Rabah Rabah
email: rabah@irccyn.ec-nantes.fr

Grigory Sklyar
email: sklar@univ.szczecin.pl

Pavel Barkhayev
email: pbarhaev@inbox.ru

Stability analysis of mixed retarded-neutral type systems in Hilbert space

Keywords: Retarded-neutral type systems, exponential stability, strong stability, infinite dimensional systems Retarded-neutral type systems, exponential stability, strong stability, infinite dimensional systems. Mathematical subject classification. 34K06, 34K20, 34K40, 49K25, 93C23

We investigate the stability property of mixed retarded-neutral type systems. Considering an operator model of the system in Hilbert space we are interesting in the critical case when the spectrum of the operator belongs to the open left half-plane and there exists a sequence of eigenvalues with real parts approaching to zero. In this case the exponential stability is not possible and we are studying the strong asymptotic stability property. The present paper extends the results obtained in [R. Rabah, G.M. Sklyar, A. V. Rezounenko, Stability analysis of neutral type systems in Hilbert space. J. of Differential Equations, 214(2005), No. 2, 391-428] in which stability of systems of neutral type was studied using the existence of a Riesz basis of invariant finite-dimensional subspaces. However, for mixed retarded-neutral type systems such a basis may not exist for the whole state space. Though the main result on stability remains the same for mixed retarded-neutral type systems, the technic of its proof had to be changed and it involves a proof of resolvent boundedness on some invariant subspace. We show that the property of asymptotic stability is determinated not only by the spectrum of the system but also depends on geometrical characteristics of its main neutral term which in our situation may be singular. We also give an explicit example of two systems having the same spectrum in the open left half-plane, but one of them is asymptotically stable while the other one is unstable.

Introduction

The interest in considering delay differential equations and corresponding infinite-dimensional dynamical systems is caused by a huge amount of applied problem which can be described by these equations. The stability theory of such type of systems was studied intensively (see e.g. [START_REF] Bellman | Differential-difference equations[END_REF][START_REF] Diekmann | Delay equations[END_REF][START_REF] Hale | Introduction to functional-differential equations[END_REF][START_REF] Kolmanovskii | Introduction to the theory and applications of functional-differential equations[END_REF][START_REF] Kolmanovskii | Stability of functional-differential equations[END_REF]). Number of results was obtained for retarded systems, however an analysis of neutral type systems is much more complicated and these systems are still studied not so deeply. In the case of neutral type systems one meets the notion of strong asymptotic (non-exponential) stability. This type of stability could happen if exponential stability is not possible. We note also that for retarded system exponential stability and strong asymptotic stability occur at the same time.

We consider systems given by the following functional differential equation:

ż(t) = A -1 ż(t -1) + 0 -1 A 2 (θ) ż(t + θ) dθ + 0 -1 A 3 (θ)z(t + θ) dθ, (1.1)
where A -1 is a constant n × n-matrix and A 2 , A 3 are n × n-matrices whose elements belongs to L 2 (-1, 0; C). The term A -1 ż(t -1) is called principal.

In [START_REF] Rabah | Stability analysis of neutral type systems in hilbert space[END_REF] an analysis of strong stability of systems (1.1) has been carried out in the case when the matrix A -1 of the principal neutral term is nonsingular. In this paper we want to avoid this restriction and to perform a stability analysis in case when det A -1 is allowed to be zero, i.e. in the case of systems of mixed retarded-neutral type.

We use the general theory of C 0 -semigroups of linear bounded operators (see e.g. [START_REF] Van Neerven | The asymptotic behaviour of semigroups of linear operators[END_REF]). Following [START_REF] Rabah | Stability analysis of neutral type systems in hilbert space[END_REF], we consider the operator model of the system introduced by Burns et al. [START_REF] Burns | Linear functional-differential equations as semigroups on product spaces[END_REF] in product space:

ẋ = Ax, x(t) = y(t) z t (•) , (1.2)
where A is the generator of a C 0 -semigroup, which is defined by

Ax(t) = A y(t) z t (•) = 0 -1 A 2 (θ) żt (θ) dθ + 0 -1 A 3 (θ)z t (θ) dθ dz t (θ)/ dθ , (1.3)
with the domain

D(A) = {(y, z(•)) T : z ∈ H 1 (-1, 0; C n ), y = z(0) -A -1 z(-1)} ⊂ M 2 , (1.4)
where M 2 def = C n × L 2 (-1, 0; C n ) is the state space. In the case when A 2 (θ) ≡ A 3 (θ) ≡ 0 we use the notation A for A.

The dynamical system (e tA , M 2 ) is said to be strongly asymptotically stable if for all x ∈ M 2 : lim t→+∞ e tA x = 0.

In the paper [START_REF] Rabah | Stability analysis of neutral type systems in hilbert space[END_REF] the following result on strong stability of neutral type systems of the form (1.1) was obtained.

Theorem (R. Rabah, G.M. Sklyar, A.V. [START_REF] Rabah | Stability analysis of neutral type systems in hilbert space[END_REF]. Let us assume that det A -1 = 0. And we assume also that σ(A) ⊂ {λ : Reλ < 0} and max{|µ| : µ ∈ σ 1 (A -1 )} = 1. Let us put σ 1 = σ(A -1 ) ∩ {µ : |µ| = 1}. Then the following three mutually exclusive possibilities hold true:

1. σ 1 consists of simple eigenvalues only, i.e. an one-dimensional eigenspace corresponds to each eigenvalue and there are no root vectors. Then system (1.2) is asymptotically stable.

2. The matrix A -1 has a Jordan block, corresponding to an eigenvalue µ ∈ σ 1 . In this case system (1.2) is unstable.

3. There are no Jordan blocks, corresponding to eigenvalues from σ 1 , but there exists an eigenvalue µ ∈ σ 1 whose eigenspace is at least two dimensional. In this case system (1.2) can be either stable or unstable. Moreover, there exist two systems with the same spectrum, such that one of them is stable while the other one is unstable.

The present paper extends the last result on the case of mixed retarded-neutral type systems, i.e. on the case when det A -1 is allowed to be zero. Let us analyze the propositions of the theorem above in this case.

If the matrix A -1 has a Jordan block, corresponding to an eigenvalue µ ∈ σ 1 (item 2), then the result remains the same: system (1.2) is unstable. The proof of this fact given in [START_REF] Rabah | Stability analysis of neutral type systems in hilbert space[END_REF] does not use the assumption det A -1 = 0 (this proof does not involve the Riesz basis technic).

The example given in [START_REF] Rabah | Stability analysis of neutral type systems in hilbert space[END_REF] for illustrating the item 3 does not essentially need the assumption det A -1 = 0. In this paper we give also an explicit method of construction of such examples (two systems with the same spectrum and with the same matrix A -1 , and one of these systems is stable while the other one is unstable) which does not use the requirement det A -1 = 0. Thus, the dilemma of the item 3 holds also for mixed retarded-neutral type systems.

However, an extension of the result on stability (item 1 of the theorem) is not obvious at all. The reason of this is that the proof of stability in the case det A -1 = 0 is essentially based on the results on existence of a Riesz basis of invariant finite-dimensional subspaces. However, for mixed retarded-neutral type systems such a basis may not exist for the whole state space M 2 .

Let us describe this problem in more details. A motivation for involving Riesz basis notion in stability analysis could be found in [START_REF] Rabah | On strong stability and stabilizability of linear systems of neutral type[END_REF][START_REF] Sklyar | Stability of a strongly stabilizing control for systems with a skew-adjoint operator in Hilbert space[END_REF][START_REF] Sjoerd | A functional model approach to linear neutral functional-differential equations[END_REF]. It is noted in [START_REF] Sjoerd | A functional model approach to linear neutral functional-differential equations[END_REF] that if the spectrum of the system is contained in a vertical strip and inf{|λ -λ |, λ, λ ∈ σ(A), λ = λ } > 0, then the generalized eigenspaces form a Riesz basis [START_REF] Gohberg | Introduction to the theory of linear nonselfadjoint operators[END_REF]. On the other hand, in [START_REF] Rabah | Generalized riesz basis property in the analysis of neutral type systems[END_REF] the example was given which shows that the neutral type system (1.2) may not possess such a basis since its eigenvalues may be not separated.

It was shown in [START_REF] Rabah | Stability analysis of neutral type systems in hilbert space[END_REF] that the state space M 2 possesses a Riesz basis of finite-dimensional subspaces which are A-invariant (and e tA -invariant). The proof of stability is essentially based on this fundamental result which requires the condition det A -1 = 0.

However, if det A -1 = 0 then an infinite part of the spectrum of the operator A is not situated in vertical strips. Thus, we are not able to use the results on existence of the Riesz basis in the whole space and we need to involve another technic to investigate the stability property. Namely, we combine the Riesz basis technic with analysis of resolvent's boundedness on some invariant space.

We decompose the state space M 2 onto a direct sum of A-invariant subspaces, one of which possesses the Riesz basis of finite-dimensional subspaces and, therefore, the Riesz basis technic may be applied in this subspace. To prove stability property on the second subspace we use the following criterion of exponential stability which could be found in [START_REF] Van Neerven | The asymptotic behaviour of semigroups of linear operators[END_REF] (Corollary 4.2.8 from Theorem 4.2.8, p.119) or in [START_REF] Luo | Stability and stabilization of infinite dimensional systems with applications[END_REF] (Theorem 3.35, p.139). Theorem (on exponential stability). Let T (t) be a C 0 -semigroup on a Hilbert space H with a generator A. Then T (t) is exponentially stable if and only if the following conditions hold:

1. {λ : Reλ ≥ 0} ⊂ ρ(A); 2. R(λ, A) ≤ M for all {λ : Reλ ≥ 0} and for some constant M > 0.

Combining two technics, we prove the following theorem:

Theorem 1.1 (on stability). If σ(A) ⊂ {λ : Reλ < 0} and σ 1 = σ(A -1 ) ∩ {µ : |µ| = 1} consists only of simple eigenvalues, then system (1.1) is strongly asymptotically stable.

We emphasize that we require no other conditions. In particular, we do not require the spectrum of the operator A * to coincide with the spectrum of the operator A.

We emphasize also that though the formulation of the main result on stability remains the same for mixed retarded-neutral type systems, the technic of its proof was essentially changed and it involves a proof of resolvent boundedness on some invariant subspace.

In [START_REF] Rabah | Stability analysis of neutral type systems in hilbert space[END_REF] it was given an example of two systems having the same spectrum in the open left half-plane but one of them is asymptotically stable while the other one is unstable. That example was constructed mostly in infinite dimensional space M 2 . In this paper we give an example of the same situation but in terms of systems of the form (1.1). The analysis of the spectrum being carried out in this example is essentially based on deep results on transcendental equations obtained in the paper of L. Pontryagin [START_REF] Pontryagin | On the zeros of some elementary transcendental functions[END_REF]. Such an example helps to understand better the situation that may occur in the case when there are no Jordan blocks, corresponding to eigenvalues in σ 1 , but there exists an eigenvalue µ ∈ σ 1 whose eigenspace is at least two-dimensional.

The paper is organized as follows. Sections 2-5 are devoted to a proof of Theorem 1.1 on stability of mixed neutral-retarded type systems. In Section 2 we construct a direct decomposition of the state space onto invariant subspaces and we give a proof of the main result on stability. In Section 3 we prove validity of the direct decomposition and in Section 4 we give a proof of resolvent boundedness on the invariant subspace. Section 5 is devoted to a proof of some auxiliary results which was used in proofs of the main results. In Section 6 we give an explicit example of two systems having the same spectrum in the open left half-plane but one of these systems is asymptotically stable while the other one is unstable.

Stability analysis

Notations and assumptions

Following the notation given in [START_REF] Rabah | Stability analysis of neutral type systems in hilbert space[END_REF][START_REF] Rabah | On strong regular stabilizability for linear neutral type systems[END_REF], we introduce matrix-functions

(λ) = A (λ) = -λI + λe -λ A -1 + λ 0 -1 e λs A 2 (s) ds + 0 -1 e λs A 3 (s) ds, (2.5) * (λ) = A * (λ) = -λI + λe -λ A * -1 + λ 0 -1 e λs A * 2 (s) ds + 0 -1 e λs A * 3 (s) ds, (2.6)
which are in the relation ( (λ)) * = * (λ) and the eigenvalues of the operators A and A * are roots of the equations det (λ) = 0 and det * (λ) = 0, respectively.

As it has been shown in [START_REF] Rabah | Stability analysis of neutral type systems in hilbert space[END_REF]Proposition 1], the resolvent of the operator A has the following form:

R(λ, A) z ξ(•) ≡ e -λ A -1 0 -1 e -λs ξ(s) ds + (I -e -λ A -1 ) -1 (λ)D(z, ξ, λ) θ 0 e λ(θ-s) ξ(s) ds + e λθ -1 (λ)D(z, ξ, λ)
, where (λ) is defined by (2.5) and D(z, ξ, λ) is the following vector-function:

D(z, ξ) = D(z, ξ, λ) = z + λe -λ A -1 0 -1 e -λθ ξ(θ) dθ - 0 -1 A 2 (θ)ξ(θ) dθ - 0 -1 e λθ [λA 2 (θ) + A 3 (θ)] θ 0 e -λs ξ(s) ds dθ, z ∈ C n , ξ(•) ∈ L 2 (-1, 0; C n ). (2.7)
We denote by µ 1 , . . . , µ the set of (distinct) eigenvalues of the matrix A -1 and by p 1 , . . . , p their multiplicities; by λ

(k) m = ln |µ m | + i(arg µ m + 2πk), m = 1, , k ∈ Z we denote eigenvalues of the operator A and by L (k) m (r (k) ) -circles of the radius r (k) ≤ 1 3 | λ (k) m -λ (j) i |, (m, k) = (i, j) centered at λ (k) m .
We introduce a notation σ 1 = σ(A -1 ) ∩ {µ : |µ| = 1} and, without loss of generality, we assume that {µ 1 , . . . , µ 1 } ⊂ σ 1 and, thus, p 1 = . . . = p 1 = 1. We assume that the matrix A -1 is in Jordan form:

A -1 =          µ 1 . . . 0 0 . . . 0 . . . . . . . . . . . . . . . . . . 0 . . . µ 1 0 . . . 0 0 . . . 0 J 1 +1 . . . 0 . . . . . . . . . . . . . . . . . . 0 . . . 0 0 . . . J          , (2.8)
where J m are Jordan blocks corresponding to eigenvalues µ 1 +1 , . . . , µ . We introduce the set of eigenvalues Λ 1 = Λ 1 (N ):

Λ 1 = Λ 1 (N ) = {λ k m : 1 ≤ m ≤ 1 , |k| ≥ N }, (2.9)
and further we specify more precisely how large the number N should be.

