# Existence Results for a Coupled System of Nonlinear Singular Fractional Differential Equations with Impulse Effects

Authors: Yuji Liu; Juan J. Nieto; Óscar Otero-Zarraquiños
Journal: Mathematical Problems in Engineering
 (2013)

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## Abstract

A boundary value problem for the singular fractional differential system with impulse effects is presented. By applying Schauder's fixed point theorem in a suitably Banach space, we obtain the existence of at least one solution for this problem. Two examples are presented to illustrate the main theorem.


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## Body

## 1. Introduction

Fractional differential equations have received increasing attention during recent years since the behavior of many physical, chemical, and engineering processes can be properly described by using fractional differential equations theory; see the books [1–3], papers [4, 5] and references therein. For details on the geometric and physical interpretation of the derivatives of noninteger order, see, for example, [6–11]. For some recent works with applications to engineering we refer the reader to [12–15].For an introduction of the basic theory of impulsive differential equation, we refer the reader to [16]. Among previous research, little is concerned with differential equations with fractional order with impulses [17]. Ahmad and Sivasundaram [18, 19] gave some existence results for two-point boundary value problems involving nonlinear impulsive hybrid differential equations of fractional order 1<α≤2. Ahmad and Nieto in [20] establish sufficient conditions for the existence of solutions of the antiperiodic boundary value problem for impulsive differential equations with the Caputo derivative of order q∈(1,2]. Some recent results on impulsive initial value problems or boundary value problems for fractional differential equations on a finite interval can be found in [21–23] and references therein. The memory property of fractional calculus makes studies more complicated.This paper is motivated by [24] in which the following boundary value problem for the fractional differential equation
(1)D0+αx(t)=f(t,y(t),D0+py(t)),t∈(0,1),D0+βy(t)=g(t,x(t),D0+qx(t)),t∈(0,1),x(0)=0,y(0)=0,x(1)-γx(η)=0,y(1)-γy(η)=0
was studied, where 1<α, β<2, 0<p≤β-1 and 0<q≤α-1, γ>0, 1>γηα-1, 1>γηβ-1 and f,g:[0,1]×R2→R are continuous functions, and D0+ is the Riemann-Liouville fractional derivative. An existence result was proved for BVP (1) in [24]. The growth assumptions imposed on f and g are sublinear cases (see [24, Theorem 2.1]); that is, there exist functions a,b∈L1(0,1), nonnegative constants ϵ1,ϵ2>0, δ1,δ2≥0 and ρ1,ρ2,σ1,σ2∈(0,1) such that
(2)|f(t,x,y)|≤a(t)+ϵ1|x|ρ1+ϵ2|y|ρ2,|g(t,x,y)|≤b(t)+δ1|x|σ1+δ2|y|σ2.In [25], the following boundary value problem for the fractional differential equation
(3)D0+αx(t)=f(t,y(t),D0+py(t)),t∈(0,1),D0+βy(t)=g(t,x(t),D0+qx(t)),t∈(0,1),x(0)=0,y(0)=0,x(1)=0,y(1)=0
was studied, where 1<α, β<2, 0<p≤β-1 and 0<q≤α-1, and f,g:[0,1]×R2→R are continuous functions, and D0+ is the Riemann-Liouville fractional derivative. The growth assumptions imposed on f and g are sublinear cases (see [25, Theorem 3.1]), that is, there exist functions a,b∈L1(0,1), nonnegative constants ϵ1,ϵ2>0, δ1,δ2≥0, and ρ1,ρ2,σ1,σ2∈(0,1] such that
(4)|f(t,x,y)|≤a(t)+ϵ1|x|ρ1+ϵ2|y|ρ2,|g(t,x,y)|≤b(t)+δ1|x|σ1+δ2|y|σ2,or sublinear cases, that is, there exist nonnegative constants ϵ1,ϵ2>0, δ1,δ2≥0 and ρ1,ρ2,σ1,σ2∈(1,∞) such that
(5)|f(t,x,y)|≤ϵ1|x|ρ1+ϵ2|y|ρ2,|g(t,x,y)|≤δ1|x|σ1+δ2|y|σ2.
We find that in the superlinear cases, BVP (3) has a pair of solutions (x,y)=(0,0) without needing any other assumptions. Hence, these cases are trivial ones discussed in [25].It is interesting to consider the solvability of BVP (1) when the growth assumptions imposed on f,g are superlinear cases. Furthermore, the solvability of BVP (1) is not studied when q>α-1 or p>β-1.In this paper we consider the following nonlinear boundary value problem for the singular multiterm fractional differential equation with impulse effects whose boundary conditions are of integral form(6)D0+αx(t)=ϕ(t)f(t,y(t),D0+py(t)),mmmmmmmmimt∈(0,1),t≠t1,D0+βy(t)=ψ(t)g(t,x(t),D0+qx(t)),mmmmmmmmimit∈(0,1),t≠t1,limt→0t2-αx(t)=∫01‍u(s)G(s,y(s),D0+py(s))ds,limt→0t2-βy(t)=∫01‍v(s)H(s,x(s),D0+qx(s))ds,x(1)=∫01‍m(s)M(s,y(s),D0+py(s))ds,(7)y(1)=∫01‍n(s)N(s,x(s),D0+qx(s))ds,Δx(t1)=limt→t1+x(t)-limt→t1-x(t)=I(t1,y(t1),D0+py(t1)),Δy(t1)=limt→t1+y(t)-limt→t1-y(t)=J(t1,x(t1),D0+qx(t1)),ΔD0+qx(t1)=limt→t1+D0+qx(t)-limt→t1-D0+qx(t)=I1(t1,y(t1),D0+py(t1)),ΔD0+py(t1)=limt→t1+D0+py(t)-limt→t1-D0+py(t)=J1(t1,x(t1),D0+qx(t1)),
where(a)
1<α, β≤2, 0<p<β and 0<q<α, D0+ is the Riemann-Liouville fractional derivative,(b)
ϕ,ψ:(0,1)→R,f,g defined on (0,1)×R2,(c)
m,n,u,v:(0,1)→R with m,n,u,v∈L1(0,1), G,H,M,N defined on (0,1)×R2,(d)
0=t0<t1<t2=1,(e)
I,I1,J,J1:(0,1)×R2→R.A pair of functions(x,y) defined on (0,1) is called a solution of BVP (1) and BVP (3), if x|(tk,tk+1], D0+qx|(tk,tk+1] and y|(tk,tk+1], D0+py|(tk,tk+1](k=0,1) are continuous, there exists the limits
(8)limt→tk+t2-αx(t),limt→tk+t2-βy(t),limt→tk+t2+q-αD0+qx(t),limt→tk+t2+p-βD0+py(t),mmmmmmmmmmmmmmmmmmk=0,1,D0+αx, D0+βy∈L1(0,1) and (x,y) satisfies all equations in (6) and (7).The novelty of this paper is as follows: first, the fractional differential equations in (6) are multiterm ones and their nonlinearities f,g depend on the lower fractional derivatives; second, both ϕ and ψ may be singular at t=0 and t=1, that is, ϕ(t)f(t,x,y) and ψ(t)g(t,x,y) may be not continuous functions on [0,1]×R2, the boundary conditions are integral boundary conditions, and we obtain the results on the existence of at least one solution of BVP (6)-(7); third, 0<p<β and 0<q<α are supposed; the growth assumptions imposed on f,g,G,H,M,N and I,I1,J,J1 are allowed to be sublinear cases. Finally, two examples are given to illustrate the efficiency of the main theorem.The remainder of this paper is as follows: in Section2, we present preliminary results. In Section 3, the main theorem and its proof are given. In Section 4, two examples are given to illustrate the main results.


## 2. Preliminaries

In this section, we present some background definitions and preliminary results.Definition 1 (see [1]).
The Riemann-Liouville fractional integral of orderα>0 of a function g:(0,∞)→R is given by
(9)I0+αg(t)=1Γ(α)∫0t‍(t-s)α-1g(s)ds,
provided that the right-hand side exists.Definition 2 (see [1]).
The Riemann-Liouville fractional derivative of orderα>0 of a continuous function g:(0,∞)→R is given by
(10)D0+αg(t)=1Γ(n-α)dndtn∫0t‍g(s)(t-s)α-n+1ds,
where n-1≤α<n, provided that the right-hand side is pointwise defined on (0,∞).Definition 3.
K
:
(
0,1
)
×
R
2
→
R is called a β-Caratheodory function if K satisfies that(i)
t→K(t,tβ-2U,tβ-p-2V) is continuous on (tk,tk+1](k=0,1) for every (U,V)∈R2;(ii)
(U,V)→K(t,tβ-2U,tβ-p-2V) is continuous on R2 for every t∈(0,1);(iii)
for eachr>0 there exists a constant Ar>0 such that |K(t,tβ-2U,tβ-p-2V)|≤Ar, t∈(0,1), |U|,|V|≤r.Definition 4.
Q
:
(
0,1
)
×
R
2
→
R is called a α-Caratheodory function if Q satisfies that(i)
t→Q(t,tα-2U,tα-q-2V) is continuous on (tk,tk+1](k=0,1) for every (U,V)∈R2;(ii)
(U,V)→Q(t,tα-2U,tα-q-2V) is continuous on R2 for every t∈(0,1);(iii)
for eachr>0 there exists a constant Br>0 such that |Q(t,tα-2U,tα-q-2V)|≤Br, t∈(0,1), |U|,|V|≤r.Lemma 5 (the Leray-Schauder nonlinear alternative [23]).
