Open Access

Eigenparameter dependent inverse boundary value problem for a class of Sturm-Liouville operator

Boundary Value Problems20142014:194

https://doi.org/10.1186/s13661-014-0194-3

Received: 10 March 2014

Accepted: 29 July 2014

Published: 25 September 2014

Abstract

In this work a Sturm-Liouville operator with piecewise continuous coefficient and spectral parameter in the boundary conditions is considered. The eigenvalue problem is investigated; it is shown that the eigenfunctions form a complete system and an expansion formula with respect to the eigenfunctions is obtained. Uniqueness theorems for the solution of the inverse problem with a Weyl function and spectral data are proved.

MSC: 34L10, 34L40, 34A55.

Keywords

Sturm-Liouville operatorexpansion formulainverse problemWeyl function

1 Introduction

We consider the boundary value problem
y + q ( x ) y = λ 2 ρ ( x ) y , 0 x π ,
(1)
U ( y ) : = y ( 0 ) + ( α 1 λ 2 α 2 ) y ( 0 ) = 0 ,
(2)
V ( y ) : = λ 2 ( β 4 y ( π ) + β 2 y ( π ) ) β 1 y ( π ) β 3 y ( π ) = 0 ,
(3)
where q ( x ) L 2 ( 0 , π ) is a real valued function, λ is a complex parameter, α i , β j , i = 1 , 2 , j = 1 , 4 ¯ are positive real numbers and
ρ ( x ) = { 1 , 0 x < a , γ 2 , a < x π ,

where 0 < γ 1 .

Physical applications of the eigenparameter dependent Sturm-Liouville problems, i.e. the eigenparameter appears not only in the differential equation of the Sturm-Liouville problem but also in the boundary conditions, are given in [1]–[4]. Spectral analyses of these problems are examined as regards different aspects (eigenvalue problems, expansion problems with respect to eigenvalues, etc.) in [5]–[13]. Similar problems for discontinuous Sturm-Liouville problems are examined in [14]–[18].

Inverse problems for differential operators with boundary conditions dependent on the spectral parameter on a finite interval have been studied in [19]–[23]. In particular, such problems with discontinuous coefficient are studied in [24]–[27].

We investigate a Sturm-Liouville operator with discontinuous coefficient and a spectral parameter in boundary conditions. The theoretic formulation of the operator for the problem is given in a suitable Hilbert space in Section 2. In Section 3, an asymptotic formula for the eigenvalues is given. In Section 4, an expansion formula with respect to the eigenfunctions is obtained and Section 5 contains uniqueness theorems for the solution of the inverse problem with a Weyl function and spectral data.

2 Operator formulation

Let φ ( x , λ ) and ψ ( x , λ ) be the solutions of (1) satisfying the initial conditions
φ ( 0 , λ ) = 1 , φ ( 0 , λ ) = λ 2 α 2 α 1 ,
(4)
ψ ( π , λ ) = β 1 λ 2 β 4 , ψ ( π , λ ) = λ 2 β 2 β 3 .
(5)
For the solution of (1), the following integral representation as μ ± ( x ) = ± x ρ ( x ) + a ( 1 ρ ( x ) ) is obtained similar to [28] for all λ:
e ( x , λ ) = 1 2 ( 1 + 1 ρ ( x ) ) e i λ μ + ( x ) + 1 2 ( 1 1 ρ ( x ) ) e i λ μ ( x ) + μ + ( x ) μ + ( x ) K ( x , t ) e i λ t d t ,
where K ( x , ) L 1 ( μ + ( x ) , μ + ( x ) ) . The following properties hold for the kernel K ( x , t ) which has the partial derivative K x belonging to the space L 1 ( μ + ( x ) , μ + ( x ) ) for every x [ 0 , π ] :
d d x K ( x , μ + ( x ) ) = 1 4 ρ ( x ) ( 1 + 1 ρ ( x ) ) q ( x ) ,
(6)
d d x K ( x , μ ( x ) + 0 ) d d x K ( x , μ ( x ) 0 ) = 1 4 ρ ( x ) ( 1 1 ρ ( x ) ) q ( x ) .
(7)
We obtain the integral representation of the solution φ ( x , λ ) :
φ ( x , λ ) = φ 0 ( x , λ ) + 0 μ + ( x ) A ( x , t ) cos λ t d t + ( λ 2 α 2 α 1 ) 0 μ + ( x ) A ˜ ( x , t ) sin λ t λ d t ,
(8)
where
A ( x , t ) = K ( x , t ) K ( x , t ) , A ˜ ( x , t ) = K ( x , t ) + K ( x , t )

