# On Some Existence and Uniqueness Results for a Class of Equations of Order0

Authors: Abdourazek Souahi; Assia Guezane-Lakoud; Rabah Khaldi
Journal: International Journal of Differential Equations
 (2016)

---

## Abstract

This paper investigates the existence and uniqueness of solution for a class of nonlinear fractional differential equations of fractional order0
<
α
≤
1 in arbitrary time scales. The results are established using extensions of Krasnoselskii-Krein, Rogers, and Kooi conditions.


---

## Body

## 1. Introduction

This work concerns the investigation of sufficient conditions for the existence and uniqueness of the solution of the following initial value problem with fractional derivative up to the first order on arbitrary time scales:(1)
D
t
T
α
u
t
=
f
t
,
u
t
,
t
∈
t
0
,
t
0
+
a
T
,
0
<
α
≤
1
,
I
t
T
1
-
α
u
t
0
=
0
,where D
t
T
α is the (left) Riemann-Liouville fractional derivative of order α on time scales T, I
t
T is the Riemann-Liouville fractional integral on time scales, and [
t
0
,
t
0
+
a
]
T is an interval on T. We assume that f is a right-dense continuous function.The theory of time scales calculus allows us to study the dynamic equations, which include both difference and differential equations, both of which are very important in implementing applications; for further information about the theoretical and potential applications of the theory of time scales, we refer the reader to [1–8] and the survey [9].The quantitative behaviour of solutions to ordinary differential equations on time scales is currently undergoing active investigations. Many authors studied the existence and the uniqueness of the solutions of initial and boundary differential equations; see [8, 10–20] and the references cited therein. In the papers [21–25], several authors were interested by the existence and uniqueness of the first-order differential equations on time scales with initial or boundary conditions using diverse techniques and conditions. On the other hand, some existence results for the fractional order differential equations were obtained in [10].Our ideas arise from the papers [26–34], especially [30, 31], where the authors used Nagumo and Krasnoselskii-Krein conditions on the nonlinear term f, without satisfying Lipschitz assumption. Motivated greatly by the above works, under appropriate time scales versions of the Krasnoselskii-Krein conditions, we obtain the uniqueness and existence of solution for the following two classes of differential equations, namely, the first-order ODE (2)
u
Δ
t
=
f
t
,
u
t
,
t
∈
t
0
,
t
0
+
a
T
,
u
t
0
=
0
,and the fractional order FDE:(3)
D
t
T
α
u
t
=
f
t
,
u
t
,
t
∈
t
0
,
t
0
+
a
T
,
0
<
α
≤
1
,
I
t
T
1
-
α
u
t
0
=
0
.The rest of the paper is organized as follows. In Section2, we give some definitions and lemmas that will be used in our work. Section 3 is devoted to the main results; we first establish the uniqueness of the solution under Krasnoselskii-Krein conditions for the first-order problem; then we establish the convergence of the successive approximations to the unique solution. Later, we prove the uniqueness for the fractional order problem under some other conditions.


## 2. Preliminaries

In this section, we recall basic results and definitions in time scales calculus.A time scaleT is a nonempty closed subset of R. We assume that c
a
r
d
(
T
)
≥
2. The forward and backward jump operators σ
,
ρ
:
T
→
T, are, respectively, defined by (4)
σ
t
=
inf
⁡
s
∈
T
:
s
>
t
,
ρ
t
=
sup
⁡
s
∈
T
:
s
<
t
.The point t
∈
T is left-dense, left-scattered, right-dense, and right scattered if ρ
(
t
)
=
t, ρ
(
t
)
<
t, σ
(
t
)
=
t, and σ
(
t
)
>
t, respectively.We setT
K
=
T
∖
{
max
⁡
T
} whenever T admits a left-scattered maximum, and T
K
=
T otherwise. We denote A
T
=
A
∩
T. An interval of T is defined by I
T, where I is an interval of R.Definition 1 (delta derivative [1]).
Assumef
:
T
→
R and let t
∈
T
K. We define(5)
f
Δ
t
=
lim
s
→
t
⁡
f
σ
s
-
f
t
σ
s
-
t
,
t
≠
σ
s
,provided the limit exists. We call f
Δ
(
t
) the delta derivative (or Hilger derivative) of f at t. Moreover, we say that f is delta differentiable on T
K provided f
Δ exists for all t
∈
T
K. The function f
Δ
:
T
K
→
R is called the (delta) derivative of f on T
K.Definition 2 (see [10]).
A functionf
:
T
→
R is called rd-continuous provided it is continuous at right-dense points in T and its left-sided limits exist (finite) at left-dense points in T. The set of rd-continuous function f
:
T
→
R is denoted by C
r
d. Similarly, a function f
:
T
→
R is called ld-continuous provided it is continuous at left-dense points in T and its right-sided limits exist (finite) at right-dense points in T. The set of ld-continuous function f
:
T
→
R is denoted by C
l
d. For f
∈
C
r
d define f
=
sup
t
∈
T
⁡
f
t. It is easy to see that C
r
d is a Banach space with this norm.Definition 3 (delta antiderivative [10]).
A functionF
:
[
a
,
b
]
T
→
R is called a delta antiderivative of a function f
:
[
a
,
b
)
T
→
R provided F is continuous on [
a
,
b
]
T, delta differentiable on [
a
,
b
)
T, and F
δ
(
t
)
=
f
(
t
) for all t
∈
[
a
,
b
)
T. Then, we define the Δ-integral of f from a to b by(6)
∫
a
b
f
t
Δ
t
≜
F
b
-
F
a
.Lemma 4.
Letf be an increasing continuous function on the [
a
,
b
]
T. We define the extension f
~ of f to the real interval [
a
,
b
] by(7)
f
~
s
≜
f
s
i
f
s
∈
T
,
0
i
f
s
∈
t
,
σ
t
∉
T
.Then(8)
∫
a
b
f
t
Δ
t
≤
∫
a
b
f
~
t
d
t
,
f
~
Δ
t
=
f
Δ
t
,
f
o
r
e
v
e
r
y
t
∈
a
,
b
T
.Lemma 5.
Lety
:
[
t
0
,
t
0
+
a
]
T
→
R be continuous. Then the general solution of the differential equation(9)
u
Δ
t
=
y
tis given by(10)
u
t
=
u
t
0
+
∫
t
0
t
y
s
Δ
s
,
t
∈
t
0
,
t
0
+
a
T
.Proof.
Lemma5 is an immediate consequence of Theorem  4.1 [5].Definition 6 (fractional integral on time scales [10]).
SupposeT is a time scale, [
a
,
b
] is an interval of T, and h is an integrable function on [
a
,
b
]. Let 0
<
α
<
1. Then the (left) fractional integral of order α of h is defined by(11)
I
a
T
t
α
h
t
=
∫
a
t
t
-
s
α
-
1
Γ
α
h
s
Δ
s
,where Γ is the gamma function.Definition 7 (fractional Riemann-Liouville derivative on time scales [10]).
LetT be a time scale, t
∈
T, 0
<
α
<
1, and h
:
T
→
R. Then the (left) Riemann-Liouville fractional derivative of order α of h is defined by(12)
D
a
T
t
α
h
t
=
1
Γ
1
-
α
∫
a
t
t
-
s
-
α
h
s
Δ
s
Δ
.For the sake of simplicity, we use the following notationI
t
T
α and D
t
T
α instead of I
t
0
T
t
α and D
t
0
T
t
α, respectively, whenever a
=
t
0.Lemma 8 (see [10]).
For any functionf integrable on [
t
0
,
t
0
+
a
]
T one has the following:(13)
D
t
T
α
∘
I
t
T
α
f
=
f
.Lemma 9 (see [10]).
Letf
∈
C
(
[
t
0
,
t
0
+
a
]
T
) and 0
<
α
<
1. If I
t
T
1
-
α
f
(
t
)
t
=
t
0
=
0, then(14)
I
t
T
α
∘
D
t
T
α
f
=
f
.Lemma 10 (see [10]).
Let0
<
α
<
1 and f
:
[
t
0
,
t
0
+
a
]
T
×
R
→
R. The function u is a solution of problem (2) if and only if it is a solution of the following integral equation:(15)
u
t
=
1
Γ
α
∫
t
0
t
t
-
s
α
-
1
f
s
,
u
s
Δ
s
,
t
∈
t
0
,
t
0
+
a
T
.Lemma 11 (see [31]).
The solution of the equation(16)
D
R
L
t
0
α
R
t
=
R
t
δis given by(17)
R
t
=
L
t
-
t
0
σ
,where L
=
Γ
(
1
-
α
)
1
/
(
1
-
δ
) and σ
=
α
/
(
1
-
δ
) and D
R
L
t
0
α is the fractional Riemann-Liouville derivative of order α
∈
(
0,1
) on the interval [
t
0
,
t
0
+
a
]; see [35].


