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On periodic one-parameter groups of linear operators in a Banach space and applications

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 [420]).

Let T be the one-dimensional torus { e i t :πt<π}. Further we consider T as the additive group Q/2πZ{t:πt<π} with its Euclidean topology, where Q is the field of real numbers. Let α(t) (tT) 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 α(t). The evident form of the resolvent operator R(μ,D) of D is known (see [7], Lemma 2.25)

R(μ,D)= ( 1 e μ t ) 1 0 2 π e μ t α(t)xdt,

where μ is an element of the resolvent set of D. For the element μ of the resolvent set of D and arbitrary element aH, the element R(μ,D)a is the evident form of the unique solution of the equation Dxμx=a.

In the present paper, we obtain conditions of the existence of a solution of the equation Dxμx=a 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 P(D)x=a 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 H(D) of the definition of D. For D and any λC, we introduce the operator R λ :HH 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 R λ is bounded and has properties (3) and (4) below.

In Section 4 we prove that H(D)= R λ (H) for all λC, the spectrum σ(D) of the operator D is a point spectrum and σ(D)={imSpec(H)}, where Spec(H) is the spectrum of the linear representation α. It is proved that R λ 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 A:HH of a complex Banach space H by GL(H). The following Definitions 1-4 are known [21].

Definition 1 A homomorphism α:TGL(H) is called a linear representation of T on a Banach space H.

Definition 2 Linear representations α:TGL(H) and β:TGL(V) are called equivalent if there exists a bounded invertible linear operator B:HV such that Bα(t)=β(t)B for all tT.

Definition 3 A linear representation α of T on a Banach space H is called isometric if α(t)x=x for all tT and xH.

Definition 4 A linear representation α of T on a Banach space H is called strongly continuous if lim t 0 α(t)x=x for all xH.

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 nZ. Put

H n := { x H : α ( t ) x = e i n t x t T } .

Let xH and nZ. By the theorem in ([22], p.314), Riemann’s integral

F n (x):= 1 2 π π π e i n t α(t)xdt

exists and F n (x) H n .

Proposition 1 Let α be a strongly continuous isometric linear representation of T on a Banach space H. Then

  1. 1.
    α(t) F n (x)= F n (α(t)x)= e i n t F n (x)

    for all nZ, xH and tT;

  2. 2.
    F n F m =0

    and F n 2 = F n for all m,nZ, mn;

  3. 3.
    F n (x)x

    for all nZ and xH.

Proof It is easy, so it is omitted. □

A series in the form k = x k , x k H k , is called the Fourier series of an element xH, if x k = F k (x) for all kZ. It is written in the form

x k = x k orx k = k = F k (x).

For xH and for an integer number n0, let us put

H f

is a subspace of H.

In the present paper, we assume that Spec(H) is infinite. The case of the finite Spec(H) 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 lim n ψ n (x)=x for every xH and H f ¯ =H.

Proof

In a standard manner, we obtain the equality

ψ n (x)= 1 2 π π π K n (t)α(t)xdt.

Since lim t 0 α(t)x=x, for every ε>0, there exists δ>0 such that 0<δ < π 2 and α(t)xx< ε 2 for all t(δ,δ). We have

K n (t)= 1 n + 1 ( sin ( n + 1 ) 2 t sin t 2 ) 2 1 ( n + 1 ) sin 2 δ 2

for all t[δ,π]. Since α is a strongly continuous isometric linear representation, K n (t)= K n (t) for all t[π,π], 1 2 π π π K n (t)dt=1 and sin δ 2 sin t 2 for all t[δ,π), it follows that

ψ n ( x ) x = 1 2 π π π K n ( t ) α ( t ) x d t 1 2 π π π K n ( t ) x d t 1 2 π π δ K n ( t ) ( α ( t ) x x ) d t + 1 2 π δ δ K n ( t ) ( α ( t ) x x ) d t + 1 2 π δ π K n ( t ) ( α ( t ) x x ) d t 1 π δ π K n ( t ) α ( t ) x x d t + 1 2 π δ δ K n ( t ) α ( t ) x x d t 1 π ( n + 1 ) sin 2 δ 2 δ π α ( t ) x x d t + 1 2 π δ δ K n ( t ) ε d t 2 x ( n + 1 ) sin 2 δ 2 + ε 2 π .

