Abstract
This work aims to examine a SturmLiouville operator with a piecewise continuous coefficient and a spectral parameter in boundary condition. The orthogonality of the eigenfunctions, realness and simplicity of the eigenvalues are investigated. The asymptotic formula of the eigenvalues is found, and the resolvent operator is constructed. It is shown that the eigenfunctions form a complete system and the expansion formula with respect to eigenfunctions is obtained. Also, the evolution of the Weyl solution and Weyl function is discussed. Uniqueness theorems for the solution of the inverse problem with Weyl function and spectral data are proved.
MSC: 34L10, 34L40, 34A55.
Keywords:
SturmLiouville operator; expansion formula; inverse problem; Weyl function1 Introduction
In recent years, there has been a growing interest in physical applications of boundary value problems with a spectral parameter, contained in the boundary conditions. The relationship between diffusion processes and SturmLiouville problem with eigenparameter in the boundary conditions has been shown in [1]. Another example of this relationship between the same problem and the wave equation has been examined in [2,3]. SturmLiouville problems with a discontinuous coefficient arise upon nonhomogeneous material properties.
In a finite interval, inverse problems for the SturmLiouville operator with spectral parameter, contained in the boundary conditions, have been investigated, and the uniqueness of the solution of these problems has been shown in [49]. The inverse problem has been analyzed by zeros of the eigenfunctions in [6], by numerical methods in [9] and by two spectra, consisting of sequences of eigenvalues and the normed constants in [10]. In [11,12], eigenvaluedependent inverse problem with the discontinuities inside the interval was examined by the Weyl function. In a finite interval, discontinuous and no eigenvalue parameter containing direct problem and inverse problem with the Weyl function were discussed in [13,14]. The similar problem was investigated in the half line by scattering data in [15,16].
We consider the boundary value problem
where is a real valued function, λ is a complex parameter, , , , are real numbers and
2 Special solutions
Let and be the solutions of equation (1) satisfying the initial conditions
For the solution of equation (1), the following integral representation is obtained in [13] for all λ:
The kernel has the partial derivative belonging to the space for every , and the properties below hold:
Moreover, if is differentiable, then the following are valid
Using the representation of the solution and formula
we obtain the integral representation of the solution
where . The kernel can be represented with the coefficients and
With the help of equation (6), we have a representation for the function
Denote
which is independent of . Substituting and into (8), we get
is an entire function of λ and is called the characteristic function of the boundary value problem (1)(3).
3 Some spectral properties
Lemma 1The square values of the rootsof the characteristic function coincide with the eigenvalues of the boundary value problem (1)(3), and for every, there exists a sequencesuch that
whereandare the eigenfunctions of the boundary value problem (1)(3), corresponding to the eigenvalue.
Proof The proof can be done in a similar way to [8]. Indeed, let us assume that is an eigenvalue of the function . Then
holds, i.e., the functions and are linearly dependent (), and they satisfy the boundary conditions (2), (3). Hence is an eigenvalue, and are eigenfunctions, related to this eigenvalue. Conversely, let be an eigenvalue of the operator A, and let , be the corresponding eigenfunctions. Then the boundary conditions (2), (3) hold both for the eigenfunctions and . Additionally, if the functions and satisfy the conditions , , then , . According to boundary conditions (2), (3), we have
Similarly, if we assume that , , then , . Again from the boundary conditions (2), (3), it is obvious that
Therefore, we have proved that for each eigenvalue , there exists only one (up to a multiplicative constant) eigenfunction. □
In the Hilbert space an inner product defined by
where
Let us define
with
where
The boundary value problem (1)(3) is equivalent to the equation .
The eigenfunctions of operator A are in the form of
Lemma 2The eigenfunctionsand, corresponding to different eigenvalues, are orthogonal.
Proof Since and are the solutions of the boundary value problem (1)(3), the equations below are valid
Multiplying the first equation by and the second equation by and adding together, we have
Integrating it from 0 to π, and using the boundary condition (3), we obtain
Since , the lemma is proved. □
Corollary 3The eigenvalues of the boundary value problem (1)(3) are real.
The values
are called the norming constants of the boundary value problem (1)(3).
Now, let us agree to denote differentiation with respect to λ with a dot ().
