# Henry-Gronwall Integral Inequalities with “Maxima” and Their Applications to Fractional Differential Equations

Authors: Phollakrit Thiramanus; Jessada Tariboon; Sotiris K. Ntouyas
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

Some new weakly singular Henry-Gronwall type integral inequalities with “maxima”
are established in this paper. Applications to Caputo fractional differential
equations with “maxima” are also presented.


---

## Body

## 1. Introduction

It is well known that Gronwall-Bellman type integral inequalities play a dominant role in the study of quantitative properties of solutions of differential and integral equations [1–5]. Usually, the integrals concerning these type inequalities have regular or continuous kernels, but some problems of theory and practicality require us to solve integral inequalities with singular kernels. For example, Henry [6] proposed a method to find solutions and proved some results concerning linear integral inequalities with weakly singular kernel. Moreover, Medved’ [7, 8] presented a new approach to solve integral inequalities of Henry-Gronwall type and their Bihari version and obtained global solutions of semilinear evolution equations. Ye and Gao [9] considered the integral inequalities of Henry-Gronwall type and their applications to fractional differential equations with delay. Ma and Pečarić [10] established some weakly singular integral inequalities of Gronwall-Bellman type and used them in the analysis of various problems in the theory of certain classes of differential equations, integral equations, and evolution equations. Shao and Meng [11] studied a certain class of nonlinear inequalities of Gronwall-Bellman type, which is used to a qualitative analysis to certain fractional differential equations. For other results on the subject we refer to [12–18] and references cited therein.Differential equations with “maxima” are a special type of differential equations that contain the maximum of the unknown function over a previous interval. Several integral inequalities have been established in the case when maxima of the unknown scalar function are involved in the integral; see [19, 20] and references cited therein.Recently in [21] some new types of integral inequalities on time scales with “maxima” are established, which can be used as a handy tool in the investigation of making estimates for bounds of solutions of dynamic equations on time scales with “maxima.” In this paper we establish some Henry-Gronwall type integral inequalities with “maxima.” The significance of our work lies in the fact that “maxima” are taken on intervals [
β
t
,
t
] which have nonconstant length, where 0
<
β
<
1. Most of the papers take the “maxima” on [
t
-
h
,
t
], where h
>
0 is a given constant. We apply our results to demonstrate the bound of solutions and the dependence of solutions on the orders with initial conditions for Caputo fractional differential equations with “maxima”
(1)
D
α
x
(
t
)
=
f
(
t
,
x
(
t
)
,
max
⁡
s
∈
[
β
t
,
t
]
x
(
s
)
)
,
t
∈
I
=
[
t
0
,
T
)
,
x
(
t
)
=
ϕ
(
t
)
,
t
∈
[
β
t
0
,
t
0
]
.The paper is organized as follows. In Section2 we recall some results from [21] in the special case T
=