LetX be a Banach space and T:X→X be a completely continuous operator. Suppose Ω is a nonempty open subset of X centered at zero. Then either there exists x∈∂Ω and λ∈(0,1) such that x=λTx or there exists x∈Ω¯ such that x=Tx.Let the gamma and beta functionsΓ(α) and B(p,q) be defined by
(11)Γ(α)=∫0+∞‍xα-1e-xdx,B(p,q)=∫01‍xp-1(1-x)q-1dx,∥m∥1=∫01‍|m(s)|dsform∈L1(0,1).
Choose
(12)X={x|(tk,tk+1]∈C0(tk,tk+1](k=0,1),D0+qx|(tk,tk+1]∈C0(tk,tk+1](k=0,1),x:(0,1]⟶Rthere  exist  the  limitslimt→tk+t2-αx(t),limt→tk+t2+q-αD0+qx(t)},Y={y|(tk,tk+1]∈C0(tk,tk+1](k=0,1),D0+px|(tk,tk+1]∈C0(tk,tk+1](k=0,1),y:(0,1]⟶Rthere  exist  the  limitslimt→tk+t2-βy(t),limt→tk+t2+p-βD0+py(t)}.
For x∈X, define the norm by
(13)∥x∥=∥x∥X=max{supt∈(0,1)t2-α|x(t)|,supt∈(0,1)t2+q-α|D0+qx(t)|}.
It is easy to show that X is a real Banach space. For y∈Y, define the norm by
(14)∥y∥=∥y∥Y=max{supt∈(0,1)t2-β|y(t)|,supt∈(0,1)t2+p-β|D0+py(t)|}.
It is easy to show that Y is a real Banach space. Thus, (X×Y,||·||) is a Banach space with the norm defined by ||(x,y)||=max{||x||X,||y||Y}for(x,y)∈X×Y.In this paper, we suppose the following:(A)
ϕ satisfies that there exist constants L1>0, k>-1, δ∈(q-α,0] such that α+2δ-q>0, α+k+δ-q≥0, and |ϕ(t)|≤L1tk(1-t)δ for all t∈(0,1); ψ satisfies that there exist constants L2>0, l>-1, θ∈(p-β,0] such that β+2θ-p>0, β+l+θ-p≥0, and |ψ(t)|≤L2tl(1-t)θ for all t∈(0,1).(B)
f,G,M,I,I1 are β-Caratheodory functions and g,H,N,J,J1 are α-Caratheodory functions.Remark 6.
Suppose thatf is a β-Caratheodory function. For example, α=7/4, q=1/8, choose k=-1/2, δ=-3/4 and ϕ(t)=tk(1-t)δ, then k>-1, δ∈(-α,0] such that α+2δ-q>0, α+k+δ-q≥0, and |ϕ(t)|≤tk(1-t)δ for all t∈(0,1). It is easy to see that ϕ is singular at t=0 and t=1.Lemma 7.
Suppose thaty∈Y, and (a)–(e), (A)-(B) hold. Then x∈X is a solution of
(15)D0+αx(t)=ϕ(t)f(t,y(t),D0+py(t)),t∈(0,1),t≠t1,limt→0t2-αx(t)=∫01‍u(s)G(s,y(s),D0+py(s))ds,x(1)=∫01‍m(s)M(s,y(s),D0+py(s))ds,Δx(t1)=I(t1,y(t1),D0+py(t1)),ΔD0+qx(t1)=I1(t1,y(t1),D0+py(t1)),
if and only if x∈X satisfies the integral equation
(16)x(t)={∫0t(t-s)α-1Γ(α)ϕ(u)f(u,y(u),D0+py(u))du-tα-1Γ(α)×∫01(1-s)α-1ϕ(s)f(s,y(s),D0+py(s))ds+tα-2∫01u(s)G(s,y(s),D0+py(s))ds+tα-1∫01m(s)M(s,y(s),D0+py(s))ds+tα-1Π×(Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2)×I(t1,y(t1),D0+py(t1))+tα-1(t1α-2-t1α-1)Π×I1(t1,y(t1),D0+py(t1)),t∈(0,t1],∫0t(t-s)α-1Γ(α)ϕ(s)f(s,y(s),D0+py(s))ds-tα-1Γ(α)×∫01(1-s)α-1ϕ(s)f(s,y(s),D0+py(s))ds+(tα-2-tα-1)×∫01‍u(s)G(s,y(s),D0+py(s))ds+tα-1∫01‍m(s)M(s,y(s),D0+py(s))ds+tα-1-tα-2ΠΓ(α)Γ(α-q)t1α-q-1×I(t1,y(t1),D0+py(t1))+tα-2-tα-1Πt1α-1I1(t1,y(t1),D0+py(t1)),mmmmmmmmmimmmmmmmt∈(t1,1],
where
(17)Π=(Γ(α-1)Γ(α-q-1)-Γ(α)Γ(α-q))t12α-q-3.Proof.
Ify∈Y is a solution of BVP (15), then
(18)∥y∥=max{supt∈(0,1)t2-β|y(t)|,supt∈(0,1)t2+p-β|D0+py(t)|}=r<+∞,
and x satisfies all equations in (31) From (B), f is a β-Caratheodory function, then there exists Ar>0 such that
(19)|f(t,y(t),D0+py(t))|=|f(t,tβ-2t2-βy(t),tβ-p-2t2+p-βD0+py(t))|≤Ar.
Similarly we get that there exist constants Ar′,Ar′′,Br′,Br′′>0 such that
(20)|G(t,y(t),D0+py(t))|≤Ar′,|M(t,y(t),D0+py(t))|≤Ar′′,mmmmmmmmmmt∈(0,1),|I(t1,y(t1),D0+py(t1))|≤Br′,|I1(t1,y(t1),D0+py(t1))|≤Br′′.
It follows from (15) that, for t∈(tk,tk+1](k=0,1), there exist constants ck,dk∈R such that
(21)x(t)=1Γ(α)∫0t‍(t-s)α-1ϕ(s)f(s,y(s),D0+py(s))ds+cktα-1+dktα-2,t∈(tk,tk+1],k=0,1.
Fromlimt→0t2-αx(t)=∫01‍u(s)G(s,y(s),D0+py(s))ds, we get
(22)d0=∫01‍u(s)G(s,y(s),D0+py(s))ds.
Fromx(1)=∫01‍m(s)M(s,y(s),D0+py(s))ds, we get
(23)1Γ(α)∫01‍(1-s)α-1ϕ(s)f(s,y(s),D0+py(s))ds+c1+d1=∫01‍m(s)M(s,y(s),D0+py(s))ds.
FromΔx(t1)=I(t1,y(t1),D0+py(t1)), we get
(24)(c1-c0)t1α-1+(d1-d0)t1α-2=I(t1,y(t1),D0+py(t1)).
FromΔD0+qx(t1)=I1(t1,y(t1),D0+py(t1)), we get
(25)(c1-c0)Γ(α)Γ(α-q)t1α-q-1+(d1-d0)Γ(α-1)Γ(α-q-2)t1α-q-2=I1(t1,y(t1),D0+py(t1)).
It follows that(26)c1-c0=(Γ(α-1)Γ(α-q-2)t1α-q-2I(t1,y(t1),D0+py(t1))-t1α-2I1(t1,y(t1),D0+py(t1)){Γ(α-1)Γ(α-q-2)})×(Π)-1,d1-d0=({Γ(α)Γ(α-q)}t1α-1I1(t1,y(t1),D0+py(t1))-Γ(α)Γ(α-q)t1α-q-1I(t1,y(t1),D0+py(t1)))×(Π)-1.
Then
(27)d1=({Γ(α)Γ(α-q)}t1α-1I1(t1,y(t1),D0+py(t1))-Γ(α)Γ(α-q)t1α-q-1I(t1,y(t1),D0+py(t1)))×(Π)-1+∫01‍u(s)G(s,y(s),D0+py(s))ds.
So
(28)c1=∫01‍m(s)M(s,y(s),D0+py(s))ds-1Γ(α)∫01‍(1-s)α-1ϕ(s)f(s,y(s),D0+py(s))ds-({Γ(α)Γ(α-q)}t1α-1I1(t1,y(t1),D0+py(t1))-Γ(α)Γ(α-q)t1α-q-1I(t1,y(t1),D0+py(t1)))×(Π)-1-∫01‍u(s)G(s,y(s),D0+py(s))ds,c0=∫01‍m(s)M(s,y(s),D0+py(s))ds-1Γ(α)∫01‍(1-s)α-1ϕ(s)f(s,y(s),D0+py(s))ds-({Γ(α)Γ(α-q)}t1α-1I1(t1,y(t1),D0+py(t1))-Γ(α)Γ(α-q)t1α-q-1I(t1,y(t1),D0+py(t1)))×(Π)-1-∫01‍u(s)G(s,y(s),D0+py(s))ds-(Γ(α-1)Γ(α-q-2)t1α-q-2I(t1,y(t1),D0+py(t1))-t1α-2I1(t1,y(t1),D0+py(t1)){Γ(α-1)Γ(α-q-2)})×(Π)-1.
Hence, fort∈(0,t1], we have
(29)x(t)=∫0t‍(t-s)α-1Γ(α)ϕ(u)f(u,y(u),D0+py(u))du-tα-1Γ(α)∫01‍(1-s)α-1ϕ(s)f(s,y(s),D0+py(s))ds+tα-2∫01‍u(s)G(s,y(s),D0+py(s))ds+tα-1∫01‍m(s)M(s,y(s),D0+py(s))ds+tα-1Π(Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2)×I(t1,y(t1),D0+py(t1))+tα-1(t1α-2-t1α-1)ΠI1(t1,y(t1),D0+py(t1)).
