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IJNSNS 2017; aop
Jing Ren and Chengbo Zhai∗
A Fractional q-difference Equation with Integral
Boundary Conditions and Comparison Theorem
https://doi.org/10.1515/ijnsns-2017-0056
Abstract: In this article, we mainly prove the existence
of extremal solutions for a fractional q-difference equation involving Riemann–Lioville type fractional derivative
with integral boundary conditions. A comparison theorem
under weak conditions is also build, and then we apply
the comparison theorem, monotone iterative technique
and lower–upper solution method to prove the existence
of extremal solutions. Moreover, we can construct two iterative schemes approximating the extremal solutions of
the fractional q-difference equation with integral boundary conditions. In the last section, a simple example is
presented to illustrate the main result.
Keywords: fractional q-difference equation, comparison
theorem, integral boundary conditions, extremal solutions, iterative method
MSC® (2010). 34B18, 33D05
1 Introduction
We will devote to consider the following nonlinear fractional q-difference equation with integral boundary condition
D!q u(t) + h(t)f (t, u(t)) = 0, t ∈ (0, 1),
1
j
Dq u(0) = 0, 0 ≤ j ≤ n – 2, u(1) = , 0 g(s)u(s)dq s,
(1)
where 0 < q < 1, D!q denotes the Riemann–Lioville type
fractional q-derivative of order !, ! ∈ (n – 1, n] and n ≥ 3
is an integer, , > 0 is a constant, the functions g, h, f are
continuous.
Over the last 30 years, fractional differential equations have been of great interest due to the numerous applications in various disciplines, one can see [1–
24] and references therein. As we all know, fractional
*Corresponding author: Chengbo Zhai, School of Mathematical
Sciences, Shanxi University, Taiyuan 030006, Shanxi, P.R. China,
E-mail: cbzhai@sxu.edu.cn
Jing Ren, School of Mathematical Sciences, Shanxi University,
Taiyuan 030006, Shanxi, P.R. China, E-mail: 1974155068@qq.com
calculus play an important role in several fields and
bring deep influence on solving boundary value problems of q-calculus, which allowed to deal with continuous but not necessarily smooth functions. Following some initial works [1–18], several techniques have
been developed to prove the existence of solutions for
fractional q-difference equation boundary value problems. Now people pay more attention to the practicability
of mathematical tools such as Banach fixed-point theorem, Krasnoselskii’s fixed-point theorem, the monotone
iterative method, Leray–Schauder alternative, the lower–
upper solution method, etc. Recently, nonlinear fractional
q-difference equation boundary value problems have been
addressed more extensively, and a variety of new results
related to the existence and uniqueness of the solutions
were obtained based on a great diversity of methods, see
[8–11, 24] and references therein. For instance, Liang and
Zhang [8] investigated the existence and uniqueness of
positive solutions for three-point boundary value problems of nonlinear fractional q-difference equations by
means of Krasnoselskii’s fixed-point theorem on cones.
In [9], Miao and Liang obtained the uniqueness of positive solutions for fractional q-difference equation boundary value problem with p-Laplacian operator by using
a fixed-point theorem in partially ordered set. Especially, many researchers can flexibly combine different
methods to achieve the goal. Yang [17] obtained the
existence of positive solution for q-fractional boundary
value problems with I-Laplacian operator via the lower–
upper solution method combined with the Schauder fixedpoint theorem. By using the Leray–Schauder alternative combined with Krasnoselskii’s fixed-point theorem,
Graef and Kong [13] discussed the existence of positive
solutions for fractional q-derivatives equation boundary
value problem. In [10], by applying monotone iterative method and some inequalities associated with the
Green’s function, Li and Yang discussed the existence
of positive solutions for nonlinear fractional q-difference
equation. Zhao et al. [18] discussed the existence of positive solutions for nonlocal q-integral boundary value
problems of fractional q-difference equations by using
the generalized Banach contraction principle, the monotone iterative method and Krasnoselskii’s fixed-point
theorem.
