Open Access

Inverse problem for a class of Sturm-Liouville operator with spectral parameter in boundary condition

Boundary Value Problems20132013:183

DOI: 10.1186/1687-2770-2013-183

Received: 5 April 2013

Accepted: 29 July 2013

Published: 14 August 2013

Abstract

This work aims to examine a Sturm-Liouville operator with a piece-wise continuous coefficient and a spectral parameter in boundary condition. The orthogonality of the eigenfunctions, realness and simplicity of the eigenvalues are investigated. The asymptotic formula of the eigenvalues is found, and the resolvent operator is constructed. It is shown that the eigenfunctions form a complete system and the expansion formula with respect to eigenfunctions is obtained. Also, the evolution of the Weyl solution and Weyl function is discussed. Uniqueness theorems for the solution of the inverse problem with Weyl function and spectral data are proved.

MSC:34L10, 34L40, 34A55.

Keywords

Sturm-Liouville operator expansion formula inverse problem Weyl function

1 Introduction

In recent years, there has been a growing interest in physical applications of boundary value problems with a spectral parameter, contained in the boundary conditions. The relationship between diffusion processes and Sturm-Liouville problem with eigen-parameter in the boundary conditions has been shown in [1]. Another example of this relationship between the same problem and the wave equation has been examined in [2, 3]. Sturm-Liouville problems with a discontinuous coefficient arise upon non-homogeneous material properties.

In a finite interval, inverse problems for the Sturm-Liouville operator with spectral parameter, contained in the boundary conditions, have been investigated, and the uniqueness of the solution of these problems has been shown in [49]. The inverse problem has been analyzed by zeros of the eigenfunctions in [6], by numerical methods in [9] and by two spectra, consisting of sequences of eigenvalues and the normed constants in [10]. In [11, 12], eigenvalue-dependent inverse problem with the discontinuities inside the interval was examined by the Weyl function. In a finite interval, discontinuous and no eigenvalue parameter containing direct problem and inverse problem with the Weyl function were discussed in [13, 14]. The similar problem was investigated in the half line by scattering data in [15, 16].

We consider the boundary value problem
y + q ( x ) y = λ 2 ρ ( x ) y , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ1_HTML.gif
(1)
U 1 ( y ) : = y ( 0 ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ2_HTML.gif
(2)
U 2 ( y ) : = λ 2 [ β 1 y ( π ) + β 2 y ( π ) ] + α 1 y ( π ) + α 2 y ( π ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ3_HTML.gif
(3)
where q ( x ) L 2 ( 0 , π ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq1_HTML.gif is a real valued function, λ is a complex parameter, α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq2_HTML.gif, α 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq3_HTML.gif, β 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq4_HTML.gif, β 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq5_HTML.gif are real numbers and
ρ ( x ) = { 1 , 0 x < a , α 2 , a < x π , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equa_HTML.gif

where 0 < α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq6_HTML.gif.

2 Special solutions

Let φ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq7_HTML.gif and ψ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq8_HTML.gif be the solutions of equation (1) satisfying the initial conditions
φ ( 0 , λ ) = 0 , φ ( 0 , λ ) = 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ4_HTML.gif
(4)
ψ ( π , λ ) = λ 2 β 1 α 2 , ψ ( π , λ ) = λ 2 β 2 + α 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ5_HTML.gif
(5)
For the solution of equation (1), the following integral representation is obtained in [13] for all λ:
e ( x , λ ) = 1 2 ( 1 + 1 ρ ( x ) ) e i λ μ + ( x ) + 1 2 ( 1 1 ρ ( x ) ) e i λ μ ( x ) + μ + ( x ) μ + ( x ) K ( x , t ) e i λ t d t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equb_HTML.gif

where K ( x , ) L 1 ( μ + ( x ) , μ + ( x ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq9_HTML.gif.

The kernel K ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq10_HTML.gif has the partial derivative K x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq11_HTML.gif belonging to the space L 1 ( μ + ( x ) , μ + ( x ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq12_HTML.gif for every x [ 0 , π ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq13_HTML.gif, and the properties below hold:
d d x K ( x , μ + ( x ) ) = 1 4 ρ ( x ) ( 1 + 1 ρ ( x ) ) q ( x ) , d d x { K ( x , μ ( x ) + 0 ) K ( x , μ ( x ) 0 ) } = 1 4 ρ ( x ) ( 1 1 ρ ( x ) ) q ( x ) , K ( x , μ + ( x ) ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equc_HTML.gif
Moreover, if q ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq14_HTML.gif is differentiable, then the following are valid
ρ ( x ) K t t K x x + q ( x ) K = 0 , | t | < μ + ( x ) , μ + ( x ) μ + ( x ) | K ( x , t ) d t | C ( exp { 0 x | q ( t ) | d t } 1 ) , 0 < C = const. https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equd_HTML.gif
Using the representation of the solution e ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq15_HTML.gif and formula
φ ( x , λ ) = e ( x , λ ) e ( x , λ ) 2 i λ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Eque_HTML.gif
we obtain the integral representation of the solution φ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq7_HTML.gif
φ ( x , λ ) = φ 0 ( x , λ ) + 0 μ + ( x ) A ( x , t ) sin λ t λ d t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ6_HTML.gif
(6)
where A ( x , t ) = K ( x , t ) K ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq16_HTML.gif. The kernel A ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq17_HTML.gif can be represented with the coefficients ρ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq18_HTML.gif and q ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq14_HTML.gif
d d x A ( x , μ + ( x ) ) = 1 4 ρ ( x ) ( 1 + 1 ρ ( x ) ) q ( x ) , d d x { A ( x , μ ( x ) + 0 ) A ( x , μ ( x ) 0 ) } = 1 4 ρ ( x ) ( 1 1 ρ ( x ) ) q ( x ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equf_HTML.gif
With the help of equation (6), we have a representation for the function ψ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq8_HTML.gif
ψ ( x , λ ) = 1 2 ( 1 + α ρ ( x ) ) sin λ ( μ + ( π ) μ + ( x ) ) α λ 1 2 ( 1 α ρ ( x ) ) sin λ ( μ + ( π ) μ ( x ) ) α λ + 0 μ + ( π ) μ + ( x ) A ˜ ( x , t ) sin λ t λ d t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ7_HTML.gif
(7)

where A ˜ ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq19_HTML.gif is a real function.

