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Existence and uniqueness results for mixed derivative involving fractional operators

  • In this article, we discuss the existence and uniqueness results for mix derivative involving fractional operators of order β(1,2) and γ(0,1). We prove some important results by using integro-differential equation of pantograph type. We establish the existence and uniqueness of the solutions using fixed point theorem. Furthermore, one application is likewise given to represent our fundamental results.

    Citation: Abeer Al Elaiw, Farva Hafeez, Mdi Begum Jeelani, Muath Awadalla, Kinda Abuasbeh. Existence and uniqueness results for mixed derivative involving fractional operators[J]. AIMS Mathematics, 2023, 8(3): 7377-7393. doi: 10.3934/math.2023371

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  • In this article, we discuss the existence and uniqueness results for mix derivative involving fractional operators of order β(1,2) and γ(0,1). We prove some important results by using integro-differential equation of pantograph type. We establish the existence and uniqueness of the solutions using fixed point theorem. Furthermore, one application is likewise given to represent our fundamental results.



    The pantograph equation is a special type of functional differential equations with proportional delay. The present study introduces a compound technique incorporating the perturbation method with a iteration algorithm to solve numerically the delay differential equation of pantograph type. The pantograph equations became a prime example of delay differential equation in the recent years. Over the last few years, the continuous and discrete cases of the pantograph equation have been extensively explored see [1,2,3].

    Different authors discuss linear and non-linear pantograph equations. The solution to the simplest homogeneous linear pantograph cannot be expressed in terms of elementary functions. We can't solve even the simplest non-homogeneous linear pantograph problem using standard approaches like variation of constants and Laplace transformation. The existence and uniqueness of solutions for the linear pantograph equation's initial value problems change significantly depending on the beginning locations chosen. In general, the solution to the initial value problem may or may not exist, or may not be unique. some authors discuss linear pantograph Volterra delay-integro-differential equation and the multi-terms boundary value problem of fractional pantograph differential equations [4,5].

    It is also possible to obtain additive, multiplicative, and functional separable solutions, as well as several additional precise solutions. Nonlinear pantograph-type PDEs of a more broad form, containing one or two arbitrary functions Polyanin et al. [6] examine Nonlinear pantograph-type diffusion PDEs, exact solutions and the principle of analogy. Recently, many research on fractional-order pantograph differential equations have recently been published, involving various operators [7,8,9], φ-Caputo derivative [10], Atangana-Baleanu-Caputo derivative [11,23,24].

    Furthermore, several scientific scholars have produced results regarding the existence and uniqueness of solutions for various classes of fractional pantograph equations by applying various fixed-point theorems as like Shah et al. [12] discussed the dynamics and stability of-fractional pantograph equations and Houas et al. [13] studied the existence and Ulam stability of fractional pantograph differential equations with two Caputo-Hadamard derivatives.

    Different authors like [14,15] work on the pantograph-catenary electrical contact system of high speed railways. The pantograph-catenary electrical contact system, which serves as the only power entrance, keeps the high-speed train's power transfer reliable and efficient. A pantograph-catenary system must take into account the wind, sand, rain, thunder, ice, and snow while designing it due to the rapid expansion of high-speed trains around the world. Commercialized lines are also being developed in China to cover isolated areas with severe environments. There are some recent results on the existence of solutions for fractional integro-differential and fractional differential equations [16,17].

    The following fractional integro-differential equation of pantograph type is considered in this work, along with appropriate initial conditions.

    {CDβ[CDγv(t)]=ϕ(t,v(t),v(Λt))+qt0k1(t,s,v(s))ds+t0k2(t,s,v(s))ds,tJ:=[0,M],0Dβ1tv(0)=v0v(0)=v1. (1.1)

    Where 0<γ<1 and 1<β<2 as well as Λ,q<1, ϕ:J×R×RR and ki:J×J×RR are continuous for i=1,2, and CDβ,CDγ are the Caputo fractional derivatives.

    We shall first look into the existence and uniqueness of solutions for (1.1). To do so, we turn the original problem into an equivalent integral equation, then establish the existence and uniqueness of the solutions using fixed point theorems.

