<0 for λ ∈ −∞, ν1. Since the eigenvalues μn and νn, n ∈ N, are simple zeros of functions
ayl, λ−cTyl, λandcyl, λ aTyl, λ, respectively, then by3.29the relations
−1n1cyl, μn
aTyl, μn
>0,
−1n1
a∂y
l, μn
∂λ −c
∂Tyl, μn
∂λ
<0, n∈N,
3.33
are true.
Taking into account3.33, in3.32we getcn<0,n∈N. The casesa/c <0,a0, and
c0 can be treated along similar lines. DenoteΩnε {λ∈C| |√4
λ−√4μn|< ε}whereε >0 is some small number. From the
asymptotic formula3.25, it follows that forε < π/4lthe domainsΩnεasymptotically do not intersect and contain only one poleμnof the functionGλ.
By3.11,3.12,3.23,3.24, and3.25, we see that outside of domainsΩnεthe asymptotic formulae are true:
Gλ
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
−a
c O
ρ−1 ifac /0,
ρ3zρ1Oρ−1 ifc0,
−ρ−3zρ−11Oρ−1 ifa0,
3.34
where
zρ
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
cosπ/4sgnγcosρl π/2sgnβ π/4sgnγ
sinπ/41sgnγcosρl π/2sgnβ π/41sgnγ ifβ∈
0,π 2
,
√
21−2 sgnγcosρl−1−sgnγπ/4
cosρl π/4sgnγ ifβ0.
3.35
Following the corresponding reasoning see 36, Chapter VII, Section 2, formula
27, we see that outside of domainsΩnεthe estimation
|Gλ| ≤
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
M1 ifac /0,
M24
|λ|3 ifc0,
M34
|λ|−3 ifa0,
holds; using it in3.32we get
|cn| 1
2πi
∂Ωnε
Gλdλ 2 π
|ν−√4μn|ε
ν3Gν4dν≤
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
M1n3 ifac /0,
M2n6 ifc0,
M3 ifa0,
3.37
where M1, M2,M3, M1, M2, M3 are some positive constants. By 3.37 and asymptotic
formula 3.25 the series ∞n1cn|μn|−2 converges. Then, according to Theorem 2 in 35, Chapter 6, Section 5, in formula3.31we can assumesn1,n∈N.
Let {Γn}∞n1 be a sequence of the expanding circles which are not crossing domains
Ωnε. Then, according to Formula9in37, Chapter V, Section 13, we have
Gλ−
μk∈intΓn
ck
λ−μk
1 2πi
Γn
Gξ ξ−λdξ,
G0
μk∈intΓn
ck
μk
1 2πi
Γn
Gξ
ξ dξ.
3.38
By3.38, we get
Gλ−G0
μk∈intΓn
λck
μk
λ−μk
1
2πi
Γn
λGξ
ξξ−λdξ. 3.39
From3.36, the right side of3.39tends to zero asn → ∞. Then, passing to the limit asn → ∞in3.39, we obtain
Gλ G0
∞
n1
λcn
μn
λ−μn
, 3.40
which impliesG1λ≡G0.
Differentiating the right side of the least equality, we have
Gsλ −1ss!
∞
n1
cn
λ−μn
s1, s1,2,3. 3.41
Note that the functionFλhas the following expansion:
Fλ F0
∞
n1
λcn
λn0λ−λn0, 3.42
where
cn res
λλn0
Now letμ10, that is,F0 a/c.Gλhas the following expansion:
Gλ G1λ c1
λ
∞
n2
λcn
μn
λ−μn
. 3.44
Again, according to Formula9in37, Chapter V, Section 13, we have
Gλ− c1
λ −
μk∈intΓn
k /1
ck
λ−μk
1 2πi
Γn
Gξ
ξ−λdξ. 3.45
By2.6 18and3.9, we get
c1−c−2
a2c2F0−1. 3.46
Using3.42,3.41, and3.46, we obtain
Gλ−c1
λ −
a
c c
−2a2c2
∞
n1
cn/λ2n0λ−λn0
F0∞n1cn/λn0λ−λn0
. 3.47
Passing to the limit asλ → 0 in3.47, we get
lim
λ→0
Gλ− c1 λ
−a
c c
−2a2c2∞
n1
cn
λ2n0
−1∞
n1
cn
λ3n0
−1
a
c 2c
−2a2c2F0−2F0 c
0.
