Citation: Sina Etemad, Sotiris K. Ntouyas. Application of the fixed point theorems on the existence of solutions for q-fractional boundary value problems[J]. AIMS Mathematics, 2019, 4(3): 997-1018. doi: 10.3934/math.2019.3.997
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The classical fractional calculus is one of the branches of applied mathematics which has merged in pure mathematics. In fact, the topic of fractional differential equations and inclusions, as one of the subjects of the fractional calculus, is an important and effective field of research which the basic techniques of the functional analysis and fixed point theory were used to proving the existence and uniqueness of solutions for this kind of differential equations and inclusions. On the other hand, the extensiveness and importance of this topic has been caused to publish the many works and papers by other researchers (for example, see [2,3,5,7,9,12,13,14,15,19,32,33,34].
Later, q−difference calculus or quantum calculus, as a generalization of the classical calculus, has gained considerable attention of researchers and mathematicians. The first work on the subject of q−difference calculus dates back to Jackson's works [25].
In recent years, many researchers have studied and published various and distinct papers on the existence theory of fractional q−difference equations and inclusions (for examples, see [1,4,6,17,18,20,21,22,23,26,29,30,35,37,39,41]).
In [8], Bashir Ahmad et al. studied the existence of solutions for the nonlocal boundary value problem of fractional q−difference equation
cDαqx(t)=f(t,x(t)),0≤t≤1,1<α≤2,α1x(0)−β1Dqx(0)=γ1x(η1),α2x(1)+β2Dqx(1)=γ2x(η2), |
where, cDαq is the fractional q−derivative of the Caputo type and αi,βi,γi∈R. Authors in that paper, by applying the Banach contraction principle, Krasnoselskii's fixed point theorem and the Leray-Schauder nonlinear alternative studied the existence results.
In [40], Zhao et al. dealt with the following nonlinear fractional q−difference equation with the nonlocal q−integral boundary value conditions:
Dαqx(t)+f(t,x(t))=0,0<t<1,1<α≤2,x(0)=0,x(1)=μIβqx(η),0<β≤2, |
where, Dαq is the fractional q−derivative of Riemann-Liouville type and μ>0. They studied the existence of solutions for the above problem by using the generalized Banach contraction principle, the monotone iterative method and Krasnoselskii's fixed point theorem.
In 2015, Alsaedi, Ntouyas and Ahmad investigated the fractional q−difference inclusion with nonlocal and substrip type boundary conditions
cDνqx(t)∈F(t,x(t)),0≤t≤1,1<ν≤2,x(0)=g(x),x(w)=b∫1δx(s)dqs0<w<δ<1, |
where cDνq denotes the Caputo fractional q−derivative of order ν [10].
Motivated by the above papers, in this paper, we discuss the existence of solutions for fractional q−difference equation
{cDαqu(t)=f(t,u(t),Dqu(t)),0≤t≤1,0<q<1,u(0)=0,Dqu(1)=0,D2qu(1)=0, | (1.1) |
where, cDαq denotes the fractional q−derivative of the Caputo type of order α and α∈(2,3] and f:[0,1]×R2→R is a continuous mapping.
Also, we study the existence of solutions for the following fractional q−difference inclusion
{cDαqu(t)∈F(t,u(t),Dqu(t)),0≤t≤1,0<q<1,u(0)=0,Dqu(1)=0,D2qu(1)=0, | (1.2) |
where F:[0,1]×R2→P(R) is a compact multivalued map.
The rest of the paper is organized as follows: In section 2, we state some important definitions and lemmas on the fundamental concepts of q−fractional calculus and fixed point theory. In section 3, we state main results on the existence of solutions for q−fractional boundary value problem (1.1). Section 4 contains some main theorems on the existence of solutions for q−fractional boundary value problem (1.2). Finally, in section 5, we give some illustrative examples to show the validity and applicability of our results.
In this section, we first recall some known definitions and lemmas about q−fractional calculus. For more details in this regard, see [22,25,26,27].
Let 0<q<1. For each a∈R we define [a]q=1−qa1−q. The q−analogue of the power function (a−b)n with n∈N0:={0,1,2,…} is given by
(a−b)(0)=1,(a−b)(n)=n−1∏k=0(a−bqk),n∈N,a,b∈R. |
In general, if α is real number then
(a−b)(α)=aα∞∏k=0a−bqka−bqα+k,a≠0. |
It is clear that if b=0, then a(α)=aα. The q−Gamma function is defined by
Γq(x)=(1−q)(x−1)(1−q)x−1, |
where x∈R∖{0,−1,−2,…}. Also, we have Γq(x+1)=[x]qΓq(x). The q−derivative of a real-valued function f is defined by
(Dqf)(x)=f(x)−f(qx)(1−q)x,(Dqf)(0)=limx→0(Dqf)(x). |
The q−derivative of higher order of a function f is given by
(D0qf)(x)=f(x),(Dnqf)(x)=Dq(Dn−1qf)(x),(n∈N). |
The q−integral of a function f defined in the interval [0,b] is given by
(Iqf)(x)=∫x0f(s)dqs=x(1−q)∞∑k=0f(xqk)qk,x∈[0,b] |
such that the sum is absolutely convergent. Now, if a∈[0,b], then q−integral of f from a to b is
∫baf(s)dqs=Iqf(b)−Iqf(a)=∫b0f(s)dqs−∫a0f(s)dqs=(1−q)∞∑k=0[bf(bqk)−af(aqk)]qk, |
provided that the series converges. Similar to q−derivatives, an operator Inq is given by
(I0qf)(x)=f(x),(Inqf)(x)=Iq(In−1qf)(x),(n∈N). |
Note that (DqIqf)(x)=f(x) and if f is continuous at x=0, then (IqDqf)(x)=f(x)−f(0).
Let α≥0 and f be a function defined on [0,1]. The fractional q−integral of the Riemann-Liouville type is (I0qf)(x)=f(x) and for x∈[0,1]
(Iαqf)(x)=1Γq(α)∫x0(x−qs)(α−1)f(s)dqs,α>0. |
The fractional q−derivative of the Caputo type of order α≥0 is defined by
(cDαqf)(x)=(I[α]−αqD[α]qf)(x)=1Γq([α]−α)∫x0(x−qs)([α]−α−1)(D[α]qf)(s)dqs,α>0, |
where [α] is the smallest integer greater than or equal to α [16]. In the following lemmas, we bring some important properties of these q−operators.