    The following is the outline for this paper. We look at some essential preliminaries in section 2. We discuss the existence and uniqueness of problem (1.1) in section 3. We explore a helpful application to represent our primary finding in section 4. In section 5, we present some numerical methods. In section 6, we find some numerical results to show the applicability of our results.

    We present some well-known definitions and lemma in this part.

    Lemma 2.1. [18,19] Suppose that β>ϱ>0 and ϕL1([b,d]). At that point DϱIβϕ(t)=Iβϱϕ(t),t[b,d].

    Lemma 2.2. [18,19] For β>0 and ϱ>1, we get

    Iβ[(ts)ϱ]=Γ(ϱ+1)Γ(β+ϱ+1)(ty)β+ϱ.

    For ϱ=0 and y=0, we get

    Iβ[1]=1Γ(β+1)tβ.

    Assume that [b,d]R be a finite interval as well as suppose that β,γ,ξC and R(z)=Re(z) for zC. The RL-fractional integral and derivative of order βC are defined by

    (Iβb+ϕ)(y)=1Γ(β)ybϕ(y)(ys)1βds,y>b,R(β)>0, (2.1)

    and

    (Dβb+ϕ)(y)=1Γ(mβ)dmdymybϕ(s)(ys)βm+1ds=dmdym(Imβb+ϕ)(y),y>b,R(β)0.

    On the interval [b,d], the Caputo fractional derivative of order β is defined by

    (CDβb+z)(y)=(Dβb+[z(t)m1k=0z(k)(b)k!(tb)k])(y).

    When b=0, Iβb+z and CDβb+z are denoted by Iβz and CDβz. The semi-group features of the fractional integral operator Iβb+ as well as the fractional differentiation operator Dγb+ are given by [19].

    Lemma 2.3. Suppose that R(β),R(γ)>0 as well as ϕ(y)C[b,d]. For y[b,d] the following statements are true:

    (i) (Iβb+Iγb+ϕ)(y)=(Iβ+γb+ϕ)(y).

    (ii) (Dβb+Iβb+ϕ)(y)=ϕ(y).

    (iii) If R(β)>R(γ) at that point

    (Dβb+Iβb+ϕ)(y)=(Iβγb+ϕ)(y).

    (iv) Suppose that m=[R(β)>0]+1 for R(β)N and ϕmβ(y)=(Imβb+ϕ)(y)Cm[b,d], then

    (Iβb+Dβb+ϕ)(y)=ϕ(y)mk=1ϕ(mk)mβΓ(βk+1)(ya)βk.

    Suppose that Cξ[b,d] be the space of function ϕ defined on (b,d] in such a way that (yb)ξϕ(y)C[b,d] along the norm ϕCξ=(yb)ξϕ(y)C:=supy[b,d]|(yb)ξϕ(y)|. Note that for ξ=0,Cξ[b,d]=C[b,d]. The continuity of the fractional integral operator Iβb+ from the space Cξ[b,d] into C[b,d] is discussed in the following lemma ([19] Lemma 2.8 (a)).

    Lemma 2.4. Suppose that R(β)>0 and 1R(ξ)0. If R(ξ)R(β) at that point the fractional integral operator Iβb+ is bounded from Cξ[b,d] into C[b,d]

    Iβb+ϕChϕCξ,h=(db)R(βξ)Γ(R(β))|Γ(1R(ξ))||Γ(β)|Γ(1+R(βξ)).

    According to the following ([19] Lemma 2.21, part (a)) when R(β)N0 the Riemann-Liouville fractional integral operator Iβb+ is the left inverse of the Caputo fractional differentiation operator CDβb+.

    Lemma 2.5. Suppose that βC with 0<R(β) as well as z(y)C[b,d]. If R(β)N, at that point

    (CDβb+Iβb+z)(y)=z(y).

    The fixed point theorem in [20], first presented by Krasnoselskii, is necessary to show that the existence of solution for (1.1).

    Theorem 2.6. Assume that E be a nonempty and convex closed subset of a Banach space Y. Let S as well as R be two operators such that

    (i) when v,wE then Sv+RwE,

    (ii) S is continuous and compact,

    (iii) R be a contraction mapping.