R which are used to prove our main results which are presented in Section 3. In the last Section 4 we give applications of our results for an initial value problem for a Caputo fractional differential equation with “maxima.”


## 2. Preliminaries

For convenience we let throughoutt
0
>
0. The following results in Lemmas 1 and 3 are obtained by reducing the time scalesT
=
R, f
(
t
)
=
g
(
t
)
≡
1, and a
(
t
)
=
b
(
t
)
≡
0 for all t
∈
[
t
0
,
T
) in Theorems 3.3 and 3.2 ([21], page 8 and page 6), respectively.Lemma 1 (see [21]).
Let the following conditions be satisfied:(
H
1
)
the functionsp and q
∈
C
(
[
t
0
,
T
)
,
R
+
);
(
H
2
)
the functionϕ
∈
C
(
[
β
t
0
,
T
)
,
R
+
) with max
⁡
s
∈
[
β
t
0
,
t
0
]
ϕ
(
s
)
>
0, where 0
<
β
<
1;
(
H
3
)
the functionu
∈
C
(
[
β
t
0
,
T
)
,
R
+
) and satisfies the inequalities
(2)
u
(
t
)
≤
ϕ
(
t
)
+
∫
t
0
t
[
p
(
s
)
u
(
s
)
+
q
(
s
)
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
]
d
s
,
00000000000000000000000000000000
t
∈
[
t
0
,
T
]
,
u
(
t
)
≤
ϕ
(
t
)
,
t
∈
[
β
t
0
,
t
0
]
.Then
(3)
u
(
t
)
≤
ϕ
(
t
)
+
h
(
t
)
exp
⁡
(
∫
t
0
t
[
p
(
s
)
+
q
(
s
)
]
d
s
)
,
t
∈
[
t
0
,
T
)
holds, where
(4)
h
(
t
)
=
max
⁡
s
∈
[
β
t
0
,
t
0
]
ϕ
(
s
)
+
∫
t
0
t
[
p
(
s
)
ϕ
(
s
)
+
q
(
s
)
max
⁡
ξ
∈
[
β
s
,
s
]
ϕ
(
ξ
)
]
d
s
,
000000000000000000000000000000000000000
t
∈
[
t
0
,
T
)
.By splitting the initial functionϕ to be two functions, we deduce the following corollary.Corollary 2.
Let the following conditions be satisfied:(
H
4
)
the functionsp, q, and v
∈
C
(
[
t
0
,
T
)
,
R
+
);
(
H
5
)
the functionw
∈
C
(
[
β
t
0
,
t
0
]
,
R
+
) with max
⁡
s
∈
[
β
t
0
,
t
0
]
w
(
s
)
>
0 and w
(
t
0
)
=
v
(
t
0
), where 0
<
β
<
1;
(
H
6
)
the functionu
∈
C
(
[
β
t
0
,
T
)
,
R
+
) and satisfies the inequalities
(5)
u
(
t
)
≤
v
(
t
)
+
∫
t
0
t
[
p
(
s
)
u
(
s
)
+
q
(
s
)
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
]
d
s
,
000000000000000000000000000000000
t
∈
[
t
0
,
T
)
,
u
(
t
)
≤
w
(
t
)
,
t
∈
[
β
t
0
,
t
0
]
.Then
(6)
u
(
t
)
≤
v
(
t
)
+
h
(
t
)
exp
⁡
(
∫
t
0
t
[
p
(
s
)
+
q
(
s
)
]
d
s
)
,
000000000000000000000000000000
t
∈
[
t
0
,
T
)
holds, where
(7)
h
(
t
)
=
max
⁡
s
∈
[
β
t
0
,
t
0
]
w
(
s
)
+
∫
t
0
t
[
p
(
s
)
v
(
s
)
+
q
(
s
)
max
⁡
ξ
∈
[
β
s
,
s
]
m
(
ξ
)
]
d
s
,
0000000000000000000000000000000000000000
t
∈
[
t
0
,
T
)
,
with
(8)
m
(
t
)
=
{
v
(
t
)
,
t
∈
[
t
0
,
T
)
,
w
(
t
)
,
t
∈
[
β
t
0
,
t
0
]
.Lemma 3 (see [21]).
Let the condition(
H
1
) of Lemma 1 be satisfied. In addition, assume that(
H
7
)
the functionk
∈
C
(
[
t
0
,
T
)
,
(
0
,
∞
)
) is nondecreasing;
(
H
8
)
the functionϕ
∈
C
(
[
β
t
0
,
t
0
)
,
R
+
), where 0
<
β
<
1;
(
H
9
)
the functionu
∈
C
(
[
β
t
0
,
T
)
,
R
+
) and satisfies the inequalities
(9)
u
(
t
)
≤
k
(
t
)
+
∫
t
0
t
[
p
(
s
)
u
(
s
)
+
q
(
s
)
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
]
d
s
,
000000000000000000000000000000000
t
∈
[
t
0
,
T
)
,
u
(
t
)
≤
ϕ
(
t
)
,
t
∈
[
β
t
0
,
t
0
]
.Then
(10)
u
(
t
)
≤
N
k
(
t
)
exp
⁡
(
∫
t
0
t
[
p
(
s
)
+
q
(
s
)
]
d
s
)
,
t
∈
[
t
0
,
T
)
holds, where
(11)
N
=
max
⁡
{
1
,
max
⁡
s
∈
[
β
t
0
,
t
0
]
ϕ
(
s
)
k
(
t
0
)
}
.The following lemma is a consequence of Jensen’s inequality which can be found in [22].Lemma 4 (see [22]).
Letn
∈
N, and let x
1
,
…
,
x
n be nonnegative real numbers. Then, for σ
>
1,
(12)
(
∑
i
=
1
n
‍
x
i
)
σ
≤
n
σ
-
1
∑
i
=
1
n
x
i
σ
.


