Skip to content

Advertisement

  • Research
  • Open Access

Some existence theorems for fractional integro-differential equations and inclusions with initial and non-separated boundary conditions

  • Bashir Ahmad1Email author,
  • Ahmed Alsaedi1,
  • Sayyedeh Zahra Nazemi2 and
  • Shahram Rezapour2
Boundary Value Problems20142014:249

https://doi.org/10.1186/s13661-014-0249-5

Received: 19 September 2014

Accepted: 13 November 2014

Published: 9 December 2014

Abstract

In this paper, we study the existence of solutions for a new class of boundary value problems of nonlinear fractional integro-differential equations and inclusions of arbitrary order with initial and non-separated boundary conditions. In the case of inclusion, the existence results are obtained for convex as well as non-convex multifunctions. Our results rely on the standard tools of fixed point theory and are well illustrated with the aid of examples.

1 Introduction

The subject of fractional calculus has recently been investigated in an extensive manner. The publication of several books, special issues, and a huge number of articles in journals of international repute, exploring numerous aspects of this branch of mathematics, clearly indicates the popularity of the topic. One of the key factors accounting for the utility of the subject is that fractional-order operators are nonlocal in nature in contrast to the integer-order operators and can describe the hereditary properties of many underlying phenomena and processes. Owing to this characteristic, the principles of fractional calculus have played a significant role in improving the modeling techniques for several real world problems [1]–[4].

Many researchers have focused their attention on fractional differential equations and inclusions, and a variety of interesting and important results concerning existence and uniqueness of solutions, stability properties of solutions, analytic and numerical methods of solutions of these equations have been obtained and the surge for investigating more and more results is still under way. For details and examples, we refer the reader to a series of papers [5]–[29] and the references therein. Anti-periodic boundary value problems occur in the mathematical modeling of a variety of physical processes and some works have been published in this area, for instance; see [30]–[34] and the references therein.

In this paper, for α ( n 1 , n ] , n 5 , n N , t I = [ 0 , T ] , T > 0 , we investigate the fractional integro-differential equation
D α c x ( t ) = f ( t , x ( t ) , x ( t ) , x ( t ) , x ( t ) , ϕ x ( t ) , ψ x ( t ) , c D μ 1 x ( t ) , c D μ 2 x ( t ) , , c D μ m x ( t ) , D ν 1 c x ( t ) , c D ν 2 x ( t ) , , c D ν m x ( t ) , c D ξ 1 x ( t ) , c D ξ 2 x ( t ) , , c D ξ m x ( t ) ) ,
(1.1)
and related fractional integro-differential inclusion
D α c x ( t ) F ( t , x ( t ) , x ( t ) , x ( t ) , x ( t ) , ϕ x ( t ) , ψ x ( t ) , c D μ 1 x ( t ) , c D μ 2 x ( t ) , , c D μ m x ( t ) , D ν 1 c x ( t ) , c D ν 2 x ( t ) , , c D ν m x ( t ) , c D ξ 1 x ( t ) , c D ξ 2 x ( t ) , , c D ξ m x ( t ) ) ,
(1.2)
supplemented with initial boundary conditions
x ( 4 ) ( 0 ) = = x ( n 1 ) ( 0 ) = 0 , a x ( 0 ) + b x ( T ) = 0 , D p c x ( 0 ) = c D p x ( T ) , c D q x ( 0 ) = c D q x ( T ) , D γ c x ( 0 ) = c D γ x ( T ) , 0 < p < 1 , 1 < q < 2 , 2 < γ < 3 , a + b 0 , a , b R ,
(1.3)
where c D denotes the Caputo fractional derivative, f : [ 0 , T ] × R 6 + m + m + m R is a continuous function, F : [ 0 , 1 ] × R 6 + m + m + m P ( R ) is a multifunction, P ( R ) is the family of all non-empty subsets of R , and the maps ϕ and ψ are defined by
ϕ x ( t ) = 0 t γ ( t , s ) h 1 ( t , s , x ( s ) , x ( s ) , x ( s ) , x ( s ) , c D δ 1 x ( s ) , c D β 1 x ( s ) , c D θ 1 x ( s ) ) d s
and
ψ x ( t ) = 0 t λ ( t , s ) h 2 ( t , s , x ( s ) , x ( s ) , x ( s ) , x ( s ) , c D δ 2 x ( s ) , c D β 2 x ( s ) , c D θ 2 x ( s ) ) d s ,

where γ , λ : [ 0 , T ] × [ 0 , T ] R and h 1 , h 2 : [ 0 , T ] × [ 0 , T ] × R 7 R are continuous maps, 0 < μ i < 1 ( 1 i m ), 1 < ν j < 2 ( 1 j m ), 2 < ξ k < 3 ( 1 k m ), 0 < δ i < 1 , 1 < β i < 2 , and 2 < θ i < 3 for i = 1 , 2 .

The paper is organized as follows. In Section 2, we recall some preliminary facts that we used in the sequel. Section 3 deals with the existence result for single-valued initial boundary value problem, while the results for multivalued problem are presented in Section 4. We present some examples illustrating the main results in Section 5.

2 Preliminaries

Let ( X , ) be a normed space, P ( X ) the set of all non-empty subsets of X, P c l ( X ) the set of all non-empty closed subsets of X, P b ( X ) the set of all non-empty bounded subsets of X, P c p ( X ) the set of all non-empty compact subsets of X and P c p , c ( X ) the set of all non-empty compact and convex subsets of X[35]. A multivalued map G : X P ( X ) is said to be convex (closed) valued whenever G ( x ) is convex (closed) for all x X [35]. The multifunction G is called bounded on bounded sets whenever G ( B ) = x B G ( x ) is bounded subset of X for all B P b ( X ) , that is, sup x B { sup { | y | : y G ( x ) } } < for all B P b ( X ) [35]. Also, the multifunction G : X P ( X ) is called upper semi-continuous whenever for each x 0 X the set G ( x 0 ) is a non-empty closed subset of X, and for every open set N of X containing G ( x 0 ) , there exists an open neighborhood N 0 of x 0 such that G ( N 0 ) N [36], [37]. The multifunction G : X P ( X ) is called compact whenever G ( B ) is relatively compact for all B P b ( X ) and also is called completely continuous whenever G is upper semi-continuous and compact [38], [39]. It is well known that a compact multifunction G with non-empty compact valued is upper semi-continuous if and only if G has a closed graph, that is, x n x , y n G ( x n ) for all n, and y n y imply y G ( x ) [37]. We say that x 0 X is a fixed point of a multifunction G whenever x 0 G ( x 0 ) [40]. Let T > 0 and G : [ 0 , T ] P c l ( R ) a multifunction. We say that G is measurable whenever the function t d ( y , G ( t ) ) = inf { | y z | : z G ( t ) } is measurable for all y R [38], [39].

