Spectral Expansion Series with Parenthesis for the Nonself-adjoint Periodic Differential Operators

In this paper we construct the spectral expansion for the differential operator generated in all real line by ordinary differential expression of arbitrary order with periodic complex-valued coefficients by introducing new concepts as essential spectral singularities and singular quasimomenta and using the series with parenthesis. Moreover, we find a criteria for which the spectral expansion coincides with the Gelfand expansion for the self-adjoint case.

The obtained results can be easily curry out for p k ∈ L 1 [0, 1], ∀k = 2, 3, ..., n by combining the approaches of this paper and the paper [18] , where the asymptotic formulas were obtained for operator L with arbitrary Lebesgue integrable on (0, 1) complex-valued coefficients.
Tkachenko proved in [11] that the nonself-adjoint Hill's operator (one-dimensional Schrodinger operator with a periodic potential) H can be reduced to triangular form if all eigenvalues of the operators H t for t ∈ [0, 2π) are simple, where the operators L and L t , in case l(y) = −y ′′ + q(x)y, are redenoted by H and H t respectively. McGarvey in [5,6] proved that L is spectral operator if the projections of the operator L are uniformly bounded. However, in general, the eigenvalues are not simple and the projections are not uniformly bounded, since the operator H with potential q(x) = e i2πx has infinitely many spectral singularities (see [1]). Note that the spectral singularity of L is the point of σ(L) in neighborhood on which the projections of the operator L are not uniformly bounded. The spectral expansion for the self-adjoint operator L was constructed by Gelfand [2], Titchmarsh [10] and Tkachenko [13]. The existence of the spectral singularities and the absence of the Parseval's equality for the nonself-adjoint operator L t do not allow us to apply the elegant method of Gelfand (see [2]) for construction of the spectral expansion for the nonself-adjoin operator L. These situation essentially complicate the construction of the spectral expansion for the nonself-adjoint case. Gesztezy and Tkachenko [3] proved two versions of a criterion for the Hill operator H(q) with q ∈ L 2 [0, 1] to be a spectral operator of scalar type, in sense of Dunford, one analytic and one geometric. The analytic version was stated in term of the solutions of Hill's equation. The geometric version of the criterion uses algebraic and geometric properties of the spectra of periodic/antiperiodic and Dirichlet boundary value problems. In paper [20][21] we found the conditions on the potential q such that H(q) has no spectral singularity at infinity and it is an asymptotically spectral operator and in [22] we considered the spectral expansion for the asymptotically spectral operator.
In this paper, we investigate the spectral expansion for the general and frequent case when the operator L is not an asymptotically spectral operator. We consider the effect of the spectral singularities, especially the spectral singularity at infinity, to the spectral expansion and try to write the spectral expansion in term of the quasimomentum t as precise as possible. For this we introduce new concepts as essential singular quasimomentum (ESQ) and essential spectral singularities (ESS) defined at the end of this section (see Definition 5).
Here we do not investigate the technical details as replacement of t to the spectral parameter λ and regularization which can be done in a standard way. The case n = 2 is investigated in detail. To discuss more precisely the obtained results we need some preliminary facts about: (a) eigenvalues and eigenfunction of L t , (b) spectral singularities of L and (c) spectral expansion of L (a) Eigenvalues and eigenfunctions of L t . First let us introduce some definitions, notations and results of [8] from which we use fundamentally. Recall that if n is odd, n = 2µ − 1, then the boundary conditions (4)  . . . ω n−1 n are both different from zero (see [8], p. 56-57), where the numbers ω 1 , ω 2 , . . . , ω n denote the different nth roots of −1. It is clear that where A is Vandermonde determinant for ω 1 , ω 2 , . . . , ω n . Hence the boundary conditions (4) are regular. If n is even, n = 2µ, then the boundary conditions (4) are regular if the numbers θ −1 and θ 1 defined by the identity θ−1 s + θ 0 + θ 1 s = (−e it ) µ−1 × If the roots ξ ′ and ξ ′′ of θ 1 ξ 2 + θ 0 ξ + θ −1 = 0 are simple, that is, if θ 2 0 − 4θ 1 θ −1 = 0 then the regular boundary conditions are said to be strongly regular ( see [7]). Clearly ξ ′ = e it , ξ ′′ = e −it , θ 2 0 − 4θ 1 θ −1 = 0, ∀t = 0, π.
Hence, if n = 2µ and t = 0, π, then the boundary conditions (4) are strongly regular. If n = 2µ−1, then each regular boundary condition is strongly regular and hence the conditions (4) are strongly regular for all values of t. Therefore (see [7], for the case (2) and [14] for the case (3) the root functions of L t ,with t = 0, π for n = 2µ and with arbitrary t for n = 2µ − 1, form a Riesz basis in L 2 (0, 1). The eigenvalues λ k (t) of the boundary-value problem (1), (4) are −(ρ k (t)) n where ρ k (t) are the zeroes of the characteristic determinant Here y 1 (x, ρ), y 2 (x, ρ), . . . , y n (x, ρ) are n linearly independent solutions of the equation l(y) = −ρ n y satisfying for ν = 1, 2, · · · , n. Besides we use the following result of [18] about the eigenvalues that were easily obtained by following the arguments in [8] of the proof of the asymptotic formulas for the eigenvalues and eigenfunction, and taking into consideration the uniformity with respect to t.