## 3. Main Results

Theorem 5.
Suppose that the following conditions are satisfied:(
H
10
)
the functionsp and r
∈
C
(
[
t
0
,
T
)
,
R
+
);
(
H
11
)
the functionϕ
∈
C
(
[
β
t
0
,
t
0
]
,
R
+
) with max
⁡
s
∈
[
β
t
0
,
t
0
]
ϕ
(
s
)
>
0, where 0
<
β
<
1;
(
H
12
)
the functionu
∈
C
(
[
β
t
0
,
T
)
,
R
+
) with
(13)
u
(
t
)
≤
r
(
t
)
+
∫
t
0
t
(
t
-
s
)
α
-
1
p
(
s
)
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
d
s
,
t
∈
[
t
0
,
T
)
,
(14)
u
(
t
)
≤
ϕ
(
t
)
,
t
∈
[
β
t
0
,
t
0
]
,where α
>
0.
Then the following assertions hold.(
R
1
)
Supposeα
>
1
/
2; then
(15)
u
(
t
)
≤
e
t
[
c
1
r
2
(
t
)
+
h
1
(
t
)
exp
⁡
(
K
1
∫
t
0
t
p
2
(
s
)
d
s
)
]
1
/
2
,
00000000000000000000000000000000000
t
∈
[
t
0
,
T
)
,
where(16)
c
1
=
max
⁡
{
2
e
-
2
t
0
,
e
-
2
β
t
0
}
,
(17)
K
1
=
Γ
(
2
α
-
1
)
4
α
-
1
,
(18)
h
1
(
t
)
=
c
1
max
⁡
s
∈
[
β
t
0
,
t
0
]
ϕ
2
(
s
)
+
c
1
K
1
∫
t
0
t
p
2
(
s
)
max
⁡
ξ
∈
[
β
s
,
s
]
m
1
2
(
ξ
)
d
s
,
000000000000000000000000000000000000
t
∈
[
t
0
,
T
)
,
with(19)
m
1
(
t
)
=
{
r
(
t
)
,
t
∈
[
t
0
,
T
)
,
ϕ
(
t
)
,
t
∈
[
β
t
0
,
t
0
]
.
Moreover, ifr
∈
C
(
[
t
0
,
T
)
,
(
0
,
∞
)
) is a nondecreasing function, then
(20)
u
(
t
)
≤
c
1
N
1
r
(
t
)
exp
⁡
(
t
+
1
2
K
1
∫
t
0
t
p
2
(
s
)
d
s
)
,
0000000000000000000000000000
t
∈
[
t
0
,
T
)
,
where(21)
N
1
=
max
⁡
{
1
,
max
⁡
s
∈
[
β
t
0
,
t
0
]
ϕ
2
(
s
)
r
2
(
t
0
)
}
.
(
R
2
)
Suppose0
<
α
≤
1
/
2; then
(22)
u
(
t
)
≤
e
t
[
c
2
r
b
(
t
)
+
h
2
(
t
)
exp
⁡
(
2
b
-
1
K
2
b
∫
t
0
t
p
b
(
s
)
d
s
)
]
1
/
b
,
00000000000000000000000000000000000000
t
∈
[
t
0
,
T
)
,
where(23)
a
=
α
+
1
,
(24)
b
=
1
+
1
α
,
(25)
c
2
=
max
⁡
{
2
b
-
1
e
-
b
t
0
,
e
-
b
β
t
0
}
,
(26)
K
2
=
(
Γ
(
α
2
)
a
α
2
)
1
/
a
,
(27)
h
2
(
t
)
=
c
2
max
⁡
s
∈
[
β
t
0
,
t
0
]
ϕ
b
(
s
)
+
2
b
-
1
c
2
K
2
b
∫
t
0
t
‍
p
b
(
s
)
max
⁡
ξ
∈
[
β
s
,
s
]
m
1
b
(
ξ
)
d
s
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
∈
[
t
0
,
T
)
.
Moreover, ifr
∈
C
(
[
t
0
,
T
)
,
(
0
,
∞
)
) is a nondecreasing function, then
(28)
u
(
t
)
≤
(
c
2
N
2
)
1
/
b
r
(
t
)
exp
⁡
(
t
+
2
b
-
1
b
K
2
b
∫
t
0
t
p
b
(
s
)
d
s
)
,
00000000000000000000000000000000
t
∈
[
t
0
,
T
)
,
where(29)
N
2
=
max
⁡
{
1
,
max
⁡
s
∈
[
β
t
0
,
t
0
]
ϕ
b
(
s
)
r
b
(
t
0
)
}
.Proof.
Consider(
R
1
)
α
>
1
/
2. Using the Cauchy-Schwarz inequality with (13), we get for t
∈
[
t
0
,
T
)
(30)
u
(
t
)
≤
r
(
t
)
+
∫
t
0
t
(
t
-
s
)
α
-
1
e
s
p
(
s
)
e
-
s
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
d
s
≤
r
(
t
)
+
[
∫
t
0
t
‍
(
t
-
s
)
2
α
-
2
e
2
s
d
s
]
1
/
2
×
[
∫
t
0
t
p
2
(
s
)
e
-
2
s
(
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
)
2
d
s
]
1
/
2
.
The first integral of (30) implies the estimate
(31)
∫
t
0
t
(
t
-
s
)
2
α
-
2
e
2
s
d
s
=
∫
0
t
-
t
0
τ
2
α
-
2
e
2
(
t
-
τ
)
d
τ
≤
e
2
t
∫
0
t
τ
2
α
-
2
e
-
2
τ
d
τ
=
2
e
2
t
4
α
∫
0
2
t
σ
2
α
-
2
e
-
σ
d
σ
<
2
e
2
t
4
α
Γ
(
2
α
-
1
)
.
Therefore, from (30) and (31), we obtain
(32)
u
(
t
)
≤
r
(
t
)
+
[
2
e
2
t
4
α
Γ
(
2
α
-
1
)
]
1
/
2
×
[
∫
t
0
t
p
2
(
s
)
e
-
2
s
(
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
)
2
d
s
]
1
/
2
.
Applying Lemma 4 with n
=
2, σ
=
2, we get
(33)
u
2
(
t
)
≤
2
r
2
(
t
)
+
e
2
t
4
α
-
1
Γ
(
2
α
-
1
)
∫
t
0
t
p
2
(
s
)
e
-
2
s
(
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
)
2
d
s
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
∈
[
t
0
,
T
)
.
Now, taking v
(
t
)
=
[
e
-
t
u
(
t
)
]
2, we have
(34)
v