satisfying (6), (7).

Let us define
Δ ( λ ) : = φ ( x , λ ) , ψ ( x , λ ) = φ ( x , λ ) ψ ( x , λ ) φ ( x , λ ) ψ ( x , λ ) ,
(9)
which is independent from x [ 0 , π ] . Substituting x = 0 and x = π into (9) we get
Δ ( λ ) = U ( ψ ) = V ( φ ) .

The function Δ ( λ ) is entire and has zeros at the eigenvalues of the problem (1)-(3).

In the Hilbert space H ρ = L 2 , ρ ( 0 , π ) C 2 let an inner product be defined by
( f , g ) : = 0 π f 1 ( x ) g 1 ( x ) ¯ ρ ( x ) d x + f 2 g 2 ¯ α 2 + f 3 g 3 ¯ δ 2 ,
where
f = ( f 1 ( x ) f 2 f 3 ) H ρ , g = ( g 1 ( x ) g 2 g 3 ) H ρ , δ 2 : = β 1 β 2 β 3 β 4 > 0 .
We define the operator
L ( f ) : = ( f 1 ( x ) + q ( x ) f 1 ( x ) f 1 ( 0 ) + α 1 f 1 ( 0 ) β 1 f 1 ( π ) + β 3 f 1 ( π ) )
with
D ( L ) = { f H ρ : f 1 ( x ) , f 1 ( x ) A C [ 0 , π ] , l ( f 1 ) L 2 [ 0 , π ] , f 2 = α 2 f 1 ( 0 ) , f 3 = β 4 f 1 ( π ) + β 2 f 1 ( π ) } ,
where
l ( f 1 ) = 1 ρ ( x ) { f 1 + q ( x ) f 1 } .
The boundary value problem (1)-(3) is equivalent to the equation L Y = λ 2 Y . When λ = λ n are the eigenvalues, the eigenfunctions of operator L are in the form of
Φ ( x , λ n ) = Φ n : = ( φ ( x , λ n ) α 2 φ ( 0 , λ n ) β 4 φ ( π , λ n ) + β 2 φ ( π , λ n ) ) , n = 1 , 2 .
For any eigenvalue λ n the solutions (4), (5) satisfy the relation
ψ ( x , λ n ) = k n φ ( x , λ n )
(10)
and the normalized numbers of the boundary value problem (1)-(3) are given below:
α n : = 0 π φ 2 ( x , λ n ) ρ ( x ) d x + α 2 φ 2 ( 0 , λ n ) + 1 δ 2 ( β 4 φ ( π , λ n ) + β 2 φ ( π , λ n ) ) 2 .
(11)

Lemma 1

The eigenvalues of the boundary value problem (1)-(3) are simple, i.e.
Δ ˙ ( λ ) = 2 λ n k n α n .
(12)

