We discuss a class of singular
boundary value problems of fractional q-difference equations. Some
existence and uniqueness results are obtained by a fixed point
theorem in partially ordered sets. Finally, we give an example to
illustrate the results.

1. Introduction

In recent years, many papers on fractional differential equations have appeared, because of their demonstrated applications in various fields of science and engineering; see [1–11] and the references therein. Based on the increasingly extensive application of discrete fractional calculus and the development of q-difference calculus or quantum calculus (see [12–19] and the references therein), fractional q-difference equations have attracted the attention of researchers for the numerous applications in a number of fields such as physics, chemistry, aerodynamics, biology, economics, control theory, mechanics, electricity, signal and image processing, biophysics, blood flow phenomena, aerodynamics, and fitting of experimental data; see [20–23]. Some recent work on the existence theory of fractional q-difference equations can be found in [24–29]. However, the study of singular boundary value problems (BVPs) with fractional q-difference equations is at its infancy and lots of work on the topic should be done.

Recently, in [25], Ferreira has investigated the existence of positive solution for the following fractional q-difference equations BVP (1)Dqαyx+fx,yt=0,0<x<1,y0=Dqy0=0,Dqy1=β≥0,by applying a fixed point theorem in cones.

More recently, in [30], Caballero et al. have studied positive solutions for the following BVP: (2)D0+αut+ft,ut=0,0<t<1,u(0)=u(1)=0,by fixed point theorem in partially ordered sets.

Motivated by the work above, in this paper, we discuss the existence and uniqueness of solutions for the singular BVPs of factional q-difference equations given by (3)(Dqαu)(t)+f(t,u(t))=0,0<t<1,u(0)=u(1)=0,(Dqu)(0)=0,where 2<α≤3 and f:[0,1]×[0,∞)→[0,∞) is continuous with limt→0+f(t,·)=∞ (i.e., f is singular at t=0).

2. Preliminary Results

For convenience, we present some definitions and lemmas which will be used in the proofs of our results.

Let q∈(0,1) and define (4)aq=1-qa1-q,a∈R.

The q-analogue of the power function (a-b)n with n∈N0 is (5)a-b0=1,a-bn=∏k=0n-1a-bqk,n∈N,a,b∈R.More generally, if α∈R, then (6)a-b(α)=aα∏n=0n-1a-bqna-bqα+n.Note that if b=0 then a(α)=aα. The q-gamma function is define by (7)Γqx=1-q(x-1)1-qx-1,x∈R∖{0,-1,-2,…},and it satisfies Γq(x+1)=[x]qΓq(x).

Following, let us recall some basic concepts of q-calculus [12].

Definition 1.

For 0<q<1, we define the q-derivative of a real-value function f as (8)Dqfx=fx-fqx1-qx,Dqf0=limx→0Dqfx.Note that limq→1-Dqf(x)=f′(x).

Definition 2.

The higher order q-derivatives are defined inductively as (9)(Dq0f)(x)=f(x),(Dqnf)(t)=Dq(Dqn-1f)(t),n∈N.

For example, Dq(tk)=[k]qtk-1, where k is a positive integer and the bracket [k]q=(qk-1)/(q-1). In particular, Dq(t2)=(1+q)t.

Definition 3.

The q-integral of a function f in the interval [0,b] is given by (10)Iqfx=∫0xftdqt=x1-q∑n=0∞fxqnqn,x∈0,b.If a∈[0,b] and f is defined in the interval [0,b], its integral from a to b is define by (11)∫abf(t)dqt=∫0bf(t)dqt-∫0af(t)dqt.Similarly as done for derivatives, an operator Iqn can be define, namely, (12)(Iq0f)(x)=f(x),(Iqnf)(x)=Iq(Iqn-1f)(x),n∈N.

Observe that (13)DqIqf(x)=f(x),and if f is continuous at x=0, then IqDqf(x)=f(x)-f(0).

We now point out three formulas (iDq denotes the derivative with respect to variable i) (14)at-s(α)=aαt-s(α),(15)tDqt-s(α)=αqt-s(α-1),Dxq∫0xfx,tdqt=∫0xxDqfx,tdqt+fqx,x.

Remark 4.

We note that if α≥0 and a≤b≤t, then (t-a)(α)≥(t-b)(α) [24].

