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

Copyright q 2012 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 problem y 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.


Introduction
The following boundary value problem is considered: Obviously, the operator L is well defined.By immediate verification we conclude that problem 1.1 , 1.2a -1.2d is equivalent to the following spectral problem: that is, the eigenvalue λ n 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: y n y n x , m n ←→ y n x , m n ay n l − cT y n l .

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.
If σ > 0, then L is a selfadjoint discrete lower-semibounded operator in H and hence has a system of eigenvectors {{y n x , m n }} ∞ n 1 , that forms an orthogonal basis in H.In the case σ < 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 Π 1 L 2 0, l ⊕ C by means of the inner products J-metric 26 : Proof.JL is selfadjoint in H by virtue of Theorem 2.2 11 .Then, J-selfadjointness of L on Π 1 follows from 27, Section 3, Proposition 3 0 .
Lemma 2.2 see 27, Section 3, Proposition 5 0 .Let L * be an operator adjoined to the operator L in H.Then, L * JLJ.
Let λ be an eigenvalue of operator L of algebraic multiplicity ν.Let us suppose that ρ λ is equal to ν if Im λ / 0 and equal to whole part ν/2 if Im λ 0.

International Journal of Mathematics and Mathematical Sciences
Theorem 2.3 see 28 .The eigenvalues of operator L arrange symmetrically with regard to the real axis.n k 1 ρ λ k ≤ 1 for any system {λ k } n k 1 n ≤ ∞ of eigenvalues with nonnegative parts.
From Theorem 2.3 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.

Some Auxiliary Results
As in 17,19,29,30 for the analysis of the oscillation properties of eigenfunctions of the problem 1.1 , 1.2a -1.2d we will use a Pr üfer-type transformation of the following form: 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 .

3.1
Consider the boundary conditions see 29, 30 y 0 cos α − y 0 sin α 0, 1.2a * y l cos δ − Ty l sin δ 0, 1.2d * 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 Pr üfer 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 Pr üfer 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.
Let u 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 : transform 0, l into the interval 0, l and 1.1 into where p lω −1 h 3 , r l −1 ωh −1 ; h x , y x are taken as functions of t and • : d/dt.Furthermore, the following relations are useful in the sequel: 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.  (3.5c).If a is a zero of u and ü in the interval 0, l , then u t Tu t < 0 in a neighborhood of a.If a is a zero of u or Tu in 0, l , then u t ü t < 0 in a neighborhood of a. Theorem 3.5.Let u be a nontrivial solution of the problem 1.1 , 1.2a and 1.2c for λ > 0. Then the Jacobian J u r 3 cos ψ sin ψ of the transformation 3.1 does not vanish in 0, l .
Proof.Let u 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 x 0 nπ for some integer n and for some x 0 ∈ 0, l .Then, the transformation 3.1 implies that u x 0 Tu x 0 0. Using the transformation 3.2 , the solution u of 3.3 also satisfies u t 0 Tu t 0 0, where t 0 l −1 ω x 0 0 h s ds ∈ 0, l .However, it is incompatible with the conclusion of Lemma 3.4.The proof of the inequality cos ψ x / 0, x ∈ 0, l , proceeds in the same fashion as in the previous case.The proof of Theorem 3.5 is complete.
Let y 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 : 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 Obviously, the eigenvalues λ n δ , n ∈ N, of the problem 1.1 , 1.2a -1.2c , and 1.2d * are zeros of the entire function y l, λ cos δ − Ty l, λ cos δ 0. Note that the function F λ Ty l, λ /y l, λ is defined for Lemma 3.9 see 19, Lemma 5 .Let λ ∈ A. Then, the following relation holds: In 1.1 we set λ ρ 4 .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 z k x, ρ , k 1, 4, regular in ρ for sufficiently large ρ and satisfying the relations where ω k , k 1, 4, are the distinct fourth roots of unity, 1 1 O 1/ρ .For brevity, we introduce the notation s δ 1 , δ 2 sgn δ 1 sgn δ 2 .Using relation 3.10 and taking into account boundary conditions 1.2a -1.2c , we obtain International Journal of Mathematics and Mathematical Sciences 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
Furthermore, we have