According [START_REF] Rabah | On strong regular stabilizability for linear neutral type systems[END_REF]Theorem 4], there exists N 1 such that for every index m = 1, 1 and k : |k| ≥ N 1 the total multiplicity of the roots of the equation det (λ) = 0, contained in the circle L (k) m (r (k) ), equals p m = 1, where radii r (k) satisfy relation k∈Z (r (k) ) 2 ≤ ∞. We denote these roots (eigenvalues of the operator A) as (k) ) are centered on the imaginary axis and their radii are approaching to zero, then let us chose N ≥ N 1 big enough such that the strip {λ : -r (N ) < Reλ < 0} contains no eigenvalues of A except those belonging to circles

λ k m , 1 ≤ m ≤ 1 , |k| ≥ N 1 . Without loss of generality, we assume that r (k 1 ) ≥ r (k 2 ) when N 1 ≤ k 1 < k 2 . Since for m = 1, 1 all the circle L (k) m (r
L (k) m (r (k) ), m = 1, 1 , |k| ≥ N .

Direct decomposition of the state space

Our aim is to divide the system onto exponentially stable part and strongly asymptotically stable part. To do this we construct a decomposition of the state space M 2 onto the direct sum of some A-invariant subspaces. The method of construction is described below.

We split the spectrum of the operator A:

σ(A) = Λ 1 ∪ Λ 0 , (2.10)
where Λ 1 consists of eigenvalues situated "not far away" from the imaginary axis, i.e., taking some small ε > 0, we put

Λ 1 = σ(A) ∩ {λ : -ε < Reλ < 0} (2.11)
and Λ 0 = σ(A)\Λ 1 . We notice here that we take ε > 0 such small that Λ 1 consists only of eigenvalues "corresponding" to µ i ∈ σ 1 and the corresponding eigenvectors form a Riesz basis of the closure of their linear span. We give the precise way of construction the set Λ 1 in the proof of the main result.

We introduce an A-invariant subspace M 1 2 consisting of eigenvectors of the operator A:

M 1 2 = Cl Lin{ϕ : (A -λI)ϕ = 0, λ ∈ Λ 1 }. (2.12)
To construct another A-invariant subspace (which we denote by M 0 2 ), we consider the closure of the linear span of eigenvectors of the operator A * corresponding to eigenvalues from Λ 1 :

M 1 2 = Cl Lin{ψ : (A * -λI)ψ = 0, λ ∈ Λ 1 }. (2.13)
Let us denote by M 0 2 the orthogonal complement to the subspace M 1 2 in M 2 , i.e.:

M 2 = M 1 2 ⊥ ⊕M 0 2 . (2.14)
Since by the construction M 1 2 is an A * -invariant subspace, then M 0 2 is an A-invariant subspace.

Finally, we consider the subspace M 1 2 ⊕ M 0 2 and we want to prove that it coincides with the whole space M 2 :

M 2 = M 1 2 ⊕ M 0 2 . (2.15)
Our plan of proof of stability is the following: we prove the validity of the direct decomposition (2.15). Further, we prove that the resolvent of the operator A is uniformly bounded on the subspace M 0 2 and then, due to the theorem on exponential stability given above, we conclude that the restriction of the system (1.2) onto the A-invariant subspace M 0 2 is exponentially stable. Proving also that the restriction of the system (1.2) onto the subspace M 1 2 is strongly stable, we obtain strong asymptotical stability of the system on the whole space M 2 .

Proof of stability

Further we assume that the spectrum of the operator A belongs to the open left half-plane and also assume that the set σ 1 = σ(A -1 ) ∩ {µ : |µ| = 1} consists only of simple eigenvalues (i.e. the multiplicity of these eigenvalues is equal to one).

The proof of Theorem 1.1 is based on the following theorems.

Theorem 2.1 (on direct decomposition). There exists N ∈ Z such that the part of the spectrum Λ 1 = Λ 1 (N ) given by (2.9) and the subspaces M 0 2 , M 1 2 , M 1 2 , given by (2.12), (2.13) and (2.14), define the direct decomposition (2.15) of the space M 2 , i.e.

M 2 = M 1 2 ⊕ M 0 2 and M 1 2 , M 0 2 are A-invariant subspaces.
Theorem 2.2 (on resolvent boundedness). The restriction of the resolvent R(λ, A) is uniformly bounded on the subspace M 0 2 for λ : Reλ ≥ 0, i.e. there exists C > 0 such that

R(λ, A)x ≤ C x , x ∈ M 0 2 . (2.16)
Proofs of these two theorems are given below and now we prove the main result on stability.

Proof of Theorem 1.1. Let us show that for any x ∈ M 2 : e tA x → 0 when t → +∞. Due to Theorem 2.1 each x ∈ M 2 allows the following representation:

x = x 0 + x 1 , x 0 ∈ M 0 2 , x 1 ∈ M 1 2 .
It has been shown in [START_REF] Rabah | Stability analysis of neutral type systems in hilbert space[END_REF]Theorem 15] that there exists N ∈ Z such that elements

{ϕ k m : (A -λ k m I)ϕ k m = 0, λ k m ∈ Λ 1 = Λ 1 (N )} (2.17)
form a Riesz basis of the closure of their linear span. Thus, for any x 1 ∈ M 1 2 we have representations

x 1 = 1≤m≤l 1 |k|≥N c k m ϕ k m , e tA x 1 = 1≤m≤l 1 |k|≥N e λ k m t c k m ϕ k m , 1≤m≤l 1 |k|≥N |c k m | 2 < ∞.
Let us consider a norm • 1 in which the Riesz basis (2.17) is orthogonal, then we have the following estimate:

e tA x 1 1 =    1≤m≤l 1 |k|≥N e 2Reλ k m t c k m ϕ k m 2 1    1 2 ≤ x 1 1 . (2.18)
Since the series 1≤m≤l 1 |k|≥N c k m ϕ k m converges and since ϕ k m 1 is uniformly bounded then for any ε > 0 there exists N 1 ≥ N such that we have the estimate

1≤m≤l 1 |k|≥N 1 c k m ϕ k m 2 1 ≤ ε 2 8 .
Moreover, since the set {(m, k) : 1 ≤ m ≤ l 1 , N ≤ |k| ≤ N 1 } is finite and since Reλ k m < 0, then there exists t 0 > 0 such that for any t ≥ t 0 we have an estimate

1≤m≤l 1 N ≤|k|≤N 1 e 2Reλ k m t c k m ϕ k m 2 1 ≤ ε 2 8 .
Thus, we have ≤ M e -ωt . Thus, for any x 0 ∈ M 0 2 there exists t 0 > 0 such that for any t ≥ t 0 we have an estimate

1≤m≤l 1 |k|≥N e 2Reλ k m t c k m ϕ k m 2 1 ≤ 1≤m≤l 1 N ≤|k|≤N 1 e 2Re k m λt c k m ϕ k m 2 1 + 1≤m≤l 1 |k|≥N 1 c k m ϕ k m 2 1 ≤ ε 2 4 . ( 2
e tA x 0 1 ≤ M e -ωt x 0 1 ≤ ε 2 . (2.20)
Finally, from the estimates (2.18), (2.19) and (2.20) we conclude that for any x ∈ M 2 and for any ε > 0 there exists t 0 > 0 such that for any t ≥ t 0 the following estimate holds:

e tA x 1 ≤ e tA x 0 1 + e tA x 1 1 ≤ ε 2 + ε 2 = ε. (2.21)
Therefore, lim t→+∞ e tA x 1 = 0 what implies that the system (1.2) is strongly asymptotically stable. 2

Proof of the direct decomposition

As it has been shown in [START_REF] Rabah | Stability analysis of neutral type systems in hilbert space[END_REF] and [START_REF] Rabah | On strong regular stabilizability for linear neutral type systems[END_REF] an eigenvector ϕ k m of the operator A corresponding to an eigenvalue λ k m is of the form

ϕ k m = (I -e -λ k m A -1 )x k m e λ k m θ x k m , (3.22)
where x k m ∈ Ker (λ k m ); an eigenvector ψ k m of the operator A * corresponding to an eigenvalue λ k m is of the form

ψ k m = y k m λ k m e -λ k m θ -A * 2 (θ) + e -λ k m θ θ 0 e λ k m s A * 3 (s) ds + λ k m e -λ k m θ θ 0 e λ k m s A * 2 (s) ds y k m , (3.23
) where y k m ∈ Ker * (λ k m ) and * (λ) is given by (2.6).

Lemma 3.1 Let λ k m ∈ σ(A), λ j i ∈ σ(A *
) and ϕ k m , ψ j i are corresponding eigenvectors:

(A -λ k m I)ϕ k m = 0, (A * -λ j i I)ψ j i = 0.
The scalar product ϕ k m , ψ j i M 2 equals to the following value:

ϕ k m , ψ j i M 2 = 0, (m, k) = (i, j) - (λ k m )x k m , y k m C n , (m, k) = (i, j) , (3.24
m C n = 1, y k m C n = 1, are bounded, i.e. there exists a constant C > 0 such that ϕ k m ≤ C, 1 λ k m ψ k m ≤ C, 1 ≤ m ≤ l 1 , |k| ≥ N . (3.25) Lemma 3.3 Let λ k m ∈ Λ 1 then there exists matrices P m,k =      1 -p 2 . . . -p n 0 1 . . . 0 . . . . . . . . . . . . 0 0 . . . 1      , Q m,k =      1 0 . . . 0 -q 2 1 . . . 0 . . . . . . . . . . . . -q n 0 . . . 1      (3.26) such that the matrix 1 λ k m R m P m,k (λ k m )R m Q m,k has the following form: m,k (λ k m ) def = 1 λ k m P m,k R m (λ k m )R m Q m,k =      0 0 . . . 0 0 . . . 0 S m,k      , det S m,k = 0, (3.27)
where

R m = R m 0 0 I , I = I n-m ∈ C n-m×n-m , R m =        0 0 . . . 0 1 0 1 . . . 0 0 . . . . . . . . . . . . . . . 0 0 . . . 1 0 1 0 . . . 0 0        ∈ C m×m .
Moreover, for any ε > 0 there exists N ∈ Z such that for any |k| ≥ N we have estimates on p i = p i (m, k), q i = q i (m, k):

|p i | ≤ ε, |q i | ≤ ε, i = 2, n.
(3.28)

Corollary 3.1 In a neighborhood U (λ k m ) of each λ k m ∈ Λ 1 the matrix function (λ) allows a representation m,k (λ) =      (λ -λ k m )r 11 (λ) (λ -λ k m )r 12 (λ) . . . (λ -λ k m )r 1n (λ) (λ -λ k m )r 21 (λ) . . . (λ -λ k m )r n1 (λ) M m,k (λ)      , λ ∈ U (λ k m ), (3.29) where r ij (λ) = r m,k ij (λ) are analytic functions. Moreover, r m,k 11 (λ k m ) = 0, |r m,k 11 (λ k m )| → 1, k → ∞. (3.30)
Remark 3.1 Direct computations give us:

P -1 m,k =      1 p 2 . . . p n 0 1 . . . 0 . . . . . . . . . . . . 0 0 . . . 1      , Q -1 m,k =      1 0 . . . 0 q 2 1 . . . 0 . . . . . . . . . . . . q n 0 . . . 1      . (3.31) Lemma 3.4 There exist constants 0 < C 1 < C 2 , N ∈ Z such that for any λ k m ∈ Λ 1 = Λ 1 (N ) we have an estimate 0 < C 1 ≤ 1 λ k m (λ k m )x k m , y k m ≤ C 2 , λ k m ∈ Λ 1 . (3.32)
Using the propositions given above we prove the theorem on direct decomposition. Proof of Theorem 2.1. Let us prove that any element x ∈ M 2 allows the following representation:

x = x 0 + 1≤m≤l 1 |k|≥N c k m ϕ k m , x 0 ∈ M 0 2 , ϕ k m ∈ M 1 2 , 1≤m≤l 1 |k|≥N |c k m | 2 < ∞. (3.33)
It has been shown in [START_REF] Rabah | Stability analysis of neutral type systems in hilbert space[END_REF]Theorem 15] that there exists N ∈ Z such that elements

{ϕ k m : (A -λ k m I)ϕ k m = 0, λ k m ∈ Λ 1 = Λ 1 (N )} form a Riesz basis of the closure of their linear span. Also the set { 1 λ k m ψ k m : (A * -λ k m I)ψ k m = 0, λ k m ∈ Λ 1 = Λ 1 (N )
} form a Riesz basis of the closure of their linear span.