Result of [18]. If n = 2µ − 1 then for arbitrary t, if n = 2µ then for t = 0, π, the large eigenvalues of the operator L t consist of the sequence {λ k (t) : |k| > N } satisfying For any fixed h ( h ∈ (0, 1)) there exists an integer N (h) such that the eigenvalue These formulas are uniform with respect to t in Q for n = 2µ − 1 and in Q h for n = 2µ. The uniformity of (9) and (10) means that there exists a positive constant c 1 , independent on t, where t ∈ Q for n = 2µ − 1 and t ∈ Q h for n = 2µ, such that for j = n − 2 and j = −1 respectively. From (7) it is not hard to see that the characteristic determinant ∆(λ, t), where λ = −ρ n , has the form ∆(λ, t) = e int + a 1 (λ)e i(n−1)t + a 2 (λ)e i(n−2)t + ... + a n (λ), (12) that is, ∆(λ, t) is a polynomial of e it with entire coefficients a 1 (λ), a 2 (λ), .... Since da1 dλ and the resultant determinant R(λ) of the polynomials ∆ and ∂∆ ∂λ are nonzero entire functions, the set A of t for which the operator L t has a multiple eigenvalue is countable. Thus and for t / ∈ A all eigenvalues of L t are simple eigenvalues. It follows from Result of [18] (see above) that if n = 2µ − 1 then the set A ∩ Q is finite, if n = 2µ then the possible accumulation points of the set A ∩ Q are 0 and π. Denote the set (A ∩ Q) ∪ {0, π} by A.
In [19] we proved that the eigenvalues of L t can be numbered (counting the multiplicity) as λ 1 (t), λ 2 (t), ..., such that | λ p (t) |→ ∞ as p → ∞ and for each p the function λ p (t) is continuous in Q and is analytic in Q\A(p), where A(p) is a finite subset of A ∩ Q . It readily follows from the arguments of [19] (see also Remark 5 ) that they can be renumbered by elements of the set of integers Z such that (9) holds. In this paper we prefer to consider the set of eigenvalues L t for t ∈ Q as {λ k (t) : k ∈ Z} .
Replacing the ith row of the determinant ∆(λ k (t), t) (see (7)) by the row vector we get the eigenfunctions Ψ k,i,t (x) for i = 1, 2, ..., n corresponding to the eigenvalue λ k (t). It is clear that if t / ∈ A then at least one these eigenfunctions is nonzero, since the rank of the matrix in (7) is n − 1. Moreover it is analytic with respect to t for each x and its norm is continuous in some neighborhood of t / ∈ A. For t / ∈ A we denote by Ψ k,t the normalized eigenfunctions of L t corresponding to the eigenvalue λ k (t) and by X k,t the eigenfunction of the adjoint operator L * t which is orthogonal to all eigenfunctions of L t except Ψ k,t and (Ψ k,t , X k,t ) = 1. It is clear that where Ψ * k,t is the normalized eigenfunctions of L * t corresponding to the eigenvalue λ k (t) and Thus {X k,t : k ∈ Z} is a biorthogonal system of {Ψ k,t : k ∈ Z}. Is clear that Ψ * k,t , Ψ k,t and α k (t) are continuous and α k (t) = 0 for t ∈ Q\A, since the system {Ψ k,t : k ∈ Z} is complete.
In [16] we defined projection P (γ) of L for the arc γ ⊂ {λ p (t) : t ∈ (−π, π]} ⊂ σ(L) which does not contain the multiple eigenvalues of the operators L t , as follows where γ 1 ε ⊂ ρ(L) and γ 2 ε ⊂ ρ(L) are the connected curves lying in opposite sides of γ and Here ρ(L) denotes the resolvent set of L. Moreover, we proved that if γ does not contain the common roots of then where δ = {t ∈ (−π, π] : λ p (t) ∈ γ}. It gives the following definition of the spectral singularities.
Definition 1 We say that λ ∈ σ(L(q)) is a spectral singularity of L(q) if for all ε > 0 there exists a sequence {γ n } of arcs γ n ⊂ {z ∈ C :| z − λ |< ε} such that they do not contain multiple eigenvalues and common roots of (16) and In the similar way, we defined in [20] the spectral singularity at infinity.

Definition 2
We say that the operator H has a spectral singularity at infinity if there exists a sequence {γ n } of the arcs such that they do not contain multiple eigenvalues and common roots of (16) and d(0, γ n ) → ∞ as n → ∞ and (18) holds, where d(0, γ n ) is the distance from the point (0, 0) to the arc γ n .
The asymptotic spectrality of the operator H(q) was defined in [20] as follows. Let e(t, γ) be the spectral projection defined by contour integration of the resolvent of L t (q), where γ ∈ R and R is the ring consisting of all sets which are the finite union of the half closed rectangles. In [5] it was proved Theorem 3.5 which can be written in the form: e(t, γ) ) < ∞.
According to this theorem, in [20] we gave the following definition of the asymptotic spectrality.

Definition 3
The operator H(q) is said to be an asymptotically spectral operator if there exists a positive constant C such that sup γ∈R(C) (ess sup t∈(−π,π] e(t, γ) ) < ∞, where R(C) is the ring consisting of all sets which are the finite union of the half closed rectangles lying in {λ ∈ C :| λ |> C}.
In [20] the definitions 2,3 are given for the operator H. Everywhere replacing H by L we get the corresponding definitions for L.
One can readily see that if λ k (t) is a simple eigenvalue of L t , then the spectral projection e(t, γ) defined by contour integration of the resolvent of L t (q), where γ is closed contour containing only the eigenvalue λ k (t) has the form (see p. of [8]) and This with (17) and Definition 1implies that the spectral singularity can be defined as follows Definition 4 A point λ ∈ σ(L) is called a spectral singularity of L if for all ε > 0 the projections of the operators L t for t ∈ [0, 2π) corresponding to a simple eigenvalue lying in {z ∈ C :| z − λ |< ε} are not uniformly bounded.
