Existence and Numerical Simulation of Solutions for Fractional

Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2013, Article ID 268347, 15 pages http://dx.doi.org/10.1155/2013/268347

Research Article Existence and Numerical Simulation of Solutions for Fractional Equations Involving Two Fractional Orders with Nonlocal Boundary Conditions Jing Zhao, Peifen Lu, and Yiliang Liu Guangxi Key Laboratory of Hybrid Computation and IC Design Analysis and College of Sciences, Guangxi University for Nationalities, Nanning, Guangxi 530006, China Correspondence should be addressed to Yiliang Liu; [email protected] Received 26 March 2013; Accepted 6 June 2013 Academic Editor: Naseer Shahzad Copyright © 2013 Jing Zhao et al. 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 study a boundary value problem for fractional equations involving two fractional orders. By means of a fixed point theorem, we establish sufficient conditions for the existence and uniqueness of solutions for the fractional equations. In addition, we describe the dynamic behaviors of the fractional Langevin equation by using the 𝐺2 algorithm.

1. Introduction Fractional differential equations arise in many engineering and scientific disciplines as the mathematical modeling of systems and processes in various fields, such as physics, mechanics, chemical technology, population dynamics, biotechnology, and economics (see, e.g., [1–7]). As one of the important topics in the research on differential equations, the boundary value problem has attained a great deal of attention from many researchers (see [8–18]) and the references therein. As pointed out in [19], the nonlocal boundary condition can be more useful than the standard condition to describe some physical phenomena. There are several noteworthy papers (see [20–22]) dealing with nonlocal boundary value problems of fractional differential equations. In [19], Benchohra et al. investigated the existence and uniqueness of the solutions for the differential equations with nonlocal conditions: 𝑐

𝐷𝛼 𝑢 (𝑡) + 𝑓 (𝑡, 𝑢 (𝑡)) = 0, 𝑢 (0) = 𝑔 (𝑢) ,

1 < 𝛼 ≤ 2, 0 < 𝑡 < 𝑇, 𝑢 (𝑇) = 𝑢𝑇 ,

(1)

where 𝑐 𝐷𝛼 denotes Caputo’s fractional derivative of order 𝛼 with the lower limit zero.

In [22], Zhong and Lin studied the existence and uniqueness of solutions in the nonlocal and multiple-point boundary value problem for fractional differential equation: 𝑐

𝐷𝑞 𝑢 (𝑡) + 𝑓 (𝑡, 𝑢 (𝑡)) = 0,

𝑢 (0) = 𝑢0 + 𝑔 (𝑢) , 𝑐

0 < 𝑡 < 1, 1 < 𝑞 ≤ 2, 𝑚−2

𝑢󸀠 (1) = 𝑢1 + ∑ 𝑏𝑖 𝑢󸀠 (𝜉𝑖 ) ,

(2)

𝑖=1

𝑞

where 𝐷 denotes Caputo’s fractional derivative of order 𝑞 with the lower limit zero. In this paper we will study the fractional Langevin equation where the fractional derivative is in Caputo sense. In 1908 the French physicist Langevin introduced the concept of the equation of motion with a random variable, which reads as 𝑚

𝑑𝑥 (𝑡) 𝑑2 𝑥 (𝑡) = −𝛾 + 𝐹 (𝑡) + 𝜉 (𝑡) , 2 𝑑𝑡 𝑑𝑡

(3)

where 𝑚 is the mass of the particle, 𝛾 is the coefficient of viscosity, 𝐹(𝑥) is the external force, and 𝜉(𝑡) is the random force. The Langevin equation is always regarded as the first stochastic differential equation. Langevin equation has been widely used to describe the evolution of physical phenomena in fluctuating environm-

2

Journal of Applied Mathematics

ents [23–25]. For instance, Brownian motion is well described by the Langevin equation when the random fluctuation force is assumed to be white noise. In case the random fluctuation force is not white noise, the motion of the particle is described by the generalized Langevin equation [26]. For systems in complex media, ordinary Langevin equation does not provide the correct description of the dynamics. Various generalizations of Langevin equations have been proposed to describe dynamical processes in a fractal medium. One such generalization is the generalized Langevin equation [27–32] which incorporates the fractal and memory properties with a dissipative memory kernel into the Langevin equation. Fractional order models are more accurate than integerorder models as fractional order models allow more degrees of freedom. The presence of memory term in such models not only takes into account the history of the process involved but also carries its impact to present and future development of the process. Fractional differential equations are also regarded as an alternative model to nonlinear differential equations [33]. In consequence, the subject of fractional differential equations is gaining much importance and attention. For some recent work on fractional differential equations, see [1, 34–46]. In [47], Ahmad et al. studied nonlinear Langevin equation involving two fractional orders in different intervals: 𝑐

𝑥 (0) = 0,

1 < 𝛼 ≤ 2,

𝑥 (𝜂) = 0,

𝑥 (1) = 0,

0 < 𝛽 ≤ 1,

(4)

0 < 𝜂 < 1,

𝛼

where 𝑐 𝐷 and 𝑐 𝐷𝛽 denote Caputo’s fractional derivative of order 𝛼 and 𝛽 with the lower limit zero. In [48], A. Chen and Y. Chen studied existence of solutions to nonlinear Langevin equation involving two fractional orders with boundary value conditions: 𝑐

𝛼

𝐷𝛽 ( 𝑐 𝐷 + 𝜆) 𝑢 (𝑡) = 𝑓 (𝑡, 𝑢 (𝑡) , 𝑢󸀠 (𝑡)) , 0 < 𝑡 < 𝑇,

𝑢 (0) = −𝑢 (𝑇) , 𝛼

0 < 𝛼 ≤ 1,

1 < 𝛽 ≤ 2,

𝑐

𝛼

𝐷𝛽 ( 𝑐 𝐷 + 𝜆) 𝑢 (𝑡) = 𝑓 (𝑡, 𝑢 (𝑡) , 𝑢󸀠 (𝑡)) , 0 < 𝑡 < 1,

1 < 𝛼 ≤ 2,

0 < 𝛽 ≤ 1,

𝛼1 𝑢 (0) + 𝛽1 𝑢 (1) = 𝑔1 (𝑢) ,

(6)

𝛼2 𝑢󸀠 (0) + 𝛽2 𝑢󸀠 (1) = 𝑔2 (𝑢) , 𝑢 (0) = 𝜂𝑢󸀠 (0) ,

𝜂 ≠ 0,

where 𝑐 𝐷𝛼 and 𝑐 𝐷𝛽 denote Caputo’s fractional derivative of order 𝛼 and 𝛽 with the lower limit zero, 𝑓 : [0, 1] × 𝑅2 → 𝑅 is a given continuous function and 𝜆 is a real number, and 𝑔1 , 𝑔2 : 𝐶([0, 1], 𝑅) → 𝑅 are two continuous functions, 𝛼𝛽2 [(𝛼1 + 𝛽1 )𝜂 + 𝛽1 ] ≠ 𝛽1 (𝛼2 + 𝛽2 ). Evidently, problem (6) not only includes boundary value problems mentioned above but also extends them to a much wider case. The organization of this paper is as follows. In Section 2, we will give some lemmas which are essential to prove our main results. In Section 3, main results are given. In Section 4, we will give the numerical simulation for the fractional Langevin equation.

2. Preliminaries

𝛼

𝐷𝛽 ( 𝑐 𝐷 + 𝜆) 𝑥 (𝑡) = 𝑓 (𝑡, 𝑥 (𝑡) , 𝑥󸀠 (𝑡)) , 0 < 𝑡 < 1,

fixed point theorem and use 𝐺2 algorithm to describe the dynamic behaviors for the following problem:

(5)

𝑢󸀠 (0) = 𝑢󸀠 (𝑇) = 0,

where 𝑐 𝐷 and 𝑐 𝐷𝛽 denote Caputo’s fractional derivative of order 𝛼 and 𝛽 with the lower limit zero. The fractional calculus has been studied for more than three hundred years. In recent few decades, the fractional calculus has been widely used in many fields such as chaotic dynamics, viscoelasticity, acoustics, and physical chemistry. In [49], Guo studied the numerical solution of fractional partial differential equations. In [50], Guo studied the numerical simulation of the fractional Langevin equation. As far as we know, there are no papers discussing the existence and numerical simulation of solutions for fractional equations involving two fractional orders with nonlocal boundary conditions. Motivated by the works mentioned above, in this paper, we establish the existence and uniqueness of solutions by the

In this section, we introduce notations, definitions, and preliminary facts. Throughout this paper, set C = 𝐶([0, 1], 𝑅) denotes the Banach space of all continuous functions from [0, 1] → 𝑅 with the norm ‖𝑥‖C = sup𝑡∈[0,1] |𝑥(𝑡)|. We also introduce the Banach space 𝑢 ∈ 𝐶1 ([0, 1], 𝑅) endowed with the norm defined by ‖𝑢‖𝐶1 = max{sup𝑡∈[0,1] |𝑢(𝑡)|, sup𝑡∈[0,1] |𝑢󸀠 (𝑡)|}. For the convenience of the readers, let us recall the following useful definitions and fundamental facts of fractional calculus theory. Definition 1 (see [1, 6]). The Riemann-Liouville derivative of order 𝛾 with the lower limit zero for a function 𝑓 : [0, ∞) → 𝑅 can be written as 𝑓 (𝑠) 𝑑𝑛 𝑡 1 𝐿 𝛾 𝐷 𝑓 (𝑡) = 𝑑𝑠, ∫ Γ (𝑛 − 𝛾) 𝑑𝑡𝑛 0 (𝑡 − 𝑠)𝛾+1−𝑛 (7) 𝑡 > 0,

𝑛 − 1 < 𝛾 < 𝑛.

Definition 2 (see [1, 6]). The fractional (arbitrary) order integral of the function 𝑓 : [0, ∞) → 𝑅 of order 𝑝 > 0 is defined by 𝐼𝑝 𝑓 (𝑥) =

𝑥 1 ∫ (𝑥 − 𝑠)𝑝−1 𝑓 (𝑠) 𝑑𝑠. Γ (𝑝) 0

(8)

Definition 3 (see [1]). Let 𝛼 ≥ 0, 𝑛 = [𝛼] + 1. If 𝑓 ∈ 𝐴𝐶𝑛 [𝑎, 𝑏], the Caputo fractional derivative of order 𝛼 of 𝑓 is defined by 𝑐

𝐷𝛼 𝑓 (𝑡) =

𝑡 𝑓(𝑛) (𝑠) 1 𝑑𝑠, ∫ Γ (𝑛 − 𝛼) 𝑎 (𝑡 − 𝑠)𝛼+1−𝑛

𝑡 > 0,

𝑛 − 1 < 𝛼 < 𝑛.

