Wang and Liu Advances in Difference Equations (217) 217:7 DOI 1.1186/s13662-16-162-5 R E S E A R C H Open Access Positive solutions for a class of fractional 3-point boundary value problems at resonance Yongqing Wang 1,2* and Lishan Liu 2,3 * Correspondence: wyqing981@163.com 1 School of Statistics, Qufu Normal University, Qufu, Shandong 273165, P.R. China 2 School of Mathematical Sciences, Qufu Normal University, Qufu, Shandong 273165, P.R. China Full list of author information is available at the end of the article Abstract In this paper, we study the nonlocal fractional differential equation: D α + u(t)+f(t, u(t)) =, < t <1, u() =, u(1) = ηu(ξ), where 1 < α <2,<ξ <1,ηξ α 1 =1,D α + is the standard Riemann-Liouville derivative, f : [, 1] [, + ) R is continuous. The existence and uniqueness of positive solutions are obtained by means of the fixed point index theory and iterative technique. Keywords: fractional differential equation; positive solution; resonance; fixed point index 1 Introduction In this paper, we consider the following fractional differential equation: D α + u(t)+f (t, u(t)) =, < t <1, u() =, u(1) = ηu(ξ), (1.1) where 1 < α <2,<ξ <1,ηξ α 1 =1,D α + is the standard Riemann-Liouville derivative, f :[,1] [, + ) R is continuous. Problem (1.1) happens to be at resonance, since λ = is an eigenvalue of the linear problem D α + u = λu, <t <1, u() =, u(1) = ηu(ξ), (1.2) and ct α 1, c R, is the corresponding eigenfunction. Fractional differential equations occur frequently in various fields such as physics, chemistry, engineering and control of dynamical systems, etc. During the last few decades, many papers and books on fractional calculus and fractional differential equations have appeared (see [1 22] and the references therein). The Author(s) 217. This article is distributed under the terms of the Creative Commons Attribution 4. International License (http://creativecommons.org/licenses/by/4./), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Wang and Liu Advances in Difference Equations (217) 217:7 Page 2 of 13 When < ηξ α 1 <1,problem(1.1) is non-resonant. In [9], the author studied the existence of positive solutions for the non-resonant case by means of the fixed point index theory under sublinear conditions. In [18], the authors investigated the existence and multiplicity results of positive solutions by using of the fixed point theorem for the fractional differential equation given by D α + u(t)=f (t, u(t)), < t <1, u() =, D β + u(1) = adβ + u(ξ), (1.3) where 1 < α 2, β 1, < ξ <1, a 1withaξ α β 2 <1 β, α β 1. Recently, there are some papers dealing with the existence of solutions of fractional boundary value problem at resonance by using the coincidence degree theory due to Mawhin (see [19 22]). In [22], the authors investigated the following fractional threepoint boundary value problem (BVP for short) at resonance: D α + u(t)+f (t, u(t), Dα 1 + u(t)) =, < t <1, (1.4) u() =, u(1) = ηu(ξ), where 1 < α 2, < ξ <1,ηξ α 1 =1,D α + is the standard Riemann-Liouville derivative, f :[,1] R 2 R is continuous. By using the coincidence degree theory, the existence of solutions forbvp (1.4) are obtained under certain growth conditions. To the best of our knowledge, there are only very few papers dealing with the existence of positive solutions for resonant boundary value problems since the corresponding linear operator is non-reversible. For the case that α is an integer, some work has been done dealing with the existence of positive solutions for resonant boundary value problems by using Leggett-Williams norm-type theorem for coincidence (see [23 25]). Webb [26] established existence of positive solutions for second order boundary value problems at resonance by considering equivalent non-resonant perturbed problems with the same boundary conditions. Inspired by the work mentioned above, in this paper we aim to establish the existence of positive solutions for resonant problem (1.1). The paper is organized as follows. Firstly, we reduce non-perturbed boundary value problems at resonance to equivalent non-resonant perturbed problems with the same boundary conditions. Then we derive the corresponding Green s function and argue its properties. Finally, the existence and uniqueness results of positive solutions are obtained by using of the fixed point index and iterative technique. 2 Basic definitions and preliminaries In this section, we present some preliminaries and lemmas. The definitions and properties of fractional derivative can be found in the literature [1 22]. Definition 2.1 The fractional integral of order α >ofafunctionu :(,+ ) R is given by I α + u(t)= 1 Ɣ(α) t (t s) α 1 u(s) ds provided that the right-hand side is point-wise defined on (, + ).
