Proof.
Firstly we prove that T:P→P is continuous.
We will show that T:P→P is well defined and T(P)⊂P. For all u(t)∈P, by (H2), Φ(t) and f are nonnegative functions, and we have Tu(t)≥0. From (H1), (H2), we obtain
(A)A(Φf)=1Δ|T(∫0+∞G(τ,s)Φ(s)f(s,u(s),u′(s))ds)ρ-T(b(τ))-K(∫0+∞G(τ,s)Φ(s)f(s,u(s),u′(s))ds)K(b(τ))|≤maxy1∈[0,‖u‖],|y2|≤‖u‖Q(y1,y2)Δ|T(∫0+∞G(τ,s)Φ(s)q(s)ds)ρ-T(b(τ))-K(∫0+∞G(τ,s)Φ(s)q(s)ds)K(b(τ))|=A(Φq)maxy1∈[0,‖u‖],|y2|≤‖u‖Q(y1,y2).
In the same way, we have
(B)B(Φf)≤B(Φq)maxy1∈[0,‖u‖],|y2|≤‖u‖Q(y1,y2).
By Lemma 2.3(1), (A), (B), and (H1), for all u(t)∈P, we have
(3.4)(Tu)(t)ρ-1[1+a(t)b(t)]=∫0+∞G(t,s)ρ-1[1+a(t)b(t)]Φ(s)f(s,u(s),u′(s))ds+A(Φf)a(t)ρ-1[1+a(t)b(t)]+B(Φf)b(t)ρ-1[1+a(t)b(t)]≤∫0+∞Φ(s)f(s,u(s),u′(s))ds+A(Φf)ρ-1β2+B(Φf)ρ-1β1≤maxy1∈[0,‖u‖],|y2|≤‖u‖Q(y1,y2)(∫0+∞Φ(s)q(s)ds+A(Φq)ρ-1β2+B(Φq)ρ-1β1)<+∞,(3.5)|(Tu)′(t)|=1p(t)|∫0t-α2a(s)ρΦ(s)f(s,u(s),u′(s))ds1p(t) +∫t+∞α1b(s)ρΦ(s)f(s,u(s),u′(s))ds+A(Φf)α1-B(Φf)α2|≤supt∈[0,+∞)1p(t)maxy1∈[0,‖u‖],|y2|≤‖u‖Q(y1,y2)×(∫0+∞Φ(s)q(s)ds+A(Φq)α1+B(Φq)α2)<+∞.
Hence, T:P→P is well defined. By (3.1), (H1), the Lebesgue dominated convergence theorem and the continuity of p(t), for any u∈P, t1,t2∈R+, we have
(3.6)|(Tu)′(t1)-(Tu)′(t2)|≤α2a(∞)ρ|1p(t1)-1p(t2)|∫0t1Φ(s)f(s,u(s),u′(s))ds+α2a(∞)ρp(t2)∫t1t2Φ(s)f(s,x(s),x′(s))ds+α1b(0)ρ|1p(t1)-1p(t2)|∫0+∞Φ(s)f(s,u(s),u′(s))ds+α1b(0)ρp(t2)∫t1t2Φ(s)f(s,x(s),x′(s))ds+(A(Φf)α1+B(Φf)α2)|1p(t1)-1p(t2)|⟶0, as t1⟶t2.
That is, (Tu)(t)∈C1(R0+); therefore, (Tu)(t)∈E.
By Lemma 2.4, we have
(3.7)mint∈[a*,b*](Tu)(t)ρ-1[1+a(t)b(t)]=mint∈[a*,b*](∫0+∞G(t,s)ρ-1[1+a(t)b(t)]Φ(s)f(s,u(s),u′(s))ds=mint∈[a*,b*]+a(t)A(Φf)ρ-1[1+a(t)b(t)]+b(t)B(Φf)ρ-1[1+a(t)b(t)])≥c*(∫0+∞G(τ,s)ρ-1[1+a(τ)b(τ)]Φ(s)f(s,u(s),u′(s))dsc*=+a(τ)A(Φf)ρ-1[1+a(t)b(t)]+b(τ)B(Φf)ρ-1[1+a(t)b(t)])=c*(Tu)(τ)ρ-1[1+a(τ)b(τ)],
therefore T:P→P.