    At that point yE must exist in such a way that y=Sy+Ry.

    Consider C(J) be a Banach space along the norm vC=suptJ|v(t)|. We define

    (K1v)(t):=qt0k1(t,s,v(s))ds,(K2v)(t):=t0k2(t,s,v(s))ds.

    In the next lemma, we present an integral equation that corresponding to Eq (1.1).

    Lemma 3.1. Suppose that ϕ:J×R×RR as well as ki:J×J×RR are continuous functions. If and only if v is a solution of the fractional integral equation, then the function vC(J) fulfils problem (1.1).

    v(t)=tβ1Γ(β)v0+v0(β+γ)Γ(β)Γ(γ)tβ+γ+v1Γ(γ+1)tγ+Iβ+γ0+(ϕ(t,v(t),v(Λt)))+Iβ+γ0+((K1v)(t)+(K2v)(t)). (3.1)

    Proof. Suppose that vC(J) to solve the problem (1.1). Using the concept of the Caputo fractional derivative in Lemma (2.3) (d), we get

    CDγv(t)=v(0)+v(0)t+Iβ0+(ϕ(t,v(t),v(Λt)))+Iβ0+((K1v)(t)+(K2v)(t))=tβ1Γ(β)v0+v1t+Iβ0+(ϕ(t,v(t),v(Λt)))+Iβ0+((K1v)(t)+(K2v)(t)),

    apply Iγ0+ on both sides

    Iγ0+CDγv(t)=v(0)+1Γ(β)Γ(γ)t0(ts)β+γ+1v0ds+1Γ(γ)t0(ts)γ1v1ds+Iβ+γ0+(ϕ(t,v(t),v(Λt)))+Iβ+γ0+((K1v)(t)+(K2v)(t))v(t)=tβ1Γ(β)v0+v0Γ(β)Γ(γ)[(ts)β+γβ+γ]t0+v1Γ(γ)[(ts)γγ]t0+Iβ+γ0+(ϕ(t,v(t),v(Λt)))+Iβ+γ0+((K1v)(t)+(K2v)(t))=tβ1Γ(β)v0+v0(β+γ)Γ(β)Γ(γ)tβ+γ+v1γΓ(γ)tγ+Iβ+γ0+(ϕ(t,v(t),v(Λt)))+Iβ+γ0+((K1v)(t)+(K2v)(t))=tβ1Γ(β)v0+v0(β+γ)Γ(β)Γ(γ)tβ+γ+v1Γ(γ+1)tγ+Iβ+γ0+(ϕ(t,v(t),v(Λt)))+Iβ+γ0+((K1v)(t)+(K2v)(t)).

    We define

    Δ={(t,s):0st},Δq={(t,s):qts0}.

    The existence of (1.1) is investigated under the following conditions:

    H1:ϕ:J×R×R×RR is continuous and a continuous function exists, b:E[0,) in such a way that t[0,T] and vi,wiR,i=1,2.

    |ϕ(t,v1,w1)ϕ(t,v2,w2)|b(t)(|v1w1|+|v2w2|).

    H2:ki:J×J×RR,i=1,2 be continuous and there exist b1:Δq[0,) and b2:Δ[0,) in such a way that d1(t):=qt0b1(t,s)dsC(J),d2(t):=t0b2(t,s)dsC(J)

    |ki(t,s,y)|bi(t,s)(1+|y|).

    H3:

    Mβ+γΓ(β+γ)(d1+d2C+2bC)<1.

    Assume that the closed ball with radius r0 and centre at 0 is Br0C(J) as well as put

    ˜ϕ:=sup{|ϕ(t,0,0)|:tJ},

    and

    r0:=1+tβ1Γ(β)|v0|+|v1t|+Mβ+γΓ(β+γ)(d1+d2C+˜ϕ)1Mβ+γΓ(β+γ)(d1+d2C+2bC),

    and define

    (Sv)(t):=(Iβ0+Iγ0+((K1v)(s)+(K2v)(s))(t)).

