Computers and Mathematics with Applications 62 (2011) 1333–1340

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Computers and Mathematics with Applications

journal homepage: www.elsevier.com/locate/camwa

Existence and uniqueness of strictly nondecreasing and positive solutionfor a fractional three-point boundary value problemSihua Liang a,∗, Jihui Zhang b

a College of Mathematics, Changchun Normal University, Changchun 130032, Jilin, PR Chinab Institute of Mathematics, School of Mathematical Science, Nanjing Normal University, Nanjing, Jiangsu 210046, PR China

a r t i c l e i n f o

Keywords:Partially ordered setsFixed point theoremPositive solution

a b s t r a c t

In this paper, we consider the following nonlinear fractional three-point boundary valueproblem

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

u(0) = u′(0) = u′′(0) = 0, u′′(1) = βu′′(η),

where Dα0+ is the standard Riemann–Liouville fractional derivative. By using a fixed pointtheorem in partially ordered sets, we obtain sufficient conditions for the existence anduniqueness of positive and nondecreasing solution to the above boundary value problem.

Crown Copyright© 2011 Published by Elsevier Ltd. All rights reserved.

1. Introduction

Recently, an increasing interest in studying the existence of solutions for boundary value problems of fractional orderfunctional differential equations has been observed [1–6]. Fractional differential equations describe many phenomena invarious fields of science and engineering such as physics, mechanics, chemistry, control, engineering, etc. For an extensivecollection of such results, we refer the readers to the monographs by Samko et al. [7], Podlubny [8], Kilbas et al. [9] andDiethelm [10].

On the other hand, some basic theory for the initial value problems of fractional differential equations involving theRiemann–Liouville differential operator has been discussed by Lakshmikantham [11,12,2], Bai and Lü [13], El-Sayed et al.[14,15] and Bai [16,17], Zhang [18], etc.

El-Shahed [15] considered the following nonlinear fractional boundary value problem

Dα0+u(t)+ λa(t)f (u(t)) = 0, 0 < t < 1, 2 < α ≤ 3,u(0) = u′(0) = u′(1) = 0,

where Dα0+ is the standard Riemann–Liouville fractional derivative. They used Krasnoselskii’s fixed point theorem on coneexpansion and compression to show the existence and non-existence of positive solutions for the above fractional boundaryvalue problem.

Liang and Zhang [19] considered the following nonlinear fractional boundary value problem

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

∗ Corresponding author. Tel.: +86 13944114928.E-mail addresses: liangsihua@163.com (S. Liang), jihuiz@jlonline.com (J. Zhang).

0898-1221/$ – see front matter Crown Copyright© 2011 Published by Elsevier Ltd. All rights reserved.doi:10.1016/j.camwa.2011.03.073

1334 S. Liang, J. Zhang / Computers and Mathematics with Applications 62 (2011) 1333–1340

where Dα0+ is the standard Riemann–Liouville fractional derivative. By means of lower and upper solution method and fixedpoint theorems, some results on the existence of positive solutions to the above boundary value problem are obtained. Butthe uniqueness is not treated.

We note that the results dealing with the existence of positive solutions of multi-point boundary value problems offractional order differential equations are relatively scarce.

Li et al. [20] considered the following three-point boundary value problem of fractional order differential equation

Dα0+u(t)+ f (t, u(t)) = 0, 0 < t < 1, 1 < α ≤ 2,u(0) = 0, Dβ0+

u(1) = aDβ0+u(ξ),

where Dα0+ is the standard Riemann–Liouville fractional derivative. They proved some results on existence and multiplicityof positive solutions by using some fixed point theorems. But the uniqueness is not treated.