Denote
Δ ( λ ) : = φ ( x , λ ) , ψ ( x , λ ) = φ ( x , λ ) ψ ( x , λ ) φ ( x , λ ) ψ ( x , λ ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ8_HTML.gif
(8)
which is independent of x [ 0 , π ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq20_HTML.gif. Substituting x = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq21_HTML.gif and x = π https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq22_HTML.gif into (8), we get
Δ ( λ ) = U 2 ( φ ) = U 1 ( ψ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equg_HTML.gif

Δ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq23_HTML.gif is an entire function of λ and is called the characteristic function of the boundary value problem (1)-(3).

3 Some spectral properties

Lemma 1 The square values of the roots ( λ n ) n = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq24_HTML.gif of the characteristic function coincide with the eigenvalues of the boundary value problem (1)-(3), and for every λ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq25_HTML.gif, there exists a sequence ( k n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq26_HTML.gif such that
ψ ( x , λ n ) = k n φ ( x , λ n ) ( k n 0 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ9_HTML.gif
(9)

where ψ ( x , λ n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq27_HTML.gif and φ ( x , λ n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq28_HTML.gif are the eigenfunctions of the boundary value problem (1)-(3), corresponding to the eigenvalue λ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq25_HTML.gif.

Proof The proof can be done in a similar way to [8]. Indeed, let us assume that λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq29_HTML.gif is an eigenvalue of the function Δ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq23_HTML.gif. Then
Δ ( λ ) = | φ ( x , λ 0 ) ψ ( x , λ 0 ) φ ( x , λ 0 ) ψ ( x , λ 0 ) | = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equh_HTML.gif
holds, i.e., the functions φ ( x , λ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq30_HTML.gif and ψ ( x , λ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq31_HTML.gif are linearly dependent ψ ( x , λ 0 ) = k n φ ( x , λ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq32_HTML.gif ( k n = const. https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq33_HTML.gif), and they satisfy the boundary conditions (2), (3). Hence λ 0 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq34_HTML.gif is an eigenvalue, ψ ( x , λ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq31_HTML.gif and φ ( x , λ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq30_HTML.gif are eigenfunctions, related to this eigenvalue. Conversely, let λ 0 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq34_HTML.gif be an eigenvalue of the operator A, and let y 0 ( x , λ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq35_HTML.gif, y 0 ( x , λ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq36_HTML.gif be the corresponding eigenfunctions. Then the boundary conditions (2), (3) hold both for the eigenfunctions y 0 ( x , λ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq35_HTML.gif and y 0 ( x , λ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq36_HTML.gif. Additionally, if the functions y 0 ( x , λ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq35_HTML.gif and y 0 ( x , λ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq36_HTML.gif satisfy the conditions y ( 0 , λ 0 ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq37_HTML.gif, y ( 0 , λ 0 ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq38_HTML.gif, then y 0 ( x , λ 0 ) φ ( x , λ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq39_HTML.gif, y 0 ( x , λ 0 ) φ ( x , λ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq40_HTML.gif. According to boundary conditions (2), (3), we have
Δ ( λ 0 ) = U 2 ( φ ( π , λ 0 ) ) = U 2 ( y 0 ( π , λ 0 ) ) = 0 , Δ ( λ 0 ) = U 2 ( φ ( π , λ 0 ) ) = U 2 ( y 0 ( π , λ 0 ) ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equi_HTML.gif
Similarly, if we assume that y 0 ( π , λ 0 ) = λ 0 2 β 2 + α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq41_HTML.gif, y 0 ( π , λ 0 ) = λ 0 2 β 2 + α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq42_HTML.gif, then y 0 ( x , λ 0 ) ψ ( x , λ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq43_HTML.gif, y 0 ( x , λ 0 ) ψ ( x , λ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq44_HTML.gif. Again from the boundary conditions (2), (3), it is obvious that
Δ ( λ 0 ) = U 1 ( ψ ( 0 , λ 0 ) ) = U 1 ( y 0 ( 0 , λ 0 ) ) = 0 , Δ ( λ 0 ) = U 1 ( ψ ( 0 , λ 0 ) ) = U 1 ( y 0 ( 0 , λ 0 ) ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equj_HTML.gif

Therefore, we have proved that for each eigenvalue λ 0 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq34_HTML.gif, there exists only one (up to a multiplicative constant) eigenfunction. □

In the Hilbert space H ρ = L 2 , ρ ( 0 , π ) C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq45_HTML.gif an inner product defined by
( F , G ) : = 0 π F 1 ( x ) G 1 ( x ) ¯ ρ ( x ) d x + F 2 G 2 ¯ χ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equk_HTML.gif
where
F = ( F 1 ( x ) F 2 ) H ρ , G = ( G 1 ( x ) G 2 ) H ρ , χ : = α 1 β 1 α 2 β 2 > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equl_HTML.gif
Let us define
A ( F ) : = ( F 1 ( x ) + q ( x ) F 1 ( x ) α 1 F 1 ( π ) + α 2 F 1 ( π ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equm_HTML.gif
with
D ( A ) = { F H ρ : F 1 ( x ) , F 1 ( x ) A C [ 0 , π ] , l F 1 L 2 , ρ ( 0 , π ) F 1 ( 0 ) = 0 , F 2 = β 1 F 1 ( π ) + β 2 F 1 ( π ) } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equn_HTML.gif
where
l ( F 1 ) = 1 ρ ( x ) { F 1 + q ( x ) F 1 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equo_HTML.gif

The boundary value problem (1)-(3) is equivalent to the equation A Y = λ 2 Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq46_HTML.gif.