Hence lim n ψ n (x)=x. This, in view of ψ n (x) H f for all n, implies that H f ¯ =H. □

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 x,yH. If F n (x)= F n (y) for each nZ, then x=y.

Proof Since F n (x)= F n (y) for each nZ, it follows that ψ n (x)= ψ n (y) for each nZ. Hence Theorem 1 gives x=y. □

Theorem 2 Let α be a strongly continuous isometric linear representation of T on a Banach space H. Then lim n F n (x)=0 for each xH.

Proof Let ε>0 be given. Since lim t 0 α(t)x=x, there exists a natural number N(ε) such that α( π n )xx<ε for all natural numbers nN(ε). Since

F n (x)= 1 2 π π + π n π + π n e i n t α(t)xdt= 1 2 π π π e i n t α ( t + π n ) xdt,

we have

F n (x)= 1 4 π π π e i n t α ( t + π n ) ( α ( π n ) x x ) dt.

So, F n (x)α( π n )(x)x<ε for all nN(ε). □

Remark 2 This theorem is a generalization of Riemann-Lebesque’s lemma ([23], p.13).

3 The operator R λ

Let α be a strongly continuous isometric linear representation of T on a Banach space H.

Definition 5 ([7], p.45)

A point xH is called a differentiable point of α if there exists

Dx:= lim t 0 α ( t ) x x t in H.

Denote the set of all differentiable points of α by H(D). The set Spec(H) is called the spectrum of D. The set CSpec(H) 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

  1. (i)
    H(D)

    is a linear subspace of H, H f H(D) and H ( D ) ¯ =H;

  2. (ii)
    H(D)

    is α(T)-invariant and α(t)Dx=Dα(t)x for all tT, xH(D);

  3. (iii)
    D F n (x)= F n (Dx)=in F n (x)

    for all nZ and xH(D).

Proof (i) It is obvious that H(D) is a linear subspace of H. Let x H f . Then x can be expressed in the form = m m F (x) for some m. Since F (x) H , we get

lim t 0 α ( t ) x x t = lim t 0 = m m ( e i t 1 t ) F (x)= = m m i F (x).

Hence xH(D) and Dx= = m m i F (x). Therefore H f H(D). By Theorem 1 H f ¯ = H ( D ) ¯ =H.

  1. (ii)

    Let xH(D). Since α(t) is strongly continuous, we have

    α(t)Dx=α(t) lim s 0 α ( s ) x x s = lim s 0 α ( s ) α ( t ) x α ( t ) x s =Dα(t)x.

Hence α(t)xH(D) and α(t)Dx=Dα(t)x.

  1. (iii)

    Let nZ and xH(D). Using the continuity of F n and Proposition 1, we get

    D F n (x)= F n (Dx)= lim t 0 e i n t F n ( x ) F n ( x ) t =in F n (x).

 □

Remark 3 It is easily seen that H=H(D) if and only if Spec(H) is finite.

Definition 6 The operator D is called an infinitesimal generator of a linear representation α (see [7], p.45).

Proposition 3 Let xH(D). Then the function G x (t):=α(t)x is differentiable on T and G x (t)=α(t) G x (0)=α(t)Dx.

Proof Since α is strongly continuous, we have

G x (t)= lim s 0 α ( t + s ) x α ( t ) x s =α(t) lim s 0 α ( s ) x x s =α(t)Dx.

 □

Let α be a strongly continuous isometric linear representation of T on a Banach space H. For any xH and λC, there exists the following vector-valued Riemann’s integral:

0 2 π e λ t ( 0 t e λ s α ( s ) x d s ) dt.

We consider linear operators R λ on H defined by

R λ (x)= λ 1 e 2 π λ 0 2 π e λ t ( 0 t e λ s α ( s ) x d s ) dt+ 1 2 π ( 1 e 2 π λ ) 0 2 π α(t)xdt
(1)

for all λC such that λim, mZ and

R i m (x)= 1 + π 2 π 0 2 π e i m t α(t)xdt 1 2 π 0 2 π ( 0 t e i m s α ( s ) x d s ) dt
(2)

for all mZ. In Theorem 4(iv) below, we prove that R λ is equal to the resolvent operator of D for every point λ of the resolvent set of D. The operator R i m (x) is called the quasi-resolvent operator of D for the point im of the spectrum of D. The operator R λ was introduced in [9, 13].

