# Multiscale Nonconforming Finite Element Computation to Small Periodic Composite Materials of Elastic Structures on Anisotropic Meshes

Authors: Ying Hao; Shicang Song; Junfeng Guan
Journal: Mathematical Problems in Engineering
 (2016)

---

## Abstract

The small periodic elastic structures of composite materials with the multiscale asymptotic expansion and homogenized method are discussed. A nonconforming Crouzeix-Raviart finite element is applied to calculate every term of the asymptotic expansion on anisotropic meshes. The approximation scheme to the higher derivatives of the homogenized solution is also derived. Finally, the optimal error estimate in·
h for displacement vector is obtained.


---

## Body

## 1. Introduction

Because the composite materials are rapidly oscillating and the period of oscillation is very small, the calculation of the small periodic elastic structures of composite materials is rather complex. It is hard to obtain an analytical solution of most problems. The heterogeneous multiscale method (HMM) [1–4] is a general method for efficient numerical computation of problems with rapidly oscillating coefficients. The key concepts such as resonance, fast-slow scale interactions, averaging, and techniques for transformations to nonstiff forms have been discussed in [1]. There are a number of numerical examples of the HMM applied to long time integration of wave propagation problems in both periodic and nonperiodic medium found in [2]. The finite difference heterogeneous multiscale method (FD-HMM) [3] is used to solve multiscale parabolic problems. The heterogeneous multiscale finite element method (HM-FEM) is used to solve the elliptic problems in perforated domains for the first time in [4].The elastic problems of periodic composite materials can be formulated mathematically as the following boundary value problems of elliptic equations by using Einstein summation notation:(1)
-
∂
∂
x
p
A
p
k
x
ε
∂
u
ε
x
∂
x
k
=
f
x
in
Ω
,
u
ε
x
=
0
on
∂
Ω
,where A
p
k
x
/
ε is a family of n
×
n matrices. Let ξ
=
x
/
ε; its elements can be presented as a
i
j
p
k
ξ, which is 1-periodic functions of ξ. u
ε
x
=
u
1
ε
x
,
u
2
ε
x
,
…
,
u
n
ε
x
T denotes the displacement column vector and f
x
=
f
1
x
,
f
2
x
,
…
,
f
n
x
T is the body force column vector.According to the inhomogeneous anisotropic properties of the composite materials and the complexities in geometric forms, the elements ofA