Wang and Liu Advances in Difference Equations (217) 217:7 Page 3 of 13 Definition 2.2 The Riemann-Liouville fractional derivative of order α > ofafunction u :(,+ ) R is given by ( ) D α + u(t)= 1 d n t (t s) n α 1 u(s) ds, Ɣ(n α) dt where n =[α]+1,[α]denotes the integer part of number α, provided that the right-hand side is point-wise defined on (, + ). Denote g(t)= α 2 + Ɣ(α 1) + t k Ɣ((k +1)α 2). (2.1) k=1 It is easy to check that g (t)>on(,+ ), and g() = α 2 Ɣ(α 1) <, lim t + g(t)=+. Therefore, there exists a unique b >suchthat g ( b ) =. (2.2) For the convenience in presentation, we here list the assumptions to be used throughout the paper. (H 1 ) b (, b ] is a constant. (H 2 ) f :[,1] [, + ) R is continuous and f (t, x)+bx. (2.3) Set G b (t)=t α 1 E α,α ( bt α ), (2.4) where E α,α (x)= + k= x k Ɣ((k +1)α) (2.5) is the Mittag-Leffler function (see [1, 2]). Next we consider the following boundary value problem: D α + u(t)+bu(t)=f (t, u(t)) + bu(t), < t <1, u() =, u(1) = ηu(ξ). (2.6) It is clear that (1.1)isequivalentto(2.6).
Wang and Liu Advances in Difference Equations (217) 217:7 Page 4 of 13 Set K (t, s)= 1 G b (t)g b (1 s), t s 1, G b (1) G b (t)g b (1 s) G b (t s)g b (1), s t 1; (2.7) ηk (ξ, s) q(s)= G b (1) ηg b (ξ) ; (2.8) K(t, s)=k (t, s)+g b (t)q(s). (2.9) Lemma 2.1 Suppose that (H 1 ) holds, and y L[, 1]. Then the unique solution of the problem D α + u(t)+bu(t)=y(t), < t <1, (2.1) u() =, u(1) = ηu(ξ), is u(t)= K(t, s)y(s) ds. Proof By [1, 2], we know that the solution of (2.1)canbeexpressedby u(t)= t By u() =, we have c 2 =. On the other hand, we have u(1) = G b (t s)y(s) ds + c 1 G b (t)+c 2 G b (t). G b (1 s)y(s) ds + c 1 G b (1), (2.11) ξ u(ξ)= G b (ξ s)y(s) ds + c 1 G b (ξ). (2.12) Noting that ηξ α 1 =1,and<ξ <1,wehave G b (1) ηg b (ξ)= + k= Equations (2.11)and(2.12)yield b k [1 ηξ (k+1)α 1 ] Ɣ((k +1)α) = + k= c 1 = G b(1 s)y(s) ds η ξ G b(ξ s)y(s) ds. G b (1) ηg b (ξ) Therefore, the solution of (2.1)is b k (1 ξ kα ) >. (2.13) Ɣ((k +1)α) t u(t)= G b (t s)y(s) ds + G b(1 s)y(s) ds η ξ G b(ξ s)y(s) ds G b (t) G b (1) ηg b (ξ) = G b(t)g b (1 s)y(s) ds t G b(1)g b (t s)y(s) ds G b (1)
Wang and Liu Advances in Difference Equations (217) 217:7 Page 5 of 13 G b(1 s)y(s) ds G b (t)+ G b (1) G b(1 s)y(s) ds G b (t) G b (1) ηg b (ξ) η ξ G b(ξ s)y(s) ds G b (t) G b (1) ηg b (ξ) = K (t, s)y(s) ds + η G b(ξ)g b (1 s)y(s) ds G b (t) G b (1)[G b (1) ηg b (ξ)] = η ξ G b(1)g b (ξ s)y(s) ds G b (t) G b (1)[G b (1) ηg b (ξ)] K(t, s)y(s) ds. This completes the proof. Lemma 2.2 Suppose that (H 1 ) holds. The function K(t, s) has the following properties: (1) K(t, s)>, t, s (, 1); (2) ω 2 (s)t α 1 K(t, s) ω 1 (s)t α 1, t, s [, 1], where ω 1 (s)=g b (1 s)+g b (1)q(s), ω 2 (s)= q(s) Ɣ(α). (2.14) Proof Itisclearthatwejustneedtoprovethat(2)holds. By (2.4), we can get t α 1 + Ɣ(α) G b(t)=t α 1 b k t αk Ɣ((k +1)α) tα 1 G b (1), t [, 1], (2.15) G b (t)= + k= k= b k t (k+1)α 2 >, Ɣ((k +1)α 1) t (, 1], (2.16) and [ ] α 2 + b (t)=tα 3 Ɣ(α 1) + b k t kα Ɣ((k +1)α 2) G k=1 = t α 3 g ( bt α) < t α 3 g(b) t α 3 g ( b ) =, t (, 1), (2.17) which implies G b (t) is strictly increasing on [, 1], and G b (t) is strictly decreasing on (, 1]. By (2.15), we have K(t, s)=k (t, s)+g b (t)q(s) G b(t)g b (1 s) + G b (t)q(s) G b (1) G b (1 s)t α 1 + t α 1 G b (1)q(s)=ω 1 (s)t α 1. (2.18) On the other hand, when < t s < 1, noticing G b () =, and the monotonicity of G b (t), it is clear that G b (t)g b (1 s)>. (2.19)
Wang and Liu Advances in Difference Equations (217) 217:7 Page 6 of 13 When < s < t <1,wehave [ Gb (t)g b (1 s) G b (t s)g b (1) ] s = G b (t s)g b(1) G b (t)g b (1 s) [ G b (1) G b (t) ] G b (1 s). (2.2) Integrating (2.2)withrespectto s,weobtain G b (t)g b (1 s) G b (t s)g b (1) s [ Gb (1) G b (t) ] G b (1 τ) dτ = [ G b (1) G b (t) ][ G b (1) G b (1 s) ] >. (2.21) By (2.7), (2.19), and (2.21), we get K (t, s)>, t, s (, 1). Then K(t, s)=k (t, s)+g b (t)q(s) G b (t)q(s) tα 1 Ɣ(α) q(s)=ω 2(s)t α 1 >, t, s (, 1). This completes the proof. Let E = C[, 1] be endowed with the maximum norm u = max t 1 u(t), θ is the zero element of E, B r = u E : u < r}.defineaconep by P = u E : u(t), t [, 1] }. Let Au(t)= Tu(t)= K(t, s) [ f ( s, u(s) ) + bu(s) ] ds. (2.22) K(t, s)u(s) ds. (2.23) By means of Lemma 2.2 and the Arzela-Ascoli theorem, we can get A : P P is completely continuous, T : P P is completely continuous linear operator. By virtue of the Krein-Rutmann theorem and Lemma 2.2, we have the spectral radius r(t)>andt has a positive eigenfunction corresponding to its first eigenvalue λ 1 =(r(t)) 1.Sinceλ =is the eigenvalue of the linear problems (1.2), and t α 1 is the corresponding eigenfunction, we have the following lemma. Lemma 2.3 Suppose that (H 1 ) holds, then the first eigenvalue of T is λ 1 = b, and ϕ 1 (t)=t α 1 is the positive eigenfunction corresponding to λ 1, that is, ϕ 1 = bt ϕ 1. Lemma 2.4 ([27]) Let P be a cone in a Banach space E, and be a bounded open set in E. Suppose that A: P P is a completely continuous operator. If there exists u Pwith
