Positive solutions for three-point nonlinear fractional boundary value problems

Transcript

1 Electronic Journal of Qualitative Theory of Differential Equations 2, No. 2, -9; Positive solutions for three-point nonlinear fractional boundary value problems Abdelkader Saadi, Maamar Benbachir Abstract In this paper, we give sufficient conditions for the existence or the nonexistence of positive solutions of the nonlinear fractional boundary value problem D α u atfut =, < t <, 2 < α < 3, u = u =, u u η = λ, where D α is the standard Riemann-Liouville fractional differential operator of order α, η,,, are two arbitrary constants» «and η α 2 λ [, is a parameter. The proof uses the Guo-Krasnosel skii fixed point theorem and Schauder s fixed point theorem. Introduction In this paper, we are interested in the existence or non-existence of positive solutions for the nonlinear fractional boundary value problem BVP for short D α u atfut =, 2 < α < 3, < t <,. u = u =, u u η = λ,.2 where α is a real number, D [ α is the standard Riemann-Liouville differentiation of order α, η,,, are arbitrary constants and λ [, is a η α 2 parameter. A positive solution is a function ut which is positive on, and satisfies.-.2. We show, under suitable conditions on the nonlinear term f, that the fractional boundary value problem.-.2 has at least one or has non positive solutions. By employing the fixed point theorems for operators acting on cones 2 Mathematics Subject Classifications: 26A33, 34B25, 34B5. Key words and phrases: Fractional derivatives; three-point BVPs; positive solutions; fixed point theorem. EJQTDE, 2 No. 2, p.

2 in a Banach space see, for example [7, 8, 3, 4, 5]. The use of cone techniques in order to study boundary value problems has a rich and diverse history. That is, some authors have used fixed point theorems to show the existence of positive solutions to boundary value problems for ordinary differential equations, difference equations, and dynamic equations on time scales, see for example [, 2, 3]. Moreover, Delbosco and Rodino [7] considered the existence of a solution for the nonlinear fractional differential equation D α u = ft, u, where < α < and f : [, a] R R, < a is a given function, continuous in, a R. They obtained results for solutions by using the Schauder fixed point theorem and the Banach contraction principle. Bai and Lü [5] studied the existence and multiplicity of positive solutions of nonlinear fractional differential equation boundary value problem: D α u ft, ut =, < α < 2, < t <, u = u =, where D α is the standard Riemann-Liouville differential operator of order α. Recently Bai and Qiu [4]. considered the existence of positive solutions to boundary value problems of the nonlinear fractional differential equation D α u ft, ut =, 2 < α 2, < t <, u = u = u =, where D α is the Caputo s fractional differentiation, and f :, ] [, [,, with lim t ft,. =. They obtained results for solutions by using the Krasnoselskii s fixed point theorem and the nonlinear alternative of Leray-Schauder type in a cone. Lü Zhang [8] considered the existence of solutions of nonlinear fractional boundary value problem involving Caputo s derivative D α t u ft, ut =, < α < 2, < t <, u = ν, u = ρ. In another paper, by using fixed point theory on cones, Zhang [9] studied the existence and multiplicity of positive solution of nonlinear fractional boundary value problem D α t u ft, ut =, < α < 2, < t <, u u =, u u =, where Dt α is the Caputo s fractional derivative. By using the Krasnoselskii fixed point theory on cones, Benchohra, Henderson, Ntoyuas and Ouahab [6] used the Banach fixed point and the nonlinear alternative of Leray-Schauder to investigate the existence of solutions for fractional order functional and neutral functional differential equations with infinite delay D α yt = ft, y t, for each t J = [, b], < α <, yt = φt, t,], EJQTDE, 2 No. 2, p. 2

