Citation: Nichaphat Patanarapeelert, Thanin Sitthiwiratthame. On nonlocal fractional symmetric Hanh integral boundary value problems for fractional symmetric Hahn integrodifference equation[J]. AIMS Mathematics, 2020, 5(4): 3556-3572. doi: 10.3934/math.2020231
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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(q−1)+ω,t≠ω0:=ω1−q. |
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(q−1(t−ω))(q−q−1)t+(1+q−1)ωfort≠ω0. |
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),),t∈ITq,ω,u(ω0)=λ1˜Iθ1q,ωg(η1)u(η1),u(T)=λ2˜Iθ2q,ωg(η2)u(η2),η1,η2∈ITq,ω−{ω0,T}, | (1.1) |
where ITq,ω:={qkT+ω[k]q:k∈N0}∪{ω0}; α∈(1,2];β,γ,θ1,θ2∈(0,1]; ω>0;q∈(0,1); λ1,λ2∈R+; F∈C(ITq,ω×R3,R) and g1,g2∈C(ITq,ω,R+) are given functions; and for φ∈C(ITq,ω×ITq,ω,[0,∞)), we define
˜Ψγq,ωu(t):=(˜Iγq,ωφu)(t)=q(α2)˜Γq,ω(γ)∫tω0~(t−s)γ−1_q,ωφ(t,σα−1q,ω(s))u(σα−1q,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=ω1−q and [k]q=1−qk1−q, we define
~[k]q:={1−q2k1−q2=[k]q2,k∈N1, k=0, |
~[k]q!:={~[k]q~[k−1]q⋅⋅⋅~[1]q=k∏i=11−q2i1−q2,k∈N1, 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):=t−ω[k]qqk, |
where k∈N.
Let n∈N0:={0,1,2,...}, and a,b∈R. We define the power functions as follows:
● The q-analogue of the power function
(a−b)0_q:=1,(a−b)n_q:=n−1∏i=0(a−bqi). |
● The q-symmetric analogue of the power function
~(a−b)0_q:=1,~(a−b)n_q:=n−1∏i=0(a−bq2i+1). |
● The q,ω-symmetric analogue of the power function
~(a−b)0_q,ω:=1,~(a−b)n_q,ω:=n−1∏i=0[a−σ2i+1q,ω(b)]. |
Generally, for α∈R, we have
(a−b)α_q=aα∞∏i=01−(ba)qi1−(ba)qα+i,a≠0, |
~(a−b)α_q=aα∞∏i=01−(ba)q2i+11−(ba)q2(α+i)+1,a≠0, |
~(a−b))α_q,ω=~((a−ω0)−(b−ω0))α_q=(a−ω0)α∞∏i=01−(b−ω0a−ω0)q2i+11−(b−ω0a−ω0)q2(α+i)+1,a≠ω0. |
Particularly, we have aα_q=˜aα_q=aα and ~(a−ω0)α_q,ω=(a−ω0)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):={(1−q2)x−1_q(1−q2)x−1=~(1−q)x−1_q(1−q2)x−1,x∈R∖{0,−1,−2,...}~[x−1]q!,x∈N˜Bq(x,y):=∫10(q−1s)x−1~(1−s)y−1_q˜dqs=˜Γq(x)˜Γq(y)˜Γq(x+y), |
respectively.
Lemma 2.1. [21] For m,n∈N0 and α∈R,
(a) ~(x−σnq,ω(x))α_q,ω=(x−ω0)k~(1−qn)α_q,
(b) ~(σmq,ω(x))−σnq,ω(x))α_q,ω=qmα(x−ω0)α~(1−qn−m)α_q.
Definition 2.1. [20] For q∈(0,1), ω>0, and f is a function defined on ITq,ω⊆R, 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)t∈ITq,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˜DN−1q,ωf(x) where N∈N. |
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:I→R be a given function. The symmetric Hahn integral of f from a to b is defind by
∫baf(t)˜dq,ωt:=∫bω0f(t)˜dq,ωt−∫aω0f(t)˜dq,ωt, |
where
˜Iq,ωf(t)=∫xω0f(t)˜dq,ωt:=(1−q2)(x−ω0)∞∑k=0q2kf(σ2k+1q,ω(x)),x∈I. |
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,b∈I.