We show that T:P→P is continuous. In fact suppose {um}⊆P,u0∈P and um→u0(m→+∞), then there exists M>0, such that ∥um∥≤M. By (H1), we have
(3.8)∫0+∞Φ(s)|f(s,um(s),um′(s))-f(s,u0(s),u0′(s))|ds≤2∫0+∞Φ(s)f(s,u(s),u′(s))ds≤2maxy1∈[0,M],|y2|≤MQ(y1,y2)×∫0+∞Φ(s)q(s)ds<+∞.
Therefore, by Lemma 2.3(1), the continuity of f and the Lebesgue dominated convergence theorem imply that
(3.9)|(Tum)(t)-(Tu0)(t)ρ-1[1+a(t)b(t)]|=|∫0+∞G(t,s)ρ-1[1+a(t)b(t)]=×Φ(s)[f(s,um(s),um′(s))-f(s,u0(s),u0′(s))]ds∫0+∞|≤∫0+∞Φ(s)|f(s,um(s),um′(s))-f(s,u0(s),u0′(s))|ds⟶0,m⟶+∞,|(Tum)′(t)-(Tu0)′(t)|≤supt∈[0,+∞)1p(t)∫0+∞Φ(s)|f(s,um(s),um′(s))-f(s,u0(s),u0′(s))|ds⟶0, m⟶+∞.
Thus, ∥Tum-Tu0∥→0(m→+∞). Therefore T:P→P is continuous.
Secondly we show that T:P→P is compact operator.
For any bounded set B⊂P, there exists a constant L>0 such that ∥u∥≤L, for all u∈B. By Lemma 2.3(1), (A), (B), and (H1), we have
(3.10)(Tu)(t)=ρ-1[1+a(t)b(t)](Tu)(t)ρ-1[1+a(t)b(t)]≤ρ-1[1+a(∞)b(0)](∫0+∞G(t,s)ρ-1[1+a(t)b(t)]Φ(s)f(s,u(s),u′(s))ds+A(Φf)a(t)ρ-1[1+a(t)b(t)]+B(Φf)b(t)ρ-1[1+a(t)b(t)])≤ρ-1[1+a(∞)b(0)](∫0+∞Φ(s)f(s,u(s),u′(s))ds+A(Φf)ρ-1β2+B(Φf)ρ-1β1)≤ρ-1[1+a(∞)b(0)]maxy1∈[0,L],|y2|≤LQ(y1,y2)(∫0+∞Φ(s)q(s)ds+A(Φq)ρ-1β2+B(Φq)ρ-1β1)<+∞,(Tu)(∞)=∫0+∞G-(s)Φ(s)f(s,u(s),u′(s))ds+A(Φf)a(∞)+B(Φf)b(∞)=β2ρ∫0+∞a(s)Φ(s)f(s,u(s),u′(s))ds+A(Φf)a(∞)+B(Φf)b(∞)≤maxy1∈[0,L],|y2|≤LQ(y1,y2)(β2a(∞)ρ∫0+∞Φ(s)q(s)ds+A(Φq)a(∞)+B(Φq)b(∞))<+∞.
Therefore, (Tu)(t)⊆Cl(R+,R).
By (3.4) and (3.5), we have
(3.11)‖Tu‖1=supt∈[0,+∞)(Tu)(t)ρ-1[1+a(t)b(t)]≤maxy1∈[0,L],|y2|≤LQ(y1,y2)(∫0+∞Φ(s)q(s)ds+A(Φq)ρ-1β2+B(Φq)ρ-1β1)<+∞,‖(Tu)′‖∞=maxt∈[0,+∞)|(Tu)′(t)|≤supt∈[0,+∞)1p(t)maxy1∈[0,L],|y2|≤LQ(y1,y2)(∫0+∞Φ(s)q(s)ds+A(Φq)α1+B(Φq)α2)<+∞,
so TB is bounded.
Given T>0, t1, t2∈[0,T], by (H1) and Lemma 2.3(1), we have
(3.12)|G(t1,s)ρ-1[1+a(t1)b(t1)]-G(t2,s)ρ-1[1+a(t2)b(t2)]|Φ(s)f(s,u(s),u′(s))≤2maxy1∈[0,L],|y2|≤LQ(y1,y2)×Φ(s)q(s).
