Leapfrog/Finite Element Method for Fractional Diffusion Equation

Hindawi Publishing Corporation e Scientific World Journal Volume 2014, Article ID 982413, 9 pages http://dx.doi.org/10.1155/2014/982413

Research Article Leapfrog/Finite Element Method for Fractional Diffusion Equation Zhengang Zhao1 and Yunying Zheng2 1 2

Department of Fundamental Courses, Shanghai Customs College, Shanghai 201204, China School of Mathematical Sciences, Huaibei Normal University, Huaibei 235000, China

Correspondence should be addressed to Zhengang Zhao; [email protected] Received 18 January 2014; Accepted 17 February 2014; Published 3 April 2014 Academic Editors: C. Li, A. Sikorskii, and S. B. Yuste Copyright © 2014 Z. Zhao and Y. Zheng. 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 analyze a fully discrete leapfrog/Galerkin finite element method for the numerical solution of the space fractional order (fractional for simplicity) diffusion equation. The generalized fractional derivative spaces are defined in a bounded interval. And some related properties are further discussed for the following finite element analysis. Then the fractional diffusion equation is discretized in space by the finite element method and in time by the explicit leapfrog scheme. For the resulting fully discrete, conditionally stable scheme, we prove an 𝐿2 -error bound of finite element accuracy and of second order in time. Numerical examples are included to confirm our theoretical analysis.

1. Introduction Fractional calculus and fractional partial differential equations (FPDEs) have many applications in various aspects such as in viscoelastic mechanics, power-law phenomenon in fluid and complex network, allometric scaling laws in biology and ecology, colored noise, electrode-electrolyte polarization, dielectric polarization, boundary layer effects in ducts, electromagnetic waves, quantitative finance, quantum evolution of complex systems, and fractional kinetics [1]. And a lot of attention has recently been paid to the problem of the numerical approximation of FPDEs. Generally speaking, the finite difference method and the finite element method are the two main means to solve FPDEs. Recently, some typical fractional difference methods have been utilized to solve FPDEs numerically [2–4]. On the other hand, the finite element method has also been used to find the variational solution of FPDEs [5–14]. But there are still some interesting schemes that can be constructed to enhance the convergence order by using the finite difference/finite element mixed method. In this paper, we use the explicit leapfrog difference/ Galerkin finite element mixed method to numerically solve

the space fractional diffusion equation in order to get a higher convergence order. The fractional diffusion equation as a typical kind of fractional partial differential equation [15] is a generalization of the classical diffusion equation, which can be used to better characterize anomalous diffusion phenomena. Besides, the spatial fractional diffusion equation usually describes 2𝛼 the Lévy flights. The operator RL 𝐷2𝛼 𝑎,𝑥 ( RL 𝐷𝑥,𝑏 ) is commonly referred to the left (right) sided Lévy stable distribution, where the underlying stochastic process is Lévy 𝛼-stable flights; see [16–18]. And a more general form 𝜅1 ⋅ RL 𝐷2𝛼 𝑎,𝑥 + is widely used for mathematical modelling and 𝜅2 ⋅ RL 𝐷2𝛼 𝑥,𝑏 numerical computation. Here, we mainly focus on constructing and analyzing a kind of efficient numerical schemes for approximately solving space fractional diffusion equation. The considered problem reads as follows: for 1/2 < 𝛼 < 1, 𝜕𝑡 𝑢 (𝑥, 𝑡) − Δ𝛼 (𝜆 ⋅ 𝑢 (𝑥, 𝑡)) = 𝑓 (𝑥, 𝑡) , 𝑢 (𝑥, 𝑡) = 𝜑 (𝑡) ,

𝑥 ∈ Ω, 𝑡 ∈ [0, 𝑇] ,

𝑥 ∈ 𝜕Ω, 𝑡 ∈ [0, 𝑇] ,

𝑢 (𝑥, 0) = 𝜓 (𝑥) ,

𝑥 ∈ Ω, (1)

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where Ω = [𝑎, 𝑏], time 𝑇 > 0. Here the spatial fractional 2𝛼 differential operator Δ𝛼 is denoted by 𝜅1 ⋅ RL 𝐷2𝛼 𝑎,𝑥 +𝜅2 ⋅ RL 𝐷𝑥,𝑏 , where 0 ≤ 𝜅1 , 𝜅2 ≤ 1, and 𝜅1 + 𝜅2 = 1. When 𝛼 = 1, the problem models a Brownian diffusion process. And 𝑓 is a source term, 𝜆 is a positive constant. The rest of this paper is constructed as follows. In Section 2, the preliminary knowledge of fractional derivative and the generalized fractional derivative spaces are defined. And some related properties are further discussed. The approximate system of the equation, existence and uniqueness of the weak solution, and the error estimates of the fully discrete scheme for (1) are studied in Section 3. In Section 4, numerical examples are presented to demonstrate the efficiency of the theoretical results derived in Section 3.

Lemma 3 (see [5]). Let Ω = [𝑎, 𝑏] be bounded and 𝛼 > 0. Then 𝑢 ∈ 𝐿𝑝 (Ω) satisfies 󵄩󵄩 󵄩󵄩 (𝑏 − 𝑎)𝛼 󵄩󵄩 RL 𝐷−𝛼 󵄩󵄩 𝑝 ≤ 𝑢(𝑥) 𝑎,𝑥 󵄩 󵄩𝐿 (Ω) Γ (𝛼 + 1) ‖𝑢‖𝐿𝑝 (Ω) ,

(4)

󵄩󵄩 󵄩󵄩 (𝑏 − 𝑎)𝛼 󵄩󵄩 RL 𝐷−𝛼 󵄩󵄩 𝑝 ≤ 𝑢(𝑥) 𝑥,𝑏 󵄩 󵄩𝐿 (Ω) Γ (𝛼 + 1) ‖𝑢‖𝐿𝑝 (Ω) .

Lemma 4 (fractional integration by parts, see [20]). The relation ∫

𝑏

𝑎

−𝛼 RL 𝐷𝑎,𝑥 𝑢 (𝑥)

𝑏

⋅ V (𝑥) 𝑑𝑥 = ∫ 𝑢 (𝑥) ⋅ 𝑎

−𝛼 RL 𝐷𝑥,𝑏 V (𝑥) 𝑑𝑥

(5)

is valid under the assumption that 𝑢 (𝑥) ∈ 𝐿𝑝 (Ω) ,

2. Generalized Fractional Derivative Spaces In this section, we first give the definition of fractional derivatives. There are several definitions for the fractional derivatives, but Riemann-Liouville derivative is one of the most often used fractional derivatives, which is a reasonable generalization of the classical derivative [1, 19–22]. Then we define the generalized fractional derivative spaces by using Riemann-Liouville derivative, which is extended from the 𝐿2 sense to the 𝐿𝑝 sense. Definition 1. The 𝛼th order left and right Riemann-Liouville integrals of function 𝑢(𝑥) are defined in a finite interval (𝑎, 𝑏) as follows: 𝑥 𝑢 (𝑠) 1 𝑑𝑠, ∫ Γ (𝛼) 𝑎 (𝑥 − 𝑠)1−𝛼

−𝛼 RL 𝐷𝑎,𝑥 𝑢 (𝑥)

=

−𝛼 RL 𝐷𝑥,𝑏 𝑢 (𝑥)

𝑏 𝑢 (𝑠) 1 = 𝑑𝑠, ∫ Γ (𝛼) 𝑥 (𝑠 − 𝑥)1−𝛼

(2)

1 1 + ≤ 1 + 𝛼, 𝑝 𝑞

Corollary 5 (see [20]). The formula ∫

𝑛

𝑑 1 ∫ (𝑥 − 𝜏)𝑛−𝛼−1 𝑢 (𝜏) 𝑑𝜏, Γ (𝑛 − 𝛼) 𝑑𝑥𝑛 𝑎

=

𝛼 RL 𝐷𝑥,𝑏 𝑢 (𝑥)

(−1)𝑛 𝑑𝑛 𝑏 = ∫ (𝜏 − 𝑥)𝑛−𝛼−1 𝑢 (𝜏) 𝑑𝜏, Γ (𝑛 − 𝛼) 𝑑𝑥𝑛 𝑥

𝛼 𝑅𝐿 𝐷𝑎,𝑥 𝑢 (𝑥)

𝑏

⋅ V (𝑥) 𝑑𝑥 = ∫ 𝑢 (𝑥) ⋅ 𝑎

𝛼 𝑅𝐿 𝐷𝑥,𝑏 V (𝑥) 𝑑𝑥

(7)

−𝛼 𝑝 is valid under the assumption that 𝑢(𝑥) ∈ 𝑅𝐿 𝐷𝑎,𝑥 (𝐿 (Ω)), −𝛼 𝑞 V(𝑥) ∈ 𝑅𝐿 𝐷𝑥,𝑏 (𝐿 (Ω)), 1/𝑝 + 1/𝑞 ≤ 1 + 𝛼, where the function 𝑝 −𝛼 space 𝑅𝐿 𝐷−𝛼 𝑎,𝑥 (𝐿 (Ω)) = {𝑓(𝑥) | 𝑓(𝑥) = 𝑅𝐿 𝐷𝑎,𝑥 𝜙(𝑥), 𝜙(𝑥) ∈ 𝑝 −𝛼 𝑞 𝐿 (Ω)}, 𝑅𝐿 𝐷𝑥,𝑏 (𝐿 (Ω)) = {𝑓(𝑥) | 𝑓(𝑥) = 𝑅𝐿 𝐷−𝛼 𝑥,𝑏 𝜓(𝑥), 𝜓(𝑥) ∈ 𝐿𝑞 (Ω)}.

