Citation: Abdelkader Amara. Existence results for hybrid fractional differential equations with three-point boundary conditions[J]. AIMS Mathematics, 2020, 5(2): 1074-1088. doi: 10.3934/math.2020075
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The domain of fractional calculus is interested with the generalization of the classical integer order differentiation and integration to an arbitrary order. Fractional calculus has found important applications in different fields of science, especially in problems related to biology, chemistry, mathematical physics, economics, control theory, blood flow phenomena and aerodynamics, etc (see, for example, [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]. For some recent developments on the existence and uniqueness of solutions for differential equations nvolving the fractional derivatives, for more information, we advise reading these papers [16,17,18,19,20,21,22,23,24] and the references therein. In this literature, we show some contributions of researchers to the finding of the existence and uniqueness of the solution for the different fractional differential equations. Bai [16] studied the existence and uniqueness of positive solutions for the following three-point fractional boundary value problem:
{cDq0+x(t)=f(t,x(t)),t∈(0,1),q∈(1,2],x(0)=0,x(1)=βx(η),η∈(0,1), | (1.1) |
where Dq denotes the Riemann-Liouville fractional derivative, and 0<βηq−1<1.
Ahmad et al. in [17] discussed the existence and uniqueness of solutions for the following boundary value problem of fractional order differential equations with three-point integral boundary conditions:
{cDqx(t)=f(t,x(t)),t∈(0,1),q∈(1,2],x(0)=0,x(1)=α∫η0x(s)ds,η∈(0,1), | (1.2) |
where cDq denotes the Caputo fractional derivative of order q, and α∈R, 2η≠α.
In [18], the authors discussed the existence and uniqueness of solutions for the following nonlinear fractional differential equations with three-point fractional integral boundary conditions:
{cDqx(t)=f(t,x(t)),t∈(0,1),q∈(1,2],x(0)=0,x(1)=αIpx(η),η∈(0,1), | (1.3) |
where cD denotes the Caputo fractional derivative of order q, Ip is the Riemann-Liouville fractional integral of order p and α∈R, α≠Γ(p+2)ηp+1.
In [20] existence and uniqueness results are obtained for the following boundary value problem of fractional order differential equations with three-point fractional integral boundary conditions:
{cDα(x(t))=f(t,x(t),cDβx(t)),t∈[0,1],α∈(1,2]andβ∈(0,1),x(0)=0,bx(1)=c−aIγx(η),η∈(0,1), | (1.4) |
where cD denotes the Caputo fractional derivative of order q, Iγ the Riemann-Liouville fractional integral of order γ, f is a given continuous function, and a, b, c are real constants with aη1+γ≠−bΓ(β+2). We are concerned to study the hybrid fractional differential equations (also called the quadratic perturbations of nonlinear differential equations) because they have been extensively studied and have achieved a great deal of interest. First time, Dhage and Lakshmikantham in [26] proposed hybrid differential equations and showed some essential results on this kind of differential equations. In this class of the equations, the perturbations of the original differential equations are involved in many ways, for hybrid fractional differential equations, we refer to [25,26,27,28,29,30,31,32,33,34,35] and references therein.
In this paper, we discuss existence and uniqueness results for hybrid fractional differential equations with three-point boundary hybrid conditions, these results are determined, by applying fixed point theorems such as Banach’s fixed point theorem and Leray-Schauder Nonlinear Alternative. Our assumed problem will more complicated and general than the problems considered before and aforementioned above, we study the existence and uniqueness of solutions for the hybrid fractional differential equations given by with boundary hybrid conditions where t∈[0,1], γ and qi∈(0,1],i=1,....,m, with m∈N, η∈(0,1), and aη1+γΓ(γ+2)+b≠0.
cDα0+(x(t)−m∑i=1Iqi0+hi(t,x(t),cDβ0+x(t),Iq0+x(t)))=f(t,x(t),cDβ0+x(t),Iq0+x(t))α∈(1,2],βandq∈(0,1), | (1.5) |
{[x(t)−m∑i=1Iqi0+hi(t,x(t),cDβ0+x(t),Iq0+x(t))]t=0=0,aIγ0+[x(t)−m∑i=1Iqi0+hi(t,x(t),cDβ0+x(t),Iq0+x(t))]t=η+b[x(t)−m∑i=1Iqi0+hi(t,x(t),cDβ0+x(t),Iq0+x(t))]t=1=c, | (1.6) |
cDα0+ denotes the Caputo fractional derivative of order α and Iq0+ denotes Riemann-Liouville fractional integral of order q, and a, b, c are real constants with f,hi∈C([0,1]×R3,R).
