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Uniqueness criteria for initial value problem of conformable fractional differential equation

  • This paper presents four uniqueness criteria for the initial value problem of a differential equation which depends on conformable fractional derivative. Among them is the generalization of Nagumo-type uniqueness theory and Lipschitz conditional theory, and advances its development in proving fractional differential equations. Finally, we verify the main conclusions of this paper by providing four concrete examples.

    Citation: Yumei Zou, Yujun Cui. Uniqueness criteria for initial value problem of conformable fractional differential equation[J]. Electronic Research Archive, 2023, 31(7): 4077-4087. doi: 10.3934/era.2023207

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  • This paper presents four uniqueness criteria for the initial value problem of a differential equation which depends on conformable fractional derivative. Among them is the generalization of Nagumo-type uniqueness theory and Lipschitz conditional theory, and advances its development in proving fractional differential equations. Finally, we verify the main conclusions of this paper by providing four concrete examples.



    Concurrent with the development of classic calculus theory, quantum calculus (calculus without limit) have received a great deal of attention in the last three decades. Quantum calculus have been found in many problems such as particle physics, quantum mechanics, and calculus of variations. In this paper, we study on the development of Hahn calculus, which is a type of quantum calculus. Hahn difference operator was first introduced by Hahn [1] in 1949 in the form of

    Dq,ωf(t):=f(qt+ω)f(t)t(q1)+ω,tω0:=ω1q.

    This operator has been further employed in many research works such as the studies of the right inverse and its properties of Hahn difference operator [2,3], Hahn quantum variational calculus [4,5,6], the initial value problems [7,8,9], and the boundary value problems [10,11]. The approximation problems and constructing families of orthogonal polynomials [12,13,14], Hahn difference operator is an important tool used to study in these areas.

    Based on the iadea of Hahn, in 2017, Brikshavana and Sitthiwirattham [15] introduced a general case of order of Hahn's operator, the so-called fractional Hahn difference operators. This operator has been used in the study of existencne and uniqueness of solution of boundary value problems for fractional Hahn difference equations (see [16,17,18,19]).

    The symmetric Hahn difference operator ˜Dq,ω is another opertor related to Hahn's operator. It was introduced by Artur et al. in 2013 [20] where

    ˜Dq,ωf(t):=f(qt+ω)f(q1(tω))(qq1)t+(1+q1)ωfortω0.

    Recently, Patanarapeelert and Sitthiwirattham [21] introduced the fractional symmetric Hahn integral, Riemann-Liouville and Caputo fractional symmetric Hahn difference operators and their properties. To present the advantage of this newest knowledge, in this paper, we devote our attention to study the solutions of boundary value problem for fractional symmetric Hahn difference equation.

    Our problem is a nonlocal fractional symmetric Hanh integral boundary value problem for fractional symmetric Hahn integrodifference equation of the form

    ˜Dαq,ωu(t)=F(t,u(t),˜Dβq,ωu(t),˜Ψγq,ωu(t),),tITq,ω,u(ω0)=λ1˜Iθ1q,ωg(η1)u(η1),u(T)=λ2˜Iθ2q,ωg(η2)u(η2),η1,η2ITq,ω{ω0,T}, (1.1)

    where ITq,ω:={qkT+ω[k]q:kN0}{ω0}; α(1,2];β,γ,θ1,θ2(0,1]; ω>0;q(0,1); λ1,λ2R+; FC(ITq,ω×R3,R) and g1,g2C(ITq,ω,R+) are given functions; and for φC(ITq,ω×ITq,ω,[0,)), we define

    ˜Ψγq,ωu(t):=(˜Iγq,ωφu)(t)=q(α2)˜Γq,ω(γ)tω0~(ts)γ1_q,ωφ(t,σα1q,ω(s))u(σα1q,ω(s))˜dq,ωs.

    In the next section, we give some definitions and lemmas related to fractional symmetric Hahn calculus. In section 3, we analyze the existence and uniqueness of a solution of problem (1.1) by using the Banach fixed point theorem. Moreover, we show the existence of at least one solution of problem (1.1) by using the Schuader's fixed point theorem. Finally, we present an example to illustrate our results in the last section.

    In this section, we provide some notations, definitions, and lemmas related to the fractional symmetric Hahn difference calculus as follows [20,21,22,23].