Since the proof of (17) is long and technical, in order to avoid eclipsing the essence by the technical details, in [], we used Definition 4. Note that this definition is natural, since in the spectral expansion of L the eigenfunctions and eigenprojections of L t for t ∈ [0, 2π) are used.
(c) Spectral expansion of L. By Gelfand's Lemma (see [2]) every compactly supported and continuous function f can be represented in the form where f t is defined by Then the following relations hold where Ψ k,t and X k,t are extended to (−∞, ∞) by Let h ∈ (0, 1 2 )\A, and let l be a continuous curve joining the points −h and 2π − h and satisfying if n = 2µ − 1 and n = 2µ respectively. Suppose f is a compactly supposed and continuous function. Then f t (x) is an analytic function of t in a neighborhood of D for each x, where D is the closure of the domain enclosed by l ∪ [−h, 2π − h]. Hence the Cauchy's theorem and (20) give By (24), for each t ∈ l the system {Ψ k,t : k ∈ Z} is a Reisz basis of L 2 [0, 1] and hence we have a decomposition where Using (26) in (25), we get In [15,17] we proved that for a continuous curve l that is compact subset of Q h \A the series in (28) can be integrated term by term l k∈Z Therefore we have where the series converges in the norm of L 2 (a, b) for every a, b ∈ R.
To get the spectral expansion in the term of t from (30) we need to replace the integrals over l by the integral over [0, 2π) if it is possible. For each x the expression a k (t)Ψ k,t (x) is continuous on [0, 2π)\A and hence is it continuous almost everywhere in [0, 2π). The existence of the integral of a k (t)Ψ k,t over [0, 2π) depend on the behavior of α k (t) when t tends to a point of A, since (see (14) and (27)). If lim t→c α k (t) = 0 for some c ∈ A ∩ [0, 2π) then 1 α k (t) and a k (t)Ψ k,t (x) are unbounded. It follows from (17) and Definition 1 that the boundlessness of 1 α k (t) is the characterization of the spectral singularities. Since 1 α k (t) may have an integrable boundlessness, in general, its boundlessness is not a criterion for the nonexistence of the integral of a k (t)Ψ k,t over [0, 2π]. Moreover, the considerations of the spectral singularities, that is the consideration of the boundlessness of 1 α k (t) play only the crucial rule for the investigations of the spectrality of L. On the other hand, the papers [1,3,16] show that the Hill's operator H, in general, is not a spectral operator. Therefore to construct the spectral expansion for the operator L in the general case we need to introduce the new concepts connected with the integrability of a k (t)Ψ k,t over [0, 2π). Indeed, to construct the spectral expansion for compactly supported continuous function f we need to consider the existence of the integral w.r.t. t of the function for all x ∈ [0, 1] which can be reduced to the investigation of the integrability (not the boundlessness) of 1 α k (t) . In this paper, everywhere, we suppose that f is a compactly supported and continuous function and use the following obvious claim.
Proposition 1 Let f be compactly supported and continuous function and λ k (t) be a simple eigenvalue for t ∈ I, where I is a measurable subset of (−π, π]. If 1 α k is integrable on I then a k (t)Ψ k,t (x) is also integrable on I for each x ∈ [0, 1].
Thus we introduce the following notions for the construction of the spectral expansion.
As we noted above if λ k (t) is a simple eigenvalue then the function 1 α k is continuous in some neighborhood of t. Therefore, the following proposition follows from (17)  In this paper we construct the spectral expansion by using the concepts ESS and ESQ. First (in Section 2) we consider the simplest case n = 2µ − 1 in which there exists at most finite number of the multiple eigenvalues. Then, in Section 3, we investigate the complicated case n = 2µ. In Section 4, the Hill's operator is considered in detail.

Spectral Expansion for Odd Order
In this section we consider the case n = 2µ − 1 which is simpler than the case n = 2µ. Moreover, we consider the odd order case so that it helps to read the complicated even order case. As it was noted in the introduction in the case By Result of [18] (see introduction) for n = 2µ − 1 the number of multiple eigenvalues of L tj is finite. Therefore using the Definition 2 and taking into account that under condition (2) the eigenfunction Ψ * k,t of the adjoint operator L * t for t ∈ [0, 2π] satisfies (10) we obtain the following: Proposition 3 If n = 2µ − 1, then the operator L has at most finite number spectral singularities and has no spectral singularity at infinity.
the different multiple eigenvalues of the operator L tj . The set is finite since the multiplicity of the eigenvalues of L tj is finite.
where |T| denotes the number of the elements of the set T. It mean that Λ v (t j ) is the p multiple root of the equation The following proposition immediately follows from implicit function theorem and (12).
is the eigenvalue of L tj of multiplicity p then there exist a disk and positive numbers r 1 and r 2 , where r 1 < r 2 , such that the followings hold: where \{t j } and for t = t j have no other eigenvalues than (37) and Λ v (t j ) respectively. Now using this proposition we prove the following theorem that helps to replace the integrals over l by the integral over [0, 2π) in (30).
T(v, j) and ε j are defined in (34) and in Proposition 4 respectively, a k (t)Ψ k,t (x) is defined in introduction and f is a compactly supposed and continuous function.