Proof

Since
φ ( x , λ n ) + q ( x ) φ ( x , λ n ) = λ n 2 ρ ( x ) φ ( x , λ n ) , ψ ( x , λ ) + q ( x ) ψ ( x , λ ) = λ 2 ρ ( x ) ψ ( x , λ ) ,
we get
d d x [ φ ( x , λ n ) ψ ( x , λ ) φ ( x , λ n ) ψ ( x , λ ) ] = ( λ n 2 λ 2 ) ρ ( x ) φ ( x , λ n ) ψ ( x , λ ) .
With the help of (2), (3) we get
Δ ( λ n ) Δ ( λ ) = ( λ n 2 λ 2 ) 0 π φ ( x , λ n ) ψ ( x , λ ) ρ ( x ) d x .
Adding
( λ n 2 λ 2 ) α 2 φ ( 0 , λ n ) ψ ( 0 , λ ) + ( λ n 2 λ 2 ) δ 2 ( β 4 φ ( π , λ n ) + β 2 φ ( π , λ n ) ) ( β 4 ψ ( π , λ ) + β 2 ψ ( π , λ ) )
to both sides of the last equation and using the relations (10), (11) we have
Δ ( λ n ) Δ ( λ ) = ( λ n + λ ) ( λ n λ ) k n α n .

Taking λ λ n , we find (12). □

3 Asymptotic formulas of the eigenvalues

The solution of (1) satisfying the initial conditions (4) when q ( x ) 0 is in the following form:
φ 0 ( x , λ ) = c 0 ( x , λ ) + ( λ 2 α 2 α 1 ) s 0 ( x , λ ) λ ,
(13)
where
c 0 ( x , λ ) = { cos λ x , 0 x < a , 1 2 ( 1 + 1 ρ ( x ) ) cos λ μ + ( x ) + 1 2 ( 1 1 ρ ( x ) ) cos λ μ ( x ) , a < x π ,
and
s 0 ( x , λ ) = { sin λ x λ , 0 x < a , 1 2 ( 1 + 1 ρ ( x ) ) sin λ μ + ( x ) λ + 1 2 ( 1 1 ρ ( x ) ) sin λ μ ( x ) λ , a < x π .
The eigenvalues λ n 0 ( n = 0 , 1 , 2 , ) of the boundary value problem (1)-(3) when q ( x ) 0 can be found by using the equation
Δ 0 ( λ ) = ( λ 2 β 2 β 3 ) φ 0 ( π , λ ) ( β 1 λ 2 β 4 ) φ 0 ( π , λ ) = 0
and can be represented in the following way:
λ n 0 = n + ψ ( n ) , n = 0 , 1 , 2 , ,

where sup n | ψ ( n ) | < + .

Roots λ n 0 of the function Δ 0 ( λ ) are separated, i.e.,
inf n k | λ n 0 λ k 0 | = τ > 0 .

Lemma 2

The eigenvalues of the boundary value problem (1)-(3) are in the form of
λ n = λ n 0 + d n λ n 0 + η n n , λ n > 0 ,
(14)
where ( d n ) is a bounded sequence,
d n = 1 4 λ n 0 Δ ˙ ( λ n 0 ) 0 π ( 1 1 ρ ( t ) ) q ( t ) sin ( λ n 0 μ ( π ) ) ρ ( t ) d t α 1 α 2 4 λ n 0 Δ ˙ ( λ n 0 ) 0 π ( 1 + 1 ρ ( t ) ) q ( t ) cos ( λ n 0 μ ( π ) ) ρ ( t ) d t

and { η n } l 2 .