3.15
By virtue of 18, Theorem 3.1 , one has the asymptotic formulas where relation 3.17 holds uniformly for x ∈ 0, l .
International Journal of Mathematics and Mathematical Sciences 9 By 3.14 , we have lim
Let s λ be the number of zeros of the function y x, λ in the interval 0, l .
The proof is similar to that of 19, Lemma 10 using Theorems 3.6 and 3.7 and Remark 3.11.
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: has an infinite set of nonpositive eigenvalues ρ n tending to −∞ and satisfying the asymptote
Remark 3.15.By Remark 3.10 the number of zeros of the eigenfunction y δ 1 x corresponding to an eigenvalue λ 1 δ < 0 can by arbitrary.In views of 31, Corollary 2.5 , as λ 1 δ < 0 varies, International Journal of Mathematics and Mathematical Sciences new zeros of the corresponding eigenfunction y δ 1 x enter the interval 0, l only through the end point x 0 since y δ 1 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 λ 1 δ .In the case β 0 see 31, Theorem 5.3 .
Let ac / 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 A n , n 1, 2, . .., the equation F λ a/c resp., F λ −c/a has a unique solution μ n resp., ν n ; moreover, if a/c > 0 and if a/c < 0. Besides, μ 1 0 if a/c < 0 and F 0 a/c; ν 1 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 4 B n and is a finite-order meromorphic function and the eigenvalues ν n and μ n , 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.
Let λ ∈ B. Using formula 3.9 , we get

3.29
Lemma 3.17.The expansion holds, where c n , 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 μ n allows the representation where G 1 λ is an entire function, and integers s n , 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 .

3.35
Following the corresponding reasoning see 36, Chapter VII, Section 2, formula 27 , we see that outside of domains Ω n ε the estimation

3.36
International Journal of Mathematics and Mathematical Sciences 13 holds; using it in 3.32 we get Let {Γ n } ∞ n 1 be a sequence of the expanding circles which are not crossing domains Ω n ε .Then, according to Formula 9 in 37, Chapter V, Section 13 , we have

3.38
By 3.38 , we get

3.39
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 Differentiating the right side of the least equality, we have

International Journal of Mathematics and Mathematical Sciences
Now let μ 1 0, that is, F 0 a/c.G λ has the following expansion: λc n μ n λ − μ n .

3.44
Again, according to Formula 9 in 37, Chapter V, Section 13 , we have

3.48
Using 3.48 in 3.45 , we have

3.49
In view of 3.49 and 3.45 , we get

3.50
Passing to the limit as n → ∞ in 3.50 , we obtain Lemma 3.17 is proved.

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
For c / 0, we find a positive integer N from the inequality μ N−1 < −d/c ≤ μ N .
Theorem 4.1.The problem 1.1 , 1.2a -1.2d for σ > 0 has a sequence of real and simple eigenvalues including at most 1 sgn |c| number of negative ones.The corresponding eigenfunctions have the following oscillation properties.
a If c 0, then the eigenfunction y n x , n ≥ 2, has exactly n − 1 zeros in 0, l , the eigenfunction y 1 x has no zeros in 0, l in the case λ 1 ≥ 0, and the number of zeros of y 1 x can be arbitrary in the case λ 1 < 0.
b If c / 0, then the eigenfunction y n x corresponding to the eigenvalue λ n ≥ 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 that σ < 0. Let λ, 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 Π 1 , since the operator L is J-selfadjoint in Π 1 .Hence, by 2.4 , we have

4.7
From the second relation it follows that Im y x, μ /m μ 0, which by 1.1 contradicts the condition μ ∈ C \ R. The obtained contradictions prove Lemma 4.2.

4.8
Hence, we get 4.9 From 4.9 , it follows that y 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 λ n , 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 that B 1 contains algebraically two (either two simple or one double) eigenvalues, and B n , n 2, 3, . .., contain precisely one simple eigenvalues.
2 All eigenvalues are real, at that B 1 contains no eigenvalues but, for some s ≥ 2, B s contains algebraically three (either three simple, or one simple and one double, or one triple) eigenvalues, and B n , n 2, 3, . .., n / s contain precisely one simple eigenvalue.
3 There are two nonreal eigenvalues appearing as a conjugate pair, at that B 1 contains no eigenvalues, and B n , 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 − a 2 c 2 /σ, B − ab cd \ σ see 3.28 .From 3.41 , it follows that G λ > 0 for λ ∈ B 1 ; therefore, the function G λ is convex on the interval B 1 .By virtue of 3.18 and 3.30 , we have