Therefore for any u∈B, by (3.1), the Lebesgue dominated convergence theorem and the continuity of G(t,s), a(t), and b(t), we have
(3.13)|(Tu)(t1)ρ-1[1+a(t1)b(t1)]-(Tx)(t2)ρ-1[1+a(t2)b(t2)]| ≤∫0+∞|G(t1,s)ρ-1[1+a(t1)b(t1)]-G(t2,s)ρ-1[1+a(t2)b(t2)]| ×Φ(s)f(s,u(s),u′(s))ds +A(Φf)|a(t1)ρ-1[1+a(t1)b(t1)]-a(t2)ρ-1[1+a(t2)b(t2)]| +B(Φf)|b(t1)ρ-1[1+a(t1)b(t1)]-b(t2)ρ-1[1+a(t2)b(t2)]| ⟶0, as t1⟶t2.
By a similar proof as (3.6), we obtain |(Tu)′(t1)-(Tu)′(t2)|→0, as t1→t2. Thus, TB is equicontinuous on [0,T]. Since T>0 is arbitrary, TB is locally equicontinuous on [0,+∞).
By Lemma 2.3(2), (H2) and the Lebesgue dominated convergence theorem, we obtain
(3.14)limt→+∞|(Tu)(t)ρ-1[1+a(t)b(t)]| =1ρ-1[1+a(∞)b(∞)]|∫0+∞β2a(s)Φ(s)f(s,x(s),x′(s))ds+A(Φf)a(∞)+B(Φf)b(∞)| ≤maxy1∈[0,L],|y2|≤LQ(y1,y2)ρ-1(1+β1β2)(β2a(∞)∫0+∞Φ(s)q(s)ds+A(Φq)a(∞)+B(Φq)b(∞)) <+∞,|(Tu)(t)ρ-1[1+a(t)b(t)]-(Tu)(∞)ρ-1[1+a(∞)b(∞)]| ≤∫0ta(s)b(t)|11+a(t)b(t)-11+a(∞)b(∞)|Φ(s)f(s,x(s),x′(s))ds +∫0ta(s)1+a(∞)b(∞)|b(t)-β2|Φ(s)f(s,x(s),x′(s))ds +∫t+∞b(s)|a(t)-a(s)|1+a(t)b(t)Φ(s)f(s,x(s),x′(s))ds +∫t+∞a(s)b(s)|11+a(t)b(t)-11+a(∞)b(∞)|Φ(s)f(s,x(s),x′(s))ds +∫t+∞a(s)1+a(∞)b(∞)|b(s)-β2|Φ(s)f(s,x(s),x′(s))ds+A(Φf)|a(t)-a(∞)|ρ-1[1+a(t)b(t)] +B(Φf)|b(t)-b(∞)|ρ-1[1+a(t)b(t)]+[A(Φf)a(∞)+B(Φf)b(∞)] ×|1ρ-1[1+a(t)b(t)]-1ρ-1[1+a(∞)b(∞)]| ≤maxy1∈[0,M],|y2|≤MQ(y1,y2) ×{b(0)a(∞)∫0t|11+a(t)b(t)-11+a(∞)b(∞)|Φ(s)q(s)ds +a(∞)1+a(∞)b(∞)∫0t|b(t)-β2|Φ(s)q(s)ds+b(0)1+β1β2∫t∞|a(t)-a(s)|Φ(s)q(s)ds +a(∞)b(0)∫t∞|11+a(t)b(t)-11+a(∞)b(∞)|Φ(s)q(s)ds +a(∞)1+a(∞)b(∞)∫t+∞|b(s)-β2|Φ(s)q(s)ds+A(Φq)|a(t)-a(∞)|ρ-1[1+a(t)b(t)] +B(Φq)|b(t)-b(∞)|ρ-1[1+a(t)b(t)]+(A(Φq)a(∞)+B(Φq)b(∞)) ×|1ρ-1[1+a(t)b(t)]-1ρ-1[1+a(∞)b(∞)]|}⟶0, as t⟶+∞.