p
k
x
/
ε are rapidly oscillating when 0
<
ε
<
1. The computational resources required to solve numerically for the smallest scales are prohibitive. To solve problem (1), a two-scale asymptotic analysis method is presented in [5], where the solution can be written as the form(2)
u
ε
x
=
u
0
x
+
∑
l
=
1
∞
ε
l
∑
α
=
l
N
α
ξ
D
α
u
0
x
,where α
=
α
1
,
α
2
,
…
,
α
l is multiple index and α
=
α
1
+
α
2
+
⋯
+
α
l.N
α
ξ
=
N
α
1
⋯
α
l
ξ are 1-periodic matrix functions that can be solved in the periodic cell Q. These are the solutions of the following auxiliary periodic problems:(3)
-
∂
∂
ξ
p
A
p
k
ξ
∂
N
α
1
ξ
∂
ξ
k
=
∂
∂
ξ
p
A
p
α
1
ξ
,
(4)
-
∂
∂
ξ
p
A
p
k
ξ
∂
N
α
1
α
2
ξ
∂
ξ
k
=
∂
∂
ξ
p
A
p
α
1
ξ
N
α
2
ξ
+
A
α
1
k
ξ
∂
N
α
2
ξ
∂
ξ
k
+
A
α
1
α
2
ξ
-
A
^
α
1
α
2
,where A
^
α
1
α
2
=
∫
Q
A
α
1
α
2
ξ
+
A
α
1
k
ξ
∂
N
α
2
ξ
/
∂
ξ
k
d
ξ.Generally, whenα
=
l
≥
3, it can be obtained that(5)
-
∂
∂
ξ
p
A
p
k
ξ
∂
N
α
1
⋯
α
l
ξ
∂
ξ
k
=
∂
∂
ξ
p
A
p
α
1
ξ
N
α
2
⋯
α
l
ξ
+
A
α
1
k
ξ
∂
N
α
2
⋯
α
l
ξ
∂
ξ
k
+
A
α
1
α
2
ξ
N
α
3
⋯
α
l
ξ
.Letu
0
x be the solution of homogenized problem in the whole domain Ω:(6)
∂
∂
x
p
A
^
p
k
∂
u
0
x
∂
x
k
=
f
x
in
Ω
,
u
0
x
=
0
on
∂
Ω
,where A
^
p
k
=
∫
Q
A
p
k
ξ
+
A
p
m
ξ
∂
N
k
ξ
/
∂
ξ
m
d
ξ and N
k
ξ is the solution of (3).As can be seen from (2), solving the elastic problem using the multiscale technique involves three major steps:(1)
the periodic solutionN
α
1
⋯
α
l
ξ is solved in the periodic cell Q;
(2)
the solutionu
0
x about homogenized problem is available in the whole domain Ω;
(3)
the high order derivatives ofu
0
x are calculated.The asymptotic expansion of the solution and the associated truncation errors have been deeply investigated in [6–8]. However, there is little reference focusing on how to calculate every term of the asymptotic expansion.Based on [5, 8], an achievable finite element computational scheme about the small periodic composite materials of elastic structures on anisotropic meshes by combining multiscale technique and finite element method is presented in this paper. A nonconforming Crouzeix-Raviart finite element is applied to estimate every term of the asymptotic expansion on anisotropic meshes [9]. At the same time, the approximation scheme to the higher derivative of the homogenized solution is derived. Finally, the optimal error estimate in ·
h for displacement vector is obtained.