Lemma 2.1. [17] Let α,β≥0 and f be a function defined on [0,1]. Then
(i) (IβqIαqf)(x)=(Iα+βqf)(x),
(ii) (DαqIαqf)(x)=f(x).
Lemma 2.2. [17] Let α>0 and n be a positive integer. Then the following equality holds:
(IαqDnqf)(x)=(DnqIαqf)(x)−n−1∑k=0xα−n+kΓq(α+k−n+1)(Dkqf)(0). |
Now, we recall some definitions and lemmas on the multifunctions and fixed point theory which are needed in the sequal.
Consider the set X with the metric d. Denote by P(X), 2X, Pcl(X), Pcp(X), Pb(X) and Pcv(X), the class of all subsets, the class of all nonempty subsets of X, the class of all closed subsets of X, the class of all compact subsets of X, the class of all bounded subsets of X and the class of all convex subsets of X, respectively. Let F:X→2X be a multivalued map. If u∈Fu then we say that u∈X is a fixed point of the multifunction F [16,24,38]. The fixed point set of the multivalued operator F will be denoted by Fix(F).
A multifunction F:[0,1]→Pcl(R) is said to be measurable whenever the function (t)↦d(y,F(t))=inf{‖y−v‖:v∈F(t)} is measurable for all y∈R [16,24]. The Pompeiu-Hausdorff metric Hd:2X×2X→[0,∞) is defined by
Hd(A,B)=max{supa∈Ad(a,B),supb∈Bd(A,b)}, |
where d(a,B)=infb∈Bd(a,b), d(A,b)=infa∈Ad(a,b) [16,24]. Then the space (Pcl,b(X),Hd) is a metric space and (Pcl,b(X),Hd) is a generalized metric space, where Pcl,b(X) is the set of closed and bounded subsets of X. [16,24].
A multi-valued mapping F:X→Pcl(X) is called a contraction if there exists γ∈(0,1) such that Hd(F(x),F(y))≤γd(x,y) for all x,y∈X [16].
F is called upper semi-continuous (u.s.c.) on X if for each x0∈X, the set F(x0) is a nonempty closed subset of X, and if for each open set N of X containing F(x0), there exists an open neighborhood N0 of x0 such that F(N0)⊆N [16]. The operator F is said to be completely continuous if F(B) is relatively compact for every B∈Pb(X).
An element x∈X is called an endpoint of a multifunction F:X→P(X) whenever Fx={x} [11]. Also, we say that F has an approximate endpoint property whenever infx∈Xsupy∈Fxd(x,y)=0 [11]. A real-valued function f:R→R is called upper semi-continuous whenever limsupn→∞f(λn)≤f(λ) for all sequence {λn}n≥1 with λn→λ.
We denote by Ψ, the family of nondecreasing functions ψ:[0,∞)→[0,∞) such that ∑∞n=1ψn(t)<∞ for all t>0 [36]. It is known that ψ(t)<t for all t>0 [36]. In 2012, Samet, Vetro and Vetro introduced the notion of α-ψ-contractive type mappings [36]. We say that the selfmap T:X→X is an α-ψ-contraction whenever α(u,v)d(Tu,Tv)≤ψ(d(u,v)) for all u,v∈X [36]. Also, the selfmap T is called α-admissible whenever α(u,v)≥1 implies α(Tu,Tv)≥1 [36]. We say that X has the property (B) whenever for each sequence {un} in X with α(un,un+1)≥1 for all n≥1 and un→u, we have α(un,u)≥1 for all n [36].
In 2013, Mohammadi, Rezapour and Shahzad generalized this notion to multifunctions [31]. A multifunction F:X→Pcl,b(X) is called α−ψ−contraction whenever
α(u,v)Hd(Fu,Fv)≤ψ(d(u,v)) |
for each u,v∈X [31]. Similarly, the space X has the property (Cα) whenever for each sequence {un} in X with α(un,un+1)≥1 for all n∈N, there exists a subsequence {unk} of {un} such that α(unk,u)≥1 for all k∈N. The multi-valued map F is α−admissible whenever for each u∈X and v∈Fu with α(u,v)≥1, we have α(v,w)≥1 for all w∈Fv [31].
Our main results based on the following fixed point theorems.
Theorem 2.1. ([36]) Let (X,d) be a complete metric space, ψ∈Ψ, α:X×X→R a map and T an α−admissible and α−ψ−contractive selfmap on X such that α(x0,Tx0)≥1, for some x0∈X. If X has the property (B), then T has a fixed point.
Theorem 2.2. ([28,38], Krasnoselskii) Let M be a closed, bounded, convex and nonempty subset of a Banach space X. Let A and B be two operators mapping M to X such that:
(i) Ax+By∈M whenever x,y∈M;
(ii) A is compact and continuous;
(iii) B is a contraction mapping.
Then there exists z∈M such that z=Az+Bz.
Theorem 2.3. ([31]) Let (X,d) be a complete metric space, α:X×X→[0,∞) a map, ψ∈Ψ a strictly increasing map, F:X→Pcl,b(X) an α-admissible α-ψ-contractive multifunction and α(u0,u1)≥1 for some u0∈X and u1∈Fu0. If the space X has the property (Cα), then F has a fixed point.
Theorem 2.4. ([11]) Let (X,d) be a complete metric space and ψ:[0,∞)→[0,∞) be an upper semi-continuous function such that ψ(t)<t and lim inft→∞(t−ψ(t))>0 for all t>0. Suppose that T:X→Pcl,b(X) is a multifunction such that Hd(Tx,Ty)≤ψ(d(x,y)) for all x,y∈X. Then T has a unique endpoint if and only if T has approximate endpoint property.