## 3. Main Results

In the following, we denoteS
0
=
{
(
t
,
x
)
:
t
∈
[
t
0
,
t
0
+
a
]
T, x
≤
b
,
a
,
b
∈
R
+
}.


### 3.1. Uniqueness Results for First-Order ODE

Theorem 12 (Krasnoselskii-Krein conditions).
Letf
(
t
,
x
) be continuous in S
0 and for all (
t
,
x
)
,
(
t
,
x
¯
)
∈
S
0 satisfying(
H1)
f
t
,
x
-
f
t
,
x
¯
≤
k
t
-
t
0
-
1
x
-
x
¯, t
≠
t
0,
(H2)
f
t
,
x
-
f
t
,
x
¯
≤
c
x
-
x
¯
δ, where c and k are positive constants; the real number δ is such that 0
<
δ
<
1, and k
(
1
-
δ
)
<
1.Then, the first-order initial value problem (2) has at most one solution on [
t
0
,
t
0
+
a
]
T.Proof.
Supposeu and v are two solutions of (2) in [
t
0
,
t
0
+
a
]
T. We will show that u
≡
v. Let us define ϕ
(
t
) and R
(
t
) by(18)
ϕ
t
=
u
t
-
v
t
,
for  every
t
∈
t
0
,
t
0
+
a
T
,
R
t
=
∫
t
0
t
c
ϕ
~
δ
s
d
s
,
for  every
t
∈
t
0
,
t
0
+
a
,such that ϕ
~ is the extension of ϕ to the real interval [
t
0
,
t
0
+
a
]. It follows from condition (H2) that(19)
ϕ
t
=
∫
t
0
t
f
s
,
u
s
-
f
s
,
v
s
Δ
s
≤
∫
t
0
t
f
s
,
u
s
-
f
s
,
v
s
Δ
s
≤
∫
t
0
t
c
u
s
-
v
s
δ
Δ
s
≤
∫
t
0
t
c
u
~
s
-
v
~
s
δ
d
s
=
R
t
.On the other hand, since R
(
t
0
)
=
0, R
(
t
)
>
0 for t
>
t
0, and R
Δ
(
t
)
=
c
ϕ
~
δ
(
t
), for every t
∈
[
t
0
,
t
0
+
a
]
T we deduce from (18) and (19) that(20)
R
′
t
≤
c
R
δ
t
,
for  every
t
∈
t
0
,
t
0
+
a
.Multiplying both sides of this inequality by (
1
-
δ
)
R
1
-
δ
(
t
) and then integrating the resulting inequality, we obtain(21)
R
1
-
δ
t
≤
c
1
-
δ
t
-
t
0
.It immediately follows that(22)
ϕ
t
≤
c
1
-
δ
-
1
1
-
δ
1
-
δ
-
1
t
-
t
0
1
-
δ
-
1
.Moreover, if we define ψ
(
t
)
=
ϕ
(
t
)
/
(
t
-
t
0
)
k, we get(23)
0
≤
ψ
t
≤
c
1
-
δ
-
1
1
-
δ
1
-
δ
-
1
t
-
t
0
1
-
δ
-
1
-
k
,
for  every
t
∈
t
0
,
t
0
+
a
T
.It follows that the exponent of t in the above inequality is positive, since k
(
1
-
δ
)
<
1. Hence, lim
t
→
t
0
⁡
ψ
(
t
)
=
0. Therefore, if we define ψ
(
t
0
)
=
0, then the function is rd-continuous in [
t
0
,
t
0
+
a
]
T.
Now, to prove thatϕ
≡
0, we prove by absurdity that ψ
≡
0 on [
t
0
,
t
0
+
a
]
T. Assume that ψ does not vanish at some points t; that is, ψ
(
t
)
>
0 on t
0
,
t
0
+
a
T; then there exists a maximum m
>
0 reached when t is equal to some t
1: t
0
<
t
1
≤
t
0
+
a such that ψ
(
s
)
<
m
=
ψ
(
t
1
), for s
∈
[
t
0
,
t
1
)
T. But from condition (H1), we have(24)
m
=
ψ
t
1
=
t
1
-
t
0
-
k
ϕ
t
1
≤
t
1
-
t
0
-
k
∫
t
0
t
1
k
s
-
t
0
-
1
ϕ
s
Δ
s
≤
t
1
-
t
0
-
k
∫
t
0
t
1
k
s
-
t
0
k
-
1
ψ
s
Δ
s
<
m
t
1
-
t
0
-
k
∫
t
0
t
1
k
s
-
t
0
k
-
1
Δ
s
<
m
t
1
-
t
0
-
k
∫
t
0
t
1
k
s
-
t
0
k
-
1
d
s
<
m
,which is a contradiction. Thus, the uniqueness of the solution is established.Theorem 13 (Kooi’s conditions).
Letf
(
t
,
x
) be continuous in S
0 and satisfying for all (
t
,
x
)
,
(
t
,
x
¯
)
∈
S
0
(I1)
f
t
,
x
-
f
t
,
x
¯
≤
k
t
-
t
0
-
1
x
-
x
¯, t
≠
t
0,
(I2)
t
-
t
0
β
f
t
,
x
-
f
t
,
x
¯
≤
c
x
-
x
¯
δ, where c and k are positive constants; the real numbers β, δ are such that 0
<
β
<
δ
<
1, and k
(
1
-
δ
)
<
1
-
β.Then, the first-order initial value problem (2) has at most one solution on [
t
0
,
t
0
+
a
]
T.Proof.
The proof is similar to that of Theorem12; thus we omit it.


