Estimation of unknown function of a class of integral inequalities and applications in fractional integral equations

Applied Mathematics and Computation 268 (2015) 1029–1037

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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc

Estimation of unknown function of a class of integral inequalities and applications in fractional integral equations Wu-Sheng Wang∗ School of Mathematics and Statistics, Hechi University, Guangxi, Yizhou 546300, PR China

a r t i c l e

i n f o

a b s t r a c t

MSC: 26D10 26D15 26D20 45A99

In this paper, we establish a class of iterated integral inequalities, which includes a nonconstant term outside the integrals. The upper bound of the embedded unknown function is estimated explicitly by adopting novel analysis techniques, such as: change of variable, amplification method, differential and integration. The derived result can be applied in the study of qualitative properties of solutions of fractional integral equations.

Keywords: Gronwall-type inequality Analysis technique Explicit bound Fractional integral equation

© 2015 Published by Elsevier Inc.

1. Introduction It is well known that differential equations and integral equations are important tools to discuss the rule of natural phenomena. In the study of the existence, uniqueness, boundedness, stability, oscillation and other qualitative properties of solutions of differential equations and integral equations, one often deals with certain integral inequalities. One of the best known and widely used inequalities in the study of nonlinear differential equations is Gronwall–Bellman inequality [1,2], which can be stated as follows: If u and f are non-negative continuous functions on an interval [a, b] satisfying

u(t ) ≤ c +



t

a

f (s)u(s)ds,

for some constant c ≥ 0, then

u(t ) ≤ c exp



t a

t ∈ [a, b],

(1)



f (s)ds ,

t ∈ [a, b].

Pachpatte in [5] investigated the retarded inequality

u(t ) ≤ k +



t

a

g(s)u(s)ds +

 α(t ) a

h(s)u(s)ds,

(2)

where k is a constant. Replacing k by a nondecreasing continuous function f (t) in (1), Rashid in [12] studied the following retarded inequality

u(t ) ≤ f (t ) + ∗



t a

g(s)u(s)ds +

 α(t ) a

h(s)u(s)ds.

Tel.: +861 8677857348. E-mail address: [email protected], [email protected]

http://dx.doi.org/10.1016/j.amc.2015.07.015 0096-3003/© 2015 Published by Elsevier Inc.

(3)

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W.-S. Wang / Applied Mathematics and Computation 268 (2015) 1029–1037

In 2011, Abdeldaim and Yakout [10] studied some new integral inequalities

u(t ) ≤ u0 + u(t ) ≤ u0 +





t

g(s)u(s) u(s) +

0



t 0





s

h(τ ) u(τ ) +

0

[g(s)u(s) + q(s)]ds +

u p+1 (t ) ≤ u0 +



t

f (s)u p (s)ds

0

2





t

0







s 0



t

f (s)u p (s) u(s) +

0



r(ξ )u(ξ )dξ dτ ds,

g(s)u(s) u(s) +

0

+2

 τ

(4)



h(τ )u(τ )dτ ds, 

s 0

(5)



f (τ )u p (τ )dτ ds.

(6)

In 2014, El-Owaidy et al. [13] investigated some new retarded nonlinear integral inequalities

u(t ) ≤ f (t ) +



t a

u p (t ) ≤ f p (t ) +

g(s)u p (s)ds +

 α1 (t ) a

 α(t )

h(s)u p (s)ds,

a

g(s)u(s)ds +

 α2 (t ) a

(7)

h(s)u(s)ds.

(8)

In 2014, Zheng [16] discussed the inequalities of the following form

u(t ) ≤ C + u(t ) ≤ C +



1

 (α) 

t 0

t

(t − s)α−1 g(s)u(s)ds +

0

h(s)u p (s)ds +

1

 (α)



t 0



1

 (α)

T 0

(T − s)α−1 g(s)u(s)ds,

(t − s)α−1 g(s)uq (s) +



T 0

h(s)u p (s)ds +

(9)

1

 (α)



T 0

(T − s)α−1 g(s)uq (s)ds.

