A Generalized Henry-Type Integral Inequality and Application to

Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2014, Article ID 841718, 12 pages http://dx.doi.org/10.1155/2014/841718

Research Article A Generalized Henry-Type Integral Inequality and Application to Dependence on Orders and Known Functions for a Fractional Differential Equation Jun Zhou1,2 1 2

Department of Mathematics, Sichuan University, Chengdu, Sichuan 610064, China College of Mathematics and Software Science, Sichuan Normal University, Chengdu, Sichuan 610066, China

Correspondence should be addressed to Jun Zhou; [email protected] Received 30 March 2014; Accepted 10 June 2014; Published 7 July 2014 ¨ ¸ Academic Editor: Turgut Ozis Copyright © 2014 Jun Zhou. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We discuss on integrable solutions for a generalized Henry-type integral inequality in which weak singularity and delays are involved. Not requiring continuity or differentiability for some given functions, we use a modified iteration argument to give an estimate of the unknown function in terms of the multiple Mittag-Leffler function. We apply the result to give continuous dependence of solutions on initial data, derivative orders, and known functions for a fractional differential equation.

1. Introduction Since Gronwall [1] and Bellman [2] discussed the integral inequalities

if and only if 𝑢 = 0) on [0, ∞). In 2000 this result was generalized by Lipovan [4] to the delay case 𝑡

𝑢 (𝑡) ≤ 𝑎 + ∫ 𝑓 (𝑠) 𝑤 (𝑢 (𝑠)) 𝑑𝑠 𝑡0

𝑡

0 ≤ 𝑢 (𝑡) ≤ ∫ (𝑎 + 𝑏𝑢 (𝑠)) 𝑑𝑠, 𝛼

(1)

𝑡

𝑢 (𝑡) ≤ 𝑎 + ∫ 𝑓 (𝑠) 𝑢 (𝑠) 𝑑𝑠,

(2)

𝛼

respectively, there have been made many generalizations, some of which were applied to existence, uniqueness, boundedness, and stability of solutions and invariant manifolds for differential equations and integral equations. In 1956 Bihari [3] discussed a nonlinear version of the integral inequality 𝑡

𝑢 (𝑡) ≤ 𝑎 + ∫ 𝑓 (𝑠) 𝑤 (𝑢 (𝑠)) 𝑑𝑠, 0

𝑡 ≥ 0,

(3)

where the given function 𝑤 is continuous, nondecreasing, and positive definite (i.e., 𝑤(𝑢) ≥ 0 for all 𝑢 and 𝑤(𝑢) = 0

+∫

𝑏(𝑡)

𝑏(𝑡0 )

(4) 𝑔 (𝑠) 𝑤 (𝑢 (𝑠)) 𝑑𝑠,

𝑡0 ≤ 𝑡 < 𝑡1 ,

where 𝑏 is a continuously differentiable and nondecreasing function from [𝑡0 , 𝑡1 ) to [𝑡0 , 𝑡1 ) such that 𝑏(𝑡) ≤ 𝑡. In 2005 Agarwal et al. [5] investigated the integral inequality of a finite sum 𝑛

𝑢 (𝑡) ≤ 𝑎 (𝑡) + ∑ ∫

𝑏𝑖 (𝑡)

𝑖=1 𝑏𝑖 (𝑡0 )

𝑓𝑖 (𝑡, 𝑠) 𝑤𝑖 (𝑢 (𝑠)) 𝑑𝑠,

(5)

𝑡0 ≤ 𝑡 < 𝑡1 , where the monotonicity of 𝑎 is not required but 𝑤𝑖+1 has stronger monotonicity than 𝑤𝑖 for each 𝑖. In recent years some new generalizations were given in, for example, [6–8]. Many results on integral inequalities can be found in Pachpatte’s monograph (see [9]). Among various types of integral inequalities, integral inequalities with weakly singular kernels are important tools

2

Journal of Applied Mathematics

in the discussion of reaction-diffusion equations and fractional differential equations. As shown in [10], the integral 𝑡 ∫𝑡0 𝐾(𝑡, 𝑠) is singular on the line 𝑠 = 𝑡; it is referred to be 𝑡

weakly singular if it is singular and ∫𝑡0 |𝑓(𝑡, 𝑠)|𝑑𝑠 < ∞ for all 𝑡0 ≤ 𝑡 < 𝑇 ≤ ∞. In fractional differential equations we need to consider the following Riemann-Liouville derivative operator and integral operator (see [11]) 𝐷𝑡𝛼0 𝑓 (𝑥) :=

1 𝑑 Γ (1 − 𝛼) 𝑑𝑥 𝑥

× ∫ (𝑥 − 𝑡)−𝛼 𝑓 (𝑡) 𝑑𝑡, 𝑡0

𝛽 𝐼𝑡0 𝑓 (𝑥)

0 < 𝛼 < 1,

(6)

(7)

in a Banach space 𝑋, where 𝐴 is a sectorial operator and 𝑓 is locally H¨older continuous with H¨older index 0 < 𝜃 < 1. From the variation-constant formula (see [12]), we encounter the following: 󵄩󵄩 −𝐴(𝑡−𝑠) 󵄩󵄩 󵄩󵄩 = 𝑂 ((𝑡 − 𝑠)−1 ) , 󵄩󵄩𝐴𝑒 󵄩 󵄩 󵄩 󵄩󵄩 𝜃 󵄩󵄩𝑓 (𝑡) − 𝑓 (𝑠)󵄩󵄩󵄩 = 𝑂 ((𝑡 − 𝑠) ) ,

(8)

and therefore the singular kernel (𝑡 − 𝑠) is the estimation of its solution. For this reason the integral inequality with a weakly singular kernel 𝑡 ≥ 0,

𝑛

(𝑏Γ (𝛽)) 𝑢 (𝑡) ≤ 𝑎 (𝑡) + ∫ { ∑ (𝑡 − 𝑠)𝑛𝛽−1 𝑎 (𝑠)} 𝑑𝑠, Γ (𝑛𝛽) 0 𝑛=1

𝑡 ≥ 0,

𝑡

× (∫ 𝐹(𝑠)𝑞 𝑒−𝑞𝑠 𝑤(𝑢 (𝑠))𝑞 𝑑𝑠) 0

𝜇−1 𝜏−1

𝑢 (𝑡) ≤ 𝑎 (𝑡) + 𝑏 (𝑡) ∫ (𝑡𝜎 − 𝑠𝜎 ) 0

𝑡

+ 𝑐 (𝑡) ∫ (𝑡𝛼 − 𝑠𝛼 ) 0

𝛽−1 𝛾−1

𝑠

𝑠

0

𝑡

+ 𝑐4 ∫ (𝑡 − 𝑠) 0

𝛽−1

𝑢 (𝑠) 𝑑𝑠,

(11) 𝑡 ≥ 0,

where 𝑐1 , 𝑐2 , 𝑐3 , 𝑐4 ≥ 0, 0 < 𝛽 < 1. Inequality (9) was also extended by Ye et al. [14] in 2007 by replacing the constant

𝑔 (𝑠) 𝑢 (𝑠) 𝑑𝑠

𝑓 (𝑠) 𝑤 (𝑢 (𝑠)) 𝑑𝑠,

(14)

where 𝛼, 𝛽, 𝛾, 𝜎, 𝜇, 𝜏 ≥ 0. Recently Ma and Peˇcari´c [17] also employed the separation approach to discuss another weakly singular integral inequality 𝑡

𝛽−1 𝛾−1

𝑢𝑝 (𝑡) ≤ 𝑎 (𝑡) + 𝑏 (𝑡) ∫ (𝑡𝛼 − 𝑠𝛼 ) 0

𝑠

𝑓 (𝑠) 𝑢𝑞 (𝑠) 𝑑𝑠,

(15)

𝑡 ≥ 0, under the condition 𝑝 ≥ 𝑞 ≥ 0. In this paper we investigate the following integral inequality of finite sum: 𝑚

𝑏𝑖 (𝑡)

𝑖=1 𝑏𝑖 (𝑡0 )

𝑢 (𝑡) ≤ 𝑐1 + 𝑐2 𝑡𝛽−1 + 𝑐3 ∫ 𝑢 (𝑠) 𝑑𝑠

,

𝑡 ≥ 0,

𝑢 (𝑡) ≤ 𝑎 (𝑡) + ∑ ∫

𝑡

1/𝑞

where 𝑝, 𝑞 > 1 are certain constants such that 1/𝑝 + 1/𝑞 = 1, so that the inserted exponential factor 𝑒𝑠 makes the singular 𝑡 integral ∫0 (𝑡 − 𝑠)𝑝(𝛽−1) 𝑒𝑝𝑠 𝑑𝑠 convergent as 𝑠 → 𝑡 and the inequality is reduced to the classic Bihari’s form (3). In 2002 Ma and Yang [16] improved Medved’s method and gave an estimation to the Volterra-type integral inequality with a more general form of weakly singular kernel

(10) where Γ is the Gamma function, was given by an iteration approach. Following Henry’s idea, in 1994 Sano and Kunimatsu [13] extended Henry’s result to a more general integral inequality

(13)

0

(9)

where 𝑏 ≥ 0, 0 < 𝛽 < 1 and both 𝑎 and 𝑢 are nonnegative and locally integrable, was considered in Henry’s book [12] in 1981 and the estimate ∞

1/𝑝

≤ 𝑎 (𝑡) + (∫ (𝑡 − 𝑠)𝑝(𝛽−1) 𝑒𝑝𝑠 𝑑𝑠)

𝑡

𝜃−1

𝑡

where 0 < 𝛽 < 1. This inequality is of Bihari’s form (3) with a weakly singular kernel. He applied the well-known H¨older inequality to separate the unknown 𝑢 from the singular kernel, that is,

𝑡

𝛽−1

𝑥 (𝑡0 ) = 𝑥0 ,

0

𝑡 ≥ 0, (12)

0

are in which the singular kernels (𝑥 − 𝑡) and (𝑥 − 𝑡) included, respectively. Another example can be seen from the Cauchy problem of the evolution equation