In this paper, we deal with the following three-point boundary value problem

Dα0+u(t)+ f (t, u(t)) = 0, 0 < t < 1, 3 < α ≤ 4, (1.1)

u(0) = u′(0) = u′′(0) = 0, u′′(1) = βu′′(η), (1.2)

where Dα0+ is the standard Riemann–Liouville fractional derivative and 0 < η < 1 satisfies 0 < βηα−3 < 1. We will provethe existence and uniqueness of a positive and nondecreasing solution for the boundary value problems (1.1)–(1.2) by usinga fixed point theorem in partially ordered sets. Existence of fixed point in partially ordered sets has been considered recentlyin [21–25]. This work is motivated by papers [21,20].

2. Some definitions and fixed point theorems

We need the following definitions and lemmas which will be used to prove our main result.

Definition 2.1 ([8]). The integral

Is0+f (x) =1

Γ (s)

∫ x

0

f (t)(x − t)1−s

dt, x > 0,

where s > 0, is called Riemann–Liouville fractional integral of order s and Γ (s) is the Euler gamma function defined by

Γ (s) =

∫+∞

0ts−1e−tdt, s > 0.

Definition 2.2 ([9]). For a function f (x) given in the interval [0,∞), the expression

Ds0+f (x) =

1Γ (n − s)

ddx

n ∫ x

0

f (t)(x − t)s−n+1

dt,

where n = [s] + 1, [s] denotes the integer part of number s, is called the Riemann–Liouville fractional derivative of order s.

The following two lemmas can be found in [13,9] which are crucial in finding an integral representation of fractionalboundary value problem (1.1) and (1.2).

Lemma 2.1 ([13,9]). Let α > 0 and u ∈ C(0, 1) ∩ L(0, 1). Then fractional differential equation

Dα0+u(t) = 0

has

u(t) = c1tα−1+ c2tα−2

+ · · · + cntα−n, ci ∈ R, i = 0, 1, . . . , n, n = [α] + 1

as unique solution.

Lemma 2.2 ([13,9]). Assume that u ∈ C(0, 1) ∩ L(0, 1) with a fractional derivative of order α > 0 that belongs to C(0, 1) ∩

L(0, 1). Then

Iα0+Dα0+u(t) = u(t)+ c1tα−1

+ c2tα−2+ · · · + cntα−n,

for some ci ∈ R, i = 0, 1, . . . , n, n = [α] + 1.

Now, we present the fixed point theorems which will be used later.

S. Liang, J. Zhang / Computers and Mathematics with Applications 62 (2011) 1333–1340 1335

Theorem 2.1 ([22]). Let (E,≤) be a partially ordered set and suppose that there exists a metric d in E such that (E, d) is acomplete metric space. Assume that E satisfies the following condition

if {xn}is a nondecreasing sequence in E such thatxn → x, then xn ≤ x, ∀ n ∈ N. (2.1)

Let T : E → E be nondecreasing mapping such that

d(Tx, Ty) ≤ d(x, y)− ψ(d(x, y)), for x ≥ y,

whereψ : [0,+∞) → [0,+∞) is a continuous and nondecreasing function such that ψ is positive in (0,+∞),ψ(0) = 0 andlimt→∞ ψ(t) = ∞. If there exists x0 ∈ E with x0 ≤ T (x0), then T has a fixed point.

Remark 2.1. Notice that the condition limt→∞ ψ(t) = ∞ is superfluous.

If we consider that (E,≤) satisfies the following condition:

for x, y ∈ E there exists z ∈ E which is comparable to x and y, (2.2)

then we have the following result.

Theorem 2.2 ([22]). Adding condition (2.2) to the hypotheses of Theorem 2.1, we obtain uniqueness of the fixed point.

3. Related lemmas

The basic space used in this paper is E = C[0, 1]. Then E is a real Banach space with the norm ‖u‖ = max0≤t≤1 |u(t)|.Note that this space can be equipped with a partial order given by

x, y ∈ C[0, 1], x ≤ y ⇔ x(t) ≤ y(t), t ∈ [0, 1].

