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Article

A Class of Singular Sturm–Liouville Problems with Discontinuity and an Eigenparameter-Dependent Boundary Condition

School of Mathematical Sciences, Qufu Normal University, Qufu 273165, China
*
Author to whom correspondence should be addressed.
Mathematics 2022, 10(23), 4430; https://doi.org/10.3390/math10234430
Submission received: 12 October 2022 / Revised: 17 November 2022 / Accepted: 21 November 2022 / Published: 24 November 2022

Abstract

:
In this paper, we study a singular Sturm–Liouville problem with an eigenparameter-dependent boundary condition and transmission conditions at two interior points. Using an operator-theoretical formulation, we transfer the problem to an operator in an appropriate Hilbert space. It is proved that the operator is self-adjoint. We also give the asymptotic formulas of the eigenvalues of the problem. Moreover, Green’s function is also discussed.

1. Introduction

Discontinuous Sturm–Liouville problems have profound application backgrounds, such as in vibrating string problems when a string is additionally loaded with point masses, or in heat and mass transfer (see [1]). To solve interior discontinuities, some extra conditions are imposed on the discontinuous point; these are often called interface conditions, point interactions, or transmission conditions [2,3]. In recent years, such problems have attracted many scholars’ attention, and many important results have been obtained. In Refs. [4,5], the authors considered Sturm–Liouville problems with transmission conditions and obtained the variational principles and asymptotic formulas of eigenvalues, respectively. In Ref. [6], Kadakal and Mukhtarov considered the case of Sturm–Liouville problems with two discontinuities and gave some properties of eigenvalues. In Refs. [7,8,9,10], the eigenfunction expansions, periodic eigenvalues, and weak eigenfunctions of discontinuous Sturm–Liouville problems were investigated, respectively. For more details on discontinuous Sturm–Liouville problems, we refer to [11,12,13,14] and the references cited therein.
For the classical self-adjoint Sturm–Liouville problems, lots of work can be found (see [15,16,17] and the references therein). Regular Sturm–Liouville problems with eigenparameter-dependent boundary conditions have received much attention, since these problems are applied to engineering, physics, and electric circuits (see [18,19]). Such problems can be traced back to Feller, who considered these problems and solved the proper connections in probability theory (see [20,21]). Some examples of spectral problems appearing in mechanical engineering and containing eigenparameters in the boundary conditions were listed in a classic book [22]. In [23], Walter considered a Sturm–Liouville problem with eigenparameter-dependent boundary conditions and obtained the expansion theorem of the eigenfunctions. This problem has been studied in various fields, such as in the dependence of eigenvalues on coefficients and parameters, inverse problems, self-adjoint realization, the oscillation of eigenfunctions, and so on (see [23,24,25,26,27]).
As is commonly known, all of the above references studied the regular Sturm–Liouville problem with one discontinuous point and with eigenparameter-dependent boundary conditions. However, little is known about singular Sturm–Liouville problems with transmission conditions. In particular, the authors of [28] considered singular Sturm–Liouville problems with one discontinuous point and a boundary condition that was rationally dependent on the parameter. The authors obtained asymptotic formulas of the eigenvalues. In this paper, we study a singular Sturm–Liouville problem with a limit-circle endpoint and with an eigenparameter-dependent boundary condition; moreover, two discontinuous points are involved in the considered interval. To this end, we study the following singular discontinuous Sturm–Liouville problem:
L u : = ( p ( x ) u ( x ) ) + q ( x ) u ( x ) = λ ω ( x ) u ( x ) , x J = [ a , ξ 1 ) ( ξ 1 , ξ 2 ) ( ξ 2 , b ) ,
where the boundary conditions at the endpoints a and b are
L 1 u : = λ ( α 1 u ( a ) α 2 u ( a ) ) ( α 3 u ( a ) α 4 u ( a ) ) = 0 ,
L 2 u : = m 1 [ u , v 1 ] ( b ) + m 2 [ u , v 2 ] ( b ) = 0 ,
and the transmission conditions at the points of discontinuity x = ξ 1 and x = ξ 2 are
L 3 u : = u ( ξ 1 + 0 ) θ 1 u ( ξ 1 0 ) θ 3 u ( ξ 1 0 ) = 0 ,
L 4 u : = u ( ξ 1 + 0 ) θ 2 u ( ξ 1 0 ) θ 4 u ( ξ 1 0 ) = 0 ,
L 5 u : = u ( ξ 2 + 0 ) ρ 1 u ( ξ 2 0 ) ρ 3 u ( ξ 2 0 ) = 0 ,
L 6 u : = u ( ξ 2 + 0 ) ρ 2 u ( ξ 2 0 ) ρ 4 u ( ξ 2 0 ) = 0 ,
where < a < b + , b is assumed to be a limit -circle point, v 11 , v 21 are linearly independent real-valued solutions of the equation ( p ( x ) u ( x ) ) + q ( x ) u ( x ) = 0 on [ a , ξ 1 ) , v 12 , v 22 are linearly independent real-valued solutions of the equation ( p ( x ) u ( x ) ) + q ( x ) u ( x ) = 0 on ( ξ 1 , ξ 2 ) , and v 13 , v 23 are linearly independent real-valued solutions of the equation ( p ( x ) u ( x ) ) + q ( x ) u ( x ) = 0 on ( ξ 2 , b ) . [ v 13 , v 23 ] ( b ) 0 , where [ y , z ] ( x ) = p ( y z ¯ y z ¯ ) ( x ) is the sesquilinear form; p ( x ) = p 1 2 on x [ a , ξ 1 ) , p ( x ) = p 2 2 on x [ ξ 1 , ξ 2 ] , and p ( x ) = p 3 2 on x ( ξ 2 , b ) ; ω ( x ) = ω 1 2 on x [ a , ξ 1 ) , ω ( x ) = ω 2 2 on x [ ξ 1 , ξ 2 ] , and ω ( x ) = ω 3 2 on x ( ξ 2 , b ) ; λ is a complex eigenparameter; q ( x ) is a real-valued locally integrable function on J with finite limits q ( ± ξ i ) = lim x ( ξ i ± 0 ) q ( x ) ( i = 1 , 2 ) ; w i , p i ( i = 1 , 2 , 3 ) , α j , θ j , ρ j ( j = 1 , 2 , 3 , 4 ) , m 1 , m 2 are non-zero real numbers, and
v 1 ( x ) = v 11 ( x ) , x [ a , ξ 1 ) , v 12 ( x ) , x ( ξ 1 , ξ 2 ) , v 13 ( x ) , x ( ξ 2 , b ) , v 2 ( x ) = v 21 ( x ) , x [ a , ξ 1 ) , v 22 ( x ) , x ( ξ 1 , ξ 2 ) , v 23 ( x ) , x ( ξ 2 , b ) ,
where v 1 ( x ) and v 2 ( x ) satisfy the conditions in (4)–(7). Moreover, we assume that
α = α 1 α 2 α 3 α 4 > 0 , m 1 m 2 0 , θ = θ 1 θ 3 θ 2 θ 4 > 0 , ρ = ρ 1 ρ 3 ρ 2 ρ 4 > 0 .
For any interval I R , L 2 ( I ) denotes the set of functions that are square integrable on I, and A C ( I ) denotes the set of functions that are absolutely continuous on I.
Using the methods of classical analysis and operator theory, we define a self-adjoint operator A in a new Hilbert space such that the eigenvalues of the considered problem coincide with those of operator A. Thereby, the singular problem in (1)–(7) is transformed into an operator problem.
This paper is organized as follows: After the lemma in this section, we define the operator formulation in Section 2 and discuss the properties of the eigenvalues in Section 3. The asymptotic formulas of the eigenvalues are discussed in Section 4. Green’s function and the resolvent operator are given in Section 5.
The following lemma is useful in this paper.
Lemma 1
([15]). Let D ( L ) = { f | f L 2 ( I ) , f A C ( I ) , L f L 2 ( I ) } . Then, for any f, g, h, ι D ( L ) , we have
[ f , h ] [ f , ι ] [ g , h ] [ g , ι ] = [ f , g ¯ ] [ h ¯ , ι ] .

