Expanded Mixed Finite Element Method for the Two-Dimensional ...

Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2013, Article ID 934973, 9 pages http://dx.doi.org/10.1155/2013/934973

Research Article Expanded Mixed Finite Element Method for the Two-Dimensional Sobolev Equation Qing-li Zhao,1,2 Zong-cheng Li,2 and You-zheng Ding2 1 2

School of Mathematics, Shandong University, Jinan 250100, China School of Science, Shandong Jianzhu University, Jinan 250101, China

Correspondence should be addressed to Qing-li Zhao; [email protected] Received 23 April 2013; Revised 7 June 2013; Accepted 7 June 2013 Academic Editor: Carlos Conca Copyright © 2013 Qing-li 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. Expanded mixed finite element method is introduced to approximate the two-dimensional Sobolev equation. This formulation expands the standard mixed formulation in the sense that three unknown variables are explicitly treated. Existence and uniqueness of the numerical solution are demonstrated. Optimal order error estimates for both the scalar and two vector functions are established.

1. Introduction In this paper, we consider the following Sobolev equation: 𝑢𝑡 (x, 𝑡) − ∇ ⋅ {𝑎 (x, 𝑡) ∇𝑢𝑡 + 𝑏1 (x, 𝑡) ∇𝑢 (x, 𝑡)} = 𝑓 (x, 𝑡) ,

(x, 𝑡) ∈ Ω × (0, 𝑇] ,

𝑢 (x, 𝑡) = 0,

(x, 𝑡) ∈ 𝜕Ω × [0, 𝑇] ,

𝑢 (x, 0) = 𝑢0 (x) , 2

(1)

x ∈ Ω,

in a bounded domain Ω ⊂ 𝑅 with Lipschitz continuous boundary 𝜕Ω. Equation (1) has a wide range of applications in many mathematical and physical problems [1, 2], for example, the percolation theory when the fluid flows through the cracks, the transfer problem of the moisture in the soil, and the heat conduction problem in different materials. So there exists great and actual significance to research Sobolev equation. Up till now, there are some different schemes studied to solve this kind of equation (see [1–4] for instance). The mixed finite element method, which is a finite element method [5] with constrained conditions, plays an important role in the research of the numerical solution for partial differential equations. Its general theory was proposed by Babuska [6] and Brezzi [7]. Falk and Osborn [8] improved their theory and expanded the adaptability of the mixed finite element method. The mixed finite element method

(see [9, 10] for instance) is wildly used for the modeling of fluid flow and transport, as it provides accurate and locally mass conservative velocities. The main motivation of the expanded mixed finite element method [11–15] is to introduce three (or more) auxiliary variables for practical problems, while the traditional finite element method and mixed finite element method can only approximate one and two variables, respectively. The expanded mixed finite element method also has some other advantages except introducing more variables. It can treat individual boundary conditions. Also, this method is suitable for differential equation with small coefficient (close to zero) which does not need to be inverted. Consequently, this method works for the problems with small diffusion or low permeability terms in fluid problems. Using this method, we can get optimal order error estimates for certain nonlinear problems, while standard mixed formulation sometimes gives only suboptimal error estimates. The object of this paper is to present the expanded mixed finite element method for the Sobolev equation. We conduct theoretical analysis to study the existence and uniqueness and obtain optimal order error estimates. The rest of this paper is organized as follows. In Section 2, the mixed weak formulation and its mixed element approximation are considered. In Section 3, we prove the existence and uniqueness of approximation form. In Section 4, some lemmas are given.

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Journal of Applied Mathematics

In Section 5, optimal order semidiscrete error estimates are established. Throughout the paper, we will use 𝐶, with or without subscript, to denote a generic positive constant which does not depend on the discretization parameter ℎ. Vectors will be expressed in boldface. At the same time, we show a useful 𝜀Cauchy inequality 1 𝑎𝑏 ≤ 𝜀𝑎2 + 𝑏2 , 4𝜀

Let n be the unit exterior normal vector to the boundary of Ω. Then (8) is formulated in the following expanded mixed weak form: find (𝑢, 𝜆, 𝜎) ∈ 𝑊 × Λ × V, such that (𝑢𝑡 , 𝑤) + (div 𝜎, 𝑤) = (𝑓, 𝑤) , ∀𝑤 ∈ 𝑊, 0 < 𝑡 ≤ 𝑇, (𝜆, k) − (𝑢𝑡 + 𝑏𝑢, div k) + (c𝑢, k) = 0,

𝜀 > 0.

∀k ∈ V, 0 < 𝑡 ≤ 𝑇,

(2) (𝜎, 𝜏) − (𝑎𝜆, 𝜏) = 0,

2. Mixed Weak Form and Mixed Element Approximation

𝑊 = 𝐿2 (Ω) , (3)

denote the Sobolev spaces [16] endowed with the norm 1/𝑞

󵄩 𝛼 󵄩𝑞 󵄩󵄩 󵄩󵄩 󵄩󵄩𝑓󵄩󵄩𝑚,𝑞,Ω = ( ∑ 󵄩󵄩󵄩𝐷 𝑓󵄩󵄩󵄩𝐿𝑞 (Ω) )

(4)

|𝛼|≤𝑚

(the subscript Ω will always be omitted). Let 𝐻𝑚 (Ω) = 𝑊𝑚,2 (Ω) with the norm ‖ ⋅ ‖𝑚 = ‖ ⋅ ‖𝑚,2 . The notation ‖ ⋅ ‖ will mean ‖ ⋅ ‖𝐿2 (Ω) or ‖ ⋅ ‖𝐿2 (Ω)2 . We denote by ( , ) the inner product in either 𝐿2 (Ω) or 𝐿2 (Ω)2 ; that is,

Ω

(𝜉, 𝜂) = ∫ 𝜉 ⋅ 𝜂𝑑x. Ω

(5)

𝜕Ω

𝜎 = 𝑎𝜆 = −𝑎∇𝑢𝑡 − 𝑏1 ∇𝑢.

(𝑢ℎ,𝑡 , 𝑤) + (div 𝜎ℎ , 𝑤) = (𝑓, 𝑤) , ∀𝑤 ∈ 𝑊ℎ , 0 < 𝑡 ≤ 𝑇, (𝜆ℎ , k) − (𝑢ℎ,𝑡 + 𝑏𝑢ℎ , div k) + (c𝑢ℎ , k) = 0,

(6)

(7)

∀k ∈ Vℎ , 0 < 𝑡 ≤ 𝑇, (𝜎ℎ , 𝜏) − (𝑎𝜆ℎ , 𝜏) = 0,

(x, 𝑡) ∈ Ω × (0, 𝑇] ,

𝑢 (x, 𝑡) = 0,

(x, 𝑡) ∈ 𝜕Ω × (0, 𝑇] ,

𝑢 (x, 0) = 𝑢0 (x) ,

x ∈ Ω.

𝑢0 , 𝑤) , (𝑢 (0) , 𝑤) = (̃

∀𝑤 ∈ 𝑊ℎ ,

where

(12)

Vℎ (𝐸) = {k ∈ V : k|𝐸 ∈ Vℎ (𝐸) , ∀𝐸 ∈ 𝑇ℎ } .

