Unconditional superconvergent analysis of a linearized finite element method for Ginzburg-Landau equation

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Applied Numerical Mathematics ••• (••••) •••–•••

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Applied Numerical Mathematics www.elsevier.com/locate/apnum

Unconditional superconvergent analysis of a linearized finite element method for Ginzburg-Landau equation Shi Dongyang ∗ , Liu Qian School of Mathematics and Statistics, Zhengzhou University, Zhengzhou 450001, China

a r t i c l e

i n f o

Article history: Received 9 May 2018 Received in revised form 8 January 2019 Accepted 22 August 2019 Available online xxxx Keywords: Ginzburg-Landau equation Linearized Galerkin FEM Unconditional superclose and superconvergent estimates

a b s t r a c t In this paper, unconditional superconvergent estimate with a Galerkin finite element method (FEM) is presented for Ginzburg-Landau equation by conforming bilinear FE, while all previous works require certain time-step restrictions. First of all, a time-discrete system is introduced, with which the error function is split into a temporal error and a spatial error. On one hand, the regularity of the time-discrete system is deduced with a rigorous analysis. On the other hand, the error between the numerical solution and the solution of the time-discrete system is derived τ -independently with order O (h2 + hτ ), where h is the subdivision parameter and τ , the time step. Then, the unconditional superclose result of order O (h2 + τ ) in the H 1 -norm is deduced directly based on the above estimates. Furthermore, the global superconvergent result is obtained through the interpolated postprocessing technique. At last, numerical results are provided to confirm the theoretical analysis. © 2019 IMACS. Published by Elsevier B.V. All rights reserved.

1. Introduction The Ginzburg-Landau equation is an important nonlinear evolution equation and has important applications in many fields. It was introduced by Ginzburg and Landau in the 1950s to describe a vast of phenomena from nonlinear waves to second-order phase transitions, from superconductivity, superfluidity and Bose-Einstein condensation to liquid crystal and strings in field theory [1]. The existence and uniqueness of weak solution, the regularity and long-time behavior of the solution for the problem can be found in [5–7,10,11,13]. Numerical methods for Ginzburg-Landau equation such as spectral method [4,20,28], finite difference method [19,31,32,34] and FEM [9,33] have already been applied to the convergence analysis or optimal error estimate. In this article, we are concerned with a class of linearized finite element approximation for the following complex Ginzburg-Landau equation

⎧ ⎨ ut − (ν + i α )u + (κ + i β)|u |2 u − γ u = 0, ( X , t ) ∈ × (0, T ], u ( X , t ) = 0, ( X , t ) ∈ ∂ × [0, T ], ⎩ u ( X , 0) = u 0 ( X ), X ∈ . Where is a bounded rectangular region with boundary ∂ , X = (x, y ), u 0 ( X ) is complex-valued function. constants, α , β are real parameters and γ is the linear evolution term.

*

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

https://doi.org/10.1016/j.apnum.2019.08.023 0168-9274/© 2019 IMACS. Published by Elsevier B.V. All rights reserved.

(1.1)