(10)

During the past few years, some investigators have established a lot of useful and interesting integral inequalities in order to achieve various goals; see [3–20] and the references cited therein. In this paper, based on the works of [10,13,16], we discuss some new integral inequalities with weak singularity



1

t

1



t

(t − s)α−1 g(s)u(s)ds + (t − s)α−1 g(s)u(s)  (α) 0  (α) 0  s  τ     1 1 × u(s) + (s − τ )α−1 h(τ ) u(τ ) + (τ − ξ )α−1 q(ξ )u(ξ )dξ dτ ds,  (α) 0  (α) 0

u(t ) ≤ f (t ) +



1

t

(t − s)α−1 [g(s)u(s) + q(s)]ds  t  s   1 1 + (t − s)α−1 g(s)u(s) u(s) + (s − τ )α−1 h(τ )u(τ )dτ ds,  (α) 0  (α) 0  1 t 2 u p+1 (t ) ≤ f p+1 (t ) + (t − s)α−1 g(s) f (1−p)/2 (s)u p (s)ds  (α) 0  t  s   1 1 +2 (t − s)α−1 g(s)u p (s) u(s) + (s − τ )α−1 g(τ ) f 1−p (τ )u p (τ )dτ ds.  (α) 0  (α) 0 u(t ) ≤ f (t ) +

 (α)

(11)

0

(12)

(13)

2. Main result Throughout this paper, let R+ = (0, +∞), I = [0, +∞). The modified Riemann– Liouville fractional derivative, presented by Jumarie in [17,18] is defined by the following expression. Definition 1. The modified Riemann– Liouville derivative of order α is defined by the following expression:

Dtα f (t ) =

⎧ ⎨

1 d  (1 − α) dt



t 0

(t − ξ )−α ( f (ξ ) − f (0))dξ , 0 < α < 1,

⎩ (n) (α −n) ( f (t )) , n < α < n + 1, n ≥ 1.

(14)

Definition 2. The Riemann– Liouville fractional integral of order α on the interval I is defined by

Itα f (t ) =

1  (1 + α)



0

t

α

f (s)(ds) =

1

 (α)



t 0

(t − s)α−1 f (s)ds.

(15)

In 2014, Zheng [16] proved the property Lemma 1. Suppose that 0 < α < 1, f is a continuous function, then

Dtα (Itα f (t )) = f (t ).

(16)

W.-S. Wang / Applied Mathematics and Computation 268 (2015) 1029–1037

1031

Some important properties for the modified Riemann–Liouville derivative and fractional integral are listed as follows (see [19]):

Dtα t r =

 (1 + r) r−α t ,  (1 + r − α)

(17)

Dtα ( f (t )g(t )) = g(t )Dtα f (t ) + f (t )Dtα g(t ).

(18)

α Dtα f [g(t )] = fg [g(t )]Dtα g(t ) = Dαg f [g(t )](g (t )) ,

(19)

Itα (Dtα f (t )) = f (t ) − f (0),

(20)

Itα (g(t )Dtα f (t )) = f (t )g(t ) − f (0)g(0) − Itα ( f (t )Dtα g(t )),

(21)

Dtα C = 0, where C is a constant.

(22)

Theorem 1. Suppose that h(t ), q(t ) ∈ C (I, R+ ), f ∈ C (R+ , R+ ) is a nondecreasing function with f(t) > 0 for t > 0. Suppose that

1−



1

 (α)

t 0



(t − τ )α−1 

× exp −

τα  (1+α)

0

1

1



1

[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds g(τ ) f (τ )dτ > 0, t ∈ I.

(23)

If u(t) satisfies (11), then





tα  (1+α)



1



t

(t − τ )α−1  (α) 0   (τ1+α α)  −1 1 1 1 × exp − [g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds g(τ ) f (τ )dτ , t ∈ I. (24)

u(t ) ≤ f (t ) exp −

0

1

1

1

[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds

1−

0

Proof. Noting that f(t) is a positive and nondecreasing function, from (11) we obtain

1 u(t ) ≤ 1+ f (t )  (α) ×

 u(s)

f (s)



t

0

+

(t − s)α−1 g(s) 1

 (α)



s 0

1 u(s) ds + f (s)  (α) α −1

(s − τ )



t

(t − s)α−1 g(s) f (s)

