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Positive solutions of the threepoint boundary value problem for fractionalorder differential equations with an advanced argument
Advances in Difference Equationsvolume 2011, Article number: 2 (2011)
Abstract
In this article, we consider the existence of at least one positive solution to the threepoint boundary value problem for nonlinear fractionalorder differential equation with an advanced argument
where 2 < α ≤ 3, 0 < η < 1, , ^{C} D ^{α} is the Caputo fractional derivative. Using the wellknown GuoKrasnoselskii fixed point theorem, sufficient conditions for the existence of at least one positive solution are established.
MSC (2010): 34A08; 34B18; 34K37.
1 Introduction
The study of threepoint BVPs for nonlinear integerorder ordinary differential equations was initiated by Gupta [1]. Many authors since then considered the existence and multiplicity of solutions (or positive solutions) of threepoint BVPs for nonlinear integerorder ordinary differential equations. To identify a few, we refer the reader to [2–13] and the references therein.
Fractional differential equations arise in many engineering and scientific disciplines as the mathematical modeling of systems and processes in the fields of physics, chemistry, aerodynamics, electrodynamics of complex medium, polymer rheology, etc. [14–17]. In fact, fractionalorder models have proved to be more accurate than integerorder models, i.e., there are more degrees of freedom in the fractionalorder models. In consequence, the subject of fractional differential equations is gaining much importance and attention. For details, see [18–36] and the references therein.
Differential equations with deviated arguments are found to be important mathematical tools for the better understanding of several real world problems in physics, mechanics, engineering, economics, etc. [37, 38]. In fact, the theory of integer order differential equations with deviated arguments has found its extensive applications in realistic mathematical modelling of a wide variety of practical situations and has emerged as an important area of investigation. For the general theory and applications of integer order differential equations with deviated arguments, we refer the reader to the references [39–45].
As far as we know, fractional order differential equations with deviated arguments have not been much studied and many aspects of these equations are yet to be explored. For some recent work on equations of fractional order with deviated arguments, see [46–48] and the references therein. In this article, we consider the following threepoint BVPs for nonlinear fractionalorder differential equation with an advanced argument
where 2 < α ≤ 3, 0 < η < 1, , ^{C} D ^{α} is the Caputo fractional derivative and f : [0, ∞) → [0, ∞) is a continuous function.
By a positive solution of (1.1), one means a function u(t) that is positive on 0 < t < 1 and satisfies (1.1).
Our purpose here is to give the existence of at least one positive solution to problem (1.1), assuming that
(H _{1}): a ∈ C ([0, 1], [0, ∞)) and a does not vanish identically on any subinterval.
(H _{2}): The advanced argument θ ∈ C((0, 1), (0, 1)) and t ≤ θ(t) ≤ 1, ∀t ∈ (0, 1).
Let E = C[0, 1] be the Banach space endowed with the supnorm. Set
The main results of this paper are as follows.
Theorem 1.1 Assume that (H _{1}) and (H _{2}) hold. If f _{0} = ∞ and f _{ ∞ } = 0, then problem (1.1) has at least one positive solution.
Theorem 1.2 Assume that (H _{1}) and (H _{2}) hold. If f _{0} = ∞ and f _{ ∞ } = ∞, then problem (1.1) has at least one positive solution.
Remark 1.1 It is worth mentioning that the conditions of our theorems are easily to verify, so they are applicable to a variety of problems, see Examples 4.1 and 4.2.
The proof of our main results is based upon the following wellknown GuoKrasnoselskii fixed point theorem:
Theorem 1.3 [49] Let E be a Banach space, and let P ⊂ E be a cone. Assume that Ω_{1}, Ω_{2} are open subsets of E with 0 ∈ Ω_{1}, , and let be a completely continuous operator such that

(i)
Tu ≥ u, u ∈ P ∩ ∂ Ω_{1}, and Tu ≤ u, u ∈ P ∩ ∂ Ω_{2}; or

(ii)
Tu ≤ u, u ∈ P ∩ ∂ Ω_{1}, and Tu ≥ u, u ∈ P ∩ ∂ Ω_{2}.
Then T has a fixed point in .
2 Preliminaries
For the reader's convenience, we present some necessary definitions from fractional calculus theory and Lemmas.
Definition 2.1 For a function f : [0, ∞) → ℝ, the Caputo derivative of fractional order α is defined as
where [α] denotes the integer part of real number α.
Definition 2.2 The RiemannLiouville fractional integral of order α is defined as
provided the integral exists.
Definition 2.3 The RiemannLiouville fractional derivative of order α for a function f(t) is defined by
provided the righthand side is pointwise defined on (0, ∞).
Lemma 2.1 [15] Let α > 0, then fractional differential equation
has solution
where n is the smallest integer greater than or equal to α.
Lemma 2.2 [15] Let α > 0, then
for some C _{ i } ∈ ℝ, i = 0, 1, 2, . . . , N  1, where N is the smallest integer greater than or equal to α.
Lemma 2.3 Let 2 < α ≤ 3, 1 ≠ βη. Assume y(t) ∈ C[0, 1], then the following problem
has a unique solution
Proof. We may apply Lemma 2.2 to reduce Equation (2.1) to an equivalent integral equation
for some b _{1}, b _{2}, b _{3} ∈ ℝ.
In view of the relation ^{C} D ^{α} I ^{α} u(t) = u(t) and I ^{α} I ^{β} u(t) = I ^{α+β} u(t) for α, β > 0, we can get that
By u(0) = u″ (0) = 0, it follows b _{1} = b _{3} = 0. Then by the condition βu(η) = u(1), we have
Combine with (2.3), we get
This complete the proof.