(9)

Journal of Applied Mathematics

3

Definition 4 (see [6]). Let 𝑝 ∈ (𝑛 − 1, 𝑛], 𝑛 ∈ 𝑁 and the the Gr¨unwald-Letnikov fractional derivative of order 𝑝 of 𝑓 defined by 𝐷𝑝 𝑓 (𝑡) = lim

ℎ→0 𝑛ℎ=𝑡−𝑎

1 𝑛 𝑝 ∑(−1)𝑟 ( ) 𝑓 (𝑡 − 𝑟ℎ) , 𝑟 ℎ𝑝 𝑟=0

+

𝛽2 [𝛽1 (𝜂 + 𝑡) − 𝑡𝛼 ((𝛼1 + 𝛽1 ) 𝜂 + 𝛽1 )] Γ (𝛼 − 1) [𝛼𝛽2 ((𝛼1 + 𝛽1 ) 𝜂 + 𝛽1 ) − 𝛽1 (𝛼2 + 𝛽2 )] 1

× ∫ (1 − 𝜏)𝛼−2 0

(10) ×[

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 𝜎 (𝑠) 𝑑𝑠 − 𝜆𝑢 (𝜏)] 𝑑𝜏. Γ (𝛽) 0

where ( 𝑝𝑟 ) = 𝑝(𝑝 − 1)(𝑝 − 2) ⋅ ⋅ ⋅ (𝑝 − 𝑟 + 1)/𝑟!.

(13)

Lemma 5 (see [1]). Let 𝑝 ∈ (𝑚−1, 𝑚], 𝑚 ∈ 𝑁 and the Caputo derivative of order 𝑝 for a function 𝑓 : [0, ∞) → 𝑅. If for 𝑡 ∈ [0, 1], 𝑓 ∈ 𝐴𝐶𝑚 [0, 1] or 𝑓 ∈ 𝐶𝑚 [0, 1],

𝛽

𝑚−1 𝑘

𝑡 (𝑘) 𝑓 (0) . 𝑘! 𝑘=0

𝐼𝑝 𝑐 𝐷𝑝 𝑓 (𝑡) = 𝑓 (𝑡) − ∑

Proof. According to Lemma 5 and applying the operator 𝐼𝛽 to both sides of (12), for some constants 𝑐0 , 𝑐1 , and 𝑐2 , we get

(11)

𝛼

𝐼𝛽 𝑐 𝐷 ( 𝑐 𝐷 + 𝜆) 𝑢 (𝑡) = 𝐼𝛽 𝜎 (𝑡) ;

(14)

then the above equation can be written as We also easily prove the following lemmas. Lemma 6. Let 𝜎 ∈ 𝐿𝑞 ([0, 1], 𝑅), 𝑞 > 1/(𝛼 + 𝛽). 𝑐 𝐷𝛼 𝑢 ∈ 𝐶1 ([0, 1], 𝑅), 𝑢 ∈ 𝐶2 ([0, 1], 𝑅) satisfying the following differential equation: 𝑐

𝛼

( 𝑐 𝐷 + 𝜆) 𝑢 (𝑡) = 𝐼𝛽 𝜎 (𝑡) + 𝑐0 ,

(15)

and applying the operator 𝐼𝛼 to both sides of the above equation, we obtain

𝐷𝛽 ( 𝑐 𝐷𝛼 + 𝜆) 𝑢 (𝑡) = 𝜎 (𝑡) ,

0 < 𝑡 < 1,

1 < 𝛼 ≤ 2,

𝛼1 𝑢 (0) + 𝛽1 𝑢 (1) = 𝑔1 (𝑢) , 󸀠

𝐼𝛼 𝐷𝛼 𝑢 (𝑡) = 𝐼𝛼 [𝐼𝛽 𝜎 (𝑡) − 𝜆𝑢 (𝑡)] + 𝐼𝛼 𝑐0 ;

0 < 𝛽 ≤ 1, (12)

(16)

then the above equation can be written as

󸀠

𝛼2 𝑢 (0) + 𝛽2 𝑢 (1) = 𝑔2 (𝑢) , 𝑢 (0) = 𝜂𝑢󸀠 (0) ,

𝜂 ≠ 0,

is a solution of the following integral equation:

𝑡 1 ∫ (𝑡 − 𝜏)𝛼−1 Γ (𝛼) 0

×[ +

𝜏

1 ∫ (𝜏 − 𝑠)𝛽−1 𝜎 (𝑠) 𝑑𝑠 − 𝜆𝑢 (𝜏)] 𝑑𝜏 Γ (𝛽) 0

𝛼𝛽2 (𝜂 + 𝑡) − 𝑡𝛼 (𝛼2 + 𝛽2 ) 𝑔 (𝑢) 𝛼𝛽2 [(𝛼1 + 𝛽1 ) 𝜂 + 𝛽1 ] − 𝛽1 (𝛼2 + 𝛽2 ) 1

𝛽1 (𝜂 + 𝑡) − 𝑡𝛼 [(𝛼1 + 𝛽1 ) 𝜂 + 𝛽1 ] − 𝑔 (𝑢) 𝛼𝛽2 [(𝛼1 + 𝛽1 ) 𝜂 + 𝛽1 ] − 𝛽1 (𝛼2 + 𝛽2 ) 2 +

(17)

that can be written as (13). The proof is completed. Definition 7. The function 𝑢 ∈ 𝐶1 ([0, 1], 𝑅) satisfying (13) is a generalized solution of the nonlocal boundary value problem (6).

𝑢 (𝑡) =

𝑢 (𝑡) = 𝐼𝛼 [𝐼𝛽 𝜎 (𝑡) − 𝜆𝑢 (𝑡)] + 𝐼𝛼 𝑐0 + 𝑐1 + 𝑐2 𝑡,

𝛽1 𝑡𝛼 (𝛼2 + 𝛽2 ) − 𝛼𝛽1 𝛽2 (𝜂 + 𝑡) Γ (𝛼) [𝛼𝛽2 ((𝛼1 + 𝛽1 ) 𝜂 + 𝛽1 ) − 𝛽1 (𝛼2 + 𝛽2 )] 1

× ∫ (1 − 𝜏)𝛼−1 0

𝜏 1 ×[ ∫ (𝜏 − 𝑠)𝛽−1 𝜎 (𝑠) 𝑑𝑠 − 𝜆𝑢 (𝜏)] 𝑑𝜏 Γ (𝛽) 0

Lemma 8 (Krasnoselskii). Let B be a closed convex and nonempty subset of 𝑋. Suppose that L and N are general nonlinear operators which map B into 𝑋 such that (1) L𝑥 + N𝑦 ∈ B whenever 𝑥, 𝑦 ∈ B; (2) L is a contraction mapping; (3) N is compact and continuous. Then there exists 𝑧 ∈ B such that 𝑧 = L𝑧 + N𝑧.

3. Main Results In order to apply Lemma 8 to prove our main results, we first give 𝐹, 𝑆, 𝑇 as follows. Let Ω𝑟 = {𝑢 ∈ 𝐶1 ([0, 1], 𝑅) : ‖𝑢‖𝐶1 ≤ 𝑟}, 𝑟 > 0.

4

Journal of Applied Mathematics Define an operator 𝐹 : 𝐶1 → 𝐶1 by

and differentiating both sides of the above equation, we get

(𝐹𝑢) (𝑡) = (𝑆𝑢) (𝑡) + (𝑇𝑢) (𝑡) , 𝑢󸀠 (𝑡) = 𝐼𝛼+𝛽−1 𝑓 (𝑡, 𝑢 (𝑡) , 𝑢󸀠 (𝑡)) − 𝜆𝐼𝛼−1 𝑢 (𝑡)

(𝑆𝑢) (𝑡) =

𝑡 1 ∫ (𝑡 − 𝜏)𝛼−1 Γ (𝛼) 0

×[

+

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 𝑓 (𝑠, 𝑢 (𝑠) , 𝑢󸀠 (𝑠)) 𝑑𝑠 Γ (𝛽) 0

−𝜆𝑢 (𝜏) ] 𝑑𝜏, (𝑇𝑢) (𝑡) =

𝛼𝛽2 (𝜂 + 𝑡) − 𝑡𝛼 (𝛼2 + 𝛽2 ) 𝑔 (𝑢) 𝛼𝛽2 [(𝛼1 + 𝛽1 ) 𝜂 + 𝛽1 ] − 𝛽1 (𝛼2 + 𝛽2 ) 1

It is clear that every term of the above equation belongs to 𝐶; then 𝑢 ∈ 𝐶1 . Secondly, we show that 𝑢 is the generalized solution of the problem (6). Let 𝑢 be a generalized solution of the problem (6) and 𝑢 (𝑡) = 𝐼𝛼+𝛽 𝑓 (𝑡, 𝑢 (𝑡) , 𝑢󸀠 (𝑡)) − 𝜆𝐼𝛼 𝑢 (𝑡) +

𝛽1 (𝜂 + 𝑡) − 𝑡𝛼 [(𝛼1 + 𝛽1 ) 𝜂 + 𝛽1 ] − 𝑔 (𝑢) 𝛼𝛽2 [(𝛼1 + 𝛽1 ) 𝜂 + 𝛽1 ] − 𝛽1 (𝛼2 + 𝛽2 ) 2 +

(20)

𝑐0 𝑡𝛼−1 + 𝑐2 . Γ (𝛼)

𝑐0 𝑡𝛼 Γ (1 + 𝛼)

(21)

+ 𝑐1 + 𝑐2 𝑡.

𝛽1 𝑡𝛼 (𝛼2 + 𝛽2 ) − 𝛼𝛽1 𝛽2 (𝜂 + 𝑡) Γ (𝛼) [𝛼𝛽2 ((𝛼1 + 𝛽1 ) 𝜂 + 𝛽1 ) − 𝛽1 (𝛼2 + 𝛽2 )]

Applying the operator 𝑐 𝐷𝛼 to both sides of the above equation, we obtain

1

× ∫ (1 − 𝜏)𝛼−1 0

𝑐

×[

𝜏

1 ∫ (𝜏 − 𝑠)𝛽−1 𝑓 (𝑠, 𝑢 (𝑠) , 𝑢󸀠 (𝑠)) 𝑑𝑠 Γ (𝛽) 0

𝛼

𝛼

𝑐 𝛼+𝛽 󸀠 𝛼 𝐷 𝑢 (𝑡) = 𝐷 [𝐼 𝑓 (𝑡, 𝑢 (𝑡) , 𝑢 (𝑡)) − 𝜆𝐼 𝑢 (𝑡)

+

−𝜆𝑢 (𝜏) ] 𝑑𝜏 +

𝑐0 𝑡𝛼 + 𝑐 + 𝑐 𝑡] Γ (1 + 𝛼) 1 2

𝛼

= 𝑐 𝐷 𝐼𝛼+𝛽 𝑓 (𝑡, 𝑢 (𝑡) , 𝑢󸀠 (𝑡))

𝛽2 [𝛽1 (𝜂 + 𝑡) − 𝑡𝛼 ((𝛼1 + 𝛽1 ) 𝜂 + 𝛽1 )] Γ (𝛼 − 1) [𝛼𝛽2 ((𝛼1 + 𝛽1 ) 𝜂 + 𝛽1 ) − 𝛽1 (𝛼2 + 𝛽2 )]

𝛼

− 𝜆 𝑐 𝐷 𝐼𝛼 𝑢 (𝑡) + 𝑐 𝐷

1

× ∫ (1 − 𝜏)𝛼−2

𝛼

𝑐0 𝑡𝛼 Γ (1 + 𝛼)

𝛼

+ 𝑐 𝐷 𝑐1 + 𝑐 𝐷 𝑐2 𝑡

0

×[

𝛼

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 𝑓 (𝑠, 𝑢 (𝑠) , 𝑢󸀠 (𝑠)) 𝑑𝑠 Γ (𝛽) 0

= 𝐼𝛽 𝑓 (𝑡, 𝑢 (𝑡) , 𝑢󸀠 (𝑡)) − 𝜆𝑢 (𝑡) , ( 𝑐 𝐷𝛼 + 𝜆) 𝑢 (𝑡) = 𝐼𝛽 𝑓 (𝑡, 𝑢 (𝑡) , 𝑢󸀠 (𝑡)) ,

−𝜆𝑢 (𝜏) ] 𝑑𝜏.