## 2. Construction of the Element

LetT
h be an anisotropic triangle subdivision of the unit cube Q with Q
=
⋃
K
∈
T
h. Given K
∈
T
h, we denote the length of edges parallel to the x-axis and the y-axis by 2
h
x and 2
h
y, respectively. Given K
∈
T
h, we note that K is a triangle element of x
-
y plan, M
1
(
0,0
), M
2
(
h
x
,
0
), and M
3
(
0
,
h
y
) are the three vertices, and l
i
=
M
i
M
i
+
1
M
4
=
M
1, i
=
1,2
,
3, are the three edges of K. And T
h do not meet the regular condition; that is, h
y
≪
h
x. Let K
^ be a reference element of ξ
-
η plan, the three vertices are M
^
1
(
0,0
), M
^
2
(
1,0
), and M
^
3
(
0,1
), and the three edges are l
^
i
=
M
^
i
M
^
i
+
1. There exists an inverse mapping F
K
:
K
^
→
K:(7)
x
=
x
0
+
h
x
ξ
,
y
=
y
0
+
h
y
η
.The finite elementK
^
,
P
^
,
Σ
^ on K
^ can be defined as(8)
Σ
^
=
1
l
^
i
∫
l
^
i
v
^
i
d
s
^
,
i
=
1,2
,
3
,
P
^
=
span
⁡
1
,
ξ
,
η
.For anyv
^
∈
H
1
K
^, it can be easily checked that the interpolation function I
^
v
^ can be expressed as(9)
I
^
v
^
=
v
^
3
+
v
^
1
-
v
^
2
+
2
v
^
2
-
v
^
3
ξ
+
2
v
^
2
-
v
^
1
η
.Then, the following lemma can be obtained.Lemma 1 (see [10]).
The interpolation operator shows the anisotropic characteristic. That is to say, when the multiple indexβ
=
1, there exists a constant C which satisfies, ∀
v
^
∈
H
2
K
^,(10)
D
^
β
v
^
-
I
^
v
^
0
,
K
^
≤
C
D
^
β
v
^
1
,
K
^
.Throughout this paper, C denotes a general positive constant whose value may be different at different places but remains independent of h and h
K
/
ρ
K.The associated finite element space is defined as(11)
V
h
0
=
v
∣
v
i
j
∈
L
2
Q
2
×
2
,
v
^
i
j
=
v
i
j
K
∘
F
K
∈
P
^
,
∫
l
i
v
i
j
d
s
=
0
,
i
=
1,2
,
3
,where v
i
j denotes the jump of v
i
j across the boundary l
i and v
i
j
=
v
i
j if l
i
∈
∂