ν , κ are positive

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2

It is well known that the boundedness of numerical solution in L ∞ -norm or a stronger norm is often required for studying the nonlinear problem with FEM. If a priori estimate for numerical solution in such a norm can not be obtained, one may employ the inverse inequality to deal with this issue, which will result in some time-step restrictions, such as τ = O (hr +1 )(r = 0, 1, 2) for Ginzburg-Landau equation in superconductivity [9,33]. In fact, the restrictions of τ are also needed for other evolution equations, such as nonlinear Sobolev problems [12,24], nonlinear Schrödinger equations [2,25] and Navier-Stokes equations [3,14]. In order to overcome this difficulty, the so-called splitting technique [16] was used to solve a class of nonlinear parabolic equations so as to deduce the unconditional error estimates, where a corresponding time-discrete system was introduced to split the error into two parts, the temporal error and the spatial error. More precisely, the spatial error is τ -independent and the numerical solution can be bounded in L ∞ -norm by an inverse inequality unconditionally. Subsequently, [8,15,17,29,30] applied this technique to various nonlinear problems as incompressible miscible flow in porous media, heat and sweat transport in textile materials, joule heating equations, etc. However, their work only arrived at convergence analysis and optimal estimate. Recently, by use of low order conforming and nonconforming FEMs, the unconditional superclose or superconvergent properties for Sobolev equation [26], nonlinear Schrödinger equation [21], nonlinear parabolic equation [22,27], nonlinear hyperbolic equation [23] have been discussed deeply. As far as we know, few studies have been devoted to this kind of analysis for (1.1). The main aim of this paper is to study the unconditional superconvergent character of u in the H 1 -norm for (1.1) with bilinear element. First of all, a linearized backward Euler FE scheme is constructed. Secondly, motivated by the idea of splitting technique, we introduce a time-discrete system with solution of U n to split the error of un − U hn and deduce the correspond temporal error estimates. Thirdly, based on these estimates, we obtain the unconditional superclose result of u in the H 1 -norm with order O (h2 + τ ) by arriving at the spatial error with order O (h2 + hτ ) directly. Moreover, we obtain the global superconvergent estimate through interpolation postprocessing technique. Lastly, we carry out a numerical test to illustrate the validity and efficiency of our theoretical analysis. To our best knowledge, the results achieved herein have never been seen in the existing literature.  Throughout this paper, for two complex functions u , v, we denote the L 2 inner product as (u , v ) = u v¯ d X ( v¯ represents √ the conjugate of v), and the corresponding L 2 -norm as u 0 = (u , u ). Let H 01 ( ) = {u ∈ H 1 ( ) : u |∂ = 0}. Furthermore, we use the classical Sobolev spaces W m, p ( ), 1 ≤ p ≤ ∞, denoted by W m, p , with norm  · m, p . When p = 2, we simply

b

write  · m, p as  · m . In addition, we define the space L p (a, b; Y ) with norm  f  L p (a,b;Y ) = ( the integral is replaced by the essential supremum.

a

p

1

 f (·, t )Y dt ) p , and if p = ∞,

2. A linearized finite element approximation scheme Let h be a regular rectangular subdivision of . The bilinear finite element space is defined by V h = { v h : v h | K ∈ span{1, x, y , xy }, v h |∂ = 0, ∀ K ∈ h }. Let I h be the associated interpolation operator over V h , then we have the following important lemma which has been proved in [18]. Lemma 1. If u ∈ H 3 ( ), then for v h ∈ V h , there holds

(∇(u − I h u ), ∇ v h ) = O (h2 )u 3  v h 1 . Let {tn : tn = nτ ; 0 ≤ n ≤ N } be a uniform partition of [0, T ] with the time step define

ϕ n = ϕ (tn ), ∂t ϕ n =

(2.1)

τ = T / N. For a smooth function ϕ , we

ϕ n − ϕ n −1 . τ

(2.2)

Then with these notations, we develop the linearized Galerkin finite element scheme to problem (1.1) as: find U hn ∈ V h , such that for tn ∈ (0, T ]



(∂t U hn , v h ) + (ν + i α )(∇ U hn , ∇ v h ) + (κ + i β)(|U hn−1 |2 U hn , v h ) − (γ U hn , v h ) = 0, ∀ v h ∈ V h , U h0 = I h u 0 .

(2.3)

3. Error estimates for time-discrete system We introduce the following time-discrete scheme for n ≥ 1

⎧ ⎨ ∂t U n − (ν + i α )U n + (κ + i β)|U n−1 |2 U n − γ U n = 0, ( X , t ) ∈ , U n = 0, X ∈ ∂ . ⎩ 0 U = u 0 , X ∈ .

(3.1)

The existence and uniqueness of the solution U n follow immediately from the fact that (3.1) is a system of linear elliptic equation. In what follows, we set en = un − U n (n = 0, 1, 2, ..., N ), analyze un − U n i (i = 0, 1, 2) and give the related estimates.

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Theorem 1. Let um and U m (m = 0, 1, 2, ..., N ) be the solutions of (1.1) and (3.1) respectively, assume u ∈ L 2 (0, T ; H 3 ( )), ut ∈ L ∞ (0, T ; H 2 ( )), utt ∈ L 2 (0, T ; L 2 ( )), then for m = 0, 1, ..., N, there exists τ0 such that when τ < τ0 ,

em 2 + ∇ em 0 + ∂t em 0 ≤ C 0 τ .

(3.2)

Here and later, C (with or without subscript) denotes a positive constant independent of m, h and

τ.