(22) (18)

Lemma 9. The function 𝑢 ∈ 𝐶1 ([0, 1], 𝑅) is a generalized solution of the nonlocal boundary value problem (6) if 𝐹𝑢(𝑡) = 𝑢(𝑡), for all 𝑡 ∈ [0, 1]. Proof. Firstly, we show that 𝑢 ∈ 𝐶1 . Assuming 𝑢 ∈ 𝐶1 is a generalized solution of the problem (6), there exist three constants 𝑐0 , 𝑐1 , and 𝑐2 . Equation (13) can be written as 𝑢 (𝑡) = 𝐼𝛼+𝛽 𝑓 (𝑡, 𝑢 (𝑡) , 𝑢󸀠 (𝑡)) − 𝜆𝐼𝛼 𝑢 (𝑡) 𝑐0 𝑡𝛼 + + 𝑐 + 𝑐 𝑡, Γ (1 + 𝛼) 1 2

(19)

and then applying the operator 𝑐 𝐷𝛽 to both sides of the above equation, we obtain 𝑐

𝛽

𝛼

𝛽

𝑐 𝑐 𝛽 󸀠 𝐷 ( 𝐷 + 𝜆) 𝑢 (𝑡) = 𝐷 𝐼 𝑓 (𝑡, 𝑢 (𝑡) , 𝑢 (𝑡)) 󸀠

(23)

= 𝑓 (𝑡, 𝑢 (𝑡) , 𝑢 (𝑡)) . By simple calculations, it is clear that 𝑢 satisfies conditions (6); then it is a generalized solution for the problem (6). The proof is completed.

Journal of Applied Mathematics

5 (𝑇𝑢)󸀠 (𝑡)

For convenience, let us set

= (𝛼𝛽2 − 𝛼𝑡𝛼−1 (𝛼2 + 𝛽2 )) Λ 2 𝑔1 (𝑢)

Λ 1 = (𝛼1 + 𝛽1 ) 𝜂 + 𝛽1 ,

+ (−𝛽1 + 𝛼𝑡𝛼−1 Λ 1 ) Λ 2 𝑔2 (𝑢)

1 , Λ2 = 𝛼𝛽2 [(𝛼1 + 𝛽1 ) 𝜂 + 𝛽1 ] − 𝛽1 (𝛼2 + 𝛽2 ) 1

+

𝛼−1

Λ 3 = ∫ (1 − 𝜏) 0

+

𝜏 1 ×[ ∫ (𝜏 − 𝑠)𝛽−1 𝑓 (𝑠, 𝑢 (𝑠) , 𝑢󸀠 (𝑠)) 𝑑𝑠 Γ (𝛽) 0

𝛽2 (𝛽1 − 𝛼𝑡𝛼−1 Λ 1 ) Λ 2 Λ 4 Γ (𝛼 − 1)

.

Now, we make the following hypotheses. (H1) There exist two real-valued functions 𝑔 ∈ 𝐿1/𝑟 ([0, 1], 𝑅) for some 𝑟 ∈ (0, 1), such that

1

Λ 4 = ∫ (1 − 𝜏)𝛼−2

󵄨 󵄨󵄨 󵄨󵄨𝑓 (𝑡, 𝑢, 𝑢󸀠 ) − 𝑓 (𝑡, V, V󸀠 )󵄨󵄨󵄨 󵄨 󵄨

0

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 𝑓 (𝑠, 𝑢 (𝑠) , 𝑢󸀠 (𝑠)) 𝑑𝑠 Γ (𝛽) 0

󵄨 󵄨 ≤ 2𝑔 (𝑡) max {|𝑢 − V| , 󵄨󵄨󵄨󵄨𝑢󸀠 − V󸀠 󵄨󵄨󵄨󵄨} ,

−𝜆𝑢 (𝜏) ] 𝑑𝜏,

(26)

for almost all 𝑡 ∈ [0, 1], 𝑢, V ∈ 𝑅. (H2) There exist two positive constants 𝑙1 , 𝑙2 such that 𝑙1 + 𝑙2 = 𝐿 < 1. Moreover, 𝑔1 (0) = 0, 𝑔2 (0) = 0 and

1

Λ 5 = ∫ (1 − 𝜏)𝛼−1 0

󵄨 󵄨󵄨 󵄨󵄨𝑔1 (𝑢) − 𝑔1 (V)󵄨󵄨󵄨 ≤ 𝑙1 ‖𝑢 − V‖𝐶1 , 󵄨󵄨 󵄨 󵄨󵄨𝑔2 (𝑢) − 𝑔2 (V)󵄨󵄨󵄨 ≤ 𝑙2 ‖𝑢 − V‖𝐶1 ,

𝜏 1 ×[ ∫ (𝜏 − 𝑠)𝛽−1 𝑓 (𝑠, V (𝑠) , V󸀠 (𝑠)) 𝑑𝑠 Γ (𝛽) 0

(27)

∀𝑢, V ∈ 𝐶1 ([0, 1] , 𝑅) .

−𝜆V (𝜏) ] 𝑑𝜏, 1

Theorem 10. Let 𝑓 : [0, 1]×𝑅×𝑅 → 𝑅 be a jointly continuous function and the assumptions (H1) and (H2) hold. In addition, assume that

Λ 6 = ∫ (1 − 𝜏)𝛼−2 0

×[

Γ (𝛼)

(25)

−𝜆𝑢 (𝜏) ] 𝑑𝜏,

×[

[𝛼𝛽1 𝑡𝛼−1 (𝛼2 + 𝛽2 ) − 𝛼𝛽1 𝛽2 ] Λ 2 Λ 3

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 𝑓 (𝑠, V (𝑠) , V󸀠 (𝑠)) 𝑑𝑠 Γ (𝛽) 0

Λ ≜ max {Υ1 + (𝜂 + 1) × (𝛼𝛽2 Λ 2 𝐿 + 𝛽1 Λ 2 𝐿 + 𝛽1 𝛽2 Λ 2 Υ2 )

−𝜆V (𝜏) ] 𝑑𝜏. (24)

+ 𝛽1 (𝛼2 + 𝛽2 ) Λ 2 Υ1 , Υ2

(28)

+ (𝛼2 + 𝛽2 ) 𝛼𝛽1 Λ 2 Υ1 + 𝛼𝛽2 Λ 2 𝐿 Clearly, for any 𝑡 ∈ [0, 1], (𝑆𝑢)󸀠 (𝑡) =

𝑡 1 ∫ (𝑡 − 𝜏)𝛼−2 Γ (𝛼 − 1) 0

×[

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 𝑓 (𝑠, 𝑢 (𝑠) , 𝑢󸀠 (𝑠)) 𝑑𝑠 Γ (𝛽) 0

−𝜆𝑢 (𝜏) ] 𝑑𝜏,

+𝛼Λ 1 Λ 2 𝐿 + 𝛽1 𝛽2 Λ 2 Υ2 } < 1, where 𝑝 ∈ (0, 1), 1 < 𝛼 ≤ 2, 0 < 𝛽 ≤ 1, 𝜂 ≠ 0, 𝑔∗ = 1 (∫0 (𝑔(𝑠))1/(1−𝑝) 𝑑𝑠)1−𝑝 . Then the problem (6) has at most one solution. Proof. The proof will be given in two steps. Step 1. 𝐹 is bounded. Now we show that 𝐹Ω𝑟 ⊂ Ω𝑟 .

6

Journal of Applied Mathematics 𝜏

Let 𝑀 = sup𝑠∈[0,1] |𝑓(𝑠, 0, 0)|. For any 𝑢 ∈ Ω𝑟 , we have

0

󵄨 1 󵄨󵄨 󵄨󵄨 󵄨󵄨󵄨󵄨 𝛼−1 󵄨󵄨Λ 3 󵄨󵄨 = 󵄨󵄨∫ (1 − 𝜏) 󵄨󵄨 0 ×[

+

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 𝑓 (𝑠, 𝑢 (𝑠) , 𝑢󸀠 (𝑠)) 𝑑𝑠 Γ (𝛽) 0

󵄨󵄨 󵄨 −𝜆𝑢 (𝜏) ] 𝑑𝜏󵄨󵄨󵄨󵄨 󵄨󵄨

𝜏 𝑀 1 ∫ (1 − 𝜏)𝛼−1 ∫ (𝜏 − 𝑠)𝛽−1 𝑑𝑠 𝑑𝜏 Γ (𝛽) 0 0

𝑝

2𝑟𝑔∗ 1 𝑝𝜏(𝛽+𝑝−1)/𝑝 ≤ ) 𝑑𝜏 ∫ (1 − 𝜏)𝛼−1 ( 𝛽+𝑝−1 Γ (𝛽) 0 +

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 Γ (𝛽) 0

≤

󵄨 󵄨 × (󵄨󵄨󵄨󵄨𝑓 (𝑠, 𝑢 (𝑠) , 𝑢󸀠 (𝑠)) − 𝑓 (𝑠, 0, 0)󵄨󵄨󵄨󵄨 󵄨 󵄨 + 󵄨󵄨󵄨𝑓 (𝑠, 0, 0)󵄨󵄨󵄨 ) 𝑑𝑠

≤ ∫ (1 − 𝜏)

2𝑟𝑔∗ 𝑝𝑝 𝐵 (𝛽 + 𝑝, 𝛼) Γ (𝛽) (𝛽 + 𝑝 − 1)

𝑝

+

2𝑟𝑔∗ 𝑝𝑝 𝐵 (𝛽 + 𝑝, 𝛼 − 1) 𝑝

Γ (𝛽) (𝛽 + 𝑝 − 1)

