Unconditional superconvergence analysis of a two-grid finite element method for nonlinear wave equations

Unconditional superconvergence analysis of a two-grid finite element method for nonlinear wave equations

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

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

Unconditional superconvergence analysis of a two-grid finite element method for nonlinear wave equations Dongyang Shi, Ran Wang ∗ School of Mathematics and Statistics, Zhengzhou University, 450001, China

a r t i c l e

i n f o

Article history: Received 3 July 2019 Received in revised form 5 September 2019 Accepted 18 September 2019 Available online xxxx Keywords: Nonlinear wave equation TGM Nonconforming E Q 1rot element Supercloseness and superconvergence Unconditionally

a b s t r a c t This paper aims to present a two-grid method (TGM) with low order nonconforming E Q 1rot finite element for solving a class of nonlinear wave equations, and to give the superconvergent error analysis unconditionally. Firstly, for the Crank-Nicolson fully discrete scheme, the existence and uniqueness of the numerical solutions are proved, and based on the special characters of this element as well as the priori estimates, the supercloseness and superconvergence analysis of both the original variable u and the auxiliary variable q = ut in the broken H 1 -norm are deduced on the coarse mesh for the Galerkin finite element method (FEM). Then, by employing the interpolation postprocessing approach and the boundness of the numerical solution in the broken H 1 -norm on the coarse mesh, the corresponding superconvergence results of order O (τ 2 + H 4 + h2 ) for the TGM are obtained unconditionally, here τ , H and h denote time step, coarse and fine grid sizes, respectively. Finally, numerical experiments are provided to confirm our theoretical results and effectiveness of the proposed algorithm. © 2019 IMACS. Published by Elsevier B.V. All rights reserved.

1. Introduction We discuss the following nonlinear wave equations with dispersive term:

⎧ u − utt − ut − u = f (u ), (x, t ) ∈  × J , ⎪ ⎨ tt u (x, t ) = 0, (x, t ) ∈ ∂  × J , ⎪ ⎩ u (x, 0) = φ(x), ut (x, 0) = ψ(x), (x, t ) ∈  × {t = 0},

(1.1)

where  ⊂ R2 is a rectangle with boundary ∂  parallel to the coordinate axis, J = (0, T ], 0 < T < ∞, x = (x, y ), f (u ) = −|u |2 u, φ(x), ψ(x) are given smooth functions. The nonlinear wave equation (1.1) describes the spread problems of longitudinal strain waves in the nonlinear elastic rods [26], and there have been a lot of studies about this problem in [25,22,2,16,15]. More precisely, when φ(x), ψ(x) ∈ H 2 () ∩ H 01 (), the existence and uniqueness of global strong solution u ∈ W 2,∞ (0, T ; H 2 ∩ H 01 ()) were proved in [25] u under the condition f (u )u ≤ 0 f (s)ds ≤ 0, ∀u ∈ R. The existence of global attractors in the Hilbert space H 01 () × H 01 () was investigated in [22] when the nonlinear term f (u ) satisfies a critical exponential growth condition lim sup|s|→∞

*

Corresponding author. E-mail address: [email protected] (R. Wang).

https://doi.org/10.1016/j.apnum.2019.09.012 0168-9274/© 2019 IMACS. Published by Elsevier B.V. All rights reserved.

f (s) s



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2

λ, | f (s)| ≤ C 0 (1 + |s| p ), 1 ≤ p ≤ +∞, where C 0 , λ are positive constants and λ < λ1 , λ1 is the first eigenvalue of − in 1, p H 01 () with the Dirichlet boundary condition. A local well posedness result was obtained in the Banach spaces W 0 () × 1, p ρ − 1 ρ − 1 W 0 () for p ∈ (1, ∞) in [2] when f (u ) satisfies | f (s1 ) − f (s2 )| ≤ c |s1 − s2 |(1 + |s1 | + |s2 | ), s1 , s2 ∈ R, for some c > 0 and ρ ∈ (1, ∞). In [16] and [15], the superconvergence results of original variable u in the broken H 1 -norm and the real stress variable q = ∇ ut in H (di v ; ) norm were deduced for a new H 1 -Galerkin mixed FEM when f = f (x, y ) and f (u ) is global Lipschtiz continuous with respect to the variable u, respectively. As we know, the TGM is an efficient algorithm for solving the nonlinear problems [23,24]. The convergence analysis and optimal error estimates of this method have been investigated for the elliptic eigenvalue equations [3,1], the Navier-Stokes equations [10,8,4,5], the hyperbolic equations [21], and the semi-linear integro-differential equations [6], etc. Recently, TGM also has been applied to study the unconditional optimal error estimates [19] and superconvergence behavior of semilinear parabolic equations [13]. In this article, we consider the unconditional superconvergence of the FEM with a TGM for problem (1.1). Through the special characters of nonconforming E Q 1rot element [14], i.e., its interpolation and Ritz projection are equal to each other, and its consistency error can reach O (h2 ), which is one order higher than its interpolation error when the exact solution belongs to H 3 (), the superconvergence estimates for u and q = ut in the broken H 1 -norm are obtained without the requirement of the numerical solutions in L ∞ () norm. In addition, weaker regularities of utt ∈ L 2 ( J ; H 3 ()) and ut ∈ L 2 ( J ; H 3 ()) are required instead of utt ∈ L ∞ ( J ; H 3 ()) ∩ L 2 ( J ; H 5 ()) and ut ∈ L 2 ( J ; H 5 ()) in [15]. The rest of the paper is organized as follows. In section 2, the existence and uniqueness of numerical solutions are proved for the Crank-Nicolson fully discrete scheme, and the superclose estimates of order O (h2 + τ 2 ) in the broken H 1 -norm for u and q are derived without the requirement of numerical solutions U hn , Q hn ∈ L ∞ (). In section 3, we set up the TGM and gain the superclose estimates for u and q in the broken H 1 -norm of order O ( H 2 + τ 2 ) on the coarse grid and of order O (h2 + H 4 + τ 2 ) on the fine grid, respectively. In addition, the global superconvergence of order O (h2 + H 4 + τ 2 ) for the TGM is deduced unconditionally through interpolation post-processing approach. In the last section, a numerical example is given to verify the theoretical analysis and show that the TGM can save much computing cost compared with the traditional Galerkin mixed FEM. Throughout this paper, we equip the classical Sobolev spaces W m, p (), 1 ≤ p ≤ ∞ with the norm · m, p , especially, we simply write W m,2 () as H m () and · m,2 as · m , respectively. At the same time, we equip the space L p ( J ; Y ) with the