(x, 𝑡) ∈ Ω × (0, 𝑇] ,

𝜎 − 𝑎𝜆 = 0,

∀𝜏 ∈ Λℎ , 0 < 𝑡 ≤ 𝑇,

Λℎ (𝐸) = {𝜏 ∈ Λ : 𝜏|𝐸 ∈ Vℎ (𝐸) , ∀𝐸 ∈ 𝑇ℎ } ,

(x, 𝑡) ∈ Ω × (0, 𝑇] ,

𝜆 + ∇𝑢𝑡 − ∇ (𝑏𝑢) + c𝑢 = 0,

(11)

𝑊ℎ (𝐸) = {𝜔 ∈ 𝑊 : 𝜔|𝐸 ∈ 𝑊ℎ (𝐸) , ∀𝐸 ∈ 𝑇ℎ } ,

Letting c = −∇𝑏, we rewrite (1) as the following system: 𝑢𝑡 + ∇ ⋅ 𝜎 = 𝑓,

(10)

Let 𝑇ℎ be a quasiregular polygonalization of Ω (by triangles or rectangles), with ℎ being the maximum diameter of the elements of the polygonalization. Let 𝑊ℎ × Λℎ × Vℎ be a conforming mixed element space with index 𝑘 and discretization parameter ℎ. 𝑊ℎ × Λℎ × Vℎ is an approximation to 𝑊 × Λ × V. There are many conforming (or compatible) mixed element function spaces such as Raviart-Thomas elements [17], BDFM elements [18, 19]. Some RT type mixed elements are listed in Table 1. Here, 𝑃𝑘 (𝑇) is the polynomial up to 𝑘 order in two-dimensional domain used in triangle, while 𝑄𝑘,𝑙 (𝑇) is the polynomial up to 𝑘, 𝑙 in each dimension used in rectangle. Replacing the three variables by their approximation, we get the expanded mixed finite element approximation problem: find (𝑢ℎ , 𝜆ℎ , 𝜎ℎ ) ∈ 𝑊ℎ × Λℎ × Vℎ , such that

To formulate the weak form, let 𝑏 = 𝑏1 /𝑎; we introduce two vector variables 𝜆 = −∇𝑢𝑡 − 𝑏∇𝑢,

Λ = 𝐿2 (Ω)2 ,

V = 𝐻 (div, Ω) = {k ∈ 𝐿2 (Ω)2 | ∇ ⋅ k ∈ 𝐿2 (Ω)} .

The notation ⟨, ⟩ denotes the 𝐿2 -inner product on the boundary of Ω ⟨𝜉, 𝜂⟩ = ∫ 𝜉𝜂𝑑𝑠.

∀𝑤 ∈ 𝑊,

where

For 1 ≤ 𝑞 ≤ +∞ and 𝑚 any nonnegative integer, let

(𝜉, 𝜂) = ∫ 𝜉𝜂𝑑x ,

∀𝜏 ∈ Λ, 0 < 𝑡 ≤ 𝑇,

(𝑢 (0) , 𝑤) = (𝑢0 , 𝑤) ,

𝑊𝑚,𝑞 (Ω) = {𝑓 ∈ 𝐿𝑞 (Ω) | 𝐷𝛼 𝑓 ∈ 𝐿𝑞 (Ω) , |𝛼| ≤ 𝑚}

(9)

(8)

The error analysis next makes use of three projection operators. The first operator is the Raviart-Thomas projection (or Brezzi-Douglas-Marini projection) Πℎ : V → Vℎ ; Πℎ satisfies (∇ ⋅ (k − Πℎ k) , 𝜔ℎ ) = 0,

∀𝜔ℎ ∈ 𝑊ℎ .

(13)

Journal of Applied Mathematics

3 By (20), it is easy to see that

Table 1: Some RT type mixed elements. Dim 2D 2D

Element Triangle Rectangle

Vℎ (𝐸) RT𝑘 (𝑇) = 𝑃𝑘 (𝐸)2 ⊕ x𝑃𝑘 (𝐸) RT[𝑘] (𝑇) = 𝑄𝑘+1,𝑘 (𝐸) ⊕ 𝑄𝑘,𝑘+1 (𝐸)

𝑊ℎ (𝐸) 𝑃𝑘 (𝐸) 𝑄𝑘,𝑘 (𝐸)

󵄩 󵄩󵄩 𝑟 󵄩󵄩∇ ⋅ (k − Πℎ k)󵄩󵄩󵄩𝑟 ≤ 𝐶ℎ ‖∇ ⋅ k‖𝑟 ,

1 ≤ 𝑟 ≤ 𝑘 + 1, 2 0 ≤ 𝑟 ≤ 𝑘 + 1.

(14)

(𝜏 − 𝑅ℎ 𝜏, 𝜏ℎ ) = 0,

∀𝜔 ∈ 𝑊, kℎ ∈ Vℎ , ∀𝜏 ∈ Λ, 𝜏ℎ ∈ Λℎ .

Choosing 𝜔 = div 𝜎ℎ , k = 𝜎ℎ , and 𝜏 = 𝜆ℎ in (19), we get

In the third equation of (19), letting 𝜏 = 𝜎ℎ and 𝜏 = 𝜆ℎ , respectively, then 󵄩 󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩𝜎ℎ 󵄩󵄩 ≤ 𝐶 󵄩󵄩󵄩𝜆ℎ 󵄩󵄩󵄩 , 󵄩 󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩𝜆ℎ 󵄩󵄩 ≤ 𝐶 󵄩󵄩󵄩𝜎ℎ 󵄩󵄩󵄩 .

(15)

The other two operators are the standard 𝐿2 -projection [5] 𝑃ℎ and 𝑅ℎ onto 𝑊ℎ and Λℎ , respectively, (𝜔 − 𝑃ℎ 𝜔, ∇ ⋅ kℎ ) = 0,

(16)

󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 𝑎0 󵄩󵄩󵄩𝜆ℎ 󵄩󵄩󵄩 + 󵄩󵄩󵄩div 𝜎ℎ 󵄩󵄩󵄩 ≤ 𝐶󵄩󵄩󵄩𝑢ℎ 󵄩󵄩󵄩 +

(18)

In this section, we consider the existence and uniqueness of the solution of (11).

Proof. In fact, this equation is linear; it suffices to show that the associated homogeneous system (𝑢ℎ,𝑡 , 𝑤) + (div 𝜎ℎ , 𝑤) = 0,

(27)

From (27), we know 󵄩 𝑡 󵄩󵄩 𝑡 𝑡 󵄩󵄩 󵄩󵄩 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝑢ℎ 󵄩󵄩 = 󵄩󵄩∫ 𝑢ℎ,𝑡 𝑑𝑡󵄩󵄩󵄩 ≤ 𝐶 ∫ 󵄩󵄩󵄩𝑢ℎ,𝑡 󵄩󵄩󵄩 𝑑𝑡 ≤ 𝐶 ∫ 󵄩󵄩󵄩𝑢ℎ 󵄩󵄩󵄩 𝑑𝑡. 󵄩󵄩 0 0 󵄩󵄩 0

(28)

Using Gronwall’s inequality to (28), we can prove 𝑢ℎ = 0. Further, from (26) and (23), we know that 𝜆ℎ = 0, 𝜎ℎ = 0. The proof is completed.

In the study of parabolic equations, we usually introduce a mixed elliptic projection associated with our equations. ̃ ,𝜎 Define a map: find (̃ 𝑢ℎ , 𝜆 ℎ ̃ ℎ ) ∈ 𝑊ℎ × Λℎ × Vℎ , such that ̃ ℎ ) , 𝑤) = 0, (div (𝜎 − 𝜎

∀𝑤 ∈ 𝑊ℎ , 0 < 𝑡 ≤ 𝑇,

∀𝑤 ∈ 𝑊ℎ , 0 < 𝑡 ≤ 𝑇,

̃ , k) − (𝑢 − 𝑢̃ + 𝑏 (𝑢 − 𝑢̃ ) , div k) + (c (𝑢 − 𝑢̃ ) , k) (𝜆 − 𝜆 ℎ 𝑡 ℎ,𝑡 ℎ ℎ

(𝜆ℎ , k) − (𝑢ℎ,𝑡 + 𝑏𝑢ℎ , div k) + (c𝑢ℎ , k) = 0,

= 0, (19)

(𝜎ℎ , 𝜏) − (𝑎𝜆ℎ , 𝜏) = 0,

∀k ∈ Vℎ , 0 < 𝑡 ≤ 𝑇,

̃ ) , 𝜏) = 0, ̃ ℎ , 𝜏) − (𝑎 (𝜆 − 𝜆 (𝜎 − 𝜎 ℎ (𝑢 (0) − 𝑢̃ℎ (0) , 𝑤) = 0,

∀𝜏 ∈ Λℎ , 0 < 𝑡 ≤ 𝑇,

∀𝜏 ∈ Λℎ , 0 < 𝑡 ≤ 𝑇, ∀𝑤 ∈ 𝑊ℎ . (29)

∀𝑤 ∈ 𝑊ℎ ,

has only the trivial solution. In the first equation of (19), if we take 𝜔 = 𝑢ℎ,𝑡 and 𝜔 = div 𝜎ℎ , respectively, then we have that 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩 󵄩󵄩𝑢ℎ,𝑡 󵄩󵄩 ≤ 󵄩󵄩div 𝜎ℎ 󵄩󵄩󵄩 , 󵄩󵄩 󵄩 󵄩 󵄩 󵄩󵄩div 𝜎ℎ 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑢ℎ,𝑡 󵄩󵄩󵄩 .