𝑢 (𝑡) ≤ 𝑎 (𝑡) + 𝑏 ∫ (𝑡 − 𝑠)𝛽−1 𝑢 (𝑠) 𝑑𝑠,

0

𝑢 (𝑡) ≤ 𝑎 (𝑡) + ∫ (𝑡 − 𝑠)𝛽−1 𝑒𝑠 𝐹 (𝑠) 𝑒−𝑠 𝑤 (𝑢 (𝑠)) 𝑑𝑠

𝛽 > 0,

−𝛼

𝑡

𝑡

𝑢 (𝑡) ≤ 𝑎 (𝑡) + ∫ (𝑡 − 𝑠)𝛽−1 𝐹 (𝑠) 𝑤 (𝑢 (𝑠)) 𝑑𝑠,

𝑡

𝑥 1 := ∫ (𝑥 − 𝑡)𝛽−1 𝑓 (𝑡) 𝑑𝑡, Γ (𝛽) 𝑡0

𝑥̇ + 𝐴 (𝑡) 𝑥 = 𝑓 (𝑡) ,

𝑏 with a nonnegative, nondecreasing, bounded, and continuous function 𝑔(𝑡). Another idea for inequalities with a weakly singular kernel was introduced by Medved [15] in 1997 for the following Henry-Bihari type integral inequality:

𝛽

(𝑡 − 𝑓𝑖 (𝑠)) 𝑖 ℎ𝑖 (𝑡, 𝑠) 𝑢 (𝑠) 𝑑𝑠,

(16)

𝑡 ≥ 0, in which weakly singular kernels and delays are involved. Not requiring continuity of 𝑎 or differentiability of 𝑏𝑖 , we give an estimate for locally integrable 𝑢 in terms of the multiple Mittag-Leffler function (see [18]). We prefer Henry’s iteration

Journal of Applied Mathematics

3

approach in our proof because the approach does not reduce the problem to the classic one so that no more continuity and differentiability are required. Our result generalizes the works made in [14–17] in some sense because of the more general form (16). Finally, we apply the result to give continuous dependence of solutions on initial data, derivative orders, and known functions for a fractional differential equation.

2. Main Results For constants 𝑡0 , 𝑇 such that 𝑡0 < 𝑇 ≤ +∞, consider inequality (16), where 𝑎, 𝑏𝑖 , 𝑓𝑖 and ℎ𝑖 are given nonnegative functions and satisfy the following hypotheses: (H1 ) 𝑎 is a locally integrable function on [𝑡0 , 𝑇), that is, Lebesgue integrable on every compact subset of [𝑡0 , 𝑇), and the integrations in (16) are bounded by replacing 𝑢 with 𝑎 in (16);

was proposed as an extension of the exponential function by Mittag-Leffler ([19]) in 1903. An extension of two parameters ∞

𝑧𝑘 , Γ (𝑘𝛼 + 𝛽) 𝑘=0

(19) was proposed by Wiman [20] in 1905. Later, two MittagLeffler functions with three parameters were given separately by Prabhakar [21] and Kilbas and Saigo [22]. In 1996 Hadid and Luchko [18] generalized the function into the multiple form

(H4 ) every 𝑓𝑖 , 𝑖 = 1, . . . , 𝑚, is continuously differentiable on [𝑡0 , 𝑇) such that 𝑓𝑖󸀠 (𝑡) > 0 and 𝑡 ≤ 𝑓𝑖 (𝑡) ≤ 𝑏𝑖−1 (𝑡) for all 𝑡 ∈ [𝑡0 , 𝑇). Theorem 1. Suppose that (H1 )–(H4 ) hold and 𝛽𝑖 > −1, 𝑖 = 1, . . . , 𝑚. Then, every nonnegative and locally integrable function 𝑢(𝑡) which satisfies (16) on [𝑡0 , 𝑇) and the integrations in (16) bounded has the estimate 𝑢 (𝑡) ≤ 𝑎 (𝑡)

𝑙

(𝑘; 𝑙1 , . . . , 𝑙𝑛 ) ∏𝑛𝑖=1 𝑧𝑖𝑖 , 𝐸(𝛼1 ,...,𝛼𝑛 ),𝛽 (𝑧1 , . . . , 𝑧𝑛 ) := ∑ ∑ Γ (∑𝑛𝑖=1 𝑙𝑖 𝛼𝑖 + 𝛽) 𝑘=0 𝑙 +⋅⋅⋅+𝑙 =𝑘 ∞

1

(20) where 𝛼𝑖 , 𝛽, 𝑧𝑖 ∈ C, R(𝛼𝑖 ), R(𝛽) > 0, 𝑖 = 1, . . . , 𝑛 and (𝑘; 𝑙1 , . . . , 𝑙𝑛 ) :=

⋅∫

𝑏𝑘1 (𝑡)

𝑏𝑘1 (𝑡0 )

(𝑡 − 𝑓𝑘1 (𝑠))

𝑛

𝑖=1

𝛼𝑖 , 𝑧𝑖 ∈ R+ ,

𝑧𝑘 , Γ (𝑘𝛼 + 1) 𝑘=0

𝑚

𝑚

𝑖=1

𝑖=1

𝑎𝑖 > 0, 𝑖 = 1, . . . , 𝑚. (23)

𝑡 ∈ [𝑡0 , 𝑇) , (17)

∞ 1 (1 + (1/𝑛))∑𝑖=1 𝑎𝑖 +1 . Γ (∑𝑎𝑖 + 1) = 𝑚 ∏ ∑𝑖=1 𝑎𝑖 + 1 𝑛=1 1 + (∑𝑚 𝑖=1 (𝑎𝑖 + 1) /𝑛) 𝑖=1

(24)

It follows that Γ (∑𝑚 𝑖=1 𝑎𝑖 + 1) 𝑚 ∏𝑖=1 Γ (𝑎𝑖 + 1) =

∞ ∏𝑚 ∏𝑚 𝑖=1 (𝑎𝑖 + 1) 𝑖=1 (1 + ((𝑎𝑖 + 1) /𝑛)) . ∏ 𝑚 ∑𝑖=1 𝑎𝑖 + 1 𝑛=1 (1 + (1/𝑛))𝑚−1 (1 + (∑𝑚 𝑖=1 (𝑎𝑖 + 1) /𝑛)) (25)

𝑚 It is easy to know that ∏𝑚 𝑖=1 (𝑎𝑖 + 1) ≥ ∑𝑖=1 𝑎𝑖 + 1. Then, we can prove by induction that 𝑚

(18)

1 ∞ (1 + (1/𝑛))𝑎𝑖 +1 , ∏ 𝑎𝑖 + 1 𝑛=1 1 + ((𝑎𝑖 + 1) /𝑛) 𝑚

𝑚

𝑧 ∈ C, 𝛼 > 0,

𝑖 = 1, . . . , 𝑛.

In fact, we can show that

Γ (𝑎𝑖 + 1) =

We leave the proof of this theorem to next section. In what follows, we express the estimate of series form in terms of the multiple Mittag-Leffler function (see [18]). The original Mittag-Leffler function ∞

(22)

By Euler’s definition on Gamma function (see [24]), we have

where ̃ℎ𝑖 (𝑡) = max(𝑠,𝜏)∈[𝑡0 ,𝑡]×[𝑡0 ,𝑡] ℎ𝑖 (𝑠, 𝜏)/𝑓𝑖󸀠 (𝜏), 𝑖 = 1, . . . , 𝑚.

𝐸𝛼 (𝑧) := ∑

(21)

𝐸(𝛼1 ,...,𝛼𝑛 ),1 (𝑧1 , . . . , 𝑧𝑛 ) ≤ ∏𝐸𝛼𝑖 (𝑧𝑖 ) ,

∑𝑛𝑖=1 𝛽𝑘𝑖 +𝑛−1

} ×𝑓𝑘󸀠1 (𝑠) 𝑎 (𝑠) 𝑑𝑠} , }

𝑘! . 𝑙1 ! × ⋅ ⋅ ⋅ × 𝑙𝑛 !

These generalized Mittag-Leffler functions have been treated as significant special functions since they played an important role in computing fractional calculus and solving fractional differential and integral equations modeled in physics, chemistry, biology, engineering, and applied sciences (see monographs [11, 23]). We have the following inequality:

Γ (∑𝑎𝑖 + 1) ≥ ∏Γ (𝑎𝑖 + 1) ,

∞ { ∏𝑛𝑖=1 Γ (𝛽𝑘𝑖 + 1) ̃ℎ𝑘𝑖 (𝑡) +∑{ ∑ Γ (∑𝑛𝑖=1 𝛽𝑘𝑖 + 𝑛) 𝑛=1 1≤𝑘1 ,...,𝑘𝑛 ≤𝑚 {

𝑛

𝑙1 ,...,𝑙𝑛 ≥0

(H2 ) every ℎ𝑖 , 𝑖 = 1, . . . , 𝑚, is continuous on [𝑡0 , 𝑇)×[𝑡0 , 𝑇); (H3 ) every 𝑏𝑖 : [𝑡0 , 𝑇) → [𝑡0 , 𝑇), 𝑖 = 1, . . . , 𝑚, is continuous and strictly increasing such that 𝑏𝑖 (𝑡) ≤ 𝑡 for all 𝑡 ∈ [𝑡0 , 𝑇);

𝛼, 𝛽, 𝑧 ∈ C, R (𝛼) , R (𝛽) > 0,

𝐸𝛼,𝛽 (𝑧) := ∑

∏ (1 + 𝑖=1

∑𝑚 𝑎 + 1 𝑎𝑖 + 1 1 𝑚−1 (1 + 𝑖=1 𝑖 ) ≥ (1 + ) ). 𝑛 𝑛 𝑛

(26)

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Journal of Applied Mathematics

Obviously, (26) is true for 𝑚 = 1. Assume that (26) is true for 𝑚 = 𝑙. We get 𝑙+1

∏ (1 + 𝑖=1

∑𝑙 𝑎 + 1 𝑎𝑖 + 1 1 𝑙−1 ) = (1 + ) (1 + 𝑖=1 𝑖 ) 𝑛 𝑛 𝑛 𝑎 +1 × (1 + 𝑙+1 ) 𝑛 ∑𝑙+1 𝑎 + 1 1 𝑙 ≥ (1 + ) (1 + 𝑖=1 𝑖 ), 𝑛 𝑛

∞ ∏𝑛𝑖=1 Γ (𝛽𝑘𝑖 + 1) ̃ℎ𝑘𝑖 (𝑡) { ≤ 𝑎̃ (𝑡) {1 + ∑ ( ∑ 𝑛 𝑛=1 1≤𝑘1 ,...,𝑘𝑛 ≤𝑚 Γ (∑𝑖=1 𝛽𝑘𝑖 + 𝑛 + 1) { ∑𝑛𝑖=1 (𝛽𝑘𝑖 +1)

×(𝑡 − 𝑡0 ) (27)

∞

= 𝑎̃ (𝑡) ∑

} )} }

𝑛

(∏Γ (𝛽𝑘𝑖 + 1) ̃ℎ𝑘𝑖 (𝑡)

