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21
63 α Then D( z , ˆ 0) M ˜ ||I α q || + M b ||p|| = M . Set N : U ¯ Γ(α + 1) U = { z C 0 |D 0 ( z , ˆ 0) M + 1} . C 0 is continuous and completely continuous. From the choice of U , there is no z U such that z = λN (z ) for λ (0, 1). As a consequence of the nonlinear alternativ e of Leray-Schauder ty p e [36], we deduce that N has a fixed point z in U . 5.4 Example In thi s section we give an example to illustrat e the usefulness of our main results. Let us consider the following fractional differential equation 2 ¸ t D α t u( t, x) = x 2 u( t, x) + f (t, t s, ξ , u(s, ξ ))ds, 0 < α < 1, (5.4.1) −∞ b (t, x) [0, b] × (0 , π) , t = 2 , u(t, 0) = u( t, π ), t [0, b] . (5.4.2) Let X = L 2 ([0, π ]) and consider the operator A : D(A) X X defined by D(A) = , u X : u , 2 u , X and u(0) = u( π) = 0 , 2 Au = x 2 u. x x 2 Clearly A is densely defined in X and is the infinitesimal generator of a strongly cont inuous Semigroup ( T ( t)) t 0 on X . Now let u 0 X , and f (t, .) is cont inuous function. All conditions of theorem 5.3.1 are now fulfilled so we deduce that (5.4.1) (5.4. 2) has an integral solution.

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α

Then

D(z, 0) ≤ M ||I αq|| + M b ||p||∞

= M ∗.

Set

N : U

Γ(α + 1)

U = z ∈ C0|D0(z, 0) ≤ M ∗ + 1.

→ C0 is continuous and completely continuous. From the choice of U ,

there is no z ∈ ∂U such that z = λN (z) for λ ∈ (0, 1). As a consequence of the

nonlinear alternative of Leray-Schauder type [36], we deduce that N has a fixed

point z in U .

5.4 Example

In this section we give an example to illustrate the usefulness of our main

results. Let us consider the following fractional differential equation

∂2 ¸ t

t u(t, x) = ∂x2

u(t, x) +

f (t, t − s, ξ , u(s, ξ))ds, 0 < α < 1, (5.4.1) −∞

b (t, x) ∈ [0, b] × (0, π), t =

2 ,

u(t, 0) = u(t, π), t ∈ [0, b]. (5.4.2)

Let X = L2([0, π]) and consider the operator A : D(A) ⊂ X → X defined by

D(A) = ,

u ∈ X : ∂u

, ∂

2u , ∈ X and u(0) = u(π) = 0 ,

∂2 Au = −

∂x2 u.

∂x ∂x2

Clearly A is densely defined in X and is the infinitesimal generator of a strongly

continuous Semigroup (T (t))t≥0 on X . Now let u0 ∈ X, and f (t, .) is continuous

function. All conditions of theorem 5.3.1 are now fulfilled so we deduce that

(5.4.1) − (5.4.2) has an integral solution.

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Chapter 6

EXISTENCE RESULTS FOR DELAY EVOLUTION

FRACTIONAL FUZZY INTEGRODIFFERENTIAL

EQUATIONS

6.1 Introduction

Differential equations of fractional order have recently proved valuable tools in the

modelling of many physical phenomena [20; 31; 33; 59; 60; 74]. There has been

a significant theoretical development in fractional differential equations in recent

years; see the monographs of Kilbas et al [48], Miller and Ross[62], Podlubny

[70], Samko et al[75], and the papers of Bai and Lu[8], Diethelm et al [20; 21; 22],

El-Raheem [26], El-Sayed[27; 28; 29], Kilbas and Trujillo [48], Mainardi [59],

Momani and Hadid [65], Momani et al[64], Nakhushev [66], and Yu and Gao

[82; 83].