    For 0<q<1, ω>0, ω0=ω1q and [k]q=1qk1q, we define

    ~[k]q:={1q2k1q2=[k]q2,kN1,    k=0,
    ~[k]q!:={~[k]q~[k1]q~[1]q=ki=11q2i1q2,kN1,    k=0.

    The q,ω-forward jump operator is defined by

    σkq,ω(t):=qkt+ω[k]q,

    and the q,ω-backward jump operator is defined by

    ρkq,ω(t):=tω[k]qqk,

    where kN.

    Let nN0:={0,1,2,...}, and a,bR. We define the power functions as follows:

    ● The q-analogue of the power function

    (ab)0_q:=1,(ab)n_q:=n1i=0(abqi).

    ● The q-symmetric analogue of the power function

    ~(ab)0_q:=1,~(ab)n_q:=n1i=0(abq2i+1).

    ● The q,ω-symmetric analogue of the power function

    ~(ab)0_q,ω:=1,~(ab)n_q,ω:=n1i=0[aσ2i+1q,ω(b)].

    Generally, for αR, we have

    (ab)α_q=aαi=01(ba)qi1(ba)qα+i,a0,
    ~(ab)α_q=aαi=01(ba)q2i+11(ba)q2(α+i)+1,a0,
    ~(ab))α_q,ω=~((aω0)(bω0))α_q=(aω0)αi=01(bω0aω0)q2i+11(bω0aω0)q2(α+i)+1,aω0.

    Particularly, we have aα_q=˜aα_q=aα and ~(aω0)α_q,ω=(aω0)α if b=0. If a=b, we define (0)α_q=~(0)α_q=~(ω0)α_q,ω=0 for α>0.

    The q-symmetric gamma and q-symmetric beta functions are defined as

    ˜Γq(x):={(1q2)x1_q(1q2)x1=~(1q)x1_q(1q2)x1,xR{0,1,2,...}~[x1]q!,xN˜Bq(x,y):=10(q1s)x1~(1s)y1_q˜dqs=˜Γq(x)˜Γq(y)˜Γq(x+y),

    respectively.

    Lemma 2.1. [21] For m,nN0 and αR,

    (a) ~(xσnq,ω(x))α_q,ω=(xω0)k~(1qn)α_q,

    (b) ~(σmq,ω(x))σnq,ω(x))α_q,ω=qmα(xω0)α~(1qnm)α_q.

    Definition 2.1. [20] For q(0,1), ω>0, and f is a function defined on ITq,ωR, the symmetric Hahn difference of f is defined by

    ˜Dq,ωf(t):=f(σq,ω(t))f(ρq,ω(t))σq,ω(t)ρq,ω(t)tITq,ω{ω0},˜Dq,ωf(ω0)=f(ω0)  where  f  is  differentiable  at  ω0.

    ˜Dq,ωf is called q,ω-symmetric derivative of f, and f is q,ω-symmetric differentiable on ITq,ω.

    In addition, we define

    ˜D0q,ωf(x)=f(x) and ˜DNq,ωf(x)=˜Dq,ω˜DN1q,ωf(x) where NN.

    Remarks If f and g are q,ω-symmetric differentiable on ITq,ω,

    (a) ˜Dq,ω[f(t)+g(t)]=˜Dq,ωf(t)+˜Dq,ωg(t),

    (b) ˜Dq,ω[f(t)g(t)]=f(ρq,ω(t))˜Dq,ωg(t)+g(σq,ω(t))˜Dq,ωf(t),

    (c) ˜Dq,ω[f(t)g(t)]=g(ρq,ω(t))˜Dq,ωf(t)f(ρq,ω(t))˜Dq,ωg(t)g(ρq,ω(t))g(σq,ω(t)), g(ρq,ω(t))g(σq,ω(t))0,

    (d) ˜Dq,ω[C]=0 where C is constant.

    Definition 2.2. [20] Let I be any closed interval of R containing a,b and ω0 and f:IR be a given function. The symmetric Hahn integral of f from a to b is defind by

    baf(t)˜dq,ωt:=bω0f(t)˜dq,ωtaω0f(t)˜dq,ωt,

    where

    ˜Iq,ωf(t)=xω0f(t)˜dq,ωt:=(1q2)(xω0)k=0q2kf(σ2k+1q,ω(x)),xI.