One can find basic notions of fractional calculus in [1] and [2]. We recall two necessary ones here.

The Riemann-Liouville fractional integral of order q > 0 with the lower limit zero for a function f : [ 0 , ) R is defined by I q f ( t ) = 1 Γ ( q ) 0 t f ( s ) ( t s ) 1 q d s for t > 0 provided the integral exists.

The Caputo fractional derivative of order q > 0 for a function f C n ( [ 0 , ) , R ) can be written as
D q c f ( t ) = 1 Γ ( n q ) 0 t f ( n ) ( s ) ( t s ) q + 1 n d s = I n q f ( n ) ( t ) ,

where n 1 < q n and t > 0 .

To define the solution for problems (1.1)-(1.3) and (1.2)-(1.3), we establish the following lemma.

Lemma 2.1

Let y L 1 ( [ 0 , T ] , R ) . Then the integral solution of the linear equation
D α c x ( t ) = y ( t )
(2.1)
subject to the initial boundary conditions (1.3) is given by
x ( t ) = 0 T G ( t , s ) y ( s ) d s ,
(2.2)
where
G ( t , s ) = { ( t s ) α 1 Γ ( α ) + G 1 ( t , s ) , if  s t , G 1 ( t , s ) , if  t s ,
and
G 1 ( t , s ) = b ( T s ) α 1 ( a + b ) Γ ( α ) + [ b T ( a + b ) t ] Γ ( 2 p ) ( T s ) α p 1 ( a + b ) Γ ( α p ) T 1 p [ b p T 2 ( a + b ) ( 2 T t ( 2 p ) t 2 ) ] Γ ( 3 q ) ( T s ) α q 1 2 ( a + b ) ( 2 p ) Γ ( α q ) T 2 q Γ ( 4 γ ) ( T s ) α γ 1 6 ( a + b ) ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ ) T 3 γ × ( b ( 6 ( q p ) + ( 2 p ) ( 3 p ) q ) T 3 + ( a + b ) ( 6 ( p q ) T 2 t + ( 2 p ) ( 3 p ) ( 3 T t 2 + ( 3 q ) t 3 ) ) ) .

Proof

It is well known that the solution of (2.1) can be written as
x ( t ) = I α y ( t ) b 0 b 1 t b 2 t 2 b 3 t 3 b 4 t 4 b n 1 t n 1 = 1 Γ ( α ) 0 t ( t s ) α 1 y ( s ) d s b 0 b 1 t b 2 t 2 b 3 t 3 b 4 t 4 b n 1 t n 1 ,
(2.3)
where b 0 , b 1 , b 2 , b 3 , b 4 , , b n 1 R are arbitrary constants. Using the initial conditions x ( 4 ) ( 0 ) = = x ( n 1 ) ( 0 ) = 0 , we find that b 4 = = b n 1 = 0 . Since D p c c = 0 for all constant c, D p c t = t 1 p Γ ( 2 p ) , D p c t 2 = 2 t 2 p Γ ( 3 p ) , D p c t 3 = 6 t 3 p Γ ( 4 p ) , D q c t = 0 , D q c t 2 = 2 t 2 q Γ ( 3 q ) , D q c t 3 = 6 t 3 q Γ ( 4 q ) , D γ c t = 0 , D γ c t 2 = 0 , D γ c t 3 = 6 t 3 γ Γ ( 4 γ ) , D p c I α y ( t ) = I α p y ( t ) , D q c I α y ( t ) = I α q y ( t ) , and D γ c I α y ( t ) = I α γ y ( t ) , therefore
D p c x ( t ) = 1 Γ ( α p ) 0 t ( t s ) α p 1 y ( s ) d s b 1 t 1 p Γ ( 2 p ) b 2 2 t 2 p Γ ( 3 p ) b 3 6 t 3 p Γ ( 4 p ) , D q c x ( t ) = 1 Γ ( α q ) 0 t ( t s ) α q 1 y ( s ) d s b 2 2 t 2 q Γ ( 3 q ) b 3 6 t 3 q Γ ( 4 q ) , D γ c x ( t ) = 1 Γ ( α γ ) 0 t ( t s ) α γ 1 y ( s ) d s b 3 6 t 3 γ Γ ( 4 γ ) .
Now using the conditions a x ( 0 ) + b x ( T ) = 0 , D p c x ( 0 ) = c D p x ( T ) , D q c x ( 0 ) = c D q x ( T ) , and D γ c x ( 0 ) = c D γ x ( T ) , we obtain
b 0 = b ( a + b ) [ 1 Γ ( α ) 0 T ( T s ) α 1 y ( s ) d s Γ ( 2 p ) T p Γ ( α p ) 0 T ( T s ) α p 1 y ( s ) d s + p Γ ( 3 q ) T q 2 ( 2 p ) Γ ( α q ) 0 T ( T s ) α q 1 y ( s ) d s + [ 6 ( q p ) + ( 2 p ) ( 3 p ) q ] Γ ( 4 γ ) T γ 6 ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ ) 0 T ( T s ) α γ 1 y ( s ) d s ] , b 1 = Γ ( 2 p ) Γ ( α p ) T 1 p 0 T ( T s ) α p 1 y ( s ) d s Γ ( 3 q ) ( 2 p ) Γ ( α q ) T 1 q 0 T ( T s ) α q 1 y ( s ) d s + ( q p ) Γ ( 4 γ ) ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ ) T 1 γ 0 T ( T s ) α γ 1 y ( s ) d s , b 2 = Γ ( 3 q ) 2 Γ ( α q ) T 2 q 0 T ( T s ) α q 1 y ( s ) d s Γ ( 4 γ ) 2 ( 3 q ) Γ ( α γ ) T 2 γ 0 T ( T s ) α γ 1 y ( s ) d s , b 3 = Γ ( 4 γ ) 6 Γ ( α γ ) T 3 γ 0 T ( T s ) α γ 1 y ( s ) d s .

Substituting the values of b 0 , b 1 , b 2 , b 3 , b 4 , , b n 1 in (2.3), we get the solution (2.2). □

3 Existence results for problem (1.1)-(1.3)

Consider the space X = { u : u C 3 ( I ) } endowed with the norm
u = sup t I | u ( t ) | + sup t I | u ( t ) | + sup t I | u ( t ) | + sup t I | u ( t ) | .