In addition,
˜I0q,ωf(x)=f(x),˜INq,ωf(x)=˜Iq,ω˜IN−1q,ωf(x)whereN∈N, |
˜Dq,ω˜Iq,ωf(x)=f(x),and˜Iq,ω˜Dq,ωf(x)=f(x)−f(ω0). |
Remarks [20] Let a,b∈ITq,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,c∈ITq,ω,a<c<b,
(d) ∫ba[αf(t)+βg(t)]˜dq,ωt=α∫baf(t)˜dq,ωt+β∫bag(t)˜dq,ωt,α,β∈R,
(e) ∫ba[f(ρq,ω(t))˜Dq,ωg(t)]˜dq,ωt=[f(t)g(t)]ba−∫ba[g(σq,ω(t))˜Dq,ωf(t)]˜dq,ωt.
Lemma 2. [20] [Fundamental theorem of symmetric Hahn calculus]
Let f:I→R be continuous at ω0. Then,
F(x):=∫xω0f(t)˜dq,ωt,x∈I |
is continuous at ω0 and ˜Dq,ωF(x) exists for every x∈σq,ω(I):={qt+ω:t∈I} where
˜Dq,ωF(x)=f(x). |
In addition,
∫ba˜Dq,ωf(t)˜dq,ωt=f(b)−f(a) for all a,b∈I. |
Lemma 2.3. [21] Let 0<q<1, ω>0 and f:I→R be continuous at ω0. Then,
∫tω0∫rω0f(s)˜dq,ωs˜dq,ωr=q∫tω0∫tqs+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~(t−s)α−1_q,ωf(σα−1q,ω(s))˜dq,ωs=(1−q2)q(α2)(t−ω0)˜Γq(α)∞∑k=0q2k(t−σ2k+1q,ω(t))α−1_q,ωf(σ2k+αq,ω(t))=(1−q2)q(α2)(t−ω0)α˜Γq(α)∞∑k=0q2k~(1−q2k+1)α−1_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˜IN−αq,ωf(t)=q(−α2)˜Γq(−α)∫tω0~(t−s)−α−1_q,ωf(σ−α−1q,ω(s))˜dq,ωs,˜D0q,ωf(t)=f(t) |
where N−1<α<N,N∈N.
Lemma 2.4. [21] Let α,ω>0,0<q<1 and f:ITq,ω→R. Then,
˜Iαq,ω˜Dαq,ωf(t)=f(t)+C1(t−ω0)α−1+C2(t−ω0)α−2+...+CN(t−ω0)α−N |
for some Ci∈R,i=1,2,...,N and N−1<α<N for N∈N.
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:D→D be the map such that the set Tu:u∈U is relatively compact in D. Then the operator T has at least one fixed point u∗∈U: 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=t−ω0 and ∫tω0(s−ω0)n˜dq,ωs=qn~[n+1]q(t−ω0)n+1. |
Proof. Using the definition of symmetric Hahn integral, we have
∫tω0˜dq,ωs=(1−q2)(t−ω0)∞∑k=0q2k=(1−q2)(t−ω0)[11−q2]=t−ω0, |
and
∫tω0(s−ω0)n˜dq,ωs=(1−q2)(t−ω0)∞∑k=0q2k(σ2k+1q,ω(t)−ω0)n=qn(1−q2)(t−ω0)n+1∞∑k=0q(n+1)2k=qn(1−q2)(t−ω0)n+1[11−q2(n+1)]=qn~[n+1]q(t−ω0)n+1. |
The proof is complete.
Lemma 2.9. Let α,β>0,q∈(0,1) and ω>0. Then,