Wang and Liu Advances in Difference Equations (217) 217:7 Page 7 of 13 u θ such that u Au λu, λ, x P, then i(a, P, P)=. Lemma 2.5 ([27]) Let P be a cone in a Banach space E, and be a bounded open set in E. Suppose that A: P P is a completely continuous operator. If Au λu, λ 1, u P, then i(a, P, P)=1. 3 The uniqueness result Theorem 3.1 Assumethatthereexists λ (,b) such that f (t, u)+bu f (t, v) bv λ u v, for t [, 1], u, v [, ), then (1.1) has a unique nonnegative solution. Proof Firstly, we will prove A has fixed point in P. Set Q = u P : l 1, l 2 >suchthatl 1 t α 1 u(t) l 2 t α 1}. (3.1) For any u P \θ},let l i (u)= ω i (s)u(s) ds, i =1,2. (3.2) By Lemma 2.2,itisobviousthatl 1 (u), l 2 (u)>,and l 1 (u)t α 1 (Tu)(t) l 2 (u)t α 1, (3.3) that is, T : P \θ} Q. For any u P \θ},let u n = A(u n 1 ), n =1,2,... (3.4) We may suppose that u 1 u θ (otherwise, the proof is finished). Then there exists l 2 ( u 1 u )>,suchthat T ( u 1 u ) l 2 ( u1 u ) t α 1 = l 2 ( u1 u ) ϕ 1. (3.5)
Wang and Liu Advances in Difference Equations (217) 217:7 Page 8 of 13 Thus u 2 u 1 = λ K(t, s) [ f ( s, u 1 (s) ) + bu 1 (s) f ( s, u (s) ) bu (s) ] ds K(t, s) f ( s, u 1 (s) ) + bu 1 (s) f ( s, u (s) ) bu (s) ds K(t, s) u1 (s) u (s) ds = λt ( u 1 u ) ( λl 2 u1 u ) ϕ 1, (3.6) u 3 u 2 = K(t, s) [ f ( s, u 2 (s) ) + bu 2 (s) f ( s, u 1 (s) ) bu 1 (s) ] ds λ K(t, s) f ( s, u 2 (s) ) + bu 2 (s) f ( s, u 1 (s) ) bu 1 (s) ds K(t, s) u2 (s) u 1 (s) ds λ 2 ( l 2 u1 u ) Tϕ 1 = λ2 b l ( 2 u1 u ) ϕ 1, (3.7). By induction, we can get u n+1 u n ( ) λ n 1 ( λl 2 u1 u ) ϕ 1. (3.8) b Then, for any n, m N,wehave u n+m u m u n+m u n+m 1 + + u n+1 u n [( ) λ n+m 1 ( ) λ n ] ( + + λl 2 u1 u ) ϕ 1 b b ( λ b )n 1 λ b ( λl 2 u1 u ) ϕ 1 = λn+1 l 2 ( u 1 u ) ϕ b n 1 1. (3.9) (b λ) So, u n+m u m λn+1 l 2 ( u 1 u ) b n 1 (b λ), n, (3.1) which implies u n } is a Cauchy sequence. Therefore, there exists a u P, suchthatu n } converges to u. Clearly, u is a fixed point of A. In the following, we will prove the fixed point of A is unique. Suppose v u is a fixed point of A. Then there exists l 2 ( u v )>,suchthat T ( u v ) l 2 ( u v ) ϕ 1. (3.11)
Wang and Liu Advances in Difference Equations (217) 217:7 Page 9 of 13 Then Au Av 1 = K(t, s) [ f ( s, u (s) ) + bu (s) f ( s, v(s) ) bv(s) ] ds λ K(t, s) f ( s, u (s) ) + bu (s) f ( s, v(s) ) bv(s) ds K(t, s) u (s) v(s) ds = λt ( u v ) λl 2 ( u v ) ϕ 1. (3.12) By induction, we can get A n u A n v ( λ b ) n 1 λl 2 ( u v ) ϕ 1. (3.13) So, u v = A n u A n v ( λ b ) n 1 λl 2 ( u v ), n. (3.14) Consequently, the fixed point of A is unique. This completes the proof. Remark 3.1 The unique nonnegative solution u of (1.1) can be approximated by the iterative schemes: for any u P \θ},let u n = A(u n 1 ), n =1,2,..., then u n u. Remark 3.2 If f (t,) on [, 1], then θ istheuniquesolutionof(1.1)inp;iff (t,) on [, 1], then the unique solution u is a positive solution. 4 Existence of positive solutions Theorem 4.1 Assume that (H 1 ), (H 2 ), and the following assumptions hold: lim inf x + lim sup x + min t [,1] max t [,1] f (t, x) x f (t, x) x Then (1.1) has at least one positive solution. >, (4.1) <. (4.2) Proof It follows from (4.1) that there exists r 1 >suchthat f (t, x), (t, x) [, 1] [, r 1 ]. (4.3)