∑

𝑛=0 1≤𝑘1 ,...,𝑘𝑛 ≤𝑚

𝑖=1

𝛽𝑘𝑖 +1

×(𝑡 − 𝑡0 )

since ∑𝑙𝑖=1 𝑎𝑖

(1 +

+1

𝑛

) (1 +

𝑎𝑙+1 + 1 ) 𝑛

−1

𝑛

∑𝑙+1 𝑎 + 1 1 ), ≥ (1 + ) (1 + 𝑖=1 𝑖 𝑛 𝑛

(28)

× (Γ (∑𝛽𝑘𝑖 + 𝑛 + 1)) 𝑖=1

= 𝑎̃ (𝑡) ∞

implying that (26) holds for 𝑚 = 𝑙 + 1. Thus, from (26) we see that (23) holds. By (23), we have

×∑

∑

𝑘=0 𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑘 𝑘1 ,...,𝑘𝑚 ≥0

( (𝑘; 𝑘1 , . . . , 𝑘𝑚 ) 𝑚

𝐸(𝛼1 ,...,𝛼𝑛 ),1 (𝑧1 , . . . , 𝑧𝑛 ) ∞

= ∑

∑

∞

∞

× ∏ (Γ (𝛽𝑖 + 1) ̃ℎ𝑖 (𝑡) 𝑖=1

∏𝑚 𝑖=1 𝑧𝑘𝑖

𝑚 𝑚=0 1≤𝑘1 ,...,𝑘𝑚 ≤𝑛 Γ (∑𝑖=1

𝛼𝑘𝑖 + 1)

𝑘𝑛

𝛽𝑖 +1 𝑘𝑖

×(𝑡 − 𝑡0 )

(29)

𝑘

𝑛 ∏𝑛𝑖=1 𝑧𝑖 𝑖 = ∏𝐸𝛼𝑖 (𝑧𝑖 ) . 𝑛 ∏ Γ (𝑘𝑖 𝛼𝑖 + 1) 𝑖=1 =0 𝑖=1

𝑚

≤ ∑ ⋅⋅⋅ ∑ 𝑘1 =0

)

) )

−1

× (Γ (∑𝑘𝑖 (𝛽𝑖 + 1) + 1)) 𝑖=1

Therefore, (22) is proved.

= 𝑎̃ (𝑡) 𝐸(𝛽1 +1,...,𝛽𝑚 +1),1

Corollary 2. Suppose that the hypotheses of Theorem 1 hold and additionally that 𝑎(𝑡) is continuous on [𝑡0 , 𝑇). Then,

𝛽 +1 × (̃ℎ1 (𝑡) Γ (𝛽1 + 1) (𝑡 − 𝑡0 ) 1 ,

. . . , ̃ℎ𝑚 (𝑡) Γ (𝛽𝑚 + 1) (𝑡 − 𝑡0 )

𝑢 (𝑡) ≤ 𝑎̃ (𝑡) 𝐸(𝛽1 +1,...,𝛽𝑚 +1),1

𝛽𝑚 +1

). (31)

𝛽 +1 × (̃ℎ1 (𝑡) Γ (𝛽1 + 1) (𝑡 − 𝑡0 ) 1 , 𝛽𝑚 +1

. . . , ̃ℎ𝑚 (𝑡) Γ (𝛽𝑚 + 1) (𝑡 − 𝑡0 )

)

(30)

𝑚

𝛽 +1 ≤ 𝑎̃ (𝑡) ∏𝐸𝛽𝑖 +1 (̃ℎ𝑖 (𝑡) Γ (𝛽𝑖 + 1) (𝑡 − 𝑡0 ) 𝑖 ) , 𝑖=1

where 𝑎̃(𝑡) = max𝑡0 ≤𝜏≤𝑡 𝑎(𝜏) and 𝐸𝛼 (𝑧) and 𝐸(𝛼1 ,...,𝛼𝑛 ),1 are defined in (18) and (20).

𝑏𝑘1 (𝑡0 )

𝛽 +1 × (̃ℎ1 (𝑡) Γ (𝛽1 + 1) (𝑡 − 𝑡0 ) 1 , 𝛽 +1 . . . , ̃ℎ𝑚 (𝑡) Γ (𝛽𝑚 + 1) (𝑡 − 𝑡0 ) 𝑚 )

(32)

𝑚

∞ { ∏𝑛𝑖=1 Γ (𝛽𝑘𝑖 + 1) ̃ℎ𝑘𝑖 (𝑡) { 𝑢 (𝑡) ≤ 𝑎̃ (𝑡) {1 + ∑ { ∑ Γ (∑𝑛𝑖=1 𝛽𝑘𝑖 + 𝑛) 𝑛=1 1≤𝑘1 ,...,𝑘𝑛 ≤𝑚 { { 𝑏𝑘1 (𝑡)

𝑎̃ (𝑡) 𝐸(𝛽1 +1,...,𝛽𝑚 +1),1

𝛽 +1 ≤ 𝑎̃ (𝑡) ∏𝐸𝛽𝑖 +1 (̃ℎ𝑖 (𝑡) Γ (𝛽𝑖 + 1) (𝑡 − 𝑡0 ) 𝑖 ) .

Proof. Starting from (17), we have

×∫

It follows from (22) that

(𝑡 − 𝑓𝑘1 (𝑠))

∑𝑛𝑖=1 𝛽𝑘𝑖 +𝑛−1

}} ×𝑓𝑘󸀠1 (𝑠) 𝑑𝑠}} }}

𝑖=1

The corollary is proved.

3. Proof of Theorem Let 𝐿1loc,𝑤 [𝑡0 , 𝑇) consist of all locally integrable nonnegative 𝑏 (𝑡)

𝑖 𝛽𝑖 functions 𝜙 on [𝑡0 , 𝑇) such that ∑𝑚 𝑖=1 ∫𝑏𝑖 (𝑡0 ) (𝑡 − 𝑓𝑖 (𝑠)) ℎ𝑖 (𝑡, 𝑠)𝜙(𝑠)𝑑𝑠 < ∞ for arbitrarily given 𝑡 ∈ [𝑡0 , 𝑇). We can

Journal of Applied Mathematics

5

verify that 𝐿1loc,𝑤 [𝑡0 , 𝑇) is a linear space, due to the linearity of integration. Define a linear operator 𝐵 on 𝐿1loc,𝑤 [𝑡0 , 𝑇) as 𝑚

𝐵𝜙 (𝑡) := ∑ ∫

𝑏𝑖 (𝑡)

𝛽

𝑖=1 𝑏𝑖 (𝑡0 )

(𝑡 − 𝑓𝑖 (𝑠)) 𝑖 ℎ𝑖 (𝑡, 𝑠) 𝜙 (𝑠) 𝑑𝑠,

It follows that 𝐵𝑙+1 𝑢 (𝑡) ≤

𝑡 ∈ [𝑡0 , 𝑇) ,

𝑏 (𝑡)

𝑖 𝛽𝑖 [𝑡0 , 𝑇), ∑𝑚 𝑖=1 ∫𝑏𝑖 (𝑡0 ) (𝑡 − 𝑓𝑖 (𝑠)) ℎ𝑖 (𝑡, 𝑠)𝜙(𝑠)𝑑𝑠 < ∞, implying that 𝐵𝜙 ∈ 𝐶+ [𝑡0 , 𝑇), the set of all continuous and nonnegative functions on [𝑡0 , 𝑇). We know that 𝐶+ [𝑡0 , 𝑇) ⊂ 𝐿1loc [𝑡0 , 𝑇). For all 𝜓 ∈ 𝐶+ [𝑡0 , 𝑇), from the continuity of 𝜓 and ℎ𝑖 , we know that they are all locally bounded. We also know that (𝑡 − 𝑓𝑖 (𝑠))𝛽𝑖 are integrable on 𝑠 ∈ [𝑏𝑖 (𝑡0 ), 𝑏𝑖 (𝑡)] by (H3 ), (H4 ) and 𝛽𝑖 > −1. It follows that

𝑏 (𝑡)

𝑖 𝛽𝑖 ∑𝑚 𝑖=1 ∫𝑏 (𝑡 ) (𝑡 − 𝑓𝑖 (𝑠)) ℎ𝑖 (𝑡, 𝑠)𝜓(𝑠)𝑑𝑠 𝑖


0 on [𝑡0 , 𝑇). In fact, (35) is true for 𝑛 = 1. Assume that (35) is true for 𝑛 = 𝑙. Then

≤

𝑓𝑘𝑙 (𝑏𝑘𝑙 (𝑡))

𝑓𝑘𝑙 (𝑏𝑘𝑙 (𝑡0 ))

(35)

∏𝑙𝑖=1 Γ (𝛽𝑘𝑖 ∑ Γ (∑𝑙𝑖=1 1≤𝑘1 ,...,𝑘𝑙 ≤𝑚

(𝑠 − 𝑓𝑘𝑙+1 (𝜏))

(𝑠) 𝜑 (𝑠) 𝑑𝑠,

Note that 𝑓𝑘𝑖 (𝑏𝑘𝑖 (𝑡)) ≤ 𝑡 by (H4 ). It implies that 𝑠 − 𝑓𝑘𝑙+1 (𝜏) ≥ 0 for all 𝜏 ∈ [𝑏𝑘𝑙+1 (𝑡0 ), 𝑏𝑘𝑙+1 (𝑠)] by the monotonicity of 𝑓𝑘𝑙+1 . It follows from (39) that 𝜑 is a nondecreasing function. Hence, 𝜑(𝑓𝑘−1 (𝑡)) ≤ 𝜑(𝑡) by (H4 ). On the other hand, from (H3 ) 𝑙 and (H4 ) we see that 𝑏𝑖 (𝑡0 ) = 𝑓𝑖 (𝑡0 ) = 𝑡0 , implying that 𝑓𝑘𝑙 (𝑏𝑘𝑙 (𝑡0 )) = 𝑡0 . With the change of variables 𝜉 = 𝑓𝑘𝑙 (𝑠), it follows from (38) that

𝛽𝑘𝑖 + 𝑛) ∑𝑛𝑖=1

∑𝑙𝑖=1 𝛽𝑘𝑖 +𝑙−1 󸀠 𝑓𝑘𝑙

(𝑡 − 𝑓𝑘𝑙 (𝑠))

𝑏𝑘𝑙+1 (𝑡0 )

∏𝑛𝑖=1 Γ (𝛽𝑘𝑖 + 1) ̃ℎ𝑘𝑖 (𝑡)

∑

(37)

𝑚

𝑙=0

We claim that for all integers 𝑛 ≥ 1,

Γ (∑𝑙𝑖=1 𝛽𝑘𝑖 + 𝑙)

1≤𝑘1 ,...,𝑘𝑙 ≤𝑚

(33) where 𝜙 ∈ 𝐿1loc,𝑤 [𝑡0 , 𝑇). We claim that 𝐵 is self-mapping on 𝐿1loc,𝑤 [𝑡0 , 𝑇). In fact, for all 𝜙 ∈ 𝐿1loc,𝑤 [𝑡0 , 𝑇) and given 𝑡 ∈

∏𝑙𝑖=1 Γ (𝛽𝑘𝑖 + 1) ̃ℎ𝑘𝑖 (𝑡)

∑

(𝑡 − 𝑠)∑𝑖=1 𝛽𝑘𝑖 +𝑙−1 × (𝑠 − 𝑓𝑘𝑙+1 (𝜏))

(𝑠)

(41) 𝛽𝑘𝑙+1

𝑑𝑠

× 𝑓𝑘󸀠𝑙+1 (𝜏) 𝑢 (𝜏) 𝑑𝜏, (36)

ℎ𝑘𝑙+1 (𝑠, 𝜏)

× (𝑠 − 𝑓𝑘𝑙+1 (𝜏)) ×𝑢 (𝜏) 𝑑𝜏} 𝑑𝑠.