Let X={u:u,Dqu∈C([0,1],R)}. Then X is a Banach space via the norm
‖u‖=supt∈[0,1]|u(t)|+supt∈[0,1]|Dqu(t)|. |
Put
Λ1=1Γq(α+1)+1Γq(α)+2+q(1+q)Γq(α−1),Λ2=2Γq(α)+2Γq(α−1),Δ1=1Γq(α)+2+q(1+q)Γq(α−1),Δ2=1Γq(α)+2Γq(α−1),Φ1=‖m‖Λ1,Φ2=‖m‖Λ2. | (3.1) |
Lemma 3.1. Let y∈C([0,1],R). Then the integral solution of the q−fractional boundary value problem
{cDαqu(t)=y(t),u(0)=0,Dqu(1)=0,D2qu(1)=0 | (3.2) |
is given by
u(t)=∫t0(t−qs)(α−1)Γq(α)y(s)dqs−t∫10(1−qs)(α−2)Γq(α−1)y(s)dqs−t2−t(1+q)1+q∫10(1−qs)(α−3)Γq(α−2)y(s)dqs. | (3.3) |
Proof. Choose the constants c0, c1 and c2∈R such that
u(t)=∫t0(t−qs)(α−1)Γq(α)y(s)dqs+c0+c1t+c2t2. | (3.4) |
Thus, we have
Dqu(t)=∫t0(t−qs)(α−2)Γq(α−1)y(s)dqs+c1+c2(1+q)t, |
D2qu(t)=∫t0(t−qs)(α−3)Γq(α−2)y(s)dqs+c2(1+q). |
By using the boundary value conditions, we find that c0=0 and
c1=−∫10(1−qs)(α−2)Γq(α−1)y(s)dqs+∫10(1−qs)(α−3)Γq(α−2)y(s)dqs, |
and
c2=−11+q∫10(1−qs)(α−3)Γq(α−2)y(s)dqs. |
By substituting the values of ci's in Eq (3.3), we obtain the q−integral equation (3.2). The converse follows by direct computation. The proof is completed.
In view of the above lemma, we define an operator S:X→X as follows:
(Su)(t)=∫t0(t−qs)(α−1)Γq(α)f(s,u(s),Dqu(s))dqs−t∫10(1−qs)(α−2)Γq(α−1)f(s,u(s),Dqu(s))dqs−t2−t(1+q)1+q∫10(1−qs)(α−3)Γq(α−2)f(s,u(s),Dqu(s))dqs. | (3.5) |
It is evident that the solution of the problem (1.1) is a fixed point of an operator S; that is Su=u.
Now, we are ready to prove our main results.
Theorem 3.1. Let ψ∈Ψ, ξ:R2×R2→R be a map and f:[0,1]×R2→R a continuous function. Suppose that:
(H1) For all u1,u2,v1,v2∈R,
|f(t,u1,v1)−f(t,u2,v2)|≤ψ(|u1−u2|+|v1−v2|)(1Λ1+Λ2), |
with ξ((u1(t),v1(t)),(u2(t),v2(t)))≥0 for t∈[0,1].
(H2) There exists u0∈R such that ξ((u0(t),Dqu0(t)),(Fu0(t),Dq(Fu0(t))))≥0 for all t∈[0,1] and ξ((u(t),Dqu(t)),(v(t),Dqv(t)))≥0 implies
ξ((Fu(t),Dq(Fu(t))),(Fv(t),Dq(Fv(t))))≥0 |
for all t∈[0,1] and u,v∈R.
(H3) For each convergent sequence {un}n≥1 in R with un→u and
ξ((un(t),Dqun(t)),(un+1(t),Dqun+1(t)))≥0 |
for all n and t∈[0,1], we have ξ((un(t),Dqun(t)),(u(t),Dqu(t)))≥0.
Then the fractional q−difference Eq (1.1) has at least one solution.
Proof. Let u,v∈R with ξ((u(t),Dqu(t)),((v(t),Dqv(t)))≥0 for all t∈[0,1]. Then, we get
|Su(t)−Sv(t)|≤∫t0(t−qs)(α−1)Γq(α)|f(s,u(s),Dqu(s))−f(s,v(s),Dqv(s))|dqs+t∫10(1−qs)(α−2)Γq(α−1)|f(s,u(s),Dqu(s))−f(s,v(s),Dqv(s))|dqs+|t2−t(1+q)|(1+q)∫10(1−qs)(α−3)Γq(α−2)|f(s,u(s),Dqu(s))−f(s,v(s),Dqv(s))|dqs≤1Γq(α+1)ψ(|u(s)−v(s)|+|Dqu(s)−Dqv(s)|)(1Λ1+Λ2)+1Γq(α)ψ(|u(s)−v(s)|+|Dqu(s)−Dqv(s)|)(1Λ1+Λ2)+2+q(1+q)Γq(α−1)ψ(|u(s)−v(s)|+|Dqu(s)−Dqv(s)|)(1Λ1+Λ2)≤1Γq(α+1)ψ(‖u−v‖)(1Λ1+Λ2)+1Γq(α)ψ(‖u−v‖)(1Λ1+Λ2)+2+q(1+q)Γq(α−1)ψ(‖u−v‖)(1Λ1+Λ2)=Λ1ψ(‖u−v‖)(1Λ1+Λ2), |
and
|DqSu(t)−DqSv(t)|≤∫t0(t−qs)(α−2)Γq(α−1)|f(s,u(s),Dqu(s))−f(s,v(s),Dqv(s))|dqs+∫10(1−qs)(α−2)Γq(α−1)|f(s,u(s),Dqu(s))−f(s,v(s),Dqv(s))|dqs+|t−1|∫10(1−qs)(α−3)Γq(α−2)|f(s,u(s),Dqu(s))−f(s,v(s),Dqv(s))|dqs≤1Γq(α)ψ(|u(s)−v(s)|+|Dqu(s)−Dqv(s)|)(1Λ1+Λ2)+1Γq(α)ψ(|u(s)−v(s)|+|Dqu(s)−Dqv(s)|)(1Λ1+Λ2)≤1Γq(α)ψ(‖u−v‖)(1Λ1+Λ2)+1Γq(α)ψ(‖u−v‖)(1Λ1+Λ2)+1Γq(α−1)ψ(‖u−v‖)(1Λ1+Λ2)=Λ2ψ(‖u−v‖)(1Λ1+Λ2). |
Hence
‖Su−Sv‖≤(Λ1+Λ2)ψ(‖u−v‖)(1Λ1+Λ2)=ψ(‖u−v‖). |
Now, define the function α:R×R→[0,∞) as follows
α(u,v)={1,ifξ((u(t),Dqu(t)),(v(t),Dqv(t)))≥00,otherwise. |
By definition of the above function, it is clear that
α(u,v)d(Su,Sv)≤ψ(d(u,v)) |
for each u,v∈R. This means that S is an α−ψ−contractive operator. Also, it is easy to see that S is an α−admissible and α(u0,Su0)≥1. Suppose that {un}n≥1 is a sequence in R with un→u and α(un,un+1)≥1 for all n. By definition of the function α, we have
ξ((un(t),Dqun(t)),(un+1(t),Dqun+1(t)))≥0. |
Thus by the hypothesis, ξ((un(t),Dqun(t)),(u(t),Dqu(t)))≥0. This shows that for all n, α(un,u)≥1. So R has the property (B). Finally, Theorem 2.1 implies that the operator S has fixed point u∗∈R which is the solution for the q−fractional problem (1.1). This completes the proof.
