Unique positive solutions for fractional differential equation boundary value problems

Applied Mathematics Letters 23 (2010) 1095–1098

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Unique positive solutions for fractional differential equation boundary value problemsI Liu Yang a,b , Haibo Chen a,∗ a

Department of Mathematics, Central South University, Changsha, Hunan 410075, PR China

b

Department of Mathematics, Hengyang Normal University, Hengyang, Hunan 421008, PR China

article

info

Article history: Received 26 October 2009 Received in revised form 29 April 2010 Accepted 29 April 2010 Keywords: Riemann–Liouville differentiation Fractional differential equation Positive solution Iterative schemes Boundary value problem

abstract In this work, we consider the uniqueness of positive solutions for fractional differential equation boundary value problems. Our results can not only guarantee the existence of a unique positive solution, but also be applied to construct an iterative scheme for approximating it. © 2010 Elsevier Ltd. All rights reserved.

1. Introduction Fractional differential equations arise in many fields, such as physics, mechanics, chemistry, engineering and biological sciences, etc; see [1–3]. In applications one is interested in showing the existence of positive solutions. In consequence, many authors have investigated the existence of positive solutions for fractional differential equation boundary value problems; see [4–11]. In particular, by means of a mixed monotone method, Xu, Jiang and Yuan [10] considered the uniqueness of the solution for the following singular boundary value problem:



Dα0+ u(t ) − f (t , u(t )) = 0, 0 < t < 1, 3 < α ≤ 4, u(0) = u0 (0) = u(1) = u0 (1) = 0,

(1.1)

where f (t , u) = q(t )[g (u) + h(u)], g : [0, +∞) → [0, +∞) is continuous and nondecreasing, h : (0, +∞) → (0, +∞) R1 is continuous and nonincreasing, q ∈ C ((0, 1), (0, +∞)) satisfies 0 s2−η(2−α) (1 − s)α−2−2η q(s)ds < +∞, η ∈ (0, 1). By a similar method, in [11], Zhang considered a unique positive solution for the singular boundary value problem



Dα0+ u(t ) + q(t )f (u, u0 , . . . , u(n−2) ) = 0, 0 < t < 1, n − 1 < α ≤ n, n ≥ 2, u(0) = u0 (0) = · · · = u(n−2) (0) = u(n−2) (1) = 0,

where f = g + h, and g , h have different monotone properties.

I This work was supported by the National Natural Science Foundation of China (no. 10871206).

∗

Corresponding author. E-mail addresses: [email protected] (L. Yang), [email protected] (H. Chen).

0893-9659/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.aml.2010.04.042

(1.2)

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L. Yang, H. Chen / Applied Mathematics Letters 23 (2010) 1095–1098

Motivated by the works mentioned above, in this work, by means of fixed point theory for the u0 concave operator, we obtain the uniqueness of the solution for the following nonsingular boundary value problem without making the assumption f = q(t )(g + h) as in [10,11]:



Dα0+ u(t ) + f (t , u, u0 , . . . , u(n−2) ) = 0, 0 < t < 1, n − 1 < α ≤ n, n ≥ 2, u(0) = u0 (0) = · · · = u(n−2) (0) = u(n−2) (1) = 0.

(1.3)

Moreover, we can construct an iterative scheme to help us to find the solution, which is important for evaluation and application. 2. Preliminaries In this section, we will present some definitions and lemmas that will be used in the proof of our main results. Definition 2.1 ([3]). The Riemann–Liouville fractional integral of order α > 0 of a function y : (0, ∞) → R is given by I0α+ y(t )

=

t

Z

1

Γ (α)

(t − s)α−1 y(s)ds,

0

where the right side is defined pointwise on (0, ∞). Definition 2.2 ([3]). The Riemann–Liouville fractional derivative of order α > 0 of a continuous function y : (0, ∞) → R is given by Dα0+ y(t ) =



1

Γ (n − α)

d dt

n Z

t

(t − s)α−1 y(s)ds,

0

where n = [α] + 1, [α] denotes the integer part of the number α , and the right side is pointwise defined on (0, ∞). Like in the method in [11], we can transform (1.3) into the following problem:



n +2 Dα− v(t ) + f (t , I0n+−2 v(t ), . . . , I01+ v(t ), v(t )) = 0, 0+ v(0) = v(1) = 0.