𝛽𝑘𝑙+1

by interchanging the order of integration. In (41) we observe that ∫

𝑡

𝑙

𝑏𝑘−1 (𝜏) 𝑙+1

(𝑡 − 𝑠)∑𝑖=1 𝛽𝑘𝑖 +𝑙−1 (𝑠 − 𝑓𝑘𝑙+1 (𝜏)) ∑𝑙+1 𝑖=1 𝛽𝑘𝑖 +𝑙

= (𝑡 − 𝑓𝑘𝑙+1 (𝜏))

𝛽𝑘𝑙+1

𝑑𝑠

6

Journal of Applied Mathematics 𝑡

×∫

𝑏𝑘−1

𝑙+1

𝑡−𝑠 ) ( (𝜏) 𝑡 − 𝑓𝑘𝑙+1 (𝜏) ×(

𝑠 − 𝑓𝑘𝑙+1 (𝜏) 𝑡 − 𝑓𝑘𝑙+1 (𝜏)

∑𝑙𝑖=1 𝛽𝑘𝑖 +𝑙−1

𝛽𝑘𝑙+1

)

𝑑(

where 𝑥 ∈ (0, ∞) and 𝑎 > 0 is a constant. By Euler’s definition on Gamma function (see [24]), Γ(𝑥) = (1/𝑥)∏∞ 𝑛=1 (1 + (1/𝑛))𝑥 /(1 + (𝑥/𝑛)), we know that

𝑠 − 𝑓𝑘𝑙+1 (𝜏) 𝑡 − 𝑓𝑘𝑙+1 (𝜏)

)

Ψ𝑎 (𝑥) = (1 +

∑𝑙+1 𝑖=1 𝛽𝑘𝑖 +𝑙

= (𝑡 − 𝑓𝑘𝑙+1 (𝜏)) 1

𝑙

𝑧1

∑𝑙+1 𝑖=1 𝛽𝑘𝑖 +𝑙

≤ (𝑡 − 𝑓𝑘𝑙+1 (𝜏)) × ∫ (1 − 𝑧)

𝑧

0

∑𝑙+1 𝑖=1 𝛽𝑘𝑖 +𝑙

= (𝑡 − 𝑓𝑘𝑙+1 (𝜏)) =

Γ (𝛽𝑘𝑗 + 1)

∑𝑙𝑖=1 𝛽𝑘𝑖 +𝑙−1 𝛽𝑘𝑙+1

Γ (∑𝑛𝑖=1

𝑑𝑧

𝐵 (∑𝛽𝑘𝑖 + 𝑙, 𝛽𝑘𝑙+1 + 1) 𝑖=1

,

≤

where 𝑧 := {𝑠 − 𝑓𝑘𝑙+1 (𝜏)}/{𝑡 − 𝑓𝑘𝑙+1 (𝜏)} and 𝑧1 := − 𝑓𝑘𝑙+1 (𝜏)}/{𝑡 − 𝑓𝑘𝑙+1 (𝜏)} ≥ 0 by (H4 ). Thus, from (37) we see that inequality (35) holds for 𝑛 = 𝑙 + 1 and the claimed (35) is proved. For every 𝜙(𝑡) ∈ 𝐿1loc,𝑤 [𝑡0 , 𝑇) and each 𝑛 = 0, 1, . . . , ∞, define a sequence of linear operators 𝐴 𝑛 : 𝐿1loc,𝑤 [𝑡0 , 𝑇) → 𝐶[𝑡0 , 𝑇) by

𝑏𝑘𝑛 (𝑡)

⋅∫

𝑏𝑘𝑛 (𝑡0 )

∏𝑛𝑖=1 Γ (𝛽𝑘𝑖 + 1) ̃ℎ𝑘𝑖 (𝑡) Γ (∑𝑛𝑖=1 𝛽𝑘𝑖 + 𝑛) ∑𝑛𝑖=1

(𝑡 − 𝑓𝑘𝑛 (𝑠)) ×

𝑏𝑘𝑛 (𝑡)

where ∫𝑏 𝑏𝑘𝑛 (𝑡)

∫𝑏

𝑘𝑛 (𝑡0 )

𝑘𝑛 (𝑡0 )

(𝑡 −

𝑓𝑘󸀠𝑛

Γ (∑𝑛𝑖=1 𝛽𝑘𝑖 − 𝛿 + 𝑛)

Let 𝛽∗ := min1≤𝑖≤𝑚 𝛽𝑖 , 𝛽∗ := max1≤𝑖≤𝑚 𝛽𝑖 and 𝑐 := (𝛽∗ + 1)/(𝛽∗ + 1) ≥ 1. Then (48) implies that ∏𝑛𝑖=1 Γ (𝛽𝑘𝑖 + 1) Γ (∑𝑛𝑖=1 𝛽𝑘𝑖 + 𝑛)

Theorem 1 and (𝑡 − [𝑏𝑘𝑛 (𝑡0 ), 𝑏𝑘𝑛 (𝑡)). It implies from (35) that 𝑛

𝐵 𝜙 (𝑡) ≤ 𝐴 𝑛 𝜙 (𝑡) .

are continuous on

(44)

In what follows we will prove that for all 𝜙 ∈ 𝐿1loc,𝑤 [𝑡0 , 𝑇) and arbitrarily given 𝑡 ∈ [𝑡0 , 𝑇), ∑∞ 𝑛=1 𝐴 𝑛 𝜙(𝑡) converges. Define Γ (𝑥) , Γ (𝑥 + 𝑎)

(45)

(49)

=

Γ (∑𝑛𝑖=1 𝛽𝑘𝑖 + 𝑛)

1≤𝑘1 ,...,𝑘𝑛 ≤𝑚 𝑏𝑘𝑛 (𝑡)

𝑏𝑘𝑛 (𝑡0 )

(𝑠) 𝜙 (𝑠) 𝑑𝑠,

𝑛−1 𝑓𝑘𝑛 (𝑠))∑𝑖=1 (𝛽𝑘𝑖 +1) 𝑓𝑘󸀠𝑛 (𝑠)

(Γ (𝛽∗ + 1)) . Γ (𝑛 (𝛽∗ + 1))

∏𝑛𝑖=1 Γ (𝛽𝑘𝑖 + 1) ̃ℎ𝑘𝑖 (𝑡)

∑ ×∫

(𝑡 − 𝑓𝑘𝑛 (𝑠))𝛽𝑘𝑛 𝜙(𝑠)𝑑𝑠 < ∞ from the supposition of

𝑛

≤

When 𝑡 ∈ (𝑡0 + 1, 𝑇), we have

(43)

< ∞ since

.

(48)

𝐴 𝑛 𝜙 (𝑡) =

𝛽𝑘𝑖 +𝑛−1

𝑛 𝑓𝑘𝑛 (𝑠))∑𝑖=1 𝛽𝑘𝑖 +𝑛−1 𝑓𝑘󸀠𝑛 (𝑠)𝜙(𝑠)𝑑𝑠

Ψ𝑎 (𝑥) :=

(47)

𝑗−1

{𝑏𝑘−1 (𝜏) 𝑙+1

∑

.

∏𝑖=1 Γ (𝛽𝑘𝑖 + 1) Γ (𝛽𝑘𝑗 − 𝛿 + 1) ∏𝑛𝑖=𝑗+1 Γ (𝛽𝑘𝑖 + 1)

(42)

1≤𝑘1 ,...,𝑘𝑛 ≤𝑚

Γ (∑𝑛𝑖=1 𝛽𝑘𝑖 − 𝛿 + 𝑛)

Γ (∑𝑛𝑖=1 𝛽𝑘𝑖 + 𝑛)

Γ (∑𝑙+1 𝑖=1 𝛽𝑘𝑖 + 𝑙 + 1)

𝐴 𝑛 𝜙 (𝑡) :=

Γ (𝛽𝑘𝑗 − 𝛿 + 1)

∏𝑛𝑖=1 Γ (𝛽𝑘𝑖 + 1)

Γ (∑𝑙𝑖=1 𝛽𝑘𝑖 + 𝑙) Γ (𝛽𝑘𝑙+1 + 1)

× (𝑡 − 𝑓𝑘𝑙+1 (𝜏))

𝛽𝑘𝑖 + 𝑛)

≤

Multiplying the above inequality by (∏𝑛𝑖=1 Γ(𝛽𝑘𝑖 + 1))/Γ(𝛽𝑘𝑗 + 1), we get that

𝑙

∑𝑙+1 𝑖=1 𝛽𝑘𝑖 +𝑙

(46)

and therefore Ψ𝑎 is strictly decreasing on (0, ∞). Thus, for all integers 1 ≤ 𝑘𝑖 , 𝑘𝑗 ≤ 𝑚, 1 ≤ 𝑖, 𝑗 ≤ 𝑛 and 𝛿 ≥ 0 satisfying that −1 < 𝛽𝑘𝑗 − 𝛿 < 0, we have Ψ𝑎 (𝛽𝑘𝑗 + 1) ≤ Ψ𝑎 (𝛽𝑘𝑗 − 𝛿 + 1), where 𝑎 := (∑𝑛𝑖=1 𝛽𝑘𝑖 ) + 𝑛 − 𝛽𝑘𝑗 − 1, implying that

