R E S E A R C H
Open Access
On periodic one-parameter groups of linear
operators in a Banach space and applications
Abdullah Çavu¸s
*, Djavvat Khadjiev and Mehmet Kunt
*Correspondence: [email protected]
Karadeniz Technical University, Trabzon, Turkey
Abstract
LetDbe 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 ofD) is defined for points of the spectrum ofD and its evident form is given. The theorem on integral for the operatorD, theorems on the existence of periodic solutions of a linear differential equation of thenth 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 nth 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 ofD. MSC: 42A; 43; 47D
Keywords: Fourier series; infinitesimal generator; resolvent operator; periodic solution; theorem on integral
1 Introduction
One-parameter groups of linear operators and periodic one-parameter groups of linear operators in topological vector spaces were investigated by Stone [], Dunford [], Gelfand [] and others (see [–]).
LetTbe the one-dimensional torus{eit: –π≤t<π}. Further we considerTas the addi-tive groupQ/πZ {t: –π≤t<π}with its Euclidean topology, whereQis the field of real numbers. Letα(t) (t∈T) be a strongly continuous one-parameter group of bounded linear operators in a Banach spaceH, and letDbe an infinitesimal generator of the groupα(t). The evident form of the resolvent operatorR(μ,D) ofDis known (see [], Lemma .)
R(μ,D) = –e–μt–
π
e–μtα(t)x dt,
whereμis an element of the resolvent set ofD. For the elementμof the resolvent set of
Dand arbitrary elementa∈H, the elementR(μ,D)ais the evident form of the unique solution of the equationDx–μx=a.
In the present paper, we obtain conditions of the existence of a solution of the equation
Dx–μx=afor points of the spectrum ofD. We define an analog of the resolvent oper-ator (= quasi-resolvent operoper-ator) for points of the spectrum ofDand, in the case of the existence of a solution, we give the evident form of all solutions by using a quasi-resolvent
operator ofD. We apply resolvent and quasi-resolvent operators to a solution of a linear differential equationP(D)x=aof thenth order with constant coefficients and to a sys-tem of linear differential equations of the first order with constant coefficients in a Banach spaceH.
Contents of the present paper is the following. In Section we give generalizations of Fejer’s theorem and Riemann-Lebesque’s lemma for a strongly continuous linear repre-sentation ofTin a Banach space. These results are used in the next sections.
In Section we give a definition of the infinitesimal generatorDof a strongly continuous linear representation ofTin a Banach spaceHand the domainH(D) of the definition ofD. ForDand anyλ∈C, we introduce the operatorRλ:H→Hby formula () below for the pointλof the resolvent set ofDand by () for the point of the spectrum ofD. We show that the linear operatorRλis bounded and has properties () and () below.
In Section we prove thatH(D) =Rλ(H) for allλ∈C, the spectrumσ(D) of the operator
Dis a point spectrum andσ(D) ={im∈Spec(H)}, whereSpec(H) is the spectrum of the linear representationα. It is proved thatRλis equal to the resolvent operator ofDfor all pointsλof the resolvent set ofD. We obtain the theorem on an integral forD.
In Section we give conditions of the existence of a periodic solution of a linear differ-ential equation of thenth order with constant coefficients. In the case of existence, the evident form of all periodic solutions is given.
In Section 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 ofTin a Banach space
Denote the group of all invertible bounded linear operatorsA:H→Hof a complex Ba-nach spaceHbyGL(H). The following Definitions - are known [].
Definition A homomorphismα:T→GL(H) is called a linear representation ofT on
a Banach spaceH.
Definition Linear representationsα:T→GL(H) andβ:T→GL(V) are called
equiv-alent if there exists a bounded invertible linear operatorB:H→Vsuch thatBα(t) =β(t)B
for allt∈T.
Definition A linear representationα ofT on a Banach spaceHis called isometric if
α(t)x=xfor allt∈T andx∈H.
Definition A linear representationαofT on a Banach spaceHis called strongly
con-tinuous iflimt→α(t)x=xfor allx∈H.
Letαbe a strongly continuous linear representation ofTon a Banach spaceH, and let
Zbe the ring of all integers andn∈Z. Put
Hn:=
x∈H:α(t)x=eintx∀t∈T.
Letx∈Handn∈Z. By the theorem in ([], p.), Riemann’s integral
Fn(x) := π
π
–π
e–intα(t)x dt
exists andFn(x)∈Hn.