3.48
Using3.48in3.45, we have
c0
μk∈intΓn
k /1
ck
λ−μk
1 2πi
Γn
Gξ
ξ dξ. 3.49
In view of3.49and3.45, we get
Gλ−c0−c1
λ −
μk∈intΓn
k /1
λck
μk
λ−μk
1
2πi
Γn
λGξ
ξξ−λdξ. 3.50
Passing to the limit asn → ∞in3.50, we obtain
Gλ c0c1
λ
∞
n2
λcn
μn
λ−μn
. 3.51
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 integerNfrom the inequalityμN−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
λ1< λ2<· · ·< λn−→∞, 4.1
including at most 1sgn|c|number of negative ones. The corresponding eigenfunctions have the
following oscillation properties.
aIf c 0, then the eigenfunction ynx, n ≥ 2, has exactly n− 1 zeros in 0, l, the
eigenfunction y1x has no zeros in0, lin the case λ1 ≥ 0, and the number of zeros
ofy1xcan be arbitrary in the caseλ1<0.
bIfc /0, then the eigenfunctionynxcorresponding to the eigenvalueλn ≥ 0 has exactly
n−1 simple zeros forn ≤ Nand exactlyn−2 simple zeros for n > Nin 0, land the eigenfunctions associated with the negative eigenvalues may have an arbitrary number of
simple zeros in0, l.
The proof of this theorem is similar to that of18, Theorem 2.2usingRemark 3.15. Throughout the following, we assume thatσ <0.
Letλ, μλ /μbe the eigenvalue of the operatorL. The eigenvectorsyλ {yx, λ, mλ}andyμ {yx, μ, mμ}corresponding to the eigenvaluesλandμ, respectively, are orthogonal inΠ1, since the operatorLisJ-selfadjoint inΠ1. Hence, by2.4, we have
l
0
yx, λyx, μdx−σ−1mλmμ. 4.2
Lemma 4.2. Let λ∗ ∈ R be an eigenvalue of boundary value problem 1.1, 1.2a–1.2d and
Gλ∗≤A, whereA−a2c2/σ. Then, problem1.1,1.2a–1.2dhas no nonreal eigenvalues.
Proof. Letμ∈C\Rbe an eigenvalue of problem1.1,1.2a–1.2d. Then, fromRemark 3.16
and equality4.2, we obtain
l
0
yx, λ∗ mλ∗
⎛
⎝yx, μ
mμ
⎞
⎠dx−σ−1, l
0
yx, λ∗ mλ∗
yx, μ
mμ dx−σ
−1, 4.3
l
0
yx, μ
mμ
2
dx−σ−1. 4.4
In view of formula3.29, the inequality
l
0
yx, λ∗ mλ∗
2
dx≤ −σ−1 4.5
By4.3,
l
0
yx, λ∗
mλ∗ Re
yx, μ
mμ dx−σ
−1. 4.6
From4.4–4.6, we get
l
0 ⎧ ⎨ ⎩
yx, λ∗
mλ∗ −Re
yx, μ
mμ
2
Im2y
x, μ
mμ
⎫ ⎬
⎭dx <0 ifGλ∗< A,
l
0 ⎧ ⎨ ⎩
yx, λ∗
mλ∗ −Re
yx, μ
mμ
2
Im2y
x, μ
mμ
⎫ ⎬
⎭dx0 ifGλ∗ A.
4.7
From the second relation it follows that Imyx, μ/mμ 0, which by 1.1
contradicts the conditionμ∈C\R. The obtained contradictions proveLemma 4.2.
Lemma 4.3. Letλ∗1, λ∗2∈R,λ∗1/λ∗2be eigenvalues of problem1.1,1.2a–1.2dandGλ∗1≤A.
Then,Gλ∗2> A.
Proof. LetGλ∗2≤A. By3.29and4.2, we have
l
0
yx, λ∗1 mλ∗1
2
dx≤−σ−1,
l
0
yx, λ∗2 mλ∗2
2
dx≤−σ−1,
l
0
yx, λ∗1 mλ∗1
yx, λ∗2
mλ∗2 dx−σ
−1.