Proof. Let C be the circle with center Λ v (t j ) and radius r, where r 1 < r < r 2 and the numbers r i for i = 1, 2 are defined in Proposition 4. Consider the total projections T (x, t) =: f t is defined in (21) and G(x, ξ, λ, t) is the Green function of the operator L t . It is well-known that the Green function G(x, ξ, λ, t) of L t is defined by formulas (see [8] pages 36 and 37) where H(x, ξ, λ, t) is the (n + 1) × (n + 1) determinant determined as follows g(x, ξ) = ± 1 2W (ξ) .
and W (ξ) is the Wronskian of the solutions y 1 , y 2 , . . . , y n . In (44) the positive sign being taken if x > ξ, and the negative sign if x < ξ. By Proposition 4 the circle C encloses only the eigenvalues (37) and lies in the resolvent sets of L t for t ∈ U (t j ). By (12), ∆(λ, t) is continuous in the compact C × U (t j ) and hence there exists a positive constant c such that Therefore from (40)-(45) we readily obtain the following results about the functions A(x, λ, t) and T (x, t), where f t (x) is defined by (21) and f is a compactly supported and continuous function which implies that there are finite number of summands in (21): (a) There exists a constant M v,j such that Note that the proof of (a) − (c) follows immediately from (40)-(45) and the proof of (d) follows from (a) − (c).
Since inside of the circle C the operator L t for t ∈ U (t j )\{t j } has p simple eigenvalues (37) we have where Ψ k,t and X k,t are defined in the introduction. On the other hand by (d) for each x ∈ [0, 1] the function T (x, t) is analytic on the disk U (t j ). Therefore we have Finally taking into account that, by Proposition 4, for t ∈ γ(t j , ε) the eigenvalues are simple and hence Thus (38) follows from (47)-(49). Note that by (d) for each x the function T (x, t) is a bounded function in U (t j )\{t j }, but, in general, the summands of (47) for same values of k may became unbounded and nonintegrable. This situation agree with the well-known argument of the general perturbation theory in finite dimensional spaces (see Chapter 2 of [4]).
Let h and ε be positive numbers such that −h / ∈ A, t s < 2π − h < 2π and ε < 1 2 min j=1,2,...,s−1 that is, ε is less than half of the length of the intervals (−h, t 1 ), (t 1 , t 2 ), ..., (t s−1 , t s ) and (t s , 2π − h) (see (32) for t j ). Besides we assume that ε satisfies the inequality in (39). Let l(ε) be a curve joining the points −h and 2π − h, passing over the points t 1 < t 2 < ... < t s and consisting of the intervals and semicircles γ(t j , ε) for j = 1, 2, ..., s defined in (39). Thus Since A ∩ Q is a finite set, we choose l(ε) so that it and the open domain D(ε) enclosed by the curve l(ε) ∪ [−h, 2π − h] does not contain the point of the set A. Now we try to replace the semicircle γ(t j , ε) by the interval [t j − ε, t j + ε], that is, prove the equality for some value of k if it is possible. More precisely we determine whether (53) holds or not in order to replace in and get a spectral expansion for L.
is the simple eigenvalue of L t then it readily follows from the proof of (d) and (47) that a k (t)Ψ k,t (x) for each x is analytic in some neighborhood of t.
We say that the set is not ESS of the operator L then it follows from Definition 5 that all elements a k (t)Ψ k,t (x) of the bundle and by Theorem 1 since T(v, j) consist of the finite number of indices and (56) hold whenever Now using (30), (52) and (57) and taking into account that if n = 2µ − 1 then the set of multiple eigenvalues of L t for t ∈ Q is finite we obtain Theorem 2 If n = 2µ − 1 and L has no ESS, then for each compactly supported and continuous function f the following spectral expansion holds where the series converges in the norm of L 2 (a, b) for every a, b ∈ R.
Now consider the case when L has the ESS. By Proposition 2 set of the ESS is the subset of the set of multiple eigenvalues and (59) is a finite set. Therefore, for the simplicity of the notation and without loss of generality, we renumare the elements of (59) so that where m ≤ s and m j ≤ s j , denotes the set of ESS of L. Then due to Definition 5 the set of is an ESS of the operator L then for some values of k ∈ T(v, j) the function a k (t)Ψ k,t (x) for almost all x is nonintegrable on D(t j , ε), while some of elements of the bundle (55) may be integrable and the total sum of elements of (55) is bounded due to the cancellations of the nonintegrable terms. At least two element of the bundle must be nonintegrable in order to do the cancellations. In fact, we may to huddle together only the nonintegrable elements of the bundle (55). To do the same handling for all compactly supported and continuous f we use the following is called an essential singular part of the bundle (55) corresponding to the ESS Λ v (t j ).
Let us stress the following.
Remark 3 By the definition of S(v, j) for each k ∈ S(v, j), a k (t)Ψ k,t (x) may become nonintegrable function for almost all x, while is a an integrable function in D(t j , ε) for all v and x, since the total sum of elements of (55) and a k (t)Ψ k,t (x) for k ∈ T(v, j)\S(v, j) are integrable (see Definition 6 and Proposition 1) Now using (38) we obtain where by absolute continuity of the Lebesque integral we have Introduce the notations Here S and T are finite subsets of Z, since S ⊂ T and the number of elements of T(v, j) is equal to the multiplicity of the eigenvalue Λ v (t j ) of the operator L tj . The definitions of S j , S and T immediately imply the following.