We choose x k m C n = 1 and y k m C n = 1 and, due to Lemma 3.2, we obtain that there exists

C > 0 such that ϕ k m M 2 ≤ C and ψ k m M 2 = 1 λ k m ψ k m M 2 ≤ C for all λ k m ∈ Λ 1 . Applying decomposition (2.14) to vectors ϕ k m we obtain ϕ k m = γ k m + 1≤i≤l 1 |j|≥N a j i ψ j i , γ k m ∈ M 0 2 .
Since, due to Lemma 3.1, ϕ k m , ψ j i = 0 for (m, k) = (i, j), then the last representation can be rewritten as follows:

ϕ k m = γ k m + a k m ψ k m , γ k m ∈ M 0 2 , (3.34)
and, moreover, due to (3.24) we have

a k m = ϕ k m , ψ k m M 2 ψ k m 2 M 2 = 1 λ k m ϕ k m , ψ k m M 2 ψ k m 2 M 2 = -1 λ k m (λ k m )x k m , y k m C n ψ k m 2 M 2 . (3.35)
We note also that

γ k m ≤ ϕ k m + |a k m | ψ k m ≤ C + | 1 λ k m (λ k m )x k m , y k m |. (3.36)
Using decomposition (2.14) and the relation (3.34) we represent each vector x ∈ M 2 as follows:

x = x 0 + 1≤m≤l 1 |k|≥N b k m ψ k m = x 0 - 1≤m≤l 1 |k|≥N b k m a k m γ k m + 1≤m≤l 1 |k|≥N b k m a k m ϕ k m = x 0 + 1≤m≤l 1 |k|≥N c k m ϕ k m , (3.37)
where

x 0 ∈ M 0 2 , 1≤m≤l 1 |k|≥N |b k m | 2 < ∞, x 0 = x 0 - 1≤m≤l 1 |k|≥N b k m a k m γ k m ∈ M 0 2 , c k m = b k m a k m .
To prove the validity of the decomposition (3.37) it is enough to show that 1

a k m ≤ C 1 and γ k m ≤ C 2 .
Taking into account (3.35) and (3.36), the last means to give an estimate

0 < C 1 ≤ 1 λ k m (λ k m )x k m , y k m ≤ C 2 , λ k m ∈ Λ 1 , (3.38)
which is proved by Lemma 3.4. Therefore, the representation (3.37) holds for any x ∈ M 2 and this completes the proof of the theorem. 2

Now we prove the auxiliary lemmas. Proof of Lemma 3.1. 1. Let us compute directly the scalar product ϕ k m , ψ j i M 2 using the representations (3.22) and (3.23):

ϕ k m , ψ j i M 2 = (I-e -λ k m A -1 )x k m , y j i C n + 0 -1 e λ k m θ x k m , λ j i e -λ j i θ y j i C n dθ- 0 -1 e λ k m θ x k m , A * 2 (θ)y j i C n dθ + 0 -1 e λ k m θ x k m , e -λ j i θ θ 0 e λ j i s A * 3 (s) ds•y j i C n dθ+ 0 -1 e λ k m θ x k m , λ j i e -λ j i θ θ 0 e λ j i s A * 2 (s) ds•y j i C n dθ = (I -e -λ k m A -1 )x k m , y j i + 0 -1 λ j i e (λ k m -λ j i )θ dθ • x k m , y j i - 0 -1 e λ k m θ A 2 (θ) dθ • x k m , y j i + 0 -1 e (λ k m -λ j i )θ θ 0 e λ j i s A 3 (s) ds dθ • x k m , y j i + λ j i 0 -1 e (λ k m -λ j i )θ θ 0 e λ j i s A 2 (s) ds dθ • x k m , y j i = Γ(λ k m , λ j i )x k m , y j i , (3.39)
where

Γ(λ k m , λ j i ) = I -e -λ k m A -1 + λ j i 0 -1 e (λ k m -λ j i )θ dθ - 0 -1 e λ k m θ A 2 (θ) dθ + 0 -1 e (λ k m -λ j i )θ θ 0 e λ j i s A 3 (s) ds dθ + λ j i 0 -1 e (λ k m -λ j i )θ θ 0 e λ j i s A 2 (s) ds dθ. (3.40)
The last two terms of Γ(λ k m , λ j i ) are integrals with the domain -1 ≤ θ ≤ s ≤ 0 and we integrate first by s and after that by θ. Let us change the order of integrating:

0 -1 ( θ 0 ds) dθ = - 0 -1 ( 0 θ ds) dθ = - 0 -1 ( s -1 dθ) ds. Since also 0 -1 e (λ k m -λ j i )θ dθ = 1 λ k m -λ j i (1 -e λ j i -λ k m ), we obtain 0 -1 e (λ k m -λ j i )θ θ 0 e λ j i s A i (s) ds dθ = - 0 -1 e λ j i s A i (s) s -1 e (λ k m -λ j i )θ dθ ds = 1 λ k m -λ j i e λ j i -λ k m 0 -1 e λ j i s A i (s) ds - 0 -1 e λ k m s A i (s) ds .
Finally, we have

Γ(λ k m , λ j i ) = 1 λ k m -λ j i (λ k m -λ j i )I -(λ k m -λ j i )e -λ k m A -1 + λ j i (1 -e λ j i -λ k m )I -(λ k m -λ j i ) 0 -1 e λ k m θ A 2 (θ) dθ - 0 -1 e λ k m s A 3 (s) ds -λ j i 0 -1 e λ k m s A 2 (s) ds +e λ j i -λ k m 0 -1 e λ j i s A 3 (s) ds + λ j i e λ j i -λ k m 0 -1 e λ j i s A 2 (s) ds = 1 λ k m -λ j i λ k m I -λ k m e -λ k m A -1 - 0 -1 e λ k m s A 3 (s) ds -λ k m 0 -1 e λ k m s A 2 (s) ds -λ j i e λ j i -λ k m I + λ j i e -λ k m A -1 + e λ j i -λ k m 0 -1 e λ j i s A 3 (s) ds + λ j i e λ j i -λ k m 0 -1 e λ j i s A 2 (s) ds = 1 λ k m -λ j i -(λ k m ) + e λ j i -λ k m (λ j i ) .
Taking into account that

x k m ∈ Ker (λ k m ), y j i ∈ Ker * (λ j i ) and ( (λ j i )) * = * (λ j i ), we conclude that ϕ k m , ψ j i M 2 = 1 λ k m -λ j i -(λ k m ) + e λ j i -λ k m (λ j i ) x k m , y j i C n = e λ j i -λ k m λ k m -λ j i x k m , * (λ j i )y j i C n = 0. (3.41) 2.
Let (m, k) = (i, j), then from (3.39), (3.40) we have:

ϕ k m , ψ k m = Γ(λ k m )x k m , y k m ,
where

Γ(λ k m ) = I -e -λ k m A -1 + λ k m I - 0 -1 e λ k m θ A 2 (θ) dθ + 0 -1 θ 0 e λ k m s A 3 (s) ds dθ + λ k m 0 -1 θ 0 e λ k m s A 2 (s) ds dθ. (3.42)
The last two terms of Γ(λ k m ) are integrals with the domain -1 ≤ θ ≤ s ≤ 0 and we integrate first by s and after that by θ. Let us change the order of integrating:

0 -1 ( θ 0 ds) dθ = - 0 -1 ( 0 θ ds) dθ = - 0 -1 ( s -1 dθ) ds. Thus, we obtain Γ(λ k m ) = I -e -λ k m A -1 + λ k m I - 0 -1 e λ k m θ A 2 (θ) dθ - 0 -1 e λ k m s A 3 (s) s -1 dθ ds -λ k m 0 -1 e λ k m s A 2 (s) s -1 dθ ds = I -e -λ k m A -1 - 0 -1 e λ k m θ A 2 (θ) dθ - 0 -1 e λ k m s sA 3 (s) ds -λ k m 0 -1 e λ k m s sA 2 (s) ds + λ k m I - 0 -1 e λ k m s sA 3 (s) ds -λ k m 0 -1 e λ k m s sA 2 (s) ds = Γ 1 (λ k m ) + Γ 2 (λ k m ).
It is easy to see that

Γ 1 (λ k m ) + λ k m e -λ k m A -1 = -(λ k m ) Γ 2 (λ k m ) -λ k m e -λ k m A -1 = -(λ k m ). Taking into account that x k m ∈ Ker (λ k m ), we conclude that ϕ k m , ψ k m M 2 = - (λ k m )x k m , y k m C n . (3.43)
The last completes the proof of the lemma. 2

Proof of Lemma 3.2. We choose x k m C n = 1 and y k m C n = 1 and obtain the following estimates.

ϕ k m 2 = (I -e -λ k m A -1 )x k m 2 + 0 -1 e λ k m θ x k m 2 dθ ≤ I -e -λ k m A -1 2 + 0 -1 e 2Reλ k m θ dθ ≤ 1 + e 2Reλ k m A -1 2 + 1 -e -2Reλ k m 2Reλ k m ≤ 1 + A -1 2 + 1 -e 2r (N ) 2r (N ) ≤ C.
We have used here the fact that the real function 1-e -y y decreases monotone from ∞ to 1 when y → -0.

Moreover,

1 λ k m ψ k m 2 = y k m 2 + 0 -1 (e -λ k m θ - 1 λ k m A * 2 (θ) + 1 λ k m e -λ k m θ θ 0 e λ k m s A * 3 (s) ds +e -λ k m θ θ 0 e λ k m s A * 2 (s) ds)y k m 2 dθ ≤ y k m 2 (1 + e 2Reλ k m -1 2Reλ k m + 1 |λ k m | 2 A * 2 (θ) 2 L 2 (-1,0;C n×n ) + 1 |λ k m | 2 0 -1 e -2Reλ k m θ dθ 0 -1 e 2Reλ k m s A * 3 (s) ds + 0 -1 e -2Reλ k m θ dθ 0 -1 e 2Reλ k m s A * 2 (s) ds) ≤ 1+ 1 |λ k m | 2 A * 2 (θ) 2 L 2 (-1,0;C n×n ) + e 2Reλ k m -1 2Reλ k m 1 + 1 |λ k m | 2 A * 3 (θ) 2 L 2 (-1,0;C n×n ) + A * 2 (θ) 2 L 2 (-1,0;C n×n ) ≤ 2 + 1 |λ k m | 2 + 1 A * 1 |λ k m | 2 A * 3 (θ) 2 L 2 (-1,0;C n×n ) ≤ C.
Here we used the fact that the real function e y -1 y increases monotone from 0 to 1 when y → -0. 2 Remark 3.2 We note that the norm of eigenvectors ψ k m (with y k m = 1) increases infinitely when k → ∞. This could be seen on example of eigenvectors ψ k m of the operator

A * (A 2 (θ) = A 2 (θ) ≡ 0): ψ k m 2 = y k m 2 + 0 -1 λ k m e -λ k m θ y k m 2 dθ = y k m 2 1 + |λ k m | 2 e 2Reλ k m -1 2Reλ k m ≥ (1+C|λ k m | 2 ) → ∞.
Proof of Lemma 3.3. Let λ k m ∈ Λ 1 and we prove the representation (3.27). From the form of the matrices R m we conclude that

R -1 m = R * m = R m , 1 ≤ m ≤ 1 and R 1 = I.
Let us analyze the matrix

1 λ k m R m (λ k m )R m = -I+e -λ k m R m A -1 R m + 0 -1 e λ k m s R m A 2 (s)R m ds+ 1 λ k m 0 -1 e λ k m s R m A 3 (s)R m ds.
Since the matrix A -1 is in Jordan form (2.8), then, multiplying A -1 on R m from the left and from the right we change places of one-dimensional Jordan blocks µ 1 and µ m :

R m A -1 R m = µ m 0 0 S ,
where S ∈ C (n-1)×(n-1) . Let us consider the matrix

-I + e -λ (k) m R m A -1 R m , where λ k m = i(arg µ m + 2πk
) is an eigenvalue of the operator A, i.e. root of the equation e λ = µ m . This matrix is of the form

-I + e -λ (k) m R m A -1 R m = 0 0 0 -I n-1 + e -λ (k) m S ,
where I n-1 is the identity matrix in C (n-1)×(n-1) . Thus, det(-I + e -λ (k) m R m A -1 R m ) = 0 and since the multiplicity of the eigenvalue µ m ∈ σ 1 equals 1, then rg(-

I + e -λ (k) m R m A -1 R m ) = n -1, which means that det(-I n-1 + e -λ (k) m S) = 0, Since |λ k m -λ (k) m | → 0 when k → ∞,
I n-1 + e -λ k m S) = det(-I n-1 + e -λ (k) m S + (e -λ k m -e -λ (k) m )S) = 0. (3.44)
According to Proposition 4.3, elements of the matrices

0 -1 e λ k m s R m A i (s)R m ds, i = 2
, 3 are small when k → ∞. We introduce the following notation:

0 -1 e λs R m A 2 (s)R m ds + 1 λ 0 -1 e λs R m A 3 (s)R m ds =    ε 11 (λ) . . . ε 1n (λ) . . . . . . . . . ε n1 (λ) . . . ε nn (λ)    .
Let us choose N > 0 such that for any k

: |k| ≥ N the estimate |ε ij (λ k m )| ≤ ε holds. Therefore, due to Proposition 4.2, we obtain that det S m,k ≡ det    -I n-1 + e -λ k m S +    ε 22 (λ k m ) . . . ε 2n (λ k m ) . . . . . . . . . ε n2 (λ k m ) . . . ε nn (λ k m )       = 0. (3.45)
In the notation introduced the singular matrix 1

λ k m R m (λ k m )R m has the following struc- ture: 1 λ k m R m (λ k m )R m =      -1 + e -λ k m µ m + ε 11 (λ k m ) ε 12 (λ k m ) . . . ε 1n (λ k m ) ε 21 (λ k m ) . . . ε n1 (λ k m ) S m,k      . (3.46)
We denote elements of S m,k as S m,k = {s m,k ij } 2≤i,j≤n = {s ij } 2≤i,j≤n . Since det S m,k = 0 then the first row of the matrix (3.46) is a linear combination of all other rows and also the first column is a linear combination of all other columns:

ε 1i (λ k m ) = p 2 s 2i + . . . + p n s ni , i = 2, n (3.47)
and ε j1 (λ k m ) = q 2 s j2 + . . . + q n s jn , j = 2, n.

(3.48)

The last relations defines two matrices given by (3.26), i.e.

P m,k =      1 -p 2 . . . -p n 0 1 . . . 0 . . . . . . . . . . . . 0 0 . . . 1      , Q m,k =      1 0 . . . 0 -q 2 1 . . . 0 . . . . . . . . . . . . -q n 0 . . . 1     
, and direct computations gives us that the matrix 1

λ k m P m,k R m (λ k m )R m Q m,k has the form (3.27), i.e.: 1 λ k m P m,k R m (λ k m )R m Q m,k =      0 0 . . . 0 0 . . . 0 S m,k     
. Now let us give the estimate (3.28), i.e. an estimate on numbers p i and q i . The equations (3.47), (3.48) can be rewritten in the form:

v 1 = (S m,k ) T w 1 , v 2 = S m,k w 2 ,
where

v 1 = (ε 12 (λ k m ), . . . , ε 1n (λ k m )) T , w 1 = (p 2 , . . . , p n ) T , v 2 = (ε 21 (λ k m ), . . . , ε n1 (λ k m ))
T , w 2 = (q 2 , . . . , q n ) T . Since det S m,k = 0 we have:

w 1 = (S -1 m,k ) T v 1 , w 2 = S -1 m,k v 2
and let us estimate S -1 m,k . From (3.44) and (3.45) we see that S m,k = -I n-1 + 1 µm S -B m,k , where elements of the matrix B m,k are small: |b ij | < ε. Thus, there exists N ∈ Z such that the norm of the operator B m,k ≡ (-I n-1 + 1 µm S) -1 B m,k is small: B m,k < 1 (and, moreover, it is less then any ε > 0) for any |k| > N . Thus, there exists an inverse of the matrix I n-1 -B m,k (see e.g. [13, p.233, Theorem 5]), which is uniformly bounded for any k (for example, we take

B m,k < 1 2 ): (I n-1 -B m,k ) -1 = ∞ i=0 ( B m,k ) i ≤ C 1 .
Thus, we have

S -1 m,k = (I n-1 -B m,k ) -1 (-I n-1 + 1 µ m S) -1 ≤ C 1 (-I n-1 + 1 µ m S) -1 ≤ C.
Since ε 1i (λ k m ) and ε j1 (λ k m ) are small then v 1 ≤ ε and v 2 ≤ ε, therefore, w 1 ≤ εC, w 2 ≤ εC and we conclude that the estimate (3.28) holds.

