On sampling theories and discontinuous Dirac systems with eigenparameter in the boundary conditions

Boundary Value Problems20132013:65

DOI: 10.1186/1687-2770-2013-65

Received: 8 November 2012

Accepted: 11 March 2013

Published: 29 March 2013

Abstract

The sampling theory says that a function may be determined by its sampled values at some certain points provided the function satisfies some certain conditions. In this paper we consider a Dirac system which contains an eigenparameter appearing linearly in one condition in addition to an internal point of discontinuity. We closely follow the analysis derived by Annaby and Tharwat (J. Appl. Math. Comput. 2010, doi:10.1007/s12190-010-0404-9) to establish the needed relations for the derivations of the sampling theorems including the construction of Green’s matrix as well as the eigen-vector-function expansion theorem. We derive sampling representations for transforms whose kernels are either solutions or Green’s matrix of the problem. In the special case, when our problem is continuous, the obtained results coincide with the corresponding results in Annaby and Tharwat (J. Appl. Math. Comput. 2010, doi:10.1007/s12190-010-0404-9).

MSC:34L16, 94A20, 65L15.

Keywords

Dirac systems transmission conditions eigenvalue parameter in the boundary conditions discontinuous boundary value problems

1 Introduction

Sampling theory is one of the most powerful results in signal analysis. It is of great need in signal processing to reconstruct (recover) a signal (function) from its values at a discrete sequence of points (samples). If this aim is achieved, then an analog (continuous) signal can be transformed into a digital (discrete) one and then it can be recovered by the receiver. If the signal is band-limited, the sampling process can be done via the celebrated Whittaker, Shannon and Kotel’nikov (WSK) sampling theorem [13]. By a band-limited signal with band width σ, σ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq1_HTML.gif, i.e., the signal contains no frequencies higher than σ / 2 π http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq2_HTML.gif cycles per second (cps), we mean a function in the Paley-Wiener space B σ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq3_HTML.gif of entire functions of exponential type at most σ which are L 2 ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq4_HTML.gif-functions when restricted to ℝ. In other words, f ( t ) B σ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq5_HTML.gif if there exists g ( ) L 2 ( σ , σ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq6_HTML.gif such that, cf. [4, 5],
f ( t ) = 1 2 π σ σ e i x t g ( x ) d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ1_HTML.gif
(1.1)
Now WSK sampling theorem states [6, 7]: If f ( t ) B σ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq5_HTML.gif, then it is completely determined from its values at the points t k = k π / σ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq7_HTML.gif, k Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq8_HTML.gif, by means of the formula
f ( t ) = k = f ( t k ) sinc σ ( t t k ) , t C , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ2_HTML.gif
(1.2)
where
sinc t = { sin t t , t 0 , 1 , t = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ3_HTML.gif
(1.3)

The sampling series (1.2) is absolutely and uniformly convergent on compact subsets of ℂ.

The WSK sampling theorem has been generalized in many different ways. Here we are interested in two extensions. The first is concerned with replacing the equidistant sampling points by more general ones, which is very important from the practical point of view. The following theorem which is known in some literature as the Paley-Wiener theorem [5] gives a sampling theorem with a more general class of sampling points. Although the theorem in its final form may be attributed to Levinson [8] and Kadec [9], it could be named after Paley and Wiener who first derived the theorem in a more restrictive form; see [6, 7, 10] for more details.

The Paley-Wiener theorem states that if { t k } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq9_HTML.gif, k Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq8_HTML.gif, is a sequence of real numbers such that
D : = sup k Z | t k k π σ | < π 4 σ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ4_HTML.gif
(1.4)
and G is the entire function defined by
G ( t ) : = ( t t 0 ) k = 1 ( 1 t t k ) ( 1 t t k ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ5_HTML.gif
(1.5)
then, for any function of the form (1.1), we have
f ( t ) = k Z f ( t k ) G ( t ) G ( t k ) ( t t k ) , t C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ6_HTML.gif
(1.6)

The series (1.6) converges uniformly on compact subsets of ℂ.

The WSK sampling theorem is a special case of this theorem because if we choose t k = k π / σ = t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq10_HTML.gif, then
G ( t ) = t k = 1 ( 1 t t k ) ( 1 + t t k ) = t k = 1 ( 1 ( t σ / π ) 2 k 2 ) = sin t σ σ , G ( t k ) = ( 1 ) k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equa_HTML.gif

The sampling series (1.6) can be regarded as an extension of the classical Lagrange interpolation formula to ℝ for functions of exponential type. Therefore, (1.6) is called a Lagrange-type interpolation expansion.

The second extension of WSK sampling theorem is the theorem of Kramer, [11] which states that if I is a finite closed interval, K ( , t ) : I × C C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq11_HTML.gif is a function continuous in t such that K ( , t ) L 2 ( I ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq12_HTML.gif for all t C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq13_HTML.gif. Let { t k } k Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq14_HTML.gif be a sequence of real numbers such that { K ( , t k ) } k Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq15_HTML.gif is a complete orthogonal set in L 2 ( I ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq16_HTML.gif. Suppose that
f ( t ) = I K ( x , t ) g ( x ) d x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equb_HTML.gif
where g ( ) L 2 ( I ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq17_HTML.gif. Then
f ( t ) = k Z f ( t k ) I K ( x , t ) K ( x , t k ) ¯ d x K ( , t k ) L 2 ( I ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ7_HTML.gif
(1.7)

Series (1.7) converges uniformly wherever K ( , t ) L 2 ( I ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq18_HTML.gif, as a function of t, is bounded. In this theorem, sampling representations were given for integral transforms whose kernels are more general than exp ( i x t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq19_HTML.gif. Also Kramer’s theorem is a generalization of WSK theorem. If we take K ( x , t ) = e i t x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq20_HTML.gif, I = [ σ , σ ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq21_HTML.gif, t k = k π σ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq22_HTML.gif, then (1.7) is (1.2).

The relationship between both extensions of WSK sampling theorem has been investigated extensively. Starting from a function theory approach, cf. [12], it was proved in [13] that if K ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq23_HTML.gif, x I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq24_HTML.gif, t C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq25_HTML.gif, satisfies some analyticity conditions, then Kramer’s sampling formula (1.7) turns out to be a Lagrange interpolation one; see also [1416]. In another direction, it was shown that Kramer’s expansion (1.7) could be written as a Lagrange-type interpolation formula if K ( , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq26_HTML.gif and t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq27_HTML.gif were extracted from ordinary differential operators; see the survey [17] and the references cited therein. The present work is a continuation of the second direction mentioned above. We prove that integral transforms associated with Dirac systems with an internal point of discontinuity can also be reconstructed in a sampling form of Lagrange interpolation type. We would like to mention that works in direction of sampling associated with eigenproblems with an eigenparameter in the boundary conditions are few; see, e.g., [1820]. Also, papers in sampling with discontinuous eigenproblems are few; see [2124]. However, sampling theories associated with Dirac systems which contain eigenparameter in the boundary conditions and have at the same time discontinuity conditions, do not exist as far as we know. Our investigation is be the first in that direction, introducing a good example. To achieve our aim we briefly study the spectral analysis of the problem. Then we derive two sampling theorems using solutions and Green’s matrix respectively.

2 The eigenvalue problem

In this section, we define our boundary value problem and state some of its properties. We consider the Dirac system
u 2 ( x ) p 1 ( x ) u 1 ( x ) = λ u 1 ( x ) , u 1 ( x ) + p 2 ( x ) u 2 ( x ) = λ u 2 ( x ) , x [ 1 , 0 ) ( 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ8_HTML.gif
(2.1)
U 1 ( u ) : = sin α u 1 ( 1 ) cos α u 2 ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ9_HTML.gif
(2.2)
U 2 ( u ) : = ( a 1 + λ sin β ) u 1 ( 1 ) ( a 2 + λ cos β ) u 2 ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ10_HTML.gif
(2.3)
and transmission conditions
U 3 ( u ) : = u 1 ( 0 ) δ u 1 ( 0 + ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ11_HTML.gif
(2.4)
U 4 ( u ) : = u 2 ( 0 ) δ u 2 ( 0 + ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ12_HTML.gif
(2.5)

where λ C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq28_HTML.gif; the real-valued functions p 1 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq29_HTML.gif and p 2 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq30_HTML.gif are continuous in [ 1 , 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq31_HTML.gif and ( 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq32_HTML.gif and have finite limits p 1 ( 0 ± ) : = lim x 0 ± p 1 ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq33_HTML.gif and p 2 ( 0 ± ) : = lim x 0 ± p 2 ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq34_HTML.gif; a 1 , a 2 , δ R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq35_HTML.gif, α , β [ 0 , π ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq36_HTML.gif; δ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq37_HTML.gif and ρ : = a 1 cos β a 2 sin β > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq38_HTML.gif.