Corollary 6 (see [13]). One can further give the following corollary: 𝑏

2𝛼 𝑅𝐿 𝐷𝑎,𝑥 𝑢 (𝑥)

𝑏

⋅ V (𝑥) 𝑑𝑥 = ∫

𝑎

𝛼 𝑅𝐿 𝐷𝑎,𝑥 𝑢 (𝑥)

⋅

𝛼 𝑅𝐿 𝐷𝑥,𝑏 V (𝑥) 𝑑𝑥

(8)

under the assumption that 𝑢(𝑥) ∈ −𝛼 𝑞 𝑅𝐿 𝐷𝑥,𝑏 (𝐿 (Ω)), 1/𝑝 + 1/𝑞 ≤ 1 + 𝛼.

−2𝛼 𝑝 𝑅𝐿 𝐷𝑎,𝑥 (𝐿 (Ω)),

V(𝑥) ∈

𝑝 Note that the above assumption 𝑢(𝑥) ∈ RL 𝐷−2𝛼 𝑎,𝑥 (𝐿 (Ω)) −𝛼 𝑝 implies 𝑢(𝑥) ∈ RL 𝐷𝑎,𝑥 (𝐿 (Ω)) one can prove that by using Lemma 3.

𝑥

𝛼 RL 𝐷𝑎,𝑥 𝑢 (𝑥)

𝑏

𝑎

𝑎

Definition 2. The 𝛼th order left and right Riemann-Liouville derivatives of function 𝑢(𝑥) defined in a finite interval (𝑎, 𝑏) are given as

(6)

𝑝 ≥ 1, 𝑞 ≥ 1,

with 𝑝 ≠ 1, 𝑞 ≠ 1 in the case 1/𝑝 + 1/𝑞 = 1 + 𝛼.

∫ where 𝛼 > 0.

V (𝑥) ∈ 𝐿𝑞 (Ω) ,

(3)

Corollary 7 (see [13]). Consider 𝑏

∫ in which 𝑛 − 1 < 𝛼 < 𝑛 ∈ 𝑍+ . Obviously, they are the integer derivatives of the left and right fractional integrals, respectively. Now, we give some lemmas and corollaries which are necessary to define the generalized fractional derivative spaces.

𝑎

2𝛼 𝑅𝐿 𝐷𝑥,𝑏 𝑢 (𝑥)

⋅ V (𝑥) 𝑑𝑥 = ∫

𝑏

𝑎

𝛼 𝑅𝐿 𝐷𝑥,𝑏 𝑢 (𝑥)

under the assumption that 𝑢(𝑥) ∈ −𝛼 𝑞 𝑅𝐿 𝐷𝑎,𝑥 (𝐿 (Ω)), 1/𝑝 + 1/𝑞 ≤ 1 + 𝛼.

⋅

𝛼 RL 𝐷𝑎,𝑥 V (𝑥) 𝑑𝑥

(9) −2𝛼 𝑝 𝑅𝐿 𝐷𝑥,𝑏 (𝐿 (Ω)),

V(𝑥) ∈

Note that, from the definition of the function space 𝑝 we can get that if 𝑢(𝑥) ∈ RL 𝐷−𝛼 𝑎,𝑥 (𝐿 (Ω)),

−𝛼 𝑝 RL 𝐷𝑎,𝑥 (𝐿 (Ω)),


3

𝛼 then 𝑢(𝑥) = RL 𝐷−𝛼 𝑎,𝑥 𝜙(𝑥), and RL 𝐷𝑎,𝑥 𝑢(𝑥) = 𝜙(𝑥), where 𝑝 𝑝 𝜙(𝑥) ∈ 𝐿 (Ω), such that 𝑢 ∈ 𝐿 (Ω), which is obtained by Lemma 3. And RL 𝐷𝛼𝑎,𝑥 𝑢(𝑥) ∈ 𝐿𝑝 (Ω) naturally holds. So, by the above idea, we define the following fractional derivative spaces from the 𝐿2 sense to the 𝐿𝑝 sense, which will be proved to be equivalent with the fractional Sobolev spaces under some certain conditions.

Definition 8. Define the following norms of the left (with 𝛼,𝑝 symbol 𝑊𝐿 ) fractional derivative space and the right (with 𝛼,𝑝 symbol 𝑊𝑅 ) fractional derivative space in a bounded interval Ω = [𝑎, 𝑏] as follows correspondingly, where 1 < 𝑝 < +∞: 𝛼,𝑝

𝑊𝐿 (Ω) ≡ {𝑢 ∈ 𝐿𝑝 (Ω) :

𝛼 RL 𝐷𝑎,𝑥 𝑢 (𝑥)

∈ 𝐿𝑝 (Ω)}

(10)

equipped with seminorm 󵄩 󵄩 𝛼 𝑢(𝑥)󵄩󵄩󵄩󵄩𝐿𝑝 (Ω) |𝑢|𝑊𝐿𝛼,𝑝 (Ω) = 󵄩󵄩󵄩󵄩 RL 𝐷𝑎,𝑥

(11)

1/𝑝

[𝛼]

,

𝐿

𝑘=0

𝛼,𝑝

𝑊𝑅 (Ω) ≡ {𝑢 ∈ 𝐿𝑝 (Ω) :

RL

From [6], we can get the following lemma, which is true in the 𝐿2 sense. 𝛼,2 𝛼,2 𝛼 Lemma 11. The spaces 𝑊𝐿,0 (Ω), 𝑊𝑅,0 (Ω), 𝐻𝑆,0 (Ω), and 𝛼 𝐻0 (Ω) are equal to equivalent seminorms and norms, where 𝐻𝛼 (Ω) is the fractional Sobolev space in terms of the Fourier transform.

Therefore, in this paper we always use 𝐻0𝛼 when 𝑝 = 2, to denote the fractional derivative space equipped with the norm ‖ ⋅ ‖𝛼 which can be any one of (12), (15), and (18), and 𝐻−𝛼 (Ω) is denoted as the dual space of 𝐻0𝛼 (Ω), with norm ‖ ⋅ ‖−𝛼 . Moreover, we can present some new properties about norms for the above left and right fractional derivative spaces in the 𝐿𝑝 sense.

(12) (1)

𝛼 𝐷𝑥,𝑏 𝑢 ∈ 𝐿𝑝 (Ω)}

−𝛼 𝑅𝐿 𝐷𝑎,𝑥 𝑢(𝑥)

: 𝐿𝑝 (Ω) → 𝐿𝑝 (Ω) is a bounded linear

−𝛼 𝑅𝐿 𝐷𝑥,𝑏 𝑢(𝑥)

: 𝐿𝑝 (Ω) → 𝐿𝑝 (Ω) is a bounded linear

𝛼 𝑅𝐿 𝐷𝑎,𝑥 𝑢(𝑥)

: 𝑊𝐿 (Ω) → 𝐿𝑝 (Ω) is a bounded linear

𝛼 𝑅𝐿 𝐷𝑥,𝑏 𝑢(𝑥)

: 𝑊𝑅 (Ω) → 𝐿𝑝 (Ω) is a bounded linear

−𝛼 𝑅𝐿 𝐷𝑎,𝑥 𝑢(𝑥)

: 𝐿𝑝 (Ω) → 𝑊𝐿 (Ω) is a bounded linear

−𝛼 𝑅𝐿 𝐷𝑥,𝑏 𝑢(𝑥)

: 𝐿𝑝 (Ω) → 𝑊𝑅 (Ω) is a bounded linear

operator;

(13)

equipped with seminorm

(2)

operator;

󵄩 󵄩 𝛼 𝑢(𝑥)󵄩󵄩󵄩󵄩𝐿𝑝 (Ω) |𝑢|𝑊𝑅𝛼,𝑝 (Ω) = 󵄩󵄩󵄩󵄩 RL 𝐷𝑥,𝑏

(14) (3)

and norm

operator; [𝛼]

.

