tional Differential Equations with Infinite Delay Next, we consider the case where (3.26) is linear, that is, (H5) F(t, φ) is linear inφwith supt≥0|F(t,ã)|X≤L.
Theorem 3.15 Assume that the conditions(H2) and(H5)hold. LetBbe a fading memory space. Then the following statements hold:
i) If the null solution of (3.26)isBC-TS, then it isBC-UAS .
ii) Assume that B is a uniform fading memory space. If the null solution of (3.26) isB-TS then it isB- UAS.
Proof. Claim i). Letσ∈R+ andφ∈BC with|φ|BC <min(1, δ(1)),where δ(ã) is the one given for theBC-TS of the null solution of (3.26). Thenu(t) :=u(t, σ, φ) satisfies|u(t)|X<1 for allt∈R. Now, for anyε >0, 0< ε <1,andα >0,we set
a(t) :=a(t, α, ε) =
(1 + 2αt)(1 +εαt), t≥0 1, t <0, and definev(t) andh(t) by
v(t) =a(t−σ)u(t), t∈R and
h(t) = ˙a(t−σ)u(t) +a(t−σ)F(t, ut)−F(t, vt), t≥σ, (3.51) respectively, where ˙adenotes the right hand derivative of a. Clearlyh(t) is contin- uous int≥σ and it satisfies
|h(t)|X ≤ 2α|u(t)|X+|F(t, a(t−σ)ut−vt)|X
≤ 2α+L|a(t−σ)ut−vt|B
fort≥σ by (H2), because of|a(t)| ≤˙ 2αfort≥0. We first assert that v(t) is the (mild) solution of (3.34) withh(t) given by (3.51); that is,v(t) satisfies the relation
v(t) =T(t−σ)v(σ) + Z t
σ
T(t−s){a(s˙ −σ)u(s) +a(s−σ)F(s, us)}ds, (3.52) for all τ ≥ t > σ. Indeed, one can take sequences {xn} ⊂ D(A) and {hn} ⊂ BC([σ, τ];X) such thathn is continuously differentiable on [σ, τ] and|xn−φ(0)|X+ supσ≤t≤τ|hn(t)−F(t, ut)|X → 0 as n → ∞. Set un(t) = T(t−σ)xn +Rt
σT(t− s)hn(s)dsandvn(t) =a(t−σ)un(t) fort∈[σ, τ].Then un(t)→u(t) andvn(t)→ v(t) inX as n→ ∞ uniformly fort∈ [σ, τ]. From [179, Theorem 4.2.4] it follows thatun is continuously differentiable on (σ, τ] with (dun(t))(dt) =Aun(t) +hn(t).
Hencevn is continuously differentiable on (σ, τ] with (dvn(t))(dt) =Avn(t) + ˙a(t− σ)un(t) +a(t−σ)hn(t).Then, from [179, Corollary 4.2.2] it follows that
vn(t) =T(t−σ)vn(σ) + Z t
σ
T(t−s){a(s˙ −σ)un(s) +a(s−σ)hn(s)}ds fort∈[σ, τ].Lettingn→ ∞in the above, we get (3.52) as required.
Now, for each positive integer n we set Sn := sup{|φ|B : φ ∈ BC,|φ|BC ≤ 1 and suppφ⊂(−∞,−n]}, where supp φdenotes the support ofφ.Observing that Sn→0,by (A2), asn→ ∞, we take so largen:=n(ε) thatSn < εδ(1)/(8L),and then chooseα:=α(ε) such that 2α(1 +LJ n)< δ(1)/2. We claim that|h|[σ,∞)<
δ(1). To see this, for anyt≥σ we consider functionsq, qn andwn in BC defined by
q(θ) = a(t−σ)u(t+θ)−v(t+θ)
= [a(t−σ)−a(t+θ−σ)]u(t+θ), θ≤0,
qn(θ) =
q(θ) if −n≤θ≤0, linear if −n−1≤θ≤ −n,
0 if θ≤ −n−1,
andwn=q−qn.Then the support ofwn is contained in (−∞,−n], and|wn|BC ≤ 2 sups∈R|a(s)| ≤ 4/ε. Consequently, |wn|B ≤ (4/ε)Sn < δ(1)/(2L). Also, since
|qn|BC ≤ |q|[−n,0] ≤2αn, we get |qn|B ≤J|qn|BC ≤2αJ n. Then |h(t)|X ≤2α+ L|q|B < δ(1) or |h|[σ,∞) < δ(1) as required. Since the null solution of (3.26) is BC-TS, we get|v(t)|X < 1 for all t ≥ σ. Hence, if t ≥ σ+ (1−ε)/(εα(ε)), then
|u(t)|X<1/a(t−σ)< ε,which proves the first claim of the theorem.