The study of fuzzy differential equations has been initiated as an independent

subject in conjunction with fuzzy valued analysis [18; 67] and set-valued differ-

ential equations [51]. Using the Hukuhara derivative of muti-valued functions,

Puri and Ralescu [72] have introduced the concept of H-differentiability for fuzzy

functions. This concept has been studied and applied in context of fuzzy differ-

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t

ential equations by Seikkala [77] and Kaleva [46] in time dependent form. Kaleva

showed in [46; 47] that if f is continuous and satisfies the Lipschitz condition

with respect to u, then there exists a unique local solution for the fuzzy initial

value problem ut(t) = f (t, u), u(0) = u0 on (En , D). The existence theorem of

solutions for fuzzy initial value problem under different sets of assumptions are

given in Ding et.al [19] and Song et.al [78]. Park et.al [69] studied the approxi-

mate solutions of the fuzzy functional integral equations. The local existence and

uniqueness are investigated by Balasubramaniam and Muralisankar [7] for non-

linear fuzzy neutral functional differential equations. Guo et.al [38] established

some existence results for the fuzzy impulsive functional differential equations.

Recently Lupulescu [58] has established the local and global existence and unique-

ness results for fuzzy functional differential equations. Our aim in this paper is

to study the existence and the uniqueness of the solution for the delay evolution

fractional fuzzy integrodifferential equation.

¸ t

t x(t) = A(t)x(t) + K(t, s, xs)ds + f (t, xt), f or each t ∈ J = [0, b], 0

x(t) = φ(t), t ∈ (−∞, 0], (6.1.1)

where 0 < α < 1 and A(t), t ∈ J is a family of closed densely defined linear

unbounded operators on X . Dα

is the caputo fractional derivative, f : J ×

C∞ → X, K : J × J × C∞ → X are given fuzzy functions satisfying appropriate

1

conditions, φ ∈ C∞, φ(0) = 0 and C∞ = C((−∞], E ).

For any mapping x defined on (−∞, b] and t ∈ J , we denote by xt the element

of C∞ defined by

xt(θ) = x(t + θ), θ ∈ (−∞, 0].

This chapter is organized as follows: In section 6.2, we introduce the mild solution

of (6.1.1) − (6.1.3) and recall some Lemmas which are used in the sequel. In

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section 6.3, we study the existence and uniqueness of mild solution of system

(6.1.1) − (6.1.3) under some suitable conditions. An example to illustrate our

results is given at last.

6.2 Preliminaries

In this section, we introduce notations, definitions and preliminary facts which

are used throughout this chapter.

By C(J, R) we denote the Banach space of all continuous functions from J

into R with the norm

||y||∞ = sup||y(t)||t ∈ J .

Let Kc(Rn) denote the collection of all nonempty compact convex subsets of Rn.

We define the Hausdorff distance between A, B ∈ KC (Rn) by

dH (A, B) = maxsup inf ||x − y||, sup inf ||x − y||. x∈A y∈B y∈B x∈A

Denote

En = u : Rn → [0, 1]| u satisfies (i) − (iv)

(i) u is normal, that is, there exists an x0 ∈ Rn such that u(x0) = 1,

(ii) u is fuzzy convex, that is, for x, y ∈ Rn and 0 ≤ λ ≤ 1,

u(λx + (1 − λ)y) ≥ minu(x), u(y);

(iii) u is upper semicontinuous;

(iv) [u]0 = clx ∈ Rn : u(x) > 0 is compact.

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For 0 < α ≤ 1, denote [u]α = x ∈ Rn : u(x) ≥ α. Then from (i) − (iv), it

folows that the α- level set [u]α ∈ KC ((R\)) for all 0 < α ≤ 1. For later purpose,

we define 0 ∈ En as 0(x) = 1 if x = 0 and 0(x) = 0 if x = 0. If we define

D[u, v] = sup 0≤α≤1

dH ([u]α, [v]α),

then it is well known that D is a metric in En and that (En , D) is a complete

metric space [46; 71]. We list the following properties of D[u, v];

(i) D[u + w, v + w] = D[u, v], D[u, v] = D[v, u],

(ii) D[λu, λv] = |λ|D[u, v],

(iii) D[u, v] ≤ D[u, w] + D[w, v]

for all u, v, w ∈ En and λ ∈ R.

