Positive solutions for a class of fractional boundary value problems

Nonlinear Analysis: Modelling and Control, Vol. 21, No. 1, 1 17 ISSN 1392-5113 http://dx.doi.org/1.15388/na.216.1.1 Positive solutions for a class of fractional boundary value problems Jiafa Xu a, Zhongli Wei b,c a School of Mathematical Science Chongqing Normal University, Chongqing 41331, China xujiafa292@sina.com b School of Mathematic Shandong University, Jinan 251, Shandong, China c Department of Mathematic Shandong Jianzhu University, Jinan 2511, Shandong, China jnwzl32@163.com Received: November 11, 213 / Revised: August 22, 214 / Published online: November 16, 215 Abstract. In this work, by virtue of the Krasnoselskii Zabreiko fixed point theorem, we investigate the existence of positive solutions for a class of fractional boundary value problems under some appropriate conditions concerning the first eigenvalue of the relevant linear operator. Moreover, we utilize the method of lower and upper solutions to discuss the unique positive solution when the nonlinear term grows sublinearly. Keywor: fractional boundary value problem, Krasnoselskii Zabreiko fixed point theorem, positive solution, uniqueness. 1 Introduction In this paper we consider the existence of positive solutions for the boundary value problem of fractional order involving Riemann Liouville s derivative D+D α +u α = f t, u, u, D+u α, t [, 1], u = u = u 1 = D+u α = D+ α+1 u = Dα+1 + u1 =, 1 where α 2, 3] is a real number, D α + is the standard Riemann Liouville fractional derivative of order α and f C[, 1] R 3 +, R + R + := [, +. Recently, the fractional differential calculus and fractional differential equation have drawn more and more attention due to the applications of such constructions in various This research was supported by the NNSF-China 11371117, Shandong Provincial Natural Science Foundation ZR213AM9. c Vilnius University, 216

2 J. Xu, Z. Wei sciences such as physic mechanic chemistry, engineering, etc. Many books on fractional calculu fractional differential equations have appeared, for instance, see [7,1,11]. This may explain the reason that the last two decades have witnessed an overgrowing interest in the research of such problem with many papers in this direction published. We refer the interested reader to [1, 2, 4, 5, 6, 12, 13, 14, 15] and the references therein. In [4,6], by using the fixed point index theory and Krein Rutman theorem, Jiang et al. obtained the existence of positive solutions for the multi-point boundary value problems of fractional differential equations and D α +ut + f t, ut =, < t < 1, 1 < α 2, m 2 u =, D β + u1 = i=1 a i D β + uξ i, D α u Mu = λf t, ut, t [, 1], < α < 1, n u = β i uξ i. i=1 In this paper, we first construct an integral operator for the corresponding linear boundary value problem and find out its first eigenvalue and eigenfunction. Then we establish a special cone associated with the Green s function of 1. Finally, by employing the Krasnoselskii Zabreiko fixed point theorem, combined with a priori estimates of positive solution we obtain the existence of positive solutions for 1. Note that our nonlinear term f involves the fractional derivatives of the dependent variable this is seldom studied in the literature and most research articles on boundary value problems consider nonlinear terms that involve the unknown function u only, for example, [1, 2, 4, 5,6,12,13,15]. Moreover, we adopt the method of lower and upper solutions to discuss the uniqueness of positive solutions for 1, and prove that the unique positive solution can be uniformly approximated by an iterative sequence beginning with any function which is continuou nonnegative and not identically vanishing on [, 1] Thi together with the fact that our nonlinearity may be of distinct growth, means that our methodology and main results here are entirely different from those in the above papers. 2 Preliminaries For convenience, we give some background materials from fractional calculus theory to facilitate analysis of problem 1. These materials can be found in the recent book see [7, 1, 11]. Definition 1. See [7,1], [11, pp. 36 37]. The Riemann Liouville fractional derivative of order α > of a continuous function f :, +, + is given by D α +ft = n t 1 d t s n α 1 fs d Γn α dt 2 3 http://www.mii.lt/na