### 3.2. Existence of the Solution under Krasnoselskii-Krein Conditions on Time Scales

Theorem 14.
Assume that conditions (H1) and (H2) are satisfied; then the successive approximations given by(25)
u
n
+
1
t
=
∫
t
0
t
f
s
,
u
n
s
Δ
s
,
u
0
t
=
0
,
n
=
0,1
,
…converge uniformly to the unique solution u of (2) on [
t
0
,
t
0
+
η
]
T, where η
=
min
⁡
a
,
b
/
M, and M is the bound for f on S
0.Proof.
With the uniqueness of the solution being proved in Theorem12, we prove the existence of the solution using Arzela-Ascoli Theorem.
Step
1. The successive approximations {
u
n
+
1
}, n
=
0,1
,
2
,
… given by (25) are well defined and continuous. Indeed,(26)
u
n
+
1
t
=
∫
t
0
t
f
s
,
u
n
s
Δ
s
≤
∫
t
0
t
f
s
,
u
n
s
Δ
s
.This yields for n
=
0
(27)
u
1
t
≤
∫
t
0
t
f
s
,
u
0
s
Δ
s
≤
M
t
≤
b
.By induction, the sequence {
u
j
+
1
(
t
)
} is well defined and uniformly bounded on [
t
0
,
t
0
+
η
]
T.
Step
2. We prove that y is a continuous function in [
t
0
,
t
0
+
η
]
T, where y is defined by(28)
y
t
=
lim
sup
j
→
∞
⁡
u
j
t
-
u
j
-
1
t
.For t
1
,
t
2
∈
[
t
0
,
t
0
+
η
]
T, we have(29)
u
j
+
1
t
1
-
u
j
t
1
≤
u
j
+
1
t
2
-
u
j
t
2
+
2
M
t
2
-
t
1
.In fact, (30)
u
j
+
1
t
1
-
u
j
t
1
-
u
j
+
1
t
2
-
u
j
t
2
≤
u
j
+
1
t
1
-
u
j
t
1
-
u
j
+
1
t
2
+
u
j
t
2
≤
∫
t
0
t
1
f
s
,
u
j
s
-
f
s
,
u
j
-
1
s
Δ
s
-
∫
t
0
t
2
f
s
,
u
j
s
-
f
s
,
u
j
-
1
s
Δ
s
≤
2
M
∫
t
1
t
2
Δ
s
≤
2
M
t
2
-
t
1
.The right-hand side in inequality (29) is at most y
(
t
2
)
+
ϵ
+
2
M
(
t
2
-
t
1
) for large n if ϵ
>
0 provided that |
t
2
-
t
1
|
≤
ϵ
/
2
M. Since ϵ is arbitrary and t
1, t
2 can be interchangeable, we get(31)
y
t
1
-
y
t
2
≤
2
M
t
2
-
t
1
.This implies that y is continuous on [
t
0
,
t
0
+
η
]
T. Using condition (H2) and the definition of successive approximations, we obtain(32)
u
j
+
1
t
-
u
j
t
≤
c
∫
t
0
t
u
j
s
-
u
j
-
1
s
α
Δ
s
.The sequence {
u
n
} is equicontinuous: that is, for each function u
n and any ϵ
>
0, t
1
,
t
2
∈
[
t
0
,
t
0
+
η
]
T if there exists τ
=
ϵ
/
M such that t
2
-
t
1
≤
τ, then(33)
u
n
+
1
t
1
-
u
n
+
1
t
2
=
∫
t
1
t
2
f
s
,
u
n
s
Δ
s
≤
∫
t
1
t
2
f
~
s
,
u
n
s
d
s
≤
M
t
1
-
t
2
≤
ϵ
.
All of the Arzela-Ascoli Theorem conditions are fulfilled for the family{
u
j
} in C
r
d
[
t
0
,
t
0
+
η
]
T. Hence, there exists a subsequence {
u
j
k
} converging uniformly on [
t
0
,
t
0
+
η
]
T as j
k
→
∞.
Let us note(34)
m
∗
t
=
lim
k
→
∞
⁡
u
j
k
t
-
u
j
k
-
1
t
.Further, if {
u
j
-
u
j
-
1
}
→
0 as j
→
∞, then the limit of any subsequence is the unique solution u of (25). It follows that a selection of subsequences is unnecessary and that the entire sequence {
u
j
} converges uniformly to u. For that, it suffices to show that y
≡
0 which will lead to m
∗
(
t
) being null.
Setting(35)
R
t
=
∫
t
0
t
y
s
α
d
s
,and by defining ψ
∗
(
t
)
=
t
-
k
y
(
t
), we show that lim
t
→
0
+
⁡
ψ
∗
(
t
)
=
0.
We prove by absurdity thatψ
∗
≡
0. Assume that ψ
∗
(
t
)
>
0 at any point in ]
t
0
,
t
0
+
η
]
T; then there exists t
1 such that 0
<
m
¯
=
ψ
∗
(
t
1
)
=
max
t
∈
[
t
0
,
t
0
+
η
]
T
⁡
ψ
∗
(
t
). Hence, from condition (H1), we obtain(36)
m
¯
=
ψ
t
1
=
t
1
-
k
y
t
1
≤
m
¯
t
1
<
m
¯
.We end up with a contradiction. So ψ
∗
≡
0. Therefore, the Picard iterates (25) converge uniformly to the unique solution u of (2) on [
t
0
,
t
0
+
η
]
T.


