Unconditional optimal error estimates of a two-grid method for semilinear parabolic equation

Unconditional optimal error estimates of a two-grid method for semilinear parabolic equation

Applied Mathematics and Computation 310 (2017) 40–47 Contents lists available at ScienceDirect Applied Mathematics and Computation journal homepage:...

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Applied Mathematics and Computation 310 (2017) 40–47

Contents lists available at ScienceDirect

Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc

Unconditional optimal error estimates of a two-grid method for semilinear parabolic equation Dongyang Shi∗, Huaijun Yang School of Mathematics and Statistics, Zhengzhou University, Zhengzhou 450001, China

a r t i c l e

i n f o

a b s t r a c t

Keywords: Unconditional Optimal error estimates Two-grid method Semilinear parabolic equation,

In this paper, the error analysis of a two-grid method (TGM) with backward Euler scheme is discussed for semilinear parabolic equation. Contrary to the conventional finite element analysis, the error between exact solution and finite element solution is split into two parts (temporal error and spatial error) by introducing a corresponding time-discrete system. This can lead to the spatial error independent of τ (time step). Secondly, based on the above technique, optimal error estimates in L2 and H1 -norms of TGM solution are deduced unconditionally, while previous works always require a certain time step size condition. Finally, a numerical experiment is provided to confirm the theoretical analysis. © 2017 Elsevier Inc. All rights reserved.

1. Introduction We consider the following semilinear parabolic equation:

⎧ ⎨ut − u = f (u ), u = 0,



u ( 0 ) = u0 ,

(X, t ) ∈  × J, (X, t ) ∈ ∂  × J, (X, t ) ∈  × {t = 0},

(1.1)

where  ⊂ R2 is a bounded domain with Lipschitz boundary ∂ , X = (x, y ), J = (0, T ] and ut = ∂∂ut . f(·) is twice continuously differentiable and u0 (X) is a given smooth function. As we know, some works about problem (1.1) have been studied. Such as, the mathematical theory of Galerkin finite element methods (FEMs) for parabolic PDE discussed in monograph [1]. Both the semi-discrete and fully-discrete finite element approximations considered under the hypothesis that the exact solution is asymptotically stable as t → ∞ in [2]. An efficient finite element discretization algorithm developed for nonlinear reduced-order model in [3]. Two-grid finite element and finite volume methods studied for two-dimensional semilinear parabolic problems in [4]. In addition, a priori error estimate for a space-time FEM was proved for (1.1) in [5]. Our goal is to solve problem (1.1) by two grid method with bilinear element unconditionally. This method was introduced by Xu [6–8] mainly for nonlinear or linear but not symmetric positive definite elliptic problems based on a coarse mesh with size H and a finer mesh with size h (h  H). More precisely, on the coarse mesh, a nonlinear or nonsymmetric problem is solved to reduce the computational cost. Then the solution obtained from coarse grid is used as a initial guess to solve a linearized problem (one Newton like iteration) on the finer mesh. The two grid method also has been applied to nonlinear parabolic equation [9–13], eigenvalue problems [14–19], Cahn–Hilliard equations [20], Stokes and Navier–Stokes equations [21–25], Maxwell’s equations [26] and so on. ∗

Corresponding author. E-mail addresses: [email protected], [email protected] (D. Shi).

http://dx.doi.org/10.1016/j.amc.2017.04.010 0 096-30 03/© 2017 Elsevier Inc. All rights reserved.

