Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2012, Article ID 808216, 35 pages
doi:10.1155/2012/808216
Research Article
The Spectral Method for the Cahn-Hilliard
Equation with Concentration-Dependent Mobility
Shimin Chai and Yongkui Zou
Department of Mathematics, Jilin University, Changchun 130012, China
Correspondence should be addressed to Yongkui Zou, zouyk@jlu.edu.cn
Received 14 April 2012; Accepted 9 July 2012
Academic Editor: Ram N. Mohapatra
Copyright q 2012 S. Chai and Y. Zou. 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.
This paper is concerned with the numerical approximations of the Cahn-Hilliard-type equation
with concentration-dependent mobility. Convergence analysis and error estimates are presented
for the numerical solutions based on the spectral method for the space and the implicit Euler
method for the time. Numerical experiments are carried out to illustrate the theoretical analysis.
1. Introduction
In this paper, we apply the spectral method to approximate the solutions of Cahn-Hilliard
equation, which is a typical class of nonlinear fourth-order diffusion equations. Diffusion
phenomena is widespread in the nature. Therefore, the study of the diffusion equation
caught wide concern. Cahn-Hilliard equation was proposed by Cahn and Hilliard in
1958 as a mathematical model describing the diffusion phenomena of phase transition in
thermodynamics. Later, such equations were suggested as mathematical models of physical
problems in many fields such as competition and exclusion of biological groups 1, moving
process of river basin 2, and diffusion of oil film over a solid surface 3. Due to the
important application in chemistry, material science, and other fields, there were many
investigations on the Cahn-Hilliard equations, and abundant results are already brought
about.
The systematic study of Cahn-Hilliard equations started from the 1980s. It was Elliott
and Zheng 4 who first study the following so-called standard Cahn-Hilliard equation with
constant mobility:
∂u
divk∇Δu − ∇Au 0.
∂t
1.1
2
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Basing on global energy estimates, they proved the global existence and uniqueness of
classical solution of the initial boundary problem. They also discussed the blow-up property
of classical solutions. Since then, there were many remarkable studies on the Cahn-Hilliard
equations, for example, the asymptotic behavior of solutions 5–8, perturbation of solutions
9, 10, stability of solutions 11, 12, and the properties of the solutions for the Cahn-Hilliard
equations with dynamic boundary conditions 13–16. In the mean time, a number of the
numerical techniques for Cahn-Hilliard equations were produced and developed. These
techniques include the finite element method 4, 17–24, the finite difference method 25–
29, the spectral method, and the pseudospectral method 30–35. The finite element method
for the Cahn-Hilliard equation is well investigated by many researchers. For example, in
23, 36, 1.1 was discreted by conforming finite element method with an implicit time
discretization. In 22, semidiscrete schemes which can define a Lyapunov functional and
remain mass constant were used for a mixed formulation of the governing equation. In 37,
a mixed finite element formulation with an implicit time discretization was presented for the
Cahn-Hilliard equation 1.1 with Dirichlet boundary conditions. The conventional strategy
to obtain numerical solutions by the finite differece method is to choose appropriate mesh
size based on the linear stability analysis for different schemes. However, this conventional
strategy does not work well for the Cahn-Hilliard equation due to the bad numerical stability.
Therefore, an alternative strategy is proposed for general problems, for example, in 26, 38
the strategy was to design such that schemes inherit the energy dissipation property and the
mass conservation by Furihata. In 27, 28, a conservative multigrid method was developed
by Kim.
The advantage of the spectral method is the infinite order convergence; that is, if
the exact solution of the Cahn-Hilliard equation is Ck smooth, the approximate solution
will be convergent to the exact solution with power for exponent N −k , where N is the
number of the basis function. This method is superior to the finite element methods and
finite difference methods, and a lot of practice and experiments convince the validity of
the spectral method 39. Many authors have studied the solution of the Cahn-Hilliard
equation which has constant mobility by using spectral method. For example, in 33–35,
Ye studied the solution of the Cahn-Hilliard equation by Fourier collocation spectral method
and Legendre collocation spectral method under different boundary conditions. In 30, the
author studied a class of the Cahn-Hilliard equation with pseudospectral method. However,
the Cahn-Hilliard equation with varying mobility can depict the physical phenomena more
accurately; therefore, there is practical meaning to study the numerical solution for the CahnHilliard equation with varying mobility. Yin 40, 41 studied the Cahn-Hilliard equation
with concentration-dependent mobility in one dimension and obtained the existence and
uniqueness of the classical solution. Recently, Yin and Liu 42, 43 investigate the regularity
for the solution in two dimensions. Some numerical techniques for the Cahn-Hilliard
equation with concentration-dependent mobility are already studied with the finite element
method 28 and with finite difference method 44.
In this paper, we consider an initial-boundary value problem for Cahn-Hilliard
equation of the following form:
∂u
D mu D3 u − DAu 0, x, t ∈ 0, 1 × 0, T ,
∂t
Du|x0,1 D3 u
0, t ∈ 0, T ,
x0,1
ux, 0 u0 ,
x ∈ 0, 1,
1.2
1.3
1.4
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3
where D ∂/∂x and
As −s γ1 s2 γ2 s3 ,
1.5
γ2 > 0.
Here, ux, t represents a relative concentration of one component in binary mixture. The
function mu is the mobility which depends on the unknown function u, which restricts
diffusion of both components to the interfacial region only. Denote QT 0, 1 × 0, T .
Throughout this paper, we assume that
0 < m0 ≤ ms ≤ M0 ,
m s ≤ M1 ,
∀s ∈ R,
1.6
where m0 , M0 , and M1 are positive constants. The existence and uniqueness of the classical
solution of the problems 1.2–1.4 were proved by Yin 41.
In this paper, we will apply the spectral method to discretize the spatial variables
of 1.2 to construct a semidiscrete system. We prove the existence and boundedness of
the solutions of this semidiscrete system. Then, we apply implicit midpoint Euler scheme
to discretize the time variable and obtain a full-discrete scheme, which inherits the energy
dissipation property. The property of the mobility depending on the solution of 1.2 causes
much troubles for the numerical analysis. Furthermore, with the aid of Nirenberg inequality
we investigate the boundedness and convergence of the numerical solutions of the fulldiscrete equations. We also obtain the error estimation for the numerical solutions to the exact
ones.
This paper is organized as follows. In Section 2, we study the spectral method for 1.2–
1.4 and obtain the error estimate between the exact solution u and the spectral approximate
solution uN . In Section 3, we use the implicit Euler method to discretize the time variable
and obtain the error estimate between the exact solution u and the full-discrete approximate
j
solution UN . Finally in Section 4, we present a numerical computation to illustrate the
theoretical analysis.
2. Semidiscretization with Spectral Method
In this section, we apply the spectral method to discretize 1.2–1.4 and study the error
estimate between the exact solution and the semidiscretization solution.
Denote by ·k and |·|k the norm and seminorm of the Sobolev spaces H k 0, 1k ∈ N,
respectively. Let ·, · be the standard L2 inner product over 0, 1. Define
HE2 0, 1 v ∈ H 2 0, 1; Dv|x0,1 0 .
2.1
A function u is said to be a weak solution of the problems 1.2–1.4, if u ∈
L∞ 0, T ; HE2 0, 1 and satisfies the following equations:
∂u
,v
∂t
D2 u − Au, DmuDv 0,
u·, 0, v − u0 , v 0,
∀v ∈ HE2 .
∀v ∈ HE2 ,
2.2
2.3
4
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For any integer N > 0, let SN span{cos kπx, k 0, 1, 2, . . . , N}. Define a projection
operator PN : HE2 → SN by
1
PN uxvxdx 0
1
uxvxdx,
∀v ∈ SN .
