GLOBAL EXISTENCE OF SOLUTIONS FOR A SYSTEM OF SINGULAR FRACTIONAL DIFFERENTIAL EQUATIONS WITH IMPULSE EFFECTS

Abstract

By employing a fixed point theorem in a weighted Banach space, we establish the existence of a solution for a system of impulsive singular fractional differential equations. Some examples are presented to illustrate the efficiency of the results obtained.

1. Introduction

Fractional differential equation is a generalization of ordinary differential equation to arbitrary non-integer orders. The origin of fractional calculus goes back to Newton and Leibniz in the seventeenth century. Recent investigations have shown that many physical systems can be represented more accurately through fractional derivative formulation [14]. Fractional differential equations therefore find numerous applications in different branches of physics, chemistry and biological sciences such as visco-elasticity, feed back amplifiers, electrical circuits, electro analytical chemistry, fractional multipoles and neuron modelling [16]. The reader may refer to the books and monographs [15,6,8] for fractional calculus and developments on fractional differential and fractional integrodifferential equations with applications.

On the other hand, the theory of impulsive differential equations describes processes which experience a sudden change of their state at certain moments. Processes with such characteristics arise naturally and often, for example, phenomena studied in physics, chemical technology, population dynamics, biotechnology and economics. For an introduction of the basic theory of impulsive differential equation, we refer the reader to [11].

On the other hand, in recently, there have been many papers [1,2,3,4,5,7,9,12,19,17,18] concerned with the existence of solutions for different initial value problems involving impulsive fractional differential equations. However, there has been few papers discussed the Global existence of solutions of initial value problems for impulsive fractional differential systems on half line.

Motivated by the above reason, we discuss the following initial value problem of singular fractional differential system on the half line

where

A pair of functions (x, y) with x : (0, ∞) → R and y : (0, ∞) → R is said to be a solution of (1.1) if

x|(ts,ts+1], y|(ts,ts+1] ∈ C0(ts,ts+1], s ∈ N0, s ∈ N0, exist and (x, y) satisfies all equations in (1.1).

We construct a weighted Banach space and apply the Leray-Schauder nonlinear alternative to obtain the existence of at least one solution of (1.1). Our results are new and naturally complement the literature on fractional differential equations.

The paper is outlined as follows. Section 2 contains some preliminary results. The main results are presented in Section 3. Finally, in Section 4 we give two examples to illustrate the efficiency of the results obtained.

 

2. Preliminaries

For the convenience of the readers, we shall state the necessary definitions from fractional calculus theory.

For h ∈ L1 (0, ∞), denote For r > 0, p > 0, q > 0, let the Gamma and beta functions Γ(r) and B(p, q) be defined by

Definition 2.1 ([6]). Let a ∈ R. The Riemann-Liouville fractional integral of order α > 0 of a function h : (a, ∞) → R is given by

provided that the right-hand side exists.

Definition 2.2 ([6]). Let a ∈ R. The Caputo fractional derivative of order α > 0 of a function h : (a, ∞) → R is given by

where n − 1 ≤ α < n, provided that the right-hand side exists.

Remark 2.1. Let a, b ∈ R with a < b. From Theorem 2.14 in [6], we know that almost all t ∈ (a, b) for every h ∈ L1 (a, b). From Theorem 2.23 in [6], we have which is the set of all absolutely continuous functions see ([6], page 10), i.e. the functions h for which there exists (almost everywhere) a function H ∈ L1 (a, b) such that

Definition 2.3 ([10]). An odd homeomorphism Φ of the real line R onto itself is called a sup-multiplicative-like function if there exists a homeomorphism ω of [0, +∞) onto itself which supports Φ in the sense that for all v1, v2 ≥ 0,

ω is called the supporting function of Φ.

Remark 2.2. Note that any sup-multiplicative function is sup-multiplicativelike function. Also any function of the form is sup-multiplicative-like, provided that cj ≥ 0. Here a supporting function is defined by ω(u) := min{uk+1, u}, u ≥ 0.

Remark 2.3. It is clear that a sup-multiplicative-like function Φ and any corresponding supporting function ω are increasing functions vanishing at zero. Moreover, their inverses Φ−1 and ν respectively are increasing and such that for all w1, w2 ≥ 0,

ν is called the supporting function of Φ−1.

In this paper we always suppose that Φ is a sup-multiplicative-like function with its supporting function ω. The inverse function Φ−1 has its supporting function ν.

Definition 2.4. Let σ > k + α and δ > l + β. We say K : (0, +∞) × R2 → R is a Carathéodory function if it satisfies the following:

Definition 2.5. {G : {ts : s ∈ N } × R2} is called a Carathéodory sequence if

To obtain the main results, we need the Leray-Schauder nonlinear alternative.