D. Shi, H. Yang / Applied Mathematics and Computation 310 (2017) 40–47

41

On the other hand, for nonlinear evolution equation, a certain restriction condition between spatial mesh size and temporal step size is usual needed to obtain optimal error estimates. For example, in [11], a two grid algorithm with expanded mixed FEM for nonlinear reaction diffusion equation was analyzed in detail, however, optimal error estimates were obtained under the condition τ · h−1 = O (1 ). Moreover, the restriction τ · H −1 = O (1 ) was also required for two-scale product approximation of (1.1) in the mixed FEM in [13]. The reason why the standard FEMs need time restriction is due to the error estimate of quadratic term arising from linearization by using inverse inequality. Recently, it has been shown that this restriction is not necessary and can be removed when the error splitting technique developed in [27,28] is employed. In this case, different from the conventional finite element analysis, the error is split into two parts (temporal error and spatial error) by introducing a corresponding time-discrete system with certain regularity. The error between Rh Un (Rh is Ritz projection operator and Un is the solution of time-discrete system) and Uhn (Uhn is the finite element solution) is proved to be independent of τ . This so-called splitting technique also has been applied to deal with other problems. Such as, nonlinear Joule heating equations [29–31], nonlinear Schrödinger equation [32], Ginzburg–Landau equations [33], miscible displacement in porous media [34] and Navier–Stokes equations [35], nonlinear parabolic equation [36] and nonlinear Sobolev equation [37]. The main aim of this paper is to solve problem (1.1) by a two grid algorithm with bilinear FEM and to obtain optimal error estimate, in which the time step restriction required in [11,13] is removed. The paper is organized as follows. In Section 2, we derive optimal error estimates of finite element solution in both temporal and spatial direction by the splitting technique, respectively, Then, in Section 3, we present the two grid algorithm and derive optimal error estimate of TGM solution unconditionally. A numerical example is provided in Section 4 to confirm the theoretical analysis. Throughout this paper, let Wm, p () be the standard Sobolev space with norm ·m, p and semi-norm|·|m, p . (see [38] for details). (·, ·) denote the standard L2 -inner product. The generic constant C > 0 (with or without subscript) is independent of n (time level), h (spatial size), τ (time step size) and may be different in different places. 2. Error estimates for FEM solution Let  be a rectangle and Th be a regular partition of  into rectangles Tj and h = max{diam T j } be the mesh size. Let Vh be bilinear finite element space and Vh0 = {v ∈ Vh , v|∂  = 0}. Moreover, we also let Ih be the associated interpolation operator over Vh and Rh : H01 () → Vh0 be the Ritz projection operator satisfying

(∇ (u − Rh u ), ∇ω ) = 0,

∀ ω ∈ Vh0 .

Then it can be found in [1,39] that

u − Rh u0 + hu − Rh u1 ≤ Ch2 u2 , ∀u ∈ H 2 ().

(2.1)

The weak formula of (1.1) is to find u : J → H01 (), such that

(ut , v ) + (∇ u, ∇ v ) = ( f (u ), v ), ∀v ∈ H01 (). (2.2) n Let {tn | tn = nτ ; 0 ≤ n ≤ N} be a uniform partition in time with time step τ and u = u(X, tn ). For a sequence of of functions { f n }Nn=0 , we denote Dτ f n = ( f n − f n−1 )/τ . Then the backward Euler scheme of (1.1) is to find Uhn ∈ Vh0 for n = 1, . . . , N, such that



(Dτ Uhn , vh ) + (∇ Uhn , ∇ vh ) = ( f (Uhn ), vh ), ∀vh ∈ Vh0 ,

Uh0 = Ih u0 ,

(2.3)

where Ih is the nodal Lagrange interpolation operator. Throughout this paper, we assume that the solution to problem (1.1) exists and satisfies

u0 2 + uL∞ (J;H2 ∩W 2,4 ) + ut L∞ (J;H2 ) + utt L∞ (J;L2 ) ≤ C.

(2.4)

Now, we present the first main result of this paper as follows. Theorem 2.1. Suppose that (1.1) has a unique solution u and satisfies (2.4). Thus, (2.3) also has a unique solution Uhn . Then there exist two positive constants h0 and τ 0 such that when h < h0 , τ < τ 0 ,

un − Uhn 0 ≤ C0 (h2 + τ ),

un − Uhn 1 ≤ C0 (h + τ ).

(2.5)

In order to prove Theorem 2.1, we present the error estimates of temporal direction and spatial direction in the next two subsections, respectively. 2.1. Temporal error estimate To discuss the error estimate of temporal direction, we introduce a corresponding time-discrete system to (1.1) as:

⎧ n n n ⎨Dτ U − U = f (U ), U n = 0,



U 0 = u0 .

n = 1, . . . , N, X ∈ ∂ ,

42

D. Shi, H. Yang / Applied Mathematics and Computation 310 (2017) 40–47

Theorem 2.2. Let un and Un be the solutions of (1.1) and (2.6), respectively. Then, for n = 0, 1, . . . , N, there exists a constant τ 1 such that when τ < τ 1 ,

en 1 + τ 2 en 2 ≤ C1 τ , 1

(2.7)

and

U n 2 + τ

N 



Dτ U n 22 ≤ C ,

(2.8)

n=1



where en = un − U n , C1 ≥ C utt 2∞

L (J;L2 )

and C ≥ C (u2∞

L (J;H 2 )

+ ut 2∞

L (J;H 2 )

+ utt 2∞

L (J;H 2 )

).

