The Spectral Method for the Cahn-Hilliard Equation with … · 2019. 7. 31. · the spectral method...

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2012, Article ID 808216, 35 pagesdoi:10.1155/2012/808216

Research ArticleThe Spectral Method for the Cahn-HilliardEquation with Concentration-Dependent Mobility

Shimin Chai and Yongkui Zou

Department of Mathematics, Jilin University, Changchun 130012, China

Correspondence should be addressed to Yongkui Zou, [email protected]

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 CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

This paper is concerned with the numerical approximations of the Cahn-Hilliard-type equationwith concentration-dependent mobility. Convergence analysis and error estimates are presentedfor the numerical solutions based on the spectral method for the space and the implicit Eulermethod 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-Hilliardequation, which is a typical class of nonlinear fourth-order diffusion equations. Diffusionphenomena is widespread in the nature. Therefore, the study of the diffusion equationcaught wide concern. Cahn-Hilliard equation was proposed by Cahn and Hilliard in1958 as a mathematical model describing the diffusion phenomena of phase transition inthermodynamics. Later, such equations were suggested as mathematical models of physicalproblems in many fields such as competition and exclusion of biological groups [1], movingprocess of river basin [2], and diffusion of oil film over a solid surface [3]. Due to theimportant application in chemistry, material science, and other fields, there were manyinvestigations on the Cahn-Hilliard equations, and abundant results are already broughtabout.

The systematic study of Cahn-Hilliard equations started from the 1980s. It was Elliottand Zheng [4] who first study the following so-called standard Cahn-Hilliard equation withconstant mobility:

∂u

∂t+ div[k∇Δu − ∇A(u)] = 0. (1.1)

2 Journal of Applied Mathematics

Basing on global energy estimates, they proved the global existence and uniqueness ofclassical solution of the initial boundary problem. They also discussed the blow-up propertyof classical solutions. Since then, there were many remarkable studies on the Cahn-Hilliardequations, 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-Hilliardequations with dynamic boundary conditions [13–16]. In the mean time, a number of thenumerical techniques for Cahn-Hilliard equations were produced and developed. Thesetechniques 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 methodfor 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 timediscretization. In [22], semidiscrete schemes which can define a Lyapunov functional andremain 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 theCahn-Hilliard equation (1.1) with Dirichlet boundary conditions. The conventional strategyto obtain numerical solutions by the finite differece method is to choose appropriate meshsize based on the linear stability analysis for different schemes. However, this conventionalstrategy 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 themass conservation by Furihata. In [27, 28], a conservative multigrid method was developedby Kim.

The advantage of the spectral method is the infinite order convergence; that is, ifthe exact solution of the Cahn-Hilliard equation is Ck smooth, the approximate solutionwill be convergent to the exact solution with power for exponent N−k, where N is thenumber of the basis function. This method is superior to the finite element methods andfinite difference methods, and a lot of practice and experiments convince the validity ofthe spectral method [39]. Many authors have studied the solution of the Cahn-Hilliardequation 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 methodand Legendre collocation spectral method under different boundary conditions. In [30], theauthor studied a class of the Cahn-Hilliard equation with pseudospectral method. However,the Cahn-Hilliard equation with varying mobility can depict the physical phenomena moreaccurately; therefore, there is practical meaning to study the numerical solution for the Cahn-Hilliard equation with varying mobility. Yin [40, 41] studied the Cahn-Hilliard equationwith concentration-dependent mobility in one dimension and obtained the existence anduniqueness of the classical solution. Recently, Yin and Liu [42, 43] investigate the regularityfor the solution in two dimensions. Some numerical techniques for the Cahn-Hilliardequation with concentration-dependent mobility are already studied with the finite elementmethod [28] and with finite difference method [44].

In this paper, we consider an initial-boundary value problem for Cahn-Hilliardequation of the following form:

∂u

∂t+D[m(u)

(D3u −DA(u)

)]= 0, (x, t) ∈ (0, 1) × (0, T), (1.2)

Du|x=0,1 = D3u∣∣∣x=0,1

= 0, t ∈ (0, T), (1.3)

u(x, 0) = u0, x ∈ (0, 1), (1.4)

Journal of Applied Mathematics 3

where D = ∂/∂x and

A(s) = −s + γ1s2 + γ2s

3, γ2 > 0. (1.5)

Here, u(x, t) represents a relative concentration of one component in binary mixture. Thefunction m(u) is the mobility which depends on the unknown function u, which restrictsdiffusion of both components to the interfacial region only. Denote QT = (0, 1) × (0, T).Throughout this paper, we assume that

0 < m0 ≤ m(s) ≤ M0,∣∣m′(s)

∣∣ ≤ M1, ∀s ∈ R, (1.6)

where m0, M0, and M1 are positive constants. The existence and uniqueness of the classicalsolution 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 variablesof (1.2) to construct a semidiscrete system. We prove the existence and boundedness ofthe solutions of this semidiscrete system. Then, we apply implicit midpoint Euler schemeto discretize the time variable and obtain a full-discrete scheme, which inherits the energydissipation property. The property of the mobility depending on the solution of (1.2) causesmuch troubles for the numerical analysis. Furthermore, with the aid of Nirenberg inequalitywe investigate the boundedness and convergence of the numerical solutions of the full-discrete equations. We also obtain the error estimation for the numerical solutions to the exactones.

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 approximatesolution uN . In Section 3, we use the implicit Euler method to discretize the time variableand obtain the error estimate between the exact solution u and the full-discrete approximatesolution U

j

N . Finally in Section 4, we present a numerical computation to illustrate thetheoretical analysis.

2. Semidiscretization with Spectral Method

In this section, we apply the spectral method to discretize (1.2)–(1.4) and study the errorestimate between the exact solution and the semidiscretization solution.

Denote by ‖·‖k and | · |k the norm and seminorm of the Sobolev spacesHk(0, 1)(k ∈ N),respectively. Let (·, ·) be the standard L2 inner product over (0, 1). Define

H2E(0, 1) =

{v ∈ H2(0, 1); Dv|x=0,1 = 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 ;H2

E(0, 1)) and satisfies the following equations:

(∂u

∂t, v

)+(D2u −A(u), D(m(u)Dv)

)= 0, ∀v ∈ H2

E, (2.2)

(u(·, 0), v) − (u0, v) = 0, ∀v ∈ H2E. (2.3)


For any integer N > 0, let SN = span{cos kπx, k = 0, 1, 2, . . . ,N}. Define a projectionoperator PN : H2

E → SN by

∫1

0(PN)u(x)v(x)dx =

∫1

0u(x)v(x)dx, ∀v ∈ SN. (2.4)

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 H2E(I), that is,

PND2v = D2PNv, ∀v ∈ H2E(I). (2.5)

(ii) For any 0 ≤ μ ≤ σ, there exists a positive constant C such that

‖v − PNv‖μ ≤ CNμ−σ |v|σ, v ∈ Hσ(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 ∈ Wm,r(Ω), then we have

∥∥∥Dju∥∥∥Lp

� C1‖Dmu‖aLr‖u‖1−aLq + C2‖u‖Lq , (2.7)

where

j

m� a < 1,

1p=

j

n+ a

(1r− m

n

)+ (1 − a)

1q. (2.8)

By [41], we have the following.

Lemma 2.3. Assume that m(s) ∈ C1+α(R), u0 ∈ C4+α(I), Diu0(0) = Diu0(1) = 0 (i =0, 1, 2, 3, 4), m(s) > 0, then there exists a unique solution of the problems (1.2)–(1.4) such that

u ∈ C1+α/4, 4+α(QT

). (2.9)

The spectral approximation to (2.2) is to find an element uN(·, t) ∈ SN such that

(∂uN

∂t, vN

)+(D2uN − PNA(uN), D(m(uN)DvN)

)= 0, ∀vN ∈ SN, (2.10)

(uN(·, 0), vN) − (u0, vN) = 0, ∀vN ∈ SN. (2.11)

Now we study the L∞ norm estimates of the function uN(·, t) and DuN(·, t) for 0 ≤t ≤ T .


Theorem 2.4. Assume (1.6) and u0 ∈ H2E. 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 = C(u0, T, γ1, γ2) is a positive constant.

Proof. From (2.11) it follows that uN(·, 0) = PNu0(·). The existence and uniqueness of theinitial problem follow from the standard ODE theory. Now we study the estimate.

Define an energy function:

F(t) = (H(uN), 1) +12‖DuN‖2, (2.13)

where H(s) =∫s0 A(t)dt = (1/4)γ2s4 + (1/3)γ1s3 − (1/2)s2. Direct computation gives

dF

dt=(A(uN),

∂uN

∂t

)+(DuN,D

∂uN

∂t

)=(PNA(uN),

∂uN

∂t

)−(D2uN,

∂uN

∂t

)

=(∂uN

∂t, PNA(uN) −D2uN

).

