On the Basis Property of the Root Functions of Some Class of Non-self-adjoint Sturm--Liouville Operators

We obtain the asymptotic formulas for the eigenvalues and eigenfunctions of the Sturm-Liouville operators with some 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.

In conditions (2), (3), (4) and (5) if β = 1, β = −1, α = 1 and α = −1 respectively, then any λ ∈ C is an eigenvalue of infinite multiplicity. In (2) and (4) if β = −1 and α = −1 then they are periodic boundary conditions; In (3) and (5) if β = 1 and α = 1 then they are antiperiodic boundary conditions. These boundary conditions are regular but not strongly regular. Note that, if the boundary conditions are strongly regular, then the root functions form a Riesz basis (this result was proved independently in [6], [10] 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.
In the regular but not strongly regular boundary conditions, periodic and antiperiodic 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 q ∈ C 4 [0, 1], q(1) = q(0), then the root functions of the operator L 0 (q) form a Riesz basis in L 2 [0, 1], where L 0 (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]. They proved that if the conditions lim n→∞ ln |n| nq 2n = 0, q 2n ∼ q −2n (7) hold, then the root functions of L 0 (q) form a Riesz basis in L 2 [0, 1], where q n =: (q, e i2πnx ) is the Fourier coefficient of q and everywhere, without loss of generality, it is assumed that q 0 = 0. Here (., .) denotes the inner product in L 2 [0, 1] and a n ∼ b n means that a n = O(b n ) and b n = O(a n ) as n → ∞. Makin [11] improved this result. Using another method he proved that the assertion on the Riesz basis property remains valid if condition (7) holds, but condition (6) is replaced by a less restrictive one: q ∈ W s 1 [0, 1], q (k) (0) = q (k) (1), ∀ k = 0, 1, ..., s − 1 holds and | q 2n |> cn −s−1 with some c > 0 for sufficiently large n, where s is a nonnegative integer. Besides, some conditions which imply the absence of the Riesz basis property were presented in [11]. Shkalilov and Veliev obtained in [22] more general results which cover all results discussed above.
The basis properties of regular but not strongly regular other some problems are studied in [9,12,13]. It was proved in [12] that the system of the root functions of the operator generated by (1) and the boundary conditions forms an unconditional basis of the space L 2 [0, 1], where q (x) is an arbitrary complex-valued function from the class L 1 [0, 1], γ is an arbitrary nonzero complex constant and σ = 0, 1.
In this paper we prove that if where s k = (q, sin 2πkx) , then the large eigenvalues of the operators T 1 and T 3 are simple. Moreover, if there exists a sequence {n k } such that (8) holds when n is replaced by n k , then the root functions of these operators do not form a Riesz basis. Similarly, if lim n→∞ ln |n| ns 2n+1 = 0, then the large eigenvalues of the operators T 2 and T 4 are simple and if there exists a sequence {n k } such that (9) holds when n is replaced by n k , then the root functions of these operators do not form a Riesz basis.
Moreover we obtain asymptotic formulas of arbitrary order for the eigenvalues and eigenfunctions of the operators T 1 ,T 2 , T 3 and T 4 .

Main Results
We will focus only on the operator T 1 . The investigations of the operators T 2 , T 3 and T 4 are similar. It is well-known that ( see formulas (47a), (47b)) in page 65 of [18] ) the eigenvalues of the operators T 1 (q) consist of the sequences {λ n,1 }, {λ n,2 } satisfying for j = 1, 2. From this formula one can easily obtain the following inequality for j = 1, 2; k = n; k = 0, 1, ...; and n ≥ N, where we denote by N a sufficiently large positive integer, that is, N ≫ 1.
It is easy to verify that if q(x) = 0 then the eigenvalues of the operator T 1 , denoted by T 1 (0), are λ n = (2πn) 2 for n = 0, 1, . . . The eigenvalue 0 is simple and the corresponding eigenfunction is 1. The eigenvalues λ n = (2πn) 2 for n = 1, 2, . . . are double and the corresponding eigenfunctions and associated functions are respectively. Note that for any constant c, φ n (x) + cy n (x) is also an associated function. It can be shown that the adjoint operator T * 1 (0) is associated with the boundary conditions: It is easy to see that, 0 is a simple eigenvalue of T * 1 (0) and the corresponding eigenfunction is y * 0 (x) = x − 1 1 + β . The other eigenvalues λ * n = (2πn) 2 for n = 1, 2, . . ., are double and the corresponding eigenfunctions and associated functions are respectively. Let and The system of the root functions of T * 1 (0) can be written as {f n : n ∈ Z}, where One can easily verify that it forms a basis in L 2 [0, 1] and the biorthogonal system {g n : n ∈ Z} is the system of the root functions of T 1 (0), where To obtain the asymptotic formulas for the eigenvalues λ n,j and the corresponding normalized eigenfunctions Ψ n,j (x) of T 1 (q) we use (11) and the well-known relations (λ N,j − (2πn) 2 )(Ψ N,j , sin 2πnx) = (qΨ N,j , sin 2πnx) (18) and where which can be obtained by multiplying both sides of the equality by sin 2πnx and ϕ * n respectively. It follows from (18) and (19) that Moreover, we use the following relations for N ≫ 1,where M = sup |q n | . These relations are obvious for q ∈ L 2 (0, 1), since to obtain (22) and (23) we can use the decomposition of q sin 2πnx and qϕ * n by basis (16). For q ∈ L 1 (0, 1) see Lemma 1 of [23].
In these notations we have