(
t
)
≤
2
e
-
2
t
r
2
(
t
)
+
Γ
(
2
α
-
1
)
4
α
-
1
×
∫
t
0
t
p
2
(
s
)
e
-
2
s
(
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
)
2
d
s
≤
2
e
-
2
t
0
r
2
(
t
)
+
Γ
(
2
α
-
1
)
4
α
-
1
×
∫
t
0
t
p
2
(
s
)
max
⁡
ξ
∈
[
β
s
,
s
]
(
e
-
2
ξ
u
2
(
ξ
)
)
d
s
≤
c
1
r
2
(
t
)
+
K
1
∫
t
0
t
p
2
(
s
)
max
⁡
ξ
∈
[
β
s
,
s
]
v
(
ξ
)
d
s
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
∈
[
t
0
,
T
)
,
and, for t
∈
[
β
t
0
,
t
0
],
(35)
v
(
t
)
≤
e
-
2
t
ϕ
2
(
t
)
≤
e
-
2
β
t
0
ϕ
2
(
t
)
≤
c
1
ϕ
2
(
t
)
,
where c
1 and K
1 are defined by (16) and (17), respectively.
Applying Corollary2 for (34) and (35), we obtain
(36)
v
(
t
)
≤
c
1
r
2
(
t
)
+
h
1
(
t
)
exp
⁡
(
K
1
∫
t
0
t
p
2
(
s
)
d
s
)
,
0000000000000000000000000000
t
∈
[
t
0
,
T
)
,
where h
1 is defined by (18). Therefore, we get the required inequality in (15).
Moreover, ifr
∈
C
(
[
t
0
,
T
)
,
(
0
,
∞
)
) is a nondecreasing function, then, by applying Lemma 3 for (34) and (35), we obtain the estimate
(37)
v
(
t
)
≤
c
1
N
1
r
2
(
t
)
exp
⁡
(
K
1
∫
t
0
t
p
2
(
s
)
d
s
)
,
t
∈
[
t
0
,
T
)
,
where N
1 is defined by (21). Thus, we get the desired inequality in (20). This completes the proof of the first part.
Consider(
R
2
)
0
<
α
≤
1
/
2. Let a, b be defined by (23) and (24), respectively. It is obvious that (
1
/
a
)  +  (
1
/
b
)
=
1. Using the Hölder inequality in (13), for t
∈
[
t
0
,
T
), we have
(38)
u
(
t
)
≤
r
(
t
)
+
∫
t
0
t
(
t
-
s
)
α
-
1
e
s
p
(
s
)
e
-
s
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
d
s
≤
r
(
t
)
+
[
∫
t
0
t
‍
(
t
-
s
)
a
(
α
-
1
)
e
a
s
d
s
]
1
/
a
×
[
∫
t
0
t
p
b
(
s
)
e
-
b
s
(
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
)
b
d
s
]
1
/
b
.
Repeating the process to get (31), the first integral of (38) implies the estimate
(39)
∫
t
0
t
(
t
-
s
)
a
(
α
-
1
)
e
a
s
d
s
<
e
a
t
a
1
-
a
(
1
-
α
)
Γ
(
1
-
a
(
1
-
α
)
)
.
Obviously, 1
-
a
(
1
-
α
)
=
α
2
>
0 and Γ
(
1
-
a
(
1
-
α
)
)
∈
R. From (38) and (39), it follows that
(40)
u
(
t
)
≤
r
(
t
)
+
K
2
e
t
[
∫
t
0
t
p
b
(
s
)
e
-
b
s
(
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
)
b
d
s
]
1
/
b
,
where K
2 is defined by (26). Applying Lemma 4 with n
=
2, σ
=
b, we have
(41)
u
b
(
t
)
≤
2
b
-
1
[
r
b
(
t
)
+
K
2
b
e
b
t
∫
t
0
t
p
b
(
s
)
e
-
b
s
(
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
)
b
d
s
]
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
∈
[
t
0
,
T
)
.
By setting v
(
t
)
=
[
e
-
t
u
(
t
)
]
b, we get
(42)
v
(
t
)
≤
c
2
r
b
(
t
)
+
2
b
-
1
K
2
b
∫
t
0
t
p
b
(
s
)
max
⁡
ξ
∈
[
β
s
,
s
]
v
(
ξ
)
d
s
,
00000000000000000000000000000
t
∈
[
t
0
,
T
)
,
and, for t
∈
[
β
t
0
,
t
0
],
(43)
v
(
t
)
≤
e
-
b
β
t
0
ϕ
b
(
t
)
≤
c
2
ϕ
b
(
t
)
,
where c
2 is defined by (25). Consequently, applying Corollary 2 with (42) and (43), we have
(44)
v
(
t
)
≤
c
2
r
b
(
t
)
+
h
2
(
t
)
exp
⁡
(
2
b
-
1
K
2
b
∫
t
0
t
p
b
(
s
)
d
s
)
,
00000000000000000000000000000000
t
∈
[
t
0
,
T
)
,
where h
2 is defined by (27). Therefore, the desired inequality (22) is established.
Furthermore, ifr
∈
C
(
[
t
0
,
T
)
,
(
0
,
∞
)
) is a nondecreasing function, then by applying Lemma 3 for (42) and (43) we deduce that
(45)
v
(
t
)
≤
c
2
N
2
r
b
(
t
)
exp
⁡
(
2
b
-
1
K
2
b
∫
t
0
t
p
b
(
s
)
d
s
)
,
000000000000000000000000000000
t
∈
[
t
0
,
T
)
.
Thus, inequality (28) is proved. This completes the proof.Theorem 6.
Assume that(
H
13
)
the conditions(
H
10
), (
H
