Research Article Existence of Positive Solutions for a Kind of Fractional Boundary Value Problems

Abstract and Applied Analysis Volume 24 Article ID 6264 8 pages http://dx.doi.org/.55/24/6264 Research Article Existence of Positive Solutions for a Kind of Fractional Boundary Value Problems Hongjie Liu Xiao Fu 2 and Liangping Qi 2 School of Land Science and Technology China University of Geosciences Beijing 83 China 2 School of Science China University of Geosciences Beijing 83 China Correspondence should be addressed to Hongjie Liu; l hongjie@63.com Received 8 March 24; Accepted 26 April 24; Published 2 May 24 Academic Editor: Yonghui Xia Copyright 24 Hongjie Liu et al. This is an open access article distributed under the Creative Commons Attribution License which permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited. We are concerned with the following nonlinear three-point fractional boundary value problem: D α + u (t) +λa(t) f (t u (t)) <t< u () andu () βu(η)where<α 2 <β< <η< D α + is the standard Riemann-Liouville fractional derivative a (t) >is continuous for t andf is continuous on [ ] [ ). By using Krasnoesel skii s fixed-point theorem and the corresponding Green function we obtain some results for the existence of positive solutions. At the end of this paper we give an example to illustrate our main results.. Introduction Fractional differential equations have been of great interest recently. With the development of nonlinear science the researchers found that nonlinear fractional differential equations could describe something s changing rules more accurately. Therefore it is significant to study nonlinear fractional differential equations to solve the nonlinear problems. Recently many researchers paid attention to existence and multiplicity of solution of the boundary value problem for fractional differential equations with different boundary conditions such as [ ]. Bai and Lü [] investigated the existence and multiplicity of positive solutions for nonlinear fractional boundary value problem: D α + u (t) +f(t u (t)) <t< u () u() where < α 2 is a real number D α + is the standard Riemann-Liouville fractional derivative and f : [] [ ) [ ) is continuous. By means of some fixedpoint theorems on cone some existence and multiplicity results of positive solutions are obtained. Agarwal et al. [2] () investigated the existence of positive solutions for the singular fractional boundary value problem: D α + u (t) +f(tu(t) Dμ u (t)) < t < u () u() where <α 2μ>are real numbers f is positive and D α + is the standard Riemann-Liouville fractional derivative. By means of a fixed point theorem on cone the existence of positive solutions is obtained. Delbosco and Rodino [3] investigated the nonlinear Dirichlet-type problem: x α (D α + y) (x) f(y(x)) y () y(). <α<2 x They proved that if f(y) is a Lipschitzian function then the problem has at least one solution y(x) in a certain subspace of C[ ]. In this paper we study the following three-point fractional boundary value problem. Consider D α + u (t) +λa(t) f (t u (t)) <t< u () βu(η)u() (2) (3) (4)

