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J. Nonlinear Sci. Appl. 1 (2008), no. 3, 123–131 The Journal of Nonlinear Sciences and Applications http://www.tjnsa.com POSITIVE SOLUTIONS FOR BOUNDARY VALUE PROBLEM OF NONLINEAR FRACTIONAL DIFFERENTIAL EQUATION TINGTING QIU1∗ AND ZHANBING BAI2 Abstract. We investigate the positive solution of nonlinear fractional differential equation with semi-positive nonlinearity ½ α D0+ u(t) + f (t, u(t)) = 0, 0 < t < 1, u(0) = u0 (1) = u00 (0) = 0 where 2 < α ≤ 3 is a real number, D0α+ is the Caputo’s differentiation, and f : [0, 1] × [0, ∞) → (−∞, ∞). By use of Krasnosel’skii fixed point theorem, the existence results of positive solution are obtained. 1. Introduction and preliminaries Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications of such constructions in various sciences such as physics, mechanics, chemistry, engineering, etc. For details, see [4, 6, 7, 8] and the references therein. It should be noted that most of papers and books on fractional calculus are devoted to the solvability of linear initial fractional differential equations in terms of special functions [9]. Recently, there are some papers deal with the existence and multiplicity of solution (or positive solution) of nonlinear initial fractional differential equation by the use of techniques of nonlinear analysis (fixed-point theorems, Leray-Shauder theory, etc.), see [2, 3, 5, 11, 12]. However, there are few papers consider the semipositone nonlinearity differential equations of fractional order. No contributions exist, as far as we know, Date: Received: 4 September 2008; Revised: 28 September 2008. Corresponding author. 2000 Mathematics Subject Classification. Primary 34B15; Secondary 34B18. Key words and phrases. Fractional differential equation; Positive solution; Fixed-point theorem. ∗ 123 124 TINGTING QIU AND ZHANBING BAI concerning the positive solution with semipositone nonlinearity of the following problem: ½ α D0+ u(t) + f (t, u(t)) = 0, 0 < t < 1, (1.1) u(0) = u0 (1) = u00 (0) = 0 where 2 < α ≤ 3 is a real number, D0α+ is the Caputo’s differentiation, and f : [0, 1] × [0, ∞) → (−∞, ∞) is continuous. In this paper, we firstly derive the corresponding Green’s function,then we give the properties of Green’s function,finally,the existence of positive solution with semipositone nonlinearity are obtained by Krasnosel’skii fixed point theorem. Here, a positive solution u∗ of (1.1) will mean a solution u∗ of (1.1) satisfying u∗ (t) > 0, 0 < t < 1. For the convenience of the reader, we present here the necessary definitions from fractional calculus theory. These definitions can be found in the recent literature. Definition 1.1. The Riemann-Liouville fractional integral of order α > 0 of a function f : [0, 1] → R is given by Z t 1 α I0+ f (t) = (t − s)α−1 f (s)ds Γ(α) 0 provided that the right side is pointwise defined on (0, ∞). Definition 1.2. The Caputo’s fractional derivative of order α > 0 of a function f ∈ AC n [0, 1] is given by Z t 1 f (n) (s) α D0+ f (t) = ds