Proof.
We define the operator ℱ:C([-r,T],X)→C([-r,T],X) in the following way:
(3.6)(Fu)(t)={ϕ(t),t∈[-r,0],Sq(t)ϕ(0)+∫0t(t-s)q-1Pq(t-s)f(s,us)ds +∫0t(t-s)q-1Pq(t-s)a(u)(s)ds,t∈[0,T].
It is clear that the operator ℱ is well defined.
We define
(3.7)ϕ^(t)={ϕ(t),t∈[-r,0],Sq(t)ϕ(0),t∈[0,T].
Let u(t)=v(t)+ϕ^(t). It is easy to see that v satisfies v0=0 and
(3.8)v(t)=∫0t(t-s)q-1Pq(t-s)f(s,vs+ϕ^s)ds+∫0t(t-s)q-1Pq(t-s)a(v+ϕ^)(s)ds, t∈[0,T]
if and only if u satisfies
(3.9)u(t)=Sq(t)ϕ(0)+∫0t(t-s)q-1Pq(t-s)f(s,us)ds+∫0t(t-s)q-1Pq(t-s)a(u)(s)ds, t∈[0,T]
and u(t)=ϕ(t), t∈[-r,0].
Let ℱ^:C([-r,T],X)→C([-r,T],X) be an operator defined by (ℱ^v)(t)=0 for t∈[-r,0] and
(3.10)(F^v)(t)=∫0t(t-s)q-1Pq(t-s)f(s,vs+ϕ^s)ds+∫0t(t-s)q-1Pq(t-s)a(v+ϕ^)(s)ds, t∈[0,T].
Clearly the operator ℱ^ has a fixed point is equivalent to ℱ having one.
We define 𝒞:={v∈C([-r,T],X):v0=0}⊂C([-r,T],X). Next we will prove that ℱ^ has a fixed point on 𝒞.
Let {vn}n∈N be a sequence such that vn→v in 𝒞 as n→∞. Since f satisfies (Hf)(1) and g satisfies (Hg)(1), for almost every t∈[0,T] and (t,s)∈Λ, we get
(3.11)f(t,vtn+ϕ^t)⟶f(t,vt+ϕ^t), as n⟶∞,g(t,s,vsn+ϕ^s)⟶g(t,s,vs+ϕ^s), as n⟶∞.
Noting that
(3.12)‖vt+ϕ^t‖[-r,0]≤supt∈[-r,0]‖v(t)‖+supt∈[0,T]‖v(t)‖+supt∈[-r,0]‖ϕ^(t)‖+supt∈[0,T]‖ϕ^(t)‖=supt∈[0,T]‖v(t)‖+supt∈[-r,0]‖ϕ(t)‖+supt∈[0,T]‖Sq(t)ϕ(0)‖:=α.
Therefore,
(3.13)‖f(t,vt+ϕ^t)‖≤μ(t)‖vt+ϕ^t‖[-r,0]≤αμ(t),
and
(3.14)‖a(v+ϕ^)(t)‖≤∫0t‖g(t,s,vs+ϕ^s)‖ds≤∫0tm(t,s)‖vs+ϕ^s‖[-r,0]ds≤m*α.
Since vn→v in 𝒞, it follows that there exists ε>0 such that ∥vn-v∥[0,T]≤ε for n sufficiently large. Moreover, noting that ∥vtn-vt∥[-r,0]≤∥vn-v∥[0,T], we have
(3.15)‖f(t,vtn+ϕ^t)-f(t,vt+ϕ^t)‖≤μ(t)‖vtn+ϕ^t‖[-r,0]+αμ(t)≤μ(t)‖vtn-vt‖[-r,0]+2αμ(t)≤(ε+2α)μ(t).
Similarly,
(3.16)‖g(t,s,vsn+ϕ^s)-g(t,s,vs+ϕ^s)‖≤(ε+2α)m(t,s),‖∫0tg(t,s,vsn+ϕ^s)ds-∫0tg(t,s,vs+ϕ^s)ds‖≤m*(ε+2α).