Ω. Let h
0
=
max
⁡
h
x
,
h
y.The interpolation operator is defined asI
h
0
:
H
1
Q
2
×
2
→
V
h
0:(12)
I
h
0
K
=
I
K
,
I
K
ϕ
=
I
^
ϕ
^
∘
F
K
-
1
.We note that V
h
0 is the nonconforming finite element space since V
h
0
⊄
H
0
1
Q. Define ·
0
=
∑
K
·
0
,
K
2, ·
h
=
∑
K
·
1
,
K
2, and ·
2
=
∑
K
·
2
,
K
2.Lemma 2 (see [10]).
∀
v
∈
V
h
0, one has v
0
≤
C
v
h.Lemma 3.
Ifu
∈
H
0
1
Q
∩
H
2
Q, for any v
∈
V
h
0on anisotropic meshes, one has(13)
∑
K
∫
∂
K
∂
u
∂
n
v
d
s
≤
C
h
0
u
2
v
h
.Proof.
LetP
0
l
i
v
=
1
/
l
i
∫
l
i
v
d
s; then one has (14)
∑
K
∫
∂
K
∂
u
∂
n
P
0
l
i
v
d
s
=
∑
K
P
0
l
i
v
∫
∂
K
∂
u
∂
n
d
s
=
0
,
∑
K
∑
i
=
1
3
∫
l
i
P
0
l
i
∂
u
∂
n
v
-
P
0
l
i
v
d
s
=
∑
K
∑
i
=
1
3
P
0
l
i
∂
u
∂
n
∫
l
i
v
d
s
-
l
i
P
0
l
i
v
=
0
.
Now we show the affine equivalence property ofP
0
l
i:(15)
P
0
l
i
v
=
1
l
i
∫
l
i
v
d
s
=
1
l
i
∫
l
^
i
v
^
l
i
l
^
i
d
s
^
=
1
l
^
i
∫
l
^
i
v
^
d
s
^
=
P
0
l
^
i
v
^
.Therefore (16)
∑
K
∫
∂
K
∂
u
∂
n
v
d
s
=
∑
K
∑
i
=
1
3
∫
l
i
∂
u
∂
n
-
P
0
l
i
∂
u
∂
n
v
-
P
0
l
i
v
d
s
=
∑
K
∑
i
=
1
3
∂
u
∂
n
-
P
0
l
i
∂
u
∂
n
0
,
l
i
v
-
P
0
l
i
v
0
,
l
i
≤
C
h
0
∑
K
∑
i
=
1
3
∂
u
^
∂
n
-
P
^
0
l
i
∂
u
^
∂
n
0
,
l
^
i
v
^
-
P
^
0
l
i
v
^
0
,
l
^
i
≤
C
h
0
∑
K
∑
i
=
1
3
∂
u
^
∂
n
0
,
l
^
i
v
^
-
P
^
0
l
i
v
^
0
,
l
^
i
≤
C
h
0
∑
K
u
^
1
,
∂
K
^
v
^
-
P
^
0
l
i
v
^
0
,
K
^
≤
C
h
0
∑
K
u
^
1
,
K
^
+
u
^
2
,
K
^
1
/
2
v
^
-
P
^
0
l
i
v
^
0
,
K
^
≤
C
h
0
∑
K
u
2
v
h
.Lemma 4.
Letw
=
u
-
I
h
u; one has(17)
w
0
+
h
0
w
h
≤
C
h
0
2
u
2
,
(18)
∇
w
,
∇
v
h
=
0
.Proof.
We only prove (18).
It can be seen thatΔ
v
K
=
0 and ∂
v
/
∂
n
∂
K is a constant about the above nonconforming Crouzeix-Raviart triangle element. Using the interpolation condition, we have(19)
∇
w
,
∇
v
h
=
∑
K
∫
K
∇
w
∇
v
d
x
d
y
=
∑
K
∫
∂
K
∂
u
∂
n
v
d
s
-
∑
K
∫
K
w
Δ
v
d
x
d
y
=
0
.