3 where D α is the standard Riemman-Liouville fractional derivative, f : J B R is a given function satisfying some suitable assumptions, φ B, φ = and B is called a phase space. By using the Krasnoselskii fixed point theory on cones, El-Shahed [8] Studied the existence and nonexistence of positive solutions to nonlinear fractional boundary value problem D α u λatfut =, 2 < α < 3, < t <, u = u = u =, where D α is the standard Riemann-Liouville differential operator of order α. Some existence results were given for the problem.-.2 with α = 3 by Sun [7].The BVP.-.2 arises in many different areas of applied mathematics and physics, and only its positive solution is significant in some practice. For existence theorems of fractional differential equation and application, the definitions of fractional integral and derivative and related proprieties we refer the reader to [7,, 2, 6]. The rest of this paper is organized as follows: In section 2, we present some preliminaries and lemmas. Section 3 is devoted to prove the existence and nonexistence of positive solutions for BVP Elementary Background and Preliminary lemmas In this section, we will give the necessary notations, definitions and basic lemmas that will be used in the proofs of our main results. We also present a fixed point theorem due to Guo and Krasnosel skii. Definition [,, 5]. The fractional arbitrary order integral of the function h L [a, b],r of order α R is defined by I α a ht = t a t s α hsds, where Γ is the gamma function. When a =, we write I α ht = h ϕ α t, where ϕ α t = tα for t >, and ϕ αt = for t, and ϕ α δt as α, where δ is the delta function. Definition 2 [,, 5]. For a function h given on the interval [a, b], the αth Riemann-Liouville fractional-order derivative of h, is defined by n Da α ht = d t fs ds, n = [α]. Γn α dt t s α n Lemma 3 [4] Let α >. If u C, L,, then the fractional differential equation D α ut = 2.2 EJQTDE, 2 No. 2, p. 3

4 has solution ut = c t α c 2 t α 2... c n t α n, c i R, i =, 2,..., n, n = [α]. Lemma 4 [4] Assume that u C, L, with a fractional derivative of order α >. Then I α Dα ut = ut c t α c 2 t α 2... c n t α n, 2.3 for some c i R, i =, 2,..., n, n = [α]. Lemma 5 Let y C [, ] = y C [, ], yt, t [, ] }, then the BVP D α ut yt =, 2 < α < 3, < t <, 2.4 has a unique solution where and ut = u = u =, u u η = λ, 2.5 Gt, sysds tα G η, sysds λtα 2 η 2.6 t Gt, s = α s α 2 t s α, s t t α s α 2, t s η G η, s = α 2 s α 2 η s α 2, s η η α 2 s α 2, η s Proof. By applying Lemmas 3 and 4, the equation 2.4 is equivalent to the following integral equation ut = c t α c 2 t α 2 c 3 t α 3 t t s α ysds. 2.9 for some arbitrary constants c, c 2, c 3 R. Boundary conditions 2.5, permit us to deduce there exacts values c = η α 2 c 2 = c 3 = [ s α 2 ysds η then, the unique solution of is given by the formula ut = t t α t = t s α ysds t α η ] } η s α 2 ysds λ α η s α 2 ysds λt α α t s α ysds tα s α 2 ysds s α 2 ysds EJQTDE, 2 No. 2, p. 4

5 ηα 2 t α λt α α = tα t tα η α 2 t s α 2 ysds t s α ysds tα t s α 2 ysds tα η α 2 t α η α 2 η α 2 = where, η s α 2 ysds η Gt, sysds Gt, s = G η, s = η t α s α 2 ysds η s α 2 ysds η s α 2 ysds λt α α t α G η, sysds η s α 2 ysds t α s α 2 t s α, s t t α s α 2, t s η α 2 s α 2 η s α 2, s η η α 2 s α 2, η s λt α α This ends the proof. In order to check the existence of positive solutions, we give some properties of the functions Gt, s and G t, s. Lemma 6 For all t, s [, ] [, ], we have P Gt, s t = α G t, s. P2 G η, s ηα 2 s α 2, G η, sds = ηηα 2 α P3 γg, s Gt, s G, s, t, s [τ, ] [, ]. Where G, s = s sα 2, γ = τ α 2, and τ satisfies τ s s α 2 asds >. 2. Proof. P and P2 are obvious. We prove that P3 holds. For all t, s [, ] [, ], P and P2 imply that, Gt, s G, s. If s t, we have Gt, s G, s = tα s α 2 t s α s s α 2 tt tsα 2 t ts α s s α 2 = tt tsα 2 t tst ts α 2 s s α 2 = tst tsα 2 s s α 2 = tα. EJQTDE, 2 No. 2, p. 5