Obviously ( X , ) is a Banach space.

We need the following result [40] in the sequel.

Theorem 3.1

Let E be a Banach space, S : E E a completely continuous operator and
V = { x E : x = μ S x , 0 μ 1 }

a bounded set. Then S has a fixed point in E.

Let us define the operator T : X X by
( T x ) ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 f ˜ ( s , x ( s ) ) d s b ( a + b ) Γ ( α ) 0 T ( T s ) α 1 f ˜ ( s , x ( s ) ) d s + [ b T ( a + b ) t ] Γ ( 2 p ) ( a + b ) Γ ( α p ) T 1 p 0 T ( T s ) α p 1 f ˜ ( s , x ( s ) ) d s [ b p T 2 ( a + b ) ( 2 T t ( 2 p ) t 2 ) ] Γ ( 3 q ) 2 ( a + b ) ( 2 p ) Γ ( α q ) T 2 q 0 T ( T s ) α q 1 f ˜ ( s , x ( s ) ) d s ( [ b ( 6 ( q p ) + ( 2 p ) ( 3 p ) q ) T 3 6 ( a + b ) ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ ) T 3 γ + ( a + b ) ( 6 ( q p ) T 2 t + ( 2 p ) ( 3 p ) ( 3 T t 2 + ( 3 q ) t 3 ) ) ] Γ ( 4 γ ) 6 ( a + b ) ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ ) T 3 γ ) × 0 T ( T s ) α γ 1 f ˜ ( s , x ( s ) ) d s ,
where
f ˜ ( s , x ( s ) ) = f ( s , x ( s ) , x ( s ) , x ( s ) , x ( s ) , ϕ x ( s ) , ψ x ( s ) , c D μ 1 x ( s ) , c D μ 2 x ( s ) , , c D μ m x ( s ) , D ν 1 c x ( s ) , c D ν 2 x ( s ) , , c D ν m x ( s ) , c D ξ 1 x ( s ) , c D ξ 2 x ( s ) , , c D ξ m x ( s ) ) .
For the sake of convenience, we set
M 1 = [ | a | + 2 | b | | a + b | Γ ( α + 1 ) + ( | a | + 2 | b | ) Γ ( 2 p ) | a + b | Γ ( α p + 1 ) + ( | b | p + | a + b | ( 4 p ) ) Γ ( 3 q ) 2 | a + b | ( 2 p ) Γ ( α q + 1 ) + ( [ | b | ( 6 ( q p ) + ( 2 p ) ( 3 p ) q ) 6 | a + b | ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ + 1 ) + | a + b | ( 6 ( q p ) + ( 2 p ) ( 3 p ) ( 6 q ) ) ] Γ ( 4 γ ) 6 | a + b | ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ + 1 ) ) ] T α + [ 1 Γ ( α ) + Γ ( 2 p ) Γ ( α p + 1 ) + ( 3 p ) Γ ( 3 q ) ( 2 p ) Γ ( α q + 1 ) + [ 2 ( q p ) + ( 2 p ) ( 3 p ) ( 5 q ) ] Γ ( 4 γ ) 2 ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ + 1 ) ] T α 1 + [ 1 Γ ( α 1 ) + Γ ( 3 q ) Γ ( α q + 1 ) + ( 4 q ) Γ ( 4 γ ) ( 3 q ) Γ ( α γ + 1 ) ] T α 2 + [ 1 Γ ( α 2 ) + Γ ( 4 γ ) Γ ( α γ + 1 ) ] T α 3 .

Theorem 3.2

The operator T : X X is completely continuous.