And for t∈(t1,1], we have
(30)x(t)=∫0t‍(t-s)α-1Γ(α)ϕ(s)f(s,y(s),D0+py(s))ds-tα-1Γ(α)∫01(1-s)α-1ϕ(s)f(s,y(s),D0+py(s))ds+(tα-2-tα-1)∫01‍u(s)G(s,y(s),D0+py(s))ds+tα-1∫01‍m(s)M(s,y(s),D0+py(s))ds+tα-1-tα-2ΠΓ(α)Γ(α-q)t1α-q-1×I(t1,y(t1),D0+py(t1))+tα-2-tα-1Πt1α-1I1(t1,y(t1),D0+py(t1)).
Hence, x∈X satisfies (16).
On the other hand, ify∈Y and x∈X is a solution of (16), then we can prove that x∈X is a solution of BVP (6)-(7). The proof is completed.Lemma 8.
Suppose thatx∈X, and (a)–(e), (A)-(B) hold. Then y∈Y is a solution of
(31)D0+βy(t)=ψ(t)g(t,x(t),D0+qx(t)),t∈(0,1),t≠t1,limt→0t2-βy(t)=∫01‍v(s)H(s,x(s),D0+qx(s))ds,y(1)=∫01‍n(s)N(s,x(s),D0+qx(s))ds,Δy(t1)=J(t1,x(t1),D0+qx(t1)),ΔD0+py(t1)=J1(t1,x(t1),D0+qx(t1)),
if and only if y∈Y satisfies the integral equation
(32)y(t)={∫0t‍(t-s)β-1Γ(β)ψ(u)g(u,x(u),D0+qx(u))du-tβ-1Γ(β)×∫01‍(1-s)β-1ψ(s)g(s,x(s),D0+qx(s))ds+tβ-2×∫01‍v(s)H(s,y(s),D0+py(s))ds+tβ-1×∫01‍n(s)N(s,x(s),D0+qx(s))ds+tβ-1Ξ×(Γ(β)Γ(β-p)t1β-p-1-Γ(β-1)Γ(β-p-2)t1β-p-2)×J(t1,x(t1),D0+qx(t1))+tβ-1(t1β-2-t1β-1)Ξ×J1(t1,x(t1),D0+qx(t1)),t∈(0,t1],∫0t‍(t-s)β-1Γ(β)ψ(s)g(s,x(s),D0+qx(s))ds-tβ-1Γ(β)×∫01‍(1-s)β-1ψ(s)g(s,x(s),D0+qx(s))ds+(tβ-2-tβ-1)×∫01‍v(s)H(s,y(s),D0+py(s))ds+tβ-1×∫01‍n(s)N(s,x(s),D0+qx(s))ds+tβ-1-tβ-2ΞΓ(β)Γ(β-p)t1β-p-1×J(t1,x(t1),D0+qx(t1))+tβ-2-tβ-1Ξt1β-1×J1(t1,x(t1),D0+qx(t1)),t∈(t1,1],
where
(33)Ξ=(Γ(β-1)Γ(β-p-1)-Γ(β)Γ(β-p))t12β-p-3.Proof.
The proof is similar to that of the proof of Lemma7 and is omitted.
Now, we define the operatorT on X×Y by T(x,y)(t)=((T1y)(t),(T2x)(t)) with
(34)(T1y)(t)={∫0t‍(t-s)α-1Γ(α)ϕ(u)f(u,y(u),D0+py(u))du-tα-1Γ(α)∫01‍(1-s)α-1ϕ(s)f(s,y(s),D0+py(s))ds+tα-2∫01‍u(s)G(s,y(s),D0+py(s))ds+tα-1∫01‍m(s)M(s,y(s),D0+py(s))ds+tα-1Π(Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2)×I(t1,y(t1),D0+py(t1))+tα-1(t1α-2-t1α-1)Π×I1(t1,y(t1),D0+py(t1)),t∈(0,t1],∫0t‍(t-s)α-1Γ(α)ϕ(s)f(s,y(s),D0+py(s))ds-tα-1Γ(α)∫01(1-s)α-1ϕ(s)f(s,y(s),D0+py(s))ds+(tα-2-tα-1)∫01‍u(s)G(s,y(s),D0+py(s))ds+tα-1∫01‍m(s)M(s,y(s),D0+py(s))ds+tα-1-tα-2ΠΓ(α)Γ(α-q)t1α-q-1I(t1,y(t1),D0+py(t1))+tα-2-tα-1Πt1α-1I1(t1,y(t1),D0+py(t1)),mmmmmmmmmmmmmmmmt∈(t1,1],(35)(T2x)(t)={∫0t‍(t-s)β-1Γ(β)ψ(u)g(u,x(u),D0+qx(u))du-tβ-1Γ(β)∫01‍(1-s)β-1ψ(s)g(s,x(s),D0+qx(s))ds+tβ-2∫01‍v(s)H(s,y(s),D0+py(s))ds+tβ-1∫01‍n(s)N(s,x(s),D0+qx(s))ds+tβ-1Ξ(Γ(β)Γ(β-p)t1β-p-1-Γ(β-1)Γ(β-p-2)t1β-p-2)×J(t1,x(t1),D0+qx(t1))+tβ-1(t1β-2-t1β-1)ΞJ1(t1,x(t1),D0+qx(t1)),mmmmmmmmmmmmmmmmmt∈(0,t1],∫0t‍(t-s)β-1Γ(β)ψ(s)g(s,x(s),D0+qx(s))ds-tβ-1Γ(β)∫01‍(1-s)β-1ψ(s)g(s,x(s),D0+qx(s))ds+(tβ-2-tβ-1)∫01‍v(s)H(s,y(s),D0+py(s))ds+tβ-1∫01‍n(s)N(s,x(s),D0+qx(s))ds+tβ-1-tβ-2ΞΓ(β)Γ(β-p)t1β-p-1×J(t1,x(t1),D0+qx(t1))+tβ-2-tβ-1Ξt1β-1J1(t1,x(t1),D0+qx(t1)),+t1β-1J1(t1,x(t1)mmmmm)t∈(t1,1].Remark 9.
By Lemmas7 and 8, (x,y)∈X×Y is a solution of BVP (6)-(7) if and only if (x,y)∈X×Y is a fixed point of the operator T.Lemma 10.
Suppose that (a)–(e) and (A)-(B) hold. ThenT:X×Y→X×Y is well defined and is completely continuous.Proof.
The proof is very long, so we list the steps. First, we prove thatT is well defined; second, we prove that T is continuous, and, finally, we prove that T is compact. So T is completely continuous. Thus, the proof is divided into three steps.
Step  1. Prove that T:X×Y→X×Y is well defined.
For(x,y)∈X×Y, we have ||(x,y)||=r>0. Then
(36)max{supt∈(0,1)t2-α|x(t)|,supt∈(0,1)t2+q-α|D0+qx(t)|}≤r<+∞,max{supt∈(0,1)t2-β|y(t)|,supt∈(0,1)t2+p-β|D0+py(t)|}≤r<+∞.
From (B), f,G,M,I,I1 are β-Caratheodory functions, then there exist constants Ar>0 such that
(37)|f(t,y(t),D0+py(t))|≤Ar,t∈(0,1),|G(t,y(t),D0+py(t))|≤Ar,t∈(0,1),|M(t,y(t),D0+py(t))|≤Ar,t∈(0,1),|I(t1,y(t1),D0+py(t1))|≤Ar,t∈(0,1),|I1(t1,y(t1),D0+py(t1))|≤Ar,t∈(0,1).
Hence,
(38)|∫0t‍(t-s)α-1Γ(α)ϕ(u)f(u,y(u),D0+py(u))du|≤∫0t‍(t-s)α-1Γ(α)|ϕ(u)f(u,y(u),D0+py(u))|du≤ArL1B(α+δ,k+1)Γ(α)<∞.
From (34), (37), and (38), we see that (T1y)(t) is defined on (0,1], continuous on (0,t1] and (t1,1], respectively. One sees that
(39)limt→0t2-α(T1y)(t)=limt→0[t2-α∫0t‍(t-s)α-1Γ(α)ϕ(u)f(u,y(u),D0+py(u))du-tΓ(α)∫01‍(1-s)α-1ϕ(s)f(s,y(s),D0+py(s))ds+∫01‍u(s)G(s,y(s),D0+py(s))ds+t∫01‍m(s)M(s,y(s),D0+py(s))ds+tΠ(Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2)×I(t1,y(t1),D0+py(t1))+t(t1α-2-t1α-1)ΠI1(t1,y(t1),D0+py(t1))(t-s)α-1Γ(α)]=∫01‍u(s)G(s,y(s),D0+py(s))ds,
and there exits the limit limt→t1+(T1y)(t).