𝑅

𝑘=0

(15)

(4)

operator;

Definition 9. Define the symmetric fractional derivative space (with symbol 𝐻𝑆𝛼 ) in a bounded interval Ω = [𝑎, 𝑏] in the 𝐿2 sense 𝐻𝑆𝛼 (Ω) ≡ { 𝑢 ∈ 𝐿2 (Ω)

(5)

operator; (6)

operator. :∫

RL

𝑎

𝛼,𝑝

1/𝑝

󵄩 󵄩𝑝 𝑝 ‖𝑢‖𝑊𝑅𝛼,𝑝 (Ω) = ( ∑ 󵄩󵄩󵄩󵄩𝐷𝑘 𝑢󵄩󵄩󵄩󵄩𝐿𝑝 (Ω) + |𝑢|𝑊𝛼,𝑝 (Ω) )

𝑏

𝛼,𝑝

Lemma 12. Let 𝛼 > 0 and Ω = [𝑎, 𝑏] ⊂ R be bounded. Then the following mapping properties hold:

and norm 󵄩 󵄩𝑝 𝑝 ‖𝑢‖𝑊𝐿𝛼,𝑝 (Ω) = ( ∑ 󵄩󵄩󵄩󵄩𝐷𝑘 𝑢󵄩󵄩󵄩󵄩𝐿𝑝 (Ω) + |𝑢|𝑊𝛼,𝑝 (Ω) )

𝛼,𝑝

Definition 10. Define the spaces 𝑊𝐿,0 (Ω), 𝑊𝑅,0 (Ω), and 𝛼 𝐻𝑆,0 (Ω) as the closures of 𝐶0∞ (Ω) under their respective norms.

𝛼 𝐷𝑎,𝑥 𝑢 (𝑥) ⋅

𝛼 RL 𝐷𝑥,𝑏 𝑢 (𝑥) 𝑑𝑥

∈ 𝐿2 (Ω)} (16)

𝛼,𝑝

𝛼,𝑝

𝛼,𝑝

Proof. Properties (1) and (2) follow directly from Lemma 3. Property (3) follows directly from the definition of 𝛼,𝑝 𝛼,𝑝 𝑊𝐿 (Ω) and 𝑊𝑅 (Ω) as

equipped with seminorm |𝑢|𝐻𝑆𝛼 (Ω)

󵄨󵄨 𝑏 󵄨 = 󵄨󵄨󵄨󵄨∫ 󵄨󵄨 𝑎

𝐷𝛼 RL 𝑎,𝑥

𝑢 (𝑥) ⋅

𝐷𝛼 RL 𝑥,𝑏

󵄨󵄨1/2 󵄨 𝑢 (𝑥) 𝑑𝑥󵄨󵄨󵄨󵄨 󵄨󵄨

(17)

[𝛼]

󵄩𝑝 󵄩 󵄩𝑝 󵄩 ≤ ( ∑ 󵄩󵄩󵄩󵄩𝐷𝑘 𝑢󵄩󵄩󵄩󵄩𝐿𝑝 (Ω) + 󵄩󵄩󵄩󵄩 RL 𝐷𝛼𝑎,𝑥 𝑢 (𝑥)󵄩󵄩󵄩󵄩𝐿𝑝 (Ω) )

and norm ‖𝑢‖𝐻𝑆𝛼 (Ω)

󵄩󵄩 󵄩 𝛼 󵄩󵄩 𝐷𝑎,𝑥 𝑢 (𝑥)󵄩󵄩󵄩󵄩𝐿𝑝 (Ω) 󵄩 RL

𝑘=0

[𝛼]

1/2

󵄩 󵄩2 = ( ∑ 󵄩󵄩󵄩󵄩𝐷𝑘 𝑢󵄩󵄩󵄩󵄩𝐿2 (Ω) + |𝑢|2𝐻𝛼 (Ω) ) 𝑆 𝑘=0

.

(18)

Property (4) follows similarly.

1/𝑝

(19) .

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Property (5) follows from the definition of 𝑊𝐿 (Ω) and the semigroup property of fractional operator, 󵄩󵄩 󵄩󵄩 󵄩 𝛼,𝑝 󵄩󵄩 RL 𝐷−𝛼 𝑎,𝑥 𝑢 (𝑥)󵄩 󵄩𝑊𝐿 (Ω) 󵄩 [𝛼]

󵄩 = ( ∑ 󵄩󵄩󵄩󵄩𝐷𝑘 ⋅

𝐷−𝛼 RL 𝑎,𝑥

𝑘=0

󵄩 + 󵄩󵄩󵄩󵄩 RL 𝐷𝛼𝑎,𝑥 ⋅

Therefore, (25) is true. And the following inequality certainly holds: 󵄩 󵄩󵄩 󵄩𝑝 . ‖𝑢‖𝐿𝑝 (Ω) ≤ 𝐶󵄩󵄩󵄩󵄩 RL 𝐷𝛼−𝑠 (28) 𝑎,𝑥 𝑢(𝑥)󵄩 󵄩𝐿 (Ω) So, we get that

󵄩𝑝 𝑢 (𝑥)󵄩󵄩󵄩󵄩𝐿𝑝 (Ω)

󵄩󵄩𝑝 −𝛼 󵄩𝑝 ) RL 𝐷𝑎,𝑥 𝑢 (𝑥)󵄩 󵄩𝐿 (Ω)

1/𝑝

(20)

󵄩 󵄩󵄩 󵄩󵄩 󵄩 󵄩𝑝 󵄩󵄩 RL 𝐷𝑠𝑎,𝑥 𝑢 (𝑥)󵄩󵄩󵄩 𝑝 ≤ 󵄩󵄩󵄩 RL 𝐷𝑠𝑎,𝑥 RL 𝐷𝛼−𝑠 𝑎,𝑥 𝑢 (𝑥)󵄩 󵄩𝐿 (Ω) 󵄩 󵄩𝐿 (Ω) 󵄩 󵄩󵄩 󵄩 = 󵄩󵄩󵄩 RL 𝐷𝛼𝑎,𝑥 𝑢(𝑥)󵄩󵄩󵄩󵄩𝐿𝑝 (Ω) .

(29)

Therefore, (26) holds. [𝛼]

󵄩 󵄩𝑝 𝑝 𝑘−𝛼 = ( ∑ 󵄩󵄩󵄩󵄩 RL 𝐷𝑎,𝑥 𝑢(𝑥)󵄩󵄩󵄩󵄩𝐿𝑝 (Ω) + ‖𝑢‖𝐿𝑝 (Ω) )

1/𝑝

.

3. Error Estimates of the Leapfrog/Finite Element Scheme

𝑘=0

Using Lemma 3, there exist constants 𝑐𝑘 , 𝑘 = 0, . . . , [𝛼] such that 󵄩𝑝 󵄩󵄩 𝑝 𝑘−𝛼 󵄩󵄩 𝐷𝑎,𝑥 𝑢(𝑥)󵄩󵄩󵄩󵄩𝐿𝑝 (Ω) ≤ 𝐶𝑘 ‖𝑢‖𝐿𝑝 (Ω) . (21) 󵄩 RL Therefore, we obtain the bound [𝛼]

1/𝑝

󵄩󵄩 󵄩 −𝛼 󵄩󵄩 𝐷𝑎,𝑥 𝑢(𝑥)󵄩󵄩󵄩󵄩𝑊𝛼,𝑝 (Ω) ≤ (1 + ∑ 𝑐𝑘 ) 󵄩 RL 𝐿 𝑘=0

‖𝑢‖𝐿𝑝 (Ω) .