Proposition Letα be a strongly continuous isometric linear representation of T on a
Banach space H.Then
. α(t)Fn(x) =Fn(α(t)x) =eintFn(x)for alln∈Z,x∈Handt∈T; . Fn·Fm= andFn=Fnfor allm,n∈Z,m=n;
. Fn(x) ≤ xfor alln∈Zandx∈H.
Proof It is easy, so it is omitted.
A series in the form∞k=–∞xk,xk∈Hk, is called the Fourier series of an elementx∈H, ifxk=Fk(x) for allk∈Z. It is written in the form
x∼ ∞
k=–∞
xk or x∼ k=∞
k=–∞ Fk(x).
Forx∈Hand for an integer numbern≥, let us put
sn(x) := k=–n
k=n
Fk(x), ψn(x) :=s(x) +s(x) +· · ·+sn(x)
n+ ,
Kn(t) =
n+
sin(n+)t
sint
, Spec(x) :=in:n∈Z,Fn(x)=
forx∈H,
Spec(H) := x=θ,x∈H
Spec(x), Hf:=
x∈H:Spec(x) is finite.
Hf is a subspace ofH.
In the present paper, we assume thatSpec(H) is infinite. The case of the finiteSpec(H) is investigated easy and it is omitted.
Theorem Letαbe a strongly continuous isometric linear representation of T on a Banach
space H.Thenlimn→∞ψn(x) =x for every x∈H and Hf =H.
Proof In a standard manner, we obtain the equality
ψn(x) = π
π
–π
Sincelimt→α(t)x=x, for everyε> , there existsδ> such that <δ<π andα(t)x–
x<εfor allt∈(–δ,δ). We have
Kn(t) =
n+
sin(n+)t
sint
≤
(n+ )sinδ
for allt∈[δ,π]. Sinceαis a strongly continuous isometric linear representation,Kn(t) =
Kn(–t) for allt∈[–π,π], π
π
–πKn(t)dt= andsin δ
≤sin
t
for allt∈[δ,π), it follows
that
ψn(x) –x= π π
–π
Kn(t)α(t)x dt– π
π
–π
Kn(t)x dt
≤ π
–δ
–π
Kn(t)α(t)x–xdt+ π
δ
–δ
Kn(t)α(t)x–xdt
+ π
π
δ
Kn(t)α(t)x–xdt
≤
π
π
δ
Kn(t)α(t)x–xdt+ π
δ
–δ
Kn(t)α(t)x–xdt
≤
π(n+ )sinδ π
δ
α(t)x–xdt+ π
δ
–δ
Kn(t)εdt
≤ x (n+ )sinδ +
ε
π.
Hencelimn→∞ψn(x) =x. This, in view ofψn(x)∈Hf for alln, implies thatHf=H.
Remark Theorem is known for the homogeneous Banach spaces onT([], p.; [],
pp.-). For a strongly continuous linear representation ofT in a locally convex space, it is obtained in [].
Corollary Letαbe a strongly continuous isometric linear representation of T on a
Ba-nach space H and x,y∈H.If Fn(x) =Fn(y)for each n∈Z,then x=y.
Proof SinceFn(x) =Fn(y) for eachn∈Z, it follows thatψn(x) =ψn(y) for eachn∈Z. Hence
Theorem givesx=y.
Theorem Letαbe a strongly continuous isometric linear representation of T on a
Ba-nach space H.Thenlimn→∞Fn(x) = for each x∈H.
Proof Letε> be given. Sincelimt→α(t)x=x, there exists a natural numberN(ε) such
thatα(–π
n)x–x<εfor all natural numbersn≥N(ε). Since
Fn(x) = π
π+πn –π+πn
e–intα(t)x dt= – π
π
–π
e–intα t+π
n
x dt,
we have
Fn(x) = π
π
–π
e–intα t+π
n α –
π n
x–x
dt.
Remark This theorem is a generalization of Riemann-Lebesque’s lemma ([], p.).
3 The operatorRλ
Letαbe a strongly continuous isometric linear representation ofTon a Banach spaceH.
Definition ([], p.) A pointx∈His called a differentiable point ofαif there exists
Dx:=lim t→
α(t)x–x t inH.
Denote the set of all differentiable points ofαbyH(D). The setSpec(H) is called the spectrum ofD. The setC\Spec(H) is called the resolvent set ofD.
Proposition Letα be a strongly continuous isometric linear representation of T on a
Banach space H.Then
(i) H(D)is a linear subspace ofH,Hf⊂H(D)andH(D) =H;
(ii) H(D)isα(T)-invariant andα(t)Dx=Dα(t)xfor allt∈T,x∈H(D); (iii) DFn(x) =Fn(Dx) =inFn(x)for alln∈Zandx∈H(D).
