Citation: Dehong Ji, Weigao Ge. A nonlocal boundary value problems for hybrid ϕ-Caputo fractional integro-differential equations[J]. AIMS Mathematics, 2020, 5(6): 7175-7190. doi: 10.3934/math.2020459
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Hybrid differential equations have been considered more important and served as special cases of dynamical systems. Dhage and Lakshmikantham [1] were the first to study ordinary hybrid differential equation and studied the existence of solutions for this boundary value problem. In recent years, with the wide study of fractional differential equations, the theory of hybrid fractional differential equations were also studied by several researchers, see [2,3,4,5,6,7,8,9,10] and the references therein.
Zhao et al. [2] studied existence and uniqueness results for the following hybrid differential equations involving Riemann-Liouville fractional derivative
Dq0+(x(t)f(t,x(t)))=g(t,x(t)), a.e.t∈J=[0,T] |
x(0)=0, |
where 0<q<1,f∈C(J×R→R∖{0}) and g∈C(J×R,R).
Zidane Baitiche et al. [11] considered the following boundary value problem of nonlinear fractional hybrid differential equations involving Caputo's derivative
CDα0+(x(t)f(t,x(μ(t))))=g(t,x(μ(t))), t∈I=[0,1] |
a[x(t)f(t,x(μ(t)))]|t=0+b[x(t)f(t,x(μ(t)))]|t=1=c, |
where 0<α≤1,CDα0+ is the Caputo fractional derivative. f∈C(I×R→R∖{0}),g∈C(I×R,R).
As we all known, the hadamard fractional differential equations are also popular in the literature, see [12,13,14,15,16], so some authors began to study the theory of fractional hybrid differential equation of hadamard type.
Zidane Baitiche et al. [17] studied the existence of solutions for fractional hybrid differential equation of hadamard type with dirichlet boundary conditions
HDα(x(t)f(t,x(t)))=g(t,x(t)), 1<t<e, 1<α≤2, |
x(1)=0, x(e)=0, |
where 1<α≤2, HDα is the Hadamard fractional derivative, f∈C([1,e]×R→R∖{0}) and g∈C([1,e]×R,R).
In [18], M. Jamil et al. discussed the existence result for the boundary value problem of hybrid fractional integro-differential equations involving Caputo's derivative given by
CDα(CDωu(t)−∑mi=1Iβifi(t,u(t))g(t,u(t)))=h(t,u(t),Iγu(t)), t∈J=[0,1], |
u(0)=0, Dωu(0)=0, u(1)=δu(η), 0<δ<1, 0<η<1, |
where CDα is the Caputo fractional derivative of order α, CDω is the Caputo fractional derivative of order ω, 0<α≤1, 1<ω≤2.
In order to analyze fractional differential equations in a generic way, a fractional derivative with respect to another function called φ-Caputo derivative was proposed [19].
By mixing idea of the above works, we derived an existence result for the nonlocal boundary value problems of hybrid φ-Caputo fractional integro-differential equations
CDα φ(CDβ φu(t)−∑mi=1Iωi φfi(t,u(t),Iμ1 φu(t),⋅⋅⋅,Iμn φu(t))g(t,u(t),Iγ1 φu(t),⋅⋅⋅,Iγp φu(t)))=h(t,u(t)),t∈J=[0,1], | (1.1) |
u(0)=0, CDβ φu(0)=0, u(1)=k∑j=1δju(ξj), | (1.2) |
where 0<α≤1, 1<β≤2, CDα φ is the φ-Caputo fractional derivative of order α, CDβ φ is the φ-Caputo fractional derivative of order β, the function φ: [0,1]→R is a strictly increasing function such that φ∈C2[0,1] with φ′(x)>0 for all x∈[0,1], Iμ φ denote the φ-Riemann-Liouville fractional integral of order μ, g∈C(J×Rp+1,R∖{0}), h∈C(J×R,R) and fi∈C(J×Rn+1,R) with fi(0,0,⋅⋅⋅,0⏟n+1)=0, wi>0, i=1,2,⋅⋅⋅,m, μ1,⋅⋅⋅,μn>0 and γ1,⋅⋅⋅,γp>0, 0<δj<1, j=1,2,⋅⋅⋅,k, 0<ξ1<ξ2<⋅⋅⋅<ξk<1.