(22)

In this section, we firstly give a fully discrete scheme, where we use the leapfrog difference method in the temporal direction and the finite element method in the spatial direction and then analyze the error estimate. Let 𝑆ℎ denote a uniform partition on Ω, with grid parameter ℎ. For 𝑘 ∈ 𝑁, let 𝑃𝑘 (Ω) denote the space of polynomials on Ω with degree not greater than 𝑘. Then we define 𝑋ℎ as the finite element space on 𝑆ℎ with the basis of the piecewise polynomials of order 𝑘 ∈ 𝑍+ ; that is, 𝑋ℎ = {V ∈ 𝑋 ∩ 𝐶 (Ω) : V|𝐷 ∈ 𝑃𝑘 (𝐷) , ∀𝐷 ∈ 𝑆ℎ } ,

Property (6) follows similarly. Corollary 13. Consider 𝛼,𝑝

𝛼−[𝛼],𝑝

(Ω) ,

𝛼,𝑝

𝛼−[𝛼],𝑝

(Ω)

𝑊𝐿,0 (Ω) 󳨀→ 𝑊𝐿,0

𝑊𝑅,0 (Ω) 󳨀→ 𝑊𝑅,0

(23)

in which 𝐷 is the unit of 𝑆ℎ . The following property of finite element spaces is necessary for our subsequent analysis [23]: for 𝑢 ∈ 𝐻𝑘+1 (Ω), 0 ≤ 𝜇 ≤ 𝑘 + 1, there exists V ∈ 𝑋ℎ such that ‖𝑢 − V‖𝜇 ≤ 𝐶ℎ𝑘+1−𝜇 ‖𝑢‖𝑘+1 .

for 1 ≤ 𝑝 < ∞. And if 1 ≤ 𝑝 ≤ 𝑞 < ∞, one has 𝛼,𝑝

𝛼,𝑝

𝑊𝑅,0 (Ω) 󳨀→ 𝐿𝑞 (Ω) .

(24)

Lemma 15 (discrete Gronwall’s lemma, see [24]). Let Δ𝑡, 𝐻 and 𝑎𝑛 , 𝑏𝑛 , 𝑐𝑛 , 𝛾𝑛 (for integer 𝑛 ≥ 0) be nonnegative numbers such that

It is obviously true by using the norms of fractional derivative spaces and imbedding theorems for 𝐿𝑝 (Ω). Lemma 14. Let Ω = [𝑎, 𝑏] ⊂ R be bounded. Then for 𝑢 ∈ 𝛼,𝑝 𝑊𝐿,0 (Ω), one has 𝛼,𝑝 ‖𝑢‖𝐿𝑝 (Ω) ≤ 𝐶‖𝑢‖𝑊𝐿,0 (Ω) ,

(25)

𝛼,𝑝

𝑁

𝑁

𝑁

𝑛=0

𝑛=0

𝑛=0

𝑎𝑁 + Δ𝑡 ∑ 𝑏𝑛 ≤ Δ𝑡 ∑ 𝛾𝑛 𝑎𝑛 + Δ𝑡 ∑ 𝑐𝑛 + 𝐻,

(26)

Proof. If 𝑢 ∈ 𝑊𝐿,0 (Ω) by using Lemmas 3 and 12, we have that 󵄩 󵄩󵄩 −𝛼 𝛼 󵄩𝑝 ‖𝑢‖𝐿𝑝 (Ω) = 󵄩󵄩󵄩󵄩 RL 𝐷𝑎,𝑥 RL 𝐷𝑎,𝑥 𝑢 (𝑥)󵄩 󵄩𝐿 (Ω) (27) 󵄩 (𝑏 − 𝑎)𝛼 󵄩󵄩 󵄩󵄩 RL 𝐷𝛼𝑎,𝑥 𝑢 (𝑥)󵄩󵄩󵄩 𝑝 . ≤ 󵄩𝐿 (Ω) Γ (𝛼 + 1) 󵄩

(32)

for 𝑁 ≥ 0. Suppose that Δ𝑡𝛾𝑛 < 1 for all 𝑛, and set 𝜎𝑛 = (1 − Δ𝑡𝛾𝑛 )−1 ; then 𝑁

𝑁

𝑁

𝑛=0

𝑛=0

𝑛=0

𝑎𝑁 + Δ𝑡 ∑ 𝑏𝑛 ≤ exp (Δ𝑡 ∑ 𝜎𝑛 𝛾𝑛 ) {Δ𝑡 ∑ 𝑐𝑛 + 𝐻}

and for 0 < 𝑠 < 𝛼, one has 𝑠,𝑝 𝛼,𝑝 ‖𝑢‖𝑊𝐿,0 (Ω) ≤ 𝐶‖𝑢‖𝑊𝐿,0 (Ω) .

(31)

The Gronwall’s lemma is also needed for the error analysis.

𝛼,𝑞

𝑊𝐿,0 (Ω) 󳨀→ 𝑊𝐿,0 (Ω) ,

(30)

(33)

for 𝑁 ≥ 0. In the following, we give the fully discrete scheme of (1). Let Δ𝑡 denote the step size for 𝑡 so that 𝑡𝑛 = 𝑛Δ𝑡, 𝑛 = 1, 2, . . . , 𝑁 − 1. For notational convenience, we denote 𝑢𝑛 := 𝑢(⋅, 𝑡𝑛 ) and 𝑑𝑡 𝑢𝑛 :=

𝑢𝑛+1 − 𝑢𝑛−1 . 2Δ𝑡

(34)


5

Let 𝑢ℎ𝑛 of (1) be the finite element solution at time 𝑡 = 𝑡𝑛 of the following fully discrete scheme: (𝑑𝑡 𝑢ℎ𝑛 , V) − (Δ𝛼 (𝜆 ⋅ 𝑢ℎ𝑛 ) , V) = ⟨𝑓𝑛 , V⟩ ,

∀V ∈ 𝑋ℎ ;

(35)

that is, (𝑢ℎ𝑛+1 − 𝑢ℎ𝑛−1 , V) − 2Δ𝑡 (Δ𝛼 (𝜆 ⋅ 𝑢ℎ𝑛 ) , V) = 2Δ𝑡⟨𝑓𝑛 , V⟩,

(36)

∀V ∈ 𝑋ℎ ,

where (⋅, ⋅) is denoted by an 𝐿2 inner product and (Δ𝛼 (𝜆 ⋅ 𝛼 𝛼 (𝜆 ⋅ 𝑢ℎ𝑛 ), RL 𝐷𝛼𝑥,𝑏 V) + 𝜅2 ⋅ ( RL 𝐷𝑥,𝑏 (𝜆 ⋅ 𝑢ℎ𝑛 ), V) = 𝜅1 ⋅ ( RL 𝐷𝑎,𝑥 𝑛 𝛼 𝛼 𝑛 𝑢ℎ ), RL 𝐷𝑎,𝑥 V). For brevity, we always use (Δ (𝜆⋅𝑢ℎ ), V) instead of the right hand side of this equation. Lemma 16. For a sufficient small step size Δ𝑡 > 0, there exists a unique solution 𝑢ℎ𝑛 ∈ 𝑋ℎ satisfying (36). (𝑢ℎ𝑛 , 𝑢ℎ𝑛 )/2Δ𝑡

𝛼

𝑢ℎ𝑛 ), 𝑢ℎ𝑛 )

− (Δ (𝜆 ⋅ Proof. Firstly, we prove that is positive, which is one of the sufficient conditions for the existence and uniqueness of 𝑢ℎ𝑛 . For Δ𝑡 > 0 chosen sufficiently small, we have that

(𝑢ℎ𝑛 , 𝑢ℎ𝑛 ) 󵄩 󵄩2 (37) − (Δ𝛼 (𝜆 ⋅ 𝑢ℎ𝑛 ) , 𝑢ℎ𝑛 ) ≥ 𝐶󵄩󵄩󵄩𝑢ℎ𝑛 󵄩󵄩󵄩𝛼 . 2Δ𝑡 Besides, by using the fractional Poincare-Friedrichs formula, we can easily get the continuity of (𝑢ℎ𝑛 , 𝑢ℎ𝑛 )/2Δ𝑡 − (Δ𝛼 (𝜆 ⋅ 𝑢ℎ𝑛 ), 𝑢ℎ𝑛 ). Hence, by using the Lax-Milgram theorem, we have that (36) is uniquely solvable for 𝑢ℎ𝑛 . Now, we carry out the error analysis for the fully discrete problem. The following norms are also used in the analysis: 󵄩 󵄩 ‖𝑢‖∞,𝑘 = max 󵄩󵄩󵄩𝑢𝑛 󵄩󵄩󵄩𝑘 , 1≤𝑛≤𝑁 ‖𝑢‖0,𝛼

𝑁

󵄩 󵄩2 = ( ∑ 󵄩󵄩󵄩𝑢𝑛 󵄩󵄩󵄩𝛼 𝑑𝑡)

(38)

1/2

.