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 Γ (𝛽) 0

󵄨 󵄨 × (2𝑔 (𝑠) max {|𝑢 (𝑠)| , 󵄨󵄨󵄨󵄨𝑢󸀠 (𝑠)󵄨󵄨󵄨󵄨}+𝑀) 𝑑𝑠

1

×[

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 Γ (𝛽) 0

× 𝑓 (𝑠, 𝑢 (𝑠) , 𝑢󸀠 (𝑠)) 𝑑𝑠

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 2𝑔 (𝑠) 𝑟𝑑𝑠 𝑑𝜏 Γ (𝛽) 0

+ ∫ (1 − 𝜏)𝛼−1 0

−𝜆𝑢 (𝜏) ] 𝑑𝜏 + (𝛼𝛽2 (𝜂 + 𝑡) − 𝑡𝛼 (𝛼2 + 𝛽2 )) Λ 2 𝑔1 (𝑢)

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 𝑀𝑑𝑠 𝑑𝜏 Γ (𝛽) 0

− (𝛽1 (𝜂 + 𝑡) − 𝑡𝛼 Λ 1 ) Λ 2 𝑔2 (𝑢)

1

+ ∫ (1 − 𝜏)𝛼−1 |𝜆| 𝑟𝑑𝜏

(𝛽1 𝑡𝛼 (𝛼2 + 𝛽2 ) − 𝛼𝛽1 𝛽2 (𝜂 + 𝑡)) Λ 2 Λ 3 Γ (𝛼) 󵄨󵄨 𝛼 𝛽 [𝛽 (𝜂 + 𝑡) − 𝑡 Λ 1 ] 󵄨 + 2 1 Λ 2 Λ 4 󵄨󵄨󵄨󵄨 Γ (𝛼 − 1) 󵄨󵄨

+

0

≤

1 𝜏 2𝑟 ∫ (1 − 𝜏)𝛼−1 ∫ (𝜏 − 𝑠)𝛽−1 𝑔 (𝑠) 𝑑𝑠 𝑑𝜏 Γ (𝛽) 0 0

+

𝜏 𝑀 1 ∫ (1 − 𝜏)𝛼−1 ∫ (𝜏 − 𝑠)𝛽−1 𝑑𝑠 𝑑𝜏 Γ (𝛽) 0 0 1

+ |𝜆| 𝑟 ∫ (1 − 𝜏)𝛼−1 𝑑𝜏 1 2𝑟 ∫ (1 − 𝜏)𝛼−1 Γ (𝛽) 0 𝜏

1 1 ∫ (1 − 𝜏)𝛼−1 Γ (𝛼) 0 𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 Γ (𝛽) 0

× 𝑓 (𝑠, 𝑢 (𝑠) , 𝑢󸀠 (𝑠)) 𝑑𝑠 𝑝

× [(∫ (𝜏 − 𝑠)(𝛽−1)/𝑝 𝑑𝑠) 0

≤

×[

0

≤

𝑀Γ (𝛼 − 1) |𝜆| 𝑟 + ≜ Φ4 , 𝛼−1 Γ (𝛼 + 𝛽)

󵄨󵄨 𝑡 󵄨 1 = 󵄨󵄨󵄨󵄨 ∫ (𝑡 − 𝜏)𝛼−1 󵄨󵄨 Γ (𝛼) 0

+ |𝜆| 𝑟] 𝑑𝜏

0

+

|(𝐹𝑢) (𝑡)|

0

1

𝑀Γ (𝛼) |𝜆| 𝑟 + ≜ Φ3 . 𝛼 Γ (𝛼 + 𝛽 + 1)

Clearly, we also can get

Λ4 ≤ 𝛼−1

𝑀𝐵 (𝛽 + 1, 𝛼) |𝜆| 𝑟 + 𝛼 𝛽Γ (𝛽)

(29)

+ |𝜆𝑢 (𝜏)| ] 𝑑𝜏

≤ ∫ (1 − 𝜏)𝛼−1

] 𝑑𝜏

0

0

×[

1−𝑝

1

1

1

𝑑𝑠)

+ |𝜆| 𝑟 ∫ (1 − 𝜏)𝛼−1 𝑑𝜏

≤ ∫ (1 − 𝜏)𝛼−1 ×[

1/(1−𝑝)

× (∫ (𝑔 (𝑠))

−𝜆𝑢 (𝜏) ] 𝑑𝜏

Journal of Applied Mathematics

7

+ 𝛼𝛽2 (𝜂 + 1) Λ 2 𝑙1 ‖𝑢‖ + Λ 1 Λ 2 𝑙2 ‖𝑢‖ 𝛽1 (𝛼2 + 𝛽2 ) Λ 2 Λ 3 𝛽2 𝛽1 (𝜂 + 1) + Λ 2Λ 4 Γ (𝛼) Γ (𝛼 − 1)

+ ≤

Λ3 + 𝛼𝛽2 (𝜂 + 1) Λ 2 𝑙1 𝑟 + Λ 1 Λ 2 𝑙2 𝑟 Γ (𝛼) 𝛽1 (𝛼2 + 𝛽2 ) Λ 2 Λ 3 𝛽2 𝛽1 (𝜂 + 1) + Λ 2Λ 4 Γ (𝛼) Γ (𝛼 − 1)

+ ≤

Λ4 + 𝛼𝛽2 Λ 2 𝑙1 𝑟 + 𝛼Λ 1 Λ 2 𝑙2 𝑟 Γ (𝛼 − 1) +

≤

𝛼𝛽1 (𝛼2 + 𝛽2 ) Λ 2 Λ 3 𝛽2 𝛽1 Λ 2 Λ 4 + Γ (𝛼) Γ (𝛼 − 1)

Φ4 + 𝛼 (𝛽2 + Λ 1 ) Λ 2 𝐿𝑟 Γ (𝛼 − 1) +

𝛼𝛽1 (𝛼2 + 𝛽2 ) Λ 2 Φ3 𝛽2 𝛽1 Λ 2 Φ4 + . Γ (𝛼) Γ (𝛼 − 1) (30)

Φ3 + (𝛼𝛽2 (𝜂 + 1) + Λ 1 ) Λ 2 𝐿𝑟 Γ (𝛼) 𝛽1 (𝛼2 + 𝛽2 ) Λ 2 Φ3 𝛽2 𝛽1 (𝜂 + 1) + Λ 2 Φ4 , Γ (𝛼) Γ (𝛼 − 1)

+

󵄨󵄨 󵄨 󵄨󵄨(𝐹𝑢)󸀠 (𝑡)󵄨󵄨󵄨 󵄨 󵄨

×[

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 Γ (𝛽) 0

× 𝑓 (𝑠, 𝑢 (𝑠) , 𝑢󸀠 (𝑠)) 𝑑𝑠

× (𝛼𝛽2 − 𝛼𝑡𝛼−1 (𝛼2 + 𝛽2 )) Λ 2 𝑔1 (𝑢) 𝛼−1

+ (−𝛽1 + 𝛼𝑡

+

Λ 1 ) Λ 2 𝑔2 (𝑢)

[𝛼𝛽1 𝑡𝛼−1 (𝛼2 + 𝛽2 ) − 𝛼𝛽1 𝛽2 ] Λ 2 Λ 3 Γ (𝛼)

𝛽1 (𝛼2 + 𝛽2 ) Λ 2 Φ3 Γ (𝛼)

+

𝛽2 𝛽1 (𝜂 + 1) Φ4 Λ 2 Φ4 , Γ (𝛼 − 1) Γ (𝛼 − 1)

1 ∫ (1 − 𝜏)𝛼−2 Γ (𝛼 − 1) 0

𝛼𝛽1 (𝛼2 + 𝛽2 ) Λ 2 Φ3 𝛽2 𝛽1 Λ 2 Φ4 + }, Γ (𝛼) Γ (𝛼 − 1)

where we have used the H¨older inequality and the following equalities:

1

𝐵 (𝛽 + 1, 𝛼) = ∫ (1 − 𝜏)𝛼−1 𝜏𝛽 𝑑𝜏 =

Γ (𝛽 + 1) Γ (𝛼) . Γ (𝛼 + 𝛽 + 1)

Step 2. 𝐹 is a contraction operator. For convenience, we get

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 Γ (𝛽) 0 󸀠

× 𝑓 (𝑠, 𝑢 (𝑠) , 𝑢 (𝑠)) 𝑑𝑠 −𝜆𝑢 (𝜏) ] 𝑑𝜏 + 𝛼𝛽2 Λ 2 𝑙1 ‖𝑢‖ + 𝛼Λ 1 Λ 2 𝑙2 ‖𝑢‖ 𝛼𝛽1 (𝛼2 + 𝛽2 ) Λ 2 Λ 3 𝛽2 𝛽1 Λ 2 Λ 4 + Γ (𝛼) Γ (𝛼 − 1)

(31)

+ 𝛼 (𝛽2 + Λ 1 ) Λ 2 𝐿𝑟

Therefore, ‖(𝐹𝑢)(𝑡)‖ ≤ 𝜓.

1

+

+

0

󵄨 𝛽2 [𝛽1 − 𝛼𝑡𝛼−1 Λ 1 ] Λ 2 Λ 4 󵄨󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 Γ (𝛼 − 1) 󵄨󵄨

×[

Φ3 + (𝛼𝛽2 (𝜂 + 1) + Λ 1 ) Λ 2 𝐿𝑟 Γ (𝛼)

+

−𝜆𝑢 (𝜏) ] 𝑑𝜏

+

For convenience, we let

𝜓 = max {

󵄨󵄨 𝑡 1 󵄨 = 󵄨󵄨󵄨󵄨 ∫ (𝑡 − 𝜏)𝛼−2 󵄨󵄨 Γ (𝛼 − 1) 0

≤

≤

|(𝑆𝑢) (𝑡) − (𝑆V) (𝑡)| 󵄨󵄨 1 𝑡 󵄨 = 󵄨󵄨󵄨 ∫ (𝑡 − 𝜏)𝛼−1 󵄨󵄨 Γ (𝛼) 0 ×[

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 𝑓 (𝑠, 𝑢 (𝑠) , 𝑢󸀠 (𝑠)) 𝑑𝑠 Γ (𝛽) 0

−𝜆𝑢 (𝜏) ] 𝑑𝜏

(32)

8

Journal of Applied Mathematics −

𝑡 1 ∫ (𝑡 − 𝜏)𝛼−1 Γ (𝛼) 0

×[

For 𝑢, V ∈ 𝐶1 ([0, 1], 𝑅) and for each 𝑡 ∈ [0, 1], we obtain

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 𝑓 (𝑠, V (𝑠) , V󸀠 (𝑠)) 𝑑𝑠 Γ (𝛽) 0