(2.14)

Noticing that PNA(uN) −D2uN ∈ SN and setting vN = PNA(uN) −D2uN in (2.10), applyingintegration by part, we obtain

dF

dt=(m(uN)

(D3uN −DPNA(uN)

), D(PNA(uN) −D2uN

))

= −(m(uN)

(D3uN −DPNA(uN)

), D3uN −DPNA(uN)

)

≤ −m0

∥∥∥D(PNA(uN) −D2uN

)∥∥∥2 ≤ 0.

(2.15)

Hence,

F(t) ≤ F(0) =∫1

0H(PNu0)ds +

12‖DPNu0‖2, ∀0 ≤ t ≤ T. (2.16)

Applying Young inequality, we obtain

u2N ≤ εu4

N + C1ε, u3N ≤ εu4

N + C2ε, (2.17)

where C1ε and C2ε are positive constants. Letting ε = 3γ2/(8|γ1| + 12), then for all 0 ≤ t ≤ T ,

∫1

0H(uN)dx ≥

∫1

0

(14γ2u

4N − 1

3

∣∣∣γ1u3N

∣∣∣ − 12u2N

)dx ≥ γ2

8

∫1

0(uN)4dx −K1, (2.18)


where K1 is a positive constant depending only on γ1 and γ2. Therefore, we get

12‖DuN(·, t)‖2 + γ2

8

∫1

0(uN(·, t))4dx −K1 ≤ 1

2‖DuN(·, t)‖2 +

∫1

0H(uN(·, t))dx

= F(t) ≤ F(0) =∫1

0H(PNu0)dx + ‖DPNu0‖2.

(2.19)

Thus,

‖DuN(·, t)‖2 ≤ 2K1 + 2F(0) = 2K1 + 2∫1

0H(PNu0)dx + ‖DPNu0‖2 � C,

∫1

0(uN(x, t))4dx ≤ 8

γ2(K1 + F(0)) =

8γ2

(K1 +

∫1

0H(PNu0)dx +

12‖DPNu0‖2

)� C,

(2.20)

where C = C(u0, γ1, γ2) is a constant. By Holder inequality, we obtain

‖uN‖2 =∫1

0u2Ndx ≤

(∫1

0u4Ndx

)1/2(∫1

012dx

)1/2

=

(∫1

0u4Ndx

)1/2

. (2.21)

Therefore,

‖uN‖ ≤ C, ‖DuN‖ ≤ C. (2.22)

From the embedding theorem it follows that

‖uN‖∞ ≤ C, ∀0 ≤ t ≤ T. (2.23)

Theorem 2.5. Assume (1.6) and let uN(·, t) be the solution of (2.10) and (2.11). Then there is apositive constant C = C(u0, m, T, γ1, γ2) > 0 such that

‖DuN‖∞ ≤ C,∥∥∥D2uN

∥∥∥ ≤ C. (2.24)

Proof. Setting vN = D4uN in (2.10) and integrating by parts, we get

12d

dt

∥∥∥D2uN

∥∥∥2+(m(uN)

(D4uN −D2PNA(uN)

), D4uN

)

+(m′(uN)DuN

(D3uN −DPNA(uN)

), D4uN

)= 0.

(2.25)


Consequently,

12d

dt

∥∥∥D2uN

∥∥∥2+(m(uN)D4uN,D4uN

)= I1 + I2 + I3, (2.26)

where

I1 =(m(uN)D2PNA(uN), D4uN

),

I2 = −(m′(uN)DuND3uN,D4uN

),

I3 =(m′(uN)DuNDPNA(uN), D4uN

).

(2.27)

Noticing (1.6), we have

12d

dt

∥∥∥D2uN

∥∥∥2+m0

∥∥∥D4uN

∥∥∥2 ≤ 1

2d

dt

∥∥∥D2uN

∥∥∥2+(m(uN)D4uN,D4uN

). (2.28)

In terms of the Nirenberg inequality (2.7), there is a constant C > 0 such that

‖DuN‖∞ ≤ C

(∥∥∥D4uN

∥∥∥3/8

‖u‖5/8 + ‖u‖),

‖DuN‖∞ ≤ C

(∥∥∥D2uN

∥∥∥3/4

‖u‖1/4 + ‖u‖),

∥∥∥D3uN

∥∥∥∞≤ C

(∥∥∥D4uN

∥∥∥7/8

‖u‖1/8 + ‖u‖).

(2.29)

Noticing the definition of the function A and the estimates in (2.12), we have

‖A(uN)‖∞ ≤ C,∥∥A′(uN)

∥∥∞ ≤ C,

∥∥A′′(uN)∥∥∞ ≤ C, (2.30)


for some constant C = C(u0, T, γ1, γ2) > 0. Applying Holder inequality and Young inequality,for any ε > 0,

|I1| ≤ M0

(∥∥∥A′(uN)D2uN

∥∥∥ +∥∥∥A′′(uN)(DuN)2

∥∥∥)·∥∥∥D4uN

∥∥∥

≤ M0∥∥A′(uN)

∥∥∞ ·∥∥∥D2uN

∥∥∥∥∥∥D4uN

∥∥∥ +M0∥∥A′′(uN)

∥∥∞ · ‖DuN‖‖DuN‖∞

∥∥∥D4uN

∥∥∥

≤ ε∥∥∥D4uN

∥∥∥2+ C∥∥∥D2uN

∥∥∥2+ C

(∥∥∥D4uN

∥∥∥3/8

+ 1)∥∥∥D4uN

∥∥∥

≤ ε∥∥∥D4uN

∥∥∥2+ C∥∥∥D2uN

∥∥∥2+ C∥∥∥D4uN

∥∥∥11/8

+ C∥∥∥D4uN

∥∥∥

≤ ε∥∥∥D4uN

∥∥∥2+ C∥∥∥D2uN

∥∥∥2+ε

2

∥∥∥D4uN

∥∥∥2+ C +

ε

2

∥∥∥D4uN

∥∥∥2+ C

≤ 2ε∥∥∥D4uN

∥∥∥2+ C∥∥∥D2uN

∥∥∥2+ C,

(2.31)

where C = C(u0, m, T, γ1, γ2, ε) > 0 is a positive constant. Similarly, we obtain

|I2| ≤ M1‖DuN‖∥∥∥D3uN

∥∥∥∞

∥∥∥D4uN

∥∥∥

≤ C

(∥∥∥D4uN

∥∥∥7/8

+ 1)∥∥∥D4uN

∥∥∥

≤ C

(∥∥∥D4uN

∥∥∥15/8

+∥∥∥D4uN

∥∥∥)

≤ ε

2

∥∥∥D4uN

∥∥∥2+ C +

ε

2

∥∥∥D4uN

∥∥∥2+ C

≤ ε∥∥∥D4uN

∥∥∥2+ C,

|I3| ≤ M1‖DPNA(uN)‖∞ · ‖DuN‖ ·∥∥∥D4uN

∥∥∥

≤ C‖DPNA(uN)‖∞∥∥∥D4uN

∥∥∥

≤ C

(∥∥∥D2PNA(uN)∥∥∥3/4

‖PNA(uN)‖1/4 + ‖PNA(uN)‖)∥∥∥D4uN

∥∥∥

≤ C

(∥∥∥D2PNA(uN)∥∥∥3/4

‖A(uN)‖1/4 + ‖A(uN)‖)∥∥∥D4uN

∥∥∥

≤ C

(∥∥∥D2PNA(uN)∥∥∥3/4

‖A(uN)‖1/4∞ + ‖A(uN)‖∞)∥∥∥D4uN

∥∥∥

≤ C

(∥∥∥D2PNA(uN)∥∥∥3/4

+ 1)∥∥∥D4uN

∥∥∥


≤ C(∥∥∥D2PNA(uN)

∥∥∥ + 1)∥∥∥D4uN

∥∥∥

≤ C(∥∥∥D2A(uN)

∥∥∥ + 1)∥∥∥D4uN

∥∥∥

≤ C(∥∥∥A′(uN)D2uN

∥∥∥ +∥∥∥A′′(uN)(DuN)2

∥∥∥)·∥∥∥D4uN

∥∥∥

≤ ε∥∥∥D4uN

∥∥∥2+ C.

(2.32)

Hence,

12d

dt

∥∥∥D2uN

∥∥∥2+m0

∥∥∥D4uN

∥∥∥2 ≤ 4ε

∥∥∥D4uN

∥∥∥2+ C∥∥∥D2uN

∥∥∥2+ C. (2.33)

Taking ε = m0/8, we have

12d

dt

∥∥∥D2uN

∥∥∥2+m0

2

∥∥∥D4uN

∥∥∥2 ≤ C

∥∥∥D2uN

∥∥∥2+ C, (2.34)

where C = C(u0, m, T, γ1, γ2) > 0 is a positive constant. Therefore,

12d

dt

∥∥∥D2uN

∥∥∥2 ≤ C

∥∥∥D2uN

∥∥∥2+ C. (2.35)

From Gronwall inequality it follows that

∥∥∥D2uN

∥∥∥2 ≤ expCt

(∥∥∥D2u0

∥∥∥2+ 1)

≤ expCT

(∥∥∥D2u0

∥∥∥2+ 1)

≤ C, (2.36)

where C = C(u0, 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 = C(u0, m, T, γ1, γ2) > 0 is a positive constant.