2. Operator Formulation

In the following, we define a self-adjoint operator A in a new Hilbert space such that the eigenvalues of this problem coincide with those of operator A. Firstly, we introduce the Hilbert spaces:
H 1 = L 2 [ a , ξ 1 ) L 2 ( ξ 1 , ξ 2 ) L 2 ( ξ 2 , b ) ,
and
H = H 1 C ,
with the inner product
F , G = ω 1 2 θ ρ p 1 2 a ξ 1 f g ¯ d x + ω 2 2 ρ p 2 2 ξ 1 ξ 2 f g ¯ d x + ω 3 2 p 3 2 ξ 2 b f g ¯ d x + θ ρ α f 0 g 0 ¯ ,
for F = ( f ( x ) , f 0 ) H , G = ( g ( x ) , g 0 ) H .
For convenience, we use the following notations:
R ( u ) = α 3 u ( a ) α 4 u ( a ) ,
R 1 ( u ) = α 1 u ( a ) α 2 u ( a ) .
Now, we introduce the operator A in the Hilbert space H as follows:
D ( A ) = { F = ( f ( x ) , f 0 ) H f ( x ) , f ( x ) are absolutely continuous on [ a , ξ 1 ) ( ξ 1 , ξ 2 ) ( ξ 2 , b ) , with finite limits f ( ξ 1 ± 0 ) , f ( ξ 2 ± 0 ) , f ( ξ 1 ± 0 ) , f ( ξ 2 ± 0 ) and satisfing L f H 1 , L 2 f = L 3 f = L 4 f = L 5 f = L 6 f = 0 , f 0 = R 1 ( f ) } ,
which acts by the rule
A F = ( 1 ω ( x ) [ ( p ( x ) f ) + q ( x ) f ] , R ( f ) ) ,
with F = ( f , R 1 ( f ) ) D ( A ) . Now, the singular Sturm–Liouville problem in (1)–(7) can be rewritten in the operator-theoretic formulation as A F = λ F for F = ( f ( x ) , f 0 ) D ( A ) .
Corollary 1.
(c.f. [15]) The eigenvalues and eigenfunctions of the singular Sturm–Liouville problem in (1)–(7) correspond to the eigenvalues and the first elements of the corresponding eigenfunctions of operator A, respectively.
Lemma 2.
The domain D ( A ) is dense in H.
Proof. 
Let F = ( f ( x ) , f 0 ) H , F D ( A ) , and let C 0 be a class of infinitely many continuous differentiable functions with compact support: φ 1 ( x ) C 0 [ a , ξ 1 ) , φ 2 ( x ) C 0 ( ξ 1 , ξ 2 ) and φ 3 ( x ) C 0 ( ξ 2 , b ) . Since C 0 0 D ( A ) , for all V = ( v ( x ) , 0 ) C 0 0 , it is orthogonal to F, namely,
F , V = ω 1 2 θ ρ p 1 2 a ξ 1 f v ¯ d x + ω 2 2 ρ p 2 2 ξ 1 ξ 2 f v ¯ d x + ω 3 2 p 3 2 ξ 2 b f v ¯ d x = 0 ,
so f ( x ) = 0 . For all G = ( g ( x ) , g 0 ) D ( A ) , F , G = 1 α f 0 g 0 ¯ = 0 . Thus, f 0 = 0 , since g 0 = R 1 ( g ) is arbitrary. Therefore, F = ( 0 , 0 ) , which proves the assertion. □
Definition 1.
We define the Wronskians of functions f ( x ) and g ( x ) as follows: W ( f , g ; x ) = f ( x ) g ( x ) f ( x ) g ( x ) . One has
lim x b W ( f , g ; x ) = 1 p 3 2 [ f , g ¯ ] ( b ) .
Lemma 3.
A is a symmetric operator in H.
Proof. 
For each F , G D ( A ) , F = ( f , R 1 ( f ) ) , G = ( g , R 1 ( g ) ) , integration by parts yields
A F , G F , A G = θ ρ W ( f , g ¯ ) | a ξ 1 + ρ W ( f , g ¯ ) | ξ 1 ξ 2 + W ( f , g ¯ ) | ξ 2 b + ρ θ α [ R ( f ) R 1 ( g ) ¯ R 1 ( f ) R ( g ) ¯ ] .
From (4)–(7), we have
W ( f , g ¯ ; ξ 1 + 0 ) = θ W ( f , g ¯ ; ξ 1 0 ) ,
W ( f , g ¯ ; ξ 2 + 0 ) = ρ W ( f , g ¯ ; ξ 2 0 ) .
From boundary conditions (2) and (3), we have
R ( f ) R 1 ( g ¯ ) R 1 ( f ) R ( g ¯ ) = α W ( f , g ¯ ; a ) ,
W ( f , g ¯ ; b ) = 0 .
Substituting (10)–(13) into (9) yields A F , G = F , A G . This completes the proof. □
Taking the above lemmas into account, we have the following theorem.
Theorem 1.
A is a self-adjoint operator.
Proof. 
From Lemmas 2 and 3, it is sufficient to prove that if A F , G = F , W for each F D ( A ) , then G D ( A ) and A G = W , where G = ( g ( x ) , r ) T , W = ( w ( x ) , s ) T . Precisely, we need to prove that the following properties hold:
(i)
g ( x ) , g ( x ) are absolutely continuous on J , L g H 1 ;
(ii)
r = R 1 ( g ) = α 1 g ( a ) α 2 g ( a ) ;
(iii)
L 2 g = L 3 g = L 4 g = L 5 g = L 6 g = 0 ;
(iv)
w ( x ) = L g ;
(v)
s = R ( g ) = α 3 g ( a ) α 4 g ( a ) .
For an arbitrary point F C 0 0 D ( A ) (where 0 C ), we have
L f , g H 1 = f , w H 1 ,
namely,
ω 1 2 ρ θ p 1 2 a ξ 1 ( L f ) g ¯ d x + ω 2 2 p 2 2 ξ 1 ξ 2 ( L f ) g ¯ d x + ω 3 2 p 3 2 ξ 2 b ( L f ) g ¯ d x = ω 1 2 ρ θ p 1 2 a ξ 1 f w ¯ d x + ω 2 2 p 2 2 ξ 1 ξ 2 f w ¯ d x + ω 3 2 p 3 2 ξ 2 b f w ¯ d x .
By virtue of the classical Sturm–Liouville theory, (i) and (iv) hold. For all F D ( A ) , A F , G = F , W implies that
L f , g + ρ θ α R ( f ) r ¯ = f , L g + ρ θ α R 1 ( f ) s ¯ .
In addition, integration by parts yields
L f , g = f , L g + θ ρ ( W ( f , g ¯ ; ξ 1 0 ) W ( f , g ¯ ; a ) ) + ρ ( W ( f , g ¯ ; ξ 2 0 ) W ( f , g ¯ ; ξ 1 + 0 ) ) + ( W ( f , g ¯ ; b ) W ( f , g ¯ ; ξ 2 + 0 ) ) .
Combining (14) with (15), we have
ρ θ α R 1 ( f ) s ¯ = ρ θ α R ( f ) r ¯ + θ ρ ( W ( f , g ¯ ; ξ 1 0 ) W ( f , g ¯ ; a ) ) + ρ ( W ( f , g ¯ ; ξ 2 0 ) W ( f , g ¯ ; ξ 1 + 0 ) ) + ( W ( f , g ¯ ; b ) W ( f , g ¯ ; ξ 2 + 0 ) ) .
From Naimark’s Patching Lemma [16,17], there exists a F D ( A ) such that
[ f , v 1 ] ( b ) = [ f , v 2 ] ( b ) = f ( ξ 1 0 ) = f ( ξ 1 0 ) = f ( ξ 1 + 0 ) = f ( ξ 1 + 0 ) = f ( ξ 2 0 ) = f ( ξ 2 0 ) = f ( ξ 2 + 0 ) = f ( ξ 2 + 0 ) = 0 , f ( a ) = α 2 , f ( a ) = α 1 .
Substituting the above equations into (16), we get (ii). Similarly, we have (v). Next, let F D ( A ) such that
f ( a ) = f ( a ) = f ( ξ 1 ± 0 ) = f ( ξ 1 ± 0 ) = f ( ξ 2 ± 0 ) = f ( ξ 2 ± 0 ) = 0 , [ f , v 1 ] ( b ) = m 2 [ v 1 , v 2 ] , [ f , v 2 ] ( b ) = m 1 [ v 1 , v 2 ¯ ] .
Then, from (16), we obtain L 2 g = 0 . Further, let F D ( A ) satisfy
f ( a ) = f ( a ) = f ( ξ 1 + 0 ) = f ( ξ 2 + 0 ) = f ( ξ 2 0 ) = f ( ξ 2 + 0 ) = f ( ξ 2 0 ) = [ f , v 1 ] ( b ) = [ f , v 2 ] ( b ) = 0 , f ( ξ 1 0 ) = θ 3 , f ( ξ 1 0 ) = θ 1 , f ( ξ 1 + 0 ) = θ .
From (16), we find that g ( ξ 1 + 0 ) = θ 1 g ( ξ 1 0 ) + θ 3 g ( ξ 1 0 ) , namely, L 3 g = 0 . Analogously, we get L 4 g = L 5 g = L 6 g = 0 , which means that (iii) holds. Hence, A is a self-adjoint operator. □
By the properties of self-adjoint operator, we have the following results.
Corollary 2.
All eigenvalues of the problem in (1)–(7) are real-valued.
Corollary 3.
If λ 1 and λ 2 are two different eigenvalues of the considered problem in (1)–(7), then the corresponding eigenfunctions f ( x ) and g ( x ) are orthogonal in the sense of
ω 1 2 θ ρ p 1 2 a ξ 1 f ( x ) g ¯ ( x ) d x + ω 2 2 ρ p 2 2 ξ 1 ξ 2 f ( x ) g ¯ ( x ) d x + ω 3 2 p 3 2 ξ 2 b f ( x ) g ¯ ( x ) d x + θ ρ α [ ( α 1 f ( a ) α 2 f ( a ) ) ( α 3 g ¯ ( a ) α 4 g ¯ ( a ) ) ] = 0 .