󵄨󵄨 󵄨 −𝜆V (𝜏) ] 𝑑𝜏󵄨󵄨󵄨󵄨 󵄨󵄨 ≤

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 2𝑔 (𝑠) Γ (𝛽) 0

1 𝜏 1 ∫ (1 − 𝜏)𝛼−1 ∫ (𝜏 − 𝑠)𝛽−1 2𝑔 (𝑠) 𝑑𝑠 𝑑𝜏 Γ (𝛼) Γ (𝛽) 0 0 1

1 𝜏 2 ∫ (1 − 𝜏)𝛼−1 (∫ (𝜏 − 𝑠)(𝛽−1)/𝑝 𝑑𝑠) Γ (𝛼) Γ (𝛽) 0 0 1/(1−𝑝)

0

≤ ‖𝑢 − V‖ (

𝑑𝑠)

1−𝑝

(𝛽1 𝑡𝛼 (𝛼2 + 𝛽2 ) − 𝛼𝛽1 𝛽2 (𝜂 + 𝑡)) Λ 2 Λ 3 Γ (𝛼) 󵄨󵄨 𝛽 [𝛽 (𝜂 + 𝑡) − 𝑡𝛼 Λ 1 ] 󵄨 Λ 2 Λ 4 ]󵄨󵄨󵄨󵄨 + 2 1 Γ (𝛼 − 1) 󵄨󵄨

≤ |(𝑆𝑢) (𝑡) − (𝑆V) (𝑡)| 󵄨 󵄨 + 󵄨󵄨󵄨(𝛽1 (𝜂 + 𝑡) − 𝑡𝛼 Λ 1 )󵄨󵄨󵄨 Λ 2 𝑙2 ‖𝑢 − V‖

≤ ‖𝑢 − V‖

𝜏

𝛽2 [𝛽1 (𝜂 + 𝑡) − 𝑡𝛼 Λ 1 ] Λ 2Λ 4 Γ (𝛼 − 1)

󵄨 󵄨 + 󵄨󵄨󵄨(𝛼𝛽2 (𝜂 + 𝑡) − 𝑡𝛼 (𝛼2 + 𝛽2 ))󵄨󵄨󵄨 Λ 2 𝑙1 ‖𝑢 − V‖

|𝜆| ∫ (1 − 𝜏)𝛼−1 𝑑𝜏) Γ (𝛼) 0

× (∫ (𝑔 (𝑠))

+

+

≤ ‖𝑢 − V‖

×[

(𝛽1 𝑡𝛼 (𝛼2 + 𝛽2 ) − 𝛼𝛽1 𝛽2 (𝜂 + 𝑡)) Λ 2 Λ 3 Γ (𝛼)

− (𝛽1 (𝜂 + 𝑡) − 𝑡𝛼 Λ 1 ) Λ 2 𝑔2 (V)

|𝜆| 1 + ∫ (1 − 𝜏)𝛼−1 ‖𝑢 − V‖ 𝑑𝜏 Γ (𝛼) 0

+

+

− [𝑆V (𝑡) + (𝛼𝛽2 (𝜂 + 𝑡) − 𝑡𝛼 (𝛼2 + 𝛽2 )) Λ 2 𝑔1 (V)

󵄨 󵄨 × max {|𝑢 − V| , 󵄨󵄨󵄨󵄨𝑢󸀠 − V󸀠 󵄨󵄨󵄨󵄨} 𝑑𝑠] 𝑑𝜏

×(

󵄨󵄨 󵄨 = 󵄨󵄨󵄨󵄨𝑆𝑢 (𝑡) + (𝛼𝛽2 (𝜂 + 𝑡) − 𝑡𝛼 (𝛼2 + 𝛽2 )) Λ 2 𝑔1 (𝑢) 󵄨󵄨 − (𝛽1 (𝜂 + 𝑡) − 𝑡𝛼 Λ 1 ) Λ 2 𝑔2 (𝑢)

1 1 ∫ (1 − 𝜏)𝛼−1 Γ (𝛼) 0

×[

|(𝐹𝑢) (𝑡) − (𝐹V) (𝑡)|

𝑑𝜏 +

𝑝

|𝜆| ] Γ (𝛼 + 1)

2𝑔∗ 𝑝𝑝 Γ (𝛽 + 𝑝) Γ (𝛽) Γ (𝛼 + 𝛽 + 𝑝) (𝛽 + 𝑝 − 1)

𝑝

+

󵄨󵄨 (𝛽 𝑡𝛼 (𝛼 + 𝛽 ) − 𝛼𝛽 𝛽 (𝜂 + 𝑡)) Λ (Λ − Λ ) 󵄨 2 2 1 2 2 3 5 + 󵄨󵄨󵄨󵄨 1 Γ (𝛼) 󵄨󵄨 󵄨󵄨 𝛽 [𝛽 (𝜂 + 𝑡) − 𝑡𝛼 Λ 1 ] 󵄨 + 2 1 Λ 2 (Λ 4 − Λ 6 ) 󵄨󵄨󵄨󵄨 Γ (𝛼 − 1) 󵄨󵄨 ≤ |(𝑆𝑢) (𝑡) − (𝑆V) (𝑡)|

|𝜆| ) Γ (𝛼 + 1)

+ 𝛼𝛽2 (𝜂 + 𝑡) Λ 2 𝑙1 ‖𝑢 − V‖ + 𝛽1 (𝜂 + 𝑡) Λ 2 𝑙2 ‖𝑢 − V‖ + 𝛽1 𝑡𝛼 (𝛼2 + 𝛽2 ) Λ 2 Υ1 ‖𝑢 − V‖

≜ Υ1 ‖𝑢 − V‖ . (33)

+ 𝛽2 𝛽1 (𝜂 + 𝑡) Λ 2 Υ2 ‖𝑢 − V‖ ≤ ‖𝑢 − V‖ (Υ1 + (𝜂 + 1)

Clearly, we can also get

× (𝛼𝛽2 Λ 2 𝐿 + 𝛽1 Λ 2 𝐿 + 𝛽1 𝛽2 Λ 2 Υ2 ) + 𝛽1 (𝛼2 + 𝛽2 ) Λ 2 Υ1 ) ,

󵄨󵄨 󵄨 󵄨󵄨(𝑆𝑢)󸀠 (𝑡) − (𝑆V)󸀠 (𝑡)󵄨󵄨󵄨 󵄨 󵄨 ≤ ‖𝑢 − V‖ (

2𝑔∗ 𝑝𝑝 Γ (𝛽 + 𝑝)

𝑝

Γ (𝛽) Γ (𝛼 + 𝛽 + 𝑝 − 1) (𝛽 + 𝑝 − 1) +

|𝜆| ) ≜ Υ2 ‖𝑢 − V‖ . Γ (𝛼)

(34)

󵄨 󵄨󵄨 󵄨󵄨(𝐹𝑢)󸀠 (𝑡) − (𝐹V)󸀠 (𝑡)󵄨󵄨󵄨 󵄨 󵄨 󵄨󵄨 󵄨 󸀠 ≤ 󵄨󵄨󵄨(𝑆𝑢) (𝑡) − (𝑆𝑢)󸀠 (𝑡)󵄨󵄨󵄨󵄨 󵄨 + 󵄨󵄨󵄨󵄨(𝛼𝛽2 − 𝛼𝑡𝛼−1 (𝛼2 + 𝛽2 )) Λ 2 (𝑔1 (𝑢) − 𝑔1 (V)) 󵄨 + (−𝛽1 + 𝛼𝑡𝛼−1 Λ 1 ) Λ 2 (𝑔2 (𝑢) − 𝑔2 (V)) 󵄨󵄨󵄨󵄨

Journal of Applied Mathematics

9

󵄨󵄨 󵄨󵄨 [𝛼𝛽1 𝑡𝛼−1 (𝛼2 + 𝛽2 ) − 𝛼𝛽1 𝛽2 ] Λ 2 (Λ 3 − Λ 5 ) + 󵄨󵄨󵄨󵄨 Γ (𝛼) 󵄨󵄨 󵄨 󵄨 𝛽2 [𝛽1 − 𝛼𝑡𝛼−1 Λ 1 ] Λ 2 (Λ 4 − Λ 6 ) 󵄨󵄨󵄨󵄨 󵄨󵄨 + 󵄨󵄨 Γ (𝛼 − 1) 󵄨󵄨

For 𝑢, V ∈ 𝐶1 ([0, 1], 𝑅) and for each 𝑡 ∈ [0, 1], we obtain |(𝑇𝑢) (𝑡) − (𝑇V) (𝑡)| ≤ ‖𝑢 − V‖ ((𝜂 + 1) (𝛼𝛽2 Λ 2 𝐿 + 𝛽1 Λ 2 𝐿 + 𝛽1 𝛽2 Λ 2 Υ2 ) +𝛽1 (𝛼2 + 𝛽2 ) Λ 2 Υ1 ) , 󵄨 󵄨󵄨 󵄨󵄨(𝑇𝑢)󸀠 (𝑡) − (𝑇V)󸀠 (𝑡)󵄨󵄨󵄨 󵄨 󵄨

≤ Υ2 ‖𝑢 − V‖ + 𝛼𝛽2 Λ 2 𝑙1 ‖𝑢 − V‖ + 𝛼Λ 1 Λ 2 𝑙2 ‖𝑢 − V‖ 𝛼−1

+ 𝛼𝛽1 𝑡

≤ ‖𝑢 − V‖ ((𝛼2 + 𝛽2 ) 𝛼𝛽1 Λ 2 Υ1

(𝛼2 + 𝛽2 ) Λ 2 Υ1 ‖𝑢 − V‖

+ 𝛼𝛽2 Λ 2 𝐿 + 𝛼Λ 1 Λ 2 𝐿 + 𝛽1 𝛽2 Λ 2 Υ2 ) . (40)

+ 𝛽2 𝛽1 Λ 2 Υ2 ‖𝑢 − V‖

Since 𝜉 < 1, we have ‖𝑇(𝑢) − 𝑇(V)‖ ≤ 𝜉‖𝑢 − V‖; that is, 𝑇 is a nonlinear contraction.

≤ ‖𝑢 − V‖ (Υ2 + (𝛼2 + 𝛽2 ) 𝛼𝛽1 Λ 2 Υ1 + 𝛼𝛽2 Λ 2 𝐿 + 𝛼Λ 1 Λ 2 𝐿 + 𝛽1 𝛽2 Λ 2 Υ2 ) . (35) Since Λ < 1, we have ‖𝐹(𝑢) − 𝐹(V)‖ ≤ Λ‖𝑢 − V‖; that is, 𝐹 is a nonlinear contraction. Hence, by using Lemma 8, the conclusion of the theorem holds by Banach fixed point theorem. The proof is completed.