The eigenfunctions of operator A are in the form of
Φ ( x , λ n ) = Φ n : = ( φ ( x , λ n ) β 1 φ ( π , λ n ) + β 2 φ ( π , λ n ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equp_HTML.gif

Lemma 2 The eigenfunctions Φ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq47_HTML.gif and Φ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq48_HTML.gif, corresponding to different eigenvalues λ 1 λ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq49_HTML.gif, are orthogonal.

Proof Since Φ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq47_HTML.gif and Φ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq48_HTML.gif are the solutions of the boundary value problem (1)-(3), the equations below are valid
Φ ( x , λ 1 ) + q ( x ) Φ ( x , λ 1 ) = λ 1 2 ρ ( x ) Φ ( x , λ 1 ) , Φ ( x , λ 2 ) + q ( x ) Φ ( x , λ 2 ) = λ 2 2 ρ ( x ) Φ ( x , λ 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equq_HTML.gif
Multiplying the first equation by Φ ( x , λ 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq50_HTML.gif and the second equation by Φ ( x , λ 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq51_HTML.gif and adding together, we have
d d x { Φ ( x , λ 1 ) Φ ( x , λ 2 ) , Φ ( x , λ 1 ) Φ ( x , λ 2 ) } = ( λ 1 2 λ 2 2 ) ρ ( x ) Φ ( x , λ 1 ) Φ ( x , λ 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equr_HTML.gif
Integrating it from 0 to π, and using the boundary condition (3), we obtain
( λ 1 2 λ 2 2 ) [ 0 π Φ ( x , λ 1 ) Φ ( x , λ 2 ) ρ ( x ) d x ] + ( λ 1 2 λ 2 2 ) [ 1 k ( β 1 φ ( x , λ 1 ) + β 2 φ ( x , λ 1 ) ) ( β 1 φ ( x , λ 2 ) + β 2 φ ( x , λ 2 ) ) ] = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equs_HTML.gif

Since λ 1 λ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq49_HTML.gif, the lemma is proved. □

Corollary 3 The eigenvalues of the boundary value problem (1)-(3) are real.

The values
γ n = 0 π φ 2 ( x , λ n ) d x + ( β 1 φ ( π , λ n ) + β 2 φ ( π , λ n ) ) 2 χ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ10_HTML.gif
(10)

are called the norming constants of the boundary value problem (1)-(3).

Now, let us agree to denote differentiation with respect to λ with a dot Δ ˙ = d d λ Δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq52_HTML.gif ( Δ ˙ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq53_HTML.gif).

Lemma 4 The following equality holds
Δ ˙ ( λ n ) = 2 λ n k n γ n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ11_HTML.gif
(11)
Proof Since
φ ( x , λ n ) + q ( x ) φ ( x , λ n ) = λ n 2 φ ( x , λ n ) , ψ ( x , λ ) + q ( x ) ψ ( x , λ ) = λ 2 ψ ( x , λ ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equt_HTML.gif
we get
d d x Δ ( λ ) = ( λ n 2 λ 2 ) φ ( x , λ n ) ψ ( x , λ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equu_HTML.gif
With the help of (2) and (3),
Δ ( λ ) = ( λ n λ ) ( λ n + λ ) k n [ 0 π φ 2 ( x , λ n ) ρ ( x ) d x + χ ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equv_HTML.gif

Taking into consideration (9) and (10), for λ λ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq54_HTML.gif, we arrive (11). □

Corollary 5 All zeros of Δ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq23_HTML.gif are simple, i.e., Δ ˙ ( λ n ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq55_HTML.gif.

4 Asymptotic formulas of eigenvalues

Let φ 0 ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq56_HTML.gif be the solution of equation (1) satisfying the initial conditions (4) when q ( x ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq57_HTML.gif
φ 0 ( x , λ ) = 1 2 ( 1 + 1 ρ ( x ) ) sin λ ( μ + ( x ) ) λ + 1 2 ( 1 1 ρ ( x ) ) sin λ ( μ ( x ) ) λ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ12_HTML.gif
(12)
The eigenvalues λ n 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq58_HTML.gif ( n = 0 , ± 1 , ± 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq59_HTML.gif) of the boundary value problem (1)-(3) when q ( x ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq57_HTML.gif can be found using the equation
Δ 0 ( λ ) = λ 2 [ β 1 φ 0 ( π , λ ) + β 2 φ 0 ( π , λ ) ] + α 1 φ 0 ( π , λ ) + α 2 φ 0 ( π , λ ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equw_HTML.gif
from [17] (see also [18]) and can be represented in the following way
λ n 0 = n π μ + ( π ) + h n , n = 0 , ± 1 , ± 2 , , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ13_HTML.gif
(13)

where sup n | h n | = h < + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq60_HTML.gif.