p

1

norm ς L p ( J ;Y ) = ( J ς Y dτ ) p and the space H (di v ; ) with the norm ζ H (di v ;) = ( ζ 20 + di v ζ 20 )1/2 . Remarkably, the generic constant C > 0 is independent of the mesh parameters like H and h (coarse and fine mesh sizes), τ (time step) and may be different in different places. 2. Superconvergence analysis for Crank-Nicolson scheme Let Th be a regular rectangular partition of . For any element K ∈ Th , its four vertices and four edges are A i , li = A i A i +1 (mod 4) (i = 1, 2, 3, 4), respectively, h K = diam( K ), h = max h K . K ∈Th

Define the nonconforming E Q 1rot finite element space V h as follows [14]:

 2

2

V h = { v h : v h | K ∈ span{1, x, y , x , y },

[ v h ]ds = 0, F ⊂ ∂ K , ∀ K ∈ Th }, F

where [ v h ] denotes the jump of v h across the edge F if F is an internal edge, and it is equal to v h itself if F belongs to ∂ . The associated interpolation operator I h : v ∈ V = H 1 () → I h v ∈ V h , I h | K = I K , satisfies

⎧ ⎪ ⎪ ⎪ ( I K v − v )ds = 0, i = 1, 2, 3, 4, ⎪ ⎪ ⎨ li  ⎪ ⎪ ⎪ ( I K v − v )dx = 0, ∀ K ∈ Th . ⎪ ⎪ ⎩ K

Then the following important lemma has been proved in [14,17]. Lemma 2.1. For w ∈ H 1 () and v h ∈ V h , there holds

(∇h ( w − I h w ), ∇h v h ) = 0,

(2.1)

v h 0,2k ≤ C v h h , ∀k = 1, 2, · · · .

(2.2)

Moreover, for w ∈ H 3 () and v h ∈ V h , there holds

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|



3

∇ w · nv h ds| = O (h2 )| w |3 v h h ,

(2.3)

K ∂K

where ∇h denotes the gradient operator defined piecewisely, ( 1 , 2 )h =

 K ∈Th

K

1 · 2 dx and the broken H 1 norm · h = (

K ∈Th



1

|21, K ) 2 is a norm on V h , n is the outward unit normal vector on ∂ K . By setting q = ut as a new unknown variable, we can rewrite the problem (1.1) as:

⎧ qt − qt − q − u = f (u ), ⎪ ⎪ ⎪ ⎪ ⎨ q − u t = 0,

(x, t ) ∈  × J , (x, t ) ∈  × J ,

(2.4)

⎪ u (x, t ) = 0, (x, t ) ∈ ∂  × J , ⎪ ⎪ ⎪ ⎩ u (x, 0) = φ(x), q(x, 0) = ψ(x), (x, t ) ∈  × {t = 0}. The weak formulation of (2.4) is to find (u , q) : J → H 01 () × H 01 (), such that for all ( v , w ) ∈ H 01 () × H 01 (),

⎧ (q , v ) + a(qt , v ) + a(q, v ) + a(u , v ) = ( f (u ), v ), (x, t ) ∈  × J , ⎪ ⎨ t (x, t ) ∈  × J , (q, w ) = (ut , w ), ⎪ ⎩ x ∈ , (u (x, 0) − φ(x), v ) = 0, (q(x, 0) − ψ(x), w ) = 0,

(2.5)



where a( w , v ) =  ∇ w ∇ vdx. Let {tn |tn = nτ ; 0 ≤ n ≤ N } be a uniform partition in time with time step

1

1

1

τ , for any function ϕ (x, t ), we denote that

1

ϕ n = ϕ (x, tn ), tn− 1 = (n − )τ , ϕ n− 2 = (ϕ n + ϕ n−1 ), ∂t ϕ n− 2 = (ϕ n − ϕ n−1 )/τ . 2

2

2

The Crank-Nicolson fully discrete scheme of (2.5) is to find (U hn , Q hn ) ∈ V h × V h , (n = 1, 2, · · · , N), such that for all

(vh , wh ) ∈ V h × V h ,

⎧ n− 1 n− 1 n− 1 n− 1 n− 1 ⎪ ⎪ (∂t Q h 2 , v h )h + ah (∂t Q h 2 , v h ) + ah ( Q h 2 , v h ) + ah (U h 2 , v h ) = ( f (U h 2 ), v h )h , ⎪ ⎨ n− 1

n− 1

(2.6)

( Q h 2 , w h )h = (∂t U h 2 , w h )h , ⎪ ⎪ ⎪ ⎩ 0 U h = I h φ(x), Q h0 = I h ψ(x), x ∈ , where ah ( w h , v h ) =



K ∈Th

K

∇h w h ∇h v h dx.