Proof

First, we show that the operator T : X X is continuous. Let { x n } be a sequence in X with x n x 0 and 0 < μ 1 , , μ m < 1 . Then we have
sup t I | c D μ i x n ( t ) c D μ i x 0 ( t ) | = sup t I | 1 Γ ( 1 μ i ) 0 t ( t s ) μ i x n ( s ) d s 1 Γ ( 1 μ i ) 0 t ( t s ) μ i x 0 ( s ) d s | = sup t I | 1 Γ ( 1 μ i ) 0 t ( t s ) μ i [ x n ( s ) x 0 ( s ) ] d s | T 1 μ i Γ ( 2 μ i ) sup t I | x n ( t ) x 0 ( t ) | T 1 μ i Γ ( 2 μ i ) x n x 0 .
Since x n x 0 , lim n c D μ i x n ( t ) = c D μ i x 0 ( t ) uniformly on I. Similarly, lim n c D ν j x n ( t ) = c D ν j x 0 ( t ) uniformly on I for 1 j m , lim n c D ξ k x n ( t ) = c D ξ k x 0 ( t ) uniformly on I for 1 k m . Also, we get lim n c D δ i x n ( t ) = c D δ i x 0 ( t ) , lim n c D β i x n ( t ) = c D β i x 0 ( t ) , and lim n c D θ i x n ( t ) = c D θ i x 0 ( t ) uniformly on I for i = 1 , 2 . Since
T x n T x 0 = sup t I | T x n ( t ) T x 0 ( t ) | + sup t I | ( T x n ) ( t ) ( T x 0 ) ( t ) | + sup t I | ( T x n ) ( t ) ( T x 0 ) ( t ) | + sup t I | ( T x n ) ( t ) ( T x 0 ) ( t ) | ,
using the continuity of f, h 1 , h 2 , we get T x n T x 0 . Thus, T is continuous on X. Now, let Ω X be a bounded subset. Then there exists a positive constant L > 0 such that | f ˜ ( t , x ( t ) ) | L for all t I and x Ω . We show that T Ω is a bounded set. We have
| ( T x ) ( t ) | 1 Γ ( α ) 0 t ( t s ) α 1 | f ˜ ( s , x ( s ) ) | d s + | b | | a + b | Γ ( α ) 0 T ( T s ) α 1 | f ˜ ( s , x ( s ) ) | d s + | b T ( a + b ) t | Γ ( 2 p ) | a + b | Γ ( α p ) T 1 p 0 T ( T s ) α p 1 | f ˜ ( s , x ( s ) ) | d s + | b p T 2 ( a + b ) ( 2 T t ( 2 p ) t 2 ) | Γ ( 3 q ) 2 | a + b | ( 2 p ) Γ ( α q ) T 2 q 0 T ( T s ) α q 1 | f ˜ ( s , x ( s ) ) | d s + ( | b ( 6 ( q p ) + ( 2 p ) ( 3 p ) q ) T 3 6 | a + b | ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ ) T 3 γ + ( a + b ) ( 6 ( q p ) T 2 t + ( 2 p ) ( 3 p ) ( 3 T t 2 + ( 3 q ) t 3 ) ) | Γ ( 4 γ ) 6 | a + b | ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ ) T 3 γ ) × 0 T ( T s ) α γ 1 | f ˜ ( s , x ( s ) ) | d s ( | a | + 2 | b | ) L T α | a + b | Γ ( α + 1 ) + ( | a | + 2 | b | ) Γ ( 2 p ) L T α | a + b | Γ ( α p + 1 ) + ( | b | p + | a + b | ( 4 p ) ) Γ ( 3 q ) L T α 2 | a + b | ( 2 p ) Γ ( α q + 1 ) + ( [ | b | ( 6 ( q p ) + ( 2 p ) ( 3 p ) q ) 6 | a + b | ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ + 1 ) + | a + b | ( 6 ( q p ) + ( 2 p ) ( 3 p ) ( 6 q ) ) ] Γ ( 4 γ ) L T α 6 | a + b | ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ + 1 ) ) , | ( T x ) ( t ) | 1 Γ ( α 1 ) 0 t ( t s ) α 2 | f ˜ ( s , x ( s ) ) | d s + Γ ( 2 p ) Γ ( α p ) T 1 p 0 T ( T s ) α p 1 | f ˜ ( s , x ( s ) ) | d s + | T ( 2 p ) t | Γ ( 3 q ) ( 2 p ) Γ ( α q ) T 2 q 0 T ( T s ) α q 1 | f ˜ ( s , x ( s ) ) | d s + | 2 ( q p ) T 2 + ( 2 p ) ( 3 p ) ( 2 T t + ( 3 q ) t 2 ) | Γ ( 4 γ ) 2 ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ ) T 3 γ × 0 T ( T s ) α γ 1 | f ˜ ( s , x ( s ) ) | d s T α 1 L Γ ( α ) + Γ ( 2 p ) L T α 1 Γ ( α p + 1 ) + ( 3 p ) Γ ( 3 q ) L T α 1 ( 2 p ) Γ ( α q + 1 ) + [ 2 ( q p ) + ( 2 p ) ( 3 p ) ( 5 q ) ] Γ ( 4 γ ) L T α 1 2 ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ + 1 ) , | ( T x ) ( t ) | 1 Γ ( α 2 ) 0 t ( t s ) α 3 | f ˜ ( s , x ( s ) ) | d s + Γ ( 3 q ) Γ ( α q ) T 2 q 0 T ( T s ) α q 1 | f ˜ ( s , x ( s ) ) | d s + | T + ( 3 q ) t | Γ ( 4 γ ) ( 3 q ) Γ ( α γ ) T 3 γ 0 T ( T s ) α γ 1 | f ˜ ( s , x ( s ) ) | d s L T α 2 Γ ( α 1 ) + Γ ( 3 q ) L T α 2 Γ ( α q + 1 ) + ( 4 q ) Γ ( 4 γ ) L T α 2 ( 3 q ) Γ ( α γ + 1 ) ,
and
| ( T x ) ( t ) | 1 Γ ( α 3 ) 0 t ( t s ) α 4 | f ˜ ( s , x ( s ) ) | d s + Γ ( 4 γ ) Γ ( α γ ) T 3 γ 0 T ( T s ) α γ 1 | f ˜ ( s , x ( s ) ) | d s L T α 3 Γ ( α 2 ) + Γ ( 4 γ ) L T α 3 Γ ( α γ + 1 )
for all x Ω . Hence, we get
T x = sup t I | ( T x ) ( t ) | + sup t I | ( T x ) ( t ) | + sup t I | ( T x ) ( t ) | + sup t I | ( T x ) ( t ) | [ | a | + 2 | b | | a + b | Γ ( α + 1 ) + ( | a | + 2 | b | ) Γ ( 2 p ) | a + b | Γ ( α p + 1 ) + ( | b | p + | a + b | ( 4 p ) ) Γ ( 3 q ) 2 | a + b | ( 2 p ) Γ ( α q + 1 ) + ( [ | b | ( 6 ( q p ) + ( 2 p ) ( 3 p ) q ) 6 | a + b | ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ + 1 ) + | a + b | ( 6 ( q p ) + ( 2 p ) ( 3 p ) ( 6 q ) ) ] Γ ( 4 γ ) 6 | a + b | ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ + 1 ) ) ] L T α + [ 1 Γ ( α ) + Γ ( 2 p ) Γ ( α p + 1 ) + ( 3 p ) Γ ( 3 q ) ( 2 p ) Γ ( α q + 1 ) + [ 2 ( q p ) + ( 2 p ) ( 3 p ) ( 5 q ) ] Γ ( 4 γ ) 2 ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ + 1 ) ] L T α 1 + [ 1 Γ ( α 1 ) + Γ ( 3 q ) Γ ( α q + 1 ) + ( 4 q ) Γ ( 4 γ ) ( 3 q ) Γ ( α γ + 1 ) ] L T α 2 + [ 1 Γ ( α 2 ) + Γ ( 4 γ ) Γ ( α γ + 1 ) ] L T α 3 = M 1 L .
This implies that the operator T maps bounded sets of X into bounded sets. Now, we prove that the sets { T x : x Ω } , { ( T x ) : x Ω } , { ( T x ) : x Ω } , { ( T x ) : x Ω } are equicontinuous on I. For 0 t 1 < t 2 T , we have
| ( T x ) ( t 2 ) ( T x ) ( t 1 ) | = | 1 Γ ( α ) 0 t 2 ( t 2 s ) α 1 f ˜ ( s , x ( s ) ) d s 1 Γ ( α ) 0 t 1 ( t 1 s ) α 1 f ˜ ( s , x ( s ) ) d s ( t 2 t 1 ) Γ ( 2 p ) Γ ( α p ) T 1 p 0 T ( T s ) α p 1 f ˜ ( s , x ( s ) ) d s + [ 2 T ( t 2 t 1 ) ( 2 p ) ( t 2 2 t 1 2 ) ] Γ ( 3 q ) 2 ( 2 p ) Γ ( α q ) T 2 q 0 T ( T s ) α q 1 f ˜ ( s , x ( s ) ) d s [ 6 ( q p ) T 2 ( t 2 t 1 ) + ( 2 p ) ( 3 p ) ( 3 T ( t 2 2 t 1 2 ) + ( 3 q ) ( t 2 3 t 1 3 ) ) ] Γ ( 4 γ ) 6 ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ ) T 3 γ × 0 T ( T s ) α γ 1 f ˜ ( s , x ( s ) ) d s | L Γ ( α ) 0 t 1 [ ( t 2 s ) α 1 ( t 1 s ) α 1 ] d s + L Γ ( α ) t 1 t 2 ( t 2 s ) α 1 d s + ( t 2 t 1 ) Γ ( 2 p ) L T α 1 Γ ( α p + 1 ) + [ 2 T ( t 2 t 1 ) + ( 2 p ) ( t 2 2 t 1 2 ) ] Γ ( 3 q ) L T α 2 2 ( 2 p ) Γ ( α q + 1 ) + [ 6 ( q p ) T 2 ( t 2 t 1 ) + ( 2 p ) ( 3 p ) ( 3 T ( t 2 2 t 1 2 ) + ( 3 q ) ( t 2 3 t 1 3 ) ) ] Γ ( 4 γ ) L T α 3 6 ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ + 1 ) L Γ ( α + 1 ) ( t 2 α t 1 α ) + ( t 2 t 1 ) Γ ( 2 p ) L T α 1 Γ ( α p + 1 ) + [ 2 T ( t 2 t 1 ) + ( 2 p ) ( t 2 2 t 1 2 ) ] Γ ( 3 q ) L T α 2 2 ( 2 p ) Γ ( α q + 1 ) + [ 6 ( q p ) T 2 ( t 2 t 1 ) + ( 2 p ) ( 3 p ) ( 3 T ( t 2 2 t 1 2 ) + ( 3 q ) ( t 2 3 t 1 3 ) ) ] Γ ( 4 γ ) L T α 3 6 ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ + 1 ) .
In a similar manner, one can find that
| ( T x ) ( t 2 ) ( T x ) ( t 1 ) | L Γ ( α ) ( t 2 α 1 t 1 α 1 ) + ( t 2 t 1 ) Γ ( 3 q ) L T α 2 Γ ( α q + 1 ) + [ 2 T ( t 2 t 1 ) + ( 3 q ) ( t 2 2 t 1 2 ) ] Γ ( 4 γ ) L T α 3 2 ( 3 q ) Γ ( α γ + 1 ) , | ( T x ) ( t 2 ) ( T x ) ( t 1 ) | L Γ ( α 1 ) ( t 2 α 2 t 1 α 2 ) + ( t 2 t 1 ) Γ ( 4 γ ) L T α 3 Γ ( α γ + 1 ) ,
and
| ( T x ) ( t 2 ) ( T x ) ( t 1 ) | L Γ ( α 2 ) ( t 2 α 3 t 1 α 3 ) .