(i)∫tω0~(t−s)α−1_q,ω˜dq,ωs=(t−ω0)α~[α]q,(ii)∫tω0~(t−s)α−1_q,ω(σα−1q,ω(s)−ω0)β˜dq,ωs=qαβ(t−ω0)α+βBq(β+1,α),(iii)∫tω0∫σα−1q,ω(s)ω0~(t−s)α−1_q,ω~(σα−1q,ω(s)−r)β−1_q,ω˜dq,ωr˜dp,ωs=qαβ~[β]q(t−ω0)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~(t−s)α−1_q,ω˜dq,ωs=(1−q2)(t−ω0)∞∑k=0q2k~(t−σ2k+1q,ω(t))α−1_q,ω=(1−q2)(t−ω0)α∞∑k=0q2k~(1−q2k+1)α−1_q=(1−q2)(t−ω0)α∞∑k=0q2k[∞∏i=01−q2k+2i+21−q2k+2i+2α]=(t−ω0)α~[α]q,(ii)∫tω0~(t−s)α−1_q,ω(σα−1q,ω(s)−ω0)β˜dq,ωs=(1−q2)(t−ω0)∞∑k=0q2k~(t−σ2k+1q,ω(t))α−1_q,ω(qα−1(σ2k+1q,ω(t)−ω0))β=qαβ(1−q2)(t−ω0)α+β∞∑k=0q2k~(1−q2k+1)α−1_q(q−1q2k+1)β=qαβ(t−ω0)α+β∫1ω0~(1−s)α−1_q,ω(q−1s)β˜dq,ωs=qαβ(t−ω0)α+β˜Bq(β+1,α). |
Using (ⅰ) and (ⅱ), we have
∫tω0∫σα−1q,ω(s)ω0~(t−s)α−1_q,ω~(σα−1q,ω(s)−r)β−1_q,ω˜dq,ωr˜dp,ωs=∫tω0~(t−s)α−1_q,ω[∫σα−1q,ω(s)ω0~(σα−1q,ω(s)−r)β−1_q,ω˜dq,ωr]˜dp,ωs=1~[β]q∫tω0~(t−s)α−1_q,ω(σα−1q,ω(s)−ω0)β˜dp,ωs=qαβ~[β]q(t−ω0)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 Λ≠0;ω>0;q∈(0,1); α∈(1,2];θ1,θ2∈(0,1]; λ1,λ2∈R+; h∈C(ITq,ω,R) and g1,g2∈C(ITq,ω,R+) be given functions. Then the linear variant form
˜Dαq,ωu(t)=h(t),t∈ITq,ω,u(ω0)=λ1˜Iθ1q,ωg(η1)u(η1),u(T)=λ2˜Iθ2q,ωg(η2)u(η2),η1,η2∈ITq,ω−{ω0,T}, | (2.1) |
has the unique solution which is
u(t)=q(α2)˜Γq(α)∫tω0~(t−s)α−1_q,ωh(σα−1q,ω(s))˜dq,ωs+(t−ω0)α−1Λ{B2Φ1[h]+A2Φ2[h]}−(t−ω0)α−2A2{(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)∫η1ω0∫σθ1−1q,ω(s)ω0~(η1−s)θ1−1_q,ω~(σθ1−1q,ω(s)−r)α−1_q,ωg1(σθ1−1q,ω(s))×h(σα−1q,ω(r))˜dq,ωr˜dq,ωs, | (2.3) |
Φ2[h]:=λ2q(θ22)+(α2)˜Γq(θ2)˜Γq(α)∫η2ω0∫σθ2−1q,ω(s)ω0~(η2−s)θ2−1_q,ω~(σθ2−1q,ω(s)−r)α−1_q,ωg2(σθ2−1q,ω(s))×h(σα−1q,ω(r))˜dq,ωr˜dq,ωs−q(α2)˜Γq(α)∫Tω0~(T−s)α−1_q,ωh(σα−1q,ω(s))˜dq,ωs, | (2.4) |
and the constants A1,A2,B1,B2 and Λ are defined by
A1:=λ1q(θ12)˜Γq(θ1)∫η1ω0~(η1−s)θ1−1_q,ωg1(σθ1−1q,ω(s))(σθ1−1q,ω(s)−ω0)α−1˜dq,ωs, | (2.5) |
A2:=λ1q(θ12)˜Γq(θ1)∫η1ω0~(η1−s)θ1−1_q,ωg1(σθ1−1q,ω(s))(σθ1−1q,ω(s)−ω0)α−2˜dq,ωs, | (2.6) |
B1:=(T−ω0)α−1−λ2q(θ22)˜Γq(θ2)∫η2ω0~(η2−s)θ2−1_q,ωg2(σθ2−1q,ω(s))(σθ2−1q,ω(s)−ω0)α−1˜dq,ωs, | (2.7) |
B2:=(T−ω0)α−2−λ2q(θ22)˜Γq(θ2)∫η2ω0~(η2−s)θ2−1_q,ωg2(σθ2−1q,ω(s))(σθ2−1q,ω(s)−ω0)α−2˜dq,ωs, | (2.8) |
Λ:=A2B1−A1B2. | (2.9) |
Proof. Taking fractional symmetric Hahn integral of order α for (2.1), we obtain
