Research Article | Open Access
Josefa Caballero, Mohamed Abdalla Darwish, Kishin Sadarangani, Wafa M. Shammakh, "Existence Results for a Coupled System of Nonlinear Fractional Hybrid Differential Equations with Homogeneous Boundary Conditions", Abstract and Applied Analysis, vol. 2014, Article ID 672167, 10 pages, 2014. https://doi.org/10.1155/2014/672167
Existence Results for a Coupled System of Nonlinear Fractional Hybrid Differential Equations with Homogeneous Boundary Conditions
We study an existence result for the following coupled system of nonlinear fractional hybrid differential equations with homogeneous boundary conditions and where and denotes the Riemann-Liouville fractional derivative. The main tools in our study are the techniques associated to measures of noncompactness in the Banach algebras and a fixed point theorem of Darbo type.
Fractional differential equations arise in many engineering and scientific disciplines as the mathematical modelling of a great number of processes which appear in physics, chemistry, aerodynamics, and so forth and involve also derivatives of fractional order. For details, see [1–5] and the references therein.
On the other hand, about the theory of hybrid differential equations, we refer to the paper  where the authors studied the hybrid differential equation of first order: where and .
Recently, in , the authors studied the following fractional hybrid initial value problem with supremum: where , , and .
In this paper, we consider the following coupled system: where and is the standard Riemann-Liouville fractional derivative.
The main tool in our study is a fixed point theorem of Darbo type associated to measures of noncompactness.
We begin this section with some definitions and results about fractional calculus.
Let and , where denotes the integer part of . For a function , the Riemann-Liouville fractional integral of order of is defined as provided that the right side is pointwise defined on .
The Riemann-Liouville fractional derivative of order of a continuous function is defined by provided that the right side is pointwise defined on .
The following lemma will be useful for our study, .
Lemma 1. Let and . Then,
Lemma 2. Let and suppose that and . Then, the unique solution of the fractional hybrid initial value problem is given by
Proof. Suppose that is a solution of problem (9). Using the operator and taking into account Lemma 1, we get or, equivalently, Since (because ), we have This means that Conversely, suppose that is given by This means that Applying and taking into account Lemma 1 and that for , we obtain Moreover, for in (16), we have . This completes the proof.
In the sequel, we recall some definitions and basic facts about measures of noncompactness.
Assume that is a real Banach space with norm and zero element . By we denote the closed ball in centered at with radius . By we denote the ball . If is a nonempty subset of , by the symbols and we denote the closure and the convex closure of , respectively. By we denote the quantity . Finally, by we will denote the family of all nonempty and bounded subsets of and by we denote its subfamily consisting of all relatively compact subsets of .
Definition 3. A mapping is said to be a measure of noncompactness in if it satisfies the following conditions.(a)The family is nonempty and .(b).(c).(d) for .(e)If is a sequence of closed subsets from such that and , then .
The family appearing in (a) is called the kernel of the measure of noncompactness . Notice that the set appearing in (e) is an element of ker. Indeed, since for , we infer that and this says that .
An important theorem about fixed point theorem in the context of measures of noncompactness is the following Darbo’s fixed point theorem .
Theorem 4. Let be a nonempty, bounded, closed, and convex subset of a Banach space and let be a continuous mapping. Suppose that there exists a constant such that
for any nonempty subset of .
Then, has a fixed point.
Theorem 5. Let be a nonempty, bounded, closed, and convex subset of a Banach space and let be a continuous operator satisfying
for any nonempty subset of with .
Then, has a fixed point.
Next, we will assume that the space has structure of Banach algebra. By we will denote the product of two elements and by we will denote the set defined by .
Definition 6. Let be a Banach algebra. We will say that a measure of noncompactness defined on satisfies condition if for any , .
This definition appears in .
In this paper, we will work in the space consisting of all real functions defined and continuous on with the standard supremum norm for . It is clear that is a Banach algebra, where the multiplication is defined as the usual product of real functions.
Next, we present the measure of noncompactness in which will be used later. Let us fix and . For , we denote by the modulus of continuity of ; that is, Put In , it is proved that is a measure of noncompactness in .
Proposition 7. The measure of noncompactness on satisfies condition .
Proof. Fix , , , and , with . Then, we have This means that and, therefore, Taking , we get This completes the proof.