### 3.3. Uniqueness Results for Fractional Order ODE

In this section, we denoteC
p
(
[
t
0
,
t
0
+
a
]
T
,
R
)
=
{
u
∣
u
∈
C
(
[
t
0
,
t
0
+
a
]
T
,
R
) and (
t
-
t
0
)
1
-
q
u
∈
C
(
[
t
0
,
t
0
+
a
]
T
,
R
)
}.Theorem 15 (Krasnoselskii-Krein conditions).
Letf
(
t
,
x
) be continuous in S
0 and satisfying for all (
t
,
x
)
,
(
t
,
x
¯
)
∈
S
0
(J1)
f
t
,
x
-
f
t
,
x
¯
≤
k
r
Γ
α
t
-
t
0
-
α
x
-
x
¯, t
≠
t
0,
(J2)
f
t
,
x
-
f
t
,
x
¯
≤
c
x
-
x
¯
δ, where c, r, k are positive constants such that k
>
1, k
r
≤
α, and k
(
1
-
δ
)
<
1, and the real number δ is such that 0
<
δ
<
1.Then, the fractional order initial value problem (3) has at most one solution on [
t
0
,
t
0
+
a
]
T.Proof.
Supposeu and v are two solutions of (3) in [
t
0
,
t
0
+
a
]
T. We will show that u
≡
v. Let us define ϕ
(
t
) and R
(
t
) by(37)
ϕ
t
=
u
t
-
v
t
,
for  every
t
∈
t
0
,
t
0
+
a
T
,
R
t
=
c
Γ
α
∫
t
0
t
t
-
s
α
-
1
ϕ
~
δ
s
d
s
,
for  every
t
∈
t
0
,
t
0
+
a
,such that ϕ
~ is the extension of ϕ to the real interval [
t
0
,
t
0
+
a
]. It follows from condition (J2) that(38)
ϕ
t
=
1
Γ
α
∫
t
0
t
t
-
s
α
-
1
f
s
,
u
s
-
f
s
,
v
s
Δ
s
≤
1
Γ
α
∫
t
0
t
t
-
s
α
-
1
f
s
,
u
s
-
f
s
,
v
s
Δ
s
≤
1
Γ
α
∫
t
0
t
c
t
-
s
α
-
1
u
s
-
v
s
δ
Δ
s
≤
∫
t
0
t
c
Γ
α
t
-
s
α
-
1
u
~
s
-
v
~
s
δ
d
s
=
R
t
.On the other hand, R
(
t
0
)
=
0, R
(
t
)
>
0 for t
>
t
0, and D
t
T
α
R
(
t
)
=
ϕ
~
δ
(
t
)
=
R
δ
(
t
), for every t
∈
[
t
0
,
t
0
+
a
]
T. Now from relations (37) and (38) and using Lemma 11, we obtain for every t
∈
[
t
0
,
t
0
+
a
]
T
(39)
ϕ
t
≤
R
t
=
L
t
-
t
0
σ
,where L and σ are defined as in Lemma 11. Moreover, if we define ψ
(
t
)
=
ϕ
(
t
)
/
(
t
-
t
0
)
k, we get(40)
0
≤
ψ
t
≤
L
t
-
t
0
σ
-
k
α
,
for  every
t
∈
t
0
,
t
0
+
a
T
.It follows that the exponent of t in the above inequality is positive, since k
(
1
-
δ
)
<
1. Hence, l
i
m
t
→
t
0
ψ
(
t
)
=
0. Therefore, if we define ψ
(
t
0
)
=
0, then the function is rd-continuous in [
t
0
,
t
0
+
a
]
T.
Now, to show thatϕ
≡
0, we prove by absurdity that ψ
≡
0 on [
t
0
,
t
0
+
a
]
T. Assume that ψ does not vanish at some points t; that is, ψ
(
t
)
>
0 on ]
t
0
,
t
0
+
a
]
T; then there exists a maximum m
>
0 reached when t is equal to some t
1
:
t
0
<
t
1
≤
t
0
+
a such that ψ
(
s
)
<
m
=
ψ
(
t
1
), for s
∈
[
t
0
,
t
1
)
T. But from condition (J1), we have (41)
m
=
ψ
t
1
=
t
1
-
t
0
-
k
ϕ
t
1
<
t
1
-
t
0
-
k
α
∫
t
0
t
1
k
r
t
-
s
α
-
1
f
s
,
u
s
-
f
s
,
v
s
Δ
s
≤
t
1
-
t
0
-
k
α
∫
t
0
t
1
k
r
t
-
s
α
-
1
ϕ
s
s
-
t
0
-
α
Δ
s
≤
t
1
-
t
0
-
k
α
∫
t
0
t
1
k
r
t
-
s
α
-
1
s
-
t
0
k
α
-
α
ψ
s
Δ
s
≤
m
k
r
t
1
-
t
0
-
α
∫
t
0
t
1
t
1
-
s
α
-
1
Δ
s
≤
m
k
r
t
1
-
t
0
-
α
∫
t
0
t
1
t
1
-
s
α
-
1
d
s
≤
m
k
r
α
<
m
,which is a contradiction. Thus, the uniqueness of the solution is established.Theorem 16 (Kooi’s conditions).
Letf
(
t
,
x
) be continuous in S
0 and satisfying for all (
t
,
x
)
,
(
t
,
x
¯
)
∈
S
0
(K1)
f
t
,
x
-
f
t
,
x
¯
≤
k
r
Γ
α
t
-
t
0
-
α
x
-
x
¯, t
≠
t
0,
(K2)
t
-
t
0
β
f
t
,
x
-
f
t
,
x
¯
≤
c
x
-
x
¯
δ, where c, r, and k are positive constants; the positive real numbers β, δ, k, r are such that 0
<
β
<
δ
<
1, and k
(
1
-
δ
)
<
1
-
β, and k
r
≤
α.Then, the first-order initial value problem (3) has at most one solution on [
t
0
,
t
0
+
a
]
T.Proof.
The proof is similar to that of Theorem15; thus, we omit it.