    Providing that the above series converges at x=a and x=b, f is symmetric Hahn integrable on [a,b]. In addition, f is symmetric Hahn integrable on I if it is symmetric Hahn integrable on [a,b] for all a,bI.

    In addition,

    ˜I0q,ωf(x)=f(x),˜INq,ωf(x)=˜Iq,ω˜IN1q,ωf(x)whereNN,
    ˜Dq,ω˜Iq,ωf(x)=f(x),and˜Iq,ω˜Dq,ωf(x)=f(x)f(ω0).

    Remarks [20] Let a,bITq,ω and f,g be symmetric Hahn integrable on ITq,ω. Then,

    (a) aaf(t)˜dq,ωt=0,

    (b) baf(t)˜dq,ωt=abf(t)˜dq,ωt,

    (c) baf(t)˜dq,ωt=bcf(t)˜dq,ωt+caf(t)˜dq,ωt,cITq,ω,a<c<b,

    (d) ba[αf(t)+βg(t)]˜dq,ωt=αbaf(t)˜dq,ωt+βbag(t)˜dq,ωt,α,βR,

    (e) ba[f(ρq,ω(t))˜Dq,ωg(t)]˜dq,ωt=[f(t)g(t)]baba[g(σq,ω(t))˜Dq,ωf(t)]˜dq,ωt.

    Lemma 2. [20] [Fundamental theorem of symmetric Hahn calculus]

    Let f:IR be continuous at ω0. Then,

    F(x):=xω0f(t)˜dq,ωt,xI

    is continuous at ω0 and ˜Dq,ωF(x) exists for every xσq,ω(I):={qt+ω:tI} where

    ˜Dq,ωF(x)=f(x).

    In addition,

    ba˜Dq,ωf(t)˜dq,ωt=f(b)f(a)  for  all  a,bI.

    Lemma 2.3. [21] Let 0<q<1, ω>0 and f:IR be continuous at ω0. Then,

    tω0rω0f(s)˜dq,ωs˜dq,ωr=qtω0tqs+ωf(qs+ω)˜dq,ωr˜dq,ωs.

    Definition 2.3. [21] Let α,ω>0,0<q<1, and f be a function defined on ITq,ω. The fractional symmetric Hahn integral is defined by

    ˜Iαq,ωf(t):=q(α2)˜Γq(α)tω0~(ts)α1_q,ωf(σα1q,ω(s))˜dq,ωs=(1q2)q(α2)(tω0)˜Γq(α)k=0q2k(tσ2k+1q,ω(t))α1_q,ωf(σ2k+αq,ω(t))=(1q2)q(α2)(tω0)α˜Γq(α)k=0q2k~(1q2k+1)α1_qf(σ2k+αq,ω(t))

    and ˜I0q,ωf)(t)=f(t).

    Definition 2.4. [21] For α,ω>0,0<q<1 and f defined on ITq,ω, the fractional symmetric Hahn difference operator of Riemann-Liouville type of order α is defined by

    ˜Dαq,ωf(t):=˜DNq,ω˜INαq,ωf(t)=q(α2)˜Γq(α)tω0~(ts)α1_q,ωf(σα1q,ω(s))˜dq,ωs,˜D0q,ωf(t)=f(t)

    where N1<α<N,NN.

    Lemma 2.4. [21] Let α,ω>0,0<q<1 and f:ITq,ωR. Then,

    ˜Iαq,ω˜Dαq,ωf(t)=f(t)+C1(tω0)α1+C2(tω0)α2+...+CN(tω0)αN

    for some CiR,i=1,2,...,N and N1<α<N for NN.

    Lemma 2.5. [24] (Arzelá-Ascoli theorem) A set of function in C[a,b] with the sup norm, is relatively compact if and only if it is uniformly bounded and equicontinuous on [a,b].

    Lemma 2.6. [24] If a set is closed and relatively compact then it is compact.

    Lemma 2.7. [25] (Schauder's fixed point theorem) Let (D,d) be a complete metric space, U be a closed convex subset of D, and T:DD be the map such that the set Tu:uU is relatively compact in D. Then the operator T has at least one fixed point uU: Tu=u.

    In this section, we formulate some lemmas that will be used as a tool for our calculations as follows.