In the following we recall some main concepts and properties of differentia-

bility and integrability for fuzzy functions [46; 71; 72].

If there exists w ∈ En such that u = v + w, then w is called the H-difference

of u and v and is denoted by u − v. Let I be an interval in R. A mapping

F : I → En is differentiable at t0 ∈ I if there exists a F t(t0) ∈ En such that the

limits

lim h→0+

F (t0 + h) − F (t0)

h

and lim h→0+

F (t0) − F (t0 − h)

h

exist and equal to F t(t0). If F : I → E1 is a fuzzy function such that

[F (t)]α = [F α(t), F α(t)], α ∈ [0, 1], 1 2

and there exists F t(t0) for xsome t0 ∈ I , then

[F t(t0)]α = [(F α)t(t0), (F α)t(t0)], α ∈ [0, 1]. 1 2

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I

I

I

I I

I I

The fuzzy function F : I → En is called strongly measurable if for each α ∈ [0, 1]

the set-valued function Fα : I → KC (Rn) defined by Fα(t) = [F (t)]α is Lebesgue

measurable. A function F : I → En is called integrable bounded if there exists

an integrable function h : I → R+ such that D[F0(t), 0] ≤ h(t) for all t ∈ I . The

integral of F over I , denoted ¸

F (t)dt, is defined levelwise by the equation

. ¸

F (t)dt.α

=

¸

I I

Fα(t)dt

= . ¸

I

f (t)dt|

f : I → Rn is measurableselection f or Fα

.

for all 0 ≤ α ≤ 1. A strongly measurable and integrable function F : I → En is

said to be integrable over I if ¸

F (t)dt ∈ En.

Proposition 6.2.1 If F : I → En is continuous, then it is integrable over I .

Moreover, in this case, the function G(t) = ¸ t

F (s)ds, t , t ∈ I , is differentiable

and Gt(t) = F (t).

t0 0

Proposition 6.2.2Let F ; I → En be differentiable and assume that the

derivative F t(t) is integrable over I . Then, for each t ∈ I , we have

¸ t

F (t) = F (t0) + F t(s)ds, t0 ∈ I t0

It is well known that F, G : I → En are integrable, then the followings hold

[46]

(i) ¸

(F (t) + G(t))dt = ¸ F (t)dt +

¸ G(t)dt;

(ii) ¸ λF (t)dt = λ

¸ F (t)dt, λ ∈ R;

(iii) D[F, G] is integrable;

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I α

I I

α

(iv) D[¸ F (t)dt,

¸ G(t)dt] ≤

¸I D[F (t), G(t)]dt;

(v) ¸ t2 F (t)dt =

¸ t1 F (t)dt + ¸ t2 F (t)dt for t , t , t

∈ I .

t0 t0 t1 0 1 2

If I is a compact interval of R, then C(I , En) denotes the set of all fuzzy contin-

uous functions from I into En. On the space C(I , En) we consider the following

metric:

D(u, v) = sup D[u(t), v(t)]. t∈I

We denote by C∞ the space C((−∞, 0], E 1). Also, we denote by

D∞(u, v) = sup

t∈(−∞,0]

D[u(t), v(t)]

the metric on the space C∞. Let u(.) ∈ C∞. Then for each t ∈ [0, b] we denote

by ut the element of C∞ defined by ut(s) = u(t + s), s ∈ (−∞, 0].

Definition 6.2.1. The fractional primitive of order α > 0 of a function h :

R+ → R is defined by

0 h(t) =

¸ t (t − s)α−1

h(s)ds 0 Γ(α)

provided the right side exists pointwise on R+ and Γ is the gamma function.