This article is structured as follows. In part 2, we introduce notations, definitions, and lemmas. Next, in part 3, we prove the existence results for problems 1.5 and 1.6 using the fixed point theorem. Finally, we illustrate the results with examples.
In this section, we present the notation, definitions to be used throughout this article.
Definition 2.1. [25] The Riemann-Liouville fractional integral of order q for 0 continuous function f:[0,+∞)⟶R is defined as
Iq0+f(t)=1Γ(q)∫t0(t−s)α−1f(s)ds, |
where Γ is the Euler gamma function.
Definition 2.2. [34] Let α>0 and n=[α]+1. If f∈Cn([0,1]), then the Caputo fractional derivative of order α defined by
cDα0+f(t)=1Γ(n−α)∫t0(t−s)n−α−1f(n)(s)ds, |
exists almost everywhere on [0,1], [α] is the integer part of α.
Lemma 2.3. [4] Let α>β>0 and f∈L1([0,1]). Then for all t∈[0,1], we have:
Iα0+Iβ0+f(t)=Iα+β0+f(t),
cDα0+Iα0+f(t)=f(t),
cDα0+Iβ0+f(t)=Iα−β0+f(t).
Lemma 2.4. Let α>0. Then the differential equation
(cDα0+f)(t)=0, |
has a solution
f(t)=m−1∑j=0citi,ci∈R,j=0,...,m−1, |
where m–1<α<m.
Theorem 2.5. [19] Let X be a Banach space, let B be a closed, convex subset of X, let U be an open subset of B and 0∈U. Suppose that
P:U⟶B is a continuous and compact map. Then either
(a) P has a fixed point in U, or
(b) there exist an x∈∂U (the boundary of U) and λ∈(0,1) with x=λP(x).
In this section, we show the existence results for the boundary value problems on the interval [0,1].
Lemma 3.1. Let A(t) be continuous function on [0,1]. Then the solution of the boundary value problem
{cDα0+(x(t)−m∑i=1Iqi0+hi(t,x(t),cDβ0+x(t),Iq0+x(t)))=A(t),α∈(1,2],βandq∈(0,1),[x(t)−m∑i=1Iqi0+hi(t,x(t),cDβ0+x(t),Iq0+x(t))]t=0=0,aIγ0+[x(t)−m∑i=1Iqi0+hi(t,x(t),cDβ0+x(t),Iq0+x(t))]t=η+b[x(t)−m∑i=1Iqi0+hi(t,x(t),cDβ0+x(t),Iq0+x(t))]t=1=c, | (3.1) |
where t∈[0,1], γ,η and qi∈(0,1),i=1,....,m,m∈N, is given by
x(t)=Iα0+A(t)+t(c−bIα0+A(1)−aIγ+α0+A(η))aη1+γΓ(γ+2)+b+m∑i=1Iqi0+hi(t,x(t),cDβ0+x(t),Iq0+x(t)). | (3.2) |
Proof. For 1<α≤2 and some constants c0, c1∈R, the general solution of the equation
cDα0+(x(t)−m∑i=1Iqi0+hi(t,x(t),cDβ0+x(t),Iq0+x(t)))=A(t), |
can be written as
x(t)=Iα0+A(t)+c0+c1t+m∑i=1Iqi0+hi(t,x(t),cDβ0+x(t),Iq0+x(t)), | (3.3) |
applying the boundary conditions, we find that
c0=0, |
and
c1=c−bIα0+A(1)−aIα+γ0+A(η)aη1+γΓ(γ+2)+b. | (3.4) |
Substituting the values of c0, c1, we obtain the result, this completes the proof.
Now we list the following hypotheses
(H1) The functions f,hi:[0,1]×R3→R are continuous.