6 If t s, we have Gt, s G, s = tα s α 2 s s α 2 tα s α 2 s α 2 = t α. Thus, Therefore, t α G, s Gt, s G, s, t, s [, ] [, ]. τ α G, s Gt, s G, s, t, s [τ, ] [, ]. This completes the proof. Lemma 7 If y C [, ], then the unique solution ut of the BVP is nonnegative and satisfies min ut γ u. t [τ,] Proof. Let y C [, ]; it is obvious that ut is nonnegative. For any t [, ], by 2.6 and Lemma 6, it follows that ut = and thus u Gt, sysds G, sysds G, sysds tα G η, sysds G η, sysds G η, sysds More that, 2.6 and Lemma 6 imply that, for any t [τ, ], ut = Hence γ Gt, sysds G, sysds = γ G, sysds This completes the proof. tα α τ G η, sysds min ut γ u. t [τ,] G η, sysds G η, sysds λt α α λ α, λ α. λt α α λτ α α λ α Definition 8 Let E be a real Banach space. A nonempty closed convex set K E is called cone of E if it satisfies the following conditions. EJQTDE, 2 No. 2, p. 6

7 A x K, σ implies σx K; A2 x K, x K implies x =. Definition 9 An operator is called completely continuous if it continuous and maps bounded sets into precompact sets To establish the existence or nonexistence of positive solutions of BVP.-.2, we will employ the following Guo-Krasnosel skii fixed point theorem: Theorem [2] Let E be a Banach space and let K E be a cone in E. Assume that Ω and Ω 2 are open subsets of E with Ω and Ω Ω 2. Let T : K Ω 2 \Ω K be completely continuous operator. In addition, suppose either H Tu u, u K Ω and Tu u, u K Ω 2 or H2 Tu u, u K Ω 2 and Tu u, u K Ω, holds. Then T has a fixed point in K Ω 2 \Ω. 3 Existence of solutions In this section, we will apply Krasnosel skii s fixed point theorem to the problem.-.2. We note that ut is a solution of.-.2 if and only if ut = tα Gt, sasfusds η α 2 G η, sasfusds Let us consider the Banach space of the form λt α α. 3. E=C [, ] = u C [, ], ut, t [, ] }, equipped with standard norm We define a cone K by and an integral operator T : E E by u = max ut : t [, ]}. } K= u E : min ut γ u, t [τ,] Tut = tα Gt, sasfusds η α 2 G η, sasfusds λt α α. 3.2 It is not difficult see that, fixed points of T are solutions of.-.2. Our aim is to show that T : K K is completely continuous, in order to use Theorem. EJQTDE, 2 No. 2, p. 7

8 Lemma Let f : [, [, continuous. Assume the following condition C a C [, ], [,. Then operator T : K K is completely continuous. Proof. Since Gt, s, G η, s, then Tut for all u K. We first prove that TK K. In fact, Tut = tα Gt, sasfusds } G η, sasfusds λ α } so, G, sasfusds t [, ] Tu G, sasfusds τ α G η, sasfusds λ α } G η, sasfusds λ α, on the other hand, Lemma 6 imply that, for any t [τ, ], Tut = tα Gt, sasfusds G η, sasfusds λ γ G, sasfusds G η, sasfusds = γ and, for u K G, sasfusds γ min Tut γ Tu. t [τ,] α G η, sasfusds Consequently, we have TK K. Next, we prove that T is continuous. In fact, let N = 2 s sα 2 asds sα 2 asds, assume that u n, u K and u n u, then u n c <, for every n. Since f is continuous on [, c], it is uniformly continuous. Therefore, for any ǫ >, there exists δ > such that u u 2 < δ implies that fu fu 2 < ε 2N. Since u n u, there exists n N such that u n u < δ for n n. Thus we have fu n t fu t < ε 2N, for n n and t [, ]. This implies that } λ α λ α } EJQTDE, 2 No. 2, p. 8