u(t)=C1(t−ω0)α−1+C2(t−ω0)α−2+q(α2)˜Γq(α)∫tω0~(t−s)α−1_q,ωh(σα−1q,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~(t−s)θi−1_q,ω[C1(σθi−1q,ω(s)−ω0)α−1+C2(σθi−1q,ω(s)−ω0)α−2]˜dq,ωs+q(θi2)+(α2)˜Γq(θi)˜Γq(α)∫tω0∫σθi−1q,ω(s)ω0~(t−s)θi−1_q,ω~(σθi−1q,ω(s)−r)α−1_q,ωh(σα−1q,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
‖u‖C=maxt∈ITq,ω{|u(t)|,|˜Dβq,ωu(t)|}, |
where α∈(1,2];β,γ,θ1,θ2∈(0,1]; ω>0;q∈(0,1); λ1,λ2∈R+. We define an operator F:C→C as
(Fu)(t):=q(α2)˜Γq(α)∫tω0~(t−s)α−1_q,ω×F(σα−1q,ω(s),u(σα−1q,ω(s)),˜Dβq,ωu(σα−1q,ω(s)),˜Ψγq,ωu(σα−1q,ω(s)))˜dq,ωs+(t−ω0)α−1Λ{B2Φ1[f(u)]+A2Φ2[f(u)]}−(t−ω0)α−2A2{(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)∫η1ω0∫σθ1−1q,ω(s)ω0~(η1−s)θ1−1_q,ω~(σθ1−1q,ω(s)−r)α−1_q,ωg1(σθ1−1q,ω(s))×F(σα−1q,ω(r),u(σα−1q,ω(r)),˜Dβq,ωu(σα−1q,ω(r)),˜Ψγq,ωu(σα−1q,ω(r)))˜dq,ωr˜dq,ωs, | (3.2) |
Φ2[F(u)]:=λ2q(θ22)+(α2)˜Γq(θ2)˜Γq(α)∫η2ω0∫σθ2−1q,ω(s)ω0~(η2−s)θ2−1_q,ω~(σθ2−1q,ω(s)−r)α−1_q,ωg2(σθ2−1q,ω(s))×F(σα−1q,ω(r),u(σα−1q,ω(r)),˜Dβq,ωu(σα−1q,ω(r)),˜Ψγq,ωu(σα−1q,ω(r)))˜dq,ωr˜dq,ωs−q(α2)˜Γq(α)∫Tω0~(T−s)α−1_q,ω×F(σα−1q,ω(r),u(σα−1q,ω(r)),˜Dβq,ωu(σα−1q,ω(r)),˜Ψγq,ωu(σα−1q,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×R3→R, and g1,g2:ITq,ω→R+ 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 ℓ1,ℓ2,ℓ3>0 such that for each t∈ITq,ω and u,v∈R,
|F(t,u,˜Dβq,ωu,˜Ψγq,ωu)−F(t,v,˜Dβq,ωv,˜Ψγq,ωv)|≤ℓ1|u−v|+ℓ2|˜Dβq,ωu−˜Dβq,ωv|+ℓ3|˜Ψγq,ωu−˜Ψγq,ωv|. |
(H2) There exist constants gi,Gi>0 where i=1,2 such that for each t∈ITq,ω,
0<gi<gi(t)<Gi. |
(H3) Θ<1,
where
Ω1:=λ1G1q(θ12)+(α2)+θ1+α˜Γq(θ1+α+1)(η1−ω0)θ1+α | (3.4) |
Ω2:=λ2G2q(θ22)+(α2)+θ2+α˜Γq(θ2+α+1)(η2−ω0)θ2+α+q(α2)˜Γq(α+1)(T−ω0)α | (3.5) |
¯A1:=λ1G1q(θ12)+θ1(α−1)˜Γq(α)˜Γq(θ1+α)(η1−ω0)θ1+α−1>A1 | (3.6) |
¯A2:=λ1G1q(θ12)+θ1(α−2)˜Γq(α−1)˜Γq(θ1+α−1)(η1−ω0)θ1+α−2>A2 | (3.7) |
A∗2:=λ1g1q(θ12)+θ1(α−2)˜Γq(α−1)˜Γq(θ1+α−1)(η1−ω0)θ1+α−2<A2 | (3.8) |
¯B1:=(T−ω0)α−1+λ2G2q(θ22)+θ2(α−1)˜Γq(α)˜Γq(θ2+α)(η2−ω0)θ2+α−1>B1 | (3.9) |
¯B2:=(T−ω0)α−2+λ2G2q(θ22)+θ2(α−2)˜Γq(α−1)˜Γq(θ2+α−1)(η2−ω0)θ2+α−2>B2 | (3.10) |
Λ∗:=g1g2|¯A2¯B1−¯A1¯B2|<|Λ| | (3.11) |
and