Theorem 3 Let α be a strongly continuous isometric linear representation of T on a Banach space H. Then

  1. (i)

    the operator R λ , defined by (1) and (2), is bounded for all λC;

  2. (ii)
    R λ ( F n ( x ) ) = F n ( R λ ( x ) ) = 1 i n λ F n (x)
    (3)

for all λin, λC, nZ;

  1. (iii)
    R i n ( F n ( x ) ) = F n ( R i n ( x ) ) = F n (x)
    (4)

for all nZ and xH;

  1. (iv)
    R λ R μ (x)= R μ R λ (x)

    for all xH.

Proof (i) Let L:= 0 2 π e λ t ( 0 t e λ s ds)dt. Then it is obvious that

R λ ( x ) { ( λ L + 1 ) 1 e 2 π λ x if  λ C  and  λ i m , m Z , ( 1 + 2 π ) x if  λ = i m , m Z .

Therefore R λ is bounded in H for all λC.

  1. (ii)

    Let x H f , λC and λim for all mZ. Then x= = k k F (x) for some k. Since F (x) H , using Proposition 1, we get

    R λ ( x ) = λ 1 e 2 π λ 0 2 π e λ t ( 0 t e λ s ( = k k α ( s ) F ( x ) ) d s ) d t + 1 2 π ( 1 e 2 π λ ) 0 2 π ( = k k α ( t ) F ( x ) ) d t = λ 1 e 2 π λ 0 2 π e λ t ( = k k 0 t e ( i λ ) s F ( x ) d s ) d t + 1 2 π ( 1 e 2 π λ ) 0 2 π = k k e i t F ( x ) d t = λ 1 e 2 π λ = k k 0 2 π e i t i λ F ( x ) d t λ 1 e 2 π λ = k k 0 2 π e λ t i λ F ( x ) d t + 1 2 π ( 1 e 2 π λ ) = k k 0 2 π e i t F ( x ) d t = 1 e 2 π λ 1 F 0 ( x ) + = k k 1 i λ F ( x ) 1 e 2 π λ 1 F 0 ( x ) = = k k 1 i λ F ( x ) .

Hence, using F m F l =0 for lm (Proposition 1), we get F n ( R λ (x))= 1 i n λ F n (x)= R λ ( F n (x)).

Let x be an arbitrary element of H. By Theorem 1, lim p ψ p (x)=x. Since ψ p (x) H f , we have F n ( R λ ( ψ p (x)))= 1 i n λ F n ( ψ p (x)) for all nZ and pN. On the other hand, from lim p ψ p (x)=x, using the continuity of operators F n and R λ , we get F n ( R λ (x))= 1 i n λ F n (x)= R λ ( F n (x)).

Now we prove equality (3) for λ=im, mZ, nm. Let x H f and F m (x)=0. Then x= m , = k k F (x) for some kN and α(t)x= m , = k k e i t F (x). Hence

R i m ( x ) = 1 + π 2 π m , = k k 0 2 π e i ( m ) t F ( x ) d t 1 2 π 0 2 π ( m , = k k 0 t e i ( m ) s F ( x ) ) d s = m , = 1 k 1 i i m F ( x ) .

It implies that

R i m ( F n ( x ) ) = F n ( R i m ( x ) ) = m , = k k 1 i i m F n ( F ( x ) ) = { 1 i n i m F n ( x ) , n m , n k , 0 , n = m  or  n > k .
(5)

Let F m (x)0. Then x= F m (x)+ m , = k k F (x) for some kN. Since F m (x F m (x))=0, using equalities (5) and (2), we have

Hence

m , = k k 1 i i m F ( x ) + F m ( x ) = ( 1 + π ) F m ( x ) 1 2 π 0 2 π ( 0 t e i m s α ( s ) ( x ) d s ) d t = R i m ( x ) .

This equality implies that

R i m ( F n ( x ) ) = F n ( R i m ( x ) ) = 1 i n i m F n (x)
(6)

for all n,mZ, nm, and R i m ( F m (x))= F m ( R i m (x))= F m (x) for all mZ, x H f .

