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; zhaoqingliabc@163.com
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.
2
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)
4
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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