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On the Basis Property of the Root Functions of Sturm-Liouville Operators with General Regular Boundary Conditions

We obtain the asymptotic formulas for the eigenvalues and eigenfunctions of the Sturm-Liouville operators with general regular boundary conditions. Using these formulas, we find sufficient conditions on the potential q such that the root functions of these operators do not form a Riesz basis.


Introduction and Preliminary Facts
In this paper we consider the operators generated in L 2 [0, 1] by the differential expression l (y) = −y ′′ + q(x)y (1) and regular boundary conditions that are not strongly regular. Note that, if the boundary conditions are strongly regular, then the root functions (eigenfunctions and associated functions) form a Riesz basis (this result was proved independently in [6], [9] and [17]). In the case when an operator is regular but not strongly regular, the root functions generally do not form even usual basis. However, Shkalikov [20,21] proved that they can be combined in pairs, so that the corresponding 2-dimensional subspaces form a Riesz basis of subspaces.
Case (a) The cases β 2 , β 4 = 0 , β 1 , β 3 = (−1) σ in (7), (9) for σ = 1 and σ = 0 coincide with the periodic and antiperiodic boundary conditions respectively. These boundary conditions are the ones more commonly studied. Therefore, let us briefly describe some historical developments related to the Riesz basis property of the root functions of the periodic and antiperiodic boundary value problems. First results were obtained by Kerimov and Mamedov [8]. They established that, if then the root functions of the operator L(q) form a Riesz basis in L 2 [0, 1], where L(q) denotes the operator generated by (1) and the periodic boundary conditions.
The first result in terms of the Fourier coefficients of the potential q was obtained by Dernek and Veliev [1]. Makin [11] extended this result for the larger class of functions. Shkalilov and Veliev obtained in [22] more general results which cover all results about periodic and antiperiodic boundary conditions discussed above.
Case (c) The cases β 2 , β 4 = 0 and β 1 , β 3 = (−1) σ are investigated in [12,13] and [19]. We call the boundary conditions (7) and (9) for β 2 , β 4 = 0 and β 1 , β 3 = (−1) σ which are different from the special cases (a) , (b) and (c) as general regular boundary conditions that are not strongly regular. Note that in any case β 1 , β 3 = − (−1) σ by (8) and (10). For the case (c) and general boundary conditions Makin [12,13] proved that the systems of the root functions of the Sturm-Liouville operators corresponding to these cases are Riesz bases in L 2 (0, 1) if and only if all large eigenvalues are multiple. Note that this result is not effective, since the conditions are given in implicit form and can not be verified for concrete potentials. In [19] we find explicit conditions on potential such that the system of the root functions of the Sturm-Liouville operator corresponding to the case (c) does not form a Reisz basis. Namely we proved that if lim n→∞ ln |n| ns 2n = 0, where s n = (q, sin 2πnt) and (., .) is the inner product in L 2 [0, 1] , then the large eigenvalues of each of the operators corresponding to these cases are simple for σ = 1. Moreover, if there exists a sequence {n k } such that (11) holds when n is replaced by n k , then the root functions of these operators do not form a Riesz basis. Similarly, if the condition lim n→∞ ln |n| ns 2n+1 = 0 holds instead of (11), then the same statements continue to hold for σ = 0. In this paper we find explicit conditions on potential q such that the system of the root functions of the Sturm-Liouville operator generated by (1) and the general regular boundary conditions does not form a Reisz basis. Now let us describe briefly the main results of this paper. Let T σ 1 (q) and T σ 2 (q) be the Sturm-Liouville operators associated by the boundary conditions (7) and (9) respectively. Without loss of generality we assume that then the large eigenvalues of T σ 1 (q) and T σ 2 (q) for σ = 1, are simple. Moreover if there exists a subsequence {n k } such that (12) holds whenever n is replaced by n k , then the system of the root functions of each operators T σ 1 (q) and T σ 2 (q) for σ = 1, does not form a Riesz basis. The same results continue to hold for T σ 1 (q) and T σ 2 (q) for σ = 0, if instead of (12) the condition holds.