Let x be an arbitrary element of H. By Theorem 1, ψ p (x) H f and equality (6), we obtain

R i m ( F n ( x ) ) = F n ( R i m ( x ) ) = lim p F n ( R i m ( ψ p ( x ) ) ) = lim p 1 i n i m F n ( ψ p ( x ) ) = 1 i n i m F n ( x )

for all n,mZ, nm, xH.

  1. (iii)

    The proof of equality (4) is similar to the proof of equality (3).

  2. (iv)

    Using (3) and (4), we obtain F n R λ R μ (x)= F n R μ R λ (x) for xH, nZ, λC, μC. Hence, by Corollary 1, we have R λ R μ (x)= R μ R λ (x) for all xH, λC, μC. □

Corollary 2 Let aH, imSpec(H) and λC. Then F m R λ (a)=0 if and only if F m (a)=0.

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

  1. (i)
    R λ R 0 =λ( R λ R 0 ) 1 λ F 0

    for all λC, λim, mZ;

  2. (ii)
    R i m R 0 =im( R i m R 0 ) 1 i m F 0 1 i m F m

    for all mZ, m0.

Proof (i) Let xH, λC and λim, mZ. For n0, using Theorem 4 and F n F 0 =0, we obtain

F n ( λ R λ R 0 ( x ) 1 λ F 0 ( x ) ) = F n ( λ R λ R 0 ( x ) ) 1 λ F n ( F 0 ( x ) ) = λ i n ( i n λ ) F n ( x ) = 1 i n λ F n ( x ) 1 i n F n ( x ) = F n ( R λ ( x ) R 0 ( x ) ) .

Similarly,

F 0 ( R λ ( x ) R 0 ( x ) ) = 1 λ F 0 ( x ) F 0 ( R 0 ( x ) ) = 1 λ F 0 ( F 0 ( x ) ) F 0 ( R 0 ( x ) ) = 1 λ F 0 ( F 0 ( x ) ) + F 0 ( λ R λ R 0 ( x ) ) = F 0 ( λ R λ R 0 ( x ) 1 λ F 0 ( x ) ) .

Hence F n ( R λ (x) R 0 (x))= F n (λ R λ R 0 (x) 1 λ F 0 (x)) for every nZ. By Corollary 1 we have R λ (x) R 0 (x)=λ R λ R 0 (x) 1 λ F 0 (x) for all xH, λC and λim, mZ.

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

  1. (i)
    H(D)= R λ (H)

    for all λC;

  2. (ii)
    R i m (Dim)x=x

    and (Dim) R i m (y)=y for all imSpec(H) and xH(D), yH such that F m (x)= F m (y)=0;

  3. (iii)
    R λ (Dλ)x=x

    and (Dλ) R λ (y)=y for all λCSpec(H) and xH(D), yH;

  4. (iv)
    R λ (x)= ( 1 e 2 π λ ) 1 0 2 π e λ s α(s)xds

    for all λCSpec(H);

  5. (v)

    the spectrum σ(D) of D is a point spectrum and σ(D)=Spec(H).

Proof (i) We need the following two lemmas.

Lemma 1 D R 0 (x)=x F 0 (x) for all x H f .

Proof An element x H f has the form x= = k k F (x) for some k. Using Theorem 3 and Proposition 2, we obtain F n (D R 0 (x))= F n (x) for all n0 and F 0 (D R 0 (x))=0. Hence D R 0 (x)=x F 0 (x) for any x H f . Lemma is proved. □

Lemma 2 Let xH such that F 0 (x)=0. Then

α(t) R 0 (x)= 0 t α(s)xds+ R 0 (x).
(7)

Proof Let us define the functions f n ,f:[0,2π)H by f n (t):=α(t) R 0 ( ψ n (x)) and f(t):=α(t) R 0 (x). Since α is a strongly continuous isometric linear representation, we have

α ( t ) x α ( t ) ψ n ( x ) = x ψ n ( x )
(8)

and

f ( t ) f n ( t ) = R 0 ( x ) R 0 ( ψ n ( x ) ) .
(9)

Equality (8) implies that

0 t α ( s ) x d s 0 t α ( s ) ψ n ( x ) d s 2π x ψ n ( x ) .
(10)

Using F 0 (x)=0, Proposition 3, Theorem 3, R 0 ( ψ n (x)) H f , F 0 ( ψ n (x))= F 0 (x) and Lemma 1, we get

f n (t)=(α(t) R 0 ( ψ n ( x ) ) =α(t)D R 0 ( ψ n ( x ) ) =α(t) ( ψ n ( x ) F 0 ( x ) ) =α(t) ψ n (x).

