R E S E A R C H
Open Access
Existence and uniqueness results to
positive solutions of integral boundary value
problem for fractional
q
-derivatives
Furi Guo
1, Shugui Kang
1and Fu Chen
1**Correspondence:
chenfu-88@163.com
1School of Mathematics and
Computer Science, Shanxi Datong University, Datong, P.R. China
Abstract
In this paper,we are interested in the existence and uniqueness of positive solutions for integral boundary value problem with fractionalq-derivative:
Dαqu(t) +f
(
t,u(t),u(t))
+g(
t,u(t))
= 0, 0 <t< 1, u(0) =Dqu(0) = 0, u(1) =μ
1
0
u(s)dqs,
whereDαq is the fractionalq-derivative of Riemann–Liouville type, 0 <q< 1, 2 <
α
≤3, andμ
is a parameter with 0 <μ
< [α
]q. By virtue of fixed point theorems for mixedmonotone operators, we obtain some results on the existence and uniqueness of positive solutions.
Keywords: Positive solution; Mixed monotone operator; Fractionalq-difference equation; Existence and uniqueness
1 Introduction
The theory that fractional differential equations arise in the fields of science and engineer-ing such as physics, chemistry, mechanics, economics, and biological sciences, etc.; see, for example, [1–6]. Theq-difference calculus or quantum calculus is an old subject that was put forward by Jackson [7,8]. The essential definitions and properties ofq-difference calculus can be found in [9,10]. Early development forq-fractional calculus can be seen in the papers by Al-Salam [11] and Agarwal [12] on the existence theory of fractional q-difference. These days the fractionalq-difference equation have given fire to increasing scholars’ imaginations. Some works considered the existence of positive solutions for non-linearq-fractional boundary value problem [13–32]. For example, Ferreira [13] studied the existence of positive solutions to the fractionalq-difference equation
⎧ ⎨ ⎩
Dα
qu(t) +f(t,u(t)) = 0, 0 <t< 1, 1 <α≤2,
u(0) =u(1) = 0. (1.1)
Ferreira [14] also considered the existence of positive solutions to the nonlinear q -difference boundary value problem
⎧ ⎨ ⎩
Dα
qu(t) +f(t,u(t)) = 0, 0 <t< 1, 1 <α≤3,
u(0) =Dqu(0) = 0, Dqu(1) =β≥0.
(1.2)
EI-Shahed and AI-Askar [15] studied the existence of a positive solution to the fractional q-difference equation
⎧ ⎨ ⎩
cDαqu(t) +a(t)f(t) = 0, 0≤t≤1, 2 <α≤3, u(0) =D2
q(0) = 0, γDqu(1) +βD2qu(1) = 0,
(1.3)
whereγ,β≤0, andcDαqis the fractionalq-derivative of Caputo type.
Darzi and Agheli [16] studied the existence of a positive solution to the fractionalq -difference equation
⎧ ⎨ ⎩
Dα
qu(t) +a(t)f(t) = 0, 0≤t≤1, 3 <α≤4,
u(0) =Dqu(0) =D2qu(0) = 0, D2qu(1) =βD2qu(η),
(1.4)
where 0 <η< 1 and 1 –βηα–3> 0.
The methods used in the papers mentioned are mainly the Krasnoselskii fixed point theorem, the Schauder fixed point theorem, the Leggett–Williams fixed point theorem, and so on. Differently from methods used in the literature mentioned, on the basis of the enlightenment of the works [17,18,26], we will use fixed point theorems for mixed monotone operators to demonstrate the existence and uniqueness of positive solutions for integral boundary value problems of the form
⎧ ⎨ ⎩
Dα
qu(t) +f(t,u(t),u(t)) +g(t,u(t)) = 0, 0 <t< 1,
u(0) =Dqu(0) = 0, u(1) =μ
1 0 u(s)dqs,
(1.5)
whereDα
q is the fractionalq-derivative of Riemann–Liouville type, 0 <q< 1, 2 <α≤3, 0 <μ< [α]q. Our results ensure the existence of a unique positive solution. Moreover, an iterative scheme is constructed for approximating the solution. As far as we know, there are still very few works utilizing the fixed point results for mixed monotone operators to study the existence and uniqueness of a positive solution for fractional q-derivative integral boundary value problems.