    Lemma 2.8. Let q(0,1),ω>0 and n>0. Then,

    tω0˜dq,ωs=tω0  and  tω0(sω0)n˜dq,ωs=qn~[n+1]q(tω0)n+1.

    Proof. Using the definition of symmetric Hahn integral, we have

    tω0˜dq,ωs=(1q2)(tω0)k=0q2k=(1q2)(tω0)[11q2]=tω0,

    and

    tω0(sω0)n˜dq,ωs=(1q2)(tω0)k=0q2k(σ2k+1q,ω(t)ω0)n=qn(1q2)(tω0)n+1k=0q(n+1)2k=qn(1q2)(tω0)n+1[11q2(n+1)]=qn~[n+1]q(tω0)n+1.

    The proof is complete.

    Lemma 2.9. Let α,β>0,q(0,1) and ω>0. Then,

    (i)tω0~(ts)α1_q,ω˜dq,ωs=(tω0)α~[α]q,(ii)tω0~(ts)α1_q,ω(σα1q,ω(s)ω0)β˜dq,ωs=qαβ(tω0)α+βBq(β+1,α),(iii)tω0σα1q,ω(s)ω0~(ts)α1_q,ω~(σα1q,ω(s)r)β1_q,ω˜dq,ωr˜dp,ωs=qαβ~[β]q(tω0)α+βBq(β+1,α).

    Proof. From the definition of q,ω-symmetric analogue of the power function, Lemma 2.1 and Definition 2.2, we obtain

    (i)tω0~(ts)α1_q,ω˜dq,ωs=(1q2)(tω0)k=0q2k~(tσ2k+1q,ω(t))α1_q,ω=(1q2)(tω0)αk=0q2k~(1q2k+1)α1_q=(1q2)(tω0)αk=0q2k[i=01q2k+2i+21q2k+2i+2α]=(tω0)α~[α]q,(ii)tω0~(ts)α1_q,ω(σα1q,ω(s)ω0)β˜dq,ωs=(1q2)(tω0)k=0q2k~(tσ2k+1q,ω(t))α1_q,ω(qα1(σ2k+1q,ω(t)ω0))β=qαβ(1q2)(tω0)α+βk=0q2k~(1q2k+1)α1_q(q1q2k+1)β=qαβ(tω0)α+β1ω0~(1s)α1_q,ω(q1s)β˜dq,ωs=qαβ(tω0)α+β˜Bq(β+1,α).

    Using (ⅰ) and (ⅱ), we have

    tω0σα1q,ω(s)ω0~(ts)α1_q,ω~(σα1q,ω(s)r)β1_q,ω˜dq,ωr˜dp,ωs=tω0~(ts)α1_q,ω[σα1q,ω(s)ω0~(σα1q,ω(s)r)β1_q,ω˜dq,ωr]˜dp,ωs=1~[β]qtω0~(ts)α1_q,ω(σα1q,ω(s)ω0)β˜dp,ωs=qαβ~[β]q(tω0)α+β˜Bq(β+1,α).

    The following lemma present a solution of a linear variant form of the problem (1.1).

    Lemma 2.10. Let Λ0;ω>0;q(0,1); α(1,2];θ1,θ2(0,1]; λ1,λ2R+; hC(ITq,ω,R) and g1,g2C(ITq,ω,R+) be given functions. Then the linear variant form

    ˜Dαq,ωu(t)=h(t),tITq,ω,u(ω0)=λ1˜Iθ1q,ωg(η1)u(η1),u(T)=λ2˜Iθ2q,ωg(η2)u(η2),η1,η2ITq,ω{ω0,T}, (2.1)

    has the unique solution which is

    u(t)=q(α2)˜Γq(α)tω0~(ts)α1_q,ωh(σα1q,ω(s))˜dq,ωs+(tω0)α1Λ{B2Φ1[h]+A2Φ2[h]}(tω0)α2A2{(1+A1B2Λ)Φ1[h]+A1A2ΛΦ2[h]} (2.2)