Now, we study the error estimates between the exact solution u and the semidiscretespectral approximation solution uN . Set the following decomposition:

u − uN = η + e, η = u − PNu, e = PNu − uN. (2.38)

From the inequality (2.6) it follows that

∥∥η∥∥ ≤ CN−4. (2.39)

Hence, it remains to obtain the approximate bounds of e.


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), andm is smooth and satisfies (1.6), then there exists a constant C = C(u0, m, γ1, γ2) such that

‖e‖ ≤ C(N−2 + ‖e(0)‖

). (2.40)

Before we prove this theorem, we study some useful approximation properties.

Lemma 2.7. For any ε > 0, we have

−(D2η +D2e,D(m(uN)De)

)≤ −m0

∥∥∥D2e∥∥∥2+ 3ε∥∥∥D2e

∥∥∥2+ C(‖e‖2 +N−4

), (2.41)

where C = C(u0, m, T, γ1, γ2, ε) > 0 is a positive constant.

Proof. Direct computation gives

−(D2η +D2e,D(m(uN)De)

)

= −(D2e,m(uN)D2e

)−(D2η,m(uN)D2e

)

−(D2η,m′(uN)DuNDe

)−(D2e,m′(uN)DuNDe

)

≤ −m0

∥∥∥D2e∥∥∥2+M0

∥∥∥D2e∥∥∥∥∥∥D2η

∥∥∥2

+M1

∥∥∥D2η∥∥∥‖DuN‖∞‖De‖ +M1

∥∥∥D2e∥∥∥‖DuN‖∞‖De‖

≤ −m0

∥∥∥D2e∥∥∥2+ ε∥∥∥D2e

∥∥∥2+M2

0

4ε

∥∥∥D2η∥∥∥2

+M1‖DuN‖∞(‖De‖2 +

∥∥∥D2η∥∥∥2)+ ε∥∥∥D2e

∥∥∥2+M2

1‖DuN‖2∞4ε

‖De‖2

≤ −m0

∥∥∥D2e∥∥∥2+ 2ε∥∥∥D2e

∥∥∥2+

(M2

0

4ε+M1‖DuN‖∞

)∥∥∥D2η∥∥∥2

+

(M1‖DuN‖∞ +

M21‖DuN‖2∞

4ε

)‖De‖2

≤ −m0

∥∥∥D2e∥∥∥2+ 2ε∥∥∥D2e

∥∥∥2+

(M2

0

4ε+M1‖DuN‖∞

)∥∥∥D2η∥∥∥2


+

(M1‖DuN‖∞ +

M21‖DuN‖2∞

4ε

)∥∥∥D2e∥∥∥ · ‖e‖

≤ −m0

∥∥∥D2e∥∥∥2+ 2ε∥∥∥D2e

∥∥∥2+ C∥∥∥D2η

∥∥∥2+ ε∥∥∥D2e

∥∥∥2+ C‖e‖2

≤ −m0

∥∥∥D2e∥∥∥2+ 3ε∥∥∥D2e

∥∥∥2+ C

(‖e‖2 +

∥∥∥D2η∥∥∥2)

≤ −m0

∥∥∥D2e∥∥∥2+ 3ε∥∥∥D2e

∥∥∥2+ C(‖e‖2 +N−4

),

(2.42)


Lemma 2.8. For any ε > 0, we have

(A(u) − PNA(uN), D(m(uN)De)) ≤ 3εM20

∥∥∥D2e∥∥∥2+ C(‖e‖2 +N−4

), (2.43)


Proof. Noticing that

(A(u) − PNA(uN), D(m(uN)De))

= (A(u) − PNA(u), D(m(uN)De)) + (PNA(u) − PNA(uN), D(m(uN)De))

≤ ‖A(u) − PNA(u)‖ · ‖D(m(uN)De)‖ +∥∥∥PN

(A(u, uN)(u − uN)

)∥∥∥ · ‖D(m(uN)De)‖

≤ ‖A(u) − PNA(u)‖ · ‖D(m(uN)De)‖ +∥∥∥A(u, uN)

(η + e

)∥∥∥ · ‖D(m(uN)De)‖,(2.44)

where

A(s, τ) = γ2(s2 + sτ + τ2

)+ 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 followsthat

∥∥∥A(u, uN)∥∥∥∞≤ C. (2.46)


Then we obtain

‖A(u) − PNA(u)‖2 ≤ C

(∥∥∥u3 − PNu3∥∥∥2+∥∥∥u2 − PNu2

∥∥∥2+ ‖u − PNu‖2

)

≤ CN−8(∣∣∣u3∣∣∣4+∣∣∣u2∣∣∣4+ |u|4

)2 ≤ CN−8,

∥∥∥A(u, uN)(η + e

)∥∥∥2 ≤∥∥∥A(u, uN)

∥∥∥2

∞

(‖e‖2 + ∥∥η∥∥2

)≤ C(N−8 + ‖e‖2

),

‖D(m(uN)De)‖2 ≤ 2∥∥∥m(uN)D2e

∥∥∥2+ 2∥∥m′(uN)DuNDe

∥∥2

≤ 2M20

∥∥∥D2e∥∥∥2+ 2M2

1‖DuN‖2∞(∥∥∥D2e

∥∥∥2‖e‖2)

≤ 2M20

∥∥∥D2e∥∥∥2+M2

0

∥∥∥D2e∥∥∥2+M4

1‖DuN‖4∞M2

0

‖e‖2

≤ 3M20

∥∥∥D2e∥∥∥2+ C‖e‖2,

(2.47)

where C = C(u0, m, T, γ1, γ2) > 0 is a positive constant. By Cauchy inequality, for any ε > 0,we have

(A(u) − PNA(uN), D(m(uN)De))

≤ ‖A(u) − PNA(u)‖ · ‖D(m(uN)De)‖ +∥∥∥A(u, uN)

(η + e

)∥∥∥ · ‖D(m(uN)De)‖

≤ ε

2‖D(m(u)De)‖2 + 1

2ε‖A(u) − PNA(u)‖2 + ε

2‖D(m(u)De)‖2 + 1

2ε

∥∥∥A(u, uN)(η + e

)∥∥∥2

≤ 3εM20

∥∥∥D2e∥∥∥2+ C(‖e‖2 +N−8

),

(2.48)


Lemma 2.9. Assume that u is the solution of (1.2)–(1.4), there exists a positive constant C =C(u0, m, γ1, γ2) such that

(D2u −A(u), D(m(u) −m(uN))De

)≤ m0

2

∥∥∥D2e∥∥∥2+ C(‖e‖2 +N−4

), (2.49)

where C = C(u0, m, T, γ1, γ2) > 0 is a positive constant.

Proof. By Lemma 2.3, we have

∥∥∥D3u −DA(u)∥∥∥∞≤∥∥∥D3u

∥∥∥∞+∥∥A′(u)Du

∥∥∞ ≤ C. (2.50)


In the other hand, it follows that

‖m(u) −m(uN)‖2 ≤ M21‖u − uN‖2 ≤ M2

1

(‖e‖2 + ∥∥η∥∥2

). (2.51)

By the Young inequality,

(D2u −A(u), D(m(u) −m(uN)De)

)= −

(D3u −DA(u), (m(u) −m(uN))De

)

≤∥∥∥D3u −DA(u)

∥∥∥∞‖m(u) −m(uN)‖ · ‖De‖

≤ C‖m(u) −m(uN)‖‖De‖

≤ ε‖De‖2 + Cε

(‖e‖2 + ∥∥η∥∥2

)

≤ ε∥∥∥D2e

∥∥∥2+ Cε

(‖e‖2 + ∥∥η∥∥2

).

(2.52)

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

∂t, e

)+(D2u −A(u), D(m(u)De)

)= 0. (2.53)

Setting vN = e in (2.10), we get

(∂uN

∂t, e

)+(D2uN − PNA(uN), D(m(uN)De)

)= 0. (2.54)

(2.53)minus (2.54) gives

(∂e

∂t, e

)= −

(Du2 −A(u), D(m(u) −m(uN)De)

)−(D2u −D2uN,D(m(uN)De)

)

+ (A(u) − PNA(uN), D(m(uN)De)).

(2.55)

According to Lemmas 2.7, 2.8 and 2.9, we have

12d

dt‖e‖2 ≤ −m0

∥∥∥D2e∥∥∥2+ 3ε∥∥∥D2e

∥∥∥2+ C(‖e‖2 +N−4

)

+ 3εM20

∥∥∥D2e∥∥∥2+ C(N−8 + ‖e‖2

)

≤(−m0 +

(4 + 3M2

0

)ε)∥∥∥D2e

∥∥∥2+ C(N−4 + ‖e‖2

).

(2.56)


Set ε = m0/(4 + 3M20), then there exists a positive constant C = C(u0, m, γ1, γ2) such that

d

dt‖e‖2 ≤ C

(‖e‖2 +N−4

). (2.57)

By Gronwall inequality, we have

‖e‖ ≤ expCt(‖e0‖ +N−2

)≤ expCT

(‖e0‖ +N−2

)≤ C(‖e0‖ +N−2

), (2.58)

where C = C(u0, m, T, γ1, γ2) > 0 is a constant.