2 Preliminaries
For convenience of the reader, on one hand, we recall some well-known facts onq-calculus and, on the other hand, some notations and lemmas that will be used in the proofs of our theorems.
A nonempty closed convex setP⊂E is a cone if (1)x∈P,r≥0⇒rx∈Pand (2)x∈ P, –x∈P⇒x=θ (θ is the zero element ofE), where (E, · ) is a real Banach space. For allx,y∈E, if there existμ,ν> 0 such thatμx≤y≤νx, then we writex∼y. Obviously,∼ is an equivalence relation. LetPh={x∈E|x∼h,h>θ}.
Letq∈(0, 1). Then theq-number is given by
[a]q=1 –q a
1 –q, a∈R.
Theq-analogue of the power function (a–b)(n)withn∈N 0is
(a–b)(0)= 1, (a–b)(n)= n–1
k=0
a–bqk , n∈N,a,b∈R.
More generally, ifα∈R, then
(a–b)(α)=aα
∞
k=0
a–bqk
a–bqα+k, α= 0.
Note that ifb= 0, thena(α)=aα. Theq-gamma function is defined by
q(x) =(1 –q)
(x–1)
1 –qx–1 , x∈R
+,
and satisfiesq(x+ 1) = [x]qq(x).
Theq-derivative of a functionf is defined by
(Dqf)(x) =
f(qx) –f(x)
(q– 1)x , (Dqf)(0) =xlim→0(Dqf)(x),
andq-derivatives of higher order by
D0qf (x) =f(x), Dnqf (x) =Dq
Dnq–1f (x), n∈N.
Theq-integral of a functionf defined in the interval [0,b] is given by
(Iqf)(x) =
x
0
f(s)dqs=x(1 –q)
∞
k=0
fxqk qk, x∈[0,b].
Ifa∈[0,b] andf is defined in the interval [0,b], then its integral fromatobis defined by
b
a
f(s)dqs=
b
0
f(s)dqs–
a
0
Similarly to the derivatives, the operatorIn
q is given by
Iq0f (x) =f(x), In
qf (x) =Iq
In–1
q f (x), n∈N.
The fundamental theorem of calculus applies to the operatorsIqandDq, that is,
(DqIqf)(x) =f(x),
and iff is continuous atx= 0, then
(IqDqf)(x) =f(x) –f(0).
The following formulas will be used later (tDqdenotes the derivative with respect to vari-ablet):
tDq(t–s)(α)= [α]q(t–s)(α–1),
xDq
x
0
f(x,t)dqt
(x) =
x
0
xDqf(x,t)dqt+f(qx,x).
Definition 2.1(see [4]) Letα≥0, and letf be a function defined on [0, 1]. The fractional q-integral of the Riemann–Liouville type is defined by (I0
qf)(x) =f(x) and
Iqαf (x) = 1
q(α)
x
0
(x–qt)(α–1)f(t)dqt, α> 0,x∈[0, 1].
Definition 2.2(see [10]) The fractionalq-derivative of the Riemann–Liouville type is de-fined by
D0qf (x) =f(x), Dαqf (x) =DpqIpq–αf (x), α> 0,
wherepis the smallest integer greater than or equal toα.
Lemma 2.1(see [10]) Letα,β≥0,and let f be a function defined on[0, 1].Then the fol-lowing formulas hold:
(i) (IqβIqαf)(x) = (I
β+α
q f)(x),
(ii) (Dα
qIqαf)(x) =f(x).
Lemma 2.2(see [10]) Letα> 0,and let be p be a positive integer.Then the following equal-ity holds:
IqαDpqf (x) =DpqIqαf (x) – p–1
k=0
xα–p+k
q(α+k–p+ 1)
Dkqf (0).