Proof (i) It is obvious thatH(D) is a linear subspace ofH. Letx∈Hf. Thenxcan be ex-pressed in the formm=–mF(x) for somem. SinceF(x)∈H, we get
lim t→
α(t)x–x t =limt→
m
=–m
eit–
t
F(x) = m
=–m
iF(x).
Hence x∈H(D) andDx=m=–miF(x). Therefore Hf ⊂H(D). By Theorem Hf =
H(D) =H.
(ii) Letx∈H(D). Sinceα(t) is strongly continuous, we have
α(t)Dx=α(t)lim s→
α(s)x–x s =lims→
α(s)α(t)x–α(t)x
s =Dα(t)x.
Henceα(t)x∈H(D) andα(t)Dx=Dα(t)x.
(iii) Letn∈Zandx∈H(D). Using the continuity ofFnand Proposition , we get
DFn(x) =Fn(Dx) =lim t→
eintF
n(x) –Fn(x)
t =inFn(x).
Remark It is easily seen thatH=H(D) if and only ifSpec(H) is finite.
Definition The operatorDis called an infinitesimal generator of a linear representation
α(see [], p.).
Proposition Let x∈H(D).Then the function Gx(t) :=α(t)x is differentiable on T and
Gx(t) =α(t)Gx() =α(t)Dx.
Proof Sinceαis strongly continuous, we have
Gx(t) =lim s→
α(t+s)x–α(t)x
s =α(t)lims→
α(s)x–x
Letαbe a strongly continuous isometric linear representation ofTon a Banach spaceH. For anyx∈Handλ∈C, there exists the following vector-valued Riemann’s integral:
π
eλt
t
e–λsα(s)x ds
dt.
We consider linear operatorsRλonHdefined by
Rλ(x) = λ –eπ λ
π
eλt
t
e–λsα(s)x ds
dt+ π( –eπ λ)
π
α(t)x dt ()
for allλ∈Csuch thatλ=im,m∈Zand
Rim(x) = +π π
π
e–imtα(t)x dt– π
π
t
e–imsα(s)x ds
dt ()
for allm∈Z. In Theorem (iv) below, we prove thatRλis equal to the resolvent oper-ator ofDfor every pointλof the resolvent set ofD. The operatorRim(x) is called the quasi-resolvent operator ofDfor the pointimof the spectrum ofD. The operatorRλwas introduced in [, ].
Theorem Letαbe a strongly continuous isometric linear representation of T on a
Ba-nach space H.Then
(i) the operatorRλ,defined by()and(),is bounded for allλ∈C; (ii)
RλFn(x)
=Fn
Rλ(x)=
in–λFn(x) ()
for allλ=in,λ∈C,n∈Z; (iii)
Rin
Fn(x)=Fn
Rin(x)=Fn(x) ()
for alln∈Zandx∈H;
(iv) RλRμ(x) =RμRλ(x)for allx∈H.
Proof (i) LetL:=πeλt(t
e
–λsds)dt. Then it is obvious that Rλ(x)≤
⎧ ⎨ ⎩
(λL+)
–eπ λx ifλ∈Candλ=im,m∈Z, ( + π)x ifλ=im,m∈Z.
ThereforeRλis bounded inHfor allλ∈C.
(ii) Letx∈Hf,λ∈Candλ=imfor allm∈Z. Thenx=k=–kF(x) for somek. Since
F(x)∈H, using Proposition , we get
Rλ(x) = λ –eπ λ
π
eλt t
e–λs k
=–k
α(s)F(x)
ds
dt
+
π( –eπ λ)
π
k
=–k
α(t)F(x)
= λ –eπ λ
π
eλt
k
=–k t
e(i–λ)sF(x)ds
dt
+
π( –eπ λ)
π
k
=–k
eitF(x)dt
= λ
–eπ λ
k
=–k π
eit
i–λF(x)dt– λ
–eπ λ
k
=–k π
eλt
i–λF(x)dt
+
π( –eπ λ)
k
=–k π
eitF(x)dt
=
eπ λ– F(x) +
k
=–k
i–λF(x) –
eπ λ– F(x) =
k
=–k
i–λF(x).
Hence, usingFm◦Fl= forl=m(Proposition ), we getFn(Rλ(x)) =in–λFn(x) =Rλ(Fn(x)).
Let xbe an arbitrary element of H. By Theorem ,limp→∞ψp(x) =x. Since ψp(x)∈
Hf, we haveFn(Rλ(ψp(x))) = in–λFn(ψp(x)) for all n∈Z andp∈N. On the other hand,
fromlimp→∞ψp(x) =x, using the continuity of operatorsFnandRλ, we getFn(Rλ(x)) =
in–λFn(x) =Rλ(Fn(x)).