× ∫ (1 − 𝑧)∑𝑖=1 𝛽𝑘𝑖 +𝑙−1 𝑧𝛽𝑘𝑙+1 𝑑𝑧

1

1 𝑎 𝑎 ∞ )∏ ), 𝑎 (1 + 𝑥 𝑛=1 (1 + (1/𝑛)) 𝑛+𝑥

(𝑡 − 𝑓𝑘𝑛 (𝑠))

(𝑠) 𝜙 (𝑠) 𝑑𝑠

∏𝑛𝑖=1 Γ (𝛽𝑘𝑖 + 1) ̃ℎ𝑘𝑖 (𝑡)

∑ 1≤𝑘1 ,...,𝑘𝑛 ≤𝑚

× {∫

∑𝑛𝑖=1 𝛽𝑘𝑖 +𝑛−1 󸀠 𝑓𝑘𝑛

𝑓𝑘−1 (𝑡−1) 𝑛

𝑏𝑘𝑛 (𝑡0 )

Γ (∑𝑛𝑖=1 𝛽𝑘𝑖 + 𝑛) 𝛽𝑘𝑛

(𝑡 − 𝑓𝑘𝑛 (𝑠))

𝜙 (𝑠)

∑𝑛−1 𝑖=1 𝛽𝑘𝑖 +𝑛−1 󸀠 𝑓𝑘𝑛

× (𝑡 − 𝑓𝑘𝑛 (𝑠)) +∫

𝑏𝑘𝑛 (𝑡)

𝑓𝑘−1 (𝑡−1) 𝑛

𝛽𝑘𝑛

(𝑡 − 𝑓𝑘𝑛 (𝑠))

×(𝑡 − 𝑓𝑘𝑛 (𝑠))

(𝑠) 𝑑𝑠

𝜙 (𝑠)

∑𝑛−1 𝑖=1 𝛽𝑘𝑖 +𝑛−1 󸀠 𝑓𝑘𝑛

(𝑠) 𝑑𝑠}

Journal of Applied Mathematics

7 𝑛

Then ∑∞ 𝑛=1 𝑐𝑛 (𝑡) is convergent on (𝑡0 + 1, 𝑇) by the ratio test ([25, pages 66-67]) because

𝑛 (Γ (𝛽 + 1)) ∗ ≤ 𝑚𝑛−1 ( max {̃ℎ𝑖 (𝑡)}) 𝑖=1,...,𝑚 Γ (𝑛 (𝛽∗ + 1)) 𝑚

× ∑ {∫

𝑓𝑖−1 (𝑡−1)

𝑏𝑖 (𝑡0 )

𝑖=1

(𝑡 − 𝑓𝑖 (𝑠)) 𝜙 (𝑠)

× (𝑡 − 𝑏𝑖 (𝑡)

+∫

lim

𝛽𝑖

𝑛 → +∞

(𝑛−1)𝑐(𝛽∗ +1) 󸀠 𝑓𝑖 (𝑠)) 𝑓𝑖

= lim 𝑚 max {̃ℎ𝑖 (𝑡)} 𝑛 → +∞

(𝑠) 𝑑𝑠

×

𝛽𝑖

𝑓𝑖−1 (𝑡−1)

𝑐𝑛+1 (𝑡) 𝑐𝑛 (𝑡)

(𝑡 − 𝑓𝑖 (𝑠)) 𝜙 (𝑠)

𝑖=1,...,𝑚

Γ (𝛽∗ + 1) Γ (𝑛 (𝛽∗ + 1)) 𝑐(𝛽 +1) (𝑡 − 𝑡0 ) ∗ Γ ((𝑛 + 1) (𝛽∗ + 1))

= lim 𝐵 (𝛽∗ + 1, 𝑛 (𝛽∗ + 1)) 𝑚

(53)

𝑛 → +∞

(𝑛−1)(𝛽∗ +1) 󸀠 𝑓𝑖

×(𝑡 − 𝑓𝑖 (𝑠))

(𝑠) 𝑑𝑠}

𝑐(𝛽 +1) × max {̃ℎ𝑖 (𝑡)} (𝑡 − 𝑡0 ) ∗ 𝑖=1,...,𝑚

𝑛

𝑛 (Γ (𝛽 + 1)) ∗ ≤ (𝑚 max {̃ℎ𝑖 (𝑡)}) 𝑖=1,...,𝑚 Γ (𝑛 (𝛽∗ + 1))

= lim

× {(𝑡 − 𝑡0 ) ×∫

𝑏𝑗 (𝑡0 )

+∫

𝑏𝑗 (𝑡)

where we note that 𝑚 max𝑖=1,...,𝑚 {̃ℎ𝑖 (𝑡)}(𝑡 − 𝑡0 )𝑐(𝛽∗ +1) is a continuous function on [𝑡0 , 𝑇) and Γ(𝛽∗ + 1)/(𝑛(𝛽∗ + 1))𝛽∗ +1 is a Stirling’s approximation of 𝐵(𝛽∗ + 1, 𝑛(𝛽∗ + 1)) as 𝑛 → +∞, as known in [26, pages 59]. It implies that ∑∞ 𝑛=1 𝐴 𝑛 𝜙(𝑡) is convergent on (𝑡0 + 1, 𝑇). The case of 𝑡 ∈ [𝑡0 , 𝑡0 + 1] can be proved similarly. Then, ∑∞ 𝑛=1 𝐴 𝑛 𝜙(𝑡) is convergent for 𝜙 ∈ 𝐿1loc,𝑤 [𝑡0 , 𝑇) and arbitrarily given 𝑡 ∈ [𝑡0 , 𝑇). Passing to the limit as 𝑛 → +∞ in (34), by (44) we get

𝛽𝑗

(𝑡 − 𝑓𝑗 (𝑠)) 𝜙 (𝑠) 𝑓𝑗󸀠 (𝑠) 𝑑𝑠 (𝑡 − 𝑓𝑗 (𝑠)) 𝜙 (𝑠) 𝑓𝑗󸀠 (𝑠) 𝑑𝑠} 𝑛

𝑛 (Γ (𝛽 + 1)) ∗ ≤ (𝑚 max {̃ℎ𝑖 (𝑡)}) 𝑖=1,...,𝑚 Γ (𝑛 (𝛽∗ + 1))

∞

𝑢 (𝑡) ≤ ∑ 𝐴 𝑛 𝑎 (𝑡) 𝑛=0

(𝑛−1)𝑐(𝛽∗ +1)

× (𝑡 − 𝑡0 ) 𝑏𝑗 (𝑡)

∏𝑛𝑖=1 Γ (𝛽𝑘𝑖 + 1) ̃ℎ𝑘𝑖 (𝑡) { = 𝑎 (𝑡) + ∑ { ∑ Γ (∑𝑛𝑖=1 𝛽𝑘𝑖 + 𝑛) 𝑛=1 1≤𝑘1 ,...,𝑘𝑛 ≤𝑚 { ∞

𝛽𝑗

×∫

𝑏𝑗 (𝑡0 )

(𝑡 − 𝑓𝑗 (𝑠)) 𝜙 (𝑠) 𝑓𝑗󸀠 (𝑠) 𝑑𝑠, (50)

×∫

𝑏𝑘𝑛 (𝑡)

𝑏𝑘𝑛 (𝑡0 )

where ∫

𝑏𝑗 (𝑡)

(𝑡 − 𝑓𝑗 (𝑠)) 𝜙 (𝑠) 𝑓𝑗󸀠 (𝑠) 𝑑𝑠

= max {∫ 𝑖=1,...,𝑚

(51)

𝑏𝑖 (𝑡)

𝛽

𝑏𝑖 (𝑡0 )

(𝑡 − 𝑓𝑘𝑛 (𝑠))

∑𝑛𝑖=1 𝛽𝑘𝑖 +𝑛−1

(54)

} ×𝑓𝑘󸀠𝑛 (𝑠) 𝑎 (𝑠) 𝑑𝑠} . }

𝛽𝑗

𝑏𝑗 (𝑡0 )

𝑚

𝑖=1,...,𝑚

𝛽𝑗

𝑓𝑗−1 (𝑡−1)

𝛽∗ +1

(𝑛 (𝛽∗ + 1))

𝑐(𝛽 +1) × max {̃ℎ𝑖 (𝑡)} (𝑡 − 𝑡0 ) ∗ = 0,

(𝑛−1)𝑐(𝛽∗ +1)

𝑓𝑗−1 (𝑡−1)

Γ (𝛽∗ + 1)

𝑛 → +∞

(𝑡 − 𝑓𝑖 (𝑠)) 𝑖 𝜙 (𝑠) 𝑓𝑖󸀠 (𝑠) 𝑑𝑠} < ∞.

Since 𝑘𝑖 ’s are chosen arbitrarily, by interchanging 𝑘𝑛 with 𝑘1 in (54), we obtain the estimate (17) and complete the proof of the theorem.

4. Application to Dependence

Let

Recently, increasing interest was given to fractional differential equations (see monographs [11, 23]). In this section we consider the Cauchy problem of the general fractional differential equation

𝑛

(Γ (𝛽∗ + 1)) 𝑐𝑛 (𝑡) := (𝑚 max {̃ℎ𝑖 (𝑡)}) 𝑖=1,...,𝑚 Γ (𝑛 (𝛽∗ + 1)) 𝑛

(𝑛−1)𝑐(𝛽∗ +1)

× (𝑡 − 𝑡0 ) ×∫

𝑏𝑗 (𝑡)

𝑏𝑗 (𝑡0 )

(52) 𝛽𝑗

(𝑡 − 𝑓𝑗 (𝑠)) 𝜙 (𝑠) 𝑓𝑗󸀠 (𝑠) 𝑑𝑠.