Θ:=(ℓ1+ℓ2+ℓ3φ0(T+ω0)γ˜Γq(γ+1)){(T−ω0)α−1Λ∗[¯B2Ω1+¯A2Ω2]+(t−ω0)α−2A∗2[(1+¯A1¯B2Λ∗)Ω1+¯A1¯A2Λ∗Ω2]+q(α2)(T−ω0)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|u−v|(t):=|F(t,u(t),˜Dβq,ωu(t),˜Ψγq,ωu(t))−F(t,v(t),˜Dβq,ωv(t),˜Ψγq,ωv(t))|, |
for each t∈ITq,ω and u,v∈C. We find that
|Φ1[F(u)]−Φ1[F(v)]|≤λ1G1q(θ12)+(α2)˜Γq(θ1)˜Γq(α)∫η1ω0∫σθ1−1q,ω(s)ω0~(η1−s)θ1−1_q,ω~(σθ1−1q,ω(s)−r)α−1_q,ωH|u−v|(r)˜dq,ωr˜dq,ωs≤(ℓ1|u−v|+ℓ2|˜Dβq,ωu−˜Dβq,ωv|+ℓ3|˜Ψγq,ωu−˜Ψγq,ωv|)×λ1G1q(θ12)+(α2)˜Γq(θ1)˜Γq(α)∫η1ω0∫σθ1−1q,ω(s)ω0~(η1−s)θ1−1_q,ω~(σθ1−1q,ω(s)−r)α−1_q,ω˜dq,ωr˜dq,ωs≤[(ℓ1+ℓ3φ0(T+ω0)γ˜Γq(γ+1))|u−v|+ℓ2|˜Dβq,ωu−˜Dβq,ωv|]Ω1≤‖u−v‖C(ℓ1+ℓ2+ℓ3φ0(T+ω0)γ˜Γq(γ+1))Ω1. |
Similary,
|Φ2[F(u)]−Φ2[F(v)]|≤‖u−v‖C(ℓ1+ℓ2+ℓ3φ0(T+ω0)γ˜Γq(γ+1))Ω2. |
In addition, we find that
|(Fu)(t)−(Fv)(t)|≤q(α2)˜Γq(α)∫tω0~(t−s)α−1_q,ωH|u−v|(s)˜dq,ωs+‖u−v‖C(ℓ1+ℓ2+ℓ3φ0(T+ω0)γ˜Γq(γ+1))(T−ω0)α−1|Λ|{B2Ω1+A2Ω2}+‖u−v‖C(ℓ1+ℓ2+ℓ3φ0(T+ω0)γ˜Γq(γ+1))(T−ω0)α−2A2{(1+A1B2|Λ|)Ω1+A1A2|Λ|Ω2}≤‖u−v‖C(ℓ1+ℓ2+ℓ3φ0(T+ω0)γ˜Γq(γ+1)){q(α2)(T−ω0)α˜Γq(α+1)+(T−ω0)α−1Λ∗[¯B2Ω1+¯A2Ω2]+(t−ω0)α−2A∗2[(1+¯A1¯B2Λ∗)Ω1+¯A1¯A2Λ∗Ω2]}=‖u−v‖CΘ. | (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)∫tω0∫σ−β−1q,ω(s)ω0~(t−s)−β−1_q,ω~(σ−β−1q,ω(s)−r)α−1_q,ω×F(σα−1q,ω(r),u(σα−1q,ω(r)),˜Dβq,ωu(σα−1q,ω(r)),˜Ψγq,ωu(σα−1q,ω(r)))˜dq,ωr˜dq,ωs+[B2Φ1[f(u)]+A2Φ2[f(u)]Λ]q(−β2)˜Γq(−β)∫tω0~(t−s)−β−1_q,ω(σ−β−1q,ω(s)−ω0)α−1˜dq,ωs−1A2[(1+A1B2Λ)Φ1[f(u)]+A1A2ΛΦ2[f(u)]]×q(−β2)˜Γq(−β)∫tω0~(t−s)−β−1_q,ω(σ−β−1q,ω(s)−ω0)α−2˜dq,ωs. | (3.14) |
Similary, we have
\begin{align} \big|(\tilde{D}_{q, \omega}^\beta{\mathcal{F}}u)(t) -(\tilde{D}_{q, \omega}^\beta{\mathcal{F}}v)(t) \big| \lt \|u-v\|_{\mathcal{C}}\Theta. \end{align} | (3.15) |
From (3.13) and (3.15), we get
\|{\mathcal{F}}u-{\mathcal{F}}v\|_{\mathcal{C}}\leq\|u-v\|_{\mathcal{C}}\Theta. |
Using (H_3) we can conclude that {\mathcal{F}} is a contraction. Based on Banach fixed point theorem, {\mathcal{F}} has a fixed point which is a unique solution of problem (1.1) on I^T_{q, \omega} .
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 (H_1) and (H_3) defined in Theorem 3.1 hold. Then, problem (1.1) has at least one solution on I^T_{q, \omega} .
Proof. The proof is established as the following structures.