Now we prove equality () forλ=im,m∈Z,n=m. Letx∈Hf andFm(x) = . Then
x=k=m,=–kF(x) for somek∈Nandα(t)x= k
=m,=–keitF(x). Hence
Rim(x) = +π π
k
=m,=–k π
ei(–m)tF(x)dt– π
π
k
=m,=–k t
ei(–m)sF(x)
ds
= k
=m,=
i–imF(x).
It implies that
Rim
Fn(x)=Fn
Rim(x)= k
=m,=–k
i–imFn
F(x)
= ⎧ ⎨ ⎩
in–imFn(x), n=m,n ≤k,
, n=morn>k. ()
LetFm(x)= . Thenx=Fm(x) +k=m,=–kF(x) for somek∈N. SinceFm(x–Fm(x)) = , using equalities () and (), we have
k
=m,=–k
i–imF
x–Fm(x)=Rim
x–Fm(x),
k
=m,=–k
i–imF(x)
= +π π
π
e–imtα(t)x–Fm(t)
dt– π
π
t
e–imsα(s)x–Fm(x)
ds
= +π π
π
e–imtα(t)x dt– +π π
π
e–imtα(t)Fm(x)dt
+ π π t
e–imsα(s)Fm(x)ds
dt– π
π
t
e–imsα(s)x ds
dt
= ( +π)Fm(x) – ( +π)Fm(x) + π
π
tFm(x)dt– π
π
t
e–imsα(s)x ds
dt
=πFm(x) – π
π
t
e–imsα(s)x ds
dt.
Hence
k
=m,=–k
i–imF(x) +Fm(x) = ( +π)Fm(x) –
π π t
e–imsα(s)(x)ds
dt
=Rim(x).
This equality implies that
Rim
Fn(x)=Fn
Rim(x)=
in–imFn(x) ()
for alln,m∈Z,n=m, andRim(Fm(x)) =Fm(Rim(x)) =Fm(x) for allm∈Z,x∈Hf.
Letxbe an arbitrary element ofH. By Theorem ,ψp(x)∈Hfand equality (), we obtain
Rim
Fn(x)
=Fn
Rim(x)
= lim p→∞Fn
Rim
ψp(x)
= lim p→∞
in–imFn
ψp(x)
=
in–imFn(x)
for alln,m∈Z,n=m,x∈H.
(iii) The proof of equality () is similar to the proof of equality ().
(iv) Using () and (), we obtainFnRλRμ(x) =FnRμRλ(x) forx∈H,n∈Z,λ∈C,μ∈C. Hence, by Corollary , we haveRλRμ(x) =RμRλ(x) for allx∈H,λ∈C,μ∈C.
Corollary Let a∈H,im∈Spec(H)andλ∈C.Then FmRλ(a) = if and only if Fm(a) = .
Proof It follows easily from Theorem .
Proposition Letα be a strongly continuous isometric linear representation of T on a
Banach space H.Then
(i) Rλ–R=λ(Rλ◦R) –λFfor allλ∈C,λ=im,m∈Z;
(ii) Rim–R=im(Rim◦R) –im F–im Fmfor allm∈Z,m= .
Proof (i) Letx∈H,λ∈Candλ=im,m∈Z. Forn= , using Theorem andFnF= ,
we obtain
Fn λRλ◦R(x) –
λF(x)
=Fn
λRλ◦R(x)
–
λFn
F(x)
= λ
in(in–λ)Fn(x)
=
in–λFn(x) –
inFn(x) =Fn
Rλ(x) –R(x)
Similarly,
F
Rλ(x) –R(x)
= –
λF(x) –F
R(x)
= –
λF
F(x)
–F
R(x)
= –
λF
F(x)
+F
λRλ◦R(x)
=F λRλ◦R(x) –
λF(x)
.
HenceFn(Rλ(x) –R(x)) =Fn(λRλ◦R(x) –λF(x)) for everyn∈Z. By Corollary we have Rλ(x) –R(x) =λRλ◦R(x) –λF(x) for allx∈H,λ∈Candλ=im,m∈Z.
A proof of (ii) is similar.