9 for n n Tu n Tu = Gt, sas[fu ns fu s] ds tα η α 2 G η, sas[fu n s fu s] ds ε [ ] 2N G, sasds η α 2 G η, sasds ε [ ] 2N s sα 2 asds ηα 2 sα 2 asds ε [ 2N s sα 2 asds η ] α 2 sα 2 asds ε 2N = ε. 2N That is, T : K K is continuous. Finally, let B K be bounded, we claim that TB K is uniformly bounded. Indeed, since B is bounded, there exists some m > such that u m, for all u B. Let then C = max fut : u m} Tu C N for all u B. λ α N such that C = C At last, we prove TB is equicontinuous. Hence TB is bounded, for all ε >, each } u B, t, t 2 [, ], t < t 2, let ε δ = min 6C a, ηα 2 ε 3C a, ηα 2 ε 3λ, this allows us to show that, One has Tut 2 Tut < ε when t 2 t < δ. Tut 2 Tut = Gt 2, s Gt, sasfusds tα 2 t α } G η, sasfusds λ α C a Gt 2, s Gt, sds tα 2 t α λtα 2 t α α G η, sds C a t Gt 2, s Gt, sds t 2 t Gt 2, s Gt, s ds C a t 2 Gt 2, s Gt, sds tα 2 t α C a G η, sds λtα 2 t α α } EJQTDE, 2 No. 2, p. 9

10 C a = C a = C a = C a I I 2 I 3 tα 2 t α C a G η, sds λ α I 4 I 5 I 6 tα 2 t α I 7 I 8 I 9 tα 2 t α I I I 2 tα 2 t α C a G η, sds λ α C a G η, sds λ α C a G η, sds λ α = C a I 3 t α 2 t α η η α 2 C a α λt α 2 t α C a I 3 t α 2 t α C a η α 2 α λt α α 2 t α α, where I = t [ s α 2 t2 α t α t 2 s α t s α ] ds I 2 = t 2 [ t s α 2 t2 α t α t 2 s α ] ds I 3 = t 2 s α 2 t α 2 t α ds I 4 = t α 2 t α t sα 2 ds t I 5 = t α 2 t α t sα 2 ds t t 2 s α ds t 2 s α ds I 6 = t 2 t t 2 s α ds t α 2 t α t 2 s α 2 ds I 7 = α tα 2 t α [ t α ] α [t 2 t α t α 2 ] α t α I 8 = α tα 2 t α [ t 2 α t α ] I 9 = α t 2 t α α tα 2 t α t 2 α I = α tα 2 t α α tα 2 t α t α I = α t 2 t α α t 2 α α t α α tα 2 t α t 2 α I 2 = α tα 2 t α t α α t 2 t α α tα 2 t α t 2 α I 3 = α tα 2 t α α tα 2 tα. In order to estimate t 2 α t α and t 2 α t α, we can apply a method used in [4, 8]; by means value theorem of differentiation, we have Thus, we obtain t 2 α t α αt 2 t < αδ 3δ, t 2 α t α α t 2 t < α δ 2δ. Tut 2 Tut < C a < C a < 2C a [ α δ α αδ α I t α 2 t α C a ] α δ C a C a λ η α 2 α λt α 2 t α α η α 2 α λα δ α δ < ε 3 ε 3 ε 3 = ε, EJQTDE, 2 No. 2, p.

11 where I = α tα 2 t α α t 2 α t α. By means of the Arzela-Ascoli theorem, T : K K is completely continuous. The proof is achieved. In all what follow, we assume that the next conditions are satisfied. C f : [, [, is continuous; C2 a :, [, is continuous, < and < s s α 2 asds < s α 2 asds <. that is a is singular at t =, t = Lemma 2 [5] Suppose that E is a Banach space, T n : E E n =, 2, 3,... are completely continuous operators, T : E E, and lim max T nu Tu = for all r >. n u <r Then T is completely continuous. For any natural number n n 2, we set inf t<s as, t n n, a n t = at, n t n, 3.3 inf n <s<t as, n t. Then a n : [, ] [, is continuous and a n t at, t,. Let T n ut = Gt, sa nsfusds tα G η, sa n sfusds λt α α. Lemma 3 If C, C2 hold. Then T : K K is completely continuous. Proof. By a similar as in the proof of Lemma it is obvious that T n : E E is completely continuous. Since < tα Gt, sasds < < η α 2 G η, sasds G η, sasds G, sasds [ ] s sα 2 asds ηα 2 η α 2 sα 2 asds <, and by the absolute continuity of the integral, we have [ en s ] sα 2 asds ηα 2 t α en sα 2 asds =, lim n EJQTDE, 2 No. 2, p.