In [23] it is proved that (C[0, 1],≤)with the classic metric given by

d(x, y) = sup0≤t≤1

{|x(t)− y(t)|}

satisfied condition (2.1) of Theorem 2.1. Moreover, for x, y ∈ C[0, 1] as the function max{x, y} ∈ C[0, 1], (C[0, 1],≤)satisfies condition (2.2).

Lemma 3.1. Let 0 < η < 1 and β =1

ηα−3 . If h ∈ C[0, 1], then the boundary value problem

Dα0+u(t)+ h(t) = 0, 0 < t < 1, 3 < α ≤ 4, (3.1)

u(0) = u′(0) = u′′(0) = 0, u′′(1) = βu′′(η), (3.2)

has a unique solution

u(t) =

∫ 1

0G(t, s)h(s)ds +

βtα−1

(α − 1)(α − 2)(1 − βηα−3)

∫ 1

0H(η, s)h(s)ds, (3.3)

where

G(t, s) =1

Γ (α)

tα−1(1 − s)α−3

− (t − s)α−1, 0 ≤ s ≤ t ≤ 1,tα−1(1 − s)α−3, 0 ≤ t ≤ s ≤ 1,

(3.4)

H(t, s) =∂2G(t, s)∂t2

=(α − 1)(α − 2)

Γ (α)

tα−3(1 − s)α−3

− (t − s)α−3, 0 ≤ s ≤ t ≤ 1,tα−3(1 − s)α−3, 0 ≤ t ≤ s ≤ 1.

(3.5)

Proof. By Lemma 2.2, the solution of (3.1) can be written as

u(t) = c1tα−1+ c2tα−2

+ c3tα−3+ c4tα−4

−

∫ t

0

(t − s)α−1

Γ (α)h(s)ds.

From (3.2), we know that c2 = c3 = c4 = 0 and

c1 =1

Γ (α)(1 − βηα−3)

[∫ 1

0(1 − s)α−3h(s)ds −

∫ η

0β(η − s)α−3h(s)ds

].

1336 S. Liang, J. Zhang / Computers and Mathematics with Applications 62 (2011) 1333–1340

Therefore, the unique solution of boundary value problem (3.1) and (3.2) is

u(t) = −

∫ t

0

(t − s)α−1

Γ (α)h(s)ds +

tα−1

Γ (α)(1 − βηα−3)

[∫ 1

0(1 − s)α−3h(s)ds −

∫ η

0β(η − s)α−3h(s)ds

]= −

∫ t

0

(t − s)α−1

Γ (α)h(s)ds +

tα−1

Γ (α)+

βηα−3tα−1

Γ (α)(1 − βηα−3)

∫ 1

0(1 − s)α−3h(s)ds

−βtα−1

Γ (α)(1 − βηα−3)

∫ η

0(η − s)α−3h(s)ds

=1

Γ (α)

∫ t

0(tα−1(1 − s)α−3

− (t − s)α−1)h(s)ds +1

Γ (α)

∫ 1

ttα−1(1 − s)α−3h(s)ds

+βηα−3tα−1

Γ (α)(1 − βηα−3)

∫ 1

0(1 − s)α−3h(s)ds −

βtα−1

Γ (α)(1 − βηα−3)

∫ η

0(η − s)α−3h(s)ds

=

∫ 1

0G(t, s)h(s)ds +

βtα−1

(α − 1)(α − 2)(1 − βηα−3)

∫ 1

0H(η, s)h(s)ds.

The proof is complete. �

Lemma 3.2. G is a continuous function and G(t, s) ≥ 0.

Proof. The continuity of G is easily checked. On the other hand, for 0 ≤ t ≤ s ≤ 1 it is obvious that

G(t, s) =tα−1(1 − s)α−3

Γ (α)≥ 0.

In the case 0 ≤ s ≤ t ≤ 1(s = 1), we have

G(t, s) =1

Γ (α)

[tα−1(1 − s)α−1

(1 − s)2− (t − s)α−1

]≥

1Γ (α)

tα−1(1 − s)α−1

− (t − s)α−1=

1Γ (α)

(t − ts)α−1

− (t − s)α−1≥ 0.