Definition 6.2.2. The fractional derivative of order α > 0 of a fuzzy continuous

function h : R+ → E1 is givn by

dαh(t) 1 d ¸ t

dtα =

=

Γ(1 − α) dt d

I 1−α

(t − s) a

h(s)ds

dt a h(t).

Definition 6.2.3. Let f : R+ → R. The Caputo fractional derivative of order

α of f is defined by

1 ¸ t

n−α−1

(n)

t f (t) = Γ(n − α)

wh ere α ∈ (n − 1, n), n ∈ N .

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(t − s) 0

f

(s)d

s, t >

0,

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∂t

Definition 6.2.4. Let f : R+ → R be a continuous function on R+ and α ≥ 0.

Then the expression

1 ¸ t

J α

α−1

t f (t) = Γ(α)

(t − s) 0

f (s)ds, t > 0

is cal led the Riemann-Liouville integral of order α.

In this chapter, we assume that the space (C∞, D∞) is a metric space of fuzzy

functions from (−∞, 0] to E1 and satisfies the following axioms:

If y : (−∞, b] → E1 and y0 ∈ C

conditions hold:

, then for every t ∈ [0, b] the following

(A1) yt is in C∞;

(A2) D∞(yt, 0) ≤ K(t) supD[y(s), 0]| 0 ≤ s ≤ t + M (t)D∞(y0, 0);

Lemma 6.2.1. (1) U (t, 0) ∈ Lb(X ) for 0 ≤ θ ≤ t ≤ T . Denotes M =

supt∈J ¦U (t, s)¦Lb (X ), which is a finite number

(2) U (t, r)U (r, θ) = U (t, θ) for 0 ≤ θ ≤ r ≤ t ≤ T .

(3) U (·, ·)x ∈ C(Λ, X) for x ∈ X, Λ = (t, 0) ∈ J × J | 0 ≤ θ ≤ t ≤ T .

(4) For 0 ≤ θ ≤ t ≤ T , U (t, θ) : X → D and t → U (t, θ) is strongly differen-

tiable on X. The derivative ∂ U (t, θ) ∈ X and it is strongly continuous on

0 ≤ θ ≤ t ≤ T . Moreover,

∂t U (t, θ) = −A(t)U (t, θ) for 0 ≤ θ ≤ t ≤ T .

∂ C

∂t U (t, θ)

X = ¦A(t)U (t, θ)¦X ≤

t − θ ,

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¦A(t)U (t, θ)A(θ)−1¦X ≤ C for 0 ≤ θ ≤ t ≤ T .

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Γ(α+1)

−∞

(5) For every v ∈ D and t ∈ (0, T ], U (t, θ)v is differentiable with respect to θ

on 0 ≤ θ ≤ t ≤ T

∂ U (t, θ)v = U (t, θ)A(θ)v.

∂θ

(6) U (t, θ) is compact operator for 0 ≤ θ ≤ t ≤ b. And, for each x0 ∈ X, the

Cauchy problem (..) has a unique classical solution x ∈ C1(J, X) given by

x(t) = U (t, 0)x0, t ∈ J.

6.3 Existence Results

Let the space

Ω = .

x : (−∞, b) → E1|x|( ,0] ∈ C∞ and x| [0,b] is continuous

..

Definition 6.3.1. A fuzzy function x ∈ Ω is said to be a solution to be a solution

of (6.1.1) if y satisfies the equation Dαx(t) = A(t)x(t) + ¸ t

K(t, s, x )ds + f (t, x )

0 s t

on [0, b] and the condition x(t) = φ(t) on (−∞, 0].

Theorem 6.3.1. Let f : [0, b] × C∞ → E1. Assume that there exists l, l1 > 0

such that

D[f (t, φ), f (t, ψ)] ≤ lD∞(φ, ψ)

and

D.

t ¸

K(t, τ , φ)dτ ,

t . K(t, τ , ψ)dτ

l D (φ, ψ),

0 0

M bα Kb l

1 ∞

M l1 b

α Kb

for t ∈ [0, b] and for every φ, ψ ∈ C∞. If .