Γ(n − α) 0 (t − s)α−n+1 where n−1 < α ≤ n, provided that the right side is pointwise defined on (0, ∞). From the definition we can obtain the following lemma: Lemma 1.3. Assume that u ∈ C n [0, 1], then α α I0+ D0+ u(t) = u(t) − C1 − C2 t − · · · − Cn tn−1 where Ci ∈ R, i = 1, 2, . . . , n, n = [α] + 1. Lemma 1.4. [6] The relation α+β α β Ia+ Ia+ ϕ = Ia+ ϕ is valid in following case Reβ > 0, Re(α + β) > 0, ϕ(x) ∈ L1 (a, b). Lemma 1.5. Given f ∈ C[0, 1], and 2 < α ≤ 3, the unique solution of 0 < t < 1, D0α+ u(t) + f (t) = 0, 0 00 u(0) = u (1) = u (0) = 0. is Z 1 G(t, s)f (s)ds u(t) = 0 (1.2) BOUNDARY VALUE PROBLEM where ( G(t, s) = (α−1)t(1−s)α−2 −(t−s)α−1 , Γ(α) α−2 t(1−s) , Γ(α−1) 0 ≤ s ≤ t ≤ 1, 0 ≤ t ≤ s ≤ 1. 125 (1.3) Proof. We may apply Lemma1.3 to reduce Eq.(1.2) to an equivalent integral equation α u(t) = −I0+ f (t) + C1 + C2 t + C3 t2 for some Ci ∈ R, i = 1, 2, 3. By Lemma1.4 we have α−1 1 1 α 1 f (t) + C2 + 2C3 t I0+ I0+ f (t) + C2 + 2C3 t = −D0+ I0+ u0 (t) = −D0+ α−1 f (t) + C2 + 2C3 t = −I0+ 1 α−1 1 1 α−2 α−2 u00 (t) = −D0+ I0+ f (t) + 2C3 = −D0+ I0+ I0+ f (t) + 2C3 = −I0+ f (t) + 2C3 . From u(0) = u0 (1) = u00 (0) = 0, one has Z 1 1 C1 = 0, C2 = (1 − s)α−2 f (s)ds, C3 = 0. Γ(α − 1) 0 Therefore, the unique solution of problem (1.2) is Z t Z 1 1 1 α−1 (t − s) f (s)ds + t(1 − s)α−2 f (s)ds u(t) = − Γ(α) 0 Γ(α − 1) 0 ¸ Z t· Z 1 t(1 − s)α−2 (t − s)α−1 t(1 − s)α−2 = − f (s)ds + f (s)ds Γ(α − 1) Γ(α) Γ(α − 1) 0 t Z 1 = G(t, s)f (s)ds 0 The proof is complete. ¤ Lemma 1.6. The function G(t, s) defined by Eq. (1.3) satisfies (1) G(t, s) > 0, for t, s ∈ (0, 1); (2) min 1/4≤t≤3/4 G(t, s) ≥ 1 1 max G(t, s) = G(1, s), for 0 < s < 1. 4 0≤t≤1 4 (1.4) Proof. Setting (α − 1)t(1 − s)α−2 − (t − s)α−1 t(1 − s)α−2 , g2 (t, s) = . Γ(α) Γ(α − 1) (1) For g1 (t, s), since 2 < α ≤ 3, 0 ≤ s ≤ t ≤ 1 we can obtain g1 (t, s) = (α − 1)t(1 − s)α−2 ≥ t(1 − s)α−2 ≥ t(t − s)α−2 ≥ (t − s)α−1 we get g1 (t, s) > 0. It is clearly that g2 (t, s) > 0, therefore G(t, s) > 0, for t, s ∈ (0, 1). 126 TINGTING QIU AND ZHANBING BAI (2) Since (α − 1)(1 − s)α−2 − (α − 1)(t − s)α−2 ∂g1 (t, s) = > 0, ∂t Γ(α) ∂g2 (t, s) (1 − s)α−2 = > 0. ∂t Γ(α − 1) we know G(t, s) is increasing with respect to t. Consequently, 0 ≤ G(t, s) ≤ max G(t, s) = G(1, s), for t, s ∈ [0, 1]. 0≤t≤1 One has 1 min G(t, s) = G( , s) = 1/4≤t≤3/4 4 max G(t, s) = G(1, s) = 0≤t≤1 1/4≤t≤3/4 1 (1−s)α−2 4 Γ(α−1) Γ(α) , s ∈ (0, 14 ], , s ∈ [ 14 , 1). (α − 1)(1 − s)α−2 − (1 − s)α−1 , s ∈ (0, 1). Γ(α) we will prove min 1 (α−1)(1−s)α−2 −( 14 −s)α−1 4 G(t, s) ≥ 1 1 max G(t, s) = G(1, s) 0≤t≤1 4 4 As follows, 10 : For 0 < s ≤ 14 , (α − 1)(1 − s)α−2 ( 14 − s)α−1 − min G(t, s) = 1/4≤t≤3/4 4Γ(α) Γ(α) 1 1 (α − 1)(1 − s)α−2 (1 − s)α−1 max G(t, s) = G(1, s) = − 4 0≤t≤1 4 4Γ(α) 4Γ(α) 1 Since 2 < α ≤ 3, 0 < s ≤ 4 we can obtain 1 1 1 1 ( − s)α−1 = ( )α−1 (1 − 4s)α−1 ≤ (1 − 4s)α−1 < (1 − s)α−1 4 4 4 4 Thus, min 1/4≤t≤3/4 0 2 : For 1 4 G(t, s) ≥ 1 1 max G(t, s) = G(1, s). 