Proposition 5 (a)The relations k ∈ S and k ∈ T\S hold respectively if and only if 1 α k is nonintegrable and integrable in s j=1 D(t j , ε). (65) To replace the circles γ(t j , ε) by the intervals [t j − ε, t j + ε] for j = 1, 2, ..., s and hence to get integrals over [0, 2π) instead of l(ε) (see (52)), that is, to obtain a real spectral expansion (see (30)), we divide the set Z into three pairwise disjoint subsets Z\T, T\S, and S and consider separately the integrals of a k (t)Ψ k,t (x) over l(ε) when the index k varies through these subsets. For this first we prove the following we obtain (66).
(b) It follows from Proposition 1 that to prove (b) it is enough to show that 1 α k (t) is integrable in the set (67) if k ∈ T\S. By Proposition 5(a) it is integrable in (65). On the other hand, as we noted in introduction, 1 α k is continuous on the set (67) and hence is integrable in the compact Now the proof follows from (68), (67) and (65).
(c) Since S(x, t) is integrable in the compact (68), it is enough to show that S(x, t) is a integrable function on D(t j , ε) for arbitrary j. Using Remark 1 and taking into account and by Remark 3, S j (x, t) is a integrable function on D(t j , ε). Now consider the function where the summation is taken over k ∈ S\S j . By Proposition 5(b), if k / ∈ S j , then 1 α k (t) and hence by Proposition Now we ready to prove the following.
Theorem 3 If L has the ESS Λ v (t j ) for j = 1, 2, ...m and v = 1, 2, ..., m j then for each continuous and compactly supported function f the following spectral expansion holds where I δ is obtained from [−h, 2π − h] by deleting the δ ∈ (0, ε) neighborhood of the ESQ t j for j = 1, 2, ..., m (see Definition 5): The series in (69) converges in the norm of L 2 (a, b) for every a, b ∈ R. If L has no ESQ then (58) holds.
Proof. Using Lemma 1(a) and (30), we obtain where T consist of finite number of indices. By Theorem 1 and Remark 1 we have On the other hand, by Proposition 5(c) if k ∈ T\T j then λ k (t) is a simple eigenvalue in U (t j ) and hence The last 2 equalities yield for arbitrary j = 1, 2, ..., s. Therefore from (52) we obtain where S(x, t) is defined in (64). It is clear that for each k ∈ S, where |S| < ∞, and for each fixed δ ∈ (0, ε) the function a k (t)Ψ k,t is integrable on I(δ). Therefore for any δ ∈ (0, ε) we have

Remark 4
In the case n = 2µ, l(y) = y (n) (x) + p 1 (x) y (n−1) + p 2 (x) y (n−2) + p 3 (x) y (n−3) + ... + p n (x)y, is finite (see [16]). Therefore this case is similar to the case n = 2µ − 1 and the results of this section can be carry out to this case without any changes in the proofs.

Spectral Expansion for Even Order
The case n = 2µ is more complicated than the case n = 2µ − 1 due to the following. By Proposition 3, in the case of odd order the number of the spectral singularities is finite and the operator L has no spectral singularity at infinity that easify the investigations. In the big contrary of the odd order case, in the case of even order we meet with the both complexities: (a) The existence of the infinite number of the spectral singularities in σ(L) (b) The existence of the spectral singularity at infinity. The complexity (a) occurs due the existence of the infinite number of the multiple eigenvalues in σ(L) which means that the set A ∩ Q is not finite. Fortunately, it follows from Result of [18] (see introduction) that the possible accumulation points of the set A ∩ Q are 0 and π. Hence the set A ∩ Q h is finite and the sets A ∩ {t : |t| ≤ h} and A ∩ {t : |t − π| ≤ h} are not finite, in general. Therefore we need to investigate in detail the eigenvalues of L t for the cases |t| ≤ h and |t − π| ≤ h. Moreover, the complexity (b) is connected with the fact there exists a sequence of pairs {n k , t k } such that |n k | → ∞ and α n k (t k ) → 0 as k → ∞. It is possible if either t k → 0 or t n → π which again shows the importance of the detail consideration about 0 and π.
Thus, first of all, let us consider the cases |t| ≤ h and |t − π| ≤ h. For this we follow the arguments in [8] of the proof of the asymptotic formulas for the eigenvalues and eigenfunction, and take into consideration the uniformity with respect to t. First, let us introduce some notations. Let T be a domain of the complex plane such that if ρ ∈ T, then the inequalities hold, for a suitable ordering of the nth roots ω 1 , ω 2 , . . . , ω n of −1 (see p. 45 of [8]). Let h and r be constant satisfying Let the constant c in (75) be chosen so that 1 ωµ ρ ∈ T if ρ belongs to the disks for the large positive values of k. From the arguments of [8] by the simple estimations we obtain the following contain only two eigenvalues (counting multiplicities) denoted by λ k (t) and λ −k (t) of the operators L t for |t| ≤ h. Moreover, the washer for k > N h (0) does not contain the eigenvalues of L t for |t| ≤ h.
Proof. It was proved in [8] (see (69) in p.71 of [8]) that to investigate the roots of the characteristic equations ∆(ρ, t) = 0 it is enough to consider the equation (5)). Therefore (80) has the form where z = ρω µ and there exists a constant c 2 independent of t for |t| ≤ h such that Using the Taylor series of e z at 2πki we get if z belongs to the circle γ r =: {z ∈ C : |z − (i2kπ)| = r} since r ≤ 1 2 (see (76)). If |t| ≤ h, then using (76) and the Maclaurin's series of e x we obtain This last two inequality imply that and hence we have if z ∈ γ r . It immediately yields the claim (Claim 1) that (81) has no roots on the washer for large values of k. Moreover taking into account that for |t| ≤ h < 1 32 the equation and using Rouche's theorem we get the claim (Claim 2) that the disk (84) contains only 2 roots of (81). Thus repeating the arguments of the proof of Theorem 2 in p. 70-74 of [8] from the Claim 1 and Claim 2 we obtain the proof of the theorem.