    Lemma 3.2. Suppose that (H1)(H3) be satisfied, at that point the operator S maps Br0 into itself, and S:Br0Br0 is continuous and compact.

    Proof. Step 1: We prove that S(Br0)Br0 where Br0={vW:vr0}. For vBr0, for assumption (H1)

    |ϕ(t,v(t),v(Λt)||ϕ(t,v(t),v(Λt))ϕ(s,0,0)|+|ϕ(s,0,0)|2b(t)v+L12b(t)r0+L1|ϕi(t,v(t),v(Λt)||ϕi(t,v(t),v(Λt))ϕi(s,0,0)|+|ϕi(s,0,0)|2bi(t)v+Li2bi(t)r0+Li,i=1,2,...m.

    For tJ using assumptions (H2) and (H3) we have

    |(Sv)(t)|(Iβ+γ0+(|(K1v)(s)|+|(K2v)(s)|))(t)(Iβ+γ0+(qs0b1(s,ξ)dξ+s0b2(s,ξ)dξ))Mβ+γΓ(β+γ+1)(d1+d2C)+r0Mβ+γΓ(β+γ+1)(d1+d2C)r0. (3.2)

    Step 2: S:Br0Br0 is continuous. ϵ>0 be a fixed point, choose an arbitrary v,wBr0 in such a way that vwϵ. For tJ we get

    |(Kiv)(t)(Kiw)(t)|t0|ki(t,s,v(s))ki(t,s,w(s))ωr0(ki,ϵ)M,

    where

    ωr0(ki,ϵ)=sup{|ki(t,s,v1)ki(t,s,v2)|:t,sJ,v1,v2[r0,r0],|v1v2|ϵ}, (3.3)

    for i=1,2 using (3.3) we have

    |(Sv)(t)(Sw)(t)|(Iβ+γ0+(|(K1v)(s)(K1w)(s)|+|(K2v)(s)(K2w)(s)|))(t)(ωr0(k1,ξ)+ωr0(k2,ξ))Mβ+γ+1Γ(β+γ+1). (3.4)

    We see that ωr0(ki,ϵ)0,asϵ0 from the uniform continuity of ki,i=1,2 on bounded subsets of J×R×R. As a result of the inequality (3.4) S:Br0Br0 is continuous.

    Step 3: An equi-continuous subset of C(J) is S(Br0). Supposition (H2) states that for any vBr0 as well as sJ we have

    |(K1v)(s)|qs0|k1(s,ξ,v(ξ))|dξqs0b1(s,ξ)(1+|v(ξ)|)dξ(1+r0)d1(s), (3.5)

    and similarly

    |(K2v)(s)|s0|k2(s,ξ,v(ξ))|dξs0b2(s,ξ)(1+|v(ξ)|)dξ(1+r0)d2(s). (3.6)

    Now, let t1,t2J and t1t2.

    By Eqs (3.5) and (3.6), we get

    |(Sv)(t1)(Sv)(t2)||(Iβ+γ0+(K1v)(s))(t1)(Iβ+γ0+(K1v)(s))(t2)|+|(Iβ+γ0+(K2v)(s))(t1)(Iβ+γ0+(K2v)(s))(t2)|r0+1Γ(β+γ)t0(d1+d2)(1(t1s)1(β+γ)1(t2s)1(β+γ))ds+r0+1Γ(β+γ)t2t1(d1+d2)(t2s)1(β+γ)dsd1+d2CΓ(β+γ+1)(2(t2t1)β+γ+tβ+γ2tβ+γ1)(r0+1). (3.7)

    As t1t2, the right hand side of inequality (3.7) tends to zero. We can see from Steps 1–3 and the Arzela-Ascoli theorem that S:Br0Br0 is continuous and compact.

    Theorem 3.3. In the space C(J), problem (1.1) has at least one solution with assumptions (H1)(H3).

    Proof. Define the R operator on C(J) as follows:

    (Rv)(t):=tβ1Γ(β)v0+v1t+(Iβ+γ0+ϕ(s,v(s),v(Λs)))(t).

    The operator R is clearly defined and RvC(J) for some vC(J) due to the continuity of ϕ and Lemma (2.3).