Proof

From (8), it follows that
φ ( π , λ ) = φ 0 ( π , λ ) + 0 μ + ( π ) A ( π , t ) cos λ t d t + ( λ 2 α 2 α 1 ) 0 μ + ( π ) A ˜ ( π , t ) sin λ t λ d t .
(15)
The expressions of Δ ( λ ) and Δ 0 ( λ ) let us calculate Δ ( λ ) Δ 0 ( λ ) :
Δ ( λ ) Δ 0 ( λ ) = λ A ˜ ( π , μ + ( π ) ) ( α + a + π 1 2 α ) sin λ μ + ( π ) + ( α + a + π 1 2 α ) A ( π , μ + ( π ) ) cos λ μ + ( π ) + I ( λ ) λ 3 ,
where
I ( λ ) = α 2 β 4 0 μ + ( π ) x A ˜ ( π , t ) sin λ t d t + O ( e | Im λ | μ + ( π ) λ 2 ) .
Therefore, for sufficiently large n, on the contours
Γ n = { λ : | λ | = | λ n 0 | + τ 2 } ,
we have
| Δ ( λ ) Δ 0 ( λ ) | < | Δ 0 ( λ ) | .
By the Rouche theorem, we obtain the result that the number of zeros of the function
{ Δ ( λ ) Δ 0 ( λ ) } + Δ 0 ( λ ) = Δ ( λ )
inside the contour Γ n coincides with the number of zeros of the function Δ 0 ( λ ) . Moreover, applying the Rouche theorem to the circle γ n ( δ ) = { λ : | λ λ n 0 | δ } we find, for sufficiently large n, that there exists one zero λ n of the function Δ ( λ ) in γ n ( δ ) . Owing to the arbitrariness of δ > 0 we have
λ n = λ n 0 + ϵ n , ϵ n = o ( 1 ) , n .
(16)
Substituting (16) into (15), as n taking into account the equality Δ 0 ( λ n 0 ) = 0 and the relations sin ϵ n μ + ( π ) ϵ n μ + ( π ) , cos ϵ n μ + ( π ) 1 , integrating by parts and using the properties of the kernels A ( x , t ) and A ˜ ( x , t ) we have
ϵ n d n λ n 0 + ϵ n + η n λ n 0 ,
where
η n = 0 μ + ( π ) A t ( π , t ) sin λ n 0 t d t + ( α 1 α 2 ) 0 μ + ( π ) A t ( π , t ) cos λ n 0 t d t .
Let us show that η n l 2 . It is obvious that η n can be reduced to the integral
μ + ( π ) μ + ( π ) R ( t ) e i λ t d t ,
where R ( t ) L 2 ( μ + ( π ) , μ + ( π ) ) . Now, take
ζ ( λ ) : = μ + ( π ) μ + ( π ) R ( t ) e i λ t d t .

It is clear from [28] (p.66) that { ζ n } = ζ ( λ n ) l 2 . By virtue of this we have { η n } l 2 . The lemma is proved. □

4 Expansion formula with respect to eigenfunctions

Denote
G ( x , t ; λ ) : = 1 Δ ( λ ) { φ ( t , λ ) ψ ( x , λ ) , t x , ψ ( t , λ ) φ ( x , λ ) , t x
(17)
and consider the function
y ( x , λ ) : = 0 π G ( x , t ; λ ) f ( t ) ρ ( t ) d t f 1 Δ ( λ ) ψ ( x , λ ) + f 2 Δ ( λ ) φ ( x , λ ) .
(18)

Theorem 3

The eigenfunctions Φ ( x , λ n ) of the boundary value problem (1)-(3) form a complete system in L 2 , ρ ( 0 , π ) C 2 .

Proof

With the help of (10) and (12), we can write
ψ ( x , λ n ) = Δ ˙ ( λ n ) 2 λ n α n φ ( x , λ n ) .
(19)
Using (17) and (18) we get
Res λ = λ n y ( x , λ ) = 1 2 λ n α n φ ( x , λ n ) 0 π φ ( t , λ n ) f ( t ) ρ ( t ) d t 1 2 λ n α n φ ( x , λ n ) ( f 1 f 2 k n ) .
(20)
Now let f ( x ) L 2 , ρ ( 0 , π ) C 2 and assume
( Φ ( x , λ n ) , f ( x ) ) = 0 π φ ( x , λ n ) f 1 ( x ) ¯ ρ ( x ) d x + φ ( 0 , λ n ) f 2 ¯ + ( β 4 φ ( π , λ n ) + β 2 φ ( π , λ n ) ) f 3 ¯ δ 2 = 0 .
(21)
Then from (20), we have Res λ = λ n y ( x , λ ) = 0 . Consequently, for fixed x [ 0 , π ] the function y ( x , λ ) is entire with respect to λ. Let us denote
G δ : = { λ : | λ λ n 0 | δ , n = 0 , 1 , 2 , } ,
where δ is sufficiently small positive number. It is clear that the relation below holds:
| Δ ( λ ) | C | λ | 3 e | Im λ | μ + ( π ) , λ G δ , C = cons .
(22)
From (18) it follows that for fixed δ > 0 and sufficiently large λ > 0 we have
| y ( x , λ ) | C | λ | , λ G δ , | λ | λ , C = cons .