It is notable that the fractional hybrid integro-differential equation presented in this paper is the novel in the sense that the fractional derivative with respect to another function called φ-Caputo fractional derivative. Note that the hybrid fractional integro-differential equations involving Caputo's derivative in [18] is a special case of our hybrid φ-Caputo fractional integro-differential equations with φ(t)=t. Moreover, all dependent functions fi and g in our paper are in the form of multi-term. Furthermore, our problem is more general than the work in [8], as we consider the problem with multi-point boundary conditions, while the authors in [8] only investigated two-point boundary condition.
The organization of this work is as follows. Section 2 contains some preliminary facts that we need in the sequel. In section 3, we present the solution for the hybrid fractional integro-differential equation (1.1), (1.2) and then prove our main existence results. Finally, we illustrate the obtained results by an example.
In the following and throughtout the text, a>0 is a real, x:[a,b]→R an integrable function and φ∈C2[a,b] an increasing function such that with φ′(t)≠0 for all t∈[a,b].
Definition 2.1 The φ-Riemann-Liouville fractional integral of x of order α is defined as follows
Iα φa+x(t):=1Γ(α)∫taφ′(s)(φ(t)−φ(s))α−1x(s)ds. |
Definition 2.2 The φ-Riemann-Liouville fractional derivative of x of order α is defined as follows
Dα φa+x(t):=(1φ′(t)ddt)nIn−α φa+x(t)=1Γ(n−α)(1φ′(t)ddt)n∫taφ′(s)(φ(t)−φ(s))n−α−1x(s)ds, |
here n=[α]+1.
Remark 2.1 Let α,β>0, then the relation holds
Iα φa+Iβ φa+x(t)=Iα+β φa+x(t). |
Definition 2.3 Let α>0 and x∈Cn−1[a,b], the φ-Caputo fractional derivative of x of order α is defined as follows
CDα φa+x(t):=Dα φa+[x(t)−n−1∑k=0x[k]φ(a)k!(φ(t)−φ(a))k], n=[α]+1 for α∉N, n=α for α∈N, |
where x[k]φ(t):=(1φ′(t)ddt\bigamma)kx(t).
Theorem 2.1 [20] Let x:[a,b]→R. The following results hold:
1. If x∈C[a,b], then CDα φa+Iα φa+x(t)=x(t);
2. If x∈Cn−1[a,b], then
Iα φ Ca+Dα φa+x(t)=x(t)−n−1∑k=0x[k]φ(a)k!(φ(t)−φ(a))k. |
Lemma 2.2 [18] Let S be a nonempty, convex, closed, and bounded set such that S⊆E, and let A:E→E and B:S→E be two operators which satisfy the following :
(H1)A is contraction;
(H2)B is compact and continuous, and
(H3)u=Au+Bv, ∀v∈S⇒u∈S.
Then there exists a solution of the operator equation u=Au+Bu.
Let E=C(J,R) be a Banach space equipped with the norm
‖u‖=supt∈J|u(t)| and (uv)(t)=u(t)v(t), ∀ t∈J. |
Then E is a Banach algebra with the above norm and multiplication.