Proof. Note K  max1≤m≤ N um 2 + 1 and we begin to prove (3.2) by mathematical induction. In fact, when m = 1, with (1.1) and (3.1), we have

∂t e 1 − (ν + i α )e 1 = (κ + i β)(|U 0 |2 U 1 − |u 0 |2 u 1 ) + γ e 1 + R 11 + (κ + i β) R 12 ,  R 11 0 1

(3.3)

employing Taylor’s expansion yields = O (τ ). On one hand, multiplying (3.3) by e and integrating it over lead to

(

e1 − e0

τ

+  R 12 0

, e 1 ) + (ν + i α )∇ e 1 20 = (κ + i β)(|U 0 |2 U 1 − |u 0 |2 u 1 , e 1 ) + γ (e 1 , e 1 )

(3.4)

+ ( R 11 , e 1 ) + (κ + i β)( R 12 , e 1 ). Comparing the real parts of (3.4), we have

e 1 20

τ

+ ν ∇ e 1 20 = (κ + i β)(|U 0 |2 U 1 − |u 0 |2 u 1 , e 1 ) + γ (e 1 , e 1 ) (3.5)

+ ( R 11 , e 1 ) + (κ + i β)( R 12 , e 1 ) ≤ C τ 2 + C e 1 20 . Since e 1 ∈ H 2 ( ) ∩ H 01 ( ), there exists

τ1 , such that for τ ≤ τ1 , it follows that

e 1 0 √ + ∇ e 1 0 ≤ C 1 τ .

(3.6)

τ

On the other hand, multiplying e 1 of (3.3) yields

(

∇ e1

τ

, ∇ e 1 ) + (ν + i α )e 1 20 = −(κ + i β)(|U 0 |2 U 1 − |u 0 |2 u 1 , e 1 ) + γ (∇ e 1 , ∇ e 1 )

(3.7)

− ( R 11 , e 1 ) + (κ + i β)( R 12 , e 1 ). Then we compare the imaginary part of (3.7) to give

e 1 0 ≤ C 2 τ ,

(3.8)

which together with e ∈ H ( ) ∩ follows that U 2 ≤ e 2 + u 2 ≤ C C 2 τ + u 2 ≤ K , where By mathematical induction, we assume that (3.2) holds for m ≤ n − 1. Thus we have 1

2

H 01 ( )

1

1

1

1

τ ≤ τ2 ≤ 1/C C 2 .

em 2 ≤ C 0 τ , U m 2 ≤ um 2 + em 2 ≤ um 2 + C 0 τ ≤ K , where τ ≤ τ3 ≤ 1/C 0 . Now, we start to discuss the case of m = n. From (1.1) and (3.1), we derive the following error equation

∂t en − (ν + i α )en = (κ + i β)(|U n−1 |2 U n − |un−1 |2 un ) + γ en + R n1 + (κ + i β) R n2 .  R n1 0

(3.9)

+  R n2 0 n

Obviously, = O (τ ). Firstly, multiplying e on both sides of (3.9) yields

(∂t en , en ) + (ν + i α )(∇ en , ∇ en ) = (κ + i β)(|U n−1 |2 U n − |un−1 |2 un , en ) + γ (en , en ) + ( R n1 , en ) + (κ + i β)( R n2 , en ),

(3.10)

and comparing the real parts gives

en 20 − en−1 20 2τ

+ ν ∇ en 20 ≤ C en−1 20 + C en 20 + C τ 2 .

Then, multiplying 2τ on the two sides of (3.11) and summing from i = 2 to n, we have

(3.11)

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en 20 + τ

n 

∇ e i 20 ≤ C τ 2 + C τ

n 

e i 20 .

i =2

i =2

By Gronwall’s inequality, there exists

τ4 such that

en 0 + (τ

n 

(3.12)

∇ e i 20 )1/2 ≤ C 3 τ ,

(3.13)

i =2

when τ ≤ τ4 . In what follows, we give the result of ∇ en 0 ≤ C 0 τ . In fact, multiplying ∂t en on both sides of (3.9) and comparing the real parts result in

(∂t en , ∂t en ) + (ν + i α )(∇ en , ∂t ∇ en ) = ((κ + i β)(|U n−1 |2 U n − |un−1 |2 un , ∂t en ) + γ (en , ∂t en ) + ( R n1 , ∂t en ) + (κ + i β)( R n2 , ∂t en ). Similarly with the estimates of (3.13), there exists

∇ en 0 + (τ

n 

(3.14)

τ5 such that for τ ≤ τ5 , we have

∂t e i 20 )1/2 ≤ C 4 τ .

