2.4. FIXED POINT THEOREMS AND FREDHOLM OPERATORS
2.4.4. Existence of Periodic Solutions: Bounded Perturbations
In this subsection we will apply Theorem 2.28 to study the existence of periodic solutions of Eq.(2.155). To this end, we first consider the operatorTb(ω).
Proposition 2.14 Let (T(t))t≥0 be aC0-semigroup onE. Thenφ∈N(I−T(ω))b if and only ifφ(0)∈N(I−T(ω))andφ(θ) =T(nω+θ)φ(0),−r≤θ≤0,whenever nω≥r ;in particular φ∈N(I−Tb(ω))is a restriction to[−r,0]of an ω-periodic continuous function. Furthermore,
dimN(I−T(ω)) = dimb N(I−T(ω)).
Proof. Ifx∈N(I−T(ω)), thenx=T(ω)xandT(t)x=T(t)T(ω)x=T(t+ω)x fort≥0; and vice versa.
Suppose thatφ∈N(I−Tb(ω)). Sinceφ=Tb(ω)φ, it follows thatφ(0) =T(ω)φ(0) or φ(0) ∈ N(I −T(ω)). Since Tb(t) is a C0-semigroup, we see that Tb(ω)n = Tb(nω), n= 1,2,ã ã ã.Ifnω > r, thennω+θ≥0 for θ∈[−r,0] ; and hence
(Tb(ω)nψ)(θ) =T(nω+θ)ψ(0), θ∈[−r,0].
Thusφ =Tb(ω)nφ for n≥ 1 ; in particular, if nω > r, φ(θ) = T(nω+θ)φ(0) for θ∈[−r,0].
Conversely, suppose thatx∈N(I−T(ω)) and set (φn)(θ) =T(nω+θ)x, θ ∈ [−r,0] fornsuch thatnω≥r. SinceT(t+ω)x=T(t)xfort≥0,φn is independent ofn > r/ω. Denote byφthis independent function. Thenφ(0) =T(nω)x=x. Let θ∈[−r,0]. Ifω+θ≥0, then
Tb(ω)φ(θ) =T(ω+θ)φ(0) =T(ω+θ)x=T(nω+θ)x=φ(θ).
Forω+θ <0, one has
Tb(ω)φ(θ) = φ(ω+θ) =φn(ω+θ)
= T(nω+ω+θ)x=T(nω+θ)x
= φn(θ) =φ(θ).
Hence,Tb(ω)φ=φ. It is obvious that this map from N(I−T(ω)) to N(I−T(ω))b is injective. Its surjectiveness follows from the first half of the proof.
To see that the rangeR(I−T(ω)) is a closed subspace, we solve the equationb (I−T(ω))φb =ψ. Letpbe a positive integer such that
(p−1)ω < r≤pω. (2.157)
SetIp= [−r,−(p−1)ω) andIk= [−kω,−(k−1)ω) fork= 1,2,ã ã ã, p−1 provided p≥2.
Proposition 2.15 The functionsφ, ψ∈ C satisfy the equation(I−Tb(ω))φ=ψ if and only if
i) (I−T(ω))φ(0) =ψ(0), ii) φ(θ) =Pk−1
j=0ψ(θ+jω) +T(θ+kω)φ(0), θ∈Ik, k= 1,2,ã ã ã, p.
Proof. Suppose that (I−Tb(ω))φ=ψ. Then ψ(θ) =
φ(θ)−T(θ+ω)φ(0) θ∈I1
φ(θ)−φ(θ+ω) θ∈Ik, k≥2.
Puttingθ= 0 in the first equation, we obtain the condition i). Solving this equation with respect toφ(θ) onIk successively fork= 1,2,ã ã ã, pwe have the representation ofφ(θ) in the condition ii) as mentioned above. The valueφ(−kω) is well defined fork≥1 because of the condition (I−T(ω))φ(0) =ψ(0).
