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. Brought to you by | National University of Singapore - NUS Libraries Authenticated Download Date | 10/28/17 12:24 PM 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 Brought to you by | National University of Singapore - NUS Libraries Authenticated Download Date | 10/28/17 12:24 PM 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 (!) Brought to you by | National University of Singapore - NUS Libraries Authenticated Download Date | 10/28/17 12:24 PM 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 Brought to you by | National University of Singapore - NUS Libraries Authenticated Download Date | 10/28/17 12:24 PM 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. Brought to you by | National University of Singapore - NUS Libraries Authenticated Download Date | 10/28/17 12:24 PM 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. Brought to you by | National University of Singapore - NUS Libraries Authenticated Download Date | 10/28/17 12:24 PM 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) Brought to you by | National University of Singapore - NUS Libraries Authenticated Download Date | 10/28/17 12:24 PM 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. 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