### 3.4. Existence of Solutions under Krasnoselskii-Krein Conditions on Time Scales

Theorem 17.
Assume that conditions (J1) and (J2) are satisfied; then the successive approximations given by(42)
u
n
+
1
t
=
∫
t
0
t
f
s
,
u
n
s
Δ
s
,
u
0
t
=
0
,
n
=
0,1
,
…converge uniformly to the unique solution u of (3) on [
t
0
,
t
0
+
η
]
T, where(43)
η
=
min
⁡
a
,
b
Γ
1
+
α
M
1
/
αand M is the bound for f on S
0.Proof.
With the uniqueness of the solution being proved in Theorem15, we prove the existence of the solution using Arzela-Ascoli Theorem.
Step
1. The successive approximations {
u
n
+
1
}, n
=
0,1
,
2
,
… given by (42) are well defined and continuous. Indeed,(44)
u
n
+
1
t
=
1
Γ
α
∫
t
0
t
t
-
s
α
-
1
f
s
,
u
n
s
Δ
s
≤
1
Γ
α
∫
t
0
t
t
-
s
α
-
1
f
s
,
u
n
s
Δ
s
.This yields, for n
=
0,(45)
u
1
t
≤
M
Γ
α
∫
t
0
t
t
-
s
α
-
1
Δ
s
≤
M
Γ
α
∫
t
0
t
t
-
s
α
-
1
d
s
≤
M
a
α
Γ
α
+
1
≤
b
.By induction, the sequence {
u
j
+
1
(
t
)
} is well defined and uniformly bounded on [
t
0
,
t
0
+
η
]
T.
Step
2. We prove that y is a continuous function in [
t
0
,
t
0
+
η
]
T, where y is defined by(46)
y
t
=
lim
sup
j
→
∞
⁡
u
j
t
-
u
j
-
1
t
.For t
1
,
t
2
∈
[
t
0
,
t
0
+
η
]
T, we have(47)
u
j
+
1
t
1
-
u
j
t
1
≤
u
j
+
1
t
2
-
u
j
t
2
+
4
M
Γ
α
+
1
t
2
-
t
1
α
.In fact,(48)
u
j
+
1
t
1
-
u
j
t
1
-
u
j
+
1
t
2
-
u
j
t
2
≤
u
j
+
1
t
1
-
u
j
t
1
-
u
j
+
1
t
2
+
u
j
t
2
≤
1
Γ
α
∫
t
0
t
1
t
1
-
s
α
-
1
f
s
,
u
j
s
-
f
s
,
u
j
-
1
s
Δ
s
-
∫
0
t
2
t
2
-
s
α
-
1
f
s
,
u
j
s
-
f
s
,
u
j
-
1
s
Δ
s
≤
2
M
Γ
α
∫
0
t
1
t
1
-
s
α
-
1
-
t
2
-
s
α
-
1
Δ
s
-
∫
t
1
t
2
t
2
-
s
α
-
1
Δ
s
≤
2
M
Γ
α
∫
0
t
1
t
1
-
s
α
-
1
-
t
2
-
s
α
-
1
d
s
-
∫
t
1
t
2
t
2
-
s
α
-
1
d
s
≤
2
M
α
Γ
α
t
1
α
-
t
2
α
+
2
t
2
-
t
1
α
≤
4
M
Γ
α
+
1
t
2
-
t
1
α
.The right-hand side in inequality (47) is at most y
(
t
2
)
+
ϵ
+
(
4
M
/
Γ
α
+
1
)
(
t
2
-
t
1
)
α for large n if ϵ
>
0 provided that |
t
2
-
t
1
|
≤
ϵ
Γ
(
α
+
1
)
/
4
M
1
/
α. Since ϵ is arbitrary and t
1, t
2 can be interchangeable, we get(49)
y
t
1
-
y
t
2
≤
4
M
Γ
α
+
1
t
2
-
t
1
.This implies that y is continuous on [
t
0
,
t
0
+
η
]
T. Using condition (J2) and the definition of successive approximations, we obtain(50)
u
j
+
1
t
-
u
j
t
≤
c
Γ
α
∫
t
0
t
u
j
s
-
u
j
-
1
s
α
Δ
s
.The sequence {
u
n
} is equicontinuous: that is, for each function u
n and any ϵ
>
0, t
1
,
t
2
∈
[
t
0
,
t
0
+
η
]
T if there exists τ
=
ϵ
-
α
Γ
(
α
+
1
)
/
M such that t
2
-
t
1
≤
τ; then (51)
u
n
+
1
t
1
-
u
n
+
1
t
2
=
1
Γ
α
∫
t
0
t
2
t
2
-
s
α
-
1
f
s
,
u
n
s
Δ
s
-
∫
t
0
t
1
t
1
-
s
α
-
1
f
s
,
u
n
s
Δ
s
≤
2
M
Γ
α
+
1
t
1
-
t
2
α
≤
ϵ
,where we used a similar argument as in (48).
All of the Arzela-Ascoli Theorem conditions are fulfilled for the family{
u
j
} in C
r
d
[
t
0
,
t
0
+
a
]
T. Hence, there exists a subsequence {
u
j
k
} converging uniformly on [
t
0
,
t
0
+
a
]
T as j
k
→
∞.
Let us note(52)
m
∗
t
=
lim
k
→
∞
⁡
u
j
k
t
-
u
j
k
-
1
t
.Further, if {
|
u
j
-
u
j
-
1
|
}
→
0 as j
→
∞, then the limit of any subsequence is the unique solution u of (42). It follows that a selection of subsequences is unnecessary and that the entire sequence {
u
j
} converges uniformly to u. For that, it is sufficient to show that y
≡
0 which will lead to m
∗
(
t
) being null.
Setting(53)
R
t
=
c
Γ
α
∫
t
0
t
t
-
s
α
-
1
y
s
α
d
s
,and defining ψ
∗
(
t
)
=
t
-
k
y
(
t
) and then using Lemma 11, we obtain that ψ
(
t
)
≤
L
(
t
1
-
t
0
)
J
-
k
α. Which yields that l
i
m
t
→
0
+
ψ
∗
(
t
)
=
0.
We prove by absurdity thatψ
∗
≡
0. Assume that ψ
∗
(
t
)
>
0 at any point in ]
t
0
,
t
0
+
η
]
T; then there exists t
1 such that 0
<
m
¯
=
ψ
∗
(
t
1
)
=
m
a
x
t
∈
[
t
0
,
t
0
+
η
]
T
ψ
∗
(
t
). Hence, from condition (J1), we obtain (54)
m
=
ψ
t
1
=
t
1
-
t
0
-
k
α
ϕ
t
1
≤
t
1
-
t
0
-
k
α
∫
t
0
t
1
k
r
t
1
-
s
α
-
1
s
-
t
0
-
α
ϕ
s
d
s
≤
k
r
t
1
-
t
0
-
k
α
∫
t
0
t
1
t
1
-
s
α
-
1
s
-
t
0
k
α
-
α
ψ
s
d
s
<
k
r
m
t
1
-
t
0
-
α
∫
t
0
t
1
t
1
-
t
0
α
-
1
d
s
<
k
r
m
α
<
m
.We end up with a contradiction. So ψ
∗
≡
0. Therefore, the Picard iterates (42) converge uniformly to the unique solution u of (2) on [
t
0
,
t
0
+
η
]
T.Remark 18.
For the caseT
=
R, Theorem 15 is reduced to [31, Theorem  2.1].