Γ(α+1) +

. < 1, where Kb =

sup|K(t)|, t ∈ [0, b], then there exists a unique solution for the fuzzy fractional

integrodifferential equation (6.1.1) on the interval (−∞, b].

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0

Proof. Transform the problem (6.1.1) into a fixed point problem. Consider the

operator N : Ω → Ω defined by

φ(t) if t ∈ (−∞, 0],

(N x)(t) =

1 ¸ t

α−1

Γ(α)

0 (t − s) U (t, s)f (s, xs)ds

s

+ 1 ¸ t α−1

Γ(α) 0 (t − s) U (t, s)

¸ K(t, τ , xτ )dτ ds if t ∈ [0, b].

Let y(.) : (−∞, b] → E1 be the fuzzy function defined by

0 if t ∈ [0, b], y(t) =

φ(t) if t ∈ (−∞, 0].

Then y0(t) = y(t + 0) = φ(t) for every t ∈ (−∞, 0]. For each z ∈ C([0, b], E1)

with z(0) = 0, we denote by z the fuzzy function defined by

z(t) if t ∈ [0, b] z (t) =

0 if t ∈ (−∞, 0]

If x(.) satisfies the fuzzy integral equation

x(t) = 1

Γ(α)

¸ t

(t − s)α−1U (t, s)f (s, xs)ds 0

1 + Γ(α)

¸ t ¸ s

(t − s)α−1U (t, s) 0 0

K(t, τ , xτ )dτ ds,

we can decompose x(.) as x(t) = z(t) + y(t), 0 ≤ t ≤ b, which implies

xt = zt + yt, 0 ≤ t ≤ b and the fuzzy function z(.) satisfies

z(t) =

1

Γ(α)

¸ t

(t − s)α−1U (t, s)f (s, zs + ys)ds 0

1 + Γ(α)

¸ t ¸ s

(t − s)α−1U (t, s) 0 0

K(t, τ , zτ + yτ )dτ ds.

Set

C0 = z ∈ C([0, b], E1) : z0 = 0

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s

τ

∞ s

s ∞

Γ(α) 0

≤ 0

and let D0 be the metric in C0 defined by

D0(z, 0) = D∞(z0, 0) + supD[z(t), 0], 0 ≤ t ≤ b

= supD[z(t), 0], 0 ≤ t ≤ b, z ∈ C0.

Then the space C0 is a complete metric space. Let the operator N : C0 → C0 be

defined by

(N z)(t) =

1

Γ(α)

¸ t

(t − s)α−1U (t, s)f (s, zs + ys)ds 0

1 + Γ(α)

¸ t ¸ s

(t − s)α−1U (t, s) 0 0

K(t, τ , zτ + yτ )dτ ds, t ∈ [0, b].

Then the operator N has a fixed point is equivalent to p has a fixed point, and

so we turn to proving that N has a fixed point. We shall show that N : C0 → C0

is a contractive mapping. Indeed, consider z, z∗ ∈ C0. Then we have for each

t ∈ [0, b],

D[pz(t), pz∗(t)] ≤

1

Γ(α)

¸ t

(t − s) 0

α−1

U (t, s)D[f (s, zs + xs), f (s, z∗ + xs)]ds,

1 + Γ(α)

¸ t

(t − s)α−1U (t, s) 0

×D.

1

¸ s ¸ s

K(t, τ , zτ + xτ )dτ , 0 0

¸ t

K(t, τ , z∗ + xτ )dτ .ds

≤ Γ(α)

(t − s)α−1U (t, s)lD 0

(Zs, Z ∗)ds

1 + Γ(α)

¸ t

(t − s)α−1U (t, s)l1D 0

(Zs, Z ∗)ds

1 ¸ t

(t − s)α−1U (t, s)lKa sup

D[Z(s), Z∗(s)]ds

≤ Γ(α) 0

1 ¸ t

s∈[0,t]

+ Γ(α)

(t − s)α−1U (t, s)l1Kb sup 0 s∈[0,t]

D[Z(s), Z∗(s)]ds

≤ M lKb

D (z, z∗)

¸ t

(t − s) 0

α−1ds

+ M l1Kb

Γ(α)

D0(z, z∗)

¸ t

(t − s)α−1ds 0

M bαlKb D (z, z∗) +

Γ(α + 1)

M l1bαlKb D0(z, z∗)

Γ(α + 1)

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76

.