(26)

4. Some Lemmas

Lemma 1. Equation (11) has a unique solution.

(𝑢 (0) , 𝑤) = 0,

1 󵄩󵄩 󵄩2 𝑎 󵄩 󵄩2 󵄩󵄩div 𝜎ℎ 󵄩󵄩󵄩 + 0 󵄩󵄩󵄩𝜆ℎ 󵄩󵄩󵄩 . 2 2 (25)

󵄩 󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩𝜆ℎ 󵄩󵄩 ≤ 𝐶 󵄩󵄩󵄩𝑢ℎ 󵄩󵄩󵄩 , 󵄩󵄩󵄩𝑢ℎ,𝑡 󵄩󵄩󵄩 ≤ 𝐶 󵄩󵄩󵄩𝑢ℎ 󵄩󵄩󵄩 . 󵄩 󵄩 󵄩 󵄩

(17)

3. Existence and Uniqueness of Approximation Form

∀k ∈ Vℎ , 0 < 𝑡 ≤ 𝑇,

(24)

Note that (21), we deserve

The two projections Πℎ and 𝑃ℎ preserve the commuting property div ∘ Πℎ = 𝑃ℎ ∘ div : 𝐻1 (Ω)2 󳨀→ 𝑊ℎ .

(23)

Using the 𝜀-Cauchy inequality to (22), we have

They have the approximation properties 󵄩 󵄩󵄩 𝑟 󵄩󵄩𝜔 − 𝑃ℎ 𝜔󵄩󵄩󵄩 ≤ 𝐶ℎ ‖𝜔‖𝑟 , 0 ≤ 𝑟 ≤ 𝑘 + 1, 󵄩󵄩 󵄩󵄩 𝑟󵄩 󵄩 󵄩󵄩 󵄩󵄩𝜏 − 𝑅ℎ 𝜏󵄩󵄩 ≤ 𝐶ℎ 󵄩󵄩𝜇󵄩󵄩𝑟 , 0 ≤ 𝑟 ≤ 𝑘 + 1.

(21)

(𝑎𝜆ℎ , 𝜆ℎ ) + (div 𝜎ℎ , div 𝜎ℎ ) = (𝑏𝑢ℎ , div 𝜎ℎ ) − (c𝑢ℎ , 𝜎ℎ ) . (22)

The following approximation properties are well known. 󵄩 󵄩󵄩 𝑟 󵄩󵄩k − Πℎ k󵄩󵄩󵄩 ≤ 𝐶ℎ ‖k‖𝑟 ,

󵄩 󵄩 󵄩 󵄩󵄩 󵄩󵄩div 𝜎ℎ 󵄩󵄩󵄩 = 󵄩󵄩󵄩𝑢ℎ,𝑡 󵄩󵄩󵄩 .

(20)

Similarly to Lemma 1, we can prove that system (29) has a unique solution. ̃ ,𝜎 Now we give some error estimates of (̃ 𝑢ℎ , 𝜆 ℎ ̃ ℎ ). Define 𝑢 − 𝑢̃ℎ = (𝑝ℎ 𝑢 − 𝑢̃ℎ ) + (𝑢 − 𝑝ℎ 𝑢) = 𝑢1 + 𝑢2 , ̃ , 𝜆1 = 𝜆 − 𝜆 ℎ

̃ℎ. 𝜎1 = 𝜎 − 𝜎

(30)

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Journal of Applied Mathematics

System (29) can be rewritten as follows:

It is easy to see that 𝐼1 = (𝜆1 , Πℎ (𝑎∇𝜙))

(div 𝜎1 , 𝑤) = 0,

= (𝜆1 , Πℎ (𝑎∇𝜙) − 𝑎∇𝜙) + (𝜆1 , 𝑎∇𝜙)

∀𝑤 ∈ 𝑊ℎ ,

(𝜆1 , k) − (𝑢1,𝑡 + 𝑏 (𝑢1 + 𝑢2 ) , div k) + (c (𝑢1 + 𝑢2 ) , k) = 0, ∀k ∈ Vℎ , (𝜎1 , 𝜏) − (𝑎𝜆1 , 𝜏) = 0,

∀𝜏 ∈ Λℎ .

(𝑢 (0) − 𝑢̃ℎ (0) , 𝑤) = 0,

∀𝑤 ∈ 𝑊ℎ .

𝐸2 = (𝑎𝜆1 , ∇ (𝜙 − 𝑝ℎ 𝜙)) + (𝑎𝜆1 , ∇ (𝑝ℎ 𝜙))

Lemma 2. Let (𝑢1 , 𝜆1 , 𝜎1 ) be the solution of system (31). If 𝑢, 𝜆, and 𝜎 are sufficiently smooth, then there exist positive constants 𝐶 such that 𝑡

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩 󵄩󵄩 1−𝛿 󵄩 󵄩 󵄩󵄩𝑢1 󵄩󵄩 ≤ 𝐶 ∫ (ℎ 󵄩󵄩󵄩𝜆1 󵄩󵄩󵄩 + ℎ 𝑘0 󵄩󵄩󵄩𝜆1 󵄩󵄩󵄩 + ℎ 󵄩󵄩󵄩𝑢2 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢2 󵄩󵄩󵄩−1 ) 𝑑𝑡, 0 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩 󵄩󵄩 1−𝛿 󵄩 󵄩 󵄩󵄩𝑢1,𝑡 󵄩󵄩 ≤ 𝐶 (ℎ 󵄩󵄩󵄩𝜆1 󵄩󵄩󵄩 + ℎ 𝑘0 󵄩󵄩󵄩𝜆1 󵄩󵄩󵄩 + ℎ 󵄩󵄩󵄩𝑢2 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢2 󵄩󵄩󵄩−1 + 󵄩󵄩󵄩𝑢1 󵄩󵄩󵄩) , (32)

where 𝛿𝑘0 = 0 for 𝑘 = 0, and 𝛿𝑘0 = 0 for 𝑘 ≥ 1. Proof. Let 𝜙 ∈ 𝐻 following problem:

= (𝑎𝜆1 , ∇ (𝜙 − 𝑝ℎ 𝜙)) + (𝜎1 , ∇ (𝑝ℎ 𝜙))

(36)

= (𝑎𝜆1 , ∇ (𝜙 − 𝑝ℎ 𝜙)) − (div 𝜎1 , (𝑝ℎ 𝜙))

Now we consider the estimates of 𝑢1 and 𝑢1,𝑡 .

(Ω) ⋂ 𝐻01 (Ω)

= 𝐸1 + 𝐸2 , 󵄩 󵄩󵄩 󵄩 󵄩 󵄩󵄩 󵄩 𝐸1 ≤ 𝐶ℎ 󵄩󵄩󵄩𝜆1 󵄩󵄩󵄩 󵄩󵄩󵄩𝜙󵄩󵄩󵄩2 ≤ 𝐶ℎ 󵄩󵄩󵄩𝜆1 󵄩󵄩󵄩 󵄩󵄩󵄩𝜓󵄩󵄩󵄩 ,

(31)

2

= (𝜆1 , Πℎ (𝑎∇𝜙) − 𝑎∇𝜙) + (𝑎𝜆1 , ∇𝜙)

be the solution of the

= (𝑎𝜆1 , ∇ (𝜙 − 𝑝ℎ 𝜙)) 󵄩 󵄩 󵄩󵄩 ≤ 𝐶 󵄩󵄩󵄩𝜆1 󵄩󵄩󵄩 󵄩󵄩󵄩∇ (𝜙 − 𝑝ℎ 𝜙)󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝐶ℎ1−𝛿𝑘𝑜 󵄩󵄩󵄩𝜙󵄩󵄩󵄩2 󵄩󵄩󵄩𝜆1 󵄩󵄩󵄩 󵄩 󵄩󵄩 󵄩 ≤ 𝐶ℎ1−𝛿𝑘𝑜 󵄩󵄩󵄩𝜓󵄩󵄩󵄩 󵄩󵄩󵄩𝜆1 󵄩󵄩󵄩 . Now, we estimate 𝐼2 , (c𝑢1 , Πℎ (𝑎∇𝜙)) 󵄩 󵄩 󵄩󵄩 ≤ 𝐶 󵄩󵄩󵄩𝑢1 󵄩󵄩󵄩 󵄩󵄩󵄩Πℎ (𝑎∇𝜙) − 𝑎∇𝜙 + 𝑎∇𝜙󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝐶 󵄩󵄩󵄩𝑢1 󵄩󵄩󵄩 (ℎ 󵄩󵄩󵄩∇𝜙󵄩󵄩󵄩 + 󵄩󵄩󵄩∇𝜙󵄩󵄩󵄩) 󵄩 󵄩󵄩 󵄩 ≤ 𝐶 󵄩󵄩󵄩𝑢1 󵄩󵄩󵄩 󵄩󵄩󵄩𝜓󵄩󵄩󵄩 ,