Lemma 6 Roots λ n 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq58_HTML.gif of the function Δ 0 ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq61_HTML.gif are separated, i.e.,
inf n k | λ n 0 λ k 0 | = τ > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equx_HTML.gif
Proof Assume the contrary. Then there are sequences ( λ k 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq62_HTML.gif, ( λ k 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq63_HTML.gif of zeros of functions λ 2 [ β 1 y ( π ) + β 2 y ( π ) ] + α 1 y ( π ) + α 1 y ( π ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq64_HTML.gif such that
λ k 0 λ k 0 , λ k 0 , λ k 0 , lim k ( λ k 0 λ k 0 ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equy_HTML.gif
Since the eigenfunctions φ 0 ( x , λ k 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq65_HTML.gif, φ 0 ( x , λ k 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq66_HTML.gif are orthogonal, we have
0 = λ k 0 λ k 0 0 π ρ ( x ) φ 0 ( x , λ k 0 ) φ 0 ( x , λ k 0 ) d x = 0 π ρ ( x ) λ k 0 φ 0 ( x , λ k 0 ) [ λ k 0 φ 0 ( x , λ k 0 ) λ k 0 φ 0 ( x , λ k 0 ) ] d x + 0 π ρ ( x ) ( λ k 0 φ 0 ( x , λ k 0 ) ) 2 d x = I k + 0 π ρ ( x ) ( λ k 0 φ 0 ( x , λ k 0 ) ) 2 d x I k + 0 a ρ ( x ) ( λ k 0 φ 0 ( x , λ k 0 ) ) 2 d x = I k + 0 a sin 2 λ k 0 x d x = I k + a 2 sin 2 λ k 0 a 4 λ k 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equz_HTML.gif
where
I k = 0 π ρ ( x ) λ k 0 φ 0 ( x , λ k 0 ) [ λ k 0 φ 0 ( x , λ k 0 ) λ k 0 φ 0 ( x , λ k 0 ) ] d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equaa_HTML.gif
Thus,
0 I k + a 2 sin 2 λ k 0 a 4 λ k 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ14_HTML.gif
(14)
Let us prove that I k 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq67_HTML.gif as k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq68_HTML.gif. In fact, (12) implies the following estimate
| λ k 0 φ ( x , λ k 0 ) λ k 0 φ ( x , λ k 0 ) | C | λ k 0 λ k 0 | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equab_HTML.gif

Consequently, lim k [ λ k 0 φ ( x , λ k 0 ) λ k 0 φ ( x , λ k 0 ) ] = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq69_HTML.gif holds uniformly on x [ 0 , π ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq70_HTML.gif. Now, passing to the limit in equality (14), as k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq71_HTML.gif, we have 0 a 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq72_HTML.gif. This is a contradiction, and it proves the validity of lemma’s statement. □

Lemma 7 The eigenvalues of the boundary value problem (1)-(3) are in the form of
λ n = λ n 0 + d n λ n 0 + η n n , λ n 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ15_HTML.gif
(15)
where d n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq73_HTML.gif is a bounded sequence
d n = 1 4 λ n 0 Δ ˙ 0 ( λ n 0 ) { 0 π 1 ρ ( t ) ( 1 1 ρ ( t ) ) q ( t ) cos ( λ n 0 μ ( π ) ) d t } + 1 4 λ n 0 Δ ˙ 0 ( λ n 0 ) { 0 π 1 ρ ( t ) ( 1 + 1 ρ ( t ) ) q ( t ) cos ( λ n 0 μ + ( π ) ) d t } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equac_HTML.gif

and { η n } l 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq74_HTML.gif.

Proof From (6), it follows that
Δ ( λ ) = Δ 0 ( λ ) + 0 μ + ( π ) A ( π , t ) sin λ t λ d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ16_HTML.gif
(16)
Lemma 1.3.1 in [19] and from [17], we get
Δ ( λ ) Δ 0 ( λ ) = O ( 1 | λ | ) e | Im λ | μ + ( π ) , | λ | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ17_HTML.gif
(17)
Therefore, for sufficiently large n, on the contours
Γ n = { λ : | λ | = | λ n 0 | + τ 2 } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equad_HTML.gif
we have
| Δ ( λ ) Δ 0 ( λ ) | < | Δ 0 ( λ ) | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equae_HTML.gif
By the Rouche theorem, we obtain that the number of zeros of the function { Δ ( λ ) Δ 0 ( λ ) } + Δ 0 ( λ ) = Δ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq75_HTML.gif inside the contour Γ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq76_HTML.gif coincides with the number of zeros of the function Δ 0 ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq61_HTML.gif. Furthermore, applying the Rouche theorem to the circle σ n ( δ ) = { λ : | λ λ n 0 | δ } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq77_HTML.gif, we get that, for sufficiently large n there exists only one zero λ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq25_HTML.gif of the function Δ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq23_HTML.gif in σ n ( δ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq78_HTML.gif. Owing to the arbitrariness of δ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq79_HTML.gif we have
λ n = λ n 0 + ϵ n , ϵ n = o ( 1 ) , n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ18_HTML.gif
(18)
Substituting (18) into (16), we get
Δ ( λ n 0 + ϵ n ) = Δ 0 ( λ n 0 + ϵ n ) + 0 μ + ( π ) A ( π , t ) sin ( λ n 0 + ϵ n ) λ n 0 + ϵ n d t = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equaf_HTML.gif
Hence, as n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq80_HTML.gif, taking into account the equality Δ 0 ( λ n 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq81_HTML.gif and relations sin ϵ n μ + ( π ) ϵ n μ + ( π ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq82_HTML.gif, cos ϵ n μ + ( π ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq83_HTML.gif, integrating by parts and using the properties of the kernel A ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq17_HTML.gif, we have
ϵ n 1 λ n 0 Δ ˙ ( λ n 0 ) { A ( π , μ + ( π ) ) cos ( λ n 0 + ϵ n ) λ n 0 + ϵ n μ + ( π ) } 1 λ n 0 Δ ˙ ( λ n 0 ) { [ A ( π , μ ( π ) + 0 ) A ( π , μ ( π ) + 0 ) ] cos ( λ n 0 + ϵ n ) λ n 0 + ϵ n μ ( π ) } 1 λ n 0 Δ ˙ ( λ n 0 ) { 0 μ + ( π ) A t ( π , t ) cos ( λ n 0 + ϵ n ) λ n 0 + ϵ n t d t } = d n λ n 0 + ϵ n + η ˜ n λ n 0 + ϵ n , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equag_HTML.gif
where
{ η ˜ n } : = { 0 μ + ( π ) A t ( π , t ) cos ( λ n 0 + ϵ n ) t d t } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equah_HTML.gif
Let us show that { η ˜ n } l 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq84_HTML.gif. It is obvious that
0 μ + ( π ) A t ( π , t ) cos λ t d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equai_HTML.gif
can be reduced to
μ + ( π ) μ + ( π ) R ( t ) e i λ t d t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equaj_HTML.gif
where R ( t ) L 2 ( μ + ( π ) , μ + ( π ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq85_HTML.gif. Now, take
ζ ( λ ) : = μ + ( π ) μ + ( π ) R ( t ) e i λ t d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equak_HTML.gif
It is clear from [19] (p.66) that { ζ n } = { ζ ( λ n ) } l 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq86_HTML.gif. By virtue of this we have η ˜ n l 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq87_HTML.gif (see [20, 21]). Therefore, as
η n = η ˜ n λ n 0 + ϵ n , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equal_HTML.gif