4 The theorem on resolvent and quasi-resolvent operators
Theorem Letαbe a strongly continuous isometric linear representation of T on a
Ba-nach space H.Then
(i) H(D) =Rλ(H)for allλ∈C;
(ii) Rim(D–im)x=xand(D–im)Rim(y) =yfor allim∈Spec(H)andx∈H(D),y∈H
such thatFm(x) =Fm(y) = ;
(iii) Rλ(D–λ)x=xand(D–λ)Rλ(y) =yfor allλ∈C\Spec(H)andx∈H(D),y∈H; (iv) Rλ(x) = ( –e–π λ)–π
e–
λsα(s)x dsfor allλ∈C\Spec(H);
(v) the spectrumσ(D)ofDis a point spectrum andσ(D) =Spec(H).
Proof (i) We need the following two lemmas.
Lemma DR(x) =x–F(x)for all x∈Hf.
Proof An elementx∈Hf has the form x= k
=–kF(x) for somek. Using Theorem and Proposition , we obtainFn(DR(x)) =Fn(x) for alln= andF(DR(x)) = . Hence
DR(x) =x–F(x) for anyx∈Hf. Lemma is proved.
Lemma Let x∈H such that F(x) = .Then
α(t)R(x) =
t
α(s)x ds+R(x). ()
Proof Let us define the functionsfn,f : [, π)→H byfn(t) :=α(t)R(ψn(x)) andf(t) :=
α(t)R(x). Sinceαis a strongly continuous isometric linear representation, we have
α(t)x–α(t)ψn(x)=x–ψn(x) ()
and
f(t) –fn(t)=R(x) –R
ψn(x). ()
Equality () implies that
t
α(s)x ds– t
α(s)ψn(x)ds
Using F(x) = , Proposition , Theorem , R(ψn(x))∈ Hf, F(ψn(x)) = F(x) and
Lemma , we get
fn(t) = (α(t)R
ψn(x)=α(t)DR
ψn(x)=α(t)ψn(x) –F(x)
=α(t)ψn(x).
Hence
fn(t) = t
α(s)ψn(x)ds+C,
whereCn∈H. Puttingt= , we obtainCn=fn() =α()R(ψn(x)) =R(ψn(x)) and
fn(t) = t
α(s)ψn(x)ds+R
ψn(x)
. ()
Using equalities (), (), () and inequality (), we obtain
f(t) – t
α(s)x ds+R(x)
=f(t) –fn(t) + fn(t) – t
α(s)x ds–R(x)
≤f(t) –fn(t)+ t
α(s)ψn(x) –xds
+R
ψn(x) –x≤π+Rψn(x) –x
for allt∈T. Sincelimn→∞ψn(x) =x, it follows that
f(t) = t
α(s)x ds+R(x)
and Lemma is proved.
We continue the proof of the theorem.
(i) From equality () we obtain that the functionf(t) is differentiable andf(t) =α(t)x. Using Proposition , we have
α(t)x=f(t) =α(t)f() =α(t)lim s→
α(s)R(x) –R(x)
s =α(t)D
R(x)
.
HenceR(x)∈H(D) for allx∈Hsuch thatF(x) = . Now letx∈Hbe an arbitrary
ele-ment. SinceF(x–F(x)) = , we haveR(x–F(x))∈H(D). On the other hand, by
The-orem and Proposition ,RF(x) =F(x)∈H⊂H(D). Sincex= (x–F(x)) +F(x) and
H(D) is a linear subspace ofH, we getR(x)∈H(D). HenceR(H)⊂H(D).
Conversely, let x∈H(D). By Proposition and Theorem ,Dx∼∞n=–∞inFn(x) and
R(Dx)∼
∞
n=,n=–∞Fn(x). HenceR(Dx) +F(x)∼
∞
n=–∞Fn(x). Using Corollary and
equality (), we getR(Dx) +F(x) =x,R(F(x)) =F(x) andR(Dx+F(x)) =x. Let us put
a:=Dx+F(x). ThenR(a) =x. This means thatx∈R(H), that is,H(D)⊂R(H); and
consequentlyH(D) =R(H).
Now we prove that R(H) =Rim(H) for allm∈Z. By claim (ii) of Proposition and
claim (iv) of Theorem ,Rim(x) =R(x) +imRRim(x) –imF(x) –im Fm(x). SinceF(x)∈H,
Hf ⊂Rim(H). Claim (ii) of Proposition implies that R(x) =Rim(x) –imRimR(x) +
imF(x) +
imFm(x). Hence R(H)⊂Rim(H) and H(D) =Rim(H). Similarly, using claims (i) and (ii) of Proposition , we obtainH(D) =Rλ(H) for allλ∈C,λ=im,m∈Z. Thus
H(D) =Rλ(H) for allλ∈C.
