SpringerOpen Newsletter

Receive periodic news and updates relating to SpringerOpen.

Open Access Research

The asymptpotics of eigenvalues and trace formula of operator associated with one singular problem

Nigar M Aslanova

Author Affiliations

Institute of Mathematics and Mechanics of NAS of Azerbaijan, Baku, Azerbaijan

Mathematics Department, Khazar University, Baku, Azerbaijan

Boundary Value Problems 2012, 2012:8  doi:10.1186/1687-2770-2012-8


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2012/1/8


Received:14 September 2011
Accepted:23 January 2012
Published:23 January 2012

© 2012 Aslanova; licensee Springer.

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In the article, spectrum of operator generated by differential operator expression given on semi axis is investigated and proved formula for regularized trace of this operator.

Introduction

Let H be a separable Hilbert space with a scalar product (.,.) and norm ∥.∥. Consider in L2((0, ∞), H) the problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M1">View MathML</a>

(1)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M2">View MathML</a>

(2)

where A is a self-adjoint positive-definite operator in H which has a compact inverse operator and A > E (E is an identity operator in H). Denote the eigenvalues and eigenvectors of the operator A by γ1 γ2 ≤ ..., and φ1, φ2, ..., respectively.

Suppose that operator-valued function q(x) is weakly measurable, ∥q(x)∥ is bounded on [0, ∞), q*(x) = q(x)∀x ∈ [o, ∞). The following properties hold:

(1) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M3','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M3">View MathML</a>, ∀x ∈ [0, ∞).

(2) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M4','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M4">View MathML</a> is summable on (0, ∞), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M5','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M5">View MathML</a> for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M6">View MathML</a>

(3) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M7">View MathML</a>.

In the case q(x) ≡ 0 in L2(H, (0, ∞)) associate with problems (1), (2) a self-adjoint operator L0 whose domain is

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M8">View MathML</a>

In the case q(x) ≠ 0 denote the corresponding operator by L, so L = L0 + q.

In this article the asymptotics of eigenvalues and the trace formula of operator L will be studied.

In [1] the regularized traces of all orders of the operator generated by the expression

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M9">View MathML</a>

and the boundary conditions

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M10','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M10">View MathML</a>

are obtained.

In [2] the sum of eigenvalue differences of two singular Sturm-Liouville operators is studied.

The asymptotics of eigenvalues and trace formulas for operators generated by differential expressions with operator coefficients are studied, for example, in [3-7]. We could also refer to papers [8-10] where trace formulas for abstract operators are obtained. Trace formulas are used for evaluation of first eigenvalues, they have application to inverse problems, index theory of operators and so forth. For further detailed discussions of the subject refer to [11].

1 The asymptotic formula for eigenvalues of L0 and L

One could easily show that under conditions A > E, A-1 σ, the spectrum of L0 is discrete.

Suppose that γk ~ akα (k → ∞, a > 0, α > 0). Denote yk(x) = (y(x), φk). Then by virtue of the spectral expansion of the self-adjoint operator A we get the following boundary-value problem for the coefficients yk(x):

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M11','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M11">View MathML</a>

(1.1)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M12','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M12">View MathML</a>

(1.2)

In the case x + γk > λ solution of problem (1.1) from L2(0, ∞) is

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M13','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M13">View MathML</a>

(1.3)

and in the case x + γk< λ we can write it as a function of real argument as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M14">View MathML</a>

(1.4)

For this solution to satisfy (1.2) it is necessary and sufficient to hold

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M15','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M15">View MathML</a>

(1.5)

at least for one γk(λ γk). Therefore, the spectrum of the operator L0 consists of those real values of λ γk such that at least for one k

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M16','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M16">View MathML</a>

(1.6)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M17">View MathML</a>.

Prove the following two lemmas which we will need further.

Lemma 1.1. Equation (1.6) has only real roots.