(3.15)

i =2

To derive the estimate of en 0 , we take difference between two time levels n and n − 1 of (3.9) and multiply it by τ1 on both sides, then there holds

∂tt en − (ν + i α )∂t en + (κ + i β)∂t P 1n − γ ∂t en = ∂t R n1 + (κ + i β)∂t R n2 ,

(3.16)

where P 1n = |U n−1 |2 U n − |un−1 |2 un . We multiply (3.16) by ∂t en , integrate it over and compare the real parts to get

ν ∂t en 20 + ν

∇∂t en 20 − ∇∂t en−1 20 2τ

= −(κ + i β)(∂t P 1n , ∂t en ) + γ (∂t en , ∂t en ) + (∂t R n1 , ∂t en ) + (κ + i β)(∂t R n2 , ∂t en ) 

4 

(3.17)

Ai .

i =1

For A 1 , it is not hard to check that

A 1 ≤ C (∂t (|U n−1 |2 U n − |un−1 |2 un ), ∂t en ) ≤ C ∂t en 20 + C ∂t en−1 20 +

ν 4

∂t en 20 .

(3.18)

With Green’s formula, we obtain

A 2 = |γ (∇∂t en , ∇∂t en )| ≤ C ∇∂t en 20 . By Taylor’s expansion, we have

A3 + A4 ≤ C τ 2 +

ν 2

∂t R n1 0

(3.19)

+ ∂t R n2 0

= O (τ ) which follows by

∂t en 20 .

(3.20)

Combining the above estimates and summing from i = 2...n, we have

∇∂t en 20 + τ

n 

∂t e i 20 ≤ C τ 2 + τ

i =2

n 

∂t en 20 + τ

i =2

Applying Gronwall’s lemma and noticing that en 20 ≤ C τ

∇∂t en 0 + en 2 ≤ C 5 τ ,

n 

∇∂t en 20 ≤ C τ 2 .

(3.21)

i =2

n

i =1 ∂t e

i 2 0

and en ∈ H 2 ( ) ∩ H 01 ( ), we get

(3.22)

where C 5 is independent of C 0 . n n n Therefore, 5 when τ ≤ τ6 ≤ 1/C 5 , we get U 2 ≤ e 2 + u 2 ≤ K . As a result, (3.2) holds for all m = 0, 1, 2... N if we take C 0 ≥ i =1 C i and τ0 ≤ min1≤i ≤6 τi . The induction is closed and the proof is completed. 2 Remark 1. The result of en 2 = O (τ ) in (3.2) which may lead to ∂t U n 2 ≤ C 0 is very important to the superclose analysis in next section.

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4. Error estimates for spatial-discrete scheme and unconditional superconvergent result In this section, we will firstly establish a estimate for  I h U n − U hn 1 = O (h2 + hτ ) which results in the final estimate of  I h un − U n 1 . In order to do this, for n = 0, 1, 2, ..., N, we let

U n − U hn = U n − I h U n + I h U n − U hn  ηn + ξ n ,

(4.1)

un − U hn = un − I h un + I h un − U hn  ρ n + θ n .

Theorem 2. Let U m and U hm be the solutions of (3.1) and (2.3) respectively, for m = 0, 1, 2, ..., N. Under the conditions in Theorem 1, there exist h0 , τ0 , such that when h < h0 , τ < τ0 , there holds

ξ m 1 ≤ C 0 (h2 + hτ ).

(4.2)

Proof. Since  I h U m 0,∞ ≤ C U m 2 ≤ C , we can define K = max1≤m≤ N  I h U m 0,∞ + 1. Then, we prove (4.2) by mathematical induction. For m = 1, we have the following equation from (2.3) and (3.1)

(∂t ξ 1 , v h ) + (ν + i α )(∇ξ 1 , ∇ v h ) = −(∂t η1 , v h ) − (ν + i α )(∇ η1 , ∇ v h ) + (κ + i β)(|U h0 |2 U h1 − |U 0 |2 U 1 , v h ) + γ (U h1 − U 1 , v h ).