Lemma 3.1 Suppose that α,β,ωi,i=1,2,⋅⋅⋅,m,γi,i=1,2,⋅⋅⋅,p,μi,i=1,2,⋅⋅⋅,n,δj,ξj,j=1,2,⋅⋅⋅,k and functions g,h,fi,i=1,2,⋅⋅⋅,m satisfy problem (1.1), (1.2). Then the unique solution of (1.1), (1.2) is given by
u(t)=∫t0(φ(t)−φ(s))β−1Γ(β)φ′(s)g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,u(τ))dτds+m∑i=1Iωi+β φfi(t,u(t),Iμ1 φu(t),⋅⋅⋅,Iμn φu(t))+φ(t)−φ(0)k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))[∫10(φ(1)−φ(s))β−1Γ(β)φ′(s)g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,u(τ))dτds+m∑i=1Iωi+β φfi(1,u(1),Iμ1 φu(1),⋅⋅⋅,Iμn φu(1))−k∑j=1δj∫ξj0(φ(ξj)−φ(s))β−1Γ(β)φ′(s)g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,u(τ))dτds−k∑j=1δjm∑i=1Iωi+β φfi(ξj,u(ξj),Iμ1 φu(ξj),⋅⋅⋅,Iμn φu(ξj))], | (3.1) |
where
Iωi+β φfi(t,u(t),Iμ1 φu(t),⋅⋅⋅,Iμn φu(t))=∫t0(φ(t)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,u(s),Iμ1 φu(s),⋅⋅⋅,Iμn φu(s))ds; |
Iωi+β φfi(1,u(1),Iμ1 φu(1),⋅⋅⋅,Iμn φu(1))=∫10(φ(1)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,u(s),Iμ1 φu(s),⋅⋅⋅,Iμn φu(s))ds; |
Iωi+β φfi(ξj,u(ξj),Iμ1 φu(ξj),⋅⋅⋅,Iμn φu(ξj))=∫ξj0(φ(ξj)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,u(s),Iμ1 φu(s),⋅⋅⋅,Iμn φu(s))ds. |
Proof. We apply φ-Riemann-Liouville fractional integral Iα φ on both sides of (1.1), by Theorem 2.1, we have
CDβ φu(t)−∑mi=1Iωi φfi(t,u(t),Iμ1 φu(t),⋅⋅⋅,Iμn φu(t))g(t,u(t),Iγ1 φu(t),⋅⋅⋅,Iγp φu(t))=Iα φh(t,u(t))+c0, |
then by u(0)=0, CDβ φu(0)=0, fi(0,0,⋅⋅⋅,0⏟n+1)=0, we get c0=0. i.e,
CDβ φu(t)=g(t,u(t),Iγ1 φu(t),⋅⋅⋅,Iγp φu(t))∫t0(φ(t)−φ(s))α−1Γ(α)φ′(s)h(s,u(s))ds+m∑i=1Iωi φfi(t,u(t),Iμ1 φu(t),⋅⋅⋅,Iμn φu(t)). | (3.2) |
Apply again fractional integral Iβ φ on both sides of (3.2) and by Theorem 2.1, we get
u(t)=∫t0(φ(t)−φ(s))β−1Γ(β)φ′(s)g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,u(τ))dτds+m∑i=1Iωi+β φfi(t,u(t),Iμ1 φu(t),⋅⋅⋅,Iμn φu(t))+c1+c2(φ(t)−φ(0)), | (3.3) |
u(0)=0, fi(0,0,⋅⋅⋅,0⏟n+1)=0 yield c1=0, thus equation (3.3) is reduced to
u(t)=∫t0(φ(t)−φ(s))β−1Γ(β)φ′(s)g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,u(τ))dτds+m∑i=1Iωi+β φfi(t,u(t),Iμ1 φu(t),⋅⋅⋅,Iμn φu(t))+c2(φ(t)−φ(0)), | (3.4) |
specially.