Wang and Liu Advances in Difference Equations (217) 217:7 Page 1 of 13 Thus, for any u B r1 P,wehave Au(t)= K(t, s) [ f ( s, u(s) ) + bu(s) ] ds btu(t). (4.4) We may suppose that A has no fixed points on B r1 P (otherwise, the proof is finished). Now we show that u Au μϕ 1, u B r1 P, μ >. (4.5) If otherwise, there exist u 1 B r1 P and μ >suchthat u 1 Au 1 = μ ϕ 1. Then u 1 = Au 1 + μ ϕ 1 μ ϕ 1. Denote μ = supμ : u 1 μϕ 1 }. (4.6) It is clear that μ μ and u 1 μ ϕ 1.SinceT(P) P, wehavebtu 1 μ bt ϕ 1 = μ ϕ 1. Then u 1 = Au 1 + μ ϕ 1 btu 1 + μ ϕ 1 ( μ + μ ) ϕ1, contradicts the definition of μ.hence(4.5) holds and we see from Lemma 2.4 that i(a, B r1 P, P)=. (4.7) On the other hand, it follows from (4.2) that there exist < σ <1andr 2 > r 1 such that f (t, x) (σ 1)bx, t [, 1], x r 2. (4.8) Let T 1 u = σ btu.thent 1 is a bounded linear operator and T 1 (P) P.Set W = u P u = μau, μ 1}. (4.9) In the following, we will prove that W is bounded. For any u W,setũ(t)=minu(t), r 2 }.Then f ( t, u(t) ) + bu(t) σ bu(t)+f ( t, ũ(t) ) + bũ(t). Therefore, u(t)=μau(t) Au(t) σ btu(t)+aũ(t) T 1 u(t)+m,
Wang and Liu Advances in Difference Equations (217) 217:7 Page 11 of 13 where M = max (t,x) [,1] [,r 2 ] } f (t, x)+bx ω 1 (s) ds. (4.1) Thus (I T 1 )u(t) M, t [, 1]. Noticing b is the first eigenvalue of T and < σ <1,we have (r(t 1 )) 1 = σ 1 > 1. So the inverse operator of I T 1 exists, and (I T 1 ) 1 = I + T 1 + T 2 1 + + T n 1 +. It follows from T 1 (P) P that (I T 1 ) 1 (P) P.Wehave u(t) (I T 1 ) 1 M M (I T 1 ) 1, t [, 1], which implies W is bounded. Select r 3 > maxr 2, M (I T 1 ) 1 }. Then by Lemma 2.5,wehave i(a, B r3 P, P) = 1. (4.11) By (4.7)and(4.11), we have i ( A,(B r3 \ B r1 ) P, P ) = i(a, B r3 P, P) i(a, B r1 P, P)=1, which implies that A has at least one fixed point on (B r3 \ B r1 ) P. ThismeansthatBVP (1.1) has at least one positive solution. This completes the proof. Theorem 4.2 Assume that (H 1 ), (H 2 ), and the following assumptions hold: lim sup x + max t [,1] f (t, x) x <, (4.12) and f (t,) on [,1].Then (1.1) has at least one positive solution. Proof It follows from (4.12) that there exists r 1 >suchthat f (t, x), (t, x) [, 1] [, r 1 ]. (4.13) Denote T 2 u = btu.obviously,r(t 2 )=1. We may suppose that A hasnofixedpointson B r1 P (otherwise, the proof is finished). In the following, we prove that Au μu, u B r1 P, μ >1. (4.14) If otherwise, there exist u 1 B r1 P, μ >1,suchthatAu 1 = μ u 1.Itisclearthatμ u 1 = Au 1 T 2 u 1,and μ n u 1 T n 2 u 1, n =1,2,...