Hence

f n (t)= 0 t α(s) ψ n (x)ds+C,

where C n H. Putting t=0, we obtain C n = f n (0)=α(0) R 0 ( ψ n (x))= R 0 ( ψ n (x)) and

f n (t)= 0 t α(s) ψ n (x)ds+ R 0 ( ψ n ( x ) ) .
(11)

Using equalities (8), (9), (11) and inequality (10), we obtain

f ( t ) ( 0 t α ( s ) x d s + R 0 ( x ) ) = f ( t ) f n ( t ) + ( f n ( t ) 0 t α ( s ) x d s R 0 ( x ) ) f ( t ) f n ( t ) + 0 t α ( s ) ( ψ n ( x ) x ) d s + R 0 ( ψ n ( x ) x ) 2 ( π + R 0 ) ψ n ( x ) x

for all tT. Since lim n ψ n (x)=x, it follows that

f(t)= 0 t α(s)xds+ R 0 (x)

and Lemma 2 is proved. □

We continue the proof of the theorem.

  1. (i)

    From equality (7) we obtain that the function f(t) is differentiable and f (t)=α(t)x. Using Proposition 3, we have

    α(t)x= f (t)=α(t) f (0)=α(t) lim s 0 α ( s ) R 0 ( x ) R 0 ( x ) s =α(t)D ( R 0 ( x ) ) .

Hence R 0 (x)H(D) for all xH such that F 0 (x)=0. Now let xH be an arbitrary element. Since F 0 (x F 0 (x))=0, we have R 0 (x F 0 (x))H(D). On the other hand, by Theorem 3 and Proposition 2, R 0 F 0 (x)= F 0 (x) H 0 H(D). Since x=(x F 0 (x))+ F 0 (x) and H(D) is a linear subspace of H, we get R 0 (x)H(D). Hence R 0 (H)H(D).

Conversely, let xH(D). By Proposition 2 and Theorem 3, Dx n = in F n (x) and R 0 (Dx) n 0 , n = F n (x). Hence R 0 (Dx)+ F 0 (x) n = F n (x). Using Corollary 1 and equality (4), we get R 0 (Dx)+ F 0 (x)=x, R 0 ( F 0 (x))= F 0 (x) and R 0 (Dx+ F 0 (x))=x. Let us put a:=Dx+ F 0 (x). Then R 0 (a)=x. This means that x R 0 (H), that is, H(D) R 0 (H); and consequently H(D)= R 0 (H).

Now we prove that R 0 (H)= R i m (H) for all mZ. By claim (ii) of Proposition 4 and claim (iv) of Theorem 3, R i m (x)= R 0 (x)+im R 0 R i m (x) 1 i m F 0 (x) 1 i m F m (x). Since F 0 (x) H 0 , F m (x) H m and H f H(D)= R 0 (H), we get R i m (H) R 0 (H). By Theorem 3, R i m ( H f )= H f R i m (H). Claim (ii) of Proposition 4 implies that R 0 (x)= R i m (x)im R i m R 0 (x)+ 1 i m F 0 (x)+ 1 i m F m (x). Hence R 0 (H) R i m (H) and H(D)= R i m (H). Similarly, using claims (i) and (ii) of Proposition 4, we obtain H(D)= R λ (H) for all λC, λim, mZ. Thus H(D)= R λ (H) for all λC.

  1. (ii)

    Let xH(D), F m (x)=0. By Proposition 2, Dx k = ik F k (x). Hence Dximx k = (ikim) F k (x). Using F m (x)=0 and Theorem 3, we get R i m (Dim)x k m , k = F k (x). By Corollary 1, R i m (Dim)x=x. Thus R i m (Dim)x=x for all xH(D) such that F m (x)=0.