0

u(s) f (s)

 τ  u(τ )   1 u(ξ ) h(τ ) + (τ − ξ )α−1 q(ξ ) dξ dτ ds. f (τ )  (α) 0 f (ξ )

Let z1 (t ) = u(t )/ f (t ). From (25) we see

1



t

1



t

(t − s)α−1 g(s)z1 (s)ds + (t − s)α−1 g(s) f (s)z1 (s)  (α) 0  (α) 0  s  τ     1 1 × z1 (s) + (s − τ )α−1 h(τ ) z1 (τ ) + (τ − ξ )α−1 q(ξ )z1 (ξ )dξ dτ ds.  (α) 0  (α) 0

z1 (t ) ≤ 1 +

(25)

(26)

Defining the function z2 (t) by the right hand side of the inequality (26), i. e.

1



t

1



t

(t − s)α−1 g(s)z1 (s)ds + (t − s)α−1 g(s) f (s)z1 (s)  (α) 0  (α) 0  s  τ     1 1 × z1 (s) + (s − τ )α−1 h(τ ) z1 (τ ) + (τ − ξ )α−1 q(ξ )z1 (ξ )dξ dτ ds  (α) 0  (α) 0  t

 1 (t − τ )α−1 h(τ ) = 1 + Itα (g(t )z1 (t )) + Itα g(t ) f (t )z1 (t ) z1 (t ) +  (α) 0  τ    1 × z1 (τ ) + (τ − ξ )α−1 q(ξ )z1 (ξ )dξ dτ , t ∈ I.  (α) 0

z2 (t ) = 1 +

(27)

We observe that z2 (t) is a positive and nondecreasing function on I. From (26) and (27) we have

z1 (t ) ≤ z2 (t ), u(t ) ≤ z2 (t ) f (t ), t ∈ I,

(28)

z2 (0) = 1.

(29)

Using Lemma 1, (22), (28) and by the definition of fractional integral, we get



Dtα z2 (t ) = g(t )z1 (t ) + g(t ) f (t )z1 (t ) z1 (t ) +

1

 (α)



t 0

 (t−τ )α−1 h(τ ) z1 (τ ) +

1

 τ

 (α)

0

  (τ −ξ )α−1 q(ξ )z1 (ξ )dξ dτ

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W.-S. Wang / Applied Mathematics and Computation 268 (2015) 1029–1037







≤ g(t )z2 (t ) + g(t ) f (t )z2 (t ) z2 (t ) + Itα h(t ) z2 (t ) + = g(t )z2 (t )[1 + f (t )z3 (t )], t ∈ I, where





z3 (t ) = z2 (t ) + Itα h(t ) z2 (t ) +



1

 (α)



1

 (α)

t

0

(t − ξ )α−1 q(ξ )z2 (ξ )dξ

 (30)

t

(t − ξ )α−1 q(ξ )z2 (ξ )dξ

0



, t ∈ I,

(31)

which is a positive and nondecreasing function on I. From (29) and (31) we have

z2 (t ) ≤ z3 (t ), t ∈ I,

(32)

z3 (0) = z2 (0) = 1.

(33)

Using (30) and (32), we have



Dtα z3 (t ) = Dtα z2 (t ) + h(t ) z2 (t ) +



1

 (α)

0

t

(t − ξ )α−1 q(ξ )z2 (ξ )dξ



 t   1 ≤ g(t )z2 (t )[1 + f (t )z3 (t )] + h(t ) z2 (t ) + (t − ξ )α−1 q(ξ )z2 (ξ )dξ  (α) 0  t   1 ≤ g(t )z3 (t )[1 + f (t )z3 (t )] + h(t ) z3 (t ) + (t − ξ )α−1 q(ξ )z3 (ξ )dξ  (α) 0 ≤ g(t )z3 (t )[1 + f (t )z3 (t )] + h(t )z4 (t ), t ∈ I,

(34)

where

z4 (t ) = z3 (t ) + Itα q(t )z3 (t ),

(35)

which is a positive and nondecreasing function on I. From (33) and (35) we have

z3 (t ) ≤ z4 (t ), t ∈ I,

(36)

z4 (0) = z3 (0) = 1.