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2
J. Ren and C. Zhai: A Fractional q-difference Equation
Furthermore, the method of lower–upper solution
associated with the monotone iterative technique is
regarded as an excellent tool for dealing with boundary
value problems of nonlinear difference equations. In this
way, Wang et al. [15] obtained the existence of extremal
solutions for a class of nonlinear fractional differential
equation. Wang and Zhang [12] investigated the existence
of extremal solutions for Caputo type q-fractional initial
value problems. Motivated by the works mentioned above,
we will discuss the existence of extremal solutions for the
problem (1). Different from the above works mentioned, in
this paper, we first establish a comparison theorem and
then use monotone iterative technique coupled with the
lower–upper solution method to prove the existence of
extremal solutions. In Section 2, we shall introduce some
definitions and lemmas to prove our main results. In Section 3, the main theorems are proved. At last, an example
is given to illustrate the main result.
f (qt)–f (t)
,
(q–1)t
′
f (0),
(Iq! Dq f )(t) = (Dq Iq! f )(t)–
p
p
n=0
Definition 2.2: (See [27]). The fractional q-integral of
Riemann–Liouville type of order ! ≥ 0 is defined by
(Iq! f )(t) =
0
∞
f (tqk )tqk , t ∈ [0, b]. (2)
k=0
Similar to the q-derivative, an operator Iqn is given by
(Iq0 f )(t)
= f (t),
(Iqn f )(t)
=
Iq (Iqn–1 f )(t),
t
(t – qs)(!–1) f (s)dq s, ! > 0, t ∈ [0, 1],
0
(1 – q)(!–1)
, Aq (! + 1) = [!]q Aq (!), q ∈ (0, 1),
(1 – q)!–1
here
1 – q!
, (a – b)(0) = 1 and
1–q
∞
1 – (b/a)qn
(a – b)(!) = a!
, a, b, ! ∈ R.
1 – (b/a)qn+!
n=0
[!]q =
0
f (s)dq s = (1 – q)
1
Aq (!)
Aq (+ + 1)
(t – a)(!++) , 0 < a < t.
Aq (! + + + 1)
In particular, for + = 0, a = 0, we have Iq! (1)(t) =
conclusion, we obtain
The q-integral of a function f in the interval [0, b] is
defined by
t
t!–p+n
(Dn f )(0), p ∈ N.
Aq (! – p + n + 1) q
(3)
Iq! (t – a)(+) =
(D0q f )(t) = f (t), (Dnq f )(t) = Dq (Dn–1
q f )(t), n ∈ N.
(Iq f )(t) =
p–1
Remark 2.1: (See [25]). For + ∈ (–1, ∞) and ! ≥ 0, the
following equality holds
t ≠ 0,
t = 0.
And the q-derivatives of higher order by
f )(t), ! > 0, t ∈ [0, 1],
where ! is the smallest integer greater than or equal to
!.
As we all know, (D!q f )(t) = Dq f (t), when ! = 1. Further
analysis showed that
Aq (!) =
For the convenience of the readers, we list some known
background materials of fractional q-calculus. For details,
the readers can find them in [25, 26] and references
therein.
The q-derivative (0 < q < 1) of a function f is defined
by
(Dq f )(t) =
! !–!
(D!q f )(t) = (Dq Iq
where
2 Preliminaries and previous
results
Definition 2.1: (See [27]). The fractional q-derivative of
Riemann–Liouville type of order ! ≥ 0 is defined by
t
(t – qs)(!–1) dq s = Aq (!)Iq! (1)(t) =
t!
.
Aq (!+1)
1 !
t .
[!]q
In
(4)
Lemma 2.1: Assume that t ≥ b ≥ a, if ! ≥ 0, then we have
(t – a)(!) ≥ (t – b)(!) , further, we have (t – a)(–!) ≤ (t – b)(–!) ,
for a, b ∈ R.
Proof: We only prove the last inequality. If (t – a)(–!) ≤
(t – b)(–!) is not true, then there exist a1 , b1 ∈ R with a1 ≤ b1
such that
n ∈ N.
The fundamental theorem of calculus applies to Iq and Dq ,
(Dq Iq f )(t) = f (t). Suppose that f is continuous at t = 0, then
(Iq Dq f )(t) = f (t) – f (0).
t–!
∞
∞
t – a1 qk
t – b1 qk
–!
>
t
, 0 < q < 1.
k–!
t – a1 q
t – b1 qk–!
k=0
k=0
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J. Ren and C. Zhai: A Fractional q-difference Equation
Then, for k ∈ N, we calculate
1
H(t, s) =
Aq (!)