In [24] the authors discussed problem (2.1)-(2.5) but with the condition sin β u 1 ( 1 ) cos β u 2 ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq39_HTML.gif instead of (2.3). To formulate a theoretic approach to problem (2.1)-(2.5), we define the Hilbert space H = H C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq40_HTML.gif with an inner product, see [19, 20],
U ( ) , V ( ) H : = 1 0 u ( x ) v ¯ ( x ) d x + δ 2 0 1 u ( x ) v ¯ ( x ) d x + δ 2 ρ z w ¯ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ13_HTML.gif
(2.6)
where ⊤ denotes the matrix transpose,
U ( x ) = ( u ( x ) z ) , V ( x ) = ( v ( x ) w ) H , u ( x ) = ( u 1 ( x ) u 2 ( x ) ) , v ( x ) = ( v 1 ( x ) v 2 ( x ) ) H , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equc_HTML.gif
u i ( ) , v i ( ) L 2 ( 1 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq41_HTML.gif, i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq42_HTML.gif, z , w C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq43_HTML.gif. For convenience, we put
( R a ( u ( x ) ) R β ( u ( x ) ) ) : = ( a 1 u 1 ( 1 ) a 2 u 2 ( 1 ) sin β u 1 ( 1 ) cos β u 2 ( 1 ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ14_HTML.gif
(2.7)
Equation (2.1) can be written as
( u ) : = A u ( x ) P ( x ) u ( x ) = λ u ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ15_HTML.gif
(2.8)
where
A = ( 0 1 1 0 ) , P ( x ) = ( p 1 ( x ) 0 0 p 2 ( x ) ) , u ( x ) = ( u 1 ( x ) u 2 ( x ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ16_HTML.gif
(2.9)
For functions u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq44_HTML.gif, which are defined on [ 1 , 0 ) ( 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq45_HTML.gif and have finite limit u ( ± 0 ) : = lim x ± 0 u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq46_HTML.gif, by u ( 1 ) ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq47_HTML.gif and u ( 2 ) ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq48_HTML.gif, we denote the functions
u ( 1 ) ( x ) = { u ( x ) , x [ 1 , 0 ) , u ( 0 ) , x = 0 , u ( 2 ) ( x ) = { u ( x ) , x ( 0 , 1 ] , u ( 0 + ) , x = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ17_HTML.gif
(2.10)

which are defined on Γ 1 : = [ 1 , 0 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq49_HTML.gif and Γ 2 : = [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq50_HTML.gif, respectively.

In the following lemma, we prove that the eigenvalues of problem (2.1)-(2.5) are real.

Lemma 2.1 The eigenvalues of problem (2.1)-(2.5) are real.

Proof Assume the contrary that λ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq51_HTML.gif is a nonreal eigenvalue of problem (2.1)-(2.5). Let ( u 1 ( x ) u 2 ( x ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq52_HTML.gif be a corresponding (non-trivial) eigenfunction. By (2.1), we have
d d x { u 1 ( x ) u ¯ 2 ( x ) u ¯ 1 ( x ) u 2 ( x ) } = ( λ ¯ 0 λ 0 ) { | u 1 ( x ) | 2 + | u 2 ( x ) | 2 } , x [ 1 , 0 ) ( 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equd_HTML.gif
Integrating the above equation through [ 1 , 0 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq53_HTML.gif and [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq54_HTML.gif, we obtain
( λ ¯ 0 λ 0 ) [ 1 0 ( | u 1 ( x ) | 2 + | u 2 ( x ) | 2 ) d x ] = u 1 ( 0 ) u ¯ 2 ( 0 ) u ¯ 1 ( 0 ) u 2 ( 0 ) [ u 1 ( 1 ) u ¯ 2 ( 1 ) u ¯ 1 ( 1 ) u 2 ( 1 ) ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ18_HTML.gif
(2.11)
( λ ¯ 0 λ 0 ) [ 0 1 ( | u 1 ( x ) | 2 + | u 2 ( x ) | 2 ) d x ] = u 1 ( 1 ) u ¯ 2 ( 1 ) u ¯ 1 ( 1 ) u 2 ( 1 ) [ u 1 ( 0 + ) u ¯ 2 ( 0 + ) u ¯ 1 ( 0 + ) u 2 ( 0 + ) ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ19_HTML.gif
(2.12)
Then from (2.2), (2.3) and transmission conditions, we have, respectively,
u 1 ( 1 ) u ¯ 2 ( 1 ) u ¯ 1 ( 1 ) u 2 ( 1 ) = 0 , u 1 ( 1 ) u ¯ 2 ( 1 ) u ¯ 1 ( 1 ) u 2 ( 1 ) = ρ ( λ ¯ 0 λ 0 ) | u 2 ( 1 ) | 2 | a 1 + λ 0 sin β | 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Eque_HTML.gif
and
u 1 ( 0 ) u ¯ 2 ( 0 ) u ¯ 1 ( 0 ) u 2 ( 0 ) = δ 2 [ u 1 ( 0 + ) u ¯ 2 ( 0 + ) u ¯ 1 ( 0 + ) u 2 ( 0 + ) ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equf_HTML.gif
Since λ 0 λ ¯ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq55_HTML.gif, it follows from the last three equations and (2.11), (2.12) that
1 0 ( | u 1 ( x ) | 2 + | u 2 ( x ) | 2 ) d x + δ 2 0 1 ( | u 1 ( x ) | 2 + | u 2 ( x ) | 2 ) d x = ρ δ 2 | u 2 ( 1 ) | 2 | a 1 + λ 0 sin β | 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ20_HTML.gif
(2.13)

This contradicts the conditions 1 0 ( | u 1 ( x ) | 2 + | u 2 ( x ) | 2 ) d x + δ 2 0 1 ( | u 1 ( x ) | 2 + | u 2 ( x ) | 2 ) d x > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq56_HTML.gif and ρ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq57_HTML.gif. Consequently, λ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq51_HTML.gif must be real. □

Let D ( A ) H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq58_HTML.gif be the set of all U ( x ) = ( u ( x ) R β ( u ( x ) ) ) H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq59_HTML.gif such that u 1 ( i ) ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq60_HTML.gif, u 2 ( i ) ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq61_HTML.gif are absolutely continuous on Γ i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq62_HTML.gif, i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq42_HTML.gif, and ( u ) H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq63_HTML.gif, sin α u 1 ( 1 ) cos α u 2 ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq64_HTML.gif, u i ( 0 ) δ u i ( 0 + ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq65_HTML.gif, i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq42_HTML.gif. Define the operator A : D ( A ) H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq66_HTML.gif by
A ( u ( x ) R β ( u ( x ) ) ) = ( ( u ) R a ( u ( x ) ) ) , ( u ( x ) R β ( u ( x ) ) ) D ( A ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ21_HTML.gif
(2.14)
Thus, the operator A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq67_HTML.gif is symmetric in H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq68_HTML.gif. Indeed, for U ( ) , V ( ) D ( A ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq69_HTML.gif,
A U ( ) , U ( ) H = 1 0 ( ( u ( x ) ) ) v ¯ ( x ) d x + δ 2 0 1 ( ( u ( x ) ) ) v ¯ ( x ) d x δ 2 ρ R a ( u ( x ) ) R ¯ β ( v ( x ) ) = 1 0 ( u 2 ( x ) p 1 ( x ) u 1 ( x ) ) v ¯ 1 ( x ) d x 1 0 ( u 1 ( x ) + p 2 ( x ) u 2 ( x ) ) v ¯ 2 ( x ) d x + δ 2 0 1 ( u 2 ( x ) p 1 ( x ) u 1 ( x ) ) v ¯ 1 ( x ) d x δ 2 0 1 ( u 1 ( x ) + p 2 ( x ) u 2 ( x ) ) v ¯ 2 ( x ) d x δ 2 ρ R a ( u ( x ) ) R ¯ β ( v ( x ) ) = 1 0 u 2 ( x ) ( v ¯ 1 ( x ) + p 2 ( x ) v ¯ 2 ( x ) ) d x + 1 0 u 1 ( x ) ( v ¯ 2 ( x ) p 1 ( x ) v ¯ 1 ( x ) ) d x δ 2 0 1 u 2 ( x ) ( v ¯ 1 ( x ) + p 2 ( x ) v ¯ 2 ( x ) ) d x + δ 2 0 1 u 1 ( x ) ( v ¯ 2 ( x ) p 1 ( x ) v ¯ 1 ( x ) ) d x + ( u 2 ( 0 ) v ¯ 1 ( 0 ) u 1 ( 0 ) v ¯ 2 ( 0 ) ) ( u 2 ( 1 ) v ¯ 1 ( 1 ) u 1 ( 1 ) v ¯ 2 ( 1 ) ) + δ 2 ( u 2 ( 1 ) v ¯ 1 ( 1 ) u 1 ( 1 ) v ¯ 2 ( 1 ) ) δ 2 ( u 2 ( 0 + ) v ¯ 1 ( 0 + ) u 1 ( 0 + ) v ¯ 2 ( 0 + ) ) δ 2 ρ R a ( u ( x ) ) R ¯ β ( v ( x ) ) = 0 1 u ( x ) ¯ v ( x ) d x δ 2 ρ R β ( u ( x ) ) R ¯ a ( v ( x ) ) = U ( ) , A V ( ) H . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equg_HTML.gif

The operator A : D ( A ) H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq70_HTML.gif and the eigenvalue problem (2.1)-(2.5) have the same eigenvalues. Therefore they are equivalent in terms of this aspect.