2

Proof of Corollary 3.1. The matrix function m,k (λ)

def = 1 λ P m,k R m (λ)R m Q m,k is analytic in a neighborhood of the point λ k m since (λ) is analytic. Moreover, since also m,k (λ k m ) is of the form (3.27), then we conclude that (λ) is of the form (3.29), i.e. m,k (λ) =      (λ -λ k m )r 11 (λ) (λ -λ k m )r 12 (λ) . . . (λ -λ k m )r 1n (λ) (λ -λ k m )r 21 (λ) . . . (λ -λ k m )r n1 (λ) M m,k (λ)      , λ ∈ U (λ k m ),
where r ij (λ) = r m,k ij (λ) are analytic functions. Let us prove the relation (3.30). If we suppose that r m,k

11 (λ k m ) = 0 then (λ-λ k m )r m,k 1i (λ) = (λ -λ k m ) 2 r m,k 11 (λ).
Decomposing det m,k (λ) by the elements of the first row, we see that all the term of this decomposition have the multiplier (λ-

λ k m ) 2 , i.e. det m,k (λ) = (λ-λ k m ) 2 r(λ),
where r(λ) is an analytic function. The last contradicts to the assumption that λ k m is an eigenvalue of the multiplicity one of the operator A.

Moreover, taking into account (3.46) and the form of the transformations (3.27) we see that

(λ -λ k m )r k 11 (λ) = (-1 + e -λ µ m + ε 11 (λ)) - n i=2 p i ε i1 (λ) - n j=2 q j ε 1j (λ)
and, therefore,

r 11 (λ k m ) = ((λ -λ k m )r 11 (λ)) λ=λ k m = -e -λ k m µ m + ε 11 (λ) - n i=2 p i ε i1 (λ) - n j=2 q j ε 1j (λ) λ=λ k m .
The terms (ε ij (λ)) are of the form

ε ij (λ)) = 0 -1 e λs s(A 2 (s)) ij ds + 1 λ 0 -1 e λs s(A 3 (s)) ij ds - 1 λ 2 0 -1
e λs (A 3 (s)) ij ds, therefore, using Proposition 4.3, we conclude that

ε 11 (λ) - n i=2 p i ε i1 (λ) - n j=2 q j ε 1j (λ) λ=λ k m → 0, k → ∞.
Since -e -λ k m µ m → -1 when k → ∞, then we obtain the relation (3.30) and, in particular, there exists a constant C > 0 and an integer N such that for |k| > N we have

0 < C ≤ |r k 11 (λ k )|.
The last completes the proof of the proposition. 2

Proof of Lemma 3.4. Let us first prove the estimate (3.32) for eigenvalues λ k 1 ∈ Λ 1 . We use Lemma 3.3: since x k 1 ∈ Ker (λ k 1 ), then

0 = 1 λ k 1 P -1 1,k P 1,k (λ k 1 )Q 1,k Q -1 1,k x k 1 = P -1 1,k (λ k 1 )Q -1 1,k x k 1 , (3.49)
where P 1,k , Q 1,k , (λ k 1 ) are defined by (3.26), (3.27). Taking into account the form (3.27) of the matrix (λ k 1 ) and, namely, det S 1,k = 0, we conclude that Q -1 1,k x k 1 = ( x 1 , 0, . . . , 0) T . On the other hand, from (3.31) and multiplying directly

Q -1 1,k on x k 1 , we conclude that x 1 = (x k 1 ) 1 and (x k 1 ) i = -q i (x k 1 ) 1 , i = 2, n.
Thus, taking into account (3.28), we have:

1 = x k 1 2 = |(x k 1 ) 1 | 2 (1 + |q 2 | 2 + . . . + |q n | 2 ) ≤ |(x k 1 ) 1 | 2 (1 + (n -1)ε 2 )
and, therefore, we conclude that

|(x k 1 ) 1 | ≥ 1 √ 1+(n-1)ε 2 → 1 when k → ∞. Summarizing, we have Q -1 1,k x k 1 = ((x k 1 ) 1 , 0, . . . , 0) T , 0 < C ≤ |(x k 1 ) 1 | ≤ 1, λ k 1 ∈ Λ 1 . (3.50) Conjugating (3.27), we obtain      0 0 . . . 0 0 . . . 0 S * 1,k      = ( 1 λ k 1 P 1,k (λ k 1 )Q 1,k ) * = 1 λ k 1 Q * 1,k * (λ k 1 )P * 1,k = * (λ k 1 ). (3.51) Since y k 1 ∈ Ker * (λ k 1 ), then 0 = 1 λ k 1 (Q * 1,k ) -1 Q * 1,k (λ k 1 )P * 1,k (P * 1,k ) -1 y k 1 = (Q * 1,k ) -1 * (λ k 1 )(P * 1,k ) -1 y k 1 (3.52)
and, taking into account the form (3.51) of the matrix * (λ k 1 ), we conclude that (P * 1,k ) -1 y k 1 = ( y 1 , 0, . . . , 0) T . On the other hand, from the form of (P * 1,k ) -1 we conclude that y 1 = (y k 1 ) 1 and (y k 1 ) i = -p i (y k 1 ) 1 , i = 2, n. Thus, taking into account (3.28), we have:

1 = y k 1 2 = |(y k 1 ) 1 | 2 (1 + |p 2 | 2 + . . . + |p n | 2 ) ≤ |(y k 1 ) 1 | 2 (1 + (n -1)ε 2 )
and, therefore, we conclude that

|(y k 1 ) 1 | ≥ 1 √ 1+(n-1)ε 2 → 1 when k → ∞. (P * 1,k ) -1 y k 1 = ((y k 1 ) 1 , 0, . . . , 0) T , 0 < C ≤ |(y k 1 ) 1 | ≤ 1, λ k 1 ∈ Λ 1 . ( 3
     r 11 (λ k 1 ) r 12 (λ k 1 ) . . . r 1n (λ k 1 ) r 21 (λ k 1 ) . . . r n1 (λ k 1 ) M m,k (λ k 1 )      = (λ k 1 ) = ( 1 λ k 1 P 1,k (λ k 1 )Q 1,k ) λ=λ k 1 = P 1,k 1 λ k 1 (λ k 1 ) - 1 (λ k 1 ) 2 (λ k 1 ) Q 1,k . (3.54)
Finally, we have

- 1 λ k 1 (λ k 1 )x k 1 , y k 1 = -P -1 1,k P 1,k 1 λ k 1 (λ k 1 ) - 1 (λ k 1 ) 2 (λ k 1 ) Q 1,k Q -1 1,k x k 1 , y k 1 = -P 1,k 1 λ k 1 (λ k 1 ) - 1 (λ k 1 ) 2 (λ k 1 ) Q 1,k Q -1 1,k x k 1 , (P -1 1,k ) * y k 1 = -      r 11 (λ k 1 ) r 12 (λ k 1 ) . . . r 1n (λ k 1 ) r 21 (λ k 1 ) . . . r n1 (λ k 1 ) M m,k (λ k 1 )           (x k 1 ) 1 0 . . . 0      ,      (y k 1 ) 1 0 . . . 0      = -r 11 (λ k 1 )(x k 1 ) 1 (y k 1 ) 1 → 1, k → ∞. (3.55)
From the last the estimate (3.32) follows and the lemma is proved in the case of eigenvalue

λ k 1 ∈ L (m) 1 . 2 Remark 3.3 For λ k m ∈ Λ 1 , 1 ≤ m ≤ 1 we have that - 1 λ k m (λ k m )x k m , y k m = -r 11 (λ k m )(x k m ) m (y k m ) m → 1, k → ∞. (3.56)
Proof of Remark 3.3. The ideas of our arguing remain quite the same as in the case when we consider λ k 1 ∈ Λ 1 but now arguing becomes more complicated. As we have noted in the proof of Lemma 3.3, multiplying A -1 on R m from the left and from the right we change places of one-dimensional Jordan blocks µ 1 and µ j . Therefore, all previous arguing could be applying to the matrix R m (λ k m )R m instead of (λ k m ). Also, we note that

R -1 m = R * m = R m , 1 ≤ m ≤ 1 and R 1 = I.
Using Lemma 3.3 and since x k m ∈ Ker (λ k m ), we have

0 = 1 λ k m R m P -1 m,k P m,k R m (λ k m )R m Q m,k Q -1 m,k R m x k m = R m P -1 m,k (λ k m )Q -1 m,k R m x k m , (3.57)
where P m,k , Q m,k , (λ k m ) are of the form (3.26), (3.27). From the form (3.27) of the matrix (λ k m ) we conclude that

Q -1 m,k R m x k m = ( x 1 , 0, . . . , 0) T . Multiplication Q -1
m,k on R m from the right changes places the first and the j-th column of Q -1 m,k , therefore, multiplying Q -1 m,k R m on x k m , we obtain:

x 1 = (x k m ) m and (x k m ) i = -q m (x k m ) m , i ∈ {1, .
. . , n}\{m}. Thus, taking into account (3.28), we have:

1 = x k m 2 ≤ |(x k m ) m | 2 (1 + (n -1)ε 2 )
and, therefore, we conclude that

|(x k m ) m | ≥ 1 √ 1+(n-1)ε 2 → 1 when k → ∞. Summarizing, we have Q -1 m,k R m x k m = ((x k m ) m , 0, . . . , 0) T , 0 < C ≤ |(x k m ) m | ≤ 1, λ k m ∈ Λ 1 . (3.58)
The same arguing gives us that

(P * m,k ) -1 R m y k m = ((y k m ) m , 0, . . . , 0) T , 0 < C ≤ |(y k m ) m | ≤ 1, λ k m ∈ Λ 1 . (3.59)
and also

     r 11 (λ k m ) r 12 (λ k m ) . . . r 1n (λ k m ) r 21 (λ k m ) . . . r n1 (λ k m ) M m,k (λ k m )      = P m,k R m 1 λ k m (λ k m ) - 1 (λ k m ) 2 (λ k m ) R m Q m,k .
(3.60) Finally, we obtain

- 1 λ k m (λ k m )x k m , y k m = -R m P -1 m,k P m,k R m 1 λ k m (λ k m ) - 1 (λ k m ) 2 (λ k m ) R m Q m,k Q -1 m,k R m x k m , y k m = -P m,k R m 1 λ k m (λ k m ) - 1 (λ k m ) 2 (λ k m ) R m Q m,k Q -1 m,k x k m , (P -1 m,k ) * R m y k m = -      r 11 (λ k m ) r 12 (λ k m ) . . . r 1n (λ k m ) r 21 (λ k m ) . . . r n1 (λ k m ) M m,k (λ k m )           (x k m ) m 0 . . . 0      ,      (y k m ) m 0 . . . 0      = -r 11 (λ k m )(x k m ) m (y k m ) m → 1, k → ∞. (3.61)
The last completes the proof of the lemma. 2

Resolvent boundedness

Our aim is to prove that the resolvent R(λ, A) is uniformly bounded on the subspace M 0 2 for λ : Reλ ≥ 0. We give the proof of this proposition by the following steps: 1. We prove that the relation D(z, ξ, λ k m ) ∈ Im (λ k m ) holds for any (z, ξ(θ)) T ∈ M 0 2 and for any λ k m ∈ Λ 1 . 2. Using the previous fact we prove that the resolvent is uniformly bounded on the subspace M 0 2 in the neighborhood of each λ k m ∈ Λ 1 . 3. Finally, we prove that the resolvent is uniformly bounded on the subspace M 0 2 in the whole right half-plane.

We need the following propositions. Lemma 4.1 For any vector g = (z, ξ(θ)) T ∈ M 0 2 and for any eigenvalue λ k m ∈ Λ 1 the following relation holds:

D(z, ξ, λ k m ) ∈ Im (λ k m ). (4.62) Lemma 4.2 The vector-function -1 (λ)D(z, ξ, λ) is uniformly bounded in neighborhoods U δ (λ k m ) of eigenvalues λ k m ∈ Λ 1 for some fixed δ > 0, i.e.: 1. for any 1 ≤ m ≤ 1 , |k| > N the estimate -1 (λ)D(z, ξ, λ) ≤ C m,k holds in a neighborhood U δ (λ k m ); 2. there exists C > 0 such that C m,k ≤ C for all 1 ≤ m ≤ 1 , |k| > N .
{λ k } is such that Imλ k → ∞ and -∞ < a ≤ Reλ k ≤ b < ∞ then for any f (s) ∈ L 2 (0, 1; C n×m ) we have: 0 -1 e λ k s f (s) ds → 0 when k → ∞.
Proof of Theorem 2.2. We introduce two sets K 1 (δ) = {λ : Reλ ≥ 0} ∩ U δ ( λ k m ) and K 2 (δ) = {λ : Reλ ≥ 0}\K 1 for small enough δ > 0 such that Lemma 4.2 holds.

First, let us prove that R(λ, A)x ≤ C x for any λ ∈ K 1 (δ) and x ∈ M 0 2 . Due to Lemma 4.2 we have:

-1 (λ)D(z, ξ, λ) ≤ C 1 . Due to Corollary 4.1 we obtain that 0 -1 e -λs ξ(s) ds ≤ C 2 . Thus, for any x = (z, ξ(•)) ∈ M 0 2 we obtain:

R(λ, A)x = e -λ A -1 0 -1 e -λs ξ(s) ds + (I -e -λ A -1 ) -1 (λ)D(z, ξ, λ) C n + θ 0 e λ(θ-s) ξ(s) ds + e λθ -1 (λ)D(z, ξ, λ) L 2 ≤ e δ A -1 C 2 + (1 + e δ A -1 )C 1 + ( 0 -1 θ 0 e λ(θ-s) ξ(s) ds + e λθ -1 (λ)D(z, ξ, λ) 2 C n dθ) 1 2 ≤ e δ A -1 C 2 + (1 + e δ A -1 )C 1 + (e δ C 2 + C 2 1 )
1 2 ≤ C. Thus, for any x =∈ M 0 2 the family R(λ, A)x is bounded and due to Banach-Steinhaus theorem we have that R(λ, A) uniformly bounded on M 0 2 . Now we prove that R(λ, A)x ≤ C x for any λ ∈ K 2 (δ) and x ∈ M 2 . We begin with proof that | 1 λ det (λ)| ≥ ε for some ε > 0 and for any λ ∈ K 2 (δ)\U (0). Let us suppose the contrary then there exists a sequence {λ i } such that | 1 λ i det (λ i )| → 0. If we suppose that this sequence is bounded then its subsequence converges: λ i j → λ and | 1 λ det ( λ)| = 0. However, closure of λ ∈ K 2 (δ)\U (0) does not contains zeroes of det (λ) and we obtain a contradiction. Thus, |λ i | → ∞ when i → ∞. Analyzing the matrix Let ψ k m ∈ M 1 2 be an eigenvector of A * corresponding to the eigenvalue λ k m . Thus, this eigenvector is of the form (3.23), i.e.