Summig up the properties above, we obtain the following.

Theorem 2.10. Assume m(s) 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 =C(u0, 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 geta full-discrete form. Furthermore, we investigate the boundedness of numerical solution andthe convergence of the numerical solutions of the full-discrete system. We also obtain theerror 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 andh = T/Λ is time-step size. Then the full-discretization spectral method for (1.2)–(1.4) reads:∀v ∈ SN , findU

j

N ∈ SN (j = 0, 1, 2, . . . ,Λ) such that

⎛⎝U

j+1N −U

j

N

h, v

⎞⎠ +

(D2U

j+1/2N − PNA

(U

j+1N ,U

j

N

), D

(m

(U

j+1/2N

)Dv

))= 0, (3.1)

(U0

N, v)− (u0, v) = 0, (3.2)

where Uj+1/2N = (Uj+1

N +Uj

N)/2 and

A(φ, ϕ)=

γ24

(φ3 + ϕ3 + φ2ϕ + φϕ2

)+γ13

(φ2 + φϕ + ϕ2

)− 12(φ + ϕ

). (3.3)

The solution Uj

N has the following property.


Lemma 3.1. Assume thatUj

N ∈ SN (j = 1, 2, . . . ,Λ) is the solution of (3.1)-(3.2). Then there existsa constant C = C(u0, m, γ1, γ2) > 0 such that

∥∥∥Uj

N

∥∥∥∞≤ C,

∥∥∥DUj

N

∥∥∥ ≤ C. (3.4)

Proof. Define a discrete energy function at time tj by

F(j)=

12

∥∥∥DUj

N

∥∥∥2+(H(U

j

N

), 1). (3.5)

Notice that

1h

(F(j + 1) − F

(j))

=

⎛⎝DU

j+1/2N ,D

Uj+1N −U

j

N

h

⎞⎠ +

⎛⎝PNA

(U

j+1N ,U

j

N

),U

j+1N −U

j

N

h

⎞⎠

= −⎛⎝D2U

j+1/2N − PNA

(U

j+1N ,U

j

N

),U

j+1N −U

j

N

h

⎞⎠.

(3.6)

Setting v = D2Uj+1/2N − PNA(Uj+1

N ,Uj

N) ∈ SN in (3.1), we obtain

1h

(F(j + 1) − F

(j))

= −(m

(U

j+1/2N

)(D3U

j+1/2N −DPNA

(U

j+1N ,U

j

N

)), D3U

j+1/2N −DPNA

(U

j+1N ,U

j

N

))

≤ −m0

∥∥∥∥D3Uj+1/2N −DPNA

(U

j+1N ,U

j

N

)∥∥∥∥2

≤ 0,

(3.7)

which implies

F(j) ≤ F(0) =

12

∥∥∥DU0N

∥∥∥2+(H(U0

N

), 1). (3.8)

By (2.18), we have

∫1

0H(U

j

N

)dx ≥ γ2

8

∫1

0

(U

j

N

)4dx −K1. (3.9)


Then

12

∥∥∥DUj

N

∥∥∥2+γ28

∫1

0

(U

j

N

)4dx −K1 ≤ 1

2

∥∥∥DUj

N

∥∥∥2+∫1

0H(U

j

N

)dx = F

(j)

≤ F(0) =12

∥∥∥DU0N

∥∥∥2+(H(U0

N

), 1).

(3.10)

So we obtain

∥∥∥DUj

N

∥∥∥2 ≤ 2K1 +

∥∥∥DU0N

∥∥∥2+ 2(H(U0

N

), 1)≡ C1,

∫1

0

(U

j

N

)4dx ≤ 8

γ2

(K1 +

12

∥∥∥DU0N

∥∥∥2+(H(U0

N

), 1))

≡ C2,

(3.11)

where C1 = C1(u0, m, γ1, γ2) and C2 = C2(u0, m, γ1, γ2) are positive constants. By Holderinequality, we get

∥∥∥Uj

N

∥∥∥2=∫1

0

(U

j

N

)2dx ≤

(∫1

0

(U

j

N

)4dx

)1/2

≤ C1/22 . (3.12)

Therefore,

∥∥∥Uj

N

∥∥∥ ≤ C1/42 . (3.13)

By the embedding theorem, we obtain

∥∥∥Uj

N

∥∥∥∞≤ C, (3.14)

where C = C(u0, m, γ1, γ2) > 0 is a constant.

Lemma 3.2. Assume thatUj

N ∈ SN (j = 1, 2, . . . ,Λ) is the solution of the full-discretization scheme(3.1)-(3.2), then there is a constant C = C(u0, m, T, γ1, γ2) > 0 such that

∥∥∥DUj

N

∥∥∥∞≤ C,

∥∥∥D2Uj

N

∥∥∥ ≤ C. (3.15)


Proof. Setting v = 2D4Uj+1/2N ∈ SN in (3.1), we have

⎛⎝U

j+1N −U

j

N

h, 2D4U

j+1/2N

⎞⎠

+(D

(m

(U

j+1/2N

)D3U

j+1/2N −DPNA

(U

j+1N ,U

j

N

))2D4U

j+1/2N

)

=

⎛⎝D2U

j+1N −D2U

j

N

h,D2U

j+1N +D2U

j

N

⎞⎠ +

(m

(U

j+1/2N

)D4U

j+1/2N , 2D4U

j+1/2N

)

+(m′(U

j+1/2N

)DU

j+1/2N D3U

j+1/2N , 2D4U

j+1/2N

)

−(m

(U

j+1/2N

)D2PNA

(U

j+1N +U

j

N

), 2D4U

j+1/2N

)

−(m′(U

j+1/2N

)DU

j+1/2N DPNA

(U

j+1N ,U

j

N

), 2D4U

j+1/2N

)= 0.

(3.16)

Therefore,

1h

(∥∥∥D2Uj+1N

∥∥∥2 −∥∥∥D2U

j

N

∥∥∥2)+ 2(m

(U

j+1/2N

)D4U

j+1/2N ,D4U

j+1/2N

)

= 2(m′(U

j+1/2N

)D3U

j+1/2N DU

j+1/2N ,D4U

j+1/2N

)

+ 2(m

(U

j+1/2N

)D2PNA

(U

j+1N ,U

j

N

), D4U

j+1/2N

)

+ 2(m′(U

j+1/2N

)DU

j+1/2N DPNA

(U

j+1N ,U

j+1N

), D4U

j+1/2N

)

= Ij

1 + Ij

2 + Ij

3.

(3.17)

By Nirenberg inequality (2.7), we have

∥∥∥DUj

N

∥∥∥∞≤ C

(∥∥∥D4Uj

N

∥∥∥3/8∥∥∥Uj

N

∥∥∥5/8

+∥∥∥Uj

N

∥∥∥),

∥∥∥DUj

N

∥∥∥∞≤ C

(∥∥∥D2Uj

N

∥∥∥3/4∥∥∥Uj

N

∥∥∥1/4

+∥∥∥Uj

N

∥∥∥),

∥∥∥D3Uj

N

∥∥∥∞≤ C

(∥∥∥D4Uj

N

∥∥∥7/8∥∥∥Uj

N

∥∥∥1/8

+∥∥∥Uj

N

∥∥∥).

(3.18)


According to (3.14), we obtain

∥∥∥A′1

(U

j+1N ,U

j

N

)∥∥∥∞≤ C,

∥∥∥A′2

(U

j+1N ,U

j

N

)∥∥∥∞≤ C,

∥∥∥A′′11

(U

j+1N ,U

j

N

)∥∥∥∞≤ C,

∥∥∥A′′12

(U

j+1N ,U

j

N

)∥∥∥∞≤ C,

∥∥∥A′′22

(U

j+1N ,U

j

N

)∥∥∥∞≤ C,

(3.19)

whereC = C(u0, m, γ1, γ2) is a positive constant. By Young inequality, for any positive constantε > 0, it follows that