Using maximum principle for module of analytic functions and Liouville theorem, we get y ( x , λ ) 0 . From this we obtain f ( x ) 0 a.e. on [ 0 , π ] . Thus we conclude the completeness of the eigenfunctions Φ ( x , λ n ) in L 2 , ρ ( 0 , π ) C 2 . □

Theorem 4

If f ( x ) D ( L ) , then the expansion formula
f ( x ) = n = 1 a n φ ( x , λ n )
(23)
is valid, where
a n = 1 2 α n 0 π φ ( t , λ n ) f ( t ) ρ ( t ) d t ,
and the series converges uniformly with respect to x [ 0 , π ] . For f ( x ) L 2 , ρ ( 0 , π ) , the series converges in L 2 , ρ ( 0 , π ) , moreover, the Parseval equality holds:
0 π | f ( x ) | 2 ρ ( x ) d x = n = 1 α n | a n | 2 .

Proof

Since φ ( x , λ ) and ψ ( x , λ ) are the solutions of the boundary value problem (1)-(3), we have
y ( x , λ ) = ψ ( x , λ ) Δ ( λ ) { 0 π [ φ ( t , λ ) + q ( t ) φ ( t , λ ) ] f ( t ) λ 2 d t } φ ( x , λ ) Δ ( λ ) { π x [ ψ ( t , λ ) + q ( t ) ψ ( t , λ ) ] f ( t ) λ 2 d t } f 1 Δ ( λ ) ψ ( x , λ ) + f 2 Δ ( λ ) φ ( x , λ ) .
(24)
Integrating by parts and taking into account the boundary conditions (2), (3) we obtain
y ( x , λ ) = 1 λ 2 f ( x ) 1 λ 2 [ Z 1 ( x , λ ) + Z 2 ( x , λ ) ] f 1 Δ ( λ ) ψ ( x , λ ) + f 2 Δ ( λ ) φ ( x , λ ) ,
(25)
where
Z 1 ( x , λ ) = 1 Δ ( λ ) ψ ( x , λ ) 0 x φ ( t , λ ) f ( t ) d t + 1 Δ ( λ ) φ ( x , λ ) x π ψ ( t , λ ) f ( t ) d t , Z 2 ( x , λ ) = 1 Δ ( λ ) [ ( λ 2 α 2 α 1 ) ψ ( x , λ ) f ( 0 ) ] 1 Δ ( λ ) [ ( λ 2 β 2 β 3 ) φ ( x , λ ) f ( π ) ] + 1 Δ ( λ ) ψ ( x , λ ) 0 x φ ( t , λ ) q ( t ) f ( t ) d t + 1 Δ ( λ ) φ ( x , λ ) x π ψ ( t , λ ) q ( t ) f ( t ) d t .
If we consider the following contour integral where Γ n is a counter-clockwise oriented contour:
I n ( x ) = 1 2 π i Γ n λ y ( x , λ ) d λ ,
and then taking into consideration (20) we get
I n ( x ) = n = 1 Res λ = λ n [ λ y ( x , λ ) ] = n = 1 a n φ ( x , λ n ) + n = 1 λ n f 1 Δ ˙ ( λ n ) ψ ( x , λ n ) n = 1 λ n f 2 Δ ˙ ( λ n ) φ ( x , λ n ) ,
(26)
where
a n = 1 α n 0 π φ ( t , λ n ) f ( t ) ρ ( t ) d t .
On the other hand, with the help of (25) we get
I n ( x ) = f ( x ) 1 2 π i Γ n [ Z 1 ( x , λ ) + Z 2 ( x , λ ) ] d λ + n = 1 λ n f 1 Δ ˙ ( λ n ) ψ ( x , λ n ) n = 1 λ n f 2 Δ ˙ ( λ n ) φ ( x , λ n ) .
(27)
Comparing (26) and (27) we obtain
n = 1 a n φ ( x , λ n ) = f ( x ) + ϵ n ( x ) ,
where
ϵ n ( x ) = 1 2 π i Γ n [ Z 1 ( x , λ ) + Z 2 ( x , λ ) ] d λ .
The relations below hold for sufficiently large λ > 0
max x [ 0 , π ] | Z 2 ( x , λ ) | C 2 | λ | 2 , λ G δ , | λ | λ ,
(28)
max x [ 0 , π ] | Z 1 ( x , λ ) | C 1 | λ | 2 , λ G δ , | λ | λ .
(29)
The validity of
lim n max x [ 0 , π ] | ϵ n ( x ) | = 0
can easily be seen from (28) and (29). The last equation gives us the expansion formula
f ( x ) = n = 1 a n φ ( x , λ n ) .
Since the system of Φ ( x , λ n ) is complete and orthogonal in L 2 , ρ ( 0 , π ) C 2 , the Parseval equality
0 π | f ( x ) | 2 ρ ( x ) d x = n = 1 α n | a n | 2