Theorem 3.2. Let f:[0,1]×R2→R be a continuous function. Suppose that:
(H4) There exists a continuous function L:[0,1]→R such that for each t∈[0,1] and for all ui,vi∈R, i=1,2, we have
|f(t,u1,u2)−f(t,v1,v2)|≤L(t)(|u1−v1|+|u2−v2|). |
(H5) There exist a continuous function μ:[0,1]→R+ and a non-decreasing continuous function ψ:[0,1]→R+ such that
|f(t,u1,u2)|≤μ(t)ψ(|u1|+|u2|),t∈[0,1],ui∈R,i=1,2. |
Then, the fractional q−difference equation (1.1) has at least one solution on [0,1] if
k:=‖L‖(Λ1+Λ2)<1, |
where ‖L‖=supt∈[0,1]|L(t)| and Λ1,Λ2 are given by Eq (3.1).
Proof. Define ‖μ‖=supt∈[0,1]|μ(t)| and choose a suitable constant r such that
r≥ψ(‖u‖)‖μ‖{Δ1+Δ2}, | (3.6) |
where Δi's are given by Eq (3.1). We consider the set Br={u∈X:‖u‖≤r}, where r is given in Eq (3.6). It is clear that Br is a closed, bounded, convex and nonempty subset of the Banach space X. Now, define two operators S1 and S2 on Br as follows:
(S1u)(t)=∫t0(t−qs)(α−1)Γq(α)f(s,u(s),Dqu(s))dqs, | (3.7) |
and
(S2u)(t)=−t∫10(1−qs)(α−2)Γq(α−1)f(s,u(s),Dqu(s))dqs−t2−t(1+q)1+q∫10(1−qs)(α−3)Γq(α−2)f(s,u(s),Dqu(s))dqs, | (3.8) |
for each t∈[0,1]. Put a=supu∈Xψ(‖u‖). For u,v∈Br, we have
|(S1u+S2v)(t)|≤∫t0(t−qs)(α−1)Γq(α)|f(s,u(s),Dqu(s))|dqs+t∫10(1−qs)(α−2)Γq(α−1)|f(s,v(s),Dqv(s))|dqs+|t2−t(1+q)|(1+q)∫10(1−qs)(α−3)Γq(α−2)|f(s,v(s),Dqv(s))|dqs≤1Γq(α)∫t0(t−qs)(α−1)μ(s)ψ(|u(s)|+|Dqu(s)|)dqs+tΓq(α−1)∫10(1−qs)(α−2)μ(s)ψ(|v(s)|+|Dqv(s)|)dqs+|t2−t(1+q)|(1+q)Γq(α−2)∫10(1−qs)(α−3)μ(s)ψ(|v(s)|+|Dqv(s)|)dqs≤a‖μ‖[1Γq(α+1)+1Γq(α)+2+q(1+q)Γq(α−1)]=a‖μ‖Λ1. |
Also,
|(DqS1u+DqS2v)(t)|≤∫t0(t−qs)(α−2)Γq(α−1)|f(s,u(s),Dqu(s))|dqs+∫10(1−qs)(α−2)Γq(α−1)|f(s,v(s),Dqv(s))|dqs+|t−1|∫10(1−qs)(α−3)Γq(α−2)|f(s,v(s),Dqv(s))|dqs≤1Γq(α−1)∫t0(t−qs)(α−2)μ(s)ψ(|u(s)|+|Dqu(s)|)dqs+1Γq(α−1)∫10(1−qs)(α−2)μ(s)ψ(|v(s)|+|Dqv(s)|)dqs+|t−1|Γq(α−2)∫10(1−qs)(α−3)Γq(α−2)μ(s)ψ(|v(s)|+|Dqv(s)|)dqs≤a‖μ‖[2Γq(α)+2Γq(α−1)]=a‖μ‖Λ2. |
Hence ‖S1u+S2v‖≤r and so, S1u+S2v∈Br.
Clearly, the continuity of S1 is follows from the function f. Also, for each u∈Br, we have
|(S1u)(t)|≤∫t0(t−qs)(α−1)Γq(α)|f(s,u(s),Dqu(s))|ds≤1Γq(α+1)‖μ‖ψ(‖u‖), |
and
|(DqS1u)(t)|≤∫t0(t−qs)(α−2)Γq(α−1)|f(s,u(s),Dqu(s))|ds≤1Γq(α)‖μ‖ψ(‖u‖). |
Thus ‖S1u‖≤{1Γq(α+1)+1Γq(α)}‖μ‖ψ(‖u‖). This proves that the operator S1 is uniformly bounded on Br.