Positive solutions for a class of fractional boundary value problems 3 where n = [α] + 1, [α] denotes the integer part of number α, provided that the right side is pointwise defined on, +. Definition 2. See [11, Def. 2.1]. The Riemann Liouville fractional integral of order α > of a function f :, +, + is given by I α +ft = 1 Γα t t s α 1 fs d provided that the right side is pointwise defined on, +. From the definition of the Riemann Liouville derivative, we can obtain the following statement. Lemma 1. See [1]. Let α >. If we assume u C, 1 L, 1, then the fractional differential equation D α +ut = has a unique solution ut = c 1 t α 1 + c 2 t α 2 + + c N t α N, c i R, i = 1, 2,..., N, where N is the smallest integer greater than or equal to α. Lemma 2. See [1]. Assume that u C, 1 L, 1 with a fractional derivative of order α > that belongs to C, 1 L, 1. Then I α +D α +ut = ut + c 1 t α 1 + c 2 t α 2 + + c N t α N, c i R, i = 1, 2,..., N, where N is the smallest integer greater than or equal to α. In what follow we shall discuss some properties of the Green s function for fractional boundary value problem 1. Let G 1 t, s := 1 Γα { t α 1 1 s α 2 t s α 1, s t 1, t α 1 1 s α 2, t s 1. 4 Then we can easily obtain that G 2 t, s := t G 1t, s = α 1 Γα { t α 2 1 s α 2 t s α 2, s t 1, t α 2 1 s α 2, t s 1. 5 Lemma 3. See [2, Lemma 2.7]. Let f be as in 1 and D α +u := v. Then 1 is equivalent to vt = G 1 t, sf G 1 τvτ dτ, G 2 τvτ dτ, vs. 6 Nonlinear Anal. Model. Control, 211:1 17

4 J. Xu, Z. Wei Lemma 4. See [2, Lemma 2.8] and [5, Thms. 1.1, 1.2]. The functions G i t, s C[, 1] [, 1], R + i = 1, 2, moreover, the following two inequalities hold: t α 1 s1 s α 2 ΓαG 1 t, s s1 s α 2 t, s [, 1]. 7 α 1α 2t α 2 1 ts1 s α 2 ΓαG 2 t, s α 1t α 3 s1 s α 2 t, s [, 1]. 8 In what follow we shall define two extra functions by G 1, G 2. Let G 3 t, s := G 4 t, s := G 1 t, τg 1 τ, s dτ t, s [, 1], G 1 t, τg 2 τ, s dτ t, s [, 1]. Then G i t, s C[, 1] [, 1], R + i = 3, 4. Moreover, by Lemma 4, we easily have Similarly, Let α α 1Γ2α tα 1 s1 s α 2 = t α 1 τ1 τ α 2 Γα τ α 1 s1 s α 2 Γα s1 s α 2 τ1 τ α 2 Γ 2 dτ = α α 1α 2 t α 1 s1 s α 2 Γ2α = t α 1 τ1 τ α 2 G 4 t, s = E := C[, 1], Γα dτ G 3 t, s 9 s1 sα 2 αα 1Γ 2 α. 1 α 1α 2τ α 2 1 τs1 s α 2 Γα α 1τ α 3 s1 s α 2 τ1 τ α 2 Γ 2 dτ α s1 s α 2 α 1Γ2α 2. 11 { } v := max vt, P := v E: vt t [, 1]. t [,1] dτ http://www.mii.lt/na