Step Ⅰ. Verify {\mathcal{F}} map bounded sets into bounded sets in B_R . Let B_R = \{u\in C(I^T_{q, \omega}) :\|u\|_{\mathcal{C}}\leq R\} , \; \max_{t\in I^T_{q, \omega}}|F(t, 0, 0, 0)| = M and choose a constant
\begin{equation} R\geq \frac{ M \Bigg\{ \frac{(T-\omega_0)^{\alpha-1}}{\Lambda^*}\big[\overline{\textbf{B}}_2\Omega_1+\overline{{\textbf{A}}}_{2}\Omega_2 \big] +\frac{(t-\omega_0)^{\alpha-2}}{\textbf{A}^*_2} \Big[\left(1+ \frac{\overline{\textbf{A}}_1\overline{\textbf{B}}_2 }{\Lambda^*} \right) \Omega_1 +\frac{\overline{{\textbf{A}}}_{1} \overline{{\textbf{A}}}_{2}}{\Lambda^*} \Omega_2\Big] +\frac{ q^{ \alpha \choose 2} (T-\omega_0)^{\alpha} }{ \tilde{\Gamma}_q(\alpha+1) } \Bigg\} }{1-\Theta}. \end{equation} | (3.16) |
Denote that
\big|\mathcal{K}(t, u, 0)\big| = \bigg|F\big(t, u(t), \tilde{D}^\beta_{q, \omega}u(t), \tilde{\Psi}^\gamma_{q, \omega}u(t)\big)-F\big(t, 0, 0, 0\big)\bigg|+\big|F\big(t, 0, 0, 0\big)\big|. |
We find that
\begin{align} &\Big|{\Phi_{1}[F(u)]} \Big|\\ \leq\; &\frac{\lambda_1G_1q^{ {\theta_1 \choose 2 }+{\alpha \choose 2 } } }{\tilde{\Gamma}_q(\theta_1) \tilde{\Gamma}_q(\alpha)} \int_{\omega_0}^{\eta_1} \int_{\omega_0}^{ \sigma_{q, \omega}^{\theta_1-1}(s) } \widetilde{\left(\eta_1-s \right)}_{q, \omega}^{\underline{\theta_1-1}} \widetilde{\left(\sigma_{q, \omega}^{\theta_1-1}(s)-r\right)}_{q, \omega}^{\underline{\alpha-1}} \big|\mathcal{K}(t, u, 0)\big|\tilde{d}_{q, \omega}r\tilde{d}_{q, \omega}s\\ \leq\; &\left[ \Big( \ell_1|u|+\ell_2\Big|\tilde{D}^\beta_{q, \omega}u\Big| +\ell_3\big|\tilde{\Psi}^\gamma_{q, \omega}u\big|\Big)+M\right]\Omega_1 \\ \leq\; & \left[ \left( \ell_1+\ell_2+\ell_3 \varphi_0 \frac{(T+\omega_0)^\gamma}{\tilde{\Gamma}_q(\gamma+1)} \right) \|u-v\|_{\mathcal{C}}+M\right] \Omega_1 \\ \leq\; & \left[ \left( \ell_1+\ell_2+\ell_3 \varphi_0 \frac{(T+\omega_0)^\gamma}{\tilde{\Gamma}_q(\gamma+1)} \right)R+M\right] \Omega_1, \end{align} | (3.17) |
where t\in I_{q, \omega}^T and u\in B_{R} .
Similary,
\begin{align} \Big|{\Phi_{2}[F(u)]} \Big| \leq \left[ \left( \ell_1+\ell_2+\ell_3 \varphi_0 \frac{(T+\omega_0)^\gamma}{\tilde{\Gamma}_q(\gamma+1)} \right) R+M\right] \Omega_2. \end{align} | (3.18) |
Employing (3.17) and (3.18), we find that
\begin{align} \big| ({\mathcal{F}}u)(t)\big| \leq\; &\left[ \left( \ell_1+\ell_2+\ell_3 \varphi_0 \frac{(T+\omega_0)^\gamma}{\tilde{\Gamma}_q(\gamma+1)} \right) R+M\right] \Bigg\{ \frac{(T-\omega_0)^{\alpha-1}}{\Lambda^*}\big[\overline{\textbf{B}}_2\Omega_1+\overline{{\textbf{A}}}_{2}\Omega_2 \big] \\ & +\frac{(t-\omega_0)^{\alpha-2}}{\textbf{A}^*_2} \Big[\left(1+ \frac{\overline{\textbf{A}}_1\overline{\textbf{B}}_2 }{\Lambda^*} \right) \Omega_1 +\frac{\overline{{\textbf{A}}}_{1} \overline{{\textbf{A}}}_{2}}{\Lambda^*} \Omega_2\Big] +\frac{ q^{ \alpha \choose 2} (T-\omega_0)^{\alpha} }{ \tilde{\Gamma}_q(\alpha+1) } \Bigg\} \\ \leq\; &R. \end{align} | (3.19) |
Since
\begin{align} \bigg|\left(\tilde{D}_{q, \omega}^\beta{\mathcal{F}}u\right) (t)\bigg| \lt R. \end{align} | (3.20) |
Therefore, \|{\mathcal{F}}u\|_{\mathcal{C}}\leq R . Hence, {\mathcal{F}} is uniformly bounded.