Moreover, as G(t, 1) = 0, then we conclude that G(t, s) ≥ 0 for all (t, s) ∈ [0, 1] × [0, 1]. The proof is complete. �

Lemma 3.3.

supt∈[0,1]

∫ 1

0G(t, s)ds =

2(α − 2)Γ (α + 1)

,

∫ 1

0H(η, s)ds =

ηα−3(α − 1)(1 − η)

Γ (α).

Proof. Since∫ 1

0G(t, s)ds =

∫ t

0G(t, s)ds +

∫ 1

tG(t, s)ds

=1

Γ (α)

∫ t

0(tα−1(1 − s)α−3

− (t − s)α−1)ds +1

Γ (α)

∫ 1

ttα−1(1 − s)α−3ds

=1

Γ (α)

tα−1

α − 2−

1αtα

.

On the other hand, let

ρ(t) =

∫ 1

0G(t, s)ds =

1Γ (α)

tα−1

α − 2−

1αtα

,

then, as

ρ ′(t) =1

Γ (α)

α − 1α − 2

tα−2− tα−1

> 0, for t > 0,

S. Liang, J. Zhang / Computers and Mathematics with Applications 62 (2011) 1333–1340 1337

the function ρ(t) is strictly increasing and, consequently,

supt∈[0,1]

ρ(t) = supt∈[0,1]

∫ 1

0G(t, s)ds = ρ(1) =

1Γ (α)

1

α − 2−

1α

=

2α(α − 2)Γ (α)

=2

(α − 2)Γ (α + 1).

By direct computation, we have∫ 1

0H(η, s)ds =

ηα−3(α − 1)(1 − η)

Γ (α).

The proof is complete. �

Lemma 3.4. G(t, s) is strictly increasing in the first variable.

Proof. For s fixed, we let

g1(t) =1

Γ (α)

tα−1(1 − s)α−3

− (t − s)α−1 , for s ≤ t,

g2(t) =1

Γ (α)tα−1(1 − s)α−3, for t ≤ s.

It is easy to check that g1(t) is strictly increasing on [s, 1] and g2(t) is strictly increasing on [0, s]. Thenwe have the followingcases:Case 1: t1, t2 ≤ s and t1 < t2. In this case, we have g2(t1) < g2(t2), i.e. G(t1, s) < G(t2, s).Case 2: s ≤ t1, t2 and t1 < t2. In this case, we have g1(t1) < g1(t2), i.e. G(t1, s) < G(t2, s).Case 3: t1 ≤ s ≤ t2 and t1 < t2. In this case, we have g2(t1) ≤ g2(s) = g1(s) ≤ g1(t2). We claim that g2(t1) < g1(t2). Infact, if g2(t1) = g1(t2), then g2(t1) = g2(s) = g1(s) = g1(t2), from the monotone of g1 and g2, we have t1 = s = t2, whichcontradicts with t1 < t2. This fact implies that G(t1, s) < G(t2, s). The proof is complete. �

4. Main result

The main result of this paper is the following.

Theorem 4.1. The boundary value problem (1.1)–(1.2) has a unique positive and strictly increasing solution u(t) if the followingconditions are satisfied:(i) f : [0, 1] × [0,+∞) → [0,+∞) is continuous and nondecreasing with respect to the second variable and f (t, u(t)) ≡ 0

for t ∈ Z ⊂ [0, 1] with µ(Z) > 0 (µ denotes the Lebesgue measure);

(ii) There exists 0 < λ <

2(α−2)Γ (α+1) +

βηα−3(1−η)(α−2)(1−βηα−3)Γ (α)

−1such that for u, v ∈ [0,+∞) with u ≥ v and t ∈ [0, 1]

f (t, u)− f (t, v) ≤ λ · ln(u − v + 1).

Proof. Consider the cone

K = {u ∈ C[0, 1] : u(t) ≥ 0} .