the validity of η n l 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq88_HTML.gif can be seen directly. Lemma is proved. □

5 Expansion formula

Assume that λ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq89_HTML.gif is not a spectrum point of operator A. Then, there exists resolvent operator R λ 2 ( A ) = ( A λ 2 I ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq90_HTML.gif. Let us find the expression of the operator R λ 2 ( A ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq91_HTML.gif.

Lemma 8 The resolvent R λ 2 ( A ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq91_HTML.gif is the integral operator with the kernel
G ( x , t ; λ ) = 1 Δ ( λ ) { φ ( t , λ ) ψ ( x , λ ) , t x , ψ ( t , λ ) φ ( x , λ ) , t x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ19_HTML.gif
(19)
Proof To construct the resolvent operator of A, we need to solve the boundary value problem
y + q ( x ) y = λ 2 ρ ( x ) y + ρ ( x ) f ( x ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ20_HTML.gif
(20)
y ( 0 ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ21_HTML.gif
(21)
λ 2 [ β 1 y ( π ) + β 2 y ( π ) ] + α 1 y ( π ) + α 2 y ( π ) = f 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ22_HTML.gif
(22)
where f ( x ) D ( A ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq92_HTML.gif. By applying the method of variation of constants, we seek the solution of the problem (20)-(22) in the following form
y ( x , λ ) = c 1 ( x , λ ) ψ ( x , λ ) + c 2 ( x , λ ) φ ( x , λ ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ23_HTML.gif
(23)
and we get the coefficients c 1 ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq93_HTML.gif and c 2 ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq94_HTML.gif as
c 1 ( x , λ ) = c 1 ( 0 , λ ) 1 Δ ( λ ) 0 x φ ( t , λ ) f ( t ) ρ ( t ) d t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ24_HTML.gif
(24)
c 2 ( x , λ ) = 1 Δ ( λ ) x π ψ ( t , λ ) f ( t ) ρ ( t ) d t + c 2 ( π , λ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ25_HTML.gif
(25)
Substituting (24) and (25) into (23) and taking into account the boundary conditions (21) and (22), we have
y ( x , λ ) = 0 π G ( x , t ; λ ) f ( t ) ρ ( t ) d t f 2 Δ ( λ ) φ ( x , λ ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ26_HTML.gif
(26)

where G ( x , t ; λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq95_HTML.gif is as in (19). □

Theorem 9 The eigenfunctions φ ( x , λ n ) n 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq96_HTML.gif of the boundary value problem (1)-(3) form a complete system in L 2 , ρ ( 0 , π ) C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq97_HTML.gif.

Proof With the help of (9) and (11), we can write
ψ ( x , λ n ) = Δ ˙ ( λ n ) 2 λ n γ n φ ( x , λ n ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ27_HTML.gif
(27)
Using (19) and (26), we get
Res λ = λ n y ( x , λ ) = 1 2 λ n γ n φ ( x , λ n ) { 0 π φ ( t , λ n ) f ( t ) ρ ( t ) d t + f 2 k n } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ28_HTML.gif
(28)
Now, let f ( x ) L 2 , ρ ( 0 , π ) C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq98_HTML.gif and assume
( φ ( x , λ n ) , f ( x ) ) = 0 π φ ( x , λ n ) f 1 ( x ) ¯ ρ ( x ) d x + f 2 ( β 1 φ ( π , λ n ) + β 2 φ ( π , λ n ) ) χ = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ29_HTML.gif
(29)

Then from (28), we have Res λ = λ n y ( x , λ ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq99_HTML.gif. Consequently, for fixed x [ 0 , π ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq100_HTML.gif the function y ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq101_HTML.gif is entire with respect to λ.

Let us denote that
G δ : = { λ : | λ λ n 0 | δ , n = 0 , 1 , 2 , } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equam_HTML.gif

where δ is a sufficiently small positive number. (16) is valid from Theorem 12.4 in [22] for λ G δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq102_HTML.gif.