𝑛=1

Theorem 17. Assume that (1) has a solution 𝑢 satisfying 𝑢 ∈ 𝐿2 (0, 𝑇; 𝐻𝛼 ∩ 𝐻𝑘+1 (Ω)), 𝑢𝑡 ∈ 𝐿2 (0, 𝑇; 𝐻𝑘+1 (Ω)), and 𝑢𝑡𝑡𝑡 ∈ 𝐿2 (0, 𝑇; 𝐿2 (Ω)), with 𝑢0 ∈ 𝐻𝑘+1 (Ω). 𝑢ℎ𝑛 is the solution of (36), and 𝑢ℎ1 is computed in such a way that 󵄩 󵄩󵄩 1 󵄩󵄩𝑢 (𝑥) − 𝑢 (𝑥, Δ𝑡)󵄩󵄩󵄩 ≤ 𝐶(Δ𝑡)2 . (39) 󵄩 󵄩 Then, there exists a constant 𝐶0 independent of ℎ and Δ𝑡, such that if Δ𝑡 ⋅ ℎ−2𝛼 ≤ 𝐶0 ,

(40)

then the finite element approximation (36) is convergent to the solution of (1) on the interval (0, 𝑇) as Δ𝑡, ℎ → 0. And the approximation solution 𝑢ℎ satisfies the following error estimates: 󵄩 󵄩 󵄩󵄩 𝑘+1 󵄩 󵄩 2󵄩 󵄩󵄩𝑢 − 𝑢ℎ 󵄩󵄩󵄩0,𝛼 ≤ 𝐶 (ℎ 󵄩󵄩󵄩𝑢𝑡 󵄩󵄩󵄩0,𝑘+1 + (Δ𝑡) 󵄩󵄩󵄩𝑢𝑡𝑡𝑡 󵄩󵄩󵄩0,0 (41) + ℎ𝑘+1−𝛼 ‖𝑢‖0,𝑘+1 ) ; 󵄩󵄩 󵄩 󵄩 𝑘+1 󵄩 󵄩 2󵄩 󵄩󵄩𝑢 − 𝑢ℎ 󵄩󵄩󵄩∞,0 ≤ 𝐶 (ℎ 󵄩󵄩󵄩𝑢𝑡 󵄩󵄩󵄩0,𝑘+1 + (Δ𝑡) 󵄩󵄩󵄩𝑢𝑡𝑡𝑡 󵄩󵄩󵄩0,0 +ℎ𝑘+1−𝛼 ‖𝑢‖0,𝑘+1 + ℎ𝑘+1 ‖𝑢‖∞,𝑘+1 ) .

Proof. In order to estimate (41) and (42), we first discuss the error at 𝑡 = 𝑡𝑛 , 𝑛 = 1, 2, . . . , 𝑁 − 1. Let 𝑢𝑛 = 𝑢(⋅, 𝑡𝑛 ) represent the solution of (1), define 𝜀𝑛 = 𝑢𝑛 −𝑢ℎ𝑛 , and for 𝑈𝑛 ∈ 𝑋ℎ , define Λ𝑛 and 𝐸𝑛 as Λ𝑛 = 𝑢𝑛 − 𝑈𝑛 , 𝐸𝑛 = 𝑈𝑛 − 𝑢ℎ𝑛 . So, we have 𝜀𝑛 = Λ𝑛 + 𝐸𝑛 . Obviously the true solution of this problem (1) 𝑢𝑛 also satisfies (𝑑𝑡 𝑢𝑛 , V) − (Δ𝛼 (𝜆 ⋅ 𝑢𝑛 ) , V) (43) = ⟨𝑓𝑛 , V⟩ − (𝑢𝑡𝑛 − 𝑑𝑡 𝑢𝑛 , V) , ∀V ∈ 𝑋ℎ . Therefore, subtracting (36) from (43) gives (𝑑𝑡 𝜖𝑛 , V) − (Δ𝛼 (𝜆 ⋅ 𝜖𝑛 ) , V) = (𝑑𝑡 𝑢𝑛 − 𝑢𝑡𝑛 , V) ,

∀V ∈ 𝑋ℎ ; (44)

that is, (𝜖𝑛+1 − 𝜖𝑛−1 , V) − 2Δ𝑡 (Δ𝛼 (𝜆 ⋅ 𝜖𝑛 ) , V) = 2Δ𝑡 (𝑑𝑡 𝑢𝑛 − 𝑢𝑡𝑛 , V) ,

∀V ∈ 𝑋ℎ .

Substituting 𝜖𝑛+1 = Λ𝑛+1 + 𝐸𝑛+1 , V = 𝐸𝑛+1 + 𝐸𝑛−1 into (45) leads to (𝐸𝑛+1 − 𝐸𝑛−1 , 𝐸𝑛+1 + 𝐸𝑛−1 ) − 2Δ𝑡 (Δ𝛼 (𝜆 ⋅ 𝐸𝑛 ) , 𝐸𝑛+1 + 𝐸𝑛−1 ) = − (Λ𝑛+1 − Λ𝑛−1 , 𝐸𝑛+1 + 𝐸𝑛−1 )

(46)

+ 2Δ𝑡 (Δ𝛼 (𝜆 ⋅ Λ𝑛 ) , 𝐸𝑛+1 + 𝐸𝑛−1 ) + 2Δ𝑡 (𝑑𝑡 𝑢𝑛 − 𝑢𝑡𝑛 , 𝐸𝑛+1 + 𝐸𝑛−1 ) . 2

After adding ‖𝐸𝑛 ‖ to both sides of (46), we obtain the identity 󵄩󵄩󵄩𝐸𝑛+1 󵄩󵄩󵄩2 + 󵄩󵄩󵄩𝐸𝑛 󵄩󵄩󵄩2 − 2Δ𝑡 (Δ𝛼 (𝜆 ⋅ 𝐸𝑛 ) , 𝐸𝑛+1 ) 󵄩󵄩 󵄩 󵄩 󵄩󵄩 󵄩2 󵄩 󵄩2 󵄩 = 󵄩󵄩󵄩𝐸𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝐸𝑛−1 󵄩󵄩󵄩󵄩 + 2Δ𝑡 (Δ𝛼 (𝜆 ⋅ 𝐸𝑛 ) , 𝐸𝑛−1 ) − (Λ𝑛+1 − Λ𝑛−1 , 𝐸𝑛+1 + 𝐸𝑛−1 )

(47)

+ 2Δ𝑡 (Δ𝛼 (𝜆 ⋅ Λ𝑛 ) , 𝐸𝑛+1 + 𝐸𝑛−1 ) + 2Δ𝑡 (𝑑𝑡 𝑢𝑛 − 𝑢𝑡𝑛 , 𝐸𝑛+1 + 𝐸𝑛−1 ) . Define now the quantity 𝐴𝑛+1 , for 1 ≤ 𝑛 ≤ 𝑁 − 1, by 󵄩2 󵄩 󵄩2 󵄩 𝐴𝑛+1 = 󵄩󵄩󵄩󵄩𝐸𝑛+1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐸𝑛 󵄩󵄩󵄩 − 2Δ𝑡 (Δ𝛼 (𝜆 ⋅ 𝐸𝑛 ) , 𝐸𝑛+1 ) . We can rewrite (47) as

(48)

𝐴𝑛+1 = 𝐴𝑛 − (Λ𝑛+1 − Λ𝑛−1 , 𝐸𝑛+1 + 𝐸𝑛−1 ) + 2Δ𝑡 (Δ𝛼 (𝜆 ⋅ 𝐸𝑛 ) , 𝐸𝑛−1 ) + 2Δ𝑡 (Δ𝛼 (𝜆 ⋅ Λ𝑛 ) , 𝐸𝑛+1 + 𝐸𝑛−1 ) + 2Δ𝑡 (Δ𝛼 (𝜆 ⋅ 𝐸𝑛−1 ) , 𝐸𝑛 )

(42)

(45)

+ 2Δ𝑡 (𝑑𝑡 𝑢𝑛 − 𝑢𝑡𝑛 , 𝐸𝑛+1 + 𝐸𝑛−1 ) .

