Let B be a fading memory space which is separable. We shall discuss some rela- tionships between BC-stabilities andρ-stabilities, and extend some results due to Murakami and Yoshizawa [161] forX=Rn to the case where dimX=∞.
The solution ¯u(t) of (3.26) in (H4) is said to be BC-totally stable (BC-TS) if for any ε > 0 there exists a δ(ε) > 0 with the property that σ ∈ R+, φ ∈ BC with|¯uσ−φ|BC< δ(ε) andh∈BC([σ,∞);X) with supt∈[σ,∞)|h(t)|X< δ(ε) imply
|¯u(t)−u(t, σ, φ, F+h)|X< εfort≥σ, whereu(ã, σ, φ, F+h) denotes the solution of
du
dt =Au(t) +F(t, ut) +h(t), t≥σ, (3.34) through (σ, φ).
The other BC-stabilities for ¯u(t) are given in a similar way; we omit the details.
Theorem 3.11 Assume that conditions (H1)–(H4) hold. If u(t)¯ of (3.26) is BC- UAS, then it is BC-UASinΩ(F)andBC-TS.
Proof. First, we shall show that ¯u(t) is BC-UAS in Ω(F).
Let (δ(ã), δ0, t0(ã)) be the triple for BC-UAS of ¯u(t), where we may assume δ0< δ(1). We first establish that
σ∈R+, (¯v, G)∈Ω(¯u, F) and |φ−v¯σ|BC < δ(η 2)
imply |x(t, σ, φ, G)−v(t)|X< η for t≥σ. (3.35) Select a sequence{tn}withtn→ ∞asn→ ∞such that (¯utn, Ftn)→(¯v(t), G(t, φ)) compactly, and consider any solutionx(ã, σ+tn, φ−v¯σ+ ¯uσ+tn, F). For anyn∈N, set xn(t) =x(t+tn, σ+tn, φ−¯vσ+ ¯uσ+tn, F), t ∈ R. Since the solution ¯u(t) of (3.26) is BC-UAS, from the fact that |xnσ−u¯σ+tn|BC = |φ−vσ|BC < δ(η/2) it follows that
|xn(t)−u(t¯ +tn)|X< η
2 for all t≥σ and n∈N. (3.36) Observe that the set{xn(σ) :n∈N} is relatively compact inX. In virtue of this fact and (3.36), repeating almost the same argument as in the proof of Lemma 3.5 we can see that the set{xn(t) :t≥σ, n∈N} is relatively compact inX and that the sequence{xn}is uniformly equicontinuous on [σ,∞). Thus we may assume that xn(t)→y(t) compactly on [σ,∞) for some functiony(t) : [σ,∞)7→X. Since xn(σ) =φ(0)−v(σ) + ¯¯ u(σ+tn), we obtainy(σ) =φ(0). Hence, if we extend the functiony by settingyσ=φ, theny∈C(R,X) and|xnt −yt|B→0 compactly on [σ,∞). Lettingn→ ∞in the relation
xn(t) =T(t−σ){φ(0)−¯v(σ) + ¯u(σ+tn)}+ Z t
σ
T(t−s)F(s+tn, xns)ds, t≥σ, we obtain
y(t) =T(t−σ)φ(0) + Z t
σ
T(t−s)G(s, ys)ds, t≥σ,
which means thaty(t)≡x(t, σ, φ, G) fort≥σ by the regularity assumption (H3).
Then (3.35) follows from (3.36) by lettingn→ ∞.
Repeating the above arguments withη= 2, we can establish that σ∈R+, (¯v, G)∈Ω(¯u, F) and |φ−vσ|BC < δ0
imply |x(t, σ, φ, G)−¯v(t)|X< ε for t≥σ+t0(ε
2). (3.37) Next, we suppose that the solution ¯u(t) is not BC-TS. Then there exist anε,0<
ε < δ0, sequences{τn} ⊂R+,{rn}, rn >0,{φn} ⊂BC,{hn}, hn∈BC([τn,∞);X), and solutions{x(ã, τn, φn, F +hn)} such that, for alln∈N,
|φn−u¯τn|BC < 1
n and sup
t≥τn
|hn(t)|< 1
n (3.38)
and
|zn(τn+rn)−u(τ¯ n+rn)|X=εand|zn(t)−u(t)|¯ X< ε∀t∈(−∞, τn+rn), (3.39) herezn(t) :=x(t, τn, φn, F+hn). We first consider the case that{rn}is unbounded.