Positive solutions for a class of fractional boundary value problems 5 Then E, becomes a real Banach space and P is a cone on E. Define B ρ := {v E: v < ρ} for ρ > in the sequel. Let Avt := G 1 t, sf G 1 τvτ dτ, G 2 τvτ dτ, vs 12 for all v E. The continuity of G 1, G 2 and f implies that A : E E is a completely continuous nonlinear operator. As mentioned in Lemma 3, D α +u = v, together with the boundary conditions u = u = u 1 =, we have ut = G 1 t, svs d 13 where G 1 is determined by 4. Therefore, we find the existence of solutions of 1 is equivalent to that of fixed points of A. For a, b, c with a 2 + b 2 + c 2, let G a,b,c t, s := ag 3 t, s + bg 4 t, s + cg 1 t, s t, s [, 1], and define a linear operator L a,b,c as follows: L a,b,c vt = G a,b,c t, svs v E. 14 Obviously, L a,b,c is positive, i.e., L a,b,c P P. The continuity of G 1, G 3, G 4 implies that L a,b,c is a completely continuous operator. From now on, we utilize rl a,b,c to denote the spectral radius of L a,b,c. Furthermore, Gelfand s theorem enables us to obtain the following result. Lemma 5. Let Then aα bα 1α 2 ξ a,b,c := + α 1Γ2α Γ2α a η a,b,c := αα 1Γ 2 α + ξ a,b,c Γα + 1Γα 1 Γ2α Proof. By 7, 1, and 11, we obtain L a,b,c = max t [,1] G a,b,c t, s η a,b,c + c Γα, b α 1Γ2α 2 + c Γα. rl a,b,c η a,b,c αα 1. s1 s α 2 = η a,b,c αα 1. Nonlinear Anal. Model. Control, 211:1 17

6 J. Xu, Z. Wei Similarly, we find, for all n N +, L n a,b,c = max t [,1] }{{} n [ ] n ηa,b,c. αα 1 Gelfand s theorem implies that On the other hand, L a,b,c = max G a,b,c t, s n 2 G a,b,c s 2, s 1 G a,b,c s 1, s G a,b,c τ n 2 1 dτ n rl a,b,c = lim L n n a,b,c η a,b,c αα 1. t [,1] = ξ a,b,c αα 1. Similarly, we also obtain and L 2 a,b,c = max t [,1] max G a,b,c t, s max t [,1] = ξ 2 a,b,c t [,1] G a,b,c t, sg a,b,c τ dτ ξ a,b,c t α 1 s1 s α 2 ξ 2 a,b,ct α 1 s1 s α 2 s α 1 τ1 τ α 2 dτ s α 1 s α 2 1 L 3 a,b,c ξ 3 a,b,c s α 1 s α 2 Therefore, for all n N +, 1 L n a,b,c ξ n a,b,c s α 1 s α 2 τ1 τ α 2 dτ 2 1 n 1 1 τ1 τ α 2 dτ. τ1 τ α 2 dτ. http://www.mii.lt/na

Positive solutions for a class of fractional boundary value problems 7 By Gelfand s theorem, we see This completes the proof. n rl a,b,c = lim L n n a,b,c ξ a,b,c = ξ a,b,cγα + 1Γα 1. Γ2α s α 1 s α 2 By Lemma 5, we see rl a,b,c >, and thus the Krein Rutman theorem [9] asserts that there are ϕ a,b,c P \ {} and ψ a,b,c P \ {} such that G a,b,c t, sϕ a,b,c s = rl a,b,c ϕ a,b,c t, G a,b,c t, sψ a,b,c t dt = rl a,b,c ψ a,b,c s. 15 Note that we can normalize ψ a,b,c such that Let ω a,b,c = ξ a,b,c η 1 a,b,c tα 1 ψ a,b,c t dt and define P := { v P : Clearly, P is also a cone of E. Lemma 6. L a,b,c P P. Proof. We easily have the following inequality: ψ a,b,c t dt = 1. 16 vtψ a,b,c t dt ω a,b,c v G a,b,c t, s ξ a,b,c η 1 a,b,c tα 1 G a,b,c τ, s t, τ [, 1]. For vt, t [, 1], we have L a,b,c vtψ a,b,c t dt = } G a,b,c t, svsψ a,b,c t dt. Nonlinear Anal. Model. Control, 211:1 17