𝑚

𝛼

𝐷𝑡10 𝑥 (𝑡) = ∑𝐷𝑡0𝑖 𝑓𝑖 (𝑡, 𝑥 (𝑏𝑖 (𝑡))) ,

(55)

𝑥 (𝑡0 ) = 𝑎,

(56)

𝑖=1

8

Journal of Applied Mathematics 𝛼

where 0 ≤ 𝛼𝑖 < 1, 𝑡0 ≤ 𝑡 ≤ 𝑇 < +∞, 𝐷𝑡0𝑖 is defined as in the Introduction, particularly, 𝐷𝑡10 denotes 𝑑/𝑑𝑡 and 𝐷𝑡00 denotes identity map, 𝑏𝑖 : [𝑡0 , 𝑇] → [𝑡0 , 𝑇] and 𝑓𝑖 : [𝑡0 , 𝑇] × R → R, 𝑖 = 1, . . . , 𝑚. This system includes the system 𝐷01 𝑥 (𝑡)

+

𝑝𝐷0𝛼 𝑥 (𝑡)

+

𝐷00 𝑥 (𝑡)

= 𝑞 (𝑡) ,

+

𝑥 (0 ) = 𝑎,

Proposition 3. Suppose that each 𝑏𝑖 ∈ 𝐶1 ([𝑡0 , 𝑇], [𝑡0 , 𝑇]) is strictly increasing such that 𝑏𝑖 (𝑡) ≤ 𝑡 and each 𝑓𝑖 ∈ 𝐶([𝑡0 , 𝑇] × [𝑎−𝜌, 𝑎+𝜌], R) satisfies that |𝑓𝑖 (𝑡, 𝑥1 )−𝑓𝑖 (𝑡, 𝑥2 )| ≤ 𝐿 𝑖 |𝑥1 −𝑥2 |, 𝑖 = 1, . . . , 𝑚, where 𝑇 > 0, 𝜌 > 0, and 𝐿 𝑖 > 0 are constant. Then, the Cauchy problem (55)-(56) has a unique solution on [𝑡0 , 𝑡0 + ℎ] for a certain constant 0 < ℎ < 𝑇 and the solution depends continuously on 𝑎, 𝛼𝑖 ’s, 𝑓𝑖 ’s, and 𝑏𝑖 ’s for all 𝑡 ∈ [𝑡0 , 𝑡0 + ℎ]. Proof. The Cauchy problem (55)-(56) is equivalent to the integral equation 𝑥 (𝑡) 𝑚

𝑡

𝑖=1 𝑡0 𝑚

𝑀 :=

𝑚

𝑀𝜎1−𝛼𝑖 = 𝜌. 𝑖=1 Γ (2 − 𝛼𝑖 )

󵄨 󵄨󵄨 󵄨󵄨𝜑𝑙 (𝑡) − 𝜑𝑙−1 (𝑡)󵄨󵄨󵄨 ≤

𝑀∏𝑙𝑖=2 𝐿 𝑘𝑖

∑

1≤𝑘1 ,...,𝑘𝑙 ≤𝑚 Γ (1

Define a sequence {𝜑𝑛 (𝑡)} such that 𝜑0 (𝑡) ≡ 𝑎 and 1 𝑖=1 Γ (1 − 𝛼𝑖 )

𝜑𝑛 (𝑡) = 𝑎 + ∑

≤ (59)

𝑀ℎ ≤ 𝜌, 𝑖=1 Γ (2 − 𝛼𝑖 )

≤∑

1≤𝑘1 ,...,𝑘𝑙 ≤𝑚 Γ (1

∑𝑙𝑖=1 (1−𝛼𝑘𝑖 )

𝑀∏𝑙𝑖=2 𝐿 𝑘𝑖

∑

1≤𝑘1 ,...,𝑘𝑙 ≤𝑚 Γ (1

+ ∑𝑙𝑖=1 (1 − 𝛼𝑘𝑖 )) .

Combining with (59), we have 󵄨 󵄨󵄨 󵄨󵄨𝜑𝑙+1 (𝑡) − 𝜑𝑙 (𝑡)󵄨󵄨󵄨 𝑚

1

≤ ∑

𝑘𝑙+1 =1 Γ (1

− 𝛼𝑘𝑙+1 )

𝑡

× ∫ (𝑡 − 𝑠)−𝛼𝑘𝑙+1 (60)

(63)

+ ∑𝑙𝑖=1 (1 − 𝛼𝑘𝑖 ))

∑𝑙𝑖=1 (1−𝛼𝑘𝑖 )

𝑚

,

𝑀∏𝑙𝑖=2 𝐿 𝑘𝑖

∑

× (𝑠 − 𝑡0 )

𝑡0

1−𝛼𝑖

(1 − 𝛼𝑘𝑖 ))

× (𝑏𝑘𝑙+1 (𝑠) − 𝑡0 )

𝑚

𝑚

∑𝑙𝑖=1 (1−𝛼𝑘𝑖 )

(𝑡 − 𝑡0 )

󵄨󵄨 󵄨 󵄨󵄨𝜑𝑙 (𝑏𝑘𝑙+1 (𝑠)) − 𝜑𝑙−1 (𝑏𝑘𝑙+1 (𝑠))󵄨󵄨󵄨 󵄨 󵄨

(58)

𝑡 1 󵄨 󵄨 ≤∑ ∫ (𝑡 − 𝑠)−𝛼𝑖 󵄨󵄨󵄨𝑓𝑖 (𝑠, 𝜑𝑛 (𝑏𝑖 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑖=1 Γ (1 − 𝛼𝑖 ) 𝑡0

+

∑𝑙𝑖=1

for all 𝑡 ∈ [𝑡0 , 𝑡0 + ℎ] and all integers 𝑙. It can be checked easily for 𝑙 = 1. Suppose that it is true for some 𝑙. By changing 𝑡 into 𝑏𝑘𝑙+1 (𝑠) in (63), we get from 𝑏𝑘𝑙+1 (𝑠) ≤ 𝑠 that

≤

where 𝑡 ∈ [𝑡0 , 𝑡0 + ℎ]. We first claim that all 𝜑𝑛 (𝑡) are welldefined and continuous in [𝑡0 , 𝑡0 +ℎ], such that |𝜑𝑛 (𝑡)−𝑎| ≤ 𝜌. In fact, it is true for 𝑛 = 0. Suppose that it is true for some 𝑛. Then, 𝜑𝑛+1 is also well defined by (59) and continuous in [𝑡0 , 𝑡0 + ℎ]. Moreover, 󵄨󵄨 󵄨 󵄨󵄨𝜑𝑛+1 (𝑡) − 𝑎󵄨󵄨󵄨

(62)

The existence of 𝜎 is guaranteed by the intermediate theorem for continuous functions. Thus, the claim is proved by induction for all 𝑛. The convergence of the sequence {𝜑𝑛 (𝑡)} is equivalent to the convergence of the series 𝜑0 (𝑡) + ∑∞ 𝑙=1 (𝜑𝑙 (𝑡) − 𝜑𝑙−1 (𝑡)). We claim that

𝑡 1 ∫ (𝑡 − 𝑠)−𝛼𝑖 𝑓𝑖 (𝑠, 𝑥 (𝑏𝑖 (𝑠))) 𝑑𝑠. 𝑖=1 Γ (1 − 𝛼𝑖 ) 𝑡0

𝑛 = 1, 2, . . . ,

(61)

∑

=𝑎+∑

× ∫ (𝑡 − 𝑠)−𝛼𝑖 𝑓𝑖 (𝑠, 𝜑𝑛−1 (𝑏𝑖 (𝑠))) 𝑑𝑠,

󵄨 󵄨󵄨 󵄨󵄨𝑓𝑖 (𝑡, 𝑥)󵄨󵄨󵄨 ,

and 𝜎 is a positive constant such that

𝑠 1 𝑑 (∫ (𝑠 − 𝜏)−𝛼𝑖 𝑓𝑖 (𝜏, 𝑥 (𝑏𝑖 (𝜏))) 𝑑𝜏) Γ (1 − 𝛼𝑖 ) 𝑡0

𝑡

max

(𝑡,𝑥)∈[𝑡0 ,𝑇]×[𝑎−𝜌,𝑎+𝜌] 𝑖=1,...,𝑚

ℎ := min {𝑇 − 𝑡0 , 𝜎} ,

(57)

as a special case, which was considered in [27] and corresponds to the Basset problem when 𝛼 = 1/2, a classical problem in fluid dynamics concerning the unsteady motion of a particle accelerating in a viscous fluid under the action of gravity (see [28]). We will give continuous dependence for solutions of (55) associated with the initial condition (56) on the derivative orders 𝛼𝑖 ’s, the initial data 𝑎, and the functions 𝑓𝑖 and 𝑏𝑖 .

= 𝑎 + ∑∫

where

𝑡0

󵄨 × 󵄨󵄨󵄨󵄨𝑓𝑘𝑙+1 (𝑠, 𝜑𝑙 (𝑏𝑘𝑙+1 (𝑠)))

󵄨 −𝑓𝑘𝑙+1 (𝑠, 𝜑𝑙−1 (𝑏𝑘𝑙+1 (𝑠)))󵄨󵄨󵄨󵄨 𝑑𝑠

(64)

Journal of Applied Mathematics 𝑚

9

𝐿 𝑘𝑙+1

≤ ∑

Next, we prove the continuous dependence of the solution. Consider the Cauchy problem

𝑘𝑙+1 =1 Γ (1 − 𝛼𝑘𝑙+1 )

𝑚

𝑡 󵄨 󵄨 × ∫ (𝑡 − 𝑠)−𝛼𝑘𝑙+1 󵄨󵄨󵄨󵄨𝜑𝑙 (𝑏𝑘𝑙+1 (𝑠)) − 𝜑𝑙−1 (𝑏𝑘𝑙+1 (𝑠))󵄨󵄨󵄨󵄨 𝑑𝑠 𝑡

1≤𝑘1 ,...,𝑘𝑙+1 ≤𝑚 Γ (1

+ ∑𝑙𝑖=1 (1 − 𝛼𝑘𝑖 )) Γ (1 − 𝛼𝑘𝑙+1 )

𝑡

∑𝑙𝑖=1 (1−𝛼𝑘𝑖 )

𝑡0

(68)

where 0 ≤ 𝛼̃𝑖 < 1, 𝑡0 ≤ 𝑡 ≤ 𝑇 < +∞, ̃𝑏𝑖 ∈ 𝐶1 ([𝑡0 , 𝑇], [𝑡0 , 𝑇]) is strictly increasing such that 𝑏𝑖 (𝑡) ≤ 𝑡, and 𝑓̃𝑖 ∈ 𝐶([𝑡0 , 𝑇] × [̃ 𝑎 − 𝜌, 𝑎̃ + 𝜌], R) is Lipschitzian in the second variable with ̃ 𝑖 , 𝑖 = 1, . . . , 𝑚. As above, we similarly the Lipschitz constant 𝐿 obtain a solution 𝜓 of the Cauchy problem (67)-(68). Similar to (58), we have