(H2) There exist positive functions ϕi,i=1,...m with bounds ‖ϕi‖1τi, such that
|hi(t,x1,y1,z1)−hi(t,x2,y2,z2)|≤ϕi(t)(|x1−x2|+|y1−y2|+|z1−z2|), |
for t∈[0,1],(xk,yk,zk)∈R3,k=1,2 and ϕi(t)∈L1τi([0,1],R+) and τi∈(0,1−α),i=1,..,m.
(H3) There exist positive function ψ with bounds ‖ψ‖1μ, such that
|f(t,x1,y1,z1)−f(t,x2,y2,z2)|≤ψ(t)(|x1−x2|+|y1−y2|+|z1−z2|), |
for t∈[0,1],(xk,yk,zk)∈R3,k=1,2 and ψ(t)∈L1μ([0,1],R+) and μ∈(0,1−α).
(H4) If Δ+Λ+Θ<1, where Δ,Λ and Θ are given by
Δ=‖ψ‖1μ(1−μα−μ)1−μΓ(α)(1+|b||aη1+γΓ(γ+2)+b|)+|a|‖ψ‖1μηα+γ−μ(1−μα+γ−μ)1−μΓ(α+γ)|aη1+γΓ(γ+2)+b|+m∑i=1‖ϕi‖1τi(1−τiqi−τi)1−τiΓ(qi). |
Λ=‖ψ‖1μ(1−μα+q−μ)1−μΓ(α+q)+‖ψ‖1μΓ(q+1)(aη1+γΓ(γ+2)+b)(|b|(1−μα−μ)1−μΓ(α)+|a|ηα+γ−μ(1−μα+γ−μ)1−μΓ(α+γ)|)+m∑i=1‖ϕi‖1τi(1−τiq+qi−τi)1−τiΓ(q+qi). |
Θ=‖ψ‖1μ(1−μα−β−μ)1−μΓ(α−β)+‖ψ‖1μΓ(2−β)(aη1+γΓ(γ+2)+b)(|b|(1−μα−μ)1−μΓ(α)+|a|ηα+γ−μ(1−μα+γ−μ)1−μΓ(α+γ)|)+m∑i=1‖ϕi‖1τi(1−τiβ−qi−τi)1−τiΓ(β−qi). |
Theorem 3.2. Assume that condition (H1−H4) hold, then problems (1.5) and (1.6) have a unique solution defined on [0,1]
Proof. Define the space
X={x:x,Iq0+xandcDβ0+x∈C([0,1],R),0<q,β<1}, |
endowed with the norm
‖x‖=maxt∈[0,1]|x(t)|+maxt∈[0,1]|Iq0+x(t)|+maxt∈[0,1]|cDβ0+x(t)|. |
We put
Fx(t)=f(t,x(t),cDβ0+x(t),Iq0+x(t)).
Hix(t)=hi(t,x(t),cDβ0+x(t),Iq0+x(t)),i=1,...,m.m∈N.
Obviously, (X,‖x‖) is Banach espace. In order to obtain the existence results of problems (1.5) and (1.6), by Lemma 3.1, we define an operator S:X⟶X as follows
Sx(t)=Iα0+Fx(t)+t(c−b(Iα0+F)(1)−a(Iγ+α0+Fx)(η))aη1+γΓ(γ+2)+b+m∑i=0Iqi0+Hix(t). |
Since f,hi continuous, it is easy to see that
(Iq0+Sx)(t)=Iα+q0+F(t)+(tq(c−b(Iα0+F)(1)−a(Iγ+α0+Fx)(η))Γ(q+1)(aη1+γΓ(γ+2)+b))+m∑i=0Iq+qi0+Hix(t),
and
(cDβ0+Sx)(t)=(Iα−β0+Fx)(t)+(t1−β(c−b(Iα0+F)(1)−a(Iγ+α0+Fx)(η))Γ(2−β)(aη1+γΓ(γ+2)+b))+(m∑i=0Iqi−β0+Hix)(t).