. M bαlKb M l1bαKb

. D0(pz, pz∗) ≤ +

Γ(α + 1) Γ(α + 1) D0(z, z∗).

By hypothesis, p is a contractive mapping. Thus p has a unique fixed point by

Banach’s contraction principle.

Lemma 6.3.2. [42] Let v : [0, b] → [0, ∞) be a real function and w(·) is a

nonnegative, localy integrable function on [0, b] and there exist constants a >

0, 0 < α < 1 such that

¸ t

v(t) ≤ w(t) + a 0

v(s) ds.

(t − s)α

Then there exists a constant K = K(α) such that

for every t ∈ [0, b].

¸ t

v(t) ≤ w(t) + K(α)a 0

W (s) ds.

(t − s)α

Assume that the following hypotheses hold:

(H1) f is a continuous function,

(H2) There exist p1, q1 ∈ C([0, b], R+) such that

D[f (t, ϕ), 0] ≤ p1(t) + q1(t)D∞(ϕ, 0)

(H3) There exist p2, q2 ∈ C([0, b], R+) such that

D. t

K(t, s, ϕ), 0

p (t) + q (t)D

(ϕ, 0)

≤ 2 2 ∞ 0

for t ∈ [0, b] and each ϕ ∈ C∞ and ||I αp|| < +∞. Then the fuzzy fractional

integrodifferential equation (6.1.1) has at least one solution on (−∞, b].

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s s

∞ ∞ ≤

Proof. Let p : C0 → C0 . We shall show that the operator p is continuous and

completely continuous.

Step 1. p is continuous. Let zn be a sequence such that zn → z in C0 as n → ∞.

Then

D[pZn(t), pz(t)] ≤ 1

Γ(α)

¸ t

(t − s) 0

α−1

U (t, s)D[f (s, zns + xs), f (s, Zs + xs)]ds

1 + Γ(α)

¸ t

(t − s)α−1U (t, s) 0 ¸ s

×D.

0

¸ s

K(t, τ , znτ + xτ )dτ , 0

K(t, τ , zτ + xτ )dτ .ds.

Since f is continuous, we obtain D0(pzn , pz) → 0 as n → ∞. Thus p is continuous.

Step 2. p maps bounded sets into bounded sets in C0.

Indeed, it is enough to show that for any η > 0, there exists a positive constant

L such that for each z ∈ Bη = z ∈ C0|D0(z, 0) ≤ η one has D0(pz, 0) ≤ L. Let

z ∈ Bη . From (H1) − (H3), we have for each t ∈ [0, b],

¸ t

D[pz(t), 0] ≤ 1

(t − s)α−1U (t, s)D[f (s, z + x ), 0]ds Γ(α) 0

1 ¸ t ¸ s

+ Γ(α)

(t − s)α−1U (t, s)D.

0

K(t, τ , zτ + xτ )dτ , 0.ds

0

1 ≤

Γ(α)

¸ t

(t − s)α−1U (t, s)D[p1(s) + q1(s)D 0

(zs + xs, 0)]ds,

1 + Γ(α)

¸ t

(t − s)α−1U (t, s)D[p2(s) + q2(s)D 0

(zτ + xτ , 0)]ds,

M bα||p1|| M bα||q1|| +

Γ(α + 1)

M bα||p2||

η Γ(α + 1) M bα||q2||

+ ∞

+ Γ(α + 1)

∞ η

Γ(α + 1) ∗

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78

t1

t2

where

D∞(zs + xs, 0) ≤ D∞(zs, 0) + D∞(xs, 0)