u(1)=∫10(φ(1)−φ(s))β−1Γ(β)φ′(s)g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,u(τ))dτds+m∑i=1Iωi+β φfi(1,u(1),Iμ1 φu(1),⋅⋅⋅,Iμn φu(1))+c2(φ(1)−φ(0)), |
u(ξj)=∫ξj0(φ(ξj)−φ(s))β−1Γ(β)φ′(s)g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,u(τ))dτds+m∑i=1Iωi+β φfi(ξj,u(ξj),Iμ1 φu(ξj),⋅⋅⋅,Iμn φu(ξj))+c2(φ(ξj)−φ(0)), |
from u(1)=k∑j=1δju(ξj), we have
c2=1k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))[∫10(φ(1)−φ(s))β−1Γ(β)φ′(s)g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,u(τ))dτds+m∑i=1Iωi+β φfi(1,u(1),Iμ1 φu(1),⋅⋅⋅,Iμn φu(1))−k∑j=1δj∫ξj0(φ(ξj)−φ(s))β−1Γ(β)φ′(s)g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,u(τ))dτds−k∑j=1δjm∑i=1Iωi+β φfi(ξj,u(ξj),Iμ1 φu(ξj),⋅⋅⋅,Iμn φu(ξj))]. |
Consequently, we can get the desired result. The proof is completed.
Theorem 3.2 Suppose that functions g∈C(J×Rp+1,R∖{0}), h∈C(J×R,R) and fi∈C(J×Rn+1,R) with fi(0,0,⋅⋅⋅,0⏟n+1)=0. Furthermore, assume that
(C1) there exist bounded mapping σ:[0,1]→R+, λ:[0,1]→R+ such that
|g(t,k1,k2,⋅⋅⋅,kp+1)−g(t,k′1,k′2,⋅⋅⋅,k′p+1)|≤σ(t)p+1∑i=1|ki−k′i| |
for t∈J and (k1,k2,⋅⋅⋅,kp+1),(k′1,k′2,⋅⋅⋅,k′p+1)∈Rp+1, and
|h(t,u)−h(t,v)|≤λ(t)|u−v| for t∈J and u,v∈R;
(C2) there exist ϕi,Ω,χ∈C(J,R+),i=1,2,⋅⋅⋅,m such that
|fi(t,k1,k2,⋅⋅⋅,kn+1)|≤ϕi(t), ∀ (t,k1,k2,⋅⋅⋅,kn+1)∈J×Rn+1, |
|h(t,u)|≤Ω(t), ∀ (t,u)∈J×R, |
|g(t,k1,k2,⋅⋅⋅,kp+1)|≤χ(t), ∀ (t,k1,k2,⋅⋅⋅,kp+1)∈J×Rp+1; |
(C3) there exists r>0 such that
(1+(φ(1)−φ(0))(1+k∑j=1δj)|k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))|)(χ∗Ω∗(φ(1)−φ(0))αΓ(α+1)(φ(1)−φ(0))βΓ(β+1)+m∑i=1ϕ∗i(φ(1)−φ(0))ωi+βΓ(ωi+β+1))≤r; | (3.5) |
(χ∗λ∗+Ω∗σ∗p+1∑i=1(φ(1)−φ(0))γiΓ(γi+1))(φ(1)−φ(0))αΓ(α+1)(1+(φ(1)−φ(0))(1+k∑j=1δj)|k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))|)(φ(1)−φ(0))βΓ(β+1)<1, | (3.6) |
where Ω∗=sup0≤t≤1|Ω(t)|, ϕ∗i=sup0≤t≤1|ϕi(t)|, i=1,2,⋅⋅⋅,p, χ∗=sup0≤t≤1|χ(t)|, λ∗=sup0≤t≤1|λ(t)|, σ∗=sup0≤t≤1|σ(t)|.
Then the hybrid problem (1.1), (1.2) has at least one solution.