Before deriving the error estimate for this approximate solution in the broken H 1 -norm, we will propose the following three lemmas. Lemma 2.2. For n = 1, 2, · · · , N, system (2.6) has a unique solution (U hn , Q hn ) ∈ V h × V h . Proof. First, we define a operator T : V h × V h → V h × V h : (Y h , P h ) = T ( X h , R h ), for given U hn−1 , Q hn−1 , X h , R h ∈ V h ,

⎧ ∇h P h − ∇h Q hn−1 ∇h P h + ∇h Q hn−1 P h − Q hn−1 ⎪ ⎪ ⎪ , v h )h + ( , ∇h v h )h + ( , ∇h v h )h ( ⎪ ⎪ τ τ 2 ⎪ ⎪ ⎨ ∇h Y h + ∇h U hn−1 X h + U hn−1 + ( , ∇ v ) = ( f ( ), v h )h , h h h ⎪ ⎪ 2 2 ⎪ ⎪ ⎪ n − 1 n − 1 ⎪ ⎪ ⎩ ( Y h − U h , w ) = ( P h + Q h , w ) , ∀( v , w ) ∈ V × V . h h h h h h h h

τ

We set up Y h =

(2.7)

2

k i =1

y i ϕi , P h =

k i =1

p i ϕi . Then (2.7) can be written as

⎧ − → − → τ − → τ − → − → ⎪ ⎨A P +B P + B P + BY = F , ⎪ ⎩

2

− →

AY =

2

τ − → 2

(2.8)

− →

AP + G,

− →

− →

− →

where A = (ϕi , ϕ j ), B = (∇h ϕi , ∇h ϕ j ), Y = ( y 1 , y 2 , · · · , yk )T , P = ( p 1 , p 2 , · · · , pk )T , F = ( F j ), F j = ( f ( qnh−1 ,

ϕ

n−1 j ) + (∇ qh

− → − τ2 (∇ unh−1 + ∇ qnh−1 ), ∇ ϕ j ), G = (G j ), G j = 2t (qnh−1 , ϕ j ) + (unh−1 , ϕ j ), i , j = 1, 2, · · · k.

X h +unh−1 ) 2

τ+

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Arranging (2.8), it follows that

⎧ τ τ2 − − → → − → τ ⎪ ⎨ (A + B + B + B ) P = F − B A −1 G , 2

⎪ ⎩

4

− →

Y =

τ− → 2

2

(2.9)

− → P + A −1 G .

It can be seen that A , B are positive definite matrices, so for given ( X h , R h ), we can solve (Y h , P h ) uniquely. It shows that the operator T is reasonable. Next, we define M 1 = {ω1 | ω1 − U hn h ≤ d1 }, M 2 = {ω2 | ω2 − Q hn h ≤ d2 }, d1 , d2 are two positive constants to be determined. Now we prove that the operator T is mapping to M 1 × M 2 . In fact, the second equation in (2.7) is equivalent to

(

∇h Y h − ∇h U hn−1

, ∇h w h )h = (

τ

∇h P h + ∇h Q hn−1 2

, ∇h w h )h .

(2.10)

Let v h = P h − Q hn−1 in (2.7), w h = τ2 ( P h − Q hn−1 ) in (2.10), then by Schwarz’s inequality and (2.2), we get

τ

P h − Q hn−1 20 + (1 + ≤ τ

X h + U hn−1 2

2

+

τ2 4

) P h − Q hn−1 h2

30,6 P h − Q hn−1 0 + τ ( Q hn−1 h + U hn−1 h ) P h − Q hn−1 h

≤ C τ ( X h − U hn−1 h6 + Q hn−1 h2 + U hn−1 h2 ) +

τ 2

P h − Q hn−1 h2 .

(2.11)

On the other hand, let w h = (Y h − U hn−1 ) in (2.10), we obtain

Y h − U hn−1 h ≤

τ 2

P h − Q hn−1 h + τ Q hn−1 h .

(2.12)

So, when τ is small enough, we can determine d1 , d2 so that (Y h , P h ) ∈ M 1 × M 2 . Finally, we prove that the operator T is a contractive mapping. In fact, for ( X h1 , R h1 ), ( X h2 , R h2 ) ∈ M 1 × M 2 , we put (Y h1 , P h1 ) = T ( X h1 , R h1 ), (Y h2 , P h2 ) = T ( X h2 , R h2 ) into (2.7), and make additive operation, to yield for all ( v h , w h ) ∈ V h × V h ,

⎧ 2 ∇h P h2 − ∇h P h1 ∇h P h2 − ∇h P h1 P − P h1 ⎪ ⎪ ⎪ , v h )h + ( , ∇h v h )h + ( , ∇h v h )h ( h ⎪ ⎪ τ τ 2 ⎪ ⎪ ⎨ ∇h Y h2 − ∇h Y h1 X 2 + U hn−1 X 1 + U hn−1 +( , ∇h v h )h = ( f ( h )− f( h ), v h )h , ⎪ ⎪ 2 2 2 ⎪ ⎪ ⎪ 2 1 2 1 ⎪ ⎪ ⎩ ( Y h − Y h , w h )h = ( P h − P h , w h )h , ∀( v h , w h ) ∈ V h × V h .