The other main result is the following: If the potential q is an absolutely continuous function and then the large eigenvalues of T σ 1 (q) for σ = 0, 1 are simple and the system of the root functions of T σ 1 (q) does not form a Riesz basis. Similarly, if the condition holds instead of (13), then the same results remain valid for T σ 2 (q) for σ = 0, 1. Moreover we obtain subtle asymptotic formulas for the eigenvalues and eigenfunctions for the operators T σ 1 (q) and T σ 2 (q) for both cases q ∈ L 1 [0, 1] and q is an absolutely continuous function. Note that the general cases we investigate in this paper are essentially different from the case (c) as the method of investigations and obtained results.

Main Results
We will focus only on the operator T 1 1 (q). The investigations of the operators T 0 1 (q) , T 0 2 (q) and T 1 2 (q) are similar. First let us prove the following simple proposition about T 1 1 (0). Note that the simplest case q(x) ≡ 0 was completely solved in [10]. Here we write the asymptotic formulas for the eigenvalues of T 1 1 (0) in the form we need.

Proposition 1
The square roots (with nonnegative real part) of the eigenvalues of the operator T 1 1 (0) consist of the sequences {µ n,1 (0)} and {µ n,2 (0)} satisfying Proof. Using the fundamental solutions e iµx and e −iµx of −y ′′ = λy where µ = √ λ, one can readily see that the characteristic determinant ∆ 0 (µ) of T 1 1 (0) has the form After simplifying this equation, we have The solution of the first equation in (18) is µ n,1 (0) = 2πn for n ∈ Z, that is, (15) is proved.
To prove (16), we estimate the roots of (19). Using Rouche's theorem on the circle µ : |µ − 2πn| = c n for some constant c, one can easily see that, the roots of (19) has the form Now we prove that For this, let us consider the roots of (19) in detail. By (20) and (19) we have On the other hand, using Maclaurin expansion of e iξ and taking into account the second equality of (20) we see that This with (22) gives us (21). Now (16) follows from (20) and (21). Lemma is proved. For q = 0 it is known that (see (21) of [13]) the characteristic polynomial of T 1 1 (q) has the form where ∆ 0 (µ) is defined in (17) and After some arrangements (23) can be written in the form Using (17) in this formula we obtain Therefore the characteristic determinant ∆ (µ), can be written as where To obtain the asymptotic formulas for the eigenvalues of T 1 1 (q) first let us consider the roots of ∆ 1 (µ) .
Since c µ = o(1), there exists a sequence δ n such that δ n = o(1) and Now to estimate the zeros of g (µ), we use Rouche's theorem for the functions f (µ) and g (µ) on the circle where µ n,2 (0) is defined in (16) and ε n is chosen so that For this let us estimate |f (µ)| on γ n by using the Taylor series of f (µ) about µ n,2 (0) : there exist a constant c > 0 such that |f ′ (µ)| > c and for µ ∈ γ n . Thus by (30)-(33) and Rouche's theorem, there exists a root µ 1 n,2 of g (µ) inside the circle (31). Therefore (29), follows from (16). Now using (26), (27) and Lemma 1, we get one of the main results of this paper.
Theorem 1 (a) If (12) holds, then the large eigenvalues of T 1 1 (q) are simple and the square roots (with nonnegative real part) of these eigenvalues consist of two sequences {µ n,1 (q)} and {µ n,2 (q)} satisfying the asymptotic formulas µ n,2 (q) = 2πn + β 2 β 1 − 1 Moreover the normalized eigenfunctions ϕ n,1 (x) and ϕ n,2 (x) corresponding to the eigenvalues (µ n,1 (q)) 2 and (µ n,2 (q)) 2 satisfy the same asymptotic formula for j = 1, 2 (b) If there exists a subsequence {n k } such that (12) holds whenever n is replaced by n k , then the system of the root functions of T 1 1 (q) does not form a Riesz basis.