Proof. Define a subset S of E as
S={u∈E: ‖u‖≤r}, |
where r satisfies inequality (3.5). Clearly S is closed, convex and bounded subset of the Banach space E. Define two operators A:E→E by
Au(t)=∫t0(φ(t)−φ(s))β−1Γ(β)φ′(s)g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,u(τ))dτds+φ(t)−φ(0)k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))∫10(φ(1)−φ(s))β−1Γ(β)φ′(s)g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,u(τ))dτds−(φ(t)−φ(0))k∑j=1δjk∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))∫ξj0(φ(ξj)−φ(s))β−1Γ(β)φ′(s)g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,u(τ))dτds, | (3.7) |
Bu(t)=m∑i=1∫t0(φ(t)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,u(s),Iμ1 φu(s),⋅⋅⋅,Iμn φu(s))ds+φ(t)−φ(0)k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))m∑i=1∫10(φ(1)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,u(s),Iμ1 φu(s),⋅⋅⋅,Iμn φu(s))ds−(φ(t)−φ(0))k∑j=1δjk∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))m∑i=1∫ξj0(φ(ξj)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,u(s),Iμ1 φu(s),⋅⋅⋅,Iμn φu(s))ds. | (3.8) |
Then u(t) is a solution of problem (1.1), (1.2) if and only if u(t)=Au(t)+Bu(t). We shall show that the operators A and B satisfy all the conditions of Lemma 2.2. We split the proof into several steps.
Step 1. We first show that A is a contraction mapping. Let u(t),v(t)∈S, we write
G(s)=g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,u(τ))dτ−g(s,v(s),Iγ1 φv(s),⋅⋅⋅,Iγp φv(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,v(τ))dτ, |
then by (C1) we have
|G(s)|=|g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,u(τ))dτ−g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,v(τ))dτ+g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,v(τ))dτ−g(s,v(s),Iγ1 φv(s),⋅⋅⋅,Iγp φv(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,v(τ))dτ|≤|g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))|∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)|h(τ,u(τ))−h(τ,v(τ))|dτ+∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)|h(τ,v(τ))|dτ|g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))−g(s,v(s),Iγ1 φv(s),⋅⋅⋅,Iγp φv(s))|≤χ∗λ∗‖u−v‖(φ(s)−φ(0))αΓ(α+1)+Ω∗(φ(s)−φ(0))αΓ(α+1)σ∗p+1∑i=1(φ(s)−φ(0))γiΓ(γi+1)‖u−v‖≤(χ∗λ∗+Ω∗σ∗p+1∑i=1(φ(1)−φ(0))γiΓ(γi+1))(φ(1)−φ(0))αΓ(α+1)‖u−v‖, |
thus we have
|Au(t)−Av(t)|≤∫t0(φ(t)−φ(s))β−1Γ(β)φ′(s)G(s)ds+φ(t)−φ(0)|k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))|∫10(φ(1)−φ(s))β−1Γ(β)φ′(s)G(s)ds+φ(t)−φ(0)|k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))|k∑j=1δj∫ξj0(φ(ξj)−φ(s))β−1Γ(β)φ′(s)G(s)ds≤(χ∗λ∗+Ω∗σ∗p+1∑i=1(φ(1)−φ(0))γiΓ(γi+1))(φ(1)−φ(0))αΓ(α+1)(1+(φ(1)−φ(0))(1+k∑j=1δj)|k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))|)(φ(1)−φ(0))βΓ(β+1)‖u−v‖, |
which implies
‖Au(t)−Av(t)‖≤[(χ∗λ∗+Ω∗σ∗p+1∑i=1(φ(1)−φ(0))γiΓ(γi+1))(φ(1)−φ(0))αΓ(α+1)(1+(φ(1)−φ(0))(1+k∑j=1δj)|k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))|)(φ(1)−φ(0))βΓ(β+1)]‖u−v‖, |
in view of (3.6), this shows that A is a contraction mapping.
Step 2. The operator B is compact and continuous on S.