    where the functionals Φ1[h],Φ2[h] are defined by

    Φ1[h]:=λ1q(θ12)+(α2)˜Γq(θ1)˜Γq(α)η1ω0σθ11q,ω(s)ω0~(η1s)θ11_q,ω~(σθ11q,ω(s)r)α1_q,ωg1(σθ11q,ω(s))×h(σα1q,ω(r))˜dq,ωr˜dq,ωs, (2.3)
    Φ2[h]:=λ2q(θ22)+(α2)˜Γq(θ2)˜Γq(α)η2ω0σθ21q,ω(s)ω0~(η2s)θ21_q,ω~(σθ21q,ω(s)r)α1_q,ωg2(σθ21q,ω(s))×h(σα1q,ω(r))˜dq,ωr˜dq,ωsq(α2)˜Γq(α)Tω0~(Ts)α1_q,ωh(σα1q,ω(s))˜dq,ωs, (2.4)

    and the constants A1,A2,B1,B2 and Λ are defined by

    A1:=λ1q(θ12)˜Γq(θ1)η1ω0~(η1s)θ11_q,ωg1(σθ11q,ω(s))(σθ11q,ω(s)ω0)α1˜dq,ωs, (2.5)
    A2:=λ1q(θ12)˜Γq(θ1)η1ω0~(η1s)θ11_q,ωg1(σθ11q,ω(s))(σθ11q,ω(s)ω0)α2˜dq,ωs, (2.6)
    B1:=(Tω0)α1λ2q(θ22)˜Γq(θ2)η2ω0~(η2s)θ21_q,ωg2(σθ21q,ω(s))(σθ21q,ω(s)ω0)α1˜dq,ωs, (2.7)
    B2:=(Tω0)α2λ2q(θ22)˜Γq(θ2)η2ω0~(η2s)θ21_q,ωg2(σθ21q,ω(s))(σθ21q,ω(s)ω0)α2˜dq,ωs, (2.8)
    Λ:=A2B1A1B2. (2.9)

    Proof. Taking fractional symmetric Hahn integral of order α for (2.1), we obtain

    u(t)=C1(tω0)α1+C2(tω0)α2+q(α2)˜Γq(α)tω0~(ts)α1_q,ωh(σα1q,ω(s))˜dq,ωs. (2.10)

    Taking fractional symmetric Hahn integral of order θi,i=1,2 for (2.10), we get

    Iθiq,ωu(t)=q(θi2)˜Γq(θi)tω0~(ts)θi1_q,ω[C1(σθi1q,ω(s)ω0)α1+C2(σθi1q,ω(s)ω0)α2]˜dq,ωs+q(θi2)+(α2)˜Γq(θi)˜Γq(α)tω0σθi1q,ω(s)ω0~(ts)θi1_q,ω~(σθi1q,ω(s)r)α1_q,ωh(σα1q,ω(r))˜dq,ωr˜dq,ωs. (2.11)

    After substituting i=1 into (2.11) and employing the first condition of (2.1), we have

    A1C1+A2C2=Φ1[h]. (2.12)

    Taking i=2 into (2.11) and employing the second condition of (2.1), we have

    B1C1+B2C2=Φ2[h], (2.13)

    where Φ1[h],Φ2[h],A1,A2,B1 and B2 are defined as (2.3)(2.8), respectively.

    Solving the system of Eqs (2.12)(2.13), we have

    C1=B2Φ1[h]+A2Φ2[h]ΛandC2=1A2{(1+A1B2Λ)Φ1[h]+A1A2ΛΦ2[h]},

    where Λ is defined as (2.9). Substituting the constants C1 and C2 into (2.10), we obtain (2.2).

    In this section, we prove the existence and uniqueness of solution of the problem (1.1). Furthermore, we show the existence of at least one solution of problem (1.1).

    In this section, we consider the existence and uniqueness result for the problem (1.1). Let C=C(ITq,ω,R) be a Banach space of all function u with the norm defined by

    uC=maxtITq,ω{|u(t)|,|˜Dβq,ωu(t)|},

    where α(1,2];β,γ,θ1,θ2(0,1]; ω>0;q(0,1); λ1,λ2R+. We define an operator F:CC as

    (Fu)(t):=q(α2)˜Γq(α)tω0~(ts)α1_q,ω×F(σα1q,ω(s),u(σα1q,ω(s)),˜Dβq,ωu(σα1q,ω(s)),˜Ψγq,ωu(σα1q,ω(s)))˜dq,ωs+(tω0)α1Λ{B2Φ1[f(u)]+A2Φ2[f(u)]}(tω0)α2A2{(1+A1B2Λ)Φ1[f(u)]+A1A2ΛΦ2[f(u)]} (3.1)