3. Fundamental Solutions and Properties of Eigenvalues

In this section, we rebuild the fundamental solutions of the singular discontinuous problem in (1)–(7) and give the properties of the eigenvalues.
Lemma 4.
(c.f. [15]) Let q ( x ) be a real-valued continuous function on I = [ a , b ) and let f ( λ ) , g ( λ ) be given entire functions. Then, for any λ C , the equation
L u : = ( p ( x ) u ) + q ( x ) u = λ ω ( x ) u , x [ a , b )
has a unique solution u = u ( x , λ ) satisfying the initial conditions
[ u , v 1 ] ( b , λ ) = f ( λ ) , [ u , v 2 ] ( b , λ ) = g ( λ ) .
For each fixed x [ a , b ) , u ( x , λ ) is an entire function of λ.
Now, we define fundamental solutions φ ( x , λ ) and χ ( x , λ ) of Equation (1) with the following procedure, where
φ ( x , λ ) = φ 1 ( x , λ ) , x [ a , ξ 1 ) , φ 2 ( x , λ ) , x ( ξ 1 , ξ 2 ) , φ 3 ( x , λ ) , x ( ξ 2 , b ) ,
and
χ ( x , λ ) = χ 1 ( x , λ ) , x [ a , ξ 1 ) , χ 2 ( x , λ ) , x ( ξ 1 , ξ 2 ) , χ 3 ( x , λ ) , x ( ξ 2 , b ) .
Let φ 1 ( x , λ ) be the solution of Equation (1) on the interval [ a , ξ 1 ) satisfying the initial conditions
u ( a , λ ) = λ α 2 α 4 ,
u ( a , λ ) = λ α 1 α 3 .
In accordance with Lemma 4, we can define the solution φ 2 ( x , λ ) of Equation (1) on ( ξ 1 , ξ 2 ) with the initial conditions
φ 2 ( ξ 1 + 0 ) φ 2 ( ξ 1 + 0 ) = θ 1 φ 1 ( ξ 1 0 ) + θ 3 φ 1 ( ξ 1 0 ) θ 2 φ 1 ( ξ 1 0 ) + θ 4 φ 1 ( ξ 1 0 )
and we define the solution φ 3 ( x , λ ) of Equation (1) on ( ξ 2 , b ) by the initial conditions
φ 3 ( ξ 1 + 0 ) φ 3 ( ξ 1 + 0 ) = ρ 1 φ 2 ( ξ 2 0 ) + ρ 3 φ 2 ( ξ 2 0 ) ρ 2 φ 2 ( ξ 2 0 ) + ρ 4 φ 2 ( ξ 2 0 ) .
Similarly, we define the solutions χ 3 ( x , λ ) , χ 2 ( x , λ ) , and χ 1 ( x , λ ) of Equation (1) by the initial conditions
χ 3 ( b , λ ) : = [ χ 3 , v 1 ] ( b ) = m 2 ,
χ 3 ( b , λ ) : = [ χ 3 , v 2 ] ( b ) = m 1 ,
χ 2 ( ξ 2 0 ) χ 2 ( ξ 2 0 ) = ρ 4 χ 3 ( ξ 2 + 0 ) ρ 3 χ 3 ( ξ 2 + 0 ) ρ ρ 1 χ 3 ( ξ 2 + 0 ) ρ 2 χ 3 ( ξ 2 + 0 ) ρ ,
χ 1 ( ξ 1 0 ) χ 1 ( ξ 1 0 ) = θ 4 χ 2 ( ξ 1 + 0 ) θ 3 χ 2 ( ξ 1 + 0 ) θ θ 1 χ 2 ( ξ 1 + 0 ) θ 2 χ 2 ( ξ 1 + 0 ) θ .
Now, we consider Wronskians:
W i ( λ ) : = W ( φ i , χ i ; x ) = φ i ( x , λ ) χ i ( x , λ ) φ i ( x , λ ) χ i ( x , λ ) ( i = 1 , 2 , 3 ) .
From the dependence of the solutions of the initial value problems on the parameter, we get that W i ( λ ) ( i = 1 , 2 , 3 ) are independent of x.
Lemma 5.
For each λ C , W 3 ( λ ) = ρ W 2 ( λ ) = ρ θ W 1 ( λ ) .
Proof. 
From the definitions of W i ( λ ) ( i = 1 , 2 , 3 ) , we have
W 1 ( λ ) = φ 1 ( ξ 1 0 , λ ) χ 1 ( ξ 1 0 , λ ) φ 1 ( ξ 1 0 , λ ) χ 1 ( ξ 1 0 , λ ) ,
W 2 ( λ ) = φ 2 ( ξ 1 + 0 , λ ) χ 2 ( ξ 1 + 0 , λ ) φ 2 ( ξ 1 + 0 , λ ) χ 2 ( ξ 1 + 0 , λ ) ,
W 3 ( λ ) = φ 3 ( ξ 2 + 0 ) χ 3 ( ξ 2 + 0 ) φ 3 ( ξ 2 + 0 ) χ 3 ( ξ 2 + 0 ) .
From the transmission conditions (4) and (5), simple calculation gives
W 2 ( λ ) = θ W 1 ( λ ) .
According to (6) and (7), we get W 3 ( λ ) = ρ W 2 ( λ ) . Thus, for each λ C , we have W 3 ( λ ) = ρ W 2 ( λ ) = ρ θ W 1 ( λ ) . This completes the proof. □
In addition, we set W ( λ ) : = W 1 ( λ ) = 1 θ W 2 ( λ ) = 1 θ ρ W 3 ( λ ) .
Theorem 2.
All of the eigenvalues of the singular problem in (1)–(7) coincide with the roots of W ( λ ) = 0 .
Proof. 
Let ν 0 ( x , λ 0 ) be any eigenfunction corresponding to the eigenvalue λ 0 ; then, the function ν 0 ( x , λ 0 ) can be written as
ν 0 ( x , λ 0 ) = a 1 φ 1 ( x , λ 0 ) + a 2 χ 1 ( x , λ 0 ) , x [ a , ξ 1 ) , a 3 φ 2 ( x , λ 0 ) + a 4 χ 2 ( x , λ 0 ) , x ( ξ 1 , ξ 2 ) , a 5 φ 3 ( x , λ 0 ) + a 6 χ 3 ( x , λ 0 ) , x ( ξ 2 , b ) ,
where at least one of the constants a i 0 ( i = 1 , 2 , , 6 ) . It is sufficient to prove that W ( λ 0 ) = 0 . Suppose, to the contrary, that there exists a λ 0 R satisfying W ( λ 0 ) = W 1 ( λ 0 ) = 1 θ W 2 ( λ 0 ) = 1 θ ρ W 3 ( λ 0 ) 0 . Since the eigenfunction ν 0 ( x , λ 0 ) satisfies (2)–(7), we have L i ν 0 ( x , λ 0 ) = 0 ( i = 1 , 2 , 3 , 4 , 5 , 6 ) , while the determinant of the coefficient matrix is
w ( λ 0 ) = p 3 2 W 1 ( λ 0 ) W 3 ( λ 0 ) W 2 ( λ 0 ) W 3 ( λ 0 ) = p 3 2 ρ 2 θ 3 W 4 ( λ 0 ) 0 ,
so we have a i = 0 ( i = 1 , 2 , 3 , 4 , 5 , 6 ) , which is a contradiction; then, W ( λ 0 ) = 0 . Conversely, if W ( λ 0 ) = 0 , then W ( λ 0 ) = W 1 ( λ 0 ) = 1 θ W 2 ( λ 0 ) = 1 θ ρ W 3 ( λ 0 ) = 0 . Therefore, χ i ( x , λ 0 ) = k φ i ( x , λ 0 ) ( i = 1 , 2 , 3 ) , for some k 0 . Since both φ 3 ( x , λ 0 ) and χ 3 ( x , λ 0 ) satisfy the boundary condition (3), thus,
φ ( x , λ 0 ) = φ 1 ( x , λ 0 ) , x [ a , ξ 1 ) , φ 2 ( x , λ 0 ) , x ( ξ 1 , ξ 2 ) , φ 3 ( x , λ 0 ) , x ( ξ 2 , b ) ,
satisfies (1)–(7). So, φ ( x , λ 0 ) is an eigenfunction of the problem in (1)–(7) corresponding to the eigenvalue λ 0 . This completes the proof. □
Theorem 3.
The eigenvalues of the singular problem in (1)–(7) are algebraically single.
Proof. 
Let λ = s + i t ; we use the following notations for convenience: φ = φ ( x , λ ) , φ 1 λ = φ 1 λ , and φ 1 λ = φ 1 λ . Differentiating the equation A χ = λ χ with respect to λ yields
A χ λ = χ + λ χ λ .
Then,
λ χ λ , φ H 1 χ λ , λ φ H 1 = χ , φ H 1 .
Through integration by parts, we obtain
A χ λ , φ χ λ , A φ = θ ρ ( χ 1 λ φ 1 ¯ χ 1 λ φ 1 ¯ ) | a ξ 1 + ρ ( χ 2 λ φ 2 ¯ χ 2 λ φ 2 ¯ ) | ξ 1 ξ 2 + ( χ 3 λ φ 3 ¯ χ 3 λ φ 3 ¯ ) | ξ 2 b .
Further, from the initial conditions, we have
θ ρ ( χ 1 λ φ 1 ¯ χ 1 λ φ 1 ¯ ) | a ξ 1 + ρ ( χ 2 λ φ 2 ¯ χ 2 λ φ 2 ¯ ) | ξ 1 ξ 2 + ( χ 3 λ φ 3 ¯ χ 3 λ φ 3 ¯ ) | ξ 2 b = θ ρ [ χ 1 λ ( a ) φ 1 ( a ) ¯ χ 1 λ ( a ) φ 1 ( a ) ¯ ] + χ 3 λ ( b ) φ 3 ( b ) ¯ χ 3 λ ( b ) φ 3 ( b ) ¯ .
By virtue of the definitions of W ( λ ) , (17), and (18), we observe that
W ( λ ) = W 1 ( λ ) = φ 1 χ 1 φ 1 χ 1 | x = a = φ 1 ( a ) χ 1 ( a ) φ 1 ( a ) χ 1 ( a ) = ( λ α 2 α 4 ) χ 1 ( a ) ( λ α 1 α 3 ) χ 1 ( a ) .
Differentiating the above identity yields
W ( λ ) = α 1 χ 1 ( a ) + α 2 χ 1 ( a ) + ( λ α 2 α 4 ) χ 1 λ ( a ) ( λ α 1 α 3 ) χ 1 λ ( a ) .
Next, let λ 0 be a zero of the function W ( λ ) . Then, we get φ i ( x , λ 0 ) = k χ i ( x , λ 0 ) ( i = 1 , 2 , 3 ) ( k R , k 0 ) . Noting that λ 0 R , by a direct calculation, (26) becomes
θ ρ W ( λ 0 ) = χ , φ H 1 = k ( θ ρ ω 1 2 p 1 2 a ξ 1 | φ 1 ¯ | 2 d x + ρ ω 2 2 p 2 2 ξ 1 ξ 2 | φ 2 ¯ | 2 d x + ω 3 2 p 3 2 ξ 2 b | φ 3 ¯ | 2 d x ) .
Since k 0 , θ , ρ , ω i 2 and p i 2 > 0 ( i = 1 , 2 , 3 ) , we get W ( λ 0 ) 0 . Hence, the eigenvalues are algebraically single. □