As K is a closed set of C[0, 1], K is a complete metric space with the distance given by d(u, v) = supt∈[0,1] |u(t)− v(t)|.Now, we consider the operator T defined by

Tu(t) =

∫ 1

0G(t, s)f (s, u(s))ds +

βtα−1

(α − 1)(α − 2)(1 − βηα−3)

∫ 1

0H(η, s)f (s, u(s))ds,

By Lemma 3.2 and condition (i), we have that T (K) ⊂ K .We now show that all the conditions of Theorems 2.1 and 2.2 are satisfied.Firstly, by condition (i), for u, v ∈ K and u ≥ v, we have

Tu(t) =

∫ 1

0G(t, s)f (s, u(s))ds +

βtα−1

(α − 1)(α − 2)(1 − βηα−3)

∫ 1

0H(η, s)f (s, u(s))ds

≥

∫ 1

0G(t, s)f (s, v(s))ds +

βtα−1

(α − 1)(α − 2)(1 − βηα−3)

∫ 1

0H(η, s)f (s, v(s))ds

= Tv(t).

This proves that T is a nondecreasing operator.

1338 S. Liang, J. Zhang / Computers and Mathematics with Applications 62 (2011) 1333–1340

On the other hand, for u ≥ v and by condition (ii) we have

d(Tu, Tv) = sup0≤t≤1

|(Tu)(t)− (Tv)(t)| = sup0≤t≤1

((Tu)(t)− (Tv)(t))

≤ sup0≤t≤1

∫ 1

0G(t, s)(f (s, u(s))− f (s, v(s)))ds

+β

(α − 1)(α − 2)(1 − βηα−3)

∫ 1

0H(η, s)(f (s, u(s))− f (s, v(s)))ds

≤ sup0≤t≤1

∫ 1

0G(t, s)λ · ln(u(s)− v(s)+ 1)ds

+β

(α − 1)(α − 2)(1 − βηα−3)

∫ 1

0H(η, s)λ · ln(u(s)− v(s)+ 1)ds.

Since the function h(x) = ln(x + 1) is nondecreasing, by Lemma 3.3 and condition (ii), we have

d(Tu, Tv) ≤ λ ln(‖u − v‖ + 1)

sup0≤t≤1

∫ 1

0G(t, s)ds +

β

(α − 1)(α − 2)(1 − βηα−3)

∫ 1

0H(η, s)ds

= λ ln(‖u − v‖ + 1)

2

(α − 2)Γ (α + 1)+

βηα−3(1 − η)

(α − 2)(1 − βηα−3)Γ (α)

< ‖u − v‖ − (‖u − v‖ − ln(‖u − v‖ + 1)).

Letψ(x) = x− ln(x+1). Obviouslyψ : [0,+∞) → [0,+∞) is continuous, nondecreasing, positive in (0,+∞),ψ(0) = 0.Thus, for u ≥ v, we have

d(Tu, Tv) ≤ d(u, v)− ψ(d(u, v)).

As G(t, s) ≥ 0 and f ≥ 0, (T0)(t) = 10 G(t, s)f (s, 0)ds ≥ 0 and by Theorem 2.1 we know that problem (1.1)–(1.2) has at

least one nonnegative solution.Obviously, as (K ,≤) satisfies condition (2.2), Theorem 2.2 implies the uniqueness of the solution.Finally, we will prove that this solution u(t) is strictly increasing function. As u(0) =

10 G(0, s)f (s, u(s))ds and G(0, s) =

0 we have u(0) = 0.Moreover, if we take t1, t2 ∈ [0, 1] with t1 < t2, we can consider the following cases.

Case 1: t1 = 0, in this case, u(t1) = 0 and, as u(t) ≥ 0, suppose that u(t2) = 0. Then 0 = u(t2) = 10 G(t2, s)f (s, u(s))ds.