Lemma 4The following equality holds
Proof Since
we get
With the help of (2) and (3),
Taking into consideration (9) and (10), for , we arrive (11). □
4 Asymptotic formulas of eigenvalues
Let be the solution of equation (1) satisfying the initial conditions (4) when
The eigenvalues () of the boundary value problem (1)(3) when can be found using the equation
from [17] (see also [18]) and can be represented in the following way
Lemma 6Rootsof the functionare separated, i.e.,
Proof Assume the contrary. Then there are sequences , of zeros of functions such that
Since the eigenfunctions , are orthogonal, we have
where
Thus,
Let us prove that as . In fact, (12) implies the following estimate
Consequently, holds uniformly on . Now, passing to the limit in equality (14), as , we have . This is a contradiction, and it proves the validity of lemma’s statement. □
Lemma 7The eigenvalues of the boundary value problem (1)(3) are in the form of
Proof From (6), it follows that
Lemma 1.3.1 in [19] and from [17], we get
Therefore, for sufficiently large n, on the contours
we have
By the Rouche theorem, we obtain that the number of zeros of the function inside the contour coincides with the number of zeros of the function . Furthermore, applying the Rouche theorem to the circle , we get that, for sufficiently large n there exists only one zero of the function in . Owing to the arbitrariness of we have
Substituting (18) into (16), we get
Hence, as , taking into account the equality and relations , , integrating by parts and using the properties of the kernel , we have
where
Let us show that . It is obvious that
can be reduced to
It is clear from [19] (p.66) that . By virtue of this we have (see [20,21]). Therefore, as
5 Expansion formula
Assume that is not a spectrum point of operator A. Then, there exists resolvent operator . Let us find the expression of the operator .
Lemma 8The resolventis the integral operator with the kernel
Proof To construct the resolvent operator of A, we need to solve the boundary value problem
where . By applying the method of variation of constants, we seek the solution of the problem (20)(22) in the following form
and we get the coefficients and as
Substituting (24) and (25) into (23) and taking into account the boundary conditions (21) and (22), we have
Theorem 9The eigenfunctionsof the boundary value problem (1)(3) form a complete system in.
Proof With the help of (9) and (11), we can write
Using (19) and (26), we get
Then from (28), we have . Consequently, for fixed the function is entire with respect to λ.
Let us denote that
where δ is a sufficiently small positive number. (16) is valid from Theorem 12.4 in [22] for .
On the other hand, we can say from Lemma 1.3.1 in [19] that for every , the following relation holds
Also, for , the relations below hold
From the estimates (31)(34), it is obvious that
From (26), it follows that for fixed and sufficiently large , we have
Using maximum principle for module of analytic functions and Liouville theorem, we get . From this and the expression of the boundary value problem (20)(22), we obtain that a.e. on . Thus, we reach the completeness of the eigenfunctions in . □
Theorem 10If, then the expansion formula
is valid, where
and the series converges uniformly with respect to. For, the series converges in, moreover, the Parseval equality holds
Proof Since and are the solutions of the boundary value problem (1)(3), we have
Integrating by parts and taking into account the boundary conditions (2), (3), we obtain
where
as . Let us consider the following contour integral
where is a contour oriented counterclockwise, and n is a sufficiently large natural number. With the help of Residue theorem, we get
where
On the other hand, taking into account (38), we have
Comparing (39) and (40), we obtain
where
Thus, we obtain
Now, let us show that
From estimates (31)(34) of solutions , and the inequality (35) for the function , it follows, for fixed and sufficiently large
Let us show that . If we suppose that , and by then integrate by parts the expression of , we obtain
In general case, let us take an arbitrary fixed number and assume that , such that . Then we can find a that for and . Also, using the equation below,
and with the help of the estimates of functions , and , we get
From the arbitrariness of ϵ, we reach
The validity of (42) can be easily seen from (43) and (44). Thus, we obtain
If we take
the last equation gives us the expansion formula
Since the system of is complete and orthogonal in , the Parseval equality
holds. Extension of the Parseval equality to an arbitrary vectorfunction of the class can be carried out by usual methods. □
6 Weyl solution, Weyl function
Let be the solution of equation (1) that satisfies the conditions
Denote by the solution of equation (1), which satisfies the initial conditions , . Then the solution can be represented as follows
or
Denote
It is clear that
The functions and are respectively called the Weyl solution and the Weyl function of the boundary value problem (1)(3). The Weyl function is a meromorphic function having simple poles at points eigenvalues of boundary value problem (1)(3). Relations (46), (48) yield
It can be shown that
Let us take into consideration a boundary value problem with the coefficient similar to (1)(3) and assume that if an element α belongs to boundary value problem (1)(3), then belongs to one with .
Validity of the equation below can be shown analogously to [8]
Theorem 11The boundary value problem (1)(3) is identically denoted by the Weyl function. (If, then.)
Proof Let us identify the matrix as
From (50) and (52), we have
or
Taking equation (49) into consideration in (54), we get
Now, from the estimates
and
we have from equation (55)
for . Now, if we take consideration equation (48) into (53), we get
Therefore, if , then and are entire functions for every fixed x. It can easily be seen from equation (56) that and . Consequently, we get and for every x and λ. Hence, we arrive at . □
Theorem 12The spectral data identically define the boundary value problem (1)(3).
Proof From (51), it is clear that the function can be constructed by . Since for every , from Theorem 10, we can say that . Then from Theorem 11, it is obvious that . □
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the manuscript and read and approved the final manuscript.
Acknowledgements
This work is supported by the Scientific and Technological Research Council of Turkey (TÜBİTAK).
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