Lemma 2.6 (Leray-Schauder Nonlinear Alternative [13]). Let X be a Banach space and T : X → X be a completely continuous operator. Suppose Ω is a nonempty open subset of X centered at zero. Then, either there exists x ∈ ∂Ω and λ ∈ (0, 1) such that x = λT x, or there exists such that x = Tx.

 

3. Main Results

In this section we shall establish the existence of at least one solution of system (1.1). Throughout, we assume that the functions and parameters in (1.1) satisfy (a)–(e) (stated in Section 1) and the following:

Let

It is easy to show that X is a real Banach space. Thus, (X × X, ∥ · ∥) is a Banach space with the norm defined by ||(x, y)|| = max {||x||X, ||y||X}, (x, y) ∈ X×X.

Let x ∈ X and y ∈ X. Then, there exists r > 0 such that ||(x, y)|| = r < +∞ From (A), f is a Carathéodory function, thus there exists Ar ≥ 0 such that

Similarly, there exist positive constants A′r and Ar,s (s = 1, 2, · · · ) such that

Likewise, g, G and Js are also Carathéodory functions, so there exist positive constants Br, B′r and Br,s (s = 1, 2, · · · ) such that

Lemma 3.1. Suppose that x, y ∈ X. Then, u ∈ X is a solution of

if and only if u ∈ X satisfies the integral equation

Proof. Let u ∈ X be a solution of (3.4). Then, it follows from (3.4) that for t ∈ (ti, ti+1] (i = 0, 1, 2, · · · ), there exist constants ci ∈ R such that

From we get From ∆u(ti) = I(ti, x(ti), y(ti)), we have ci − ci−1 = I(ti, x(ti), y(ti)), which leads to On substituting ci into (3.6), we obtain for t ∈ (ti, ti+1] (i = 0, 1, 2, · · · ),

which is simply the same as (3.5). Moreover, since x ∈ X and y ∈ Y, we have (3.1)–(3.3) which will lead to the expression of u in (3.5) is indeed in X.

On the other hand, if x ∈ X, y ∈ Y and u ∈ X satisfies (3.5), then we can prove that u ∈ X is a solution of (3.4). The proof is complete. ΢

Lemma 3.2. Suppose that x, y ∈ X. Then, v ∈ Y is a solution of

if and only if v ∈ Y satisfies the integral equation

Proof. The proof is similar to that of Lemma 3.1. ΢

Now, we define the operator T on X × Y by T(x, y)(t) = (T1(x, y)(t), T2(x, y)(t)) where

and

Remark 3.1. By Lemmas 3.1 and 3.2, (x, y) ∈ X × Y is a solution of system (1.1) if and only if (x, y) ∈ X × Y is a fixed point of the operator T.

Lemma 3.3. The operator T : X × X → X × X is well defined and is completely continuous.

Proof. The proof is long and will be divided into parts. First, we prove that T is well defined. Next, we show that T is continuous, and finally we prove that T is compact. Hence, T is completely continuous.

Step 1. We shall prove that T : X × Y → X × Y is well defined. For (x, y) ∈ X × Y, we have ∥(x, y)∥ = r > 0. Then, (3.1)–(3.4) hold. Hence,

It follows that

and

We see that T1 (x, y)(t) is defined on (0, +∞) and is continuous on (ts, ts+1] (s = 0, 1, 2, · · · ). Next, we have exist.

Also, in view of (3.1)–(3.3) and (3.12), we find for t ∈ (ti, ti+1]

With this we have shown that T1 (x, y) ∈ X. Similarly we can show that T2 (x, y) ∈ X. Hence, (T1 (x, y), T2 (x, y) ∈ X × X and T : X × X → X × X is well defined.

Step 2. We shall prove that T is continuous. Let (xn, yn) ∈ X × X with (xn, yn) → (x0, y0) as n → ∞. We shall show that T (xn, yn) → T (x0, y0) as n → ∞, i.e., T1 (xn, yn) → T1 (x0, y0) and T2 (xn, yn) → T2 (x0, y0) as n → ∞. In fact, there exists r > 0 such that ∥(xn, yn)∥ ≤ r, (n = 0, 1, 2, · · ·). Then, (3.1)–(3.4) hold for (x, y) = (xn, yn). Also,

as n → +∞. Noting

from the Lebesgue dominated convergence theorem, we get

as n → +∞. Similarly,

as n → +∞. Hence, T is continuous.

Step 3. We shall prove that T is compact, i.e., for each nonempty open bounded subset Ω of X × Y, we shall prove that is relatively compact. For this, we shall show that is uniformly bounded, equicontinuous on each (ti, ti+1] (i = 0, 1, 2, · · · ), both are equiconvergent as t → +∞.