4. Asymptotic Formulas of the Fundamental Solutions and Eigenvalues

In this section, we give the asymptotic formulas of the eigenvalues of the considered problem in (1)–(7).
Lemma 6.
Let λ = s 2 , s = σ + i t ; then, the following equalities hold for k = 0 and k = 1 :
d k d x k φ 1 ( x , λ ) = ( s 2 α 2 α 4 ) d k d x k cos [ ω 1 s p 1 ( x a ) ] + p 1 ω 1 s ( s 2 α 1 α 3 ) d k d x k sin [ ω 1 s p 1 ( x a ) ] + 1 p 1 ω 1 s a x d k d x k sin [ ω 1 s p 1 ( x τ ) ] q ( τ ) φ 1 ( τ ) d τ ,
d k d x k φ 2 ( x , λ ) = p 2 ω 2 s [ θ 2 φ 1 ( ξ 1 0 ) + θ 4 φ 1 ( ξ 1 0 ) ] d k d x k sin [ ω 2 s p 2 ( x ξ 1 ) ] + [ θ 1 φ 1 ( ξ 1 0 ) + θ 3 φ 1 ( ξ 1 0 ) ] d k d x k cos [ ω 2 s p 2 ( x ξ 1 ) ] + 1 p 2 ω 2 s ξ 1 x d k d x k sin [ ω 2 s p 2 ( x τ ) ] q ( τ ) φ 2 ( τ ) d τ ,
d k d x k φ 3 ( x , λ ) = p 3 ω 3 s [ ρ 2 φ 2 ( ξ 2 0 ) + ρ 4 φ 2 ( ξ 2 0 ) ] d k d x k sin [ ω 3 s p 3 ( x ξ 2 ) ] + [ ρ 1 φ 2 ( ξ 2 0 ) + ρ 3 φ 2 ( ξ 2 0 ) ] d k d x k cos [ ω 3 s p 3 ( x ξ 2 ) ] + 1 p 3 ω 3 s ξ 2 x d k d x k sin [ ω 3 s p 3 ( x τ ) ] q ( τ ) φ 3 ( τ ) d τ .
Proof. 
For the case of k = 0 , since p φ 1 + q φ 1 = s 2 ω φ 1 , we obtain q φ 1 = p φ 1 + s 2 ω φ 1 and
a x sin [ ω 1 s p 1 ( x τ ) ] q ( τ ) φ 1 ( τ ) d τ = s 2 ω 1 2 a x φ 1 ( τ ) sin [ ω 1 s p 1 ( x τ ) ] d τ + p 1 2 a x sin [ ω 1 s p 1 ( x τ ) ] φ 1 ( τ ) d τ .
Integration by parts and the initial conditions yield
φ 1 ( 0 , λ ) = λ α 2 α 4 , φ 1 ( 0 , λ ) = λ α 1 α 3 ,
and we get
φ 1 ( x , λ ) = ( s 2 α 2 α 4 ) cos [ ω 1 s p 1 ( x a ) ] + p 1 ω 1 s ( s 2 α 1 α 3 ) sin [ ω 1 s p 1 ( x a ) ] + 1 p 1 ω 1 s a x sin [ ω 1 s p 1 ( x τ ) ] q ( τ ) φ 1 ( τ ) d τ .
Then, (27) can be obtained by differentiating (30) with respect to x. The proofs for (28) and (29) are similar; hence, we omit the details. □
Similarly, we have the following lemma.
Lemma 7.
Let λ = s 2 , s = σ + i t ; then, the following equalities hold for k = 0 and k = 1 :
d k d x k χ 1 ( x , λ ) = p 1 ω 1 s θ [ θ 1 χ 2 ( ξ 1 + 0 ) θ 2 χ 2 ( ξ 1 + 0 ) ] d k d x k sin [ ω 1 s p 1 ( x ξ 1 ) ] + 1 θ [ θ 4 χ 2 ( ξ 1 + 0 ) θ 3 χ 2 ( ξ 1 + 0 ) ] d k d x k cos [ ω 1 s p 1 ( x ξ 1 ) ] 1 p 1 ω 1 s x ξ 1 d k d x k sin [ ω 1 s p 1 ( x τ ) ] q ( τ ) χ 1 ( τ ) d τ ,
d k d x k χ 2 ( x , λ ) = p 2 ω 2 s ρ [ ρ 1 χ 3 ( ξ 2 + 0 ) ρ 2 χ 3 ( ξ 2 + 0 ) ] d k d x k sin [ ω 2 s p 2 ( x ξ 2 ) ] + 1 ρ [ ρ 4 χ 3 ( ξ 2 + 0 ) ρ 3 χ 3 ( ξ 2 + 0 ) ] d k d x k cos [ ω 2 s p 2 ( x ξ 2 ) ] 1 p 2 ω 2 s x ξ 2 d k d x k sin [ ω 2 s p 2 ( x τ ) ] q ( τ ) χ 2 ( τ ) d τ ,
d k d x k χ 3 ( x , λ ) = m 2 d k d x k cos [ ω 3 s p 3 ( x b ) ] m 1 p 3 ω 3 s d k d x k sin [ ω 3 s p 3 ( x b ) ] 1 p 3 ω 3 s x b d k d x k sin [ ω 3 s p 3 ( x τ ) ] q ( τ ) χ 3 ( τ ) d τ .
Lemma 8.
Let λ = s 2 , s = σ + i t . For k = 0 , 1 , φ ( x , λ ) has the following asymptotic representations:
Case 1: If α 2 0 ,
d k d x k φ 1 ( x , λ ) = α 2 s 2 d k d x k cos [ ω 1 s p 1 ( x a ) ] + O [ | s | k + 1 e | t | ω 1 p 1 ( x a ) ] ,
d k d x k φ 2 ( x , λ ) = θ 3 ω 1 α 2 s 3 p 1 sin [ ω 1 s p 1 ( ξ 1 a ) ] d k d x k cos [ ω 2 s p 2 ( x ξ 1 ) ] + O [ | s | k + 2 e | t | [ ω 2 p 2 ( x ξ 1 ) + ω 1 p 1 ( ξ 1 a ) ] ] ,
d k d x k φ 3 ( x , λ ) = ρ 3 ω 1 ω 2 α 2 θ 3 s 4 p 1 p 2 sin [ ω 1 s p 1 ( ξ 1 a ) ] sin [ ω 2 s p 2 ( ξ 2 ξ 1 ) ] d k d x k cos [ ω 3 s p 3 ( x ξ 2 ) ] + O [ | s | k + 3 e | t | [ ω 2 p 2 ( ξ 2 ξ 1 ) + ω 1 p 1 ( ξ 1 a ) + ω 3 p 3 ( x ξ 2 ) ] ] .
Case 2: If α 2 = 0 ,
d k d x k φ 1 ( x , λ ) = p 1 α 1 s ω 1 d k d x k sin [ ω 1 s p 1 ( x a ) ] + O ( | s | k e | t | ω 1 p 1 ( x a ) ) ,
d k d x k φ 2 ( x , λ ) = α 1 θ 3 s 2 cos [ ω 1 s p 1 ( ξ 1 a ) ] d k d x k cos [ ω 2 s p 2 ( x ξ 1 ) ] + O ( | s | k + 1 e | t | [ ω 2 p 2 ( x ξ 1 ) + ω 1 p 1 ( ξ 1 a ) ] ) ,
d k d x k φ 3 ( x , λ ) = α 1 ρ 3 θ 3 s 3 ω 2 p 2 cos [ ω 1 s p 1 ( ξ 1 a ) ] sin [ ω 2 s p 2 ( ξ 2 ξ 1 ) ] d k d x k cos [ ω 3 s p 3 ( x ξ 2 ) ] + O ( | s | k + 2 e | t | [ ω 2 p 2 ( ξ 2 ξ 1 ) + ω 1 p 1 ( ξ 1 a ) + ω 3 p 3 ( x ξ 2 ) ] ) .
Each of these estimations holds uniformly for x as | λ | .
Proof. 
The proofs of the asymptotic equalities for φ 1 ( x , λ ) and φ 2 ( x , λ ) are similar to those of Titchmarsh’s proof for φ ( x , λ ) (see [29]), so we only prove (31).
When α 2 0 , let
F 3 ( x , λ ) | s | 4 e | t | [ ω 2 p 2 ( ξ 2 ξ 1 ) + ω 1 p 1 ( ξ 1 a ) + ω 3 p 3 ( x ξ 2 ) ] = φ 3 ( x , λ ) .
It is easy to show that φ 3 ( x , λ ) = O ( | s | 4 e | t | [ ω 2 p 2 ( ξ 2 ξ 1 ) + ω 1 p 1 ( ξ 1 a ) + ω 3 p 3 ( x ξ 2 ) ] ) for k = 0 . Then, differentiating it with respect to x, we have (31). The proofs for the others are similar. □
Lemma 9.
Let λ = s 2 , s = σ + i t . Then, for k = 0 , 1 , χ ( x , λ ) has the following asymptotic representations:
d k d x k χ 1 ( x , λ ) = θ 3 p 3 m 2 ω 2 ω 3 s 2 p 2 p 3 sin [ ω 3 s p 3 ( ξ 2 b ) ] sin [ ω 2 s p 2 ( ξ 1 ξ 2 ) ] d k d x k cos [ ω 1 s p 1 ( x ξ 1 ) ] + O ( | s | k + 1 e | t | [ ω 3 p 3 ( ξ 2 b ) + ω 2 p 2 ( ξ 1 ξ 2 ) + ω 1 p 1 ( x ξ 1 ) ] ) ,
d k d x k χ 2 ( x , λ ) = ρ 3 m 2 ω 3 s p 3 sin [ ω 3 s p 3 ( ξ 2 b ) ] d k d x k cos [ ω 2 s p 2 ( x ξ 2 ) ] + O ( | s | k e | t | [ ω 3 p 3 ( ξ 2 b ) + ω 2 p 2 ( x ξ 2 ) ] ) ,
d k d x k χ 3 ( x , λ ) = m 2 d k d x k cos [ ω 3 s p 3 ( x b ) ] + O ( | s | k 1 e | t | ω 3 p 3 ( x b ) ) .
Each of these asymptotic formulas holds uniformly for x as | λ | .
According to the definition of W ( λ ) and the estimations of φ 1 ( x , λ ) and χ 1 ( x , λ ) in Lemmas 8 and 9, we have the following theorem.
Theorem 4.
Let λ = s 2 , s = σ + i t . Then, the characteristic function W ( λ ) has the following asymptotic representations:
if α 2 0 ,
W ( λ ) = m 2 ρ 3 ω 1 ω 2 α 2 θ 3 s 5 p 1 p 2 p 3 sin [ ω 1 s p 1 ( ξ 1 a ) ] sin [ ω 2 s p 2 ( ξ 2 ξ 1 ) ] sin [ ω 3 s p 3 ( b ξ 2 ) ] + O ( | s | 4 e | t | [ ω 1 p 1 ( ξ 1 a ) + ω 2 p 2 ( ξ 2 ξ 1 ) + ω 3 p 3 ( b ξ 2 ) ] ) ;
if α 2 = 0 ,
W ( λ ) = m 2 ρ 3 ω 2 α 1 θ 3 s 4 p 2 p 3 cos [ ω 1 s p 1 ( ξ 1 a ) ] sin [ ω 2 s p 2 ( ξ 2 ξ 1 ) ] sin [ ω 3 s p 3 ( b ξ 2 ) ] + O ( | s | 3 e | t | [ ω 1 p 1 ( ξ 1 a ) + ω 2 p 2 ( ξ 2 ξ 1 ) + ω 3 p 3 ( b ξ 2 ) ] ) .
Theorem 5.
The eigenvalues λ n = s n 2 ( n = 0 , 1 , 2 , ) of the singular problem in (1)–(7) have the following asymptotic representations as n :
if α 2 0 ,
λ n = p 1 ( n 1 ) π ω 1 ( ξ 1 a ) + O ( 1 n ) , λ n = p 2 ( n 1 ) π ω 2 ( ξ 2 ξ 1 ) + O ( 1 n ) , λ n = p 3 ( n 1 ) π ω 3 ( b ξ 2 ) + O ( 1 n ) ;
if α 2 = 0 ,
λ n = p 1 ( n 1 2 ) π ω 1 ( ξ 1 a ) + O ( 1 n ) , λ n = p 2 ( n 1 ) π ω 2 ( ξ 2 ξ 1 ) + O ( 1 n ) , λ n = p 3 ( n 1 ) π ω 3 ( b ξ 2 ) + O ( 1 n ) .
Proof. 
By applying the well-known Rouche theorem, we can obtain these conclusions (c.f.Theorem 2.3 of [16]). □
Corollary 4.
The eigenvalues of the singular Sturm–Liouville problem in (1)–(7) are bounded below.
Proof. 
Setting s = i t ( t > 0 ) in Theorem 4, we get W ( λ ) = W ( t 2 ) ( t ). Then, W ( t 2 ) 0 for a negative and sufficiently large λ . This completes the proof. □

