On Fractional Order Hybrid Differential Equations

and Applied Analysis 3 3. Fractional Hybrid Differential Equation of the First Type Consider the fractional hybrid differential equation in the form D α [ x (t) f (t, x (t)) ] = h (t) , a.e. t ∈ J = [0, T] , x (0) = x0 ∈ R; (10) we have the following Lemma. Lemma 7. Any function satisfies (10) with h ∈ L(J,R) will also satisfy the integral equation

Recently, the quadratic perturbation of nonlinear differential equations (called hybrid differential equations) has captured much attention.The importance of the investigations of hybrid differential equation lies in the fact that they include several dynamic systems as special cases.Dhage and Lakshmikantham [12] discussed the existence and uniqueness theorems of the solution to the ordinary first-order hybrid differential equation with perturbation of first type   [  ()  (,  ()) ] =  (,  ()) , a.e. ∈ ,  ( 0 ) =  0 ∈ R, where  ∈ ( × R, R − {0}) and  ∈ C( × R, R), where  = [ 0 ,  0 + ] is a bounded interval in R for some  0 and  ∈ R with  > 0, ( × R, R) is the class of continuous functions and C( × R, R) is called the Caratheodory class of functions  :  × R → R which are Lebesgue integrable bounded by a Lebesgue integrable function on .Moreover (i) the map   → (, ) is measurable for each  ∈ R; (ii) the map   → (, ) is continuous for each  ∈ .
Lu et al. [15] discussed a generalization of (2) by replacing the classical differentiation by fractional derivative in the Riemann-Liouvile sense as subject to  ( 0 ) =  0 .
They proved that () is a solution to (5) if and only if it is a solution to the integral equation then they studied this integral equation.Again, we recall that this concept of a solution is not realistic because of neglecting the initial condition.In fact, consider the following trivial example with (, ()) = 1, (, ()) = 0,  0 = 0; we get that the function ) is a solution to (5) but not for (6).
Here we discuss the existence of solutions to hybrid fractional differential equations in both types using the Caputo fractional derivative instead of the classical one in both (1) and (2).
Our paper is organized as follows.In Section 2 some basic definitions and theorems are given.Fractional hybrid differential equation of type 1 is discussed in Section 3 while in Section 4 we discuss the fractional hybrid differential equation of type 2. Our conclusion is presented in Section 5.

Preliminaries
Below we present some definitions and theorems used in the rest of this paper.

Fractional Hybrid Differential Equation of the First Type
Consider the fractional hybrid differential equation in the form we have the following Lemma.
Lemma 7. Any function satisfies (10) with ℎ ∈  1 (, R + ) will also satisfy the integral equation In addition if the function   → /(0, ) is injective, and   ℎ() is an absolutely continuous function, then the converse is true.

Definition 8. (1)
The function  ∈ (, R) is called a mild solution of the hybrid differential equation of first type (14) if it satisfies the integral equation (2) The function  ∈ (, R), the space of absolutely continuous real-valued functions defined on , is called a strong solution of ( 14 ( Now we divide our proof in several steps. Step Taking the supremum over  ∈  we get that  is Lipschitz on  with Lipschitz constant . Step 2. The operator  is continuous operator on B  (0).
Let {  } be a convergent sequence in B  (0) converging to  ∈ B  (0).Then, by the Lebesgue dominated converging theorem, lim for all  ∈  with prove the continuity of the operator .
Step 3. The operator  is compact operator on B  (0).
Let  be arbitrary in B  (0).By hypothesis (H2) and using Young's inequality for convolutions we get Hence, for  > 0, there exists a  > 0 such that This shows that (B  (0)) is an equicontinuous set in .By the Arzela Ascoli Theorem we get that the operator  is a compact operator. Step Taking supremum over  and using (H4) and  ∈ (0, 1) we get which contradicts ‖‖ = ; thus (ii) of Theorem 4 is not possible; hence the operator  =  has a solution in B  (0).As a result problem ( 14) has a mild solution on  which completes the proof of our theorem.
We finish this section by the following example.
Definition 12.One has the following functions.
Now we prove our theorem by proving that the conditions of Theorem 6 are satisfied.
(a) Using the hypothesis (A1) we get We finish this section with the following example.

Conclusions
In this paper we gave definitions of both strong and mild solutions to the fractional hybrid boundary value problems in two types using the Caputo fractional derivative of order  ∈ (0, 1) and then we discussed the existence of at least one mild solution for each type.We gave examples proving the importance of taking into account of the initial conditions, therefore our results are realistic.We mention that in [14,15] the initial conditions were neglected.

Theorem 4 .Definition 5 .Theorem 6 .
Let B  (0) and B  (0) be open and closed balls in a Banach algebra  centered at origin 0 of radius , for some real number  > 0, and let ,  : B  (0) →  be two operators satisfying the following.(a)  is Lipschitz with Lipschitz constant ; (b)  is continuous and compact; (c)  < 1, where  = ‖(B  (0))‖ = sup{‖()‖ :  ∈ B  (0)}.Then, either (i) the equation  =  has a solution in B  (0), or (ii) there is an element  ∈  such that ‖‖ =  satisfying  = , for some 0 <  < 1.Let  be a Banach space.A mapping  :  →  is called D-Lipschitzian if there exists a continuous and nondecreasing function  : R + → R + such that ‖ − ‖ ≤ (‖ − ‖) for all ,  ∈ , where (0) = 0.If  is not necessarily nondecreasing and satisfies () < , for  > 0, the mapping  is called a nonlinear contraction with a contraction function .Let  be a closed convex and bounded subset of the Banach space  and let  :  →  and  :  →  be two operators such that (a)  is nonlinear contraction, (b)  is continuous and compact, and (c)  =  +  for all  ∈  ⇒  ∈ .
c) Let  ∈  be fixed and  ∈  be arbitrary such that  =  + .Then we get Thus the conditions of Theorem 6 are satisfied; then the operator equation () + () = () has a solution in  which proves the existence of a mild solution to problem (33) in .
thus the operator  is a nonlinear contraction with the function  defined by () = /( + ).(b)Similarly, in proving Theorem 9 we can prove that  is continuous and compact.(