On the other hand, we have(40)D0+q(T1y)(t)={∫0t‍(t-s)α-q-1Γ(α-q)ϕ(u)f(u,y(u),D0+py(u))du-tα-q-1Γ(α-q)×∫01‍(1-s)α-1ϕ(s)f(s,y(s),D0+py(s))ds+tα-q-2Γ(α-1)Γ(α-q-1)×∫01‍u(s)G(s,y(s),D0+py(s))ds+tα-q-1Γ(α)Γ(α-q)×∫01‍m(s)M(s,y(s),D0+py(s))ds+tα-q-1ΠΓ(α)Γ(α-q)×(Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2)×I(t1,y(t1),D0+py(t1))+Γ(α)Γ(α-q)tα-q-1(t1α-2-t1α-1)Π×I1(t1,y(t1),D0+py(t1)),t∈(0,t1],∫0t‍(t-s)α-q-1Γ(α-q)ϕ(s)f(s,y(s),D0+py(s))ds-tα-q-1Γ(α-q)×∫01‍(1-s)α-1ϕ(s)f(s,y(s),D0+py(s))ds+(tα-q-2Γ(α-1)Γ(α-q-1)-tα-q-1Γ(α)Γ(α-q))×∫01‍u(s)G(s,y(s),D0+py(s))ds+tα-q-1Γ(α)Γ(α-q)∫01‍m(s)M(s,y(s),D0+py(s))ds+1Π(tα-q-1Γ(α)Γ(α-q)-tα-q-2Γ(α-1)Γ(α-q-1))×Γ(α)Γ(α-q)t1α-q-1I(t1,y(t1),D0+py(t1))+1Π(tα-q-2Γ(α-1)Γ(α-q-1)-tα-q-1Γ(α)Γ(α-q))×t1α-1I1(t1,y(t1),D0+py(t1)),t∈(t1,1],D0+p(T2x)(t)={∫0t‍(t-s)β-p-1Γ(β-p)ψ(u)g(u,x(u),D0+qx(u))du-tβ-p-1Γ(β-p)×∫01‍(1-s)β-1ψ(s)g(s,x(s),D0+qx(s))ds+tβ-p-2Γ(β-1)Γ(β-p-1)×∫01‍v(s)H(s,y(s),D0+py(s))ds+tβ-p-1Γ(β)Γ(β-p)×∫01‍n(s)N(s,x(s),D0+qx(s))ds+tβ-p-1ΞΓ(β)Γ(β-p)×(Γ(β)Γ(β-p)t1β-p-1-Γ(β-1)Γ(β-p-2)t1β-p-2)×J(t1,x(t1),D0+qx(t1))+tβ-p-1(t1β-2-t1β-1)ΞΓ(β)Γ(β-p)×J1(t1,x(t1),D0+qx(t1)),t∈(0,t1],∫0t‍(t-s)β-p-1Γ(β-p)ψ(s)g(s,x(s),D0+qx(s))ds-tβ-p-1Γ(β)Γ(β)Γ(β-p)×∫01‍(1-s)β-1ψ(s)g(s,x(s),D0+qx(s))ds+(tβ-p-2Γ(β-1)Γ(β-p-1)-tβ-p-1Γ(β)Γ(β-p))×∫01‍v(s)H(s,y(s),D0+py(s))ds+tβ-p-1Γ(β)Γ(β-p)∫01‍n(s)N(s,x(s),D0+qx(s))ds+1Ξ(tβ-p-1Γ(β)Γ(β-p)-tβ-p-2Γ(β-1)Γ(β-p-1))×Γ(β)Γ(β-p)t1β-p-1J(t1,x(t1),D0+qx(t1))+1Ξ(tβ-p-2Γ(β-1)Γ(β-p-1)-tβ-p-1Γ(β)Γ(β-p))×t1β-1J1(t1,x(t1),D0+qx(t1)),t∈(t1,1].
It is easy to see that(41)|∫0t‍(t-s)α-q-1Γ(α-q)ϕ(u)f(u,y(u),D0+py(u))du|≤ArL1B(α+δ-q,k+1)Γ(α-q)<∞.
From (37) and (41), we see that D0+q(T1y)(t) is defined on (0,1], continuous on (0,t1] and (t1,1], respectively. One sees that
(42)limt→0t2+q-αD0+q(T1y)(t)=limt→0[{(t1,y(t1),D0+py(t1))}{∫0t‍(Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2)}t2+q-α×∫0t‍(t-s)α-q-1Γ(α-q)ϕ(u)f(u,y(u),D0+py(u))dulimt→0-tΓ(α-q)×∫01‍(1-s)α-1ϕ(s)f(s,y(s),D0+py(s))dslimt→0+Γ(α-1)Γ(α-q-1)∫01‍u(s)G(s,y(s),D0+py(s))dslimt→0+tΓ(α)Γ(α-q)∫01‍m(s)M(s,y(s),D0+py(s))dslimt→0+tΠΓ(α)Γ(α-q)×(Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2)limt→0×I(t1,y(t1),D0+py(t1))+Γ(α)Γ(α-q)t(t1α-2-t1α-1)Πlimt→0×I1(t1,y(t1),D0+py(t1)){∫01‍(Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2)}]=Γ(α-1)Γ(α-q-1)∫01‍u(s)G(s,y(s),D0+py(s))ds,
and there exits the limit limt→t1+D0+q(T1y)(t).
From the above discussion, we have(T1y)∈X. Similarly, we can show that (T2x)∈Y. Hence, ((T1y),(T2x))∈X×Y. Then T:X×Y→X×Y is well defined.
Step  2. We prove that T is continuous. Let (xn,yn)∈X×Y with (xn,yn)→(x0,y0) as n→∞. We will show that T(xn,yn)→T(x0,y0) as n→∞, that is, prove that T1yn→T1y0 and T2xn→T2x0 as n→∞.
In fact, we haver>0 such that ||(xn,yn)||=r>0. Then
(43)max{supt∈(0,1)t2-α|xn(t)|,supt∈(0,1)t2+q-α|D0+qxn(t)|}≤r<+∞,n=0,1,2,…,max{supt∈(0,1)t2-β|yn(t)|,supt∈(0,1)t2+p-β|D0+pyn(t)|}≤r<+∞,n=0,1,2,….
From (B),f,G,M,I,I1 are β-Caratheodory functions, then there exist constants Ar>0 such that
(44)|f(t,yn(t),D0+pyn(t))|≤Ar,mmt∈(0,1),n=0,1,2,…,|G(t,yn(t),D0+pyn(t))|≤Ar,mmt∈(0,1),n=0,1,2,…,|M(t,yn(t),D0+pyn(t))|≤Ar,mmt∈(0,1),n=0,1,2,…,|I(t1,yn(t1),D0+pyn(t1))|≤Ar,mmmt∈(0,1),n=0,1,2,…,|I1(t1,yn(t1),D0+pyn(t1))|≤Ar,mmimt∈(0,1),n=0,1,2,…,supt∈(0,1)t2-α|xn(t)-x0(t)|⟶0,supt∈(0,1)t2-β|yn(t)-y0(t)|,supt∈(0,1)t2+q-α|D0+qxn(t)-D0+qx0(t)|⟶0,supt∈(0,1)t2+p-β|D0+pyn(t)-D0+py0(t)|⟶0,
as n→∞. We have
(45)D0+q(T1yn)(t)={∫0t‍(t-s)α-q-1Γ(α-q)ϕ(u)f(u,yn(u),D0+pyn(u))du-tα-q-1Γ(α-q)×∫01‍(1-s)α-1ϕ(s)f(s,yn(s),D0+pyn(s))ds+tα-q-2Γ(α-1)Γ(α-q-1)×∫01‍u(s)G(s,yn(s),D0+pyn(s))ds+tα-q-1Γ(α)Γ(α-q)×∫01‍m(s)M(s,yn(s),D0+pyn(s))ds+tα-q-1ΠΓ(α)Γ(α-q)×(Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2)×I(t1,yn(t1),D0+pyn(t1))+Γ(α)Γ(α-q)tα-q-1(t1α-2-t1α-1)Π×I1(t1,yn(t1),D0+pyn(t1)),t∈(0,t1],∫0t‍(t-s)α-q-1Γ(α-q)ϕ(s)f(s,yn(s),D0+pyn(s))ds-tα-q-1Γ(α-q)×∫01‍(1-s)α-1ϕ(s)f(s,y(s),D0+py(s))ds+(tα-q-2Γ(α-1)Γ(α-q-1)-tα-q-1Γ(α)Γ(α-q))×∫01‍u(s)G(s,yn(s),D0+pyn(s))ds+tα-q-1Γ(α)Γ(α-q)×∫01‍m(s)M(s,yn(s),D0+pyn(s))ds+1Π(tα-q-1Γ(α)Γ(α-q)-tα-q-2Γ(α-1)Γ(α-q-1))×Γ(α)Γ(α-q)t1α-q-1I(t1,yn(t1),D0+pyn(t1))+1Π(tα-q-2Γ(α-1)Γ(α-q-1)-tα-q-1Γ(α)Γ(α-q))×t1α-1I1(t1,yn(t1),D0+pyn(t1)),t∈(t1,1].
From the Lebesgue dominated convergence theorem, we get
(46)supt∈(0,1)t2-β|(T1yn)(t)-(T1y0)(t)|,supt∈(0,1)t2+p-β|D0+p(T1yn)(t)-D0+p(T1y0)(t)|⟶0,
as n→∞. Similarly, we can show that
(47)supt∈(0,1)t2-α|(T2xn)(t)-(T2x0)(t)|⟶0,supt∈(0,1)t2+q-α|D0+q(T2xn)(t)-D0+q(T2x0)(t)|⟶0,
as n→∞. It follows from (46) and (47) that T is continuous.
Step  3. We prove that T is compact, that is, for each nonempty open bounded subset Ω of X×Y, prove that T(Ω¯) is relatively compact. We must prove that T(Ω¯) is uniformly bounded, equicontinuous on each subinterval [a,b]⊆(tk,tk+1](k=0,1), T(Ω¯) is equiconvergent as t→0, and equiconvergent as t→t1.
LetΩ be a bounded open subset of Y. We have r>0 such that
(48)max{supt∈(0,1)t2-β|y(t)|,supt∈(0,1)t2+p-β|D0+py(t)|}≤r<+∞,y∈Ω¯.
From (B),f,G,M,I,I1 are β-Caratheodory functions, then there exist constants Ar>0 such that
(49)|f(t,y(t),D0+py(t))|≤Ar,t∈(0,1),|G(t,y(t),D0+py(t))|≤Ar,t∈(0,1),|M(t,y(t),D0+py(t))|≤Ar,t∈(0,1),|I(t1,y(t1),D0+py(t1))|≤Ar,t∈(0,1),|I1(t1,y(t1),D0+py(t1))|≤Ar,t∈(0,1).