5. Green’s Function and the Resolvent Operator

In this section, we give Green’s function and the resolvent operator of the singular Sturm–Liouville problem in (1)–(7).
We consider the following differential equation:
( p ( x ) u ( x ) ) + q ( x ) u ( x ) λ ω ( x ) u ( x ) = f ( x ) ω ( x ) , x J ,
together with the eigenparameter-dependent boundary and transmission conditions in (2)–(7); p ( x ) and ω ( x ) are defined in Section 1. By applying the method of variation of constants, we give the general solution U ( x , λ ) of the non-homogeneous differential Equation (36) in the form
U ( x , λ ) = a 1 ( x , λ ) φ 1 ( x , λ ) + a 2 ( x , λ ) χ 1 ( x , λ ) , x [ a , ξ 1 ) , a 3 ( x , λ ) φ 2 ( x , λ ) + a 4 ( x , λ ) χ 2 ( x , λ ) , x ( ξ 1 , ξ 2 ) , a 5 ( x , λ ) φ 3 ( x , λ ) + a 6 ( x , λ ) χ 3 ( x , λ ) , x ( ξ 2 , b ) .
Simple calculation implies that
a 1 ( x , λ ) = ω 1 2 p 1 2 W ( λ ) x ξ 1 f ( t ) χ 1 ( t ) d t + a 1 , a 2 ( x , λ ) = ω 1 2 p 1 2 W ( λ ) a x f ( t ) φ 1 ( t ) d t + a 2
for x [ a , ξ 1 ) ,
a 3 ( x , λ ) = ω 2 2 p 2 2 W 2 ( λ ) x ξ 2 f ( t ) χ 2 ( t ) d t + a 3 , a 4 ( x , λ ) = ω 2 2 p 2 2 W 2 ( λ ) ξ 1 x f ( t ) φ 2 ( t ) d t + a 4
for x ( ξ 1 , ξ 2 ) , and
a 5 ( x , λ ) = ω 3 2 p 3 2 W 3 ( λ ) x b f ( t ) χ 3 ( t ) d t + a 5 , a 6 ( x , λ ) = ω 3 2 p 3 2 W 3 ( λ ) ξ 2 x f ( t ) φ 3 ( t ) d t + a 6
for x ( ξ 2 , b ) , where a i ( i = 1 , 2 , 3 , 4 , 5 , 6 ) are arbitrary constants. Substituting them into (37), the solution of the non-homogeneous differential Equation (36) has the following representation:
U ( x , λ ) = ω 1 2 φ 1 ( x , λ ) p 1 2 W ( λ ) x ξ 1 f ( t ) χ 1 ( t ) d t + ω 1 2 χ 1 ( x , λ ) p 1 2 W ( λ ) a x f ( t ) φ 1 ( t ) d t + a 1 φ 1 ( x , λ ) + a 2 χ 1 ( x , λ ) , x [ a , ξ 1 ) , ω 2 2 φ 2 ( x , λ ) p 2 2 W 2 ( λ ) x ξ 2 f ( t ) χ 2 ( t ) d t + ω 2 2 χ 2 ( x , λ ) p 2 2 W 2 ( λ ) ξ 1 x f ( t ) φ 2 ( t ) d t + a 3 φ 2 ( x , λ ) + a 4 χ 2 ( x , λ ) , x ( ξ 1 , ξ 2 ) , ω 3 2 φ 3 ( x , λ ) p 3 2 W 3 ( λ ) x b f ( t ) χ 3 ( t ) d t + ω 3 2 χ 3 ( x , λ ) p 3 2 W 3 ( λ ) ξ 2 x f ( t ) φ 3 ( t ) d t + a 5 φ 3 ( x , λ ) + a 6 χ 3 ( x , λ ) , x ( ξ 2 , b ) .
Differentiating it with respect to x, we get
U ( x , λ ) = ω 1 2 φ 1 ( x , λ ) p 1 2 W ( λ ) x ξ 1 f ( t ) χ 1 ( t ) d t + ω 1 2 χ 1 ( x , λ ) p 1 2 W ( λ ) a x f ( t ) φ 1 ( t ) d t + a 1 φ 1 ( x , λ ) + a 2 χ 1 ( x , λ ) , x [ a , ξ 1 ) , ω 2 2 φ 2 ( x , λ ) p 2 2 W 2 ( λ ) x ξ 2 f ( t ) χ 2 ( t ) d t + ω 2 2 χ 2 ( x , λ ) p 2 2 W 2 ( λ ) ξ 1 x f ( t ) φ 2 ( t ) d t + a 3 φ 2 ( x , λ ) + a 4 χ 2 ( x , λ ) , x ( ξ 1 , ξ 2 ) , ω 3 2 φ 3 ( x , λ ) p 3 2 W 3 ( λ ) x b f ( t ) χ 3 ( t ) d t + ω 3 2 χ 3 ( x , λ ) p 3 2 W 3 ( λ ) ξ 2 x f ( t ) φ 3 ( t ) d t + a 5 φ 3 ( x , λ ) + a 6 χ 3 ( x , λ ) , x ( ξ 2 , b ) .
Then, from the boundary conditions and transmission conditions (2)–(7), we have
L 1 U : = a 2 W ( λ ) ,
L 2 U : = a 5 p 3 2 W 3 ( λ ) ,
L 3 U : = ω 2 2 φ 2 ( ξ 1 + 0 , λ ) p 2 2 W 2 ( λ ) ξ 1 ξ 2 f ( t ) χ 2 ( t ) d t ω 1 2 χ 2 ( ξ 1 + 0 , λ ) p 1 2 W ( λ ) a ξ 1 f ( t ) φ 1 ( t ) d t a 1 φ 2 ( ξ 1 + 0 , λ ) a 2 χ 2 ( ξ 1 + 0 , λ ) + a 3 φ 2 ( ξ 1 + 0 , λ ) + a 4 χ 2 ( ξ 1 + 0 , λ ) ,
L 4 U : = ω 2 2 φ 2 ( ξ 1 + 0 , λ ) p 2 2 W 2 ( λ ) ξ 1 ξ 2 f ( t ) χ 2 ( t ) d t ω 1 2 χ 2 ( ξ 1 + 0 , λ ) p 1 2 W ( λ ) a ξ 1 f ( t ) φ 1 ( t ) d t a 1 φ 2 ( ξ 1 + 0 , λ ) a 2 χ 2 ( ξ 1 + 0 , λ ) + a 3 φ 2 ( ξ 1 + 0 , λ ) + a 4 χ 2 ( ξ 1 + 0 , λ ) ,
L 5 U : = ω 3 2 φ 3 ( ξ 2 + 0 , λ ) p 3 2 W 3 ( λ ) ξ 2 b f ( t ) χ 3 ( t ) d t ω 2 2 χ 3 ( ξ 2 + 0 , λ ) p 2 2 W 2 ( λ ) ξ 1 ξ 2 f ( t ) φ 2 ( t ) d t a 3 φ 3 ( ξ 2 + 0 , λ ) a 4 χ 3 ( ξ 2 + 0 , λ ) + a 5 φ 3 ( ξ 2 + 0 , λ ) + a 6 χ 3 ( ξ 2 + 0 , λ ) ,