It follows from the Lebesgue's Dominated Convergence Theorem that
(3.17)‖∫0tg(t,s,vsn+ϕ^s)ds-∫0tg(t,s,vs+ϕ^s)ds‖⟶0, as n⟶∞,
and from (2.23), we have
(3.18)‖∫0t(t-s)q-1Pq(t-s)[f(s,vsn+ϕ^s)-f(s,vs+ϕ^s)]ds‖ +‖∫0t(t-s)q-1Pq(t-s)[∫0sg(s,τ,vτn+ϕ^τ)dτ-∫0sg(s,τ,vτ+ϕ^τ)dτ]ds‖ ≤M∫0t(t-s)-qγ-1[‖∫0sg(s,τ,vτn+ϕ^τ)dτ-∫0sg(s,τ,vτ+ϕ^τ)dτ‖‖f(s,vsn+ϕ^s)-f(s,vs+ϕ^s)‖ +‖∫0sg(s,τ,vτn+ϕ^τ)dτ-∫0sg(s,τ,vτ+ϕ^τ)dτ‖]ds⟶0, as n⟶∞.
Therefore, we obtain that
(3.19)limn→∞‖F^vn-F^v‖[0,T]=0,
then we see that ℱ^ is continuous.
Let us consider the MNC ν in the space 𝒞 with values in the cone R+3 of the following way: for every bounded subset Ω⊂𝒞,
(3.20)ν(Ω)=(supt∈[-r,0]χ(Ω(t)),Φ(Ω),modc(Ω)),
where mod c(Ω) is the module of equicontinuity of Ω given by
(3.21)modc(Ω)=limδ→0supv∈Ωmax|t1-t2|≤δ‖v(t1)-v(t2)‖,Φ(Ω)=supt∈[0,T](e-Ltsups∈[0,t]χ(Ω(s))),
where L>0 is a constant chosen so that
(3.22)Msupt∈[0,T]∫0t(t-s)-qγ-1η(s)e-L(t-s)ds=L1<1,(3.23)Mξ*supt∈[0,T]∫0t(t-s)-qγ-1e-L(t-s)ds=L2<1.
Noting that for any ψ∈L1([0,T],X), we have
(3.24)limL→+∞supt∈[0,T]∫0te-L(t-s)ψ(s)ds=0,
so, we can take the appropriate L to satisfy (3.22) and (3.23).
Next, we show that the operator ℱ^ is ν-condensing on every bounded subset of 𝒞.
Let Ω⊂𝒞 be a nonempty, bounded set for which
(3.25)ν(F^(Ω))≥ν(Ω).
Noting that
(3.26)supt∈[-r,0]χ(F^(Ω)(t))=0
and (3.25), we can see that sup t∈[-r,0] χ(Ω(t))=0.
Next, we estimate Φ(Ω). For any t∈[0,T], we set
(3.27)F^1(Ω)(t)={∫0t(t-s)q-1Pq(t-s)f(s,vs+ϕ^s)ds:v∈Ω}.
We consider the multifunction s∈[0,t]⊸H(s):
(3.28)H(s)={(t-s)q-1Pq(t-s)f(s,vs+ϕ^s):v∈Ω}.
Obviously, H is integrable, and from (2.23), (Hf)(1), and (3.13), it follows that H is integrably bounded. Moreover, noting that (Hf)(2), we have the following estimate for a.e. s∈[0,t]:
(3.29)χ(H(s))=χ({(t-s)q-1Pq(t-s)f(s,vs+ϕ^s):v∈Ω})=χ((t-s)q-1Pq(t-s)f(s,Ωs+ϕ^s))≤M(t-s)-qγ-1η(s)supτ∈[0,s]χ(Ω(τ))=M(t-s)-qγ-1η(s)eLse-Lssupτ∈[0,s]χ(Ω(τ))≤M(t-s)-qγ-1η(s)eLsΦ(Ω).
Applying Proposition 2.6, we have
(3.30)χ(F^1(Ω)(t))=χ(∫0tH(s)ds)≤M∫0t(t-s)-qγ-1η(s)eLsds⋅Φ(Ω).
Therefore, from (3.22), we have
(3.31)supt∈[0,T](e-Ltsups∈[0,t]χ(F^1(Ω)(s)))≤Msupt∈[0,T]∫0t(t-s)-qγ-1η(s)e-L(t-s)ds⋅Φ(Ω)=L1Φ(Ω).