In the same way we prove the following.

Remark 5 Consider the family of operators
where L t (0) denotes the case when all coefficients of (1) are zero. Following [8] and repeating the proof of Theorem 4 one can readily see that there exist N h (0), independent of z ∈ [0, 1] and |t| ≤ h, so that Theorems 4 continues to hold for the operators (86). Therefore, there exists a closed curve Γ(0) such that: (a) The curve Γ(0) lies in the resolvent set of the operator L t,z for z ∈ [0, 1] and |t| ≤ h.
(b) All eigenvalues of L t,z for z ∈ [0, 1] and |t| ≤ h that do not lie in (78) for |k| > N h (0) belong to the set enclosed by Γ(0). Therefore, taking into account that the family L t,z is halomorphic with respect to z, we obtain that the number of eigenvalues of the operators L t,0 = L t (0) and L t,1 = L t lying inside of Γ(0) are the same. It means that apart from the eigenvalues λ k (t), where |k| > N h (0), there exist (2N h (0) + 1) eigenvalues of the operator L t for |t| ≤ h.denoted by and lying in Γ(0), where N(0) = {0, ±1, ..., ±N h (0)} By the same arguments we obtain that apart from the eigenvalues λ k (t) and λ −(k+1) (t) lying in (85) for k > N h (π) there exist (2N h (π)+2) eigenvalues the operator L t for |t−π| ≤ h denoted by where N(π) = {0, ±1, ..., ±N h (π), −(N h (π) + 1)} , and there exist a closed curve Γ(π) which contains inside only the eigenvalues (88). Thus in any case we numerate the eigenvalues of L t by the elements of Z.
Theorem 4 shows that for the large values of k the eigenvalue λ k (t) of L t for |t| ≤ h is close to the eigenvalue (±2kπi + it) n of L t (0) and far from the other eigenvalues L t (0). More precisely, Theorem 4 implies that for p = ±k, |t| ≤ h and k ≫ 1. Using this one can easily verify that the formula holds uniformly with respect to |t| ≤ h from which we obtain the following Theorem 6 If t / ∈ A, then the normalized eigenfunction Ψ k,t (x) of L t corresponding to λ k (t) satisfies the following, uniform with respect to |t| ≤ h, asymptotic formula Proof. We use the formula which can be obtained from by multiplying by e (2pπi+it)x and using where, L * t (0) is the adjoint operator to L t (0). Using the integration by part and (2) one can easily verify that the right-hand side of (90) is O(p n−2 ). Therefore by (89) we have Now decomposing Ψ k,t by basis {e (2kπi+it)x : k ∈ Z}, we get the proof of the theorem.
Instead of Theorem 4 using Theorem 5 and repeating the proof of Theorem 6 we obtain Theorem 7 If t / ∈ A, then the normalized eigenfunction Ψ k,t (x) of L t corresponding to λ k (t) satisfies the following, uniform with respect to |t − π| ≤ h, asymptotic formula Now we are ready to construct the curve of integration l (see (30)) for the even order case. Since in the case n = 2µ the numbers 0 and π are accumulation points of A ∩ Q, we consider the set (A ∩ Q) ∪ {0, π} denoted by A and choose the curve of integration so that it pass over the points of A . Namely, we construct l as follows. Let h be positive number satisfying (76) and such that ±h / ∈ A, (π ± h) / ∈ A. By Result of [18] and h < t 1 < t 2 < · · · < t p < π − h < π + h < t p+1 < t p+2 < · · · < t s < 2π − h.
Let ε be positive number satisfying ε < 1 2 min j∈{1,2,...,s−1}\p that is, ε is less than half of the distance between the neighboring points of (92). Let C(h, ε) be a curve lying in Q h , joining the points −h and 2π − h and consisting of the intervals for j ∈ {1, 2, ..., s − 1} \ {p} and semicircles for j = 1, 2, ..., s. Since A ∩ Q δ is a finite set for any δ > 0, the numbers h and ε can be chosen so that the curve C(h, ε) does not contain the point of the set A. Divide C(h, ε) into three parts: γ(0, h), γ(π, h) and l(h, ε) = C(h, ε)\(γ(0, h) ∪ γ(π, h)) Thus l(h, ε) consist of the intervals (94) and semicircles (96) and lies in the domain Q h \A. Since A ∩ Q h is a finite set, ε can be chosen so that l(ε) ∩ A = ∅, and the open domain D(ε) enclosed above by the curve l(h, ε) and below by B(h) does not contain the point of the set A. The semicircles γ(0, h) and γ(π, h) also lie in Q h \A.
In (28) instead of l taking C(h, ε) and then using C(h, ε) = l(h, ε) ∪ γ(0, h) ∪ γ(π, h) and (26) we obtain In [17] we proved that the series in (97) can be integrated term by term for any continuous curve lying in Q h \A. Therefore we have The investigation of Σ(l(h, ε)) is the repetition of the investigation of (54), since in the cases Case 1 n = 2µ − 1, t ∈ (−π, π] Case 2 n = 2µ, t ∈ B(h) we have the same situation. Namely the followings that were used for the investigations of the Case 1 are the same.
1. In the both cases the set of the multiple eigenvalues, the set of the ESS and the set of the ESQ are the finite sets. In Case 2 we also use the notations (32) and (33) and Definition 5 for these notions, where in Case 2, t 1 , t 2 , ..., t s belong to B(h) (see above).
