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On periodic one-parameter groups of linear operators in a Banach space and applications
Journal of Inequalities and Applications volume 2013, Article number: 172 (2013)
Abstract
Let D be the infinitesimal generator of a strongly continuous periodic one-parameter group of linear operators in a Banach space. Main results: An analog of the resolvent operator (= quasi-resolvent operator of D) is defined for points of the spectrum of D and its evident form is given. The theorem on integral for the operator D, theorems on the existence of periodic solutions of a linear differential equation of the n th order with constant coefficients and systems of linear differential equations with constant coefficients in Banach spaces are obtained. In the case of the existence of periodic solutions, evident forms of all periodic solutions of a linear differential equation of the n th order with constant coefficients and systems of linear differential equations with constant coefficients in Banach spaces are given in terms of resolvent and quasi-resolvent operators of D.
MSC:42A, 43, 47D.
1 Introduction
One-parameter groups of linear operators and periodic one-parameter groups of linear operators in topological vector spaces were investigated by Stone [1], Dunford [2], Gelfand [3] and others (see [4–20]).
Let T be the one-dimensional torus . Further we consider T as the additive group with its Euclidean topology, where Q is the field of real numbers. Let () be a strongly continuous one-parameter group of bounded linear operators in a Banach space H, and let D be an infinitesimal generator of the group . The evident form of the resolvent operator of D is known (see [7], Lemma 2.25)
where μ is an element of the resolvent set of D. For the element μ of the resolvent set of D and arbitrary element , the element is the evident form of the unique solution of the equation .
In the present paper, we obtain conditions of the existence of a solution of the equation for points of the spectrum of D. We define an analog of the resolvent operator (= quasi-resolvent operator) for points of the spectrum of D and, in the case of the existence of a solution, we give the evident form of all solutions by using a quasi-resolvent operator of D. We apply resolvent and quasi-resolvent operators to a solution of a linear differential equation of the n th order with constant coefficients and to a system of linear differential equations of the first order with constant coefficients in a Banach space H.
Contents of the present paper is the following. In Section 2 we give generalizations of Fejer’s theorem and Riemann-Lebesque’s lemma for a strongly continuous linear representation of T in a Banach space. These results are used in the next sections.
In Section 3 we give a definition of the infinitesimal generator D of a strongly continuous linear representation of T in a Banach space H and the domain of the definition of D. For D and any , we introduce the operator by formula (1) below for the point λ of the resolvent set of D and by (2) for the point of the spectrum of D. We show that the linear operator is bounded and has properties (3) and (4) below.
In Section 4 we prove that for all , the spectrum of the operator D is a point spectrum and , where is the spectrum of the linear representation α. It is proved that is equal to the resolvent operator of D for all points λ of the resolvent set of D. We obtain the theorem on an integral for D.
In Section 5 we give conditions of the existence of a periodic solution of a linear differential equation of the n th order with constant coefficients. In the case of existence, the evident form of all periodic solutions is given.
In Section 6 we give conditions of the existence of a periodic solution of a system of linear differential equations of the first order with constant coefficients. In the case of existence, the evident form of all periodic solutions is given.
For simplicity, we prove our main results for an isometric strongly continuous linear representation. But they are true for any strongly continuous linear representation.
2 Fejer’s theorem and Riemann-Lebesque’s lemma for a strongly continuous linear representation of T in a Banach space
Denote the group of all invertible bounded linear operators of a complex Banach space H by . The following Definitions 1-4 are known [21].
Definition 1 A homomorphism is called a linear representation of T on a Banach space H.
Definition 2 Linear representations and are called equivalent if there exists a bounded invertible linear operator such that for all .
Definition 3 A linear representation α of T on a Banach space H is called isometric if for all and .
Definition 4 A linear representation α of T on a Banach space H is called strongly continuous if for all .
It is known that every strongly continuous linear representation of T on a Banach space is equivalent to a strongly continuous isometric linear representation of T on a Banach space [21].
Let α be a strongly continuous linear representation of T on a Banach space H, and let Z be the ring of all integers and . Put
Let and . By the theorem in ([22], p.314), Riemann’s integral
exists and .
Proposition 1 Let α be a strongly continuous isometric linear representation of T on a Banach space H. Then
-
1.
for all , and ;
-
2.
and for all , ;
-
3.
for all and .