    where the functionals Φ1[F(u)],Φ2[F(u)] are given by

    Φ1[F(u)]:=λ1q(θ12)+(α2)˜Γq(θ1)˜Γq(α)η1ω0σθ11q,ω(s)ω0~(η1s)θ11_q,ω~(σθ11q,ω(s)r)α1_q,ωg1(σθ11q,ω(s))×F(σα1q,ω(r),u(σα1q,ω(r)),˜Dβq,ωu(σα1q,ω(r)),˜Ψγq,ωu(σα1q,ω(r)))˜dq,ωr˜dq,ωs, (3.2)
    Φ2[F(u)]:=λ2q(θ22)+(α2)˜Γq(θ2)˜Γq(α)η2ω0σθ21q,ω(s)ω0~(η2s)θ21_q,ω~(σθ21q,ω(s)r)α1_q,ωg2(σθ21q,ω(s))×F(σα1q,ω(r),u(σα1q,ω(r)),˜Dβq,ωu(σα1q,ω(r)),˜Ψγq,ωu(σα1q,ω(r)))˜dq,ωr˜dq,ωsq(α2)˜Γq(α)Tω0~(Ts)α1_q,ω×F(σα1q,ω(r),u(σα1q,ω(r)),˜Dβq,ωu(σα1q,ω(r)),˜Ψγq,ωu(σα1q,ω(r)))˜dq,ωs, (3.3)

    and the constants A1,A2,B1,B2 and Λ are defined by (2.5)-(2.9), respectively.

    We find that the problem (1.1) has solution if and only if the operator F has fixed point.

    Theorem 3.1. Assume that F:ITq,ω×R3R, and g1,g2:ITq,ωR+ are continuous, and φ:ITq,ω×ITq,ω[0,) is continuous with φ0=max{φ(t,s):(t,s)ITq,ω×ITq,ω}. Suppose that the following conditions hold:

    (H1) There exist constants 1,2,3>0 such that for each tITq,ω and u,vR,

    |F(t,u,˜Dβq,ωu,˜Ψγq,ωu)F(t,v,˜Dβq,ωv,˜Ψγq,ωv)|1|uv|+2|˜Dβq,ωu˜Dβq,ωv|+3|˜Ψγq,ωu˜Ψγq,ωv|.

    (H2) There exist constants gi,Gi>0 where i=1,2 such that for each tITq,ω,

    0<gi<gi(t)<Gi.

    (H3) Θ<1,

    where

    Ω1:=λ1G1q(θ12)+(α2)+θ1+α˜Γq(θ1+α+1)(η1ω0)θ1+α (3.4)
    Ω2:=λ2G2q(θ22)+(α2)+θ2+α˜Γq(θ2+α+1)(η2ω0)θ2+α+q(α2)˜Γq(α+1)(Tω0)α (3.5)
    ¯A1:=λ1G1q(θ12)+θ1(α1)˜Γq(α)˜Γq(θ1+α)(η1ω0)θ1+α1>A1 (3.6)
    ¯A2:=λ1G1q(θ12)+θ1(α2)˜Γq(α1)˜Γq(θ1+α1)(η1ω0)θ1+α2>A2 (3.7)
    A2:=λ1g1q(θ12)+θ1(α2)˜Γq(α1)˜Γq(θ1+α1)(η1ω0)θ1+α2<A2 (3.8)
    ¯B1:=(Tω0)α1+λ2G2q(θ22)+θ2(α1)˜Γq(α)˜Γq(θ2+α)(η2ω0)θ2+α1>B1 (3.9)
    ¯B2:=(Tω0)α2+λ2G2q(θ22)+θ2(α2)˜Γq(α1)˜Γq(θ2+α1)(η2ω0)θ2+α2>B2 (3.10)
    Λ:=g1g2|¯A2¯B1¯A1¯B2|<|Λ| (3.11)

    and

    Θ:=(1+2+3φ0(T+ω0)γ˜Γq(γ+1)){(Tω0)α1Λ[¯B2Ω1+¯A2Ω2]+(tω0)α2A2[(1+¯A1¯B2Λ)Ω1+¯A1¯A2ΛΩ2]+q(α2)(Tω0)α˜Γq(α+1)}. (3.12)

    Then the problem (1.1) has a unique solution in ITq,ω.