Let xH, F m (x)=0. Since x k m , k = F k (x), we have R i m (x) k m , k = 1 i k i m F k (x). According to the statement (i) of this theorem, R i m (x)H(D). Using Proposition 2, we get (Dim) R i m (x) k m , k = F k (x), and hence (Dim) R i m (x)=x. Thus (Dim) R i m (x)=x for all xH such that F m (x)=0.

  1. (iii)

    The proof of claim (iii) is similar to the one of (ii).

  2. (iv)

    Equalities (iii) mean that R λ is the resolvent operator for all λCSpec(H). Hence equality (iv) follows from the form of the resolvent operator in ([7], Lemma 2.25).

  3. (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, R λ 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, aH and mZ. Then the equation Dximx=a has a solution if and only if F m (a)=0. In the case F m (a)=0, a general solution of the equation Dximx=a has the form x= R i m (a)+c, where c is an arbitrary element of H m .

Proof Suppose that the equation Dximx=a has a solution x. By Proposition 2 we have Dx k = ik F k (x) and Dximx k = (ikim) F k (x). Hence F m (a)= F m (Dximx)=0.

For the converse, we assume that F m (a)=0. By Theorem 4 we have (Dim) R i m (a)=a. Therefore x= R i m (a) is a solution of the equation Dximx=a. Let yH be a solution of the equation Dyimy=0. Then Dyimy k = (ikim) F k (y)= k m , k = (ikim) F k (y)=0. Hence F k (y)=0 for all km, that is, y= F m (y) H m . On the other hand, Dyimy=0 for all y H m . Thus a general solution of the equation Dximx=a has the form x= R i m (a)+c, where c is an arbitrary element of H m . □

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 aH(D). Then the Fourier series of the element a is convergent to a in H.

Proof

In a standard manner, we have the equality

s n (a)= 1 2 π π π D n (t)α(t)adt,where  D n (t):= sin ( n + 1 2 ) t sin 1 2 t .

Using 1 2 π π π D n (t)dt=1, we obtain s n (a)a= 1 2 π π π D n (t)(α(t)aa)dt. Let us consider the function

g(t):= 1 2 t g t 2 1 t .

Since lim t 0 g(t)=0, defining g(0)=0, we obtain that g(t) is a continuous function on [π,π]. Using the equality

D n (t)= sin n t t +g(t)sinnt+ 1 2 cosnt,

we get

s n ( a ) a = 1 2 π π π ψ ( t ) sin n t d t + 1 2 π π π g ( t ) sin n t ( α ( t ) a a ) d t + 1 4 π π π cos n t ( α ( t ) a a ) d t ,
(12)

where ψ(t):= α ( t ) a a t for t0 and ψ(0):=D(a). Functions ψ(t), α(t)aa and g(t)(α(t)aa) are continuous vector-valued functions on T. So, using Theorem 2 and equality (12), we obtain lim n s n =a. □

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 x k = x k . Then the series k = 1 i k λ x k is convergent to R λ (x) in Hfor all λCSpec(H) and the series x m + k m , k = 1 i k m x k is convergent to R i m (x) in H for all imSpec(H).

Proof Let λCSpec(H). According to Theorem 4, we have R λ (x)H(D). By Theorem 6 and R λ (x) k = 1 i k λ x k , the series k = 1 i k λ x k is convergent to R λ (x) 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 xH.

  1. (i)

    Let λCSpec(H). Then xH(D) if and only if there exists yH such that x= k = 1 i k λ F k (y);

  2. (ii)

    Let λSpec(H). Then xH(D) if and only if there exists yH such that x= y m + k m , k = 1 i k i m F k (y).

The proof follows from Theorem 4, Corollary 2 and Proposition 2. □

5 Periodic solutions of the linear differential equation P(D)x=aof the n th order with constant coefficients

Let α(t) be a strongly continuous linear representation of T in a Banach space H and D be the infinitesimal generator of α. For aH, we consider a solution of the linear differential equation

P(D)x=a
(13)

in H, where

P(D)= c 0 I+ c 1 D++ c n 1 D n 1 + D n ,
(14)
c i C

, i=0,,n1, I is the unit operator in H.