(37)

Using (34) and (36), we have

Dtα z4 (t ) = Dtα z3 (t ) + q(t )z3 (t ) ≤ g(t )z3 (t )[1 + f (t )z3 (t )] + h(t )z4 (t ) + q(t )z3 (t ) ≤ g(t ) f (t )z42 (t ) + [g(t ) + h(t ) + q(t )]z4 (t ), t ∈ I.

(38)

Using the properties (19) and z4 (t) > 0, from (38) we have

Dtα ( − (z4 (t )) Let x(t ) = −(z4 (t ))

−1

−1

) ≤ [g(t ) + h(t ) + q(t )](z4 (t ))−1 + g(t ) f (t ).

, then x(0) = −(z4 (0))

−1

(39)

= −1, from (39) we get

Dtα x(t ) − [g(t ) + h(t ) + q(t )]x(t ) ≤ g(t ) f (t ).

(40)

On other hand, using the properties (17)–(19), we have









tα  (1+α)

Dtα x(t ) exp −



= exp −

tα  (1+α)

0

0





1



= exp −

tα  (1+α)

0

 

= exp −



≤ exp −

tα  (1+α)

0



1



tα  (1+α)





1

0

1

1

1

[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds 1

tα  (1+α)

0

1





1

1

1



1

[g((s (1+α)) α )+h((s (1+α)) α )+q((s (1 + α)) α )]ds [g(t )+h(t )+q(t )]Dtα 1

1

1

[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds

tα  (1+α)

0

1

[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds Dtα x(t )

 

− x(t ) exp −

1

[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds Dtα x(t )

+ x(t )Dtα exp −



1

[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds

1

1

1





tα  (1 + α)



Dtα x(t )−[g(t ) + h(t ) + q(t )]x(t )



[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds g(t ) f (t ), t ∈ I.

(41)

W.-S. Wang / Applied Mathematics and Computation 268 (2015) 1029–1037

1033

Substituting t with τ in (41), making a fractional integral of order α for (41) with respect to τ from 0 to t and using the properties (20), we obtain that



x(t ) exp −



tα  (1+α)

0





× exp −

1

1



1

[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds − x(0) ≤

τα  (1+α)

1

1



1



1

 (α)

t 0

(t − τ )α−1

[g((s (1 + α)) α ) + h((s (1 + α)) α )]ds + q((s (1 + α)) α ) g(τ ) f (τ )dτ , t ∈ I.

0

(42)

From (42), we have



x(t ) ≤ exp

tα  (1+α)

0





× exp − By x(t ) = −(z4 (t ))

−1

1

1

1

[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds τα  (1+α)

0

1

1



−1+



1

1



 (α)

t

0

(t − τ )α−1



[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds g(τ ) f (τ )dτ , t ∈ I.

(43)

and (43), we have



z4 (t ) ≤ exp −



tα  (1+α)

0





× exp −

1

1

1

[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds

τα  (1+α)

0

1

1



1

1−

 (α)



1



t

0

[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds g(τ ) f (τ )dτ

(t − τ )α−1

−1

, t ∈ I.

(44)

From (28), (32), (36), we get

u(t )/ f (t ) = z1 (t ) ≤ z2 (t ) ≤ z3 (t ) ≤ z4 (t ).

(45)

From (44) and (45), we have





tα  (1+α)



1



t

(t − τ )α−1  (α) 0 −1   (τ1+α α)  1 1 1 × exp − [g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds g(τ ) f (τ )dτ , t ∈ I, (46)

u(t ) ≤ f (t ) exp −

0

1

1

1

[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds

1−

0

that is the required estimation (24). The proof is complete.  Theorem 2. Suppose that h(t ), q(t ) ∈ C (I, R+ ), f ∈ C (R+ , R+ ) is a nondecreasing function with f(t) > 0 for t > 0. Suppose that

1−



1

 (α)

t 0

  (t − τ )α−1 exp −

1

[g((s (1 + α)) α )

 1 1 ) + q((s (1 + α)) α )/ f ((s (1 + α)) α )]ds g(τ ) f (τ )dτ > 0, t ∈ I. 0

+ h((s (1 + α))

τα  (1+α)

1

α

(47)

If u(t) satisfies (12), then



u(t ) ≤ f (t ) exp −





tα  (1+α)

0

1

1

1

1

[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )/ f ((s (1 + α)) α )]ds



  (τ1+α α) 1 1 1 (t − τ )α−1 exp − [g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )/  (α) 0 0  −1 1 f ((s (1 + α)) α )]ds g(τ ) f (τ )dτ , t ∈ I. 