Lemma 2.3(see [27]) Let2 <α≤3and0 <μ< [α]q.Let x∈C[0, 1].Then the boundary value problem
Dαqu(t) +x(t) = 0, 0 <t< 1, (2.1)
u(0) =Dqu(0) = 0, u(1) =μ
1
0
has a unique solution
u(t) =
1
0
G(t,qs)x(s)dqs,
where
G(t,s) =
⎧ ⎨ ⎩
tα–1(1–s)(α–1)([α]
q–μ+μqα–1s)–([α]q–μ)(t–s)α–1
([α]q–μ)q(α) , 0≤s≤t≤1,
tα–1(1–s)(α–1)([α]q–μ+μqα–1s)
([α]q–μ)q(α) , 0≤t≤s≤1.
(2.3)
Lemma 2.4(see [27]) The function G(t,qs)defined by(2.3)has the following properties:
(i) G(t,qs)is a continuous function andG(t,qs)≥0; (ii) μqα([tαα]–1(1–qs)(α–1)s
q–μ)q(α) ≤G(t,qs)≤
M0tα–1
([α]q–μ)q(α),t,s∈[0, 1],
where M0=max{[α– 1]q([α]q–μ) +μqα,qα–1[α]q}.
Definition 2.3 (see [17]) An operator A:P×P→P is said to be a mixed monotone
operator ifA(x,y) is increasing inxand decreasing iny, that is,xi,yi∈P(i= 1, 2),x1≤x2,
y1≥y2implyA(x1,y1)≤A(x2,y2). An elementx∈Pis called a fixed point ofAifA(x,x) =x.
Definition 2.4(see [18]) An operatorA:P→Pis said to be subhomogeneous if
A(tx)≥tA(x) for anyt∈(0, 1),x∈P. (2.4)
Definition 2.5(see [18]) LetD=P, and letγbe a real number with 0≤γ < 1. An operator A:D→Dis said to beγ-concave if it satisfies
A(tx)≥tγA(x) for anyt∈(0, 1),x∈D. (2.5)
Lemma 2.5(see [17]) Let h>θ andγ ∈(0, 1).
Let A:P×P→P be a mixed monotone operator satisfying
Atx,t–1y ≥tγA(x,y) for any t∈(0, 1),x,y∈P, (2.6)
and let B:P→P be an increasing subhomogeneous operator.Assume that
(i) there ish0∈Phsuch thatA(h0,h0)∈PhandBh0∈Ph;
(ii) there exists a constantδ0such thatA(x,y)≥δ0Bxfor anyx,y∈P.
Then:
(1) A:Ph×Ph→PhandB:Ph→Ph;
(2) there existu0,v0∈Phandr∈(0, 1)such that
rv0≤u0<v0, u0≤A(u0,v0) +Bu0≤A(v0,u0) +Bv0≤v0;
(3) the operator equationA(x,x) +Bx=xhas a unique solutionx∗∈Ph;
(4) for any initial valuesx0,y0∈Ph,constructing successively the sequences
xn=A(xn–1,yn–1) +Bxn–1, yn=A(yn–1,xn–1) +Byn–1, n= 1, 2, . . . ,
Remark2.1 WhenB=θ in Lemma2.5, then the corresponding conclusion still holds.
Lemma 2.6(see [18]) Let h>θ andγ ∈(0, 1).
Let A:P×P→P be a mixed monotone operator satisfying
Atx,t–1y ≥tA(x,y), for any t∈(0, 1),x,y∈P, (2.7)
and let B:P→P be an increasingγ-concave operator.Assume that
(i) there ish0∈Phsuch thatA(h0,h0)∈PhandBh0∈Ph;
(ii) there exists a constantδ0such thatA(x,y)≤δ0Bxfor anyx,y∈P.