(4.3)

Taking v h = ξ 1 and comparing the real parts to get

ξ 1 20

τ

+ ν ∇ξ 1 20 = −(∂t η1 , ξ 1 ) − (ν + i α )(∇ η1 , ∇ξ 1 ) + (κ

+ i β)(|U h0 |2 U h1

0 2

1

1

− | U | U , ξ ) + γ

(U h1

1

1

− U ,ξ ) 

4 

(4.4) Hi.

i =1

Thanks to Theorem 1, we can easily get

H 1 + H 4 ≤ Ch4 (U 1 22 + ∂t U 1 22 ) + C ξ 1 20 ,

(4.5)

H 2 ≤ C (∇(e 1 − I h e 1 ) + ∇(u 1 − I h u 1 ), ∇ξ 1 ) ≤ Ch2 τ 2 + Ch4 u 1 23 + C ∇ξ 1 20

(4.6)

and

For the term of H 3 , we have

H 3 ≤ (|U h0 |2 U h1 − |U h0 |2 U 1 + |U h0 |2 U 1 − |U 0 |2 U 1 , ξ 1 ) ≤ Ch4 (U 1 22 + U 0 22 ) + C ξ 1 20 .

(4.7)

Altogether, we obtain

ξ 1 0 √ + ∇ξ 1 0 ≤ C 1 (h2 + hτ ),

(4.8)

τ

which also implies that √1 ξ 1 0 ≤ C 1 h, ∇ξ 1 0 ≤ C 1 h(h + τ ),ξ 1 1 ≤ C 1 (h2 + τ 2 ). τ

Thus, it follows that

U h1 0,∞ ≤ Ch−1 U h1 − I h U 1 0 +  I h U 1 0,∞ ≤ C C 1 (h + τ ) +  I h U 1 0,∞ ≤ K , where h ≤ h1 ≤ 1/2C C 1 , τ

≤ τ

1

(4.9)

≤ 1/2C C . 1

By mathematical induction, we assume (4.2) holds for m ≤ n − 1. Then, there yields

U hm 0,∞ ≤ Ch−1 U hm − I h U m 0 +  I h U m 0,∞ ≤ C C 0 (h + τ ) +  I h U m 0,∞ ≤ K , when h ≤ h2 ≤ 1/2C C 0 , τ

≤ τ

2

(4.10)

≤ 1/2C C . 0

Now, we prove that (4.2) holds true for m = n. Subtracting (2.3) from (3.1), we have

(∂t ξ n , v h ) + (ν + i α )(∇ξ n , ∇ v h ) = −(∂t ηn , v h ) − (ν + i α )(∇ ηn , ∇ v h ) + (κ + i β)(|U hn−1 |2 U hn − |U n−1 |2 U n , v h ) + γ (ξ n + ηn , v h ). Choosing v h = ξ n and comparing the real parts to find

(4.11)

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ξ n 20 − ξ n−1 20 2τ

+ ν ∇ξ n 20 = −(∂t ηn , ξ n ) − ν (∇ ηn , ∇ξ n ) − α (∇ ηn , ∇ξ n ) + κ (|U hn−1 |2 U hn − |U n−1 |2 U n , ξ n ) + β(|U hn−1 |2 U hn − |U n−1 |2 U n , ξ n ) + γ (ξ n + ηn , ξ n ) 

6 

(4.12)

Bi.

i =1

Firstly, with interpolation theory, we get

B 1 + B 6 ≤ Ch4 (∂t U n 22 + U n 22 ) + C ξ n 20 .

(4.13)

Then, for B 2 and B 3 , we have

B 2 + B 3 ≤ C (∇(U n − un + un − I h un + I h un − I h U n ), ∇ξ n )

≤ C (∇(un − I h un ), ∇ξ n ) + C (∇(en − I h en ), ∇ξ n ) ≤ Ch4 un 23 + C ∇ξ n 20 + Ch2 en 22

(4.14)

≤ Ch4 + Ch2 τ 2 + C ∇ξ n 20 . As to B 4 , B 5 , it is not difficult to check that

B 4 + B 5 ≤ C (|U hn−1 |2 U hn − |U hn−1 |2 U n + |U hn−1 |2 U n − |U n−1 |2 U n , ξ n )

≤ C U hn−1 0,∞ ((ξ n + ηn ), ξ n ) + C U n 0,∞ ((ξ n−1 + ηn−1 ), ξ n ) ≤ Ch4 (U n 22 + U n−1 22 ) + C ξ n−1 20 + C ξ n 20

(4.15)

≤ Ch4 + C ξ n−1 20 + C ξ n 20 . Recalling (4.13)-(4.15), it follows that

ξ n 20 − ξ n−1 20 2τ

+ ν ∇ξ n 20 ≤ Ch4 + Ch2 τ 2 + C ξ n−1 20 + C ξ n 20 + C ∇ξ n 20 .