Step 3. 𝑆 is continuous and compact. Firstly, we show that the operator 𝑆 is continuous. For {𝑢𝑛 } ⊂ Ω𝑟 , 𝑢0 ∈ Ω𝑟 such that 𝑢𝑛 → 𝑢0 in Ω𝑟 ; then 󵄨 󵄨󵄨 󵄨󵄨𝑆𝑢𝑛 (𝑡) − 𝑆𝑢0 (𝑡)󵄨󵄨󵄨 󵄨󵄨 1 𝑡 󵄨 = 󵄨󵄨󵄨 ∫ (𝑡 − 𝜏)𝛼−1 󵄨󵄨 Γ (𝛼) 0

Theorem 11. Let 𝑓 : [0, 1]×𝑅×𝑅 → 𝑅 be a jointly continuous function and the assumptions (H1) and (H2) hold. In addition,

×[

(H3) assume that there exist a constant 𝑙 ∈ (0, 1) and a realvalued function 𝑚 ∈ 𝐿1/𝑙 ([0, 1], 𝑅+ ) such that 󵄨󵄨 󵄨 󵄨󵄨𝑓 (𝑡, 𝑢, 𝑢󸀠 )󵄨󵄨󵄨 ≤ 𝑚 (𝑡) , ∀ (𝑡, 𝑢, 𝑢󸀠 ) ∈ [0, 1] × 𝑅 × 𝑅, 󵄨 󵄨 (36) with sup |𝑚 (𝑡)| = ‖𝑚‖ .

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 𝑓 (𝑠, 𝑢𝑛 (𝑠) , 𝑢𝑛󸀠 (𝑠)) 𝑑𝑠 Γ (𝛽) 0

−𝜆𝑢𝑛 (𝜏) ] 𝑑𝜏 −

𝑡∈[0,1]

𝑡 1 ∫ (𝑡 − 𝜏)𝛼−1 Γ (𝛼) 0

×[

Then the problem (6) has at least one solution on [0, 1] if

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 Γ (𝛽) 0

𝜉 ≜ max {(𝜂 + 1) (𝛼𝛽2 Λ 2 𝐿 + 𝛽1 Λ 2 𝐿 + 𝛽1 𝛽2 Λ 2 Υ2 ) + 𝛽1 (𝛼2 + 𝛽2 ) Λ 2 Υ1 , (𝛼2 + 𝛽2 ) 𝛼𝛽1 Λ 2 Υ1 + 𝛼𝛽2 Λ 2 𝐿

× 𝑓 (𝑠, 𝑢0 (𝑠) , 𝑢0󸀠 (𝑠)) 𝑑𝑠

+ 𝛼Λ 1 Λ 2 𝐿 + 𝛽1 𝛽2 Λ 2 Υ2 } < 1. Proof. Step 1. There exists a positive constant 𝑟 > 0 such that 𝑆𝑢 + 𝑇𝑢 ∈ Ω𝑟 . For 𝑢 ∈ Ω𝑟 , by the same arguments of the first step of the proof in Theorem 10, we have ‖𝑆𝑢 + 𝑇𝑢‖ ≤ 𝜓. In virtue of the definition of 𝜓 and a simple calculation, we obtain 𝜓 ≤ 𝑀 + 𝜉𝑟,

󵄨󵄨 󵄨 −𝜆𝑢0 (𝜏) ] 𝑑𝜏 󵄨󵄨󵄨󵄨 󵄨󵄨

(37)

󵄨󵄨 1 𝑡 󵄨 ≤ 󵄨󵄨󵄨 ∫ (𝑡 − 𝜏)𝛼−1 󵄨󵄨 Γ (𝛼) 0 ×

× [𝑓 (𝑠, 𝑢𝑛 (𝑠) , 𝑢𝑛󸀠 (𝑠))

(38)

−𝑓 (𝑠, 𝑢0 (𝑠) , 𝑢0󸀠 (𝑠))] 𝑑𝑠 𝑑𝜏

where 𝑀 is a constant. By the assumptions, 𝜉 < 1. Therefore, there exists a positive constant 𝑟 large enough such that 𝜓 ≤ 𝑀 + 𝜉𝑟 < 𝑟.

+

(39)

Hence, there exists a positive constant 𝑟 such that 𝑆𝑢 + 𝑇𝑢 ∈ Ω𝑟 . Step 2. 𝑇 is a contraction operator.

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 Γ (𝛽) 0

≤

󵄨󵄨 𝑡 𝜆 󵄨 ∫ (𝑡 − 𝜏)𝛼−1 (𝑢𝑛 (𝜏) − 𝑢0 (𝜏)) 𝑑𝜏󵄨󵄨󵄨 󵄨󵄨 Γ (𝛼) 0

𝑡 1 ∫ (𝑡 − 𝜏)𝛼−1 Γ (𝛼) 0

×

𝜏 1 ∫ (𝜏−𝑠)𝛽−1 Γ (𝛽) 0

10

Journal of Applied Mathematics 󵄨 󵄨 × 2𝑔 (𝑠) max {󵄨󵄨󵄨𝑢𝑛 (𝑠) − 𝑢0 (𝑠)󵄨󵄨󵄨 ,

−𝜆𝑢 (𝜏) ] 𝑑𝜏

󵄨󵄨 󸀠 󵄨 󵄨󵄨𝑢𝑛 (𝑠) − 𝑢0󸀠 (𝑠)󵄨󵄨󵄨} 𝑑𝑠 𝑑𝜏 󵄨 󵄨

+

−

𝑡

|𝜆| 󵄨 󵄨 ∫ (𝑡 − 𝜏)𝛼−1 󵄨󵄨󵄨𝑢𝑛 (𝜏) − 𝑢0 (𝜏)󵄨󵄨󵄨 𝑑𝜏 Γ (𝛼) 0

𝑡 1 ≤ ∫ (𝑡 − 𝜏)𝛼−1 Γ (𝛼) 0

× +

𝑠1 1 𝛼−1 ∫ (𝑠1 − 𝜏) Γ (𝛼) 0

×[

󵄨󵄨 󵄨 −𝜆𝑢 (𝜏) ] 𝑑𝜏󵄨󵄨󵄨󵄨 󵄨󵄨

𝜏 1 󵄩 󵄩 ∫ (𝜏 − 𝑠)𝛽−1 2𝑔 (𝑠) 󵄩󵄩󵄩𝑢𝑛 − 𝑢0 󵄩󵄩󵄩 𝑑𝑠 𝑑𝜏 Γ (𝛽) 0

󵄨󵄨 1 𝑠1 𝛼−1 𝛼−1 󵄨 ≤ 󵄨󵄨󵄨 ∫ [(𝑠2 − 𝜏) − (𝑠1 − 𝜏) ] 󵄨󵄨 Γ (𝛼) 0

|𝜆| 𝑡 󵄩 󵄩 ∫ (𝑡 − 𝜏)𝛼−1 󵄩󵄩󵄩𝑢𝑛 − 𝑢0 󵄩󵄩󵄩 𝑑𝜏 Γ (𝛼) 0

1 2 1 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑢𝑛 − 𝑢0 󵄩󵄩󵄩 ( ∫ (1 − 𝜏)𝛼−1 Γ (𝛼) 0 Γ (𝛽)

×[

𝜏

× ∫ (𝜏 − 𝑠)𝛽−1 𝑔 (𝑠) 𝑑𝑠 𝑑𝜏 |𝜆| ) Γ (𝛼 + 1)

+

󵄩 󵄩 ≤ Υ1 󵄩󵄩󵄩𝑢𝑛 − 𝑢0 󵄩󵄩󵄩 .

𝑠2 1 𝛼−1 ∫ (𝑠1 − 𝜏) Γ (𝛼) 𝑠1

×[

(41)

(42)

≤

𝑠1 1 𝛼−1 𝛼−1 ∫ [(𝑠2 − 𝜏) − (𝑠1 − 𝜏) ] Γ (𝛼) 0

we get sequences 𝑢𝑛 (𝑡) and 𝑢0 (𝑡), which converge on [0, 1] with lim𝑛 → ∞ 𝑢𝑛 (𝑡) = 𝑢0 (𝑡) and lim𝑛 → ∞ 𝑢𝑛󸀠 (𝑡) = 𝑢0󸀠 (𝑡). Since

×(

󵄩 󵄩󵄩 󵄩󵄩𝑆𝑢𝑛 − 𝑆𝑢0 󵄩󵄩󵄩 󵄨 󵄨 = max { sup 󵄨󵄨󵄨𝑆𝑢𝑛 (𝑡) − 𝑆𝑢0 (𝑡)󵄨󵄨󵄨 , 𝑡∈[0,1]

(43)

󵄨 󵄨󵄨 󵄨󵄨(𝑆𝑢𝑛 )󸀠 (𝑡) − (𝑆𝑢0 )󸀠 (𝑡)󵄨󵄨󵄨} . 󵄨 󵄨 𝑡∈[0,1] sup

Combining (41) and (42), we can get ‖𝑆𝑢𝑛 − 𝑆𝑢0 ‖ → 0. Thus 𝑆 is continuous in 𝐶1 ([0, 1], 𝑅). Secondly, we show that the operator 𝑆 is equicontinuous. Let 𝑀𝑟 = max𝑠∈[0,1] {|𝑓(𝑠, 𝑢, 𝑢󸀠 )|, 𝑢 ∈ Ω𝑟 }. For any 𝑢 ∈ Ω𝑟 , for all 𝑠1 , 𝑠2 ∈ [0, 1], 0 ≤ 𝑠1 < 𝑠2 ≤ 1, we obtain 󵄨󵄨 󵄨 󵄨󵄨(𝑆𝑢) (𝑠2 ) − (𝑆𝑢) (𝑠1 )󵄨󵄨󵄨 󵄨󵄨 1 𝑠2 𝛼−1 󵄨 = 󵄨󵄨󵄨 ∫ (𝑠2 − 𝜏) 󵄨󵄨 Γ (𝛼) 0 ×[

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 𝑓 (𝑠, 𝑢 (𝑠) , 𝑢󸀠 (𝑠)) 𝑑𝑠 Γ (𝛽) 0

𝜏 1 󵄨 󵄨 ∫ (𝜏 − 𝑠)𝛽−1 󵄨󵄨󵄨󵄨𝑓 (𝑠, 𝑢 (𝑠) , 𝑢󸀠 (𝑠))󵄨󵄨󵄨󵄨 𝑑𝑠 Γ (𝛽) 0

󵄨󵄨 󵄨 + |𝜆| |𝑢 (𝜏)| ] 𝑑𝜏󵄨󵄨󵄨󵄨 󵄨󵄨

Similarly, we get 󵄨󵄨 󸀠 󵄨 󵄨󵄨𝑆𝑢𝑛 (𝑡) − 𝑆𝑢0󸀠 (𝑡)󵄨󵄨󵄨 ≤ Υ2 󵄩󵄩󵄩󵄩𝑢𝑛 − 𝑢0 󵄩󵄩󵄩󵄩 , 󵄨 󵄨

𝜏 1 󵄨 󵄨 ∫ (𝜏 − 𝑠)𝛽−1 󵄨󵄨󵄨󵄨𝑓 (𝑠, 𝑢 (𝑠) , 𝑢󸀠 (𝑠))󵄨󵄨󵄨󵄨 𝑑𝑠 Γ (𝛽) 0

+ |𝜆| |𝑢 (𝜏)| ] 𝑑𝜏

0

+

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 𝑓 (𝑠, 𝑢 (𝑠) , 𝑢󸀠 (𝑠)) 𝑑𝑠 Γ (𝛽) 0