Clearly the right-hand sides of the above inequalities tend to zero as t 2 t 1 . So T is completely continuous. This completes the proof. □

Theorem 3.3

Assume that there exist positive constants d 0 > 0 , d i 0 ( 1 i 6 ), ζ i 0 ( 1 i m ), η j 0 ( 1 j m ), τ k 0 ( 1 k m ), l i 1 , l i 2 0 ( 1 i 7 ), c 01 , c 02 > 0 such that
| f ( t , x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , y 1 , y 2 , , y m , z 1 , z 2 , , z m , w 1 , w 2 , , w m ) | d 0 + d 1 | x 1 | + d 2 | x 2 | + d 3 | x 3 | + d 4 | x 4 | + d 5 | x 5 | + d 6 | x 6 | + i = 1 m ζ i | y i | + j = 1 m η j | z j | + k = 1 m τ k | w k |
for all t I and x 1 , , x 6 , y 1 , , y m , z 1 , , z m , τ 1 , , τ m R and
| h j ( t , s , u 1 , u 2 , u 3 , u 4 , u 5 , u 6 , u 7 ) | c 0 j + i = 1 7 l i j | u i |
for j = 1 , 2 , all t , s I , and all u 1 , , u 7 R . In addition, assume that
M 1 = M 1 [ d 1 + d 2 + d 3 + d 4 + d 5 γ 0 ( l 11 + l 21 + l 31 + l 41 + l 51 T 1 δ 1 Γ ( 2 δ 1 ) + l 61 T 2 β 1 Γ ( 3 β 1 ) + l 71 T 3 θ 1 Γ ( 4 θ 1 ) ) + d 6 λ 0 ( l 12 + l 22 + l 32 + l 42 + l 52 T 1 δ 2 Γ ( 2 δ 2 ) + l 62 T 2 β 2 Γ ( 3 β 2 ) + l 72 T 3 θ 2 Γ ( 4 θ 2 ) ) + i = 1 m ζ i T 1 μ i Γ ( 2 μ i ) + j = 1 m η j T 2 ν j Γ ( 3 ν j ) + k = 1 m τ k T 3 ξ k Γ ( 4 ξ k ) ] < 1 ,

where γ 0 = sup t I 0 t | γ ( t , s ) | d s and λ 0 = sup t I 0 t | λ ( t , s ) | d s . Then problem (1.1)-(1.3) has at least one solution.

