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Research article

Numerical approximation of a variable-order time fractional advection-reaction-diffusion model via shifted Gegenbauer polynomials

  • The fractional advection-reaction-diffusion equation plays a key role in describing the processes of multiple species transported by a fluid. Different numerical methods have been proposed for the case of fixed-order derivatives, while there are no such methods for the generalization of variable-order cases. In this paper, a numerical treatment is given to solve a variable-order model with time fractional derivative defined in the Atangana-Baleanu-Caputo sense. By using shifted Gegenbauer cardinal function, this approach is based on the application of spectral collocation method and operator matrices. Then the desired problem is transformed into solving a nonlinear system, which can greatly simplifies the solution process. Numerical experiments are presented to illustrate the effectiveness and accuracy of the proposed method.

    Citation: Yumei Chen, Jiajie Zhang, Chao Pan. Numerical approximation of a variable-order time fractional advection-reaction-diffusion model via shifted Gegenbauer polynomials[J]. AIMS Mathematics, 2022, 7(8): 15612-15632. doi: 10.3934/math.2022855

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  • The fractional advection-reaction-diffusion equation plays a key role in describing the processes of multiple species transported by a fluid. Different numerical methods have been proposed for the case of fixed-order derivatives, while there are no such methods for the generalization of variable-order cases. In this paper, a numerical treatment is given to solve a variable-order model with time fractional derivative defined in the Atangana-Baleanu-Caputo sense. By using shifted Gegenbauer cardinal function, this approach is based on the application of spectral collocation method and operator matrices. Then the desired problem is transformed into solving a nonlinear system, which can greatly simplifies the solution process. Numerical experiments are presented to illustrate the effectiveness and accuracy of the proposed method.



    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.tJ=[0,T]
    x(0)=0,

    where 0<q<1,fC(J×RR{0}) and gC(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))),  tI=[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<α1,CDα0+ is the Caputo fractional derivative. fC(I×RR{0}),gC(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<α2,
    x(1)=0,   x(e)=0,

    where 1<α2, HDα is the Hadamard fractional derivative, fC([1,e]×RR{0}) and gC([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)),  tJ=[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<α1, 1<ω2.

    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)),tJ=[0,1], (1.1)
    u(0)=0, CDβ φu(0)=0, u(1)=kj=1δju(ξj), (1.2)

    where 0<α1, 1<β2, 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 φC2[0,1] with φ(x)>0 for all x[0,1], Iμ φ denote the φ-Riemann-Liouville fractional integral of order μ, gC(J×Rp+1,R{0}), hC(J×R,R) and fiC(J×Rn+1,R) with fi(0,0,,0n+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 φC2[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))α1x(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)nInα φa+x(t)=1Γ(nα)(1φ(t)ddt)ntaφ(s)(φ(t)φ(s))nα1x(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 xCn1[a,b], the φ-Caputo fractional derivative of x of order α is defined as follows

    CDα φa+x(t):=Dα φa+[x(t)n1k=0x[k]φ(a)k!(φ(t)φ(a))k], n=[α]+1 for αN, n=α for αN,

    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 xC[a,b], then CDα φa+Iα φa+x(t)=x(t);

    2. If xCn1[a,b], then

    Iα φ Ca+Dα φa+x(t)=x(t)n1k=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 SE, and let A:EE and B:SE be two operators which satisfy the following :

    (H1)A is contraction;

    (H2)B is compact and continuous, and

    (H3)u=Au+Bv, vSuS.

    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=suptJ|u(t)|   and  (uv)(t)=u(t)v(t),   tJ.

    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))β1Γ(β)φ(s)g(s,u(s),Iγ1 φu(s),,Iγp φu(s))s0(φ(s)φ(τ))α1Γ(α)φ(τ)h(τ,u(τ))dτds+mi=1Iωi+β φfi(t,u(t),Iμ1 φu(t),,Iμn φu(t))+φ(t)φ(0)kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))[10(φ(1)φ(s))β1Γ(β)φ(s)g(s,u(s),Iγ1 φu(s),,Iγp φu(s))s0(φ(s)φ(τ))α1Γ(α)φ(τ)h(τ,u(τ))dτds+mi=1Iωi+β φfi(1,u(1),Iμ1 φu(1),,Iμn φu(1))kj=1δjξj0(φ(ξj)φ(s))β1Γ(β)φ(s)g(s,u(s),Iγ1 φu(s),,Iγp φu(s))s0(φ(s)φ(τ))α1Γ(α)φ(τ)h(τ,u(τ))dτdskj=1δjmi=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+β1Γ(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+β1Γ(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+β1Γ(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,,0n+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))α1Γ(α)φ(s)h(s,u(s))ds+mi=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))β1Γ(β)φ(s)g(s,u(s),Iγ1 φu(s),,Iγp φu(s))s0(φ(s)φ(τ))α1Γ(α)φ(τ)h(τ,u(τ))dτds+mi=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,,0n+1)=0 yield c1=0, thus equation (3.3) is reduced to