4.8
Hence, we get
l
0
yx, λ∗1 mλ∗1
yx, λ∗2 mλ∗2
dx <0 ifGλ∗1
< AorGλ∗2
< A,
l
0
yx, λ∗1 mλ∗1
yx, λ∗2
mλ∗2 dx0 ifG
λ∗
1
Gλ∗2A.
4.9
From4.9, it follows thatyx, λ∗1/mλ∗1 yx, λ∗2/mλ∗2forx ∈ 0, l. Therefore, mλ∗2yx, λ∗1 mλ∗1yx, λ∗2.
Sinceλ1/λ2, then by1.1yx, λ1≡0. The obtained contradictions proveLemma 4.3.
By Lemmas4.2and4.3problem1.1,1.2a–1.2dcan have only one multiple real eigenvalue. From3.41, we getG3λ>0,λ∈B, whence it follows that the multiplicity of real eigenvalue of problem1.1,1.2a–1.2ddoes not exceed three.
Theorem 4.4. The boundary value problem1.1,1.2a–1.2dforσ <0 has only point spectrum,
repeated according to algebraic multiplicity and ordered so as to have increasing real parts. Moreover, one of the following occurs.
1All eigenvalues are real, at that B1 contains algebraically two (either two simple or one
double) eigenvalues, andBn,n2,3, . . ., contain precisely one simple eigenvalues.
2All eigenvalues are real, at that B1contains no eigenvalues but, for some s ≥ 2, Bs
contains algebraically three (either three simple, or one simple and one double, or one triple)
eigenvalues, andBn,n2,3, . . .,n /scontain precisely one simple eigenvalue.
3There are two nonreal eigenvalues appearing as a conjugate pair, at that B1 contains no
eigenvalues, andBn,n2,3, . . ., contain precisely one simple eigenvalue.
Proof. Remember that the eigenvalues of problem 1.1, 1.2a–1.2d are the roots of the
equationGλ AλB, whereA−a2c2/σ,B−abcd\σsee3.28. From3.41,
it follows thatGλ>0 forλ∈B1; therefore, the functionGλis convex on the intervalB1.
By virtue of3.18and3.30, we have
lim
λ→ −∞Gλ
⎧ ⎨ ⎩
−a
c ifc /0,
−∞ ifc0,
lim
λ→μn−0
Gλ ∞.
4.10
That is why for each fixed numberAthere exists numberBAsuch that the linesAλBA,
λ∈R, touch the graph of functionGλat some pointλ∈B1. Hence, in the intervalB1,3.28
has two simple rootsλ1 < λ2 ifB > BA, one double rootλ1 λif B BA, and no roots if
B < BA.
By 3.29 and 3.30 we have limλ→μn0Gλ −∞, limλ→μn−0Gλ ∞, n ∈ N.
Therefore,3.28has at least one solution in the intervalBn,n2,3, . . ..
LetB≥BA. IfB > BA, thenGλ1< A,Gλ2< A; ifBBAthenGλ1 A. By3.29, 3.28has only simple rootλn1forB > BA, λnforBBAin the intervalBn,n2,3, . . ..