Then:
(1) A:Ph×Ph→PhandB:Ph→Ph;
(2) there existu0,v0∈Phandr∈(0, 1)such that
rv0≤u0<v0, u0≤A(u0,v0) +Bu0≤A(v0,u0) +Bv0≤v0;
(3) the operator equationA(x,x) +Bx=xhas a unique solutionx∗∈Ph;
(4) for any initial valuesx0,y0∈Ph,constructing successively the sequences
xn=A(xn–1,yn–1) +Bxn–1, yn=A(yn–1,xn–1) +Byn–1, n= 1, 2, . . . ,
we havexn→x∗andyn→x∗asn→ ∞.
Remark2.2 WhenA=θ in Lemma2.6, then the corresponding conclusion still holds.
3 Main results
In this section, we give and prove our main results by applying Lemmas2.5and2.6. We consider the Banach space X=C[0, 1] endowed with standard norm x=sup{|x(t)|: t∈[0, 1]}. Clearly, this space can be equipped with a partial order given by
x,y∈C[0, 1], x≤y ⇔ x(t)≤y(t) fort∈[0, 1].
We define the coneP={x∈X:x(t)≥0,t∈[0, 1]}. Notice thatPis a normal cone inC[0, 1] and the normality constant is 1.
Theorem 3.1 Suppose that
(F1) a functionf(t,x,y) : [0, 1]×[0, +∞)×[0, +∞)→[0, +∞)is continuous,increasing with respect to the second variable,and decreasing with respect to the third variable;
(F2) a functiong(t,x) : [0, 1]×[0, +∞)→[0, +∞)is continuous and increasing with re-spect to the second variable;
(F3) there exists a constantγ ∈(0, 1)such thatf(t,λx,λ–1y)≥λγf(t,x,y)for anyt∈[0, 1], λ∈(0, 1),x,y∈[0, +∞),andg(t,λx)≥λg(t,x)forλ∈(0, 1),t∈[0, 1],u∈[0, +∞), andg(t, 0)≡0;
(F4) there exists a constantδ0> 0such thatf(t,x,y)≥δ0g(t,x),t∈[0, 1],x,y≥0.
(1) there existx0,y0∈Phandr∈(0, 1)such thatry0≤x0<y0and
x0≤ 1
0
G(t,qs)fs,x0(s),y0(s) +g
s,x0(s) dqs, t∈[0, 1],
y0≥ 1
0
G(t,qs)fs,y0(s),y0(s) +g
s,y0(s) dqs, t∈[0, 1],
whereG(t,qs)is defined by(2.3),andh(t) =tα–1,t∈[0, 1];
(2) the boundary value problem(1.5)has a unique positive solutionu∗inPh,and for any x0,y0∈Ph,constructing successively the sequences
xn+1= 1
0
G(t,qs)fs,xn(s),yn(s) +gs,xn(s) dqs, n= 0, 1, 2, . . . ,
yn+1= 1
0
G(t,qs)fs,yn(s),xn(s) +g
s,yn(s) dqs, n= 0, 1, 2, . . . ,
we havexn–u∗ →0andyn–u∗ →0asn→ ∞.
Proof We note that ifuis a solution of boundary value problem (1.5), then
u(t) =
1
0
G(t,qs)fs,u(s),u(s) +gs,u(s) dqs, 0≤t≤1. (3.1)
Define two operatorsT1:P×P→EandT2:P→Eby
T1(u,v)(t) = 1
0
G(t,qs)fs,u(s),v(s) dqs,
(T2u)(t) = 1
0
G(t,qs)gs,u(s) dqs.
(3.2)
We transform the boundary value problem (1.5) into a fixed point problemu=T1(u,u) +
T2u. From (F1), (F2), and Lemma2.4it is easy to see thatT1:P×P→PandT2:P→P.
Next, we want to prove thatT1andT2satisfy the conditions of Lemma2.5.