Step Ⅱ. That the operator {\mathcal{F}} is continuous on B_R since the continuity of F .
Step Ⅲ. Examine that {\mathcal{F}} is equicontinuous on B_R .
For any t_1, t_2\in I^T_{q, \omega} with t_1 < t_2 , by Lemma 2.9 we have
\begin{align} \big| ({\mathcal{F}}u)(t_2)&- ({\mathcal{F}}u)(t_1) \big|\, \\ \leq\; &\frac{ q^{ \alpha \choose 2} \|F\| }{ \tilde{\Gamma}_q(\alpha+1) } \Big| (t_2-\omega_0)^{\alpha} - (t_1-\omega_0)^{\alpha} \Big|\\ +& \frac{ \Big(\overline{\textbf{B}}_2\Omega_1+\overline{{\textbf{A}}}_{2}\Omega_2 \Big) \|F\|}{\Lambda^*} \Big| (t_2-\omega_0)^{\alpha-1} - (t_1-\omega_0)^{\alpha-1} \Big|\\ +& \frac{ \|F\| }{\textbf{A}^*_2} \Bigg[\left(1+ \frac{\overline{\textbf{A}}_1\overline{\textbf{B}}_2 }{\Lambda^*} \right) \Omega_1 +\frac{\overline{{\textbf{A}}}_{1} \overline{{\textbf{A}}}_{2}}{\Lambda^*} \Omega_2\Bigg] \Big| (t_2-\omega_0)^{\alpha-2} - (t_1-\omega_0)^{\alpha-2} \Big| \end{align} | (3.21) |
and
\begin{align} &\bigg|\left(\tilde{D}_{q, \omega}^\beta{\mathcal{F}}u\right) (t_1) -\left(\tilde{D}_{q, \omega}^\beta{\mathcal{F}}u\right)(t_2) \bigg|\\ \leq\, &\frac{ q^{ {\alpha \choose 2} +{-\beta \choose 2} - \alpha \beta } \|F\| }{ \tilde{\Gamma}_q(\alpha-\beta+1) } \Big| (t_2-\omega_0)^{\alpha-\beta} - (t_1-\omega_0)^{\alpha-\beta} \Big|\\ +& \frac{ q^{ {-\beta \choose 2} - \alpha \beta } \, \tilde{\Gamma}_q(\alpha) \|F\|}{\Lambda^*\tilde{\Gamma}_q(\alpha-\beta) }\Big(\overline{\textbf{B}}_2\Omega_1+\overline{{\textbf{A}}}_{2}\Omega_2 \Big)\Big| (t_2-\omega_0)^{\alpha-\beta-1} - (t_1-\omega_0)^{\alpha-\beta-1} \Big|\\ +& \frac{q^{ {-\beta \choose 2} - \alpha \beta }\, \tilde{\Gamma}_q(\alpha-1) \|F\| }{\textbf{A}^*_2\tilde{\Gamma}_q(\alpha-\beta-1)} \Bigg[\left(1+ \frac{\overline{\textbf{A}}_1\overline{\textbf{B}}_2 }{\Lambda^* } \right) \Omega_1 +\frac{\overline{{\textbf{A}}}_{1} \overline{{\textbf{A}}}_{2}}{\Lambda^*} \Omega_2\Bigg] \times\\ &\Big| (t_2-\omega_0)^{\alpha-\beta-2} - (t_1-\omega_0)^{\alpha-\beta-2} \Big|. \end{align} | (3.22) |
Since the right-hand side of (3.22) tends to be zero when |t_2-t_1|\rightarrow 0 , {\mathcal{F}} is relatively compact on B_R . Therefore, the set {\mathcal{F}}(B_R) is an equicontinuous set. From Steps I to III together with the Arzelá-Ascoli theorem, \mathcal{F}:{\mathcal{C}}\rightarrow {\mathcal{C}} 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
