Weighted Cauchy-type problem of a functional differ-integral equation

In this work, we are concerned with a nonlinear weighted Cauchy type problem of a differ-integral equation of fractional order. We will prove some local and global existence theorems for this problem, also we will study the uniqueness and stability of its solution.


Preliminaries
Let L 1 (I) be the class of Lebesgue integrable functions on the interval I = [a, b], where 0 ≤ a < b < ∞ and let Γ(.) be the gamma function.Recall that the operator T is compact if it is continuous and maps bounded sets into relatively compact ones.The set of all compact operators from the subspace U ⊂ X into the Banach space X is denoted by C(U, X).Moreover, we set B r = {u ∈ L 1 (I) : ||u|| < r, r > 0} and ||u|| 1 = 1 0 e −N t |u(t)|dt.Definition 1.1 The fractional integral of the function f (.) ∈ L 1 (I) of order β ∈ R + is defined by (see [5] - [8]) Definition 1.2 The Riemann-Liouville fractional-order derivative of f (t) of order α ∈ (0, 1) is defined as (see [5] - [8])

Introduction
We deal with the nonlinear weighted Cauchy-type problem: EJQTDE, 2007 No. 30, p. 1 This problem has been intensively studied by many authors (see [2], for instance).In comparison with earlier results of this type we get more general assumptions.In [2], the function f (t, u) is assumed to be continuous on In this work, we investigate the behavior of solutions for problem (1) with certain nonlinearities.Using the equivalence of the fractional differ-integral problem with the corresponding Volterra integral equation, we prove the existence of L 1 -solution such that the function f satisfies the Caratheodory conditions and the growth condition.Moreover, we will study the uniqueness and the stability of the solution.
Now, let us recall some results which will be needed in the sequel.
Theorem 2.1 (Rothe Fixed Point Theorem) [4] Let U be an open and bounded subset of a Banach space E, let T ∈ C( Ū , E).Then T has a fixed point if the following condition holds Theorem 2.2 (Nonlinear alternative of Laray-Schauder type) [4] Let U be an open subset of a convex set D in a Banach space E. Assume 0 ∈ U and T ∈ C( Ū , E).Then either (A1) T has a fixed point in Ū , or (A2) there exists γ ∈ (0, 1) and x ∈ ∂U such that x = γ T x.

Main results
We begin this section by proving the equivalence of problem (1) with the corresponding Volterra integral equation: Indeed: Let u(t) be a solution of (2), multiply both sides of (2) by t 1−α , we get EJQTDE, 2007 No. 30, p. 2 which gives Now, operating by I 1−α on both sides of (2), then Differentiating both sides we get Conversely, let u(t) be a solution of (1), integrate both sides, then operating by I α on both sides of the last equation, then from the initial condition, we find that C 1 = b, then we obtain (2), i.e.Problem (1) and equation ( 2) are equivalent to each other.Now define the operator T as To solve equation ( 2) it is necessary to find a fixed point of the operator T .Now, we present our main result by proving some local and global existence theorems for the integral equation ( 2) in L 1 .To facilitate our discussion, let us first state the following assumptions: (i) f : (0, 1) × R → R be a function with the following properties: (a) for each t ∈ (0, 1), f (t, .) is continuous, where a(.) ∈ L 1 (0, 1) and b(.) is measurable and bounded.
EJQTDE, 2007 No. 30, p. 3 Now, for the local existence of the solutions we have the following theorem: Let the assumptions (i) and (ii) are satisfied.
then the fractional order integral equation ( 2) has a solution u ∈ B r , where Proof.Let u be an arbitrary element in B r .Then from the assumptions (i) -(ii), we have The last estimate shows that the operator T maps L 1 into itself.Now, let u ∈ ∂B r , that is, ||u|| = r, then the last inequality implies Using inequality (3) we deduce that r > 0.Moreover, we have This estimation shows that f in L 1 (0, 1).Further, f is continuous in u (assumption (a)) and I α maps L 1 (0, 1) continuously into itself, I α f (t, u(φ(t))) is continuous in u.Since u is an arbitrary element in B r , T maps B r continuously into L 1 (0, 1).Now, we will show that T is compact, to achieve this goal we will apply Theorem 2.3.So, let Ω be a bounded subset of B r .Then T (Ω) is bounded in L 1 (0, 1), i.e. condition (i) of Theorem 2.3 is satisfied.It remains to show that (T u) h → T u in L 1 (0, 1) as h → 0, uniformly with respect to T u ∈ T Ω.We have the following estimation: Since f ∈ L 1 (0, 1) we get that I α f (.) ∈ L 1 (0, 1).Moreover t α−1 ∈ L 1 (0, 1).So, we have (see [1]) for a.e.t ∈ (0, 1).Therefore, by Theorem 2.3, we have that T (Ω) is relatively compact, that is, T is a compact operator.Therefore, Theorem 2.1 with U = B r and E = L 1 (0, 1) implies that T has a fixed point.This complete the proof.Now we prove the existence of global solution: Let the conditions (i) -(ii) be satisfied in addition to the following condition: (iii) Assume that every solution u(.) ∈ L 1 (0, 1) to the equation a.e. on (0, 1), 0 < α < 1 satisfies ||u|| = r (r is arbitrary but fixed).
Then the fractional order integral equation ( 2) has at least one solution u ∈ L 1 (0, 1).
Proof.Let u be an arbitrary element in the open set B r = {u : ||u|| < r, r > 0}.Then from the assumptions (i) -(ii), we have The above inequality means that the operator T maps B r into L 1 .Moreover, we have This estimation shows that f in L 1 (0, 1).As a consequence of Theorem 3.1 we get that T maps B r continuously into L 1 (0, 1) and T is compact.Set U = B r and D = E = L 1 (0, 1), then in the view of assumption (iii) the condition A2 of Theorem 2.2 does not hold.Therefore, Theorem 2.2 implies that T has a fixed point.This complete the proof.

Uniqueness of the solution
For the uniqueness of the solution we have the following theorem: Let the assumptions of Theorem 3.1 be satisfied, but instead of assumption (i) consider the following conditions: EJQTDE, 2007 No. 30, p. 6 then the fractional order integral equation ( 2) has a unique solution.
Proof.Let u 1 (t) and u 2 (t) be any two solutions of equation ( 2), then We choose N such that M N α > L, therefore EJQTDE, 2007 No. 30, p. 7 Now we study the stability of the Weighted Cauchy-type problem (1).
Theorem 5.1 Let the assumptions of Theorem 4.1 be satisfied, then the solution of the Weighted Cauchytype problem (1) is uniformly stable.
Proof.Let u(t) be a solution of and let u(t) be a solution of the above equation such that t