## 3. Finite Element Approximations for the Auxiliary Periodic Problems

The equivalent variational formulation for auxiliary periodic problem (3) is to find N
α
1 such that(20)
a
N
α
1
,
v
=
F
α
1
v
,
∀
v
∈
H
per
1
Q
2
×
2
,where a
w
,
v
=
∫
Q
A
p
k
ξ
∂
w
/
∂
ξ
k
∂
v
/
∂
ξ
p
d
ξ is the bilinear form and F
α
1
v
=
∫
Q
∂
/
∂
ξ
p
A
p
α
1
ξ
v
d
ξ is the linear function.The discrete problem of variational formulation (20) consists of finding N
α
1
h
0 such that(21)
a
h
0
N
α
1
h
0
,
v
h
0
=
F
α
1
h
0
v
h
0
∀
v
h
0
∈
V
h
0
,where the bilinear form a
h
0 and the linear function F
α
1
h
0 are defined on elements.It is known that (21) is the linearly elastic system [11]. Using the Lax-Milgram Lemma, discrete problem (21) presents unique solution in V
h
0.Theorem 5.
IfN
α
1
∈
H
2
Q
2
×
2 is the weak solution for (20) and N
α
1
h
0 is the corresponding finite element solution, one has(22)
N
α
1
h
0
-
N
α
1
h
≤
C
h
0
N
α
1
2
.Proof.
It can be proved by the Strong Lemma and Lemma3.The equivalent variational formulation of auxiliary periodic problem (4) is to find N
α
1
α
2 such that(23)
a
N
α
1
α
2
,
v
=
F
α
1
α
2
v
,
∀
v
∈
H
per
1
Q
2
×
2
,where the linear function is (24)
F
α
1
α
2
v
=
∫
Q
A
p
α
1
ξ
N
α
2
ξ
∂
v
∂
ξ
p
d
ξ
+
∫
Q
A
α
1
k
ξ
∂
N
α
2
ξ
∂
ξ
k
+
A
α
1
α
2
ξ
-
A
^
α
1
α
2
v
d
ξ
.We solve the following modified variational problem in practical calculation:(25)
a
N
~
α
1
α
2
,
v
=
F
~
α
1
α
2
v
,where the corresponding linear function is (26)
F
~
α
1
α
2
v
=
∫
Q
A
p
α
1
ξ
N
α
2
h
0
ξ
∂
v
∂
ξ
p
d
ξ
+
∫
Q
A
α
1
k
ξ
∂
N
α
2
h
0
ξ
∂
ξ
k
+
A
α
1
α
2
ξ
-
A
^
α
1
α
2
v
d
ξ
.Theorem 6.
IfN
α
1
α
2 and N
~
α
1
α
2 are the weak solutions for (23) and (25), respectively, one has(27)
N
α
1
α
2
-
N
~
α
1
α
2
h
≤
C
h
0
N
α
2
2
.Proof.
One has(28)
N
α
1
α
2
-
N
~
α
1
α
2
h
2
≤
a
N
α
1
α
2
-
N
~
α
1
α
2
,
N
α
1
α
2
-
N
~
α
1
α
2
=
∫
Q
A
p
α
1
ξ
N
α
2
h
0
-
N
α
2
∂
N
α
1
α
2
-
N
~
α
1
α
2
∂
ξ
p
d
ξ
+
∫
Q
A
α
1
k
ξ
∂
N
α
2
h
0
-
N
α
2
∂
ξ
k
N
α
1
α
2
-
N
~
α
1
α
2
d
ξ
=
C
N
α
2
h
0
-
N
α
2
0
N
α
1
α
2
-
N
~
α
1
α
2
h
+
C
N
α
2
h
0
-
N
α
2
h
N
α
1
α
2
-
N
~
α
1
α
2
0
≤
C
h
0
N
α
2
2
N
α
1
α
2
-
N
~
α
1
α
2
h
.
The discrete problem of variational formulation (25) consists of finding N
α
1
α
2
h
0 such that(29)
a
h
0
N
α
1
α
2
h
0
,
v
h
0
=
F
~
α
1
α
2
h
0
v
h
0
,
∀
v
h
0
∈
V
h
0
.Theorem 7.
IfN
~
α
1
α
2 is the weak solution for (25) and N
~
α
1
α
2
h
0 is the finite element solution for (25), one has(30)
N
~
α
1
α
2
h
0
-
N
~
α
1
α
2
h
≤
C
h
0
N
α
1
α
2
2
.Proof.
It can be proved by the Strong Lemma.Theorem 8.
IfN
α
1
α
2 is the weak solution for (23) and N
~
α
1
α
2
h
0 is the finite element solution for (25), one has(31)
N
~
α
1
α
2
h
0
-
N
α
1
α
2
h
≤
C
h
0
N
α
2
2
+
N
α
1
α
2
2
.Proof.
It can be proved using Theorems6 and 7 and the triangle inequality.Using the same procedure we also obtain the following.Theorem 9.
IfN
α
1
⋯
α
l is the weak solution for (5) and N
~
α
1
⋯
α
l
h
0 is the finite element solution, one has(32)
N
~
α
1
⋯
α
l
h
0
-
N
α
1
⋯
α
l
h
≤
C
h
0
∑
j
=
1
l
N
α
j
2
,
Q
.