Proof

In view of Theorem 3.2, the operator T : X X is completely continuous. Next we show that the set V = { x X : x = μ T x , 0 μ 1 } is bounded. Let x V and t I . Then we have
x ( t ) = 1 Γ ( α ) 0 t μ ( t s ) α 1 f ˜ ( s , x ( s ) ) d s b ( a + b ) Γ ( α ) 0 T μ ( T s ) α 1 f ˜ ( s , x ( s ) ) d s + [ b T ( a + b ) t ] Γ ( 2 p ) ( a + b ) Γ ( α p ) T 1 p 0 T μ ( T s ) α p 1 f ˜ ( s , x ( s ) ) d s [ b p T 2 ( a + b ) ( 2 T t ( 2 p ) t 2 ) ] Γ ( 3 q ) 2 ( a + b ) ( 2 p ) Γ ( α q ) T 2 q 0 T μ ( T s ) α q 1 f ˜ ( s , x ( s ) ) d s ( [ b ( 6 ( q p ) + ( 2 p ) ( 3 p ) q ) T 3 6 ( a + b ) ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ ) T 3 γ + ( a + b ) ( 6 ( q p ) T 2 t + ( 2 p ) ( 3 p ) ( 3 T t 2 + ( 3 q ) t 3 ) ) ] Γ ( 4 γ ) 6 ( a + b ) ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ ) T 3 γ ) × 0 T μ ( T s ) α γ 1 f ˜ ( s , x ( s ) ) d s , x ( t ) = 1 Γ ( α 1 ) 0 t μ ( t s ) α 2 f ˜ ( s , x ( s ) ) d s Γ ( 2 p ) Γ ( α p ) T 1 p 0 T μ ( T s ) α p 1 f ˜ ( s , x ( s ) ) d s + [ T ( 2 p ) t ] Γ ( 3 q ) ( 2 p ) Γ ( α q ) T 2 q 0 T μ ( T s ) α q 1 f ˜ ( s , x ( s ) ) d s [ 2 ( q p ) T 2 + ( 2 p ) ( 3 p ) ( 2 T t + ( 3 q ) t 2 ) ] Γ ( 4 γ ) 2 ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ ) T 3 γ × 0 T μ ( T s ) α γ 1 f ˜ ( s , x ( s ) ) d s , x ( t ) = 1 Γ ( α 2 ) 0 t μ ( t s ) α 3 f ˜ ( s , x ( s ) ) d s Γ ( 3 q ) Γ ( α q ) T 2 q 0 T μ ( T s ) α q 1 f ˜ ( s , x ( s ) ) d s [ T + ( 3 q ) t ] Γ ( 4 γ ) ( 3 q ) Γ ( α γ ) T 3 γ 0 T μ ( T s ) α γ 1 f ˜ ( s , x ( s ) ) d s ,
and x ( t ) = 1 Γ ( α 3 ) 0 t μ ( t s ) α 4 f ˜ ( s , x ( s ) ) d s Γ ( 4 γ ) Γ ( α γ ) T 3 γ 0 T μ ( T s ) α γ 1 f ˜ ( s , x ( s ) ) d s . Thus, we get
| x ( t ) | = μ | T x ( t ) | [ d 0 + d 1 x + d 2 x + d 3 x + d 4 x + d 5 γ 0 ( c 01 + l 11 x + l 21 x + l 31 x + l 41 x + l 51 T 1 δ 1 Γ ( 2 δ 1 ) x + l 61 T 2 β 1 Γ ( 3 β 1 ) x + l 71 T 3 θ 1 Γ ( 4 θ 1 ) x ) + d 6 λ 0 ( c 02 + l 12 x + l 22 x + l 32 x + l 42 x + l 52 T 1 δ 2 Γ ( 2 δ 2 ) x + l 62 T 2 β 2 Γ ( 3 β 2 ) x + l 72 T 3 θ 2 Γ ( 4 θ 2 ) x ) + i = 1 m ζ i T 1 μ i Γ ( 2 μ i ) x + j = 1 m η j T 2 ν j Γ ( 3 ν j ) x + k = 1 m τ k T 3 ξ k Γ ( 4 ξ k ) x ] × [ ( | a | + 2 | b | ) T α | a + b | Γ ( α + 1 ) + ( | a | + 2 | b | ) Γ ( 2 p ) T α | a + b | Γ ( α p + 1 ) + ( | b | p + | a + b | ( 4 p ) ) Γ ( 3 q ) T α 2 | a + b | ( 2 p ) Γ ( α q + 1 ) + ( [ | b | ( 6 ( q p ) + ( 2 p ) ( 3 p ) q ) 6 | a + b | ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ + 1 ) + | a + b | ( 6 ( q p ) + ( 2 p ) ( 3 p ) ( 6 q ) ) ] Γ ( 4 γ ) T α 6 | a + b | ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ + 1 ) ) ] , | x ( t ) | = μ | ( T x ) ( t ) | [ d 0 + d 1 x + d 2 x + d 3 x + d 4 x + d 5 γ 0 ( c 01 + l 11 x + l 21 x + l 31 x + l 41 x + l 51 T 1 δ 1 Γ ( 2 δ 1 ) x + l 61 T 2 β 1 Γ ( 3 β 1 ) x + l 71 T 3 θ 1 Γ ( 4 θ 1 ) x ) + d 6 λ 0 ( c 02 + l 12 x + l 22 x + l 32 x + l 42 x + l 52 T 1 δ 2 Γ ( 2 δ 2 ) x + l 62 T 2 β 2 Γ ( 3 β 2 ) x + l 72 T 3 θ 2 Γ ( 4 θ 2 ) x ) + i = 1 m ζ i T 1 μ i Γ ( 2 μ i ) x + j = 1 m η j T 2 ν j Γ ( 3 ν j ) x + k = 1 m τ k T 3 ξ k Γ ( 4 ξ k ) x ] × [ T α 1 Γ ( α ) + Γ ( 2 p ) T α 1 Γ ( α p + 1 ) + ( 3 p ) Γ ( 3 q ) T α 1 ( 2 p ) Γ ( α q + 1 ) + [ 2 ( q p ) + ( 2 p ) ( 3 p ) ( 5 q ) ] Γ ( 4 γ ) T α 1 2 ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ + 1 ) ] , | x ( t ) | = μ | ( T x ) ( t ) | [ d 0 + d 1 x + d 2 x + d 3 x + d 4 x + d 5 γ 0 ( c 01 + l 11 x + l 21 x + l 31 x + l 41 x + l 51 T 1 δ 1 Γ ( 2 δ 1 ) x + l 61 T 2 β 1 Γ ( 3 β 1 ) x + l 71 T 3 θ 1 Γ ( 4 θ 1 ) x ) + d 6 λ 0 ( c 02 + l 12 x + l 22 x + l 32 x + l 42 x + l 52 T 1 δ 2 Γ ( 2 δ 2 ) x + l 62 T 2 β 2 Γ ( 3 β 2 ) x + l 72 T 3 θ 2 Γ ( 4 θ 2 ) x ) + i = 1 m ζ i T 1 μ i Γ ( 2 μ i ) x + j = 1 m η j T 2 ν j Γ ( 3 ν j ) x + k = 1 m τ k T 3 ξ k Γ ( 4 ξ k ) x ] × [ T α 2 Γ ( α 1 ) + Γ ( 3 q ) T α 2 Γ ( α q + 1 ) + ( 4 q ) Γ ( 4 γ ) T α 2 ( 3 q ) Γ ( α γ + 1 ) ] ,
and
| x ( t ) | = μ | ( T x ) ( t ) | [ d 0 + d 1 x + d 2 x + d 3 x + d 4 x + d 5 γ 0 ( c 01 + l 11 x + l 21 x + l 31 x + l 41 x + l 51 T 1 δ 1 Γ ( 2 δ 1 ) x + l 61 T 2 β 1 Γ ( 3 β 1 ) x + l 71 T 3 θ 1 Γ ( 4 θ 1 ) x ) + d 6 λ 0 ( c 02 + l 12 x + l 22 x + l 32 x + l 42 x + l 52 T 1 δ 2 Γ ( 2 δ 2 ) x + l 62 T 2 β 2 Γ ( 3 β 2 ) x + l 72 T 3 θ 2 Γ ( 4 θ 2 ) x ) + i = 1 m ζ i T 1 μ i Γ ( 2 μ i ) x + j = 1 m η j T 2 ν j Γ ( 3 ν j ) x + k = 1 m τ k T 3 ξ k Γ ( 4 ξ k ) x ] × [ T α 3 Γ ( α 2 ) + Γ ( 4 γ ) T α 3 Γ ( α γ + 1 ) ] .
This implies that
x M 1 [ d 1 + d 2 + d 3 + d 4 + d 5 γ 0 ( l 11 + l 21 + l 31 + l 41 + l 51 T 1 δ 1 Γ ( 2 δ 1 ) + l 61 T 2 β 1 Γ ( 3 β 1 ) + l 71 T 3 θ 1 Γ ( 4 θ 1 ) ) + d 6 λ 0 ( l 12 + l 22 + l 32 + l 42 + l 52 T 1 δ 2 Γ ( 2 δ 2 ) + l 62 T 2 β 2 Γ ( 3 β 2 ) + l 72 T 3 θ 2 Γ ( 4 θ 2 ) ) + i = 1 m ζ i T 1 μ i Γ ( 2 μ i ) + j = 1 m η j T 2 ν j Γ ( 3 ν j ) + k = 1 m τ k T 3 ξ k Γ ( 4 ξ k ) ] x + M 1 ( d 0 + d 5 γ 0 c 01 + d 6 λ 0 c 02 )