Proof. Suppose that z = , α R, α ≠ 0. Then the operator associated with problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M18','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M18">View MathML</a>

(1.7)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M19','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M19">View MathML</a>

(1.8)

is positive and its eigenvalues are squares of the roots of Equation (1.6). So,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M20','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M20">View MathML</a>

But

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M21','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M21">View MathML</a>

which is contradiction. Then z can be only real, otherwise, the selfadjoint operator corresponding to (1.7), (1.8) will have nonreal eigenvalues, which is impossible. The lemma is proved.

Now, find the asymptotics of the solutions of Equation (1.6). By virtue of the asymptotics for large |z| [[12], p. 975]

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M22','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M22">View MathML</a>

we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M23','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M23">View MathML</a>

(1.9)

Hence

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M24','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M24">View MathML</a>

(1.10)

where m is a large integer. Therefore, the statement of the following lemma is true.

Lemma 1.2. For the eigenvalues of L0 the following asymptotic is true

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M25','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M25">View MathML</a>

(1.11)

For large |z| consider the rectangular contour l with vertices at the points

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M26">View MathML</a>

which bypasses the origin along the small semicircle on the right side of the imaginary axis.

The following lemma is true.

Lemma 1.3. For a sufficiently large integer N the number of the roots of the equation inside l is N + O(1).

Proof. For large |z| we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M27','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M27">View MathML</a>

(1.12)

Denote the function in braces on the right hand side of (1.12) by F(z). Then for large |z| by Rouches' theorem the number of the zeros of F(z) inside the contour equals the number of the zeros <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M28','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M28">View MathML</a>. Therefore, the number of the zeros of function

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M29','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M29">View MathML</a>

inside l is N + O(1).

Now, by using the above results, derive the asymptotic formula for the eigenvalue distribution of L0.

Denote the distribution function of L0 by N(λ). Then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M30','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M30">View MathML</a>

So, N(λ) is a number of positive integer pairs (m,k) for which

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M31','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M31">View MathML</a>

By Lemma 1.2 for the great values of m

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M32','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M32">View MathML</a>

From the asymptotics of γk we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M33','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M33">View MathML</a>

Hence, by virtue of Lemmas 1.1 and 1.3

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M34','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M34">View MathML</a>

(1.13)

where N"(λ) is the number of the positive integer pairs for which

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M35','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M35">View MathML</a>

(1.14)

N'(λ) is the number of the positive integer pairs (m, k) satisfying the inequality

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M36','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M36">View MathML</a>

(1.15)

Thus by using (1.14), (1.15) in (1.13) as in [[13], Lemma 2] we come to the following statement.

Lemma 1.4. If γk ~ akα, (0 < a, α > 0) then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M37','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M37">View MathML</a>

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M38','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M38">View MathML</a>

(1.16)

2 Trace formula

The following lemma is true.

Lemma 2.1. Let the conditions of Lemma 1.4 hold. Then for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M39','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M39">View MathML</a>there exists such a subsequence {nm} of natural numbers that the relation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M40','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M40">View MathML</a>

holds.

Proof. In virtue of Lemma 1.4 for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M41','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M41">View MathML</a>, from which it follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M42','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M42">View MathML</a>

That is why one could choose a subsequence n1 < n2 < ....nm < ..., that for each k nm holds <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M43','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M43">View MathML</a>, or <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M44','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M44">View MathML</a>. The lemma is proved.

We will call <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M45','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M45">View MathML</a> a regularized trace of the operator L. It will be shown later it is independent of the choice of {nm} satisfying the hypothesis of Lemma 2.1.

From (1.16) it is obvious that for α > 2 resolvents R(L0) and R(L) are trace class operators. By using Lemma 2.1 for α > 2 one can prove the following lemma.

Lemma 2.2. Let q(x)∥ < const on the interval [0, ∞) and also the conditions of Lemma 1.6 hold. Then for α > 2

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M46','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M46">View MathML</a>

(2.1)

where {ψn} are orthonormal eigenvectors of the operator L0.

The proof of this lemma is analogous to the proof of Lemma 2 and Theorem 2 from [8]. For this reason we will not derive it here.