    u(t)=t0(φ(t)φ(s))β1Γ(β)φ(s)g(s,u(s),Iγ1 φu(s),,Iγp φu(s))s0(φ(s)φ(τ))α1Γ(α)φ(τ)h(τ,u(τ))dτds+mi=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))β1Γ(β)φ(s)g(s,u(s),Iγ1 φu(s),,Iγp φu(s))s0(φ(s)φ(τ))α1Γ(α)φ(τ)h(τ,u(τ))dτds+mi=1Iωi+β φfi(1,u(1),Iμ1 φu(1),,Iμn φu(1))+c2(φ(1)φ(0)),
    u(ξj)=ξj0(φ(ξj)φ(s))β1Γ(β)φ(s)g(s,u(s),Iγ1 φu(s),,Iγp φu(s))s0(φ(s)φ(τ))α1Γ(α)φ(τ)h(τ,u(τ))dτds+mi=1Iωi+β φfi(ξj,u(ξj),Iμ1 φu(ξj),,Iμn φu(ξj))+c2(φ(ξj)φ(0)),

    from u(1)=kj=1δju(ξj), we have

    c2=1kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))[10(φ(1)φ(s))β1Γ(β)φ(s)g(s,u(s),Iγ1 φu(s),,Iγp φu(s))s0(φ(s)φ(τ))α1Γ(α)φ(τ)h(τ,u(τ))dτds+mi=1Iωi+β φfi(1,u(1),Iμ1 φu(1),,Iμn φu(1))kj=1δjξj0(φ(ξj)φ(s))β1Γ(β)φ(s)g(s,u(s),Iγ1 φu(s),,Iγp φu(s))s0(φ(s)φ(τ))α1Γ(α)φ(τ)h(τ,u(τ))dτdskj=1δjmi=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 gC(J×Rp+1,R{0}), hC(J×R,R) and fiC(J×Rn+1,R) with fi(0,0,,0n+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,k1,k2,,kp+1)|σ(t)p+1i=1|kiki|

    for tJ and (k1,k2,,kp+1),(k1,k2,,kp+1)Rp+1, and

    |h(t,u)h(t,v)|λ(t)|uv| for tJ and u,vR;

    (C2) there exist ϕi,Ω,χC(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+kj=1δj)|kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))|)(χΩ(φ(1)φ(0))αΓ(α+1)(φ(1)φ(0))βΓ(β+1)+mi=1ϕi(φ(1)φ(0))ωi+βΓ(ωi+β+1))r; (3.5)
    (χλ+Ωσp+1i=1(φ(1)φ(0))γiΓ(γi+1))(φ(1)φ(0))αΓ(α+1)(1+(φ(1)φ(0))(1+kj=1δj)|kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))|)(φ(1)φ(0))βΓ(β+1)<1, (3.6)

    where Ω=sup0t1|Ω(t)|, ϕi=sup0t1|ϕi(t)|, i=1,2,,p, χ=sup0t1|χ(t)|, λ=sup0t1|λ(t)|, σ=sup0t1|σ(t)|.

    Then the hybrid problem (1.1), (1.2) has at least one solution.

    Proof. Define a subset S of E as

    S={uE: ur},

    where r satisfies inequality (3.5). Clearly S is closed, convex and bounded subset of the Banach space E. Define two operators A:EE by