## 3.1. Uniqueness Results for First-Order ODE

Theorem 12 (Krasnoselskii-Krein conditions).
Letf
(
t
,
x
) be continuous in S
0 and for all (
t
,
x
)
,
(
t
,
x
¯
)
∈
S
0 satisfying(
H1)
f
t
,
x
-
f
t
,
x
¯
≤
k
t
-
t
0
-
1
x
-
x
¯, t
≠
t
0,
(H2)
f
t
,
x
-
f
t
,
x
¯
≤
c
x
-
x
¯
δ, where c and k are positive constants; the real number δ is such that 0
<
δ
<
1, and k
(
1
-
δ
)
<
1.Then, the first-order initial value problem (2) has at most one solution on [
t
0
,
t
0
+
a
]
T.Proof.
Supposeu and v are two solutions of (2) in [
t
0
,
t
0
+
a
]
T. We will show that u
≡
v. Let us define ϕ
(
t
) and R
(
t
) by(18)
ϕ
t
=
u
t
-
v
t
,
for  every
t
∈
t
0
,
t
0
+
a
T
,
R
t
=
∫
t
0
t
c
ϕ
~
δ
s
d
s
,
for  every
t
∈
t
0
,
t
0
+
a
,such that ϕ
~ is the extension of ϕ to the real interval [
t
0
,
t
0
+
a
]. It follows from condition (H2) that(19)
ϕ
t
=
∫
t
0
t
f
s
,
u
s
-
f
s
,
v
s
Δ
s
≤
∫
t
0
t
f
s
,
u
s
-
f
s
,
v
s
Δ
s
≤
∫
t
0
t
c
u
s
-
v
s
δ
Δ
s
≤
∫
t
0
t
c
u
~
s
-
v
~
s
δ
d
s
=
R
t
.On the other hand, since R
(
t
0
)
=
0, R
(
t
)
>
0 for t
>
t
0, and R
Δ
(
t
)
=
c
ϕ
~
δ
(
t
), for every t
∈
[
t
0
,
t
0
+
a
]
T we deduce from (18) and (19) that(20)
R
′
t
≤
c
R
δ
t
,
for  every
t
∈
t
0
,
t
0
+
a
.Multiplying both sides of this inequality by (
1
-
δ
)
R
1
-
δ
(
t
) and then integrating the resulting inequality, we obtain(21)
R
1
-
δ
t
≤
c
1
-
δ
t
-
t
0
.It immediately follows that(22)
ϕ
t
≤
c
1
-
δ
-
1
1
-
δ
1
-
δ
-
1
t
-
t
0
1
-
δ
-
1
.Moreover, if we define ψ
(
t
)
=
ϕ
(
t
)
/
(
t
-
t
0
)
k, we get(23)
0
≤
ψ
t
≤
c
1
-
δ
-
1
1
-
δ
1
-
δ
-
1
t
-
t
0
1
-
δ
-
1
-
k
,
for  every
t
∈
t
0
,
t
0
+
a
T
.It follows that the exponent of t in the above inequality is positive, since k
(
1
-
δ
)
<
1. Hence, lim
t
→
t
0
⁡
ψ
(
t
)
=
0. Therefore, if we define ψ
(
t
0
)
=
0, then the function is rd-continuous in [
t
0
,
t
0
+
a
]
T.
Now, to prove thatϕ
≡
0, we prove by absurdity that ψ
≡
0 on [
t
0
,
t
0
+
a
]
T. Assume that ψ does not vanish at some points t; that is, ψ
(
t
)
>
0 on t
0
,
t
0
+
a
T; then there exists a maximum m
>
0 reached when t is equal to some t
1: t
0
<
t
1
≤
t
0
+
a such that ψ
(
s
)
<
m
=
ψ
(
t
1
), for s
∈
[
t
0
,
t
1
)
T. But from condition (H1), we have(24)
m
=
ψ
t
1
=
t
1
-
t
0
-
k
ϕ
t
1
≤
t
1
-
t
0
-
k
∫
t
0
t
1
k
s
-
t
0
-
1
ϕ
s
Δ
s
≤
t
1
-
t
0
-
k
∫
t
0
t
1
k
s
-
t
0
k
-
1
ψ
s
Δ
s
<
m
t
1
-
t
0
-
k
∫
t
0
t
1
k
s
-
t
0
k
-
1
Δ
s
<
m
t
1
-
t
0
-
k
∫
t
0
t
1
k
s
-
t
0
k
-
1
d
s
<
m
,which is a contradiction. Thus, the uniqueness of the solution is established.Theorem 13 (Kooi’s conditions).
Letf
(
t
,
x
) be continuous in S
0 and satisfying for all (
t
,
x
)
,
(
t
,
x
¯
)
∈
S
0
(I1)
f
t
,
x
-
f
t
,
x
¯
≤
k
t
-
t
0
-
1
x
-
x
¯, t
≠
t
0,
(I2)
t
-
t
0
β
f
t
,
x
-
f
t
,
x
¯
≤
c
x
-
x
¯
δ, where c and k are positive constants; the real numbers β, δ are such that 0
<
β
<
δ
<
1, and k
(
1
-
δ
)
<
1
-
β.Then, the first-order initial value problem (2) has at most one solution on [
t
0
,
t
0
+
a
]
T.Proof.
The proof is similar to that of Theorem12; thus we omit it.


## 3.2. Existence of the Solution under Krasnoselskii-Krein Conditions on Time Scales

Theorem 14.
Assume that conditions (H1) and (H2) are satisfied; then the successive approximations given by(25)
u
n
+
1
t
=
∫
t
0
t
f
s
,
u
n
s
Δ
s
,
u
0
t
=
0
,
n
=
0,1
,
…converge uniformly to the unique solution u of (2) on [
t
0
,
t
0
+
η
]
T, where η
=
min
⁡
a
,
b
/
M, and M is the bound for f on S
0.Proof.
With the uniqueness of the solution being proved in Theorem12, we prove the existence of the solution using Arzela-Ascoli Theorem.
Step
1. The successive approximations {
u
n
+
1
}, n
=
0,1
,
2
,
… given by (25) are well defined and continuous. Indeed,(26)
u
n
+
1
t
=
∫
t
0
t
f
s
,
u
n
s
Δ
s
≤
∫
t
0
t
f
s
,
u
n
s
Δ
s
.This yields for n
=
0
(27)
u
1
t
≤
∫
t
0
t
f
s
,
u
0
s
Δ
s
≤
M
t
≤
b
.By induction, the sequence {
u
j
+
1
(
t
)
} is well defined and uniformly bounded on [
t
0
,
t
0
+
η
]
T.
Step
2. We prove that y is a continuous function in [
t
0
,
t
0
+
η
]
T, where y is defined by(28)
y
t
=
lim
sup
j
→
∞
⁡
u
j
t
-
u
j
-
1
t
.For t
1
,
t
2
∈
[
t
0
,
t
0
+
η
]
T, we have(29)
u
j
+
1
t
1
-
u
j
t
1
≤
u
j
+
1
t
2
-
u
j
t
2
+
2
M
t
2
-
t
1
.In fact, (30)
u
j
+
1
t
1
-
u
j
t
1
-
u
j
+
1
t
2
-
u
j
t
2
≤
u
j
+
1
t
1
-
u
j
t
1
-
u
j
+
1
t
2
+
u
j
t
2
≤
∫
t
0
t
1
f
s
,
u
j
s
-
f
s
,
u
j
-
1
s
Δ
s
-
∫
t
0
t
2
f
s
,
u
j
s
-
f
s
,
u
j
-
1
s
Δ
s
≤
2
M
∫
t
1
t
2
Δ
s
≤
2
M
t
2
-
t
1
.The right-hand side in inequality (29) is at most y
(
t
2
)
+
ϵ
+
2
M
(
t
2
-
t
1
) for large n if ϵ
>
0 provided that |
t
2
-
t
1
|
≤
ϵ
/
2
M. Since ϵ is arbitrary and t
1, t
2 can be interchangeable, we get(31)
y
t
1
-
y
t
2
≤
2
M
t
2
-
t
1
.This implies that y is continuous on [
t
0
,
t
0
+
η
]
T. Using condition (H2) and the definition of successive approximations, we obtain(32)
u
j
+
1
t
-
u
j
t
≤
c
∫
t
0
t
u
j
s
-
u
j
-
1
s
α
Δ
s
.The sequence {
u
n
} is equicontinuous: that is, for each function u
n and any ϵ
>
0, t
1
,
t
2
∈
[
t
0
,
t
0
+
η
]
T if there exists τ
=
ϵ
/
M such that t
2
-
t
1
≤
τ, then(33)
u
n
+
1
t
1
-
u
n
+
1
t
2
=
∫
t
1
t
2
f
s
,
u
n
s
Δ
s
≤
∫
t
1
t
2
f
~
s
,
u
n
s
d
s
≤
M
t
1
-
t
2
≤
ϵ
.
All of the Arzela-Ascoli Theorem conditions are fulfilled for the family{
u
j
} in C
r
d
[
t
0
,
t
0
+
η
]
T. Hence, there exists a subsequence {
u
j
k
} converging uniformly on [
t
0
,
t
0
+
η
]
T as j
k
→
∞.
Let us note(34)
m
∗
t
=
lim
k
→
∞
⁡
u
j
k
t
-
u
j
k
-
1
t
.Further, if {
u
j
-
u
j
-
1
}
→
0 as j
→
∞, then the limit of any subsequence is the unique solution u of (25). It follows that a selection of subsequences is unnecessary and that the entire sequence {
u
j
} converges uniformly to u. For that, it suffices to show that y
≡
0 which will lead to m
∗
(
t
) being null.
Setting(35)
R
t
=
∫
t
0
t
y
s
α
d
s
,and by defining ψ
∗
(
t
)
=
t
-
k
y
(
t
), we show that lim
t
→
0
+
⁡
ψ
∗
(
t
)
=
0.
We prove by absurdity thatψ
∗
≡
0. Assume that ψ
∗
(
t
)
>
0 at any point in ]
t
0
,
t
0
+
η
]
T; then there exists t
1 such that 0
<
m
¯
=
ψ
∗
(
t
1
)
=
max
t
∈
[
t
0
,
t
0
+
η
]
T
⁡
ψ
∗
(
t
). Hence, from condition (H1), we obtain(36)
m
¯
=
ψ
t
1
=
t
1
-
k
y
t
1
≤
m
¯
t
1
<
m
¯
.We end up with a contradiction. So ψ
∗
≡
0. Therefore, the Picard iterates (25) converge uniformly to the unique solution u of (2) on [
t
0
,
t
0
+
η
]
T.