    For any v,wBr0, and tJ based on assumptions (H1)(H3) and inequality (3.2).

    |(Sv)(t)+(Rw)(t)||(Sv)(t)|+tβ1Γ(β)|v0|+|v1t|+(Iβ+γ0+|ϕ(s,v(s),v(Λs))ϕ(s,0,0)|)(t)+(Iβ+γ0+a|ϕ(s,0,0)|)(t)tβ1Γ(β)|v0|+|v1t|+(Iβ+γ0+[d1+d2])(t)+r0(Iβ+γ0+[d1+d2])(t)+2r0(Iβ+γ0+b)(t)+(Iβ+γ0+|ϕ(s,0,0)|)(t)1+tβ1Γ(β)|v0|+|v1t|+Mβ+γΓ(β+γ+1)(d1+d2C+˜ϕ)+Mβ+γΓ(β+γ+1)(d1+d2C+2bC)r0r0.

    As a result, Sv+RwBr0 for every v,wBr0 We can also using (H1) for some v,wC(J) we get

    |(Rv)(t)(Rw)(t)|(Iβ+γ0+b(s)[|v(s)w(s)||v(Λs)w(qs)|])(t)2Mβ+γΓ(β+γ+1)bCvwC,tJ. (3.8)

    B is a contraction mapping, based on assumption (H3) and inequality (3.8). The assumptions of Theorem (2.6) are thus satisfied by Lemma (3.2).

    Theorem 3.4. If (H1) and (H3) are true, then the following assumption is true.

    H4 : ki:J×J×RR,i=1,2 is continuous as well as bi:J×J[0,),i=1,2 exist in such a way that

    d1(t):=qt0b1(t,s)dsC(J),d2(t):=t0b2(t,s)dsC(J),

    and

    |ki(t,s,y)ki(t,s,z)|ki(t,s)|yz|.

    Then, for J, problem (1.1) has a unique solution.

    Proof. It is sufficient to prove that the integral equation (3.1) has a unique solution using Lemma (3.1). Define the F operator on C(J) as follows:

    (Fv)(t):=tβ1Γ(β)v0+v1t+(Iβ+γ0+ϕ(s,v(s),v(Λs)))(t)+(Iβ+γ0+((K1v)(s)+(K2v)(s)))(t). (3.9)

    FvC(J) for any vC(J) is simply found using the continuity of ϕ,k1,k2, and Lemma (2.2). F be the fixed point are the solution of (3.1). In the next section, we show that F is a contraction mapping, and F has a specific fixed point according to the Banach contraction principle. Suppose that v,wC(J). According to (H1) and (H4), for any tJ we get

    |(Fv)(t)(Fw)(t)|1Γ(β+γ)t0|ϕ(s,v(s),v(Λs))ϕ(s,w(s),w(Λs))|(ts)1(β+γ)ds+1Γ(β+γ)t0|(K1v)(s)(K1w)(s)|+|(K1v)(s)(K2w)(s)|(ts)1(β+γ)ds2vwC(Iβ+γ0+b)(t)+vwC(Iβ+γ0+(d1+d2))(t).

    Hence

    FvFwMβ+γΓ(β+γ+1)(2bC+d1+d2C)vw.

    By assumption (H3) it prove that F is a contraction mapping.

    Example 4.1. Consider the integro-differential equation given below

    {CD1.5[CD0.2v(t)]=tanh(v(t)+v(12t))16(1+t2)+t80v(s)1+32tsds+t0(v(s)sin(ts)8+ts8)ds,t[1,2],0Dβ1tv(0)=v0(0)=2,v(0)=0.

    Put

    ϕ(t,y,z)=tanh(y+z)16(1+t2),M=2,β=1.5,Λ=12,q=18,
    k1(t,s,y)=y1+32ts,k2(t,s,y)=ysin(ts)8+ts8,b(t)=116(1+t2),b1(t,s)=11+32ts,b2(t,s)=ts8.