2 Abstract and Applied Analysis where < α 2 < β < < η < D α + is the standard Riemann-Liouville fractional derivative a(t) > is continuous and f is continuous on [ ] [ ). By using Krasnoesel skii s fixed-point theorem we get the existence of at least one positive solution. 2. Background Materials and Preliminaries For convenience of the readers we present here the necessary definitions from fractional calculus theory. These definitions canbefoundintherecentliterature[ 5]. Definition. The fractional integral of order α > of a function y : ( ) R is given by I α t + y (t) Γ (α) (t s) α y (s) ds (5) provided that the right side is pointwise defined on ( ). Definition 2. The fractional derivative of order α > of a function y : ( ) R is given by D α + y (t) Γ (n α) ( d dt ) n t y (s) ds n α n+ [α] + (t s) (6) where n[α]+providedthattherightsideispointwise defined on ( ). Lemma 3. Let α> if we assume u C( ) L( )then the fractional differential equation D α + u(t) has as unique solutions. u (t) C t α +C 2 t α 2 + C N t α N C i R i2...n Lemma 4. Assume that u C( ) L( ) with a fractional derivative of order α>that belongs to C( ) L( ).Then (7) I α + Dα + u (t) u(t) +C t α +C 2 t α 2 + C N t α N (8) for some C i R i2...n. Definition 5. Let E bearealbanachspace.anonemptyclosed convex set K Eis called cone of E if it satisfies the following conditions: () x Kσ>implies σx K; (2) x K x K implies x. Lemma 6. Let E be a Banach space and let K Ebe a cone in E. Assume that Ω and Ω 2 are open subsets of E with Ω Ω Ω 2.LetT:K (Ω 2 \Ω ) K be a completely continuous operator. In addition suppose that either (H) Tu u u K Ω and Tu u u K Ω 2 or (H2) Tu u u K Ω and Tu u u K Ω 2 holds. Then T has a fixed point in K (Ω 2 \Ω ). In the following we present the Green function of fractional differential equation boundary value problem. Lemma 7 (see [4]). Let y C[ ] then the boundary value problem D α + u (t) +y(t) <t< u () βu(η)u() (9) has a unique solution u (t) G (t s) y (s) ds () where ([t( s)] α βt α (η s) α (t s) α ( βη α )) (( βη α )Γ(α)) s min t η} ; [t( s)] α (t s) α ( βη α ) ( βη α )Γ(α) G (t s) <η s t ; () [t( s)] α βt α (η s) α ( βη α )Γ(α) t s η<; [t( s)] α ( βη α )Γ(α) max t η} s. Lemma 8. The function G(t s) defined by () satisfies G(t s) > for t s ( ). Proof. () For s mint η} G (t s) [t( s)]α βt α (η s) α (t s) α ( βη α ). ( βη α )Γ(α) (2) Let F(t s η) [t( s)] α βt α (η s) α (t s) α ( βη α ).Thereis (t s η) η βt α (α ) (η s) α 2 + (α ) (t s) α βη α 2 β(α ) [(t s) α η α 2 t α (η s) α 2 ] β(α ) [t α η α 2 t α (η s) α 2 ] β(α ) t α [η α 2 (η s) α 2 ]. (3)