First, we show that B is continuous on S. Let {un} be a sequence of functions in S converging to a function u∈S. Then by Lebesgue dominated convergence theorem,
limn→∞Bun(t)=limn→∞[m∑i=1∫t0(φ(t)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,un(s),Iμ1 φun(s),⋅⋅⋅,Iμn φun(s))ds+φ(t)−φ(0)k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))m∑i=1∫10(φ(1)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,un(s),Iμ1 φun(s),⋅⋅⋅,Iμn φun(s))ds−(φ(t)−φ(0))k∑j=1δjk∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))m∑i=1∫ξj0(φ(ξj)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,un(s),Iμ1 φun(s),⋅⋅⋅,Iμn φun(s))ds].=m∑i=1∫t0(φ(t)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)limn→∞fi(s,un(s),Iμ1 φun(s),⋅⋅⋅,Iμn φun(s))ds+φ(t)−φ(0)k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))m∑i=1∫10(φ(1)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)limn→∞fi(s,un(s),Iμ1 φun(s),⋅⋅⋅,Iμn φun(s))ds−(φ(t)−φ(0))k∑j=1δjk∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))m∑i=1∫ξj0(φ(ξj)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)limn→∞fi(s,un(s),Iμ1 φun(s),⋅⋅⋅,Iμn φun(s))ds=m∑i=1∫t0(φ(t)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,u(s),Iμ1 φ(s),⋅⋅⋅,Iμn φu(s))ds+φ(t)−φ(0)k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))m∑i=1∫10(φ(1)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,u(s),Iμ1 φu(s),⋅⋅⋅,Iμn φu(s))ds−(φ(t)−φ(0))k∑j=1δjk∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))m∑i=1∫ξj0(φ(ξj)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,u(s),Iμ1 φu(s),⋅⋅⋅,Iμn φu(s))ds=Bu(t). |
This shows that B is continuous on S. It is sufficient to show that B(S) is a uniformly bounded and equicontinuous set in E.
First, we note that
|Bu(t)|≤m∑i=1∫t0(φ(t)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)|fi(s,u(s),Iμ1 φu(s),⋅⋅⋅,Iμn φu(s))|ds+φ(t)−φ(0)|k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))|m∑i=1∫10(φ(1)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)|fi(s,u(s),Iμ1 φu(s),⋅⋅⋅,Iμn φu(s))|ds+(φ(t)−φ(0))k∑j=1δj|k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))|m∑i=1∫ξj0(φ(ξj)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)|fi(s,u(s),Iμ1 φu(s),⋅⋅⋅,Iμn φu(s))|ds≤m∑i=1ϕ∗i(φ(1)−φ(0))ωi+βΓ(ωi+β+1)+(φ(1)−φ(0))(1+k∑j=1δj)|k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))|m∑i=1ϕ∗i(φ(1)−φ(0))ωi+βΓ(ωi+β+1)=(1+(φ(1)−φ(0))(1+k∑j=1δj)|k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))|)m∑i=1ϕ∗i(φ(1)−φ(0))ωi+βΓ(ωi+β+1). |
This shows that B is uniformly bounded on S.