Now, we show that the operator S1 is compact on Br. For each t1,t2∈[0,1] with t1<t2, one can write
|(S1u)(t2)−(S1u)(t1)|=|∫t20(t2−qs)(α−1)Γq(α)f(s,u(s),Dqu(s))ds−∫t10(t1−qs)(α−1)Γq(α)f(s,u(s),Dqu(s))ds|≤|∫t10(t2−qs)(α−1)−(t1−qs)(α−1)Γq(α)f(s,u(s),Dqu(s))ds|+|∫t2t1(t2−qs)(α−1)Γq(α)f(s,u(s),Dqu(s))ds|≤∫t10(t2−qs)(α−1)−(t1−qs)(α−1)Γq(α)|f(s,u(s),Dqu(s))|ds+∫t2t1(t2−qs)(α−1)Γq(α)|f(s,u(s),Dqu(s))|ds≤{tα2−tα1−(t2−t1)αΓq(α+1)+(t2−t1)αΓq(α+1)}‖μ‖ψ(‖u‖). |
It is seen that |(S1u)(t2)−(S1u)(t1)|→0 as t2→t1. Also, we have
|(DqS1u)(t2)−(DqS1u)(t1)|=|∫t20(t2−qs)(α−2)Γq(α−1)f(s,u(s),Dqu(s))ds−∫t10(t1−qs)(α−2)Γq(α−1)f(s,u(s),Dqu(s))ds|≤|∫t10(t2−qs)(α−2)−(t1−qs)(α−2)Γq(α−1)f(s,u(s),Dqu(s))ds|+|∫t2t1(t2−qs)(α−2)Γq(α−1)f(s,u(s),Dqu(s))ds|≤∫t10(t2−qs)(α−2)−(t1−qs)(α−2)Γq(α−1)|f(s,u(s),Dqu(s))|ds+∫t2t1(t2−qs)(α−2)Γq(α−1)|f(s,u(s),Dqu(s))|ds≤{tα−12−tα−11−(t2−t1)α−1Γq(α)+(t2−t1)α−1Γq(α)}‖μ‖ψ(‖u‖). |
Again, we see that |(DqS1u)(t2)−(DqS1u)(t1)|→0 as t2→t1. Hence ‖(S1u)(t2)−(S1u)(t1)‖ tends to zero as t2→t1. Thus, S1 is equicontinuous and so S1 is relatively compact on Br. Consequently, the Arzelá-Ascoli theorem implies that S1 is a compact operator on Br.
Finally, we prove that S2 is a contraction mapping. For each u,v∈Br, we have
|(S2u)(t)−(S2v)(t)|≤t∫10(1−qs)(α−2)Γq(α−1)|f(s,u(s),Dqu(s))−f(s,v(s),Dqv(s))|dqs+|t2−t(1+q)|(1+q)∫10(1−qs)(α−3)Γq(α−2)|f(s,u(s),Dqu(s))−f(s,v(s),Dqv(s))|dqs≤t∫10(1−qs)(α−2)Γq(α−1)L(s)(|u(s)−v(s)|+|Dqu(s)−Dqv(s)|)dqs+|t2−t(1+q)|(1+q)∫10(1−qs)(α−3)Γq(α−2)L(s)(|u(s)−v(s)|+|Dqu(s)−Dqv(s)|)dqs. |
Also,
|(DqS2u)(t)−(DqS2v)(t)|≤∫10(1−qs)(α−2)Γq(α−1)|f(s,u(s),Dqu(s))−f(s,v(s),Dqv(s))|dqs+|t−1|∫10(1−qs)(α−3)Γq(α−2)|f(s,u(s),Dqu(s))−f(s,v(s),Dqv(s))|dqs≤∫10(1−qs)(α−2)Γq(α−1)L(s)(|u(s)−v(s)|+|Dqu(s)−Dqv(s)|)dqs+|t−1|∫10(1−qs)(α−3)Γq(α−2)L(s)(|u(s)−v(s)|+|Dqu(s)−Dqv(s)|)dqs. |
Hence, we obtain
supt∈[0,1]|(S2u)(t)−(S2v)(t)|≤‖L‖Δ1‖u−v‖, |
supt∈[0,1]|(DqS2u)(t)−(DqS2v)(t)|≤‖L‖Δ2‖u−v‖. |
Thus, ‖S2u−S2v‖≤‖L‖(Δ1+Δ2)‖u−v‖ or ‖S2u−S2v‖≤k‖u−v‖. Thus S2 is contraction on Br as k<1. Therefore, all the assumptions of Theorem 2.2 are satisfied. Hence, Theorem 2.2 implies that the q−fractional boundary value problem (1.1) has at least one solution on [0,1].
In this section, we prove our main results about the existence of solutions for fractional q−difference inclusion (1.2).
Definition 4.1. A function u∈C([0,1],R) is called a solution for the fractional q−difference inclusion (1.2) whenever it satisfies the boundary value conditions and there exists a function v∈L1([0,1]) such that v(t)∈F(t,u(t),Dqu(t)) for almost all t∈[0,1] and
u(t)=∫t0(t−qs)(α−1)Γq(α)v(s)dqs−t∫10(1−qs)(α−2)Γq(α−1)v(s)dqs−t2−t(1+q)1+q∫10(1−qs)(α−3)Γq(α−2)v(s)dqs. |
for all t∈[0,1].
Let X be a Banach space with the norm defined in the last section. For each u∈X, the set of selections of the operator F is defined by
SF,u={v∈L1([0,1]):v(t)∈F(t,u(t),Dqu(t))for almost allt∈[0,1]}. |
Also, we define the operator N:X→P(X) by
N(u)={h∈X:there existsv∈SF,usuch thath(t)=w(t)for allt∈[0,1]}, | (4.1) |
where
w(t)=∫t0(t−qs)(α−1)Γq(α)v(s)dqs−t∫10(1−qs)(α−2)Γq(α−1)v(s)dqs−t2−t(1+q)1+q∫10(1−qs)(α−3)Γq(α−2)v(s)dqs. |
Theorem 4.1. Let F:[0,1]×R2→Pcp(R) be a multifunction. Suppose that:
(H6) The operator F is integrably bounded and F(⋅,u,v):[0,1]→Pcp(R) is measurable for all u,v∈R.
(H7) Assume that there exists m∈C([0,1],[0,∞)) such that
Hd(F(t,u1,u′1),F(t,u2,u′2))≤m(t)ψ(|u1−u2|+|u′1−u′2|)(1‖m‖(Λ1+Λ2)) | (4.2) |
for all t∈[0,1] and ui,u′i∈R(i=1,2) and ψ∈Ψ. The constants Λ1 and Λ2 are given by Eq (3.1).