and so x M 1 ( d 0 + d 5 γ 0 c 01 + d 6 λ 0 c 02 ) 1 M 1 . Thus, the set V is bounded. Hence it follows by Theorem 3.1 that the operator T has at least one fixed point, which in turn implies that problem (1.1)-(1.3) has a solution. □

Theorem 3.4

Assume that f : I × R 6 + m + m + m R and h 1 , h 2 : I × I × R 7 R are continuous functions and there exist constants n i 0 ( 1 i 6 ), k i 0 ( 1 i m ), k j 0 ( 1 j m ), k k 0 ( 1 k m ), e i 1 , e i 2 0 ( 1 i 7 ) such that
| f ( t , x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , y 1 , y 2 , , y m , z 1 , z 2 , , z m , w 1 , w 2 , , w m ) f ( t , x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , y 1 , y 2 , , y m , z 1 , z 2 , , z m , w 1 , w 2 , , w m ) | n 1 | x 1 x 1 | + n 2 | x 2 x 2 | + n 3 | x 3 x 3 | + n 4 | x 4 x 4 | + n 5 | x 5 x 5 | + n 6 | x 6 x 6 | + i = 1 m k i | y i y i | + j = 1 m k j | z j z j | + k = 1 m k k | w k w k |
for all x 1 , , x 6 , x 1 , , x 6 , y 1 , , y m , y 1 , , y m , z 1 , , z m , z 1 , , z m , w 1 , , w m , w 1 , , w m R , and t I , and also
| h j ( t , s , u 1 , u 2 , u 3 , u 4 , u 5 , u 6 , u 7 ) h j ( t , s , u 1 , u 2 , u 3 , u 4 , u 5 , u 6 , u 7 ) | i = 1 7 e i j | u i u i |
for j = 1 , 2 , t , s I , and u 1 , , u 7 , u 1 , , u 7 R . In addition, suppose that
Δ = M 1 [ n 1 + n 2 + n 3 + n 4 + n 5 γ 0 ( e 11 + e 21 + e 31 + e 41 + e 51 T 1 δ 1 Γ ( 2 δ 1 ) + e 61 T 2 β 1 Γ ( 3 β 1 ) + e 71 T 3 θ 1 Γ ( 4 θ 1 ) ) + n 6 λ 0 ( e 12 + e 22 + e 32 + e 42 + e 52 T 1 δ 2 Γ ( 2 δ 2 ) + e 62 T 2 β 2 Γ ( 3 β 2 ) + e 72 T 3 θ 2 Γ ( 4 θ 2 ) ) + i = 1 m k i T 1 μ i Γ ( 2 μ i ) + j = 1 m k j T 2 ν j Γ ( 3 ν j ) + k = 1 m k k T 3 ξ k Γ ( 4 ξ k ) ] < 1 .

Then problem (1.1)-(1.3) has a unique solution.