≤

𝜏 1 ∫ (𝜏 − 𝑠)𝛽−1 𝑀𝑟 𝑑𝑠) 𝑑𝜏 Γ (𝛽) 0

+

𝑠1 1 𝛼−1 𝛼−1 ∫ [(𝑠2 − 𝜏) − (𝑠1 − 𝜏) ] |𝜆| 𝑟 𝑑𝜏 Γ (𝛼) 0

+

𝑠2 𝜏 1 1 𝛼−1 ∫ (𝑠1 − 𝜏) ( ∫ (𝜏 − 𝑠)𝛽−1 𝑀𝑟 𝑑𝑠) 𝑑𝜏 Γ (𝛼) 𝑠1 Γ (𝛽) 0

+

𝑠2 1 𝛼−1 ∫ (𝑠1 − 𝜏) |𝜆| 𝑟 𝑑𝜏 Γ (𝛼) 𝑠1

𝑠1 1 𝛼−1 𝛼−1 ∫ [(𝑠2 − 𝜏) − (𝑠1 − 𝜏) ] Γ (𝛼) 0

×(

𝑀𝑟 𝜏 ∫ (𝜏 − 𝑠)𝛽−1 𝑑𝑠) 𝑑𝜏 Γ (𝛽) 0

+

|𝜆| 𝑟 𝑠1 𝛼−1 𝛼−1 ∫ [(𝑠2 − 𝜏) − (𝑠1 − 𝜏) ] 𝑑𝜏 Γ (𝛼) 0

+

𝑠2 𝜏 𝑀 1 𝛼−1 ∫ (𝑠1 − 𝜏) ( 𝑟 ∫ (𝜏 − 𝑠)𝛽−1 𝑑𝑠) 𝑑𝜏 Γ (𝛼) 𝑠1 Γ (𝛽) 0

+

|𝜆| 𝑟 𝑠2 𝛼−1 ∫ (𝑠 − 𝜏) 𝑑𝜏 Γ (𝛼) 𝑠1 1

Journal of Applied Mathematics ≤

11

𝑠1 𝑀𝑟 𝛼−1 ∫ (𝑠2 − 𝑠1 ) 𝜏𝛽 𝑑𝜏 Γ (𝛼) Γ (𝛽 + 1) 0

get that 𝑆 is relatively compact on Ω𝑟 . Hence, by the Arzel´aAscoli theorem, 𝑆 is compact on Ω𝑟 . Thus, all the assumptions of Lemma 8 are satisfied and the conclusion of Lemma 8 implies that the boundary value problem (6) has at least one solution on [0, 1]. The proof is completed.

𝛼

+ +

|𝜆| 𝑟 (𝑠2𝛼 − 𝑠1𝛼 − (𝑠2 − 𝑠1 ) ) Γ (𝛼 + 1) 𝑠2 𝑀𝑟 𝛼−1 ∫ (𝑠1 − 𝜏) 𝜏𝛽 𝑑𝜏 Γ (𝛼) Γ (𝛽 + 1) 𝑠1

4. Algorithm for the Fractional Langevin Equation and Examples

𝛼

+

|𝜆| 𝑟(𝑠2 − 𝑠1 ) Γ (𝛼 + 1)

In this paper, we will give the numerical simulation for the fractional Langevin equation. The definition of fractional order has many kinds; the different definitions will bring different algorithm forms and will cause different proof of the algorithm stability and different method of accuracy analysis. In the practical application, there are three kinds of fractional derivative definitions, such as Gr¨unwald-Letnikov, Riemann-Liouvlle, and Caputo Fractional derivatives.

𝛼−1

≤

1 𝑀𝑟 (𝑠2 − 𝑠1 ) ∫ 𝜏𝛽 𝑑𝜏 Γ (𝛼) Γ (𝛽 + 1) 0 𝛼

+ +

𝛼

|𝜆| 𝑟 (𝑠2𝛼 − 𝑠1𝛼 − (𝑠2 − 𝑠1 ) + (𝑠1 − 𝑠2 ) ) Γ (𝛼 + 1) 𝑝 𝑠2 𝑀𝑟 (𝛼−1)/𝑝 𝑑𝜏) (∫ (𝑠1 − 𝜏) Γ (𝛼) Γ (𝛽 + 1) 𝑠1 𝑠2

× (∫ 𝜏

𝛽/(1−𝑝)

𝑠1

Remark 12 (see [51]). For 𝑚 − 1 < 𝛼 ≤ 𝑚, 𝑚 ∈ 𝑁, 𝑓(𝑡) ∈ 𝐶𝑚 [𝑎, 𝑏],

1−𝑝

𝑑𝜏)

GL 𝛼−1

≤

𝑀𝑟 (𝑠2 − 𝑠1 ) Γ (𝛼) Γ (𝛽 + 2)

1−𝑝

× [(1 − 𝑝)

𝛼+𝑝−1

GL

]

(𝛽+1−𝑝)/(1−𝑝) (𝑠2

−

𝑝

𝛼

1−𝑝 −1

(46)

Γ (𝛼 + 1)

𝑚−1

𝑓(𝑘) (𝑎) (𝑡 − 𝑎)𝑘−𝛼 Γ − 𝑎 + 1) (𝑘 𝑘=0

𝛼

𝛼

+ 𝑐𝑎 𝐷 𝑓 (𝑡) =

(47)

RL 𝛼 𝑎 𝐷 𝑓 (𝑡) .

In [52], shifted Gr¨unwald-Letnikov formula is defined by

)

GL

𝛼

|𝜆| 𝑟 (𝑠2𝛼 − 𝑠1𝛼 − (𝑠2 − 𝑠1 ) + (𝑠1 − 𝑠2 ) )

. (44)

Clearly, we also easily get

𝛼

𝐷 𝑓 (𝑡) 1 = lim 𝛼 ℎ→0ℎ

[(𝑡−𝑎)/ℎ+𝑝]

∑ 𝑘=0

𝑤𝑘(𝛼) 𝑓 (𝑡 − (𝑘 − 𝑝) ℎ) ,

𝛼 > 0.

(48)

We get the following approximation:

󵄨 󵄨󵄨 󵄨󵄨(𝑆𝑢)󸀠 (𝑠2 ) − (𝑆𝑢)󸀠 (𝑠1 )󵄨󵄨󵄨 󵄨 󵄨

𝛼 𝑎 𝐷 𝑓 (𝑡) ≈

𝑠1 1 𝛼−2 𝛼−2 ∫ [(𝑠2 − 𝜏) − (𝑠1 − 𝜏) ] Γ (𝛼 − 1) 0

×(

∑ 𝑘=0

𝑤𝑘(𝛼) 𝑓 (𝑡 − (𝑘 − 𝑝) ℎ)

(49)

We put a call shifted Gr¨unwald discrete format, simply “𝐺𝑆(𝑝) algorithm” for short. In addition, Oldham and Spanier [53] found the following approximation format in 1974:

𝛼−1 𝛽

+

[(𝑡−𝑎)/ℎ+𝑝]

:= ( 𝑎 𝐷 𝑓 (𝑡))𝐺𝑆(𝑝) .

𝑀𝑟 𝜏 + |𝜆| 𝑟) 𝑑𝜏 Γ (𝛽 + 1)

𝑀𝑟 (𝑠2 − 𝑠1 ) Γ (𝛼 + 𝛽)

1 ℎ𝛼

𝛼

𝛽

𝛼+𝛽−1

+

𝛼

𝐷 𝑓 (𝑡) .

𝐷 𝑓 (𝑡) = ∑

(𝛽+1−𝑝)/(1−𝑝) 1−𝑝 𝑠1 ) ])

× (Γ(𝛼)Γ(𝛽 + 1)(𝛼 + 𝑝 − 1) (𝛽 + 1 − 𝑝)

≤

RL

Remark 13 (see [49]). For 𝑓(𝑘) (𝑎) = 0, 𝑘 = 0, 1, . . . , 𝑚,

+ (𝑀𝑟 [𝑝𝑝 (𝑠1 − 𝑠2 )

+

𝛼

𝐷 𝑓 (𝑡) =

𝑀𝑟 (𝑠2 − 𝑠1 ) 𝑠1 Γ (𝛼) Γ (𝛽 + 1)

[(𝑡−𝑎)/ℎ−1/2]

−1 𝑎 𝐷 𝑓 (𝑡) = lim ℎ

𝛼

|𝜆| 𝑟(𝑠2 − 𝑠1 ) . + 𝛼Γ (𝛼 − 1)

ℎ→0

(45) Obviously the right hand side of the above inequality tends to zero independently of 𝑢 ∈ Ω𝑟 , as 𝑡2 − 𝑡1 → 0; we

1 −1 𝑎 𝐷 𝑓 (𝑡) = lim ℎ ℎ→0

∑

𝑗=0

[(𝑡−𝑎)/ℎ+1/2]

∑

𝑗=0

1 𝑓 (𝑡 − (𝑗 + ) ℎ) , 2

1 (−1)𝑗 𝑓 (𝑡 − (𝑗 − ) ℎ) . 2 (50)

12

Journal of Applied Mathematics

The approximation format has the rapid convergence properties. So they put forward an improved Gr¨unwaldLetnikov fractional derivative definition (take 𝑝 = 𝛼/2 to (48)): 𝑎𝐷

𝛼

According to 𝐺2 algorithm, the Caputo fractional derivatives above can be written as 𝑐

𝐷𝛽 𝑢 (𝑡) ℎ−𝛽 ℎ → 0 Γ (−𝛽)

ℎ−𝛼 ℎ → 0 Γ (−𝛼) ×

[(𝑡−𝑎)/ℎ+𝛼/2]

∑

𝑗=0

Γ (𝑗 − 𝛼) 1 𝑓 (𝑡 − (𝑗 − 𝛼) ℎ) . 2 Γ (𝑗 + 1)

𝑐

For 𝑎 = 0, the above equation can be written as

∑

ℎ−𝛼 ℎ → 0 Γ (−𝛼)

= lim

𝑗=0

𝑐

Γ (𝑗 − 𝛼) 1 𝑓 (𝑡 − (𝑗 − 𝛼) ℎ) . 2 Γ (𝑗 + 1)

(52)

𝛼 𝛼2 1 𝑓 (𝑡 − (𝑗 − 𝛼) ℎ) ≈ ( + ) 𝑓 (𝑡 − (𝑗 − 1) ℎ) 2 4 8 𝛼2 ) 𝑓 (𝑡 − 𝑗ℎ) 4

𝑐

𝐷𝛼 V (𝑡𝑛 )

1 + 𝛼2 (𝑢𝑛−𝑗+1 − 2𝑢𝑛−𝑗 + 𝑢𝑛−𝑗−1 )} , 8 where 𝛽 (𝛽) 𝑤𝑗 = (−1)𝑗 ( ) , 𝑗

𝑤𝑗(𝛼)

(54)

Remark 14 (see [49]). 𝐺2 algorithm is based on Gr¨unwaldLetnikov definition, not only used for numerical calculation of fractional derivative (𝛼 ≥ 0), but also used for numerical calculation of fractional integral (𝛼 ≤ 0). As we all know, the fractional Langevin equation form is

(59)

(𝑗 = 0, 1, 2, . . .) ,

With the above algorithm we will give some examples. Example 1. Consider the following fractional differential equations: 𝑐

𝐷0.8 ( 𝑐 𝐷1.5 + 0.125) 𝑢 (𝑡) = 1.1121𝑢 (𝑡) + 𝑡2 , 0 ≤ 𝑡 ≤ 1,

(55)

where 0 ≤ 𝑡 ≤ 1, 1 < 𝛼 ≤ 2, 0 < 𝛽 ≤ 1, 𝜆 is a constant, 𝑓(𝑡) is an external force, and 𝜉(𝑡) is a random force. The above equation can be written as

𝐷𝛼 V (𝑡) = 𝑢 (𝑡) − 𝜆V (𝑡) .