3
t!–1 (1 – s)(!–1) – (t – s)(!–1) , 0 ≤ s ≤ t ≤ 1,
t!–1 (1 – s)(!–1) , 0 ≤ t ≤ s ≤ 1,
(t – a1 qk )(t – b1 qk–! ) > (t – b1 qk )(t – a1 qk–! )
here
a1 tqk–! – b1 tqk–! – a1 tqk + b1 tqk > 0
⇒
⇒
(a1 t – b1
t)q–!
– (a1 t – b1 t) > 0
Next we give some properties on q-integral which will
be always used in the proofs of main results.
Lemma 2.2: (See [24]). If f , g are q-integral on the interval
[0, s] and f (t) ≤ g(t), for all t ∈ [0, s], then we have
s
s
(i)
0 f (t)dq t ≤ 0 g(t)dq t. Consequently, if ! > 1, we
have I ! f (s) ≤ I ! g(s), t ∈ [0, s];
s q qs
(ii) 0 f (t)dq t ≤ 0 |f (t)|dq t, for t ∈ [0, s];
(iii) If f is q-integral on the closed interval [a, b], one
has that
b
f (t)dq t =
a
c
f (t)dq t +
a
b
b
f (t)dq t, for c ∈ [a, b].
c
b
+f (t)dq t = +
a
f (t)dq t, + ∈ R.
a
If ! ≥ 0 and am → a as m → ∞, then we have
m→∞
(am – b)(!) → (a – b)(!) , a, b ∈ R.
(iv)
Throughout this paper, we always assume that , >
0, n ≥ 3, ! ∈ (n – 1, n] and
1
,
s!–1 g(s)dq s < 1, 0 <
0
1
Lemma 2.4: (See [10]). The function G(t, s) defined by
eq. (5) satisfies the following inequality
G(t, qs) ≤
t!–1 (1 – qs)(!–1)
, t, s ∈ [0, 1].
(1 – 3)Aq (!)
In the following, we give a comparison theorem which will
be very important for our proof of main results.
Lemma 2.5: (Comparison principle) If w ∈ C([0, 1], R) and
satisfies the following relations
D!q w(t) ≥ –+w(t), t ∈ [0, 1],
j
Dq w(0) = 0, 0 ≤ j ≤ n – 2, w(1) ≥ 0,
where + > –Aq (! + 1) is a constant, then w(t) ≥ 0 for all
t ∈ [0, 1].
Proof: Suppose w(t) ≥ 0 is not true, then there exist t0 , t0 ′ ∈
[0, 1) such that w(t0 ) = 0, w(t0 ′ ) < 0. In addition, if t0 > t0 ′ ,
then w(t) < 0 for t ∈ [t0 ′ , t0 ). Further, let t1 be the first
minimal point on [0, t0 ], then t1 < t0 , we will show two
cases + > 0 and + ≤ 0.
◻
Case 1: If + > 0, w(t) < 0 for t ∈ [t0 ′ , t0 ), then we have
w(t) – w(1) ≤ 0, it follows from the inequality eq. (6) that
D!q [w(t) – w(1)] ≥ –+[w(t) – w(1)] ≥ 0,
D!q w(t) – D!q w(1) ≥ 0 ⇐⇒ Iq! Dnq Iqn–! w(t) – Iq! Dnq Iqn–! w(1) ≥ 0
⇐⇒ w(t) –
D!q u(t) + x(t) = 0, t ∈ (0, 1),
1
j
Dq u(0) = 0, 0 ≤ j ≤ n – 2, u(1) = , 0 g(s)u(s)dq s,
+
has an unique solution given by
⇐⇒ w(t) +
1
u(t) =
G(t, qs)x(s)dq s,
where
,t!–1
1–3
0
1
H(4, s)g(4)dq 4,
Dn–1
q w(0)
Aq (!)
Dn–1
q w(0)
Aq (!)
Dn–1
q w(0)
Aq (!)
t!–1 – w(1)
≥0
(1 – t!–1 ) ≥ w(1).
Since w(1) ≥ 0, then we have Dn–1
q w(0) ≥ 0. Further, in view
of the inequality eq. (6), we have D!q w(t) ≥ 0. In the same
way, one has
0
G(t, s) = H(t, s) +
(6)
thus according to the Definition 2.1 and eqs. (3), (6), one
can see that
(1 – qs)(!–1) h(s)dq s < ∞.
0
Lemma 2.3: (See [10]). For any x ∈ [0, 1], then the boundary value problem
s!–1 g(s)dq s.