τ

(2.13)

2

Let v h = P h2 − P h1 , w h = Y h2 − Y h1 in (2.13), and use similar argument to (2.11) to have

Y h2 − Y h1 h ≤

τ 2

P h2 − P h1 h .

P h2 − P h1 20 + (1 + = τ( f ( ≤ τ ((|

X h2 + U hn−1 2

X h2 + U hn−1 2

τ 2

+

)− f(

|+|

τ2 4

(2.14)

) P h2 − P h1 h2

X h1 + U hn−1 2

X h1 + U hn−1 2

), P h2 − P h1 )h

|)2 |

X h2 − X h1 2

|, | P h2 − P h1 |)

(2.15)

≤ C τ ( X h2 − U hn−1 20,8 + X h1 − U hn−1 20,8 + U hn−1 20,8 ) X h2 − X h1 0,4 P h2 − P h1 0 ≤ C τ X h2 − X h1 h2 +

τ 2

P h2 − P h1 20 .

Arranging (2.14) and (2.15), we get



P h2 − P h1 h2 ≤ C τ X h2 − X h1 h2 , Y h2 − Y h1 h2 ≤ C τ 3 X h2 − X h1 h2 .

(2.16)

It shows that the operator T is a contractive mapping when τ is small enough. According to Brouwer fixed point theorem, there exists a unique solution of system (2.6). The proof is complete. 2

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Lemma 2.3. Let (U hn , Q hn ) ∈ V h × V h (n = 0, 1, 2, · · · , N ) be the solution of system (2.6), then there holds

Q hn h + U hn h ≤ C . n− 12

Proof. Let v h = ∂t U h

1 2τ

(2.17)

in (2.6), we have

( Q hn 20 − Q hn−1 20 + Q hn h2 − Q hn−1 h2 + U hn h2 − U hn−1 h2 ) +

1 8τ

n+ 12 2 h .

((U hn + U hn−1 )2 , |U hn |2 − |U hn−1 |2 ) = − Q h

(2.18)

Let D = 81τ ((U hn + U hn−1 )2 , |U hn |2 − |U hn−1 |2 ), then, when |U hn | ≥ |U hn−1 |, it’s obvious that D ≥ 0. Otherwise, for |U hn | < |U hn−1 |, by employing Young’s inequality, we have

D=

> =

1 8τ 1 4τ 1

(|U hn |2 + |U hn−1 |2 + 2U hn U hn−1 , |U hn |2 − |U hn−1 |2 ) (|U hn |2 + |U hn−1 |2 , |U hn |2 − |U hn−1 |2 ) ( U hn 40,4 − U hn−1 40,4 ).



Therefore, whether |U hn | ≥ |U hn−1 | or |U hn | < |U hn−1 |, summing up (2.18) over n from 1 to j (1 ≤ j ≤ N ), and multiplying by 2τ , we can get j

j

j

Q h 20 + Q h h2 + U h h2 ≤ Q h0 20 + Q h0 h2 + U h0 h2 + U h0 40,4 ,

(2.19)

which together (2.2) yields the desired result. The proof is complete.

2

Lemma 2.4. Let (u , q) be the solution of (2.4), u ∈ L ∞ ( J ; L 8 ()), then for all v h ∈ V h , we have 1

1

1

(∂t qn− 2 , v h )h + ah (∂t qn− 2 , v h ) + ah (qn− 2 , v h ) 1

1

+ah (un− 2 , v h ) = ( f (un− 2 ), v h ) + E n ( v h ),

(2.20)

with

tn 3

| E n ( v h )| ≤ C τ (

tn qtt 21 d

t n −1

tn + Cτ 3

utt 21 dτ ) + C v h h2

τ+ t n −1

( ut 40,8 + utt 20,4 )dτ + Ch4 τ −1

t n −1

tn u 23 dτ .

(2.21)

t n −1

Proof. Integrating (2.5) from tn−1 to tn with respect to time t and dividing by time step integrating over , we get

n− 12

(∂t q

n− 12

, v h ) − (∂t q

, vh) − (

1

tn

q(τ )dτ , v h ) − (

τ t n −1

=(

1

1

tn

u (τ )dτ , v h )

τ t n −1

tn f (u (τ ))dτ , v h ),

τ t n −1

.