ψ k m = y k m λ k m e -λ k m θ -A * 2 (θ) + e -λ k m θ θ 0 e λ k m s A * 3 (s) ds + λ k m e -λ k m θ θ 0 e λ k m s A * 2 (s) ds y k m ,
where y k m ∈ Ker * (λ k m ). We use the orthogonality of ψ k m and g:

0 = g, ψ k m = z, y k m + 0 -1 ξ(θ), λ k m e -λ k m θ y k m dθ - 0 -1 ξ(θ), A * 2 (θ)y k m dθ + 0 -1 ξ(θ), e -λ k m θ θ 0 e λ k m s A * 3 (s) ds • y k m dθ + 0 -1 ξ(θ), λ k m e -λ k m θ θ 0 e λ k m s A * 2 (s) ds • y k m dθ = z, y k m + 0 -1 λ k m e -λ k m θ ξ(θ) dθ, y k m - 0 -1 A 2 (θ)ξ(θ) dθ, y k m + 0 -1 e -λ k m θ θ 0 e λ k m s A 3 (s) ds ξ(θ) dθ, y k m + 0 -1 λ k m e -λ k m θ θ 0 e λ k m s A 2 (s) ds ξ(θ) dθ, y k m .
The last two terms are integrals with the domain -1 ≤ θ ≤ s ≤ 0 and we integrate first by s and after that by θ. Let us change the order of integrating:

0 -1 ( θ 0 ds) dθ = - 0 -1 ( 0 θ ds) dθ = - 0 -1 ( s -1 dθ) ds. Thus, we obtain g, ψ k m = z + 0 -1 λ k m e -λ k m θ ξ(θ) dθ + 0 -1 A 2 (θ)ξ(θ) dθ - 0 -1 e λ k m s A 3 (s) s -1 e -λ k m θ ξ(θ) dθ ds - 0 -1 λe λ k m s A 2 (s) s -1 e -λ k m θ ξ(θ) dθ ds, y k m . (4.65)
Since y k m ∈ Ker * (λ k m ), then for any x ∈ C n we have:

0 = x, * (λ k m )y k m = (λ k m )x, y k m .
Therefore, for any θ the relation e -λ k m θ (λ k m )ξ(θ), y k m = 0 holds, and, integrating it by θ from -1 to 0, we obtain:

0 = 0 -1 e -λ k m θ (λ k m )ξ(θ) dθ, y k m = - 0 -1 λ k m e -λ k m θ ξ(θ) dθ + 0 -1 λ k m e -λ k m θ A -1 ξ(θ) dθ +λ k m 0 -1 e λ k m s A 2 (s) ds 0 -1 e -λ k m θ ξ(θ) dθ + 0 -1 e λ k m s A 3 (s) ds 0 -1 e -λ k m θ ξ(θ) dθ, y k m . (4.66)
Adding the relation (4.66) to (4.65) we see that the term 0 -1 λ k m e -λ k m θ ξ(θ) dθ is reducing and in the last two terms we have -

0 -1 ( s -1 dθ) ds + 0 -1 ( 0 -1 dθ) ds = 0 -1 ( 0 s dθ) ds = - 0 -1 ( s 0 dθ) ds.
Finally, making a change of the variables: s = θ, θ = s in these two terms, we obtain:

0 = z + λ k m e -λ k m A -1 0 -1 e -λ k m θ ξ(θ) dθ - 0 -1 A 2 (θ)ξ(θ) dθ - 0 -1 e λ k m θ λ k m A 2 (θ) + A 3 (θ) θ 0 e -λ k m s ξ(s) ds dθ, y k m = D(z, ξ, λ k m ), y k m . Since, y k m ∈ Ker * (λ k m ) we conclude that D(z, ξ, λ k m )⊥Ker * (λ k m
) and this is equivalent to the inclusion (4.62), i.e. D(z, ξ, λ k m ) ∈ Im (λ k m ). In other words, there exists the inverse image of the vector D(z, ξ, λ k m ) though, and we emphasize this fact, the matrix -1 (λ k m ) does not exists (det (λ k m ) = 0). The last completes the proof of the lemma.

Proof of Lemma 4.2. We analyze the behavior of the vector-function -1 (λ)D(z, ξ, λ) near the imaginary axis. The matrix -1 (λ) does not exists for the values λ = λ k m , where λ k m ∈ Λ 1 are eigenvalues of the operator A and these eigenvalues approach to the imaginary axis when k → ∞. On the other hand, and we prove this fact below, the limit lim

λ→λ k m -1 (λ)D(z, ξ, λ) exists.
We introduce a notation:

f (λ) = -1 (λ)D(z, ξ, λ) = 1 λ (λ) -1 1 λ D(z, ξ, λ) . (4.67) 1.
In the first part of our proof we find a transformation that "separates" the singularity of the matrix 1 λ (λ) -1 . Using this representation we will show that in a neighborhood U (λ k m ) of each eigenvalues λ k m ∈ Λ 1 the vector-function f (λ) is analytic, i.e. we show that in the representation

f (λ) = 1 λ -λ k m f -1 + f 0 + (λ -λ k m )f 1 + ... the coefficient f -1 = lim λ→λ k m (λ -λ k m )f (λ) is equal to zero.
(We note that in the case when λ k m would be a pole of order more than 1 we would obtain lim

λ→λ k m (λ -λ k m )f (λ) = ∞.) According to Lemma 3.2 there exist matrices P m,k , Q m,k such that the matrix-function m,k (λ) def = 1 λ P m,k R m (λ)R m Q m,k at the point λ = λ k m has the form (3.27), i.e. m,k (λ k m ) def = 1 λ k m P m,k R m (λ k m )R m Q m,k =      0 0 . . . 0 0 . . . 0 S m,k      , det S m,k = 0,
We rewrite the representation (4.67) of the function f (λ) in a neighborhood U (λ k m ) as follows:

f (λ) = 1 λ R m P -1 m,k P m,k R m (λ)R m Q m,k Q -1 m,k R m -1 1 λ D(z, ξ, λ) = R m Q m,k 1 λ P m,k R m (λ)R m Q m,k -1 P m,k R m 1 λ D(z, ξ, λ) = R m Q m,k m,k (λ) -1 P m,k R m 1 λ D(z, ξ, λ) . (4.68)
The matrix-function m,k (λ) is analytic a neighborhood of the λ k m and we write down its Taylor expansion:

m,k (λ) = m,k (λ k m ) + (λ -λ k m ) m,k (λ k m ) + 1 2 (λ -λ k m ) 2 m,k (λ k m ) + . . .
On the other hand, due to Corollary (3.1), the matrix-function m,k (λ) allows the representation (3.29), i.e.

m,k (λ) =      (λ -λ k m )r 11 (λ) (λ -λ k m )r 12 (λ) . . . (λ -λ k m )r 1n (λ) (λ -λ k m )r 21 (λ) . . . (λ -λ k m )r n1 (λ) M m,k (λ)      , λ ∈ U (λ k m ),
where r ij (λ) = r m,k ij (λ) are analytic functions, and differentiating this representation by λ at the point λ = λ k m , we obtain:

m,k (λ k m ) =      r 11 (λ k m ) r 12 (λ k m ) . . . r 1n (λ k m ) r 21 (λ k m ) . . . r n1 (λ k m ) M m,k (λ k m )      =      r 11 (λ k m ) r 12 (λ k m ) . . . r 1n (λ k m ) 0 . . . 0 0      +      0 0 . . . 0 r 21 (λ k m ) . . . r n1 (λ k m ) M m,k (λ k m )      = Γ 0 + Γ 1 .
We introduce the matrix

F (λ) = m,k (λ k m ) + (λ -λ k m )Γ 0 , i.e. F (λ) =      r 11 (λ k m )(λ -λ k m ) r 12 (λ k m )(λ -λ k m ) . . . r 1n (λ k m )(λ -λ k m ) 0 . . . 0 S m,k      . ( 4.69)
The matrix

F (λ) is non-singular in a neighborhood U (λ k m ). Indeed, det F (λ) = r 11 (λ k m )(λ - λ k m ) det S m,k
, where det S m,k = 0 and r 11 (λ k m ) = 0 due to Corollary (3.1). Therefore, there exists the matrix F -1 (λ), which has the following structure:

F -1 (λ) = 1 r 11 (λ k m )(λ -λ k m ) det S m,k      det S m,k F 21 . . . F n1 0 F 22 . . . F n2 . . . . . . . . . . . . 0 F 2n . . . F nn      , (4.70)
where main minors F ij of the matrix F (λ) are of the form

F ij = (λ -λ k m )f ij , f ij ∈ C
. Thus, we have the following representation of the matrix m,k (λ):

m,k (λ) = F (λ) + (λ -λ k m )Γ 1 + 1 2 (λ -λ k m ) 2 m,k (λ k m ) + . . . = F (λ)(I + (λ -λ k m )F -1 (λ)Γ 1 + 1 2 (λ -λ k m ) 2 F -1 (λ) m,k (λ k m ) + . . .) (4.71)
We introduce the notation Υ

(λ) = (λ-λ k m )F -1 (λ)Γ 1 + 1 2 (λ-λ k m ) 2 F -1 (λ) m,k (λ k m )+.
. .. Our next goal is to prove that there exists δ > 0 such that in any neighborhood the following estimate holds:

Υ(λ) ≤ 1, λ ∈ U δ ( λ k m ). (4.72)
From (4.71) we have that Υ(λ) = F -1 m,k (λ) m,k (λ)-I, i.e. we want to prove that ∀ε > 0, ∃δ > 0, ∃N such that ∀|k| > N , ∀λ ∈ U δ ( λ k m ) we have an estimate

F -1 m,k (λ) m,k (λ) -I ≤ 1. (4.73)
Let us introduce a notation for components of the matrices Υ(λ) and M m,k (λ):

Υ(λ) = {Υ ij (λ)} n i,j=1 , M m,k (λ) = {m ij (λ)} n i,j=2 .
Direct computations gives us that

F i1 = (-1) j+1 (λ -λ k m ) r 12 (λ k m ) . . . r 1n (λ k m ) m 22 (λ k m ) . . . m 2n (λ k m ) . . . . . . . . . m i-12 (λ k m ) . . . m i-1n (λ k m ) m i+12 (λ k m ) . . . m i+1n (λ k m ) . . . . . . . . . m n2 (λ k m ) . . . m nn (λ k m ) , i = 2, n, F ij = r 11 (λ k m )(λ -λ k m )M ij (λ k m ), i, j = 2, n,
where M ij (λ) are main minors of the matrix M m,k (λ). We conclude that F i1 converges to zero when k → ∞. Let us compute the elements of the matrix Υ(λ

) = F -1 m,k (λ) m,k (λ) -I. Υ 11 (λ) = (λ -λ k m )r 11 (λ) det M m,k (λ k m ) + (λ -λ k m ) n j=2 r j1 (λ)F j1 (λ -λ k m )r 11 (λ k m ) det M m,k (λ k m ) -1 = r 11 (λ) r 11 (λ k m ) + n j=2 r j1 (λ)F j1 r 11 (λ k m ) det M m,k (λ k m ) -1 → 0, k → ∞. Υ ii (λ) = n j=2 m ji (λ)F ji (λ -λ k m )r 11 (λ k m ) det M m,k (λ k m ) -1 = n j=2 m ji (λ)M ji (λ k m ) det M m,k (λ k m ) -1 = n j=2 (m ji (λ) -m ji (λ k m ))M ji (λ k m ) det M m,k (λ k m ) → 0, k → ∞, i = 2, n. Υ 1i (λ) = (λ -λ k m )r 1i (λ) det M m,k (λ k m ) + n j=2 m ji (λ)F j1 (λ -λ k m )r 11 (λ k m ) det M m,k (λ k m ) = r 1i (λ) r 11 (λ k m ) + n j=2 m ji (λ)F j1 (λ -λ k m )r 11 (λ k m ) det M m,k (λ k m ) → 0, k → ∞, i = 2, n. Υ ij (λ) = n s=2 m sj (λ)F sj (λ -λ k m )r 11 (λ k m ) det M m,k (λ k m ) = n s=2 m sj (λ)M sj (λ k m ) det M m,k (λ k m ) = n s=2 (m sj (λ) -m sj (λ k m ))M sj (λ k m ) det M m,k (λ k m ) → 0, k → ∞, i, j = 2, n, i = j.
Therefore, the matrix I + Υ(λ) has an inverse for any λ ∈ U δ ( λ k m ):

(I + Υ(λ)) -1 = I + (λ -λ k m )Γ(λ), (4.74)
where Γ(λ) is analytic in a neighborhood U (λ k m ). We note also that the matrix F -1 (λ)Γ 1 does not depend on λ. Indeed, denoting by

f ij = f ij r 11 (λ k m ) det S m,k we obtain: F -1 (λ)Γ 1 =       1 r 11 (λ k m )(λ-λ k m ) f 21 . . . f n1 0 f 22 . . . f n2 . . . . . . . . . . . . 0 f 2n . . . f nn            0 0 . . . 0 r 21 (λ k m ) . . . r n1 (λ k m ) M m,k (λ k m )      = Γ 1 .
Therefore, the term (λ -

λ k m )F -1 (λ)Γ 1 = (λ -λ k m ) Γ 1 is small with respect to (λ -λ k m ) when λ → λ k m .
It is also easy to see that for any s ≥ 2:

(λ -λ k m )F -1 (λ) (s) m,k (λ k m ) = (λ -λ k m ) Γ s (λ)
, where Γ s (λ) are analytic matrices.