Proof. Let t = tn in (1.1), then we get

Dτ un − un = f (un ) + Rn , n = 1, . . . , N,

(2.9)

 tn

un −un−1

− utn = τ1 t (tn−1 − s )utt ds and Rn 0 ≤ C τ utt L∞ (J;L2 ) ≤ C τ . n−1 Subtracting (1.1) from (2.9) results in

where Rn =

τ

Dτ en − en = f (un ) − f (U n ) + Rn . We multiply both side of (2.10) by

1 τ



−e n

(2.10) and use Green’s formula to get

∇ (en − en−1 ), ∇ en + en 20 = ( f (un ) − f (U n ), −en ) + (Rn , −en ) := A1 + A2 .

(2.11)

Noting that

(∇ (en − en−1 ), ∇ en ) ≥

1 (∇ en 20 − ∇ en−1 20 ), 2

(2.12)

and the terms on the right hand side of (2.11) can be estimated as follows

A1 ≤ C en 0 en 0 ≤ C ∇ en 0 en 0 ≤ C ∇ en 20 + A 2 ≤  R n  0  e n  0 ≤ C τ 2 +

1 en 20 , 4

(2.13)

1 en 20 , 4

(2.14)

we have

∇ en 20 − ∇ en−1 20 + τ en 20 ≤ C ∇ en 20 + C τ 3 .

(2.15)

Then summing above inequality from n = 1, . . . , , (1 ≤  ≤ N ) and noting that

∇ e 20 + τ

 

en 20 ≤ C τ

n=1

 

e0

= 0 lead to

∇ en 20 + C τ 2 .

(2.16)

n=1

Thus, by discrete Gronwall’s lemma, there exists a small constant τ 1 > 0, such that when τ < τ 1

∇ e 20 + τ

 

en 20 ≤ C τ 2 ,

(2.17)

n=1

which together with the fact en 2 ≤ Cen 0 yields

∇ e 20 + τ

 

en 22 ≤ C1 τ 2 ,

(2.18)

n=1

this leads to the desired result of (2.7). On the other hand, it is not difficult to check that

U n 2 ≤ en 2 + un 2 ≤ C τ 2 + un 2 ≤ C. 1

Similarly

τ

 

Dτ U n 22 ≤ C τ

n=1

≤C

τ

 

Dτ en 22 + τ

n=1 −2

·τ

 

(2.19)

Dτ un 22

n=1   n=1

  +τ en 22

 





Dτ un 22

≤ C,

(2.20)

n=1

which is the desired result of (2.8). We point out that the regularity requirement Dτ un 22 = τ −1

C ut 2L∞ (J;H 2 )

≤ C is used in the last step in (2.20). The proof is complete.



 tn tn−1

ut ds22 ≤

D. Shi, H. Yang / Applied Mathematics and Computation 310 (2017) 40–47

43

2.2. Spatial error estimate The weak form of time-discrete system (2.6) is

(Dτ U n , vh ) + (∇ U n , ∇ vh ) = ( f (U n ), vh ), ∀vh ∈ Vh .

(2.21)

Then we have the following spatial error estimate. Theorem 2.3. Let Un and Uhn be the solutions of (2.21) and (2.3), respectively. Then, for n = 0, 1, 2, . . . , N, there exist two constants τ 2 > 0 and h0 > 0 such that when τ < τ 2 and h ≤ h0 ,

U n − Uhn 0 + hU n − Uhn 1 ≤ C2 h2 , where C2 ≥

C (u0 22

+

u2L∞ (J;H 2 )

+ ut 2∞ 2 L (J;H )

(2.22) +

utt 2L∞ (J;H 2 ) ).

Proof. Let U n − Uhn = U n − RhU n + RhU n − Uhn := θ n + ηn . Then we can get spatial error equation from (2.3) and (2.21) that

(Dτ ηn , vh ) + (∇ ηn , ∇ vh ) = −(Dτ θ n , vh ) − (∇ θ n , ∇ vh ) + ( f (U n ) − f (Uhn ), vh ) :=

3 

Ai .