0 < t < 1,

(2.1)

−2 Hence, if v(t ) ∈ C ([0, 1], [0, +∞)) is a solution of problem (2.1), then u(t ) = I0n+ v(t ) is a positive solution of problem (1.3). On the other hand, if v(t ) ∈ C ([0, 1], [0, +∞)) is a solution of problem (2.1), then v(t ) satisfies

v(t ) = Av :=

1

Z

−2 G(t , s)f (s, I0n+ v(s), . . . , I01+ v(s), v(s))ds,

(2.2)

0

where

 (t (1 − s))α−n+1 − (t − s)α−n+1    , Γ (α − n + 2) G(t , s) = α−n+1   (t (1 − s))  , Γ (α − n + 2)

0 ≤ s ≤ t ≤ 1, (2.3) 0 ≤ t ≤ s ≤ 1.

From [11], we know that 0 ≤ G(t , s) ≤ G(s, s). To complete our work, we first present the following new properties of G(t , s). Lemma 2.3. ∀t , s ∈ (0, 1), the function G(t , s) defined by (2.3) satisfies

α − n + 1 α−n+1 t α−n+1 (1 − t )(1 − s)α−n t (1 − t )(1 − s)α−n+1 s ≤ G(t , s) ≤ . Γ (α − n + 2) Γ (α − n + 2)

(2.4)

The proof of Lemma 2.3 is very similar to that of Theorem 1.1 in [5], so we omit it here. Lemma 2.3 generalizes Theorem 1.1 in [5]. Let E = 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 = max0≤t ≤1 |u(t )|. Define the cone P ⊂ E by P = {u ∈ E : u(t ) ≥ 0} . Assume that u0 ∈ P , u0 > 0, i.e., u0 (t ) is not identically vanishing. Let Pu0 = {u : u ∈ E , ∃ λ(u) > 0, µ(u) > 0 s.t. λ(u)u0 ≤ u ≤ µ(u)u0 } . Definition 2.4 ([12]). Assume that A : P → P , u0 > 0. A is said to be a u0 concave operator if A satisfies: (i) ∀u > 0, Au ∈ Pu0 ; (ii) there exists η(r , u) > 0 such that A(ru) ≥ r (1 + η(r , u))Au, ∀u ∈ Pu0 , 0 < r < 1.

L. Yang, H. Chen / Applied Mathematics Letters 23 (2010) 1095–1098

1097

Lemma 2.5 ([12]). Assume that P is a normal cone of E, u0 > 0, A : Pu0 → Pu0 is increasing and there exists η(r ) > 0 such that A(ru) ≥ r (1 + η(r ))Au,

∀u ∈ Pu0 , 0 < r < 1.

Then A has a unique fixed point u ∈ Pu0 if and only if there exist w0 , v0 ∈ Pu0 such that w0 ≤ Aw0 ≤ Av0 ≤ v0 . ∗

Remark 2.6 ([12]). The above fixed point of A can be approximated by using the following iterative schemes: for any x0 ∈ [w0 , v0 ], letting xn = Axn−1 , n = 1, 2, . . ., one always obtains xn → u∗ . 3. Main results Here we make the following hypotheses: (A) f ∈ C [0, 1] × [0, ∞) × Rn−2 → [0, ∞) , f (t , y1 , y2 , . . . , yn−1 ) is increasing for yi ≥ 0, i = 1, 2, . . . , n − 1, and f is not identically vanishing. (B) For any t ∈ [0, 1], yi ≥ 0, i = 1, 2, . . . , n − 1, there exist constant m1 , m2 , m1 ≤ 0 < m2 < 1 such that (i) c m2 f (t , y1 , y2 , . . . , yn−1 ) ≤ f (t , cy1 , cy2 , . . . , cyn−1 ) ≤ c m1 f (t , y1 , y2 , . . . , yn−1 ), ∀0 < c ≤ 1.




Remark 3.1. Assume that (B) holds. Applying u = ≤ c m2 f (t , y1 , y2 , . . . , yn−1 ), ∀c ≥ 1.