Conversely, ifφ, ψ∈ Chave the properties i) and ii), then it follows immediately that (I−Tb(ω))φ=ψ. The proof is complete.
Let the null space N(I−T(ω)) be of finite dimension. Then it follows from Theorem 4.6 that there exists a closed subspace M of E such that E =N ⊕M, whereN =N(I−T(ω)). LetSM be the restriction ofI−T(ω) toM. Then
SM := (I−T(ω))|M :M →R(I−T(ω))
is a continuous, bijective and linear operator. Thus there is the inverse operator SM−1 of SM. If I−T(ω)∈Φ+(E), M can be taken so that SM−1 is continuous and that
kSM−1k ≤c0(1 +√
n), (2.158)
wheren= dimN(I−T(ω)), andc0 is the constant such that
|[x]| ≤c0|(I−T(ω))x| (2.159) forx∈E (see Theorem 4.6).
Put D = {ψ ∈ C : ψ(0) ∈ R(I −T(ω))} and let ψ ∈ D. Since R(SM) = R(I−T(ω)),SM−1ψ(0) is well defined and (I−T(ω))SM−1ψ(0) =ψ(0). We define a function (V ψ)(ã) : [−r,0]→E pointwise by
[V ψ](θ) =
k−1
X
j=0
ψ(θ+jω) +T(θ+kω)SM−1ψ(0), θ∈Ik, (2.160)
fork= 1,2,ã ã ã, p,and [V ψ](0) =SM−1ψ(0). Notice thatD(V) =D.
Lemma 2.28 The operatorV defined by(2.160)is a linear operator fromD(V)to C.
Proof. It is sufficient to prove that V ψ ∈ C for ψ ∈ D(V). Clearly,V ψ is con- tinuous in each interval Ik.Thus we prove that (V ψ)(−kω) = (V ψ)(−kω−0) for k= 1,2,ã ã ãp−1.Notice thatSM−1ψ(0) =T(ω)SM−1ψ(0) +ψ(0). From the definition (2.160) ofV,we have,
(V ψ)(−kω) =
k−1
X
j=0
ψ(−kω+jω) +T(0)SM−1ψ(0)
=
k−1
X
j=0
ψ(−kω+jω) +ψ(0) +T(ω)SM−1ψ(0)
=
k
X
j=0
ψ(−kω+jω) +T(−kω+ (k+ 1)ω)SM−1ψ(0)
= lim
θ→−kω−0{
k
X
j=0
ψ(θ+jω) +T(θ+ (k+ 1)ω)SM−1ψ(0)}
as required.
Lemma 2.29
R(I−Tb(ω)) =D(V).
Proof. Letψ∈ R(I−T(ω)). Then there is ab φ∈ C such that [I−Tb(ω)]φ=ψ.
From Proposition 2.15 we see that (I−T(ω))φ(0) = ψ(0) ; and hence, ψ(0) ∈ R(I−T(ω)). This implies thatψ∈ D(V).
Conversely, if ψ ∈ D(V), then ψ(0) ∈ R(I−T(ω)). Lemma 2.28 means that V ψ∈ C. Hence it follows from Proposition 2.15 that [I−Tb(ω)]V ψ=ψ. Therefore the proof is completed.
Proposition 2.16 I−T(ω)∈Φ+(E)if and only if I−T(ω)b ∈Φ+(C).
Proof. If R(I−T(ω)) is closed in E, then D(V) is closed in C. Indeed, ifφn ∈ D(V)→φasn→ ∞, then φn(0)→φ(0) asn→ ∞. This implies that φ∈ D(V).
From this fact and Lemma 2.29 it follows thatR(I−Tb(ω)) is closed inC.
Conversely, we assume that R(I−Tb(ω)) is closed in C. Then it follows from Lemma 2.29 that D(V) is closed in C. Let xn ∈ R(I−T(ω)) → x as n → ∞.