(37)

(c𝑢2 , Πℎ (𝑎∇𝜙)) = (c𝑢2 , Πℎ (𝑎∇𝜙) − 𝑎∇𝜙) + (c𝑢2 , 𝑎∇𝜙)

∇ ⋅ (𝑎∇𝜙) = 𝜓,

∀𝑥 ∈ Ω,

𝜙|𝜕Ω = 0.

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝐶 (ℎ 󵄩󵄩󵄩𝑢2 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢2 󵄩󵄩󵄩−1 ) 󵄩󵄩󵄩𝜓󵄩󵄩󵄩 . (33)

We turn to consider 𝐼3 , (−𝑏𝑢1 , div Πℎ (𝑎∇𝜙)) 󵄩 󵄩󵄩 󵄩 󵄩 󵄩󵄩 󵄩 ≤ 𝐶 󵄩󵄩󵄩𝑢1 󵄩󵄩󵄩 󵄩󵄩󵄩∇𝜙󵄩󵄩󵄩 ≤ 𝐶 󵄩󵄩󵄩𝑢1 󵄩󵄩󵄩 󵄩󵄩󵄩𝜓󵄩󵄩󵄩 ,

Then we know

(−𝑏𝑢2 , div Πℎ (𝑎∇𝜙)) 󵄩 󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩𝜙󵄩󵄩2 ≤ 𝐶 󵄩󵄩󵄩𝜓󵄩󵄩󵄩 .

(34)

≤ − ((𝑏 − 𝑝ℎ0 𝑏) 𝑢2 , div Πℎ (𝑎∇𝜙)) −

For 0 < 𝑡 ≤ 𝑇, from the second equation of (31), we have that (𝑢1,𝑡 , 𝜓) = (𝑢1,𝑡 , div (𝑎∇𝜙)) = (𝑢1,𝑡 , div Πℎ (𝑎∇𝜙)) = (𝜆 1 , Πℎ (𝑎∇𝜙)) + (c (𝑢1 + 𝑢2 ) , Πℎ (𝑎∇𝜙)) (35) − (𝑏 (𝑢1 + 𝑢2 ) , div Πℎ (𝑎∇𝜙)) = 𝐼1 + 𝐼2 + 𝐼3 .

∑ ((𝑝ℎ0 𝑏) 𝑢2 , div Πℎ 𝑒∈𝑇ℎ

(38)

(𝑎∇𝜙))𝑒

= − ((𝑏 − 𝑝ℎ0 𝑏) 𝑢2 , div Πℎ (𝑎∇𝜙)) 󵄩 󵄩󵄩 󵄩 ≤ 𝐶ℎ 󵄩󵄩󵄩𝑢2 󵄩󵄩󵄩 󵄩󵄩󵄩𝜓󵄩󵄩󵄩 , where 𝑝ℎ0 𝑏 is the piecewise constant interpolation of function 𝑏. Using the previous estimates, we obtain 󵄩󵄩 󵄩󵄩 󵄩󵄩𝑢1,𝑡 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝐶 (ℎ 󵄩󵄩󵄩𝜆1 󵄩󵄩󵄩 + ℎ1−𝛿𝑘0 󵄩󵄩󵄩𝜆1 󵄩󵄩󵄩 + ℎ 󵄩󵄩󵄩𝑢2 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢2 󵄩󵄩󵄩−1 + 󵄩󵄩󵄩𝑢1 󵄩󵄩󵄩) . (39)

Journal of Applied Mathematics

5 Proof. For 0 < 𝑡 ≤ 𝑇, the proof proceeds in three steps as follows. (I) In the first equation of (29), taking 𝜔 = div 𝜎1 , we know

So we deserve 󵄩 𝑡 󵄩󵄩 󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩󵄩 󵄩󵄩𝑢1 󵄩󵄩 = 󵄩󵄩∫ 𝑢1,𝑡 𝑑𝑡󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 0 𝑡

(div 𝜎1 , div 𝜎1 )

󵄩 󵄩 ≤ 𝐶 ∫ 󵄩󵄩󵄩𝑢1,𝑡 󵄩󵄩󵄩 𝑑𝑡 0

(40)

= (div 𝜎1 , div (𝜎 − Πℎ 𝜎))

𝑡

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝐶 ∫ (ℎ 󵄩󵄩󵄩𝜆1 󵄩󵄩󵄩 + ℎ1−𝛿𝑘0 󵄩󵄩󵄩𝜆1 󵄩󵄩󵄩 + ℎ 󵄩󵄩󵄩𝑢2 󵄩󵄩󵄩 0

󵄩󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩div 𝜎1 󵄩󵄩󵄩 󵄩󵄩󵄩div (𝜎 − Πℎ 𝜎)󵄩󵄩󵄩 ,

󵄩 󵄩 󵄩 󵄩 +󵄩󵄩󵄩𝑢2 󵄩󵄩󵄩−1 + 󵄩󵄩󵄩𝑢1 󵄩󵄩󵄩 ) 𝑑𝑡.

which implies that 󵄩󵄩 󵄩 󵄩 󵄩 󵄩󵄩div 𝜎1 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩div (𝜎 − Πℎ 𝜎)󵄩󵄩󵄩

Applying Gronwall’s inequality to (40), we have

≤ 𝐶ℎ𝑟 ‖𝜎‖𝑟+1 ,

󵄩󵄩 󵄩󵄩 󵄩󵄩𝑢1 󵄩󵄩 𝑡

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝐶 ∫ (ℎ 󵄩󵄩󵄩𝜆1 󵄩󵄩󵄩 + ℎ1−𝛿𝑘0 󵄩󵄩󵄩𝜆1 󵄩󵄩󵄩 + ℎ 󵄩󵄩󵄩𝑢2 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢2 󵄩󵄩󵄩−1 ) 𝑑𝑡. 0

(54)

(41)

0 ≤ 𝑟 ≤ 𝑘 + 1.

(55)

̃ ℎ ), 𝜔) = 0 and (div(𝜎 − Choosing 𝜔 ∈ 𝑊ℎ , note that (div(𝜎 − 𝜎 Πℎ 𝜎), 𝜔) = 0, we get ̃ ℎ ) = 0. div (Πℎ 𝜎 − 𝜎

Noting the estimates of 𝐼1 , 𝐼2 , and 𝐼3 , the proof is completed.

(56)

Combing (15) with (56), we can prove (43). ̃ ℎ , together with (43), we (II) In (29), letting k = Πℎ 𝜎 − 𝜎 obtain

Define ‖𝑢, 𝜆, 𝜎‖𝑟 = ‖𝑢‖𝑟 + ‖𝜆‖𝑟 + ‖𝜎‖𝑟 , 𝑡

|‖𝑢, 𝜆, 𝜎‖|𝑟 = ∫ (‖𝑢‖𝑟 + ‖𝜆‖𝑟 + ‖𝜎‖𝑟 ) 𝑑𝑡.