Lemma 2.2 Let λ and μ be two different eigenvalues of problem (2.1)-(2.5). Then the corresponding eigenfunctions u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq71_HTML.gif and v ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq72_HTML.gif of this problem satisfy the following equality:
1 0 u ( x ) v ( x ) d x + δ 2 0 1 u ( x ) v ( x ) d x = δ 2 ρ R a ( u ( x ) ) R β ( v ( x ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ22_HTML.gif
(2.15)
Proof Equation (2.15) follows immediately from the orthogonality of the corresponding eigenelements:
U ( x ) = ( u ( x ) z ) , V ( x ) = ( v ( x ) w ) H , u ( x ) = ( u 1 ( x ) u 2 ( x ) ) , v ( x ) = ( v 1 ( x ) v 2 ( x ) ) H . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equh_HTML.gif

 □

Now, we construct a special fundamental system of solutions of equation (2.1) for λ not being an eigenvalue. Let us consider the next initial value problem:
u 2 ( x ) p 1 ( x ) u 1 ( x ) = λ u 1 ( x ) , u 1 ( x ) + p 2 ( x ) u 2 ( x ) = λ u 2 ( x ) , x ( 1 , 0 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ23_HTML.gif
(2.16)
u 1 ( 1 ) = cos α , u 2 ( 1 ) = sin α . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ24_HTML.gif
(2.17)
By virtue of Theorem 1.1 in [25], this problem has a unique solution u = ( ϕ 11 ( x , λ ) ϕ 21 ( x , λ ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq73_HTML.gif, which is an entire function of λ C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq28_HTML.gif for each fixed x [ 1 , 0 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq74_HTML.gif. Similarly, employing the same method as in the proof of Theorem 1.1 in [25], we see that the problem
u 2 ( x ) p 1 ( x ) u 1 ( x ) = λ u 1 ( x ) , u 1 ( x ) + p 2 ( x ) u 2 ( x ) = λ u 2 ( x ) , x ( 0 , 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ25_HTML.gif
(2.18)
u 1 ( 1 ) = a 2 + λ cos β , u 2 ( 1 ) = a 1 + λ sin β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ26_HTML.gif
(2.19)

has a unique solution u = ( χ 12 ( x , λ ) χ 22 ( x , λ ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq75_HTML.gif, which is an entire function of parameter λ for each fixed x [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq76_HTML.gif.

Now the functions ϕ i 2 ( x , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq77_HTML.gif and χ i 1 ( x , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq78_HTML.gif are defined in terms of ϕ i 1 ( x , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq79_HTML.gif and χ i 2 ( x , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq80_HTML.gif, i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq42_HTML.gif, respectively, as follows: The initial-value problem,
u 2 ( x ) p 1 ( x ) u 1 ( x ) = λ u 1 ( x ) , u 1 ( x ) + p 2 ( x ) u 2 ( x ) = λ u 2 ( x ) , x ( 0 , 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ27_HTML.gif
(2.20)
u 1 ( 0 ) = 1 δ ϕ 11 ( 0 , λ ) , u 2 ( 0 ) = 1 δ ϕ 21 ( 0 , λ ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ28_HTML.gif
(2.21)

has a unique solution u = ( ϕ 12 ( x , λ ) ϕ 22 ( x , λ ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq81_HTML.gif for each λ C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq82_HTML.gif.

Similarly, the following problem also has a unique solution u = ( χ 11 ( x , λ ) χ 21 ( x , λ ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq83_HTML.gif:
u 2 ( x ) p 1 ( x ) u 1 ( x ) = λ u 1 ( x ) , u 1 ( x ) + p 2 ( x ) u 2 ( x ) = λ u 2 ( x ) , x ( 1 , 0 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ29_HTML.gif
(2.22)
u 1 ( 0 ) = δ χ 12 ( 0 , λ ) , u 2 ( 0 ) = δ χ 22 ( 0 , λ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ30_HTML.gif
(2.23)
Let us construct two basic solutions of equation (2.1) as follows:
ϕ ( , λ ) = ( ϕ 1 ( , λ ) ϕ 2 ( , λ ) ) , χ ( , λ ) = ( χ 1 ( , λ ) χ 2 ( , λ ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equi_HTML.gif
where
ϕ 1 ( x , λ ) = { ϕ 11 ( x , λ ) , x [ 1 , 0 ) , ϕ 12 ( x , λ ) , x ( 0 , 1 ] , ϕ 2 ( x , λ ) = { ϕ 21 ( x , λ ) , x [ 1 , 0 ) , ϕ 22 ( x , λ ) , x ( 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ31_HTML.gif
(2.24)
χ 1 ( x , λ ) = { χ 11 ( x , λ ) , x [ 1 , 0 ) , χ 12 ( x , λ ) , x ( 0 , 1 ] , χ 2 ( x , λ ) = { χ 21 ( x , λ ) , x [ 1 , 0 ) , χ 22 ( x , λ ) , x ( 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ32_HTML.gif
(2.25)
Then
R β ( χ ( x , λ ) ) = ρ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ33_HTML.gif
(2.26)

By virtue of equations (2.21) and (2.23), these solutions satisfy both transmission conditions (2.4) and (2.5). These functions are entire in λ for all x [ 1 , 0 ) ( 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq84_HTML.gif.

Let W ( ϕ , χ ) ( , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq85_HTML.gif denote the Wronskian of ϕ ( , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq86_HTML.gif and χ ( , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq87_HTML.gif defined in [[26], p.194], i.e.,
W ( ϕ , χ ) ( , λ ) : = | ϕ 1 ( , λ ) ϕ 2 ( , λ ) χ 1 ( , λ ) χ 2 ( , λ ) | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equj_HTML.gif
It is obvious that the Wronskian
ω i ( λ ) : = W ( ϕ , χ ) ( x , λ ) = ϕ 1 i ( x , λ ) χ 2 i ( x , λ ) ϕ 2 i ( x , λ ) χ 1 i ( x , λ ) , x Γ i , i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ34_HTML.gif
(2.27)
are independent of x Γ i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq88_HTML.gif and are entire functions. Taking into account (2.21) and (2.23), a short calculation gives
ω 1 ( λ ) = δ 2 ω 2 ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equk_HTML.gif

for each λ C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq28_HTML.gif.

Corollary 2.3 The zeros of the functions ω 1 ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq89_HTML.gif and ω 2 ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq90_HTML.gif coincide.

Then we may take into consideration the characteristic function ω ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq91_HTML.gif as
ω ( λ ) : = ω 1 ( λ ) = δ 2 ω 2 ( λ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ35_HTML.gif
(2.28)

In the following lemma, we show that all eigenvalues of problem (2.1)-(2.5) are simple.

Lemma 2.4 All eigenvalues of problem (2.1)-(2.5) are just zeros of the function ω ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq91_HTML.gif. Moreover, every zero of ω ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq91_HTML.gif has multiplicity one.

Proof Since the functions ϕ 1 ( x , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq92_HTML.gif and ϕ 2 ( x , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq93_HTML.gif satisfy the boundary condition (2.2) and both transmission conditions (2.4) and (2.5), to find the eigenvalues of the (2.1)-(2.5), we have to insert the functions ϕ 1 ( x , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq92_HTML.gif and ϕ 2 ( x , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq93_HTML.gif in the boundary condition (2.3) and find the roots of this equation.