11
) of Theorem 5 are satisfied;
(
H
14
)
the functionq
∈
C
(
[
t
0
,
T
)
,
R
+
);
(
H
15
)
the functionu
∈
C
(
[
β
t
0
,
T
)
,
R
+
) with
(46)
u
(
t
)
≤
r
(
t
)
+
∫
t
0
t
(
t
-
s
)
α
-
1
[
p
(
s
)
u
(
s
)
+
q
(
s
)
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
]
d
s
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
∈
[
t
0
,
T
)
,
(47)
u
(
t
)
≤
ϕ
(
t
)
,
t
∈
[
β
t
0
,
t
0
]
,where α
>
0.
Then the following assertions hold.(
R
3
)
Supposeα
>
1
/
2; then
(48)
u
(
t
)
≤
e
t
[
c
3
r
2
(
t
)
+
h
3
(
t
)
exp
⁡
(
K
3
∫
t
0
t
[
p
2
(
s
)
+
q
2
(
s
)
]
d
s
)
]
1
/
2
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
∈
[
t
0
,
T
)
,
where(49)
c
3
=
max
⁡
{
3
e
-
2
t
0
,
e
-
2
β
t
0
}
,
(50)
K
3
=
6
4
α
Γ
(
2
α
-
1
)
,
(51)
h
3
(
t
)
=
c
3
max
⁡
s
∈
[
β
t
0
,
t
0
]
ϕ
2
(
s
)
+
c
3
K
3
×
∫
t
0
t
[
p
2
(
s
)
r
2
(
s
)
+
q
2
(
s
)
max
⁡
ξ
∈
[
β
s
,
s
]
m
1
2
(
ξ
)
]
d
s
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
∈
[
t
0
,
T
)
,
withm
1 being defined by (19).
Furthermore, ifr
∈
C
(
[
t
0
,
T
)
,
(
0
,
∞
)
) is a nondecreasing function, then
(52)
u
(
t
)
≤
c
3
N
1
r
(
t
)
exp
⁡
(
t
+
1
2
K
3
∫
t
0
t
[
p
2
(
s
)
+
q
2
(
s
)
]
d
s
)
,
0000000000000000000000000000000000000
t
∈
[
t
0
,
T
)
,
whereN
1 is defined by (21).
(
R
4
)
Suppose0
<
α
≤
1
/
2; then
(53)
u
(
t
)
≤
e
t
[
(
3
b
-
1
K
2
b
∫
t
0
t
‍
[
p
b
(
s
)
+
q
b
(
s
)
]
d
s
)
c
4
r
b
(
t
)
h
h
h
h
h
+
h
4
(
t
)
exp
⁡
(
3
b
-
1
K
2
b
∫
t
0
t
‍
[
p
b
(
s
)
+
q
b
(
s
)
]
d
s
)
]
1
/
b
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
∈
[
t
0
,
T
)
,
wherea, b, and K
2 are defined by (23), (24), and (26), respectively,
(54)
c
4
=
max
⁡
{
3
b
-
1
e
-
b
t
0
,
e
-
b
β
t
0
}
,
(55)
h
4
(
t
)
=
c
4
max
⁡
s
∈
[
β
t
0
,
t
0
]
ϕ
b
(
s
)
+
c
4
3
b
-
1
K
2
b
×
∫
t
0
t
[
p
b
(
s
)
r
b
(
s
)
+
q
b
(
s
)
max
⁡
ξ
∈
[
β
s
,
s
]
m
1
b
(
ξ
)
]
d
s
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
∈
[
t
0
,
T
)
.
Furthermore, ifr
∈
C
(
[
t
0
,
T
)
,
(
0
,
∞
)
) is a nondecreasing function, then
(56)
u
(
t
)
≤
(
c
4
N
2
)
1
/
b
r
(
t
)
exp
⁡
(
t
+
3
b
-
1
b
K
2
b
∫
t
0
t
[
p
b
(
s
)
+
q
b
(
s
)
]
d
s
)
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
∈
[
t
0
,
T
)
,
whereN
2 is defined by (29).Proof.
Consider(
R
3
)
α
>
1
/
2. By using the Cauchy-Schwarz inequality in (46), for t
∈
[
t
0
,
T
), we have
(57)
u
(
t
)
≤
r
(
t
)
+
∫
t
0
t
(
t
-
s
)
α
-
1
e
s
p
(
s
)
e
-
s
u
(
s
)
d
s
+
∫
t
0
t
(
t
-
s
)
α
-
1
e
s
q
(
s
)
e
-
s
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
d
s
≤
r
(
t
)
+
[
∫
t
0
t
(
t
-
s
)
2
α
-
2
e
2
s
d
s
]
1
/
2
×
[
∫
t
0
t
p
2
(
s
)
e
-
2
s
u
2
(
s
)
d
s
]
1
/
2
+
[
∫
t
0
t
(
t
-
s
)
2
α
-
2
e
2
s
d
s
]
1
/
2
×
[
∫
t
0
t
q
2
(
s
)
e
-
2
s
(
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
)
2
d
s
]
1
/
2
≤
r
(
t
)
+
[
2
e
2
t
4
α
Γ
(
2
α
-
1
)
]
1
/
2
×
{
[
∫
t
0
t
‍
q
2
(
s
)
e
-
2
s
(
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
)
2
d
s
]
1
/
2
[
∫
t
0
t
‍
p
2
(
s
)
e
-
2
s
u
2
(
s
)
d
s
]
1
/
2
h
h
h
h
h
+
[
∫
t
0
t
‍
q
2
(
s
)
e
-
2
s
(
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
)
2
d
s
]
1
/
2
[
∫
t
0
t
‍
q
2
(
s
)
e
-
2
s
(
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
)
2
d
s
]
1
/
2
[
∫
t
0
t
‍
p
2
(
s
)
e
-
2
s
u
2
(
s
)
d
s
]
1
/
2
}
.
Applying Lemma 4 with n
=
3, σ
=
2, we get
(58)
u
2
(
t
)
≤
3
r
2
(
t
)
+
6
e
2
t
4
α
Γ
(
2
α
-
1
)
×
[
(
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