1 cu, we can have (ii) c m1 f c

(t , y1 , y2 , . . . , yn−1 ) ≤ f (t , cy1 , cy2 , . . . , cyn−1 )

Theorem 3.1. Assume that (A), (B) are satisfied; then the problem (1.3) has a unique positive solution when m2 <

2m1 +1 . 3

Proof. Let e(t ) = t α−n+1 (1 − t ). By Lemma 2.3, for any given v ∈ Pe , t ∈ [0, 1], we have 1

Z

A v ≤ e( t )

0

(1 − s)α−n −2 f (s, I0n+ v(s), . . . , I01+ v(s), v(s))ds, Γ (α − n + 2)

(3.1)

α−n+1 −2 s(1 − s)α−n+1 f (s, I0n+ v(s), . . . , I01+ v(s), v(s))ds. Γ (α − n + 2)

(3.2)

and 1

Z

A v ≥ e( t )

0

From (A), for any v ∈ Pe , there exists M (v) > 0 such that

|f (s, I0n+−2 v(s), . . . , I01+ v(s), v(s))| ≤ M (v). (3.3) R R 1 (1−s)α−n α− n 1 −s ) −2 −2 v(s), . . . , I01+ v(s), v(s))ds is well defined. Choosing λ(v) = 0 Γ(1(α− v(s), . . . , I01+ Hence, 0 Γ (α−n+2) f (s, I0n+ f (s, I0n+ n+2) R 1 α−n+1 v(s), v(s))ds, and µ(v) = 0 Γ (α−n+2) s(1 − s)α−n+1 f (s, I0n+−2 v(s), . . . , I01+ v(s), v(s))ds, we have λ(v)e ≤ u ≤ µ(v)e, i.e. A : Pe → Pe . From (A), A is an increasing operator. From (B), we obtain Z 1 −2 A(c v) = G(t , s)f (s, I0n+ c v(s), . . . , I01+ c v(s), c v(s))ds 0 1

Z

−2 G(t , s)f (s, cI0n+ v(s), . . . , cI01+ v(s), c v(s))ds

= 0

≥ c m2

1

Z

−2 G(t , s)f (s, I0n+ v(s), . . . , I01+ v(s), v(s))ds 0

= c [1 + (c m2 −1 − 1)]Av,

0 < c < 1.

By 0 < m2 < 1, 0 < c < 1, we have η(c ) = c m2 −1 − 1 > 0. Let h(t ) =

R1 0

n R 1 I2 = max 1, 0

α−n+1 s( 1 Γ (α−n+2)

− s)α−n+1 f (s,

−2 I0n+ e

(1−s)α−n −2 f (s, I0n+ e(s), . . . , I01+ e(s), e(s))ds 0 Γ (α−n+2)

n R 1

−2 G(t , s)f (s, I0n+ e(s), . . . , I01+ e(s), e(s))ds, I1 = min 1,

Since

(m2 −m1 ) 1−m2 4I2

Moreover,

dI1 k1 k2

(s), . . . , I01+ e(s), e(s))ds . Like for (3.1) and (3.2), we can easily prove that (3.4)

2(m2 −m1 ) 1− 1− m2

−(m2 −m1 ) 2(m2 −m1 ) 1−m2 1 1− 1−m2 d I I 1 2 4

m2 −m1 −1 1−m2

≥ max{4I1 I2

≥ 1, we get d ≥ 1. If we let k1 = =

,

o

I1 e(t ) ≤ h(t ) ≤ I2 e(t ). Let d > 0 be a constant such that d

o

1 2

m

m2 −m1 1−m2

, 4I2 1

}. From m2 <

2m1 +1 , we have 1 3 m

1

− 2(m1−2 −mm2 1 ) > 0.

(dm1 −m2 I2 1 ) 1−m2 , k2 = 2(dm2 −m1 I2 2 ) 1−m2 , then k1 ≤

≥ 1. Hence, dI1 ≥ 1.

1 2

, k2 ≥ 2.

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L. Yang, H. Chen / Applied Mathematics Letters 23 (2010) 1095–1098

Let w0 (t ) = k1 h(t ), v0 (t ) = k2 h(t ); then w0 , v0 ∈ Pe , and w0 < v0 . Hence Aw0 < Av0 . Moreover, from (A), (B), (3.4) and the definitions of k1 , d, I1 , I2 , we have −2 f (t , I0n+ w0 , . . . , I01+ w0 , w0 ) = f (s, I0n+−2 k1 h, . . . , I01+ k1 h, k1 h)

 ≥

m2

d

 ≥

k1

m2

d

 ≥

k1

k1

m2

d

−2 f (t , I0n+ dh, . . . , I01+ dh, dh) −2 f (t , I0n+ dI1 e(t ), . . . , I01+ dI1 e(t ), dI1 e(t ))