Definition 5 (see [<xref ref-type="bibr" rid="B21">21</xref>]).

Let α≥0 and f be a function defined on [0,1]. The fractional q-integral of the Riemann-Liuville type is (Iq0f)(x)=f(x) and (16)Iqαfx=1Γqα∫0xx-qtα-1ftdqt,α>0,x∈0,1.

Definition 6 (see [<xref ref-type="bibr" rid="B23">23</xref>]).

The fractional q-derivative of the Riemann-Liuville type of α≥0 is defined by (Dq0f)(x)=f(x) and (17)(Dqαf)(x)=(DqmIqm-αf)(x),α>0,where m is the smallest integer greater than or equal to α.

Lemma 7 (see [<xref ref-type="bibr" rid="B21">21</xref>, <xref ref-type="bibr" rid="B23">23</xref>]).

Let α,β≥0 and let f be a function define on [0,1]. Then, the next formulas hold:

(IqβIqαf)(x)=(Iqα+βf)(x),

(DqαIqαf)(x)=f(x).

Lemma 8 (see [<xref ref-type="bibr" rid="B24">24</xref>]).

Let α>0 and let p be a positive integer. Then, the following equality holds: (18)IqαDqpfx=DqpIqαfx-∑k=0p-1xα-p+kΓqα+k-p+1Dqkf0.

Lemma 9.

Let y(t)∈C[0,1]⋂L1[0,1] and 2<α≤3; then the BVP (19)Dqαut+yt=0,0<t<1,u0=u1=0,Dqu0=0,has a unique solution(20)u(t)=∫01G(t,qs)y(s)dqs,where(21)Gt,s=1Γq(α)1-s(α-1)tα-1-t-s(α-1),0≤s≤t≤1,1-s(α-1)tα-1,0≤t≤s≤1.

Proof.

By Lemmas 7 and 8, we see that(22)Dqαut=-ytDqαut⟺IqαDq3Iq3-αut=-IqαytDqαut⟺ut=c1tα-1+c2tα-2Dqαut⟺ut=+c3tα-3-1Γqα∫0tt-qsα-1Dqαut⟺ut=c3tα-31Γqα×ysdqs,where c1,c2, and c3 are some constants to be determined. Since u(0)=0, we must have c3=0. Now, differentiating both sides of (22) and using (15), we obtain (23)Dqut=α-1qc1tα-2+α-2qc2tα-3Dqut-1Γq(α)∫0tα-1qt-qs(α-2)y(s)dqs.Using (Dqu)(0)=0 and u(1)=0, we must set c2=0, and (24)c1=1Γq(α)∫011-qs(α-1)y(s)dqs.

Finally, we obtain (25)ut=tα-1Γqα∫011-qsα-1ysdqsut-1Γqα∫0tt-qsα-1ysdqsut=∫01G(t,qs)y(s)dqs.The proof is complete.

Lemma 10.

Function G defined above satisfies the following conditions:

G(t,qs) is a continuous function on [0,1]×[0,1];

G(t,qs)≥0 for t,s∈[0,1].

Proof.

(i) Obviously, G(t,qs) is continuous on [0,1]×[0,1].

(ii) Let (26)g1t,s=1-sα-1tα-1-t-sα-1,0≤s≤t≤1,g2(t,s)=1-s(α-1)tα-1,0≤t≤s≤1.It is clear that g2(t,qs)≥0, for t,s∈[0,1]. Now, in view of Remark 4, for t≠0(27)g1t,qs=1-qsα-1tα-1-t-qsα-1g1t,qs=tα-11-qsα-1-1-qstα-1g1t,qs≥tα-11-qsα-1-1-qsα-1=0.Therefore, G(t,qs)≥0. This proof is complete.

By J we denote the class of those functions β:[0,∞)→[0,1) satisfying the following condition; β(tn)→1 implies tn→0.

Theorem 11 (see [<xref ref-type="bibr" rid="B31">31</xref>]).