(49)

6


Denoting

For the third term of the right hand side, one has 2Δ𝑡 (Δ𝛼 (𝜆 ⋅ Λ𝑛 ) , 𝐸𝑛+1 + 𝐸𝑛−1 )

𝐹 (𝐸𝑛−1 , 𝐸𝑛 , 𝐸𝑛+1 ) = − (Λ𝑛+1 − Λ𝑛−1 , 𝐸𝑛+1 + 𝐸𝑛−1 )

= 2Δ𝑡 ⋅ 𝜅1 ( RL 𝐷𝛼𝑎,𝑥 (𝜆 ⋅ Λ𝑛 ) ,

+ 2Δ𝑡 (Δ𝛼 (𝜆 ⋅ 𝐸𝑛 ) , 𝐸𝑛−1 )

+ 2Δ𝑡 (Δ (𝜆 ⋅ 𝐸

𝑛−1

𝑛

𝛼 RL 𝐷𝑎,𝑥

(𝐸𝑛+1 + 𝐸𝑛−1 ))

(54)

󵄩 󵄩 󵄩 󵄩 ≤ 𝐶4 Δ𝑡󵄩󵄩󵄩Λ𝑛 󵄩󵄩󵄩𝛼 ⋅ 󵄩󵄩󵄩󵄩𝐸𝑛+1 + 𝐸𝑛−1 󵄩󵄩󵄩󵄩𝛼

),𝐸 )

+ 2Δ𝑡 (𝑑𝑡 𝑢𝑛 − 𝑢𝑡𝑛 , 𝐸𝑛+1 + 𝐸𝑛−1 ) ,

󵄩 󵄩2 ≤ 𝐶5 Δ𝑡 ⋅ ℎ2(𝑘+1−𝛼) 󵄩󵄩󵄩𝑢𝑛 󵄩󵄩󵄩𝑘+1 󵄩2 󵄩 󵄩2 󵄩 + 𝐶6 Δ𝑡 (󵄩󵄩󵄩󵄩𝐸𝑛+1 󵄩󵄩󵄩󵄩𝛼 + 󵄩󵄩󵄩󵄩𝐸𝑛−1 󵄩󵄩󵄩󵄩𝛼 ) ,

then (49) can be abbreviated as 𝐴𝑛+1 = 𝐴𝑛 + 𝐹 (𝐸𝑛−1 , 𝐸𝑛 , 𝐸𝑛+1 ) .

(51)

in which 󵄩󵄩 󵄩 𝑛󵄩 𝑛󵄩 𝑘+1−𝛼 󵄩 󵄩󵄩𝑢𝑛 󵄩󵄩󵄩 . 󵄩󵄩𝜆 ⋅ Λ 󵄩󵄩󵄩𝛼 ≤ ‖𝜆‖∞ ⋅ 󵄩󵄩󵄩Λ 󵄩󵄩󵄩𝛼 ≤ 𝐶5 ℎ 󵄩 󵄩𝑘+1 For the fourth term of the right hand side, one has

= 2Δ𝑡 ⋅ 𝜅1 ( RL 𝐷𝛼𝑎,𝑥 𝜆 ⋅ 𝐸𝑛−1 ,

𝛼 𝑛 RL 𝐷𝑥,𝑏 𝐸 )

+ 2Δ𝑡 ⋅ 𝜅2 ( RL 𝐷𝛼𝑥,𝑏 𝜆 ⋅ 𝐸𝑛−1 ,

2Δ𝑡 (Δ𝛼 (𝜆 ⋅ 𝐸𝑛 ) , 𝐸𝑛−1 )

(56)

And for the term 2Δ𝑡(𝑑𝑡 𝑢𝑛 − 𝑢𝑡𝑛 , 𝐸𝑛+1 + 𝐸𝑛−1 ), by using the Cauchy-Schwarz inequality, we obtain

𝛼 𝑛−1 )) RL 𝐷𝑎,𝑥 𝐸

󵄩2 󵄩 󵄩 󵄩2 ≤ 𝐶1 Δ𝑡󵄩󵄩󵄩𝐸𝑛 󵄩󵄩󵄩𝛼 + 𝐶2 Δ𝑡󵄩󵄩󵄩󵄩𝐸𝑛−1 󵄩󵄩󵄩󵄩𝛼 , (Λ𝑛+1 − Λ𝑛−1 , 𝐸𝑛+1 + 𝐸𝑛−1 )

𝛼 𝑛 RL 𝐷𝑎,𝑥 𝐸 )

󵄩2 󵄩 󵄩 󵄩2 ≤ 𝐶7 Δ𝑡 ⋅ 󵄩󵄩󵄩󵄩𝐸𝑛−1 󵄩󵄩󵄩󵄩𝛼 + 𝐶8 Δ𝑡 ⋅ 󵄩󵄩󵄩𝐸𝑛 󵄩󵄩󵄩𝛼 .

𝛼 𝑛−1 ) RL 𝐷𝑥,𝑏 𝐸

+ 𝜅2 ( RL 𝐷𝛼𝑥,𝑏 (𝜆 ⋅ 𝐸𝑛 ) ,

(55)

2Δ𝑡 (Δ𝛼 (𝜆 ⋅ 𝐸𝑛−1 ) , 𝐸𝑛 )

We now estimate each term in 𝐹(𝐸𝑛−1 , 𝐸𝑛 , 𝐸𝑛+1 ). For the second term of the right hand side, one has

= 2Δ𝑡 (𝜅1 ( RL 𝐷𝛼𝑎,𝑥 (𝜆 ⋅ 𝐸𝑛 ) ,

(𝐸𝑛+1 + 𝐸𝑛−1 ))

+ 2Δ𝑡 ⋅ 𝜅2 ( RL 𝐷𝛼𝑥,𝑏 (𝜆 ⋅ Λ𝑛 ) ,

+ 2Δ𝑡 (Δ𝛼 (𝜆 ⋅ Λ𝑛 ) , 𝐸𝑛+1 + 𝐸𝑛−1 ) (50) 𝛼

𝛼 RL 𝐷𝑥,𝑏

2Δ𝑡 (𝑑𝑡 𝑢𝑛 − 𝑢𝑡𝑛 , 𝐸𝑛+1 + 𝐸𝑛−1 ) = (𝑢𝑛+1 − 𝑢𝑛−1 − 2Δ𝑡 ⋅ 𝑢𝑡𝑛 , 𝐸𝑛+1 + 𝐸𝑛−1 )

(52)

󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩󵄩Λ𝑛+1 − Λ𝑛−1 󵄩󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩󵄩𝐸𝑛+1 + 𝐸𝑛−1 󵄩󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 = 2Δ𝑡 󵄩󵄩󵄩𝑑𝑡 Λ𝑛 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩󵄩𝐸𝑛+1 + 𝐸𝑛−1 󵄩󵄩󵄩󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩 󵄩2 ≤ Δ𝑡󵄩󵄩󵄩𝑑𝑡 Λ𝑛 󵄩󵄩󵄩 + Δ𝑡 (󵄩󵄩󵄩󵄩𝐸𝑛+1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝐸𝑛−1 󵄩󵄩󵄩󵄩 )

(57)

󵄩 𝑛+1 󵄩2 󵄩 𝑛−1 󵄩2 󵄩 𝑛 󵄩󵄩2 ≤ 𝐶9 (Δ𝑡)5 󵄩󵄩󵄩𝑢𝑡𝑡𝑡 󵄩󵄩 + Δ𝑡 (󵄩󵄩󵄩󵄩𝐸 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝐸 󵄩󵄩󵄩󵄩 ) , where by Taylor’s theorem 󵄩󵄩 𝑛+1 󵄩 𝑛 󵄩 󵄩󵄩 . 󵄩󵄩𝑢 − 𝑢𝑛−1 − 2Δ𝑡 ⋅ 𝑢𝑡𝑛 󵄩󵄩󵄩 ≤ 𝐶9 (Δ𝑡)3 󵄩󵄩󵄩󵄩𝑢𝑡𝑡𝑡 󵄩 󵄩 󵄩 Hence, summing from 𝑛 = 1 to 𝑁 − 1, one has

󵄩2 󵄩 󵄩2 󵄩 󵄩 󵄩2 ≤ 𝐶3 Δ𝑡 ⋅ ℎ2𝑘+2 󵄩󵄩󵄩𝑢𝑡𝑛 󵄩󵄩󵄩𝑘+1 + Δ𝑡 (󵄩󵄩󵄩󵄩𝐸𝑛+1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝐸𝑛−1 󵄩󵄩󵄩󵄩 ) ,

(58)

𝑁−1

𝐴𝑁 − 𝐴𝑁−1 ≤ ∑ 𝐹 (𝐸𝑛−1 , 𝐸𝑛 , 𝐸𝑛+1 ) ;

(59)

𝑛=1

where

that is, 𝐴𝑁 ≤ 𝐴𝑁−1

𝑁

󵄩2 󵄩 ∑ Δ𝑡󵄩󵄩󵄩𝑑𝑡 Λ𝑛 󵄩󵄩󵄩

𝑁−1

󵄩 󵄩2 󵄩 󵄩2 + 𝐶10 ∑ Δ𝑡 (ℎ2𝑘+2 󵄩󵄩󵄩𝑢𝑡𝑛 󵄩󵄩󵄩𝑘+1 + ℎ2𝑘+2−2𝛼 󵄩󵄩󵄩𝑢𝑛 󵄩󵄩󵄩𝑘+1

𝑛=0

󵄩󵄩 󵄩2 󵄩 1 𝑡𝑛 𝜕Λ 󵄩󵄩 = ∑ Δ𝑡󵄩󵄩󵄩󵄩 ∫ 1 𝑑𝑡󵄩󵄩󵄩󵄩 󵄩 Δ𝑡 𝑡𝑛−1 𝜕𝑡 󵄩󵄩 𝑛=1 󵄩

𝑛=1

𝑁

𝑁

𝑡𝑛 𝑡𝑛 𝜕Λ 1 2 ≤ ∑ Δ𝑡( ) ∫ (∫ 1𝑑𝑡) (∫ 𝑑𝑡) 𝑑𝑥 Δ𝑡 Ω 𝑡𝑛−1 𝑡𝑛−1 𝜕𝑡 𝑛=1

󵄩 󵄩2 ≤ 𝐶3 ℎ2𝑘+2 󵄩󵄩󵄩𝑢𝑡 󵄩󵄩󵄩0,𝑘+1 .