0
Since (b1 t – a1 t) > 0 and 1 – q–! ≤ 0. This contradicts the
abovementioned, then we complete the proof.
◻
1
3=,
(b1 t – a1 t)(1 – q–! ) > 0.
⇒
(5)
w(t) ≥
Dn–1
q w(0)
Aq (!)
t!–1 +
Dn–2
q w(0)
Aq (!)
t!–2 + . . . +
w(0) !–n
t
≥ 0,
Aq (!)
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4
J. Ren and C. Zhai: A Fractional q-difference Equation
which contradicts with w(t) < 0, t ∈ [t0 ′ , t0 ). Hence, for
+ > 0, the result is valid.
where + > – h10 Aq (! + 1) is a constant, then v(t) ≥ 0 for all
t ∈ [0, 1].
Case 2: If 0 ≥ + > –Aq (! + 1), in view of the monotonicity of
the Riemann–Liouville fractional q-integral Iq! , we apply
the Iq! on both sides of problem (6). From Definitions 2.1
and 2.2, we conclude
Proof: Situation like Lemma 2.5, we still discuss two cases
+ > 0 and + ≤ 0.
Case 1: If + > 0, v(t) < 0 for t ∈ [t0 ′ , t0 ), it follows from the
inequality eq. (10) that D!q v(t) ≥ 0; thus by using eqs. (3),
(10) and Definition 2.1, one has v(t) ≥ 0, this contradicts
with v(t) < 0, t ∈ [t0 ′ , t0 ). Hence, for + > 0, the result is
valid.
–1 !–1
w(t) – Dn–1
+ +Iq! w(t) ≥ 0.
q w(0)Aq (!) t
Let t = t1 , one can see that
w(t1 ) +
+Iq! w(t1 )
≥ 0.
(7)
Hence, from Remark 2.1 and t1 < t0 , we can obtain that
Iq! w(t1 ) =
1
Aq (!)
1
=
Aq (!)
–
1
Aq (!)
1
≥
Aq (!)
1
≥
Aq (!)
t1
Case 2: If 0 ≥ + > – h10 Aq (! + 1), in view of the monotonicity of the Riemann–Liouville fractional q-integral Iq! ,
we apply the Iq! on both sides of problem (10). From
Definitions 2.1 and 2.2, we conclude
–1
!–1
v(t) – Dn–1
+ +h0 Iq! v(t) ≥ 0.
q v(0)Aq (!)t
(t1 – qs)(!–1) w(s)dq s
0
t0
Let t = t1 , one can see that
(t1 – qs)(!–1) w(s)dq s
0
t0
t1
t0
v(t1 ) + +h0 Iq! v(t1 ) ≥ 0.
(t1 – qs)(!–1) w(s)dq s
Hence, from the Remark 2.1 and the proof of Lemma 2.5,
1)
, that is
we have Iq! v(t1 ) ≥ Aqv(t
(!+1)
(t1 – qs)(!–1) w(s)dq s
0
t0
(11)
(t1 – qs)(!–1) w(t1 )dq s
0
t0
t0!
1
≥
(t0 – qs)(!–1) w(t1 )dq s = w(t1 )
Aq (!) 0
Aq (! + 1)
w(t1 )
≥
.
Aq (! + 1)
+h0 Iq! v(t1 ) ≤
+h0 v(t1 )
.
Aq (! + 1)
(12)
Thus, in view of eqs. (11) and (12), we have
+h0
v(t1 ) ≥ 0.
1+
Aq (! + 1)
(13)
That is,
+w(t1 )
+Iq! w(t1 ) ≤
.
Aq (! + 1)
(8)
Since v(t1 ) < 0 and + > – h10 Aq (! + 1), this contradicts
eq. (13). Therefore, for 0 ≥ + > – h10 Aq (! + 1), the result is
valid. By using the same argument, if t0 < t0 ′ , we can draw
the same conclusion.
◻
(9)
3 Main results
Thus, in view of eqs. (7) and (8), we have
1+
+
w(t1 ) ≥ 0.
Aq (! + 1)
Since w(t1 ) < 0 and + > –Aq (! + 1), this contradicts the
eq. (9). Therefore, for 0 ≥ + > –Aq (! + 1), the result is valid.
By using the same argument, if t0 < t0 ′ , we can draw the
same conclusion.