On the other hand, we can say from Lemma 1.3.1 in [19] that for every f ( x ) L 1 ( 0 , π ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq103_HTML.gif, the following relation holds
lim | λ | max 0 x π { e | Im λ | x | 0 x f ( t ) cos λ t d t | } = lim | λ | max 0 x π { e | Im λ | x | 0 x f ( t ) sin λ t d t | } = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ30_HTML.gif
(30)
Also, for | λ | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq104_HTML.gif, the relations below hold
φ ( x , λ ) = O ( 1 | λ | e | Im λ | μ + ( x ) ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ31_HTML.gif
(31)
φ ( x , λ ) = φ 0 ( x , λ ) + O ( 1 | λ | e | Im λ | μ + ( x ) ) = O ( e | Im λ | μ + ( x ) ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ32_HTML.gif
(32)
ψ ( x , λ ) = O ( 1 | λ | e | Im λ | ( μ + ( π ) μ + ( x ) ) ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ33_HTML.gif
(33)
ψ ( x , λ ) = ψ 0 ( x , λ ) O ( 1 | λ | e | Im λ | ( μ + ( π ) μ + ( x ) ) ) = O ( e | Im λ | ( μ + ( π ) μ + ( x ) ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ34_HTML.gif
(34)
From the estimates (31)-(34), it is obvious that
| Δ ( λ ) | C δ 1 | λ | e | Im λ | μ + ( π ) , λ G δ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ35_HTML.gif
(35)
From (26), it follows that for fixed δ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq79_HTML.gif and sufficiently large λ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq105_HTML.gif, we have
| y ( x , λ ) | C δ | λ | , λ G δ , | λ | λ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equan_HTML.gif

Using maximum principle for module of analytic functions and Liouville theorem, we get y ( x , λ ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq106_HTML.gif. From this and the expression of the boundary value problem (20)-(22), we obtain that f ( x ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq107_HTML.gif a.e. on [ 0 , π ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq108_HTML.gif. Thus, we reach the completeness of the eigenfunctions φ ( x , λ n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq28_HTML.gif in L 2 , ρ ( 0 , π ) C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq109_HTML.gif. □