∣∣∣Ij1∣∣∣ ≤ 2M1

∥∥∥∥D3Uj+1/2N

∥∥∥∥∞

∥∥∥∥DUj+1/2N

∥∥∥∥∥∥∥∥D4U

j+1/2N

∥∥∥∥

≤ C

(∥∥∥∥D4Uj+1/2N

∥∥∥∥7/8

+ 1

)∥∥∥∥D4Uj+1/2N

∥∥∥∥

≤ C

∥∥∥∥D4Uj+1/2N

∥∥∥∥15/8

+ C

∥∥∥∥D4Uj+1/2N

∥∥∥∥

≤ ε

2

∥∥∥∥D4Uj+1/2N

∥∥∥∥2

+ Cε +ε

2

∥∥∥∥D4Uj+1/2N

∥∥∥∥2

+ Cε

≤ ε

∥∥∥∥D4Uj+1/2N

∥∥∥∥2

+ Cε,

∣∣∣Ij2∣∣∣ ≤ 2M0

∥∥∥D2A(U

j+1N ,U

j

N

)∥∥∥∥∥∥∥D4U

j+1/2N

∥∥∥∥

≤ 2M0

∥∥∥∥A′′11

(DU

j+1N

)2+ 2A′′

12DUj+1N DU

j

N +A′′22

(DU

j

N

)2

+A′1D

2Uj+1N +A′

2D2U

j

N

∥∥∥∥∥∥∥∥D4U

j+1/2N

∥∥∥∥

≤ C

(∥∥∥D2Uj+1N

∥∥∥ +∥∥∥D2U

j

N

∥∥∥ +∥∥∥∥(DU

j+1N

)2∥∥∥∥ +∥∥∥∥(DU

j

N

)2∥∥∥∥)∥∥∥∥D4U

j+1/2N

∥∥∥∥

≤ C(∥∥∥D2U

j+1N

∥∥∥ +∥∥∥D2U

j

N

∥∥∥ +∥∥∥DU

j+1N

∥∥∥∥∥∥DU

j+1N

∥∥∥∞+∥∥∥DU

j

N

∥∥∥∥∥∥DU

j

N

∥∥∥∞

)∥∥∥∥D4Uj+1/2N

∥∥∥∥

≤ C(∥∥∥D2U

j+1N

∥∥∥ +∥∥∥D2U

j

N

∥∥∥ +∥∥∥DU

j+1N

∥∥∥∞+∥∥∥DU

j

N

∥∥∥∞

)∥∥∥∥D4Uj+1/2N

∥∥∥∥

≤ C

(∥∥∥D2Uj+1N

∥∥∥ +∥∥∥D2U

j

N

∥∥∥ +∥∥∥D2U

j+1N

∥∥∥3/4

+∥∥∥D2U

j

N

∥∥∥3/4

+ 1)∥∥∥∥D4U

j+1/2N

∥∥∥∥

≤ C(∥∥∥D2U

j+1N

∥∥∥ +∥∥∥D2U

j

N

∥∥∥ + 1)∥∥∥∥D4U

j+1/2N

∥∥∥∥

≤ ε

∥∥∥∥D4Uj+1/2N

∥∥∥∥2

+ Cε

(∥∥∥D2Uj+1N

∥∥∥2+∥∥∥D2U

j

N

∥∥∥2+ 1),


∣∣∣Ij3∣∣∣ ≤ M1

∥∥∥∥DUj+1/2N

∥∥∥∥∥∥∥DPNA

(U

j+1N ,U

j

N

)∥∥∥∞

∥∥∥∥D4Uj+1/2N

∥∥∥∥

≤ C

(∥∥∥D2PNA(U

j+1N ,U

j

N

)∥∥∥3/4

+ 1)∥∥∥∥D4U

j+1/2N

∥∥∥∥

≤ C(∥∥∥D2PNA

(U

j+1N ,U

j

N

)∥∥∥ + 1)∥∥∥∥D4U

j+1/2N

∥∥∥∥

≤ C

(∥∥∥D2A(U

j+1N ,U

j

N

)∥∥∥∥∥∥∥D4U

j+1/2N

∥∥∥∥ +∥∥∥∥D4U

j+1/2N

∥∥∥∥)

≤ ε

∥∥∥∥D4Uj+1/2N

∥∥∥∥2

+ Cε

(∥∥∥D2Uj+1N

∥∥∥2+∥∥∥D2U

j

N

∥∥∥2+ 1)+ ε

∥∥∥∥D4Uj+1/2N

∥∥∥∥2

+ Cε

≤ 2ε∥∥∥∥D4U

j+1/2N

∥∥∥∥2

+ Cε

(∥∥∥D2Uj+1N

∥∥∥2+∥∥∥D2U

j

N

∥∥∥2+ 1),

(3.20)

where Cε = Cε(u0, m, γ1, γ2) > 0 is a constant. Therefore,

1h

(∥∥∥D2Uj+1N

∥∥∥2 −∥∥∥D2U

j

N

∥∥∥2)+ 2m0

∥∥∥∥D4Uj+1/2N

∥∥∥∥2

≤ 1h

(∥∥∥D2Uj+1N

∥∥∥2 −∥∥∥D2U

j

N

∥∥∥2)+ 2(m

(U

j+1/2N

)D4U

j+1/2N ,D4U

j+1/2N

)

≤ 4ε∥∥∥∥D2U

j+1/2N

∥∥∥∥2

+ Cε

(∥∥∥D2Uj+1N

∥∥∥2+∥∥∥D2U

j

N

∥∥∥2+ 1).

(3.21)

Setting ε = m0/2, there is a positive constant C = C(u0, m, γ1, γ2) such that

∥∥∥D2Uj+1N

∥∥∥2 ≤(1 +

2Ch1 − Ch

)∥∥∥D2Uj

N

∥∥∥2+

C

1 − Chh. (3.22)

Denoted by C = C/(1 − Ch), if h is sufficiently small such that Ch ≤ 1/2, we have

∥∥∥D2Uj+1N

∥∥∥2 ≤(1 + Ch

)∥∥∥D2Uj

N

∥∥∥2+C

2h

≤(1 + Ch

)j+1∥∥∥D2U0N

∥∥∥2+C

2h

j∑i=1

(1 + Ch

)i

≤ exp(j+1)Ch(∥∥∥D2U0

N

∥∥∥2+ Cjh

)

≤ expCT

(∥∥∥D2U0N

∥∥∥2+ CT

)≡ C′,

(3.23)


where C′ = C′(u0, m, T, γ1, γ2) > 0 is a positive constant. By the embedding theorem, theestimate (3.15) holds.

Next, we investigate the error estimates for the numerical solution Uj

N to the exactsolution u(tj). Our analysis is based on the error decomposition denoted by

u(tj) −U

j

N = ηj + ej , ηj = u(tj) − PNu

(tj), ej = PNu

(tj) −U

j

N. (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 = C(u0, m, γ1, γ2) such that

∥∥∥ηj∥∥∥ ≤ CN−4. (3.25)

Hence, it remains to obtain the approximate bounds of ej . If no confusion occurs, we denotethe average of the two instant errors ej and ej+1 by ej+1/2:

ej+1/2 =ej + ej+1

2, ηj+1/2 =

ηj + ηj+1

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

∥∥∥ej+1∥∥∥2 ≤∥∥∥ej∥∥∥2+ 2h

⎛⎝ut

(tj+1/2

) − Uj+1N −U

j

N

h, ej+1/2

⎞⎠ +

1320

h4∫ tj+1

tj

‖uttt‖2dt + h∥∥∥ej+1/2

∥∥∥2.

(3.27)

Proof. Applying Taylor expansion about tj+1/2, we have

uj = uj+1/2 − h

2uj+1/2t +

h2

8uj+1/2tt − 1

2

∫ tj+1/2

tj

(t − tj

)2utttdt,

uj+1 = uj+1/2 +h

2uj+1/2t +

h2

8uj+1/2tt +

12

∫ tj+1

tj+1/2

(tj+1 − t

)2utttdt.

(3.28)

Then

uj+1/2t − uj+1 − uj

h= − 1

2h

(∫ tj+1

tj+1/2

(tj+1 − t

)2utttdt +

∫ tj+1/2

tj

(t − tj

)2utttdt

). (3.29)


From Holder inequality it follows that

∥∥∥∥∥uj+1/2t − uj+1 − uj

h

∥∥∥∥∥2

≤ 12h2

⎛⎝∥∥∥∥∥∫ tj+1

tj+1/2

(tj+1 − t

)2utttdt

∥∥∥∥∥2

+

∥∥∥∥∥∫ tj+1/2

tj

(t − tj

)2utttdt

∥∥∥∥∥2⎞⎠

≤ h3

320

∫ tj+1

tj

‖uttt‖2dt.

(3.30)

Noticing that for any v ∈ SN , we have

⎛⎝u

j+1/2t − U

j+1N −U

j

N

h, v

⎞⎠ =

(uj+1/2t − uj+1 − uj

h, v

)+1h

(ej+1 − ej , v

). (3.31)

Taking v = 2ej+1/2 in (3.31), we obtain

∥∥∥ej+1∥∥∥2=∥∥∥ej∥∥∥2+ 2h

⎛⎝u

j+1/2t − U

j+1N −U

j

N

h, ej+1/2

⎞⎠ − 2h

(uj+1/2t − uj+1 − uj

Δt, ej+1/2

)

≤∥∥∥ej∥∥∥2+ 2h

⎛⎝u

j+1/2t − U

j+1N −U

j

N

h, ej+1/2

⎞⎠

+1

320h4∫ tj+1

tj


∥∥∥2.

(3.32)

Taking v = ej+1/2 in (2.2) and (3.1), respectively, we have

(∂u(tj+1/2

)

∂t, ej+1/2

)+(D2uj+1/2 −A

(uj+1/2

), D(m(uj+1/2

)Dej+1/2

))= 0, (3.33)

⎛⎝U

j+1N −U

j

N

h, ej+1/2

⎞⎠ +

(D2U

j+1/2N − PNA

(U

j+1N ,U

j

N

), D

(m

(U

j+1/2N

)Dej+1/2

))= 0.