Let (X,≤) be a partially ordered set and suppose that there exists a metric d in X such that (X,d) is a complete metric space. Let T:X→X be a nondecreasing mapping such that there exists an element x0∈X with x0≤Tx0. Suppose that there exists β∈J such that(28)dTx,Ty≤βdx,y·dx,yforx,y∈Xwithx≥y.Assume that either T is continuous or X is such that (29)ifxnisanondecreasingsequenceinXsuchthatxn⟶xthenxn≤x∀n∈N.Besides if (30)foreachx,y∈Xthereexistsz∈Xwhichiscomparabletoxandy,then T has a unique fixed point.

Let C[0,1]={x:[0,1]→R,continuous} be the Banach space with the classic metric given by d(x,y)=sup0≤t≤1{|x(t)-y(t)|}.

Notice that this space can be equipped with a partial order given by (31)x,y∈C[0,1],x≤y⟺x(t)≤y(t),fort∈[0,1].In [32], it is proved that (C[0,1],≤) satisfies condition (29) of Theorem 11. Moreover, for x,y∈C[0,1], as the function max(x,y)∈C[0,1],(C[0,1],≤) satisfies condition (30).

3. Main Result

In this section, we will consider the question of positive solutions for BVP (3). At first, we prove some lemmas required for the main result.

Lemma 12.

Let 0<σ<1, 2<α≤3 and F:(0,1]→R is a continuous function with limt→0+F(t)=∞. Suppose that tσF(t) is a continuous function on [0,1]. Then the function defined by (32)H(t)=∫01G(t,qs)F(s)dqsis continuous on [0,1], where G(t,s) is Green function be given in Lemma 9.

Proof.

We will divide the proof into three parts.

Case 1 (t0=0). First, H(0)=0. Since tσF(t) is continuous on [0,1], we can find a positive constant M such that |tσF(t)|≤M for any t∈[0,1]. Thus, (33)Ht-H0=Ht=∫01Gt,qsFsdqs=∫01Gt,qss-σsσFsdqs=∫0t1-qsα-1tα-1-t-qsα-1Γqαs-σsσF(s)dqs+∫t11-qsα-1tα-1Γqαs-σsσF(s)dqs=∫011-qsα-1tα-1Γqαs-σsσF(s)dqs-∫0tt-qsα-1Γqαs-σsσF(s)dqs≤M∫011-qsα-1tα-1Γqαs-σdqs+M∫0tt-qsα-1Γqαs-σdqs=Mtα-1Γqα∫011-qsα-1s-σdqs+∫0t1-qst(α-1)s-σdqs.For ∫0t(1-(qs/t))(α-1)s-σdqs, let u=s/t; then we obtain(34)∫0t1-qst(α-1)s-σdqs=t1-σ∫011-qu(α-1)u-σdqu.Hence,(35)Ht≤Mtα-1Γqα∫011-qsα-1s-σdqsHt+Mtα-σΓqα∫011-quα-1u-σdquHt=Mtα-1Γqα+Mtα-σΓqαβq(1-σ,α),where βq denotes the q-beta function.

When t→0, we see that H(t)→H(0); that is H(t) is continuous at t0=0.

Case 2 (t0∈(0,1)). We should prove H(tn)→H(t0) when tn→t0. Without loss of generality, we consider tn>t0 (it is the same argument for tn<t0). In fact, (36)Htn-Ht0=∫01tnα-11-qsα-1Γqαs-σsσF(s)dqs-∫0tntn-qsα-1Γqαs-σsσFsdqs-∫01t0α-11-qsα-1Γqαs-σsσFsdqs+∫0t0t0-qsα-1Γqαs-σsσF(s)dqs=∫01(tnα-1-t0α-1)1-qsα-1Γqαs-σsσF(s)dqs-∫0t0tn-qsα-1-t0-qsα-1Γqαs-σsσFsdqs-∫t0tntn-qsα-1Γqαs-σsσF(s)dqs≤Mtnα-1-t0α-1Γqα∫011-qsα-1s-σdqs+MΓqα∫0t0tn-qsα-1-t0-qsα-1s-σdqs+MΓqα∫t0tntn-qsα-1s-σdqs=MΓq(α)βq(1-σ,α)(tnα-1-t0α-1)+MΓq(α)(an+bn),where (37)an=∫0t0tn-qsα-1-t0-qsα-1s-σdqs,bn=∫t0tntn-qsα-1s-σdqs.When n→∞, we verify an→0.