12 where en = [, n] [ n, ]. Let r > and u B r = u E : u r} and M r = max fut : t, u [, ] [, r]} <, by 3.3, Lemma 6P3, and the absolute continuity of the integral, we have lim T nu Tu n lim max Gt, sa ns as fusds n t< tα η α 2 G η, sa n s as fusds [ M ] r lim s sα 2 as a n s ds n η α 2 sα 2 as a n s ds [ M r en lim s ] sα 2 as a n s ds n en sα 2 as a n sds n = M s sα 2 as a n s ds r lim η n α 2 n sα 2 as a n s ds s s α 2 as a n sds n η α 2 s α 2 as a n sds n M [ r lim n en s sα 2 asds ηα 2 t ] α en sα 2 asds =. Then by Lemma 2, T : K K is completely continuous. Throughout this section, we shall use the following notations: Λ := s sα 2 asds γ Λ 2 := τ s sα 2 asds It is obvious that Λ 2 > Λ >. Also we define G η, sasds, γ τ G η, sasds. fr f = lim r r, f fr = lim r r. Theorem 4 Suppose that f is superlinear, i.e. f =, f =. Then BVP.-.2 has at least one positive solution for λ small enough and has no positive solution for λ large enough. Proof. We divide the proof into two steps. Step. We prove that BVP.-.2 has at least one positive solution for sufficiently small λ >.since f =, for Λ >, there exists R > such that fr r Λ 2, r [, R ]. Therefore, fr rλ 2, for r [, R ]. 3.4 EJQTDE, 2 No. 2, p. 2

13 Let Ω = u C [, ] : u R } and let λ satisfies < λ α η α 2 R Then, for any u K Ω, it follows from Lemma 6, 3.2, 3.4 and 3.5 that Tut = tα Gt, sasfusds G η, sasfusds λ α s sα 2 asfusds G η, sasfusds λ α Λ 2 α ηα 2 R 2α Λ 2 And thus = u 2 u 2 = u. s sα 2 asusds s sα 2 asds η α 2 G η, sasusds η α 2 G η, sasds u R 2 Tut u, for u K Ω. 3.6 On the other hand, since f =, for Λ 2 >, there exists R 2 > R such that Λ 2, r [γr 2,.Thus we have fr r fr rλ 2, for r [γr 2, ]. 3.7 Set Ω 2 = u C [, ] : u R 2 }. For any u K Ω 2, by Lemma 6 one has min s [τ,] us γ u = γr 2. Thus, from 3.6 we can conclude that Tu = G, sasfusds G η, sasfusds λ α G, sasfusds Λ 2 Λ 2 γ = u, τ sα 2 sasfusds τ sα 2 sasusds τ sα 2 sasds G η, sasfusds τ G η, sasfusds τ G η, sasusds u γ τ G η, sasds which implies that Tu u, for u K Ω EJQTDE, 2 No. 2, p. 3

14 Therefore, by 3.6, 3.8 and the first part of Theorem we know that the operator T has at least one fixed point u K Ω 2 \Ω, which is a positive solution of BVP.-.2. Step 2. We verify that BVP.-.2 has no positive solution for λ large enough. Otherwise, there exist < λ < λ 2 <... < λ n <..., with lim n λ n =, such that for any positive integer n, the BVP D α u atfu =, 2 < α < 3, < t <, u = u =, u u η = λ n, has a positive solution u n t, by 3., we have u n = G, sasfu nsds η α 2 G η, sasfu n sds λ n α λ n α η α 2, n. Thus u, n. Since [ f =, for 4Λ 2 >, there exists R > such that fr r γ R,, which implies that fr 2Λ 2 r, for r [ γ R,. 4Λ 2, r Let n be large enough that u n R, then u n u n = G, sasfu nsds η α 2 G η, sasfu n sds λ n α 2Λ G, sasu nsds η α 2 G η, sasu n sds 2Λ τ G, sasu nsds τ G η, sasu n sds 2Λ 2 γ τ sα 2 γ sasds τ G η, sasds u n = 2 u n, which is contradiction. The proof is complete. Moreover, if the function f is nondecreasing, the following theorem holds. Theorem 5 Suppose that f is superlinear. If f is nondecreasing, then there exists a positive constant λ such that BVP.-.2 has at least one positive solution for λ, λ and has no positive solution for λ λ,. EJQTDE, 2 No. 2, p. 4