Next, we show that B is an equicontinuous set in E. Let t1,t2∈J with t1<t2 and u∈S. Then we have
|Bu(t2)−Bu(t1)|=|m∑i=1∫t20(φ(t2)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,u(s),Iμ1 φu(s),⋅⋅⋅,Iμn φu(s))ds−m∑i=1∫t10(φ(t1)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,u(s),Iμ1 φu(s),⋅⋅⋅,Iμn φu(s))ds+φ(t2)−φ(t1)k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))m∑i=1∫10(φ(1)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,u(s),Iμ1 φu(s),⋅⋅⋅,Iμn φu(s))ds−(φ(t2)−φ(t1))k∑j=1δjk∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))m∑i=1∫ξj0(φ(ξj)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,u(s),Iμ1 φu(s),⋅⋅⋅,Iμn φu(s))ds|≤m∑i=1ϕ∗iΓ(ωi+β)[|∫t10[(φ(t2)−φ(s))ωi+β−1−(φ(t1)−φ(s))ωi+β−1]φ′(s)ds+∫t2t1[(φ(t2)−φ(s))ωi+β−1φ′(s)ds|+φ(t2)−φ(t1)|k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))|∫10(φ(1)−φ(s))ωi+β−1φ′(s)ds+(φ(t2)−φ(t1))k∑j=1δj|k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))|∫ξj0(φ(ξj)−φ(s))ωi+β−1φ′(s)ds]≤m∑i=1ϕ∗iΓ(ωi+β+1)[|(φ(t2)−φ(0))ωi+β−(φ(t1)−φ(0))ωi+β|+φ(t2)−φ(t1)|k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))|(φ(1)−φ(0))ωi+β+(φ(t2)−φ(t1))k∑j=1δj|k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))|(φ(ξj)−φ(0))ωi+β]. |
Let h(t)=(φ(t)−φ(0))ωi+β. Then h is continuously differentiable function. Consequently, for all t1,t2∈[0,1], without loss of generality, let t1<t2, then there exist positive constants M such that
|h(t2)−h(t1)|=|h′(ξ)||t2−t1|≤M|t2−t1|, ξ∈(t1,t2). |
On the other hand, for φ∈C′[0,1], thus there exist positive constants N such that |φ(t2)−φ(t1)|=|φ′(ξ)||t2−t1|≤N|t2−t1|, ξ∈(t1,t2), from which we deduce
|Bu(t2)−Bu(t1)|→0 as t2−t1→0. |
Therefore, it follows from the Arzela-Ascoli theorem that B is a compact operator on S.
Step 3. Next we show that hypothesis (H3) of Lemma 2.2 is satisfied. Let v∈S, then we have
|u(t)|=|Au(t)+Bv(t)|≤|Au(t)|+|Bv(t)|≤|∫t0(φ(t)−φ(s))β−1Γ(β)φ′(s)g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,u(τ))dτds+φ(t)−φ(0)k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))∫10(φ(1)−φ(s))β−1Γ(β)φ′(s)g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,u(τ))dτds−(φ(t)−φ(0))k∑j=1δjk∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))∫ξj0(φ(ξj)−φ(s))β−1Γ(β)φ′(s)g(s,u(s),Iγ1 φu(s),⋅⋅⋅,Iγp φu(s))∫s0(φ(s)−φ(τ))α−1Γ(α)φ′(τ)h(τ,u(τ))dτds|+|m∑i=1∫t0(φ(t)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,v(s),Iμ1 φv(s),⋅⋅⋅,Iμn φv(s))ds+φ(t)−φ(0)k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))m∑i=1∫10(φ(1)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,v(s),Iμ1 φv(s),⋅⋅⋅,Iμn φv(s))ds−(φ(t)−φ(0))k∑j=1δjk∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))m∑i=1∫ξj0(φ(ξj)−φ(s))ωi+β−1Γ(ωi+β)φ′(s)fi(s,v(s),Iμ1 φv(s),⋅⋅⋅,Iμn φv(s))ds|≤(1+(φ(1)−φ(0))(1+k∑j=1δj)|k∑j=1δj(φ(ξj)−φ(0))−(φ(1)−φ(0))|)(χ∗Ω∗(φ(1)−φ(0))αΓ(α+1)(φ(1)−φ(0))βΓ(β+1)+m∑i=1ϕ∗i(φ(1)−φ(0))ωi+βΓ(ωi+β+1))≤r, |
which implies ‖u‖≤r and so u∈S.
Thus all the conditions of Lemma 2.2 are satisfied and hence the operator equation u=Au+Bu has a solution in S. In consequence, the problem (1.1), (1.2) has a solution on J. This completes the proof.
In this section, we provide an example to illustrate our main result.