8 J. Xu, Z. Wei = ξ a,b,c η 1 a,b,c Consequently, we see ξ a,b,c η 1 a,b,c tα 1 G a,b,c τ, svsψ a,b,c t dt This completes the proof. t α 1 ψ a,b,c t dt G a,b,c τ, svs τ [, 1]. L a,b,c vtψ a,b,c t dt ω a,b,c L a,b,c v. Lemma 7. See [8]. Let E be a real Banach space and W a cone of E. Suppose that A : B R \ B r W W is a completely continuous operator with < r < R. If either i Au u for each B r W and Au u for each B R W or ii Au u for each B r W and Au u for each B R W, then A has at least one fixed point on B R \ B r W. Lemma 8. See [3]. Let E be a partial order Banach space, and x, y E with x y, D = [x, y ]. Suppose that A : D E satisfies the following conditions: i A is an increasing operator; ii x Ax, y Ay, i.e., x and y is a subsolution and a supersolution of A; iii A is a completely continuous operator. Then A has the smallest fixed point x and the largest fixed point y in [x, y ], respectively. Moreover, x = lim n A n x and y = lim n A n y. 3 Main results We first offer twelve fixed numbers α i, β i, γ i which are not all zero and let r 1 L αi,β i,γ i = λ αi,β i,γ i for i = 1, 2, 3, 4. Now, we list our assumptions on f: H1 f C[, 1] R 3 +, R + ; H1 f C[, 1] R 3 +,, +. H2 ft, x 1, x 2, x 3 lim inf > λ α1,β α 1x 1+β 1x 2+γ 1x 3 + α 1 x 1 + β 1 x 2 + γ 1 x 1,γ 1 17 3 uniformly for t [, 1]. http://www.mii.lt/na

Positive solutions for a class of fractional boundary value problems 9 H3 ft, x 1, x 2, x 3 lim sup < λ α2,β α 2x 1+β 2x 2+γ 2x 3 α + 2 x 1 + β 2 x 2 + γ 2 x 2,γ 2 18 3 uniformly for t [, 1]. H4 ft, x 1, x 2, x 3 lim inf > λ α3,β α 3x 1+β 3x 2+γ 3x 3 + α 3 x 1 + β 3 x 2 + γ 3 x 3,γ 3 19 3 uniformly for t [, 1]. H5 ft, x 1, x 2, x 3 lim sup < λ α4,β α 4x 1+β 4x 2+γ 4x 3 + α 4 x 1 + β 4 x 2 + γ 4 x 4,γ 4 2 3 uniformly for t [, 1]. H6 There exists a positive constant µ < 1 such that κ µ ft, x 1, x 2, x 3 ft, κx 1, κx 2, κx 3 κ, 1. H7 ft, x 1, x 2, x 3 is increasing in x 1, x 2, x 3, that i the inequality hol for x 1 x 1, x 2 x 2, x 3 x 3. ft, x 1, x 2, x 3 ft, x 1, x 2, x 3 3.1 Existence of positive solutions Theorem 1. Assume that H1 H3 hold. Then 1 has at least one positive solution. Proof. H2 implies that there are ε > and c 1 > such that ft, x 1, x 2, x 3 λ α1,β 1,γ 1 +εα 1 x 1 +β 1 x 2 +γ 1 x 3 c 1 x i R +, t [, 1]. 21 Let M 1 := {v P : v Av}. We claim that M 1 is bounded in P. Indeed, if v M 1, by 12 and 21, we can obtain vt G 1 t, sf λ α1,β 1,γ 1 + ε + [ 1 γ 1 G 1 t, svs = λ α1,β 1,γ 1 + ε G 1 τvτ dτ, α 1 G 3 t, τvτ dτ + ] c 1 αα 1Γα G 2 τvτ dτ, vs G α1,β 1,γ 1 t, svs β 1 G 4 t, τvτ dτ c 1 αα 1Γα. 22 Nonlinear Anal. Model. Control, 211:1 17