Theorem 7 Let P(D) be a linear differential operator (14), aH, let λ 1 ,, λ k be the roots of the polynomial P(z), and let m i be the multiplicity of λ i , m 1 ++ m k =n. Then

  1. (i)

    In the case λ 1 ,, λ n Spec(H), for any aH, there exists the unique solution of equation (13) in H, and it is x= R λ 1 m 1 R λ n m k (a).

  2. (ii)

    In the case λ 1 ,, λ r Spec(H) (r>0) and λ r + 1 ,, λ k Spec(H), a solution of equation (13) exists if and only if

    F i λ 1 (a)== F i λ r (a)=0.
    (15)

For aH, satisfying condition (15), a general solution of equation (13) has the form

x= R λ 1 m 1 R λ r m r R λ r + 1 m r + 1 R λ n m k + b 1 ++ b r ,
(16)

where b i is an arbitrary element of H i λ i , i=1,,r.

Proof (i) P(D) may be written in the form P(D)=(D λ 1 I)(D λ n I).

Then equation (13) has the form

(D λ 1 I)(D λ n I)x=a.
(17)

By Theorem 4(iii),

R λ (DλI)x=x
(18)

for all xH(D) and λSpec(H). Using equality (18) to equation (17), we obtain x= R λ 1 R λ n a.

  1. (ii)

    Let λ 1 ,, λ r Spec(H) (r>0) and λ r + 1 ,, λ n Spec(H). Using R λ r + 1 ,, R λ n to (17), we obtain

    (D λ 1 I)(D λ r I)x= R λ r + 1 R λ n a.
    (19)

This equation we can be written in the form

( D λ 1 I ) m 1 ( D λ k I ) m k x= R λ r + 1 R λ n a,
(20)

where λ i λ j for ij.

Lemma 3 Let aH, λSpec(H) and m>1. Then the equation

( D λ I ) m x=a
(21)

has a solution if and only if F i λ (a)=0. In the case F i λ (a)=0, a general solution of this equation is x= R λ m (a)+b, where b is an arbitrary element of H i λ .

Proof Put y= ( D λ I ) m 1 x. Then equation (21) has the form (DλI)y=a. By Theorem 5 this equation has a solution if and only if F i λ (a)=0. In the case F i λ (a)=0, a general solution is y= R λ (a)+ b 1 , where b 1 is an arbitrary element of H i λ . Let F i λ (a)=0. Then equation (21) reduces to the equation

( D λ I ) m 1 x= R λ (a)+ b 1 .
(22)

Put z= ( D λ I ) m 2 x. Then this equation has the form

(DλI)z= R λ (a)+ b 1 .
(23)

By Theorem 5, this equation is solvable if and only if F i λ ( R λ (a)+ b 1 )= F i λ ( R λ (a))+ F i λ ( b 1 )= F i λ (a)+ F i λ ( b 1 )= F i λ ( b 1 )=0. Therefore a general solution of equation (23) has the form z= R λ 2 (a)+ b 2 , where b 2 is an arbitrary element of H i λ . By induction, we obtain that a general solution of equation (21) has the form x= R λ m (a)+b, where b is an arbitrary element of H i λ . The lemma is proved. □

By this lemma, a solution of equation (20) reduces to a solution of the equation

( D λ 2 I ) m 2 ( D λ k I ) m k x= R λ 1 m 1 R λ r + 1 R λ n a+ b 1 ,
(24)

where b 1 is an arbitrary element of H i λ 1 . 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 ( a 1 ,, a n ) be a vector-line, where a i H, i=1,,n. Denote by the operator of transposition of a matrix. The ( a 1 , , a n ) denotes a vector-column of a i H, i=1,,n. Denote by H n the set of all vectors ( a 1 , , a n ) , where a i H, i=1,,n. For a= ( a 1 , , a n ) H n put Da= ( D a 1 , , D a n ) .

We consider the following system of linear differential equations:

Dx=Ax+a,
(25)

where a H n , x= ( x 1 , , x n ) and A is a complex n×n matrix.

Definition 7 Systems Dx=Ax+a and Dy=By+b, where A, B are complex n×n-matrices, a,b H n are called equivalent, if there exists a complex n×n-matrix K such that detK0, A= K 1 BK and a= K 1 b.

In this case y=Kx.

Proposition 5 Every system (25) is equivalent to the system of the form Dy=By+b, where the matrix B has the Jordan form.