󵄩 𝑛 󵄩󵄩2 +(Δ𝑡)4 󵄩󵄩󵄩𝑢𝑡𝑡𝑡 󵄩󵄩 )

(53)

𝑁−1 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 + 𝐶11 ∑ Δ𝑡 (󵄩󵄩󵄩󵄩𝐸𝑛+1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝐸𝑛−1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐸𝑛 󵄩󵄩󵄩𝛼 𝑛=1

󵄩2 󵄩 󵄩2 󵄩 + 󵄩󵄩󵄩󵄩𝐸𝑛−1 󵄩󵄩󵄩󵄩𝛼 + 󵄩󵄩󵄩󵄩𝐸𝑛+1 󵄩󵄩󵄩󵄩𝛼 ) .

(60)


7

We now show that, under our stability assumption (40), 2 2 𝐴𝑛+1 is positive and comparable to ‖𝐸𝑛 ‖ + ‖𝐸𝑛+1 ‖ . To this −𝛼 end, we use the inverse inequality ‖V‖𝛼 ≤ 𝐶12 ℎ ‖V‖, V ∈ 𝑋ℎ , and this yields 󵄨 󵄨 2Δ𝑡 󵄨󵄨󵄨󵄨− (Δ𝛼 (𝜆 1 𝐸𝑛 , 𝐸𝑛+1 ))󵄨󵄨󵄨󵄨 󵄩2 󵄩 󵄩2 󵄩 ≤ Δ𝑡 (󵄩󵄩󵄩𝐸𝑛 󵄩󵄩󵄩𝛼 + 󵄩󵄩󵄩󵄩𝐸𝑛+1 󵄩󵄩󵄩󵄩𝛼 ) (61) 󵄩2 󵄩 󵄩2 󵄩 ≤ 𝐶12 Δ𝑡 ⋅ ℎ−2𝛼 (󵄩󵄩󵄩𝐸𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝐸𝑛+1 󵄩󵄩󵄩󵄩 ) .

By using the discrete Gronwall’s Lemma 15, we have 𝑁 󵄩󵄩 𝑁󵄩󵄩2 󵄩󵄩𝐸 󵄩󵄩 + 𝐶14 ∑ Δ𝑡󵄩󵄩󵄩󵄩𝐸𝑛 󵄩󵄩󵄩󵄩2𝛼 󵄩 󵄩 𝑛=1

󵄩 󵄩2 ≤ 𝐶17 󵄩󵄩󵄩󵄩𝐸1 󵄩󵄩󵄩󵄩

(66)

󵄩 󵄩2 + 𝐶15 (ℎ2𝑘+2 󵄩󵄩󵄩𝑢𝑡 󵄩󵄩󵄩0,𝑘+1 + ℎ2𝑘+2−2𝛼 ‖𝑢‖20,𝑘+1 󵄩 󵄩2 + (Δ𝑡)4 󵄩󵄩󵄩𝑢𝑡𝑡𝑡 󵄩󵄩󵄩0,0 ) ,

Hence, if Δ𝑡 ⋅ ℎ−2𝛼 is sufficiently small such that 𝐶12 Δ𝑡 ⋅ ℎ−2𝛼 ≤ 𝐶13 ≤ 1, we get 󵄩2 󵄩 󵄩2 󵄩 (1 − 𝐶13 ) (󵄩󵄩󵄩󵄩𝐸𝑁−1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝐸𝑁󵄩󵄩󵄩󵄩 ) (62) 󵄩2 󵄩 󵄩2 󵄩 ≤ 𝐴𝑁 ≤ (1 + 𝐶13 ) (󵄩󵄩󵄩󵄩𝐸𝑁−1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝐸𝑁󵄩󵄩󵄩󵄩 ) . So we have that

2

2

where denoting 𝑎𝑛 = ‖𝐸𝑁‖ , 𝑏𝑛 = ‖𝐸𝑛 ‖𝛼 , 𝐻 2 𝐶15 (ℎ2𝑘+2 ‖𝑢𝑡 ‖20,𝑘+1 + ℎ2𝑘+2−2𝛼 ‖𝑢𝑛 ‖𝑘+1 + (Δ𝑡)4 ‖𝑢𝑡𝑡𝑡 ‖20,0 ).

=

Now denoting 𝐺(Δ𝑡, ℎ) = ℎ2𝑘+2 ‖𝑢𝑡 ‖20,𝑘+1 +ℎ2𝑘+2−2𝛼 ‖𝑢‖20,𝑘+1 +(Δ𝑡)4 ‖𝑢𝑡𝑡𝑡 ‖20,0 and using the condition (39), we get that 𝑁

󵄩 󵄩2 ‖𝐸‖20,𝛼 = ∑ Δ𝑡󵄩󵄩󵄩𝐸𝑛 󵄩󵄩󵄩𝛼 ≤ 𝐶18 (𝑇 + 1) 𝐺 (Δ𝑡, ℎ) .

𝑁

󵄩󵄩 𝑁−1 󵄩󵄩2 󵄩󵄩 𝑁󵄩󵄩2 󵄩󵄩𝐸 󵄩󵄩 + 󵄩󵄩𝐸 󵄩󵄩 + 𝐶14 ∑ Δ𝑡󵄩󵄩󵄩󵄩𝐸𝑛 󵄩󵄩󵄩󵄩2𝛼 󵄩 󵄩 󵄩 󵄩

(67)

𝑛=1

𝑛=1

By using the interpolation property and the following result

󵄩 󵄩2 󵄩 󵄩2 ≤ 󵄩󵄩󵄩󵄩𝐸1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝐸0 󵄩󵄩󵄩󵄩

󵄩󵄩 󵄩 󵄩󵄩𝑢 − 𝑢ℎ 󵄩󵄩󵄩0,𝛼 ≤ ‖𝐸‖0,𝛼 + ‖Λ‖0,𝛼 ,

𝑁−1

󵄩 󵄩2 󵄩 󵄩2 + 𝐶15 ∑ Δ𝑡 (ℎ2𝑘+2 󵄩󵄩󵄩𝑢𝑡𝑛 󵄩󵄩󵄩𝑘+1 + ℎ2𝑘+2−2𝛼 󵄩󵄩󵄩𝑢𝑛 󵄩󵄩󵄩𝑘+1

(63)

𝑛=1

󵄩 𝑛 󵄩󵄩2 + (Δ𝑡)4 󵄩󵄩󵄩𝑢𝑡𝑡𝑡 󵄩󵄩 ) 𝑁−1

󵄩 + 𝐶16 ∑ Δ𝑡 (󵄩󵄩󵄩󵄩𝐸

𝑛+1 󵄩 󵄩2

𝑛=1

󵄩 󵄩󵄩 + 󵄩󵄩󵄩𝐸 󵄩 󵄩

󵄩󵄩 ) . 󵄩

≤ 𝐺 (Δ𝑡, ℎ) + ℎ2𝑘+2 ‖𝑢‖2∞,𝑘+1 ,

Therefore, we obtain

4. Numerical Examples for Piecewise Linear Polynomials

𝑛=1

󵄩 󵄩2 󵄩 󵄩2 ≤ 󵄩󵄩󵄩󵄩𝐸1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝐸0 󵄩󵄩󵄩󵄩 󵄩2 2𝑘+2−2𝛼 󵄩 󵄩󵄩𝑢𝑛 󵄩󵄩󵄩2 󵄩󵄩𝑢𝑡 󵄩󵄩󵄩0,𝑘+1 + ℎ 󵄩 󵄩𝑘+1

+ 𝐶15 (ℎ

(64)