By (2.1) we obtain for λ , μ C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq94_HTML.gif, λ μ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq95_HTML.gif,
d d x { ϕ 1 ( x , λ ) ϕ 2 ( x , μ ) ϕ 1 ( x , μ ) ϕ 2 ( x , λ ) } = ( μ λ ) { ϕ 1 ( x , λ ) ϕ 1 ( x , μ ) + ϕ 2 ( x , λ ) ϕ 2 ( x , μ ) } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equl_HTML.gif
Integrating the above equation through [ 1 , 0 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq53_HTML.gif and [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq54_HTML.gif, and taking into account the initial conditions (2.17), (2.21) and (2.23), we obtain
δ 2 ( ϕ 12 ( 1 , λ ) ϕ 22 ( 1 , μ ) ϕ 12 ( 1 , μ ) ϕ 22 ( 1 , λ ) ) = ( μ λ ) ( 1 0 ( ϕ 11 ( x , λ ) ϕ 11 ( x , μ ) + ϕ 21 ( x , λ ) ϕ 21 ( x , μ ) ) d x + δ 2 0 1 ( ϕ 12 ( x , λ ) ϕ 12 ( x , μ ) + ϕ 22 ( x , λ ) ϕ 22 ( x , μ ) ) d x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ36_HTML.gif
(2.29)
Dividing both sides of (2.29) by ( λ μ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq96_HTML.gif and by letting μ λ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq97_HTML.gif, we arrive to the relation
δ 2 ( ϕ 22 ( 1 , λ ) ϕ 12 ( 1 , λ ) λ ϕ 12 ( 1 , λ ) ϕ 22 ( 1 , λ ) λ ) = ( 1 0 ( | ϕ 11 ( x , λ ) | 2 + | ϕ 21 ( x , λ ) | 2 ) d x + δ 2 0 1 ( | ϕ 12 ( x , λ ) | 2 + | ϕ 22 ( x , λ ) | 2 ) d x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ37_HTML.gif
(2.30)
We show that equation
ω ( λ ) = W ( ϕ , χ ) ( 1 , λ ) = δ 2 ( ( a 1 + λ sin β ) ϕ 12 ( 1 , λ ) ( a 2 + λ cos β ) ϕ 22 ( 1 , λ ) ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ38_HTML.gif
(2.31)
has only simple roots. Assume the converse, i.e., equation (2.31) has a double root λ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq98_HTML.gif, say. Then the following two equations hold:
( a 1 + λ sin β ) ϕ 12 ( 1 , λ ) ( a 2 + λ cos β ) ϕ 22 ( 1 , λ ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ39_HTML.gif
(2.32)
sin β ϕ 12 ( 1 , λ ) + ( a 1 + λ sin β ) ϕ 12 ( 1 , λ ) λ cos β ϕ 22 ( 1 , λ ) ( a 2 + λ cos β ) ϕ 22 ( 1 , λ ) λ = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ40_HTML.gif
(2.33)
Since ρ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq99_HTML.gif and λ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq98_HTML.gif is real, then ( a 1 + λ sin β ) 2 + ( a 2 + λ cos β ) 2 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq100_HTML.gif. Let a 1 + λ sin β 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq101_HTML.gif. From (2.32) and (2.33),
ϕ 12 ( 1 , λ ) = ( a 2 + λ cos β ) ( a 1 + λ sin β ) ϕ 22 ( 1 , λ ) , ϕ 12 ( 1 , λ ) λ = ρ ϕ 22 ( 1 , λ ) ( a 1 + λ sin β ) 2 + ( a 2 + λ cos β ) ( a 1 + λ sin β ) ϕ 22 ( 1 , λ ) λ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ41_HTML.gif
(2.34)
Combining (2.34) and (2.30) with λ = λ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq102_HTML.gif, we obtain
ρ δ 2 ( ϕ 22 ( 1 , λ ) ) 2 ( a 1 + λ sin β ) 2 = ( 1 0 ( | ϕ 11 ( x , λ ) | 2 + | ϕ 21 ( x , λ ) | 2 ) d x + δ 2 0 1 ( | ϕ 12 ( x , λ ) | 2 + | ϕ 22 ( x , λ ) | 2 ) d x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ42_HTML.gif
(2.35)

contradicting the assumption ρ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq57_HTML.gif. The other case, when a 2 + λ cos β 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq103_HTML.gif, can be treated similarly and the proof is complete. □

Let { λ n } n = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq104_HTML.gif denote the sequence of zeros of ω ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq91_HTML.gif. Then
Φ ( x , λ n ) : = ( ϕ ( x , λ n ) R β ( ϕ ( x , λ n ) ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ43_HTML.gif
(2.36)
are the corresponding eigenvectors of the operator A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq67_HTML.gif. Since A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq67_HTML.gif is symmetric, then it is easy to show that the following orthogonality relation holds:
Φ ( , λ n ) , Φ ( , λ m ) H = 0 for  n m . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ44_HTML.gif
(2.37)
Here { ϕ ( , λ n ) } n = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq105_HTML.gif is a sequence of eigen-vector-functions of (2.1)-(2.5) corresponding to the eigenvalues { λ n } n = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq104_HTML.gif. We denote by Ψ ( x , λ n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq106_HTML.gif the normalized eigenvectors of A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq67_HTML.gif, i.e.,
Ψ ( x , λ n ) : = Φ ( x , λ n ) Φ ( , λ n ) H = ( ψ ( x , λ n ) R β ( ψ ( x , λ n ) ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ45_HTML.gif
(2.38)
Since χ ( , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq107_HTML.gif satisfies (2.3)-(2.5), then the eigenvalues are also determined via
sin α χ 11 ( 1 , λ ) cos α χ 21 ( 1 , λ ) = ω ( λ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ46_HTML.gif
(2.39)
Therefore { χ ( , λ n ) } n = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq108_HTML.gif is another set of eigen-vector-functions which is related by { ϕ ( , λ n ) } n = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq105_HTML.gif with
χ ( x , λ n ) = c n ϕ ( x , λ n ) , x [ 1 , 0 ) ( 0 , 1 ] , n Z , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ47_HTML.gif
(2.40)

where c n 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq109_HTML.gif are non-zero constants, since all eigenvalues are simple. Since the eigenvalues are all real, we can take the eigen-vector-functions to be real-valued.

Now we derive the asymptotic formulae of the eigenvalues { λ n } n = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq104_HTML.gif and the eigen-vector-functions { ϕ ( , λ n ) } n = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq105_HTML.gif. We transform equations (2.1), (2.17), (2.21) and (2.24) into the integral equations, see [26], as follows:
ϕ 11 ( x , λ ) = cos ( λ ( x + 1 ) + α ) 1 x sin λ ( x t ) p 1 ( t ) ϕ 11 ( t , λ ) d t 1 x cos λ ( x t ) p 2 ( t ) ϕ 21 ( t , λ ) d t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ48_HTML.gif
(2.41)
ϕ 21 ( x , λ ) = sin ( λ ( x + 1 ) + α ) + 1 x cos λ ( x t ) p 1 ( t ) ϕ 11 ( t , λ ) d t 1 x sin λ ( x t ) p 2 ( t ) ϕ 21 ( t , λ ) d t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ49_HTML.gif
(2.42)
ϕ 12 ( x , λ ) = 1 δ ϕ 21 ( 0 , λ ) sin ( λ x ) + 1 δ ϕ 11 ( 0 , λ ) cos ( λ x ) 0 x sin λ ( x t ) p 1 ( t ) ϕ 12 ( t , λ ) d t 0 x cos λ ( x t ) p 2 ( t ) ϕ 22 ( t , λ ) d t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ50_HTML.gif
(2.43)
ϕ 22 ( x , λ ) = 1 δ ϕ 11 ( 0 , λ ) sin ( λ x ) + 1 δ ϕ 21 ( 0 , λ ) cos ( λ x ) + 0 x cos λ ( x t ) p 1 ( t ) ϕ 12 ( t , λ ) d t ϕ 22 ( x , λ ) = 0 x sin λ ( x t ) p 2 ( t ) ϕ 22 ( t , λ ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ51_HTML.gif
(2.44)
For | λ | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq110_HTML.gif the following estimates hold uniformly with respect to x, x [ 1 , 0 ) ( 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq84_HTML.gif, cf. [[25], p.55],
ϕ 11 ( x , λ ) = cos ( λ ( x + 1 ) + α ) + O ( 1 λ ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ52_HTML.gif
(2.45)
ϕ 21 ( x , λ ) = sin ( λ ( x + 1 ) + α ) + O ( 1 λ ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ53_HTML.gif
(2.46)
ϕ 12 ( x , λ ) = 1 δ ϕ 21 ( 0 , λ ) sin ( λ x ) + 1 δ ϕ 11 ( 0 , λ ) cos ( λ x ) + O ( 1 λ ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ54_HTML.gif
(2.47)
ϕ 22 ( x , λ ) = 1 δ ϕ 11 ( 0 , λ ) sin ( λ x ) + 1 δ ϕ 21 ( 0 , λ ) cos ( λ x ) + O ( 1 λ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ55_HTML.gif
(2.48)
Now we find an asymptotic formula of the eigenvalues. Since the eigenvalues of the boundary value problem (2.1)-(2.5) coincide with the roots of the equation
( a 1 + λ sin β ) ϕ 12 ( 1 , λ ) ( a 2 + λ cos β ) ϕ 22 ( 1 , λ ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ56_HTML.gif
(2.49)
then from the estimates (2.47), (2.48) and (2.49), we get
λ sin β [ 1 δ ϕ 21 ( 0 , λ ) sin λ + 1 δ ϕ 11 ( 0 , λ ) cos λ ] λ cos β [ 1 δ ϕ 11 ( 0 , λ ) sin λ + 1 δ ϕ 21 ( 0 , λ ) cos λ ] + O ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equm_HTML.gif
which can be written as
1 δ ϕ 11 ( 0 , λ ) sin ( λ β ) + 1 δ ϕ 21 ( 0 , λ ) cos ( λ β ) + O ( 1 λ ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ57_HTML.gif
(2.50)
Then, from (2.45) and (2.46), equation (2.50) has the form
sin ( 2 λ + α β ) + O ( 1 λ ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ58_HTML.gif
(2.51)
For large | λ | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq111_HTML.gif, equation (2.51) obviously has solutions which, as is not hard to see, have the form
2 λ n + α β = n π + δ n , n = 0 , ± 1 , ± 2 , . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ59_HTML.gif
(2.52)
Inserting these values in (2.51), we find that sin δ n = O ( 1 n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq112_HTML.gif, i.e., δ n = O ( 1 n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq113_HTML.gif. Thus we obtain the following asymptotic formula for the eigenvalues:
λ n = n π + β α 2 + O ( 1 n ) , n = 0 , ± 1 , ± 2 , . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ60_HTML.gif
(2.53)
Using the formulae (2.53), we obtain the following asymptotic formulae for the eigen-vector-functions ϕ ( , λ n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq114_HTML.gif:
ϕ ( x , λ n ) = { ( cos ( λ n ( x + 1 ) + α ) + O ( 1 n ) sin ( λ n ( x + 1 ) + α ) + O ( 1 n ) ) , x [ 1 , 0 ) , ( 1 δ cos ( λ n ( x + 1 ) + α ) + O ( 1 n ) 1 δ sin ( λ n ( x + 1 ) + α ) + O ( 1 n ) ) , x ( 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ61_HTML.gif
(2.54)
where
ϕ ( x , λ n ) = { ( ϕ 11 ( x , λ n ) ϕ 21 ( x , λ n ) ) , x [ 1 , 0 ) , ( ϕ 12 ( x , λ n ) ϕ 22 ( x , λ n ) ) , x ( 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ62_HTML.gif
(2.55)