Wang and Liu Advances in Difference Equations (217) 217:7 Page 12 of 13 Therefore, μ n u 1 T n 2 u 1 T n 2 u1. Thus r(t 2 )=lim n + n T n 2 μ > 1, which contradicts r(t 2 )=1.Wehavefrom Lemma 2.5 i(a, B r1 P, P)=1. (4.15) Since f (t,) on [, 1], clearly we have Aθ θ, hereθ is the zero element of E. So (4.15) implies that the problem (1.1) has at least one positive solution. Remark 4.1 Suppose u is a positive solution of (1.1), then there exist l 1, l 2 >,suchthat l 1 t α 1 u(t) l 2 t α 1. Example 4.1 (A 3-point boundary value problem at resonance) Consider the following problem: D 3 2 + u(t)+f (t, u(t)) =, < t <1, u() =, u(1) = 2u( 1 4 ). (4.16) Since Ɣ( ) is strictly increasing on [2, + ), forany t [, + ), we have g(t)= 1 + 2 π + t k Ɣ( 3 2 k 1 2 ) = 1 + 2 π + t + k=1 [ 1 + 2 π + t + t k Ɣ(k) = 1 2 π + t = 1 2 π + tet. k=2 1+ k=2 + k=1 t k Ɣ( 3 2 k 1 2 ) ] t k k! Noticing 1 2 π.282, 1 5 e 1 5.243, we have g( 1 5 ) <. Therefore b > b 1 := 1 5. Let f (t, x)= 1 (1 + t)(1 2b)x, (t, x) [, 1] [, 4), 4 1 2 (1 + t)( x bx), (t, x) [, 1] [4, + ), where b (, b 1 ]. It is clear that (H 1 )and(h 2 ) hold. Moreover, lim inf u + lim sup x + min t [,1] max t [,1] f (t, u) u f (t, x) x = 1 2b 4 = b 2 <. >, Therefore the assumptions of Theorem 4.1 are satisfied. Thus Theorem 4.1 ensures that (4.16) has at least one positive solution.
Wang and Liu Advances in Difference Equations (217) 217:7 Page 13 of 13 Competing interests The authors declare that they have no competing interests. Authors contributions All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript. Author details 1 School of Statistics, Qufu Normal University, Qufu, Shandong 273165, P.R. China. 2 School of Mathematical Sciences, Qufu Normal University, Qufu, Shandong 273165, P.R. China. 3 Department of Mathematics and Statistics, Curtin University, Perth, WA 6845, Australia. Acknowledgements The authors would like to thank the referee for his/her very important comments, which improved the results and the quality of the paper. This work was supported financially by the National Natural Science Foundation of China (11371221, 11571296), the Natural Science Foundation of Shandong Province of China (ZR213AQ14, ZR215AL2), the Specialized Research Fund for the Doctoral Program of Higher Education (213375123), and the Project of Shandong Province Higher Educational Science and Technology Program (J13LI8, J15LI16). Received: 21 October 216 Accepted: 16 December 216 References 1. Podlubny, I: Fractional Differential Equations, Mathematics in Science and Engineering. Academic Press, New York (1999) 2. Kilbas, AA, Srivastava, HM, Trujillo, JJ: Theory and Applications of Differential Equations. Elsevier, Amsterdam (26) 3. Su, X, Zhang, S: Unbounded solutions to a boundary value problem of fractional order on the half-line. Comput. Math. Appl. 61, 179-187 (211) 4. 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