Without loss of generality, we may assume that
(¯uτn+rn−t0, Fτn+rn−t0)→(¯v, G)
compactly, for some (¯v, G)∈Ω(¯u, F) and thatzn(t+τn+rn−t0)→z(t) compactly on (−∞, t0] for some function z, where t0 = t0(ε/2). Repeating almost the same argument as in the proof of the claim (3.35), we see by (3.38) thatzsatisfies (3.28) on [0, t0]. Letn→ ∞in (3.39) to obtain|z(t)−v(t)|¯ X≤εon (−∞, t0] and|z(t0)−
¯
v(t0)|X =ε. This is a contradiction, because |z0−v¯0|BC ≤ε < δ0 implies|z(t0)−
¯
v(t0)|X< εby (3.37). Therefore the sequence{rn}must be bounded. Thus we may assume that{rn}converges to some r,0≤r <∞. Moreover, we may assume that {zn(τn+t)} →ξcompactly on (−∞, r]. Consider the case where the sequence{τn} is unbounded; hence we may assume that (¯uτn, Fτn)→(w, H) compactly, for some (w, H)∈Ω(¯u, F). Thenξ(t) satisfies
ξ(t) =T(t)ξ(0) + Z t
0
T(t−s)H(s, ξs)ds
on [0, r], and moreover we have|ξ0−w0|BC = 0 and|ξ(r)−w(r)|X=εby letting n→ ∞in (3.38) and (3.39). This is a contradiction, because we must haveξ≡won [0, r] by (H3). Thus the sequence{τn}must be bounded, too. Hence we may assume that limn→∞τn =τ for someτ <∞. Thenξ(t−τ) satisfies (3.26) on [τ, τ+r], and moreover we have|ξ0−u¯τ|BC= 0 and |ξ(r)−u(τ¯ +r)|X=εby (3.38) and (3.39).
This again contradicts the fact that the solution ¯u(t) of (3.26) is BC-UAS.
3.4.2. Equivalent Relationship between BC-Uniform Asymptotic Sta- bility and ρ-Uniform Asymptotic Stability
Theorem 3.12 Assume that conditions (H1)–(H4) hold. If the solution u(t)¯ of (3.26) is BC-UAS, then it is ρ-US w.r.t.U for any closed bounded set U in X such thatUi⊃Ou¯.
Proof. We assume that the solution ¯u(t) of (3.26) is BC-UAS but not ρ-US w.r.t.U; here U ⊂ {x∈X : |x|X ≤c} for some c > 0. Since the solution ¯u(t) of (3.26) is not (ρ,X)-US w.r.t.U as noted in Proposition 3.2 there exist anε∈(0,1), sequences{τm} ⊂R+, {tm} (tm> τm), {φm} ⊂BC withφm(s)∈U fors∈R− and solutions{u(t, τm, φm, F) =: ˆum(t)} of (3.26) such that
ρ(φm,u¯τm)<1/m (3.40)
and that
|ˆum(tm)−u(t¯ m)|X=ε and |ˆum(t)−u(t)|¯ X< ε on [τm, tm) (3.41) form∈N. For eachm∈N andr∈R+, we defineφm,r∈BC by
φm,r(θ) =
φm(θ) if −r≤θ≤0, φm(−r) + ¯u(τm+θ)−u(τ¯ m−r) if θ <−r.
We note that|φm,r−u¯τm|BC=|φm−u¯τm|[−r,0] and
sup{|φm,r−φm|B:m∈N} →0 as r→ ∞. (3.42) For, if (3.42) is false, then there exist an ε > 0 and sequences {mk} ⊂ N and {rk}, rk → ∞ask→ ∞, such that|φmk,rk−φmk|B≥εfork= 1,2,ã ã ã. Putψk :=
φmk,rk−φmk. Clearly,{ψk} is a sequence in BC which converges to zero function compactly onR− and supk|ψk|BC<∞. Then Axiom (A2) yields that |ψk|B→0 as k → ∞, a contradiction. Furthermore, the set {φm, φm,r :m ∈ N, r ∈ R+} is relatively compact inB. Indeed, since the set{u¯t :t ∈R+} is relatively compact in B, (3.40) and Axiom (A2) yield that any sequence {φmj}∞j=1 (mj ∈ N) has a convergent subsequence inB.
Therefore, it sufficies to show that any sequence {φmj,rj}∞j=1(mj ∈ N, rj ∈ R+) has a convergent subsequence inB. We assert that the sequence of functions {φmj,rj(θ)}∞j=1contains a subsequence which is equicontinuous on any compact set in R−. If this is the case, then the sequence{φmj,rj}∞j=1 would have a convergent subsequence inBby the Ascoli-Arz´ela theorem and Axiom (A2), as required. Now, notice that the sequence of functions{u(τ¯ mj+θ)}is equicontinuous on any compact set in R−. Then the assertion obviously holds true when the sequence {mj} is bounded. Taking a subsequence if necessary, it is thus sufficient to consider the case mj→ ∞asj→ ∞. In this case, it follows from (3.40) thatφmj(θ)−u(τ¯ mj+θ) =:
wj(θ) → 0 compactly on R−. Concequently, {wj(θ)} is equicontinuous on any
compact set inR−, and so is{φmj(θ)}. Therefore the assertion immediately follows from this observation.