1 J. Xu, Z. Wei Multiply 22 by ψ α1,β 1,γ 1 t on both sides and integrate over [, 1] and use 15, 16 to obtain vtψ α1,β 1,γ 1 t dt λ α 1,β 1,γ 1 + ε λ α1,β 1,γ 1 Therefore, we have Consequently, Lemma 6 implies that vtψ α1,β 1,γ 1 t dt c 1 αα 1Γα. 23 vtψ α1,β 1,γ 1 t dt ε 1 λ α1,β 1,γ 1 c 1 αα 1Γα. 24 ω α1,β 1,γ 1 v ε 1 λ α1,β 1,γ 1 c 1 αα 1Γα, 25 and hence, v ε 1 ω 1 α 1,β 1,γ 1 λ α1,β 1,γ 1 c 1 αα 1Γα for all v M 1. Taking R > sup{ v : v M 1 }, we obtain 26 v Av v B R P. 27 On the other hand, by H3, there exist r, R and ε, λ α2,β 2,γ 2 such that ft, x 1, x 2, x 3 λ α2,β 2,γ 2 εα 2 x 1 + β 2 x 2 + γ 2 x 3 28 for all x i [, r] and t [, 1]. This implies that Avt λ α2,β 2,γ 2 ε α 2 = λ α2,β 2,γ 2 ε G 1 t, s G 1 τvτ dτ + β 2 G 2 τvτ dτ + γ 2 vs G α2,β 2,γ 2 t, svs 29 for all v B r P. Let M 2 := {v B r P : v Av}. Now, we claim M 2 = {}. Indeed, if v M 2, by 29, we have vt λ α2,β 2,γ 2 ε G α2,β 2,γ 2 t, svs. http://www.mii.lt/na

Positive solutions for a class of fractional boundary value problems 11 Multiply 22 by ψ α2,β 2,γ 2 t on both sides and integrate over [, 1] and use 15, 16 to obtain vtψ α2,β 2,γ 2 t dt λ α2,β 2,γ 2 ελ 1 α 2,β 2,γ 2 vtψ α2,β 2,γ 2 t dt and thus vtψ α 2,β 2,γ 2 t dt =. Consequently, we have vt, i.e., M 2 = {}. Therefore, v Av v B r P. 3 Now Lemma 7 indicates that the operator A has at least one fixed point on B R \B r P. That i 1 has at least one positive solution. This completes the proof. Theorem 2. Assume that H1, H4 and H5 hold. Then 1 has at least one positive solution. Proof. By H4, there exist r > and ε > such that ft, x 1, x 2, x 3 λ α3,β 3,γ 3 + εα 3 x 1 + β 3 x 2 + γ 3 x 3 x i [, r], t [, 1]. 31 This implies Avt λ α3,β 3,γ 3 + ε G α3,β 3,γ 3 t, svs 32 for all v B r P. Let M 3 := {v B r P : v Av}. We claim that M 3 = {}. Indeed, if v M 3, combining with 32, we find vt λ α3,β 3,γ 3 + ε G α3,β 3,γ 3 t, svs. 33 Multiply 33 by ψ α3,β 3,γ 3 t on both sides and integrate over [, 1] and use 15, 16 to obtain vtψ α3,β 3,γ 3 t dt λ α3,β 3,γ 3 + ελ 1 α 3,β 3,γ 3 vtψ α3,β 3,γ 3 t dt and thus vtψ α 3,β 3,γ 3 t dt =. Hence, we see vt, i.e., M 3 = {}. Consequently, v Av v B r P. 34 In addition, by H5, there exist ε, λ α4,β 4,γ 4 and c 2 > such that ft, x 1, x 2, x 3 λ α4,β 4,γ 4 εα 4 x 1 +β 4 x 2 +γ 4 x 3 +c 2 x i, t [, 1]. 35 Nonlinear Anal. Model. Control, 211:1 17