Abstract and Applied Analysis 3 With < β < < α 2thereis/ η <.Hence function F(t s η) is monotonically decreasing in η.with < η< F(t s η) > F(t s η) min F(t s ).Letηthereis F (t s ) [t( s)] α βt α ( s) α (t s) α ( β) ( β) [t( s)] α (t s) α } ( β) [(t ts) α (t s) α ]. (4) With <s t<<α 2wehave(t ts) α (t s) α. With β > thereisf(t s ) thenf(t s η) > F(t s ) thereisg(t s) >. (2) For <η s t< G (t s) [t( s)]α (t s) α ( βη α ). (5) ( βη α )Γ(α) Let F(t s η) [t( s)] α (t s) α ( βη α );itisobvious that function F(t s η) is monotonically increasing in ηwhen η F(t s η) min F(t s ) [t( s)] α (t s) α (t ts) α (t s) α. (6) With s t < < α thereis(t ts) α (t s) α ;wehavef(t s η) > F(t s ) theng(t s) >. (3) For <t s η< G (t s) [t( s)]α βt α (η s) α. (7) ( βη α )Γ(α) Let F(t s η) [t( s)] α βt α (η s) α thenf(t s η) is monotonically decreasing in η F(t s η) min F(t s ) [t( s)] α βt α ( s) α ( β) [t( s)] α. (8) With <β<<t s<thereisf(t s η) > then F(t s η) > F(t s ) henceg(t s) >. (4) For maxtη} s G (t s) [t( s)] α >. (9) ( βη α )Γ(α) Therefore G(t s) > for t s ( ). Theproofiscomplete. Lemma 9. G(t s) G(s s) for t s [ ]. Proof. To get G(t s) G(s s) firstly we prove the following inequality: t a s a (t s) a sa ( s) a (2) where aα s<t.letf(t) (t a s a )/(t s) a then f (t) ata (t s) a a(t a s a )(t s) a (t s) 2a a(t s)a (t s) 2a [t a (t s) (t a s a )] a(t s)a (t s) 2a s(s a t a ). (2) With <α 2thereisa ;withs<tthereiss a > t a ;thenf (t) >. Itmeansthatf(t) is monotonically increasing in tthen f(t) f();we have t a s a (t s) a sa ( s) a. (22) Here we prove G(t s) G(s s). () For s mint η} G (t s) [t( s)]α βt α (η s) α (t s) α ( βη α ) ( βη α )Γ(α) G (s s) [s( s)]α βs α (η s) α. ( βη α )Γ(α) With ( βη α )Γ(α) > G(t s) G(s s) is equivalent to or (23) t a s a (t s) a [( s) a β(η s) a ] βη a (24) (t a s a )[( s) a β(η s) a ] ( βη a ) (t s) a. (25) (a) For st (t a s a )[( s) a β(η s) a ] ( βη a )(t s) a ; (b) for stη (t a s a )[( s) a β(η s) a ] ( βη a )(t s) a ; (c) for sη t (t a s a )[( s) a β(η s) a ](t a η a )( η a ) ( βη a ) (t s) a ( βη a )(t η) a (26) if t η (t a s a )[( s) a β(η s) a ] ( βη a )(t s) a ;ift>ηfrom(22) we have (t a η a )/ (t η) a ( η a )/( η) a ( βη a )/( η) a ;

4 Abstract and Applied Analysis (d) for s<t s<η if( s) a β(η s) a < ; itisobviousthat(t a s a )[( s) a β(η s) a ] ( βη a )(t s) a.if( s) a β(η s) a >by(22) there is (η a s a )/(η s) a ( s a )/( s) a ;then (η s) a sa ( s) a (η a s a ). (27) Consider t a s a (t s) a [( s) a β(η s) a ] sa ( s) a [( s) a β(η s) a ] s a β sa ( s) a (η s) a s a β sa ( s)a ( s) a s a (η a s a ) βη a ( β)s a βη a. It means that G(t s) G(s s) holds. (2) For <η s t G (t s) [t( s)]α (t s) α ( βη α ) ( βη α )Γ(α) G (s s) [s( s)] α ( βη α )Γ(α). With ( βη α )Γ(α) > G(t s) G(s s) is equivalent to or (28) (29) [t( s)] a (t s) a ( βη a ) [s( s)] a (3) (t a s a )( s) a (t s) a ( βη a ). (3) From (22) we have ((t a s a )/(t s) a )( s) a s a ;with η s<β<wehave s a η a βη a ;then ((t a s a )/(t s) a )( s) a βη a ;withs tthereis (t a s a )( s) a (t s) a ( βη a ). (32) It means that G(t s) G(s s) holds. (3) For t s η< G (t s) [t( s)]α βt α (η s) α ( βη α )Γ(α) G (s s) [s( s)]α βs α (η s) α. ( βη α )Γ(α) With ( βη α )Γ(α) > G(t s) G(s s) is equivalent to (33) [t( s)] a βt a (η s) a [s ( s)] a βs a (η s) a (34) or (t a s a ) ( s) a β(t a s a ) (η s) a. (35) With <β< s ηthereisβ(η s) a (η s) a <( s) a ; with t sthereis (t a s a ) ( s) a