This implies that

G(t2, s) · f (s, u(s)) = 0, a.e.(s)

and as G(t2, s) = 0 a.e.(s)we get f (s, u(s)) = 0 a.e.(s).On the other hand, f is nondecreasing with respect to the second variable, then we have

f (s, 0) ≤ f (s, u(s)) = 0, a.e.(s)

which contradicts the condition (i) f (t, 0) = 0 for t ∈ Z ⊂ [0, 1](µ(Z) = 0). Thus u(t1) = 0 < u(t2).Case 2: 0 < t1. In this case, let us take t2 ∈ [0, 1] with t1 < t2, then

u(t2)− u(t1) = (Tu)(t2)− (Tu)(t1)

=

∫ 1

0(G(t2, s)− G(t1, s))f (s, u(s))ds +

β(tα−12 − tα−1

1 )

(α − 1)(α − 2)(1 − βηα−3)

∫ 1

0H(η, s)f (s, u(s))ds.

Taking into account Lemma 3.4 and the fact that f ≥ 0, we get u(t2)− u(t1) ≥ 0.Suppose that u(t2) = u(t1) then∫ 1

0(G(t2, s)− G(t1, s))f (s, u(s))ds = 0

and this implies

(G(t2, s)− G(t1, s))f (s, u(s)) = 0 a.e.(s).

Again, Lemma 3.4 gives us

f (s, u(s)) = 0 a.e.(s)

and using the same reasoning as above we have that this contradicts condition (i) f (t, 0) = 0 for t ∈ Z ⊂ [0, 1](µ(Z) = 0).Thus u(t1) < u(t2). The proof is complete. �

S. Liang, J. Zhang / Computers and Mathematics with Applications 62 (2011) 1333–1340 1339

Finally, we can replace the condition (i) of Theorem 4.1 by the following condition:(i)∗f : [0, 1] × [0,∞) → [0,∞) is continuous and nondecreasing with respect to the second variable and there existt0 ∈ [0, 1] such that f (t0, 0) > 0.

Remark 4.1. The conclusion of Theorem 4.1 is true if conditions (i)∗ and (ii) of Theorem 4.1 are satisfied.

Proof. The proof is similar to what we have done above, so we omit it here. �

5. Example

Example 5.1. The fractional boundary value problemD

720+u(t)+ (t2 + 1) ln(2 + u(t)) = 0, 0 < t < 1,

u(0) = u′(0) = u′′(0) = 0, u′′(1) = u′′

14

(5.1)

has a unique and strictly increasing solution.

Proof. In this case, f (t, u) = (t2 + 1) ln(2 + u(t)) for (t, u) ∈ [0, 1] × [0,∞). Note that f is a continuous function andf (t, u) = 0 for t ∈ [0, 1]. Moreover, f is nondecreasing with respect to the second variable since ∂ f

∂u =1

u+2 (t2+ 1) > 0. On

the other hand, for u ≥ v and t ∈ [0, 1], we have

f (t, u)− f (t, v) = (t2 + 1) ln(2 + u)− (t2 + 1) ln(2 + v) = (t2 + 1) ln2 + u2 + v

= (t2 + 1) ln

2 + v + u − v

2 + v

= (t2 + 1) ln

1 +

u − v

2 + v

≤ (t2 + 1) ln (1 + (u − v)) ≤ 2 ln(1 + u − v).

In this case, λ = 2 and, moreover, we have[2

(α − 2)Γ (α + 1)+

βηα−3(1 − η)

(α − 2)(1 − βηα−3)Γ (α)

]−1

=42Γ ( 72 )

37> 2 = λ.

Thus Theorem 4.1 implies that boundary value problem (1.1)–(1.2) has a unique and strictly increasing solution. �

Acknowledgements

The Project is supported by NSFC (10871096), Ji Jiao Ke He Zi [2011] No. 196, Natural Science Foundation of ChangchunNormal University.

The authors are indebted to the referees’ suggestions. These have greatly improved this paper.

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