12 J. Xu, Z. Wei Let M 4 := {v P : v Av}. We shall prove that M 4 is bounded in P. Indeed, if v M 4, then we have vt λ α4,β 4,γ 4 ε G α4,β 4,γ 4 t, svs + c 2 αα 1Γα. 36 Multiply 36 by ψ α4,β 4,γ 4 t on both sides and integrate over [, 1] and use 15, 16 to obtain vtψ α4,β 4,γ 4 t dt λ α4,β 4,γ 4 ελ 1 α 4,β 4,γ 4 and then It follows from Lemma 6 that vtψ α4,β 4,γ 4 t dt ε 1 λ α4,β 4,γ 4 c 2 αα 1Γα. c 2 vtψ α4,β 4,γ 4 t dt+ αα 1Γα v ε 1 ω 1 α 4,β 4,γ 4 λ α4,β 4,γ 4 c 2 αα 1Γα 37 for all v M 4. Choosing R > sup{ v : v M 4 } and R > r, we have v Av v B R P. 38 Now Lemma 7 implies that A has at least one fixed point on B R \B r P. Equivalently, 1 has at least one positive solution. This completes the proof. 3.2 Uniqueness of positive solutions In order to obtain our main results in this subsection, we first offer some lemmas. From now on, we always assume that H1 hol. Lemma 9. If vt C[, 1] is a positive fixed point of A in 12, then there exist two positive constants a v and b v such that a v ρt vt b v ρt, where ρt = G 1t, s. Proof. The continuity of G 1, G 2 and v implies that there exists M > such that vt M and G it, svs M for all t [, 1]. Taking a v = b v = min ft, x 1, x 2, x 3 >, t,x 1,x 2,x 3 [,1] [,M] 3 max ft, x 1, x 2, x 3 >. t,x 1,x 2,x 3 [,1] [,M] 3 http://www.mii.lt/na

Positive solutions for a class of fractional boundary value problems 13 Then we have a v ρt vt = Avt = b v ρt. G 1 t, sf This completes the proof. G 1 τvτ dτ, G 2 τvτ dτ, vs Lemma 1. Suppose that H1, H4 H7 hold. Then the operator A has exactly one positive fixed point. Proof. By Theorem 2, A has at least one positive fixed point. It then remains to prove that A has at most one positive fixed point. Indeed, if v 1 and v 2 are two positive fixed points of A, then v i t = G 1 t, sf G 1 τv i τ dτ, G 2 τv i τ dτ, v i s where i = 1, 2. By Lemma 9, there exist four positive numbers a i, b i for which a i ρt v i t b i ρt for t [, 1] and i = 1, 2. Clearly, v 2 a 2 /b 1 v 1. Let γ := sup{γ > : v 2 γv 1 }. Then γ > and v 2 γ v 1. We shall claim that γ 1. Suppose the contrary. Then γ < 1 and v 2 t where = G 1 t, sf G 1 t, sgs + γ µ v 1t, gs = f γ µ f G 1 τγ v 1 τ dτ, G 1 τγ v 1 τ dτ, G 1 τv 1 τ dτ, d G 2 τγ v 1 τ dτ, γ v 1 s G 2 τγ v 1 τ dτ, γ v 1 s G 2 τv 1 τ dτ, v 1 s H6 implies that g P \{} and there is a a 3 > such that G 1t, sgs a 3 ρt by Lemma 9. Consequently, v 2 t a 3 ρt + γ µ v 1t a 3 /b 1 v 1 t + γ v 1 t, which contradicts the definition of γ. As a result, γ 1 and v 2 v 1. Similarly, v 1 v 2. Hence, v 1 = v 2. This completes the proof.. Nonlinear Anal. Model. Control, 211:1 17

14 J. Xu, Z. Wei Theorem 3. Let all the conditions in Lemma 1 hold and v t be the unique positive solution of A. Then for any v P \ {}, we have A n v v n uniformly in t [, 1]. Proof. Clearly, ρt = G 1t, s is a bounded function on [, 1]. Then by Lemma 9, there exist a ρ, b ρ > such that a ρ ρt G 1 t, sf := ηt b ρ ρt. G 1 τρτ dτ, G 2 τρτ dτ, ρs Let β 1 t = δηt with < δ < min{1/b ρ, a µ/1 µ ρ }. Then we can choose < ε < min{1, a ρ }, and Aεβ 1 t = = G 1 t, sf G 1 t, sf ε µ δa ρ µ G 1 τεβ 1 τ dτ, G 1 τ εβ 1τ ρτ dτ, ρτ G 2 τ εβ 1τ ρτ ρτ dτ, εβ 1s ρs ρs G 1 t, sf G 1 τρτ dτ, = ε µ δa ρ µ ηt ε µ δηt εδηt = εβ 1 t. G 2 τεβ 1 τ dτ, εβ 1 s G 2 τρτ dτ, ρs Thus we have Aεβ 1 εβ 1. On the other hand, let β 2 t = ξηt with ξ > max{1/a ρ, b µ/1 µ ρ }. Taking ε > max{1, b ρ }, we find εβ 2 t ε µ ξηt = ε µ ξ ε µ ξ G 1 t, sf G 1 t, sf G 1 τρτ dτ, G 1 τρτεβ 2 τ εβ 2 τ G 2 τρτεβ 2 τ εβ 2 τ G 2 τρτ dτ, ρs dτ, dτ, ρsεβ 2s εβ 2 s http://www.mii.lt/na