4 0≤t≤1 4 ≤ s < 1, min 1/4≤t≤3/4 G(t, s) = 1 (1 4 − s)α−2 (α − 1)(1 − s)α−2 = Γ(α − 1) 4Γ(α) 1 (α − 1)(1 − s)α−2 (1 − s)α−1 1 max G(t, s) = G(1, s) = − 4 0≤t≤1 4 4Γ(α) 4Γ(α) It is clearly that 1 1 min G(t, s) ≥ max G(t, s) = G(1, s). 0≤t≤1 1/4≤t≤3/4 4 4 The proof is complete. ¤ BOUNDARY VALUE PROBLEM 127 Lemma 1.7. If f ∈ C[0, 1], and f ≥ 0, then,the unique solution u of problem (1.1) satisfies 1 min3 u(t) ≥ kuk, 1 4 ≤t≤ 4 4 Proof. By Lemma 1.5,u can be expressed by Z 1 Z 1 u(t) = G(t, s)f (s, u(s))ds ≤ max G(t, s)f (s, u(s))ds 0 0≤t≤1 0 so, Z max 0≤t≤1 Z 1 G(t, s)f (s, u(s))ds ≤ 0 then, Z kuk = max |u(t)| = max 0≤t≤1 0≤t≤1 1 max G(t, s)f (s, u(s))ds 0 0≤t≤1 Z 1 G(t, s)f (s, u(s))ds ≤ 0 1 max G(t, s)f (s, u(s))ds 0 0≤t≤1 Taking into account(1.4), we obtain Z min u(t) = 1min3 1 ≤t≤ 34 4 4 ≤t≤ 4 1 G(t, s)f (s, u(s))ds 0 Z Z 1 1 1 1 1 max G(t, s)f (s, u(s))ds ≥ max G(t, s)f (s, u(s))ds = kuk ≥ 4 0 0≤t≤1 4 0≤t≤1 0 4 The proof is complete. ¤ Lemma 1.8. [5] Let X be a Banach space, P ⊆ X a cone, and Ω1 , Ω2 are two bounded open balls of X centered at the origin with Ω1 ⊂ Ω2 . Suppose that A : P ∩ (Ω2 \ Ω1 ) → P is a completely continuous operator such that either (i) kAxk ≤ kxk, x ∈ P ∩ ∂Ω1 and kAxk ≥ kxk, x ∈ P ∩ ∂Ω2 , or (ii) kAxk ≥ kxk, x ∈ P ∩ ∂Ω1 and kAxk ≤ kxk, x ∈ P ∩ ∂Ω2 holds. Then A has a fixed point in P ∩ (Ω2 \ Ω1 ). 2. Main results In this section, by uses of Lemma 1.5, Lemma 1.6,Lemma 1.7 and Lemma 1.8, we will obtain the existence of positive solution with semipositone nonlinearity for Problem (1.1). Let X = C[0, 1] be endowed with the ordering u ≤ v if u(t) ≤ v(t) for all t ∈ [0, 1], and the maximum norm kuk = max |u(t)| 0≤t≤1 Define the cone K ⊂ X by ( ) ¯ 1 1 K = u ∈ X ¯u(t) ≥ 0, and 1min3 u(t) ≥ max |u(t)| = kuk, 4 0≤t≤1 4 ≤t≤ 4 4 128 TINGTING QIU AND ZHANBING BAI Define operator T : K → K Z T u(t) = 1 G(t, s)f (s, u(s))ds 0 Lemma 2.1. T : K → K is completely continuous. Proof. Let u ∈ K, we have Z 1 Z T u(t) = G(t, s)f (s, u(s))ds ≤ Z 0≤t≤1 Z 1 max T u(t) = max 0≤t≤1 G(t, s)f (s, u(s))ds ≤ 0 Taking into account (1.4), we get Z min3 T u(t) = 1min3 1 4 max G(t, s)f (s, u(s))ds 0 0≤t≤1 0 So, 1 ≤t≤ 4 4 ≤t≤ 4 1 max G(t, s)f (s, u(s))ds 0 0≤t≤1 1 G(t, s)f (s, u(s))ds 0 Z Z 1 1 1 1 1 ≥ max G(t, s)f (s, u(s))ds ≥ max G(t, s)f (s, u(s))ds = kT uk 0≤t≤1 0≤t≤1 4 0 4 4 0 Also, by Lemma1.6, for u ∈ K, we have T u(t) ≥ 0, 0 ≤ t ≤ 1. Consequently, T : K → K. Further, with the nonnegativeness and continuity of G(t, s) and f (t, u), by means of Arzela-Ascoli theorem we can easily obtain T is completely continuous. The proof is complete. ¤ Theorem 2.2. Suppose (H1 ) f ∈ C([0, 1] × [0, ∞), (−∞, +∞)), ∃M > 0, L > −M, such that − M ≤ f (t, x) ≤ L, , 0 ≤ t ≤ 1, 0 ≤ x ≤ 1; (H2 ) lim f (t,x) = +∞, t ∈ [α, β] ⊂ [0, 1], uniformly hold. x x→∞ Then Problem (1.1) has at least one positive solution. Proof. Let Z w(t) = G(t, s)ds = 0 and set 1 αt − tα 1 < Γ(α + 1) Γ(α) 1 M 1 v(t) = M w(t) < 4 , u(t) = u(t) + v(t) 4 Γ(α) then u(t) is the positive solution of problem(1.1) if and only if u(t) is the solution of the following problem: ½ α D0+ u(t) + [f (t, u(t) − v(t)) + M ] = 0, 0 < t < 1, (2.1) u(0) = u0 (1) = u00 (0) = 0 and satisfies u(t) − v(t) > 0, (0 < t < 1). Define the operator T ∗ : K → K Z 1 ∗ T x(t) = G(t, s)[f (s, x(s) − v(s)) + M ]ds 0 BOUNDARY VALUE PROBLEM 129 From Lemma2.1, we know T ∗ : K → K is completely continuous operator. Clearly, the fixed point of operator T ∗ is the solution of problem (2.1). Let ½ ¾ M +L K1 = x ∈ K : kXk < for u ∈ ∂K1 Γ(α) we get Z kT xk = max 0≤t≤1 1 G(t, s)[f (s, x(s) − v(s)) + M ]ds 0 Z ≤ (M + L) max 0≤t≤1 1 G(t, s)ds < 0 M +L = kxk Γ(α) Thus, kT xk = kxk, for u ∈ ∂K1 . By condition (H2 ) exists R > 0 is large enough such that R > 2M Γ(α) satisfies f (t, x) + M R ≥ N, for t ∈ [ξ, η] ⊂ [0, 1] , x ≥ x 8 where N is chosen so that Z 1 N η ξ+η 4 G( , s)ds > 1 2 ξ 2 Set KR = {x ∈ K : kxk < R} , for x ∈ ∂KR . we have 1 1 M x(t) > kxk = R > > 2v(t) 4 4 2Γ(α) so 1 1 R x(t) − v(t) > x(t) − x(t) = u(t) > , t ∈ [0, 1] 2 2 8 x(t) − v(t) > R , t ∈ [ξ, η] 8 Thus, we obtain f (t, x(t) − v(t)) + M ≥ N (x(t) − v(t)) ≥ Consequently, Z R , t ∈ [ξ, η]. 8 1 ξ+η , s)[f (s, x(s) − v(s)) + M ]ds 2 0 Z η Z N R 41 N R 14 η ξ + η ξ+η G( , s) ds ≥ G( , s)ds > R = kxk ≥ 2 2 2 2 ξ ξ T x(t) = G( Therefore, kT xk = kxk, for x ∈ ∂KR . An application of Lemma1.8 yields that exists u ∈ KR \ K1 , such that u = T u. Namely, u = u(t) satisfies 130 TINGTING QIU AND ZHANBING BAI ½ D0α+ u(t) + [f (t, u(t) − v(t)) + M ] = 0, 0 < t < 1, u(0) = u0 (1) = u00 (0) = 0 thus, 1 1 u(t) ≥ kuk = max |u(t)| 4 4 0≤t≤1 ¯Z 1 ¯ ¯ ¯ 1 = max ¯¯ G(t, s)[f (s, x(s) − v(s)) + M ]ds¯¯ 4 0≤t≤1 0 Z 1 1 M 1 ≥ · M max G(t, s)ds = 4 > v(t) 0≤t≤1 0 4 Γ(α) Consequently, u(t) = u(t) − v(t) > 0, 0 < t ≤ 1. and satisfies ½ D0α+ u(t) + f (t, u(t)) = 0, 0 < t < 1, u(0) = u0 (1) = u00 (0) = 0, Obviously, u(t) 0 < t < 1 is positive solution of problem (1.1). The proof is complete. ¤ References 1. Z.B. Bai, Positive solutions for boundary value problem of nonlinear fractional differential equation, J. Math. Anal. Appl. 311 (2005) 495–505. 2. A. Babakhani and V.D. Gejji, Existence of positive solutions of nonlinear fractional differential equations, J. Math. Anal. Appl. 278 (2003) 434–442. 1 3. D. Delbosco and L. Rodino, Existence and uniqueness for a nonlinear fractional differential equation, J. Math. Anal. 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Zhang, The existence of a positive solution for a nonlinear fractional differential equation, J. Math. Anal. Appl. 252 (2000) 804–812. 1 12. S.Q. Zhang, Existence of positive solution for some class of nonlinear fractional differential equations, J. Math. Anal. Appl. 278 (2003) 136–148. 1 BOUNDARY VALUE PROBLEM 131 13. Q.L. Yao, Existence of positive solution for a third order three point boundary value problem with semipositone nonlinearity, Journal of Mathematical Research Exposition. 23 (2003) 591-596. 1 Department of Mathematics, Shandong University of Science and Technology,Qingdao, 266510, PRC. E-mail address: qiutingting19833@163.com 2 Department of Mathematics, Shandong University of Science and Technology, Qingdao, 266510, PRC. E-mail address: zhanbingbai@163.com