1

× 1−

t

(48)

Proof. Noting that f(t) is a positive and nondecreasing function, from (12) we obtain

 t   1 q(s) u(t ) u(s) ≤ 1+ (t − s)α−1 g(s) + ds f (t )  (α) 0 f (s) f (s)  t  s   1 1 u(s) u(s) u(τ ) + (t − s)α−1 g(s) f (s) + (s − τ )α−1 h(τ ) dτ ds.  (α) 0 f (s) f (s)  (α) 0 f (τ )

Let w1 (t ) = u(t )/ f (t ). From (49) we see

w1 (t ) ≤ 1 + +



1

 (α) 1

 (α)



0

t 0

t

(t − s)α−1

1 q(s) ds + f (s)  (α)



t 0

(49)

(t − s)α−1 g(s)w1 (s)ds

 (t − s)α−1 g(s) f (s)w1 (s) w1 (s) +

1

 (α)

 0

s

 (s − τ )α−1 h(τ )w1 (τ )dτ ds, t ∈ I.

(50)

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W.-S. Wang / Applied Mathematics and Computation 268 (2015) 1029–1037

Define a function w2 (t) by the right hand side of the inequality (50), i. e.

w2 (t ) = 1 +

 (α)

 (α)

t 0



1

+



1

(t − s)α−1

1 q(s) ds + f (s)  (α)



t 0

(t − s)α−1 g(s)w1 (s)ds

 (t − s)α−1 g(s) f (s)w1 (s) w1 (s) +

t 0

1



 (α)

0

s

 (s − τ )α−1 h(τ )w1 (τ )dτ ds, t ∈ I.

(51)

We observe that w2 (t) is a positive and nondecreasing function on I. From (50) and (51) we have

w1 (t ) ≤ w2 (t ), u(t ) ≤ w2 (t ) f (t ), t ∈ I,

(52)

w2 (0) = 1.

(53)

Using Lemma 1, the properties (22), and the relation (52). By the definition of fractional integral, we get



q(t ) 1 + g(t )w1 (t ) + g(t ) f (t )w1 (t ) w1 (t ) + f (t )  (α)



t

(t − τ )α−1 h(τ )w1 (τ )dτ 0 

 q(t ) ≤ + g(t )w2 (t ) + g(t ) f (t )w2 (t ) w2 (t ) + Itα h(t )w2 (t ) f (t ) q(t ) = + g(t )w2 (t )[1 + f (t )w3 (t )], t ∈ I, f (t )

Dtα w2 (t ) =



(54)

where

w3 (t ) = w2 (t ) + Itα (h(t )w2 (t )),

(55)

which is a positive and nondecreasing function on I. From (53) and (55) we have

w2 (t ) ≤ w3 (t ), t ∈ I,

(56)

w3 (0) = w2 (0) = 1.

(57)

Using (54) and (56), we have

Dtα w3 (t ) = Dtα w2 (t ) + h(t )w2 (t ) ≤ ≤

q(t ) + g(t )w2 (t )[1 + f (t )w3 (t )] + h(t )w2 (t ) f (t )

q(t ) w3 (t ) + g(t ) f (t )w23 (t ) + [g(t ) + h(t )]w3 (t ) f (t )

≤ g(t ) f (t )w23 (t ) + [g(t ) + h(t ) + q(t )/ f (t )]w3 (t ), t ∈ I.