LetB < BA. ByLemma 4.3eitherGλn> Afor anyλn ∈R, or there existsk∈Nsuch
thatFλk≤AandFλn> A,n∈N\ {k}. Assume thatλk ∈Bs. Obviously,s≥2. Choose
natural numbern0such that the inequalities
ARn0B >0,
|Gλ−AλBA|>|BA−B|, λ∈SRn0,
4.11
are fulfilled; whereRnτnδ0,
τn ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
λn
π
2
ifc0,
νn ifc /0, a
c >0, λn0−1 ifc /0, a0,
νn−1 ifc /0, a
c <0,
4.12
We have
ΔSRn
0argGλ−AλBΔSRn0 argGλ−AλBAΔSRn0arg
1 BA−B
Gλ−AλBA
,
4.13
where
ΔSRn0argfz
1 i
SRn0
fz fz
dz 4.14
see37, Chapter IV, Section 10. By4.11
BA−B
Gλ−AλBA
<1, λ∈SRn0, 4.15
hence, the point
ω BA−B
Gλ−AλBA 4.16
does not go out of circle{|ω|<1}. Therefore, vectorw1ωcannot turn around the point
w0, and the second summand in4.13equals zero. Thus,
ΔSRn
0argGλ−AλB ΔSRn0argGλ−AλBA. 4.17
By the argument principlesee37, Chapter IV, Section 10, Theorem 1we have
1
2πΔSRn0argGλ−AλBA
λnBA∈intSRn0
λBA
n
−
μn∈intSRn0
μn
, 4.18
whereρλBA
n andμnare multiplicity of zeroλnBAand poleμnof the functionGλ−Aλ
BA, respectivelyλBA
1 λ
BA
2 . Obviously,
λnBA∈intSRn0
ρλBA
n n0and
μn∈intSRn0ρμn
n0−1. Then, by4.18we obtain
2π−1ΔSRn0argGλ−AλBA 1. 4.19
From4.17and4.19follows the validity of the equality
Using the argument principle again, by4.20we get
λn∈intSRn0
ρλn−
μn∈intSRn0
μn
1, 4.21
whence it follows that
λn∈intSRn0
ρλn n0, 4.22
where λn, n ∈ N, are roots of the equation Gλ Aλ B. From the above-mentioned
reasoning, by4.22we have
λm∈intSRn
ρλm n, nn0, n01, . . . , 4.23
and, therefore, problem1.1,1.2a–1.2din the intervalBn forn n0,n01, . . ., has only
one simple eigenvalue.
Consider the following two cases.
Case 1. For all real eigenvaluesλn of problem 1.1,1.2a–1.2dthe inequalities Gλn >
A,λn ∈
∞
m2Bm, are fulfilled. The problem1.1,1.2a–1.2din every interval Bm,m
2,3, . . . , n0−1, has one simple eigenvalue. Hence, problem1.1,1.2a–1.2din the interval −∞, SRn,n ≥ n0, hasn−2 simple eigenvalues, and hence, by 4.23, this problem in the
circleSRn ⊂ Chas 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 λ1, Imλ1 > 0,λn ∈ Bn−1,
n3,4, . . ..
Case 2. LetGλk≤ A,Gλn> A,n∈N | {k}andλk ∈Bs,s≥ 2. ByLemma 4.2problem
1.1, 1.2a–1.2d has no nonreal eigenvalues. From the above-mentioned reasoning it
follows that in each intervalBn,n /k,n2,3, . . ., problem1.1,1.2a–1.2dhas one simple
eigenvalue.
Subcase 1. LetGλk A,Gλk/0, that is, the eigenvalueλkis a double oneby thisλk
λk1. Then, from4.23it follows that the intervalBsbesides the eigenvalueλkcontains one
more simple eigenvalue: at that it is eitherλk−1 by thisk sorλk2 by thisk s−1.
Hence,λn ∈ Bn1,n 1,2, . . . , s−2, λs−1, λs, λs1 ∈ Bs by this eitherλs−1 < λs λs1 or
λs−1λs< λs1,λn∈Bn−1,ns2, s3, . . ..
Subcase 2. LetGλk A,Gλk 0. By3.41,Gλk/0. Hence,λkis a triple eigenvalue
of the problem1.1,1.2a–1.2d by thisλkλk1λk2. Then, from4.23it follows that
in the intervalBsproblem1.1,1.2a–1.2dhas unique triple eigenvalueλk, and therefore,
ks−1. At thisλn∈Bn1,n1,2, . . . , s−2,λs−1λsλs1∈Bs,λn∈Bn−1,ns2, s3, . . ..
Subcase 3. LetGλk< A, that is, the eigenvalueλkis simple. Then, by4.23, in the interval
Bsproblem1.1,1.2has an eigenvalueλk as well as two more simple eigenvalues, which,
byLemma 4.3, are λk−1 and λk1 and hencek s. In this case, we haveλn ∈ Bn1,n
1,2, . . . , s−2,λs−1, λs, λs1∈Bsλs−1< λs< λs1,λn∈Bn−1,ns2, s3, . . ..
ByTheorem 4.4we haveλn 2, that is,λnλn1ifns−1 orns;λn 3, that
is,λnλn1λn2ifns−1If assertion2inTheorem 4.4holds, then we sets1.