Theorem 10 If f ( x ) D ( A ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq92_HTML.gif, then the expansion formula
f ( x ) = n = 1 a n φ ( x , λ n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ36_HTML.gif
(36)
is valid, where
a n = 1 2 γ n 0 π φ ( t , λ n ) f ( t ) ρ ( t ) d t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equao_HTML.gif
and the series converges uniformly with respect to x [ 0 , π ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq13_HTML.gif. For f ( x ) L 2 , ρ ( 0 , π ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq110_HTML.gif, the series converges in L 2 , ρ ( 0 , π ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq111_HTML.gif, moreover, the Parseval equality holds
0 π | f ( x ) | 2 ρ ( x ) d x = n = 1 γ n | a n | 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equap_HTML.gif
Proof Since φ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq7_HTML.gif and ψ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq8_HTML.gif are the solutions of the boundary value problem (1)-(3), we have
y ( x , λ ) = 1 λ 2 Δ ( λ ) { ψ ( x , λ ) 0 x [ φ ( t , λ ) + q ( t ) φ ( t , λ ) ] f ( t ) d t } 1 λ 2 Δ ( λ ) { φ ( x , λ ) x π [ ψ ( t , λ ) + q ( t ) ψ ( t , λ ) ] f ( t ) d t } f 2 Δ ( λ ) φ ( x , λ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ37_HTML.gif
(37)
Integrating by parts and taking into account the boundary conditions (2), (3), we obtain
y ( x , λ ) = 1 λ 2 f ( x ) 1 λ 2 { Z 1 ( x , λ ) + Z 2 ( x , λ ) } f 2 Δ ( λ ) φ ( x , λ ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ38_HTML.gif
(38)
where
Z 1 ( x , λ ) = 1 Δ ( λ ) { ψ ( x , λ ) 0 x φ ( t , λ ) g ( t ) d t + φ ( x , λ ) x π ψ ( t , λ ) g ( t ) d t } , Z 2 ( x , λ ) = 1 Δ ( λ ) { ψ ( x , λ ) 0 x φ ( t , λ ) q ( t ) f ( t ) d t } Z 2 ( x , λ ) = + 1 Δ ( λ ) { φ ( x , λ ) x π ψ ( t , λ ) q ( t ) f ( t ) d t + φ ( x , λ ) ψ ( π , λ ) f ( π ) } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equaq_HTML.gif
as g ( t ) = f ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq112_HTML.gif. Let us consider the following contour integral
I n ( x ) = 1 2 π i Γ n λ y ( x , λ ) d λ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equar_HTML.gif
where Γ n = { λ : | λ | = | λ n 0 | + τ 2 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq113_HTML.gif is a contour oriented counter-clockwise, and n is a sufficiently large natural number. With the help of Residue theorem, we get
I n ( x ) = 2 n = 1 Res λ = λ n [ λ y ( x , λ ) ] = n = 1 a n φ ( x , λ n ) + n = 1 λ f 2 Δ ˙ ( λ n ) φ ( x , λ n ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ39_HTML.gif
(39)
where
a n = 1 γ n 0 π φ ( t , λ n ) f ( t ) ρ ( t ) d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equas_HTML.gif
On the other hand, taking into account (38), we have
I n ( x ) = f ( x ) + 1 2 π i Γ n ( Z 1 ( x , λ ) + Z 2 ( x , λ ) ) d λ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ40_HTML.gif
(40)
Comparing (39) and (40), we obtain
n = 1 a n φ ( x , λ n ) + n = 1 λ f 2 Δ ˙ ( λ n ) φ ( x , λ n ) = f ( x ) + ϵ n ( x ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equat_HTML.gif
where
ϵ n ( x ) = 1 2 π i Γ n ( Z 1 ( x , λ ) + Z 2 ( x , λ ) ) d λ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equau_HTML.gif
Thus, we obtain
f ( x ) = n = 1 1 γ n φ ( x , λ n ) { 0 π φ ( t , λ n ) f ( t ) ρ ( t ) d t } + ϵ n ( x ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ41_HTML.gif
(41)
Now, let us show that
lim n max x [ 0 , π ] | ϵ n ( x ) | = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ42_HTML.gif
(42)
From estimates (31)-(34) of solutions φ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq7_HTML.gif, ψ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq8_HTML.gif and the inequality (35) for the function Δ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq23_HTML.gif, it follows, for fixed δ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq79_HTML.gif and sufficiently large λ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq105_HTML.gif
max x [ 0 , π ] | Z 2 ( x , λ ) | C 2 | λ | , λ G δ , | λ | λ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ43_HTML.gif
(43)
Let us show that lim | λ | max 0 x π | Z 1 ( x , λ ) | = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq114_HTML.gif. If we suppose that g ( t ) A C [ 0 , π ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq115_HTML.gif, and by then integrate by parts the expression of Z 1 ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq116_HTML.gif, we obtain
Z 1 ( x , λ ) = 1 Δ ( λ ) { ψ ( x , λ ) 0 x φ ( t , λ ) g ( t ) d t + φ ( x , λ ) x π ψ ( t , λ ) g ( t ) d t } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equav_HTML.gif
Hence, similar to Z 2 ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq117_HTML.gif, we have
max x [ 0 , π ] | Z 1 ( x , λ ) | C 1 | λ | , λ G δ , | λ | λ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ44_HTML.gif
(44)
In general case, let us take an arbitrary fixed number ϵ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq118_HTML.gif and assume that g ϵ ( t ) A C [ 0 , π ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq119_HTML.gif, such that 0 π | g ( t ) g ϵ ( t ) | d t < ϵ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq120_HTML.gif. Then we can find a λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq121_HTML.gif that for λ G δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq102_HTML.gif and | λ | > λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq122_HTML.gif. Also, using the equation below,
Z 1 ( x , λ ) = 1 Δ ( λ ) { ψ ( x , λ ) 0 x ( g ( t ) g ϵ ( t ) ) φ ( t , λ ) d t } + 1 Δ ( λ ) { φ ( x , λ ) x π ( g ( t ) g ϵ ( t ) ) ψ ( t , λ ) d t } + 1 Δ ( λ ) { ψ ( x , λ ) 0 x ( g ϵ ( t ) ) φ ( t , λ ) d t + φ ( x , λ ) x π ( g ϵ ( t ) ) ψ ( t , λ ) d t } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equaw_HTML.gif
and with the help of the estimates of functions φ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq7_HTML.gif, ψ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq8_HTML.gif and Δ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq23_HTML.gif, we get
max x [ 0 , π ] | Z 1 ( x , λ ) | C 0 π | g ( t ) g ϵ ( t ) | d t + C ˜ ( ϵ ) | λ | < C ϵ + C ˜ ( ϵ ) | λ | , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equax_HTML.gif
as λ G δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq102_HTML.gif, | λ | λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq123_HTML.gif. Hence we have
lim ¯ | λ | max | Z 1 ( x , λ ) | C ϵ ( λ G δ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equay_HTML.gif
From the arbitrariness of ϵ, we reach
lim | λ | max x [ 0 , π ] | Z 1 ( x , λ ) | = 0 , λ G δ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ45_HTML.gif
(45)
The validity of (42) can be easily seen from (43) and (44). Thus, we obtain
f ( x ) = n = 1 1 γ n φ ( x , λ n ) { 0 π φ ( t , λ n ) f ( t ) ρ ( t ) d t } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equaz_HTML.gif
If we take
a n = 1 γ n 0 π φ ( t , λ n ) f ( t ) ρ ( t ) d t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equba_HTML.gif
the last equation gives us the expansion formula
f ( x ) = n = 1 a n φ ( x , λ n ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equbb_HTML.gif
Since the system of φ ( x , λ n ) n 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq124_HTML.gif is complete and orthogonal in L 2 , ρ ( 0 , π ) C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq125_HTML.gif, the Parseval equality
0 π | f ( x ) | 2 ρ ( x ) d x = n = 1 γ n | a n | 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equbc_HTML.gif

holds. Extension of the Parseval equality to an arbitrary vector-function of the class L 2 ( a , b ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq126_HTML.gif can be carried out by usual methods. □

6 Weyl solution, Weyl function

Let Φ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq127_HTML.gif be the solution of equation (1) that satisfies the conditions
Φ ( 0 , λ ) = 1 , λ 2 [ β 1 Φ ( π , λ ) + β 2 Φ ( π , λ ) ] + α 1 Φ ( π , λ ) + α 2 Φ ( π , λ ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equbd_HTML.gif
Denote by c ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq128_HTML.gif the solution of equation (1), which satisfies the initial conditions c ( 0 , λ ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq129_HTML.gif, c ( 0 , λ ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq130_HTML.gif. Then the solution ψ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq8_HTML.gif can be represented as follows
ψ ( x , λ ) = ψ ( 0 , λ ) φ ( x , λ ) Δ ( λ ) c ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Eqube_HTML.gif
or
ψ ( x , λ ) Δ ( λ ) = c ( x , λ ) ψ ( 0 , λ ) Δ ( λ ) φ ( x , λ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ46_HTML.gif
(46)
Denote
M ( λ ) : = ψ ( 0 , λ ) Δ ( λ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ47_HTML.gif
(47)
It is clear that
Φ ( x , λ ) = c ( x , λ ) + M ( λ ) φ ( x , λ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ48_HTML.gif
(48)
The functions Φ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq127_HTML.gif and M ( λ ) = Φ ( 0 , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq131_HTML.gif are respectively called the Weyl solution and the Weyl function of the boundary value problem (1)-(3). The Weyl function is a meromorphic function having simple poles at points λ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq25_HTML.gif eigenvalues of boundary value problem (1)-(3). Relations (46), (48) yield
Φ ( x , λ ) = ψ ( x , λ ) Δ ( λ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ49_HTML.gif
(49)
It can be shown that
Φ ( x , λ ) , φ ( x , λ ) 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ50_HTML.gif
(50)

Let us take into consideration a boundary value problem with the coefficient q ˜ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq132_HTML.gif similar to (1)-(3) and assume that if an element α belongs to boundary value problem (1)-(3), then α ˜ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq133_HTML.gif belongs to one with q ˜ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq132_HTML.gif.