Substep  3.1. Prove that T(Ω¯) is uniformly bounded.
In fact, fort∈(0,t1], use (49), we have
(50)t2-α|(T1y)(t)|≤t2-α∫0t‍(t-s)α-1Γ(α)|ϕ(u)f(u,y(u),D0+py(u))|du+tΓ(α)∫01(1-s)α-1|ϕ(s)f(s,y(s),D0+py(s))|ds+∫01‍|u(s)G(s,y(s),D0+py(s))|ds+t∫01‍|m(s)M(s,y(s),D0+py(s))|ds+tΠ|Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2|×|I(t1,y(t1),D0+py(t1))|+t(t1α-2-t1α-1)Π|I1(t1,y(t1),D0+py(t1))|≤ArL1B(α+δ,k+1)Γ(α)+ArL1Γ(α)B(α+δ,k+1)+Ar∥u∥1+Ar∥m∥1+ArΠ|Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2|+(t1α-2-t1α-1)ΠAr<∞.
Similarly, we can get for t∈(t1,1] that
(51)t2-α|(T1y)(t)|≤ArL1B(α+δ,k+1)Γ(α)+ArL1Γ(α)B(α+δ,k+1)+Ar∥u∥1+Ar∥m∥1+ArΠΓ(α)Γ(α-q)t1α-q-1+ArΠt1α-1<∞.
Furthermore, we have for t∈(0,t1] that
(52)t2+q-α|D0+q(T1y)(t)|≤ArL1B(α+δ-q,k+1)Γ(α-q)+ArL1Γ(α-q)B(α+δ,k+1)+ArΓ(α-1)Γ(α-q-1)∥u∥1+ArΓ(α)Γ(α-q)∥m∥1+ArΠΓ(α)Γ(α-q)×|Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2|+Γ(α)Γ(α-q)Ar(t1α-2-t1α-1)Π<∞,
and for t∈(t1,1] that
(53)t2+q-α|D0+q(T1y)(t)|≤ArL1B(α+δ-q,k+1)Γ(α-q)+ArL1Γ(α-q)B(α+δ,k+1)+(Γ(α-1)Γ(α-q-1)+Γ(α)Γ(α-q))Ar∥u∥1+Γ(α)Γ(α-q)Ar∥m∥1+1Π(Γ(α)Γ(α-q)+Γ(α-1)Γ(α-q-1))Γ(α)Γ(α-q)t1α-q-1Ar+1Π(Γ(α-1)Γ(α-q-1)+Γ(α)Γ(α-q))t1α-1Ar<∞.
Hence,
(54)max{supt∈(0,1)t2-α|(Ty)(t)|,supt∈(0,1)t2+q-α|D0+q(Ty)(t)|}<+∞,y∈Ω¯.
Similarly, we can show that
(55)max{supt∈(0,1)t2-β|(Tx)(t)|,supt∈(0,1)t2+p-β|D0+p(Tx)(t)|}<+∞,y∈Ω¯.
It is easy to see that T(Ω¯) is uniformly bounded.
Substep  3.2. Prove that T(Ω¯) is equicontinuous on each subinterval [a,b]⊆(tk,tk+1](k=0,1).
For each[a,b]⊆(t0,t1], and s1,s2∈[a,b] with s2<s1, we have
(56)|s12-α(Ty)(s1)-s22-α(Ty)(s2)|≤|s12-α∫0s1‍(s1-s)α-1Γ(α)ϕ(u)f(u,y(u),D0+py(u))du-s22-α∫0s2‍(s2-s)α-1Γ(α)ϕ(u)f(u,y(u),D0+py(u))du|+|s1-s2|Ar×[L1B(α+δ,k+1)Γ(α)+∥m∥1+1Π|Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2|+t1α-2-t1α-1Π].
Note that |τ1ϱ-τ2ϱ|≤|τ1-τ2|ϱ for all τ1,τ2≥0 and ϱ∈(0,1).
Since(57)|s12-α∫0s1‍(s1-s)α-1Γ(α)ϕ(u)f(u,y(u),D0+py(u))du-s22-α∫0s2‍(s2-s)α-1Γ(α)ϕ(u)f(u,y(u),D0+py(u))du|≤|s12-α-s22-α|ArL1B(α+δ,k+1)Γ(α)+b2-αArL1s1α+k+δ∫s2/s11‍(1-w)α+δ-1wkdwΓ(α)+|s1-s2|α-1ArL1b2-αΓ(α)∫0b‍sk(1-s)δds⟶0mmmmmmmmmmmmuniformly  ass1⟶s2.
It follows that
(58)|s12-α(Ty)(s1)-s22-α(Ty)(s2)|⟶0mmmmimmmuniformly  ass1⟶s2.
For [a,b]⊆(t1,1], and s1,s2∈[a,b] with s2<s1, we can prove similarly that
(59)|s12-α(Ty)(s1)-s22-α(Tyx)(s2)|⟶0mmmmmmimuniformly  ass1⟶s2.
On the other hand, for [a,b]⊆(t0,t1], and s1,s2∈[a,b] with s2<s1, we have
(60)|s12+q-αD0+q(T1y)(s1)-s22+q-αD0+q(T1y)(s2)|≤|s12+q-α∫0s1‍(s1-s)α-q-1Γ(α-q)ϕ(u)f(u,y(u),D0+py(u))du-s22+q-α∫0s2‍(s2-s)α-q-1Γ(α-q)ϕ(u)f(u,y(u),D0+py(u))du|+|s2-s1|Ar[L1B(α+δ,k+1)Γ(α-q)+Γ(α)∥m∥1Γ(α-q)+1ΠΓ(α)Γ(α-q)×|Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2|+Γ(α)Γ(α-q)|t1α-2-t1α-1|Π].
It is easy to see that
(61)|s12+q-α-s22+q-α|ArL1∫01‍(1-w)α+δ-q-1wkdsΓ(α-q)⟶0mmmmmmmmmmmmmuniformly  ass1⟶s2,(62)b2+q-αArL1∫s2/s11‍(1-w)α-q-1wkdwΓ(α-q)⟶0mmmmmmmmmuniformly  ass1⟶s2.
For the third term, if α-q-1≥0, use |τ1ϱ-τ2ϱ|≤|τ1-τ2|ϱ, then
(63)∫0s2‍|(s1-s)α-q-1-(s2-s)α-q-1|Γ(α-q)sk(1-s)δds≤|s1-s2|α-q-1∫0b‍1Γ(α-q)sk(1-s)δds⟶0mmmmmmmmmmmimuniformly  ass1⟶s2.
If α-q-1<0, use |τ1ϱ-τ2ϱ|≤|τ1-τ2|ϱ, then
(64)∫0s2‍|(s1-s)α-q-1-(s2-s)α-q-1|Γ(α-q)sk(1-s)δds=∫0s2‍(s2-s)α-q-1-(s1-s)α-q-1Γ(α-q)sk(s1-s)δds≤|s2α+δ+k-q-s1α+δ+k-q|B(α+δ+k-q,k+1)γ(α-q)+bα+k+δ-qB(α+2δ-q,k+1)Γ(α-q)|1-s1s2|-δ⟶0mmmmmmmmmmmmmuniformly  ass1⟶s2.
For [a,b]⊆(t1,1], and s1,s2∈[a,b] with s2<s1, we can prove similarly that
(65)|s12+q-αD0+q(T1y)(s1)-s22+q-αD0+q(T1y)(s2)|⟶0mmmmmmmmmmmmimuniformly  ass1⟶s2.
Similarly, we can show that for each [a,b]⊆(t0,t1], and s1,s2∈[a,b] with s2<s1, we have
(66)|s12-β(Tx)(s1)-s22-β(Tx)(s2)|⟶0mmmmmiimuniformly  ass1⟶s2.
For [a,b]⊆(t1,1], and s1,s2∈[a,b] with s2<s1, we can prove similarly that
(67)|s12-β(Tx)(s1)-s22-β(Tx)(s2)|⟶0mmmmmiimuniformly  ass1⟶s2.
For each [a,b]⊆(t0,t1], and s1,s2∈[a,b] with s2<s1, we have
(68)|s12+p-βD0+p(Tx)(s1)-s22+p-βD0+p(Tx)(s2)|⟶0mmmmmmmmiimmimuniformly  ass1⟶s2.
For [a,b]⊆(t1,1], and s1,s2∈[a,b] with s2<s1, we can prove similarly that
(69)|s12+p-βD0+p(Tx)(s1)-s22+p-βD0+p(Tx)(s2)|⟶0mmmmmmmmmiiiimmuniformly  ass1⟶s2.
So T(Ω¯) is equicontinuous on each subinterval [a,b]⊆(tk,tk+1](k=0,1).
Substep  3.3. Prove that T(Ω¯) is equiconvergent as t→0, and equiconvergent as t→t1.
We have(70)|t2-α(Ty)(t)-∫01‍u(s)G(s,y(s),D0+py(s))ds|≤ArL1t2+k+δB(α+δ,k+1)Γ(α)+tΓ(α)ArL1B(α+δ,k+1)+tAr∥m∥1+tΠ|Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2|Ar+t|t1α-2-t1α-1|ΠAr.
It follows that
(71)|t2-α(Ty)(t)-∫01‍u(s)G(s,y(s),D0+py(s))ds|⟶0mmmmmmmmmmmmmmiimuniformly  ast⟶0.