(42)

̃ ,𝜎 Lemma 3. Let (𝑢, 𝜆, 𝜎) and (̃ 𝑢ℎ , 𝜆 ℎ ̃ ℎ ) be the solution of (19) and (29), respectively. If 𝑢, 𝜆, and 𝜎 are sufficiently smooth, then there exist positive constants 𝐶 such that (I) if 𝑘 ≥ 0, then 0 ≤ 𝑟 ≤ 𝑘 + 1,

(43)

󵄩 󵄩󵄩 𝑟+1−𝛿𝑘0 |‖𝑢, 𝜆, 𝜎‖|𝑟 , 1 ≤ 𝑟 ≤ 𝑘 + 1, (44) 󵄩󵄩𝑝ℎ 𝑢 − 𝑢̃ℎ 󵄩󵄩󵄩 ≤ 𝐶ℎ 󵄩 󵄩󵄩 𝑟+1−𝛿𝑘0 ‖𝑢, 𝜆, 𝜎‖𝑟 , 1 ≤ 𝑟 ≤ 𝑘 + 1, (45) 󵄩󵄩𝑝ℎ 𝑢𝑡 − 𝑢̃ℎ,𝑡 󵄩󵄩󵄩 ≤ 𝐶ℎ

󵄩 󵄩 󵄩 󵄩󵄩 󵄩󵄩𝑢𝑡 − 𝑢̃ℎ,𝑡 󵄩󵄩󵄩 ≤ 𝐶ℎ (󵄩󵄩󵄩𝑢𝑡 󵄩󵄩󵄩1 + ‖𝑢, 𝜆, 𝜎‖1 + |‖𝑢, 𝜆, 𝜎‖|1 ) , 󵄩󵄩󵄩𝜆 − 𝜆 ̃ 󵄩󵄩󵄩󵄩 ≤ 𝐶ℎ (‖𝑢, 𝜆, 𝜎‖ + |‖𝑢, 𝜆, 𝜎‖| ) , 󵄩󵄩 ℎ󵄩 1 1 󵄩󵄩 󵄩 ̃ ℎ 󵄩󵄩󵄩 ≤ 𝐶ℎ (‖𝑢, 𝜆, 𝜎‖1 + |‖𝑢, 𝜆, 𝜎‖|1 ) , 󵄩󵄩𝜎 − 𝜎

(46) (47) (48) (49)

̃ , we get In the third equation of (29), choosing 𝜏 = 𝑅ℎ 𝜆 − 𝜆 ℎ ̃ ) = (𝑎 (𝜆 − 𝜆 ̃ ). ̃ ),𝑅 𝜆 − 𝜆 ̃ ℎ , 𝑅ℎ 𝜆 − 𝜆 (𝜎 − 𝜎 ℎ ℎ ℎ ℎ

(59)

So we know ̃ ) + (Π 𝜎 − 𝜎 ̃ ) ̃ ℎ , 𝑅ℎ 𝜆 − 𝜆 (𝜎 − Πℎ 𝜎, 𝑅ℎ 𝜆 − 𝜆 ℎ ℎ ℎ ̃ ),𝑅 𝜆 − 𝜆 ̃ ). = (𝑎 (𝜆 − 𝜆 ℎ ℎ ℎ

(60)

̃ ) ̃ ℎ , 𝑅ℎ 𝜆 − 𝜆 (Πℎ 𝜎 − 𝜎 ℎ ̃ℎ) = (c (𝑢ℎ − 𝑢̃ℎ ) , Πℎ 𝜎 − 𝜎 ̃ ),𝑅 𝜆 − 𝜆 ̃ ) − (𝜎 − Π 𝜎, 𝑅 𝜆 − 𝜆 ̃ ), = (𝑎 (𝜆 − 𝜆 ℎ ℎ ℎ ℎ ℎ ℎ (61) which implies that ̃ ) ̃ ℎ , 𝑅ℎ 𝜆 − 𝜆 (Πℎ 𝜎 − 𝜎 ℎ

(III) if 𝑘 ≥ 1, 2 ≤ 𝑟 ≤ 𝑘 + 1, then 󵄩󵄩 󵄩 𝑟 󵄩󵄩𝑢 − 𝑢̃ℎ 󵄩󵄩󵄩 ≤ 𝐶ℎ (‖𝑢‖𝑟 + |‖𝑢, 𝜆, 𝜎‖|𝑟−1 ) ,

̃ ,Π 𝜎 − 𝜎 ̃ ℎ ) − (c (𝑢 − 𝑢̃ℎ ) , Πℎ 𝜎 − 𝜎 ̃ ℎ ) = 0. (58) (𝑅ℎ 𝜆 − 𝜆 ℎ ℎ

Note that (44) and (45), we obtain

(II) if 𝑘 = 0, then 󵄩 󵄩󵄩 󵄩󵄩𝑢 − 𝑢̃ℎ 󵄩󵄩󵄩 ≤ 𝐶ℎ (‖𝑢‖1 + |‖𝑢, 𝜆, 𝜎‖|1 ) ,

(57)

Hence

0

󵄩󵄩 ̃ ℎ )󵄩󵄩󵄩󵄩 ≤ 𝐶ℎ𝑟 ‖𝜎‖𝑟+1 , 󵄩󵄩div (𝜎 − 𝜎

̃ ,Π 𝜎 − 𝜎 ̃ ℎ ) − (c (𝑢 − 𝑢̃ℎ ) , Πℎ 𝜎 − 𝜎 ̃ ℎ ) = 0. (𝜆 − 𝜆 ℎ ℎ

(50)

󵄩 󵄩󵄩 𝑟 󵄩 󵄩 󵄩󵄩𝑢𝑡 − 𝑢̃ℎ,𝑡 󵄩󵄩󵄩 ≤ 𝐶ℎ (󵄩󵄩󵄩𝑢𝑡 󵄩󵄩󵄩𝑟 + ‖𝑢, 𝜆, 𝜎‖𝑟 + |‖𝑢, 𝜆, 𝜎‖|𝑟−1 ) , (51) 󵄩󵄩󵄩𝜆 − 𝜆 ̃ 󵄩󵄩󵄩󵄩 ≤ 𝐶ℎ𝑟 (‖𝑢, 𝜆, 𝜎‖ + |‖𝑢, 𝜆, 𝜎‖| ) , (52) 󵄩󵄩 ℎ󵄩 𝑟 𝑟−1 󵄩󵄩 ̃ ℎ 󵄩󵄩󵄩󵄩 ≤ 𝐶ℎ𝑟 (‖𝑢, 𝜆, 𝜎‖𝑟 + |‖𝑢, 𝜆, 𝜎‖|𝑟−1 ) . (53) 󵄩󵄩𝜎 − 𝜎

󵄩 󵄩󵄩 ̃ ℎ 󵄩󵄩󵄩󵄩 ≤ 𝐶 󵄩󵄩󵄩𝑢 − 𝑢̃ℎ 󵄩󵄩󵄩 󵄩󵄩󵄩Πℎ 𝜎 − 𝜎

(62)

󵄩 󵄩 󵄩 󵄩 󵄩 ̃ ℎ 󵄩󵄩󵄩󵄩 . ≤ 𝐶 (󵄩󵄩󵄩𝑢1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢2 󵄩󵄩󵄩) 󵄩󵄩󵄩Πℎ 𝜎 − 𝜎

̃ ℎ ‖, Now we estimate ‖Πℎ 𝜎 − 𝜎 ̃ ),Π 𝜎 − 𝜎 ̃ ℎ , Πℎ 𝜎 − 𝜎 ̃ ℎ ) = (𝑎 (𝜆 − 𝜆 ̃ℎ) . (𝜎 − 𝜎 ℎ ℎ

(63)

6

Journal of Applied Mathematics By Lemma 2, if 𝑘 = 0, we have that

It is easy to see that

󵄩󵄩 ̃ 󵄩󵄩󵄩󵄩 = 󵄩󵄩󵄩𝜆 󵄩󵄩󵄩 󵄩󵄩𝜆 − 𝜆 ℎ󵄩 󵄩 1󵄩 󵄩

2 󵄩󵄩 ̃ ℎ 󵄩󵄩󵄩󵄩 󵄩󵄩Πℎ 𝜎 − 𝜎

̃ ),Π 𝜎 − 𝜎 ̃ ℎ ) + (𝑎 (𝜆 − 𝜆 ̃ℎ) = (Πℎ 𝜎 − 𝜎, Πℎ 𝜎 − 𝜎 ℎ ℎ

≤ 𝐶ℎ (‖𝑢‖1 + ‖𝜆‖1 + ‖𝜎‖1

̃ ℎ ) + (𝑎 (𝜆 − 𝑅ℎ 𝜆) , Πℎ 𝜎 − 𝜎 ̃ℎ) = (Πℎ 𝜎 − 𝜎, Πℎ 𝜎 − 𝜎

𝑡

+ ∫ (‖𝑢‖1 + ‖𝜆‖1 + ‖𝜎‖1 ) 𝑑𝜏) .