3 Green’s matrix and expansion theorem

Let F ( ) = ( f ( ) w ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq115_HTML.gif, where f ( ) = ( f 1 ( ) f 2 ( ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq116_HTML.gif, be a continuous vector-valued function. To study the completeness of the eigenvectors of A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq67_HTML.gif, and hence the completeness of the eigen-vector-functions of (2.1)-(2.5), we derive Green’s function of problem (2.1)-(2.5) as well as the resolvent of A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq67_HTML.gif. Indeed, let λ be not an eigenvalue of A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq67_HTML.gif and consider the inhomogeneous problem
( A λ I ) U ( x ) = F ( x ) , U ( x ) = ( u ( x ) R β ( u ( x ) ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equn_HTML.gif
where I is the identity operator. Since
( A λ I ) U ( x ) = ( ( u ) R a ( u ( x ) ) ) λ ( u ( x ) R β ( u ( x ) ) ) = ( f ( x ) w ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equo_HTML.gif
then we have
u 2 ( x ) { p 1 ( x ) + λ } u 1 ( x ) = f 1 ( x ) , u 1 ( x ) + { p 2 ( x ) + λ } u 2 ( x ) = f 2 ( x ) , x [ 1 , 0 ) ( 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ63_HTML.gif
(3.1)
w = R a ( u ( x ) ) λ R β ( u ( x ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ64_HTML.gif
(3.2)

and the boundary conditions (2.2), (2.4) and (2.5) with λ is not an eigenvalue of problem (2.1)-(2.5).

Now, we can represent the general solution of (3.1) in the following form:
u ( x , λ ) = { A 1 ( ϕ 11 ( x , λ ) ϕ 21 ( x , λ ) ) + B 1 ( χ 11 ( x , λ ) χ 21 ( x , λ ) ) , x [ 1 , 0 ) , A 2 ( ϕ 12 ( x , λ ) ϕ 22 ( x , λ ) ) + B 2 ( χ 12 ( x , λ ) χ 22 ( x , λ ) ) , x ( 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ65_HTML.gif
(3.3)
We applied the standard method of variation of the constants to (3.3), and thus the functions A 1 ( x , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq117_HTML.gif, B 1 ( x , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq118_HTML.gif and A 2 ( x , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq119_HTML.gif, B 2 ( x , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq120_HTML.gif satisfy the linear system of equations
A 1 ( x , λ ) ϕ 21 ( x , λ ) + B 1 ( x , λ ) χ 21 ( x , λ ) = f 1 ( x ) , A 1 ( x , λ ) ϕ 11 ( x , λ ) + B 1 ( x , λ ) χ 11 ( x , λ ) = f 2 ( x ) , x [ 1 , 0 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ66_HTML.gif
(3.4)
and
A 2 ( x , λ ) ϕ 22 ( x , λ ) + B 2 ( x , λ ) χ 22 ( x , λ ) = f 1 ( x ) , A 2 ( x , λ ) ϕ 12 ( x , λ ) + B 2 ( x , λ ) χ 12 ( x , λ ) = f 2 ( x ) , x ( 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ67_HTML.gif
(3.5)
Since λ is not an eigenvalue and ω ( λ ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq121_HTML.gif, each of the linear system in (3.4) and (3.5) has a unique solution which leads to
A 1 ( x , λ ) = 1 ω ( λ ) x 0 χ ( ξ , λ ) f ( ξ ) d ξ + A 1 , B 1 ( x , λ ) = 1 ω ( λ ) 1 x ϕ ( ξ , λ ) f ( ξ ) d ξ + B 1 , x [ 1 , 0 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ68_HTML.gif
(3.6)
A 2 ( x , λ ) = δ 2 ω ( λ ) x 1 χ ( ξ , λ ) f ( ξ ) d ξ + A 2 , B 2 ( x , λ ) = δ 2 ω ( λ ) 0 x ϕ ( ξ , λ ) f ( ξ ) d ξ + B 2 , x ( 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ69_HTML.gif
(3.7)
where A 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq122_HTML.gif, A 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq123_HTML.gif, B 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq124_HTML.gif and B 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq125_HTML.gif are arbitrary constants, and
ϕ ( ξ , λ ) = { ( ϕ 11 ( ξ , λ ) ϕ 21 ( ξ , λ ) ) , ξ [ 1 , 0 ) , ( ϕ 12 ( ξ , λ ) ϕ 22 ( ξ , λ ) ) , ξ ( 0 , 1 ] , χ ( ξ , λ ) = { ( χ 11 ( ξ , λ ) χ 21 ( ξ , λ ) ) , ξ [ 1 , 0 ) , ( χ 12 ( ξ , λ ) χ 22 ( ξ , λ ) ) , ξ ( 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equp_HTML.gif
Substituting equations (3.6) and (3.7) into (3.3), we obtain the solution of (3.1)
u ( x , λ ) = { ϕ ( x , λ ) ω ( λ ) x 0 χ ( ξ , λ ) f ( ξ ) d ξ + χ ( x , λ ) ω ( λ ) 1 x ϕ ( ξ , λ ) f ( ξ ) d ξ + A 1 ϕ ( x , λ ) + B 1 χ ( x , λ ) , x [ 1 , 0 ) , δ 2 ϕ ( x , λ ) ω ( λ ) x 1 χ ( ξ , λ ) f ( ξ ) d ξ + δ 2 χ ( x , λ ) ω ( λ ) 0 x ϕ ( ξ , λ ) f ( ξ ) d ξ + A 2 ϕ ( x , λ ) + B 2 χ ( x , λ ) , x ( 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ70_HTML.gif
(3.8)
Then from (2.2), (3.2) and the transmission conditions (2.4) and (2.5), we get
A 1 = δ 2 ω ( λ ) 0 1 χ ( ξ , λ ) f ( ξ ) d ξ w δ 2 ω ( λ ) , B 1 = 0 , A 2 = w δ 2 ω ( λ ) , B 2 = 1 ω ( λ ) 1 0 ϕ ( ξ , λ ) f ( ξ ) d ξ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equq_HTML.gif
Then (3.8) can be written as
u ( x , λ ) = w δ 2 ω ( λ ) ϕ ( x , λ ) + χ ( x , λ ) ω ( λ ) 1 x a ( ξ ) ϕ ( ξ , λ ) f ( ξ ) d ξ + ϕ ( x , λ ) ω ( λ ) x 1 a ( ξ ) χ ( ξ , λ ) f ( ξ ) d ξ , x , ξ [ 1 , 0 ) ( 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ71_HTML.gif
(3.9)
where
a ( ξ ) = { 1 , ξ [ 1 , 0 ) , δ 2 , ξ ( 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ72_HTML.gif
(3.10)
which can be written as
u ( x , λ ) = w δ 2 ω ( λ ) ϕ ( x , λ ) + 1 1 a ( ξ ) G ( x , ξ , λ ) f ( ξ ) d ξ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ73_HTML.gif
(3.11)
where
G ( x , ξ , λ ) = 1 ω ( λ ) { χ ( x , λ ) ϕ ( ξ , λ ) , 1 ξ x 1 , x 0 , ξ 0 , ϕ ( x , λ ) χ ( ξ , λ ) , 1 x ξ 1 , x 0 , ξ 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ74_HTML.gif
(3.12)
Expanding (3.12) we obtain the concrete form
G ( x , ξ , λ ) = 1 ω ( λ ) { ( ϕ 1 ( ξ , λ ) χ 1 ( x , λ ) ϕ 2 ( ξ , λ ) χ 1 ( x , λ ) ϕ 1 ( ξ , λ ) χ 2 ( x , λ ) ϕ 2 ( ξ , λ ) χ 2 ( x , λ ) ) , 1 ξ x 1 , x 0 , ξ 0 , ( ϕ 1 ( x , λ ) χ 1 ( ξ , λ ) ϕ 1 ( x , λ ) χ 2 ( ξ , λ ) ϕ 2 ( x , λ ) χ 1 ( ξ , λ ) ϕ 2 ( x , λ ) χ 2 ( ξ , λ ) ) , 1 x ξ 1 , x 0 , ξ 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ75_HTML.gif
(3.13)
The matrix G ( x , ξ , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq126_HTML.gif is called Green’s matrix of problem (2.1)-(2.5). Obviously, G ( x , ξ , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq126_HTML.gif is a meromorphic function of λ, for every ( x , ξ ) ( [ 1 , 0 ) ( 0 , 1 ] ) 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq127_HTML.gif, which has simple poles only at the eigenvalues. Although Green’s matrix looks as simple as that of Dirac systems, cf., e.g., [25, 26], it is rather complicated because of the transmission conditions (see the example at the end of this paper). Therefore
U ( x ) = ( A λ I ) 1 F ( x ) = ( w δ 2 ω ( λ ) ϕ ( x , λ ) + 1 1 a ( ξ ) G ( x , ξ , λ ) f ( ξ ) d ξ R β ( u ( x ) ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ76_HTML.gif
(3.14)

Lemma 3.1 The operator A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq67_HTML.gif is self-adjoint in H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq68_HTML.gif.