(ii) Let x∈ H(D), Fm(x) = . By Proposition , Dx∼ ∞k=–∞ikFk(x). Hence Dx–
imx∼∞k=–∞(ik–im)Fk(x). Using Fm(x) = and Theorem , we getRim(D–im)x∼ ∞
k=m,k=–∞Fk(x). By Corollary ,Rim(D–im)x=x. ThusRim(D–im)x=xfor allx∈H(D)
such thatFm(x) = .
Letx∈H,Fm(x) = . Sincex∼ ∞
k=m,k=–∞Fk(x), we haveRim(x)∼ ∞
k= m,k=–∞ik–imFk(x). According to the statement (i) of this theorem,Rim(x)∈H(D). Using Proposition , we get (D–im)Rim(x)∼
∞
k=m,k=–∞Fk(x), and hence (D–im)Rim(x) =x. Thus (D–im)Rim(x) =x for allx∈Hsuch thatFm(x) = .
(iii) The proof of claim (iii) is similar to the one of (ii).
(iv) Equalities (iii) mean thatRλis the resolvent operator for allλ∈C\Spec(H). Hence equality (iv) follows from the form of the resolvent operator in ([], Lemma .).
(v) follows from (ii) and (iii). The proof of the theorem is completed.
Remark Equality (iv) means that for pointsλof the resolvent set ofD,Rλis the other
form of the resolvent operator ofD.
Theorem Letαbe a strongly continuous isometric linear representation of T on a
Ba-nach space H,a∈H and m∈Z.Then the equation Dx–imx=a has a solution if and only if Fm(a) = .In the case Fm(a) = ,a general solution of the equation Dx–imx=a has the
form x=Rim(a) +c,where c is an arbitrary element of Hm.
Proof Suppose that the equationDx–imx=ahas a solution x. By Proposition we haveDx∼∞k=–∞ikFk(x) andDx–imx∼∞k=–∞(ik–im)Fk(x). HenceFm(a) =Fm(Dx–
imx) = .
For the converse, we assume thatFm(a) = . By Theorem we have (D–im)Rim(a) =a. Thereforex=Rim(a) is a solution of the equationDx–imx=a. Lety∈Hbe a solution of the equationDy–imy= . ThenDy–imy∼∞k=–∞(ik–im)Fk(y) =∞k= m,k=–∞(ik–
im)Fk(y) = . HenceFk(y) = for allk= m, that is,y=Fm(y)∈Hm. On the other hand,
Dy–imy= for ally∈Hm. Thus a general solution of the equationDx–imx=ahas the formx=Rim(a) +c, wherecis an arbitrary element ofHm.
Remark This theorem is the theorem on integral for periodic one-parameter groups of
operators.
The following theorem is known (see [], Theorem .)
Theorem Letαbe a strongly continuous isometric linear representation of T on a
Ba-nach space H and a∈H(D).Then the Fourier series of the element a is convergent to a in H.
Proof In a standard manner, we have the equality
sn(a) = π
π
–π
Dn(t)α(t)a dt, whereDn(t) :=sin(n+
)t
Using π–ππDn(t)dt= , we obtainsn(a) –a=π–ππDn(t)(α(t)a–a)dt. Let us consider the function
g(t) := tgt –
t.
Sincelimt→g(t) = , definingg() = , we obtain thatg(t) is a continuous function on
[–π,π]. Using the equality
Dn(t) = sinnt
t +g(t)sinnt+
cosnt,
we get
sn(a) –a= π
π
–π
ψ(t)sinnt dt+ π
π
–π
g(t)sinntα(t)a–adt
+ π
π
–π
cosntα(t)a–adt, ()
whereψ(t) :=α(t)ta–afort= andψ() :=D(a). Functionsψ(t),α(t)a–aandg(t)(α(t)a–a) are continuous vector-valued functions onT. So, using Theorem and equality (), we
obtainlimn→∞sn=a.
Remark Our proof of this theorem differs from that in ([], Theorem .).
Corollary Letα be a strongly continuous isometric linear representation of T on a
Banach space H, x be an arbitrary element of H and x∼∞k=–∞xk. Then the series
∞
k=–∞ik–λxk is convergent to Rλ(x) in Hfor all λ∈C \Spec(H) and the series xm +
∞
k=m,k=–∞ik–mxkis convergent to Rim(x)in H for all im∈Spec(H).
Proof Letλ∈C\Spec(H). According to Theorem , we haveRλ(x)∈H(D). By Theorem andRλ(x)∼k∞=–∞ik–λxk, the series∞k=–∞ik–λxkis convergent toRλ(x) inH. The proof
of the second statement is similar.
Corollary Letαbe a strongly continuous isometric linear representation of T in a
Ba-nach space H and x∈H.