Similarly, we can show that t2-α(Ty)(t) is equiconvergent at t=t1. On the other hand, we have
(72)|t2+q-αD0+q(Ty)(t)-Γ(α-1)Γ(α-q-1)×∫01‍u(s)G(s,y(s),D0+py(s))ds|≤ArL1t2+k+δB(α+δ-q,k+1)Γ(α-q)+tΓ(α-q)ArL1B(α+δ,k+1)+tΓ(α)Γ(α-q)Ar∥m∥1+tΠΓ(α)Γ(α-q)×|Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2|Ar+Γ(α)Γ(α-q)t|t1α-2-t1α-1|ΠAr.
It follows that
(73)|t2+q-αD0+q(Ty)(t)-Γ(α-1)Γ(α-q-1)×∫01‍u(s)G(s,y(s),D0+py(s))ds|⟶0mmmmmmmiiimmuniformly  ast⟶0.
Similarly, we can show that t2+q-αD0+q(Ty)(t) is equiconvergent at t=t1.
Similarly we can prove that(74)|t2-β(Tx)(t)-∫01‍v(s)H(s,x(s),D0+qx(s))ds|⟶0mmmmmmmmmmmmmmmmuniformly  ast⟶0,
and t2+p-βD0+p(Tx)(t) is equiconvergent at t=0, both t2-β(Tx)(t) and t2+p-βD0+p(Tx)(t) are equiconvergent at t=t1.
Hence,T(Ω¯) is equiconvergent as t→0 and T(Ω¯) is equiconvergent as t→t1.
SoT(Ω¯) is relatively compact. Then T is completely continuous. The proofs are completed.


## 3. Main Result

In this section, we will establish the existence of at least one solution of BVP (6)-(7).Definition 11 (see [26]).
An odd homeomorphismΦ of the real line ℝ onto itself is called a pseudo-sub-multiplicative function if there exists a homeomorphism ω of [0,∞) onto itself which supportsΦ in the sense that for all v1,v2≥0 we have Φ(v1v2)≥ω(v1)Φ(v2).  ω is called the supporting function of Φ.Remark 12.
Note that any submultiplicative function is a pseudo-submultiplicative function. Also any function of the formΦ(u):=∑j=0k‍cj|u|ju,u∈ℝ is pseudo-sup-multiplicative, provided that cj≥0. Here, a supporting function is defined by ω(u):=min{uk+1,u}, u≥0.Remark 13.
It is clear that a pseudo-submultiplicative functionΦ and any corresponding supporting function ω are increasing functions vanishing at zero; moreover, their inverses Φ-1 and ν, respectively, are increasing and for all v1,v2≥0, we have Φ-1(v1v2)≤ν(v1)Φ-1(v2).Theorem 14.
Suppose that (a)–(e) and (A)-(B) hold,Φ:R→R is a submultiplicative-like function with the supporting function ω, its inverse function is denoted by Φ-1:R→R with the supporting function ν. Furthermore, suppose that(i)
there exist nonnegative numbersCf,Bf,Af,CG,BG,AG,CM,BM, and AM such that
(75)|f(t,tα-2U,tα-q-2V)|≤Cf+BfΦ-1(|U|)+AfΦ-1(|V|),|G(t,tα-2U,tα-q-2V)|≤CG+BGΦ-1(|U|)+AGΦ-1(|V|),|M(t,tα-2U,tα-q-2V)|≤CM+BMΦ-1(|U|)+AMΦ-1(|Φ-1(|U|)|),
holds for all (U,V)∈R2, t∈(0,1].(ii)
there exist nonnegative numbersCg,Bg,Ag,CH,BH,AH,CN,BN, and AN such that
(76)|g(t,tβ-2U,tβ-p-2V)|≤Cg+BgΦ(U)+AgΦ(V),|H(t,tβ-2U,tβ-p-2V)|≤CH+BHΦ(U)+AHΦ(V),|N(t,tβ-2U,tβ-p-2V)|≤CN+BNΦ(U)+ANΦ(V),
hold for all (U,V)∈R2, t∈(0,1].(iii)
there exist the nonnegative numbersCI,BI,AI,C1,I,B1,I, and A1,I such that
(77)|I(t1,t1α-2U,t1α-q-2V)|≤CI+BIΦ-1(|U|)+AIΦ-1(|V|),|I1(t1,t1α-2U,t1α-q-2V)|≤C1,I+B1,IΦ-1(|U|)+A1,IΦ-1(|V|),
hold for all (U,V)∈R2.(iv)
there exist the nonnegative numbersCJ,BJ,AJ,C1,J,B1,J, and A1,J such that
(78)|J(t1,t1β-2U,t1β-p-2V)|≤CJ+BJΦ(U)+AJΦ(V),|J1(t1,t1β-2U,t1β-p-2V)|≤C1,J+B1,JΦ(U)+A1,JΦ(V),
hold for all (U,V)∈R2.
Then BVP (6)-(7) has at least one solution if
(79)max{Θ2,Σ2,Θ4,Σ4}ν(2max{Υ2,Λ2,Υ4,Λ4})<1ormax{Υ2,Λ2,Υ4,Λ4}w((2max{Θ2,Σ2,Θ4,Σ4})-1)<1,
where
(80)Θ1=L1B(α+δ,k+1)Γ(α)Cf+L1B(α+δ,k+1)Γ(α)Cf+∥u∥1CG+∥m∥1CM+1Π|Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2|CI+|t1α-2-t1α-1|ΠC1,I,Θ2=L1B(α+δ,k+1)Γ(α)[Bf+Af]+L1B(α+δ,k+1)Γ(α)[Bf+Af]+∥u∥1[BG+AG]+∥m∥1[BM+AM]+1Π|Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2|×[BI+AI]+|t1α-2-t1α-1|Π[B1,I+A1,I],Θ3=L1B(α+δ-q,k+1)Γ(α-q)Cf+L1Γ(α-q)B(α+δ,k+1)Cf+Γ(α-1)Γ(α-q-1)∥u∥1CG+Γ(α)Γ(α-q)∥m∥1CM+1ΠΓ(α)Γ(α-q)×|Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2|CI+Γ(α)Γ(α-q)|t1α-2-t1α-1|ΠC1,I,Θ4=L1B(α+δ-q,k+1)Γ(α-q)[Bf+Af]+L1Γ(α-q)B(α+δ,k+1)[Bf+Af]+Γ(α-1)Γ(α-q-1)∥u∥1[BG+AG]+Γ(α)Γ(α-q)∥m∥1[BM+AM]+1ΠΓ(α)Γ(α-q)×|Γ(α)Γ(α-q)t1α-q-1-Γ(α-1)Γ(α-q-2)t1α-q-2|×[BI+AI]+Γ(α)Γ(α-q)|t1α-2-t1α-1|Π[B1,I+A1,I],Σ1=L1B(α+δ,k+1)Γ(α)Cf+L1B(α+δ,k+1)Γ(α)Cf+∥u∥1CG+1ΠΓ(α)Γ(α-q)t1α-q-1CI+1Πt1α-1C1,I,Σ2=L1B(α+δ,k+1)Γ(α)[Bf+Af]+L1B(α+δ,k+1)Γ(α)[Bf+Af]+∥u∥1[BG+AG]+1ΠΓ(α)Γ(α-q)t1α-q-1[BI+AI]+1Πt1α-1[B1,I+A1,I],Σ3=L1B(α+δ-q,k+1)Γ(α-q)Cf+L1Γ(α-q)B(α+δ,k+1)Cf+(Γ(α-1)Γ(α-q-1)+Γ(α)Γ(α-q))∥u∥1CG+Γ(α)Γ(α-q)∥m∥1CM+1Π(Γ(α)Γ(α-q)+Γ(α-1)Γ(α-q-1))Γ(α)Γ(α-q)t1α-q-1CI+1Π(Γ(α-1)Γ(α-q-1)+Γ(α)Γ(α-q))t1α-1C1,I,Σ4=L1B(α+δ-q,k+1)Γ(α-q)[Bf+Af]+L1Γ(α-q)B(α+δ,k+1)[Bf+Af]+(Γ(α-1)Γ(α-q-1)+Γ(α)Γ(α-q))∥u∥1[BG+AG]+Γ(α)Γ(α-q)∥m∥1[BM+AM]+1Π(Γ(α)Γ(α-q)+Γ(α-1)Γ(α-q-1))×Γ(α)Γ(α-q)t1α-q-1[BI+AI]+1Π(Γ(α-1)Γ(α-q-1)+Γ(α)Γ(α-q))×t1α-1[B1,I+A1,I],(81)Υ1=L2B(β+θ,l+1)Γ(β)Cg+L2B(β+θ,l+1)Γ(β)Cg+∥v∥1CH+∥n∥1CN+1Ξ|Γ(β)Γ(β-p)t1β-p-1-Γ(β-1)Γ(β-p-2)t1β-p-2|CJ+|t1β-2-t1β-1|ΞC1,J,Υ2=L2B(β+θ,l+1)Γ(β)[Bg+Ag]+L2B(β+θ,l+1)Γ(β)[Bg+Ag]+∥b∥1[BH+AH]+∥n∥1[BN+AN]+1Ξ|Γ(β)Γ(β-p)t1β-p-1-Γ(β-1)Γ(β-p-2)t1β-p-2|×[BJ+AJ]+|t1β-2-t1β-1|Ξ[B1,J+A1,J],Υ3=L2B(β+θ-p,l+1)Γ(β-p)Cg+L2Γ(β-p)B(β+θ,l+1)Cg+Γ(β-1)Γ(β-p-1)∥v∥1CH+Γ(β)Γ(β-p)∥n∥1CN+1ΞΓ(β)Γ(β-p)×|Γ(β)Γ(β-p)t1β-p-1-Γ(β-1)Γ(β-p-2)t1β-p-2|CJ+Γ(β)Γ(β-p)|t1β-2-t1β-1|ΞC1,J,Υ4=L2B(β+θ-p,l+1)Γ(β-p)[Bg+Ag]+L2Γ(β-p)B(β+θ,l+1)[Bg+Ag]+Γ(β-1)Γ(β-p-1)∥v∥1[BH+AH]+Γ(β)Γ(β-p)∥n∥1[BN+AN]+1ΞΓ(β)Γ(β-p)×|Γ(β)Γ(β-p)t1β-p-1-Γ(β-1)Γ(β-p-2)t1β-p-2|[BJ+AJ]+Γ(β)Γ(β-p)|t1β-2-t1β-1|Ξ[B1,J+A1,J],Λ1=L2B(β+θ,l+1)Γ(β)Cg+L2B(β+θ,l+1)Γ(β)Cg+∥v∥1CH+1ΞΓ(β)Γ(β-p)t1β-p-1CJ+1Ξt1β-1C1,J,Λ2=L2B(β+θ,l+1)Γ(β)[Bg+Ag]+L2B(β+θ,l+1)Γ(β)[Bg+Ag]+∥v∥1[BH+AH]+1ΞΓ(β)Γ(β-p)t1β-p-1[BJ+AJ]+1Ξt1β-1[B1,J+A1,J],Λ3=L2B(β+θ-p,l+1)Γ(β-p)Cg+L2Γ(β-p)B(β+θ,l+1)Cg+(Γ(β-1)Γ(β-p-1)+Γ(β)Γ(β-p))∥v∥1CH+Γ(β)Γ(β-p)∥n∥1CN+1Ξ(Γ(β)Γ(β-p)+Γ(β-1)Γ(β-p-1))Γ(β)Γ(β-p)t1β-p-1CJ+1Ξ(Γ(β-1)Γ(β-p-1)+Γ(β)Γ(β-p))t1β-1C1,J,Λ4=L2B(β+θ-p,l+1)Γ(β-p)[Bg+Ag]+L2Γ(β-p)B(β+θ,l+1)[Bg+Ag]+(Γ(β-1)Γ(β-p-1)+Γ(β)Γ(β-p))∥v∥1[BH+AH]+Γ(β)Γ(β-p)∥n∥1[BN+AN]+1Ξ(Γ(β)Γ(β-p)+Γ(β-1)Γ(β-p-1))×Γ(β)Γ(β-p)t1β-p-1[BJ+AJ]+1Ξ(Γ(β-1)Γ(β-p-1)+Γ(β)Γ(β-p))t1β-1[B1,J+A1,J].Proof.