(2.23)

i=1

Let vh = ηn in (2.23), then the left hand side of (2.23) becomes

( Dτ η n , η n ) + ( ∇ η n , ∇ η n ) ≥

1 (ηn 20 − ηn−1 20 ) + |ηn |21 . 2τ

(2.24)

For the first term on the right hand of (2.23), by interpolation theory

A1 ≤ Ch2 Dτ U n 2 ηn 0 ≤ Ch4 Dτ U n 22 + C ηn 20 .

(2.25)

Obviously

A2 = 0,

(2.26)

and A3 can be bounded by

A3 ≤ C (θ n 0 + ηn 0 )ηn 0 ≤ C h4 U n 22 + C ηn 20 .

(2.27)

Then, substituting (2.24)–(2.27) into (2.23) yields

ηn 20 − ηn−1 20 ≤ Ch4 (τ Dτ U n 22 + τ U n 22 ) + C τ ηn 20 .

(2.28)

Summing the above inequality from n = 1, . . . , , (1 ≤  ≤ N ) leads to

η 20 ≤ (1 − C τ )η0 20 + Ch4 τ

 

(Dτ U n 22 + U n 22 ) + C τ

n=1

 

ηn 20 .

(2.29)

n=0

Noting that η0 0 = Rh u0 − Ih u0 0 ≤ Rh u0 − u0 0 + u0 − Ih u0 0 ≤ Ch2 u0 2 , we have by (2.8) that

η 20 ≤ C h4 u0 20 + C h4 τ

 

(Dτ U n 22 + U n 22 ) + C τ

n=1

 

ηn 20

n=0

≤ C (u0 22 + u2L∞ (J;H 2 ) + ut 2L∞ (J;H 2 ) + utt 2L∞ (J;H 2 ) )h4 + C τ ≤ C h4 + C τ

 

 

ηn 20

n=0

ηn 20 .

(2.30)

n=0

Thanks to discrete Gronwall’s lemma, there exists a small constant τ 2 > 0, such that when τ < τ 2

ηn 20 ≤ Ch2 .

(2.31)

Then, by inverse inequality, it follows that

ηn 21 ≤ Ch.

(2.32)

Therefore, by triangle inequality and (2.1), we have

U n − Uhn 0 + hU n − Uhn 1 ≤ C2 h2 , which is the desired result.

(2.33)



Now, we give the proof of Theorem 2.1. In fact, based on the above results of Theorems 2.2 and 2.3 together with

τ0 = min{τ1 , τ2 } and C0 = max{C1 , C2 }, then we have

un − Uhn 0 ≤ un − U n 0 + U n − Uhn 0 ≤ C0 (h2 + τ ). which is the desired result of Theorem 2.1.

(2.34)

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D. Shi, H. Yang / Applied Mathematics and Computation 310 (2017) 40–47

3. Error estimates for TGM solution In this section, we present the procedure of two-grid method and error estimate. Step I: On the coarse grid TH , for n = 1, . . . , N, solve nonlinear system for UHn ∈ VH0 , such that

(Dτ UHn , VH ) + (∇ UHn , ∇ vH ) = ( f (UHn ), vH ), ∀vH ∈ VH0 ,

(3.1)

UH0 = IH u0 .

(3.2)

Step II: On the fine grid Th , for n = 1, . . . , N, solve the following linear system for

(

v ) + (∇

Dτ U˜hn , h

U˜hn ,

∇vh ) = ( f (

UHn

) + f  (UHn )(U˜hn − UHn ), vh ),

∀vh ∈

U˜hn



Vh0 ,

such that

Vh0 ,

(3.3)

U˜h0 = Ih u0 .

(3.4)

Then, we state the second main result of TGM solution as follows. Theorem 3.1. Let un and U˜hn be the solutions of (2.2) and (3.3), then we have

un − U˜hn 0 ≤ C3 (h2 + H 3 + τ ), where C3 ≥

C (u0 22

+

u2L∞ (J;H 2 )

+

un − U˜hn 1 ≤ C3 (h + H 3 + τ ),

u2L∞ (J;W 2,4 )

+ ut 2L∞ (J;H 2 )

(3.5)

+ utt 2L∞ (J;H 2 ) ).