Then there exist a sequence{ϕn} ⊂ C of constant functions andϕ∈ C such that ϕn(0) = xn, ϕn ∈ D(V) and ϕn → ϕ as n → ∞ in C. Hence we have ϕ(0) = x and ϕ ∈ D(V). This implies that x ∈ R(I−T(ω)). The remainder follows from Proposition 2.14 ; and hence, the proof is complete.
Proposition 2.17 1∈ρ(T(ω))if and only if 1∈ρ(Tb(ω)).
Proof. It is sufficient to prove that if 1 ∈ ρ(T(ω)), then 1 ∈ ρ(Tb(ω)). Since R(I−T(ω)) = E, we have D(V) = C. Hence it follows from Lemma 2.29 that R(I−Tb(ω)) =D(V) =C. Lemma 2.14 implies that 1∈ρ(Tb(ω)).
Next, we give criteria for the existence of periodic solutions to Equation (2.155) by using Theorem 2.28.
Proposition 2.18 If there exists a positive constantc such that
|V ψ| ≤c|ψ| for all ψ∈ D(V), (2.161) whereV is defined by(2.160), then
|[φ]| ≤c|(I−Tb(ω))φ| for allφ∈ C: as a result, the range R(I−Tb(ω))is closed.
Proof. Take the quotient space C/N(I−Tb(ω)). Suppose that (I−Tb(ω))φ=ψ.
Then ψ ∈ R(I −Tb(ω)) and V ψ ∈ C, because of Lemma 2.28. Hence, we have ψ= (I−Tb(ω))V ψand [φ] = [V ψ].Using these facts and the condition (2.161), we see that
|[φ]| ≤ |V ψ| ≤c|ψ|=c|(I−Tb(ω))φ|.
We note thatR(I−T(ω)) is closed if and only if there is a positive constantb csuch that|[φ]| ≤c|(I−Tb(ω))φ|for allφ∈ C, cf. Lemma 4.1. This prove the proposition.
Theorem 2.29 Suppose thatI−T(ω)∈Φ+(E). LetdimN(I−T(ω)) =n, andc0
be a positive constant such that
|[x]| ≤c0|(I−T(ω))x|
forx∈E. ThenI−Tb(ω)∈Φ+(C),dimN(I−T(ω)) =b nand
|[φ]| ≤(p+mωc0(1 +√
n))|(I−Tb(ω))φ| forφ∈ C,
wherepis the positive integer satisfying Inequality(2.157)andmω:= sup{kT(t)k: 0≤t≤ω}.
Proof. Take the closed subspaceM so thatSM−1 is continuous and the estimate (2.158) holds. Suppose that ψ(0) ∈ R(I−T(ω)) and that V ψ ∈ C. Then, for θ∈Ik, k= 1,2,ã ã ã, p,
|V ψ(θ)| ≤k|ψ|+mωkSM−1k|ψ(0)| ≤(p+mωkSM−1k)|ψ|.
This implies that
|V ψ| ≤(p+mωkS−1M k)|ψ|. (2.162) From this, the estimate (2.158) and Proposition 2.18, we have the conclusion.
We are now in a position to prove a criterion for the existence of ω-periodic solution to Eq.(2.155)
Theorem 2.30 Assume thatI−T(ω)∈Φ+(E)satisfies the conditions in Theorem 2.29, and that Equation (2.155) has a bounded solution. If the inequality
kK(ω)k ≤1/2(1 +√
n)(p+mωc0(1 +√
n)), (2.163)
is satisfied, then SL(ω)6=∅ anddimSL(ω)≤dimN(I−T(ω)) =n.
Using Proposition 2.7, we have the following one.
Corollary 2.21 The inequality (2.163) in the theorem can be replaced by the fol- lowing inequality
Mwewω
exp Z ω
0
MwkL(r)kdr
−1
≤1/2(1 +√
n)(p+mωc0(1 +√ n)).