The orthogonal eigen-vectors of the operator L0 in L2((0, ∞), H) are

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M47','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M47">View MathML</a>

(2.2)

Calculate their norm. We have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M48','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M48">View MathML</a>

(2.3)

Take in Equation (1.7) z2 = α2 and z2 = β2. The solutions corresponding to these values denote by ψ(x, α2) and ψ(x, β2). Multiplying the first of the obtained equations by ψ (x, β2), the second by ψ (x, α2), subtracting the second one from the first one and integrating from zero to infinity we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M49','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M49">View MathML</a>

Going to limit as α β, we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M50','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M50">View MathML</a>

By making use of identities (12, p.981)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M51','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M51">View MathML</a>

(2.4)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M52','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M52">View MathML</a>

(2.5)

we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M53','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M53">View MathML</a>

(2.6)

Finally by equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M54','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M54">View MathML</a>

we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M55','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M55">View MathML</a>

(2.7)

So, the orthonormal eigenvectors of L0 are

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M56','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M56">View MathML</a>

(2.8)

Lemma 2.3. If the operator-valued function q(x) has property 1 and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M57','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M57">View MathML</a>, then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M58','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M58">View MathML</a>

(2.9)

Proof. Take (q(x)φk, φk) = qk(x). Let ε > 0 be sufficiently small number. If <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M59','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M59">View MathML</a> then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M60','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M60">View MathML</a>. For <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M61','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M61">View MathML</a> we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M62','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M62">View MathML</a> and, finally, for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M63','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M63">View MathML</a> it will be <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M64','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M64">View MathML</a>.

Consequently for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M65','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M65">View MathML</a> we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M66','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M66">View MathML</a>

and for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M67">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M68','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M68">View MathML</a>

then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M69','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M69">View MathML</a>

(2.10)

For ε → 0 we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M70','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M70">View MathML</a>

(2.11)

From asymptotic <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M71','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M71">View MathML</a> by using (2.10), (2.11) and property 1 we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M72','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M72">View MathML</a>

The lemma is proved.

By using Lemma 2.3 prove the following theorem.

Theorem 2.1. Let the conditions of Lemma 1.6 hold. If the operator-valued function q (x) has properties 1-3, then it holds the formula

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M73','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M73">View MathML</a>

Proof. In virtue of Lemma 2.1

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M74','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M74">View MathML</a>

(2.12)

Denote

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M75','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M75">View MathML</a>

Show that for each fixed value of k the m-th term of the sum TN (x) is a residue at the point αm of some function of complex variable which has poles at points <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M76','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M76">View MathML</a>.

For this purpose consider the following function

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M77','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M77">View MathML</a>

(2.13)

By taking in place of zero x in (2.6) one can show that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M78','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M78">View MathML</a>

(2.14)

Note that all zeros of the function <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M79','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M79">View MathML</a> are simple, otherwise

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M80','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M80">View MathML</a>

and by virtue of (2.7) the norm of the eigenvectors equals zero, which is contradiction.

Denote z2 - x = f (x, z) and the right hand side of (2.14) by G(f(x, z). Then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M81','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M81">View MathML</a>

(2.15)

Then from (2.14), (2.15)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M82','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M82">View MathML</a>

(2.16)

The function g(z) has poles of second order at the points αm. By using identities (2.15), (2.16) show that residues at this points equal the terms of sum TN(x). Denoting <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M83','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M83">View MathML</a>, write Taylor expansion of this function in the vicinity αm:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M84','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M84">View MathML</a>

Show that the coefficient of the expansion of function zu2 (z) at (z - αm)3 equals zero. So,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M85','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M85">View MathML</a>

(2.17)

By denoting <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M86','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M86">View MathML</a> we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M87','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M87">View MathML</a>

(2.18)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M88','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M88">View MathML</a>

(2.19)

Therefore,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M89','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M89">View MathML</a>

(2.20)

On the other hand, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M90','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M90">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M91','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M91">View MathML</a> satisfy the Bessel equation <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M92','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M92">View MathML</a> for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M93','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M93">View MathML</a>. So their difference also satisfies this equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M94','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M94">View MathML</a>

(2.21)

If <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M95','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M95">View MathML</a>, then the right hand side (2.21) vanishes. Hence,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M96','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M96">View MathML</a>

(2.22)

which shows that the coefficient at (z - αm)3 in (2.17) vanishes.