β(t a s a ) (η s) a. (36) It means that G(t s) G(s s) holds. (4) For <maxtη} s G (t s) G (s s) [t( s)] α ( βη α )Γ(α) [s( s)] α ( βη α )Γ(α). It is obvious that G(t s) G(s s). The proof is complete. (37) Lemma. G(t s) q(k)g(s s) for t s [/k (/k)] where k>2is an integer and q (k) min ( β) [( (/k))2(α ) ( (2/k)) α ] (/4) α Proof. From () we have ( β) (/k) 2(α ) (/4) α (k ) α /k 2(α ) (/4) α }. [s( s)] α βs α (η s) α s η ( βη G (s s) α )Γ(α) [s( s)] α s > η. ( βη α )Γ(α) (38) (39) It is obvious that [s( s)] α /( βη α )Γ(α) ([s( s)] α βs α (η s) α )/( βη α )Γ(α) for t s [/k (/k)] max G (s s) [s( s)] α ( βη α )Γ(α) s/2 (/4) α ( βη α )Γ(α). () For /k s mint η} (/k)let F (t s η) [t( s)] α βt α (η s) α (t s) α ( βη α ). (4) (4)

Abstract and Applied Analysis 5 By Lemma 8forη F (t s) F(t s η) min F(t s ) [t( s)] α βt α ( s) α (t s) α ( β) ( β) [t( s)] α (t s) α } ( β)[(t ts) α (t s) α ] s ( β)(α ) [(t s)α 2 t α ( s) α 2 ]. (42) With <α 2 s t then(t s) α 2 t α ( s) α 2 >;with<β< / s >.ItmeansthatF(t s) is monotonically increasing in s.similarly t ( β)(α ) [(t ts)α 2 ( s) (t s) α 2 ]<. (43) It means that F(t s) is monotonically decreasing in t. Therefore F (t s) min F (t s) F(t s) t (/k)s/k min G (t s) ( β)[( 2(α ) k ) ( 2 α ] k ) ( β) [((/k) (/k 2 )) α ( (2/k)) α ] ( βη α )Γ(α) min G (t s) max G (s s) ( β) [( (/k)) 2(α ) ( (2/k)) α ] (/4) α. (44) (2) For /k<η s t (/k)let F (t s η) [t( s)] α (t s) α ( βη α ). (45) By Lemma 8forη F (t s) F(t s η) min F(t s ) [t( s)] α (t s) α (t ts) α (t s) α s (α ) [(t s)α 2 t α ( s) α 2 ]>. It means that F(t s) is monotonically increasing in s (46) t (α ) [( s)α t α 2 (t s) α 2 ]<. (47) It means that F(t s) is monotonically decreasing in t F (t s) min F (t s) F(t s) t (/k)s(/k) ( 2(α ) k ) ( 2 α k ) min G (t s) ( β) [( (/k))2(α ) ( (2/k)) α ] ( βη α )Γ(α) min G (t s) max G (s s) ( β) [( (/k)) 2(α ) ( (2/k)) α ] (/4) α. (48) (3) For /k t s η< (/k)let F (t s η) [t( s)] α βt α (η s) α. (49) By Lemma 8forη F (t s) F(t s η) min F(t s ) [t( s)] α βt α ( s) α ( β)[t ( s)] α s (α ) tα (β ) ( s) α 2 <; then F(t s) is monotonically decreasing in s. Consider (5) t (α )( β) tα 2 ( s) α >; (5) then F(t s) is monotonically increasing in t. Therefore F (t s) min F (t s) F(t s) t/ks (/k) β k 2(α ) min G (t s) ( β) (/k2(α ) ) ( βη α )Γ(α) min G (t s) ( β) (/k)2(α ) max G (s s) (/4) α. (4) For /k maxtη} s (/k)let (52) F (t s) [t( s)] α. (53) It is obvious that F(t s) is monotonically increasing in t and monotonically decreasing in s;then we have F (t s) min F (t s) F(t s) t/ks (/k) ( k k 2 ) α ; (54) min G (t s) ((/k) (/k2 )) α ; (55) ( βη α )Γ(α)