Let {ynx}∞n1 be a system of eigen- and associated functions corresponding to the eigenvalue system{λn}∞n1of problem1.1,1.2a–1.2d, whereynx yx, λnifρλn 1;
ynx yx, λn,yn1x yn∗1x cnynx,y∗n1x ∂yx, λn/∂λ,cn is an arbitrary
constant, ifλn 2; ynx yx, λn,yn1x y∗n1x dnynx,yn2x y∗n2x
dny∗n1x hnynx,y∗n2x ∂2yx, λn/2∂λ2,dn,hn are arbitrary constants, ifρλn 3.
Here,ynxis an eigenfunction forλn and yn1xwhenρλn 2;yn1x,yn2xwhen
ρλn 3 are the associated functionssee34, Pages 16–20for more details.
We turn now to the oscillation theorem of the eigenfunctions corresponding to the positive eigenvalues of problem1.1,1.2a–1.2dsince the eigenfunctions associated with the negative eigenvalues may have an arbitrary number of simple zeros in0, l.
Theorem 4.5. For eachn < N (resp.,n > N),yn hasn−1 (resp.,n) zeros in the interval0,l.
Similarlyys,ys1both haves−1 (resp.,s) zeros ifs < N(resp.,s > N). Finally, ifc /0, then each of
yN(andys,ys1ifsN) hasN−1 orNzeros according toλN,λS,λS1≤or>−d/c, and ifc0
andsN, thenys,ys1both haveszeros.
The proof of this theorem is similar to that of11, Theorem 4.4usingLemma 3.12.
5. Asymptotic Formulae for Eigenvalues and Eigenfunctions of
the Boundary Value Problem
1.1
,
1.2a
–
1.2d
Forc /0, letKbe an integer such thatλk−1π/2 < b/a≤ λkπ/2 interpretingλ0π/2
−∞.
Lemma 5.1. The following relations hold for sufficiently largen∈N,n > n1max{s, N, K}2:
λn−20< λn< λn−1
π
2
< λn−10 if c /0, a c ≤0,
λn−20< λn−1
π
2
< λn< λn−10 if c/0, a
c >0 orc0.
5.1
Proof. Letac /0. Note that the eigenvaluesλn0 resp.,λnπ/2,n ∈ N, of problem1.1,
1.2a–1.2c, and1.2dforδ 0 resp., for δ π/2are roots of the equationGλ
−a/cresp.,Gλ c/a. The equationAλB −a/cresp.,AλB c/ahas a unique
solutionλ−d/cresp.,λ−b/a. Sincen >max{N2, K2}, in view of3.29,Gλn> max{−a/c, c/a}. Hence, by3.23,3.24, and3.29, the following relations hold forn > n1:
μn−2< λn−2
π
2
< νn−2 < λn−20< λn< μn−1 if a
c <0,
μn−2< λn−20< νn−1< λn−1
π
2
< λn< μn−1 if a
c >0.
Leta0. In this caseμn λnπ/2,νn λn0,n∈N. Sincen > N2, soGλn>0.
Then, by the equalityGλn AλnB,n∈N, we obtain
λn−2
π
2
< λn−20< λn< λn−1
π
2
< λn−10, n > n1. 5.3
Now letc 0. In this caseμn λn0,νn λnπ/2,n∈ N. Sincen > K2, soGλn >0.
Therefore, usingGλn AλnB, we have
λn−20< λn−1
π
2
< λn< λn−10, n > n1. 5.4
Relations5.1are consequences of relations5.2–5.4. The proof ofLemma 5.1is complete.
We define numbersχ,χn,n∈N, as follows:
χ
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
31sβ,|c|
4 if γ∈
0,π 2
, 5
4 − 3 8
−1sgnβ −1sgn|c| if γ0,
χn
n−χπ
l .
5.5
Using relations 5.1 and formulas 3.16, 3.17, the following corresponding
reasoning18, Theorem 3.1can be proved.
Theorem 5.2. The following asymptotic formulae hold:
4
λnχnO
n−1,
ynx zx, χn
On−1,
5.6
where relation5.6holds uniformly forx∈0, l.
6. Necessary and Sufficient Conditions of Basicity of
Root Function System of Problem
1.1
,
1.2a
–
1.2d
Note that the elementyn {ynx, mn},n ∈ N, of the system{yn}∞n1of the root vectors of
operatorLsatisfies the relation
Lyn λnynθnyn−1, 6.1
whereθnequals either 0at thatynis eigenvectoror 1at thatλnλn−1andynis associated
Theorem 6.1. The system of eigen- and associated functions of operatorLis a Riesz basis in the space
H.