(dI1 )m1 f (t , I0n+−2 e(t ), . . . , I01+ e(t ), e(t )) m

m

≥ k1 2 dm1 −m2 I2 1 f (t , I0n+−2 e(t ), . . . , I01+ e(t ), e(t )) ≥ k1 f (t , I0n+−2 e(t ), . . . , I01+ e(t ), e(t )). −2 Similarly, we can get f (t , I0n+ v0 , . . . , I01+ v0 , v0 ) ≤ k2 f (t , I0n+−2 e(t ), . . . , I01+ e(t ), e(t )). Hence, for any t ∈ [0, 1], we have

w0 (t ) = k1

1

Z

−2 G(t , s)f (s, I0n+ e(s), . . . , I01+ e(s), e(s))ds 0

1

Z

−2 G(t , s)k1 f (s, I0n+ e(s), . . . , I01+ e(s), e(s))ds

= 0 1

Z

−2 G(t , s)f (t , I0n+ w0 , . . . , I01+ w0 , w0 )ds

≤ 0

= Aw0 (t ). Similarly, we can obtain v0 ≥ Av0 . Hence, by means of Lemma 2.5, the problem (1.3) has a unique positive solution.



Remark 3.2. The unique positive solution u∗ in Theorem 3.1 can be approximated by the following iterative schemes: for any x0 ∈ [w0 , v0 ], letting xn = Axn−1 , n = 1, 2, . . . , one always obtains xn → u∗ . Example 3.1. Consider the following problem:

(

3

D02+ u(t ) + f (t , u(t )) = 0, u(0) = u(1) = 0.

It is obvious that α =

0 < t < 1,

(3.5)

t +ε

, n = 2. Let f (t , u) = u 7 , 0 < ε < 1, where ε is a constant. Letting m1 = 0, m2 = 27 , we have: m2 (i) c f (t , u) ≤ f (t , cu) ≤ c m1 f (t , x), ∀0 < c ≤ 1; (ii) c m1 f (t , u) ≤ f (t , cu) ≤ c m2 f (t , x), ∀c ≥ 1. Then (A), (B) hold true and 27 < 31 . Therefore, problem (3.5) has a unique positive solution. The nonlinearity f in Example 3.1 cannot be denoted by f (t , u) = q(t )[g (u) + h(u)], so the positive solution of problem (3.5) cannot be obtained 3 2

by virtue of [10,11]. In addition, by Remark 3.2, we can establish an iterative scheme. References [1] A. Anatoly Kilbas, M.Hari Srivastava, J. Juan Trujillo, Theory and Applications of Fractional Differential Equations, in: North-Holland Mathematics Studies, vol. 204, Elsevier Science B.V., Amsterdam, 2006. [2] B. Ross (Ed.), The Fractional Calculus and its Applications, in: Lecture Notes in Mathematics, vol. 475, Springer-Verlag, Berlin, 1975. [3] I. Podlubny, Fractional Differential Equations, in: Mathematics in Science and Engineering, vol. 198, Academic Press, New York, London, Toronto, 1999. [4] Bashir Ahamed, S. Sivasundaram, Theory of fractional differential equations with three point boundary conditions, Communication in Applied Analysis 12 (2008) 485–489. [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 Analysis 72 (2010) 710–719. [6] E.R. Kaufmann, E. Mboumi, Positive solution of a boundary value problem for a nonlinear fractional differential equation, Electronic Journal of Qualitative Theory of Differential Equations 3 (2008) 1–11. [7] Bashir Ahamed, S. Sivasundaram, Existence and uniqueness results for nonlinear boundary value problem of fractional differential equation with separated boundary conditions, Communication in Applied Analysis 13 (2009) 121–129. [8] Z. Bai, On positive solutions of a nonlocal fractional boundary value problem, Nonlinear Analysis 72 (2010) 916–924. [9] C.F. Li, X. Luo, Y. Zhou, Existence of positive solutions of the boundary value problem for nonlinear fractional differential equations, Computers and Mathematics with Applications 59 (2010) 1363–1375. [10] X. Xu, D. Jiang, Chengjun Yuan, Multiple positive solutions for the boundary value problem of a nonlinear fractional differential equation, Nonlinear Analysis 71 (2009) 4676–4688. [11] S. Zhang, Positive solutions to singular boundary value problem for nonlinear fractional differential equation, Computers and Mathematics with Applications 59 (2010) 1300–1309. [12] W. Wang, Z. Liang, Fixed point theorem of a class of nonlinear operators and applications, Acta Mathematica Sinica 48 (2005) 789–800.