(H8) There exists a function ξ:R2×R2→R with ξ((u1,u′1),(u2,u′2))≥0 for ui,u′i∈R(i=1,2).
(H9) If {un}n≥1 is a sequence in X such that un→u and ξ((un(t),Dqun(t)),(u(t),Dqu(t)))≥0 for all t∈[0,1], then there exists a subsequence {unj}j≥1 of {un} such that
ξ((unj(t),Dqunj(t)),(u(t),Dqu(t)))≥0 |
for all t∈[0,1].
(H10) There exist u0∈X and h∈N(u0) such that ξ((u0(t),Dqu0(t)),(h(t),Dqh(t)))≥0, for all t∈[0,1], where the operator N:X→P(X) is given by Eq (4.1).
(H11) For each u∈X and h∈N(u) with ξ((u(t),Dqu(t)),(h(t),Dqh(t)))≥0, there exists w∈N(u) such that ξ((h(t),Dqh(t)),(w(t),Dqw(t)))≥0.
Then the fractional q−difference inclusion (1.2) has a solution.
Proof. It is evident that the fixed point of the operator N:X→P(X) is a solution of the inclusion problem (1.2). Since the multivalued map t↦F(t,u(t),Dqu(t)) is measurable and it closed-valued for all u∈X, so F has measurable selection and the set SF,u is not empty.
We prove that N(u) is a closed subset of X for all u∈X, i.e., N(u)∈Pcl(X). For this, let {un}n≥1 be a sequence in N(u) which converges to u. We should prove that u∈N(u). For each natural number n, there exists vn∈SF,u such that
un(t)=∫t0(t−qs)(α−1)Γq(α)vn(s)dqs−t∫10(1−qs)(α−2)Γq(α−1)vn(s)dqs−t2−t(1+q)1+q∫10(1−qs)(α−3)Γq(α−2)vn(s)dqs, |
for almost all t∈[0,1].
That F has compact values, we pass into a subsequence (if necessary) to obtain that a subsequence {vn}n≥1 converges to some v∈L1([0,1]). Thus v∈SF,u and we get
un(t)→u(t)=∫t0(t−qs)(α−1)Γq(α)v(s)dqs−t∫10(1−qs)(α−2)Γq(α−1)v(s)dqs−t2−t(1+q)1+q∫10(1−qs)(α−3)Γq(α−2)v(s)dqs, |
for each t∈[0,1]. This shows that u∈N(u); that is, the operator N is closed-valued. Now, since F is a multifunction with compact values, it is easy to check that N(u) is bounded set in X for all u∈X. In the next step, we show that the operator N is an α−ψ−contractive multivalued map. For this purpose, we define a function α:X×X→[0,∞) by
α(u,u′)={1,ifξ((u(t),Dqu(t)),(u′(t),Dqu′(t)))≥0,0,otherwise |
for all u,u′∈X. Let u,u′∈X and h1∈N(u′). We choose v1∈SF,u′ such that
h1(t)=∫t0(t−qs)(α−1)Γq(α)v1(s)dqs−t∫10(1−qs)(α−2)Γq(α−1)v1(s)dqs−t2−t(1+q)1+q∫10(1−qs)(α−3)Γq(α−2)v1(s)dqs, |
for all t∈[0,1]. Since by (4.2), we have
Hd(F(t,u,Dqu),F(t,u′,Dqu′))≤m(t)ψ(|u−u′|+|Dqu−Dqu′|)(1‖m‖(Λ1+Λ2)) |
for all u,u′∈X with ξ((u(t),Dqu(t)),(u′(t),Dqu′(t)))≥0 for almost all t∈[0,1], so there exists w∈F(t,u(t),Dqu(t)) such that
|v1(t)−w|≤m(t)ψ(|u(t)−u′(t)|+|Dqu(t)−Dqu′(t)|)(1‖m‖(Λ1+Λ2)). |
Now, consider the multivalued map A:[0,1]→P(R) which is given by
A(t)={w∈R:|v1(t)−w|≤m(t)ψ(|u(t)−u′(t)|+|Dqu(t)−Dqu′(t)|)(1‖m‖(Λ1+Λ2))} |
for all t∈[0,1]. Clearly, the multifunction A(⋅)∩F(⋅,u(⋅),Dqu(⋅)) is measurable, because v1 and φ=mψ(|u−u′|+|Dqu−Dqu′|)(1Λ1+Λ2) are measurable. In this step, we can choose v2∈F(t,u(t),Dqu(t)) such that
|v1(t)−v2(t)|≤m(t)ψ(|u(t)−u′(t)|+|Dqu(t)−Dqu′(t)|)(1‖m‖(Λ1+Λ2)), |
for all t∈[0,1]. Define the element h2∈N(u) by follows:
h2(t)=∫t0(t−qs)(α−1)Γq(α)v2(s)dqs−t∫10(1−qs)(α−2)Γq(α−1)v2(s)dqs−t2−t(1+q)1+q∫10(1−qs)(α−3)Γq(α−2)v2(s)dqs |
for all t∈[0,1]. Letting supt∈[0,1]|m(t)|=‖m‖, we have