Proof Since A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq67_HTML.gif is a symmetric densely defined operator, then it is sufficient to show that the deficiency spaces are the null spaces, and hence A = A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq128_HTML.gif. Indeed, if F ( x ) = ( f ( x ) w ) H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq129_HTML.gif and λ is a non-real number, then taking
U ( x ) = ( u ( x ) z ) = ( w δ 2 ω ( λ ) ϕ ( x , λ ) + 1 1 a ( ξ ) G ( x , ξ , λ ) f ( ξ ) d ξ R β ( u ( x ) ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equr_HTML.gif
implies that U D ( A ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq130_HTML.gif. Since G ( x , ξ , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq126_HTML.gif satisfies the conditions (2.2)-(2.5), then ( A λ I ) U ( x ) = F ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq131_HTML.gif. Now we prove that the inverse of ( A λ I ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq132_HTML.gif exists. If A U ( x ) = λ U ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq133_HTML.gif, then
( λ ¯ λ ) U ( ) , U ( ) H = U ( ) , λ U ( ) H λ U ( ) , U ( ) H = U ( ) , A U ( ) H A U ( ) , U ( ) H = 0 ( since  A  is symmetric ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equs_HTML.gif
Since λ R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq134_HTML.gif, we have λ ¯ λ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq135_HTML.gif. Thus U ( ) , U ( ) H = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq136_HTML.gif, i.e., U = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq137_HTML.gif. Then R ( λ ; A ) : = ( A λ I ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq138_HTML.gif, the resolvent operator of A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq67_HTML.gif, exists. Thus
R ( λ ; A ) F = ( A λ I ) 1 F = U . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equt_HTML.gif
Take λ = ± i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq139_HTML.gif. The domains of ( A i I ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq140_HTML.gif and ( A + i I ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq141_HTML.gif are exactly H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq68_HTML.gif. Consequently, the ranges of ( A i I ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq142_HTML.gif and ( A + i I ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq143_HTML.gif are also H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq68_HTML.gif. Hence the deficiency spaces of A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq67_HTML.gif are
N i : = N ( A + i I ) = R ( A i I ) = H 1 = { 0 } , N i : = N ( A i I ) = R ( A + i I ) = H 1 = { 0 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equu_HTML.gif

Hence A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq67_HTML.gif is self-adjoint. □

The next theorem is an eigenfunction expansion theorem. The proof is exactly similar to that of Levitan and Sargsjan derived in [[25], pp.67-77]; see also [2629].

Theorem 3.2
  1. (i)
    For U ( ) H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq144_HTML.gif,
    U ( ) H 2 = n = | U ( ) , Ψ n ( ) H | 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ77_HTML.gif
    (3.15)
     
  2. (ii)
    For U ( ) D ( A ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq145_HTML.gif,
    U ( x ) = n = U ( ) , Ψ n ( ) H Ψ n ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ78_HTML.gif
    (3.16)
     

the series being absolutely and uniformly convergent in the first component for on [ 1 , 0 ) ( 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq45_HTML.gif, and absolutely convergent in the second component.

4 The sampling theorems

The first sampling theorem of this section associated with the boundary value problem (2.1)-(2.5) is the following theorem.

Theorem 4.1 Let f ( x ) = ( f 1 ( x ) f 2 ( x ) ) H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq146_HTML.gif. For λ C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq28_HTML.gif, let
F ( λ ) = 1 0 f ( x ) ϕ ( x , λ ) d x + δ 2 0 1 f ( x ) ϕ ( x , λ ) d x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ79_HTML.gif
(4.1)
where ϕ ( , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq86_HTML.gif is the solution defined above. Then F ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq147_HTML.gif is an entire function of exponential type that can be reconstructed from its values at the points { λ n } n = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq104_HTML.gif via the sampling formula
F ( λ ) = n = F ( λ n ) ω ( λ ) ( λ λ n ) ω ( λ n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ80_HTML.gif
(4.2)

The series (4.2) converges absolutely onand uniformly on any compact subset of ℂ, and ω ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq91_HTML.gif is the entire function defined in (2.28).