≤ Kbη + MbD∞(φ, 0) = η∗

D∞(zτ + xτ , 0) ≤ D∞(zτ , 0) + D∞(xτ , 0)

≤ Kbη + MbD∞(φ, 0) = η∗

and Mb = sup|M (t)| | t ∈ [0, b]. Hence D0(pz, 0) ≤ L

Step 3. p maps bounded sets into equicontinuous sets of C0. Let t1, t2 ∈ [0, b], t1 <

t2 and let Bη be a bounded set of C0 as in Step 2. Then we have

D[pz(t2), pz(t1)] ≤ 1

Γ(α)

¸ t1

0

[(t2 − s)

α−1

U (t2 − s) − (t1 − s)

α−1

U (t1 − s)]

×D[f (s, zs + ys), 0]ds

1 + Γ(α)

1 + Γ(α)

¸ t2

t1

¸ t1

0

[(t2 − s)α−1U (t2 − s)D[f (s, zs + ys), 0]ds

[(t2 − s)α−1U (t2 − s) − (t1 − s)α−1U (t1 − s)]

¸ s

×D[ 0

K(t, τ , zτ + yτ )dτ , 0]ds

1 + Γ(α)

¸ t2

t1

¸ s

(t2 − s)α−1U (t2 − s)D[ 0

t1

K(t, τ , zτ + yτ )dτ , 0]ds

||p1||∞ + ||q1||∞η∗ ¸ ≤

Γ(α) 0

[(t2 −

t2

s)α−1

U (t2 − s) − (t1 − s)

α−1

U (t1 − s)]ds

+ ||p1||∞ + ||q1||∞η∗ ¸

Γ(α) t1

+ ||p2||∞ + ||q2||∞η∗ ¸

Γ(α) 0

[(t2 −

[(t2 −

s)α−1

s)α−1

U (t2 − s)ds

U (t2 − s) − (t1 − s)

α−1

U (t1 − s)]ds

+ ||p2||∞ + ||q2||∞η∗ ¸

Γ(α) t1

[(t2 −

s)α−1

U (t2 − s)]ds

The right hand side as tends to zero t1 → t2. The equicontinuity for case t1 <

t2 ≤ 0 and t1 ≤ 0 ≤ t2 is obvious. As a consequence of Step 1 − 3, together

with Arzela-Ascoli theorem, we can conclude that p : C0 → C0 is continuous and

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79

1

+ ∞

+ ∞ .

completely continuous.

Step 4. A priori bounds. We now show there exists an open set U ⊆ C0 with

z = λp(z) for λ ∈ (0, 1) and z ∈ ∂U . Let z ∈ C0 and z = λp(z) for some

0 < λ < 1. Then for each t ∈ [0, b] we see that

z(t) = λ. 1 Γ(α)

¸ t

(t − 0

s)α−1U (t, s)f (s, zs

+ ys)ds

1 + Γ(α)

¸ t ¸ s

(t − s)α−1U (t, s) 0 0

K(t, τ , zτ + yτ )dτ ds.

This implies by (H1) − (H3).

¸ t

D[z(t), 0] ≤

(t − s)α−1U (t, s)q (s)D

(z

+ y , 0)ds

Γ(α) 0 1 ∞ s s

M bα||p1|| Γ(α + 1)

1 + Γ(α)

¸ t

(t − s)α−1U (t, s)q2(s)D 0

(zs + ys, 0)ds

M bα||p2|| Γ(α + 1)

But

D∞(zs + ys, 0) ≤ D∞(zs, 0) + D∞(ys, 0)

≤ K(t) supD[z(s), 0]| 0 ≤ s ≤ t + M (t)D∞(z0, 0)

+K(t) supD[y(s), 0]| 0 ≤ s ≤ t + M (t)D∞(y0, 0)

≤ Kb supD[z(s), 0]| 0 ≤ s ≤ t + MbD∞(φ, 0). (6.3.1)

If we take w(t) the right hand of (3.2), then we get

D∞(zs + ys, 0) ≤ w(t)

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80

∞ .