To begin with, we prove thatT1is a mixed monotone operator. In fact, foru1,u2,v1,v2∈
Pwithu1≥u2andv1≤v2, it is easy to see thatu1(t)≥u2(t),v1(t)≤v2(t),t∈[0, 1], and
by Lemma2.4and (F1),
T1(u1,v1)(t) = 1
0
G(t,qs)fs,u1(s),v1(s) dqs
≥
1
0
G(t,qs)fs,u2(s),v2(s) dqs=T1(u2,v2)(t). (3.3)
For anyλ∈(0, 1) andu,v∈P, by (F3) we have
T1
λu,λ–1v (t) =
1
0
G(t,qs)fs,λu(s),λ–1v(s) dqs
≥λγ 1
0
G(t,qs)fs,u(s),v(s) dqs≥λγT1(u,v)(t). (3.4)
Thus we haveM2h(t)≤T1(h,h)≤M1h(t),M3h(t)≤T2h≤M4h(t),t∈[0, 1]. So,T1(h,h)∈
Ph. Fromg(t, 0)≡0 it is easy to see thatT2h∈Ph. So, there ish(t) =tα–1∈Phsuch that T1(h,h)∈PhandT2h∈Ph.
Next, we prove that the operatorsT1andT2satisfy condition (ii) of Lemma2.5. In fact,
foru,v∈Pand anyt∈[0, 1], by (F4) we have
T1(u,v)(t) = 1
0
G(t,qs)fs,u(s),v(s) dqs
≥δ0 1
0
G(t,qs)gs,u(s) dqs=δ0(T2u)(t). (3.9)
Then we have T1(u,v)≥δ0T2uforu,v∈P. By Lemma2.5we can deduce: there exist
u0,v0 ∈Ph andr∈(0, 1) such thatrv0≤u0≤v0, u0≤T1(u0,v0) +T2u0≤T1(v0,u0) +
T2v0≤v0; the operator equationT1(u,u) +T2u=uhas a unique solutionu∗∈Ph; and for any initial valuesx0,y0∈Ph, constructing successively the sequences
xn=T1(xn–1,yn–1) +T2xn–1, yn=T1(yn–1,xn–1) +T2yn–1, n= 1, 2, . . . ,
we getxn→u∗andyn→u∗asn→ ∞. We have the following two inequalities:
u0(t)≤ 1
0
G(t,qs)fs,u0(s),v0(s) +g
s,u0(s) dqs, t∈[0, 1],
v0(t)≥ 1
0
G(t,qs)fs,v0(s),u0(s) +g
s,v0(s) dqs, t∈[0, 1].
Thus problem (1.5) has a unique positive solutionu∗∈Ph; for anyu0,v0∈Ph, constructing
successively the sequences
xn+1(t) = 1
0
G(t,qs)fs,xn(s),yn(s) +g
s,xn(s) dqs, n= 0, 1, 2, . . . ,
yn+1(t) = 1
0
G(t,qs)fs,yn(s),xn(s) +gs,yn(s) dqs, n= 0, 1, 2, . . . ,
we havexn–u∗ →0 andyn–u∗ →0 asn→ ∞.
Corollary 3.1 Suppose that f satisfies the conditions of Theorem 3.1 and g ≡ 0,
f(t, 0, 1)≡0.Then:
(i) there existu0,v0∈Phandr∈(0, 1)such thatrv0≤u0<v0,and
u0(t)≤ 1
0
G(t,qs)fs,u0(s),v0(s) dqs, t∈[0, 1],
v0(t)≥ 1
0
G(t,qs)fs,v0(s),u0(s) dqs, t∈[0, 1],
(ii) the BVP ⎧ ⎨ ⎩
Dα
qu(t) +f(t,u(t),u(t)) = 0, 0 <t< 1, 2 <α≤3,
u(0) =Dqu(0) = 0, u(1) =μ
1 0 u(s)dqs,
(3.10)
has a unique positive solutionu∗inPh;
(iii) for anyx0,y0∈Ph,the sequences
xn+1= 1
0
G(t,qs)fs,xn(s),yn(s) dqs, n= 0, 1, 2, . . . ,
yn+1= 1
0
G(t,qs)fs,yn(s),xn(s) dqs, n= 0, 1, 2, . . . ,
satisfyxn–u∗ →0andyn–u∗ →0asn→ ∞.