    Au(t)=t0(φ(t)φ(s))β1Γ(β)φ(s)g(s,u(s),Iγ1 φu(s),,Iγp φu(s))s0(φ(s)φ(τ))α1Γ(α)φ(τ)h(τ,u(τ))dτds+φ(t)φ(0)kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))10(φ(1)φ(s))β1Γ(β)φ(s)g(s,u(s),Iγ1 φu(s),,Iγp φu(s))s0(φ(s)φ(τ))α1Γ(α)φ(τ)h(τ,u(τ))dτds(φ(t)φ(0))kj=1δjkj=1δj(φ(ξj)φ(0))(φ(1)φ(0))ξj0(φ(ξj)φ(s))β1Γ(β)φ(s)g(s,u(s),Iγ1 φu(s),,Iγp φu(s))s0(φ(s)φ(τ))α1Γ(α)φ(τ)h(τ,u(τ))dτds, (3.7)
    Bu(t)=mi=1t0(φ(t)φ(s))ωi+β1Γ(ωi+β)φ(s)fi(s,u(s),Iμ1 φu(s),,Iμn φu(s))ds+φ(t)φ(0)kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))mi=110(φ(1)φ(s))ωi+β1Γ(ωi+β)φ(s)fi(s,u(s),Iμ1 φu(s),,Iμn φu(s))ds(φ(t)φ(0))kj=1δjkj=1δj(φ(ξj)φ(0))(φ(1)φ(0))mi=1ξj0(φ(ξj)φ(s))ωi+β1Γ(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))α1Γ(α)φ(τ)h(τ,u(τ))dτg(s,v(s),Iγ1 φv(s),,Iγp φv(s))s0(φ(s)φ(τ))α1Γ(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))α1Γ(α)φ(τ)h(τ,u(τ))dτg(s,u(s),Iγ1 φu(s),,Iγp φu(s))s0(φ(s)φ(τ))α1Γ(α)φ(τ)h(τ,v(τ))dτ+g(s,u(s),Iγ1 φu(s),,Iγp φu(s))s0(φ(s)φ(τ))α1Γ(α)φ(τ)h(τ,v(τ))dτg(s,v(s),Iγ1 φv(s),,Iγp φv(s))s0(φ(s)φ(τ))α1Γ(α)φ(τ)h(τ,v(τ))dτ||g(s,u(s),Iγ1 φu(s),,Iγp φu(s))|s0(φ(s)φ(τ))α1Γ(α)φ(τ)|h(τ,u(τ))h(τ,v(τ))|dτ+s0(φ(s)φ(τ))α1Γ(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))|χλuv(φ(s)φ(0))αΓ(α+1)+Ω(φ(s)φ(0))αΓ(α+1)σp+1i=1(φ(s)φ(0))γiΓ(γi+1)uv(χλ+Ωσp+1i=1(φ(1)φ(0))γiΓ(γi+1))(φ(1)φ(0))αΓ(α+1)uv,

    thus we have

    |Au(t)Av(t)|t0(φ(t)φ(s))β1Γ(β)φ(s)G(s)ds+φ(t)φ(0)|kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))|10(φ(1)φ(s))β1Γ(β)φ(s)G(s)ds+φ(t)φ(0)|kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))|kj=1δjξj0(φ(ξj)φ(s))β1Γ(β)φ(s)G(s)ds(χλ+Ωσp+1i=1(φ(1)φ(0))γiΓ(γi+1))(φ(1)φ(0))αΓ(α+1)(1+(φ(1)φ(0))(1+kj=1δj)|kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))|)(φ(1)φ(0))βΓ(β+1)uv,

    which implies

    Au(t)Av(t)[(χλ+Ωσp+1i=1(φ(1)φ(0))γiΓ(γi+1))(φ(1)φ(0))αΓ(α+1)(1+(φ(1)φ(0))(1+kj=1δj)|kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))|)(φ(1)φ(0))βΓ(β+1)]uv,

    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 uS. Then by Lebesgue dominated convergence theorem,