Let x,y∈X. Then for each t∈[0,1], we have
|(Sx)(t)−(Sy)(t)|≤|Iα0+(Fx−Fy)(t)|+|b||Iα0+(Fx−Fy)(1)+|a||Iγ+α0+(Fx−Fy)(η)|aη1+γΓ(γ+2)+b|+m∑i=0Iqi−β0+(Hix−Hiy)(t)≤Iα0+(ψ‖x−y‖)(t)+|b|Iα0+(ψi‖x−y‖)(1)+|a|Iγ+α0+(ψi‖x−y‖)(η)|aη1+γΓ(γ+2)+b|+m∑i=0Iqi−β0+ϕi‖x−y‖, |
by the Holder inequality, we have
|(Sx)(t)−(Sy)(t)|≤‖ψ‖1μ(1−μα−μ)1−μΓ(α)(1+|b||aη1+γΓ(γ+2)+b|)‖x−y‖+|a|‖ψ‖1μηα+γ−μ(1−μα+γ−μ)1−μΓ(α+γ)|aη1+γΓ(γ+2)+b|×‖x−y‖+m∑i=1‖ϕi‖1τi(1−τiqi−τi)1−τiΓ(qi)‖x−y‖=Δ‖x−y‖, |
similary, we have
|Iq0+(Sx)(t)−Iq0+(Sy)(t)|≤{‖ψ‖1μ(1−μα+q−μ)1−μΓ(α+q)+‖ψ1‖1μΓ(q+1)(|aη1+γΓ(γ+2)+b|)(|b|(1−μα−μ)1−μΓ(α)+|a|ηα+γ−μ(1−μα+γ−μ)1−μΓ(α+γ)|)+m∑i=1‖ϕi‖1τi(1−τiq+qi−τi)1−τiΓ(q+qi)}‖x−y‖=Λ‖x−y‖, |
and
|cDβ0+(Sx)(t)−cDβ0+(Sy)(t)|≤{‖ψ‖1μ(1−μα−β−μ)1−μΓ(α−β)+‖ψ‖1μΓ(2−β)(aη1+γΓ(γ+2)+b)(|b|(1−μα−μ)1−μΓ(α)+|a|ηα+γ−μ(1−μα+γ−μ)1−μΓ(α+γ)|)+m∑i=1‖ϕi‖1τi(1−τiβ−qi−τi)1−τiΓ(β−qi)}‖x−y‖=Θ‖x−y‖. |
Form the inequalities above, we can deduce that
‖Sx(t)−Sy(t)‖≤(Θ+Δ+Λ)‖x−y‖. |
By the contraction principale, we know that problem 3.1 has a unique solution.
Theorem 3.3. assume that
(1) We put Hi0=supt∈[0,1]hi(t,0,0,0),i=1,...,m,m∈N.
(2) There exist tree non-decreasing functions ρ1,ρ2,ρ3:[0,∞)→[0,∞) and a function ψ∈L1μ([0,1],R+) with μ∈(0,α−1)
|f(t,x,y,z)|≤ψ(t)(ρ1(|x|)+ρ2(|y|)+ρ3(|z|)), |
for t∈[0,1] and (x,y,z)∈R3.
(3) There exists a constant Z>0 such that
ZW1(Z)+‖ψ‖1μW2(ρ1(Z)+ρ2(Z)+ρ3(Z))≥1. | (3.5) |
Where
W1(Z)=(|c|(Γ(q+1)Γ(2−β)+Γ(2−β)+Γ(q+1))Γ(q+1)Γ(2−β)(|aη1+γΓ(γ+2)+b|))+m∑i=1(Hi0+Z‖ϕi‖1τi(1−τiqi−τi)1−τiΓ(qi))+m∑i=1(Hi0+Z‖ϕi‖1τi(1−τiq+qi−τi)1−τiΓ(q+qi))+m∑i=1(Hi0+Z‖ϕi‖1τi(1−τiqi−β−τi)1−τiΓ(qi−β)), |
W2=((1−μα−μ)1−μΓ(α))(1+|b||aη1+γΓ(γ+2)+b|+|b|Γ(q+1)(aη1+γΓ(γ+2)+b)+|b|Γ(2−β)(aη1+γΓ(γ+2)+b))+(|a|ηα+γ−μ(1−μα+γ−μ)1−μΓ(α+γ)|aη1+γΓ(γ+2)+b|)(1+1Γ(q+1)+1Γ(2−β))+(1−μα+q−μ)1−μΓ(α+q)+(1−μα−β−μ)1−μΓ(α−β). |
Then problem (1.5) and (1.6) have at least one solution on [0,1].