## 4. Finite Element Approximate Schemes for the Homogenized Problem

Take the two-dimensional problem; for instance, we solve the following modified homogenized problem computationally:(33)
L
h
0
u
~
0
x
=
-
∂
∂
x
p
A
^
p
k
,
h
0
∂
u
~
0
x
∂
x
k
=
f
x
in
Ω
,
u
~
0
x
=
0
on
∂
Ω
,where A
^
p
k
,
h
0
=
∫
Q
A
p
k
ξ
+
A
p
m
ξ
∂
N
k
h
0
ξ
/
∂
ξ
m
d
ξ and N
k
h
0
ξ is the finite element solution of N
k
ξ.LetJ
h
=
e be an anisotropic triangle subdivision of the region Ω, and the finite element space (34)
S
h
=
v
=
v
1
,
v
2
T
∈
L
2
Ω
2
,
v
i
∂
Ω
=
0
,
v
i
e
∈
P
1
eis still nonconforming Crouzeix-Raviart. Let h
=
max
⁡
h
e be the scale of subdivision.The equivalent variational formulation of (33) is to find u
~
0 such that (35)
b
u
~
0
,
v
=
f
v
,
∀
v
∈
H
0
1
Q
2
×
2
,where the bilinear form is(36)
b
w
,
v
=
∫
Ω
A
^
p
k
,
h
0
∂
w
∂
x
k
∂
v
∂
x
p
d
x
.The linear function is f
v
=
∫
Ω
f
·
v
d
x.It is known that (35) is the linearly elastic system [11]. Using the Lax-Milgram Lemma, variational formulation (35) presents unique solution in S
h.Theorem 10.
Ifu
0 and u
~
0 are the weak solutions of (6) and (35), respectively, one has(37)
u
0
-
u
~
0
h
≤
C
h
0
2
N
i
2
,
Q
2
f
0
,
Ω
.Proof.
It is given in Theorem 4.1 of [11].The discrete problem of variational formulation (35) consists of finding u
~
h
0 such that(38)
b
h
u
~
h
0
,
v
h
=
f
h
v
h
,
∀
v
h
∈
S
h
,where the bilinear form b
h and the linear function f
h are defined on elements.Theorem 11.
Ifu
~
0 is the weak solution for (35) and u
~
h
0 is the corresponding finite element solution, one has(39)
u
~
0
-
u
~
h
0
h
≤
C
h
f
0
,
Ω
.Proof.
One has(40)
u
~
0
-
u
~
h
0
h
2
≤
b
u
~
0
-
u
~
h
0
,
u
~
0
-
u
~
h
0
≤
b
u
~
0
-
u
~
h
0
,
u
~
0
-
I
h
u
~
0
≤
C
u
~
0
-
u
~
h
0
h
u
~
0
-
I
h
u
~
0
h
≤
C
h
u
~
0
-
u
~
h
0
h
u
~
0
2
.
Then the theorem is obtained by the reduction of a fraction and the prior error estimation.Theorem 12.
Ifu
0 is the weak solution for (6) and u
~
h
0 is the finite element solution for (35), one has(41)
u
0
-
u
~
h
0
h
≤
C
h
0
2
+
h
f
0
,
Ω
.Proof.
It can be proved using Theorems5 and 6 and the triangle inequality.Notice thath
0 and h are the scales of subdivisions of Q and Ω, respectively.