Now, for anym∈N, setum(t) = ˆum(t+τm) ift≤tm−τmandum(t) =um(tm− τm) ift > tm−τm. Moreover, setum,r(t) =φm,r(t) ift∈R− andum,r(t) =um(t) if t∈R+. Since|umt |B≤K{1+|¯u|[0,∞)}+M|φm|B≤K{1+|¯u|[0,∞)}+M J cby (3.25) and (A1-iii), by using the same arguments as in the proof of Lemma 3.5 we see that {um(t)}is uniformly equicontinuous inC(R+,X). Then it follows from Lemma 3.4 and the relative compactness of the set {φm, φm,r : m ∈ N, r ∈ R+} that the set W := {um,rt , umt :m∈N, t∈R+, r∈R+} is compact in B. Hence F(t, φ) is uniformly continuous on R+×W by (H1). Define a continuous functionqm,r on R+byqm,r(t) =F(t+τm, umt )−F(t+τm, um,rt ) if 0≤t≤tm−τm, andqm,r(t) = qm,r(tm−τm) ift > tm−τm. Since|um,rt −umt |B≤M|φm,r−φm|B (t∈R+, m∈N) by (A1-iii), it follows from (3.42) that sup{|um,rt −umt |B :t∈R+, m∈N} →0 as r→ ∞; hence one can chooser=r(ε)∈Nin such a way that
sup{|qm,r(t)|X:m∈N, t∈R+}< δ(ε/2)/2,
whereδ(ã) is the one for the BC-TS of ¯u(t). Moreover, for thisr, select anm∈N such that m > 2r(1 +δ(ε/2))/δ(ε/2). Then 2−r|φm−u¯τm|r/[1 +|φm−u¯τm|r] ≤ ρ(φm,u¯τm)<2−rδ(ε/2)/[1 +δ(ε/2)] by (3.40), which implies that
|φm−u¯τm|r< δ(ε/2) or |φm,r−u¯τm|BC< δ(ε/2).
The functionum,rsatisfiesum,r0 =φm,rand um,r(t) = um(t)
= T(t)φm(0) + Z t
0
T(t−s){F(s+τm, ums)}ds
= T(t)φm,r(0) + Z t
0
T(t−s){F(s+τm, um,rs ) +qm,r(s)}ds fort∈[0, tm−τm). Since ¯um(t) = ¯u(t+τm) is a solution of
du
dt =Au+F(t+τm, ut)
with the same δ(ã) as the one for ¯u(t), from the fact that supt≥0|qm,r(t)|X <
δ(ε/2)/2 < δ(ε/2) it follows that |um,r(t)−u(t¯ +τm)|X < ε/2 on [0, tm−τm).
In particular, we have |um,r(tm−τm)−u(t¯ m)|X < ε or |ˆum(tm)−u(t¯ m)|X < ε, which contradicts (3.41).
Theorem 3.13 Assume that conditions (H1)–(H4) are hold. Then the solution
¯
u(t)of (3.26)isBC-UASif and only if it isρ-UASw.r.t.U for any closed bounded setU in Xsuch thatUi⊃O¯u:={u(t) :¯ t∈R}.
Proof. The “if” part is easily shown by noting thatρ(φ, ξ)≤ |φ−ξ|BCforφ, ξ∈ BC. We shall establish the “ only if” part. The solution ¯u(t) of (3.26) isρ-US w.r.t.U by Theorem 3.12. Thus it suffices to establish the following assertion:
(*) For anyε > 0 there exists at∗(ε)> 0 such thatρ(φ,u¯τ) < δ1 :=δ∗(δ0/4) withφ(s)∈U, s∈R−,impliesρ(ut(τ, φ, F),u¯t)< εfor allt≥τ+t∗(ε), where δ∗(ã) is the number forρ-US of ¯u(t) andδ0 is the number for the BC-UAS in Ω(F) of ¯u(t).