Proof. Define the a ball Br as
Br={x∈X:‖x‖≤r}, |
where the constant r satisfies
r⩾W1(r)+‖ψ‖1μW2(ρ1(r)+ρ2(r)+ρ3(r)). |
Clearly, Br is a closed convex bounded subset of the Banach space X. By Lemma 3.1 the boundary value problems (1.5) and (1.6) are equivalent to the equation
Sx(t)=Iα0+Fx(t)+t(c−bIα0+Fx(1)−aIγ+α0+Fx(η))aη1+γΓ(γ+2)+b+m∑i=1Iqi0+Hix(t), | (3.6) |
|Sx(t)|≤|Iα0+Fx(t)|+(|c|+|b|Iα0+|Fx(1)|+|a|Iγ+α0+|Fx(η)|)|aη1+γΓ(γ+2)+b|+m∑i=1|Iqi0+Hix(t)|, | (3.7) |
by the Holder inequality and the hypotheses, we have
|Sx(t)|≤(‖ψ‖1μ(1−μα−μ)1−μΓ(α))(ρ1(r)+ρ2(r)+ρ3(r))(1+|b||aη1+γΓ(γ+2)+b|)+m∑i=1(Hi0+r‖ϕi‖1τi(1−τiqi−τi)1−τiΓ(qi))+(|c|Γ(α+γ)+(ρ1(r)+ρ2(r)+ρ3(r))‖ψ‖1μ|a|ηα+γ−μ(1−μα+γ−μ)1−μΓ(α+γ)|aη1+γΓ(γ+2)+b|), |
similary, we have
|(Iq0+Sx)(t)|≤(ρ1(r)+ρ2(r)+ρ3(r))‖ψ‖1μ(‖ψ‖1μ(1−μα+q−μ)1−μΓ(α+q))+(|c|Γ(q+1)(aη1+γΓ(γ+2)+b))+(|b|(ρ1(r)+ρ2(r)+ρ3(r))‖ψ‖1μ(1−μα−μ)1−μΓ(α)Γ(q+1)(aη1+γΓ(γ+2)+b))+m∑i=1(Hi0+r‖ϕi‖1τi(1−τiq+qi−τi)1−τiΓ(q+qi))+((ρ1(r)+ρ2(r)+ρ3(r))‖ψ‖1μ|a|ηα+γ−μ(1−μα+γ−μ)1−μΓ(α+γ)Γ(q+1)|aη1+γΓ(γ+2)+b|), |
and
|cDβ0+Sx(t)|≤(ρ1(r)+ρ2(r)+ρ3(r))‖ψ‖1μ((1−μα−β−μ)1−μΓ(α−β))+(|c|Γ(2−β)(aη1+γΓ(γ+2)+b))+(|b|(ρ1(r)+ρ2(r)+ρ3(r))‖ψ‖1μ(1−μα−μ)1−μΓ(α)Γ(2−β)(aη1+γΓ(γ+2)+b))m∑i=1(Hi0+r‖ϕi‖1τi(1−τiqi−β−τi)1−τiΓ(qi−β))+((ρ1(r)+ρ2(r)+ρ3(r))‖ψ‖1μ|a|ηα+γ−μ(1−μα+γ−μ)1−μΓ(α+γ)Γ(2−β)|aη1+γΓ(γ+2)+b|). |
That is to say, we have
‖Sx(t)‖≤W1(r)+‖ψ‖1μW2(ρ1(r)+ρ2(r)+ρ3(r)). | (3.8) |
Secondly, we prove that S maps bounded sets into equicontinuous sets. Let Br be any bounded set of X. Notice that f and hi are continuous, therefore, without loss of generality, we can assume that there is an Mf and Mhi, i=1,...,m, such that
supt∈[0,1]f(t,x(t),cDα0+x(t),Iq0+x(t))=Mf, |
and
supt∈[0,1]hi(t,x(t),cDα0+x(t),Iq0+x(t))=Mhi. |
Now let 0≤t1≤t2≤1.We have the following facts:
|Sx(t2)−Sx(t1)|=|Iα0+Fx(t2)+t2(c−b(Iα0+F)(1)−a(Iγ+α0+Fx)(η))aη1+γΓ(γ+2)+b+m∑i=0Iqi0+Hix(t2)−Iα0+Fx(t1)−t1(c−b(Iα0+F)(1)−a(Iγ+α0+Fx)(η))aη1+γΓ(γ+2)+b−m∑i=0Iqi0+Hix(t1)|≤t1∫0(t2−s)α+1−(t1−s)α+1Γ(α)f(s,x(s),cDα0+x(s),Iq0+x(s))ds+t2∫t1(t2−s)α+1Γ(α)f(s,x(s),cDα0+x(s),Iq0+x(s)ds|+|t2−t1|((c−b(Iα0+F)(1)−a(Iγ+α0+Fx)(η))aη1+γΓ(γ+2)+b)+m∑i=0(t1∫0(t2−s)α+1−(t1−s)qi+1Γ(qi)hi(s,x(s),cDβ0+x(s),Iq0+x(s))ds+t1∫t2(t2−s)qi+1Γ(qi)hi(s,x(s),cDβ0+x(s),Iq0+x(s)))≤Mf(t2−t1)αΓ(α+1)+Mf(tα1−tα2)+(t2−t1)αΓ(α+1)+m∑i=0(Mhi(t2−t1)qiΓ(qi+1)+Mhi(tqi1−tqi2)+(t2−t1)qiΓ(qi+1))≤2Mf(t2−t1)αΓ(α+1)+m∑i=02Mhi(t2−t1)qiΓ(qi+1), |