6 Abstract and Applied Analysis there is Let min G (t s) max G (s s) (k )α /k 2(α ) (/4) α. (56) q (k) min ( β) [( (/k))2(α ) ( (2/k)) α ] (/4) α ( β) (/k) 2(α ) (/4) α (k ) α /k 2(α ) (/4) α }. (57) Therefore G(t s)/g(s s) min G(t s)/ max G(s s) q(k). The proof is complete. We consider the Banach space XC[]equipped with standard norm u max t u(t) u X. We define a cone P by P u X:u(t) t [ ] min u (t) q(k) u }. t [/k /k] Define an integral operator T:P Xby (Tu)(t) λ G (t s) a (s) f (s u (s)) ds t u P. (58) (59) Lemma. It holds the following. () T:P Pis completely continuous. (2) u(t) is a positive solution of the fractional boundary value problem (4) if and only if u(t) is a fixed point of the operator T in cone P. Proof. For u Pt [/k (/k)]bylemma G(t s) q(k)g(s s);fort s [/k (/k)] Tu (t) λ G (t s) a (s) f (s u (s)) ds λq(k) G (s s) a (s) f (s u (s)) ds λq(k) max G (t s) a (s) f (s u (s)) ds /k t /k q(k) Tu (t). (6) It is obvious that min /k t /k (Tu)(t) q(k) Tu TP P. Thus TP P. In addition standard arguments show that T is completely continuous. (2) It is obvious that u(t) is the positive solution of BVP (4)if and only if u (t) λ G (t s) a (s) f (s u (s)) ds t. (6) It can be proved by the definition of integral operator T. 3. Main Results We denote some important constants as follows: A G (s s) a (s) q (k) ds B G (s s) a (s) ds F lim u f lim u + F lim u f lim u + t + sup max inf min t t + sup max inf min t f (t u (t)) u (t) f (t u (t)) u (t) f (t u (t)) u (t) f (t u (t)). u (t) (62) Here we assume that /Af if f /BF if F /Af if f and/bf if F. Theorem 2. Suppose that Af > BF thenforeachλ (/Af /BF )BVP(4) has at least one positive solution. Proof. We choose ε > sufficiently small such that (F + ε)λb. By the definition of F wecanseethatthereexists l >suchthatf(t u(t)) (F + ε)u(t) for u l.for u Pwith u l wehave Tu (t) λ G (s s) a (s) f (s u (s)) ds λ G (s s) a (s) (F +ε)u(s) ds λ(f +ε) u G (s s) a (s) ds λb(f +ε) u u. (63) Then we have Tu u. Thus if we let Ω u X: u < l }then Tu u for u P Ω.Wechooseδ>and c ( /4) such that λ(f δ)a. There exists l 2 >l >

Abstract and Applied Analysis 7 such that f(t u) (f δ)ufor u>l 2. Therefore for each u Pwith u l 2 wehave Tu (t) λ G (s s) a (s) f (s u (s)) ds λ G (s s) a (s) (f δ)u(s) ds c λ(f δ) u G (s s) a (s) q (k) ds c λ(f δ)a u u. (64) Thus if we let Ω 2 u E : u < l 2 }thenω Ω 2 and Tu u for u P Ω 2. Condition (H) of Krasnoesel skii s fixed point theorem is satisfied. So there exists a fixed point of T in P.Thiscompletestheproof. Theorem 3. Suppose that Af > BF thenforeachλ (/Af /BF )BVP(4) has at least one positive solution. Proof. Choose ε>sufficiently small such that (f ε)λa. From the definition of f we see that there exists l > such that f(t u) (f ε)ufor <u l.ifu Pwith u l wehave Tu (t) λ G (s s) a (s) f (s u (s)) ds λ(f ε) u A u. (65) Let Ω u X : u < l };thenwehave Tu u for u P Ω.Chooseδ > c ( /4); thenwehave λ(f +δ)a. There exists l 2 >l >suchthatf(t u) (F +δ)ufor u>l 2.Foreachu Pwith u l 2 wehave Tu (t) λ G (s s) a (s) f (s u (s)) ds λ G (s s) a (s) (F +δ)u(s) ds c λ(f +δ) u G (s s) a (s) q (k) ds c λ(f +δ)a u u. (66) Let Ω 2 u E : u < l 2 };thenω Ω 2 and we have Tu u for u P Ω 2. Condition (H2) of Krasnoesel skii s fixed-point theorem is satisfied. So there exists a fixed point of T in P.Thiscompletestheproof. 