)
2
∫
t
0
t
p
2
(
s
)
e
-
2
s
u
2
(
s
)
d
s
h
h
h
h
h
+
∫
t
0
t
q
2
(
s
)
e
-
2
s
(
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
)
2
d
s
]
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
∈
[
t
0
,
T
)
.
Taking v
(
t
)
=
[
e
-
t
u
(
t
)
]
2, we have
(59)
v
(
t
)
≤
c
3
r
2
(
t
)
+
K
3
∫
t
0
t
p
2
(
s
)
v
(
s
)
d
s
+
K
3
∫
t
0
t
q
2
(
s
)
max
⁡
ξ
∈
[
β
s
,
s
]
v
(
ξ
)
d
s
,
t
∈
[
t
0
,
T
)
,
(60)
v
(
t
)
≤
c
3
ϕ
2
(
t
)
,
t
∈
[
β
t
0
,
t
0
]
,
where c
3 and K
3 are defined by (49) and (50), respectively. Using Corollary 2 for (59) and (60), it follows that
(61)
v
(
t
)
≤
c
3
r
2
(
t
)
+
h
3
(
t
)
exp
⁡
(
K
3
∫
t
0
t
[
p
2
(
s
)
+
q
2
(
s
)
]
d
s
)
,
0000000000000000000000000000000000000
t
∈
[
t
0
,
T
)
,
where h
3 is defined by (51). Thus, we get the result in (48).
Ifr
∈
C
(
[
t
0
,
T
)
,
(
0
,
∞
)
) is a nondecreasing function, then Lemma 3 with (59) and (60) implies the estimate
(62)
v
(
t
)
≤
c
3
N
1
r
2
(
t
)
exp
⁡
(
K
3
∫
t
0
t
[
p
2
(
s
)
+
q
2
(
s
)
]
d
s
)
,
00000000000000000000000000000000000
t
∈
[
t
0
,
T
)
,
where N
1 is defined by (21). Thus, the required inequality (52) is established. This completes the proof of the first part.
Consider(
R
4
)
0
<
α
≤
1
/
2. Let a, b be defined by (23) and (24), respectively. Applying the Hölder inequality in (46), we have that for t
∈
[
t
0
,
T
)
(63)
u
(
t
)
≤
r
(
t
)
+
[
∫
t
0
t
(
t
-
s
)
a
(
α
-
1
)
e
a
s
d
s
]
1
/
a
×
[
∫
t
0
t
p
b
(
s
)
e
-
b
s
u
b
(
s
)
d
s
]
1
/
b
+
[
∫
t
0
t
(
t
-
s
)
a
(
α
-
1
)
e
a
s
d
s
]
1
/
a
×
[
∫
t
0
t
q
b
(
s
)
e
-
b
s
(
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
)
b
d
s
]
1
/
b
≤
r
(
t
)
+
K
2
e
t
{
[
∫
t
0
t
q
b
(
s
)
e
-
b
s
(
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
)
b
d
s
]
1
/
b
1
2
2
[
∫
t
0
t
p
b
(
s
)
e
-
b
s
u
b
(
s
)
d
s
]
1
/
b
h
h
h
h
h
h
h
h
h
h
+
[
∫
t
0
t
q
b
(
s
)
e
-
b
s
(
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
)
b
d
s
]
1
/
b
}
,
where K
2 is defined by (26). By using Lemma 4 with n
=
3, σ
=
b, we obtain the estimate
(64)
u
b
(
t
)
≤
3
b
-
1
r
b
(
t
)
+
3
b
-
1
K
2
b
e
b
t
×
[
(
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
)
b
∫
t
0
t
p
b
(
s
)
e
-
b
s
u
b
(
s
)
d
s
h
h
h
h
h
+
∫
t
0
t
q
b
(
s
)
e
-
b
s
(
max
⁡
ξ
∈
[
β
s
,
s
]
u
(
ξ
)
)
b
d
s
]
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
∈
[
t
0
,
T
)
.
Substituting v
(
t
)
=
[
e
-
t
u
(
t
)
]
b, we get
(65)
v
(
t
)
≤
c
4
r
b
(
t
)
+
3
b
-
1
K
2
b
×
[
max
⁡
ξ
∈
[
β
s
,
s
]
∫
t
0
t
‍
p
b
(
s
)
v
(
s
)
d
s
h
h
h
h
h
+
∫
t
0
t
‍
q
b
(
s
)
max
⁡
ξ
∈
[
β
s
,
s
]
v
(
ξ
)
d
s
]
,
t
∈
[
t
0
,
T
)
,
and, for t
∈
[
β
t
0
,
t
0
],
(66)
v
(
t
)
≤
c
4
ϕ
b
(
t
)
,
where c
4 is defined by (54). An application of Corollary 2 to (65) and (66) gives
(67)
v
(
t
)
≤
c
4
r
b
(
t
)
+
h
4
(
t
)
×
exp
⁡
(
3
b
-
1
K
2
b
∫
t
0
t
[
p
b
(
s
)
+
q
b
(
s
)
]
d
s
)
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
∈
[
t
0
,
T
)
,
where h
4 is defined by (55). Therefore, we deduce inequality (53).
As a special case, ifr
∈
C
(
[
t
0
,
T
)
,
(
0
,
∞
)
) is a nondecreasing function, then, by Lemma 3 with (65) and (66), we get
(68)
v
(
t
)
≤
c
4
N
2
r
b
(
t
)
×
exp
⁡
(
3
b
-
1
K
2
b
∫
t
0
t
‍
[
p
b
(
s
)
+
q
b
(
s
)
]
d
s
)
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
∈
[
t
0
,
T
)
.
Therefore, the desired inequality (56) is established. This completes the proof of Theorem 6.