Proof. Letμbe a regular value of operatorL, that is,Rμ L−μI−1exists and a bounded in
H. Then, problem2.4is adequate to the following problem of eigenvalues:
Rμy
λ−μ−1y, y∈DL. 6.2
By Lemma 2.1,Rμ is a completely continuousJ-selfadjoint operator inΠ1. Then, in
view of39the system of the root vectors of operatorRμhence of operatorLforms a Riesz
basis inH.Theorem 6.1is proved.
Let{υn∗}∞n1, whereυ∗n{υ∗nx, s∗n}, be a system of the root vectors of operatorL∗, that
is,
L∗υ∗nλnυ∗nθn1υ∗n1. 6.3
ByLemma 2.2and relations6.1,6.3we have the following.
Lemma 6.2. υ∗n Jynyn {ynx, mn}if λn 1; υ∗n Jy∗n1 cnJyn, υ∗n1 Jyn if
λn 2;υ∗n Jy∗n2dnJy∗n1hnJy∗n, υ∗n1 Jy∗n1dnJyn, υ∗n2 Jyn if λn 3, where
y∗n1{yn∗1x, m∗n1}, mn1mλn,yn∗2{y∗n2x, m∗n2},m∗n2 1/2mλn,cn,dn,hn are arbitrary constants.
Lemma 6.3. Let
δ
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
l
0
y2
nxdxσ−1m2n, ifρλn 1,
yn,yn∗1
Π1, ifρλn 2,
''y∗n1''2Π
1 ifρλn 3,
6.4
where · Π1is the norm inΠ1. Then,δn/0,n∈N.
Proof. By Remark 3.16,mn/0 if yn is the eigenvector of operator L. If λn 1, then
Gλn/A, whence by3.29, we getδn/0.
Letλn 2. Then,Gλn AandGλn/0. Differentiating the right-hand side of equality3.29onλ, we obtain
Gλ 2
a2c2
m2λ
(l
0
yx, λ∂yx, λ
∂λ dx−
mλ
mλ
l
0
y2x, λdx
)
. 6.5
Assumingλλnin6.5and taking into account3.29and2.6, we get
Gλn 2a2c2m−2 n
yn, y∗n1
Π1. 6.6
Now letλn 3, that is,Gλn AandGλn 0,Gλn/0. Differentiating the right-hand side of6.5onλ, we have
Gλ 2a2c2m−6λ
((l
0
∂yx, λ ∂λ
2
dx
l
0
yx, λ∂
2yx, λ
∂λ2 dx
mλ
−mλ
l
0
yx, λ∂yx, λ
∂λ dx−m
λl
0
y2x, λdx
)
mλ
−3mλ
(
mλ
l
0
yx, λ∂yx, λ
∂λ dx−m
λl
0
y2x, λdx
))
,
6.7
whencesupposing in that equalityλλn, we obtain
Gλn 2a2c2m−4n
''
y∗n1''2Π
12
yn,yn∗2
Π1
. 6.8
By6.1and6.3, we have
yn,y∗n2
Π1 ''y
∗
n1''
2
Π1; 6.9
then taking into account6.8, we get
δn''yn∗1''2Π1 1
6
a2c2−1m4nGλn/0. 6.10
Lemma 6.3is proved.
Lemma 6.4. The elementsυn{υnx, sn}of the system{υn}∞n1conjugated to the system{yn}∞n1 are defined by the equality
υnδ
−1
n υ∗n, 6.11
wherecn−cn−δn−1yn∗1
2
Π1ifλn 2;
dn−dn−δn−1yn∗1,y∗n2Π1,
hn−hn−δn−1y∗n2 2
Π1
δn−2y∗n1,yn∗2 2
Π1dndnδ
−1
Proof. On the bases of6.1,6.3,2.1,2.6, and6.9, we have
yn,υ∗n
δn if λn 1;
yn,υ∗m
0 ifλn λm 1, n /m;
yn,υ∗m
0, mk, k1, yk,υ∗k1
yk1,υ∗k
''yk∗1''2Π
1 ckckδk,
yk,υ∗k
yk1,υ∗k1
δk ifλn 1, λk 2;
yn,υ∗m
0, ym,υ∗m
δm, mk, k1, k2,
yk,υ∗k1
yk1,υ∗k
yk∗1, y∗k2Π
1
dkdk
δk,
yk,υ∗k2
yk2,υ∗k
''y∗k2''2Π
1
dkdk
yk∗1,y∗k2Π
1
dkdkhkhk
δk ifλn 1, λk 3.