2. In Case 2 the curve l(ε, h) and domain D(ε) have the same properties as the curve l(ε) and domain D(ε) in Case 1.
3. The definitions of S(v, j) and T(v, j) are the same and in the both cases they are finite sets. Therefore in Case 2 the set S(h) of indices k ∈ Z for which 1 α k is nonintegrable in B(h) is finite as the set S defined in (63) for the Case 1.
Therefore repeating the proof of Theorem 3 we obtain the following theorem.
Theorem 8 For each continuous and compactly supported function f the following equality hold, where f t (x) is defined by (21), and I(δ, h) is obtained from B(h) by deleting the δ ∈ (0, ε) neighborhood of the essential singular quasimomenta t 1 , t 1 , ..., t m lying in B(h): The series in (99) converge in the norm of L 2 (a, b) for every a, b ∈ R.
Now we consider Σ (γ(0, h)). The consideration of Σ(γ(π, h)) is similar. The complexity of the investigations of Σ(γ(0, h)) is the following. (ii) The existence of the spectral singularity at infinity is connected with the quasimomenta t lying in the intervals (−h, h) and (π − h, π + h).
That is why, in the big contrary of the Case 1 and Case 2 for the cases: and Case 4 n = 2µ, t ∈ [π − h, π + h] the set of indices k for which 1 α k is nonintegrable is not finite, in general, and may coincide with Z. This situation complicate to replace γ(0, h) with [−h, h] and γ(π, h) with [π − h, π + h]. However, now we prove that, if we huddle together the terms a k (t)Ψ k,t (x) and a −k (t)Ψ −k,t (x) for large value of k then Similarly, in the case γ(π, h), instead of C taking the circle z ∈ C : |z − (i(2kπ + π)) n | < 1 3 n(2πk + π + 1 4 ) n−1 , using Theorem 5 and repeating the proof of Theorem 1 we obtain holds.
Now let us consider the terms a k (t)Ψ k,t corresponding to the eigenvalues (87). In Remark 5 we proved that there exist a closed curve Γ(0) which contains inside only the eigenvalues (87) of L t for |t| ≤ h. Instead of the curve C using the curve Γ(0) and repeating the proof of Theorem 1 we obtain In the same way we get Introduce the notations.
Notation 1 Denote by S(0, h) and S(π, h) respectively the set of indices k ∈ N(0) and k ∈ N(π) for which 1 In the same way we define V (h, π) for S(π, h). Instead of the sets N(0) using the sets {k, −k} we construct the set V (k, h, 0) of the ESQ for each index k > N h (0) and for [−h, h]. Similarly, instead of the sets N(π) using the sets {k, −(k + 1)} we construct the corresponding set V (k, h, π) for each index k > N h (π) and for [π − h, π + h].
Since the set of the multiple eigenvalues lying in a bounded domain is finite, the sets S(0, h), V (h, 0), V (k, h, 0), S(π, h), V (h, π) and V (k, h, π) are finite. Now using (101), Theorem 6 and Proposition 6, taking into account that the sets defined in Notation 1 are finite and arguing as in the proof of Theorem 3 we obtain Theorem 9 Let f be continuous and compactly supported function. Then the following equality holds, where f t (x) is defined by (21), and I(δ, 0) is obtained from [−h, h] by deleting the δ ∈ (0, ε) neighborhood of the essential singular quasimomenta lying in V (h, 0). Moreover for |k| > N h (0) we have where I(k, δ, 0) is obtained from [−h, h] by deleting the δ ∈ (0, ε) neighborhood of the ESQ lying in V (k, h, 0). The series in (103) converge in the norm of L 2 (a, b) for every a, b ∈ R.
In the same way we obtain Theorem 10 For each continuous and compactly supported function f the following where I(k, δ, π) is obtained from [π − h, π + h] by deleting the δ ∈ (0, ε) neighborhood of the ESQ lying in V (k, h, π). The series in (105) converge in the norm of L 2 (a, b) for every a, b ∈ R. Now using (96) and the theorems 8,9,10 we obtain Theorem 11 For each continuous and compactly supported function f the following equality holds. All series in (107) converge in the norm of L 2 (a, b) for every a, b ∈ R.
The expansion (107) has a simple form if L has no spectral singularity at infinity.
Theorem 12 Let n = 2µ and L has no spectral singularities at infinity. Then (a) There exist a positive constants M and N such that where I δ is obtained from [0, 2π) by deleting the δ ∈ (0, ε) neighborhood of the ESQ t j for j = 1, 2, ..., m. Moreover, if L has no ESQ then The series in (109) and (110) converge in the norm of L 2 (a, b) for every a, b ∈ R.
Proof. The proof of (a) follows from (30) and Definition 2 and (b) is the consequence of (a). Thus the situation in case of absence of the spectral singularities at infinity is similar to the case n = 2µ − 1. Now let us prove (c). Theorems 6 and 7 continuous to hold for Ψ * k,t (x). Therefore (108) and theorems 6 and 7 immediately imply that the integrals exist for k ∈ Z\S and tend to zero as |k| → ∞. Hence, in the contrary to the general case we need not to huddle together the terms a k (t)Ψ k,t (x) and a −k (t)Ψ −k,t (x) in the case [−h, h] and the terms a k (t)Ψ k,t and a −(k+1) (t)Ψ −(k+1),t in the case [π − h, π + h] (see (107)). Hence the proof follows from (a), (b) and Theorem 11. One can readily see from the proof of Theorem 12 that we have proved the following Theorem 13 If there exists h ∈ (0, 1 32 ) such that ±h, π ± h / ∈ A and for k ∈ Z the integrals in (111) exist and tend to zero as |k| → ∞, then where J δ is obtained from B(h) by deleting the δ ∈ (0, ε) neighborhood of the ESQ t 1 , t 2 , ..., t j lying in B(h) and S being the set of all k for which 1 α k is nonintegrable in (−π, π]\A is finite. If, in addition, L has no ESQ then (110) holds.