## 4. Applications to Fractional Differential Equations with “Maxima”

In this section, we apply our results to demonstrate the bound of solutions and the dependence of solutions on the orders with initial conditions for Caputo fractional differential equations with “maxima.” We consider the following fractional differential equations (FDEs) with “maxima”(69)
D
α
x
(
t
)
=
f
(
t
,
x
(
t
)
,
max
⁡
s
∈
[
β
t
,
t
]
x
(
s
)
)
,
t
∈
I
=
[
t
0
,
T
)
,
and initial condition
(70)
x
(
t
)
=
ϕ
(
t
)
,
t
∈
[
β
t
0
,
t
0
]
,
where D
α represents the Caputo fractional derivative of order α
(
α
>
0
), f
∈
C
(
I
×
R
×
R
,
R
), ϕ is a given continuously differentiable function on [
β
t
0
,
t
0
] up to order n
(
n
=
-
[
-
α
]
), and 0
<
β
<
1. We denote ϕ
(
k
)
(
t
0
)
=
ρ
k, k
=
0,1
,
2
,
…
,
n
-
1. For more details on fractional differential equations, see [23, 24].Theorem 7.
Assume that(
H
16
)
there exist functionsy
,
z
∈
C
(
I
,
R
+
) such that, for t
∈
I, u
1
,
u
2
∈
R,
(71)
|
f
(
t
,
u
1
,
u
2
)
|
≤
y
(
t
)
|
u
1
|
+
z
(
t
)
|
u
2
|
.If x is solution of the initial value problem (69)-(70), then the following estimates hold. (
R
5
)
Suppose1
/
2
<
α
≤
1. Then
(72)
|
x
(
t
)
|
≤
c
3
M
exp
⁡
(
t
+
K
3
2
Γ
2
(
α
)
∫
t
0
t
[
y
2
(
s
)
+
z
2
(
s
)
]
d
s
)
,
0000000000000000000000000000000000000000000
t
∈
I
.
(
R
6
)
Suppose0
<
α
≤
1
/
2. Then
(73)
|
x
(
t
)
|
≤
(
c
4
)
1
/
b
M
exp
⁡
(
t
+
3
b
-
1
K
2
b
b
Γ
b
(
α
)
∫
t
0
t
[
y
b
(
s
)
+
z
b
(
s
)
]
d
s
)
,
000000000000000000000000000000000000000000000
t
∈
I
.
(
R
7
)
Supposeα
>
1. Then
(74)
|
x
(
t
)
|
≤
c
3
(
M
+
∑
j
=
1
n
-
1
|
ρ
j
|
j
!
(
t
-
t
0
)
j
)
×
exp
⁡
(
t
+
K
3
2
Γ
2
(
α
)
∫
t
0
t
[
y
2
(
s
)
+
z
2
(
s
)
]
d
s
)
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
0
h
h
h
h
h
h
h
h
h
h
h
t
∈
I
,where
(75)
M
=
max
⁡
t
∈
[
β
t
0
,
t
0
]
|
ϕ
(
t
)
|
>
0
and b, c
3, c
4, K
2, and K
3 are defined as in Theorems 5 and 6.Proof.
The solutionx of the initial value problem (69)-(70) satisfies the following equations (see [23]):
(76)
x
(
t
)
=
∑
j
=
0
n
-
1
‍
ρ
j
j
!
(
t
-
t
0
)
j
+
1
Γ
(
α
)
×
∫
t
0
t
(
t
-
s
)
α
-
1
f
(
s
,
x
(
s
)
,
max
⁡
ξ
∈
[
β
s
,
s
]
x
(
ξ
)
)
d
s
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
∈
I
,
(77)
x
(
t
)
=
ϕ
(
t
)
,
t
∈
[
β
t
0
,
t
0
]
.
For 0
<
α
≤
1, by using the assumption (
H
16
), it follows that
(78)
|
x
(
t
)
|
≤
M
+
1
Γ
(
α
)
×
∫
t
0
t
(
t
-
s
)
α
-
1
(
y
(
s
)
|
x
(
s
)
|
+
z
(
s
)
max
⁡
ξ
∈
[
β
s
,
s
]
|
x
(
ξ
)
|
)
d
s
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
0
h
h
h
h
h
h
h
t
∈
I
,
|
x
(
t
)
|
≤
M
,
t
∈
[
β
t
0
,
t
0
]
.
Hence, Theorem 6 yields the estimate inequalities (72) and (73).
Forα
>
1, by using the assumption (
H
16
) in (76), we have
(79)
|
x
(
t
)
|
≤
M
+
∑
j
=
1
n
-
1
|
ρ
j
|
j
!
(
t
-
t
0
)
j
+
1
Γ
(
α
)
∫
t
0
t
(
t
-
s
)
α
-
1
h
h
h
h
h
h
h
h
h
h
×
(
y
(
s
)
|
x
(
s
)
|
+
z
(
s
)
max
⁡
ξ
∈
[
β
s
,
s
]
|
x
(
ξ
)
|
)
d
s
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
0
h
h
h
h
h
h
h
h
h
h
t
∈
I
,
|
x
(
t
)
|
≤
M
,
t
∈
[
β
t
0
,
t
0
]
.
Since ∑
j
=
1
n
-
1
‍
(
|
ρ
j
|
/
j
!
)
(
t
-
t
0
)
j is a nondecreasing function, Theorem 6 yields the estimate inequality (74). This completes the proof.Theorem 8.
Letα
>
0 and δ
>
0 such that 0
≤
n
-
1
<
α
-
δ
<
α
≤
n. Also let f
:
I
×
R
×
R
→
R be a continuous function satisfying the following assumption:(
H
17
)
there exist constantsL
1
,
L
2
>
0 such that |
f
(
t
,
u
1
,
u
2
)
-
f
(
t
,
v
1
,
v
2
)
|
≤
L
1
|
u
1
-
v
1
|
+
L
2
|
u
2
-
v
2
|, for each t
∈
I and u
1
,
u
2
,
v
1
,
v
2
∈
R.
Ifx and y are the solutions of the initial value problem (69)-(70) and
(80)
D
α
-
δ
y
(
t
)
=
f
(
t
,
y
(
t
)
,
max
⁡
s
∈
[
β
t
,
t
]
y
(
s
)
)
,
t
∈
I
,
with initial condition
(81)
y
(
t
)
=
ϕ
¯
(
t
)
,
t
∈
[
β
t
0
,
t
0
]
,
respectively, where ϕ
¯ is a given continuous function on [
β
t
0
,
t
0
] such that ϕ
(
t
)
≢
ϕ
¯
(
t
) for all t
∈
[
β
t
0
,
t
0
) up to order n
(
n
=
-
[
-
(
α
-
δ
)
]
). we denote ϕ
¯
(
k
)
(
t
0
)
=
ρ
¯
k, k
=
0,1
,
2
,
…
,
n
-
1. Then the following estimates hold for t
0
<
t
≤
h
<
T.(
R
8
)
Supposeα
-
δ
>
1
/
2. Then for t
∈
I
(82)
|
y
(
t
)
-
x
(
t
)
|
≤
e
t
[
(
6
Γ
(
2
α
-
2
δ
-
1
)
(
L
1
2
+
L
2
2
)
(
t
-
t
0
)
4
α
-
δ
Γ
2
(
α
)
)
c
5
A
2
(
t
)
+
h
5
(
t
)
h
h
h
h
h
×
exp
⁡
(
6
Γ
(
2
α
-
2
δ
-
1
)
(
L
1
2
+
L
2
2
)
(
t
-
t
0
)
4
α
-
δ
Γ
2
(
α
)
)
]
1
/
2
.
(
R
9
)
Suppose0
<
α
-
δ
≤
1
/
2. Then for t
∈
I
(83)
|
y
(
t
)
-
x
(
t
)
|
≤
e
t
[
(
[
3
Γ
(
(
α
-
δ
)
2
)
]
1
/
α
-
δ
(
L
1
b
+
L
2
b
)
(
t
-
t
0
)
(
α
-
δ
+
1
)
α
-
δ
Γ
b
(
α
)
)
c
6
A
b
(
t
)
+
h
6
(
t
)
h
h
h
h
×

exp
⁡

(
[
3
Γ
(
(
α
-
δ
)
2
)
]
1
/
(
α
-
δ
)
(
L
1
b
+
L
2
b
)
(
t
-
t
0
)
(
α
-
δ
+
1
)
α
-
δ
Γ
b
(
α
)
)
]
1
/
b