## 3.3. Uniqueness Results for Fractional Order ODE

In this section, we denoteC
p
(
[
t
0
,
t
0
+
a
]
T
,
R
)
=
{
u
∣
u
∈
C
(
[
t
0
,
t
0
+
a
]
T
,
R
) and (
t
-
t
0
)
1
-
q
u
∈
C
(
[
t
0
,
t
0
+
a
]
T
,
R
)
}.Theorem 15 (Krasnoselskii-Krein conditions).
Letf
(
t
,
x
) be continuous in S
0 and satisfying for all (
t
,
x
)
,
(
t
,
x
¯
)
∈
S
0
(J1)
f
t
,
x
-
f
t
,
x
¯
≤
k
r
Γ
α
t
-
t
0
-
α
x
-
x
¯, t
≠
t
0,
(J2)
f
t
,
x
-
f
t
,
x
¯
≤
c
x
-
x
¯
δ, where c, r, k are positive constants such that k
>
1, k
r
≤
α, and k
(
1
-
δ
)
<
1, and the real number δ is such that 0
<
δ
<
1.Then, the fractional order initial value problem (3) has at most one solution on [
t
0
,
t
0
+
a
]
T.Proof.
Supposeu and v are two solutions of (3) in [
t
0
,
t
0
+
a
]
T. We will show that u
≡
v. Let us define ϕ
(
t
) and R
(
t
) by(37)
ϕ
t
=
u
t
-
v
t
,
for  every
t
∈
t
0
,
t
0
+
a
T
,
R
t
=
c
Γ
α
∫
t
0
t
t
-
s
α
-
1
ϕ
~
δ
s
d
s
,
for  every
t
∈
t
0
,
t
0
+
a
,such that ϕ
~ is the extension of ϕ to the real interval [
t
0
,
t
0
+
a
]. It follows from condition (J2) that(38)
ϕ
t
=
1
Γ
α
∫
t
0
t
t
-
s
α
-
1
f
s
,
u
s
-
f
s
,
v
s
Δ
s
≤
1
Γ
α
∫
t
0
t
t
-
s
α
-
1
f
s
,
u
s
-
f
s
,
v
s
Δ
s
≤
1
Γ
α
∫
t
0
t
c
t
-
s
α
-
1
u
s
-
v
s
δ
Δ
s
≤
∫
t
0
t
c
Γ
α
t
-
s
α
-
1
u
~
s
-
v
~
s
δ
d
s
=
R
t
.On the other hand, R
(
t
0
)
=
0, R
(
t
)
>
0 for t
>
t
0, and D
t
T
α
R
(
t
)
=
ϕ
~
δ
(
t
)
=
R
δ
(
t
), for every t
∈
[
t
0
,
t
0
+
a
]
T. Now from relations (37) and (38) and using Lemma 11, we obtain for every t
∈
[
t
0
,
t
0
+
a
]
T
(39)
ϕ
t
≤
R
t
=
L
t
-
t
0
σ
,where L and σ are defined as in Lemma 11. Moreover, if we define ψ
(
t
)
=
ϕ
(
t
)
/
(
t
-
t
0
)
k, we get(40)
0
≤
ψ
t
≤
L
t
-
t
0
σ
-
k
α
,
for  every
t
∈
t
0
,
t
0
+
a
T
.It follows that the exponent of t in the above inequality is positive, since k
(
1
-
δ
)
<
1. Hence, l
i
m
t
→
t
0
ψ
(
t
)
=
0. Therefore, if we define ψ
(
t
0
)
=
0, then the function is rd-continuous in [
t
0
,
t
0
+
a
]
T.
Now, to show thatϕ
≡
0, we prove by absurdity that ψ
≡
0 on [
t
0
,
t
0
+
a
]
T. Assume that ψ does not vanish at some points t; that is, ψ
(
t
)
>
0 on ]
t
0
,
t
0
+
a
]
T; then there exists a maximum m
>
0 reached when t is equal to some t
1
:
t
0
<
t
1
≤
t
0
+
a such that ψ
(
s
)
<
m
=
ψ
(
t
1
), for s
∈
[
t
0
,
t
1
)
T. But from condition (J1), we have (41)
m
=
ψ
t
1
=
t
1
-
t
0
-
k
ϕ
t
1
<
t
1
-
t
0
-
k
α
∫
t
0
t
1
k
r
t
-
s
α
-
1
f
s
,
u
s
-
f
s
,
v
s
Δ
s
≤
t
1
-
t
0
-
k
α
∫
t
0
t
1
k
r
t
-
s
α
-
1
ϕ
s
s
-
t
0
-
α
Δ
s
≤
t
1
-
t
0
-
k
α
∫
t
0
t
1
k
r
t
-
s
α
-
1
s
-
t
0
k
α
-
α
ψ
s
Δ
s
≤
m
k
r
t
1
-
t
0
-
α
∫
t
0
t
1
t
1
-
s
α
-
1
Δ
s
≤
m
k
r
t
1
-
t
0
-
α
∫
t
0
t
1
t
1
-
s
α
-
1
d
s
≤
m
k
r
α
<
m
,which is a contradiction. Thus, the uniqueness of the solution is established.Theorem 16 (Kooi’s conditions).
Letf
(
t
,
x
) be continuous in S
0 and satisfying for all (
t
,
x
)
,
(
t
,
x
¯
)
∈
S
0
(K1)
f
t
,
x
-
f
t
,
x
¯
≤
k
r
Γ
α
t
-
t
0
-
α
x
-
x
¯, t
≠
t
0,
(K2)
t
-
t
0
β
f
t
,
x
-
f
t
,
x
¯
≤
c
x
-
x
¯
δ, where c, r, and k are positive constants; the positive real numbers β, δ, k, r are such that 0
<
β
<
δ
<
1, and k
(
1
-
δ
)
<
1
-
β, and k
r
≤
α.Then, the first-order initial value problem (3) has at most one solution on [
t
0
,
t
0
+
a
]
T.Proof.
The proof is similar to that of Theorem15; thus, we omit it.