3. Main Results
We begin this section introducing the following class of functions: where denotes the -iteration of .
Remark 8. Notice that if , then , for any . Indeed, in contrary case, we can find and . Since is nondecreasing, we have
and, therefore, and this contradicts the fact that .
Moreover, the fact that for any proves that if , then is continuous at .
Theorem 9. Let be a nonempty, bounded, closed, and convex subset of a Banach space and let be a continuous operator satisfying
for any nonempty subset of , where .
Then, has a fixed point.
The following result which appears in  will be interesting in our study.
Theorem 10. Let be measures of noncompactness in the Banach spaces , respectively. Suppose that is a convex function such that if and only if for .
Then, defines a measure compactness in , where denotes the natural projection of into , for .
Remark 11. As a consequence of Theorem 10, we have that if is a measure of noncompactness on a Banach space and we consider the function defined by , then, since is convex and if and only if , defines a measure of noncompactness in the space .
Next, we present the definition of a coupled fixed point.
Definition 12. An element is said to be a coupled fixed point of a mapping if and .
The following result is crucial for our study.
Theorem 13. Let be a nonempty, bounded, closed, and convex subset of a Banach space , and let be a measure of noncompactness in . Suppose that is a continuous operator satisfying
for all nonempty subsets and of , where .
Then, has at least a coupled fixed point.
Proof. Notice that, by Remark 11, is a measure of noncompactness in the space , where and are the projections of into .
Now, we consider the mapping defined by . It is easily seen that is a nonempty, bounded, closed, and convex subset of . Since is continuous, it is clear that is also continuous.
Next, we take a nonempty of . Then, Since , by Theorem 9, the mapping has at least one fixed point. This means that there exists such that or, equivalently, and . This proves that has at least a coupled fixed point.
Now, we consider the coupled system of integral equations:
Proof. The proof is an immediate consequence of Lemma 2, so we omit it.
Next, we will study problem (4) under the following assumptions.(H1) and .(H2)The functions and satisfy respectively, for any and , , , , where , and is continuous. Notice that assumption gives us the existence of two nonnegative constants and such that and , for any .(H3)There exists satisfying the inequalities
Theorem 15. Under assumptions –, problem (4) has at least one solution in .
Proof. In virtue of Lemma 14, a solution of problem (4) satisfies
We consider the space equipped with the norm , for any .
In , we define the operator Let and be the operators given by for any and any . Then, Firstly, we will prove that applies into . To do this, it is sufficient to prove that , for any since the product of continuous functions is continuous.
In virtue of assumption , it is clear that for . In order to prove that for , we fix and we consider a sequence such that , and we have to prove that . Without loss of generality, we can suppose that . Then, we have By assumption , since , is bounded on the compact set . Denote by From the last estimate, we obtain As and , we infer that where the last inequality has been obtained by using the fact that .
Therefore, since , from the last estimate, we deduce that . This proves that . Consequently, . On the other hand, for and , we have Now, taking into account assumption , we infer that the operator applies into . Moreover, from the last estimates, it follows that
Next, we will prove that the operators and are continuous on the ball and, consequently, will be also continuous.
In fact, we fix and we take , with = = . Then, for , we have where we have used Remark 8. This proves the continuity of on .
In order to prove the continuity of on , we have Therefore, and, consequently, is a continuous operator on .
In order to prove that satisfies assumptions of Theorem 13, only we have to check the condition for any subsets and of .
To do this, we fix , , with and ; then, we have where denotes the quantity From the last estimate, we infer that Since is uniformly continuous on bounded subsets of , we deduce that as and, therefore, By assumption , since is continuous, we infer Now, we estimate the quantity .
Fix , , with and . Without loss of generality, we can suppose that ; then, we have Since is bounded on the compact subsets of , particularly, on . Put , , . Then, from the last inequality, we infer that where we have used the fact that . Therefore, From this, it follows that
Next, by Proposition 7, (46), (55), and (59), we have By assumption , since and since it is easily proved that if and , then , we deduce that where .
Finally, by Theorem 13, the operator has at least a coupled fixed point and this is the desired result. This completes the proof.
The nonoscillary character of the solutions of problem (4) seems to be an interesting question from the practical point of view. This means that the solutions of problem (4) have a constant sign on the interval