(4.16)

Summing from i = 2 to n and applying Gronwall’s lemma, it gives

ξ n 0 + (τ

n 

∇ξ n 20 )1/2 ≤ C 2 (h2 + hτ ),

(4.17)

i =2

where h ≤ h3 ≤ 1/C 2 C 0 , τ ≤ τ3 ≤ 1/C 2 C 0 and C 2 has nothing to do with C 0 . It is evident to see that

U hn 0,∞ ≤ Ch−1 U hn − I h U n 0 +  I h U n 0,∞ ≤ C C 2 (h + τ ) +  I h U m 0,∞ ≤ K . Similarly, we take C 0 ≥

2

i =1

C i , h0 ≤ min3i =1 h i and

(4.18)

τ0 ≤ min3i=1 τi , then for all m = 0, 1, 2... N, (4.2) holds true. 2

Remark 2. We control the L ∞ norm of U hm by ξ m 0 ≤ Ch(h + τ ) instead of ξ m 0 ≤ C (h2 + τ ) to avoid the restriction of h = O (τ ) and the above procedure is prepared for the unconditional superclose result. Based on the previous work, we now give the main result of this paper. Theorem 3. Let un and U hn be the solutions of (1.1) and (2.2), respectively, for n = 1, 2, ..., N. Assume that u , ut ∈ L ∞ (0, T ; H 3 ( )), utt ∈ L ∞ (0, T ; H 2 ( )), then we have

θ n 1 = O (h2 + τ ).

(4.19)

Proof. For n = 1, we have the following error equation from (1.1) and (2.3)

(∂t θ 1 , v h ) + (ν + i α )(∇θ 1 , ∇ v h ) = −(∂t ρ 1 , v h ) − (ν + i α )(∇ ρ 1 , ∇ v h ) + (κ + i β)(|U h0 |2 U h1 − |u 0 |2 u 1 , v h ) + γ (U h1 − u 1 , v h ) + ( R 11 , v h ) + (

κ

+ i β)( R 12 , v h ).

Taking v h = θ 1 and comparing the real parts of (4.20) to get

(4.20)

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(∂t θ 1 , θ 1 ) + ν (∇θ 1 , ∇θ 1 ) = (−(∂t ρ 1 , θ 1 ) − (ν + i α )(∇ ρ 1 , ∇θ 1 ) + (κ + i β)(|U h0 |2 U h1 − |u 0 |2 u 1 , θ 1 ) + γ (U h1 − u 1 , θ 1 ) + ( R 11 , θ 1 ) + (

κ

+ i β)( R 12 , θ 1 ))



6 

(4.21)

Gi.

i =1

First of all, with interpolated theory, we have

G 1 ≤ Ch4 u 1 22 + C θ 1 20 .

(4.22)

Then with (2.1) yields

G 2 ≤ Ch4 ut1 23 + C ∇θ 1 20 .

(4.23)

Moreover, it is easy to check that

G 3 ≤ C (|U h0 |2 U h1 − |U h0 |2 u 1 + |U h0 |2 u 1 − |u 0 |2 u 1 , v h ) ≤ Ch4 (u 1 22 + u 0 22 ) + C θ 1 20 ,

(4.24)

G 5 + G 6 ≤ C τ 2 + C θ 1 20 .

(4.25)

and

Combining (4.22)-(4.25), it follows that

θ 1 0 √ + ∇θ 1 0 ≤ Ch2 + C τ .

(4.26)

τ

For n ≥ 2, we have

(∂t θ n , v h ) + (ν + i α )(∇θ n , ∇ v h ) = −(∂t ρ n , v h ) − (ν + i α )(∇ ρ n , ∇ v h ) + (κ + i β)(|U hn−1 |2 U hn − |un−1 |2 un , v h ) +γ

(U hn

−u

n

, v h ) + ( R n1 , v h ) + (

κ

(4.27)

+ i β)( R n2 , v h ).