2.4
0
We collect some properties of this projection PN in the following lemma see 39.
Lemma 2.1. i PN commutes with the second derivation on HE2 I, that is,
PN D2 v D2 PN v,
2.5
∀v ∈ HE2 I.
ii For any 0 ≤ μ ≤ σ, there exists a positive constant C such that
v − PN vμ ≤ CN μ−σ |v|σ ,
v ∈ H σ I.
2.6
The following Nirenberg inequality is a key tool for our theoretical estimates.
Lemma 2.2. Assume that Ω ⊂ Rn is a bounded domain, u ∈ W m,r Ω, then we have
j D u
Lp
C1 Dm uaLr u1−a
Lq C2 uLq ,
2.7
where
j
a < 1,
m
j
1
1
1 m
a
−
1 − a .
p n
r n
q
2.8
By 41, we have the following.
Lemma 2.3. Assume that ms ∈ C1α R, u0 ∈ C4α I, Di u0 0 Di u0 1 0 i 0, 1, 2, 3, 4, ms > 0, then there exists a unique solution of the problems 1.2–1.4 such that
u ∈ C1α/4, 4α QT .
2.9
The spectral approximation to 2.2 is to find an element uN ·, t ∈ SN such that
∂uN
, vN
∂t
D2 uN − PN AuN , DmuN DvN 0,
uN ·, 0, vN − u0 , vN 0,
t ≤ T.
∀vN ∈ SN .
∀vN ∈ SN ,
2.10
2.11
Now we study the L∞ norm estimates of the function uN ·, t and DuN ·, t for 0 ≤
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5
Theorem 2.4. Assume 1.6 and u0 ∈ HE2 . Then there is a unique solution of 2.10 and 2.11 such
that
uN ∞ ≤ C,
DuN ≤ C,
0 ≤ t ≤ T,
2.12
where C Cu0 , T, γ1 , γ2 is a positive constant.
Proof. From 2.11 it follows that uN ·, 0 PN u0 ·. The existence and uniqueness of the
initial problem follow from the standard ODE theory. Now we study the estimate.
Define an energy function:
1
Ft HuN , 1 DuN 2 ,
2
where Hs s
0
2.13
Atdt 1/4γ2 s4 1/3γ1 s3 − 1/2s2 . Direct computation gives
∂uN
∂uN
∂uN
∂uN
AuN ,
DuN , D
PN AuN ,
− D 2 uN ,
∂t
∂t
∂t
∂t
∂uN
, PN AuN − D2 uN .
∂t
dF
dt
2.14
Noticing that PN AuN − D2 uN ∈ SN and setting vN PN AuN − D2 uN in 2.10, applying
integration by part, we obtain
dF muN D3 uN − DPN AuN , D PN AuN − D2 uN
dt
− muN D3 uN − DPN AuN , D3 uN − DPN AuN 2.15
2
≤ − m0 D PN AuN − D2 uN ≤ 0.
Hence,
Ft ≤ F0 1
0
1
HPN u0 ds DPN u0 2 ,
2
∀0 ≤ t ≤ T.
2.16
Applying Young inequality, we obtain
u2N ≤ εu4N C1ε ,
u3N ≤ εu4N C2ε ,
2.17
where C1ε and C2ε are positive constants. Letting ε 3γ2 /8|γ1 | 12, then for all 0 ≤ t ≤ T ,
1
0
HuN dx ≥
1
0
γ2 1
1
1 3 1 2
4
γ2 uN − γ1 uN − uN dx ≥
uN 4 dx − K1 ,
4
3
2
8 0
2.18
6
Journal of Applied Mathematics
where K1 is a positive constant depending only on γ1 and γ2 . Therefore, we get
γ2
1
DuN ·, t2 2
8
1
1
1
2
HuN ·, tdx
t
DuN ·,
2
0
2.19
1
2
HPN u0 dx DPN u0 .
Ft ≤ F0 uN ·, t4 dx − K1 ≤
0
0
Thus,
DuN ·, t2 ≤ 2K1 2F0 2K1 2
1
0
8
8
uN x, t4 dx ≤ K1 F0 γ2
γ2
1
HPN u0 dx DPN u0 2 C,
0
K1 1
0
1
HPN u0 dx DPN u0 2
2
2.20
C,
where C Cu0 , γ1 , γ2 is a constant. By Hölder inequality, we obtain
2
uN 1
0
u2N dx
≤
1
0
u4N dx
1/2 1
1/2
2
1 dx
1
0
0
1/2
u4N dx
.
2.21
Therefore,
uN ≤ C,
DuN ≤ C.
2.22
∀0 ≤ t ≤ T.
2.23
From the embedding theorem it follows that
uN ∞ ≤ C,
Theorem 2.5. Assume 1.6 and let uN ·, t be the solution of 2.10 and 2.11. Then there is a
positive constant C Cu0 , m, T, γ1 , γ2 > 0 such that
DuN ∞ ≤ C,
2 D uN ≤ C.
2.24
Proof. Setting vN D4 uN in 2.10 and integrating by parts, we get
1 d
2 2
D uN muN D4 uN − D2 PN AuN , D4 uN
2 dt
m uN DuN D3 uN − DPN AuN , D4 uN 0.
2.25
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7
Consequently,
1 d
2 2
D uN muN D4 uN , D4 uN I1 I2 I3 ,
2 dt
2.26
where
I1 muN D2 PN AuN , D4 uN ,
I2 − m uN DuN D3 uN , D4 uN ,
2.27
I3 m uN DuN DPN AuN , D4 uN .
Noticing 1.6, we have
2 1 d 1 d
2 2
2 2
D uN m0 D4 uN ≤
D uN muN D4 uN , D4 uN .
2 dt
2 dt
2.28
In terms of the Nirenberg inequality 2.7, there is a constant C > 0 such that
3/8
DuN ∞ ≤ C D4 uN u5/8 u ,
3/4
DuN ∞ ≤ C D2 uN u1/4 u ,
2.29
7/8
3 D uN ≤ C D4 uN u1/8 u .
∞
Noticing the definition of the function A and the estimates in 2.12, we have
AuN ∞ ≤ C,
A uN ≤ C,
∞
A uN ≤ C,
∞
2.30
8
Journal of Applied Mathematics
for some constant C Cu0 , T, γ1 , γ2 > 0. Applying Hölder inequality and Young inequality,
for any ε > 0,
|I1 | ≤ M0 A uN D2 uN A uN DuN 2 · D4 uN ≤ M0 A uN ∞ · D2 uN D4 uN M0 A uN ∞ · DuN DuN ∞ D4 uN 2
2
3/8
≤ εD4 uN CD2 uN C D4 uN 1 D4 uN 2
2
11/8
≤ εD4 uN CD2 uN CD4 uN CD4 uN 2
2 ε 2
≤ εD4 uN CD2 uN D4 uN C 2
2
2
≤ 2εD4 uN CD2 uN C,
ε
4 2
D uN C
2
where C Cu0 , m, T, γ1 , γ2 , ε > 0 is a positive constant. Similarly, we obtain
|I2 | ≤ M1 DuN D3 uN D4 uN ∞
4 7/8
≤ C D uN 1 D4 uN 4 15/8 4 ≤ C D uN D uN ε
4 2
D uN C 2
2
≤ εD4 uN C,
≤
ε
4 2
D uN C
2
|I3 | ≤ M1 DPN AuN ∞ · DuN · D4 uN ≤ CDPN AuN ∞ D4 uN 3/4
≤ C D2 PN AuN PN AuN 1/4 PN AuN D4 uN 3/4
≤ C D2 PN AuN AuN 1/4 AuN D4 uN 3/4
2
1/4
≤ C D PN AuN AuN ∞ AuN ∞ D4 uN 3/4
2
≤ C D PN AuN 1 D4 uN 2.31
Journal of Applied Mathematics
≤ C D2 PN AuN 1 D4 uN 9
≤ C D2 AuN 1 D4 uN ≤ C A uN D2 uN A uN DuN 2 · D4 uN 2
≤ εD4 uN C.