̃ ),Π 𝜎 − 𝜎 ̃ℎ) + (𝑎 (𝑅ℎ 𝜆 − 𝜆 ℎ ℎ 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝐶 (󵄩󵄩󵄩Πℎ 𝜎 − 𝜎󵄩󵄩󵄩 + 󵄩󵄩󵄩𝜆 − 𝑅ℎ 𝜆󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢2 󵄩󵄩󵄩) 󵄩 ̃ ℎ 󵄩󵄩󵄩󵄩 . × 󵄩󵄩󵄩Πℎ 𝜎 − 𝜎

(69)

0

If 𝑘 ≥ 1, 2 ≤ 𝑟 ≤ 𝑘 + 1, we have that (64)

󵄩󵄩 ̃ 󵄩󵄩󵄩󵄩 ≤ 𝐶ℎ𝑟 (‖𝑢‖ + ‖𝜆‖ + ‖𝜎‖ 󵄩󵄩𝜆 − 𝜆 ℎ󵄩 𝑟 𝑟 𝑟 󵄩 (70)

𝑡

By the properties of projection Πℎ , we get

+ ∫ (‖𝑢‖𝑟−1 + ‖𝜆‖𝑟−1 ‖𝜎‖𝑟−1 ) 𝑑𝑡) . 0

󵄩󵄩 ̃ ℎ 󵄩󵄩󵄩󵄩 ≤ 𝐶ℎ𝑟 (‖𝑢‖𝑟 + ‖𝜆‖𝑟 + ‖𝜎‖𝑟 + 󵄩󵄩󵄩󵄩𝑢1 󵄩󵄩󵄩󵄩) , 󵄩󵄩Πℎ 𝜎 − 𝜎

(65)

0 ≤ 𝑟 ≤ 𝑘 + 1,

̃ ℎ ‖, we obtain Using the estimate of ‖Πℎ 𝜎 − 𝜎 󵄩󵄩 ̃ ℎ 󵄩󵄩󵄩󵄩 ≤ 𝐶 (ℎ𝑟 (‖𝑢‖𝑟 + ‖𝜆‖𝑟 + ‖𝜎‖𝑟 ) + 󵄩󵄩󵄩󵄩𝑢1 󵄩󵄩󵄩󵄩) , 󵄩󵄩𝜎 − 𝜎

which proves (49) and (53). ̃ ‖, Now we estimate ‖𝑅ℎ 𝜆 − 𝜆 ℎ

0 ≤ 𝑟 ≤ 𝑘 + 1.

(71)

If 𝑘 = 0, we get

̃ ),𝑅 𝜆 − 𝜆 ̃ ) (𝑎 (𝑅ℎ 𝜆 − 𝜆 ℎ ℎ ℎ

󵄩󵄩 ̃ ℎ 󵄩󵄩󵄩󵄩 󵄩󵄩𝜎 − 𝜎

̃ ,𝑅 𝜆 − 𝜆 ̃ ) = (𝑎 (𝜆 − 𝜆 ℎ ℎ ℎ

≤ 𝐶ℎ (‖𝑢‖1 + ‖𝜆‖1 + ‖𝜎‖1

̃ ) + (𝑎 (𝑅ℎ 𝜆 − 𝜆) , 𝑅ℎ 𝜆 − 𝜆 ℎ ̃ ) + (Π 𝜎 − 𝜎 ̃ ) ̃ ℎ , 𝑅ℎ 𝜆 − 𝜆 = (𝜎 − Πℎ 𝜎, 𝑅ℎ 𝜆 − 𝜆 ℎ ℎ ℎ

(72)

𝑡

+ ∫ (‖𝑢‖1 + ‖𝜆‖1 + ‖𝜎‖1 ) 𝑑𝑡) . 0

̃ ) + (𝑎 (𝑅ℎ 𝜆 − 𝜆) , 𝑅ℎ 𝜆 − 𝜆 ℎ ̃ ) + (c (𝑢 − 𝑢̃ ) , Π 𝜎 − 𝜎 ̃ℎ) = (𝜎 − Πℎ 𝜎, 𝑅ℎ 𝜆 − 𝜆 ℎ ℎ ℎ ̃ ) + (𝑎 (𝑅ℎ 𝜆 − 𝜆) , 𝑅ℎ 𝜆 − 𝜆 ℎ 󵄩 󵄩 󵄩 󵄩 󵄩 ̃ 󵄩󵄩󵄩󵄩 ≤ (󵄩󵄩󵄩𝜎 − Πℎ 𝜎󵄩󵄩󵄩 + 𝐶 󵄩󵄩󵄩𝜆 − 𝑅ℎ 𝜆󵄩󵄩󵄩) 󵄩󵄩󵄩󵄩𝑅ℎ 𝜆 − 𝜆 ℎ󵄩 󵄩 󵄩󵄩 ̃ ℎ 󵄩󵄩󵄩󵄩 + 𝐶 󵄩󵄩󵄩𝑢1 + 𝑢2 󵄩󵄩󵄩 󵄩󵄩󵄩Πℎ 𝜎 − 𝜎 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ≤ 𝐶 (󵄩󵄩󵄩𝜎 − Πℎ 𝜎󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢2 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑅ℎ 𝜆 − 𝜆󵄩󵄩󵄩 ) 𝑐 󵄩 ̃ 󵄩󵄩󵄩󵄩2 , + 0 󵄩󵄩󵄩󵄩𝑅ℎ 𝜆 − 𝜆 ℎ󵄩 2

(66)

If 𝑘 ≥ 1, 2 ≤ 𝑟 ≤ 𝑘 + 1, we get 󵄩󵄩 ̃ ℎ 󵄩󵄩󵄩󵄩 󵄩󵄩𝜎 − 𝜎 ≤ 𝐶ℎ𝑟 (‖𝑢‖𝑟 + ‖𝜆‖𝑟 + ‖𝜎‖𝑟

(73)

𝑡

+ ∫ (‖𝑢‖𝑟−1 + ‖𝜆‖𝑟−1 + ‖𝜎‖𝑟−1 ) 𝑑𝑡) . 0

(III) By Lemma 2, we have that 󵄩󵄩 󵄩 󵄩 󵄩 𝑟+1−𝛿𝑘0 󵄩󵄩𝑝ℎ 𝑢 − 𝑢̃ℎ 󵄩󵄩󵄩 = 󵄩󵄩󵄩𝑢1 󵄩󵄩󵄩 ≤ 𝐶ℎ 𝑡

× ∫ (‖𝑢‖𝑟 + ‖𝜆‖𝑟 + ‖𝜎‖𝑟 ) 𝑑𝜏,

where

0

̃ ),𝑅 𝜆 − 𝜆 ̃ ) ≥ 𝑐 󵄩󵄩󵄩󵄩𝑅 𝜆 − 𝜆 ̃ 󵄩󵄩󵄩󵄩2 . (𝑎 (𝑅ℎ 𝜆 − 𝜆 ℎ ℎ ℎ 0󵄩 ℎ ℎ󵄩

(67)

0 ≤ 𝑟 ≤ 𝑘 + 1,

󵄩󵄩 󵄩 󵄩󵄩𝑝ℎ 𝑢𝑡 − 𝑢̃ℎ,𝑡 󵄩󵄩󵄩 ≤ 𝐶ℎ𝑟+1−𝛿𝑘0 (‖𝑢‖𝑟 + ‖𝜆‖𝑟 + ‖𝜎‖𝑟

From (66) and (67), we know 󵄩󵄩 ̃ 󵄩󵄩󵄩󵄩 ≤ 𝐶 (ℎ𝑟 (‖𝑢‖ + ‖𝜆‖ + ‖𝜎‖ ) + 󵄩󵄩󵄩𝑢 󵄩󵄩󵄩) , 󵄩󵄩𝑅ℎ 𝜆 − 𝜆 ℎ󵄩 𝑟 𝑟 𝑟 󵄩 1󵄩 󵄩 󵄩󵄩 ̃ 󵄩󵄩󵄩󵄩 ≤ 𝐶 (ℎ𝑟 (‖𝑢‖ + ‖𝜆‖ + ‖𝜎‖ ) + 󵄩󵄩󵄩𝑢 󵄩󵄩󵄩) , 󵄩󵄩𝜆 − 𝜆 ℎ󵄩 𝑟 𝑟 𝑟 󵄩 1󵄩 󵄩 0 ≤ 𝑟 ≤ 𝑘 + 1.