As tn→t0, then ((tn-qs)(α-1)-(t0-qs)(α-1))s-σ→0, when n→∞. Moreover, (38)tn-qsα-1-t0-qsα-1s-σ≤2s-σ,∫012s-σdqs=21-σqs1-σ01=21-σq<∞.

We have ((tn-qs)(α-1)-(t0-qs)(α-1))s-σ converges pointwise to the zero function and |(tn-qs)(α-1)-(t0-qs)(α-1)|s-σ is bounded by a function belonging to L1[0,1], by Lebesgue’s dominated convergence theorem an→0 when n→∞.

Now, we prove bn→0 when n→∞.

In fact, as (39)bn=∫t0tntn-qsα-1s-σdqsbn≤∫t0tns-σdqs=s1-σ1-σqt0tnbn=11-σqtn1-σ-t01-σ,and taking into account that tn→t0, we get bn→0 when n→∞.

Above all, we obtain |H(tn)-H(t0)|→0, when n→∞.

Case 3 (t0=1). It is easy to check that H(1)=0 and H(t) is continuous at t0=1. The proof is the same as the proof of Case 1.

Lemma 13.

Suppose that 0<σ<1. Then,(40)max0≤t≤1∫01G(t,qs)s-σdqs=Aα-1-Aα-σΓq(α)βq(1-σ,α),where A=((α-1)/(α-σ))1/(1-σ).

For the convenience, we denote max0≤t≤1∫01G(t,qs)s-σdqs by K.

Next, we denote the class of functions ϕ:[0,∞)→[0,∞) by A satisfying

ϕ is nondecreasing;

ϕ(x)<x for any x>0;

β(x)=ϕ(x)/x∈J, where J is the class of functions appearing in Theorem 11.

We give our main result as follows.

Theorem 14.

Let 0<σ<1, 2<α≤3,f:[0,1]×[0,∞)→[0,∞) is continuous and limt→0+f(t,·)=∞, and tσf(t,y) is a continuous function on [0,1]×[0,∞). Assume that there exists 0<λ≤1/K such that for x,y∈[0,∞) with y≥x and t∈[0,1], (43)0≤tσ(f(t,y)-f(t,x))≤λϕ(y-x),where ϕ∈A. Then the BVP (3) has a unique positive solution (i.e., x(t)>0 for t∈(0,1)).

Proof.

We define the cone P by (44)P={u∈C[0,1]:u(t)≥0}.It is clear that P is a complete metric space as P is a closed set of C[0,1]. It is also easy to check that P satisfies conditions (29) and (30) of Theorem 11.

We define the operator T by (45)Tut=∫01Gt,qsfs,usdqsTut=∫01G(t,qs)s-σsσf(s,u(s))dqs.In view of Lemma 12, Tu∈C[0,1]. Moreover, it follows from the nonnegativeness of G(t,qs) and tσf(t,y) that Tu∈P for u∈P. Thus, T:P→P.

Next, we will prove that assumptions in Theorem 11 are satisfied.

First, for u≥v, we have (46)Tut=∫01Gt,qsfs,usdqsTut=∫01Gt,qss-σsσfs,usdqsTut≥∫01Gt,qss-σsσfs,vsdqsTut=(Tv)(t).Hence, the operator T is nondecreasing. Besides, for u≥v and u≠v, (47)dTu,Tv=maxt∈0,1Tut-Tvt=maxt∈0,1Tut-Tvt=maxt∈0,1∫01Gt,qsfs,us-fs,vsdqs=maxt∈0,1∫01Gt,qss-σsσfs,us-fs,vsdqs≤maxt∈[0,1]∫01G(t,qs)s-σλϕ(u(s)-v(s))dqs.Since ϕ is nondecreasing and u(s)-v(s)≤d(u,v), (48)dTu,Tv≤maxt∈0,1∫01Gt,qss-σλϕdu,vdqsdTu,Tv=λϕdu,vmaxt∈0,1∫01Gt,qss-σdqsdTu,Tv=λϕ(d(u,v))K.Moreover, when 0<λ≤1/K, we get (49)dTu,Tv≤ϕdu,vdTu,Tv=ϕdu,vdu,v·du,vdTu,Tv=β(d(u,v))·d(u,v).Obviously, the last inequality is satisfied for u=v.