Proof It is easy, so it is omitted. □
A series in the form , , is called the Fourier series of an element , if for all . It is written in the form
For and for an integer number , let us put
is a subspace of H.
In the present paper, we assume that is infinite. The case of the finite is investigated easy and it is omitted.
Theorem 1 Let α be a strongly continuous isometric linear representation of T on a Banach space H. Then for every and .
Proof
In a standard manner, we obtain the equality
Since , for every , there exists such that and for all . We have
for all . Since α is a strongly continuous isometric linear representation, for all , and for all , it follows that
Hence . This, in view of for all n, implies that . □
Remark 1 Theorem 1 is known for the homogeneous Banach spaces on T ([6], p.87; [23], pp.14-15). For a strongly continuous linear representation of T in a locally convex space, it is obtained in [10].
Corollary 1 Let α be a strongly continuous isometric linear representation of T on a Banach space H and . If for each , then .
Proof Since for each , it follows that for each . Hence Theorem 1 gives . □
Theorem 2 Let α be a strongly continuous isometric linear representation of T on a Banach space H. Then for each .
Proof Let be given. Since , there exists a natural number such that for all natural numbers . Since
we have
So, for all . □
Remark 2 This theorem is a generalization of Riemann-Lebesque’s lemma ([23], p.13).
3 The operator
Let α be a strongly continuous isometric linear representation of T on a Banach space H.
Definition 5 ([7], p.45)
A point is called a differentiable point of α if there exists
Denote the set of all differentiable points of α by . The set is called the spectrum of D. The set is called the resolvent set of D.
Proposition 2 Let α be a strongly continuous isometric linear representation of T on a Banach space H. Then
-
(i)
is a linear subspace of H, and ;
-
(ii)
is -invariant and for all , ;
-
(iii)
for all and .
Proof (i) It is obvious that is a linear subspace of H. Let . Then x can be expressed in the form for some m. Since , we get
Hence and . Therefore . By Theorem 1 .
-
(ii)
Let . Since is strongly continuous, we have
Hence and .
-
(iii)
Let and . Using the continuity of and Proposition 1, we get
□
Remark 3 It is easily seen that if and only if is finite.
Definition 6 The operator D is called an infinitesimal generator of a linear representation α (see [7], p.45).
Proposition 3 Let . Then the function is differentiable on T and .
Proof Since α is strongly continuous, we have
□
Let α be a strongly continuous isometric linear representation of T on a Banach space H. For any and , there exists the following vector-valued Riemann’s integral:
We consider linear operators on H defined by
for all such that , and
for all . In Theorem 4(iv) below, we prove that is equal to the resolvent operator of D for every point λ of the resolvent set of D. The operator is called the quasi-resolvent operator of D for the point im of the spectrum of D. The operator was introduced in [9, 13].
Theorem 3 Let α be a strongly continuous isometric linear representation of T on a Banach space H. Then
-
(i)
the operator , defined by (1) and (2), is bounded for all ;
-
(ii)
(3)
for all , , ;
-
(iii)
(4)
for all and ;
-
(iv)
for all .
Proof (i) Let . Then it is obvious that
Therefore is bounded in H for all .
-
(ii)
Let , and for all . Then for some k. Since , using Proposition 1, we get
Hence, using for (Proposition 1), we get .
Let x be an arbitrary element of H. By Theorem 1, . Since , we have for all and . On the other hand, from , using the continuity of operators and , we get .
Now we prove equality (3) for , , . Let and . Then for some and . Hence
It implies that
Let . Then x= for some . Since , using equalities (5) and (2), we have
Hence
This equality implies that
for all , , and for all , .
Let x be an arbitrary element of H. By Theorem 1, and equality (6), we obtain
for all , , .
-
(iii)
The proof of equality (4) is similar to the proof of equality (3).
-
(iv)
Using (3) and (4), we obtain for , , , . Hence, by Corollary 1, we have for all , , . □
Corollary 2 Let , and . Then if and only if .
Proof It follows easily from Theorem 3. □
Proposition 4 Let α be a strongly continuous isometric linear representation of T on a Banach space H. Then
-
(i)
for all , , ;
-
(ii)
for all , .
Proof (i) Let , and , . For , using Theorem 4 and , we obtain
Similarly,
Hence for every . By Corollary 1 we have for all , and , .