2.32
Hence,
2
2
2
1 d
2 2
D uN m0 D4 uN ≤ 4εD4 uN CD2 uN C.
2 dt
2.33
Taking ε m0 /8, we have
2
1 d
2 2 m0 4 2
D uN D uN ≤ CD2 uN C,
2 dt
2
2.34
where C Cu0 , m, T, γ1 , γ2 > 0 is a positive constant. Therefore,
2
1 d
2 2
D uN ≤ CD2 uN C.
2 dt
2.35
From Gronwall inequality it follows that
2
2
2 2
D uN ≤ expCt D2 u0 1 ≤ expCT D2 u0 1 ≤ C,
2.36
where C Cu0 , m, T, γ1 , γ2 > 0 is a positive constant. According to the embedding theorem,
we have
DuN ∞ ≤ C,
0 ≤ t ≤ T,
2.37
where C Cu0 , m, T, γ1 , γ2 > 0 is a positive constant.
Now, we study the error estimates between the exact solution u and the semidiscrete
spectral approximation solution uN . Set the following decomposition:
u − uN η e,
η u − PN u,
e P N u − uN .
2.38
From the inequality 2.6 it follows that
η ≤ CN −4 .
Hence, it remains to obtain the approximate bounds of e.
2.39
10
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Theorem 2.6. Assume that u is the solution of 1.2–1.4, uN ∈ SN is the solution of 2.10 and
2.11, and m is smooth and satisfies 1.6, then there exists a constant C Cu0 , m, γ1 , γ2 such that
e ≤ C N −2 e0 .
2.40
Before we prove this theorem, we study some useful approximation properties.
Lemma 2.7. For any ε > 0, we have
2
2
− D2 η D2 e, DmuN De ≤ −m0 D2 e 3εD2 e C e2 N −4 ,
where C Cu0 , m, T, γ1 , γ2 , ε > 0 is a positive constant.
Proof. Direct computation gives
− D2 η D2 e, DmuN De
− D2 e, muN D2 e − D2 η, muN D2 e
− D2 η, m uN DuN De − D2 e, m uN DuN De
2
2
≤ −m0 D2 e M0 D2 eD2 η
M1 D2 ηDuN ∞ De M1 D2 eDuN ∞ De
2
2 M2 2
≤ −m0 D2 e εD2 e 0 D2 η
4ε
2
2
2 2
2 2 M1 DuN ∞
2
M1 DuN ∞ De D η εD e De2
4ε
2
2
M0
2 2
2 2
M1 DuN ∞ D2 η
≤ −m0 D e 2εD e 4ε
M12 DuN 2∞
M1 DuN ∞ De2
4ε
2
2
M0
2 2
2 2
M1 DuN ∞ D2 η
≤ −m0 D e 2εD e 4ε
2.41
Journal of Applied Mathematics
11
M2 DuN 2∞
M1 DuN ∞ 1
4ε
2 D e · e
2
2
2
2
≤ −m0 D2 e 2εD2 e CD2 η εD2 e Ce2
2
2
2 ≤ −m0 D2 e 3εD2 e C e2 D2 η
2
2
≤ −m0 D2 e 3εD2 e C e2 N −4 ,
2.42
where C Cu0 , m, T, γ1 , γ2 , ε > 0 is a positive constant.
Lemma 2.8. For any ε > 0, we have
2
Au − PN AuN , DmuN De ≤ 3εM02 D2 e C e2 N −4 ,
2.43
where C Cu0 , m, T, γ1 , γ2 , ε > 0 is a positive constant.
Proof. Noticing that
Au − PN AuN , DmuN De
Au − PN Au, DmuN De PN Au − PN AuN , DmuN De
≤ Au − PN Au · DmuN De PN Au, uN u − uN · DmuN De
≤ Au − PN Au · DmuN De Au, uN η e · DmuN De,
2.44
where
As, τ γ2 s2 sτ τ 2 γ1 s τ − 1.
2.45
From the boundedness of uN in Theorem 2.4 and the property of u in Lemma 2.3, it follows
that
Au, uN ≤ C.
∞
2.46
12
Journal of Applied Mathematics
Then we obtain
2 2
Au − PN Au2 ≤ C u3 − PN u3 u2 − PN u2 u − PN u2
2
≤ CN −8 u3 u2 |u|4 ≤ CN −8 ,
4
4
2 2 2 Au, uN η e ≤ Au, uN e2 η ≤ C N −8 e2 ,
∞
2
2
DmuN De2 ≤ 2muN D2 e 2m uN DuN De
2
2
≤ 2M02 D2 e 2M12 DuN 2∞ D2 e e2
2.47
2
2 M4 DuN 4
∞
≤ 2M02 D2 e M02 D2 e 1
e2
M02
2
≤ 3M02 D2 e Ce2 ,
where C Cu0 , m, T, γ1 , γ2 > 0 is a positive constant. By Cauchy inequality, for any ε > 0,
we have
Au − PN AuN , DmuN De
≤ Au − PN Au · DmuN De Au, uN η e · DmuN De
ε
1
ε
1
2
DmuDe2 Au − PN Au2 DmuDe2 Au, uN η e 2
2ε
2
2ε
2
≤ 3εM02 D2 e C e2 N −8 ,
≤
2.48
where C Cu0 , m, T, γ1 , γ2 , ε > 0 is a positive constant.
Lemma 2.9. Assume that u is the solution of 1.2–1.4, there exists a positive constant C Cu0 , m, γ1 , γ2 such that
m 0 2 2
D2 u − Au, Dmu − muN De ≤
D e C e2 N −4 ,
2
2.49
where C Cu0 , m, T, γ1 , γ2 > 0 is a positive constant.
Proof. By Lemma 2.3, we have
3
D u − DAu ≤ D3 u A uDu∞ ≤ C.
∞
∞
2.50
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13
In the other hand, it follows that
2 mu − muN 2 ≤ M12 u − uN 2 ≤ M12 e2 η .
2.51
By the Young inequality,
D2 u − Au, Dmu − muN De − D3 u − DAu, mu − muN De
≤ D3 u − DAu mu − muN · De
∞
≤ Cmu − muN De
2 ≤ εDe2 Cε e2 η
2.52
2
2 ≤ εD2 e Cε e2 η .
Choosing ε m0 /2 in the previous inequality, we obtain 2.49.
Proof of Theorem 2.6. Setting v e in 2.2, we obtain
∂u
, e D2 u − Au, DmuDe 0.
∂t
2.53
Setting vN e in 2.10, we get
∂uN
, e D2 uN − PN AuN , DmuN De 0.
∂t
2.54
2.53 minus 2.54 gives
∂e
,e
∂t
− Du2 − Au, Dmu − muN De − D2 u − D2 uN , DmuN De
2.55
Au − PN AuN , DmuN De.
According to Lemmas 2.7, 2.8 and 2.9, we have
2
2
1 d
e2 ≤ − m0 D2 e 3εD2 e C e2 N −4
2 dt
2
3εM02 D2 e C N −8 e2
2
≤ −m0 4 3M02 ε D2 e C N −4 e2 .