In this section, we will work in the Banach space X =
C[0, 1] with the norm u = max{|u(t)| : t ∈ [0, 1]}. We
define the standard cone P ⊂ X by
P = {u ∈ C[0, 1] : u(t) ≥ 0, t ∈ [0, 1]}.
Lemma 2.6: Let h0 > 0 and v ∈ C([0, 1], R) which satisfy
the following relations
D!q v(t) ≥ –+h0 v(t), t ∈ [0, 1],
j
Dq v(0) = 0, 0 ≤ j ≤ n – 2, v(1) ≥ 0,
(10)
Definition 3.1: A pair of functions v0 and w0 in C([0, 1], R)
are called ordered lower and upper solutions of the nonlinear BVP(1) if
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J. Ren and C. Zhai: A Fractional q-difference Equation
D!q v0 (t) + h(t)f (t, v0 (t)) ≤ 0, t ∈ (0, 1),
1
j
Dq v0 (0) = 0, 0 ≤ j ≤ n – 2, v0 (1) ≤ , 0 g(s)v0 (s)dq s,
Moreover, we obtain
(14)
and
D!q w0 (t) + h(t)f (t, w0 (t)) ≥ 0, t ∈ (0, 1),
1
j
Dq w0 (0) = 0, 0 ≤ j ≤ n – 2, w0 (1) ≥ , 0 g(s)w0 (s)dq s.
(15)
In the following, we present some existence theorems of
extremal solutions for problem (1), and three conditions
are listed below:
(H1 )
(H2 )
(H3 )
5
Assume that v0 , w0 ∈ C([0, 1], R) are lower and
upper solutions of eq. (1), respectively, and v0 ≤ w0 ;
h(t) > 0, t ∈ [0, 1];
The function f ∈ C([0, 1] × R, R) satisfies the condition
D!q v0 (t) ≤ –h(t)f (t, v0 (t)) ≤ Ir (t) – +v0 (t),
1
j
Dq v0 (0) = 0, 0 ≤ j ≤ n – 2, v0 (1) ≤ , 0 g(s)v0 (s)dq s,
and
D!q w0 (t) ≥ –h(t)f (t, w0 (t)) ≥ Ir (t) – +w0 (t),
1
j
Dq w0 (0) = 0, 0 ≤ j ≤ n – 2, w0 (1) ≥ , 0 g(s)w0 (s)dq s.
Hence v0 and w0 are lower and upper solutions of the
problem (16).
Step 2: We show the uniqueness of the solutions for problem (16).
By Lemma 2.3, the solution u(t) of the problem (16) can be
expressed as
1
G(t, qs)(+u(s) – Ir (s))dq s, ∀ t ∈ [0, 1],
u(t) =
0
f (t, v)–f (t, u) ≤
+
(v–u), v0 ≤ u ≤ v ≤ w0 , t ∈ [0, 1], where G(t, qs) is given as in eq. (5). The operator A :
h(t)
C([0, 1], R) → C([0, 1], R) is defined by
where 0 ≤ + < Aq (! + 1) is a constant.
1
Au(t) =
G(t, qs)(+u(s) – Ir (s))dq s, ∀t ∈ [0, 1].
0
Theorem 3.1: Assume that (H1 ) ∼ (H3 ) are satisfied. Then,
the problem (1) has extremal solutions in [v0 , w0 ]. And
there exist two monotone iterative sequences {vn } and {wn }
which converge uniformly to the extremal solutions.
Proof: First, we investigate the following problem
D!q u(t) = –+u(t) + Ir (t), t ∈ (0, 1),
1
(16)
j
Dq u(0) = 0, 0 ≤ j ≤ n – 2, u(1) = , 0 g(s)u(s)dq s,
where + is given in (H3 ) and Ir (t) = +r(t) – h(t)f (t, r(t)), r ∈
[v0 , w0 ]. We prove the uniqueness of solutions for eq. (16).
This part consists of three steps.