Let Ω be an open bounded subset of X × Y. There exists r > 0 such that (3.1) holds for all (x, y) ∈ . Hence, (3.2)–(3.4) also hold for all (x, y) ∈ .

Step 3a. We shall show that is uniformly bounded. Let (x, y) ∈ For t ∈ (ti, ti+1] (i = 0, 1, 2, · · ·), from (3.13) we have

Similarly, we can obtain for t ∈ (ti, ti+1] (i = 0, 1, 2, · · ·),

Hence, it is easy to see that is uniformly bounded.

Step 3b. We shall prove that s equicontinuous on each (ti, ti+1] (i = 0, 1, 2, · · ·). We define

Then is continuous on [ti, ti+1]. So is equicontinuous on [ti, ti+1]. Thus is equicontinuous on [ti, ti+1].

Similarly, we can show that is equicontinuous on [ti, ti+1].

So is equicontinuous on each (ti, ti+1] (i = 0, 1, 2, · · · ).

Step 3c. We shall show that is equiconvergent as t → +∞.

it comes from the following items:

Similarly, we can obtain for t ∈ (ti, ti+1] (i = 0, 1, 2, · · · ),

We have established that is relatively compact. So T is completely continuous. This completes the proof. ΢

We are now ready to present the main theorem.

Theorem 3.4. Let (a)–(e) and (A)–(B) hold, Φ : R → R be a sup-multiplicativelike function with supporting function ω, and its inverse function Φ−1 : R → R with supporting function ν. Furthermore, suppose that

(ii) there exist nonnegative numbers cg, bg, ag, CG, BG, AG, CJ,s, BJ,s and AJ,s such that are convergent, and the following hold for all (U, V) ∈ R2 and t ∈ (0, ∞) :

Then, the system (1.1) has at least one solution in X × X if

or

where

Proof. We shall apply Lemma 2.6. From Lemma 3.3 we note that T is completely continuous. Let us consider the operator equation

where λ ∈ (0, 1). We shall show that any solution (x, y) of (3.17) satisfies

where M is a constant independent of λ. Now, in the context of Lemma 2.1, let

In view of (3.18), it is not possible to have (x, y) ∈ ∂Ω satisfying (x, y) = λT (x, y), hence we conclude by Lemma 2.6 that there exists such that (x, y) = T (x, y), i.e., the system (1.1) has a solution in X × X. This completes the proof.

We shall now proceed to prove (3.18). Let (x, y) be a solution of the operator equation (3.17). It follows that x = λT1 (x, y) and y = λT2 (x, y), i.e.,

and

It is easy to see from condition (i) that

Similarly, we get

From (3.19), using (3.21) and (3.22), we find for t ∈ (ti , ti+1] (i = 0, 1, 2, · · · ),

where

It follows that

or equivalently

Similarly, from (3.20) we can show that

where

Case 1. Suppose (3.14) holds. If by (3.24) we have Then (3.18) holds with If

Then, using (3.24) in (3.23) as well as (3.25) and (2.2), we get

From (3.14) we have therefore it follows from (3.26) that

From the above discussion, we have either ∥x∥ ≤ W. Substituting (3.27 into (3.24) yields

Combining (3.27) and (3.28), we have proved that (3.18) holds with M = max{W, M1, M1}.

Case 2. Suppose (3.16) holds. If then (3.23) implies that Thus (3.18) holds with If

Then, using (3.23) in (3.24) and together with (3.29) and (2.1), we find

Since it is clear from (3.30) that

which on substituting into (3.23) gives

Coupling (3.31) and (3.32), we have shown that (3.18) holds with M = max{W′, M3, M4}. The proof is complete. ΢

 

4. An example

In this section, we present an example to illustrate the main theorem.

Example 4.1. Consider the following initial value problem of impulsive fractional differential system:

where c0, b0, a0, c1, b1, a1, B0 and B1 are constants.

Corresponding to system (1.1) we have

Choose Then, σ > k + 1 and δ > l + 1. It is easy to show that

Furthermore, in the context of Theorem 3.1, we have Φ−1 (x) = x3 with supporting function with supporting function ν(x) = x3. It is easy to see that conditions (i) and (ii) in Theorem 3.1 are satisfied with

By direct computation, we get

Applying Theorem 3.1, we see that system (4.1) has at least one solution if (3.14) or (3.15) holds, i.e., if

Remark 4.1. It is easy to see from (4.2) that system (4.1) has at least one solution for sufficiently small |a0|, |b0|, |a1|, |b1|, |B0| and |B1|.