    At that point

    |ϕ(t,y,z)ϕ(t,u1,u2)|b(t)(|yu1|+|zu2|),|k1(t,s,y)|b1(t,s)|y|,i=1,2,|k2(t,s,y)|b2(t,s)(1+|y|),d1(t)=qt0b1(t,s)ds=(17816)t,d2(t)=t0b1(t,s)ds=t216,
    MβΓ(β+1)(d1+d2C+2bC0.7387<1.

    All above relations shows that (H1),(H4) and (H3) are fulfilled.

    Here, we want to use the Sinc collocation method to approximation the solution of (1.1). Therefore, the Sinc basis functions must be defined. The following definition is given for the translated Sinc base functions

    S(p,e)(t)=Sinc(tpee),p=0,±1,±2,...,

    where Sinc(t) on the complete real line (,) is given below

    {Sinc(t)=sin(Πt)Πt,0t,10=t.

    Let q and m be the two integers, with the help of the previous basis function, we may approximate the following function ϕ(t) on the real line:

    ϕ(t)mp=qϕ(pe)Sinc(tpee),tR, (5.1)

    where e represent as step size. Furthermore, we can calculate the integrals on R in the following manner using the Sinc quadrature rule:

    ϕ(t)dtemp=qϕ(pe). (5.2)

    Consider the single exponential transformation

    χb,d(t)=log(tbdt).

    Consider the inverse function

    φb,d(ζ)=χ1b,d(ζ)=b+dhζhζ+1,

    creating the infinite strip

    B={ζC:|(ζ)|<l},

    the eye shaped domain

    D={w:|arg(vbdv)|<l}.

    By using φb,d, consider Γ be the image of the real line

    Γ={tC:t=φb,d(ζ),ζR}.

    We shall define the collocation points in [b,d] as the image of the equidistance points pe for some constant e in the following manner in order to employ the Sinc collocation method

    φb,d(pe)=b+dhpehpe+1,p=q,...,m.

    We will be able to approximate the function ϕ(t) at the finite interval [b,d] using the transformations χb,d as well as φb,d.

    ϕ(t)mp=qϕφb,d(pe)Sinc(χb,d(t)pee),t[b,d],

    where e is a constant, q and m be the non-negative integers. Furthermore, the Sinc quadrature rule in the finite interval can be defined as

    dbϕ(t)dt=ϕ(φb,d(ζ))φb,d(ζ)dζemp=qϕ(φb,d(pe))φb,d(pe).

    Since φb,d(ζ)=(ζb)(dζ)db and φb,d(pe)=(db)ye+b, we get

    dbϕ(y)dyemp=qϕ(φb,d(pe))φb,d(pe)=emp=qϕ(φb,d(pe))(φb,d(pe)b)(dφb,d(pe))db=e(db)mp=qϕ(φb,d(pe))ye(1ye). (5.3)

    Theorem 5.1. Suppose that ϕLρ1,σ1(E) with

    φ(t+L|ϕ(w)|dw0,ast,

    where L={aj:|a|<l} and

    limωDinfωDω|f(w)|dw<.

    Taking e=2Πlσm as well as q=[(σρ)m]+1, we get

    |dbϕ(t)dtemp=qϕ(φb,d(pe))φb,d(pe)|C1h2Πlσm,

    where C1 is a constant which depends on ϕ,l,ρ, and σ.

    The following theorems prove that for some constants ρ and σ, the relation (5.1) and (5.2) attain exponential rates of convergence when ϕ(t) belongs to Lρ,σ(E). For the solution (1.1) using Sinc collocation method, near the boundary points b and d the solution tends to 0. Define K(t)

    K(t)=v(t)μb(t)v(b)μd(t)v(d), (5.4)

    where μb(t) and μd(t) can be written as

    μb(t)=11+hχb,d(t),μd(t)=hχb,d(t)1+hχb,d(t).

    We have

    limtbμb(t)=1,limtdμb(t)=0,limtbμd(t)=0,limtdμd(t)=1.