    Proof. To show that F is contraction, we first consider

    H|uv|(t):=|F(t,u(t),˜Dβq,ωu(t),˜Ψγq,ωu(t))F(t,v(t),˜Dβq,ωv(t),˜Ψγq,ωv(t))|,

    for each tITq,ω and u,vC. We find that

    |Φ1[F(u)]Φ1[F(v)]|λ1G1q(θ12)+(α2)˜Γq(θ1)˜Γq(α)η1ω0σθ11q,ω(s)ω0~(η1s)θ11_q,ω~(σθ11q,ω(s)r)α1_q,ωH|uv|(r)˜dq,ωr˜dq,ωs(1|uv|+2|˜Dβq,ωu˜Dβq,ωv|+3|˜Ψγq,ωu˜Ψγq,ωv|)×λ1G1q(θ12)+(α2)˜Γq(θ1)˜Γq(α)η1ω0σθ11q,ω(s)ω0~(η1s)θ11_q,ω~(σθ11q,ω(s)r)α1_q,ω˜dq,ωr˜dq,ωs[(1+3φ0(T+ω0)γ˜Γq(γ+1))|uv|+2|˜Dβq,ωu˜Dβq,ωv|]Ω1uvC(1+2+3φ0(T+ω0)γ˜Γq(γ+1))Ω1.

    Similary,

    |Φ2[F(u)]Φ2[F(v)]|uvC(1+2+3φ0(T+ω0)γ˜Γq(γ+1))Ω2.

    In addition, we find that

    |(Fu)(t)(Fv)(t)|q(α2)˜Γq(α)tω0~(ts)α1_q,ωH|uv|(s)˜dq,ωs+uvC(1+2+3φ0(T+ω0)γ˜Γq(γ+1))(Tω0)α1|Λ|{B2Ω1+A2Ω2}+uvC(1+2+3φ0(T+ω0)γ˜Γq(γ+1))(Tω0)α2A2{(1+A1B2|Λ|)Ω1+A1A2|Λ|Ω2}uvC(1+2+3φ0(T+ω0)γ˜Γq(γ+1)){q(α2)(Tω0)α˜Γq(α+1)+(Tω0)α1Λ[¯B2Ω1+¯A2Ω2]+(tω0)α2A2[(1+¯A1¯B2Λ)Ω1+¯A1¯A2ΛΩ2]}=uvCΘ. (3.13)

    Taking fractional symmetric Hahn difference of order γ for (3.1), we obtain

    (˜Dβq,ωFu)(t)=q(β2)+(α2)˜Γq(β)˜Γq(α)tω0σβ1q,ω(s)ω0~(ts)β1_q,ω~(σβ1q,ω(s)r)α1_q,ω×F(σα1q,ω(r),u(σα1q,ω(r)),˜Dβq,ωu(σα1q,ω(r)),˜Ψγq,ωu(σα1q,ω(r)))˜dq,ωr˜dq,ωs+[B2Φ1[f(u)]+A2Φ2[f(u)]Λ]q(β2)˜Γq(β)tω0~(ts)β1_q,ω(σβ1q,ω(s)ω0)α1˜dq,ωs1A2[(1+A1B2Λ)Φ1[f(u)]+A1A2ΛΦ2[f(u)]]×q(β2)˜Γq(β)tω0~(ts)β1_q,ω(σβ1q,ω(s)ω0)α2˜dq,ωs. (3.14)

    Similary, we have

    |(˜Dβq,ωFu)(t)(˜Dβq,ωFv)(t)|<uvCΘ. (3.15)

    From (3.13) and (3.15), we get

    FuFvCuvCΘ.

    Using (H3) we can conclude that F is a contraction. Based on Banach fixed point theorem, F has a fixed point which is a unique solution of problem (1.1) on ITq,ω.

    In this section, we particularly study the existence of at least one solution of (1.1) by using the Schauder's fixed point theorem as follows

    Theorem 3.2. Suppose that (H1) and (H3) defined in Theorem 3.1 hold. Then, problem (1.1) has at least one solution on ITq,ω.

    Proof. The proof is established as the following structures.