Consequently, by (2.16), (2.17), (2.22) and the relation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M97','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M97">View MathML</a>

we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M98','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M98">View MathML</a>

Take as a contour of integration a rectangular contour C with vertices at the points ±AN, ±AN + +iB, which bypasses points αm above real axis, -αm below it.

Consider the right hand side of the contour with vertices at AN and AN + iB. By using the asymptotics

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M99','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M99">View MathML</a>

For x > 0, N → ∞ taking B = AN, z = u + iv we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M100','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M100">View MathML</a>

(2.23)

From condition 2

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M101','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M101">View MathML</a>

(2.24)

By conditions 2-3 as N → ∞

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M102','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M102">View MathML</a>

(2.25)

On the side of the contour with the vertices at ±AN + iB

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M103','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M103">View MathML</a>

(2.26)

In the same way as it is done in (2.25), (2.26) we get that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M104','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M104">View MathML</a>

Similarly, one may show that the integral along the left hand side of the contour converges to zero:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M105','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M105">View MathML</a>

So, by the Cauchy theorem we finally get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M106','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/8/mathml/M106">View MathML</a>

which completes the proof of the theorem.

Competing interests

The author declares that they have no competing interests.

Acknowledgements

This study was supported by the Science Development Foundation under the President of the Republic of Azerbaijan-Grant No EIF-2011-1(3)-82/18-1.

References

  1. Pechentsov, AS: Traces of one class singular differential operators: method of Lidskii-Sadovnichii. Vestnik Moscow Univ Ser I Math Mech. 5, 35–42 (1999)

  2. Gasymov, MG, Levitan, BM: About sum of differences of two singular Sturm-Liouville operators. Dokl AN SSSR. 151(5), 1014–1017 (1953)

  3. Rybak, MA: About asymptotic of eigenvalues of some boundary value problems for operator Sturm-Liouville equation. Ukr Math J. 32(2), 248–252 (1980)

  4. Qorbachuk, VI, Rybak, MA: About self-adjoint extensions of minimal operator generated by Sturm-Liouville expression with operator potentials and nonhomogeneous bondary conditions. Dokl AN URSR Ser A. 4, 300–304 (1975)

  5. Aliyev, BA: Asymptotic behavior of eigenvalues of one boundary value problem for elliptic differential operator equation of second order. Ukr Math J. 5(8), 1146–1152 (2006)

  6. Bayramoglu, M, Aslanova, NM: Distribution of eigenvalues and trace formula of operator Sturm-Liouville equation. Ukr Math J. 62(7), 867–877 (2010)

  7. Aslanova, NM: Study of the asymptotic eigenvalue distribution and trace formula of second order operator-differential equation. J Bound Value Probl. 7, 13 (2011)

  8. Maksudov, FG, Bayramoglu, M, Adygozalov, AA: On regularized trace of operator Sturm-Liouville on finite segment with unbounded operator coefficient. Dokl AN SSSR. 277(4), 795–799 (1984)

  9. Bayramoglu, M, Sahinturk, H: Higher order regularized trace formula for the regular Sturm-Liouville equation contained spectral parameter in the boundary condition. Appl Math Comput. 186(2), 1591–1599 (2007). Publisher Full Text OpenURL

  10. Aslanova, NM: Trace formula of one boundary value problem for Sturm-Liouville operator equation. Sib J Math. 49(6), 1207–1215 (2008)

  11. Sadovnichii, VA, Podolskii, VE: Traces of operators. Uspekh Math Nauk. 61:5(371), 89–156 (2006)

  12. Gradstein, IS, Ryzhik, IM: Table of Integrals, Sums, Series and Products, p. 1108. Nauka, Moscow (1971)

  13. Qorbachuk, VI, Qorbachuk, ML: On some classes of boundaryvalue problems for Sturm-Liouville equation with operator-valued potential. Ukr Math J. 24(3), 291–305 (1972)