    So, by using Sinc basis functions, the function K(t) can be approximated as shown below

    K(t)mp=qϕ(φb,d(pe))Sinc(χb,d(t)pee). (5.5)

    Using (5.4) and (5.5) we must define the approximate solution given below

    wm(t)=cqμb(t)+m1p=q+1cpSinc(χb,d(t)pee)+cmμd(t). (5.6)

    Substituting the solution wm(t) in (3.9), we get

    wm(t)=tβ1Γ(β)v0+v1t+(Iβ+γ0+ϕ(s,vm(s),vm(Λs)))(t)+(Iβ+γ0+((K1vm)(s)+(K2vm)(s)))(t). (5.7)

    Define some operators

    (Sv)(t)=(Iβ+γ0+ϕ(s,vm(s),vm(Λs))(t),(Rv)(t)=(Iβ+γ0+(K1vm)(s))(t),(Pv)(t)=(Iβ+γ0+(K2vm)(s))(t),

    since, relation (5.7) can be written as

    wm(t)=tβ1Γ(β)v0+v1t+(Sv)(t)+(Rv)(t)+(Pv)(t). (5.8)

    Define rp[0,1] as

    rp=φ0,1(pˉe)=hpˉehpˉe+1,p=q,,m,

    with ˉe2Πl(β+γ)q and m=[(β+γ)q]+1, using relation (5.3) with rp, we get

    (Sv)(t)=1Γ(β+γ)t0ϕ(s,vm(s),vm(Λs))(ts)1(β+γ)ds1Γ(β+γ)ˉetqp=mϕ(φ0,t(pˉe),vm(φ0,t(pˉe)),vm(Λφ0,t(pˉe)))t1(β+γ)(1rp)1(β+γ)rp(1rp)=1Γ(β+γ)ˉetqp=mϕ(φ0,t(pˉe),vm(φ0,t(pˉe)),vm(Λφ0,t(pˉe)))rp(1rp)β+γ, (5.9)

    for (Rv)(t), we get

    (Rv)(t)=(Iβ+γ0+(K1vm)(s))(t)=1Γ(β+γ)t0K1(vm(s))(ts)1(β+γ)ds=1Γ(β+γ)t0qs0k1(s,w,vm(w))dw(ts)1(β+γ)ds1Γ(β+γ)t0ˉeqsqa=mk1(s,φ0,qs(aˉe),wm(φ0,qs(aˉe)))ra(1ra)(ts)1(β+γ)ds1Γ(β+γ)qp=m×ˉeqφ0,t(pˉe)qa=mk1(φ0,t(pˉe),φ0,qφ0,t(pˉe)(aˉe),wm(φ0,qφ0,t(pˉe)(aˉe)))ra(1ra)t1(β+γ)(1rp)1(β+γ)rp(1rp)1Γ(β+γ)ˉetβ+γqp=mˉeqφ0,t(pˉe){qa=mk1(φ0,t(pˉe),φ0,qφ0,t(pˉe)(aˉe),wm(φ0,qφ0,t(pˉe)(aˉe))ra(1ra)}rp(1rp)β+γ. (5.10)

    For (Pv)(t), we have

    (Pv)(t)=(Iβ+γ0+(K2vm)(s))(t)1Γ(β+γ)t0s0k2(s,w,vm(w))dw(ts)1(β+γ)ds1Γ(β+γ)t0ˉesqa=mk2(s,φ0,s(aˉe),wm(φ0,s(aˉe)))ra(1ra)(ts)1(β+γ)ds1Γ(β+γ)qp=m×ˉeφ0,t(pˉe)qa=mk2(φ0,t(pˉe),φ0,φ0,t(pˉe)(aˉe),wm(φ0,φ0,t(pˉe)(aˉe)))ra(1ra)t1(β+γ)(1rp)1(β+γ)rp(1rp)1Γ(β+γ)ˉetβ+γqp=mˉeφ0,t(pˉe){qa=mk1(φ0,t(pˉe),φ0,φ0,t(pˉe)(aˉe),wm(φ0,φ0,t(pˉe)(aˉe))ra(1ra)}rp(1rp)β+γ, (5.11)