Positive solutions for a class of fractional boundary value problems 15 ε µ ξε µ ξb ρ µ G 1 t, sf G 1 t, sf = Aεβ 2 t. G 1 τεβ 2 τ dτ, G 1 τεβ 2 τ dτ, G 2 τεβ 2 τ dτ, εβ 2 s G 2 τεβ 2 τ dτ, εβ 2 s Hence, Aεβ 2 εβ 2. H7 implies that A is an increasing operator. It follows from Lemma 8 that A has the smallest fixed point v and the largest fixed point v in [εβ 1, εβ 2 ], respectively. Based on thi we first show v [εβ 1, εβ 2 ]. Indeed, for all n N +, we have εβ 1 A n εβ 1 A n εβ 2 εβ 2. 39 Let n in 39, we see εβ 1 v v v εβ 2. For all εβ 1 v εβ 2 and n N +, we have v P \ {} and A n εβ 1 A n v = v n A n εβ 2. 4 By Theorem 2 and Lemma 1, we know that A has only a positive fixed point, i.e., lim n A n εβ 1 = lim n A n εβ 2 = v, and thus lim n A n v v. This completes the proof. To facilitate computations for the following example let α 1 = α 2, β 1 = β 2, γ 1 = γ 2, α 3 = α 4, β 3 = β 4, γ 3 = γ 4 in H2 H5. Example 1. Let α = 2.5, α 1 = Γ 2 α = 9π/16 1.77, β 1 = Γ2α = 24, γ 1 = Γα = 3 π/4 1.33. Then by Lemma 5, we get.23 rl α1,β 1,γ 1 2.47, and.4 λ α1,β 1,γ 1 4.35. Let ft, x 1, x 2, x 3 = 1 sinα1 x 1 + β 1 x 2 + γ 1 x 3 + α1 x 1 + β 1 x 2 + γ 1 x 3 2. 4 Then and 1 4 lim inf sinα 1x 1 + β 1 x 2 + γ 1 x 3 + α 1 x 1 + β 1 x 2 + γ 1 x 3 2 α 1x 1+β 1x 2+γ 1x 3 + α 1 x 1 + β 1 x 2 + γ 1 x 3 = > λ α1,β 1,γ 1, lim sup α 1x 1+β 1x 2+γ 1x 3 + 1 4 sinα 1x 1 + β 1 x 2 + γ 1 x 3 + α 1 x 1 + β 1 x 2 + γ 1 x 3 2 α 1 x 1 + β 1 x 2 + γ 1 x 3 = 1 4 < λ α 1,β 1,γ 1 Nonlinear Anal. Model. Control, 211:1 17