2.56
14
Journal of Applied Mathematics
Set ε m0 /4 3M02 , then there exists a positive constant C Cu0 , m, γ1 , γ2 such that
d
e2 ≤ C e2 N −4 .
dt
2.57
By Gronwall inequality, we have
e ≤ expCt e0 N −2 ≤ expCT e0 N −2 ≤ C e0 N −2 ,
2.58
where C Cu0 , m, T, γ1 , γ2 > 0 is a constant.
Summig up the properties above, we obtain the following.
Theorem 2.10. Assume ms is sufficiently smooth and satisfies 1.6, u is the solution of 1.2–
1.4, and uN is the solution of 2.10 and 2.11. Then there exists a positive constant C Cu0 , m, T, γ1 , γ2 > 0 such that
u − uN ≤ C N −2 u0 − u0N ,
∀t ∈ 0, T .
2.59
3. Full-Discretization Spectral Scheme
In this section, we apply implicit midpoint Euler scheme to discretize time variable and get
a full-discrete form. Furthermore, we investigate the boundedness of numerical solution and
the convergence of the numerical solutions of the full-discrete system. We also obtain the
error estimates between the numerical solution and the exact ones.
Firstly, we introduce a partition of 0, T . Let 0 t0 < t1 < · · · < tΛ , where tj jh and
h T/Λ is time-step size. Then the full-discretization spectral method for 1.2–1.4 reads:
j
∀v ∈ SN , find UN ∈ SN j 0, 1, 2, . . . , Λ such that
⎛
⎞
j1
j
j1 j j1/2 UN − UN
2 j1/2
⎝
⎠
, v D UN − PN A UN , UN , D m UN
Dv
0,
h
j1/2
where UN
3.1
0
UN
, v − u0 , v 0,
3.2
φ, ϕ γ2 φ3 ϕ3 φ2 ϕ φϕ2 γ1 φ2 φϕ ϕ2 − 1 φ ϕ .
A
4
3
2
3.3
j1
j
UN UN /2 and
j
The solution UN has the following property.
Journal of Applied Mathematics
15
j
Lemma 3.1. Assume that UN ∈ SN j 1, 2, . . . , Λ is the solution of 3.1-3.2. Then there exists
a constant C Cu0 , m, γ1 , γ2 > 0 such that
j UN ≤ C,
∞
j DUN ≤ C.
3.4
Proof. Define a discrete energy function at time tj by
1
j 2
j
F j DUN H UN , 1 .
2
3.5
Notice that
⎞ ⎛
⎞
⎛
j1
j
j1 j Uj1 − Uj
UN − UN
j1/2
1 N
N
⎠ ⎝PN A
⎠
U ,U ,
F j 1 − F j ⎝DUN , D
N
N
h
h
h
⎛
−
⎞
j1 j Uj1 − Uj
N⎠
U ,U , N
.
− PN A
N
N
h
⎝D2 Uj1/2
N
3.6
j1/2
Setting v D2 UN
− PN AU
N , UN ∈ SN in 3.1, we obtain
j1
j
1 F j 1 −F j
h
j1 j j1/2
j1/2
3 j1/2
Uj1 , Uj
,
D
− m UN
U
−
DP
D3 UN − DPN A
A
UN , UN
N
N
N
N
j1 j 2
3 j1/2
≤ −m0 D UN − DPN A UN , UN ≤ 0,
3.7
which implies
2 1
0 0
,1 .
F j ≤ F0 DUN
H UN
2
3.8
By 2.18, we have
1
H
0
j
UN
γ2
dx ≥
8
1 0
j
UN
4
dx − K1 .
3.9
16
Journal of Applied Mathematics
Then
1 γ2 1 j 4
1
1
j 2
j 2
j
UN dx − K1 ≤ DUN H UN dx F j
DUN 2
8 0
2
0
2 1
0 0
≤ F0 DUN
,1 .
H UN
2
3.10
So we obtain
2
j 2
0 0
, 1 ≡ C1 ,
DUN ≤ 2K1 DUN
2 H UN
1 0
4
j
UN dx
8
≤
K1 γ2
2 1
0 0
,1
≡ C2 ,
DUN
H UN
2
3.11
where C1 C1 u0 , m, γ1 , γ2 and C2 C2 u0 , m, γ1 , γ2 are positive constants. By Hölder
inequality, we get
1 2
j
j 2
UN dx ≤
UN 1
0
0
4
j
UN dx
1/2
≤ C21/2 .
3.12
Therefore,
j UN ≤ C21/4 .
3.13
j UN ≤ C,
3.14
By the embedding theorem, we obtain
∞
where C Cu0 , m, γ1 , γ2 > 0 is a constant.
j
Lemma 3.2. Assume that UN ∈ SN j 1, 2, . . . , Λ is the solution of the full-discretization scheme
3.1-3.2, then there is a constant C Cu0 , m, T, γ1 , γ2 > 0 such that
j DUN ≤ C,
∞
2 j D UN ≤ C.
3.15
Journal of Applied Mathematics
j1/2
Proof. Setting v 2D4 UN
17
∈ SN in 3.1, we have
⎛
⎞
j1
j
UN − UN
j1/2
⎝
, 2D4 UN ⎠
h
j1/2
j1/2
4 j1/2
Uj1 , Uj
2D
D m UN
U
D3 UN − DPN A
N
N
N
⎛
⎞
j1
j
D2 UN − D2 UN 2 j1
j1/2
j
2
4 j1/2
4 j1/2
⎝
⎠
, D UN D UN m UN
D UN , 2D UN
h
j1/2
j1/2
j1/2
j1/2
DUN D3 UN , 2D4 UN
m U N
j1
j1/2
j
2
4 j1/2
− m UN
D PN A UN UN , 2D UN
− m
j1/2
UN
j1 j j1/2
4 j1/2
DUN DPN A UN , UN , 2D UN
0.
3.16
Therefore,
j1/2
j1/2
j1/2
1 2 j1 2 2 j 2
D 4 UN , D 4 UN
D UN − D UN 2 m UN
h
j1/2
j1/2
j1/2
j1/2
D3 UN DUN , D4 UN
2 m UN
j1/2
Uj1 , Uj , D4 Uj1/2
2 m UN
D 2 PN A
N
N
N
3.17
j1 j1 j1/2
j1/2
4 j1/2
2 m UN
DUN DPN A UN , UN , D UN
j
j
j
I1 I 2 I 3 .
By Nirenberg inequality 2.7, we have
j j 3/8 j 5/8
j DUN ≤ C D4 UN UN UN ,
∞
j 2 j 3/4 j 1/4 j DUN ≤ C D UN UN UN ,
∞
3 j 4 j 7/8 j 1/8 j D UN ≤ C D UN UN UN .