Claim ii). Letσ∈R+ andφ∈ B with|φ|B< δ(1), whereδ(ã) is the one given for the B-total stability of the null solution of (3.26). In what follows, we employ the same notation as in the proof of the first claim. Since B is a uniform fading memory space, we may assume that the functions K(ã) and M(ã) in (A1) satisfy supt≥0K(t) =: K < ∞, supt≥0M(t) =: M < ∞ and M(t) → 0 as t → ∞. In virtue of (A1-iii), we obtain|ut|B≤Ksupσ≤s≤t|u(s)|X+M|uσ|B≤K+M δ(1) for t≥σ.Similarly, we can get supt≥σ|vt|B ≤(2/ε)K+M δ(1).Take ant0:=t0(ε)>0 so that
L(2/ε+ 1)(2K/ε+M δ(1))M(t0)< δ(1)/2,
and then choose anα:=α(ε)>0 so that
2α{1 +Lt0(K+M δ(1))}< δ(1)/2.
Ift≥t0+σ, then
|h(t)|X ≤ 2α+L|a(t−σ)ut−vt|B
≤ 2α+L{K sup
−t0≤θ≤0
|a(t−σ)−a(t−σ+θ)||u(t+θ)|X
+M(t0)|a(t−σ)ut−t0−vt−t0|B}
≤ 2α+L{2αKt0+ (2/ε+ 1)(2K/ε+M δ(1))M(t0)}
< δ(1)
by (A1-iii). On one hand, ifσ≤t≤t0+σ,then
|h(t)|X ≤ 2α+L{K sup
σ−t≤θ≤0
|a(t−σ)−a(t−σ+θ)||u(t+θ)|X +M|a(t−σ)−1||φ|B}
≤ 2α+L{2Kα(t−σ) + 2M α(t−σ)δ(1)}
< δ(1)/2.
We thus obtain|h|[σ,∞)< δ(1).Then theB-UAS of the null solution of (3.26) follows from the same reasoning as in the proof of the first claim.
It is natural to ask if the additional assumption that B is a uniform fading memory space can be removed in the second claim of Theorem 3.15. As the following example shows, however, one cannot remove the assumption, in general.
Example 3.2 ForX=R andg(s) = 1−s, we consider the spaceCg0 constructed in Section 3.3, and define a functionalG:R+×Cg07→Rby
G(t, φ) = φ(−t)
g(−t), (t, φ)∈R+×Cg0.
For eacht∈R+,G(t,ã) :Cg07→Ris a bounded linear operator withkG(t,ã)k ≤1.
Moreover, one can see thatGis continuous onR+×Cg0.We now consider the linear functional differential equation
du
dt =−u(t) +G(t, ut), t≥0. (3.53) We first show that the null solution of (3.53) isCg0-TS. Indeed, ifσ∈R+, φ∈Cg0 with|φ|g < ε/3 andh∈BC([σ,∞);R) with |h|[σ,∞)< ε/3, then the solutionv(t) of (d/dt)u=−u(t) +G(t, ut) +h(t), t≥σ, through (σ, φ) satisfies
|v(t)| = |e−(t−σ)φ(0) + Z t
σ
e−(t−s)(G(s, vs) +h(s))ds|
≤ |φ(0)|+ Z t
σ
e−(t−s)(|φ(−σ)|/(1 +s) +|h(s)|)ds
≤ |φ|g+ Z t
σ
e−(t−s)(|φ|g+|h|[σ,∞))ds
≤ 2|φ|g+|h|[σ,∞)< ε
fort≥σ,which shows theCg0-TS of the null solution of (3.53). We claim that the null solution of (3.53) is notCg0-UAS. Indeed, if this is not true, then for anyε >0 there exists at0:=t0(ε)>1 such that supσ≥0|u(σ+t0, σ, φ)|< εwheneverφ∈Cg0 with |φ|g ≤ 1. For each σ ≥ 0, choose a nonnegative function φσ in Cg0 so that φσ(0) = 0 and|φσ|g=φσ(−σ)/g(−σ) = 1.Then
sup
σ≥0
|u(σ+t0, σ, φσ|< ε. (3.54) On the other hand,
u(σ+t0, σ, φσ) = e−t0φσ(0) + Z σ+t0
σ
e−(σ+t0−s)G(s, us)ds
= Z t0
0
eθ−t0G(θ+σ, uθ+σ)dθ
= Z t0
0
eθ−t0(1 +σ)/(1 +θ+σ)dθ
≥ Z t0
0
eθ−t0dθã(1 +σ)/(1 +t0+σ)
≥ (1−e−t0)(1 +σ)/(1 +t0+σ).
Hence we get supσ≥0|u(σ+t0, σ, φσ)| ≥1−e−t0>1−e−1,which is a contradiction to (3.54). Consequently, the null solution of (3.53) cannot beCg0-UAS.
As a direct consequence of Theorems 3.11 and 3.15, one can obtain the following result which is an extension of [93, Theorem] and [98, Theorem 3] withX=Rnto the case where dimX=∞.
Theorem 3.16 LetBbe a fading memory space, and assume(H1)-(H2)and(H5).
Then the following statements hold.
i) The null solution of (3.26)isBC-TS if and only if it isBC-UAS.
ii) Assume thatBis a uniform fading memory space. Then the null solution of (3.26) isB-TS if and only if it isB-UAS.
We note that in Claim ii) of Theorem 3.16, the additional condition thatBis a uniform fading memory space cannot necessarily be removed as Example 3.2 shows.