󵄩 󵄩2 + (Δ𝑡)4 󵄩󵄩󵄩𝑢𝑡𝑡𝑡 󵄩󵄩󵄩0,0 ) 𝑁−1 󵄩2 󵄩 󵄩2 󵄩 + 𝐶16 ∑ Δ𝑡 (󵄩󵄩󵄩󵄩𝐸𝑛+1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝐸𝑛−1 󵄩󵄩󵄩󵄩 ) . 𝑛=1

Hence, 𝑁 󵄩󵄩 𝑁󵄩󵄩2 󵄩󵄩𝐸 󵄩󵄩 + 𝐶14 ∑ Δ𝑡󵄩󵄩󵄩󵄩𝐸𝑛 󵄩󵄩󵄩󵄩2𝛼 󵄩 󵄩 𝑛=1

󵄩 󵄩2 󵄩 󵄩2 ≤ 󵄩󵄩󵄩󵄩𝐸1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝐸0 󵄩󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 + 𝐶15 (ℎ2𝑘+2 󵄩󵄩󵄩𝑢𝑡 󵄩󵄩󵄩0,𝑘+1 + ℎ2𝑘+2−2𝛼 󵄩󵄩󵄩𝑢𝑛 󵄩󵄩󵄩𝑘+1 𝑁−1 󵄩2 󵄩 󵄩 󵄩2 + (Δ𝑡) 󵄩󵄩󵄩𝑢𝑡𝑡𝑡 󵄩󵄩󵄩0,0 ) + 𝐶16 ∑ Δ𝑡󵄩󵄩󵄩󵄩𝐸𝑛+1 󵄩󵄩󵄩󵄩 . 4

𝑛=1

(69)

which yields estimate (42).

𝑁 󵄩󵄩 𝑁−1 󵄩󵄩2 󵄩󵄩 𝑁󵄩󵄩2 󵄩󵄩𝐸 󵄩󵄩 + 󵄩󵄩𝐸 󵄩󵄩 + 𝐶14 ∑ Δ𝑡󵄩󵄩󵄩󵄩𝐸𝑛 󵄩󵄩󵄩󵄩2𝛼 󵄩 󵄩 󵄩 󵄩

2𝑘+2 󵄩 󵄩

estimate (41) follows. Using estimate (66) and approximation properties, we have 󵄩2 󵄩󵄩 2 2 󵄩󵄩𝑢 − 𝑢ℎ 󵄩󵄩󵄩∞,0 ≤ ‖𝐸‖∞,0 + ‖Λ‖∞,0

𝑛−1 󵄩 󵄩2

(68)

(65)

Let 𝑆ℎ denote a uniform partition on Ω = [𝑎, 𝑏] and 𝑋ℎ the space of continuous piecewise linear functions on 𝑆ℎ ; that is, 𝑘 = 1. Then we use the Galerkin finite element method for the spatial variables. After the spatial discretization, we get classical ODEs systems with variables 𝑢ℎ𝑖 , 𝑖 = 1, 2, . . . , 𝑇/Δ𝑡. In order to satisfy the condition (39) in Theorem 17, we use the two-order Runge-Kutta method to compute the variable 𝑢ℎ1 . In this section, we present numerical calculations which support the error estimates in Theorem 17. If we suppose Δ𝑡 = 𝐶ℎ2𝛼 , then we have the convergence rate 󵄩 󵄩󵄩 󵄩󵄩𝑢(𝑡𝑛+1 ) − 𝑢ℎ𝑛+1 󵄩󵄩󵄩 ∼ O (ℎ2−𝛼 ) , 󵄩0,𝛼 󵄩 󵄩󵄩 󵄩 2−𝛼 󵄩󵄩𝑢(𝑡𝑛+1 ) − 𝑢ℎ𝑛+1 󵄩󵄩󵄩 󵄩 󵄩∞,0 ∼ O (ℎ ) .

(70)

Example 1. (i) Let 𝑢 (𝑥, 𝑡) = 𝑡2 𝑥 (1 − 𝑥)

(71)

8

The Scientific World Journal Table 1: The experiential error results and convergence rates of Example 1 (i).

ℎ 1/4 1/8 1/16 1/32 1/64

‖𝑢 − 𝑢ℎ ‖0,0 1.0569 ⋅ 10−2 4.0416 ⋅ 10−3 3.8027 ⋅ 10−4 1.0910 ⋅ 10−4 4.8599 ⋅ 10−5

‖𝑢 − 𝑢ℎ ‖∞,0 3.1077 ⋅ 10−2 6.5469 ⋅ 10−3 4.4162 ⋅ 10−4 1.4097 ⋅ 10−4 6.4241 ⋅ 10−5

cv. rate — 1.3869 3.4098 1.8014 1.1551

cv. rate — 2.2470 3.8899 1.6474 1.1338

Table 2: The experiential error results and convergence rates of Example 1 (ii). ℎ 1/4 1/8 1/16 1/32 1/64

‖𝑢 − 𝑢ℎ ‖0,0 5.6283 ⋅ 10−3 1.9379 ⋅ 10−3 7.1701 ⋅ 10−4 2.6932 ⋅ 10−4 1.0362 ⋅ 10−4

then 𝑢 is the exact solution to the problem 𝛼

𝜕𝑡 𝑢 (𝑥, 𝑡) − Δ (𝜆 ⋅ 𝑢 (𝑥, 𝑡)) = 𝑓 (𝑥, 𝑡) , 𝑢 (0, 𝑡) = 𝑢 (1, 𝑡) = 0, 𝑢 (𝑥, 0) = 0,

𝑥 ∈ Ω, 𝑡 ∈ [0, 𝑇] ,

𝑡 ∈ [0, 𝑇] , 𝑥 ∈ Ω, (72)

where 𝛼 = 0.8, 𝜅1 = 𝜅2 = 1/2, 𝜆 = 1, Ω = [0, 1], 𝑇 = 1, and 𝑓 (𝑥, 𝑡) = 2𝑡𝑥 (1 − 𝑥) − 𝑡2 (

𝑥

1−2𝛼

‖𝑢 − 𝑢ℎ ‖∞,0 9.3921 ⋅ 10−3 3.2521 ⋅ 10−3 1.2910 ⋅ 10−3 4.7917 ⋅ 10−4 1.9584 ⋅ 10−4

cv. rate — 1.5382 1.4344 1.4128 1.3780

cv. rate — 1.5301 1.4274 1.3353 1.2909

accurate explicit scheme, leapfrog difference method in time, and the finite element method in space. Under the suitably accurate initial conditions and the stability requirement that Δ𝑡 ⋅ ℎ−2𝛼 be sufficiently small, the error analysis for the fully discrete scheme is discussed, which is an 𝐿2 -error bound of finite element accuracy and of second order in time. Numerical examples are given to demonstrate the efficiency of the theoretical results.

Conflict of Interests 1−2𝛼

+ (1 − 𝑥) 2Γ (2 − 2𝛼)

+

𝑥

2−2𝛼

2−2𝛼

+ (1 − 𝑥) 2Γ (3 − 2𝛼)

).

The authors declare that there is no conflict of interests regarding the publication of this paper.

(73) The experiential error results and convergence rates are presented in Table 1. (ii) Let 𝑢 (𝑥, 𝑡) = 𝑒−𝑡 𝑥 (𝑥 + 1)

(74)

be the exact solution to the problem 𝜕𝑡 𝑢 (𝑥, 𝑡) − Δ𝛼 (𝜆 ⋅ 𝑢 (𝑥, 𝑡)) = 𝑓 (𝑥, 𝑡) , 𝑢 (0, 𝑡) = 0, 𝑢 (1, 𝑡) = 2𝑒−𝑡 𝑢 (𝑥, 0) = 𝑥 (𝑥 + 1) ,

𝑥 ∈ Ω, 𝑡 ∈ [0, 𝑇] , 𝑡 ∈ [0, 𝑇] ,

Acknowledgments This work was partially supported by the Funding Scheme for Training Young Teachers in Shanghai Colleges under Grant no. zzhg12001, the National Natural Science Foundation of China under Grant no. 11301333, the Innovation Program of Shanghai Municipal Education Commission under Grant no. 14YZ165, and Anhui Natural Science Foundation under Grant no. 1408085MA14.

References

𝑥 ∈ Ω, (75)

where 𝛼 = 0.6, 𝜅1 = 𝜅2 = 1/2, 𝜆 = 1, Ω = [0, 1], 𝑇 = 1, and 𝑓(𝑥, 𝑡) is numerically obtained. The experiential error results and convergence rates are displayed in Table 2.

5. Conclusion In this paper, we study the finite element method for fractional diffusion equation. We use the simple, second order

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