Similarly, if we set
(3.32)F^2(Ω)(t)={∫0t(t-s)q-1Pq(t-s)a(v+ϕ^)(s)ds:v∈Ω},
then we can see that the multifunction s∈[0,t]⊸H~(s),
(3.33)H~(s)={(t-s)q-1Pq(t-s)a(v+ϕ^)(s):v∈Ω}
is integrable, and from (2.23), (Hg)(1), and (3.14), it follows that H~ is integrably bounded. Moreover, noting that (Hg)(2), Proposition 2.6, and (3.23), we have the following estimate for a.e. s∈[0,t]:
(3.34)χ(H~(s))≤Mξ*(t-s)-qγ-1eLsΦ(Ω),supt∈[0,T](e-Ltsups∈[0,t]χ(F^2(Ω)(s)))≤Mξ*supt∈[0,T]∫0t(t-s)-qγ-1e-L(t-s)ds⋅Φ(Ω)=L2Φ(Ω).
Now, from (3.31) and (3.34), L>0 can be chosen so that
(3.35)Φ(F^(Ω))≤(L1+L2)Φ(Ω)=L~Φ(Ω),
where 0<L~<1.
From (3.25), we have Φ(Ω)=0. Next, we will prove that mod c(Ω)=0.
Let δ>0, t1,t2∈(0,T] such that 0<t2-t1≤δ and v∈Ω, noting that (Hf)(1) and (Hg)(1), we obtain
(3.36)‖∫0t1(t1-s)q-1Pq(t1-s)[f(s,vs+ϕ^s)+a(v+ϕ^)(s)]ds -∫0t2(t2-s)q-1Pq(t2-s)[f(s,vs+ϕ^s)+a(v+ϕ^)(s)]ds‖ ≤‖∫0t1[(t2-s)q-1-(t1-s)q-1]Pq(t2-s)[f(s,vs+ϕ^s)+a(v+ϕ^)(s)]ds‖ +‖∫t1t2(t2-s)q-1Pq(t2-s)[f(s,vs+ϕ^s)+a(v+ϕ^)(s)]ds‖ +‖∫0t1(t1-s)q-1[Pq(t2-s)-Pq(t1-s)][f(s,vs+ϕ^s)+a(v+ϕ^)(s)]ds‖ =I1+I2+I3.
Using (2.23), (3.13), and (3.14), we have
(3.37)I1≤αM∫0t1|(t2-s)q-1-(t1-s)q-1|(t2-s)-q(1+γ)[μ(s)+m*]ds.
Clearly, I1 tends to zero as t2→t1. Similarly, for I2, we have
(3.38)I2≤αM∫t1t2(t2-s)-qγ-1[μ(s)+m*]ds⟶0, as t2⟶t1.
For I3, for ε>0 small enough, noting that (2.23), (3.13), and (3.14), we have
(3.39)I3≤α∫0t1-ε(t1-s)q-1‖Pq(t2-s)-Pq(t1-s)‖L(X)[μ(s)+m*]ds+α∫t1-εt1(t1-s)q-1‖Pq(t2-s)-Pq(t1-s)‖L(X)[μ(s)+m*]ds≤αsups∈[0,t1-ε]‖Pq(t2-s)-Pq(t1-s)‖L(X)⋅∫0t1-ε(t1-s)q-1[μ(s)+m*]ds+αM∫t1-εt1((t1-s)q-1(t1-s)q(γ+1)+(t1-s)q-1(t2-s)q(γ+1))[μ(s)+m*]ds,
it follows from Theorem 2.18 that I3 tends to zero as t2→t1 and ε→0.
For the case when 0=t1<t2≤T, we can see
(3.40)‖∫0t2(t2-s)q-1Pq(t2-s)[f(s,vs+ϕ^s)+a(v+ϕ^)(s)]ds‖ ≤αM∫0t2(t2-s)-qγ-1[μ(s)+m*]ds⟶0, as t2⟶0.
Thus, the set {(ℱ^v)(·):v∈Ω} is equicontinuous, then mod c(ℱ^Ω)=0. From (3.25), we get that mod c(Ω)=0. Hence ν(Ω)=(0,0,0).
The regularity property of ν implies the relative compactness of Ω. Now, it follows from Definition 2.7 that ℱ^ is ν-condensing.