+ ∞

+

and therefore,

D[z(t), 0] 1

Γ(α)

¸ t

(t − s)α−1U (t, s)q1(s)w(s)ds 0

M bα||p1|| Γ(α + 1)

1 + Γ(α)

¸ t

(t − s)α−1)U (t, s)q2(s)w(s)ds 0

M bα||p2|| ∞ , t ∈ [0, b]. Γ(α + 1)

Using the above inequality and the definition of w we see that

M Kbbα||p1|| M K ||q || ∞ b 1 ∞

w(t) ≤ MbD∞(φ, 0) +

¸ t

+

Γ(α + 1)

Γ(α)

(t − s)α−1U (t, s)w(s)ds + 0

M Kbbα||p2||

M K ||q || MbD∞(φ, 0) + ∞

+

α + 1)

b 2 ∞

Γ( Γ(α) ¸ t

(t − s)α−1U (t, s)w(s)ds, t ∈ [0, b]. 0

Then there exists K = K(α) such that

M Kbbα||p1||

K(α)M K ||q || |w(t)| ≤ MbD∞(φ, 0) +

¸ t

∞ +

Γ(α + 1)

b 1 ∞

Γ(α)

(t − s)α−1U (t, s)Rds 0

M Kbbα||p2||

M K(α)K ||q || +MbD∞(φ, 0) +

¸ t

∞ +

Γ(α + 1)

b 2 ∞

Γ(α)

(t − s)α−1U (t, s)w(s)ds 0

where

R = 2MbD∞(φ, 0) +

M Kbbα||p1||

Γ(α + 1)

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81

+ ∞

+

∞ +

Hence,

||w||∞ ≤ R +

M RK(α)bαKb

Γ(α + 1)

M Kbbα||p2||

Γ(α + 1)

M K(α)Kb||q2||∞

Γ(α)

¸ t

(t − s)α−1w(s)ds 0

= M1.

Then,

D(z, 0) ≤ M1

.||I αq1|| + ||I αq2||

. +

M bα||p1||

M bα||p2||

= M ∗. Γ(α + 1) Γ(α + 1)

Set

p : U

U = z ∈ C0|D0(z, 0) ≤ M ∗ + 1

→ C0 is continuous and completely continuous. From the choice of U ,

there is no z ∈ ∂U such that z = λp(z) for λ ∈ (0, 1). As a consequence of the

nonlinear alternative of Leray-Schauder type [36] , we deduce that p has a fixed

point z in U .

6.4 Example

In this section we give an example to illustrate the usefulness of our main results.

Let us consider the following integrofractional differential equation

∂2 ¸ t

t u(t, x) = ∂x2

u(t, x) +

K(t, s, t − s, ξ , u(s, ξ))ds 0

+f (t, t − s, ξ , u(s, ξ)), 0 < α < 1,

b (t, x) ∈ [0, b] × (0, π), t = 2

u(t, 0) = u(t, π), t ∈ [0, b] (6.4.1)

Let X = L2([0, π]). Define D = H 2([0, π]) T

H 1([0, π]), and A(t)x = ∂2

u for

0 ∂y2

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82

u ∈ D which can determined a strongly continuous evolutionary process U (·, ·)

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83

0 0 s

in L2([0, φ]) and it is also compact and there exists a M > 0 such that ¦U (·, ·)¦ ≤

M . Define

¸ t

K(t, s, xs) = 0

¸ t

K(t, s, t − s, ξ , u(s, ξ))ds 0

and

f (t, xt) = f (t, t − s, ξ , u(s, ξ))

Clearly Now let u ∈ X , ¸ t

K(t, s, x ) and f (t, .) is continuous function. All

conditions of theorem 6.3.1 are now fulfilled so we deduce that 6.4.1 has an

integral solution.