Abstract
In this paper, we consider a discontinuous SturmLiouville operator with parameterdependent boundary conditions and two interior discontinuities. We obtain eigenvalues and eigenfunctions together with their asymptotic approximate formulas. Then, we give some uniqueness theorems by using Weyl function and spectral data, which are called eigenvalues and normalizing constants for solution of inverse problem.
MSC: 34A55, 34B24, 34L05.
Keywords:
SturmLiouville problem; eigenvalues; eigenfunctions; transmission conditions; Weyl function1 Introduction
It is well known that the theory of SturmLiouville problems is one of the most actual and extensively developing fields of theoretical and applied mathematics, since it is an important tool in solving many problems in mathematical physics (see [14]). In recent years, there has been increasing interest in spectral analysis of discontinuous SturmLiouville problems with eigenvaluelinearly and nonlinearly dependent boundary conditions [1,512]. Various physics applications of such problems can be found in [1,3,4,1319] and corresponding bibliography cited therein.
Some boundary value problems with discontinuity conditions arise in heat and mass transfer problems, mechanics, electronics, geophysics and other natural sciences (see [3] also [2029]). For instance, discontinuous inverse problems appear in electronics for building parameters of heterogeneous electronic lines with attractive technical characteristics [20,30,31]. Such discontinuity problems also appear in geophysical forms for oscillations of the earth [32,33]. Furthermore, discontinuous inverse problems appear in mathematics for exploring spectral properties of some classes of differential and integral operators.
Inverse problems of spectral analysis form recovering operators by their spectral data. The inverse problem for the classical SturmLiouville operator was studied first by Ambarsumian in 1929 [34] and then by Borg in 1945 [35]. After that, direct and inverse problems for SturmLiouville operator have been extended to so many different areas.
We consider a discontinuous SturmLiouville problem L with function
where
and , , and are given positive real numbers; ; is a complex spectral parameter; boundary conditions at the endpoints
with discontinuity conditions at two points ,
where , and , (, ) are real numbers and
In the present paper, we construct a linear operator T in a suitable Hilbert space such that problem (1)(7) and the eigenvalue problem for operator T coincide. We investigate eigenvalues and eigenfunctions together with their asymptotic behaviors of operator T. Besides, we study some uniqueness theorems according to Weyl function and spectral data, which are called eigenvalues and normalizing constants.
2 Operator formulation and spectral properties
We make known the inner product in the Hilbert space , where , ℂ denotes the Hilbert space of complex numbers and a selfadjoint operator T defined on H such that (1)(7) can be dealt with as the eigenvalue problem of operator T. We define an inner product in H by
for
Consider the operator T defined by the domain
such that for and also  are satisfied for f.
Thus, we can rewrite the considered problem (1)(7) in the operator form as .
Theorem 1The operatorTis symmetric inH.
Proof Let . By two partial integrations, we get
where by , we denote the Wronskian of the functions f and g as
Since f and g satisfy the boundary conditions (2)(3) and transmission conditions (4)(7), we obtain
Thus, we have , i.e., T is symmetric. □
Lemma 1Problem (1)(7) can be considered as the eigenvalue problem of the symmetric operatorT.
Corollary 1All eigenvalues and eigenfunctions of problem (1)(7) are real, and two eigenfunctionsand, corresponding to different eigenvaluesand, are orthogonal in the sense of
We define the solutions
of equation (1) by the initial conditions
and similarly,
respectively.
These solutions are entire functions ofλfor each fixedand satisfy the relation
Then the following integral equations and also asymptotic behaviors hold for:
Then the following integral equations and also asymptotic behaviors hold for:
The functionis called the characteristic function, and numbersare called the normalizing constants of problem (1)(7) such that
Lemma 4The following equality holds for each eigenvalue
Proof Since
we get
After that, add and subtract on the lefthand side of the last equality, and by using conditions (2)(7), we obtain
or
For , is obtained by using the equality
and (12). □
Corollary 2The eigenvalues of problemLare simple.
Lemma 5[36]
Letbe the set of real numbers satisfying the inequalities, and letbe the set of complex numbers. If, then the roots of the equation
have the form
Now, from Lemma 2 and (11), we can write
We can see that nonzero roots, namely of the equation , are real and analytically simple.
Furthermore, it can be proved by using Lemma 5 that
Theorem 2The eigenvalueshave the following asymptotic behavior for sufficiently largen:
Proof Denote
where and δ is a sufficiently small number. The relations
and
Then, by Rouche’s theorem that the number of zeros of coincides with the number of zeros of in , namely zeros, . In the annulus, between and , has accurately one positive zero, namely , for . So, it follows that . Applying to Rouche’s theorem in for sufficiently small ε and sufficiently large n, we get . Finally, we obtain the asymptotic formula
Denote
which are independent of and are entire functions such that , , .
It can be easily seen that
□
Example Let , , , , , , , , , , , , , , , .
Since
the eigenvalues of the boundary value problem (1)(7) satisfy the following asymptotic formulae:
3 Inverse problems
In this section, we study the inverse problems for the reconstruction of the boundary value problem (1)(7) by Weyl function and spectral data.
We consider the boundary value problem with the same form of L but with different coefficients , , , , , , , .
If a certain symbol α denotes an object related to L, then the symbol denotes the corresponding object related to .
The Weyl function Let be a solution of equation (1), which satisfies the condition and transmissions (4)(7).
Assume that the function is the solution of equation (1) that satisfies the conditions , and the transmission conditions (4)(7).
Since , the functions χ and φ are linearly independent. Therefore, the function can be represented by
or
that is called the Weyl solution, and
is called the Weyl function.
Theorem 3If, then, i.e., , a.e. and, , , , , , .
Proof We introduce a matrix by the formula
or
Thus, if , then the functions and are entire in λ for each fixed x.
Denote and , where w is sufficiently small number, and are square roots of the eigenvalues of the problem L and , respectively. It is easily shown that
are valid for sufficiently large , where , and . Hence, Lemma 2 and (18) yield that
According to (19), and Liouville’s theorem, and for . By virtue of (17), we get
It is obvious that
and similarly, . Thus, we have .
Otherwise, the following asymptotic expressions hold
Without loss of generality, we assume that and . From (20)(21), we get for and also
As in (22), we contradict . Therefore, , . Thus, , and . Hence, from equation (1) and transmission conditions (4)(7), , a.e., , , , and from (9) and (10), , , , . □
Lemma 6The following representation holds
Proof Weyl function is a meromorphic function with respect to λ, which has simple poles at . Therefore, we calculate
Let , where ε is a sufficiently small number. Consider the contour integral , . For ,
satisfies. Using this equality and (16), we get
Thus, . As a result, the residue theorem and (23) yield
□
Theorem 4Ifandfor alln, then, i.e., , a.e., , , , , , , . Hence, problem (1)(7) is uniquely determined by spectral data.
Proof If and for all n, then by Lemma 6. Therefore, we get by Theorem 3.
Let us consider the boundary value problem that we get the condition instead of condition (2) in L. Let be the eigenvalues of the problem . It is clear that are zeros of
□
Theorem 5Ifandfor alln, then, , , .
Hence, the problemLis uniquely determined by the sequencesand, except coefficientsand.
Proof Since the characteristic functions and are entire of order , functions and are uniquely determined up to multiplicative constant with their zeros by Hadamard’s factorization theorem [37]
where C and are constants dependent on and , respectively. Therefore, when and for all n, and . Hence, . As a result, we get by (16). So, the proof is completed by Theorem 3. □
Competing interests
The author declares that he has no competing interests.
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