4. An Example Example. Consider the following three-point fractional boundary value problem: D 3/2 + u (t) +λf(t u (t)) <t< u () u() (67) 2 u( 3 ) where f(t u(t)) ((t + )(24u 3 + u)/(u 2 + )) + u(sin u+ ) a(t). By calculations [s( s)] α βs α (η s) α ( βη α )Γ(α) st η G (s s) [s( s)] α η G (s s) ds Let kthen ( βη α )Γ(α) η ts [s ( s)] α βs α (η s) α ds ( βη α )Γ(α) [s ( s)] α + η ( βη α )Γ(α) ds [s ( s)] α ( βη α )Γ(α) ds η βs α (η s) α ( βη α )Γ(α) ds (/2)(/3) /2 Γ (3/2) [s( s)] /2 ds /3 2 s/2 (η s) /2 ds}.58832. q (k) min ( β) [( (/k))2(α ) ( (2/k)) α ] (/4) α.557 ( β) (/k) 2(α ) (/4) α (k ) α /k 2(α ) (/4) α }. (68)

8 Abstract and Applied Analysis f lim inf min [(t+) (24u2 +) u + t u 2 + sin u+] + 24 F lim sup max [(t+) (24u2 +) u + t u 2 + sin u+] + 4. Consider (69) A G (s s) a (s) q (k) ds.328 (7) B G (s s) a (s) ds.58832. (7) [7] X. Xu D. Jiang and C. Yuan Multiple positive solutions for the boundary value problem of a nonlinear fractional differential equation Nonlinear Analysis: Theory Methods & Applications vol.7no.pp.4676 468829. [8] B. Ahmad Existence of solutions for irregular boundary value problems of nonlinear fractional differential equations Applied Mathematics Lettersvol.23no.4pp.39 3942. [9]S.LiangandJ.Zhang Positivesolutionsforboundaryvalue problems of nonlinear fractional differential equation Nonlinear Analysis: Theory Methods & Applications vol. 7 no. pp. 5545 555 29. [] O. K. Jaradat A. Al-Omari and S. Momani Existence of the mild solution for fractional semilinear initial value problems Nonlinear Analysis: Theory Methods & Applicationsvol.69no. 9pp.353 35928. The condition Af > BF is obtained. From Theorem 2 we see that if λ (.538.42494) theproblem(67) has apositivesolution. Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper. Acknowledgments The authors are grateful to the reviewers for the valuable suggestions. This paper was supported by the Fundamental Research Funds for the Central Universities (no. 265224) and Beijing Higher Education Young Elite Teacher Project. References [] Z. Bai and H. Lü Positive solutions for boundary value problem of nonlinear fractional differential equation Journal of Mathematical Analysis and Applications vol.3no.2pp. 495 55 25. [2] R.P.AgarwalD.O Reganand S.Staněk Positive solutions for Dirichlet problems of singular nonlinear fractional differential equations Mathematical Analysis and Applications vol.37no.pp.57 682. [3] D. Delbosco and L. Rodino Existence and uniqueness for a nonlinear fractional differential equation Mathematical Analysis and Applications vol.24no.2pp.69 625 996. [4] Z. Bai On positive solutions of a nonlocal fractional boundary value problem Nonlinear Analysis: Theory Methods & Applicationsvol.72no.2pp.96 9242. [5]R.P.AgarwalM.BenchohraandS.Hamani Asurveyon existence results for boundary value problems of nonlinear fractional differential equations and inclusions Acta Applicandae Mathematicaevol.9no.3pp.973 332. [6] M. El-Shahed Positive solutions for boundary value problem of nonlinear fractional differential equation Abstract and Applied Analysisvol.27ArticleID3688pages27.

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