∞
3.18
18
Journal of Applied Mathematics
According to 3.14, we obtain
j1
j
A1 UN , UN ≤ C,
∞
j1 j A11 UN , UN ≤ C,
j1
j
A2 UN , UN ≤ C,
j1 j A12 UN , UN ≤ C,
∞
∞
∞
j1 j A22 UN , UN ≤ C,
∞
3.19
where C Cu0 , m, γ1 , γ2 is a positive constant. By Young inequality, for any positive constant
ε > 0, it follows that
3 j1/2 j
D
U
I1 ≤ 2M1 N
∞
4 j1/2 DUj1/2
D
U
N
N
4 j1/2 7/8
4 j1/2 ≤ C D UN 1 D UN 4 j1/2 15/8
4 j1/2 D
≤ C
D
U
C
U
N
N
2
ε
D4 Uj1/2
Cε N
2
4 j1/2 2
≤ εD UN Cε ,
≤
2
ε
D4 Uj1/2
Cε
N
2
2 j1 j j
4 j1/2 I2 ≤ 2M0 D A UN , UN D UN 2
j1 2
j1
j
j
A
≤ 2M0 DU
DU
2A
DU
DU
A
22
11
12
N
N
N
N
A 1 D
2
j1
UN
A 2D
2
j 4 j1/2 UN D UN 2 4 j1/2 j1 2 j
2 j1 2 j ≤ C D UN D UN DUN DUN D UN 4 j1/2 j1 j j1 j1 j j D
≤ C D2 UN D2 UN DUN DUN DUN DUN U
N
∞
∞ 4 j1/2 j1 j j1 j D
≤ C D2 UN D2 UN DUN DUN U
N
∞
∞ 4 j1/2 j1 j j1 3/4
j 3/4
D
≤ C D2 UN D2 UN D2 UN D2 UN 1 U
N
4 j1/2 j1 j D
≤ C D2 UN D2 UN 1 U
N
4 j1/2 2
2 j1 2 2 j 2
≤ εD UN Cε D UN D UN 1 ,
Journal of Applied Mathematics
j1/2 j
Uj1 , Uj A
DP
I3 ≤ M1 DUN N
N
N ∞
19
4 j1/2 D UN j1 j 3/4
4 j1/2 2
≤ C D PN A UN , UN 1 D UN j1 j 4 j1/2 2
≤ C D PN A UN , UN 1 D UN 4 j1/2 2 j1 j 4 j1/2 ≤ C D A UN , UN D UN D UN 4 j1/2 2
4 j1/2 2
2 j1 2 2 j 2
Cε
D
D
≤ ε
U
C
U
U
1
ε
U
D
D
ε
N
N
N
N
4 j1/2 2
2 j1 2 2 j 2
≤ 2εD UN Cε D UN D UN 1 ,
3.20
where Cε Cε u0 , m, γ1 , γ2 > 0 is a constant. Therefore,
4 j1/2 2
1 2 j1 2 2 j 2
D UN − D UN 2m0 D UN h
j1/2
1 2 j1 2 2 j 2
4 j1/2
4 j1/2
≤
D UN , D UN
D UN − D UN 2 m UN
h
2 j1/2 2
2 j1 2 2 j 2
D
U
C
U
U
1
.
≤ 4ε
D
D
ε
N
N
N
3.21
Setting ε m0 /2, there is a positive constant C Cu0 , m, γ1 , γ2 such that
2Ch C
2 j 2
2 j1 2
h.
D UN D UN ≤ 1 1 − Ch
1 − Ch
3.22
C/1 − Ch, if h is sufficiently small such that Ch ≤ 1/2, we have
Denoted by C
C
j 2
2 j1 2
D UN ≤ 1 Ch
D2 UN h
2
j j1 i
2 0 2 C
≤ 1 Ch
1 Ch
D UN h
2 i1
2 0 2 j1Ch
≤ exp
D UN Cjh
CT
≤ exp
2 0 2 D UN CT ≡ C ,
3.23
20
Journal of Applied Mathematics
where C C u0 , m, T, γ1 , γ2 > 0 is a positive constant. By the embedding theorem, the
estimate 3.15 holds.
j
Next, we investigate the error estimates for the numerical solution UN to the exact
solution utj . Our analysis is based on the error decomposition denoted by
j
u tj − UN ηj ej ,
η j u t j − PN u t j ,
j
ej PN u tj − UN .
3.24
The boundedness estimate of ηj follows from the inequality 2.6, that is, for any 0 ≤ j ≤ Λ,
there is a positive constant C Cu0 , m, γ1 , γ2 such that
j
η ≤ CN −4 .
3.25
Hence, it remains to obtain the approximate bounds of ej . If no confusion occurs, we denote
the average of the two instant errors ej and ej1 by ej1/2 :
ej1/2 ej ej1
,
2
ηj1/2 ηj ηj1
.
2
3.26
For later use, we give some estimates in the next lemmas.
Lemma 3.3. Assume that the solution of 1.2–1.4 is such that uttt ∈ L2 QT , then
⎞
⎛
j1
j
2 2
2
−
U
U
1 4 tj1
j1 N
, ej1/2 ⎠ h
uttt 2 dt hej1/2 .
e ≤ ej 2h⎝ut tj1/2 − N
h
320
tj
3.27
Proof. Applying Taylor expansion about tj1/2 , we have
u u
j
uj1
j1/2
h j1/2 h2 j1/2 1
− ut
utt
−
2
8
2
h j1/2 h2 j1/2 1
uj1/2 ut
utt
2
8
2
tj1/2
2
t − tj uttt dt,
tj
tj1
tj 1/2
3.28
2
tj1 − t uttt dt.
Then
j1/2
ut
1
uj1 − uj
−
−
h
2h
t
j1
tj 1/2
2
tj1 − t uttt dt tj1/2
tj
2
t − tj uttt dt .
3.29
Journal of Applied Mathematics
21
From Hölder inequality it follows that
⎛
2 2 ⎞
2
tj1
tj1/2
j1
j
1 ⎝
2
2
⎠
j1/2 u − u −
−
t
u
dt
u
dt
t
t
−
t
ut
≤
j1
ttt
j
ttt tj
h
2h2 tj 1/2
h3
≤
320
tj1
3.30
uttt 2 dt.
tj
Noticing that for any v ∈ SN , we have
⎛
⎝uj1/2
t
⎞ j1
j
j1
j
UN − UN
−
u
1 j1
u
j1/2
, v ⎠ ut
,v e − ej , v .
−
−
h
h
h
3.31
Taking v 2ej1/2 in 3.31, we obtain
⎞
⎛
j1
j
2 2
j1
j
U
−
U
−
u
u
j1/2
j1/2
j1 j
j1/2
j1/2
N
N
⎠ − 2h u
,e
,e
−
−
e e 2h⎝ut
t
h
Δt
⎞
⎛
j1
j
2
U
−
U
j1/2
N
, ej1/2 ⎠
− N
≤ ej 2h⎝ut
h
1 4
h
320
tj1
3.32
2
uttt 2 dt hej1/2 .
tj
Taking v ej1/2 in 2.2 and 3.1, respectively, we have
∂u tj1/2
j1/2
D2 uj1/2 − A uj1/2 , D m uj1/2 Dej1/2
,e
0,
3.33
∂t
⎞
⎛
j1
j
j1 j j1/2 UN − UN j1/2
j1/2
⎠ D2 Uj1/2
⎝
,
D
m
,e
−
P
,
U
U
De
0.
A
U
N
N
N
N
N
h
3.34
22
Journal of Applied Mathematics
Comparing 3.33 and 3.34, we have
⎛
⎝uj1/2
t
⎞
j1
j
UN − UN j1/2
⎠
,e
−
h
⎛
−⎝ D u
2 j1/2
⎞
j1
j
D2 UN D2 UN
j1/2
j1/2 ⎠
, D m UN
−
De
2
3.35
j1 j j1/2 j1/2
j1/2
A u
− PN A UN , UN , D m UN
De
D u
2 j1/2
−A u
j1/2
j1/2
j1/2
j1/2
, D m UN
De
−m u
.
Now we investigate the error estimates of the three items in the right-hand side of the
previous equation.