Taking into account that the zero function satisfies 0≤T0, in view of Theorem 11, the operator T has a unique fixed point x(t) in P.

At last, we will prove x(t) is a positive solution. We assume that there exists 0<t1<1 such that x(t1)=0. Since x(t) of problem (3) is a fixed point of the operator T, we have (50)x(t)=∫01G(t,qs)f(s,x(s))dqs,for0<t<1,x(t1)=∫01G(t1,qs)f(s,x(s))dqs=0.For the nonnegative character of G(t,qs) and f(s,x), the last relation gives (51)Gt1,qsfs,xs=0a.e.(s).f is continuous and limt→0+f(t,·)=∞; then for M>0, we can find δ>0, and, for s∈[0,1]∩(0,δ), we have f(s,0)>M. It is clear that [0,1]∩(0,δ)⊂{s∈[0,1]:f(s,x(s))>M} and μ([0,1]∩(0,δ))>0, where μ is the Lebesgue measure on [0,1]. That is to say, G(t1,qs)f(s,x(s))=0 a.e. (s). This is a contradiction because G(t1,qs) is a rational function in s.

Therefore, x(t)>0 for t∈(0,1).

The proof is complete.

4. Example

Consider the following singular BVP: (52)D1/25/2u(t)+λ(t2+1)ln(1+u(t))t1/2=0,0<t<1,λ>0,u(0)=u(1)=0,(D1/2u)(0)=0.Here, α=2.5, q=1/2, σ=1/2, and f(t,u)=λ(t2+1)ln(1+u(t))/t1/2 for (t,u)∈[0,1]×[0,∞). Notice that f is continuous in [0,1]×[0,∞) and limt→0+f(t,·)=∞.

At first, we define ϕ by (53)ϕ:[0,∞)⟶[0,∞),ϕ(x)=ln(1+x).It is clear that ϕ(x)=ln(1+x) is a nondecreasing function; for u≥v, we can get (54)ϕ(u)-ϕ(v)≥0.

Moreover, for u≥v, ϕ also satisfies (55)ϕ(u)-ϕ(v)≤ϕ(u-v).In fact, when u≥v, (56)ϕu-v-ϕu-ϕv=ln1+u-vϕu-vϕu-ϕv-ln1+u-ln1+vϕu-vϕu-ϕv=ln1+u-v1+v1+uϕu-vϕu-ϕv=ln(1+(u-v)v1+u)≥0,equivalently (57)ϕ(u)-ϕ(v)≤ϕ(u-v).Above all, 0≤ϕ(u)-ϕ(v)≤ϕ(u-v) for u≥v.

Second, for u≥v and t∈[0,1], we have (58)0≤t1/2ft,u-ft,v0=λt2+1ln1+u-ln1+v0≤λt2+1ln1+u-v0≤2λln(1+u-v);that is, f satisfies assumptions of Theorem 14.

Third, we should prove ϕ(x) belongs to A. By elemental calculus, it is easy to check that ϕ is nondecreasing and ϕ(x)<x, for x>0.

In order to prove β(x)=ϕ(x)/x∈J, we notice that if β(tn)→1, then the sequence (tn) is a bounded sequence because in contrary case, that is, tn→∞, we get (59)β(tn)=ln(1+tn)tn⟶0.Now, we assume that tn↛0, and then we find ε>0 such that for each n∈N there exists ρn≥n with tρn≥ε.

Since (tn) is a bounded sequence, we can find a subsequence (tkn) of (tρn) with tkn→a, for certain a∈[0,1). When β(tn)→1, it implies that (60)β(tkn)=ln(1+tkn)tkn⟶1.and, as the unique solution of ln(1+x)=x is x0=0, we deduce that a=0. Therefore, tkn→0 and this contradicts the fact that tkn≥ε for every n∈N.

Thus, tn→0 and this proves that β∈J.

Finally, in view of Theorem 14, (61)2λ≤1K=11/43/2-1/41/2/Γ1/23/2·β1/21/2,3/22λ≈10.96511985;that is, when λ≤5.48256, boundary value problem (52) has a unique positive solution.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This paper is supported by the Natural Science Foundation of China (11201112), the Natural Science Foundation of Hebei Province (A2013208147), (A2014208152), and (A2015208114), and the Foundation of Hebei Education Department (Z2014062).

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