Moreover λ ∈ U (n) is a double eigenvalue of T 1 if and only if it is a double root of (55) .
Proof. (a) By (10) the left hand side of (48) is O(n 1/2 ), which implies that u n,j = O(n −1/2 ). Therefore from (29) we obtain (54). Now suppose that there are two linearly independent eigenfunctions corresponding to λ n,j . Then there exists an eigenfunction satisfying Ψ n,j = √ 2 sin 2πnx + o (1) which contradicts (54).
(b) First we prove that the large eigenvalues λ n,j are the roots of the equation (55). It follows from (54), (27) and (15) that v n,j = 0. If u n,j = 0 then multiplying the equations (39) and (40) side by side and then canceling v n,j u n,j we obtain (55) . If u n,j = 0 then by (39) and (40) we have P n + B (λ n,j ) = 0 and λ n,j − (2πn) 2 − P * n − A ′ (λ n,j ) = 0 which mean that (55) holds. Thus in any case λ n,j is a root of (55). Now we prove that the roots of (55) lying in U (n) are the eigenvalues of T 1 . Let F (λ) be the left-hand side of (55) which can be written as and One can easily verify that the inequality holds for all λ from the boundary of U (n). Since the function G(λ) has two roots in the set U (n), by the Rouche's theorem we obtain that F (λ) has two roots in the same set. Thus T 1 has two eigenvalues (counting with multiplicities) lying in U (n) that are the roots of (55).
On the other hand, (55) has preciously two roots (counting with multiplicities) in U (n). Therefore λ ∈ U (n) is an eigenvalue of T 1 if and only if (55) holds. If λ ∈ U (n) is a double eigenvalue of T 1 then it has no other eigenvalues in U (n) and hence (55) has no other roots. This implies that λ is a double root of (55). By the same way one can prove that if λ is a double root of (55) then it is a double eigenvalue of T 1 .
Taking into account the last three equality and (34), (38), (50), (51), we see that (60) and (61) have the form Theorem 2 If (8) holds, then the large eigenvalues λ n,j are simple and satisfy the following asymptotic formulas for j = 1, 2. Moreover, if there exists a sequence {n k } such that (8) holds when n is replaced by n k , then the root functions of T 1 do not form a Riesz basis.
Proof. To prove that the large eigenvalues λ n,j are simple let us show that one of the eigenvalues, say λ n,1 satisfies (65) for j = 1 and the other λ n,2 satisfies (65) for j = 2. Let us prove that each of the equations (60) and (61) has a unique root in U (n) by proving that is a contraction mapping. For this we show that there exist positive real numbers K 1 , K 2 , K 3 such that where K 1 + K 2 + K 3 < 1. The proof of (66) is similar to the proof of (56) of the paper [26]. Now let us prove (67). By (62) and (8) we have On the other hand arguing as in the proof of (56) of the paper [26] we get Hence in any case we have Thus by the fixed point theorem, each of the equations (60) and (61) has a unique root λ 1 and λ 2 respectively. Clearly by (63) and (64), we have λ 1 = λ 2 which implies that the equation (55) has two simple root in U (n) . Therefore by Theorem 1(b), λ 1 and λ 2 are the eigenvalues of T 1 lying in U (n) , that is, they are λ n,1 and λ n,2 , which proves the simplicity of the large eigenvalues and the validity of (65). If there exists a sequence {n k } such that (8) holds when n is replaced by n k , then by Theorem 1(a) (Ψ n k ,1 , Ψ n k ,2 ) = 1 + O n −1/2 k . Now it follows from the theorems of [20,21] (see also Lemma 3 of [24]) that the root functions of T 1 do not form a Riesz basis. Now let us consider the operators T 2 , T 3 and T 4 . First we consider the operator T 3 . It is well-known that ( see formulas (47a), (47b)) in page 65 of [18] ) the eigenvalues of the operators T 3 (q) consist of the sequences {λ n,1,3 }, {λ n,2,3 } satisfying (10) when λ n,j is replaced by λ n,j,3 . The eigenvalues, eigenfunctions and associated functions of T 3 are λ n = (2πn) 2 ; n = 0, 1, 2, . . .
respectively. The biorthogonal systems analogous to (16), (17) are respectively. Analogous formulas to (18) and (19) are respectively, where Instead of (16)- (19) using (68)-(71) and arguing as in the proofs of Theorem 1 and Theorem 2 we obtain the following results for T 3 . (8) holds, then the large eigenvalues λ n,j,3 are simple and satisfy the following asymptotic formulas
for j = 1, 2. The eigenfunctions Ψ n,j,2 corresponding to λ n,j,2 obey Ψ n,j,2 = √ 2 cos (2n + 1) πx Moreover, if there exists a sequence {n k } such that (9) holds when n is replaced by n k , then the root functions of T 2 do not form a Riesz basis.
Moreover, if there exists a sequence {n k } such that (9) holds when n is replaced by n k , then the root functions of T 4 do not form a Riesz basis. If