Lemma 3.4. Assume that u is the solution of 1.2–1.4 such that Dutt ∈ L2 QT , then there exists
a positive constant C1 C1 m, u0 , T, γ1 , γ2 > 0 such that
⎞
2 j1
2 j
D
U
D
U
j1/2
N
N
, D m UN
Dej1/2 ⎠
− ⎝D2 uj1/2 −
2
⎛
tj1 2 m0 2 j1/2 2
2 2
j j1 2
−4
3
≤ − D e
C1 N e e h
D utt dt .
2
tj
3.36
Proof. By Taylor expansion and Hölder inequality, we obtain
u u
j
j1
u
u
j1/2
j1/2
h j1/2
− ut
2
h j1/2
ut
2
tj1/2
t − tj utt dt,
tj
tj1
tj 1/2
3.37
tj1 − t utt dt.
Therefore,
1 j
1
u uj1 − uj1/2 2
2
t
j1/2
tj
t − tj utt dt tj1
tj1/2
tj1 − t utt dt .
3.38
Journal of Applied Mathematics
23
By Hölder inequality, we have
2
2 j1/2 1 j
j1
D u
u
−
u
2
⎛ 2 ⎞
tj1/2
tj1
1 ⎝
2
t − tj utt dt tj1 − t utt dt ⎠
D
4 tj
tj1/2
⎛
2 ⎞
tj1/2
tj1/2 tj1
tj1 2
2
2
2
1 ⎝
2
2
≤
t − tj dt
tj1 − t dt
D utt dt D utt dt ⎠
4 tj
tj
tj1/2
tj1/2
⎛
2 ⎞
2
2
3 tj1/2 3 tj1 h
1 ⎝
h
D2 utt dt D2 utt dt ⎠
4 24 tj
24 tj1/2
≤
1 h3
·
4 24
tj1 2 2
D utt dt.
tj
3.39
Direct computation gives
2 h3 tj1 2 j1/2 1 j
2 2
j1
≤
D u
u
−
u
D utt dt.
2
96 tj
Then
⎛
⎞
⎞
j1
j
U
U
j1/2
N⎠
− ⎝D2 ⎝uj1/2 − N
, D m UN
Dej1/2 ⎠
2
⎛
− D
2
j1/2
u
uj1 uj
−
2
j1/2
j1/2
, D m UN
De
⎞
⎞
j1
j
j1
j
U
U
j1/2
u
u
N⎠
− N
, D m UN
− ⎝D2 ⎝
Dej1/2 ⎠
2
2
⎛
⎛
j1/2
uj1 uj 2
j1/2 j1/2
D
m
≤ D u
−
U
De
·
N
2
j1/2 j1/2
j1/2
j1/2
2
− D η
, D m UN
e
De
3.40
24
Journal of Applied Mathematics
1/2 j1/2
j1/2
j1/2
h3 tj1 2 2
j1/2 2 j1/2 m
m
· U
e
U
De
≤
D
DU
D utt dt
N
N
N
96 tj
j1/2
j1/2
j1/2
− D2 ηj1/2 , m UN
D2 ej1/2 − D2 ηj1/2 , m UN
DUN Dej1/2
j1/2
j1/2
j1/2
j1/2
2 j1/2
2 j1/2
2 j1/2
, m UN
, m UN
− D e
D e
− D e
DUN De
j
j
j
σ1 σ2 σ3 .
3.41
By Cauchy inequality, it follows that
j
σ1 ≤
h3
96
1/2 tj1 j1/2
2 2
2 j1/2 · m UN
D e
D utt dt
tj
h3
96
≤ M0
1/2 tj1 j1/2
j1/2
2 2
j1/2 DUN De
D utt dt
m UN
tj
h3
96
1/2
tj1 2 2
2 j1/2 D utt dt
D e
tj
1/2
3 tj1 j1/2 h
2 2
j1/2 M1 DUN D utt dt
De
∞ 96 tj
tj1 2
2 2
j1/2 2
j1/2 ≤ h C1ε
D utt dt εD2 e
εDe
3
tj
tj1 2
2 2
j1/2 j1/2 j1/2 3
≤ h C1ε
D utt dt εD2 e
εD2 e
e
tj
tj1 2
2 2
j1/2 2
j1/2 3
≤ h C1ε
D utt dt 2εD2 e
εe
,
tj
j1/2 j
j1/2 j1/2 σ2 ≤ M0 D2 ηj1/2 D2 e
M1 DUN D2 ηj1/2 De
∞
2
ε
2 j1/2 2
j1/2 ∞ ε
D e
De
2
2ε
2
2 ε ε
2
≤ C2ε D2 ηj1/2 D2 ej1/2 D2 ej1/2 ej1/2 2
2
2
2
ε
2
≤ C2ε D2 ηj1/2 εD2 ej1/2 ej1/2 ,
2
≤
M02 2 j1/2 2
D η
2ε
2
j1/2 2
M1 DUN Journal of Applied Mathematics
25
j1/2 2 j1/2 j
2 j1/2 2
j1/2 ·
σ3 ≤ − m0 D e
e
M1 DUN D
De
∞
2
j1/2 2
DU
M
N
2
1
2 j1/2 2 ε 2 j1/2 2
j1/2 ∞
≤ − m0 D e
De
D e
2
2ε
2
2
2
≤ − m0 D2 ej1/2 εD2 ej1/2 C3ε ej1/2 .
3.42
Then we obtain
⎞
2 j1
2 j
D
U
D
U
j1/2
N
N
− ⎝D2 uj1/2 −
, D m UN
Dej1/2 ⎠
2
⎛
tj1 j
2 j1/2 2
j j
2 2
3
≤ σ1 σ2 σ3 ≤ 4ε − m0 D e
h C1ε
D utt dt
3.43
tj
2
2
C2ε D2 ηj1/2 C3ε ej1/2 ,
where C1ε , C2ε , and C3ε are positive constants. Choosing ε m0 /8, and terms of the properties
of the projection operator PN , we complete the proof of the estimate 3.36.
Lemma 3.5. Assume that u is the solution of 1.2–1.4 such that utt ∈ L2 QT and ut ∈ L∞ QT .
Then for any positive constant ε > 0, there exists a constant C2 C2 m, u0 , T, γ1 , γ2 > 0, such that
j1/2
Uj1 , Uj , D m Uj1/2
De
A uj1/2 − PN A
N
N
N
2 2
m0 2 j1/2 2
≤
D e
C2 ej1 ej N −8
4
t
t
j1
j1/2 2 j1
2
2
3
h
.
utt dt ut
ut dt
tj
∞
3.44
tj
Proof. Firstly, we consider
3
3 2
2 3 uj1 , uj γ2 uj1/2 − γ2 uj1 uj1 uj uj1 uj uj
A uj1/2 − A
4
2 γ1 2
2 uj1 uj1 uj uj
γ1 uj1/2 −
3
1 j1
j
j
j
u uj
γ2 ρ1 γ1 ρ2 − ρ3 .