## 5. Higher-Order Derivatives and Error Estimation

In this section we provide the approximation formulae for all the partial derivatives of the vector functionu
0 and give their error estimations.Let(42)
δ
x
i
u
0
l
k
=
1
τ
l
k
∑
T
∈
σ
l
k
∂
u
~
h
0
∂
x
i
T
,where σ
l
k shows the integral of triangle unit with side of l
k, τ
l
k shows the quantity of units in σ
l
k, then u
~
h
0 is solved from problem (38), and ∂
u
~
h
0
/
∂
x
i
T shows the partial derivative value of vector function u
~
h
0 on the unit T.Define a new interpolation function∏
:
C
0
Ω
-
2
→
S
h, which satisfies(43)
∏
∂
u
0
x
∂
x
i
=
δ
x
i
u
0
l
3
+
δ
x
i
u
0
l
1
-
δ
x
i
u
0
l
2
+
2
h
x
δ
x
i
u
0
l
2
-
δ
x
i
u
0
l
3
x
+
2
h
y
δ
x
i
u
0
l
2
-
δ
x
i
u
0
l
1
y
.In the same way, let(44)
δ
x
i
x
j
u
0
l
k
=
1
τ
l
k
∑
T
∈
σ
l
k
∂
∂
x
i
∏
∂
u
0
x
∂
x
i
T
,
∏
∂
2
u
0
x
∂
x
i
∂
x
j
=
δ
x
i
x
j
u
0
l
3
+
δ
x
i
x
j
u
0
l
1
-
δ
x
i
x
j
u
0
l
2
+
2
h
x
δ
x
i
x
j
u
0
l
2
-
δ
x
i
x
j
u
0
l
3
x
+
2
h
y
δ
x
i
x
j
u
0
l
2
-
δ
x
i
x
j
u
0
l
1
y
.By that analogy, let(45)
δ
x
i
⋯
x
j
x
l
m
u
0
l
k
=
1
τ
l
k
∑
T
∈
σ
l
k
∂
∂
x
l
∏
δ
x
i
⋯
x
j
m
-
1
u
0
x
∂
x
i
⋯
∂
x
j
T
,
∏
∂
m
u
0
x
∂
x
i
⋯
∂
x
j
∂
x
l
=
δ
x
i
⋯
x
j
x
l
m
u
0
l
3
+
δ
x
i
⋯
x
j
x
l
m
u
0
l
1
-
δ
x
i
⋯
x
j
x
l
m
u
0
l
2
+
2
h
x
δ
x
i
⋯
x
j
x
l
m
u
0
l
2
-
δ
x
i
⋯
x
j
x
l
m
u
0
l
3
x
+
2
h
y
δ
x
i
⋯
x
j
x
l
m
u
0
l
2
-
δ
x
i
⋯
x
j
x
l
m
u
0
l
1
y
.Theorem 13.
One has(46)
D
l
u
0
-
∏
∂
l
u
0
x
∂
x
i
⋯
∂
x
l
h
≤
C
h
u
0
m
+
2
.Proof.
One has(47)
D
l
u
0
-
∏
∂
l
u
0
x
∂
x
i
⋯
∂
x
l
h
≤
C
h
D
l
u
0
2
≤
C
h
u
0
m
+
2
.Finally, the approximate solution of the original problem is obtained as follows:(48)
u
s
,
h
0
ε
,
h
x
=
u
~
h
0
x
+
∑
l
=
1
s
ε
l
∑
α
=
l
N
~
α
h
0
ξ
∏
∂
l
u
0
x
∂
x
i
⋯
∂
x
l
.The main conclusion of this paper is as follows.Theorem 14.
Foru
0
∈
H
s
+
2
Ω, one has the error estimation(49)
u
ε
x
-
u
s
,
h
0
ε
,
h
x
h
≤
C
ε
s
-
1
u
0
s
+
2
,
Ω
+
h
0
2
+
h
f
0
,
Ω
+
∑
l
=
1
s
ε
l
h
0
u
0
l
+
1
,
Ω
∑
j
=
1
l
N
α
j
2
,
Q
+
h
u
0
l
+
2
,
Ω
.Proof.
One has(50)
u
ε
x
-
u
s
,
h
0
ε
,
h
x
=
u
ε
-
u
s
ε
+
u
0
-
u
~
h
0
+
∑
l
=
1
s
ε
l
∑
α
=
l
N
-
N
~
α
h
0
D
α
u
0
+
∑
l
=
1
s
ε
l
∑
α
=
l
N
~
α
h
0
D
α
u
0
-
∏
∂
l
u
0
x
∂
x
i
⋯
∂
x
l
.By [5], we have(51)
u
ε
-
u
s
ε
h
≤
C
ε
s
-
1
u
0
s
+
2
,
Ω
.
The proof is completed by combining Theorems9, 12, and 13 into (50).