τ , then multiplying by v h and

where u (τ ) = u (x, τ ). Then, by means of the Green’s formula, we have

(2.22)

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6

1

1

(∂t qn− 2 , v h ) + ah (∂t qn− 2 , v h ) + ah (

tn

1

q(τ )dτ , v h ) + ah (

τ

=(

1

f (u (τ ))dτ , v h ) +

τ

 K ∂K

t n −1

+

(



(

K ∂K

1

∇ u (τ ) · n dτ ) v h ds

τ

(2.23)

t n −1

∇ q(τ ) · n dτ ) v h ds +

τ

t n −1

tn

tn

1

u (τ )dτ , v h )

τ

t n −1

tn

tn

1



(∇ ∂¯t qn · n) v h ds,

K ∂K

t n −1

which can lead to (2.20), and

1

E n ( v h ) = ah (qn− 2 −

tn

1

1

1

q(τ )dτ , v h ) + ah (un− 2 −

τ

+(

1

t n −1 1

f (u (τ ))dτ − f (un− 2 ), v h ) +

τ

 K ∂K

t n −1

+



(

K ∂K

. =

6 

tn

1

∇ q(τ ) · n dτ ) v h ds +

τ

u (τ )dτ , v h )

τ

t n −1

tn

tn



(

1

tn ∇ u (τ ) · n dτ ) v h ds

τ t n −1

(∇ ∂¯t qn · n) v h ds,

(2.24)

K ∂K

t n −1

Err i .

i =1

Moreover, by Schwarz inequality and Taylor expansion, we can obtain

3

tn

1

qtt 21 + utt 21 dτ ) 2 v h h .

Err1 + Err2 ≤ C τ 2 (( t n −1

Err4 + Err5 + Err6 ≤ Ch

2 − 12

τ

tn ((

1

u 23 + q 23 + qt 23 dτ ) 2 v h h .

t n −1

Err3 ≤ ((

1

tn

τ

1

f (u (τ ))dτ − f (u (tn− 1 ))) + ( f (u (tn− 1 )) − f (un− 2 )), v h ) 2

2

t n −1

≤ Cτ

3 2

tn u 2L ∞ ( J ; L 8 ()) (

1

utt 20,4 dτ ) 2 v h 0

t n −1

+C τ

3 2

tn u 2L ∞ ( J ; L 4 ()) (

1

ut 40,8 dτ ) 2 v h 0 .

t n −1

Hence (2.21) can be proved through Schwarz inequality. The proof is complete.

2

Now we are ready to present the superclose estimate. Theorem 2.1. Let (u , q) and (U hn , Q hn ) be the solutions of (2.5) and (2.6), respectively. Assume that u ∈ L ∞ ( J ; H 2 ()) ∩ L 2 ( J ; H 3 () ∩ H 01 ()), q ∈ L 2 ( J ; H 3 ()), ut ∈ L 2 ( J ; H 3 ()), qt ∈ L 2 ( J ; H 3 ()), utt ∈ L 2 ( J ; H 1 ()), qtt ∈ L 2 ( J ; H 1 ()),

I h un − U hn h + I h qn − Q hn h ≤ C (h2 + τ 2 ).

(2.25)

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7

Proof. Subtracting (2.20) from (2.6), and defining

.

un − U hn = (un − I h un ) + ( I h un − U hn ) = αn + β n ,

.

qn − Q hn = (qn − I h qn ) + ( I h qn − Q hn ) = ηn + ξ n , then we obtain 1

1

1

1

(∂t ξ n− 2 , v h ) + ah (∂t ξ n− 2 , v h ) + ah (ξ n− 2 , v h ) + ah (β n− 2 , v h ) n− 12

1

= ( f (u n − 2 ) − f ( U h − ah (∂t η

n− 12

1

), v h ) + E n ( v h ) − (∂t ηn− 2 , v h )

, v h ) − ah (η

n− 12

, v h ) − ah (α

n− 12

(2.26)

, v h ).

The second formula of (2.6) can be written as 1

1

1

1

ah (ξ n− 2 , w h ) = ah (∂t β n− 2 , w h ) + ah ( I h qn− 2 , w h ) − ah (∂t I h un− 2 , w h ). Thus, let v h = ξ

n− 12

(2.27)

and (2.1), which combines with (2.27), and we get 1

ξ n 20 − ξ n−1 20 + ξ n h2 − ξ n−1 h2 + β n h2 − β n−1 h2 + 2τ ξ n− 2 h2 n− 12

1

= ( f (u n − 2 ) − f ( U h n

−(η − η . =

4 

n −1

n

,ξ + ξ

1

), ξ n + ξ n−1 )τ + 2τ E n (ξ n− 2 ) n −1

) + 2τ ah (∂t I h u

n− 12

n− 12

− Ihq

(2.28)



n− 12

)

Ai .

i =1

Next we estimate the terms on the right-hand side of the equation (2.28). In fact, by Y oung  s inequality and (2.21), we have

tn A 2 + A 3 ≤ Ch4

( u 23 + q 23 + qt 23 )dτ

t n −1

tn +C τ 4

1

( utt 21 + qtt 21 )dτ + ξ n− 2 h2 τ ,

(2.29)

t n −1 1

A 4 ≤ ∇ I h (un − un−1 − τ qn− 2 ) 0 β n + β n−1 h

tn ≤ ∇

1

(ut − qn− 2 )dτ 0 β n + β n−1 h

t n −1

tn ≤ Cτ

4

qtt 21 dτ + C τ ( β n h2 + β n−1 h2 ),

(2.30)

t n −1 1

n− 12

A 1 ≤ 2τ (| f (un− 2 ) − f (U h

1

)|, |ξ n− 2 |)

1 n− 12 2 0,8 ) un− 2

≤ C τ (1 + U h

n− 12

− Uh

1

0 ξ n− 2 0,4

1

1

≤ Ch4 τ un− 2 22 + C τ ( β n−1 h2 + β n h2 ) + ξ n− 2 h2 τ .