Lemma 2.4 Let 2 < α ≤ 3, . Assume y ∈ C([0, 1], [0, ∞)), then the unique solution u of (2.1) and (2.2) satisfies u(t) ≥ 0, ∀t ∈ [0, 1].
Proof. By Lemma 2.3, we know that . It means that the graph of u(t) is concave down on (0, 1).
In addition,
Combine with u(0) = 0, it follows u(t) ≥ 0, ∀t ∈ [0, 1].
Lemma 2.5 Let 2 < α ≤ 3, . Assume y ∈ C([0, 1], [0, ∞)), then the unique solution u of (2.1) and (2.2) satisfies
where .
Proof. Note that u" (t) ≤ 0, by applying the concavity of u, the proof is easy. So we omit it.
3 Proofs of main theorems
Define the operator T : C[0, 1] → C[0, 1] as follows,
Then the problem (1.1) has a solution if and only if the operator T has a fixed point.
Define the cone , where .
Proof of Theorem 1.1. The operator T is completely continuous. Obviously, T is continuous.
Let Ω ⊂ C[0, 1] be bounded, then there exists a constant K > 0 such that a(t) f (u (θ(t)) ≤ K, ∀u ∈ Ω. Thus, we have
which implies .
On the other hand, we have
Hence, for each u ∈ Ω, let t _{1}, t _{2} ∈ [0, 1], t _{1} < t _{2}, we have
So, T is equicontinuous. The ArzelaAscoli Theorem implies that T : C[0, 1] → C[0, 1] is completely continuous.
Since t ≤ θ (t) ≤ 1, t ∈ (0, 1), then
Thus, Lemmas 2.5 and 3.2 show that TP ⊂ P. Then, T : P → P is completely continuous.
In view of f _{0} = ∞, there exists a constant ρ _{1} > 0 such that f(u) ≥ δ _{1} u for 0 < u < ρ _{1}, where δ _{1} > 0 satisfies
Take u ∈ P , such that u = ρ _{1}. Then, we have
Let Ω_{ ρ 1}= {u ∈ C 0[1]  u < ρ _{1}}. Thus, (3.4) shows Tu ≥ u, .
Next, in view of f _{∞} = 0, there exists a constant R > ρ _{1} such that f(u) ≤ δ _{2} u for u ≥ R, where δ _{2} > 0 satisfies
We consider the following two cases.
Case one. f is bounded, which implies that there exists a constant r _{1} > 0 such that f(u) ≤ r _{1} for u ∈ [0, ∞). Now, we may choose u ∈ P such that u = ρ _{2}, where ρ _{2} ≥ max {μ, R}.
Then we have
Case two. f is unbounded, which implies then there exists a constant such that f(u) ≤ f(ρ _{2}) for 0 < u ≤ ρ _{2} (note that f ∈ C([0, ∞), [0, ∞)). Let u ∈ P such that u = ρ _{2}, we have
Hence, in either case, we may always let Ω_{ ρ 2}= {u ∈ C[0, 1]  u < ρ _{2}} such that Tu ≤ u for .
Thus, by the first part of GuoKrasnoselskii fixed point theorem, we can conclude that (1.1) has at least one positive solution.
Proof of Theorem 1.2. Now, in view of f _{0} = 0, there exists a constant r _{1} > 0 such that f(u) ≤ τ _{1} u for 0 < u < r _{1}, where τ _{1} > 0 satisfies
Take u ∈ P, such that u = r _{1}. Then, we have
Let Ω_{1} = {u ∈ C[0, 1]  u < r _{1}}. Thus, (3.7) shows Tu ≤ u, u ∈ P ∩ ∂Ω _{1}.
Next, in view of f _{∞} = ∞, there exists a constant r _{2} > r _{1} such that f(u) ≥ τ _{2} u for u ≥ r _{2}, where τ _{2} > 0 satisfies
Let Ω_{2} = {u ∈ C[0, 1]  u < ρ _{2}},where , then, u ∈ P and u = ρ _{2} implies
and so
This shows that Tu ≥ u for u ∈ P ∩ ∂Ω _{2}.
Therefore, by the second part of GuoKrasnoselskii fixed point theorem, we can conclude that (1.1) has at least one positive solution .
4 Examples
Example 4.1 Consider the fractional differential equation
where 2 < α ≤ 3, 0 < η < 1, , θ(t) = t ^{v}, 0 < v < 1 and
Note that conditions (H _{1}) and (H _{2}) of Theorem 1.1 hold. Through a simple calculation we can get f _{0} = ∞ and f _{∞} = 0. Thus, by Theorem 1.1, we can get that the problem (4.1) has at least one positive solution.
Example 4.2 Consider the fractional differential equation
where 2 < α ≤ 3, 0 < η < 1, , and
Obviously, it is not difficult to verify conditions (H _{1}) and (H _{2}) of Theorem 1.2 hold. Through a simple calculation we can get f _{0} = 0 and f _{ ∞ } = ∞. Thus, by Theorem 1.2, we can get that the problem (4.2) has at least one positive solution.
Remark 4.1 In the above two examples, α, β, η could be any constants which satisfy 2 < α ≤ 3, 0 < η < 1, . For example, we can take α = 2.5, η = 0.5, β = 1.5.
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The authors declare that they have no competing interests.
Authors' contributions
GW completed the main part of this paper, SKN and LZ corrected the main theorems and gave two examples. All authors read and approved the final manuscript.
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Keywords
 Positive solution
 Threepoint boundary value problem
 Fractional differential equations
 GuoKrasnoselskii fixed point theorem
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