Finally, from (4.68), (4.71) and (4.74) we obtain:

f (λ) = R m Q m,k -1 m,k (λ)P m,k R m 1 λ D(z, ξ, λ) = R m Q m,k F (λ)(I + (λ -λ k m )F -1 (λ)Γ 1 + 1 2 (λ -λ k m ) 2 F -1 (λ) m,k (λ k m ) + . . .) -1 •R m P m,k 1 λ D(z, ξ, λ) = R m Q m,k (I +(λ-λ k m )Γ(λ))F -1 (λ)P m,k R m 1 λ D(z, ξ, λ) .
λ k m D(z, ξ, λ k m ) ∈ Im( 1 λ k m (λ k m ))
), then, due to Proposition 4.1 we have that

1 λ k m P m,k R m D(z, ξ, λ k m ) ∈ Im (λ k m ). (4.76)
Moreover, since the matrix (λ k m ) is of the form (3.27) we conclude that

1 λ k m P m,k R m D(z, ξ, λ k m ) = (0, d 0 2 , . . . , d 0 n ) T . (4.77)
Since D(z, ξ, λ) is an analytic vector-function in a neighborhood U (λ k m ) then 1 λ P m,k R m D(z, ξ, λ) is also an analytic vector-function and, due to (4.77) we conclude that

1 λ P m,k R m D(z, ξ, λ) =      (λ -λ k m )d 1 1 + . . . d 0 2 + (λ -λ k m )d 1 2 + . . . . . . d 0 n + (λ -λ k m )d 1 2 + . . .      . ( 4
(λ -λ k m )f (λ) = R m Q m,k lim λ→λ k m (λ -λ k m )F -1 (λ)P m,k R m 1 λ D(z, ξ, λ) = R m Q m,k lim λ→λ k m       1 r 11 (λ k m ) (λ -λ k m ) f 21 . . . (λ -λ k m ) f n1 0 (λ -λ k m ) f 22 . . . (λ -λ k m ) f n2 . . . . . . . . . . . . 0 (λ -λ k m ) f 2n . . . (λ -λ k m ) f nn            (λ -λ k m )d 1 1 + . . . d 0 2 + . . . . . . d 0 n + . . .      = R m Q m,k      1 r 11 (λ k m ) 0 . . . 0 0 0 . . . 0 . . . . . . . . . . . . 0 0 . . . 0           0 d 0 2 . . . d 0 n      = R m Q m,k • 0 = 0. (4.79)
Thus, we have proved that f (λ) = -1 (λ)D(z, ξ, λ) is an analytic vector-function in a neighborhood of each eigenvalue λ k m .

2. Let us prove that f (λ) is uniformly bounded in the neighborhoods of the eigenvalues λ k m ∈ Λ 1 . This means to prove that the set of vectors

f 0 = f m,k 0 = f (λ k m ) = ( -1 (λ)D(z, ξ, λ)) λ=λ k m
is bounded. Taking into account the representation (4.75) we have:

f m,k 0 = f (λ k m ) = R m Q m,k F -1 (λ)P m,k R m 1 λ D(z, ξ, λ) λ=λ k m = R m Q m,k       1 (λ-λ k m )r 11 (λ k m ) f 21 . . . f n1 0 f 22 . . . f n2 . . . . . . . . . . . . 0 f 2n . . . f nn            (λ -λ k m )d 1 1 + . . . d 0 2 + . . . . . . d 0 n + . . .      λ=λ k m = R m Q m,k           d 1 1 r 11 (λ k m ) + n i=2 f i1 d 0 i n i=2 f i2 d 0 i . . . n i=2 f in d 0 i           . ( 4
(a i,σ(i) + b i,σ(i) ) = σ∈Sn sgn(σ) c i,σ(i) =a i,σ(i) ∨b i,σ(i) n i=1 c i,σ(i) = σ∈Sn sgn(σ) n i=1 a i,σ(i) + σ∈Sn sgn(σ) condition n i=1 c i,σ(i) = det A + σ∈Sn sgn(σ) condition n i=1 c i,σ(i) ,
where condition ≡ (c i,σ(i) = a i,σ(i) ∨ b i,σ(i) and there is always at least one multiplier b i,σ(i) ). (The last notation is not too good, but the proposition is just for internal utilization :) ).

The sum σ∈Sn consists of n! terms and the sum

c i,σ(i) =a i,σ(i) ∨b i,σ(i)
consists of 2 n terms.

Therefore, the sum σ∈Sn condition consists of n!(2 n -1) terms. Thus, choosing ε

0 < | det A| 2C , where C = n!(2 n -1)(max i,j |a ij |) n-1 , we obtain | det(A+B)| ≥ | det A| -| σ∈Sn sgn(σ) condition n i=1 c i,σ(i) | ≥ | det A|- 1 2 | det A| = 1 2 | det A| > 0
what completes the proof of the proposition. 2

Proof of Proposition 4.3. Integrals a k (λ) can be considered as Fourier coefficients of the function e λs f (s), thus, they converge to zero when k → ∞. It remains to prove that they converge uniformly on the set L 0 . The last means that for any ε > 0 there exists n ∈ N such that for any |k| ≥ n and for any λ ∈ L 0 we have |a k (λ)| < ε.

Let us suppose the contrary: ∃ε > 0 such that ∀n ∈ N, ∃|k| ≥ n, ∃λ ∈ L 0 : |a k (λ)| ≥ ε. Thus, there exists a sequence k 1 < k 2 < . . . and a sequence {λ

k i } ∞ i=1 such that |a k i (λ k i )| ≥ ε.
Since L 0 is a bounded set then there exists a converging subsequence of {λ k i } ∞ i=1 which we denote by {λ j } j∈J , where J ⊂ N is a strictly increasing sequence. Moreover, since L 0 is also closed, then the limit of {λ j } j∈J belongs to L 0 : λ j → λ 0 ∈ L 0 . Let us show that the sequence {a k (λ 0 )} does not converge to zero.

Indeed, choosing big enough n ∈ N, such that for any j > n, j ∈ J and any s ∈ [-1, 0]: |e λ 0 s -e λ j s | ≤ ε/2 f (s) , we obtain

|a j (λ 0 ) -a j (λ j )| = | 0 -1 e 2πijs (e λ 0 s -e λ j s )f (s) ds| ≤ 0 -1 |e λ 0 s -e λ j s |f (s) ds ≤ ε 2 . Since |a j (λ j )| ≥ ε and assuming that |a j (λ 0 )| ≤ |a j (λ j )|, we obtain ε 2 ≥ |a j (λ 0 ) -a j (λ j )| ≥ |a j (λ j )| -|a j (λ 0 )| ≥ ε -|a j (λ 0 )|,
and, thus, |a j (λ 0 )| ≥ ε 2 for any j ∈ J, j > n. Thus, {a k (λ 0 )} does not converge to zero and we have obtained a contradiction with the fact that they are the coefficients of the Fourier series of the function e λ 0 s f (s). The last completes the proof of the proposition. 2

An example of stable and unstable situations

In this section we give an example of systems having the same spectrum but some of them are stable and some are unstable. The spectrum of these systems satisfies the following assumptions:

σ(A) = {λ : Reλ < 0}, σ(A -1 ) = {µ : |µ| ≤ 1} ⇒ σ( A) = {λ : Reλ ≤ 0},
and it satisfies the assumption that the set σ 1 = σ(A -1 ) ∩ {µ : |µ| = 1} is not empty and no eigenvalue µ ∈ σ 1 generates a Jordan block of dimension higher than one. We consider the system ż(t) = A -1 ż(t -1) + A 0 z(t), (6.81) with the state space C 2 . The matrices A -1 and A 0 are of the form:

A -1 = -1 0 0 -1 , A 0 = -b s 0 -b ,
where b is a real positive number and for the value of s we essentially distinguish two cases: s = 0 and s = 0.

Remark 6.1 The system (6.81) is a special case of the system (1.1). To show this one should choose A 2 (θ) = (θ + 1)A 0 and A 3 (θ) = A 0 . Thus,

0 -1 A 2 (θ) ż(t + θ) dθ + 0 -1 A 3 (θ)z(t + θ) dθ = 0 -1 (θ + 1)A 0 ż(t + θ) dθ + 0 -1 A 0 z(t + θ) dθ = = A 0 0 -1 ((θ + 1)z(t + θ)) dθ = A 0 z(t).
We prove the following three propositions about systems (6.81).

(-λI + λe -λ A -1 + A 0 ) = det -λ -λe -λ -b s 0 -λ -λe -λ -b = 0.
Thus, all the eigenvalues of the operator A satisfy the equation λe λ + λ + be λ = 0 (6.82) and the multiplicity of each of these eigenvalues (as zeroes of the equation det A (λ) = 0) equals two. Further, we list some results on transcendental equations obtained in the paper of L. Pontryagin [START_REF] Pontryagin | On the zeros of some elementary transcendental functions[END_REF].

Let us consider an equation H(z) = 0, where H(z) = h(z, e z ) is a polynomial with respect to z and e z .

Let us consider the equation for eigenvectors:

(A -λ k I) y z(θ) = 0 ⇔ A 0 z(0) -λ k y ż(θ) -λ k z(θ) = 0.
The solution of the second equation is given by z(θ) = e λ k θ z(0), thus, z(-1) = e -λ k z(0). Taking into account the domain of the operator A: y = z(0)-A -1 z(-1) = (I -e -λ k A -1 )z(0), we obtain from the first equation: (A 0 -λ k I + λ k e -λ k A -1 )z(0) = 0, or, in an equivalent form: A (λ k )z(0) = 0.

As we have noted above the last equation has two-dimensional solution: z(0) = z 1 k = (1, 0) T and z(0) = z 2 k = (0, 1) T , and therefore, there is a two-dimensional eigenspace of the operator A corresponding to the eigenvalue λ k . Two eigenvectors of this subspace can be chosen in the following form:

f 1 k = y 1 k z 1 k (•)
, where

y 1 k = 1 + e -λ k 0 , z 1 k (θ) = e λ k θ 0 and f 2 k = y 2 k z 2 k (•)
, where y 2 k = 0 1 + e -λ k , z 2 k (θ) = 0 e λ k θ . Thus, to any eigenvalue λ k of the operator A there corresponds the two-dimensional eigenspace.

2.

If s = 1 (or s = 0) we have that A (λ k ) = 0 e -λ k 0 0 and, therefore, the space of solutions of the equation A (λ k )z = 0 is one-dimensional: z 1 k = (1, 0) T (since, obviously, e -λ k = 0). Thus, the equation A (λ k )z = z 1 k has also one-dimensional solution which we denote by z 2 k = (0, 1) T . Arguing as above we show that the operator A possesses one eigenvector corresponding to the eigenvalue λ k :

f 1 k = y 1 k z 1 k (•)
, where y 1 k =

1 + e -λ k 0 , z 1 k (θ) = e λ k θ 0 .

Let us show that the operator A possesses one root vector. Each root vector f of the operator A satisfies the following equation: 1 + e -λ k -λ k e -λ k 0 .

Thus, if 1 + e -λ k -λ k e -λ k = 0 then z(0) is the root vector of the matrix A (λ k ) and, therefore, z(0) = z 2 k = (0, 1) T and the operator A possesses the root vector

f 2 k = y 2 k z 2 k (•)
, where y 2 k = e -λ k 1 + e -λ k and z 2 k (θ) = θe λ k θ e λ k θ . It remains to show that 1 + e -λ k -λ k e -λ k = 0. We suppose the contrary: λ k e -λ k = 1 + e -λ k and, multiplying the last expression onto e λ k , we obtain λ k = e λ k + 1. Since λ k is an eigenvalue of A, then we obtain e λ k (e λ k + 1) + e λ k + 1 + e λ k = 0 or e 2λ k + 3e λ k + 1 = 0, and we conclude that e λ k = -3± √ 5 2

. For the root e λ k = -3- > 0. We have obtained the contradiction what completes the analysis of eigenvectors and root vectors of the operator A.

Proof of Proposition 6.3. Again we consider two cases. 1. First we consider the system (6.81) when s = 0 and prove that in this case the system is stable. Let us evaluate the norm of eigenvectors. Since Reλ k → 0 when k → ∞, then lim

k→∞ 1 2Reλ k 1 -e -2Reλ k = 1. Therefore, 0 < C 1 ≤ 1 2Reλ k 1 -e -2Reλ k ≤ C 2 .
Taking into account that |e -λ k | ≤ C 3 , we obtain the following estimates:

C 1 ≤ f i k 2 ≤ (1 + C 3 ) 2 + C 2 = C 4 . (6.84)
As it has been shown in [START_REF] Rabah | Stability analysis of neutral type systems in hilbert space[END_REF], the subspaces V (k) = Lin{f 1 k , f 2 k } (and the finite-dimensional subspace W N ) form a Riesz basis of the space M 2 . Since we have proved the estimate (6.84) then the eigenvectors {f 1 k , f 2 k } (together with vectors from W N ) form a basis of M 2 . Therefore, we have a Riesz basis of eigenvectors. Further we just repeat the proof given in [START_REF] Rabah | Stability analysis of neutral type systems in hilbert space[END_REF]Theorem 23].

We consider a norm • 1 in which the eigenvectors {f 1 k , f 2 k } k∈Z are orthogonal. Let a vector x belongs to a closed span of the subspaces V (k) , then x = k∈Z (α k f 1 k + β k f 2 k ) and we have:

e At x = k∈Z e λ k t (α k f 1 k + β k f 2 k ).
Therefore,

e At x 2 1 = k∈Z e λ k t ( α k f 1 k 2 1 + β k f 2 k 2 1 ) ≤ k∈Z ( α k f 1 k 2 1 + β k f 2 k 2 1 ) = x 2 1
and, thus, the family e At is uniformly bounded in the subspace generated by the subspaces V (k) . From the last we conclude that the system is strongly asymptotically stable.

2.

Let us consider the system (6.81) when s = 1. Let an operator A has a sequence of eigenvalues {λ k } ∞ k=1 such that Reλ k < 0 and Reλ k → 0 when k → ∞ and to each λ k there corresponds one eigenvector v k and at least one root vector w k . We show that the equation ẋ = Ax is unstable. Let us suppose that v k = w k = 1. Since, for each w k we have e At w k = e λ k t (tv k + w k ) then e At w k = |e λ k t | (tv k + w k ) ≥ e Reλ k t (t -1).

For any constant C > 0 we take t ≥ 2C + 1 and for this t we take big enough k such that e Reλ k t ≥ 1 2 . Then we have:

e At w k ≥ 1 2 (2C + 1 -1) = C
and we conclude that e At ≥ C for t ≥ 2C + 1. Therefore, the family of exponents e At is not uniformly bounded and because of Banach-Steinhaus theorem there exists x ∈ D(A) such that e At x → ∞ when t → +∞. Thus, the system (6.81) is unstable when s = 1. The last completes the proof of the proposition.

2

Conclusion

We have shown that the result on stability analysis obtained in [START_REF] Rabah | Stability analysis of neutral type systems in hilbert space[END_REF] for neutral type systems can be extended on mixed retarded-neutral type systems. Though the formulation of the result remains the same, the method of its proof requires to involve resolvent boundedness technic.