Theorem 3.2 Suppose that(F1)–(F2)hold.In addition,suppose that f,g satisfy the
fol-lowing conditions:
(F5) there exists a constant γ ∈(0, 1) such thatg(t,λu)≥λγg(t,u) for anyt∈[0, 1], λ∈(0, 1),u∈[0, +∞),andf(t,λu,λ–1v)≥λf(t,u,v)forλ∈(0, 1),t∈[0, 1],u,v∈
[0, +∞);
(F6) f(t, 0, 1)≡0 fort∈[0, 1],and there exists a constantδ0> 0such that f(t,u,v)≤ δ0g(t,u),t∈[0, 1],u,v≥0.
Then:
(1) there existu0,v0∈Phandr∈(0, 1)such thatrv0≤u0<v0and
u0≤ 1
0
G(t,qs)fs,u0(s),v0(s) +g
s,u0(s) dqs, t∈[0, 1],
v0≥ 1
0
G(t,qs)fs,v0(s),u0(s) +g
s,v0(s) dqs, t∈[0, 1],
whereG(t,qs)is defined by(2.3),andh(t) =tα–1,t∈[0, 1];
(2) the boundary value problem(1.5)has a unique positive solutionu∗inPh;and for any x0,y0∈Ph,the sequences
xn+1= 1
0
G(t,qs)fs,xn(s),yn(s) +gs,xn(s) dqs, n= 0, 1, 2, . . . ,
yn+1= 1
0
G(t,qs)fs,yn(s),xn(s) +g
s,yn(s) dqs, n= 0, 1, 2, . . . ,
satisfyxn–u∗ →0andyn–u∗ →0asn→ ∞.
Proof Similarly to the proof of Theorem 3.1,T1andT2are given in (3.2). From (F1) and
(F2) we know thatT1:P×P→Pis a mixed monotone operator andT2:P→Pis
increas-ing. By (F5) we obtain
T1
The boundary value problem (1.5) has a unique positive solutionu∗∈Ph; foru0,v0∈Ph,
Now, we give two examples to illustrate our results.
Example4.1 Consider the following boundary value problem:
It is not difficult to find thatf(t,x,y) : [0, 1]×[0, +∞)×[0, +∞)→[0, +∞) is continuous, increasing with respect to the second variable, and decreasing with respect to the third variable and thatg(t,x) : [0, 1]×[0, +∞)→[0, +∞) is continuous withg(t, 0) = 2 > 0 and increasing with respect to the second variable. We also have
g(t,λu) = λu
Sofandgsatisfy the conditions of Theorem3.1. Thus by Theorem3.1the boundary value problem (4.1) has a unique positive solution inPh, whereh(t) =tα–1=t32,t∈[0, 1].
Example4.2 Consider the following boundary value problem:
If we takeδ0= 1 > 0, then we have
f(t,u,v) =
u 1 +u
1 4
+ [v+ 1]–13 + +t3≤u14+t2+ 1≤u13 +t2+ 1 =δ0g(u,t).
Sofandgsatisfy the conditions of Theorem3.2. Thus by Theorem3.2the boundary value problem (4.2) has a unique positive solution inPh, whereh(t) =tα–1=t32,t∈[0, 1].
Acknowledgements
The authors are very grateful to the reviewers for their valuable suggestions and useful comments, which led to an improvement of this paper.
Funding
This project was supported by the National Natural Science Foundation of China (Grant No. 11271235).
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
FG carried out the molecular genetic studies, participated in the sequence alignment, and drafted the manuscript. SK conceived the study and participated in its design and coordination. FC helped to draft the manuscript. All authors read and approved the final manuscript.
Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
Received: 23 April 2018 Accepted: 4 September 2018
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