Validity of the equation below can be shown analogously to [8]
M ( λ ) = M ( 0 ) + n = 1 λ 2 γ n λ n 2 ( λ 2 λ n 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ51_HTML.gif
(51)

Theorem 11 The boundary value problem (1)-(3) is identically denoted by the Weyl function M ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq134_HTML.gif. (If M ( λ ) = M ˜ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq135_HTML.gif, then q ( x ) = q ˜ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq136_HTML.gif.)

Proof Let us identify the matrix P ( x , λ ) = [ P j k ( x , λ ) ] j , k = 1 , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq137_HTML.gif as
P ( x , λ ) ( φ ˜ ( x , λ ) Φ ˜ ( x , λ ) φ ˜ ( x , λ ) Φ ˜ ( x , λ ) ) = ( φ ( x , λ ) Φ ( x , λ ) φ ( x , λ ) Φ ( x , λ ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ52_HTML.gif
(52)
From (50) and (52), we have
φ ( x , λ ) = P 11 ( x , λ ) φ ˜ ( x , λ ) + P 12 ( x , λ ) φ ˜ ( x , λ ) , Φ ( x , λ ) = P 11 ( x , λ ) Φ ˜ ( x , λ ) + P 12 ( x , λ ) Φ ˜ ( x , λ ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ53_HTML.gif
(53)
or
P 11 ( x , λ ) = φ ˜ ( x , λ ) Φ ( x , λ ) φ ( x , λ ) Φ ˜ ( x , λ ) , P 12 ( x , λ ) = φ ( x , λ ) Φ ˜ ( x , λ ) φ ˜ ( x , λ ) Φ ( x , λ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ54_HTML.gif
(54)
Taking equation (49) into consideration in (54), we get
P 11 ( x , λ ) = 1 + 1 Δ ( λ ) ψ ( x , λ ) ( φ ( x , λ ) φ ˜ ( x , λ ) ) + 1 Δ ( λ ) φ ( x , λ ) ( ψ ˜ ( x , λ ) ψ ( x , λ ) ) , P 12 ( x , λ ) = 1 Δ ( λ ) ( φ ( x , λ ) ψ ˜ ( x , λ ) + φ ˜ ( x , λ ) ψ ( x , λ ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ55_HTML.gif
(55)
Now, from the estimates
| φ ( x , λ ) φ ˜ ( x , λ ) | = O ( 1 | λ | e | Im λ | μ + ( x ) ) , | λ | , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equbf_HTML.gif
and
| ψ ( x , λ ) ψ ˜ ( x , λ ) | = O ( 1 | λ | e | Im λ | ( μ + ( π ) μ + ( x ) ) ) , | λ | , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equbg_HTML.gif
we have from equation (55)
lim | λ | max 0 x π | P 11 ( x , λ ) 1 | = lim | λ | max 0 x π | P 12 ( x , λ ) | = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equ56_HTML.gif
(56)
for λ G δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq102_HTML.gif. Now, if we take consideration equation (48) into (53), we get
P 11 ( x , λ ) = φ ˜ ( x , λ ) c ( x , λ ) φ ( x , λ ) c ˜ ( x , λ ) + φ ˜ ( x , λ ) φ ( x , λ ) [ M ( λ ) M ˜ ( λ ) ] , P 12 ( x , λ ) = φ ( x , λ ) c ˜ ( x , λ ) φ ˜ ( x , λ ) c ( x , λ ) + φ ( x , λ ) φ ˜ ( x , λ ) [ M ˜ ( λ ) M ( λ ) ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_Equbh_HTML.gif

Therefore, if M ( λ ) = M ˜ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq135_HTML.gif, then P 11 ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq138_HTML.gif and P 12 ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq139_HTML.gif are entire functions for every fixed x. It can easily be seen from equation (56) that P 11 ( x , λ ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq140_HTML.gif and P 12 ( x , λ ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq141_HTML.gif. Consequently, we get φ ( x , λ ) φ ˜ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq142_HTML.gif and Φ ( x , λ ) Φ ˜ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq143_HTML.gif for every x and λ. Hence, we arrive at q ( x ) q ˜ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq144_HTML.gif. □

Theorem 12 The spectral data identically define the boundary value problem (1)-(3).

Proof From (51), it is clear that the function M ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq134_HTML.gif can be constructed by λ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq25_HTML.gif. Since λ n = λ ˜ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq145_HTML.gif for every n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq146_HTML.gif, from Theorem 10, we can say that M ( λ ) = M ˜ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq135_HTML.gif. Then from Theorem 11, it is obvious that A = A ˜ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-183/MediaObjects/13661_2013_Article_434_IEq147_HTML.gif. □

Declarations

Acknowledgements

This work is supported by the Scientific and Technological Research Council of Turkey (TÜBİTAK).

Authors’ Affiliations

(1)
Department of Mathematics, Science and Letters Faculty, Mersin University

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