4.10
That is why for each fixed number A there exists number B A such that the lines Aλ B A , λ ∈ R, touch the graph of function G λ at some point λ ∈ B 1 .Hence, in the interval B 1 , 3.28 has two simple roots are fulfilled; where R n τ n δ 0 , eigenvalues, and hence, by 4.23 , this problem in the circle S R n ⊂ C has one pair of simple nonreal eigenvalues.In this case, the location of the eigenvalues will be in the following form: 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 B n , n / k, n 2, 3, . .., problem 1.1 , 1.2a -1.2d has one simple eigenvalue.

Subcase 1. Let G λ k
A,G λ k / 0, that is, the eigenvalue λ k is a double one by this λ k λ k 1 .Then, from 4.23 it follows that the interval B s besides the eigenvalue λ k contains one more simple eigenvalue: at that it is either λ k−1 by this k s or λ k 2 by this k s − 1 .Hence, 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 each n < N (resp., n > N), y n has n − 1 (resp., n) zeros in the interval 0, l .Similarly y s , y s 1 both have s − 1 (resp., s) zeros if s < N (resp., s > N).Finally, if c / 0, then each of y N (and y s , y s 1 if s N) has N − 1 or N zeros according to λ N , λ S , λ S 1 ≤ or > −d/c, and if c 0 and s N, then y s , y s 1 both have s zeros.
The proof of this theorem is similar to that of 11, Theorem 4.4 using Lemma 3.12.

5.2
Let a 0. In this case μ n λ n π/2 , ν n λ n 0 , n ∈ N. Since n > N 2, so G λ n > 0.Then, by the equality G λ n Aλ n B, n ∈ N, we obtain Now let c 0. In this case μ n λ n 0 , ν n λ n π/2 , n ∈ N. Since n > K 2, so G λ n > 0. Therefore, using G λ n Aλ n B, we have Relations 5.1 are consequences of relations 5.2 -5.4 .The proof of Lemma 5.1 is complete.
We define numbers χ, χ n , n ∈ N, as follows:

5.6
where relation 5.6 holds uniformly for x ∈ 0, l .Theorem 6.6.Let r be an arbitrary fixed integer.If s r / 0, then the system {y n x } ∞ n 1, n / r forms a basis in L p 0, l , p ∈ 1, ∞ , and even a Riezs basis in L 2 0, l ; if s r 0, the system {y n x } ∞ n 1,n / r is neither complete nor minimal in L p 0, l , p ∈ 1, ∞ .Proof.By Theorem 7 in 40, Chapter 1, Section 4 and Theorem 6.1, the system { υ n } ∞ n 1 is a Riesz basis in H.Then, for any vector f {f, ξ} ∈ H, the following expansion holds:

Necessary and Sufficient Conditions of Basicity of Root Function System of Problem
f, y n L 2 σ −1 ξm n {υ n , s n }, 6.14 whence it follows the equalities f, y n L 2 σ −1 ξm n s n . 6.15

y 4 x
− q x y x λy x , x ∈ 0, l , :where λ is a spectral parameter, Ty ≡ y − qy , q is absolutely continuous function on 0, l ,β, γ, a, b, c, and dare real constants such that 0 ≤ β, γ ≤ π/2 and σ bc − ad / 0.Moreover, International Journal of Mathematics and Mathematical Sciences 3 We define in the H operator L y L y, m Ty x , dTy l − by l m ∈ H/y x ∈ W 4 2 0, l , Ty x ∈ L 2 0, l , y ∈ B.C. , m ay l −cT y l , 2.3 that is dense in H 23, 25 , where B.C. denotes the set of separated boundary conditions 1.2a -1.2c .
and no roots if B < B A .By 3.29 and 3.30 we have lim λ → μ n 0 G λ −∞, lim λ → μ n −0 G λ ∞, n ∈ N. Therefore, 3.28 has at least one solution in the interval B n , n 2, 3, . ... Let B ≥ B A .If B > B A , then G λ 1 < A, G λ 2 < A; if B B A then G λ 1 A. By 3.29 , 3.28 has only simple root λ n 1 for B > B A , λ n for B B A in the interval B n , n 2, 3, . ... Let B < B A .By Lemma 4.3 either G λ n > A for any λ n ∈ R, or there exists k ∈ N such that F λ k ≤ A and F λ n > A, n ∈ N \ {k}.Assume that λ k ∈ B s .Obviously, s ≥ 2. Choose natural number n 0 such that the inequalities