If this assertion is not true, then there exist an ε > 0 and sequences {τk} ⊂ R+, {tk}, tk≥τk+ 2k, {φk} ⊂BC, and solutions{u(t, τk, φk, F)}such that
ρ(φk,u¯τk)< δ1, φk(s)∈U, s∈R− (3.43) and
ρ(utk(τk, φk, F),u¯tk)≥ε (3.44) for allk∈N. Since ¯u(t) isρ-US, (3.43) and (3.44) imply that
ρ(ut(τk, φk, F),u¯t)< δ0
4 for all t≥τk (3.45)
andρ(ut(τk, φk, F),u¯t)≥δ∗(ε) for allt∈[τk, τk+ 2k] or
ρ(ut+τk+k(τk, φk, F),u¯t+τk+k)≥δ∗(ε) for all t∈[−k, k], (3.46) respectively. Set uk(t) = u(t+τk +k, τk, φk) for t ∈ R. Since ρ(φ, ψ) ≥ 2−1|φ− ψ|1/[1 +|φ−ψ|1]≥2−1|φ(0)−ψ(0)|X/[1 +|φ(0)−ψ(0)|X], we have|φ(0)−ψ(0)|X≥ 2ρ(φ, ψ)/[1−2ρ(φ, ψ)] wheneverρ(φ, ψ)≤1/2; hence (3.45) implies that
|uk(t)−u(t¯ +τk+k)|X≤ δ0
2−δ0 for all t∈[−k, k]. (3.47) We shall show that the setO:={uk(t) :−k+1≤t≤k, k∈N}is relatively compact inXand that{uk(t)}is a family of equicontinuous on any bounded interval inR. To certify the assertion, we use the Kuratovski’s measureα(ã) of noncompactness of set inX. Let 0< ν <1. Sinceuk(t) is a mild solution of du(t)dt =Au(t)+F(t+τk+k, ut) through (−k, φk), we get
uk(t) = T(t+k)φk(0) + Z t
−k
T(t−s)F(s+τk+k, uks)ds
= T(1)[T(t+k−1)φk(0) + Z t−1
−k
T(t−1−s)F(s+τk+k, uks)ds]
+ Z t
t−1
T(t−s)F(s+τk+k, uks)ds
= T(1)[uk(t−1)] +T(ν) Z t−ν
t−1
T(t−s−ν)F(s+τk+k, uks)ds +
Z t t−ν
T(t−s)F(s+τk+k, uks)ds
fort≥ −k+1. It follows from Axiom (A1-iii), (H2) and (3.47) that sups∈[−k,∞)|F(s+
τk+k, uks)|X=:C2<∞. Then the set{Rt−ν
t−1 T(t−s−ν)F(s+τk+k, uks)ds:t≥ −k+
1}is bounded inX, and henceT(ν){Rt−ν
t−1 T(t−s−ν)F(s+τk+k, uks)ds:t≥ −k+1}
is relatively compact inXbecause of the compactness of the semigroup{T(t)}t≥0. Similarly, one can get the relative compactness of the setT(1){uk(t−1) :t≥ −k+1}.
Thus we obtain
α(O) ≤ α({
Z t t−ν
T(t−s)F(s+τk+k, uks)ds:t≥ −k+ 1})
≤ C2C3ν,
where C3 = sup0≤τ≤1kT(τ)k. Letting ν → 0 in the above, we get α(O) = 0;
consequently, O must be relatively compact in X. To certify the assertion on the equicontinuity, let−k+ 1≤s≤t≤s+ 1. Then
uk(t) =T(t−s)uk(s) + Z t
s
T(t−τ)F(τ+τk+k, ukτ)dτ. (3.48) Hence
|uk(t)−uk(s)|X ≤ |T(t−s)uk(s)−uk(s)|X
+|
Z t s
T(t−s)F(τ+τk+k, ukτk)dτ|X
≤ sup{|T(t−s)z−z|X:z∈O}+C2C3|t−s|.
Since the setOis relatively compact inX,T(τ)zis uniformly continuous inτ ∈[0,1]
uniformly forz∈O. This observation proves the assertion.
Now, applying the Ascoli-Arz´ela theorem and the diagonalization procedure, one may assume that uk(t)→ w(t) compactly on R for some w(t) ∈ BC(R;X).
Then ukt → wt compactly on R by Axiom (A1-iii). Also, we may assume that (¯uτk+k, Fτk+k)
→(¯v, G) compactly onR+× Bfor some (¯v, G)∈Ω(¯u, F). Lettingk→ ∞in (3.47) and (3.48), w(t) = u(t,0, w0, G) on R and |w0−v¯0|BC ≤δ0/[2−δ0] < δ0. Then
|w(t)−¯v(t)|X→0 ast→ ∞, by Theorem 3.11. On the other hand, lettingk→ ∞ in (3.46), we have ρ(wt,¯vt)≥δ∗(ε) for all t∈R, a contradiction. This completes the proof of Theorem 3.13.