    Step Ⅰ. Verify F map bounded sets into bounded sets in BR. Let BR={uC(ITq,ω):uCR}, maxtITq,ω|F(t,0,0,0)|=M and choose a constant

    RM{(Tω0)α1Λ[¯B2Ω1+¯A2Ω2]+(tω0)α2A2[(1+¯A1¯B2Λ)Ω1+¯A1¯A2ΛΩ2]+q(α2)(Tω0)α˜Γq(α+1)}1Θ. (3.16)

    Denote that

    |K(t,u,0)|=|F(t,u(t),˜Dβq,ωu(t),˜Ψγq,ωu(t))F(t,0,0,0)|+|F(t,0,0,0)|.

    We find that

    |Φ1[F(u)]|λ1G1q(θ12)+(α2)˜Γq(θ1)˜Γq(α)η1ω0σθ11q,ω(s)ω0~(η1s)θ11_q,ω~(σθ11q,ω(s)r)α1_q,ω|K(t,u,0)|˜dq,ωr˜dq,ωs[(1|u|+2|˜Dβq,ωu|+3|˜Ψγq,ωu|)+M]Ω1[(1+2+3φ0(T+ω0)γ˜Γq(γ+1))uvC+M]Ω1[(1+2+3φ0(T+ω0)γ˜Γq(γ+1))R+M]Ω1, (3.17)

    where tITq,ω and uBR.

    Similary,

    |Φ2[F(u)]|[(1+2+3φ0(T+ω0)γ˜Γq(γ+1))R+M]Ω2. (3.18)

    Employing (3.17) and (3.18), we find that

    |(Fu)(t)|[(1+2+3φ0(T+ω0)γ˜Γq(γ+1))R+M]{(Tω0)α1Λ[¯B2Ω1+¯A2Ω2]+(tω0)α2A2[(1+¯A1¯B2Λ)Ω1+¯A1¯A2ΛΩ2]+q(α2)(Tω0)α˜Γq(α+1)}R. (3.19)

    Since

    |(˜Dβq,ωFu)(t)|<R. (3.20)

    Therefore, FuCR. Hence, F is uniformly bounded.

    Step Ⅱ. That the operator F is continuous on BR since the continuity of F.

    Step Ⅲ. Examine that F is equicontinuous on BR.

    For any t1,t2ITq,ω with t1<t2, by Lemma 2.9 we have

    |(Fu)(t2)(Fu)(t1)|q(α2)F˜Γq(α+1)|(t2ω0)α(t1ω0)α|+(¯B2Ω1+¯A2Ω2)FΛ|(t2ω0)α1(t1ω0)α1|+FA2[(1+¯A1¯B2Λ)Ω1+¯A1¯A2ΛΩ2]|(t2ω0)α2(t1ω0)α2| (3.21)

    and

    |(˜Dβq,ωFu)(t1)(˜Dβq,ωFu)(t2)|q(α2)+(β2)αβF˜Γq(αβ+1)|(t2ω0)αβ(t1ω0)αβ|+q(β2)αβ˜Γq(α)FΛ˜Γq(αβ)(¯B2Ω1+¯A2Ω2)|(t2ω0)αβ1(t1ω0)αβ1|+q(β2)αβ˜Γq(α1)FA2˜Γq(αβ1)[(1+¯A1¯B2Λ)Ω1+¯A1¯A2ΛΩ2]×|(t2ω0)αβ2(t1ω0)αβ2|. (3.22)

    Since the right-hand side of (3.22) tends to be zero when , is relatively compact on . Therefore, the set is an equicontinuous set. From Steps I to III together with the Arzelá-Ascoli theorem, is completely continuous. By Schauder's fixed point theorem, we can conclude that problem (1.1) has at least one solution.

    Thoroughly, we provide the boundary value problem for fractional Hahn difference equation

    (4.1)

    where and .

    We let and

    For all and , we have

    and

    Thus, and hold with and .

    Since

    and

    therefore, holds with

    Hence, by Theorem problem has a unique solution.

    The new problem containing two fractional symmetric Hahn difference operators and three fractional symmetric Hahn integral with different numbers of order was proposed. The new concepts of fractional symmetric Hanh calculus were used in the study of existence results of the govern problem. The Banach fixed point and Schauder's fixed point theorems were also employed in this study.

    This research was funded by King Mongkut's University of Technology North Bangkok. Contract no.KMUTNB-61-KNOW-027. The last author of this research was supported by Suan Dusit University.

    The authors declare no conflicts of interest regarding the publication of this paper.



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