(2.31)

Then, inserting above results (2.29)-(2.31) into (2.28) leads to

ξ n 20 − ξ n−1 20 + ξ n h2 − ξ n−1 h2 + β n h2 − β n−1 h2 ≤ Cτ

( β n h2

+ β n−1 h2 ) +

tn Cτ

4

( utt h2 + qtt h2 + utt 20 )dτ

t n −1

tn +Ch4 t n −1

1

( u 23 + q 23 + qt 23 )dτ + Ch4 τ un− 2 22 .

(2.32)

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Summing up (2.32) over n from 1 to m (1 ≤ m ≤ N ) yields

(1 − C τ ) β m h2 + ξ m h2 ≤ ξ 0 20 + β 0 h2 + ξ 0 h2 +Ch4 ( u 2L 2 ( J ; H 3 ()) + q 2L 2 ( J ; H 3 ()) + qt 2L 2 ( J ; H 3 ()) + u 2L ∞ ( J ; H 2 ()) ) +C τ 4 ( utt 2L 2 ( J ; H 1 ()) + qtt 2L 2 ( J ; H 1 ()) ) + C τ

m −1 

β n h2 .

(2.33)

n =0

Noting that ξ 0 , β 0 = 0, chosing a proper τ0 so that when leads to (2.25). The proof is complete. 2

τ ≤ τ0 , 1 − C τ > 0, and applying the discrete Gronwall’s lemma

2 In order to get the global superconvergence, we use interpolated postprocessing operator I 2h constructed in [9], then we can get the following theorem easily.

Theorem 2.2. Under the assumptions of Theorem 2.1, we have 2 2 un − I 2h U hn h + qn − I 2h Q hn h ≤ C (h2 + τ 2 ).

(2.34)

3. Superconvergence analysis for the TGM In this section, we define different E Q 1rot finite element spaces V H on the coarse grid T H , which satisfies V H ⊂ V h (h  H  1). Then we consider the following TGM: 0 STEP 1: On the coarse grid T H , giving a pair (U 0H , Q H ), find (U nH , Q Hn ) ∈ V H × V H (n = 1, 2, · · · , N ), for the following nonlinear system, such that for ( v H , w H ) ∈ V H × V H

⎧ n− 1 n− 1 n− 1 ⎪ (∂t Q H 2 , v H ) H + a H (∂t Q H 2 , v H ) + a H ( Q H 2 , v H ) ⎪ ⎪ ⎪ ⎪ ⎪ n− 1 n− 1 ⎨ + a H (U H 2 , v H ) = ( f (U H 2 ), v H ) H , ⎪ n− 1 n− 1 ⎪ ⎪ ( Q H 2 , w H ) H = (∂t U H 2 , w H ) H , ⎪ ⎪ ⎪ ⎩ 0 0 U H = I H φ(x), Q H = I H ψ(x), x ∈ .

(3.1)

STEP 2: On the fine grid Th , giving a pair (U˜ h0 , Q˜ h0 ), find (U˜ hn , Q˜ hn ) ∈ V h × V h (n = 1, 2, · · · , N ), for the following linear system, such that for ( v h , w h ) ∈ V h × V h

⎧ n− 1 n− 1 n− 1 ⎪ (∂t Q˜ h 2 , v h )h + ah (∂t Q˜ h 2 , v h ) + ah ( Q˜ h 2 , v h ) ⎪ ⎪ ⎪ ⎪ ⎪ n− 1 n− 1 n− 1 n− 1 n− 1 ⎨ + ah (U˜ h 2 , v h ) = ( f (U H 2 ) + f  (U H 2 )(U˜ h 2 − U H 2 ), v h )h , ⎪ n− 1 n− 1 ⎪ ⎪ ( Q˜ h 2 , w h )h = (∂t U˜ h 2 , w h )h , ⎪ ⎪ ⎪ ⎩ 0 U˜ = I h φ(x), Q˜ 0 = I h ψ(x), x ∈ . h

(3.2)

h

Now, we will give the following unconditional superconvergent estimates of the above TGM. n ) and (U˜ hn , Q˜ hn ) be the solutions of (2.5), (3.1) and (3.2), respectively. For n = 1, 2, · · · , m (1 ≤ m ≤ Theorem 3.1. Let (u , q), (U nH , Q H N ) satisfying the conditions of Theorem 2.1, we have

I H un − U nH H + I H qn − Q Hn H ≤ C ( H 2 + τ 2 ), n

Ih u −

U˜ hn h

n

+ Ihq −

Q˜ hn h

2

4

(3.3)

2

≤ C (h + H + τ ).

(3.4)

Proof. Due to Theorem 2.1, (3.3) is true obviously. We only need to prove (3.4). In fact, by Taylor expansion, there holds 1

n− 12

f (u n − 2 ) = f (u H where

γ=

n− 1 uH 2

+ θ(u

n− 12

) + f  (u H

n− 12

n− 1 − u H 2 ),

1

n− 12

)(un− 2 − u H

(0 < θ < 1) .

1

n− 12 2

) + f  (γ )(un− 2 − u H

) /2 , (3.5)

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. n ˜n n . ˜ + β , q − Q˜ hn = qn − I h qn + I h qn − Q˜ hn = η˜ n + ξ˜ n to have Then, we split un − U˜ hn = un − I h un + I h un − U˜ hn = α 1

1

1

1

(∂t ξ˜ n− 2 , v h ) + ah (∂t ξ˜ n− 2 , v h ) + ah (ξ˜ n− 2 , v h ) + ah (β˜n− 2 , v h ) n− 12

1

= ( f (u n − 2 ) − ( f ( U H − (∂t η˜

n− 12

n− 12

) + f  (U H

, v h ) − ah (∂t η˜

n− 12

n− 12

n− 12

)(U˜ h

, v h ) − ah (η˜

− UH

n− 12

), v h ) + 2τ E n ( v h ) 1

, v h ) − ah (α˜ n− 2 , v h ).