## 6. Numerical Examples

Example 1.
Consider (1), in which Ω
=
0,1
×
0,1, and A
p
k is the identity matrix with the exact solution u
=
sin
⁡
π
x
sin
⁡
π
y
,
y
1
-
y
x
1
-
x
T. We divide the domain Ω into N uniform rectangles. The errors of displacement u under L
2-norms and h-norm with N
=
4
×
4, 8
×
8, 16
×
16, 32
×
32, 64
×
64, and 128
×
128 are obtained, respectively.Table1 shows that the optimal error estimates in L
2-norms and h-norm for displacement vector are obtained when h
→
0.Table 1
Error results inL
2-norms and h-norm for displacement vector.
u
ε
x
-
u
s
,
h
0
ε
,
h
x
0
Convergence order
u
ε
x
-
u
s
,
h
0
ε
,
h
x
h
Convergence order
4
×
4
5.4969607
e
-
02
8.0112245
e
-
01
8
×
8
1.5374911
e
-
02
1.8380561
4.2483018
e
-
01
0.91513650
16
×
16
4.0326181
e
-
03
1.9307893
2.1720831
e
-
01
0.96780695
32
×
32
1.0249727
e
-
03
1.9761313
1.0941453
e
-
01
0.98927501
64
×
64
2.5756662
e
-
04
1.9925679
5.4832867
e
-
02
0.99669150
128
×
128
6.4490254
e
-
05
1.9977926
2.7435050
e
-
02
0.99902070Example 2.
Consider the following plane stress problem:(52)
-
∂
∂
x
p
A
p
k
x
ε
∂
u
ε
x
∂
x
k
=
f
x
in
Ω
,
u
ε
x
=
0
on
∂
Ω
.The domainΩ and the periodic cell Q are shown in Figures 1 and 2.Figure 1
The domainΩ.Figure 2
The periodic cellQ.Letε
=
1
/
8. The elements in the family of 2
×
2 matrices A
p
k
x
/
ε
=
a
i
j
p
k
x
/
ε (p
,
k
,
i
,
j
=
1,2) are defined as(53)
a
i
j
,
0
p
k
=
λ
0
δ
i
p
δ
j
k
+
μ
0
δ
i
j
δ
p
k
+
μ
0
δ
i
k
δ
j
p
;
a
i
j
,
1
p
k
=
λ
1
δ
i
p
δ
j
k
+
μ
1
δ
i
j
δ
p
k
+
μ
1
δ
i
k
δ
j
p
,in which λ
m
,
μ
m (m
=
0,1) are Lame constants that can be expressed by Elasticity modulus E
m and Poisson ratio υ
m
(54)
λ
m
=
E
m
·
υ
m
1
+
υ
m
1
-
2
υ
m
;
μ
m
=
E
m
2
1
+
υ
m
.Let E
0
=
1
0
8, E
1
=
1
0
9, μ
0
=
μ
1
=
0.25, and f
x
=
2.78
×
1
0
6. The coefficient matrix family of the homogenized problem is (55)
A
^
=
0.133
×
1
0
9
0.461
×
1
0
-
7
0.545
×
1
0
-
7
0.482
×
1
0
8
0.461
×
1
0
-
7
0.355
×
1
0
8
0.482
×
1
0
8
-
0.134
×
1
0
-
7
0.545
×
1
0
-
7
0.482
×
1
0
8
0.355
×
1
0
8
-
0.274
×
1
0
-
7
0.482
×
1
0
8
-
0.134
×
1
0
-
7
-
0.274
×
1
0
-
7
0.133
×
1
0
9
.In two-scale asymptotic formulation (2), the homogenized solution u
0
x is the main part that reflects the macroscopic response of composite materials, whereas the periodic solution N
α
1
⋯
α
l
ξ and the higher derivatives of the homogenized solution D
α
u
0
x are the correction part that reflect the local micromechanical behavior of composite materials. In general, only the first-order and second-order multiscale asymptotic solution produce satisfactory results in engineering.Since it is difficult to obtain analytical solutions of (52), we take the finite element solution on the fine-mesh as the exact solution u
ε
x. The scale of fine-mesh for calculating the original problem is h
=
1
/
120, and the scale for the homogenized problem and periodic problem is h
=
1
/
48.Table2 shows that the computing resources can be saved significantly and the computational efficiency can be improved greatly by multiscale method in this paper compared with the fine-mesh finite element method. Particularly when ε is very small, the computational resources required for the finite element method scale rapidly, whereas the multiscale finite element algorithm is relatively stable.Table 2
Comparison of number of elements and nodes.
Original equation
Periodic problem
Homogenized problem
Elements number
28800
4608
4608
Nodes number
14641
2401
2401The comparison between the true solutionu
ε
x
=
w
ε
x
,
v
ε
x and the second-order multiscale asymptotic solution u
2
ε
x
=
w
2
ε
x
,
v
2
ε
x is shown in Figure 3.Figure 3
The exact solutionu
ε
x
=
w
ε
x
,
v
ε
x and the second-order asymptotic solution u
2
ε
x
=
w
2
ε
x
,
v
2
ε
x: (a) w
ε
x; (b) v
ε
x; (c) w
2
ε
x; (d) v
2
ε
x.
(a)
(b)
(c)
(d)



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*Source: 7525392-2016-07-14.xml*