we can get
|Sx(t1)−Sx(t2)|⟶0 as t2⟶t1. |
Similarly, we can obtain that
|Iq0+Sx(t1)−Iq0+Sx(t2)|⟶0ast2⟶t1,|cDα0+Sx(t1)−cDα0+Sx(t2)|⟶0ast2⟶t1. |
This implies that
‖Sx(t1)−Sx(t2)‖⟶0 as t2⟶t1. |
Finally, we let x=λSx for λ∈(0,1). Due to (3.8) and for each t∈[0,1] we have
‖x‖=‖λSx‖≤W1(‖x‖)+‖ψ‖1μW2(ρ1(‖x‖)+ρ2(‖x‖)+ρ3(‖x‖)). |
That is to say,
‖x‖W1(‖x‖)+‖ψ‖1μW2(ρ1(‖x‖)+ρ2(‖x‖)+ρ3(‖x‖))≤1. |
From (3.5), there exists Z>0 such that x≠Z. Define a set
O={y∈X:‖y‖≤Z}. |
The operator S:ˉO⟶X is continuous and completely continuous.
By the definition of the set O there is no x∈∂O such that x=λSx for some 0<λ<1.
Consequently, by Theorem 2.5, we obtain that S has a fixed point x∈O which is a solution of problems (1.5) and (1.6). This is the end of the proof.
In this section, we give two examples to illustrate the main results.
Consider the following fractional differential equation
{cD430+(x(t)−3∑i=1Iqi0+Hix(t))=esin(π(1+t2)e1(4+t)2|x|1+|x|+1(4+sin2(t))2(|cD340+x(t)|+|I780+x(t)|),[x(t)−3∑i=1Iqi0+H(t)]t=0=0,12I120+[x(t)−3∑i=1Iqi0+Hix(t)]t=12+13[x(t)−3∑i=1Iqi0+Hix(t)]t=1=14. | (4.1) |
We take
q1=9091, q2=910, q3=2930, μ=14
τ1=67, τ2=23, τ3=12
f(t,x(t),cD340+x(t),I780+x(t))=exp(sin(π(1+t2))e1(4+t)2|x|1+|x|+1(4+sin2(t))2(|cD340+x(t)|+|I780+x(t)|),h1(t,x(t),cD340+x(t),I780+x(t))=t210(x(t)+cD340+x(t)x(t)+cD340+x(t)+1+|I780+x(t)|)+exp(−t2),h2(t,x(t),cD340+x(t),I780+x(t))=e√2tsin(πt2)(√3+√5e√2t)2(cos(x(t)+cD340+x(t)x(t)+cD340+x(t)+1)+e√2tcos(πt)(√3+√5e√2t)2|I780+x(t)|),h3(t,x(t),cD340+x(t),I780+x(t))=exp(−2t)10(log(x(t)+1)+coscD340+x(t)+sin(I780+x(t))). |
We can show that
|F(x(t))−F(y(t))|≤116(|x−y|+|cD340+x−cD340+y|+|I780+x−cD780+y|),|H1(x(t))−H1(y(t))|≤t210(|x−y|+|cD340+x−cD340+y|+|I780+x−cD780+y|),|H2(x(t))−H2(y(t))|≤exp(√2t)(√3+√5exp(√2t))2(|x−y|+|cD340+x−cD340+y|+|I780+x−I780+y|),|H3(x(t))−H3(y(t))|≤exp(−2t)10(|x−y|+|cD340+x−cD340+y|+|I780+x−I780+y|), |
where
ψ=116, ϕ1=t210, ϕ2=exp(√2t)(√3+√5exp(√2t))2, ϕ3=exp(−2t)10.