(i) Letλ∈C\Spec(H).Thenx∈H(D)if and only if there existsy∈Hsuch that x=∞k=–∞ik–λFk(y);
(ii) Letλ∈Spec(H).Thenx∈H(D)if and only if there existsy∈Hsuch that x=ym+
∞
k=m,k=–∞ik–imFk(y).
The proof follows from Theorem , Corollary and Proposition .
5 Periodic solutions of the linear differential equationP(D)x=aof thenth order with constant coefficients
Letα(t) be a strongly continuous linear representation ofTin a Banach spaceHandDbe the infinitesimal generator ofα. Fora∈H, we consider a solution of the linear differential equation
inH, where
P(D) =cI+cD+· · ·+cn–Dn–+Dn, ()
ci∈C,i= , . . . ,n– ,Iis the unit operator inH.
Theorem Let P(D)be a linear differential operator(),a∈H,letλ, . . . ,λkbe the roots
of the polynomial P(z),and let mibe the multiplicity ofλi,m+· · ·+mk=n.Then
(i) In the caseλ, . . . ,λn∈/Spec(H),for anya∈H,there exists the unique solution of
equation()in H,and it isx=Rm
λ · · ·R mk
λn(a).
(ii) In the caseλ, . . . ,λr∈Spec(H)(r> )andλr+, . . . ,λk∈/Spec(H),a solution of
equation()exists if and only if
Fiλ(a) =· · ·=Fiλr(a) = . ()
Fora∈H,satisfying condition(),a general solution of equation()has the form
x=Rm
λ · · ·R mr
λrR mr+
λr+ · · ·R mk
λn +b+· · ·+br, ()
wherebiis an arbitrary element ofHiλi,i= , . . . ,r.
Proof (i)P(D) may be written in the formP(D) = (D–λI)· · ·(D–λnI). Then equation () has the form
(D–λI)· · ·(D–λnI)x=a. ()
By Theorem (iii),
Rλ(D–λI)x=x ()
for all x∈H(D) and λ∈/ Spec(H). Using equality () to equation (), we obtain x=
Rλ· · ·Rλna.
(ii) Letλ, . . . ,λr∈Spec(H) (r> ) andλr+, . . . ,λn∈/Spec(H). UsingRλr+, . . . ,Rλnto (), we obtain
(D–λI)· · ·(D–λrI)x=Rλr+· · ·Rλna. ()
This equation we can be written in the form
(D–λI)m· · ·(D–λkI)mkx=Rλr+· · ·Rλna, ()
whereλi=λjfori=j.
Lemma Let a∈H,λ∈Spec(H)and m> .Then the equation
has a solution if and only if Fiλ(a) = . In the case Fiλ(a) = , a general solution of this equation is x=Rm
λ(a) +b,where b is an arbitrary element of Hiλ.
Proof Puty= (D–λI)m–x. Then equation () has the form (D–λI)y=a. By Theorem
this equation has a solution if and only ifFiλ(a) = . In the caseFiλ(a) = , a general solution
isy=Rλ(a) +b, wherebis an arbitrary element ofHiλ. LetFiλ(a) = . Then equation ()
reduces to the equation
(D–λI)m–x=Rλ(a) +b. ()
Putz= (D–λI)m–x. Then this equation has the form
(D–λI)z=Rλ(a) +b. ()
By Theorem , this equation is solvable if and only ifFiλ(Rλ(a) +b) =Fiλ(Rλ(a)) +Fiλ(b) =
Fiλ(a) +Fiλ(b) =Fiλ(b) = . Therefore a general solution of equation () has the form
z=R
λ(a) +b, whereb is an arbitrary element of Hiλ. By induction, we obtain that a
general solution of equation () has the formx=Rmλ(a)+b, wherebis an arbitrary element
ofHiλ. The lemma is proved.
By this lemma, a solution of equation () reduces to a solution of the equation
(D–λI)m· · ·(D–λkI)mkx=RmλRλr+· · ·Rλna+b, () wherebis an arbitrary element ofHiλ. Using Lemma , by induction we obtain that a general solution of equation () has the form ().
6 Periodic solutions of the system of linear differential equations with constant coefficients
Let (a, . . . ,an) be a vector-line, whereai∈H,i= , . . . ,n. Denote by the operator of transposition of a matrix. The (a, . . . ,an)denotes a vector-column ofai∈H,i= , . . . ,n. Denote by Hn the set of all vectors (a, . . . ,an), where ai ∈H, i= , . . . ,n. For a= (a, . . . ,an)∈HnputDa= (Da, . . . ,Dan).