Proof For A there exists a complex n×n-matrix K such that A= K 1 BK, where B is the Jordan normal form of A. From Dx=Ax+a, we obtain Dy=By+b, where y=Kx and b=Ka. □

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

( B 1 0 0 0 B 2 0 0 0 B m ),
(26)

where B j is a Jordan block of the n j th order, n 1 ++ n m =n. Then a solution of system (25), where B has the form (26), reduces to a solution of the following system of equations:

{ D u 1 = B 1 u 1 + b 1 ; D u 2 = B 2 u 2 + b 2 ; D u m = B m u m + b m ,
(27)

where B j is a Jordan block of the n j th order, b j H n j , u j H n j and x= ( u 1 , u 2 , , u m ) . Therefore a solution of system (27) reduces to a solution of the equation of the form

Dx=Gx+b,
(28)

where B is a Jordan block of the q th order with eigenvalue λ: g i i =λ, i=1,,q; g i + 1 i =1, i=1,,q1; g j i =0, ji<0 and ji>1.

Theorem 8 Let Dx=Gx+b be a system of the form (28), where G is a Jordan block of the qth order with eigenvalue λ and b= ( b 1 , , b q ) H q . Then

  1. (i)

    for the case λCSpec(H), the system has the unique solution x= ( x 1 , , x q ) H q , where:

  2. (ii)

    for the case λSpec(H), λ=ip, pZ, the system has a solution if and only if F p b q =0,

  3. (iii)

    for the case λSpec(H), λ=ip, pZ and F p b q =0, a general solution x= ( x 1 , , x q ) H q of the system has the form

    { x 1 = k = 1 q 1 R i p k ( b k F p b k ) + R i p q ( b q ) + c ; x q r = k = 1 r R i p k ( b q r 1 + k F p b q r 1 + k ) + R i p r + 1 ( b q F p b q r 1 ) , r = 1 , , q 2 ; x q = R i p b q F p b q 1 ,
    (29)

where c is an arbitrary element of H p .

Proof System (28) has the form

{ D x 1 λ x 1 = x 2 + b 1 ; D x 2 λ x 2 = x 3 + b 2 ; D x q 1 λ x q 1 = x q + b q 1 ; D x q λ x q = b q .
(30)

Let λCSpec(H). From (30), using the operator R λ , we obtain

x q = R λ b q , x q 1 = R λ b q 1 + R λ 2 b q ,, x 1 = k = 1 q R λ k b k .

Let λSpec(H). Then λ=ip for some pZ. By Theorem 5, the equation

D x q ip x q = b q ,
(31)

is solvable if and only if F p b q =0.

Let F p b q =0. By Theorem 5, a general solution of (31) has the form x q = R i p b q + c q , where c q is an arbitrary element of H i p . In this case, the equation D x q 1 ip x q 1 = x q + b q 1 has the form

D x q 1 ip x q 1 = R i p b q + c q + b q 1 .
(32)

By Theorem 5, this equation has a solution if and only if

F i p ( R i p b q + c q + b q 1 )=0.
(33)

By F i p ( b q )=0 and Theorem 3(ii), (iii), we have F i p R i p b q = R i p F i p b q =0. Hence (33) is equivalent to F i p ( c q + b q 1 )=0. Then F i p c q = F i p b q 1 . By c q H i p , we obtain c q = F i p c q = F i p b q 1 . Thus equation (33) is solvable if and only if c q = F i p b q 1 . Then equation (32) has the form D x q 1 ip x q 1 =( b q 1 F i p b q 1 )+ R i p b q . By Theorem 5, a general solution of this equation is x q 1 = R i p ( b q 1 F i p b q 1 )+ R i p 2 b q + c q 1 , where c q 1 is an arbitrary element of H i p . Then the equation D x q 2 ip x q 2 = x q 1 + b q 2 has the form D x q 2 ip x q 2 =( b q 2 + c q 1 )+ R i p ( b q 1 F i p b q 1 )+ R i p 2 b q . As above, this equation is solvable if and only if c q 1 = F i p b q 2 . Similarly, by induction, we obtain a solution of system (30) in the form (29). □

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Acknowledgements

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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Keywords

  • Fourier series
  • infinitesimal generator
  • resolvent operator
  • periodic solution
  • theorem on integral