)Lemma 3 . 2 } 1≤m≤l 1 |k|≥N

 321 where x k m , y k m are defined by (3.22) and (3.23). The sets of eigenvectors {ϕ k m

  then for any ε > 0 there exists N > 0 such that for k : |k| ≥ N the estimate |e -λ k m -e -λ (k) m | S ≤ ε holds, and, thus, due to Proposition 4.2, we have det(-

  .53) Differentiating (3.29) and putting λ = λ k 1 we obtain

Lemma 4 . 3 Proposition 4 . 1 Proposition 4 . 2 Proposition 4 . 3 - 1 e

 434142431 For each (z, ξ) ∈ M 2 there exist constants C 1 , C 2 such that for any λ : Reλ > 0 If the vector y ∈ ImA, A ∈ C n×n , then for any two matrices P , Q such that det P = 0 and det Q = 0 the relation P y ∈ Im(P AQ) holds. For any matrix A ∈ C n×n with det A = 0 there exists ε 0 > 0 such that for any matrixB = {b ij } ∈ C n×n : |b ij | < ε 0 we have det(A + B) = 0. Let L 0 ⊂ C be a bounded closed set and f (s) ∈ L 2 [-1, 0]. Denote by a k (λ) = 0 2πiks e λs f (s) ds, λ ∈ L 0 , k ∈ Z. Then |a k (λ)| → 0 when k → ∞uniformly on the set L 0 . Corollary 4.1 If the sequence

1 λ 0 - 1 e 0 - 1 e- 1 e 1 λ 0 - 1 e

 101011101 (λ) = -I + e -λ A -1 + λs A 2 (s) ds + 1 λ λs A 3 (s) ds we see that 0 λs A 2 (s) ds → 0 and λs A 3 (s) ds → 0 and | det(-I + e -λ A -1 )| = | (1 + e -λ µ m )| ≥ (1 + e δ µ m ). Thus, we obtain a contradiction again.Since also 1 λ (λ) ≤ C 1 and 1 λ D(z, ξ, λ) ≤ C 2 we obtain that -1 (λ)D(z, ξ, λ) ≤ C 3 , for all λ ∈ K 2 (δ)\U (0). Since |e -λ 0 -1 s k e -λs ds| ≤ C 4 for all λ ∈ K 2 (δ)\U (0) and {s k } is a dense set in L 2 we obtain that e -λ 0 -1 e -λs ξ(s) ds ≤ C 5 . Similarly we obtain an estimate R(λ, A)x ≤ C for any λ ∈ K 2 (δ) and x ∈ M 2 and thus, due to Banach-Steinhaus theorem we have that R(λ, A), λ ∈ K 2 (δ)\U (0) uniformly bounded on M 2 . The last completes the proof of the theorem.Proof of Lemma 4.1. We show that D(z, ξ, λ k m )⊥Ker * (λ k m ) for any vector g = z ξ(θ) ∈ M 0 2 and any λ k m ∈ Λ 1 .

  (4.75) Since, due to Lemma (4.1), D(z, ξ, λ k m ) ∈ Im (λ k m ) (and, obviously 1

2 . 2 5

 22 .80) We recall that f ij = f ij r 11 (λ k m ) det S m,k and, therefore, it remains to give estimates of the following values: Q m,k , P m,k , det S m,k , r 11 (λ k m ), f ij , d 0 i and d 1 1 . An estimate on P m,k and Q m,k follows from Lemma 3.3 and, namely, from the estimate (3.28): for any ε > 0 there exists N ∈ Z such that for any |k| ≥ N : Q m,k ≤ n + (n -1)ε 2 = C, P m,k ≤ n + (n -1)ε 2 = C. An estimate 0 < C 2 ≤ |r 11 (λ k m )| follows from the relation (3.30) of the Corollary 3Estimates for det S k , f ij , d 0 i and d 1 1 follows immediately from Lemma (4.3). Thus, we conclude that f m,k 0 ≤ C for all 1 ≤ m ≤ 1 , k : |k| ≥ N what completes the proof of the Lemma. Proof of auxiliary results Proof of Proposition 4.1. The relation y ∈ ImA means that there exists a vector x such that Ax = y. Since Q is non-singular then there exists a vector x 1 such that x = Qx 1 . Therefore, AQx 1 = y and, multiplying on P from the left we obtain P AQx 1 = P y what completes the proof of the proposition. 2 Proof of Proposition 4.2. The proposition holds since the determinant is a continuous function of matrix elements. More precisely, since det(A) = σ∈Sn sgn(σ) n i=1 a i,σ(i) , where S n denotes the set of all n! permutations of the set S = {1, 2, . . . , n} we have det(A + B) = σ∈Sn sgn(σ) n i=1

Proposition 6 . 1 Proposition 6 . 2 Proposition 6 . 3 1 e 1 e

 61626311 For any b > 0 and any s ∈ C the spectrum of the corresponding operator A belongs to the open left half-plane, i.e. for any λ ∈ σ(A): Reλ < 0. If s = 0 then the operator A possesses eigenvectors only, i.e. it possesses no root vectors; if s = 0 then to any eigenvalue λ ∈ σ(A) there corresponds a pair of eigenvector and root vector of the operator A. If s = 0 then the system (6.81) is stable and, if s = 0 then it is unstable.Proof of Proposition 6.1. The eigenvalues of the operator A are determined from the equationdet A (λ) = det -λI + λe -λ A -1 + λ 0 -λs A 2 (s) ds + 0 -λs A 3 (s) ds = 0,which in our particular case has the form: det

1 k

 1 (A -λ k I) 2 f = 0, i.e. f = (A -λ k I)f ∈ Ker(A -λ k I). From the last we conclude that f = f 1 k =(1 + e -λ k , 0) T (e λ k θ , 0) T and the root vector satisfies the relationA 0 z(-1) -λ k y ż(θ) -λ k z(θ) = (1 + e -λ k , 0) T (e λ k θ , 0) T .The solution of the second equation is given byz(θ) = e λ k θ z(0) + θ 0 e λ k (θ-τ ) e λ k τ y 1 k dτ = e λ k θ z(0) + θe λ k θ y 1 k , what implies z(-1) = e -λ k z(0) -e -λ k y 1 k .Taking into account the domain of the operator A, we write down the first equation in the following form:A 0 z(0) -λ k (z(0) -A -1 (e -λ k z(0) -e -λ k y 1 k )) = (1 + e -λ k )y which is equivalent to A (λ k )z(0) = (1 + e -λ k )y 1 k + λ k A -1 e -λ k y 1 k =

√ 5 2

 2 we have that Reλ k > 0, and for the root e λ k = -

f 1 k 2 = f 2 k 2 1 e 1 e

 122211 = 1 + e -λ k , 1 + e -λ k + 0 -λ k θ e λ k θ dθ = |1 + e -λ k | 2 + 0 -2Reλ k θ dθ = |1 + e -λ k | 2 + 1 2Reλ k 1 -e -2Reλ k .

(θ) 2 L 2 (-1,0;C n×n ) +

Acknowledgments.

This work was supported in part by École Centrale de Nantes.

 Definition 6.1We say that the function H(z) = m,n a mn z m e nz possesses the principal term a rs z r e sz , if for all other terms a mn z m e nz we have that r ≥ m and s ≥ n.

We denote by F (y) and G(y) the real-valued functions of a real variable which are correspondingly the real and the imaginary parts of the function H(iy), i.e. H(iy) = F (y) + iG(y). Definition 6.2 We say that the zeroes of two real-valued functions of a real variable alternate iff a) each of these functions has no multiple roots; b) between every two zeroes of one of these functions there exists at least one zero of the other; c) the functions are never simultaneously zero.

Theorem 6.1 ([17, Pontryagin, 1955]). Let H(z) = h(z, e z ) be a polynomial with a principal term.

1. If all the zeroes of the function H(z) lie to the left side of the imaginary axis (Reλ k < 0), then the zeroes of the functions F (y) and G(y) are real, alternating and for all y ∈ R the following inequality holds G (y)F (y) -G(y)F (y) > 0. (6.83)

2. Any of the conditions below is sufficient for all the zeroes of the function H(z) to lie in the open left half-plane: a) all the zeroes of the functions F (y) and G(y) are real, alternating and the inequality (6.83) is satisfied for at least one value of y; b) all the zeroes of the function F (y) are real and for each zero y = y 0 the inequality (6.83) is satisfied, i.e. G(y 0 )F (y 0 ) < 0; c) all the zeroes of the function G(y) are real and for each zero y = y 0 the inequality (6.83) is satisfied, i.e. G (y 0 )F (y 0 ) > 0.

The following theorem allows us to check whether all zeroes of a function are real. Theorem 6.2 ([17, Pontryagin, 1955]). Let F (z) = f (z, cos z, sin z) be a polynomial with a principal term z r φ (s) m (cos z, sin z), where φ (s) m (cos z, sin z) is a homogeneous with respect to cos z and sin z polynomial.

The function F (z), z ∈ C possesses only real zeroes if and only if for all big enough k ∈ Z the function F (x), x ∈ R possesses exactly 4ks + r real roots on the interval -2πk + ε ≤ x ≤ 2πk + ε for some ε > 0.

We use the results mentioned above to analyze the equation (6.82). We divide H(iy) onto real and imaginary parts: H(iy) = iy(cos y + i sin y) + iy + b(cos y + i sin y) = (b cos y -y sin y) + i(y cos y + y + sin y) = F (y) + iG(y). and use the sufficient condition b) of Theorem 6.2. The equation F (y) = 0 can be rewritten in the form tgy = b y (since all the zeroes of the equations y = 0 and cos y = 0 are not zeroes of the equation F (y) = 0, we divide the initial equation on y cos y). Some zeroes of the equation F (y) = 0 can be seen on the picture below:

From the picture we can see that exactly 5 zeroes of the equation F (y) = 0 belongs to the interval -2π + ε ≤ y ≤ 2π + ε if we choose ε > 0 "big" enough such that the zero on the right from 2π belongs to the interval. Adding to this interval from the left and from the right 2π we add each time 4 other zeroes. Thus, on each interval -2πk + ε ≤ y ≤ 2πk + ε we have exactly 4k + 1 zeroes of the equation F (y) = 0 and, therefore, the conditions of the Theorem 6.2 are satisfied. Thus, all the zeroes of the equation F (y) = 0 are real.

Let us now prove that for each root y 0 of the equation F (y) = 0 the inequality G(y 0 )F (y 0 ) < 0 holds. We use the notation cos y = C, sin y = S and prove that -G(y 0 )F (y 0 ) > 0:

-GF = -(-bS -S -y 0 C)(y 0 C + y 0 + bS) = (bS + S + y 0 C)(y 0 C + y 0 + bS).

We rewrite tgy 0 = b y 0 as C = Sb y 0 and substitute C at the last relation:

Since S 2 > 0 and b + y 2 0 b + 1 > 0, then it remains to prove, that the third multiplier is greater then zero. From tgy 0 = b y 0 we also have b = y 0 S C and substituting we obtain:

We note also that the fact that y 0 SC > 0 can be easily seen from the picture above. Applying Theorem 6.2 we complete the proof of the fact that all the eigenvalues of the operator A belongs to the open left half-plane.

2

Proof of Proposition 6.2. We consider two cases. 1. If s = 0 then A (λ k ) ≡ 0 for each zero λ k of the equation (6.82), and, therefore, the the space of solutions of the equation A (λ k )z = 0 is two-dimensional. We choose the following basis of this space: z 1 k = (1, 0) T , z 2 k = (0, 1) T .


Rabah Rabah
email: rabah@emn.fr

Grigory Sklyar
email: sklar@sus.univ.szczecin.pl

On exact controllability of linear time delay systems of neutral type

The problem of exact null controllability is considered for linear neutral type systems with distributed delay. A characterization of this problem is given. The minimal time of controllability is precised. The results are based on the analysis of the Riesz basis property of eigenspaces in Hilbert space. Recent results on the moment problem and properties of exponential families are used.

Introduction

The problem of controllability for delay systems was considered by several authors in different framework. One approach is based on the analysis of time delay system in a module framework (space over ring, see [START_REF] Morse | Ring models for delay differential equation[END_REF]). In this case the controllability problem is considered in a formal way using different interpretations of the Kalman rank condition. Another approach is based on the analysis of time delay systems in vector spaces with finite or infinite dimension. A powerful tool is to consider a delay system as a system in a Banach functional space, this approach was developed widely in [START_REF] Hale | Theory of functional differential equations[END_REF]. Because the state space for delay systems is a functional space, the most important notion is the function space controllability. A first important contribution in the characterization of null functional controllability was given by Olbrot [START_REF] Olbrot | On degeneracy and related problems for linear time lag systems[END_REF] by using some finite dimensional tools as (A, B)-invariant subspaces for an extended system. For retarded systems one can refer to [START_REF] Manitius | Function space controllability of linear retarded systems: A derivation from abstract operator conditions[END_REF] (and references therein) for the analysis of function space controllability in abstract Banach spaces. The case of neutral type systems with discrete delay was also considered in such a framework (see O'Connor and Tarn [START_REF] O'connor | On the function space controllability of linear neutral systems[END_REF] and references therein). A general analysis of the time delay systems in infinite dimensional spaces is given in the book [START_REF] Bensoussan | Representation and control of infinite-dimensional systems[END_REF] where several methods and references are given.

The problem considered in this paper is close to that studied in [START_REF] O'connor | On the function space controllability of linear neutral systems[END_REF]. In this work the exact controllability problem was considered for neutral type systems with discrete delay using a semigroup approach in Sobolev spaces W

(1) 2 and a boundary control problem. We consider the problem of controllability for distributed delay system of neutral type in the space M 2 (-h, 0; C n ) = C n ×L 2 (-h, 0; C n ) which is natural for control problems. The semigroup theory developed here is based on the Hilbert space model introduced in [START_REF] Burns | Linear functionaldifferential equations as semigroups on product spaces[END_REF]. One of our result is a generalization of the result in [START_REF] O'connor | On the function space controllability of linear neutral systems[END_REF]. The main non trivial precision is the time of controllability. We generalize the results given [START_REF] Jacobs | Criteria for function space controllability of linear neutral systems[END_REF] for the case of a single input and one localized delay (see also [START_REF] Banks | Characterization of the controlled states in W (1) 2 of linear hereditary systems[END_REF][START_REF] Rodas Hernan Rivera | A sufficient condition for function space controllability of a linear neutral system[END_REF]). The approach developed here is different from that of [START_REF] O'connor | On the function space controllability of linear neutral systems[END_REF]. Our main results are based on the characterization of controllability as a moment problem and using some recent results on the solvability of this problem (see [START_REF] Avdonin | Families of exponentials. The method of moments in controllability problems for distributed parameter systems[END_REF] for the main tools used here). Using a precise Riesz basis in the space M 2 (-h, 0; C n ) we can give a characterization of null-controllability and of the minimal time of controllability.