L 6 U : = ω 3 2 φ 3 ( ξ 2 + 0 , λ ) p 3 2 W 3 ( λ ) ξ 2 b f ( t ) χ 3 ( t ) d t ω 2 2 χ 3 ( ξ 2 + 0 , λ ) p 2 2 W 2 ( λ ) ξ 1 ξ 2 f ( t ) φ 2 ( t ) d t a 3 φ 3 ( ξ 2 + 0 , λ ) a 4 χ 3 ( ξ 2 + 0 , λ ) + a 5 φ 3 ( ξ 2 + 0 , λ ) + a 6 χ 3 ( ξ 2 + 0 , λ ) .
It is easy to know that a 2 = a 5 = 0 , since W ( λ ) 0 . On the other hand, from (44) and (45) and the transmission conditions in (4)–(7), the following system of equations holds:
a 1 φ 2 ( ξ 1 + 0 , λ ) a 3 φ 2 ( ξ 1 + 0 , λ ) a 4 χ 2 ( ξ 1 + 0 , λ ) = ω 2 2 φ 2 ( ξ 1 + 0 , λ ) W 2 ( λ ) ξ 1 ξ 2 f ( t ) χ 2 ( t ) d t ω 1 2 χ 2 ( ξ 1 + 0 , λ ) W ( λ ) a ξ 1 f ( t ) φ 1 ( t ) d t , a 1 φ 2 ( ξ 1 + 0 , λ ) a 3 φ 2 ( ξ 1 + 0 , λ ) a 4 χ 2 ( ξ 1 + 0 , λ ) = ω 2 2 φ 2 ( ξ 1 + 0 , λ ) p 2 2 W 2 ( λ ) ξ 1 ξ 2 f ( t ) χ 2 ( t ) d t ω 1 2 χ 2 ( ξ 1 + 0 , λ ) p 1 2 W ( λ ) a ξ 1 f ( t ) φ 1 ( t ) d t , a 3 φ 3 ( ξ 2 + 0 , λ ) + a 4 χ 3 ( ξ 2 + 0 , λ ) a 6 χ 3 ( ξ 2 + 0 , λ ) = ω 3 2 φ 3 ( ξ 2 + 0 , λ ) p 3 2 W 3 ( λ ) ξ 2 b f ( t ) χ 3 ( t ) d t ω 2 2 χ 3 ( ξ 2 + 0 , λ ) p 2 2 W 2 ( λ ) ξ 1 ξ 2 f ( t ) φ 2 ( t ) d t , a 3 φ 3 ( ξ 2 + 0 , λ ) + a 4 φ 3 ( ξ 2 + 0 , λ ) a 6 χ 3 ( ξ 2 + 0 , λ ) = ω 3 2 φ 3 ( ξ 2 + 0 , λ ) p 3 2 W 3 ( λ ) ξ 2 b f ( t ) χ 3 ( t ) d t ω 2 2 χ 3 ( ξ 2 + 0 , λ ) p 2 2 W 2 ( λ ) ξ 1 ξ 2 f ( t ) φ 2 ( t ) d t .
So, we obtain the determinant of the system of equations:
φ 2 ( ξ 1 + 0 , λ ) φ 2 ( ξ 1 + 0 , λ ) χ 2 ( ξ 1 + 0 , λ ) 0 φ 2 ( ξ 1 + 0 , λ ) φ 2 ( ξ 1 + 0 , λ ) χ 2 ( ξ 1 + 0 , λ ) 0 0 φ 3 ( ξ 2 + 0 , λ ) χ 3 ( ξ 2 + 0 , λ ) χ 3 ( ξ 2 + 0 , λ ) 0 φ 3 ( ξ 2 + 0 , λ ) φ 3 ( ξ 2 + 0 , λ ) χ 3 ( ξ 2 + 0 , λ ) = W 2 ( λ ) W 3 ( λ ) 0 .
Then, the solution of (46) is unique and
a 1 = ω 2 2 p 2 2 W 2 ( λ ) ξ 1 ξ 2 f ( t ) χ 2 ( t ) d t + ω 3 2 p 3 2 W 3 ( λ ) ξ 2 b f ( t ) χ 3 ( t ) d t , a 3 = ω 3 2 p 3 2 W 3 ( λ ) ξ 2 b f ( t ) χ 3 ( t ) d t ,
a 4 = ω 1 2 p 1 2 W ( λ ) a ξ 1 f ( t ) φ 1 ( t ) d t , a 6 = ω 1 2 p 1 2 W ( λ ) a ξ 1 f ( t ) φ 1 ( t ) d t + ω 2 2 p 2 2 W 2 ( λ ) ξ 1 ξ 2 f ( t ) φ 2 ( t ) d t .
Substituting the coefficients a i ( i = 1 , 2 , 3 , 4 , 5 , 6 ) into (37), we find that the resolvent U ( x , λ ) has the following form:
U ( x , λ ) = ω 1 2 φ 1 ( x , λ ) p 1 2 W ( λ ) x ξ 1 f ( t ) χ 1 ( t ) d t + ω 1 2 χ 1 ( x , λ ) p 1 2 W ( λ ) a x f ( t ) φ 1 ( t ) d t + ω 2 2 φ 1 ( x , λ ) p 2 2 W 2 ( λ ) ξ 1 ξ 2 f ( t ) χ 2 ( t ) d t + ω 3 2 φ 1 ( x , λ ) p 3 2 W 3 ( λ ) ξ 2 b f ( t ) χ 3 ( t ) d t , x [ a , ξ 1 ) , ω 2 2 φ 2 ( x , λ ) p 2 2 W 2 ( λ ) x ξ 2 f ( t ) χ 2 ( t ) d t + ω 2 2 χ 2 ( x , λ ) p 2 2 W 2 ( λ ) ξ 1 x f ( t ) φ 2 ( t ) d t + ω 3 2 φ 2 ( x , λ ) p 3 2 W 3 ( λ ) ξ 2 b f ( t ) χ 3 ( t ) d t + ω 1 2 χ 2 ( x , λ ) p 1 2 W ( λ ) a ξ 1 f ( t ) φ 1 ( t ) d t , x ( ξ 1 , ξ 2 ) , ω 3 2 φ 3 ( x , λ ) p 3 2 W 3 ( λ ) x b f ( t ) χ 3 ( t ) d t + ω 3 2 χ 3 ( x , λ ) p 3 2 W 3 ( λ ) ξ 2 x f ( t ) φ 3 ( t ) d t + ω 1 2 χ 3 ( x , λ ) p 1 2 W ( λ ) a ξ 1 f ( t ) φ 1 ( t ) d t + ω 2 2 χ 3 ( x , λ ) p 2 2 W 2 ( λ ) ξ 1 ξ 2 f ( t ) φ 2 ( t ) d t , x ( ξ 2 , b ) .
According to the definitions of φ ( x , λ ) and χ ( x , λ ) , we find that Green’s function G ( x , t ; λ ) has the following representation:
G ( x , t ; λ ) = φ ( t , λ ) χ i ( x , λ ) W i ( λ ) , a t x b , x ξ 1 , ξ 2 , t ξ 1 , ξ 2 , φ i ( t , λ ) χ ( x , λ ) W i ( λ ) , a x t b , x ξ 1 , ξ 2 , t ξ 1 , ξ 2 .
Do not let λ be an eigenvalue of A. Obviously, the operator equation
( λ I A ) U = F , F = ( f ( x ) , f 0 ) H
is equivalent to the following problem:
p ( x ) u ( x ) + q ( x ) u ( x ) λ ω ( x ) u ( x ) = f ( x ) ω ( x ) , x J ,