− uj1/2 −
2
3.45
26
Journal of Applied Mathematics
Direct computation gives
1/2
tj1
1 tj1/2 Δt3 tj1
j
2
,
t − tj utt dt tj1 − t utt dt ≤
utt dt
ρ3 2 tj
96 tj
tj1/2
2 1 2 2 2 1 j1/2 2 j1
j j
j1/2
j1
−
u
u
−
u
− uj
2u
2
u
ρ2 6
6
1
≤ 2uj1/2 − uj1 uj
2uj1/2 uj1 uj 6
1
uj1/2 − uj1 uj1/2 uj1 uj1/2 − uj uj1/2 uj 6
1
≤ 2uj1/2 − uj1 uj · 2uj1/2 uj1 uj ∞
6
tj1
1 −Δt j1/2
ut
−
tj1 − t utt dt uj1/2 uj1
6
2
tj 1/2
tj1/2
Δt j1/2
j1/2
j u
−
u t − tj utt dt u
2 t
tj
t
t
1/2
j1
Δt3/2 j1/2 2
≤ CΔt
,
utt dt
ut dt
ut
·
∞
12
tj
tj
3 3 1 3 3 3 1
j
j1/2
j1
2uj1/2 − uj1 uj
2
u
−
u
− uj
ρ1 12
6
2
2 1
2uj1/2 − uj1 uj · 4 uj1/2 2 uj1 uj uj1/2 uj1 uj
≤
12
2
2 1 j1/2
j1
j1/2
j1 j1/2
j1
u
u
−
u
u
u
u
6
2
2 j1/2
j
j1/2
j j1/2
u
u
−u
u u
uj
⎧
1/2 ⎫
1/2
tj1
⎨ tj1
⎬
j1/2 ≤ CΔt3/2
.
ut
utt 2 dt
ut 2 dt
·
⎩ tj
⎭
∞
tj
3/2
j1
1/2
2
3.46
Then
j1/2
Uj1 , Uj , D m Uj1/2
A uj1/2 − PN A
De
N
N
N
2
3
≤ A uj1/2 − PN A uj1/2 4ε
2
ε
j1/2 D m Uj1/2
De
N
3
Journal of Applied Mathematics
27
2
ε
j1/2 D m Uj1/2
De
N
3
2
j1 j 2 ε j1/2
3
j1 j
j1/2 A u , u − A UN , UN D m UN
De
4ε
3
2
2
j1/2
j1/2 3
D
m
≤ ε
U
De
A uj1/2 − PN A uj1/2 N
4ε
2 3 2
3
j1 j
uj1 , uj Uj1 , Uj u ,u − A
A uj1/2 − A
A
N
N .
4ε
4ε
2
3
uj1 , uj A uj1/2 − A
4ε
3.47
Direct computation yields
2 2
j1/2 2
j1/2 2
2 2 j1/2 D m Uj1/2
≤
M
M
e
C
De
,
D
e
N
0
1
2
A uj1/2 − PN A uj1/2 ≤ CN −8 ,
3.48
2
j 2 j 2 j 2
uj1 , uj A uj1/2 − A
≤ C ρ1 ρ2 ρ3 ,
2 j 2 j 2 j1 j
j1 2 j 2
−8
Uj1 , Uj ,
≤
CN
A u , u − A
e
G
G
e
2
1
N
N
∞
∞
where
j
G1
j1
2 2 γ1 j1
1
γ2 j1
j1 2
j
j
j
j1
u UN
UN uj1 UN − ,
UN UN u UN
8
3
2
2 2 2
1
γ2
γ1 j1
j
j
j
j
G2 uj UN uj uj1 uj1 UN
u uj UN − .
8
3
2
3.49
Applying Lemma 2.3 and Theorem 3.7, we obtain that G1 ∞ ≤ Cu0 , m, T, γ1 , γ2 and G2 ∞ ≤
Cu0 , m, T, γ1 , γ2 . Taking ε m0 /4M02 M12 in 3.47, we have
Uj1 , Uj , D mj1/2 Dej1/2
A uj1/2 − PN A
N
N
2 2
m0 2 j1/2 2
≤
D e
C2 ej1 ej N −8
4
t
Δt3
j1
tj
3.50
t
j1/2 2 j1
utt 2 dt ut
ut 2 dt
where C2 C2 u0 , m, T, γ1 , γ2 > 0 is a constant.
∞
tj
,
28
Journal of Applied Mathematics
Lemma 3.6. Assume that u is the solution of 1.2–1.4. Then there exists a positive constant C Cu0 , m, γ1 , γ2 > 0 such that
j1/2 D2 uj1/2 − A uj1/2 , D m UN
− m uj1/2 Dej1/2
tj1
m0 2 j1/2 2
j1/2 2
−8
3
≤
utt dt .
D e
C e
N h
4
tj
3.51
Proof. By 2.9, we have
3 j1/2
− DA uj1/2 ≤ D3 uj1/2 A uj1/2 Duj1/2 D u
∞
∞
∞
≤ D3 uj1/2 A uj1/2 Duj1/2 ≤ C.
∞
∞
3.52
∞
In the other hand,
2
2
j1/2
j1/2 2
j1/2 m Uj1/2
U
−
m
u
≤
M
−
u
N
N
1
2
j1/2 uj uj1 2 uj uj1
j1/2 −u
≤ CUN −
2
2
3.53
2 2 h3 tj1
≤ Cej1/2 ηj1/2 utt dt.
96 tj
By Young inequality, we obtain
j1/2 D2 uj1/2 − A uj1/2 , D m UN
− m uj1/2 Dej1/2
j1/2 − D3 uj1/2 − DA uj1/2 , m UN
− m uj1/2 Dej1/2
j1/2 j1/2
3 j1/2
j1/2 j1/2 ≤ D u
− DA u
−m u
m U N
De
∞
2
2
j1/2
j1/2 ≤ εDej1/2 Cε m
U
−
m
u
N
2
j1/2
2 j1/2 j1/2 j1/2 ≤ εD e
−m u
e
Cε m UN
tj1
j1/2 2
2 j1/2 2
−8
3
≤ εD e
utt dt .
Cε e
N h
tj
Choosing ε m0 /4 in the previous inequality leads to 3.51.
Finally, we obtain the main theorem of this paper.
3.54
Journal of Applied Mathematics
29
Theorem 3.7. Assume that ux, t is the solution of 1.2–1.4 and satisfies that
D4 u ∈ L∞ QT ,
ut ∈ L∞ QT ,
D2 utt ∈ L2 QT ,
uttt ∈ L2 QT .
3.55
j
UN ∈ SN j 1, 2, . . . , Λ is the solution of the full-discretization 3.1 and 3.2. If h is sufficiently
small, there exists a positive constant C such that, for any j 0, 1, 2, . . . , Λ,
j1 j1 e PN u tj1 − UN ≤ C N −2 e0 Bh2 ,
where B tj1
0
3.56
D2 utt 2 utt 2 uttt 2 max0≤l≤j {ul1/2
2∞ } · ut 2 dt.
t
Proof. By 3.27, 3.36, 3.44, and 3.51, we have
tj1
2
4
j1 2 j 2 h
uttt 2 dt hej1/2 e ≤ e 320 tj
⎧ ⎛
⎞
2 j1
2 j
⎨
D
U
D
U
j1/2
N
N
, D m UN
2h −⎝D2 uj1/2 −
Dej1/2 ⎠
⎩
2
j1/2
Uj1 , Uj , D m Uj1/2
A uj1/2 − A
De
N
N
N
⎫ 3.57
⎬
j1/2 − m uj1/2 Dej1/2
D2 uj1/2 − A uj1/2 , D m UN
⎭
2
2 2 ≤ ej hC1 N −4 ej1 ej C2 h4
tj1 j1/2 2
2 2
ut 2 dt,
D utt utt 2 uttt 2 ut
∞
tj
where C1 C1 m, u0 , T, γ1 , γ2 > 0 and C2 C2 m, u0 , T, γ1 , γ2 > 0 are constants. For
2C1 C2 , we obtain
sufficiently small h such that C1 h ≤ 1/2, denoting C
2
j1 2
hN −4 h4 Bj ,
e ≤ 1 Ch
ej C
3.58
where
Bj tj1 j1/2 2
2 2
ut 2 dt.