    limnBun(t)=limn[mi=1t0(φ(t)φ(s))ωi+β1Γ(ωi+β)φ(s)fi(s,un(s),Iμ1 φun(s),,Iμn φun(s))ds+φ(t)φ(0)kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))mi=110(φ(1)φ(s))ωi+β1Γ(ωi+β)φ(s)fi(s,un(s),Iμ1 φun(s),,Iμn φun(s))ds(φ(t)φ(0))kj=1δjkj=1δj(φ(ξj)φ(0))(φ(1)φ(0))mi=1ξj0(φ(ξj)φ(s))ωi+β1Γ(ωi+β)φ(s)fi(s,un(s),Iμ1 φun(s),,Iμn φun(s))ds].=mi=1t0(φ(t)φ(s))ωi+β1Γ(ωi+β)φ(s)limnfi(s,un(s),Iμ1 φun(s),,Iμn φun(s))ds+φ(t)φ(0)kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))mi=110(φ(1)φ(s))ωi+β1Γ(ωi+β)φ(s)limnfi(s,un(s),Iμ1 φun(s),,Iμn φun(s))ds(φ(t)φ(0))kj=1δjkj=1δj(φ(ξj)φ(0))(φ(1)φ(0))mi=1ξj0(φ(ξj)φ(s))ωi+β1Γ(ωi+β)φ(s)limnfi(s,un(s),Iμ1 φun(s),,Iμn φun(s))ds=mi=1t0(φ(t)φ(s))ωi+β1Γ(ωi+β)φ(s)fi(s,u(s),Iμ1 φ(s),,Iμn φu(s))ds+φ(t)φ(0)kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))mi=110(φ(1)φ(s))ωi+β1Γ(ωi+β)φ(s)fi(s,u(s),Iμ1 φu(s),,Iμn φu(s))ds(φ(t)φ(0))kj=1δjkj=1δj(φ(ξj)φ(0))(φ(1)φ(0))mi=1ξj0(φ(ξj)φ(s))ωi+β1Γ(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)|mi=1t0(φ(t)φ(s))ωi+β1Γ(ωi+β)φ(s)|fi(s,u(s),Iμ1 φu(s),,Iμn φu(s))|ds+φ(t)φ(0)|kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))|mi=110(φ(1)φ(s))ωi+β1Γ(ωi+β)φ(s)|fi(s,u(s),Iμ1 φu(s),,Iμn φu(s))|ds+(φ(t)φ(0))kj=1δj|kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))|mi=1ξj0(φ(ξj)φ(s))ωi+β1Γ(ωi+β)φ(s)|fi(s,u(s),Iμ1 φu(s),,Iμn φu(s))|dsmi=1ϕi(φ(1)φ(0))ωi+βΓ(ωi+β+1)+(φ(1)φ(0))(1+kj=1δj)|kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))|mi=1ϕi(φ(1)φ(0))ωi+βΓ(ωi+β+1)=(1+(φ(1)φ(0))(1+kj=1δj)|kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))|)mi=1ϕi(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,t2J with t1<t2 and uS. Then we have

    |Bu(t2)Bu(t1)|=|mi=1t20(φ(t2)φ(s))ωi+β1Γ(ωi+β)φ(s)fi(s,u(s),Iμ1 φu(s),,Iμn φu(s))dsmi=1t10(φ(t1)φ(s))ωi+β1Γ(ωi+β)φ(s)fi(s,u(s),Iμ1 φu(s),,Iμn φu(s))ds+φ(t2)φ(t1)kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))mi=110(φ(1)φ(s))ωi+β1Γ(ωi+β)φ(s)fi(s,u(s),Iμ1 φu(s),,Iμn φu(s))ds(φ(t2)φ(t1))kj=1δjkj=1δj(φ(ξj)φ(0))(φ(1)φ(0))mi=1ξj0(φ(ξj)φ(s))ωi+β1Γ(ωi+β)φ(s)fi(s,u(s),Iμ1 φu(s),,Iμn φu(s))ds|mi=1ϕiΓ(ωi+β)[|t10[(φ(t2)φ(s))ωi+β1(φ(t1)φ(s))ωi+β1]φ(s)ds+t2t1[(φ(t2)φ(s))ωi+β1φ(s)ds|+φ(t2)φ(t1)|kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))|10(φ(1)φ(s))ωi+β1φ(s)ds+(φ(t2)φ(t1))kj=1δj|kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))|ξj0(φ(ξj)φ(s))ωi+β1φ(s)ds]mi=1ϕiΓ(ωi+β+1)[|(φ(t2)φ(0))ωi+β(φ(t1)φ(0))ωi+β|+φ(t2)φ(t1)|kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))|(φ(1)φ(0))ωi+β+(φ(t2)φ(t1))kj=1δj|kj=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)||t2t1|M|t2t1|,   ξ(t1,t2).

    On the other hand, for φC[0,1], thus there exist positive constants N such that |φ(t2)φ(t1)|=|φ(ξ)||t2t1|N|t2t1|,   ξ(t1,t2), from which we deduce

    |Bu(t2)Bu(t1)|0    as  t2t10.