Proof The relation (4.1) can be rewritten in the form
F ( λ ) = F ( ) , Φ ( , λ ) H = 1 0 f ( x ) ϕ ( x , λ ) d x + δ 2 0 1 f ( x ) ϕ ( x , λ ) d x , λ C , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ81_HTML.gif
(4.3)
where
F ( x ) = ( f ( x ) 0 ) , Φ ( x , λ ) = ( ϕ ( x , λ ) R β ( ϕ ( x , λ ) ) ) H . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equv_HTML.gif
Since both F ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq148_HTML.gif and Φ ( , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq149_HTML.gif are in H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq68_HTML.gif, then they have the Fourier expansions
F ( x ) = n = f ˆ ( n ) Φ ( x , λ n ) Φ ( , λ n ) H 2 , Φ ( x , λ ) = n = Φ ( , λ ) , Φ ( , λ n ) H Φ ( x , λ n ) Φ ( , λ n ) H 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ82_HTML.gif
(4.4)
where λ C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq28_HTML.gif and f ˆ ( n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq150_HTML.gif are the Fourier coefficients
f ˆ ( n ) = F ( ) , Φ ( , λ n ) H = 1 0 f ( x ) ϕ ( x , λ n ) d x + δ 2 0 1 f ( x ) ϕ ( x , λ n ) d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ83_HTML.gif
(4.5)
Applying Parseval’s identity to (4.3), we obtain
F ( λ ) = n = F ( λ n ) Φ ( , λ ) , Φ ( , λ n ) H Φ ( , λ n ) H 2 , λ C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ84_HTML.gif
(4.6)
Now we calculate Φ ( , λ ) , Φ ( , λ n ) H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq151_HTML.gif and Φ ( , λ n ) H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq152_HTML.gif of λ C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq28_HTML.gif, n Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq153_HTML.gif. To prove expansion (4.2), we need to show that
Φ ( , λ ) , Φ ( , λ n ) H Φ ( , λ n ) H 2 = ω ( λ ) ( λ λ n ) ω ( λ ) , n Z , λ C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ85_HTML.gif
(4.7)
Indeed, let λ C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq28_HTML.gif and n Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq153_HTML.gif be fixed. By the definition of the inner product of H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq68_HTML.gif, we have
Φ ( , λ ) , Φ ( , λ n ) H = 1 0 ϕ ( x , λ ) ϕ ( x , λ n ) d x + δ 2 0 1 ϕ ( x , λ ) ϕ ( x , λ n ) d x + δ 2 ρ R β ( ϕ ( x , λ ) ) R β ( ϕ ( x , λ n ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ86_HTML.gif
(4.8)
From Green’s identity, see [[25], p.51], we have
( λ n λ ) [ 1 0 ϕ ( x , λ ) ϕ ( x , λ n ) d x + δ 2 0 1 ϕ ( x , λ ) ϕ ( x , λ n ) d x ] = W ( ϕ ( 0 , λ ) , ϕ ( 0 , λ n ) ) W ( ϕ ( 1 , λ ) , ϕ ( 1 , λ n ) ) δ 2 W ( ϕ ( 0 + , λ ) , ϕ ( 0 + , λ n ) ) + δ 2 W ( ϕ ( 1 , λ ) , ϕ ( 1 , λ n ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ87_HTML.gif
(4.9)
Then (4.9) and the initial conditions (2.21) imply
1 0 ϕ ( x , λ ) ϕ ( x , λ n ) d x + δ 2 0 1 ϕ ( x , λ ) ϕ ( x , λ n ) d x = δ 2 W ( ϕ ( 1 , λ ) , ϕ ( 1 , λ n ) ) λ n λ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ88_HTML.gif
(4.10)
From (2.40), (2.19) and (2.7), we have
W ( ϕ ( 1 , λ ) , ϕ ( 1 , λ n ) ) = ϕ 12 ( 1 , λ ) ϕ 22 ( 1 , λ n ) ϕ 22 ( 1 , λ ) ϕ 12 ( 1 , λ n ) = c n 1 [ ϕ 12 ( 1 , λ ) χ 22 ( 1 , λ n ) ϕ 22 ( 1 , λ ) χ 12 ( 1 , λ n ) ] = c n 1 [ ( λ n sin β + a 1 ) ϕ 12 ( 1 , λ ) ( λ n cos β + a 2 ) ϕ 22 ( 1 , λ ) ] = c n 1 [ δ 2 ω ( λ ) + ( λ n λ ) R β ( ϕ ( x , λ ) ) ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ89_HTML.gif
(4.11)
Also, from (2.40) we have
δ 2 ρ R β ( ϕ ( x , λ ) ) R β ( ϕ ( x , λ n ) ) = δ 2 c n 1 ρ R β ( ϕ ( x , λ ) ) R β ( χ ( x , λ n ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ90_HTML.gif
(4.12)
Then from (2.26) and (4.12) we obtain
δ 2 ρ R β ( ϕ ( x , λ ) ) R β ( ϕ ( x , λ n ) ) = δ 2 c n 1 R β ( ϕ ( x , λ ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ91_HTML.gif
(4.13)
Substituting from (4.10), (4.11) and (4.13) into (4.8), we get
Φ ( , λ ) , Φ ( , λ n ) H = c n 1 ω ( λ ) λ n λ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ92_HTML.gif
(4.14)
Letting λ λ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq154_HTML.gif in (4.14), since the zeros of ω ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq91_HTML.gif are simple, we get
Φ ( , λ n ) , Φ ( , λ n ) H = Φ ( , λ n ) H 2 = c n 1 ω ( λ n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ93_HTML.gif
(4.15)
Since λ C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq28_HTML.gif and n Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq153_HTML.gif are arbitrary, then (4.14) and (4.15) hold for all λ C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq82_HTML.gif and all n Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq153_HTML.gif. Therefore, from (4.14) and (4.15), we get (4.7). Hence (4.2) is proved with a pointwise convergence on ℂ. Now we investigate the convergence of (4.2). First we prove that it is absolutely convergent on ℂ. Using Cauchy-Schwarz’ inequality for λ C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq28_HTML.gif,
k = | F ( λ k ) ω ( λ ) ( λ λ k ) ω ( λ k ) | ( k = | F ( ) , Φ ( , λ k ) H | 2 Φ ( , λ k ) H 2 ) 1 / 2 × ( k = | Φ ( , λ ) , Φ ( , λ k ) H | 2 Φ ( , λ k ) H 2 ) 1 / 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ94_HTML.gif
(4.16)
Since F ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq148_HTML.gif, Φ ( , λ ) H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq155_HTML.gif, then the two series on the right-hand side of (4.16) converge. Thus series (4.2) converges absolutely on ℂ. As for uniform convergence, let M C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq156_HTML.gif be compact. Let λ M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq157_HTML.gif and N > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq158_HTML.gif. Define ν N ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq159_HTML.gif to be
ν N ( λ ) : = | F ( λ ) k = N N F ( λ k ) ω ( λ ) ( λ λ k ) ω ( λ k ) | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ95_HTML.gif
(4.17)
Using the same method developed above, we get
ν N ( λ ) ( k = N N | F ( ) , Φ ( , λ k ) H | 2 Φ ( , λ k ) H 2 ) 1 / 2 ( k = N N | Φ ( , λ ) , Φ ( , λ k ) H | 2 Φ ( , λ k ) H 2 ) 1 / 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ96_HTML.gif
(4.18)
Therefore
ν N ( λ ) Φ ( , λ k ) H ( k = N N | F ( ) , Φ ( , λ k ) H | 2 Φ ( , λ k ) H 2 ) 1 / 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ97_HTML.gif
(4.19)
Since [ 1 , 1 ] × M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq160_HTML.gif is compact, then we can find a positive constant C M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq161_HTML.gif such that
Φ ( , λ ) H C M for all  λ M . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ98_HTML.gif
(4.20)
Then
ν N ( λ ) C M ( k = N N | F ( ) , Φ ( , λ k ) H | 2 Φ ( , λ k ) H 2 ) 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ99_HTML.gif
(4.21)
uniformly on M. In view of Parseval’s equality,
( k = N N | F ( ) , Φ ( , λ k ) H | 2 Φ ( , λ k ) H 2 ) 1 / 2 0 as  N . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equw_HTML.gif
Thus ν N ( λ ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq162_HTML.gif uniformly on M. Hence (4.2) converges uniformly on M. Thus F ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq147_HTML.gif is an entire function. From the relation
| F ( λ ) | 1 0 | f 1 ( x ) | | ϕ 11 ( x , λ ) | d x + 1 0 | f 2 ( x ) | | ϕ 21 ( x , λ ) | d x + δ 2 0 1 | f 1 ( x ) | | ϕ 12 ( x , λ ) | d x + δ 2 0 1 | f 2 ( x ) | | ϕ 22 ( x , λ ) | d x , λ C , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equx_HTML.gif

and the fact that ϕ i j ( , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq163_HTML.gif, i , j = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq164_HTML.gif, are entire functions of exponential type, we conclude that F ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq147_HTML.gif is of exponential type. □

Remark 4.2 To see that expansion (4.2) is a Lagrange-type interpolation, we may replace ω ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq91_HTML.gif by the canonical product
ω ˜ ( λ ) = ( λ λ 0 ) n = 1 ( 1 λ λ n ) ( 1 λ λ n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ100_HTML.gif
(4.22)
From Hadamard’s factorization theorem, see [4], ω ( λ ) = h ( λ ) ω ˜ ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq165_HTML.gif, where h ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq166_HTML.gif is an entire function with no zeros. Thus,
ω ( λ ) ω ( λ n ) = h ( λ ) ω ˜ ( λ ) h ( λ n ) ω ˜ ( λ n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equy_HTML.gif
and (4.1), (4.2) remain valid for the function F ( λ ) / h ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq167_HTML.gif. Hence
F ( λ ) = n = 0 F ( λ n ) h ( λ ) ω ˜ ( λ ) h ( λ n ) ω ˜ ( λ n ) ( λ λ n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ101_HTML.gif
(4.23)
We may redefine (4.1) by taking kernel ϕ ( , λ ) h ( λ ) = ϕ ˜ ( , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq168_HTML.gif to get
F ˜ ( λ ) = F ( λ ) h ( λ ) = n = 0 F ˜ ( λ n ) ω ˜ ( λ ) ( λ λ n ) ω ˜ ( λ n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ102_HTML.gif
(4.24)

The next theorem is devoted to giving vector-type interpolation sampling expansions associated with problem (2.1)-(2.5) for integral transforms whose kernels are defined in terms of Green’s matrix. As we see in (3.12), Green’s matrix G ( x , ξ , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq169_HTML.gif of problem (2.1)-(2.5) has simple poles at { λ k } k = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq170_HTML.gif. Define the function G ( x , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq171_HTML.gif to be G ( x , λ ) : = ω ( λ ) G ( x , ξ 0 , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq172_HTML.gif, where ξ 0 [ 1 , 0 ) ( 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq173_HTML.gif is a fixed point and ω ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq91_HTML.gif is the function defined in (2.28) or it is the canonical product (4.22).

Theorem 4.3 Let f ( x ) = ( f 1 ( x ) f 2 ( x ) ) H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq174_HTML.gif. Let F ( λ ) = ( F 1 ( λ ) F 2 ( λ ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq175_HTML.gif be the vector-valued transform
F ( λ ) = 1 0 G ( x , λ ) f ¯ ( x ) d x + δ 2 0 1 G ( x , λ ) f ¯ ( x ) d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ103_HTML.gif
(4.25)
Then F ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq176_HTML.gif is a vector-valued entire function of exponential type that admits the vector-valued sampling expansion
F ( λ ) = n = F ( λ n ) ω ( λ ) ( λ λ n ) ω ( λ n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ104_HTML.gif
(4.26)
The vector-valued series (4.26) converges absolutely onand uniformly on compact subsets of ℂ. Here (4.26) means
F 1 ( λ ) = n = F 1 ( λ n ) ω ( λ ) ( λ λ n ) ω ( λ n ) , F 2 ( λ ) = n = F 2 ( λ n ) ω ( λ ) ( λ λ n ) ω ( λ n ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ105_HTML.gif
(4.27)

where both series converge absolutely onand uniformly on compact sets of ℂ.