|h1−h2|≤∫t0(t−qs)(α−1)Γq(α)|v1(s)−v2(s)|dqs+t∫10(1−qs)(α−2)Γq(α−1)|v1(s)−v2(s)|dqs+|t2−t(1+q)|(1+q)∫10(1−qs)(α−3)Γq(α−2)|v1(s)−v2(s)|dqs≤∫t0(t−qs)(α−1)Γq(α)m(s)ψ(|u(s)−u′(s)|+|Dqu(s)−Dqu′(s)|)(1‖m‖(Λ1+Λ2))dqs+t∫10(1−qs)(α−2)Γq(α−1)m(s)ψ(|u(s)−u′(s)|+|Dqu(s)−Dqu′(s)|)(1‖m‖(Λ1+Λ2))dqs+|t2−t(1+q)|(1+q)∫10(1−qs)(α−3)Γq(α−2)m(s)ψ(|u(s)−u′(s)|+|Dqu(s)−Dqu′(s)|)(1‖m‖(Λ1+Λ2))dqs≤1Γq(α+1)‖m‖ψ(‖u−u′‖)(1‖m‖(Λ1+Λ2))+1Γq(α)‖m‖ψ(‖u−u′‖)(1‖m‖(Λ1+Λ2))+(2+q)(1+q)Γq(α−1)‖m‖ψ(‖u−u′‖)(1‖m‖(Λ1+Λ2))=[1Γq(α+1)+1Γq(α)+(2+q)(1+q)Γq(α−1)]‖m‖ψ(‖u−u′‖)(1‖m‖(Λ1+Λ2))=Λ1ψ(‖u−u′‖)(1(Λ1+Λ2)), |
and
|Dqh1−Dqh2|≤∫t0(t−qs)(α−2)Γq(α−1)|v1(s)−v2(s)|dqs+∫10(1−qs)(α−2)Γq(α−1)|v1(s)−v2(s)|dqs+|t−1|∫10(1−qs)(α−3)Γq(α−2)|v1(s)−v2(s)|dqs≤∫t0(t−qs)(α−2)Γq(α−1)m(s)ψ(|u(s)−u′(s)|+|Dqu(s)−Dqu′(s)|)(1‖m‖(Λ1+Λ2))dqs+∫10(1−qs)(α−2)Γq(α−1)m(s)ψ(|u(s)−u′(s)|+|Dqu(s)−Dqu′(s)|)(1‖m‖(Λ1+Λ2))dqs+|t−1|∫10(1−qs)(α−3)Γq(α−2)m(s)ψ(|u(s)−u′(s)|+|Dqu(s)−Dqu′(s)|)(1‖m‖(Λ1+Λ2))dqs≤1Γq(α)‖m‖ψ(‖u−u′‖)(1‖m‖(Λ1+Λ2))+1Γq(α)‖m‖ψ(‖u−u′‖)(1‖m‖(Λ1+Λ2))+2Γq(α−1)‖m‖ψ(‖u−u′‖)(1‖m‖(Λ1+Λ2))=[2Γq(α)+2Γq(α−1)]‖m‖ψ(‖u−u′‖)(1‖m‖(Λ1+Λ2))=Λ2ψ(‖u−u′‖)(1(Λ1+Λ2)), |
for all t∈[0,1]. Hence, we obtain
‖h1−h2‖=supt∈[0,1]|h1(t)−h2(t)|+supt∈[0,1]|Dqh1(t)−Dqh2(t)|≤(Λ1+Λ2)ψ(‖u−u′‖)(1(Λ1+Λ2))=ψ(‖u−u′‖). |
Therefore α(u,u′)Hd(N(u),N(u′))≤ψ(‖u−u′‖) for all u,u′∈X. This means that N is an α−ψ−contractive multivalued mapping. Now, let u∈X and u′∈N(u) be such that α(u,u′)≥1. Thus, by definition of α, we have ξ((u(t),Dqu(t)),(u′(t),Dqu′(t)))≥0 and by the hypothesis there exists w∈N(u′) such that ξ((u′(t),Dqu′(t)),(w(t),Dqw(t)))≥0. This implies that α(u′,w)≥1 and so, this proves that the operator N is an α−admissible.
Finally, we choose u0∈X and u′∈N(u0) such that
ξ((u0(t),Dqu0(t)),(u′(t),Dqu′(t)))≥0. |
Hence α(u0,u′)≥1. Consequently, Theorem 2.3 implies that N has a fixed point. In the other words, there exists u∗∈X such that u∗∈N(u∗) which u∗ is the solution of the fractional q−difference inclusion {(1.2)} and the proof is completed.
Theorem 4.2. Let F:[0,1]×R2→Pcp(R) be a multifunction. Suppose that:
(H12) The function ψ:[0,∞)→[0,∞) is a nondecreasing upper semi-continuous mapping such that liminft→∞(t−ψ(t))>0 and ψ(t)<t for all t>0.
(H13) The operator F:[0,1]×R2→Pcp(R) be an integrably bounded multifunction such that F(⋅,u1,u2):[0,1]→Pcp(R) is measurable for all u1,u2∈R.
(H14) There exists a function m∈C([0,1],[0,∞)) such that
Hd(F(t,u1,u2)−F(t,u′1,u′2))≤m(t)ψ(|u1−u′1|+|u2−u′2|)(1Φ1+Φ2) |
for all t∈[0,1] and ui,u′i∈R (i=1,2), where Φi's are given in Eq (3.1).
(H15) The operator N has the approximate endpoint property where N is defined in Eq (4.1).
Then the q−fractional inclusion problem (1.2) has a solution.
Proof. We show that the multifunction N:X→P(X) has an endpoint. For this, we prove that N(u) is a closed subset of P(X) for all u∈X. First of all, since the multivalued map t↦F(t,u(t),Dqu(t)) is measurable and has closed values for all u∈X, so it has measurable selection and thus SF,u is nonempty for all u∈X.
Similar to the first part of the proof of Theorem 4.1, one can see that the operator N(u) is closed-valued. Also, N(u) is a bounded set for all u∈X, because F is a compact multivalued map.