holds. □

5 Uniqueness theorems

We consider the statement of the inverse problem of the reconstruction of the boundary value problem (1)-(3) from the Weyl function.

Let the functions c ( x , λ ) and s ( x , λ ) denote the solutions of (1) satisfying the conditions c ( 0 , λ ) = 1 , c ( 0 , λ ) = 0 , s ( 0 , λ ) = 0 and s ( 0 , λ ) = 1 , respectively, and φ ( x , λ ) and ψ ( x , λ ) be the solutions of (1) under the initial conditions (4), (5).

Further, let the function Φ ( x , λ ) be the solution of (1) satisfying U ( Φ ) = 1 and V ( Φ ) = 0 . We set
M ( λ ) : = ψ ( 0 , λ ) Δ ( λ ) .
The functions Φ ( x , λ ) and M ( λ ) are called the Weyl solution and the Weyl function for the boundary value problem (1)-(3), respectively. The Weyl function is a meromorphic function having simple poles at points λ n , eigenvalues of the boundary value problem of (1)-(3). The Wronskian
W ( x ) : = φ ( x , λ ) , Φ ( x , λ )
does not depend on x. Taking x = 0 , we get
W ( 0 ) = φ ( 0 , λ ) Φ ( 0 , λ ) φ ( 0 , λ ) Φ ( 0 , λ ) = 1 .
Hence,
W ( x ) = φ ( x , λ ) , Φ ( x , λ ) = 1 .
(30)
In view of (4) and (5), we get for λ λ n
Φ ( x , λ ) = ψ ( x , λ ) Δ ( λ ) .
(31)
Using (31) we obtain
M ( λ ) = Δ 0 ( λ ) Δ ( λ ) ,
where Δ 0 ( λ ) = ψ ( 0 , λ ) is the characteristic function of the boundary value problem L 0 :
l y = λ 2 y , 0 x π , y ( 0 ) = 0 , V ( y ) = 0 .
It is clear that
Φ ( x , λ ) = s ( x , λ ) + M ( λ ) φ ( x , λ ) .
(32)

Theorem 5

The boundary value problem of (1)-(3) is identically denoted by the Weyl function M ( λ ) .