A proof of (ii) is similar. □
4 The theorem on resolvent and quasi-resolvent operators
Theorem 4 Let α be a strongly continuous isometric linear representation of T on a Banach space H. Then
-
(i)
for all ;
-
(ii)
and for all and , such that ;
-
(iii)
and for all and , ;
-
(iv)
for all ;
-
(v)
the spectrum of D is a point spectrum and .
Proof (i) We need the following two lemmas.
Lemma 1 for all .
Proof An element has the form for some k. Using Theorem 3 and Proposition 2, we obtain for all and . Hence for any . Lemma is proved. □
Lemma 2 Let such that . Then
Proof Let us define the functions by and . Since α is a strongly continuous isometric linear representation, we have
and
Equality (8) implies that
Using , Proposition 3, Theorem 3, , and Lemma 1, we get
Hence
where . Putting , we obtain and
Using equalities (8), (9), (11) and inequality (10), we obtain
for all . Since , it follows that
and Lemma 2 is proved. □
We continue the proof of the theorem.
-
(i)
From equality (7) we obtain that the function is differentiable and . Using Proposition 3, we have
Hence for all such that . Now let be an arbitrary element. Since , we have . On the other hand, by Theorem 3 and Proposition 2, . Since and is a linear subspace of H, we get . Hence .
Conversely, let . By Proposition 2 and Theorem 3, and . Hence . Using Corollary 1 and equality (4), we get , and . Let us put . Then . This means that , that is, ; and consequently .
Now we prove that for all . By claim (ii) of Proposition 4 and claim (iv) of Theorem 3, . Since , and , we get . By Theorem 3, . Claim (ii) of Proposition 4 implies that . Hence and . Similarly, using claims (i) and (ii) of Proposition 4, we obtain for all , , . Thus for all .
-
(ii)
Let , . By Proposition 2, . Hence . Using and Theorem 3, we get . By Corollary 1, . Thus for all such that .
Let , . Since , we have . According to the statement (i) of this theorem, . Using Proposition 2, we get , and hence . Thus for all such that .
-
(iii)
The proof of claim (iii) is similar to the one of (ii).
-
(iv)
Equalities (iii) mean that is the resolvent operator for all . Hence equality (iv) follows from the form of the resolvent operator in ([7], Lemma 2.25).
-
(v)
follows from (ii) and (iii). The proof of the theorem is completed. □
Remark 4 Equality (iv) means that for points λ of the resolvent set of D, is the other form of the resolvent operator of D.
Theorem 5 Let α be a strongly continuous isometric linear representation of T on a Banach space H, and . Then the equation has a solution if and only if . In the case , a general solution of the equation has the form , where c is an arbitrary element of .
Proof Suppose that the equation has a solution x. By Proposition 2 we have and . Hence .
For the converse, we assume that . By Theorem 4 we have . Therefore is a solution of the equation . Let be a solution of the equation . Then . Hence for all , that is, . On the other hand, for all . Thus a general solution of the equation has the form , where c is an arbitrary element of . □
Remark 5 This theorem is the theorem on integral for periodic one-parameter groups of operators.
The following theorem is known (see [7], Theorem 2.26)
Theorem 6 Let α be a strongly continuous isometric linear representation of T on a Banach space H and . Then the Fourier series of the element a is convergent to a in H.
Proof
In a standard manner, we have the equality
Using , we obtain . Let us consider the function
Since , defining , we obtain that is a continuous function on . Using the equality
we get
where for and . Functions , and are continuous vector-valued functions on T. So, using Theorem 2 and equality (12), we obtain . □
Remark 6 Our proof of this theorem differs from that in ([7], Theorem 2.26).
Corollary 3 Let α be a strongly continuous isometric linear representation of T on a Banach space H, x be an arbitrary element of H and . Then the series is convergent to in Hfor all and the series is convergent to in H for all .
Proof Let . According to Theorem 4, we have . By Theorem 6 and , the series is convergent to in H. The proof of the second statement is similar. □
Corollary 4 Let α be a strongly continuous isometric linear representation of T in a Banach space H and .
-
(i)
Let . Then if and only if there exists such that ;
-
(ii)
Let . Then if and only if there exists such that .