We consider the following system of linear differential equations:
Dx=Ax+a, ()
wherea∈Hn,x= (x
, . . . ,xn)andAis a complexn×nmatrix.
Definition SystemsDx=Ax+aandDy=By+b, whereA,Bare complexn×n-matrices,
a,b∈Hnare called equivalent, if there exists a complexn×n-matrixKsuch thatdetK= ,A=K–BKanda=K–b.
In this casey=Kx.
Proposition Every system()is equivalent to the system of the form Dy=By+b,where
Proof ForAthere exists a complexn×n-matrixKsuch thatA=K–BK, whereBis the
Jordan normal form ofA. FromDx=Ax+a, we obtainDy=By+b, wherey=Kxand
b=Ka.
By this proposition, a solution of system () reduces to a solution of the system of the form (), whereBhas the Jordan normal form. LetBhave the form
⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
B . . .
B . . .
..
. ... ... ...
... Bm ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠
, ()
whereBjis a Jordan block of thenjth order,n+· · ·+nm=n. Then a solution of system (), whereBhas the form (), reduces to a solution of the following system of equations:
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
Du=Bu+b;
Du=Bu+b;
.. .
Dum=Bmum+bm,
()
whereBjis a Jordan block of thenjth order,bj∈Hnj,uj∈Hnjandx= (u,u, . . . ,um).
Therefore a solution of system () reduces to a solution of the equation of the form
Dx=Gx+b, ()
whereBis a Jordan block of theqth order with eigenvalueλ:gi
i =λ,i= , . . . ,q;gii+= ,
i= , . . . ,q– ;gi
j= ,j–i< andj–i> .
Theorem Let Dx=Gx+b be a system of the form(),where G is a Jordan block of the
qth order with eigenvalueλand b= (b, . . . ,bq)∈Hq.Then
(i) for the caseλ∈C\Spec(H),the system has the unique solution x= (x, . . . ,xq)∈Hq,where:
x=Rλb+Rλb+· · ·+Rqλbq;
x=Rλb+Rλb+· · ·+Rqλ–bq; ..
.
xq=Rλbq,
(iii) for the caseλ∈Spec(H),λ=ip,p∈ZandFpbq= ,a general solution
x= (x, . . . ,xq)∈Hqof the system has the form
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
x=
q–
k=Rkip(bk–Fpbk) +Rqip(bq) +c;
xq–r=rk=Ripk(bq–r–+k–Fpbq–r–+k) +Rrip+(bq–Fpbq–r–),
r= , . . . ,q– ;
xq=Ripbq–Fpbq–,
()
wherecis an arbitrary element ofHp.
Proof System () has the form ⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
Dx–λx=x+b;
Dx–λx=x+b;
.. .
Dxq––λxq–=xq+bq–;
Dxq–λxq=bq.
()
Letλ∈C\Spec(H). From (), using the operatorRλ, we obtain
xq=Rλbq, xq–=Rλbq–+Rλbq, . . . , x=
q
k=
Rkλbk.
Letλ∈Spec(H). Thenλ=ipfor somep∈Z. By Theorem , the equation
Dxq–ipxq=bq, ()
is solvable if and only ifFpbq= .
LetFpbq= . By Theorem , a general solution of () has the formxq=Ripbq+cq, where
cqis an arbitrary element ofHip. In this case, the equationDxq––ipxq–=xq+bq–has
the form
Dxq––ipxq–=Ripbq+cq+bq–. ()
By Theorem , this equation has a solution if and only if
Fip(Ripbq+cq+bq–) = . ()
ByFip(bq) = and Theorem (ii), (iii), we haveFipRipbq=RipFipbq= . Hence () is equiv-alent toFip(cq+bq–) = . ThenFipcq= –Fipbq–. Bycq∈Hip, we obtaincq=Fipcq= –Fipbq–.
Thus equation () is solvable if and only ifcq= –Fipbq–. Then equation () has the form
Dxq––ipxq–= (bq––Fipbq–) +Ripbq. By Theorem , a general solution of this equation isxq–=Rip(bq––Fipbq–) +Ripbq+cq–, wherecq–is an arbitrary element ofHip. Then the
equationDxq––ipxq–=xq–+bq–has the formDxq––ipxq–= (bq–+cq–) +Rip(bq––
Fipbq–) +Ripbq. As above, this equation is solvable if and only ifcq–= –Fipbq–. Similarly,
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
The authors did not provide this information.
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.
Received: 12 December 2012 Accepted: 7 March 2013 Published: 15 April 2013
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doi:10.1186/1029-242X-2013-172