To apply Lemma5, we should define an open bounded subset Ω of X×Y centered at zero such that assumptions in Lemma 5 hold.
LetΩ1={(x,y)∈X×Y:(x,y)=λT(x,y)  for some λ∈(0,1)}. We prove that Ω1 is bounded. For (x,y)∈Ω1, we get (x,y)=λT(x,y). It follows that x=λT1y and y=λT2x.
Fort∈(0,t1], we obtain t2-α|x(t)|≤t2-α|(T1y)(t)|≤Θ1+Θ2Φ-1(||y||).
Fort∈(t1,1],
(82)t2-α|x(t)|≤t2-α∫0t‍(t-s)α-1Γ(α)|ϕ(s)f(s,y(s),D0+py(s))|ds+tΓ(α)∫01‍(1-s)α-1|ϕ(s)f(s,y(s),D0+py(s))|ds+(1-t)∫01‍|u(s)G(s,y(s),D0+py(s))|ds+t∫01‍|m(s)M(s,y(s),D0+py(s))|ds+1-tΠΓ(α)Γ(α-q)t1α-q-1|I(t1,y(t1),D0+py(t1))|+1-tΠt1α-1|I1(t1,y(t1),D0+py(t1))|≤Σ1+Σ2Φ-1(∥y∥).
It follows that
(83)supt∈(0,1)t2-α|x(t)|≤max{Θ1,Σ1}+max{Θ2,Σ2}Φ-1(∥y∥).
Similarly, we have for t∈(0,t1] that
(84)tq+2-α|D0+qx(t)|≤Θ3+Θ4Φ-1(∥y∥)
and for t∈(0,t1](85)tq+2-α|D0+qx(t)|≤Σ3+Σ4Φ-1(∥y∥).
It follows that(86)supt∈(0,1)t2+q-α|D0+qx(t)|≤max{Θ3,Σ3}+max{Θ4,Σ4}Φ-1(∥y∥).
Hence,(87)∥x∥≤max{Θ1,Σ1,Θ3,Σ3}mmm+max{Θ2,Σ2,Θ4,Σ4}Φ-1(∥y∥).
Similar to the above discussion we can prove that(88)∥y∥≤max{Υ1,Λ1,Υ3,Λ3}+max{Υ2,Λ2,Υ4,Λ4}Φ(∥x∥).
Case  1. Consider (max{Θ2, Σ2,Θ4,Σ4}ν  (2max{Υ2,Λ2,Υ4, Λ4})<1).
With out loss of generality, suppose that(89)∥x∥≥Φ-1(max{Υ1,Λ1,Υ3,Λ3}max{Υ2,Λ2,Υ4,Λ4}).
Then use Remark 13, and the previous inequalities to get
(90)∥x∥≤max{Θ1,Σ1,Θ3,Σ3}mmm+max{Θ2,Σ2,Θ4,Σ4}ν(2max{Υ2,Λ2,Υ4,Λ4})∥x∥.
It follows that there exists a constant W>0 such that ||x||≤W. Thus
(91)∥x∥≤max{W,Φ-1(max{Υ1,Λ1,Υ3,Λ3}max{Υ2,Λ2,Υ4,Λ4})}.
Then
(92)∥y∥≤max{Υ1,Λ1,Υ3,Λ3}mmm+max{Υ2,Λ2,Υ4,Λ4}Φmmm×(max{W,Φ-1(max{Υ1,Λ1,Υ3,Λ3}max{Υ2,Λ2,Υ4,Λ4})}).
It follows that Ω1 is bounded.
Case  2. Consider ((max{Υ2,Λ2,Υ4,Λ4}/w((2max{Θ2,Σ2,Θ4,Σ4})-1))   <1).
Without loss of generality, suppose that(93)∥y∥≥Φ(max{Θ1,Σ1,Θ3,Σ3}max{Θ2,Σ2,Θ4,Σ4}).
Then using Remark 12 and the previous inequalities, we get
(94)∥y∥≤max{Υ1,Λ1,Υ3,Λ3}mmm+max{Υ2,Λ2,Υ4,Λ4}w((2max{Θ2,Σ2,Θ4,Σ4})-1)∥y∥.
It follows that there exists a constant W>0 such that ||y||≤W. We get
(95)∥y∥≤max{W,Φ(max{Θ1,Σ1,Θ3,Σ3}max{Θ2,Σ2,Θ4,Σ4})}.
Then
(96)∥x∥≤max{Θ1,Σ1,Θ3,Σ3}mmm+max{Θ2,Σ2,Θ4,Σ4}Φ-1mmm×(max{W,Φ(max{Θ1,Σ1,Θ3,Σ3}max{Θ2,Σ2,Θ4,Σ4})}).
It follows that Ω1 is bounded.
To apply Lemma5, let Ω be a nonempty open bounded subset of X such that Ω⊃Ω1¯ centered at zero.
It is easy to see from Lemma8 that T is a completely continuous operator. One can see that
(97)(x,y)≠λT(x,y)∀(x,y)∈∂Ω,λ∈(0,1).
Thus, from Lemma 5, (x,y)=T(x,y) has at least one solution (x,y)∈Ω¯. So (x,y) is a pair of solutions of BVP (3) and BVP (6). The proof of Theorem 14 is complete.


## 4. Two Examples

To illustrate the usefulness of our main result, we present two examples that Theorem14 can readily apply.Example 15.
Consider the following impulsive boundary value problem:(98)D0+8/5x(t)=t-1/5(1-t)-1×(c+bt6/5[y(t)]3+at9/5[D0+1/5y(t)]3),mmmmmmmmmmiimt∈(0,1),t≠12,D0+9/5y(t)=t-1/5(1-t)-1×(c0+b0t1/15[x(t)]1/3+a0t2/15[D0+1/5x(t)]1/3),mmmmmmmmmmiimmmmimmt∈(0,1),t≠t1,limt→0t2/5x(t)=G,limt→0t1/5y(t)=H,x(1)=M,y(1)=N,Δx(12)=cI,Δy(12)=cJ,ΔD0+1x(12)=c1,I,ΔD0+1y(12)=c1,J,
where c,b,a,c0,b0,a0,G0,H0,M0,N0,CI,CJ,C1,I,C1,J are constants.
Corresponding to BVP (1), we have(a)
α=8/5, β=9/5, p=q=1/5,(b)
ϕ(t)=ψ(t)=t-1/5(1-t)-1/5, f(t,U,V)=c+bt6/5U3+at9/5V3 and g(t,U,V)=c0+b0t1/15U1/3 + a0t2/15V1/3 defined on (0,1)×R2,(c)
u(t)=v(t)=m(t)=n(t)≡1, G(t,U,V)=G0, H(t,U,V)=H0, M(t,U,V)=M0, N(t,U,V) = N0,(d)
0=t0<t1=(1/2)<t2=1,(e)
I(t,U,V)=cI, I1(t,U,V)=c1,I, J(t,U,V)=cJ, J1(t,U,V)=c1,J.