The proof follows from Theorem 4, Corollary 2 and Proposition 2. □
5 Periodic solutions of the linear differential equation of the n th order with constant coefficients
Let be a strongly continuous linear representation of T in a Banach space H and D be the infinitesimal generator of α. For , we consider a solution of the linear differential equation
in H, where
, , I is the unit operator in H.
Theorem 7 Let be a linear differential operator (14), , let be the roots of the polynomial , and let be the multiplicity of , . Then
-
(i)
In the case , for any , there exists the unique solution of equation (13) in H, and it is .
-
(ii)
In the case () and , a solution of equation (13) exists if and only if
(15)
For , satisfying condition (15), a general solution of equation (13) has the form
where is an arbitrary element of , .
Proof (i) may be written in the form .
Then equation (13) has the form
By Theorem 4(iii),
for all and . Using equality (18) to equation (17), we obtain .
-
(ii)
Let () and . Using to (17), we obtain
(19)
This equation we can be written in the form
where for .
Lemma 3 Let , and . Then the equation
has a solution if and only if . In the case , a general solution of this equation is , where b is an arbitrary element of .
Proof Put . Then equation (21) has the form . By Theorem 5 this equation has a solution if and only if . In the case , a general solution is , where is an arbitrary element of . Let . Then equation (21) reduces to the equation
Put . Then this equation has the form
By Theorem 5, this equation is solvable if and only if . Therefore a general solution of equation (23) has the form , where is an arbitrary element of . By induction, we obtain that a general solution of equation (21) has the form , where b is an arbitrary element of . The lemma is proved. □
By this lemma, a solution of equation (20) reduces to a solution of the equation
where is an arbitrary element of . Using Lemma 3, by induction we obtain that a general solution of equation (20) has the form (16). □
6 Periodic solutions of the system of linear differential equations with constant coefficients
Let be a vector-line, where , . Denote by ⊤ the operator of transposition of a matrix. The denotes a vector-column of , . Denote by the set of all vectors , where , . For put .
We consider the following system of linear differential equations:
where , and A is a complex matrix.
Definition 7 Systems and , where A, B are complex -matrices, are called equivalent, if there exists a complex -matrix K such that , and .
In this case .
Proposition 5 Every system (25) is equivalent to the system of the form , where the matrix B has the Jordan form.
Proof For A there exists a complex -matrix K such that , where B is the Jordan normal form of A. From , we obtain , where and . □
By this proposition, a solution of system (25) reduces to a solution of the system of the form (25), where B has the Jordan normal form. Let B have the form
where is a Jordan block of the th order, . Then a solution of system (25), where B has the form (26), reduces to a solution of the following system of equations:
where is a Jordan block of the th order, , and . Therefore a solution of system (27) reduces to a solution of the equation of the form
where B is a Jordan block of the q th order with eigenvalue λ: , ; , ; , and .
Theorem 8 Let be a system of the form (28), where G is a Jordan block of the qth order with eigenvalue λ and . Then
-
(i)
for the case , the system has the unique solution , where:
-
(ii)
for the case , , , the system has a solution if and only if ,
-
(iii)
for the case , , and , a general solution of the system has the form
(29)
where c is an arbitrary element of .
Proof System (28) has the form
Let . From (30), using the operator , we obtain
Let . Then for some . By Theorem 5, the equation
is solvable if and only if .
Let . By Theorem 5, a general solution of (31) has the form , where is an arbitrary element of . In this case, the equation has the form
By Theorem 5, this equation has a solution if and only if
By and Theorem 3(ii), (iii), we have . Hence (33) is equivalent to . Then . By , we obtain . Thus equation (33) is solvable if and only if . Then equation (32) has the form . By Theorem 5, a general solution of this equation is , where is an arbitrary element of . Then the equation has the form . As above, this equation is solvable if and only if . Similarly, by induction, we obtain a solution of system (30) in the form (29). □
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This work has been supported by the commission of Scientific Research Projects of Karadeniz Technical University, Project number: 2010.111.3.1/Faculty of Science.
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Çavuş, A., Khadjiev, D. & Kunt, M. On periodic one-parameter groups of linear operators in a Banach space and applications. J Inequal Appl 2013, 172 (2013). https://doi.org/10.1186/1029-242X-2013-172
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DOI: https://doi.org/10.1186/1029-242X-2013-172