𝛼 = (−1) ( ) , 𝑗 𝑗

𝛼 (𝛼 − 1) ⋅ ⋅ ⋅ (𝛼 − 𝑗 + 1) 𝛼 . ( )= 𝑗 𝑗!

1 + 𝛼2 (𝑓𝑛−𝑗+1 − 2𝑓𝑛−𝑗 + 𝑓𝑛−𝑗−1 )} . 8

𝑐

(𝑗 = 0, 1, 2, . . .) ,

𝛽 (𝛽 − 1) ⋅ ⋅ ⋅ (𝛽 − 𝑗 + 1) 𝛽 , ( )= 𝑗 𝑗!

𝑛−1

𝐷𝛽 𝑢 (𝑡) = 𝑓 (𝑡) ,

(58)

𝑛−1 1 = ℎ−𝛼 ∑ 𝑤𝑗(𝛼) {𝑢𝑛−𝑗 + 𝛼 (𝑢𝑛−𝑗+1 − 𝑢𝑛−𝑗−1 ) 4 𝑗=0

(𝑎 𝐷 𝑓 (𝑡𝑛 ))𝐺2

𝑐

𝑗=0

Γ (𝑗 − 𝛼) 𝛼 𝑢 (𝑡 − (𝑗 − ) ℎ) . 2 Γ (𝑗 + 1)

1 + 𝛽2 (𝑢𝑛−𝑗+1 − 2𝑢𝑛−𝑗 + 𝑢𝑛−𝑗−1 )} , 8

𝛼

𝐷𝛽 ( 𝑐 𝐷𝛼 + 𝜆) 𝑢 (𝑡) = 𝑓 (𝑡) + 𝜉 (𝑡) ,

∑

𝐷𝛽 𝑢 (𝑡𝑛 )

Then “𝐺2 algorithm” can be expressed as:

𝑐

[𝑡/ℎ+𝛼/2]

(53)

𝛼2 𝛼 + ( − ) 𝑓 (𝑡 − (𝑗 + 1) ℎ) . 8 4

1 = ℎ−𝛼 ∑ 𝑤𝑗(𝛼) {𝑓𝑛−𝑗 + 𝛼 (𝑓𝑛−𝑗+1 − 𝑓𝑛−𝑗−1 ) 4 𝑗=0

(57)

𝑛−1 1 (𝛽) = ℎ−𝛽 ∑ 𝑤𝑗 {𝑢𝑛−𝑗 + 𝛽 (𝑢𝑛−𝑗+1 − 𝑢𝑛−𝑗−1 ) 4 𝑗=0

Therefore, they put forward “fractional center difference quotient” approximation format called in general “𝐺2 algorithm.” In this paper, we use the three-point interpolation formula:

+ (1 −

𝑗=0

Γ (𝑗 − 𝛽) 𝛽 𝑢 (𝑡 − (𝑗 − ) ℎ) , 2 Γ (𝑗 + 1)

The previous equations are approximated by the three-point interpolation formula and can be written as

ℎ−𝛼 𝑎 𝐷 𝑓 (𝑡) = lim ℎ → 0 Γ (−𝛼) 𝛼

[𝑡/ℎ+𝛼/2]

∑

𝐷𝛼 𝑢 (𝑡)

(51)

×

[𝑡/ℎ+𝛽/2]

= lim

𝑓 (𝑡) = lim

(56)

𝑢 (0) = 0,

(60)

( 𝑐 𝐷1.5 + 0.125) 𝑢 (0) = 0.

Obviously, we get 󵄨 󵄨󵄨 󵄨󵄨𝑓 (𝑡, 𝑢 (𝑡) , 𝑢󸀠 (𝑡))󵄨󵄨󵄨 ≤ 1.1121 ‖𝑢‖ + 1, 󵄨 󵄨 (61) 󵄨󵄨 󵄨 󸀠 󵄨󵄨𝑓 (𝑡, 𝑢 (𝑡) , 𝑢 (𝑡)) − 𝑓 (𝑡, V (𝑡) , V󸀠 (𝑡))󵄨󵄨󵄨 ≤ 𝑔∗ ‖𝑢 − V‖ . 󵄨 󵄨

Journal of Applied Mathematics

7

13 4

×10−10

3 6

2 1

4

u(t)

u(t)

5

−1

3

−2

2

−3

1 0

0

−4 0

0.2

0.4

0.8

0.6

0

0.2

0.4

1

t

Figure 1: The displacement without a Wiener process, the numerical simulation of (60), where 𝛼 = 1.5, 𝛽 = 0.8.

0.6

0.8

1

t

Figure 2: The displacement with a Wiener process, the numerical simulation of (60), where 𝛼 = 1.5, 𝛽 = 0.8. 1

×10−9

0.9 ∗

0.8

Letting 𝑝 = 0.9, 𝑔 = 0.05, 𝐿 = 𝑙1 + 𝑙2 = 0.01, 𝛼1 = 𝛽1 = 𝛼2 = 𝛽2 = 2, 𝜂 = 1.5, and 𝑔1 (𝑢) = 𝑔2 (𝑢) = 0, we have

0.7 0.6 u(t)

𝛼𝛽2 [(𝛼1 + 𝛽1 ) 𝜂 + 𝛽1 ] = 24 ≠ 𝛽1 (𝛼2 + 𝛽2 ) = 8, Λ ≜ max {Υ1 + (𝜂 + 1)

0.5 0.4 0.3

× (𝛼𝛽2 𝐿 + 𝛽1 Λ 2 𝐿 + 𝛽1 𝛽2 Λ 2 Υ2 ) + 𝛽1 (𝛼2 + 𝛽2 ) Λ 2 Υ1 ,

(62)

Υ2 + (𝛼2 + 𝛽2 ) (𝛼Λ 2 𝐿 + 𝛼𝛽1 Λ 2 Υ1 )

0.2 0.1 0

0

0.2

0.4

+ 𝛼Λ 1 Λ 2 𝐿 + 𝛽2 𝛽1 Λ 2 Υ2 }

Thus, by Theorem 10, we can get that the problem (60) has at most one solution. With the above algorithm we get Figures 1 and 2. Example 2. Consider the following fractional differential equations:

𝑢 (0) = 0,

1

Letting 𝑝 = 0.9, 𝑔∗ = 0.05, 𝐿 = 𝑙1 + 𝑙2 = 0.01, 𝛼1 = 𝛽1 = 𝛼2 = 𝛽2 = 1, 𝜂 = 1, and 𝑔1 (𝑢) = 𝑔2 (𝑢) = 0, we have 𝛼𝛽2 [(𝛼1 + 𝛽1 ) 𝜂 + 𝛽1 ] = 4.875 ≠ 𝛽1 (𝛼2 + 𝛽2 ) = 2, Λ ≜ max {Υ1 + (𝜂 + 1) × (𝛼𝛽2 𝐿 + 𝛽1 Λ 2 𝐿 + 𝛽1 𝛽2 Λ 2 Υ2 )

𝐷0.723 ( 𝑐 𝐷1.625 + 0.351) 𝑢 (𝑡) = 1.1121𝑒𝑢(𝑡) + 1.3035𝑡3 ,

0.8

Figure 3: The displacement without a Wiener process, the numerical simulation of (63), where 𝛼 = 1.625, 𝛽 = 0.723.

= max {0.3531, 0.3975} = 0.3975 < 1.

𝑐

0.6 t

0 ≤ 𝑡 ≤ 1,

(63)

( 𝑐 𝐷1.625 + 0.351) 𝑢 (0) = 0.

Obviously, we get 󵄨 󵄨󵄨 󵄨󵄨𝑓 (𝑡, 𝑢 (𝑡) , 𝑢󸀠 (𝑡))󵄨󵄨󵄨 ≤ ‖𝑢‖2 , 󵄨 󵄨 (64) 󵄨󵄨 󵄨 󸀠 󵄨󵄨𝑓 (𝑡, 𝑢 (𝑡) , 𝑢 (𝑡)) − 𝑓 (𝑡, V (𝑡) , V󸀠 (𝑡))󵄨󵄨󵄨 ≤ 𝑔∗ ‖𝑢 − V‖ . 󵄨 󵄨

+ 𝛽1 (𝛼2 + 𝛽2 ) Λ 2 Υ1 , Υ2

(65)

+ (𝛼2 + 𝛽2 ) (𝛼Λ 2 𝐿 + 𝛼𝛽1 Λ 2 Υ1 ) + 𝛼Λ 1 Λ 2 𝐿 + 𝛽2 𝛽1 Λ 2 Υ2 } = max {0.8257, 0.9842} = 0.9842 < 1. Thus, by Theorem 10, we can get that the problem (63) has at most one solution. With the above algorithm we get Figures 3 and 4.

14

Journal of Applied Mathematics 4 3

[11]

2

u(t)

1

[12]

0 −1 −2

[13]

−3 −4

0

0.2

0.4

0.6

0.8

1

[14]

t

Figure 4: The displacement with a Wiener process, the numerical simulation of (63), where 𝛼 = 1.625, 𝛽 = 0.723.

[15]

[16]

Acknowledgments This project is supported by NNSF of China (Grants nos. 11271087 and 61263006) and Guangxi Scientific Experimental (China-ASEAN Research) Centre no. 20120116.

[17]

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[18]

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