(3.34)


Comparing (3.33) and (3.34), we have

⎛⎝u

j+1/2t − U

j+1N −U

j

N

h, ej+1/2

⎞⎠

= −⎛⎝D2uj+1/2 − D2U

j+1N +D2U

j

N

2, D

(m

(U

j+1/2N

)Dej+1/2

)⎞⎠

+(A(uj+1/2

)− PNA

(U

j+1N ,U

j

N

), D

(m

(U

j+1/2N

)Dej+1/2

))

+(D2uj+1/2 −A

(uj+1/2

), D

(m

(U

j+1/2N

)−m(uj+1/2

))Dej+1/2

).

(3.35)

Now we investigate the error estimates of the three items in the right-hand side of theprevious equation.

Lemma 3.4. Assume that u is the solution of (1.2)–(1.4) such thatDutt ∈ L2(QT ), then there existsa positive constant C1 = C1(m,u0, T, γ1, γ2) > 0 such that

−⎛⎝D2uj+1/2 − D2U

j+1N +D2U

j

N

2, D

(m

(U

j+1/2N

)Dej+1/2

)⎞⎠

≤ −m0

2

∥∥∥D2ej+1/2∥∥∥2+ C1

(N−4 +

∥∥∥ej∥∥∥2+∥∥∥ej+1

∥∥∥2+ h3

∫ tj+1

tj

∥∥∥D2utt

∥∥∥2dt

).

(3.36)

Proof. By Taylor expansion and Holder inequality, we obtain

uj = uj+1/2 − h

2uj+1/2t +

∫ tj+1/2

tj

(t − tj

)uttdt,

uj+1 = uj+1/2 +h

2uj+1/2t +

∫ tj+1

tj+1/2

(tj+1 − t

)uttdt.

(3.37)

Therefore,

12

(uj + uj+1

)− uj+1/2 =

12

(∫ tj+1/2

tj

(t − tj

)uttdt +

∫ tj+1

tj+1/2

(tj+1 − t

)uttdt

). (3.38)


By Holder inequality, we have

∥∥∥∥D2(uj+1/2 − 1

2

(uj + uj+1

))∥∥∥∥2

=14

⎛⎝∥∥∥∥∥D

2

(∫ tj+1/2

tj

(t − tj

)uttdt +

∫ tj+1

tj+1/2

(tj+1 − t

)uttdt

)∥∥∥∥∥2⎞⎠

≤ 14

⎛⎝∥∥∥∥∥

(∫ tj+1/2

tj

(t − tj

)2dt

∫ tj+1/2

tj

(D2utt

)2dt +

∫ tj+1

tj+1/2

(tj+1 − t

)2dt

∫ tj+1

tj+1/2

(D2utt

)2dt

)∥∥∥∥∥2⎞⎠

=14

⎛⎝∥∥∥∥∥

(h3

24

∫ tj+1/2

tj

(D2utt

)2dt +

h3

24

∫ tj+1

tj+1/2

(D2utt

)2dt

)∥∥∥∥∥2⎞⎠

≤ 14· h

3

24

∫ tj+1

tj

∥∥∥D2utt

∥∥∥2dt.

(3.39)

Direct computation gives

∥∥∥∥D2(uj+1/2 − 1

2

(uj + uj+1

))∥∥∥∥2

≤ h3

96

∫ tj+1

tj

∥∥∥D2utt

∥∥∥2dt. (3.40)

Then

−⎛⎝D2

⎛⎝uj+1/2 − U

j+1N +U

j

N

2

⎞⎠, D

(m

(U

j+1/2N

)Dej+1/2

)⎞⎠

= −(D2

(uj+1/2 − uj+1 + uj

2

), D

(m

(U

j+1/2N

)Dej+1/2

))

−⎛⎝D2

⎛⎝uj+1 + uj

2− U

j+1N +U

j

N

2

⎞⎠, D

(m

(U

j+1/2N

)Dej+1/2

)⎞⎠

≤∥∥∥∥∥D

2

(uj+1/2 − uj+1 + uj

2

)∥∥∥∥∥ ·∥∥∥∥D(m

(U

j+1/2N

)Dej+1/2

)∥∥∥∥

−(D2(ηj+1/2 + ej+1/2

), D

(m

(U

j+1/2N

)Dej+1/2

))


≤{h3

96

∫ tj+1

tj

∥∥∥D2utt

∥∥∥2dt

}1/2

·(∥∥∥∥m

(U

j+1/2N

)D2ej+1/2

∥∥∥∥ +∥∥∥∥m′(U

j+1/2N

)DU

j+1/2N Dej+1/2

∥∥∥∥)

−(D2ηj+1/2, m

(U

j+1/2N

)D2ej+1/2

)−(D2ηj+1/2, m′

(U

j+1/2N

)DU

j+1/2N Dej+1/2

)

−(D2ej+1/2, m

(U

j+1/2N

)D2ej+1/2

)−(D2ej+1/2, m′

(U

j+1/2N

)DU

j+1/2N Dej+1/2

)

� σj

1 + σj

2 + σj

3.

(3.41)

By Cauchy inequality, it follows that

∣∣∣σj

1

∣∣∣ ≤{h3

96

∫ tj+1

tj

∥∥∥D2utt

∥∥∥2dt

}1/2

·∥∥∥∥m(U

j+1/2N

)D2ej+1/2

∥∥∥∥

+

{h3

96

∫ tj+1

tj

∥∥∥D2utt

∥∥∥2dt

}1/2∥∥∥∥m′(U

j+1/2N

)DU

j+1/2N Dej+1/2

∥∥∥∥

≤ M0

{h3

96

∫ tj+1

tj

∥∥∥D2utt

∥∥∥2dt

}1/2∥∥∥D2ej+1/2∥∥∥

+M1

∥∥∥∥DUj+1/2N

∥∥∥∥∞

{h3

96

∫ tj+1

tj

∥∥∥D2utt

∥∥∥2dt

}1/2∥∥∥Dej+1/2∥∥∥

≤ h3C1ε

∫ tj+1

tj

∥∥∥D2utt

∥∥∥2dt + ε

∥∥∥D2ej+1/2∥∥∥2+ ε∥∥∥Dej+1/2

∥∥∥2

≤ h3C1ε

∫ tj+1

tj

∥∥∥D2utt

∥∥∥2dt + ε

∥∥∥D2ej+1/2∥∥∥2+ ε∥∥∥D2ej+1/2

∥∥∥∥∥∥ej+1/2

∥∥∥

≤ h3C1ε

∫ tj+1

tj

∥∥∥D2utt

∥∥∥2dt + 2ε

∥∥∥D2ej+1/2∥∥∥2+ ε∥∥∥ej+1/2

∥∥∥2,

∣∣∣σj

2

∣∣∣ ≤ M0

∥∥∥D2ηj+1/2∥∥∥∥∥∥D2ej+1/2

∥∥∥ +M1

∥∥∥DUj+1/2N

∥∥∥∞

∥∥∥D2ηj+1/2∥∥∥∥∥∥Dej+1/2

∥∥∥

≤ M20

2ε

∥∥∥D2ηj+1/2∥∥∥2+ε

2

∥∥∥D2ej+1/2∥∥∥2+M2

1

∥∥∥∥DUj+1/2N

∥∥∥∥2

∞2ε

ε

2

∥∥∥Dej+1/2∥∥∥2

≤ C2ε‖D2ηj+1/2‖2 + ε

2

∥∥∥D2ej+1/2∥∥∥2+ε

2

∥∥∥D2ej+1/2∥∥∥∥∥∥ej+1/2

∥∥∥

≤ C2ε‖D2ηj+1/2‖2 + ε‖D2ej+1/2‖2 + ε

2‖ej+1/2‖2,


σj

3 ≤ −m0

∥∥∥D2ej+1/2∥∥∥2+M1

∥∥∥∥DUj+1/2N

∥∥∥∥∞

∥∥∥D2ej+1/2∥∥∥ ·∥∥∥Dej+1/2

∥∥∥

≤ −m0

∥∥∥D2ej+1/2∥∥∥2+ε

2

∥∥∥D2ej+1/2∥∥∥2+M2

1

∥∥∥∥DUj+1/2N

∥∥∥∥2

∞2ε

∥∥∥Dej+1/2∥∥∥2

≤ −m0

∥∥∥D2ej+1/2∥∥∥2+ ε∥∥∥D2ej+1/2

∥∥∥2+ C3ε

∥∥∥ej+1/2∥∥∥2.

(3.42)

Then we obtain

−⎛⎝D2uj+1/2 − D2U

j+1N +D2U

j

N

2, D

(m

(U

j+1/2N

)Dej+1/2

)⎞⎠

≤∣∣∣σj

1

∣∣∣ +∣∣∣σj

2

∣∣∣ + σj

3 ≤ (4ε −m0)∥∥∥D2ej+1/2

∥∥∥2+ h3C1ε

∫ tj+1

tj

∥∥∥D2utt

∥∥∥2dt

+ C2ε

∥∥∥D2ηj+1/2∥∥∥2+ C3ε

∥∥∥ej+1/2∥∥∥2,

(3.43)

whereC1ε, C2ε, andC3ε are positive constants. Choosing ε = m0/8, and terms of the propertiesof 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

(A(uj+1/2

)− PNA

(U

j+1N ,U

j

N

), D

(m

(U

j+1/2N

)Dej+1/2

))

≤ m0

4

∥∥∥D2ej+1/2∥∥∥2+ C2

{∥∥∥ej+1∥∥∥2+∥∥∥ej∥∥∥2+N−8

+h3

(∫ tj+1

tj

‖utt‖2dt +∥∥∥uj+1/2

t

∥∥∥2

∞

∫ tj+1

tj

‖ut‖2dt)}

.