    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 vS, then we have

    |u(t)|=|Au(t)+Bv(t)||Au(t)|+|Bv(t)||t0(φ(t)φ(s))β1Γ(β)φ(s)g(s,u(s),Iγ1 φu(s),,Iγp φu(s))s0(φ(s)φ(τ))α1Γ(α)φ(τ)h(τ,u(τ))dτds+φ(t)φ(0)kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))10(φ(1)φ(s))β1Γ(β)φ(s)g(s,u(s),Iγ1 φu(s),,Iγp φu(s))s0(φ(s)φ(τ))α1Γ(α)φ(τ)h(τ,u(τ))dτds(φ(t)φ(0))kj=1δjkj=1δj(φ(ξj)φ(0))(φ(1)φ(0))ξj0(φ(ξj)φ(s))β1Γ(β)φ(s)g(s,u(s),Iγ1 φu(s),,Iγp φu(s))s0(φ(s)φ(τ))α1Γ(α)φ(τ)h(τ,u(τ))dτds|+|mi=1t0(φ(t)φ(s))ωi+β1Γ(ωi+β)φ(s)fi(s,v(s),Iμ1 φv(s),,Iμn φv(s))ds+φ(t)φ(0)kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))mi=110(φ(1)φ(s))ωi+β1Γ(ωi+β)φ(s)fi(s,v(s),Iμ1 φv(s),,Iμn φv(s))ds(φ(t)φ(0))kj=1δjkj=1δj(φ(ξj)φ(0))(φ(1)φ(0))mi=1ξj0(φ(ξj)φ(s))ωi+β1Γ(ωi+β)φ(s)fi(s,v(s),Iμ1 φv(s),,Iμn φv(s))ds|(1+(φ(1)φ(0))(1+kj=1δj)|kj=1δj(φ(ξj)φ(0))(φ(1)φ(0))|)(χΩ(φ(1)φ(0))αΓ(α+1)(φ(1)φ(0))βΓ(β+1)+mi=1ϕi(φ(1)φ(0))ωi+βΓ(ωi+β+1))r,

    which implies ur and so uS.

    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)2i=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),  tJ=[0,1], (4.1)
    u(0)=0, CD32 t4u(0)=0, u(1)=13u(13), (4.2)

    where

    2i=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=sup0t1|σ(t)|=sup0t1t24[1+t14Γ(54)+t12Γ(32)]=14(1+1Γ(54)+1Γ(32))=14(1+10.9064+10.8862)=0.8079;
    λ=sup0t1|λ(t)|=sup0t125cos(t4)=0.4;
    ϕ1=sup0t1|ϕ1(t)|=sup0t1t(1+1+1)=3;
    ϕ2=sup0t1|ϕ2(t)|=sup0t1t10(1+π2+1)=110×3.57=0.357;
    Ω=sup0t1|Ω(t)|=sup0t125cos(t4)=0.4;
    \chi^{*} = \sup\limits_{0\leq t\leq 1}|\chi(t)| = \sup\limits_{0\leq t\leq 1}\frac{t^{2}}{4}(1+1+1) = \frac{3}{4} = 0.75.

    Choose r > 0.5, then we have

    \biggl(1+\frac{\frac{1}{4}\times \frac{4}{3}}{\frac{2}{9}}\biggr)\biggl[0.75\times0.4\times\frac{(\frac{1}{4})^{\frac{1}{2}}}{\Gamma(\frac{3}{2})} \times\frac{(\frac{1}{4})^{\frac{3}{2}}}{\Gamma(\frac{5}{2})}+3\times\frac{(\frac{1}{4})^{\frac{11}{6}}}{\Gamma(\frac{17}{6})}+0.357 \times\frac{(\frac{1}{4})^{\frac{13}{6}}}{\Gamma(\frac{19}{6})}\biggr] = 0.4016\leq r.

    Moreover,

    \biggl(0.75\times 0.4+0.4\times 0.8079\times \biggl(\frac{(\frac{1}{4})^{\frac{1}{4}}}{\Gamma(\frac{5}{4})}+\frac{(\frac{1}{4})^{\frac{1}{2}}}{\Gamma(\frac{3}{2})}\biggr)\biggr) \frac{(\frac{1}{4})^{\frac{1}{2}}}{\Gamma(\frac{3}{2})}\biggl(1+\frac{\frac{1}{4}\times \frac{4}{3}}{\frac{2}{9}}\biggr)\frac{(\frac{1}{4})^{\frac{3}{2}}}{\Gamma(\frac{5}{2})} = 0.097 \lt 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 \varphi -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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