Proof

Set N = sup 0 t T | f ( t , 0 , 0 , , 0 ) | < , κ j = sup 0 t , s T | h j ( t , s , 0 , 0 , , 0 ) | < for j = 1 , 2 , and choose r ( N + n 5 γ 0 κ 1 + n 6 λ 0 κ 2 ) M 1 1 Δ . We show that T ( B r ) B r , where B r = { x X : x r } . Let x B r . Then we have
| ( T x ) ( t ) | 1 Γ ( α ) 0 t ( t s ) α 1 | f ˜ ( s , x ( s ) ) | d s + | b | | a + b | Γ ( α ) 0 T ( T s ) α 1 | f ˜ ( s , x ( s ) ) | d s + | b T ( a + b ) t | Γ ( 2 p ) | a + b | Γ ( α p ) T 1 p 0 T ( T s ) α p 1 | f ˜ ( s , x ( s ) ) | d s + | b p T 2 ( a + b ) ( 2 T t ( 2 p ) t 2 ) | Γ ( 3 q ) 2 | a + b | ( 2 p ) Γ ( α q ) T 2 q 0 T ( T s ) α q 1 | f ˜ ( s , x ( s ) ) | d s + ( | b ( 6 ( q p ) + ( 2 p ) ( 3 p ) q ) T 3 6 | a + b | ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ ) T 3 γ + ( a + b ) ( 6 ( q p ) T 2 t + ( 2 p ) ( 3 p ) ( 3 T t 2 + ( 3 q ) t 3 ) ) | Γ ( 4 γ ) 6 | a + b | ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ ) T 3 γ ) × 0 T ( T s ) α γ 1 | f ˜ ( s , x ( s ) ) | d s 1 Γ ( α ) 0 t ( t s ) α 1 [ | f ˜ ( s , x ( s ) ) f ( s , 0 , 0 , , 0 ) | + | f ( s , 0 , 0 , , 0 ) | ] d s + | b | | a + b | Γ ( α ) 0 T ( T s ) α 1 [ | f ˜ ( s , x ( s ) ) f ( s , 0 , 0 , , 0 ) | + | f ( s , 0 , 0 , , 0 ) | ] d s + | b T ( a + b ) t | Γ ( 2 p ) | a + b | Γ ( α p ) T 1 p 0 T ( T s ) α p 1 [ | f ˜ ( s , x ( s ) ) f ( s , 0 , 0 , , 0 ) | + | f ( s , 0 , 0 , , 0 ) | ] d s + | b p T 2 ( a + b ) ( 2 T t ( 2 p ) t 2 ) | Γ ( 3 q ) 2 | a + b | ( 2 p ) Γ ( α q ) T 2 q × 0 T ( T s ) α q 1 [ | f ˜ ( s , x ( s ) ) f ( s , 0 , 0 , , 0 ) | + | f ( s , 0 , 0 , , 0 ) | ] d s + ( | b ( 6 ( q p ) + ( 2 p ) ( 3 p ) q ) T 3 6 | a + b | ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ ) T 3 γ + ( a + b ) ( 6 ( q p ) T 2 t + ( 2 p ) ( 3 p ) ( 3 T t 2 + ( 3 q ) t 3 ) ) | Γ ( 4 γ ) 6 | a + b | ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ ) T 3 γ ) × 0 T ( T s ) α γ 1 [ | f ˜ ( s , x ( s ) ) f ( s , 0 , 0 , , 0 ) | + | f ( s , 0 , 0 , , 0 ) | ] d s [ ( n 1 + n 2 + n 3 + n 4 + n 5 γ 0 ( e 11 + e 21 + e 31 + e 41 + e 51 T 1 δ 1 Γ ( 2 δ 1 ) + e 61 T 2 β 1 Γ ( 3 β 1 ) + e 71 T 3 θ 1 Γ ( 4 θ 1 ) ) + n 6 λ 0 ( e 12 + e 22 + e 32 + e 42 + e 52 T 1 δ 2 Γ ( 2 δ 2 ) + e 62 T 2 β 2 Γ ( 3 β 2 ) + e 72 T 3 θ 2 Γ ( 4 θ 2 ) ) + i = 1 m k i T 1 μ i Γ ( 2 μ i ) + j = 1 m k j T 2 ν j Γ ( 3 ν j ) + k = 1 m k k T 3 ξ k Γ ( 4 ξ k ) ) r + N + n 5 γ 0 κ 1 + n 6 λ 0 κ 2 ] × [ ( | a | + 2 | b | ) T α | a + b | Γ ( α + 1 ) + ( | a | + 2 | b | ) Γ ( 2 p ) T α | a + b | Γ ( α p + 1 ) + ( | b | p + | a + b | ( 4 p ) ) Γ ( 3 q ) T α 2 | a + b | ( 2 p ) Γ ( α q + 1 ) + ( [ | b | ( 6 ( q p ) + ( 2 p ) ( 3 p ) q ) 6 | a + b | ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ + 1 ) + | a + b | ( 6 ( q p ) + ( 2 p ) ( 3 p ) ( 6 q ) ) ] Γ ( 4 γ ) T α 6 | a + b | ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ + 1 ) ) ] .
In a similar way, we can obtain
| ( T x ) ( t ) | [ ( n 1 + n 2 + n 3 + n 4 + n 5 γ 0 ( e 11 + e 21 + e 31 + e 41 + e 51 T 1 δ 1 Γ ( 2 δ 1 ) + e 61 T 2 β 1 Γ ( 3 β 1 ) + e 71 T 3 θ 1 Γ ( 4 θ 1 ) ) + n 6 λ 0 ( e 12 + e 22 + e 32 + e 42 + e 52 T 1 δ 2 Γ ( 2 δ 2 ) + e 62 T 2 β 2 Γ ( 3 β 2 ) + e 72 T 3 θ 2 Γ ( 4 θ 2 ) ) + i = 1 m k i T 1 μ i Γ ( 2 μ i ) + j = 1 m k j T 2 ν j Γ ( 3 ν j ) + k = 1 m k k T 3 ξ k Γ ( 4 ξ k ) ) r + N + n 5 γ 0 κ 1 + n 6 λ 0 κ 2 ] × [ T α 1 Γ ( α ) + Γ ( 2 p ) T α 1 Γ ( α p + 1 ) + ( 3 p ) Γ ( 3 q ) T α 1 ( 2 p ) Γ ( α q + 1 ) + [ 2 ( q p ) + ( 2 p ) ( 3 p ) ( 5 q ) ] Γ ( 4 γ ) T α 1 2 ( 2 p ) ( 3 p ) ( 3 q ) Γ ( α γ + 1 ) ] , | ( T x ) ( t ) | [ ( n 1 + n 2 + n 3 + n 4 + n 5 γ 0 ( e 11 + e 21 + e 31 + e 41 + e 51 T 1 δ 1 Γ ( 2 δ 1 ) + e 61 T 2 β 1 Γ ( 3 β 1 ) + e 71 T 3 θ 1 Γ ( 4 θ 1 ) ) + n 6 λ 0 ( e 12 + e 22 + e 32 + e 42 + e 52 T 1 δ 2 Γ ( 2 δ 2 ) + e 62 T 2 β 2 Γ ( 3 β 2 ) + e 72 T 3 θ 2 Γ ( 4 θ 2 ) ) + i = 1 m k i T 1 μ i Γ ( 2 μ i ) + j = 1 m k j T 2 ν j Γ ( 3 ν j ) + k = 1 m k k T 3 ξ k Γ ( 4 ξ k ) ) r + N + n 5 γ 0 κ 1 + n 6 λ 0 κ 2 ] × [ T α 2 Γ ( α 1 ) + Γ ( 3 q ) T α 2 Γ ( α q + 1 ) + ( 4 q ) Γ ( 4 γ ) T α 2 ( 3 q ) Γ ( α γ + 1 ) ] ,
and
| ( T x ) ( t ) | [ ( n 1 + n 2 + n 3 + n 4 + n 5 γ 0 ( e 11 + e 21 + e 31 + e 41 + e 51 T 1 δ 1 Γ ( 2 δ 1 ) + e 61 T 2 β 1 Γ ( 3 β 1 ) + e 71 T 3 θ 1 Γ ( 4 θ 1 ) ) + n 6 λ 0 ( e 12 + e 22 + e 32 + e 42 + e 52 T 1 δ 2 Γ ( 2 δ 2 ) + e 62 T 2 β 2 Γ ( 3 β 2 ) + e 72 T 3 θ 2 Γ ( 4 θ 2 ) ) + i = 1 m k i T 1 μ i Γ ( 2 μ i ) + j = 1 m