The present paper contains only the main idea of the approach and the formulations of the main results on exact controllability. A complete presentation of this approach is the subject of our extensive work which is to be published. One can also find the detailed proofs in the the preprint [START_REF] Rabah | The analysis of exact controllability of neutral type systems by the moment problem approach[END_REF].

The model and the controllability problem

We study the following neutral type system

ż(t) = A -1 ż(t -h) + 0 -h A 2 (θ) ż(t + θ)dθ + 0 -h A 3 (θ)z(t + θ)dθ + Bu, (1)
where A -1 is constant n × n-matrix, det A -1 = 0, A 2 , A 3 are n × n-matrices whose elements belong to L 2 (-h, 0), h > 0 is a constant delay. We consider the operator model of the neutral type system (1) introduced by Burns and al. in product spaces. The state space is M 2 (-h, 0;

C n ) = C n × L 2 (-h, 0; C n ), shortly M 2 , and (1) is rewritten as d dt y(t) z t (•) = A y(t) z t (•) + Bu (2)
where the operator A is given by

A y(t) z t (•) = 0 -h A 2 (θ) żt (θ)dθ + 0 -h A 3 (θ)z t (θ)dθ dz t (θ)/dθ
with the domain

D(A) = (y, z(•)) : z ∈ H 1 (-h, 0; C n ), y = z(0) -A -1 z(-h) ⊂ M 2 .
The operator A is the infinitesimal generator of a C 0 -group. The operator B is defined by Bu = (Bu, 0). The relation between the solutions of the neutral type system (1) and the system (2) is given by the substitutions

y(t) = z(t) -A -1 z(t -1), z t (θ) = z(t + θ).
The reachability set at time T is defined by

R T = T 0 e At Bu(t)dt : u(•) ∈ L 2 (0, T ; C n ) It is easy to show that R T1 ⊂ R T2 as T 1 < T 2 . An important result is that R T ⊂ D(A) ⊂ M 2 .
This non-trivial fact permits to formulate the nullcontrollability problem in the following setting: i) To find maximal possible set R T (depending on T ); ii) To find minimal T for which the set R T becomes maximal possible, i.e. R T = D(A).

Definition 1. The system (2) is said null-controllable at the time T if R T = D(A)
The main tool is to consider the null-controllability problem as a problem of moments.

The moment problem

In order to formulate the moment problem we need a Riesz basis in the Hilbert space M 2 . We recall that a Riesz basis is a basis which may be transformed to an orthogonal basis with respect to another equivalent scalar product. Each Riesz basis possesses a biorthogonal basis. Let {ϕ} be a Riesz basis in M 2 and {ψ} the corresponding biorthogonal basis. Then for each x ∈ M 2 we have x = ϕ∈{ϕ} x, ψ ϕ. In a separable Hilbert space there always exists a Riesz basis.

A state x = y z(•) ∈ M 2 is reachable at time T by a control u(•) ∈ L 2 (0, T ; C r ) iff the steering condition x = y z(•) = T 0 e At Bu(t)dt. (3)
holds. This steering condition may be expanded using the basis {ϕ}. A state x is reachable iff

Effectiveness of the proposed approach becomes obvious if we assume that the operator A possess a Riesz basis of eigenvector. This situation is characteristic, for example, for control systems of hyperbolic type when A is skew-adjoint (A * = -A) and has a compact resolvent (see, for example, [ 4), ( 5)

= iλ k ϕ k , λ k ∈ R. Assuming for simplicity r = 1, b 1 = b = k α k ϕ k , α k = 0, we have from (
x k α k = T 0 e -iλ k t u(t)dt, k ∈ N, (6)
where x = k x k ϕ k . Equalities ( 6) are a non-Fourier trigonometric moment problem whose solvability is closely connected with the property for the family of exponentials e -iλ k t , k ∈ N, to form a Riesz basis on the interval [0, T ] ( [START_REF] Avdonin | Families of exponentials. The method of moments in controllability problems for distributed parameter systems[END_REF]).

In particular, if e -iλ k t forms a Riesz basis of L 2 [0, T 0 ] then one has

R T = x : k x k α k 2 < ∞ for all T ≥ T 0 . (7)
Obviously formula [START_REF] Manitius | Function space controllability of linear retarded systems: A derivation from abstract operator conditions[END_REF] gives the complete answer to the both items of the controllability problem. Returning now to neutral type systems we observe that the operator A given in ( 2) is not skew-adjoint and, moreover, does not possess a basis even of generalized eigenvectors. So the choice of a proper Riesz basis in context of formulas ( 4), ( 5) is an essentially more complicated problem.

The choice of basis

In order to design the needed basis for our case we use spectral the properties of the operator A obtained in [START_REF] Rabah | Stability analysis of neutral type systems in Hilbert space[END_REF]. Let µ 1 , . . . , µ , µ i = µ j be eigenvalues of A -1 and let the integers p m be defined as : dim (A -1 -µ m I) n = p m , m = 1, . . . , . Denote by

λ (k) m = 1 h (ln |µ m | + i(arg µ m + 2πk)) , m = 1, . . . , ; k ∈ Z,
and let L (k) m be the circles of the fixed radius r ≤ r 0 = 1 3 min |λ

(k) m -λ (j) i | centered at λ (k) m . Let {V (k) m } k∈ Z m=1,...,
be a family of A-invariant subspaces given by

V (k) m = P (k) m M 2 , P (k) m = 1 2πi L (k) m R(A, λ)dλ.
The following theorem plays an essential role in our approach Theorem 1. [START_REF] Rabah | Generalized Riesz basis property in the analysis of neutral type systems[END_REF] There exists N 0 large enough such that for any

N ≥ N 0 i) dim V (k) m = p m , k ≥ N , ii) the family {V (k) m } |k|≥N m=1,..., ∪ V N forms a Riesz basis (of subspaces) in M 2 ,
where V N is a finite-dimensional subspace (dim V N = 2(N + 1)n) spanned by all generalized eigenvectors corresponding to all eigenvalues of A located outside of all circles

L (k) m , |k| ≥ N , m = 1, . . . , .
Using this theorem we construct a Riesz basis {ϕ} of the form

ϕ k m,j , |k| > N ; m = 1, . . . , l; j = 1, . . . , p m ∪ φN j , j = 1, . . . , 2(N + 1)n
where for any m = 1, . . . , l, and k : |k| > N the collection {ϕ k m,j } j=1,...,pm is in a special way chosen basis of V (k) m and { φN j } j=1,...,2(N +1)n is a basis of VN . In this basis equalities (4) with regard to (5) turns into a moment problem with respect to a special collection of quasipolynomials. Analyzing the mentioned moment problem by means of the methods given in [START_REF] Avdonin | Families of exponentials. The method of moments in controllability problems for distributed parameter systems[END_REF] we obtain our main results concerning the null-controllability problem.

The main results

The characterization of the null-controllability is given by the following Theorem.

Theorem 2. The system (2) is null-controllable by controls from L 2 (0, T ) for some T > 0 iff the following two conditions hold: i) rank [ ∆ A (λ) B ] = n, ∀λ ∈ C; where

∆ A (λ) = -λI + λe -λh A -1 + λ 0 -h e λs A 2 (s)ds + 0 -h e λs A 3 (s)ds. ii) rank [ B A -1 B • • • A n-1 -1 B ] = n.
The main results on the time of controllability are as follows.

Theorem 3. Let the conditions i) and ii) of Theorem 2 hold. Then i) The system (2) is null-controllable at the time T as T > nh;

ii) If the system (2) is of single control (r = 1), then the estimation of the time of controllability in i) is exact, i.e. the system is not controllable at time T = nh.

For the multivariable case, the time depends on some controllability indices. suppose that dim B = r. Let {b 1 , . . . , b r } be an arbitrary basis noted β. Let us introduce a set integers. We denote by B i = ( b i+1 , . . . , b r ) , i = 0, 1, . . . , r -1, which gives in particular B 0 = B and B r-1 = ( b r ) and we put formally B r = 0. Let us consider the integers

n β i = rank [ B i-1 A -1 B i-1 • • • A n-1 -1 B i-1 ], i = 1, .
. . , r, corresponding to the basis β. We need in fact the integers

m β i = n β i-1 -n β i , Let us denote by m min = max β m β 1 m max = min β max i m β i ,
for all possible choice of a basis β.

The main result for the multivariable case is the following Theorem.

Theorem 4. Let the conditions i) and ii) of the Theorem 2 hold, then i) The system (2) is null-controllable at the time T > m max h; ii) The system (2) is not null-controllable at the time T < m min h.

The proofs are based on the construction of a special Riesz basis of Ainvariant subspaces in the space M 2 according to [START_REF] Rabah | Generalized Riesz basis property in the analysis of neutral type systems[END_REF] and on the analysis of the properties of some quasi-exponential functions to be a Riesz basis in L 2 (0, T ) depending of the time T [START_REF] Avdonin | Families of exponentials. The method of moments in controllability problems for distributed parameter systems[END_REF].

Final Conclusions

e 0 e 4 ) 0 ee

 040 At Bu(t), ψ dtϕ, for some u(•) ∈ L 2 (-h, 0; R r ). Then the steering condition (3) can be substituted by the following system of equalities x, ψ = T At Bu(t), ψ dt, ψ ∈ {ψ}. (Let {b 1 , . . . , b r } be an arbitrary basis in ImB, the image of the matrix B and b i = b i 0 ∈ M 2 , i = 1, . . . , r. Then the right hand side of (4) takes the form T At Bu(t), ψ dt = At b i , ψ u i (t)dt.

For

  the delayed system of neutral type[START_REF] Avdonin | Families of exponentials. The method of moments in controllability problems for distributed parameter systems[END_REF] we have the following results:i) All the reachable states z(t), t ∈ [T -1, T ] from 0 are elements of H 1 [T -1, T ] (independently of T ). ii) If T > m max h then the set of reachable states on [0, T ] coincides with H 1 [T -1, T ]. iii) If T < m min h then the set of reachable states is an essential subspace of H 1 [T -1, T ]. If r = 1,this gives m max = m min = n and then iv) For T = n the reachable states form a subspace in H 1 [T -1, T ] of finite codimension; for T < n there exists an infinite-dimensional subspace in H 1 [T -1, T ] of unreachable states.

Acknowledgements. This work was realized with the partial financial support of the French-Polish grant Polonium No 07599VH.


Rabah Rabah
email: rabah@emn.fr

Grigory M Sklyar
email: sklar@szczecin.univ.pl

On Controllabilty and Stabilizability of Linear Neutral Type Systems

Keywords: Neutral type systems, Riesz basis, exact controllability, stabilizability

Linear systems of neutral type are considered using the infinite dimensional approach. Conditions for exact controllability and regular asymptotic stabilizability are given. The main tools are the moment problem approach and the existence of a Riesz basis of invariant subspaces.

In this paper we consider the problem of controllability and stabilizability for a general class of neutral systems with distributed delays given by the equation

ż(t)-A -1 ż(t-1) = Lz t (•) = 0 -1 A 2 (θ) ż(t+θ)dθ + 0 -1
A 3 (θ)z(t+θ)dθ +Bu(t), [START_REF] Burns | Linear functional-differential equations as semigroups on product spaces[END_REF] where A -1 is a constant n × n-matrix, A 2 , A 3 are n × n, L 2 valued matrices. We consider the operator model of the neutral type system (1) in the product space

M 2 = C n × L 2 (-1, 0; C n ), so (1) can be reformulated as ẋ(t) = Ax(t) + Bu(t), x(0) = y z(•) , A = 0 L 0 d dθ , B = B 0 , (2)
with

D(A) = {(y, z(•)) ∈ M 2 : z ∈ H 1 ([-1, 0]; C), y = z(0) -A -1 z(-1)
}, and A is the generator of a C 0 -semigroup. The reachability set R T is such that R T ⊂ D(A) for all T > 0, with u(•) ∈ L 2 , the solution of (2) being in D(A).

Theorem 1. The system ( 2) is exactly null-controllable, i.e. R T = D(A), iff the pair (A -1 , B) is controllable and rank ∆ A (λ) B = n for all λ ∈ C, where

∆ A (λ) = λI -λe -λ A -1 -λ 0 -1 e λs A 2 (s)ds - 0 -1 e λs A 3 (s)ds,
If these conditions hold then the system is controllable at any time T > n 1 , where n 1 is the controllability index of the pair (A -1 , B). It is not controllable at T ≤ n 1 .

The main tools of the analysis is the moment problem approach and the theory of basis of exponential families. We construct a special Riesz basis using the existence of a Riesz basis of invariant subspaces [START_REF] Rabah | Stability analysis of neutral type systems in Hilbert space[END_REF] and describe the controllability problem via a moment problem in order to get the time of controllability. See [START_REF] Jacobs | Criteria for function space controllability of linear neutral systems[END_REF] for the monovariable and discrete delay case, via a different approach, and [START_REF] Rabah | On exact controllability of linear time delay systems of neutral type[END_REF] for a preliminary result.

The same Riesz basis of subspaces allows to characterize the problem of asymptotic stabilizability by a regular feedback law. From the operator point of view, the regular feedback law

u = Fx = 0 -1 F 2 (θ) ż(t + θ)dt + 0 -1 F 3 (θ)z(t + θ)dt, (3)
where F 2 , F 3 ∈ L 2 (-1, 0; C n×n ) means a perturbation of A by the operator BF which is relatively A-bounded and verifies D(A) = D(A + BF). Such a perturbation does not mean, in general, that A + BF is the infinitesimal generator of a C 0 -semigroup. However, in our case, this fact is verified directly since after the feedback we get also a neutral type system like (1) with D(A) = D(A + BF). This feedback law is essentially different from that which use the term F ẋ(t -1) (cf. for example [START_REF] Hale | Strong stabilization of neutral functional differential equations[END_REF]) and for which D(A) = D(A + BF). Our main result is In the case when A -1 has at least one eigenvalue |µ| = 1 with a nontrivial Jordan chain, the system can not be stabilized by a control of the form (3). The same if σ(A -1 ) ⊂ {µ : |µ| ≤ 1}. This follows from the fact that any control of the form (3) leaves the system in the same form and then it remains unstable [START_REF] Rabah | Stability analysis of neutral type systems in Hilbert space[END_REF].

Theorem 2 .

 2 (Rabah, Sklyar & Rezounenko) Under the assumptions: the eigenvalues of the matrix A -1 satisfy |µ| ≤ 1, the eigenvalues µ j , |µ j | = 1 are simple, the system (1) is regularly asymptotically stabilizable if rank ∆ A (λ) B = n for all λ : Re λ ≥ 0, and rank µI -A -1 B = n for all µ : |µ| = 1.