𝑡

+ ∫ (‖𝑢‖𝑟 + ‖𝜆‖𝑟 + ‖𝜎‖𝑟 ) 𝑑𝑡) , (68)

0

0 ≤ 𝑟 ≤ 𝑘 + 1. (74)

Journal of Applied Mathematics

7

If 𝑘 = 0, we know

(I) if 𝑘 = 0, then

󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝑢 − 𝑢̃ℎ 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑢 − 𝑝ℎ 𝑢󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢1 󵄩󵄩󵄩 𝑡

≤ 𝐶ℎ (‖𝑢‖1 + ∫ (‖𝑢‖1 + ‖𝜆‖1 + ‖𝜎‖1 ) 𝑑𝑡) , 0

󵄩󵄩 󵄩 󵄩󵄩𝑢𝑡 − 𝑢̃ℎ,𝑡 󵄩󵄩󵄩

(75)

󵄩 󵄩 ≤ 𝐶ℎ (󵄩󵄩󵄩𝑢𝑡 󵄩󵄩󵄩1 + ‖𝑢‖1 + ‖𝜆‖1 + ‖𝜎‖1 𝑡

+ ∫ (‖𝑢‖1 + ‖𝜆‖1 + ‖𝜎‖1 ) 𝑑𝑡) .

󵄩 󵄨󵄩 󵄩󵄨 󵄩󵄩 󵄩󵄩𝑢 − 𝑢ℎ 󵄩󵄩󵄩 ≤ 𝐶ℎ (‖𝑢‖1 + 󵄨󵄨󵄨󵄩󵄩󵄩𝑢, 𝑢𝑡 , 𝜆, 𝜎󵄩󵄩󵄩󵄨󵄨󵄨1 ) , 󵄩 󵄩 󵄩 󵄩󵄩 󵄩󵄩𝑢𝑡 − 𝑢ℎ,𝑡 󵄩󵄩󵄩 ≤ 𝐶ℎ (󵄩󵄩󵄩𝑢, 𝑢𝑡 , 𝜆, 𝜎󵄩󵄩󵄩1 + |‖𝑢, 𝜆, 𝜎‖|1 ) , 󵄩 󵄩 󵄩 󵄨󵄩 󵄩󵄨 󵄩󵄩 󵄩󵄩𝜆 − 𝜆ℎ 󵄩󵄩󵄩 ≤ 𝐶ℎ (󵄩󵄩󵄩𝑢, 𝑢𝑡 , 𝜆, 𝜎󵄩󵄩󵄩1 + 󵄨󵄨󵄨󵄩󵄩󵄩𝑢, 𝑢𝑡 , 𝜆, 𝜎󵄩󵄩󵄩󵄨󵄨󵄨1 ) , 󵄩 󵄩󵄩 󵄩 󵄩 󵄨󵄩 󵄩󵄨 󵄩󵄩𝜎 − 𝜎ℎ 󵄩󵄩󵄩 ≤ 𝐶ℎ (󵄩󵄩󵄩𝑢, 𝑢𝑡 , 𝜆, 𝜎󵄩󵄩󵄩1 + 󵄨󵄨󵄨󵄩󵄩󵄩𝑢, 𝑢𝑡 , 𝜆, 𝜎󵄩󵄩󵄩󵄨󵄨󵄨1 ) , 󵄩 󵄩󵄩 󵄩󵄩div (𝜎 − 𝜎ℎ )󵄩󵄩󵄩 󵄩 󵄩 󵄨󵄩 󵄩󵄨 ≤ 𝐶ℎ (‖𝑢‖1 + 󵄩󵄩󵄩𝑢𝑡 󵄩󵄩󵄩1 + ‖𝜎‖2 + 󵄨󵄨󵄨󵄩󵄩󵄩𝑢, 𝑢𝑡 , 𝜆, 𝜎󵄩󵄩󵄩󵄨󵄨󵄨1 ) ,

(78) (79) (80) (81) (82)

0

(II) if 𝑘 ≥ 1 and 2 ≤ 𝑟 ≤ 𝑘 + 1, then

If 𝑘 ≥ 1, 2 ≤ 𝑟 ≤ 𝑘 + 1, we know 󵄩󵄩 󵄩 󵄩󵄩𝑢 − 𝑢̃ℎ 󵄩󵄩󵄩 𝑡

≤ 𝐶ℎ𝑟 (‖𝑢‖𝑟 + ∫ (‖𝑢‖𝑟−1 + ‖𝜆‖𝑟−1 + ‖𝜎‖𝑟−1 ) 𝑑𝑡) , 0

󵄩󵄩 󵄩 󵄩󵄩𝑢𝑡 − 𝑢̃ℎ,𝑡 󵄩󵄩󵄩

󵄩 󵄨󵄩 󵄩󵄨 󵄩󵄩 𝑟 󵄩󵄩𝑢 − 𝑢ℎ 󵄩󵄩󵄩 ≤ 𝐶ℎ (‖𝑢‖𝑟 + 󵄨󵄨󵄨󵄩󵄩󵄩𝑢, 𝑢𝑡 , 𝜆, 𝜎󵄩󵄩󵄩󵄨󵄨󵄨𝑟−1 ) , 󵄩 󵄩󵄩 󵄩󵄩𝑢𝑡 − 𝑢ℎ,𝑡 󵄩󵄩󵄩 󵄩 󵄩 ≤ 𝐶ℎ𝑟 (󵄩󵄩󵄩𝑢𝑡 󵄩󵄩󵄩𝑟 + ‖𝑢‖𝑟−1 + ‖𝜆‖𝑟−1 + ‖𝜎‖𝑟−1 𝑡

󵄩 󵄩 + ∫ (‖𝑢‖𝑟−1 + 󵄩󵄩󵄩𝑢𝑡 󵄩󵄩󵄩𝑟 + ‖𝜆‖𝑟−1 + ‖𝜎‖𝑟−1 ) 𝑑𝑡) ,

󵄩 󵄩 ≤ 𝐶ℎ𝑟 (󵄩󵄩󵄩𝑢𝑡 󵄩󵄩󵄩𝑟 + ‖𝑢‖𝑟−1 + ‖𝜆‖𝑟−1

0

󵄩 󵄩󵄩 󵄩󵄩𝜆 − 𝜆ℎ 󵄩󵄩󵄩

𝑡

+ ∫ (‖𝑢‖𝑟−1 + ‖𝜆‖𝑟−1 + ‖𝜎‖𝑟−1 ) 𝑑𝑡) . 0

(76)

≤ 𝐶ℎ𝑟 (‖𝑢, 𝜆, 𝜎‖𝑟 𝑡

󵄩 󵄩 + ∫ (󵄩󵄩󵄩𝑢𝑡 󵄩󵄩󵄩𝑟 + ‖𝑢‖𝑟−1 + ‖𝜆‖𝑟−1 + ‖𝜎‖𝑟−1 ) 𝑑𝑡) , 0

The proof is completed. Remark 4. The estimate results of (46)–(53) are optimal. If 𝑘 ≥ 1, the estimate results of (43)–(45) are superconvergent.