Proof. By Taylor expansion, we have

1  f (un ) = f (UHn ) + f  (UHn )(un − UHn ) + f (ξ )(un − UHn )2 , 2

(3.6)

where ξ is between UHn and un . Then let un − U˜hn = un − Rh un + Rh un − U˜hn := α n + β n . We have from (2.2) and (3.3) that



n

β n − β n−1 α − α n−1 , vh + ( ∇ β n , ∇ vh ) = − , vh − (∇ α n , ∇ vh ) + ( f  (UHn )(un − U˜hn ), vh ) τ τ 1 5   + f (ξ )(un − UHn )2 , vh + (Rn , vh ) := Bi . 2

(3.7)

i=1

On one hand, let vh = β n , then the left hand side of (3.7) becomes



1 β n − β n−1 n ,β + (∇ β n , ∇ β n ) ≥ (β n 20 − β n−1 20 ) + |β n |21 . τ 2τ

(3.8)

Now, we only need to estimate the terms on the right hand of (3.7). For B1 , by interpolation theory, we have

B1 ≤ Ch2 τ −1



tn

tn−1

ut 2 dt β n 0 ≤ C ut 2L∞ (J;H2 ) h4 + C β n 20 .

(3.9)

Obviously

B2 = 0.

(3.10)

Moreover, we have

B3 ≤ C ( α n 0 + β n 0 ) β n 0 ≤ C ( h2 u n 2 + β n 0 ) β n 0 ≤ C h4 un 22 + C β n 20 .

(3.11)

Similarly, it follows that

B4 ≤ C un − UHn 20,4 β n 0 .

(3.12)

Here, with the help of H1 →L4 and inverse inequality

un − UHn 0,4 ≤ un − IH un 0,4 + IH un − UHn 0,4 ≤ CH 2 un 2,4 + IH un − RH un 0,4 + RH en − en 0,4 + en 0,4 + RH U n − UHn 0,4 ≤ C H 2 + C H − 2 IH un − RH un 0 + C RH en − en 1 + C en 1 + CH − 2 RH U n − UHn 0 1

1

≤ C (u0 22 + u2L∞ (J;H 2 ) + u2L∞ (J;W 2,4 ) + ut 2L∞ (J;H 2 ) + utt 2L∞ (J;H 2 ) )(H 2 + τ ) 3

3 ≤ C˜(H 2 + τ ),

where C˜ ≥

C (u0 22

+ u2∞ 2 L (J;H )

+

(3.13)

u2L∞ (J;W 2,4 )

+ ut 2∞ 2 L (J;H )

+

utt 2L∞ (J;H 2 ) ).

D. Shi, H. Yang / Applied Mathematics and Computation 310 (2017) 40–47

45

Thus, (3.12) can be further bounded by

B4 ≤ C˜(H 6 + τ 4 ) + C β n 20 .

(3.14)

In addition, B5 can be estimated as

B5 ≤ Rn 0 β n 0 ≤ C τ 2 utt 2L∞ (J;L2 ) + C β n 20 .

(3.15)

By substituting (3.9)–(3.11) and (3.14)–(3.15) into (3.7), we have

β n 20 − β n−1 20 ≤ C τ (h4 + H 6 + τ 2 ) + C τ β n 20

(3.16)

Then, summing the above inequality from n = 1, . . . , , (1 ≤  ≤ N ) leads to

β  20 ≤ (1 − C τ )β 0 20 + C (h4 + H 6 + τ 2 ) + C τ

 

β n 20 .

(3.17)

n=0

Noting that β 0 0 = Rh u0 − Ih u0 0 ≤ Rh u0 − u0 0 + u0 − Ih u0 0 ≤ Ch2 u0 2 , when 1 − C τ > 0, by discrete Gronwall’s lemma, it follows that

β  20 ≤ C (h4 + H 6 + τ 2 ),

(3.18)

which implies un − U˜hn 0 ≤ un − Rh un 0 + Rh un − U˜hn 0 ≤ C3 (h2 + H 3 + τ ).

On the other hand, let vh = β n − β n−1 in (3.7) and follow a similar proof of un − U˜hn 0 , it is not difficult to have that

B1 ≤ Ch4 τ ut 2L∞ (J;H 2 ) +

1 β n − β n−1 20 , 4τ

(3.19)

B2 = 0,

(3.20)

B3 ≤ C h4 τ u2L∞ (J;H 2 ) + C τ |β n |21 +

B4 ≤ C˜τ (H 6 + τ 4 ) +

1 β n − β n−1 20 , 4τ

(3.21)

1 β n − β n−1 20 , 4τ

B5 ≤ C τ 3 utt 2L∞ (J;L2 ) +

(3.22)

1 β n − β n−1 20 . 4τ

(3.23)

Then, substituting the above estimates into (3.7) yields

|β n |21 − |β n−1 |21 ≤ C τ (h4 + H 6 + τ 2 ) + C τ |β n |21 .