Step 1: By using the conditions (H1 ) and (H3 ), one can see
that
+
(r(t) – v0 (t))
f (t, r(t)) – f (t, v0 (t)) ≤
h(t)
Ir (t)
+v0 (t)
=
+ f (t, r(t)) –
,
h(t)
h(t)
+
(r(t) – w0 (t))
h(t)
Ir (t)
+w0 (t)
=
+ f (t, r(t)) –
.
h(t)
h(t)
f (t, r(t)) – f (t, w0 (t)) ≥
We know that u is the solution of the problem (16) if and
only if u is the fixed point of A. By using Remark 2.1 and
Lemma 2.4, we have
1
G(t, qs)(u(s) – v(s))dq s
Au(t) – Av(t) = max +
t∈[0,1]
0
1 !–1
t (1 – qs)(!–1)
≤+
|u – v|dq s
(1 – 3)Aq (!)
0
+
≤
u – vC ,
Aq (! + 1)
where 3 = ,
1
0
s!–1 g(s)dq s < 1. By the range of +, we obtain
Au(t) – Av(t)C < u – vC .
Therefore, the operator A is a contraction. By using
Banach fixed-point theorem, it follows that A has a
unique fixed point, that is, the problem (16) has a unique
solution u.
Step 3: We show u ∈ [v0 , w0 ]. Through the abovementioned, we let z = u – v0 , by using eqs. (14), (15), and (16),
we get
D!q z(t) ≥ –+z(t), t ∈ [0, 1],
j
Dq z(0) = 0, 0 ≤ j ≤ n – 2, z(1) ≥ 0.
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6
J. Ren and C. Zhai: A Fractional q-difference Equation
In view of Lemma 2.5, z(t) ≥ 0 for t ∈ [0, 1], that is u ≥ v0 .
Reversing the argument, we similarly let z̄ = w0 – u, then
we have z̄(t) ≥ 0. Thus, we have v0 ≤ u ≤ w0 .
1
!–1
!–1
H(t1 , qs) – H(t2 , qs) + ,(t1 – t2 )
≤
1–3
0
1
H(4, s)g(4)dq 4 |+vn–1 (s) – Ir (s)|dq s
Second, we prove that there exist two monotone iterative sequences {vn } and {wn }, which satisfy
1
≤
Aq (!)
0
lim vn = v∗ , lim wn = w∗ ,
n→∞
n→∞
where v∗ , w∗ are the extremal solutions of the problem
(1). This part consists of two steps. We define an operator
A : [v0 , w0 ] → [v0 , w0 ] by Ar = u, where u is the solution of
eq. (16) for fixed r ∈ [v0 , w0 ].
Step 1: We show the monotone property of the operator
A. In fact, let 9 = v2 – v1 and v2 = Ar2 , v1 = Ar1 , for r1 , r2 ∈
[v0 , w0 ] with r1 ≤ r2 . From the condition (H3 ), we have
D!q 9(t) = D!q v2 (t) – D!q v1 (t) = –+v2 (t) + Ir2 (t) – +v1 (t) + Ir1 (t)
= –h(t)(f (t, r2 (t)) – f (t, r1 (t))) – +(v2 (t)
– v1 (t)) + +(r2 (t) – r1 (t))
≥ –+(r2 (t) – r1 (t)) – +(v2 (t) – v1 (t)) + +(r2 (t) – r1 (t))
t1 !–1 !–1
(t1 –t2 )(1–qs)(!–1) +(t2 –qs)(!–1)
0
– (t1 – qs)(!–1) |+vn–1 (s) – Ir (s)|dq s
t2 1
!–1 !–1
+
(t1 – t2 )(1 – qs)(!–1) Aq (!) t1
+ (t2 – t1 )(!–1) |+vn–1 (s) – Ir (s)|dq s
1
1
!–1 !–1
+
(t1 – t2 )(1 – qs)(!–1) Aq (!) t2
|+vn–1 (s) – Ir (s)|dq s
1
,(t1 !–1 – t2 !–1 ) 1
+
H(4, s)g(4)dq 4
1
–
3
0
0
|+vn–1 (s) – Ir (s)|dq s.
In view of Lemma 2.2 (iv), one has vn (t1 ) → vn (t2 ), as t1 →
t2 . Similarly, the sequence {wn } is also equi-continuous on
[0, 1]. By means of the Arzela–Ascoli theorem, we have
lim vn = v∗ , lim wn = w∗ .
= –+9(t),
n→∞
Finally, we show that v∗ , w∗ are the extremal solutions
of the problem (1). Let u is a solution of problem (1) with
u ∈ [v0 , w0 ], then Au = u. Since A is nondecreasing, we
have
and
j
Dq 9(0) = 0, 0 ≤ j ≤ n – 2, 9(1) ≥ 0.