(58)

Performing the same procedure as in (39)–(46), we obtain the required estimation (48). The proof is complete.  Theorem 3. Suppose that h(t ), q(t ) ∈ C (I, R+ ), f ∈ C (R+ , R+ ) is a nondecreasing function with f(t) > 0 for t > 0, p ∈ (0, 1). If u(t) satisfies (13), then

u(t ) ≤ f 1/(1−p) (t ) exp



× exp −



 1 1− p

 0

tα  (1+α)

1 2(1 − p) g((s (1 + α)) α )ds p+1



τα  (1+α)



1+

1 2(1 − p) g((s (1 + α)) α )ds 2(1 − p)g(τ )dτ p+1

0

1



 (α)

1/(1−p)

t 0

(t − τ )α−1

.

(59)

Proof. Noting that f(t) is a positive and nondecreasing function, from (13) we obtain



u p+1 (t ) 1 ≤ 1+ f p+1 (t )  (α)



(t − s)α−1 g(s)

u p (s) ds f p (s)

2



t u p (s) (t − s)α−1 g(s) p  (α) f (s) 0 0  s  u(s)  p 1 u (τ ) × + (s − τ )α−1 g(τ ) p dτ ds, t ∈ I. f (s)  (α) 0 f (τ ) t

+2

1

(60)

Let v1 (t ) = u(t )/ f (t ). From (60) we see

 t  1 t 2 1 v1p+1 (t ) ≤ 1 + (t − s)α−1 g(s)v1p (s)ds + 2 (t − s)α−1 g(s)v1p (s)  (α) 0  (α) 0  s   1 × v1 (s) + (s − τ )α−1 g(τ )v1p (τ )dτ ds, t ∈ I.  (α) 0

(61)

W.-S. Wang / Applied Mathematics and Computation 268 (2015) 1029–1037 p+1

Define a function v2

1035

(t ) by the right hand side of the inequality (61), i. e.

 t  1 t 2 1 v2p+1 (t ) = 1 + (t − s)α−1 g(s)v1p (s)ds + 2 (t − s)α−1 g(s)v1p (s)  (α) 0  (α) 0  s   1 × v1 (s) + (s − τ )α−1 g(τ )v1p (τ )dτ ds, t ∈ I.  (α) 0

(62)

We observe that v2 (t) is a positive and nondecreasing function on I. From (61) and (62) we have

v1 (t ) ≤ v2 (t ), u(t ) ≤ v2 (t ) f (t ), t ∈ I,

(63)

v2 (0) = 1.

(64)

Using Lemma 1, the properties (19), and the relation (63). By the definition of fractional integral, we get

 1 t  (t − s)α−1 g(s)v1p (s)ds g(t )v1p (t )  (α) 0  t   1 + 2g(t )v1p (t ) v1 (t ) + (t − τ )α−1 g(τ )v1p (τ )dτ  (α) 0  t   1 ≤ 2g(t )v2p (t ) v2 (t ) + 2 (t − τ )α−1 g(τ )v2p (τ )dτ  (α) 0 p = 2g(t )v2 (t )v3 (t ), t ∈ I.

( p + 1)v2p (t )Dtα v2 (t ) = 2

(65)

Since v2 (t) > 0, from (65) we have

where

Dtα v2 (t ) ≤ 2g(t )v3 (t )/( p + 1).

(66)



v3 (t ) = v2 (t ) + 2Itα g(t )v2p (t ) , t ∈ I,

(67)

which is a positive and nondecreasing function on I. From (64) and (67) we have

v2 (t ) ≤ v3 (t ), t ∈ I,

(68)

v3 (0) = v2 (0) = 1.

(69)

Using (66)–(68), we have

Dtα v3 (t ) = Dtα v2 (t ) + 2g(t )v2p (t )

≤ 2g(t )v3 (t )/( p + 1) + 2g(t )v3p (t ) ≤

2 g(t )v3 (t ) + 2g(t )v3p (t ), t ∈ I, p+1

(70)

that is

Dtα (v1−p (t )) ≤ 3 Let v4 (t ) = v3

1−p

2(1 − p) g(t )v1−p (t ) + 2(1 − p)g(t ), t ∈ I. 3 p+1

(71)

(t ), then we have

Dtα (v4 (t )) ≤

2(1 − p) g(t )v4 (t ) + 2(1 − p)g(t ) = g1 (t )v4 (t ) + g2 (t ), t ∈ I, p+1

(72)

1−p) g(t ), g2 (t ) = 2(1 − p)g(t ). Performing the same procedure as in (41)–(43), we obtain where g1 (t ) = 2(p+1

v4 (t ) ≤ exp



tα  (1+α)

0

× 1+

1

g1 ((s (1 + α)) α )ds

1

 (α)



t 0



  (t − τ )α−1 exp −

0

τα  (1+α)

1





g1 ((s (1 + α)) α )ds g2 (τ )dτ .