(3.44)

Proof. Firstly, we consider

A(uj+1/2

)− A(uj+1, uj

)= γ2(uj+1/2

)3 − γ24

[(uj+1)3

+(uj+1)2uj + uj+1

(uj)2

+(uj)3]

+ γ1(uj+1/2

)2 − γ13

[(uj+1)2

+ uj+1uj +(uj)2]

−(uj+1/2 − 1

2

(uj+1 + uj

))� γ2ρ

j

1 + γ1ρj

2 − ρj

3.

(3.45)



∥∥∥ρj3∥∥∥ =

12

∥∥∥∥∥∫ tj+1/2

tj

(t − tj

)uttdt +

∫ tj+1

tj+1/2

(tj+1 − t

)uttdt

∥∥∥∥∥ ≤(

Δt3

96

∫ tj+1

tj

‖utt‖2dt)1/2

,

∥∥∥ρj2∥∥∥ =∥∥∥∥16

[(2uj+1/2

)2 −(uj+1 + uj

)2]+16

[2(uj+1/2

)2 −(uj+1)2 −

(uj)2]∥∥∥∥

≤ 16

∥∥∥[2uj+1/2 −

(uj+1 + uj

)][2uj+1/2 +

(uj+1 + uj

)]∥∥∥

+16

∥∥∥(uj+1/2 − uj+1

)(uj+1/2 + uj+1

)+(uj+1/2 − uj

)(uj+1/2 + uj

)∥∥∥

≤ 16

∥∥∥2uj+1/2 −(uj+1 + uj

)∥∥∥ ·∥∥∥2uj+1/2 +

(uj+1 + uj

)∥∥∥∞

+16

∥∥∥∥∥

(−Δt

2uj+1/2t −

∫ tj+1

tj+1/2

(tj+1 − t

)uttdt

)(uj+1/2 + uj+1

)

+

(Δt

2uj+1/2t −

∫ tj+1/2

tj

(t − tj

)uttdt

)(uj+1/2 + uj

)∥∥∥∥∥

≤ CΔt3/2(∫ tj+1

tj

‖utt‖2dt)1/2

+Δt3/2

12

∥∥∥uj+1/2t

∥∥∥∞·(∫ tj+1

tj

‖ut‖2dt)1/2

,

∥∥∥ρj1∥∥∥ =

112

∥∥∥∥((

2uj+1/2)3 −

(uj+1 + uj

)3)+16

[2(uj+1/2

)3 −(uj+1)3 −

(uj)3]∥∥∥∥

≤ 112

∥∥∥∥[2uj+1/2 −

(uj+1 + uj

)]·[4(uj+1/2

)2+ 2(uj+1 + uj

)uj+1/2 +

(uj+1 + uj

)2]∥∥∥∥

+16

∥∥∥∥(uj+1/2 − uj+1

)((uj+1/2

)2+ uj+1uj+1/2 +

(uj+1)2)

+(uj+1/2 − uj

)((uj+1/2

)2+ ujuj+1/2 +

(uj)2)∥∥∥∥

≤ CΔt3/2

⎧⎨⎩

(∫ tj+1

tj

‖utt‖2dt)1/2

+∥∥∥uj+1/2

t

∥∥∥∞·(∫ tj+1

tj

‖ut‖2dt)1/2

⎫⎬⎭.

(3.46)

Then

(A(uj+1/2

)− PNA

(U

j+1N ,U

j

N

), D

(m

(U

j+1/2N

)Dej+1/2

))

≤ 34ε

∥∥∥A(uj+1/2

)− PNA

(uj+1/2

)∥∥∥2+ε

3

∥∥∥∥D(m

(U

j+1/2N

)Dej+1/2

)∥∥∥∥2


+34ε

∥∥∥A(uj+1/2

)− A(uj+1, uj

)∥∥∥2+ε

3

∥∥∥∥D(m

(U

j+1/2N

)Dej+1/2

)∥∥∥∥2

+34ε

∥∥∥A(uj+1, uj

)− A(U

j+1N ,U

j

N

)∥∥∥2+ε

3

∥∥∥∥D(m

(U

j+1/2N

)Dej+1/2

)∥∥∥∥2

≤ ε

∥∥∥∥D(m

(U

j+1/2N

)Dej+1/2

)∥∥∥∥2

+34ε

∥∥∥A(uj+1/2

)− PNA

(uj+1/2

)∥∥∥2

+34ε

∥∥∥A(uj+1/2

)− A(uj+1, uj

)∥∥∥2+

34ε

∥∥∥A(uj+1, uj

)− A(U

j+1N ,U

j

N

)∥∥∥2.

(3.47)

Direct computation yields

∥∥∥∥D(m

(U

j+1/2N

)Dej+1/2

)∥∥∥∥2

≤(M2

0 +M21

)∥∥∥D2ej+1/2∥∥∥2+ C∥∥∥ej+1/2

∥∥∥2,

∥∥∥A(uj+1/2

)− PNA

(uj+1/2

)∥∥∥2 ≤ CN−8,

∥∥∥A(uj+1/2

)− A(uj+1, uj

)∥∥∥2 ≤ C

(∥∥∥ρj1∥∥∥2+∥∥∥ρj2∥∥∥2+∥∥∥ρj3∥∥∥2),

∥∥∥A(uj+1, uj

)− A(U

j+1N ,U

j

N

)∥∥∥2 ≤(∥∥∥Gj

1

∥∥∥2

∞+∥∥∥Gj

2

∥∥∥2

∞

)(∥∥∥ej+1∥∥∥2+∥∥∥ej∥∥∥2+ CN−8

),

(3.48)

where

Gj

1 =γ28

((uj+1 +U

j+1N

)2+(U

j+1N +U

j

N

)2+(uj+1 +U

j

N

)2)+γ13

(U

j+1N + uj+1 +U

j

N

)− 12,

Gj

2 =γ28

((uj +U

j

N

)2+(uj + uj+1

)2+(uj+1 +U

j

N

)2)+γ13

(uj+1 + uj +U

j

N

)− 12.

(3.49)

Applying Lemma 2.3 and Theorem 3.7, we obtain that ‖G1‖∞ ≤ C(u0, m, T, γ1, γ2) and ‖G2‖∞ ≤C(u0, m, T, γ1, γ2). Taking ε = m0/4(M2

0 +M21) in (3.47), we have

(A(uj+1/2

)− PNA

(U

j+1N ,U

j

N

), D(mj+1/2Dej+1/2

))

≤ m0

4

∥∥∥D2ej+1/2∥∥∥2+ C2

{∥∥∥ej+1∥∥∥2+∥∥∥ej∥∥∥2+N−8

+Δt3(∫ tj+1

tj

‖utt‖2dt +∥∥∥uj+1/2

t

∥∥∥2

∞

∫ tj+1

tj

‖ut‖2dt)}

,

(3.50)

where C2 = C2(u0, m, T, γ1, γ2) > 0 is a constant.


Lemma 3.6. Assume that u is the solution of (1.2)–(1.4). Then there exists a positive constant C =C(u0, m, γ1, γ2) > 0 such that

(D2uj+1/2 −A

(uj+1/2

), D

(m

(U

j+1/2N

)−m(uj+1/2

))Dej+1/2

)

≤ m0

4

∥∥∥D2ej+1/2∥∥∥2+ C

(∥∥∥ej+1/2∥∥∥2+N−8 + h3

∫ tj+1

tj

‖utt‖dt).

(3.51)

Proof. By (2.9), we have

∥∥∥D3uj+1/2 −DA(uj+1/2

)∥∥∥∞≤∥∥∥D3uj+1/2

∥∥∥∞+∥∥∥A′(uj+1/2

)Duj+1/2

∥∥∥∞

≤∥∥∥D3uj+1/2

∥∥∥∞+∥∥∥A′(uj+1/2

)∥∥∥∞

∥∥∥Duj+1/2∥∥∥∞≤ C.

(3.52)

In the other hand,

∥∥∥∥m(U

j+1/2N

)−m(uj+1/2

)∥∥∥∥2

≤ M21

∥∥∥∥(U

j+1/2N − uj+1/2

)∥∥∥∥2

≤ C

∥∥∥∥∥Uj+1/2N − uj + uj+1

2

∥∥∥∥∥2

+

∥∥∥∥∥uj + uj+1

2− uj+1/2

∥∥∥∥∥2

≤ C∥∥∥ej+1/2

∥∥∥2+∥∥∥ηj+1/2

∥∥∥2+h3

96

∫ tj+1

tj

‖utt‖dt.