subject to the following boundary conditions:
λ ( α 1 u ( a ) α 2 u ( a ) ) ( α 3 u ( a ) α 4 u ( a ) ) = f 0 ,
m 1 [ u , u 1 ] ( b ) + m 2 [ u , u 2 ] ( b ) = 0 ,
and the transmission conditions in (4)–(7). The general solution V ( x , λ ) of (48) is
V ( x , λ ) = ω 1 2 φ 1 ( x , λ ) p 1 2 W ( λ ) x ξ 1 f ( t ) χ 1 d t + ω 1 2 χ 1 ( x , λ ) p 1 2 W ( λ ) a x f ( t ) φ 1 ( t ) d t + D 11 φ 1 ( x , λ ) + D 12 χ 1 ( x , λ ) , x [ a , ξ 1 ) , ω 2 2 φ 2 ( x , λ ) p 2 2 W 2 ( λ ) x ξ 2 f χ 2 d t + ω 2 2 χ 2 ( x , λ ) p 2 2 W 2 ( λ ) ξ 1 x f ( t ) φ 2 ( t ) d t + D 21 φ 2 ( x , λ ) + D 22 χ 2 ( x , λ ) , x ( ξ 1 , ξ 2 ) , ω 3 2 φ 3 ( x , λ ) p 3 2 W 3 ( λ ) x b f χ 3 d t + ω 3 2 χ 3 ( x , λ ) p 3 2 W 3 ( λ ) ξ 2 x f ( t ) φ 3 ( t ) d t + D 31 φ 3 ( x , λ ) + D 32 χ 3 ( x , λ ) , x ( ξ 2 , b ) ,
where D i j ( i , j = 1 , 2 , 3 ) are arbitrary constants. By the method of variation of constants, we get
D 11 = ω 2 2 p 2 2 W 2 ( λ ) ξ 1 ξ 2 f ( t ) χ 2 ( t ) d t + ω 3 2 p 3 2 W 3 ( λ ) ξ 2 b f ( t ) χ 3 ( t ) d t , D 12 = f 0 W 1 ( λ ) ,
D 21 = ω 3 2 p 3 2 W 3 ( λ ) ξ 2 b f ( t ) χ 3 ( t ) d t , D 22 = ω 1 2 p 1 2 W 1 ( λ ) a ξ 1 f ( t ) φ 1 ( t ) d t f 0 W 1 ( λ ) ,
D 31 = 0 , D 32 = ω 1 2 p 1 2 W 1 ( λ ) a ξ 1 f ( t ) φ 1 ( t ) d t + ω 2 2 p 2 2 W 2 ( λ ) ξ 1 ξ 2 f ( t ) φ 1 ( t ) d t f 0 W 1 ( λ ) .
Namely, the general solution V ( x , λ ) can be represented as
V ( x , λ ) = χ 1 ( x , λ ) p 1 2 W 1 ( λ ) ω 1 2 a x f ( t ) φ 1 ( t ) d t + φ 1 ( x , λ ) p 1 2 W 1 ( λ ) ω 1 2 x ξ 1 f ( t ) χ 1 ( t ) d t + φ 1 ( x , λ ) ω 2 2 p 2 2 θ W 1 ( λ ) ξ 1 ξ 2 f ( t ) χ 2 ( t ) d t + φ 1 ( x , λ ) ω 3 2 θ ρ p 3 2 W 1 ( λ ) ξ 2 b f ( t ) χ 3 ( t ) d t f 0 χ 1 ( x , λ ) W 1 ( λ ) , x [ a , ξ 1 ) , χ 2 ( x , λ ) p 1 2 W 2 ( λ ) θ ω 1 2 a x f ( t ) φ 1 ( t ) d t + χ 2 ( x , λ ) p 2 2 W 2 ( λ ) ω 2 2 ξ 1 x f ( t ) φ 2 ( t ) d t + φ 2 ( x , λ ) p 2 2 W 2 ( λ ) ω 2 2 x ξ 2 f ( t ) χ 2 ( t ) d t + φ 2 ( x , λ ) ω 3 2 p 3 2 ρ W 2 ( λ ) ξ 2 b f ( t ) χ 3 ( t ) d t θ f 0 χ 2 ( x , λ ) W 2 ( λ ) , ( ξ 1 , ξ 2 ) , χ 3 ( x , λ ) p 3 2 W 3 ( λ ) ω 3 2 ξ 2 x f φ 3 d t + χ 3 ( x , λ ) p 1 2 W 1 ( λ ) θ ρ ω 1 2 a ξ 1 f ( t ) φ 1 ( t ) d t + χ 3 ( x , λ ) p 2 2 W 3 ( λ ) ρ ω 2 2 ξ 1 ξ 2 f ( t ) φ 2 ( t ) d t + φ 3 ( x , λ ) p 3 2 W ( λ ) ω 3 2 x b f ( t ) χ 3 ( t ) d t θ ρ f 0 χ 3 ( x , λ ) W 3 ( λ ) , x ( ξ 2 , b ) .
From the definition of G ( x , t ; λ ) , we get the following equality:
V ( x , λ ) = θ ρ ω 1 2 p 1 2 a ξ 1 G ( x , t ; λ ) f ( t ) d t + ρ ω 2 2 p 2 2 ξ 1 ξ 2 G ( x , t ; λ ) f ( t ) d t + ω 3 2 p 3 2 ξ 2 b G ( x , t ; λ ) f ( t ) d t θ ρ f 0 χ ( x , λ ) W 3 ( λ ) .
On the other hand, observing that both χ ( x , λ ) and φ ( x , λ ) satisfy the initial conditions (17) and (18), we have
R 1 ( G ( x , t ; λ ) ) = α 1 G ( x , a ; λ ) α 2 G ( x , a ; λ ) t = α 1 φ ( a , λ ) W ( λ ) α 2 φ ( a , λ ) χ ( x , λ ) W ( λ ) = χ ( x , λ ) W ( λ ) [ α 1 ( λ α 2 α 4 ) ] [ α 2 ( λ α 1 α 3 ) ] = α χ ( x , λ ) W ( λ ) .
In accordance with
W 3 ( λ ) = ρ W 2 ( λ ) = ρ θ W 1 ( λ )
and (53) and (54), we obtain
V ( x , λ ) = θ ρ ω 1 2 p 1 2 a ξ 1 G ( x , t ; λ ) f ( t ) d t + ρ ω 2 2 p 2 2 ξ 1 ξ 2 G ( x , t ; λ ) f ( t ) d t + ω 3 2 p 3 2 ξ 2 b G ( x , t ; λ ) f ( t ) d t + θ ρ f 0 R 1 ( G ( x , t ; λ ) ) α .
Denoting G ˜ ( x , λ ) = ( G ( x , t ; λ ) , R 1 ( G ( x , t ; λ ) ) , F = ( f ( x ) , f 0 ) , and F ¯ = ( f ( x ) ¯ , f 0 ¯ ) ) , (55) can be rewritten as
V ( x , λ ) = G ˜ ( x , λ ) , F ¯ .
Therefore, the resolvent operator R ( λ , A ) = ( λ I A ) 1 can be represented in the form
R ( λ , A ) F = ( G ˜ ( x , λ ) , F ¯ , R 1 G ˜ ( x , λ ) , F ¯ ) .