Proof

Let us denote the matrix P ( x , λ ) = [ P j k ( x , λ ) ] j , k = 1 , 2 as
P ( x , λ ) ( φ ˜ ( x , λ ) Φ ˜ ( x , λ ) φ ˜ ( x , λ ) Φ ˜ ( x , λ ) ) = ( φ ( x , λ ) Φ ( x , λ ) φ ( x , λ ) Φ ( x , λ ) ) .
(33)
Then we have
φ ( x , λ ) = P 11 ( x , λ ) φ ˜ ( x , λ ) + P 12 ( x , λ ) φ ˜ ( x , λ ) , Φ ( x , λ ) = P 11 ( x , λ ) Φ ˜ ( x , λ ) + P 12 ( x , λ ) Φ ˜ ( x , λ )
(34)
or
P 11 ( x , λ ) = φ ( x , λ ) Φ ˜ ( x , λ ) φ ˜ ( x , λ ) Φ ( x , λ ) , P 12 ( x , λ ) = φ ˜ ( x , λ ) Φ ( x , λ ) φ ( x , λ ) Φ ˜ ( x , λ ) .
(35)
Taking (31) into consideration in (35) we get
P 11 ( x , λ ) = 1 + 1 Δ ( λ ) ψ ( x , λ ) [ φ ( x , λ ) φ ˜ ( x , λ ) ] + 1 Δ ( λ ) φ ( x , λ ) [ ψ ˜ ( x , λ ) ψ ( x , λ ) ] , P 12 ( x , λ ) = 1 Δ ( λ ) [ φ ˜ ( x , λ ) ψ ( x , λ ) φ ( x , λ ) ψ ˜ ( x , λ ) ] .
(36)
From the estimates as | λ |
| φ ( x , λ ) φ ˜ ( x , λ ) Δ ( λ ) | = O ( 1 | λ | 2 e | Im λ | μ + ( x ) ) , | ψ ˜ ( x , λ ) ψ ( x , λ ) Δ ( λ ) | = O ( 1 | λ | 2 e | Im λ | ( μ + ( π ) μ + ( x ) ) ) ,
we have from (36)
lim | λ | max x [ 0 , π ] | P 11 ( x , λ ) 1 | = lim | λ | max x [ 0 , π ] | P 12 ( x , λ ) | = 0
(37)

for λ G δ .

Now, if we take into consideration (32) and (35), we have
P 11 ( x , λ ) = φ ( x , λ ) s ˜ ( x , λ ) φ ˜ ( x , λ ) s ( x , λ ) + φ ˜ ( x , λ ) φ ( x , λ ) [ M ˜ ( λ ) M ( λ ) ] , P 12 ( x , λ ) = φ ˜ ( x , λ ) s ( x , λ ) φ ( x , λ ) s ˜ ( x , λ ) + φ ( x , λ ) φ ˜ ( x , λ ) [ M ( λ ) M ˜ ( λ ) ] .
Therefore if M ( λ ) = M ˜ ( λ ) , one has
P 11 ( x , λ ) = φ ( x , λ ) s ˜ ( x , λ ) s ( x , λ ) φ ˜ ( x , λ ) , P 12 ( x , λ ) = φ ( x , λ ) s ˜ ( x , λ ) s ( x , λ ) φ ˜ ( x , λ ) .

Thus, for every fixed x functions P 11 ( x , λ ) and P 12 ( x , λ ) are entire functions for λ. It can easily be seen from (37) that P 11 ( x , λ ) = 1 and P 12 ( x , λ ) = 0 . Consequently, we get φ ( x , λ ) φ ˜ ( x , λ ) and Φ ( x , λ ) Φ ˜ ( x , λ ) for every x and λ. Hence, we arrive at q ( x ) q ˜ ( x ) . □

The validity of the equation below can be seen analogously to [29]:
M ( λ ) = M ( 0 ) + n = 1 λ 2 α n λ n 2 ( λ 2 λ n 2 ) .
(38)

Theorem 6

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

Proof

From (38), it is clear that the function M ( λ ) can be constructed by λ n . Since λ ˜ n = λ n for every n N , we can say that M ( λ ) = M ˜ ( λ ) . Then from Theorem 5, it is obvious that L = L ˜ . □

Declarations

Acknowledgements

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

Authors’ Affiliations

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

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