,where
(84)
A
(
t
)
=
|
∑
j
=
0
n
-
1
(
ρ
¯
j
-
ρ
j
)
(
t
-
t
0
)
j
j
!
|
+
|
(
t
-
t
0
)
α
-
δ
Γ
(
α
-
δ
+
1
)
-
(
t
-
t
0
)
α
-
δ
(
α
-
δ
)
Γ
(
α
)
|
∥
f
∥
+
|
(
t
-
t
0
)
α
-
δ
(
α
-
δ
)
Γ
(
α
)
-
(
t
-
t
0
)
α
Γ
(
α
+
1
)
|
∥
f
∥
,
∥
f
∥
=
sup
⁡
t
0
≤
t
≤
h
|
f
(
t
,
y
(
t
)
,
max
⁡
s
∈
[
β
t
,
t
]
y
(
s
)
)
|
,
b
=
1
+
1
α
-
δ
,
c
5
=
max
⁡
{
3
e
-
2
t
0
,
e
-
2
β
t
0
}
,
c
6
=
max
⁡
{
3
1
/
(
α
-
δ
)
e
-
b
t
0
,
e
-
b
β
t
0
}
,
h
5
(
t
)
=
c
5
max
⁡
s
∈
[
β
t
0
,
t
0
]
|
ϕ
¯
(
s
)
-
ϕ
(
s
)
|
2
+
6
c
5
Γ
(
2
α
-
2
δ
-
1
)
4
α
-
δ
Γ
2
(
α
)
×
∫
t
0
t
(
L
1
2
A
2
(
s
)
+
L
2
2
max
⁡
ξ
∈
[
β
s
,
s
]
m
2
2
(
ξ
)
)
d
s
,
h
6
(
t
)
=
c
6
max
⁡
s
∈
[
β
t
0
,
t
0
]
|
ϕ
¯
(
s
)
-
ϕ
(
s
)
|
b
+
c
6
[
3
Γ
(
(
α
-
δ
)
2
)
]
1
/
(
α
-
δ
)
(
α
-
δ
+
1
)
α
-
δ
Γ
b
(
α
)
×
∫
t
0
t
‍
(
L
1
b
A
b
(
s
)
+
L
2
b
max
⁡
ξ
∈
[
β
s
,
s
]
m
2
b
(
ξ
)
)
d
s
,
with
(85)
m
2
(
t
)
=
{
A
(
t
)
,
t
∈
I
,
|
ϕ
¯
(
t
)
-
ϕ
(
t
)
|
,
t
∈
[
β
t
0
,
t
0
]
.Proof.
The solutionsx and y of the initial value problems (69)-(70) and (80)-(81) satisfy the following equations:
(86)
x
(
t
)
=
∑
j
=
0
n
-
1
ρ
j
j
!
(
t
-
t
0
)
j
+
1
Γ
(
α
)
×
∫
t
0
t
(
t
-
s
)
α
-
1
f
(
s
,
x
(
s
)
,
max
⁡
ξ
∈
[
β
s
,
s
]
x
(
ξ
)
)
d
s
,
y
(
t
)
=
∑
j
=
0
n
-
1
ρ
¯
j
j
!
(
t
-
t
0
)
j
+
1
Γ
(
α
-
δ
)
×
∫
t
0
t
(
t
-
s
)
α
-
δ
-
1
f
(
s
,
y
(
s
)
,
max
⁡
ξ
∈
[
β
s
,
s
]
y
(
ξ
)
)
d
s
,
respectively. So, using the assumption (
H
17
), it follows that
(87)
|
y
(
t
)
-
x
(
t
)
|
≤
|
∑
j
=
0
n
-
1
ρ
¯
j
j
!
(
t
-
t
0
)
j
-
∑
j
=
0
n
-
1
ρ
j
j
!
(
t
-
t
0
)
j
|
+
|
1
Γ
(
α
-
δ
)
∫
t
0
t
‍
(
t
-
s
)
α
-
δ
-
1
h
h
h
h
h
h
h
h
h
h
h
h
h
h
×
f
(
s
,
y
(
s
)
,
max
⁡
ξ
∈
[
β
s
,
s
]
y
(
ξ
)
)
d
s
hhhhhhh
-
1
Γ
(
α
)
∫
t
0
t
‍
(
t
-
s
)
α
-
δ
-
1
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
×
f
(
s
,
y
(
s
)
,
max
⁡
ξ
∈
[
β
s
,
s
]
y
(
ξ
)
)
d
s
|
+
|
1
Γ
(
α
)
∫
t
0
t
(
t
-
s
)
α
-
δ
-
1
h
h
h
h
h
h
h
h
h
h
h
h
×
f
(
s
,
y
(
s
)
,
max
⁡
ξ
∈
[
β
s
,
s
]
y
(
ξ
)
)
d
s
hhhhhhh
-
1
Γ
(
α
)
∫
t
0
t
(
t
-
s
)
α
-
δ
-
1
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
×
f
(
s
,
x
(
s
)
,
max
⁡
ξ
∈
[
β
s
,
s
]
x
(
ξ
)
)
d
s
|
+
|
1
Γ
(
α
)
∫
t
0
t
(
t
-
s
)
α
-
δ
-
1
h
h
h
h
h
h
h
h
h
h
h
h
×
f
(
s
,
x
(
s
)
,
max
⁡
ξ
∈
[
β
s
,
s
]
x
(
ξ
)
)
d
s
hhhhhhh
-
1
Γ
(
α
)
∫
t
0
t
(
t
-
s
)
α
-
1
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
×
f
(
s
,
x
(
s
)
,
max
⁡
ξ
∈
[
β
s
,
s
]
x
(
ξ
)
)
d
s
1
Γ
(
α
)
∫
t
0
t
(
t
-
s
)
α
-
δ
-
1
|
≤
A
(
t
)
+
1
Γ
(
α
)
∫
t
0
t
(
t
-
s
)
α
-
δ
-
1
hhhhhhhhhhhhhhh
×
(
|
max
⁡
ξ
∈
[
β
s
,
s
]
y
(
ξ
)
-
max
⁡
ξ
∈
[
β
s
,
s
]
x
(
ξ
)
|
L
1
|
y
(
s
)
-
x
(
s
)
|
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
+
L
2
|
max
⁡
ξ
∈
[
β
s
,
s
]
y
(
ξ
)
-
max
⁡
ξ
∈
[
β
s
,
s
]
x
(
ξ
)
|
)
d
s
≤
A
(
t
)
+
1
Γ
(
α
)
∫
t
0
t
(
t
-
s
)
α
-
δ
-
1
hhhhhhhhhhhhhhhh
×
(
max
⁡
ξ
∈
[
β
s
,
s
]
|
y
(
ξ
)
-
x
(
ξ
)
|
L
1
|
y
(
s
)
-
x
(
s
)
|
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
+
L
2
max
⁡
ξ
∈
[
β
s
,
s
]
|
y
(
ξ
)
-
x
(
ξ
)
|
)
d
s
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
∈
I
,
where A
(
t
) is defined by (84) and
(88)
|
y
(
t
)
-
x
(
t
)
|
=
|
ϕ
¯
(
t
)
-
ϕ
(
t
)
|
,
t
∈
[
β
t
0
,
t
0
]
.
Applying Theorem 6 yields the desired inequalities (82) and (83). This completes the proof.



---

*Source: 276316-2014-06-25.xml*