(3.6)

Due to (3.5), n− 12

1

( f (u n − 2 ) − ( f ( U H =

1 n− 1 ( f  (U H 2 )(un, 4

Thus, let v h = ξ˜

n− 12



n− 12

) + f  (U H n, 1 UH 4 ) +

n− 12

)(U˜ h

f  (γ )(U

n− 12

− UH

n− 12

)), v h )

n− 1 − u H 2 )2 /2, v h ).

(3.7)

and combine with the second equation of (3.2), and we get 1

ξ˜ n 20 − ξ˜ n−1 20 + ξ˜ n h2 − ξ˜ n−1 h2 + β˜n h2 − β˜n−1 h2 + 2τ ξ˜ n+ 2 h2 n− 12

= ( f  (U H

n− 12

)(un− 2 − U˜ h 1

n− 12 2

1

), ξ˜ n + ξ˜ n−1 )τ + ( f  (γ )(un− 2 − U H

1

1

1

) , ξ˜ n+ 2 )τ 1

1

+2τ E n (ξ˜ n− 2 ) − (η˜ n − η˜ n−1 , ξ˜ n + ξ˜ n−1 ) + 2τ ah (∂t I h un− 2 − I h qn− 2 , β˜n− 2 ) .  = Bi. 5

(3.8)

i =1

The error estimates of B i (i = 3, 4, 5) are similar to that in the proof of Theorem 2.1, i.e., 5 

tn B i ≤ Ch

4

i =3

( u 23 + q 23 + qt 23 )dτ

t n −1

tn +C τ 4

( utt h2 + qtt h2 + utt 20 )dτ

t n −1 1

+τ ξ˜ n− 2 h2 + C τ ( β˜n h2 + β˜n−1 h2 ). 1

(3.9)

4

As for B 1 , B 2 , since H () → L (), by interpolation theory, we have 1 n− 12 2 0,8 un− 2

1

n− − U˜ h 2 0 ξ˜ n + ξ˜ n−1 0,4

B 1 ≤ C τ u H

≤ Ch4 τ u 2L ∞ ( J ; H 2 ()) + C τ ( β˜n h2 + β˜n−1 h2 ) + n− 12

1

B 2 ≤ C τ ( un− 2 0,4 + u H 1

1

n− 12 2 0,4 )2

≤ C τ (H

≤ C ( H 8 + τ 8 )τ +

1 τ + H + τ ) + ξ˜ n− 2 h2

4

(3.10)

1 n− 12 2 0,4 ξ˜ n− 2 0,4

1

1

1 un− 2 22,4

2

0,4 ) un− 2 − u H

≤ C τ ( un− 2 − I H un− 2 20,4 + I H un− 2 − u H 4

τ ˜ n− 1 2 ξ 2 h .

+

τ ˜ n− 1 ξ 2 h 2

4 2

τ ˜ n− 1 2 ξ 2 h .

2

(3.11)

2

Then, substituting (3.9)-(3.11) into (3.8), yields

ξ˜ n 20 − ξ˜ n−1 20 + ξ˜ n h2 − ξ˜ n−1 h2 + β˜n h2 − β˜n−1 h2 tn ≤ Ch

4

( u 23 + q 23 + qt 23 )dτ + Ch4 τ u 2L ∞ ( J ; H 2 ())

t n −1

tn + Cτ 4

( utt h2 + qtt h2 + utt 20 )dτ

t n −1

+ C τ ( β˜n h2 + β˜n−1 h2 ) + C τ ( H 8 + τ 8 ).

(3.12)

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Summing up (3.12) over n from 1 to m (1 ≤ m ≤ N ) with ξ˜ 0 = 0 and β˜ 0 = 0, yields

(1 − C τ ) β˜m h2 + ξ˜ m h2 ≤ C τ

m −1 

β˜n h2 + C (h4 + H 8 + τ 4 ),

(3.13)

n =0

choosing a proper proof is complete.

τ0 so that when τ ≤ τ0 , 1 − C τ > 0, and then applying the discrete Gronwall’s lemma leads to (3.4). The 2

2 Based on Theorem 2.1, Theorem 3.1 and by use of interpolated postprocessing operator I 2h , we can get the following unconditional global superconvergence easily.

Theorem 3.2. Under the assumptions of Theorem 2.1, we have 2 ˜n 2 ˜n un − I 2h U h h + qn − I 2h Q h h ≤ C (h2 + H 4 + τ 2 ).