By Lemma 2.5, we have 9(t) ≥ 0 for t ∈ [0, 1]. That is
v2 ≥ v1 . Hence the operator A is nondecreasing.
Step 2: We prove that {vn } and {wn } converge uniformly to
some elements v∗ and w∗ . Let vn = Avn–1 , wn = Awn–1 , n =
1, 2, . . ., then we have
v0 ≤ v1 ≤ ⋯ ≤ vn ≤ ⋯ ≤ wn ≤ ⋯ ≤ w1 ≤ w0 .
n→∞
vn ≤ u ≤ wn , n = 1, 2, . . . .
Let n → ∞, we obtain that v∗ ≤ u ≤ w∗ . Furthermore,
in view of eq. (17), we know v∗ and w∗ are the extremal
solutions and satisfy the following relation
v0 ≤ v1 ≤ ⋯ ≤ vn ≤ v∗ ≤ u ≤ w∗ ≤ wn ≤ ⋯ ≤ w2 ≤ w1 ≤ w0 .
(17)
The proof is completed.
According to the normality of cone P, we know the
sequences {vn } and {wn } are uniformly bounded. Furthermore, for each u ∈ C[0, 1], 0 ≤ t1 < t2 ≤ 1, there
exists $ sufficiently small, such that |t1 – t2 | < $, one has
|vn (t1 ) – vn (t2 )| < %, that is to say {vn } is equi-continuous on
[0, 1]. In fact
|vn (t1 ) – vn (t2 )| = |Avn–1 (t1 ) – Avn–1 (t2 )|
1
G(t1 , qs)(+vn–1 (s) – Ir (s))dq s
= 0
1
G(t2 , qs)(+vn–1 (s) – Ir (s))dq s
–
0
◻
Next, other condition can be listed as follows:
(H3 )′ Let h0 := max{h(t) : t ∈ [0, 1]} and the function f ∈
C([0, 1] × R, R) satisfies the condition
f (t, v) – f (t, u) ≤ +(v – u), v0 ≤ u ≤ v ≤ w0 , t ∈ [0, 1],
where 0 ≤ + <
Aq (!+1)
h0
is a constant.
Corollary 3.1: Assume that (H1 ), (H2 ), and (H3 )′ are satisfied. Then, problem (1) have extremal solutions in [v0 , w0 ].
And there exist two monotone iterative sequences {vn } and
{wn } which converge uniformly to the extremal solutions.
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7
J. Ren and C. Zhai: A Fractional q-difference Equation
Proof: First, we investigate the following problem
Lemma 2.4, we have
D!q u(t) = –+h0 u(t) + Ir (t), t ∈ (0, 1),
1
(18)
j
Dq u(0) = 0, 0 ≤ j ≤ n – 2, u(1) = , 0 g(s)u(s)dq s,
where + is given in (H3 )′ and Ir (t) = +h0 r(t)–h(t)f (t, r(t)), r ∈
[v0 , w0 ]. We prove the uniqueness of solutions for eq. (18).
This part consists of three steps.
Step 1. By using the conditions (H1 ) and (H3 )′ , one can see
that
where 3 = ,
we obtain
0
s!–1 g(s)dq s < 1. From the range of + in (H3 )′ ,
Therefore, the operator A is a contraction. By using
Banach fixed-point theorem, it follows that A has a unique
fixed point, that is, the problem (18) has a unique solution
u.
Step 3. Similar to the proof of the Theorem 3.1, by means
of Lemma 2.6, we have v0 ≤ u ≤ w0 .
The rest proof is similar to the proof of Theorem
3.1, we conclude that there exist two monotone iterative sequences {vn } and {wn }, which satisfy lim vn =
+
h0 (r(t) – w0 (t))
h(t)
+h0
Ir (t)
+ f (t, r(t)) –
w0 (t).
=
h(t)
h(t)
f (t, r(t)) – f (t, w0 (t)) ≥
n→∞
Moreover, we obtain
v∗ , lim wn = w∗ , where v∗ , w∗ are the extremal solutions
n→∞
◻
of the problem (1).