It is easy to show that(A)
ϕ satisfies α+2δ-q>0, α+k+δ-q≥0, and |ϕ(t)|≤L1tk(1-t)δ for all t∈(0,1) with L1=1 and k=-(1/5)=δ;ψ satisfies η+2θ-p>0, β+l+θ-p≥0, and |ψ(t)|≤L2tl(1-t)θ for all t∈(0,1) with L2=1 and l=-(1/5)=θ;(B)
f,G,M,I,I1 are β-Caratheodory functions and g,H,N,J,J1 are α-Caratheodory functions.
Furthermore, we haveΦ-1(x)=x3 and Φ(x)=x1/3 with w(x)=x1/3 and ν(x)=x3. It is easy to see that(i)
the inequalities(99)|f(t,tα-2U,tα-q-2V)|≤Cf+BfΦ-1(|U|)+AfΦ-1(|V|),|G(t,tα-2U,tα-q-2V)|≤CG+BGΦ-1(|U|)+AGΦ-1(|V|),|M(t,tα-2U,tα-q-2V)|≤CM+BMΦ-1(|U|)+AMΦ-1(|Φ-1(|U|)|)
hold for all (U,V)∈R2, t∈(0,1] with Cf=|c|,Bf=|b|,Af=|a|, CG=|G0|,BG=0,AG=0 and CM=|M0|,BM=0,AM=0;(ii)
the inequalities(100)|g(t,tβ-2U,tβ-p-2V)|≤Cg+BgΦ(U)+AgΦ(V),|H(t,tβ-2U,tβ-p-2V)|≤CH+BHΦ(U)+AHΦ(V),|N(t,tβ-2U,tβ-p-2V)|≤CN+BNΦ(U)+ANΦ(V)
hold for all (U,V)∈R2, t∈(0,1] with Cg=|c0|, Bg=|b0|,Ag=|a0|, CH=|H0|,BH=AH=0, CN=|N0|,BN=AN=0;(iii)
the inequalities(101)|I(t1,t1α-2U,t1α-q-2V)|≤CI+BIΦ-1(|U|)+AIΦ-1(|V|),|I1(t1,t1α-2U,t1α-q-2V)|≤C1,I+B1,IΦ-1(|U|)+A1,IΦ-1(|V|)
hold for all (U,V)∈R2 with CI=|cI|, BI=AI=0, C1,I=|c1,I|, B1,I=A1,I=0;(iv)
the inequalities(102)|J(t1,t1β-2U,t1β-p-2V)|≤CJ+BJΦ(U)+AJΦ(V),|J1(t1,t1β-2U,t1β-p-2V)|≤C1,J+B1,JΦ(U)+A1,JΦ(V)
hold for all (U,V)∈R2 with CJ=|cJ|, BJ=AJ=0, C1,J=|c1,J|, B1,J=A1,J=0.By direct computation, we know that
(103)Θ2=2B(7/5,4/5)Γ(8/5)[|b|+|a|],Σ2=2B(7/5,4/5)Γ(8/5)[|b|+|a|],Θ4=(B(6/5,4/5)Γ(7/5)+B(7/5,4/5)Γ(7/5))[|b|+|a|],Σ4=(B(6/5,4/5)Γ(7/5)+B(7/5,4/5)Γ(7/5))[|b|+|a|],Υ2=2B(8/5,4/5)Γ(9/5)[|b0|+|a0|],Υ4=(B(7/5,4/5)Γ(8/5)+B(8/5,4/5)Γ(8/5))[|b0|+|a0|],Λ2=2B(8/5,4/5)Γ(9/5)[|b0|+|a0|],Λ4=(B(7/5,4/5)Γ(8/5)+B(8/5,4/5)Γ(8/5))[|b0|+|a0|].
Then Theorem 14 implies that the existence of at least one solution if
(104)max{2B(8/5,4/5)Γ(9/5),B(7/5,4/5)Γ(8/5)+B(8/5,4/5)Γ(8/5)}×(max{2B(7/5,4/5)Γ(8/5),B(6/5,4/5)Γ(7/5)+B(7/5,4/5)Γ(7/5)})1/3×[|b0|+|a0|][|b|+|a|]1/3<123.Example 16.
Consider the following boundary value problem without impulse effects:(105)D0+7/4x(t)=t-1/4(1-t)-1/4×(C+Bt3/4[y(t)]3+At15/4[D0+1y(t)]3),mmmmmmmmmmmmmmiimmt∈(0,1),D0+5/4y(t)=t-1/8(1-t)-1/8×(C0+B0t1/4[x(t)]1/3+A0t7/12[D0+1/4x(t)]1/3),mmmmmmmmmmmmmmmmmmmmimt∈(0,1),limt→0t1/4x(t)=0,limt→0t3/4y(t)=0,x(1)=0,y(1)=0,
where C, B, A, C0, B0, and A0 are constants.
Corresponding to BVP (1), we have(a)
α=7/4, β=5/4, p=1 and q=1/4,(b)
ϕ(t)=t-1/4(1-t)-1/4, ψ(t)=t-1/8(1-t)-1/8, f,g defined on (0,1)×R2, f(t,U,V)=C+Bt1/12U3 + At5/12V3 and g(t,U,V)=C0+B0t1/4U1/3+A0t7/12V1/3,(c)
m(t)=n(t)=u(t)=v(t)≡0, G(t,U,V)=H(t,U,V)=M(t,U,V)=N(t,U,V)≡0,(d)
there exists no impulse point,(e)
I(t,U,V)=I1(t,U,V)=J(t,U,V)=J1(t,U,V)≡0.
It is easy to show that(A)
ϕ satisfies α+2δ-q>0, α+k+δ-q>0, |ϕ(t)|≤L1tk(1-t)δ for all t∈(0,1) with L1=1, k=-(1/4)=δ;ψ satisfies β+2θ-p>0, β+l+θ-p≥0, and |ψ(t)|≤L2tl(1-t)θ for all t∈(0,1) with L2=1, l=-(1/8)=θ;(B)
f,G,M,I,I1 are β-Caratheodory functions and g,H,N,J,J1 are α-Caratheodory functions.Furthermore, Φ(x)=x1/3 and Φ-1(x)=x3, we have w(x)=x1/3 and ν(x)=x3, and(i)
the inequalities(106)|f(t,tα-2U,tα-q-2V)|≤Cf+BfΦ-1(|U|)+AfΦ-1(|V|),|G(t,tα-2U,tα-q-2V)|≤CG+BGΦ-1(|U|)+AGΦ-1(|V|),|M(t,tα-2U,tα-q-2V)|≤CM+BMΦ-1(|U|)+AMΦ-1(|Φ-1(|U|)|)
hold for all (U,V)∈R2, t∈(0,1) with CG=BG=AG=CM=BM=AM=0, Cf=|C|, Bf=|B| and Af=|A|;(ii)
the inequalities(107)|g(t,tβ-2U,tβ-p-2V)|≤Cg+BgΦ(U)+AgΦ(V),|H(t,tβ-2U,tβ-p-2V)|≤CH+BHΦ(U)+AHΦ(V),|N(t,tβ-2U,tβ-p-2V)|≤CN+BNΦ(U)+ANΦ(V)
hold for all (U,V)∈R2, t∈(0,1) with CH=BH=AH=CN=BN=AN=0, Cg=|C0|, Bg=|B0| and Ag=|A0|;(iii)
the inequalities(108)|I(t1,t1α-2U,t1α-q-2V)|≤CI+BIΦ-1(|U|)+AIΦ-1(|V|),|I1(t1,t1α-2U,t1α-q-2V)|≤C1,I+B1,IΦ-1(|U|)+A1,IΦ-1(|V|)
hold for all (U,V)∈R2 with CI=BI=AI=C1,I=B1,I=A1,I=0;(iv)
there exist the nonnegative numbersAi,k,Bi,k, Ci,k(i=1,2) such that
(109)|J(t1,t1β-2U,t1β-p-2V)|≤CJ+BJΦ(U)+AJΦ(V),|J1(t1,t1β-2U,t1β-p-2V)|≤C1,J+B1,JΦ(U)+A1,JΦ(V)
hold for all (U,V)∈R2 with CJ=BJ=AJ=C1,J=B1,J=A1,J=0.
By direct computation, we know that(110)Θ2=2B(3/2,3/4)Γ(7/4)[|B|+|A|],Σ2=2B(3/2,3/4)Γ(7/4)[|B|+|A|],Θ4=(B(5/4,3/4)Γ(3/2)+B(3/2,3/4)Γ(3/2))[|B|+|A|],Σ4=(B(5/4,3/4)Γ(3/2)+B(3/2,3/4)Γ(3/2))[|B|+|A|],Υ2=2B(9/8,7/8)Γ(5/4)[|B0|+|A0|],Υ4=(B(1/8,7/8)Γ(β-p)+B(9/8,7/8)Γ(1/4))[|B0|+|A0|],Λ2=2B(9/8,7/8)Γ(5/4)[|B0|+|A0|],Λ4=(B(1/8,7/8)Γ(1/4)+B(9/8,7/8)Γ(1/4))[|B0|+|A0|].
Then Theorem14 implies the existence of at least one solution if
(111)max{2B(3/2,3/4)Γ(7/4),B(5/4,3/4)Γ(3/2)+B(3/2,3/4)Γ(3/2)}mm×(max{2B(9/8,7/8)Γ(5/4),B(1/8,7/8)Γ(β-p)+B(9/8,7/8)Γ(1/4)})3mm×[|B0|+|A0|]3[|B|+|A|]<18.Remark 17.
It is easy to see that the previous boundary value problems have at least one solution for sufficiently small|B1|,|B2| and |A0|,|B0|, |a|,|b|,|a0| and |b0|. They cannot be solved by the theorems in [24, 25].



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*Source: 498781-2013-04-17.xml*