## 3.4. Existence of Solutions under Krasnoselskii-Krein Conditions on Time Scales

Theorem 17.
Assume that conditions (J1) and (J2) are satisfied; then the successive approximations given by(42)
u
n
+
1
t
=
∫
t
0
t
f
s
,
u
n
s
Δ
s
,
u
0
t
=
0
,
n
=
0,1
,
…converge uniformly to the unique solution u of (3) on [
t
0
,
t
0
+
η
]
T, where(43)
η
=
min
⁡
a
,
b
Γ
1
+
α
M
1
/
αand M is the bound for f on S
0.Proof.
With the uniqueness of the solution being proved in Theorem15, we prove the existence of the solution using Arzela-Ascoli Theorem.
Step
1. The successive approximations {
u
n
+
1
}, n
=
0,1
,
2
,
… given by (42) are well defined and continuous. Indeed,(44)
u
n
+
1
t
=
1
Γ
α
∫
t
0
t
t
-
s
α
-
1
f
s
,
u
n
s
Δ
s
≤
1
Γ
α
∫
t
0
t
t
-
s
α
-
1
f
s
,
u
n
s
Δ
s
.This yields, for n
=
0,(45)
u
1
t
≤
M
Γ
α
∫
t
0
t
t
-
s
α
-
1
Δ
s
≤
M
Γ
α
∫
t
0
t
t
-
s
α
-
1
d
s
≤
M
a
α
Γ
α
+
1
≤
b
.By induction, the sequence {
u
j
+
1
(
t
)
} is well defined and uniformly bounded on [
t
0
,
t
0
+
η
]
T.
Step
2. We prove that y is a continuous function in [
t
0
,
t
0
+
η
]
T, where y is defined by(46)
y
t
=
lim
sup
j
→
∞
⁡
u
j
t
-
u
j
-
1
t
.For t
1
,
t
2
∈
[
t
0
,
t
0
+
η
]
T, we have(47)
u
j
+
1
t
1
-
u
j
t
1
≤
u
j
+
1
t
2
-
u
j
t
2
+
4
M
Γ
α
+
1
t
2
-
t
1
α
.In fact,(48)
u
j
+
1
t
1
-
u
j
t
1
-
u
j
+
1
t
2
-
u
j
t
2
≤
u
j
+
1
t
1
-
u
j
t
1
-
u
j
+
1
t
2
+
u
j
t
2
≤
1
Γ
α
∫
t
0
t
1
t
1
-
s
α
-
1
f
s
,
u
j
s
-
f
s
,
u
j
-
1
s
Δ
s
-
∫
0
t
2
t
2
-
s
α
-
1
f
s
,
u
j
s
-
f
s
,
u
j
-
1
s
Δ
s
≤
2
M
Γ
α
∫
0
t
1
t
1
-
s
α
-
1
-
t
2
-
s
α
-
1
Δ
s
-
∫
t
1
t
2
t
2
-
s
α
-
1
Δ
s
≤
2
M
Γ
α
∫
0
t
1
t
1
-
s
α
-
1
-
t
2
-
s
α
-
1
d
s
-
∫
t
1
t
2
t
2
-
s
α
-
1
d
s
≤
2
M
α
Γ
α
t
1
α
-
t
2
α
+
2
t
2
-
t
1
α
≤
4
M
Γ
α
+
1
t
2
-
t
1
α
.The right-hand side in inequality (47) is at most y
(
t
2
)
+
ϵ
+
(
4
M
/
Γ
α
+
1
)
(
t
2
-
t
1
)
α for large n if ϵ
>
0 provided that |
t
2
-
t
1
|
≤
ϵ
Γ
(
α
+
1
)
/
4
M
1
/
α. Since ϵ is arbitrary and t
1, t
2 can be interchangeable, we get(49)
y
t
1
-
y
t
2
≤
4
M
Γ
α
+
1
t
2
-
t
1
.This implies that y is continuous on [
t
0
,
t
0
+
η
]
T. Using condition (J2) and the definition of successive approximations, we obtain(50)
u
j
+
1
t
-
u
j
t
≤
c
Γ
α
∫
t
0
t
u
j
s
-
u
j
-
1
s
α
Δ
s
.The sequence {
u
n
} is equicontinuous: that is, for each function u
n and any ϵ
>
0, t
1
,
t
2
∈
[
t
0
,
t
0
+
η
]
T if there exists τ
=
ϵ
-
α
Γ
(
α
+
1
)
/
M such that t
2
-
t
1
≤
τ; then (51)
u
n
+
1
t
1
-
u
n
+
1
t
2
=
1
Γ
α
∫
t
0
t
2
t
2
-
s
α
-
1
f
s
,
u
n
s
Δ
s
-
∫
t
0
t
1
t
1
-
s
α
-
1
f
s
,
u
n
s
Δ
s
≤
2
M
Γ
α
+
1
t
1
-
t
2
α
≤
ϵ
,where we used a similar argument as in (48).
All of the Arzela-Ascoli Theorem conditions are fulfilled for the family{
u
j
} in C
r
d
[
t
0
,
t
0
+
a
]
T. Hence, there exists a subsequence {
u
j
k
} converging uniformly on [
t
0
,
t
0
+
a
]
T as j
k
→
∞.
Let us note(52)
m
∗
t
=
lim
k
→
∞
⁡
u
j
k
t
-
u
j
k
-
1
t
.Further, if {
|
u
j
-
u
j
-
1
|
}
→
0 as j
→
∞, then the limit of any subsequence is the unique solution u of (42). It follows that a selection of subsequences is unnecessary and that the entire sequence {
u
j
} converges uniformly to u. For that, it is sufficient to show that y
≡
0 which will lead to m
∗
(
t
) being null.
Setting(53)
R
t
=
c
Γ
α
∫
t
0
t
t
-
s
α
-
1
y
s
α
d
s
,and defining ψ
∗
(
t
)
=
t
-
k
y
(
t
) and then using Lemma 11, we obtain that ψ
(
t
)
≤
L
(
t
1
-
t
0
)
J
-
k
α. Which yields that l
i
m
t
→
0
+
ψ
∗
(
t
)
=
0.
We prove by absurdity thatψ
∗
≡
0. Assume that ψ
∗
(
t
)
>
0 at any point in ]
t
0
,
t
0
+
η
]
T; then there exists t
1 such that 0
<
m
¯
=
ψ
∗
(
t
1
)
=
m
a
x
t
∈
[
t
0
,
t
0
+
η
]
T
ψ
∗
(
t
). Hence, from condition (J1), we obtain (54)
m
=
ψ
t
1
=
t
1
-
t
0
-
k
α
ϕ
t
1
≤
t
1
-
t
0
-
k
α
∫
t
0
t
1
k
r
t
1
-
s
α
-
1
s
-
t
0
-
α
ϕ
s
d
s
≤
k
r
t
1
-
t
0
-
k
α
∫
t
0
t
1
t
1
-
s
α
-
1
s
-
t
0
k
α
-
α
ψ
s
d
s
<
k
r
m
t
1
-
t
0
-
α
∫
t
0
t
1
t
1
-
t
0
α
-
1
d
s
<
k
r
m
α
<
m
.We end up with a contradiction. So ψ
∗
≡
0. Therefore, the Picard iterates (42) converge uniformly to the unique solution u of (2) on [
t
0
,
t
0
+
η
]
T.Remark 18.
For the caseT
=
R, Theorem 15 is reduced to [31, Theorem  2.1].



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*Source: 7327319-2016-07-27.xml*