15 Proof. Let Σ = λ : BVP.-.2 has at least one positive solution} and λ = sup Σ; it follows from Theorem 4 that < λ <. From the definition of λ, we know that for any λ, λ, there is a λ > λ such that BVP D α u atfut =, 2 < α < 3, < t <, u = u =, u u η = λ, has a positive solution u t. Now we prove that for any λ, λ, BVP.-.2 has a positive solution. In fact, let Ku = u K : ut u t, t [, ]} For any λ, λ, u Ku, it follows from 3.2 and the monotonicity of f that we have that Tut = tα Gt, sasfusds G η, sasfusds λ α Gt, sasfu sds tα G η, sasfu sds λ α = u t. Thus, T Ku Ku. By Shaulder s fixed point theorem we know that there exists a fixed point u Ku, which is a positive solution of BVP.-.2. The proof is complete. Now we consider the case f is sublinear. Theorem 6 Suppose that f is sublinear, i.e. f =, f =. Then BVP.-.2 has at least one positive solution for any λ,. Proof. Since f =, there exists R > such that fr Λ 2 r, for any r [, R ]. So for any u K with u = R and any λ >, we have Tu = G, sasfusds G η, sasfusds λ α sα 2 sasfusds η α 2 G η, sasfusds Λ 2 Λ 2 γ = u, τ sα 2 sasfusds τ sα 2 sasusds τ sα 2 sasds τ G η, sasfusds τ G η, sasusds u γ τ G η, sasds and consequently, Tu u. So, if we set Ω = u K : u < R }, then Tu u, for u K Ω. 3.9 EJQTDE, 2 No. 2, p. 5

16 Next we construct the set Ω 2. We consider two cases: f is bounded or f is unbounded. Case : Suppose that f is bounded, say fr M for all r [,. In this case we choose R 2 max 2R, 2M, Λ and then for u K with u = R 2, we have Tut = So, M tα Gt, sasfusds sα 2 sasds λ α M Λ R 2 2 R 2 2 R 2 2 = R 2 = u. Tu u. } 2λ α, G η, sasfusds λ G η, sasds α Case 2: When f is unbounded. Now, since f =, there exists R > such that fr Λ 2 r, for r [R,, 3. Let and be such that R 2 max 2R, R, 2λ α fr fr 2, for r [, R 2 ]. For u K with u = R 2, from 3.2 and 3., we have Tut = tα Gt, sasfusds Thus, Gt, sasfr 2ds Λ 2 sα 2 sasds = R 2 2 R 2 2 = R 2 = u. }, G η, sasfusds λ Tu u. α G η, sasfr 2 ds λ η α 2 G η, sasds Therefore, in either case we may put Ω 2 = u K : u < R 2 }, then α R 2 R 2 2 Tu u, for u K Ω So, it follows from 3.9, 3. and the second part of Theorem that T has a fixed point u K Ω 2 \Ω, Then u is a positive solution of BVP.-.2. The proof is achieved. EJQTDE, 2 No. 2, p. 6

17 4 Application In this section we give an example to illustrate the usefulness of our main results. Example 7 Let us consider the following fractional BVP D 5 2 ut t t2 u 3 2 t =, < t <, 4. u = u =, u 2 2 u = λ, We can easily show that fut = u 3 2t satisfy: fu f = lim u u = lim fu ut =, f = lim u u u = lim ut =, u obviously, for a.e. t [, ], we have s s α 2 asds = s s2 ds = ds = s s s s α 2 asds = s s s s s2 ds = ds = s So conditions C, C2 holds, then we can choose R 2 > R >, and for λ satisfies < λ 9 6 R < R 2, then we can choose Ω = u K : u < R }, Ω 2 = u K : u < R 2 } and by Theorem 4, we can show that the BVP has at least one positive solution ut K Ω 2 \Ω for λ small enough and has no positive solution for λ large enough. Example 8 Let us consider the following fractional BVP D 5 2 ut exp ut =, < t <, 4.3 t t2 u = u =, u 2 2 u = λ, We can easily show that fut = exp ut satisfy: fu f = lim u u = lim u u expu =, f fu = lim u u = lim u u expu =, EJQTDE, 2 No. 2, p. 7