𝑑𝑠

𝑙+1

∑ 1≤𝑘1 ,...,𝑘𝑙+1 ≤𝑚

(𝑀 (∏𝐿 𝑘𝑖 ) 𝑖=2

𝑚

𝑡 1 ∫ (𝑡 − 𝑠)−𝛼𝑖 𝑓𝑖 (𝑠, 𝜑 (𝑏𝑖 (𝑠))) 𝑑𝑠, 𝑖=1 Γ (1 − 𝛼𝑖 ) 𝑡0

𝜑 (𝑡) = 𝑎 + ∑

𝑙

× 𝐵 (1 + ∑ (1 − 𝛼𝑘𝑖 ) , 1 − 𝛼𝑘𝑙+1 ) 𝑖=1

𝑚

𝑡 1 ∫ (𝑡 − 𝑠)−̃𝛼𝑖 𝑓̃𝑖 (𝑠, 𝜓 (̃𝑏𝑖 (𝑠))) 𝑑𝑠. ̃𝑖 ) 𝑡0 𝑖=1 Γ (1 − 𝛼 (69)

𝜓 (𝑡) = 𝑎̃ + ∑

𝑙

× (Γ (1 + ∑ (1 − 𝛼𝑘𝑖 )) 𝑖=1

It follows that

−1

× Γ (1 − 𝛼𝑘𝑙+1 ) ) )

󵄨 󵄨󵄨 󵄨󵄨𝜑 (𝑡) − 𝜓 (𝑡)󵄨󵄨󵄨

𝑚 󵄨󵄨 1 󵄨 ≤ |𝑎 − 𝑎̃| + ∑ 󵄨󵄨󵄨󵄨 𝑖=1 󵄨󵄨 Γ (1 − 𝛼𝑖 )

∑𝑙+1 𝑖=1 (1−𝛼𝑘𝑖 )

× (𝑡 − 𝑡0 ) =

𝑥 (𝑡0 ) = 𝑎̃,

𝑀∏𝑙+1 𝑖=2 𝐿 𝑘𝑖

∑

× ∫ (𝑡 − 𝑠)−𝛼𝑘𝑙+1 (𝑠 − 𝑡0 ) =

(67)

𝑖=1

0

≤

𝛼̃ 𝐷𝑡10 𝑥 (𝑡) = ∑𝐷𝑡0𝑖 𝑓̃𝑖 (𝑡, 𝑥 (̃𝑏𝑖 (𝑡))) ,

𝑀∏𝑙+1 𝑖=2 𝐿 𝑘𝑖

∑

1≤𝑘1 ,...,𝑘𝑙+1 ≤𝑚 Γ (1

+

∑𝑙+1 𝑖=1

∑𝑙+1 𝑖=1 (1−𝛼𝑘𝑖 )

(1 − 𝛼𝑘𝑖 ))

(𝑡 − 𝑡0 )

𝑡

.

× ∫ (𝑡 − 𝑠)−𝛼𝑖 𝑓𝑖 (𝑠, 𝜑 (𝑏𝑖 (𝑠))) 𝑑𝑠 𝑡0

(65) −

󵄨󵄨 𝑡 󵄨 × ∫ (𝑡 − 𝑠)−̃𝛼𝑖 𝑓̃𝑖 (𝑠, 𝜓 (̃𝑏𝑖 (𝑠))) 𝑑𝑠󵄨󵄨󵄨󵄨 󵄨󵄨 𝑡0

Thus, (63) holds for 𝑙 + 1 and the claim is proved for all 𝑙 by induction. From (63), for 𝑡 ≤ 𝑡0 + ℎ we have

󵄨 󵄨󵄨 󵄨󵄨𝜑𝑙 (𝑡) − 𝜑𝑙−1 (𝑡)󵄨󵄨󵄨 ≤

𝑚 󵄨󵄨 1 󵄨 ≤ Ξ (𝑡) + ∑ {󵄨󵄨󵄨󵄨 Γ (1 − 𝛼𝑖 ) 󵄨 𝑖=1 󵄨

𝑙

∑

𝑀ℎ∑𝑖=1 (1−𝛼𝑘𝑖 ) ∏𝑙𝑖=2 𝐿 𝑘𝑖

1≤𝑘1 ,...,𝑘𝑙 ≤𝑚 Γ (1

+ ∑𝑙𝑖=1 (1 − 𝛼𝑘𝑖 ))

1 Γ (1 − 𝛼̃𝑖 )

𝑡

× ∫ (𝑡 − 𝑠)−𝛼𝑖 𝑓𝑖 (𝑠, 𝜑 (𝑏𝑖 (𝑠))) 𝑑𝑠

(66)

𝑡0

𝑚𝑙 𝑀ℎ𝑙(1−𝛼∗ ) 𝐿𝑙−1 , ≤ Γ (1 + 𝑙 (1 − 𝛼∗ )) where 𝐿 := max1≤𝑖≤𝑚 𝐿 𝑖 , 𝛼∗ := max1≤𝑖≤𝑚 𝛼𝑖 , and 𝛼∗ := min1≤𝑖≤𝑚 𝛼𝑖 . By the ratio test ([25, pp. 66-67]), we get that 𝑙 𝑙(1−𝛼∗ ) 𝑙−1 𝐿 /Γ(1 + 𝑙(1 − 𝛼∗ )) is convergent, implying ∑∞ 𝑙=1 𝑚 𝑀ℎ that the series 𝜑0 (𝑡) + ∑∞ 𝑙=1 (𝜑𝑙 (𝑡) − 𝜑𝑙−1 (𝑡)) and therefore the sequence {𝜑𝑛 (𝑡)} are uniformly convergent in [𝑡0 , 𝑡0 + ℎ]. Let 𝜑(𝑡) := lim𝑛 → ∞ 𝜑𝑛 (𝑡), which is well-defined and continuous in [𝑡0 , 𝑡0 + ℎ] such that |𝜑(𝑡) − 𝑎| ≤ 𝜌. One can check that 𝜑 is a continuous solution of the Cauchy problem (55)-(56) in [𝑡0 , 𝑡0 + ℎ].

−

1 Γ (1 − 𝛼𝑖 )

󵄨󵄨 𝑡 󵄨 × ∫ (𝑡 − 𝑠)−𝛼𝑖 𝑓𝑖 (𝑠, 𝜓 (𝑏𝑖 (𝑠))) 𝑑𝑠󵄨󵄨󵄨󵄨} 󵄨󵄨 𝑡0 𝑚

𝐿𝑖 𝑖=1 Γ (1 − 𝛼𝑖 )

≤ Ξ (𝑡) + ∑ 𝑏𝑖 (𝑡)

×∫

𝑏𝑖 (𝑡0 )

(𝑡 − 𝑏𝑖−1 (𝑠))

−𝛼𝑖

󵄨 󵄨󵄨 −1 󵄨󵄨𝜑 (𝑠) − 𝜓 (𝑠)󵄨󵄨󵄨 𝑑 (𝑏𝑖 (𝑠)) , (70)

10

Journal of Applied Mathematics ̃ contains |(𝑡 − 𝑡0 )1−𝛼𝑖 /Γ(2 − 𝛼𝑖 ) − Note from (71) that Ξ(𝑡) 1−̃ 𝛼𝑖 (𝑡 − 𝑡0 ) /Γ(2 − 𝛼̃𝑖 )| and |𝜓(𝑏𝑖 (𝑠)) − 𝜓(̃𝑏𝑖 (𝑠))|, which will be estimated with ‖𝑏𝑖 − ̃𝑏𝑖 ‖ := max𝑡0 ≤𝜏≤𝑡0 +ℎ |𝑏𝑖 (𝜏) − ̃𝑏𝑖 (𝜏)| and |𝛼𝑖 − 𝛼̃𝑖 |, respectively. In (71) we have ‖𝑓𝑖 ‖ ≤ 𝑀 in (61). Since (𝑡 − 𝑡0 )1−𝛼𝑖 and Γ(2 − 𝛼𝑖 ) are both 𝐶1 functions in 𝛼𝑖 ∈ [0, 1], and Γ(2−𝛼𝑖 ) > 0 for all 𝛼𝑖 ∈ [0, 1], we see that (𝑡 − 𝑡0 )1−𝛼𝑖 /Γ(2− 𝛼𝑖 ) is 𝐶1 in 𝛼𝑖 ∈ [0, 1]. By the Lagrange mean value theorem (see [25, page 108]), function (𝑡 − 𝑡0 )1−𝛼𝑖 /Γ(2 − 𝛼𝑖 ) satisfies the Lipschitz condition uniformly with respect to 𝑡 ∈ [𝑡0 , 𝑡0 + ℎ]; that is, there exists a constant 𝑁 > 0 such that

where Ξ (𝑡) := |𝑎 − 𝑎̃| 𝑚 󵄨󵄨 1 󵄨 + ∑ {󵄨󵄨󵄨󵄨 𝑖=1 󵄨󵄨 Γ (1 − 𝛼𝑖 ) 𝑡

× ∫ (𝑡 − 𝑠)−𝛼𝑖 𝑓𝑖 (𝑠, 𝜓 (𝑏𝑖 (𝑠))) 𝑑𝑠 𝑡0

󵄨󵄨 1−𝛼 1−̃ 𝛼 󵄨 󵄨󵄨 (𝑡 − 𝑡0 ) 𝑖 (𝑡 − 𝑡0 ) 𝑖 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨 Γ (2 − 𝛼 ) − Γ (2 − 𝛼̃ ) 󵄨󵄨󵄨 ≤ 𝑁 󵄨󵄨𝛼𝑖 − 𝛼̃𝑖 󵄨󵄨 , 󵄨󵄨 𝑖 𝑖 󵄨󵄨

1 − Γ (1 − 𝛼̃𝑖 ) 󵄨󵄨 󵄨 × ∫ (𝑡 − 𝑠)−̃𝛼𝑖 𝑓𝑖 (𝑠, 𝜓 (𝑏𝑖 (𝑠))) 𝑑𝑠󵄨󵄨󵄨󵄨 󵄨󵄨 𝑡0 󵄨󵄨 𝑡 1 󵄨 + 󵄨󵄨󵄨󵄨 ∫ (𝑡 − 𝑠)−̃𝛼𝑖 𝑓𝑖 (𝑠, 𝜓 (𝑏𝑖 (𝑠))) 𝑑𝑠 󵄨󵄨 Γ (1 − 𝛼̃𝑖 ) 𝑡0

∀𝛼𝑖 , 𝛼̃𝑖 ∈ [0, 1] . (73)

𝑡

On the other hand, from the equivalent integral equation of (67) 𝜓 (𝑡) = 𝜓 (𝑡0 ) 𝑚