Finally, we show that Hd(N(u),N(w))≤ψ(‖u−w‖). Let u,w∈X and h1∈N(w). Choose v1∈SF,w such that
h1(t)=∫t0(t−qs)(α−1)Γq(α)v1(s)dqs−t∫10(1−qs)(α−2)Γq(α−1)v1(s)dqs−t2−t(1+q)1+q∫10(1−qs)(α−3)Γq(α−2)v1(s)dqs, |
for almost all t∈[0,1]. Since
Hd(F(t,u(t),Dqu(t))−F(t,w(t),Dqw(t)))≤m(t)ψ(|u(t)−w(t)|+|Dqu(t)−Dqw(t)|)(1Φ1+Φ2) |
for all t∈[0,1], there exist z∈F(t,u(t),Dqu(t)) such that
|v1(t)−z|≤m(t)ψ(|u(t)−w(t)|+|Dqu(t)−Dqw(t)|)(1Φ1+Φ2) |
for all t∈[0,1]. Now, consider the multivalued map U:[0,1]→P(R) which is defined by
U(t)={z∈R:|v1(t)−z|≤m(t)ψ(|u(t)−w(t)|+|Dqu(t)−Dqw(t)|)(1Φ1+Φ2). |
Since v1 and φ=mψ(|u−w|+|Dqu−Dqw|)(1Φ1+Φ2) are measurable, the multifunction U(⋅)∩F(⋅,u(⋅),Dqu(⋅)) is measurable. Choose v2(t)∈F(t,u(t),Dqu(t)) such that
|v1(t)−v2(t)|≤m(t)ψ(|u(t)−w(t)|+|Dqu(t)−Dqw(t)|)(1Φ1+Φ2), |
for all t∈[0,1]. We define the element h2∈N(u) as follows:
h2(t)=∫t0(t−qs)(α−1)Γq(α)v2(s)dqs−t∫10(1−qs)(α−2)Γq(α−1)v2(s)dqs−t2−t(1+q)1+q∫10(1−qs)(α−3)Γq(α−2)v2(s)dqs, |
for all t∈[0,1]. Therefore, similar to the proof of Theorem 4.1, we get
‖h1−h2‖=supt∈[0,1]|h1(t)−h2(t)|+supt∈[0,1]|Dqh1(t)−Dqh2(t)|≤(Φ1+Φ2)ψ(‖u−w‖)(1Φ1+Φ2)=ψ(‖u−w‖). |
Hence Hd(N(u),N(w))≤ψ(‖u−w‖) for all u,w∈X. By hypothesis (H15), since the multifunction N has approximate endpoint property, by Theorem 2.4, there exists u∗∈X such that N(u∗)={u∗}. Consequently, the q−fractional inclusion (1.2) has the solution u∗ and the proof is completed.
Now, in this section, we present some illustrative examples to show the validity of our main results.
Example 5.1. Consider the fractional q−difference equation
cD5/21/2u(t)=t100|arcsinu|+t|arctan(D1/2u)|100+100|arctan(D1/2u)|,t∈[0,1] | (5.1) |
via the boundary value conditions
u(0)=0,D1/2u(1)=u(0),D21/2u(1)=u(0) | (5.2) |
where cD5/21/2 denotes the Caputo q−fractional derivative of order 5/2. Clearly, α=5/2 and q=1/2. We define f:[0,1]×R2→R by
f(t,x(t),y(t))=t100|arcsinx(t)|+t|arctany(t)|100+100|arctany(t)|. |
In this case, for each xi,yi∈R(i=1,2), we have
|f(t,x1(t),y1(t))−f(t,x2(t),y2(t))|≤t100|arcsinx1(t)−arcsinx2(t)|+t100|arctany1(t)−arctany2(t)|≤t100(|x1(t)−x2(t)|+|y1(t)−y2(t)|). |
Hence L(t)=t/100 and so ‖L‖=supt∈[0,1]|L(t)|=1/100. On the other hand, we define continuous and nondecreasing function ψ:R+→R by ψ(x)=x for all x∈R+. We have
|f(t,u(t),(D1/2u)(t))|≤t100(|u|+|D1/2u|)=t100ψ(|u|+|D1/2u|). |
Clearly, μ:[0,1]→R is given by μ(t)=t/100 which is continuous function. Then, we have Λ1+Λ2≈6.0085 and so k≈0.06<1. Since, all assumptions of Theorem 3.2 hold, thus the fractional q−difference equation (5.1)–(5.2) has at least one solution on [0,1].
Example 5.2. We consider the fractional q−difference inclusion
cD5/21/2u(t)∈[0,0.025t|cosu(t)|2(1+|cosu(t)|)+25t|sin(π/2)t||D1/2u(t)|2000(1+|D1/2u(t)|)],t∈[0,1] | (5.3) |
via the boundary value conditions
u(0)=0,D1/2u(1)=u(0),D21/2u(1)=u(0). | (5.4) |
Put α=5/2 and q=1/2. By these values, we get Λ1≈2.5596 and Λ2≈3.4489. We define multifunction F:[0,1]×R2→P(R) by follows:
F(t,x(t),y(t))=[0,0.025t|cosx(t)|2(1+|cosx(t)|)+25t|sin(π/2)t||y(t)|2000(1+|y(t)|)] |
for each t∈[0,1]. By above definition, there exists a continuous function m:[0,1]→[0,∞) by m(t)=5t/200 for all t. Then ‖m‖=5/200. Also, we define upper semi-continuous and nondecreasing function ψ:(0,∞)→[0,∞) by ψ(t)=t/2 for all t>0. It is clear that liminft→∞(t−ψ(t))>0 and ψ(t)<t for all t>0. On the other hand, we have Φ1≈0.06399 and Φ2≈0.08622 and 1Φ1+Φ2≈6.6577>0. For every xi,yi∈R(i=1,2), we have
Hd(F(t,x1(t),y1(t))−F(t,x2(t),y2(t)))≤5t200.12(|x1(t)−x2(t)|+|y1(t)−y2(t)|)=5t200ψ(|x1(t)−x2(t)|+|y1(t)−y2(t)|)≤m(t)ψ(|x1(t)−x2(t)|+|y1(t)−y2(t)|)(1Φ1+Φ2). |
Now, put X={u:u,D1/2u∈C([0,1],R)}. Define N:X→P(R) as follows:
N(u)={h∈X:there existsv∈SF,usuch thath(t)=w(t)for allt∈[0,1]}, |
where
w(t)=∫t0(t−12s)(52−1)Γ1/2(52)v(s)d12s−t∫10(1−12s)(52−2)Γ1/2(52−1)v(s)d12s−t2−32t32∫10(1−12s)(52−3)Γ1/2(52−2)v(s)d12s. |
Also, the operator N has the approximate endpoint property, because supu∈N(0)‖u‖=0 and so infu∈Xsupz∈N(u)‖u−z‖=0. All assumptions of Theorem 4.2 hold. Therefore, by Theorem 4.2, the fractional q−difference inclusion (5.3)–(5.4) has a solution.
All authors declare no conflicts of interest in this paper.
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