18 obviously, for a.e. t [, ], we have s α 2 asds = s s α 2 asds = ds = s s s s ds = So conditions C, C2 holds, and by Theorem 6, we can show that the BVP has at least one positive solutions ut, for any λ,. References [] R. P. Agarwal, D. O Regan and P. J. Y. Wong, Positive solutions of Differential, Difference, and Integral Equations, Kluwer Academic Publishers, Boston, 999. [2] T. S. Aleroev, The Sturm-Liouville problem for a second-order differential equation with fractional derivatives in the lower terms, Differentsial nye Uravneniya 8 982, n. 2, [3] T. S. Aleroev, On a boundary value problem for a fractional-order differential operator, Differentsial nye Uravneniya , n., 26. [4] Z. Bai and T. Qiu, Existence of positive solutions for singular fractional differential equations, Electron. J. Diff. Eqns , n 46, -9. [5] Z. Bai and H. Lü, Positive solutions for a boundary value problem of nonlinear fractional differential equations, J. Math. Anal. Appl. 3 25, [6] M. Benchohra, J. Henderson, S. K. Ntoyuas and A. Ouahab, Existence results for fractional order functional differential equations with infinite delay, J. Math. Anal. Appl , [7] D. Delbosco and L. Rodino, Existence and uniqueness for a nonlinear fractional differential equation, J. Math. Anal. Appl, 24, [8] M. El-Shahed, Positive solutions for boundary-value problem of nonlinear fractional differential equation, Abstract and Applied Analysis, vol. 27, Article ID 368, 8 pages, 27 doi:.55/27/368. [9] E. R. Kaufmann, Positive Solutions of a three-point boundary-value problems on a time scale, Electron. J. Diff. Eqns , n 82, - [] A. A. Kilbas and S. A. Marzan, Nonlinear differential equations with Caputo fractional derivative in the space of continuously differentiable functions, Diff: Uravn 4 25, n, 82-86; translation in Differ. Equ 4 25, N, EJQTDE, 2 No. 2, p. 8

19 [] A. A. Kilbas, O. I. Marichev, S. G. Samko, Fractional Integral and Derivatives Theory and Applications. Gordon and Breach, Switzerland, 993. [2] A. A. Kilbas, H. M. Srivastava, and J. J. Trujillo, Theory and Applications of Fractional Differential Equations, North-Holland Mathematics Studies, 24. Elsevier science B. V., Amsterdam, 26. [3] M. A. Krasnosel skii, Topological Methods in the Theory on Nonlinear Integral Equations, English Translated by A. H. Armstrong; A Pergamon Press Book, MacMillan, New York, 964. [4] M. A. Krasnosel skii, Positive solutions of operator equations, Noordhoff Gronigen, Netherland, 964. [5] M. A. Krasnosel skii and P. P. Zabreiko, Geometrical methods of nonlinear analysis. Springer-Verlag, New York, 984. [6] I. Podlubny, Fractional Differential Equations, Academic Press, San Diego, 999. [7] Y. Sun, Positive Solutions for third-order three-point nonhomogeneous boundary value problems, Appl. Math. Lett , 45-5 [8] S. Zhang, Existence of solution for a boundary value problems of fractional order, Acta Mathematica Scientia, vol. 26, n 2, pp , 26. [9] S. Zhang, Positive solutions for boundary value problems of nonlinear fractional differential equation, Electron. J. Diff. Eqns , n 36, -2. Received January 24, 2 Abdelkader Saadi Faculty of Science and Technology Bechar University, Bechar, Algeria address: Abdsaadi@yahoo.fr Maamar Benbachir Faculty of Science and Technology Bechar University, Bechar, Algeria address: mbenbachir2@yahoo.fr EJQTDE, 2 No. 2, p. 9