Let v h = θ and compare the real parts to derive n

θ n 20 − θ n−1 20 2τ

+ ν ∇θ n 20 = −(∂t ρ n , θ n ) − (ν + i α )(∇ ρ n , ∇θ n ) (4.28)

+ (κ + i β)(|U hn−1 |2 U hn − |un−1 |2 un , θ n ) + γ (U hn − un , θ n ) + ( R n1 , θ n ) + (κ + i β)( R n2 , θ n ).

Similarly to the estimate of (4.26), we have

θ n 20 − θ n−1 20 2τ

+

ν 2

∇θ n 20 ≤ Ch4 (unt 22 + un 23 ) + C τ 2 + C θ n 20 + C θ n−1 20 .

(4.29)

Summing (4.29) from 2 to n and applying the Gronwall’s lemma once again arrive at

θ n 0 ≤ Ch2 + C τ ,

(4.30)

here, U hn−1 0,∞ ≤ K is used. Now, we start to estimate the term of ∇θ n 0 . Taking difference between two time levels n and n − 1 of (4.20), and multiplying it by τ1 on both sides reduce to

(∂tt θ n , v h ) + (ν + i α )∇∂t θ n , ∇ v h ) = −(∂tt ρ n , v h ) + (ν + i α )∇∂t ρ n , ∇ v h ) + ((κ + i β)∂t P 2n , v h ) + γ (∂t θ n + ∂t ρ n , v h ) + (∂t R n1 , v h ) + (

κ

(4.31)

+ i β)(∂t R n2 , v h ),

|U hn−1 |2 U hn

where P 2n = − |un−1 |2 un . Then, substitute v h = ∂t θ n and compare the real parts to get

∇∂t θ n 20 + ν

∂t θ n 20 − ∂t θ n−1 20 2τ

= −(∂tt ρ n , v h ) + (ν + i α )(∇∂t ρ n , ∇ v h ) + ((κ + i β)∂t P 2n , v h ) + γ (∂t θ n + ∂t ρ n , v h ) + (∂t R n1 , v h ) + (κ + i β)(∂t R n2 , v h ) 

6  i =1

Fi.

(4.32)

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Table 1 Numerical results of u at t = 0.1. m×n

 I h un − U hn 1

Order

un − U n 1

Order

un − I 2h U hn 1

Order

4×4 4×8 16 × 16 32 × 32

1.1899e−02 3.4312e−03 9.0726e−04 2.1697e−04

– 1.7940 1.9191 2.0640

6.5516e−02 3.2922e−02 1.6486e−02 8.2454e−03

– 0.9928 0.9978 0.9996

1.8581e−01 5.8724e−02 1.5395e−02 3.8933e−03

– 1.6618 1.9315 1.9834

By interpolation theory, we have

F 1 + F 4 ≤ Ch4 (untt 22 + unt 22 ) + C ∂t θ n 20 ≤ Ch4 + C ∂t θ n 20 .

(4.33)

From (2.1), we get

F 2 ≤ Ch4 unt 23 + C ∇∂t θ n 20 . Obviously, with Taylor’ formula, we know

(4.34)

∂t R n1 0

+ ∂t R n2 0

= O (τ ), which yields

F 5 + F 6 ≤ C τ 2 + C ∂t θ n 20 .

(4.35)

With (4.10), we deduce that

F3 ≤ C(

= C(

|U hn−1 |2 U hn − |un−1 |2 un − |U hn−2 |2 U hn−1 + |un−2 |2 un−1

τ

, ∂t θ n )

|U hn−1 |2 U hn − |U hn−2 |2 U hn−1 − (|un−1 |2 un − |un−2 |2 un−1 )

τ

, ∂t θ n )

(4.36)

≤ C (∂t θ n 0 + ∂t ρ n 0 + θ n−1 0 + ρ n−1 0 )∂t θ n 0 ≤ Ch4 (unt 22 + un−1 22 ) + C θ n−1 20 + C ∂t θ n 20 . Inserting (4.33)-(4.36) into (4.32) gives

τ

n 

∇∂t θ n 20 + ∂t θ n 20 ≤ Ch4 + C τ 2 ,

(4.37)

i =2

which implies

∇θ n 0 ≤ Ch2 + C τ . 2

(4.38)

It is worthy to emphasize that the technique of taking difference between two time levels used here is vital to our final estimate, without which we can not get the result of order O (h2 ) for ∇θ n 0 . In order to get the global superconvergent result, we construct the interpolation postprocessing operator I 2h as [27] satisfying

I 2h I h u = I h u , u − I 2h u 1 ≤ Ch2 u 3 ,  I 2h v h 1 ≤ C  v h 1 , ∀u ∈ H 3 ( ), v h ∈ V h .