\begin{eqnarray} \tilde{D}^{\frac{5}{3}}_{\frac{1}{2}, \frac{2}{3}}u(t)& = &\frac{1}{\left( 100e^2+t^3\right)(1+|u(t)|)} \Bigg[ e^{-3t}\left( u^2+2|u|\right) + e^{-(\pi+\cos^2\pi t)}\left| \tilde{D}^{\frac{2}{5}}_{\frac{1}{2}, \frac{2}{3}}u(t) \right| \\ &&+ e^{-(1+\sin^2\pi t)}\left| \tilde{\Psi}_{\frac{1}{2}, \frac{2}{3}}^{\frac{3}{4}}u(t) \right| \Bigg]\\ u\left(\frac{4}{3}\right)& = &2\tilde{\mathcal{I}}_{\frac{1}{2}, \frac{2}{3}}^{\frac{3}{4}} e^{\cos\left( \frac{15}{6}\pi\right) }u\left(\frac{15}{6}\right)\\ u\left(10\right)& = &3\tilde{\mathcal{I}}_{\frac{1}{2}, \frac{2}{3}}^{\frac{1}{3}} e^{2\sin\left( \frac{47}{32}\pi\right) }u\left(\frac{47}{32}\right), \end{eqnarray} | (4.1) |
where \; t\in I_{\frac{1}{2}, \frac{2}{3}}^{10} and \varphi(t, s) = \frac{e^{-s}}{(t+10)^3} .
We let \; \alpha = \frac{5}{3}, \; \beta = \frac{2}{5}, \; \gamma = \frac{3}{4}, \; \theta_1 = \frac{3}{4}, \; \theta_2 = \frac{1}{3}, \; q = \frac{1}{2}, \; \omega = \frac{2}{3}, \; \omega_0 = \frac{\omega}{1-q} = \frac{4}{3}, \; T = 10, \; \eta_1 = 10\left(\frac{1}{2}\right)^4+\frac{2}{3}[4]_{\frac{1}{2}} = \frac{15}{8}, \; \eta_2 = 10\left(\frac{1}{2}\right)^6+\frac{2}{3}[6]_{\frac{1}{2}} = \frac{47}{32}, \; \lambda_1 = 2, \; \lambda_2 = 3, \; g_1(t) = e^{\cos\left(\pi t\right) }, \; g_2(t) = e^{2\sin\left(\pi t\right) } and \; \varphi_0 = \max\left\lbrace \varphi(t, s) \right\rbrace = \frac{27}{1156 e^{\frac{4}{3}}}.
For all \; t\in I_{\frac{1}{2}, \frac{2}{3}}^{10}\; and \; u, v\in {\mathbb{R}} , we have
\begin{align*} &\left| F\left( t, u, \tilde{D}^\beta_{q, \omega}u, \tilde{\Psi}^\gamma_{q, \omega}u\right)- F\left( t, v, \tilde{D}^\beta_{q, \omega}v, \tilde{\Psi}^\gamma_{q, \omega}v\right)\right| \\ \leq\; & \frac{1}{e^4\left( 100e^2+\frac{64}{27}\right) }|u-v| +\frac{1}{e^{\pi}\left( 100e^2+100e^2+\frac{64}{27}\right) }\left| \tilde{D}^\beta_{q, \omega}u-\tilde{D}^\beta_{q, \omega}v\right|\\ +&\frac{1}{e\left( 100e^2+100e^2+\frac{64}{27}\right) } \left|\tilde{\Psi}^\gamma_{q, \omega}u-\tilde{\Psi}^\gamma_{q, \omega}v \right|, \end{align*} |
and \; \; \frac{1}{e} < g_1(t) < e, \; \; \; \frac{1}{e^2} < g_2(t) < e^2
Thus, (H_1) and (H_2) hold with \; \ell_1 = 0.0000247, \; \ell_2 = 0.0000583, \; \ell_3 = 0.000496\; and \; g_1 = \frac{1}{e}, \; g_2 = \frac{1}{e^2}, \; G_1 = e, \; G_2 = e^2 .
Since
\Omega_1 = 0.161, \; \; \; \Omega_2 = 21.708, \; \; \; \overline{\textbf{A}}_1 = 1.518, \; \; \; \overline{\textbf{A}}_2 = 6.717, \; \; \; \textbf{A}_2^* = 0.909, |
\overline{\textbf{B}}_1 = 3.241, \; \; \; \overline{\textbf{B}}_2 = 30.841\; \; and \; \; \Lambda^* = 1.247, |
therefore, (H_3) holds with
\Theta = 0.063 \lt 1. |
Hence, by Theorem 3.1 problem (4.1) 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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