Using (73), we get the required estimation (59). The proof is complete. 

(73)

1036

W.-S. Wang / Applied Mathematics and Computation 268 (2015) 1029–1037

3. Application We use the inequalities obtained in Theorem 2 to study the boundedness and asymptotic behaviour of the solutions of a class of fractional integral equations. Furthermore, there exist many possible applications for the inequalities established in Section 2, but those presented here are sufficient to convey the importance of our results to the literature. Example. Consider the following fractional integral equation:

x(t ) = f (t ) + Itα G(t, x(t )) + Itα H[t, x(t ), Itα K (t, x(t ))],

(74)

where Itα denotes the Riemann– Liouville fractional integral of order R2 ), satisfy the following conditions:

α on the interval I as defined in (15), G, K ∈ C(I, R), H ∈ C(I,

|G(t, x)| ≤ g(t )|x| + q(s),

(75)

|H (t, x, y)| ≤ g(s)|x|[|x| + |y|],

(76)

|K (t, x)| ≤ h(t )|x|,

(77)

where f, g, q, h as defined in Theorem 2. Corollary 1. Consider the fractional integral Eq. (74) and suppose that G, H, K satisfy the conditions (75)–(77), respectively. suppose that f(t) > 1 for all t ∈ I and

1−



1

 (α)

t 0

(t − τ )α−1





× exp −

τα  (1+α)

0

1

1

1



[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds g(τ ) f (τ )dτ > 0, t ∈ I.

(78)

Then all solutions of the fractional integral Eq. (74) satisfy the following estimation





tα  (1+α)



1



t

(t − τ )α−1  (α) 0   (τ1+α α)  −1 1 1 1 × exp − [g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds g(τ ) f (τ )dτ , t ∈ I. (79)

|x(t )| ≤ f (t ) exp −

0

1

1

1

[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds

1−

0

Proof. From the conditions (75)–(77), and the fractional integral Eq. (74), we have



1

t

(t − s)α−1 [g(s)|x(s)| + q(s)]ds  t  s   1 1 (t − s)α−1 [g(s)|x(s)| |x(s)| + (s − τ )α−1 h(τ )|x(τ )|dτ ds, t ∈ I. +  (α) 0  (α) 0

|x(t )| ≤ f (t ) +

 (α)

0

(80)

Applying Theorem 2 to (80) we get the estimation (79). This completes the proof of Corollary 1.  If there are positive constants M1 , M2 such that 1 ≤ f(t) ≤ M1 for all t ∈ I, and

1−



1

 (α)

t 0



(t − τ )α−1

× exp −



τα  (1+α)

0

1

1

1



[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds g(τ ) f (τ )dτ ≥ M2 , t ∈ I.

By the estimation (79), we observe that all solutions of the fractional integral Eq. (74) satisfy

|x(t )| ≤



M1 exp − M2



tα  (1+α)

0

1

1

1

(81)



[g((s (1 + α)) α ) + h((s (1 + α)) α ) + q((s (1 + α)) α )]ds ,

(82)

for all t ∈ I. From (82) we see that the solution x(t) approaches zero as t → ∞. Author’s contributions Wu-Sheng Wang carried out the main part of this article. The author read and approved the final manuscript. Acknowledgements This research was supported by National Natural Science Foundation of China (project no. 11161018), Guangxi Natural Science Foundation (project no. 2012GXNSFAA053009), Scientific Research Foundation of the Education Department of Guangxi Autonomous Region of China (no. KY2015ZD103), and the high school specialty and curriculum integration project of Guangxi Zhuang Autonomous Region (no. GXTSZY2220).

W.-S. Wang / Applied Mathematics and Computation 268 (2015) 1029–1037

1037

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