Consider the set
(3.41)Bρ={v∈C:‖v‖[0,T]≤ρ}.
Next, we show that there exists some ρ>0 such that ℱ^Bρ⊂Bρ. Suppose on the contrary that for each ρ>0 there exist vρ(·)∈Bρ and some t∈[0,T] such that ∥(ℱ^vρ)(t)∥[0,T]>ρ.
Noting the Hölder inequality, we have
(3.42)∫0t(t-s)-1-qγμ(s)ds≤t(-(1+pqγ))/plp,q‖μ‖Lp≤T(-(1+pqγ))/plp,q‖μ‖Lp.
By (2.23), (Hf)(1), (Hg)(1), and (3.42), we have
(3.43)‖(F^vρ)(t)‖ =‖∫0t(t-s)q-1Pq(t-s)[f(s,vsρ+ϕ^s)+a(vρ+ϕ^)(s)]ds‖ ≤M[‖vρ‖[0,T]+M~]∫0t(t-s)-qγ-1[μ(s)+m*]ds ≤M(ρ+M~)⋅[lp,qT(-(1+pqγ))/p‖μ‖Lp+-m*T-qγqγ],
where M~=∥ϕ∥[-r,0]+sup t∈[0,T] ∥𝒮q(t)ϕ(0)∥.
Then,
(3.44)ρ<‖(F^vρ)(t)‖[0,T]≤M[ρ+M~]⋅[lp,qT-(1+pqγ)/p‖μ‖Lp+-m*T-qγqγ].
Dividing both sides of (3.44) by ρ, and taking ρ→∞, we have
(3.45)M(lp,qT-(1+pqγ)/p‖μ‖Lp+-m*T-qγqγ)≥1.
This contradicts (3.5). Hence for some positive number ρ, ℱ^Bρ⊂Bρ. According to Theorem 2.8, problem (1.1) has at least one mild solution.
Next, for c∈(0,1], we consider the following one-parameter family of maps:
(3.46)H:[0,1]×C⟶C(c,v)⟶H(c,v)=cF^(v).
We will demonstrate that the fixed point set of the family ℋ:
(3.47)FixH={v∈H(c,v) for some c∈(0,1]}
is a priori bounded. In fact, let v∈Fix ℋ, for t∈[0,T], we have
(3.48)‖v(t)‖≤∫0t‖(t-s)q-1Pq(t-s)[f(s,vs+ϕ^s)+a(v+ϕ^)(s)]‖ds≤M∫0t(t-s)-qγ-1[μ(s)+m*]⋅[supτ∈[0,s]‖v(τ)‖+M~]ds≤M[sups∈[0,t]‖v(s)‖∫0t(t-s)-qγ-1μ(s)ds+m*∫0t(t-s)-qγ-1supτ∈[0,s]‖v(τ)‖ds]+a1≤a1+a2sups∈[0,t]‖v(s)‖+a3∫0t(t-s)-qγ-1supτ∈[0,s]‖v(τ)‖ds,
where
(3.49)a1=MM~⋅(lp,qT-(1+pqγ)/p‖μ‖Lp+-m*T-qγqγ),a2=Mlp,qT-(1+pqγ)/p‖μ‖Lp,a3=Mm*.
We denote that x(t):=sup s∈[0,t] ∥v(s)∥. Let t~∈[0,t] such that x(t)=∥v(t~)∥. Then, by (3.48), we can see
(3.50)x(t)≤a1+a2x(t)+a3∫0t(t-s)-qγ-1x(s)ds.
By Lemma 2.10, there exists a constant κ such that
(3.51)x(t)≤a11-a2+κa1a3(1-a2)2∫0t(t-s)-qγ-1ds≤a11-a2-κa1a3T-qγqγ(1-a2)2:=ϖ.
Hence, sup t∈[0,T] ∥v(t)∥≤ϖ.
Now, we consider a closed ball as follows:(3.52)BR={v∈C:‖v‖[0,T]≤R}⊂C.
We take the radius R>0 large enough to contain the set Fix ℋ inside itself. Moreover, from the proof above-mentioned, ℱ^:BR→𝒞 is ν-condensing, and it remains to apply Theorem 2.9.