Remark 6
The obtained results can be carry out in the same way for the differential operator generated in the space L m 2 (−∞, ∞) by the differential expression where n ≥ 2, P ν = (p ν,i,j ) is a m × m matrix with the complex-valued summable entries p ν,i,j , P ν (x + 1) = P ν (x) for ν = 2, 3, ...n, and the eigenvalues µ 1 , µ 2 , ..., µ m of the matrix

Theorem 14
If t 0 = 0, π and λ k (t 0 ) is a multiple eigenvalue of H t0 , then λ k (t 0 ) is a spectral singularity of H but is not an ESS.
To prove that λ k (t 0 ) for t 0 ∈ (0, π) is not an ESS let us consider for t ∈ [b, t 0 ). Then the Floquet solution (117) for λ = λ k (t) is the nonzero eigenfunction of H t for t ∈ [b, t 0 ). Therefore for t ∈ [b, t 0 ) the normalized eigenfunction Ψ k,t (x) of H t can be defined by (119). Similarly, for t ∈ [b, t 0 ) the normalized eigenfunction Ψ * k,t (x) of H * t can be defined by Thus we have 1 The direct calculations show that (see [3] and [16]) On the other hand, using (114) and taking into account that the Wronskian of θ(x, λ) and ϕ(x, λ) is 1 (see (115)), that is, we obtain (e −it − θ(λ k (t)))(e it − θ(λ k (t))) = −ϕ(λ k (t))θ ′ (λ k (t)), and where p(λ) = 4 − F 2 (λ). Since 4 − F 2 (λ) = 0 for λ = λ k (t 0 ), p(λ) is well-defined in the neighborhood of λ k (t 0 ). Therefore we have and where Now consider two cases: Case 1: ϕ(λ 0 ) = 0. Then this inequality with p(λ 0 ) = 0 implies that the integrand in the right-hand side of (129) is a bounded function in a neighborhood of λ 0 . Since λ k (t) is continuous on [b, t 0 ], there exist positive numbers l and M independent of δ, such that the lengths of the curves λ k ([b, t 0 − δ]) for all δ > 0 are less than l and the absolute value of the integrand in the right-hand side of (129) is less than M for λ ∈ λ k ([b, t 0 )). Therefore the limit in (120) exists and hence 1 α k is integrable on the small left neighborhood [b, t 0 ) of t 0 . In the same way we prove that it is integrable on the small right neighborhood of t 0 .
Case 2: ϕ(λ 0 ) = 0. Then by (125) we have θ(λ 0 )ϕ ′ (λ 0 ) = 1. This with (126) and (127) since p(λ 0 ) = 0, and by (126) and (127) we have Now using (130) and taking into account that p(λ 0 ) = 0 and the relations (131) and (132) hold if λ 0 is replaced by λ ∈ λ k ([b, t 0 )), where b is sufficiently close to t 0 , one can readily see that the integrands in the right-hand side of (129) are uniformly (w.r.t. δ) bounded in λ k ([b, t 0 − δ]). Therefore arguing as in Case 1 we get the proof of the theorem. By the similar arguments one can find conditions on λ k (t) for t = 0, π to be or not to be the ESS. Here we prove only one criterion for large value of k which will be used essentially for the spectral expansion.
Theorem 15 There exists a positive number N such that for |k| > N the multiple eigenvalue λ k (t) for t = 0 is the spectral singularity and ESS if and only if its geometric multiplicity is 1, that is, there exists only one linearly independent eigenfunction corresponding to λ k (t).
The theorem continues to hold if t = 0 is replaced by t = π.

Conclusion 1
The series in (135) and (138) converge with parenthesis and in parenthesis is included only the integral of the functions corresponding to splitting eigenvalues. The parenthesis is necessary, due to the following. If λ k (0) is ESS then, in general, lim δ→0 δ<|t|≤h a k (t)Ψ k,t (x)dt = ∞ and hence [−h,h] a k (t)Ψ k,t (x)dt (145) does not exist. It is possible that λ k (0) is ESS for all k. Moreover, if λ k (0) is not ESS for large k then it is possible that (145) for almost all x and the norm of (145) does not tend to zero as k → ∞. Therefore the series in (135) does not converge without parenthesis. We meet with the same situation for the series (138). Note that this situation agree with the well-known result [14] that the root functions of the operators generated by a ordinary differential expression in [0, 1] with regular boundary conditions, in general, form a Riesz basis with parenthesis and in parenthesis should be included only the functions corresponding to the splitting eigenvalues. In particular, the periodic (t = 0) and antiperiodic (t = π) boundary conditions requires the parenthesis. It is natural that in the case of the operator L generated by a ordinary differential expression in (−∞, ∞) we included in parenthesis the Bloch functions Ψ k,t (x) near two t = 0 (see (135)) and t = π (see (138)).
One can readily see that if λ k (0) is ESS, that is, if λ k (0) is a double eigenvalue with geometric multiplicity 1 (see Theorem 15) then for arbitrary f the numerators of the fractions in (142) and (143) in not identically zero functions while the denominators are zero. Moreover in the neighborhood of λ k (0) we have dt dλ ∼ 1 which means that there exist the