    when (5.9)–(5.11) are substituted into (5.8), we obtain

    wm(t)=tβ1Γ(β)v0+v1t+1Γ(β+γ)ˉetqp=mϕ(φ0,t(pˉe),vm(φ0,t(pˉe)),vm(Λφ0,t(pˉe)))rp(1rp)β+γ+1Γ(β+γ)ˉetβ+γqp=mˉeqφ0,t(pˉe)×{qa=mk1(φ0,t(pˉe),φ0,qφ0,t(pˉe)(aˉe),wm(φ0,qφ0,t(pˉe)(aˉe))ra(1ra)}rp(1rp)β+γ+1Γ(β+γ)ˉetβ+γqp=mˉeφ0,t(pˉe)×{qa=mk1(φ0,t(pˉe),φ0,φ0,t(pˉe)(aˉe),wm(φ0,φ0,t(pˉe)(aˉe))ra(1ra)}rp(1rp)β+γ. (5.12)

    Define the points tk[0,T]

    tk=φ0,T(ke),k=q,,m,

    where e=Πl(β+γ)q. By definition of φb,d(t), we obtain

    {φ0,tk(pˉe)=tkrpφ0,φ0,tk(pˉe)(aˉe)=qtkrpraφ0,φ0,tk(pˉe)(aˉe)=tkrpra. (5.13)

    Using (5.13), we have

    wm(t)=tβ1Γ(β)v0+v1t+1Γ(β+γ)ˉetβ+γqp=mϕ(tkrp,wm(tkrp),wm(Λtkrp))rp(1rp)β+γ+1Γ(β+γ)ˉetβ+γqp=mˉeqtkrp×(qa=mk1(tkrp,qtkrpra,wm(qtkrpra))ra(1ra))rp(1rp)β+γ+1Γ(β+γ)ˉetβ+γqp=mˉeqtkrp×(qa=mk2(tkrp,tkrpra,wm(tkrpra))ra(1ra))rp(1rp)β+γ,k=q,...,m.

    Newton iteration method used to solve the above relationship.

    In this part, we find the numerical results for Example (4.1) to check the applicability of Sinc collocation method. In Table 1, SE means single exponential. If we indicate by E1 as well as E2 the greatest absolute errors calculated with q=q1 and q2. The practical orders of convergence can be obtained by using the following formula

    Order=log(E1/E2)log(q2/q1).
    Table 1.  Finding approximate solutions for Example (4.1).
    yq 2 4 8 16
    0.1 1.9892643553 1.9921014240 1.9915311325 1.9915647310
    0.4 1.9685783192 1.9662758325 1.9665941826 1.9666284065
    0.7 1.9349773042 1.9321150218 1.9314874599 1.9314539434
    1.0 1.8769375177 1.8767757148 1.8766451890 1.8766312223
    1.3 1.7937897759 1.7942987234 1.7942908009 1.7942769155
    1.6 1.6804986790 1.6794220904 1.6789060214 1.6788558248
    1.9 1.5285659506 1.5283296781 1.5275487916 1.5274909626

     | Show Table
    DownLoad: CSV

    In order to compare our method with other ones given in [21,22]. For different values of q=2,4,8,16, we solved Example (4.1) and tabulated the results at specific places in Table 1. Furthermore, we displayed the largest absolute errors at equidistant points in Table 2.

    Δ1={0.01,0.02,...1.99}.
    Table 2.  Maximum absolute errors at equidistant points Δ1.
    q SE Order Time
    2 3.97×103 0.572
    4 8.42×104 2.23 1.131
    8 7.35×105 3.51 7.247
    16 2.17×107 5.07 97.54
    32 1.70×109 6.99 2034

     | Show Table
    DownLoad: CSV

    Our manuscript is mainly focused on mixed derivative for fractional differential equations of order 1<β<2 and 0<γ<1. Applying the main tools from the fractional calculus, fixed point theorem, integro-differential equation, we propose the definition of α-mild solutions and obtain the existence and uniqueness dependence of the solution. Furthermore, we construct some important supposition to prove some important results. Finally, we provide an application to show the applicability of our main points.

    This work was supported by the Deanship of Scientific Research, Vice Presidency for Graduate Studies and Scientific Research, King Faisal University, Saudi Arabia [Grant No. 1388], through its KFU Research Summer initiative.

    The authors declare that they have no conflicts of interest.



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