+∑

󵄨󵄨 1 󵄨 − ∫ (𝑡 − 𝑠)−̃𝛼𝑖 𝑓̃𝑖 (𝑠, 𝜓 (𝑏𝑖 (𝑠))) 𝑑𝑠󵄨󵄨󵄨󵄨 Γ (1 − 𝛼̃𝑖 ) 𝑡0 󵄨󵄨 𝑡

󵄨󵄨 𝑡 1 󵄨 + 󵄨󵄨󵄨󵄨 ∫ (𝑡 − 𝑠)−̃𝛼𝑖 𝑓̃𝑖 (𝑠, 𝜓 (𝑏𝑖 (𝑠))) 𝑑𝑠 󵄨󵄨 Γ (1 − 𝛼̃𝑖 ) 𝑡0

𝑗=1 Γ (1

(74)

󵄨 󵄨󵄨 󵄨󵄨𝜓 (𝑡) − 𝜓 (𝑡0 )󵄨󵄨󵄨 ≤ ∑

+

𝑡 ̃𝑖 𝐿 ∫ (𝑡 − 𝑠)−̃𝛼𝑖 Γ (1 − 𝛼̃𝑖 ) 𝑡0

𝑚

‖𝑓𝑖 ‖ := max(𝑡,𝑥)∈[𝑡0 ,𝑇]×[𝑎−𝜌,𝑎+𝜌] |𝑓𝑖 (𝑡, 𝑥)|, and ‖𝑓𝑖 − 𝑓̃𝑖 ‖ := max(𝑡,𝑥)∈[𝑡0 ,𝑇]×[𝑎−𝜌,𝑎+𝜌] |𝑓𝑖 (𝑡, 𝑥) − 𝑓̃𝑖 (𝑡, 𝑥)|. Thus, applying the inequality given in Corollary 2 to (70), we obtain that 𝑚

̃ := max𝑡 ≤𝜏≤𝑡 Ξ(𝜏). where Ξ(𝑡) 0

(76)

̃ 𝑀

󵄩󵄩 󵄩1−̃𝛼 󵄩󵄩𝑏𝑖 − ̃𝑏𝑖 󵄩󵄩󵄩 𝑗 . 󵄩 󵄩 ̃𝑗 ) 𝑗=1 Γ (2 − 𝛼

≤∑

(71)

∀𝑡 ∈ [𝑡0 , 𝑡0 + ℎ] ,

󵄨 󵄨󵄨 ̃ 󵄨󵄨𝑓𝑗 (𝑡, 𝑥)󵄨󵄨󵄨 . 󵄨

max

(𝑡,𝑥)∈[𝑡0 ,𝑇]×[𝑎−𝜌,𝑎+𝜌] 󵄨 𝑗=1,...,𝑚

̃ 𝑀 󵄨1−̃𝛼 󵄨󵄨 󵄨󵄨 󵄨 𝑚 󵄨󵄨𝑏𝑖 (𝑠) − ̃𝑏𝑖 (𝑠)󵄨󵄨󵄨 𝑗 󵄨󵄨𝜓 (𝑏𝑖 (𝑠)) − 𝜓 (̃𝑏𝑖 (𝑠))󵄨󵄨󵄨 ≤ ∑ 󵄨 󵄨 󵄨 󵄨 ̃𝑗 ) 𝑗=1 Γ (2 − 𝛼

󵄨 󵄨 × 󵄨󵄨󵄨󵄨𝜓 (𝑏𝑖 (𝑠)) − 𝜓 (̃𝑏𝑖 (𝑠))󵄨󵄨󵄨󵄨 𝑑𝑠} ,

𝑖=1

(75)

Choosing 𝑡 = 𝑏𝑖 (𝑠) and 𝑡0 = ̃𝑏𝑖 (𝑠) in (74), we obtain

󵄩 󵄩󵄩 󵄩󵄩𝑓𝑖 − 𝑓̃𝑖 󵄩󵄩󵄩 󵄩 󵄩

1−𝛼 󵄨 ̃ 󵄨󵄨 (𝑡) ∏𝐸1−𝛼𝑖 (𝐿 𝑖 (𝑡 − 𝑡0 ) 𝑖 ) , 󵄨󵄨𝜑 (𝑡) − 𝜓 (𝑡)󵄨󵄨󵄨 ≤ Ξ

󵄨󵄨 󵄨1−̃𝛼 󵄨󵄨𝑡 − 𝑡0 󵄨󵄨󵄨 𝑗 ,

where ̃ := 𝑀

󵄨󵄨 1−𝛼 1−̃ 𝛼 󵄨 󵄨󵄨 (𝑡 − 𝑡0 ) 𝑖 (𝑡 − 𝑡0 ) 𝑖 󵄨󵄨󵄨 󵄩 󵄩 󵄨󵄨 󵄩󵄩𝑓 󵄩󵄩 + ∑ {󵄨󵄨󵄨 − 󵄨 Γ (2 − 𝛼̃𝑖 ) 󵄨󵄨󵄨󵄨 󵄩 𝑖 󵄩 𝑖=1 󵄨󵄨 Γ (2 − 𝛼𝑖 )

̃ 𝑀

̃𝑗 ) 𝑗=1 Γ (2 − 𝛼

𝑚

(𝑡 − 𝑡0 ) Γ (2 − 𝛼̃𝑖 )

𝑡0

𝑚

≤ |𝑎 − 𝑎̃|

+

− 𝛼̃𝑗 )

∫ (𝑡 − 𝑠)−̃𝛼𝑗 𝑓̃𝑗 (𝑠, 𝜓 (̃𝑏𝑗 (𝑠))) 𝑑𝑠,

we get

󵄨󵄨 𝑡 1 󵄨 − ∫ (𝑡 − 𝑠)−̃𝛼𝑖 𝑓̃𝑖 (𝑠, 𝜓 (̃𝑏𝑖 (𝑠))) 𝑑𝑠󵄨󵄨󵄨󵄨} Γ (1 − 𝛼̃𝑖 ) 𝑡0 󵄨󵄨

1−̃ 𝛼𝑖

𝑡

1

(72)

(77) Summarily, from (71), (73) and (75) we get Ξ (𝑡) ≤ |𝑎 − 𝑎̃| 1−̃ 𝛼

𝑚

𝑖 󵄨 (𝑡 − 𝑡0 ) 󵄨 + ∑ {𝑀𝑁 󵄨󵄨󵄨𝛼𝑖 − 𝛼̃𝑖 󵄨󵄨󵄨 + Γ (2 − 𝛼̃𝑖 ) 𝑖=1

+

󵄩 󵄩󵄩 󵄩󵄩𝑓𝑖 − 𝑓̃𝑖 󵄩󵄩󵄩 󵄩 󵄩

𝑚 𝑡 ̃𝑖 ̃ 𝐿 𝑀 ∫ (𝑡 − 𝑠)−̃𝛼𝑖 ∑ ̃ Γ (1 − 𝛼𝑖 ) 𝑡0 ̃𝑗 ) 𝑗=1 Γ (2 − 𝛼

󵄩1−̃𝛼𝑗 󵄩 ×󵄩󵄩󵄩󵄩𝑏𝑖 − ̃𝑏𝑖 󵄩󵄩󵄩󵄩 𝑑𝑠}

Journal of Applied Mathematics

11

𝑚 { ℎ1−̃𝛼𝑖 󵄩󵄩 󵄩 󵄨 󵄨 󵄩󵄩𝑓𝑖 − 𝑓̃𝑖 󵄩󵄩󵄩 ≤ |𝑎 − 𝑎̃| + ∑ {𝑀𝑁 󵄨󵄨󵄨𝛼𝑖 − 𝛼̃𝑖 󵄨󵄨󵄨 + 󵄩 󵄩 ̃ Γ (2 − 𝛼 ) 𝑖 𝑖=1 {

󵄩 󵄩1−̃𝛼𝑗 ̃𝐿 ̃ 𝑖 ℎ1−̃𝛼𝑖 𝑚 󵄩󵄩󵄩󵄩𝑏𝑖 − ̃𝑏𝑖 󵄩󵄩󵄩󵄩 } 𝑀 + . ∑ Γ (2 − 𝛼̃𝑖 ) 𝑗=1 Γ (2 − 𝛼̃𝑗 ) } } (78) It follows from (72) that 𝑚

󵄩 󵄩󵄩 1−𝛼 󵄩󵄩𝜑 − 𝜓󵄩󵄩󵄩 ≤ ∏𝐸1−𝛼𝑖 (𝐿 𝑖 ℎ 𝑖 ) |𝑎 − 𝑎̃| 𝑖=1

𝑚

𝑚

𝑖=1

𝑖=1

󵄨 󵄨 + 𝑀𝑁∏𝐸1−𝛼𝑖 (𝐿 𝑖 ℎ1−𝛼𝑖 ) ∑ 󵄨󵄨󵄨𝛼𝑖 − 𝛼̃𝑖 󵄨󵄨󵄨 𝑚

𝑚

ℎ1−̃𝛼𝑖 󵄩󵄩 󵄩 󵄩󵄩𝑓𝑖 − 𝑓̃𝑖 󵄩󵄩󵄩 󵄩 󵄩 ̃ Γ (2 − 𝛼 ) 𝑖 𝑖=1

+ ∏𝐸1−𝛼𝑖 (𝐿 𝑖 ℎ1−𝛼𝑖 ) ∑ 𝑖=1

𝑚

̃∏𝐸1−𝛼 (𝐿 𝑖 ℎ1−𝛼𝑖 ) +𝑀 𝑖 𝑖=1

̃ 𝑖 ℎ1−̃𝛼𝑖 𝐿

󵄩1−̃𝛼 󵄩󵄩 󵄩󵄩𝑏𝑖 − ̃𝑏𝑖 󵄩󵄩󵄩 𝑗 ; 󵄩 󵄩 ̃𝑖 ) Γ (2 − 𝛼̃𝑗 ) 1≤𝑖,𝑗≤𝑚 Γ (2 − 𝛼

× ∑

(79) that is, the solution 𝜑 of the Cauchy problem (55)-(56) depends continuously on 𝑎, 𝛼𝑖 , 𝑓𝑖 and 𝑏𝑖 . As a complement, we note from (79) that 𝜑 ≡ 𝜓 in [𝑡0 , 𝑡0 + ℎ] if 𝑎 = 𝑎̃, 𝛼𝑖 = 𝛼̃𝑖 , 𝑓𝑖 ≡ 𝑓̃𝑖 , and 𝑏𝑖 ≡ ̃𝑏𝑖 . This implies the uniqueness of the solution.

Conflict of Interests The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgment The author is very grateful to Professor Weinian Zhang for his valuable suggestions.

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