6. Conclusions

In this paper, we study a singular Sturm–Liouville problem with an eigenparameter-dependent boundary condition on the interval J, which involves two discontinuities, and the right endpoint is assumed to be a limit-circle endpoint. Using an operator-theoretic formulation, we transfer the problem to an operator in an appropriate Hilbert space. It is proved that the operator is a self-adjoint operator, and some properties of the eigenvalues are introduced. We also give asymptotic formulas of the eigenvalues. Moreover, the Green function of the considered problem is investigated. In future work, we will consider the singular Sturm–Liouville problem with finite discontinuities and boundary conditions that are rationally dependent on the eigenparameters.

Author Contributions

Writing—original draft preparation, J.C.; writing—review and editing, K.L.; supervision, Z.Z. All authors have read and agreed to the published version of the manuscript.

Funding

This work was supported by the Natural Science Foundation of Shandong Province (Nos. ZR2020QA009, ZR2020QA010, ZR2019MA034, and ZR2021MA047,ZR2021QA065),the National Nature Science Foundation of China (No.12101356), the Postdoctoral Foundation of China (2020M682139), and the Youth Creative Team Sci-Tech Program of Shandong Universities (No. 2019KJI007).

Data Availability Statement

Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

Acknowledgments

The authors are grateful to the referees for their careful reading and very helpful suggestions, which improved and strengthened the presentation of this manuscript.

Conflicts of Interest

The authors declare no conflict of interest.

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Cai, J.; Zheng, Z.; Li, K. A Class of Singular Sturm–Liouville Problems with Discontinuity and an Eigenparameter-Dependent Boundary Condition. Mathematics 2022, 10, 4430. https://doi.org/10.3390/math10234430

AMA Style

Cai J, Zheng Z, Li K. A Class of Singular Sturm–Liouville Problems with Discontinuity and an Eigenparameter-Dependent Boundary Condition. Mathematics. 2022; 10(23):4430. https://doi.org/10.3390/math10234430

Chicago/Turabian Style

Cai, Jinming, Zhaowen Zheng, and Kun Li. 2022. "A Class of Singular Sturm–Liouville Problems with Discontinuity and an Eigenparameter-Dependent Boundary Condition" Mathematics 10, no. 23: 4430. https://doi.org/10.3390/math10234430

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