Example 4.1 Consider the following hybrid φ-Caputo fractional integro-differential equations
CD12 t4(CD32 t4u(t)−2∑i=1Iωit4fi(t,u(t),I13t4u(t),I43t4u(t))14t2(|u(t)|1+|u(t)|+|I14t4u(t)|1+|I14t4u(t)|+sinI12t4u(t)))=25cos(t4)(|u(t)||u(t)|+1), t∈J=[0,1], | (4.1) |
u(0)=0, CD32 t4u(0)=0, u(1)=13u(13), | (4.2) |
where
2∑i=1Iωit4fi(t,u(t),I13t4u(t),I43t4u(t))=I13t4(t[|u(t)|1+|u(t)|+sin(I13t4u(t))+cos(I43t4u(t))])+I23t4(t10[|u(t)|1+|u(t)|+arctan(I13t4u(t))+sin(I43t4u(t))]). | (4.3) |
We note that α=12,β=32,m=2,n=2,p=2,k=1,δ=13,ξ=13,ω1=13,ω2=23,μ1=13,μ2=43,γ1=14,γ2=12,φ(t)=t4,
f1(t,u(t),I13t4u(t),I43t4u(t))=t[|u(t)|1+|u(t)|+sin(I13t4u(t))+cos(I43t4u(t))], |
f2(t,u(t),I13t4u(t),I43t4u(t))=t10[|u(t)|1+|u(t)|+arctan(I13t4u(t))+sin(I43t4u(t))], |
g(t,u(t),I14t4u(t),I12t4u(t))=14t2(|u(t)|1+|u(t)|+|I14t4u(t)|1+|I14t4u(t)|+sinI12t4u(t)), |
h(t,u(t))=25cos(t4)(|u(t)||u(t)|+1). |
Thus we have
|g(t,u(t),I14t4u(t),I12t4u(t))−g(t,v(t),I14t4v(t),I12t4v(t))|≤σ(t)[1+t14Γ(54)+t12Γ(32)]|u(t)−v(t)|=t24[1+t14Γ(54)+t12Γ(32)]|u(t)−v(t)|, |
|h(t,u(t))−h(t,v(t))|=25cos(t4)|u(t)−v(t)|. |
Therefore,
σ∗=sup0≤t≤1|σ(t)|=sup0≤t≤1t24[1+t14Γ(54)+t12Γ(32)]=14(1+1Γ(54)+1Γ(32))=14(1+10.9064+10.8862)=0.8079; |
λ∗=sup0≤t≤1|λ(t)|=sup0≤t≤125cos(t4)=0.4; |
ϕ∗1=sup0≤t≤1|ϕ1(t)|=sup0≤t≤1t(1+1+1)=3; |
ϕ∗2=sup0≤t≤1|ϕ2(t)|=sup0≤t≤1t10(1+π2+1)=110×3.57=0.357; |
Ω∗=sup0≤t≤1|Ω(t)|=sup0≤t≤125cos(t4)=0.4; |
χ∗=sup0≤t≤1|χ(t)|=sup0≤t≤1t24(1+1+1)=34=0.75. |
Choose r>0.5, then we have
(1+14×4329)[0.75×0.4×(14)12Γ(32)×(14)32Γ(52)+3×(14)116Γ(176)+0.357×(14)136Γ(196)]=0.4016≤r. |
Moreover,
(0.75×0.4+0.4×0.8079×((14)14Γ(54)+(14)12Γ(32)))(14)12Γ(32)(1+14×4329)(14)32Γ(52)=0.097<1. |
Now, by using Theorem 3.2, it is deduced that the fractional hybrid integro-differential problem (4.1), (4.2) has a solution.
Hybrid fractional integro-differential equations have been considered more important and served as special cases of dynamical systems. In this paper, we introduced a new class of the hybrid φ-Caputo fractional integro-differential equations. By using famous hybrid fixed point theorem due to Dhage, we have developed adequate conditions for the existence of at least one solution to the hybrid problem (1.1), (1.2). The respective results have been verified by providing a suitable example.
We express our sincere thanks to the anonymous reviewers for their valuable comments and suggestions. This work is supported by the Natural Science Foundation of Tianjin (No.(19JCYBJC30700)).
The authors declare no conflict of interest in this paper.
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