16 J. Xu, Z. Wei uniformly for t [, 1]. All conditions of Theorem 1 hold. Therefore, 1 has at least one positive solution. Example 2. Let α = 2.5, α 3 = Γ 2 α = 9π/16 1.77, β 3 = Γ2α 2 = 2, γ 3 = Γα 1 = π/2.89. Then from Lemma 5 we have.1 rl α3,β 3,γ 3.43, and 2.33 λ α3,β 3,γ 3 1. Let ft, x 1, x 2, x 3 = e t + ln 1 + α 3 x 1 + β 3 x 2 + γ 3 x 3. Then and e t + ln1 + α 3 x 1 + β 3 x 2 + γ 3 x 3 lim inf = > λ α3,β α 3x 1+β 3x 2+γ 3x 3 + α 3 x 1 + β 3 x 2 + γ 3 x 3,γ 3 3 e t + ln1 + α 3 x 1 + β 3 x 2 + γ 3 x 3 lim sup = < λ α3,β α 3x 1+β 3x 2+γ 3x 3 + α 3 x 1 + β 3 x 2 + γ 3 x 3,γ 3 3 uniformly for t [, 1]. Hence, H4, H5 hold, and Theorem 2 implies that 1 has at least one positive solution. Example 3. Let α = 2.5, α 3 = Γ 2 2α 2 = 4, β 3 = γ 3 = Γα 2 = π 1.77. By Lemma 5, we can obtain λ α3,β 3,γ 3 [1.48, 4.9]. Let ft, x 1, x 2, x 3 = e t + α 3 x 1 + β 3 x 2 + γ 3 x 3. Similar with Example 2, we can show H4 and H5 hold. On the other hand, for any κ, 1, we have κ 1 and κ [ e t + α 3 x 1 + β 3 x 2 + γ 3 x 3 ] = κe t + α 3 κx 1 + β 3 κx 2 + γ 3 κx 3 e t + α 3 κx 1 + β 3 κx 2 + γ 3 κx 3. As a result, H6 is also satisfied. In addition, H1 and H7 automatically hold. Hence, from Theorem 3, 1 has a unique positive solution. References 1. Z. Bai, H. Lü, Positive solutions for boundary value problem of nonlinear fractional differential equation, J. Math. Anal. Appl., 311:495 55, 25. 2. M. El-Shahed, Positive solutions for boundary value problems of nonlinear fractional differential equation, Abstr. Appl. Anal., 27, Article ID 1368, 27. 3. D. Guo, V. Lakshmikantham, Nonlinear Problems in Abstract Cone Notes Rep. Math. Sci. Eng., Vol. 5, Academic Pres San Diego, CA, 1988. 4. W. Jiang, Eigenvalue interval for multi-point boundary value problems of fractional differential equation Appl. Math. Comput., 219:457 4575, 213. http://www.mii.lt/na

Positive solutions for a class of fractional boundary value problems 17 5. D. Jiang, C. Yuan, The positive properties of the Green function for Dirichlet-type boundary value problems of nonlinear fractional differential equations and its application, Nonlinear Anal., Theory Metho Appl., Ser. A, 72:71 719, 21. 6. W. Jiang, B. Wang, Z. Wang, The existence of positive solutions for multi-point boundary value problems of fractional differential equation Physics Procedia, 25:958 964, 212. 7. A. Kilba H. Srivastava, J. Trujillo, Theory and Applications of Fractional Differential Equation North-Holland Math. Stud., Vol. 24, Elsevier, Amsterdam, 26. 8. M. Krasnoselski, P. Zabreiko, Geometrical Metho of Nonlinear Analysi Springer, Berlin, Heidelberg, 1984. 9. M. Krein, M. Rutman, Linear operators leaving invariant a cone in a Banach space, Trans. Am. Math. Soc., 1:199 325, 1962. 1. I. Podlubny, Fractional Differential Equation Math. Sci. Eng., Vol. 198, Academic Pres San Diego, CA, 1999. 11. S. Samko, A. Kilba O. Marichev, Fractional Integrals and Derivatives: Theory and Application Gordon and Breach, Yverdon 1993. 12. Y. Wang, L. Liu, Y. Wu, Positive solutions for a nonlocal fractional differential equation, Nonlinear Anal., Theory Metho Appl., 74:3599 365, 211. 13. J. Xu, Z. Yang, Multiple positive solutions of a singular fractional boundary value problem, Appl. Math. E-Note 1:259 267, 21. 14. J. Xu, Z. Wei, Y. Ding, Positive solutions for a boundary value problem with Riemann Liouville fractional derivative, Lith. Math. J., 524:462 476, 212. 15. J. Xu, Z. Wei, W. Dong, Uniqueness of positive solutions for a class of fractional boundary value problem, Appl. Math. Lett., 25:59 593, 212. Nonlinear Anal. Model. Control, 211:1 17