5. Main Result

󵄩 󵄩󵄩 󵄩󵄩𝜎 − 𝜎ℎ 󵄩󵄩󵄩 ≤ 𝐶ℎ𝑟 ( ‖𝑢, 𝜆, 𝜎‖𝑟 𝑡

In this section, we consider error estimates for the continuous-in-time mixed finite element approximation. Define

̃ ) + (𝜆 ̃ −𝜆 )=𝜆 +𝜆 , 𝜆 − 𝜆ℎ = (𝜆 − 𝜆 ℎ ℎ ℎ 1 2 󵄩󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝑢, 𝑢𝑡 , 𝜆, 𝜎󵄩󵄩󵄩𝑟 = ‖𝑢‖𝑟 + 󵄩󵄩󵄩𝑢𝑡 󵄩󵄩󵄩𝑟 + ‖𝜆‖𝑟 + ‖𝜎‖𝑟 ,

󵄩󵄩 󵄩 󵄩󵄩div (𝜎 − 𝜎ℎ )󵄩󵄩󵄩 󵄩 󵄩 ≤ 𝐶ℎ𝑟 (󵄩󵄩󵄩𝑢𝑡 󵄩󵄩󵄩𝑟 + ‖𝜆‖𝑟 + ‖𝜎‖𝑟+1 + ‖𝑢‖𝑟−1

𝑢ℎ − 𝑢ℎ ) = 𝑢3 + 𝑢4 , 𝑢 − 𝑢ℎ = (𝑢 − 𝑢̃ℎ ) + (̃

̃ ℎ ) + (̃ 𝜎 − 𝜎ℎ = (𝜎 − 𝜎 𝜎ℎ − 𝜎ℎ ) = 𝜎1 + 𝜎2 ,

󵄩 󵄩 + ∫ (󵄩󵄩󵄩𝑢𝑡 󵄩󵄩󵄩𝑟 + ‖𝑢‖𝑟−1 + ‖𝜆‖𝑟−1 + ‖𝜎‖𝑟−1 ) 𝑑𝑡) , 0 (83)

𝑡

(77)

󵄩 󵄩 + ∫ (󵄩󵄩󵄩𝑢𝑡 󵄩󵄩󵄩𝑟 + ‖𝑢‖𝑟−1 + ‖𝜆‖𝑟−1 + ‖𝜎‖𝑟−1 ) 𝑑𝑡) . 0 (84)

𝑡

󵄨󵄨󵄩󵄩 󵄩󵄨 󵄩 󵄩 󵄨󵄨󵄩󵄩𝑢, 𝑢𝑡 , 𝜆, 𝜎󵄩󵄩󵄩󵄨󵄨󵄨𝑟 = ∫ (‖𝑢‖𝑟 + 󵄩󵄩󵄩𝑢𝑡 󵄩󵄩󵄩𝑟 + ‖𝜆‖𝑟 + ‖𝜎‖𝑟 ) 𝑑𝑡. 0 Theorem 5. Let (𝑢, 𝜆, 𝜎) and (𝑢ℎ , 𝜆ℎ , 𝜎ℎ ) be the solution of (9) and (11), respectively. If 𝑢, 𝜆, and 𝜎 are sufficiently smooth, then there exist positive constants 𝐶 such that

Proof. From (9) and (19), we have the following error equation: (𝑢4,𝑡 + 𝑢3,𝑡 , 𝑤) + (div 𝜎2 , 𝑤) = 0, ∀𝑤 ∈ 𝑊ℎ , 0 < 𝑡 ≤ 𝑇,

8

Journal of Applied Mathematics (𝜎2 , 𝜏) − (𝑎𝜆2 , 𝜏) = 0,

Remark 6. The estimate results of (78)–(84) are optimal.

∀𝜏 ∈ Λℎ , 0 < 𝑡 ≤ 𝑇,

Acknowledgments

(𝜆2 , k) − (𝑢4,𝑡 + 𝑏𝑢4 , div k) + (c𝑢4 , k) = 0, ∀k ∈ Vℎ , 0 < 𝑡 ≤ 𝑇, 𝑢4 (0) = 0. (85) In (85), choosing 𝜔 = 𝑢4,𝑡 + 𝑝ℎ (𝑏𝑢4 ), 𝜏 = 𝜆2 , and k = 𝜎2 , we have (𝜆2 , 𝜎2 ) + (𝑢4,𝑡 , 𝑢4,𝑡 ) = − (𝑢3,𝑡 , 𝑢4,𝑡 + 𝑝ℎ (𝑏𝑢4 )) − (𝑢4,𝑡 , 𝑝ℎ (𝑏𝑢4 ))

(86)

− (c𝑢4 , 𝜎2 ) . In the second equation of (85), letting 𝜏 = 𝜎2 , we know 𝜎2 ≤ 𝐶𝜆2 .

(87)

Further, using 𝜀-inequality to (86), we get 󵄩 󵄩2 󵄩 󵄩2 󵄩󵄩 󵄩󵄩2 󵄩󵄩 󵄩󵄩2 󵄩󵄩𝑢4,𝑡 󵄩󵄩 + 󵄩󵄩𝜆2 󵄩󵄩 ≤ 𝐶 (󵄩󵄩󵄩𝑢4 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢3,𝑡 󵄩󵄩󵄩 ) . Hence, 󵄩 𝑡 󵄩󵄩 󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩󵄩 󵄩󵄩𝑢4 󵄩󵄩 = 󵄩󵄩∫ 𝑢4,𝑡 𝑑𝜏󵄩󵄩󵄩 󵄩󵄩 0 󵄩󵄩 𝑡

𝑡

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝐶 ∫ 󵄩󵄩󵄩𝑢4,𝑡 󵄩󵄩󵄩 𝑑𝜏 ≤ 𝐶 ∫ (󵄩󵄩󵄩𝑢4 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢3,𝑡 󵄩󵄩󵄩) 𝑑𝑡. 0 0

(88)

(89)

Using Gronwall’s inequality to (89), we obtain 𝑡

󵄩󵄩 󵄩󵄩 󵄩 󵄩 󵄩󵄩𝑢4 󵄩󵄩 ≤ 𝐶 ∫ 󵄩󵄩󵄩𝑢3,𝑡 󵄩󵄩󵄩 𝑑𝑡. 0

(90)

Further, we know 𝑡

󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝜎2 󵄩󵄩 + 󵄩󵄩𝑢4,𝑡 󵄩󵄩 + 󵄩󵄩𝜆2 󵄩󵄩 ≤ 𝐶 (󵄩󵄩󵄩𝑢3,𝑡 󵄩󵄩󵄩 + ∫ 󵄩󵄩󵄩𝑢3,𝑡 󵄩󵄩󵄩 𝑑𝑡) . 0

(91)

Noting the first equation of (78), we get 𝑡

󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩div 𝜎2 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑢3,𝑡 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢4,𝑡 󵄩󵄩󵄩 ≤ 𝐶 (󵄩󵄩󵄩𝑢3,𝑡 󵄩󵄩󵄩 + ∫ 󵄩󵄩󵄩𝑢3,𝑡 󵄩󵄩󵄩 𝑑𝑡) . (92) 0 So we have that 𝑡

󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝑢 − 𝑢ℎ 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑢3 󵄩󵄩󵄩 + 𝐶 ∫ 󵄩󵄩󵄩𝑢3,𝑡 󵄩󵄩󵄩 𝑑𝑡, 0 𝑡

󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝑢𝑡 − 𝑢ℎ,𝑡 󵄩󵄩󵄩 ≤ 𝐶 (󵄩󵄩󵄩𝑢3,𝑡 󵄩󵄩󵄩 + ∫ 󵄩󵄩󵄩𝑢3,𝑡 󵄩󵄩󵄩 𝑑𝑡) , 0 𝑡

󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝜆 − 𝜆ℎ 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝜆1 󵄩󵄩󵄩 + 𝐶 (󵄩󵄩󵄩𝑢3,𝑡 󵄩󵄩󵄩 + ∫ 󵄩󵄩󵄩𝑢3,𝑡 󵄩󵄩󵄩 𝑑𝑡) , 0

(93)

𝑡

󵄩 󵄩 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝜎 − 𝜎ℎ 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝜎1 󵄩󵄩󵄩 + 𝐶 (󵄩󵄩󵄩𝑢3,𝑡 󵄩󵄩󵄩 + ∫ 󵄩󵄩󵄩𝑢3,𝑡 󵄩󵄩󵄩 𝑑𝑡) , 0 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩div (𝜎 − 𝜎ℎ )󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩div 𝜎1 󵄩󵄩󵄩 + 󵄩󵄩󵄩div 𝜎2 󵄩󵄩󵄩 . Together with the results of Lemmas 2 and 3, the proof is completed.

This work is supported by the National Natural Science Foundation of China (nos. 11171190, 11101246, and 11101247) and the Natural Science Foundation of Shandong Province (ZR2011AQ020). The authors thank the anonymous referees for their constructive comments and suggestions, which led to improvements in the presentation.

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