Since the distance between the centres of the circles Γ 1 (r n ) and Γ 2 (r n ) is of order 1 n , but r n = o 1 n , the eigenvalues inside the circles Γ 1 (r n ) and Γ 2 (r n ) are different, that is, they are simple. Now let us prove (36). Since the equation −y ′′ + q(x)y = µ 2 y has the fundamental solutions of the form (see p. 52 of [18]) the eigenfunctions of T 1 1 (q) are This with the formula for j = 1, 2 (see (34) and (35)), implies (36). (b) It is clear that if (12) holds for the subsequence {n k } then (36) holds for {n k } too. Therefore the angle between the eigenfunctions ϕ n k ,1 (x) and ϕ n k ,2 (x) corresponding to µ n k ,1 (q) and µ n k ,2 (q) tends to zero. Hence the system of the root functions of T 1 1 (q) does not form a Riesz basis (see [20]). Note that (b) follows also from (a) and Theorem 2 of [12,13].
Let q be an absolutely continuous function. Then using the integration by parts formula for s µ and c µ defined in (24) we obtain and If µ ∈ U (2πn), where U (2πn) is defined in the proof of Lemma 1 , then Therefore we have and hence by (25) Now we are ready to state the second main result of this paper.
Theorem 2 Let q be an absolutely continuous function and (13) for σ = 1 hold. Then (a) the large eigenvalues of T 1 1 (q) are simple and the square roots (with nonnegative real part) of these eigenvalues consist of two sequences {µ n,1 (q)} and {µ n,2 (q)} satisfying the system of the root functions of T 1 1 (q) does not form a Riesz basis.
Proof. (a) By (46) µ n,j (q) is a root of the equation Using (17) in this equation we get By the Taylor expansions of e −iµ and e iµ at 2πn we have for µ ∈ U (2πn). Therefore (49) can be written in the form To prove the formulas (47) and (48) we consider the equation (50). In (50) substituting x = µ (µ − 2πn) and taking into account that x = O(1) for µ ∈ U (2πn) we get To solve (51) we compare the roots of the functions and on the set U (2πn), where α n = o (1). The roots of f 1 (µ) are where by the assumption (13) for σ = 1. Therefore we have two different solutions x 1 and x 2 . On the other hand the solutions of the equations µ (µ − 2πn) = x 1 and µ (µ − 2πn) = x 2 with respect to µ are (54) and (55)), we have Now consider the roots of f 2 (µ) by using Rouche's theorem on for j = 1, 2, where r n is chosen so that By (52), (53) and (58) on γ 1 (r n ) ∩ γ 2 (r n ). Since the roots of f 1 (µ) are µ ij for i, j = 1, 2, we have where A is a constant. One can easily verify by using (56)  Therefore using the Taylor series of f 1 (µ) about µ 12 for µ ∈ γ 1 (r n ) and taking into account that (µ − µ 12 ) ∼ r n we obtain On the other hand by (58) we have for µ ∈ γ 1 (r n ). Therefore on γ 1 (r n ) In the same way we prove that (60) holds on γ 2 (r n ) too. Hence inside of each of the circles γ 1 (r n ) and γ 2 (r n ), there is one root of (49) denoted by µ n,1 (q) and µ n,2 (q) respectively. Since r n = o 1 n , µ n,1 (q) and µ n,2 (q) satisfy the formulas (47) and (48). To complete the proof of (a) it is enough to note that disks enclosed by the circles γ 1 (r n ) and γ 2 (r n ) have no common points and there are only two roots of (46) in the neighborhood of 2πn. Thus (a) is proved.
(b) If there exists a subsequence {n k } such that (12a) holds whenever n is replaced by n k , then the system of the root functions of T 0 1 (q) does not form a Riesz basis.