Then, we have
‖ψ‖1μ≈0.0423, ‖ϕ1‖τ1≈0.0356, ‖ϕ2‖τ2≈0.05114, ‖ϕ3‖τ3≈0.0495,
and
Δ≈0.2109,Λ≈0.1528,Θ≈0.2276,
and
Δ+Λ+Θ≈0.5914<1.
By Theorem 3.2, we know that proplem 4.1 has a unique solution defined on [0,1].
Consider the following fractional differential equation
{cD530+(x(t)−2∑i=1Iqi0+Hix(t))=e−4t10sin(x(t)+12I890+x(t))+e−4t4D1500+x(t),[x(t)−2∑i=1Iqi0+H(t)]t=0=0,110I130+[x(t)−2∑i=1Iqi0+Hix(t)]t=14+150[x(t)−2∑i=1Iqi0+Hix(t)]t=1=12. | (4.2) |
We choose
μ=23, q1=100101, q2=1112, τ1=110, τ2=120.
f(t,x(t),cD150x(t),I890+x(t))=e−2t10sin(120x(t)+130I890+x(t))+e−4t60cD1500+x(t),
h1(t,x(t),cD150x(t),I890+x(t))=e−t2(5+t)2|x|1+|x|+116I890+x+cD150x(4+sin2(x))2+t10,
h2(t,x(t),cD150x(t),I890+x(t))=t2(sin(x(t)+cD1500+x(t))x(t)+cD1500+x(t)+1+sin(I890+x(t))+130).
We can demonstrate that
|f(t,x(t),cD150x(t),I890+x(t))|≤ψ(t)(ρ1(|x|)+ρ2(|cD150x|)+ρ3(|I890+x|)),
h1(t,x(t),cD150x(t),I890+x(t))−h1(t,y(t),cD150y(t),I890+y(t))≤116(|x−y|+|cD150x−cD150y|+|I890+x−I890+y|),
h2(t,x(t),cD150x(t),I890+x(t))−h1(t,y,cD150y,I890+y)≤t2(|x−y|+|cD150x−cD150y|+|I890+x−I890+y|),
where
ψ(t)=e−2t10, ϕ1=116, ϕ2=t2.
ρ1(|x|)=120|x|, ρ2(|cD150x|)=130|cD150x|, ρ3(|I890+x|)=160|I890+x|.
Hence we have
‖ϕ1‖τ1≈0.0625, ‖ϕ2‖τ2≈0.2714,H10=110,H10=160.
After calculation, it ensues by 3.5 that the constant Z provides the inequality Z>31,9308. Since all the stipulations of theorem 3.3 are completed, the problem 4.2 has at least one solution on [0,1].
In this article, we have presented some sufficient conditions include the existence of solutions for a kind of hybrid fractional differential equations inclose fractional Caputo derivative of order 1<α≤2. Our results depend on a fixed point theorems such as Banach’s fixed point theorem and Leray-Schauder nonlinear alternative. Our results prolong and complete those in the literature.
The author would like to express his thanks to the referees and editors for their helpful comments and suggestions.
The author declares no conflicts of interest.
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