which are basic in the sequel. Lemma 3.2. All the eigenvalues of problem
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 .Let λ > 0, and let y be a nontrivial solution of 3.3 .If y, ẏ, ÿ, and Ty are nonnegative at t a (but not all zero), they are positive for all t > a.If y, − ẏ, ÿ, and − Ty are nonnegative at t a (but not all zero), they are positive for all t < a.We also need the following results

Lemma 3.3. Let
E be the space of solution of the problem 1.1 , 1.2a -1.2c .Then, dim E 1.
It is easy to see that the eigenvalues of problem 1.1 , 1.2a -1.2d are roots of the equation Remark 3.16.Note that if λ is the eigenvalue of problem 1.1 , 1.2a -1.2d , then m λ / 0 since σ / 0. n , n ∈ N, where μ 0 −∞.We observe that the function G λ cy l, λ aT y l, λ / ay l, λ − cT y l, λ is well defined for By 3.41, G λ k / 0. Hence, λ k is a triple eigenvalue of the problem 1.1 , 1.2a -1.2d by this λ k λ k 1 λ k 2 .Then, from 4.23 it follows that in the interval B s problem 1.1 , 1.2a -1.2d has unique triple eigenvalue λ k , and therefore,k s − 1.At this λ n ∈ B n 1 , n 1, 2, ..., s − 2, λ s−1 λ s λ s 1 ∈ B s , λ n ∈ B n−1 , n s 2, s 3, .... Let G λ k < A,that is, the eigenvalue λ k is simple.Then, by 4.23 , in the interval B s problem 1.1 , 1.2 has an eigenvalue λ k as well as two more simple eigenvalues, which, by Lemma 4.3, are λ k−1 and λ k 1 and hence k s .In this case, we haveλ n ∈ B n 1 , n 1, 2, ..., s − 2, λ s−1 , λ s , λ s 1 ∈ B s λ s−1 < λ s < λ s 1 , λ n ∈ B n−1 , n s 2, s 3, ....International Journal of Mathematics and Mathematical SciencesBy Theorem 4.4 we have λ n 2, that is, λ n λ n 1 if n s − 1 or n s; λ n 3, that is, λ n λ n 1 λ n 2 if n s − 1 If assertion 2 in Theorem 4.4 holds, then we set s 1 .Let {y n x } ∞ n 1 be a system of eigen-and associated functions corresponding to the eigenvalue system {λ n } ∞ is an eigenfunction for λ n and y n 1 x when ρ λ n 2; y n 1 x , y n 2 x when ρ λ n 3 are the associated functions see 34, Pages 16-20 for more details .
1.1 , 1.2a -1.2dNote that the element y n {y n x , m n }, n ∈ N, of the system { y n } ∞ The system of eigen-and associated functions of operator L is a Riesz basis in the space H.Proof.Let μ be a regular value of operator L, that is, R μ L − μI −1 exists and a bounded in H.Then, problem 2.4 is adequate to the following problem of eigenvalues: R μ is a completely continuous J-selfadjoint operator in Π 1 .Then, in view of 39 the system of the root vectors of operator R μ hence of operator L forms a Riesz basis in H. Theorem 6.1 is proved.
n equals either 0 at that y n is eigenvector or 1 at that λ n λ n−1 and y n is associated vector see, e.g.,38 .
Since G λ n / 0, from the last equality it follows that δ n / 0.Now let λ n3, that is, G λ n A and G λ n 0, G λ n / 0. Differentiating the right-hand side of 6.5 on λ, we have The elementsυ n {υ n x , s n } of the system { υ n } ∞ n 1 conjugated to the system { y n } ∞International Journal of Mathematics and Mathematical SciencesProof.On the bases of 6.1 , 6.3 , 2.1 , 2.6 , and 6.9 , we have Using relation 6.12 and taking into account 6.11 , we get the validity of the equalityy n , υ k δ n k , 6.13where δ n k is the Kronecker delta.The proof of Lemma 6.4 is complete.