D!q v0 (t) ≤ –h(t)f (t, v0 (t)) ≤ Ir (t) – +h0 v0 (t),
1
j
Dq v0 (0) = 0, 0 ≤ j ≤ n – 2, v0 (1) ≤ , 0 g(s)v0 (s)dq s,
4 An example
and
1
Au(t) – Av(t)C < u – vC ,
+
h0 (r(t) – v0 (t))
h(t)
Ir (t)
+h0
=
+ f (t, r(t)) –
v0 (t),
h(t)
h(t)
f (t, r(t)) – f (t, v0 (t)) ≤
1 !–1
t (1
– qs)(!–1)
|u – v|dq s
(1 – 3)Aq (!)
0
+h0
u – vC .
≤ +h0 Ia! (1)(1)u – vC =
Aq (! + 1)
Au(t) – Av(t) ≤ +h0
D!q w0 (t) ≥ –h(t)f (t, w0 (t)) ≥ Ir (t) – +h0 w0 (t),
1
j
Dq w0 (0) = 0, 0 ≤ j ≤ n – 2, w0 (1) ≥ , 0 g(s)w0 (s)dq s.
In this section, we give an example to illustrate our main
result.
Hence v0 and w0 are lower and upper solutions of the
problem (18).
Step 2. We prove the uniqueness of the solutions for the
problem (18).
By Lemma 2.3, the solution u(t) of the problem (18) can be
expressed as
Example 4.1: Consider the following boundary value
problem:
1
u(t) =
(u–3t –1)5 . Let v0 (t) = 0, w0 (t) =
f (t, u) = 2 2 q 2 2
6(t sin t+2t +sin t+2)
3t + 4, by simple computation, we have
–h(t)f (t, v0 (t)) =
1
(19)
81A (!+1)
where G(t, qs) is given as in eq. (5). The operator A :
C([0, 1], R) → C([0, 1], R) is defined by
Au(t) =
81A (!+1)
D!q u(t) + 6tq2 +12 (u(t) – 3t – 1)5 = 0, t ∈ (0, 1),
1
j
Dq u(0) = 0, 0 ≤ j ≤ 1, u(1) = , 0 t2 u(s)dq s,
where ! = 52 , q = 21 , , = 1, h(t) = sin2 t + 2, g(t) = t2 , and
G(t, qs)(+h0 u(s) – Ir (s))dq s, ∀ t ∈ [0, 1],
0
81Aq (! + 1)
(3t + 1)5 ≥ D!q v0 (t),
6(t2 + 1)
G(t, qs)(+h0 u(s) – Ir (s))dq s, ∀ t ∈ [0, 1].
0
We know that u is the solution of the problem (18) if and
only if u is the fixed point of A. By using Remark 2.1 and
–h(t)f (t, w0 (t)) = –
812 Aq (! + 1)
≤ D!q w0 (t).
2(t2 + 1)
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8
J. Ren and C. Zhai: A Fractional q-difference Equation
Then, condition (H1 ) is satisfied and v0 ≤ w0 . Moreover, for
v0 ≤ u ≤ v ≤ w0 , we have
81Aq (! + 1)
[(v – 3t – 1)5
+ 2t2 + sin2 t + 2)
–(u – 3t – 1)5 ]
5Aq (! + 1)
(v – u)
≤
2
2
6(t sin t + 2t2 + sin2 t + 2)
5Aq (! + 1)
≤
(v – u)
6(sin2 t + 2)
+
(v – u).
≤
h(t)
f (t, v) – f (t, u) =
6(t2 sin2 t
5A (!+1)
Here + = q 6 . Evidently, + < Aq (!+1). Therefore, all conditions of Theorem 3.1 hold. From Theorem 3.1, we know
that eq. (19) has extremal solutions in [v0 , w0 ]; further
there exist monotone iterative sequences {vn } and {wn }
converge uniformly to the extremal solutions.
5 Conclusion
In this article, we investigate the existence of extremal
solutions for the fractional q-difference equation with
integral boundary condition (1). The main methods of
this paper are the monotone iterative technique and the
lower–upper solution method. By constructing a comparison theorem and using this method, we establish some
new existence criteria for the q-fractional boundary value
problem (1), which give some alternative answers to the
similar type of fractional q-difference equation boundary
value problem discussed in [10]. As a matter of fact, the
results on fractional q-difference equation boundary value
problem obtained by applying this method are very few in
literature. Finally, an interesting example is presented to
illustrate the main result.
Funding: This paper was supported financially by
the Youth Science Foundation of China (11201272), Shanxi
Province Science Foundation (2015011005) and 131 Talents
Project of Shanxi Province (2015).
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