Proof The integral transform (4.25) can be written as
F ( λ ) = G ( , λ ) , F ( ) H , F ( x ) = ( f ( x ) 0 ) , G ( x , λ ) = ( G ( x , λ ) R β ( G ( x , λ ) ) ) H . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ106_HTML.gif
(4.28)
Applying Parseval’s identity to (4.28) with respect to { Φ ( , λ n ) } n = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq177_HTML.gif, we obtain
F ( λ ) = n = G ( , λ ) , Φ ( , λ n ) H F ( ) , Φ ( , λ n ) ¯ H Φ ( , λ n ) H 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ107_HTML.gif
(4.29)
Let λ C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq28_HTML.gif be such that λ λ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq178_HTML.gif for n Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq153_HTML.gif. Since each Φ ( , λ n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq179_HTML.gif is an eigenvector of A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq67_HTML.gif, then
( A λ I ) Φ ( x , λ n ) = ( λ n λ ) Φ ( x , λ n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equz_HTML.gif
Thus
( A λ I ) 1 Φ ( x , λ n ) = 1 λ n λ Φ ( x , λ n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ108_HTML.gif
(4.30)
From (3.14) and (4.30) we obtain
δ 2 R β ( ϕ ( x , λ n ) ) ω ( λ ) ϕ ( ξ 0 , λ ) + 1 1 a ( x ) G ( x , ξ 0 , λ ) ϕ ( x , λ n ) d x = 1 λ n λ ϕ ( ξ 0 , λ n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ109_HTML.gif
(4.31)
Then from (2.26) and (2.40) in (4.31), we get
ρ δ 2 c n 1 ω ( λ ) ϕ ( ξ 0 , λ ) + 1 1 a ( x ) G ( x , ξ 0 , λ ) ϕ ( x , λ n ) d x = 1 λ n λ ϕ ( ξ 0 , λ n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ110_HTML.gif
(4.32)
Hence equation (4.32) can be rewritten as
ρ δ 2 c n 1 ϕ ( ξ 0 , λ ) + 1 0 G ( x , λ ) ϕ ( x , λ n ) d x + δ 2 0 1 G ( x , λ ) ϕ ( x , λ n ) d x = ω ( λ ) λ n λ ϕ ( ξ 0 , λ n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ111_HTML.gif
(4.33)
The definition of G ( , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq180_HTML.gif implies
G ( , λ ) , Φ ( , λ n ) H = 1 0 G ( x , λ ) ϕ ( x , λ n ) d x + δ 2 0 1 G ( x , λ ) ϕ ( x , λ n ) d x + 1 ρ R β ( G ( x , λ ) ) R β ( ϕ ( x , λ n ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ112_HTML.gif
(4.34)
Moreover, from (3.12) we have
R β ( G ( x , λ ) ) = ϕ ( ξ 0 , λ ) R β ( χ ( x , λ ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ113_HTML.gif
(4.35)
Then from (4.35), (2.26) and (2.40) in (4.34), we obtain
G ( , λ ) , Φ ( , λ n ) H = 1 0 G ( x , λ ) ϕ ( x , λ n ) d x + δ 2 0 1 G ( x , λ ) ϕ ( x , λ n ) d x + ρ δ 2 c n 1 ϕ ( ξ 0 , λ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ114_HTML.gif
(4.36)
Combining (4.36) and (4.33), yields
G ( , λ ) , Φ ( , λ n ) H = ω ( λ ) λ n λ ϕ ( ξ 0 , λ n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ115_HTML.gif
(4.37)
Taking the limit when λ λ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq181_HTML.gif in (4.28), we get
F ( λ n ) = lim λ λ n G ( , λ ) , F ( ) H = lim λ λ n k = G ( , λ ) , Φ ( , λ k ) H Φ ( , λ k ) , F ( ) H Φ ( , λ k ) H 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ116_HTML.gif
(4.38)
Making use of (4.37), we may rewrite (4.38) as, ξ 0 [ 1 , 0 ) ( 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq173_HTML.gif,
F ( λ n ) = ( F 1 ( λ n ) F 2 ( λ n ) ) = ( lim λ λ n k = ω ( λ ) λ k λ ϕ 1 ( ξ 0 , λ k ) Φ ( , λ k ) , F ( ) H Φ ( , λ k ) H 2 lim λ λ n k = ω ( λ ) λ k λ ϕ 2 ( ξ 0 , λ k ) Φ ( , λ k ) , F ( ) H Φ ( , λ k ) H 2 ) = ( ω ( λ n ) ϕ 1 ( ξ 0 , λ n ) Φ ( , λ n ) , F ( ) H Φ ( , λ n ) H 2 ω ( λ n ) ϕ 2 ( ξ 0 , λ n ) Φ ( , λ n ) , F ( ) H Φ ( , λ n ) H 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ117_HTML.gif
(4.39)
The interchange of the limit and summation is justified by the asymptotic behavior of Φ ( x , λ n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq182_HTML.gif and that of ω ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq91_HTML.gif. If ϕ 1 ( ξ 0 , λ n ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq183_HTML.gif and ϕ 2 ( ξ 0 , λ n ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq184_HTML.gif, then (4.39) gives
F ( ) , Φ ( , λ n ) ¯ H Φ ( , λ n ) H 2 = F 1 ( λ n ) ω ( λ n ) ϕ 1 ( ξ 0 , λ n ) F ( ) , Φ ( , λ n ) ¯ H Φ ( , λ n ) H 2 = F 2 ( λ n ) ω ( λ n ) ϕ 2 ( ξ 0 , λ n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ118_HTML.gif
(4.40)

Combining (4.37), (4.40) and (4.29), we get (4.28) under the assumption that ϕ 1 ( ξ 0 , λ n ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq183_HTML.gif and ϕ 2 ( ξ 0 , λ n ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq184_HTML.gif for all n. If ϕ i ( ξ 0 , λ n ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq185_HTML.gif, for some n , i = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq186_HTML.gif or 2, the same expansions hold with F i ( λ n ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq187_HTML.gif. The convergence properties as well as the analytic and growth properties can be established as in Theorem 4.1 above. □

Now we derive an example illustrating the previous results.

Example 4.1

Consider the system
u 2 p ( x ) u 1 = λ u 1 , u 1 + p ( x ) u 2 = λ u 2 , x [ 1 , 0 ) ( 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ119_HTML.gif
(4.41)
sin α u 1 ( 0 ) cos α u 2 ( 0 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ120_HTML.gif
(4.42)
( cos β + λ sin β ) u 1 ( 1 ) + ( sin β λ cos β ) u 2 ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ121_HTML.gif
(4.43)
and transmission conditions
u 1 ( 0 ) δ u 1 ( 0 + ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ122_HTML.gif
(4.44)
u 2 ( 0 ) δ u 2 ( 0 + ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ123_HTML.gif
(4.45)
This problem is a special case of problem (2.1)-(2.5) when p 1 ( x ) = p 2 ( x ) = p ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq188_HTML.gif and a 1 = cos β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq189_HTML.gif, a 2 = sin β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq190_HTML.gif. Then ρ = 1 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq191_HTML.gif. For simplicity, we define
P 1 ( x ) : = 1 x p ( t ) d t , P 2 ( x ) : = 1 x p ( t ) d t , x [ 1 , 0 ) ( 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equaa_HTML.gif
In the notations of the above section, the solutions ϕ ( , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq86_HTML.gif and χ ( , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_IEq107_HTML.gif are
ϕ ( x , λ ) = { ( cos [ ζ 1 ( x , λ ) ] sin [ ζ 1 ( x , λ ) ] ) , x [ 1 , 0 ) , ( 1 δ cos [ ζ 1 ( x , λ ) ] 1 δ sin [ ζ 1 ( x , λ ) ] ) , x ( 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ124_HTML.gif
(4.46)
χ ( x , λ ) = { ( δ ( sin [ ζ 2 ( x , λ ) ] + λ cos [ ζ 2 ( x , λ ) ] ) δ ( cos [ ζ 2 ( x , λ ) ] + λ sin [ ζ 2 ( x , λ ) ] ) ) , x [ 1 , 0 ) , ( sin [ ζ 2 ( x , λ ) ] + λ cos [ ζ 2 ( x , λ ) ] cos [ ζ 2 ( x , λ ) ] + λ sin [ ζ 2 ( x , λ ) ] ) , x ( 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ125_HTML.gif
(4.47)
where
ζ 1 ( x , λ ) : = λ ( x + 1 ) + P 1 ( x ) + α , ζ 2 ( x , λ ) : = λ ( x 1 ) P 2 ( x ) + β . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equab_HTML.gif
The eigenvalues are the solutions of the equation
ω ( λ ) = δ 2 [ 1 δ ( cos β + λ sin β ) cos [ ζ 1 ( 1 , λ ) ] + 1 δ ( sin β λ cos β ) sin [ ζ 1 ( 1 , λ ) ] ] = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equac_HTML.gif
which can be rewritten as
cos [ ζ 1 ( 1 , λ ) β ] λ sin [ ζ 1 ( 1 , λ ) β ] = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-65/MediaObjects/13661_2012_Article_322_Equ126_HTML.gif
(4.48)
Green’s function of problem (4.41)-(4.45) is given by
G ( x , ξ , λ ) = 1 δ [ cos [ ζ 1 ( 1 , λ ) β ] λ sin [ ζ 1 ( 1 , λ ) β ] ] × { ( δ cos [ ζ 1