By (3.5), we know that lim t→+∞|(Tu)′(t)|<+∞, then
(3.15)|(Tu)′(t)-(Tu)′(∞)| =|1p(t)∫0t-α2a(s)ρΦ(s)f(s,u(s),u′(s))ds+1p(t)∫t+∞α1b(s)ρΦ(s)f(s,u(s),u′(s))ds +1p(t)A(Φf)α1-1p(t)B(Φf)α2+1p(∞)∫0tα2a(s)ρΦ(s)f(s,u(s),u′(s))ds +1p(∞)∫t+∞α2a(s)ρΦ(s)f(s,u(s),u′(s))ds-1p(∞)A(Φf)α1+1p(∞)B(Φf)α2| ≤maxy1∈[0,L],|y2|≤LQ(y1,y2)|1p(t)-1p(∞)| ×{∫0tα2a(s)ρΦ(s)q(s)ds+1p(t)∫t+∞α1b(s)ρΦ(s)q(s)ds+1p(∞) ××∫t+∞α2a(s)ρΦ(s)q(s)ds+(A(Φq)α1+B(Φq)α2)|1p(t)-1p(∞)|}⟶0, as t⟶+∞.
Therefore, TB is equiconvergent at ∞. By Lemma 2.5, TB is completely continuous.
Finally we will show that all conditions of Theorem 2.6 hold.
From the definition of α, we can get α(u)≤∥u∥ for all u∈P. For all u∈Pr2-, we have ∥u∥≤r2; therefore 0≤y1≤r2, |y2|≤r2. By (3.4), (3.5), and (H3), we have(3.16)|(Tu)(t)|ρ-1[1+a(t)b(t)]≤maxy1∈[0,r2],|y2|≤r2Q(y1,y2)(∫0+∞Φ(s)q(s)ds+A(Φq)ρ-1β2+B(Φq)ρ-1β1)≤r2,|(Tu)′(t)|≤supt∈[0,+∞)1p(t)maxy1∈[0,r2],|y2|≤r2Q(y1,y2)×(∫0+∞Φ(s)q(s)ds+A(Φq)α1+B(Φq)α2)≤r2,
that is, ∥Tu∥≤r2 for u∈Pr2-. Thus T:Pr2-→Pr2-.
Similarly for any u∈Pr1-, we have ∥Tu∥<r1, which means that condition (c2) of Theorem 2.6 holds.
In order to apply condition (c1) of Theorem 2.6, we choose u(t)=b1ρ-1[1+a(t)b(t)]/c*, t∈R0+, then ∥u∥≤l1; this is because
(3.17)‖u‖1=b1c*≤l1, ‖u′‖∞=supt∈[0,+∞)|u′(t)|=supt∈[0,+∞)|b1ρ-1[a′(t)b(t)+a(t)b′(t)]c*|≤supt∈[0,+∞)1p(t)b1c*≤l1,
and α(u)=min t∈[a*,b*](u(t)/ρ-1[1+a(t)b(t)])=b1/c*>b1, which means that {u∈P(α,b1,l1)|α(u)>b1}≠ϕ. For all u∈P(α,b1,l1), we have α(u)≥b1 and ∥u∥≤l1, thus b1≤u(t)/ρ-1[1+a(t)b(t)]≤l1,|u′(t)|≤l1, that is, b1≤y1≤l1,|y2|≤l1. By (H4), we can get
(3.18)α(Tu(t))=mint∈[a*,b*](Tu)(t)ρ-1[1+a(t)b(t)]≥mint∈[a*,b*]1ρ-1[1+a(t)b(t)]×(∫0a*a(s)b(t)ρΦ(s)f(s,u(s),u′(s))ds=××+∫a*ta(s)b(t)ρΦ(s)f(s,u(s),u′(s))ds+∫tb*a(t)b(s)ρΦ(s)f(s,u(s),u′(s))ds=××+∫b*+∞a(t)b(s)ρΦ(s)f(s,u(s),u′(s))ds)>a(a*)b(b*)1+a(b*)b(a*)∫a*b*Φ(s)f(s,u(s),u′(s))ds>a(a*)b(b*)1+a(b*)b(a*)∫a*b*Φ(s)dsb1δ≥b1.
Consequently condition (c1) of Theorem 2.6 holds.
We will prove that condition (c3) of Theorem 2.6 holds. If u∈P(α,b1,r2), and ∥Tu(t)∥>l1, by (H4), we have
(3.19)α(Tu(t))=mint∈[a*,b*](Tu)(t)ρ-1[1+a(t)b(t)]>a(a*)b(b*)1+a(b*)b(a*)∫a*b*Φ(s)dsb1δ≥b1.
Therefore, condition (c3) of Theorem 2.6 is satisfied. Then we can complete the proof of this theorem by Leggett-Williams fixed point theorem.