(3.53)

By Young inequality, we obtain

(D2uj+1/2 −A

(uj+1/2

), D

(m

(U

j+1/2N

)−m(uj+1/2

))Dej+1/2

)

= −(D3uj+1/2 −DA

(uj+1/2

), m

(U

j+1/2N

)−m(uj+1/2

)Dej+1/2

)

≤∥∥∥D3uj+1/2 −DA

(uj+1/2

)∥∥∥∞

∥∥∥∥m(U

j+1/2N

)−m(uj+1/2

)∥∥∥∥∥∥∥Dej+1/2

∥∥∥

≤ ε∥∥∥Dej+1/2

∥∥∥2+ Cε

∥∥∥∥m(U

j+1/2N

)−m(uj+1/2

)∥∥∥∥2

≤ ε∥∥∥D2ej+1/2

∥∥∥∥∥∥ej+1/2

∥∥∥ + Cε

∥∥∥∥m(U

j+1/2N

)−m(uj+1/2

)∥∥∥∥2

≤ ε∥∥∥D2ej+1/2

∥∥∥2+ Cε

(∥∥∥ej+1/2∥∥∥2+N−8 + h3

∫ tj+1

tj

‖utt‖dt).

(3.54)

Choosing ε = m0/4 in the previous inequality leads to (3.51).

Finally, we obtain the main theorem of this paper.


Theorem 3.7. Assume that u(x, t) is the solution of (1.2)–(1.4) and satisfies that

D4u ∈ L∞(QT ), ut ∈ L∞(QT ),

D2utt ∈ L2(QT ), uttt ∈ L2(QT ).(3.55)

Uj

N ∈ SN (j = 1, 2, . . . ,Λ) is the solution of the full-discretization (3.1) and (3.2). If h is sufficientlysmall, there exists a positive constant C such that, for any j = 0, 1, 2, . . . ,Λ,

∥∥∥ej+1∥∥∥ =∥∥∥PNu

(tj+1) −U

j+1N

∥∥∥ ≤ C(N−2 +

∥∥∥e0∥∥∥ + Bh2

), (3.56)

where B =∫ tj+10 (‖D2utt‖2 + ‖utt‖2 + ‖uttt‖2 +max0≤l≤j {‖ul+1/2

t ‖2∞} · ‖ut‖2)dt.

Proof. By (3.27), (3.36), (3.44), and (3.51), we have

∥∥∥ej+1∥∥∥2 ≤∥∥∥ej∥∥∥2+

h4

320

∫ tj+1

tj


∥∥∥2

+ 2h

⎧⎨⎩−⎛⎝D2uj+1/2 − D2U

j+1N +D2U

j

N

2, D

(m

(U

j+1/2N

)Dej+1/2

)⎞⎠

+(A(uj+1/2

)− A(U

j+1N ,U

j

N

), D

(m

(U

j+1/2N

)Dej+1/2

))

+(D2uj+1/2 −A

(uj+1/2

), D

(m

(U

j+1/2N

)−m(uj+1/2

))Dej+1/2

)⎫⎬⎭

≤∥∥∥ej∥∥∥2+ hC1

(N−4 +

∥∥∥ej+1∥∥∥2+∥∥∥ej∥∥∥2)

+ C2h4∫ tj+1

tj

(∥∥∥D2utt

∥∥∥2+ ‖utt‖2 + ‖uttt‖2 +

∥∥∥uj+1/2t

∥∥∥2

∞‖ut‖2

)dt,

(3.57)

where C1 = C1(m,u0, T, γ1, γ2) > 0 and C2 = C2(m,u0, T, γ1, γ2) > 0 are constants. Forsufficiently small h such that C1h ≤ 1/2, denoting C = 2(C1 + C2), we obtain

∥∥∥ej+1∥∥∥2 ≤(1 + Ch

)∥∥∥ej∥∥∥2+ C(hN−4 + h4Bj

), (3.58)

where

Bj =∫ tj+1

tj

(∥∥∥D2utt

∥∥∥2+ ‖utt‖2 + ‖uttt‖2 +

∥∥∥uj+1/2t

∥∥∥2

∞‖ut‖2

)dt. (3.59)


By the Gronwall inequality of the discrete form, we obtain

∥∥∥ej+1∥∥∥2 ≤(1 + Ch

)j+1∥∥∥e0∥∥∥2+ C

j∑l=0

(1 + Ch

)l(htN−4 + h4Bl

)

≤ exp{C(j + 1)h}{∥∥∥e0

∥∥∥2+ C

j∑l=0

(hN−4 + h4Bl

)}

≤ exp{C(j + 1)h}{∥∥∥e0

∥∥∥2+ C

(jhN−4 + h4

j∑l=0

Bl

)}.

(3.60)


j∑l=0

Bl ≤∫ tj+1

0

(∥∥∥D2utt

∥∥∥2+ ‖utt‖2 + ‖uttt‖2 +max

0≤l≤j

{∥∥∥ul+1/2t

∥∥∥2

∞

}· ‖ut‖2

)dt. (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

D4u ∈ L∞(QT ), ut ∈ L∞(QT ),

D2utt ∈ L2(QT ), uttt ∈ L2(QT ).(3.62)

Uj

N ∈ SN (j = 1, 2, . . . ,Λ) is the solution of the full-discretization scheme (3.1)-(3.2), and U0

satisfies ‖e0‖ = ‖PNu0 − U0‖ ≤ CN−4. If h is sufficiently small, there exists a constant C =C(u0, m, T, γ1, γ2) > 0 such that

∥∥∥u(x, tj) −U

j

N

∥∥∥ ≤ 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 outnumerical computations to illustrate theoretical estimations in the previous section. Consider(2.2) with settings:

m(s) = m0 + s2, A(s) = s3 − s, (4.1)


0

1

2

3

4

5

6×10−4

00.5

1

00.05

0.1

(a)

−20

2

4

6

8

10×10−4

00.5

10

0.050.1

(b)

Figure 1: The development of the solution of the full-discrete scheme when initial value is u0(x) =x5(1 − x)5 (a) and u0(x) = x5(1 − x)5ex (b).

where m0 > 0 is a constant. The full-discretization spectral method of (2.2) and (2.3) reads:findU

j

N =∑N

l=0 αjl cos lπx (j = 1, 2, . . . ,Λ) such that

⎛⎝U

j+1N −U

j

N

h, v

⎞⎠ +

⎛⎝D2U

j+1N +D2U

j

N

2− PNA

(U

j+1N ,U

j

N

),

D

⎛⎝m

⎛⎝D2U

j+1N +D2U

j

N

2

⎞⎠Dv

⎞⎠⎞⎠ = 0,

(U0

N, v)− (u0, v) = 0.

(4.2)

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 hkΛk = T . Since we have no exact solution of (2.2) and(2.3), we takeN = 32 and h0 = 0.15625×10−4 to compute an approximating solutionUΛ0

N withh0Λ0 = T and regard this as an exact solution. we also choose five different time-step sizeshk (k = 1, 2, . . . , 5) with hkΛk = T to obtain five approximating solutions UΛk

N (k = 1, 2, . . . , 5)and compute the error estimation. Define an error function:

err(T, hk) =

(∫1

0

(UΛk

N −UΛ0N

)2dx

)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 − x)5 and

u20 = x5(1 − x)5ex 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.


00.20.40.60.810

0.050.1−1

0

1

2

3

4

5

6×10−4

(a)

−20

2

4

6

8

10×10−4

00.5

10

0.050.1

(b)

Figure 2: The development of the solution of the full-discrete scheme with initial value u0(x) = x5(1 − x)5

(a) and u0(x) = x5(1 − x)5ex (b).

00.20.40.60.81 00.05

0.1−4−20

2

4

6

8×10−3

(a)

0.015

0.01

0.005

0

−0.005−0.01−0.015

00.20.40.60.81 00.05

0.1

(b)

Figure 3: The development of the solution of the full-discrete scheme with initial value u0(x) = x5(1 − x)5

(a) and u0(x) = x5(1 − x)5ex (b).

Table 1: The error and the convergence order.

hk err1(0.1, hk) Order1 err2(0.1, hk) Order20.1 × 10−2 0.2441 × 10−6 — 0.2114 × 10−6 —0.5 × 10−3 0.6332 × 10−7 1.9466 0.5414 × 10−7 1.96510.25 × 10−3 0.1186 × 10−7 2.4159 0.1115 × 10−7 2.27990.125 × 10−3 0.1938 × 10−8 2.6142 0.1901 × 10−8 2.55150.625 × 10−4 0.2504 × 10−9 2.9518 0.2483 × 10−9 2.9368






We also choose five different time-step sizes hk to carry out numerical computationsand apply the error function in (4.3) to illustrate the estimation and convergence order intime variable t, see Table 1.

4.2. Example 2

Takem0 = 0.05 and T = 0.1. We also take two different initial functions u10(x) = x5(1− x)5 and

u20 = x5(1 − x)5ex 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 intime 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 − x)5

and u20 = x5(1 − x)5ex 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 intime variable t, see Table 3.

Acknowledgment

This work is supported by NSFC no. 11071102.

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