(4.39)

Then with similar argument as [27], we can easily get the following result: Theorem 4. Let um and U hm be the solutions of (1.1) and (2.3) respectively, for m = 1, 2, ..., N, under the conditions of Theorem 3, we have

um − I 2h U hm 1 = O (h2 + τ ).

(4.40)

5. Numerical results In this section, we present a numerical example to illustrate the theoretical analysis. Consider the following problem

ut − (i + 1)u + (i + 1)|u ( X , t )|2 u ( X , t ) − u = g ( X , t ), ∈ × (0, T ],

(5.1) i (t −2x−2 y )

with = [0, 1] × [0, 1]. Then the function g is chosen such that the exact solution is u = e xy (1 − x)(1 − y ). A uniform rectangular partition with m + 1 nodes in each direction is used in our computation. To confirm the convergence, superclose and superconvergent behavior, we choose τ = 2h2 for the fully-discrete scheme. Numerical results of u at different time levels are listed in Tables 1–4. Table 5 is given to suggest that the scheme is stable for large time steps. The error reduction results are described in Figs. 1–2. E 1 , E 2 and E 3 represent the error of  I h un − U hn 1 , un − U hn 1 and un − I 2h U hn 1 respectively. It can be seen that un − U hn 1 is convergent at order O (h) and  I h un − U hn 1 , un − I 2h U hn 1 are convergent at order O (h2 ), which coincide with the theoretical analysis.

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Table 2 Numerical results of u at t = 0.4. m×n

 I h un − U hn 1

Order

un − U hn 1

Order

un − I 2h U hn 1

Order

4×4 8×8 16 × 16 32 × 32

1.2041e−02 3.5491e−03 8.8553e−04 2.0987e−04

– 1.7624 2.0028 2.0771

6.5513e−02 3.2928e−02 1.6484e−02 8.2451e−03

– 0.9925 0.9983 0.9995

1.8632e−01 5.8739e−02 1.5395e−02 3.8930e−03

– 1.6654 1.9318 1.9835

Table 3 Numerical results of u at t = 0.7. m×n

 I h un − U hn 1

Order

un − U hn 1

Order

un − I 2h U hn 1

Order

4×4 8×8 16 × 16 32 × 32

1.2040e−02 3.4881e−03 8.9917e−04 2.1521e−04

– 1.7873 1.9558 2.0628

6.5513e−02 3.2922e−02 1.6485e−02 8.2452e−03

– 0.9928 0.9979 0.9995

1.8632e−01 5.8741e−02 1.5395e−02 3.8929e−03

– 1.6654 1.9319 1.9835

Table 4 Numerical results of u at t = 1.0. m×n

 I h un − U hn 1

Order

un − U hn 1

Order

un − I 2h U hn 1

Order

4×4 8×8 16 × 16 32 × 32

1.2043e−02 3.4816e−03 8.9016e−04 2.1410e−04

– 1.7904 1.9676 2.0557

6.5513e−02 3.2921e−02 1.6484e−02 8.2451e−03

– 0.9928 0.9979 0.9995

1.8632e−01 5.8746e−02 1.5395e−02 3.8924e−03

– 1.6652 1.9320 1.9837

Table 5 Convergence results of un − U hn 1 with h =

1 32

and h = kτ 2 .

t

k=1

k=5

k = 10

k = 20

0.1 0.4 0.7 1

8.2451e−03 8.2451e−03 8.2452e−03 8.2451e−03

8.2451e−03 8.2452e−03 8.2451e−03 8.2452e−03

8.2452e−03 8.2451e−03 8.2451e−03 8.2451e−03

8.2451e−03 8.2452e−03 8.2451e−03 8.2451e−03

Fig. 1. Error reductions for u.

Acknowledgements This research is supported by National Natural Science Foundation of China (Grant No. 11671369).

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Fig. 2. Error reductions for u.

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