6.12
Using relation6.12and taking into account6.11, we get the validity of the equality
yn,υk
δnk, 6.13
whereδnk is the Kronecker delta. The proof ofLemma 6.4is complete.
Corollary 6.5. i If λn 1, thensn/0; ii if λn 2, thensn1/0, sn/0 at cn/cn0,
sn 0 at cn cn0, where cn0 m−1n m∗n1 − δ−1n y∗n12Π1; iii if λn 3, then sn2/0,
sn1/0 at dn/dn0, sn1 0 at dn dn0; sn/0 at hn/hn0, sn 0 at hn hn0, where
d0
n m−1n m∗n1−δ−1n yn∗1,yn∗2Π1,h
0
n m−1n m∗n2−dnδ−1n y∗n1,yn∗2Π1dn−m
−1
n m∗n1−
δn−1y∗n2 2
Π1δ
−2
n yn∗1,y∗n2
2
Π1.
Theorem 6.6. Letrbe an arbitrary fixed integer. Ifsr/0, then the system{ynx}∞n1, n /r forms a
basis inLp0, l,p∈1,∞, and even a Riezs basis inL20, l; ifsr 0, the system{ynx}∞n1,n /r is
neither complete nor minimal inLp0, l,p∈1,∞.
Proof. By Theorem 7 in40, Chapter 1, Section 4andTheorem 6.1, the system {υn}∞n1 is a
Riesz basis inH. Then, for any vectorf {f, ξ} ∈H, the following expansion holds:
ff, ξ
∞
n1
f,yn
Hυn
∞
n1
f, yn
L2σ
−1ξm
n
{υn, sn}, 6.14
whence it follows the equalities
f
∞
n1
f, yn
L2σ
−1ξm
n
υn,
ξ
∞
n1
f, yn
L2σ
−1ξm
n
sn.
Ifξ0, then by6.15, we have
f
∞
n1
f, yn
L2υn, 6.16
0
∞
n1
f, yn
L2sn. 6.17
Letsr/0. Then by6.17, we obtain
f, yr
L2−s
−1 r
∞
n1
n /r
f, yn
L2sn, 6.18
considering which in6.16, we get
f
∞
n1
n /r
f, yn
L2
υn−s−1r snυr
. 6.19
By6.13and2.1, we have
yn, υk−s−1r skυr
L2
yn, υk
L2−s
−1
r sk
yn, υr
L2
yn,υk
H− |σ|−1mnsk−s
−1
r sk
yn, υr
L2
s−1r sk|σ|−1mnsr δnk, n, k /r,
6.20
that is, the system{υnx−sr−1snυrx}∞n1,n /r is conjugated to the system{ynx}∞n1,n /r. By
virtue of6.19, the system{υnx−s−1r snυrx}∞n1,n /ris a Riesz basis inL20, l. Then, on the
base of Corollary 240, Chapter 1, Section 4{ynx}∞n1,n /r is also in a Riesz basis inL20, l. The basicity of the system{ynx}∞n1,n /r in the spaceLp0, l,p∈1,∞\ {2}, can be proved by scheme of the proof of Theorem 5.1 in18usingTheorem 5.2.
Now letsr 0. Then, by2.1and6.13we have
yn, υr
L2
yn,υr
H− |σ|−1mnsr 0, n∈N, n /r. 6.21
So, the functionυτxis orthogonal to all functions of the system{ynx}∞n1, n /r, that is, the system{ynx}∞n1, n /ris incomplete inL20, l.
On the basis ofCorollary 6.5there existsk∈Nsuch thatsk/0e.g., ifλk 1. Then,
for anyfx∈L20, lthe following expansion holds:
f
∞
n1
n /k
f, υn−s−1k snυk
L2