(3.24)

Summing the above inequality from n = 1, . . . , , (1 ≤  ≤ N ) leads to

|β  |21 ≤ (1 − C τ )|β 0 |21 + C (h4 + H 6 + τ 4 ) + C τ

 

|β n |21 .

(3.25)

n=0

Similarly, noting that |β 0 |1 = |Rh u0 − Ih u0 |1 ≤ |Rh u0 − u0 |1 + |u0 − Ih u0 |1 ≤ Chu0 2 , when 1 − C τ > 0, by discrete Gronwall’s lemma, it follows that

|β  |1 ≤ C ( h + H 3 + τ ), which implies un − U˜hn 1 ≤ un − Rh un 1 + Rh un − U˜hn 1 ≤ C3 (h + H 3 + τ ). Therefore, the proof is complete.

(3.26) 

Remark 1. A striking feature of our analysis is that the condition τ · H −1 = O (1 ), which is essential in [13] for getting optimal error estimates of two-scale product approximation of semilinear parabolic problem, is removed by the error splitting approach. This means that the time step restriction is not necessary in the analysis of TGM for such kind problem. In the further study, we will apply two-grid algorithm combined with the splitting approach to other nonlinear problems (such as nonlinear Sobolev equations, nonlinear Schrödinger equation and so on) by conforming and nonconforming FEMs.

46

D. Shi, H. Yang / Applied Mathematics and Computation 310 (2017) 40–47 Table 1 The errors at t = 0.1. H

h

un − U˜hn 0

Order

un − U˜hn 1

Order

1/4 1/9 1/16

1/8 1/27 1/64

8.0714e−03 7.1155e−04 1.2673e−04

– 1.9966 1.9992

2.2781e−01 6.7513e−02 2.8483e−02

– 0.99981 0.99998

Table 2 The errors at t = 0.5. H

h

un − U˜hn 0

Order

un − U˜hn 1

Order

1/4 1/9 1/16

1/8 1/27 1/64

4.8942e−03 4.3051e−04 7.6647e−05

– 1.9984 1.9996

1.5264e−01 4.5254e−02 1.9092e−02

– 0.99949 0.99994

Table 3 The errors at t = 1. H

h

un − U˜hn 0

Order

un − U˜hn 1

Order

1/4 1/9 1/16

1/8 1/27 1/64

2.9442e−03 2.5890e−04 4.6088e−05

– 1.9986 1.9998

9.2577e−02 2.7448e−02 1.1580e−02

– 0.99947 0.99994

(a) L2 -norm errors of u.

(b) H 1 -norm errors of u.

Fig. 1. The errors of different time steps at t = 1.

4. Numerical experiment In this section, we present a numerical example to demonstrate the theoretical analysis. In the computation, we set the domain  = [0, 1] × [0, 1], the final time T = 1 and consider following problem

⎧ ⎨ut − u − sin u = g, u = 0,



u ( 0 ) = u0 ,

(X, t ) ∈  × J, (X, t ) ∈ ∂  × J, (X, t ) ∈  × {t = 0},

(4.1)

where g and u0 are computed from the exact solution

u(x, y, t ) = e−t sin(π x ) sin(π y ). In order to confirm optimal error estimates in Theorem 3.1, we choose τ = h2 and h2 = H 3 . The L2 and H1 -norms errors are listed in Tables 1–3 at t = 0.1, 0.5 , 1, respectively. Obviously, it can be seen that numerical results are in agreement with theoretical analysis, i.e., un − U˜hn 0 and un − U˜hn 1 are convergent at optimal rates of O (h2 ) and O (h ), respectively. On the other hand, in order to show the unconditional convergence behavior of scheme (3.1)–(3.4), we use four different time steps τ = 0.05, 0.1, 0.2, 0.25 on gradually refined meshes at t = 1. The corresponding numerical results are showed in Fig. 1. It is observed that for a fixed τ , numerical error tends to be a constant as h/τ → 0, which indicates that the time step restriction is not necessary.

D. Shi, H. Yang / Applied Mathematics and Computation 310 (2017) 40–47

47

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