D utt utt 2 uttt 2 ut
tj
∞
3.59
30
Journal of Applied Mathematics
By the Gronwall inequality of the discrete form, we obtain
j l j1 2
j1 2
0
1 Ch
htN −4 h4 Bl
e ≤ 1 Ch
e C
l0
j 0 2 −4
4 l
≤ exp C j 1 h e C
hN h B
3.60
l0
j
0 2 −4
4
l
.
B
≤ exp C j 1 h e C jhN h
l0
Direct computation gives
j
l0
Bl ≤
tj1 #
2 $
2 2
2
·
dt.
u
D utt utt 2 uttt 2 max ul1/2
t
t
∞
0≤l≤j
0
3.61
Then we complete the conclusion 3.56.
Furthermore, we get the following theorem.
Theorem 3.8. Assume u is the solution of 1.2–1.4 and satisfies that
D4 u ∈ L∞ QT ,
ut ∈ L∞ QT ,
D2 utt ∈ L2 QT ,
uttt ∈ L2 QT .
3.62
j
UN ∈ SN j 1, 2, . . . , Λ is the solution of the full-discretization scheme 3.1-3.2, and U0
satisfies e0 PN u0 − U0 ≤ CN −4 . If h is sufficiently small, there exists a constant C Cu0 , m, T, γ1 , γ2 > 0 such that
j u x, tj − UN ≤ C N −2 h2 ,
j 1, 2, . . . , Λ.
3.63
4. Numerical Experiments
In this section, we apply the spectral method described in 3.1 and 3.2 to carry out
numerical computations to illustrate theoretical estimations in the previous section. Consider
2.2 with settings:
ms m0 s2 ,
As s3 − s,
4.1
Journal of Applied Mathematics
31
×10−4
10
×10−4
6
5
8
4
6
3
4
2
2
1
0
0
1
−2
1
0.5
0
0
0.1
0.05
0.5
a
0
0
0.05
0.1
b
Figure 1: The development of the solution of the full-discrete scheme when initial value is u0 x x5 1 − x5 a and u0 x x5 1 − x5 ex b.
where m0 > 0 is a constant. The full-discretization spectral method of 2.2 and 2.3 reads:
%
j
find UN N
l0 αjl cos lπx j 1, 2, . . . , Λ such that
⎞ ⎛
j1
j
j1
j
D2 UN D2 UN
UN − UN
⎝
Uj1 , Uj ,
, v⎠ ⎝
− PN A
N
N
h
2
⎛
⎛ ⎛
⎞ ⎞⎞
j1
j
D2 UN D2 UN
⎠Dv⎠⎠ 0,
D ⎝ m⎝
2
4.2
0
UN
, v − u0 , v 0.
In our computations we fix N 32 and choose five different time-step sizes hk k 1, 2, . . . , 5. Let Λk be the integer with hk Λk T . Since we have no exact solution of 2.2 and
Λ0
with
2.3, we take N 32 and h0 0.15625 × 10−4 to compute an approximating solution UN
h0 Λ0 T and regard this as an exact solution. we also choose five different time-step sizes
Λk
k 1, 2, . . . , 5
hk k 1, 2, . . . , 5 with hk Λk T to obtain five approximating solutions UN
and compute the error estimation. Define an error function:
errT, hk 1
0
Λk
UN
−
Λ0 2
UN
dx
1/2
.
4.3
This function characterizes the estimations with respect to time-step size.
4.1. Example 1
Take m0 1 and T 0.1. We also take two different initial functions u10 x x5 1 − x5 and
u20 x5 1 − x5 ex to carry out numerical computations. Figure 1 shows the development of
the solutions for time t from t 0 to t 0.1 with fixed step-size h0 0.15625 × 10−4 .
32
Journal of Applied Mathematics
×10−4
10
×10−4
6
5
8
4
6
3
4
2
2
1
0
0
−1
1
0.8 0.6
0.4 0.2
0
0.05
0
0.1
−2
1
0.5
a
0.1
0.05
0
0
b
Figure 2: The development of the solution of the full-discrete scheme with initial value u0 x x5 1 − x5
a and u0 x x5 1 − x5 ex b.
×10−3
8
0.015
0.01
6
4
0.005
2
0
0
−0.005
−2
−4
1
0.8
0.6
0.4
0.2
0
0
0.05
0.1
−0.01
−0.015
1
a
0.8
0.6
0.4
0.2
0
0
0.05
0.1
b
Figure 3: The development of the solution of the full-discrete scheme with initial value u0 x x5 1 − x5
a and u0 x x5 1 − x5 ex b.
Table 1: The error and the convergence order.
hk
0.1 × 10−2
0.5 × 10−3
0.25 × 10−3
0.125 × 10−3
0.625 × 10−4
err1 0.1, hk 0.2441 × 10−6
0.6332 × 10−7
0.1186 × 10−7
0.1938 × 10−8
0.2504 × 10−9
Order1
—
1.9466
2.4159
2.6142
2.9518
err2 0.1, hk 0.2114 × 10−6
0.5414 × 10−7
0.1115 × 10−7
0.1901 × 10−8
0.2483 × 10−9
Order2
—
1.9651
2.2799
2.5515
2.9368
Table 2: The error and the convergence order.
hk
0.1 × 10−2
0.5 × 10−3
0.25 × 10−3
0.125 × 10−3
0.625 × 10−4
err1 0.1, hk 0.4748 × 10−8
0.8348 × 10−9
0.1370 × 10−9
0.2694 × 10−10
0.6398 × 10−11
Order1
—
2.5077
2.6069
2.3467
2.0741
err2 0.1, hk 0.8864 × 10−8
0.1577 × 10−8
0.2440 × 10−9
0.4457 × 10−10
0.1059 × 10−10
Order2
—
2.4907
2.6922
2.4525
2.0736
Journal of Applied Mathematics
33
Table 3: The error and the convergence order.
hk
0.1 × 10−2
0.5 × 10−3
0.25 × 10−3
0.125 × 10−3
0.625 × 10−4
err1 0.1, hk 0.1150 × 10−5
0.2872 × 10−6
0.7158 × 10−7
0.1769 × 10−7
0.4211 × 10−8
Order1
—
2.0013
2.0043
2.0170
2.0704
err2 0.1, hk 0.5224 × 10−5
0.1304 × 10−5
0.3250 × 10−6
0.8031 × 10−7
0.1912 × 10−7
Order2
—
2.0019
2.0044
2.0171
2.0704
We also choose five different time-step sizes hk to carry out numerical computations
and apply the error function in 4.3 to illustrate the estimation and convergence order in
time variable t, see Table 1.
4.2. Example 2
Take m0 0.05 and T 0.1. We also take two different initial functions u10 x x5 1 − x5 and
u20 x5 1 − x5 ex to carry out numerical computations. Figure 2 shows the development of
the solutions for time t from t 0 to t 0.1 with fixed step-sizes h0 0.15625 × 10−4 .
We also choose five different time-step sizes hk to carry out numerical computations
and apply the error function in 4.3 to illustrate the estimation and convergence order in
time variable t, see Table 2.
4.3. Example 3
Take m0 0.005 and T 0.1. We also take two different initial functions u10 x x5 1 − x5
and u20 x5 1 − x5 ex to carry out numerical computations. Figure 3 shows the development
of the solutions for time t from t 0 to t 0.1 with fixed step-sizes h0 0.15625 × 10−4 .
We also choose five different time-step sizes hk to carry out numerical computations
and apply the error function in 4.3 to illustrate the estimation and convergence order in
time variable t, see Table 3.
Acknowledgment
This work is supported by NSFC no. 11071102.
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