(3.14)

Remark 3.1. We point out that since it is hard to get priori estimates of U hn , the inverse inequality is usually used to get the boundedness of U hn 0,∞ in the previous studies of FEM analysis, e.g., when U hn − h un 0 = O (h2 + τ ) (see [20]) and U hn − h un 0 = O (h2 + τ 2 ) (see [12]), there holds

U hn 0,∞ ≤ U hn − h un 0,∞ + h un 0,∞ ≤ Ch−1 U hn − h un 0 + h un 0,∞ , where h denotes a certain interpolation operator or projection operator. In order to ensure the boundness of U hn 0,∞ , it’s inevitably to limit the ratio between the time step τ and the space parameter h, which leads to τ = O (h1+γ ) and τ = O (h1/2+γ ) (γ > 0), respectively. But in our present work, based on the boundness of the numerical solution in the broken H 1 -norm, the unconditional superconvergence is obtained. Remark 3.2. The results obtained herein are also valid to the case when f (u ) = −|u |2r u , r ∈ { Z | 0 < r < ∞}. In this situation, the term B 1 in (3.10) can be estimated as 1 n− 12 2r 0,8r un− 2

B 1 ≤ C τ u H

1

n− − U˜ h 2 0 ξ˜ n + ξ˜ n−1 0,4

≤ Ch4 τ u 2L ∞ ( J ; H 2 ()) + C τ ( β˜n h2 + β˜n−1 h2 ) +

τ ˜ n− 1 2 ξ 2 h . 2

(3.15)

Remark 3.3. Our analysis presented herein is also applicable to some other popular finite elements satisfying (2.2) such as the conforming linear triangular element [18], the nonconforming element C N Q 1rot [7] on rectangular meshes, and Q 1rot element [11] on square meshes. 4. Numerical experiments In this section, we consider the following example:

⎧ qt − qt − q − u = f (u ), ⎪ ⎪ ⎪ ⎪ ⎨ q − u t = 0,

(x, t ) ∈  × J , (x, t ) ∈  × J ,

⎪ u (x, t ) = 0, (x, t ) ∈ ∂  × J , ⎪ ⎪ ⎪ ⎩ u (x, 0) = φ(x), q(x, 0) = ψ(x), (x, t ) ∈  × {t = 0},

(4.1)

where,  is a unit square region [0, 1] × [0, 1], J = [0, 1], f (u ) = −|u |2 u, g (x, t ), φ(x) and ψ(x) could be calculated by the exact solution

u (x, y , t ) = e −t sin(π x) sin(π y ), q(x, y , t ) = −e −t sin(π x) sin(π y ). We set τ 2 = h2 , H 4 = h2 , and use Newton iterations with 1e−10 as the tolerance error to solve nonlinear system (4.1). 2 ˜n 2 ˜n Tables 4.1–4.2 show that the errors of un − I 2h U h h and qn − I 2h Q h h are convergent to order O (h2 ) at different time

level, respectively, Table 4.3 shows that the errors of I h un − U˜ hn h and I h qn − Q˜ hn h are convergent to order O (h2 ) at t = 0.5, which match with the theoretical analysis of Theorems 3.1 and 3.2. 1 In order to reveal the unconditional stability, we get the errors of un − U˜ hn h and qn − Q˜ hn h with H = 18 , h = 128 and

τ = kh in different time. From the numerical results in Table 4.4 and 4.5, we can observe that the errors of un − U˜ hn h and

qn − Q˜ hn h tend to be a constant as τh → 0, respectively, which implies the time-restrictions are not necessary. Fig. 4.1(a)

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Table 4.1 The superconvergence errors at t = 0.5. H

h

2 ˜n un − I 2h U h h

1/2 1/4 1/6 1/8

1/4 1/16 1/36 1/64

2.5990e−01 1.7250e−02 3.4182e−03 1.0821e−03

Order

2 ˜n qn − I 2h Q h h

Order

1.9566 1.9961 1.9991

2.6060e−01 1.7360e−02 3.4410e−03 1.0894e−03

1.9540 1.9957 1.9989

Order

2 ˜n qn − I 2h Q h h

Order

1.9581 1.9961 1.9990

1.5998e−01 1.0678e−02 2.1169e−03 6.7027e−04

1.9526 1.9955 1.9988

Order

I h qn − Q˜ hn h

Order

1.8660 1.9678 1.9815

3.1308e−02 2.0192e−03 4.0260e−04 1.2795e−04

1.9773 1.9885 1.9923

Table 4.2 The superconvergence errors at t = 1. H

h

2 ˜n un − I 2h U h h

1/2 1/4 1/6 1/8

1/4 1/16 1/36 1/64

1.5907e−01 1.0537e−02 2.0880e−03 6.6105e−04

Table 4.3 The supercloseness errors at t = 0.5. H

h

I h un − U˜ hn h

1/2 1/4 1/6 1/8

1/4 1/16 1/36 1/64

4.2846e−03 3.2247e−04 6.5386e−05 2.0910e−05

Table 4.4 Errors of un − U˜ hn h with H =

1 , 8

h=

1 128

and

τ = kh.

t

k=1

k=4

k=8

k = 16

0.25 0.5 0.75 1

1.2258e−02 9.5475e−03 7.4541e−03 5.9156e−03

1.2258e−02 9.5467e−03 7.4485e−03 5.8986e−03

1.2259e−02 9.5475e−03 7.4369e−03 5.8462e−03

1.2275e−02 9.6061e−03 7.5138e−03 5.8315e−03

Table 4.5 Errors of qn − Q˜ hn h with H =

1 , 8

h=

1 128

and

τ = kh.

t

k=1

k=4

k=8

k = 16

0.25 0.5 0.75 1

1.2260e−02 9.5975e−03 7.7315e−03 6.6420e−03

1.2259e−02 9.5981e−03 7.7431e−03 6.6824e−03

1.2259e−02 9.5962e−03 7.7569e−03 6.7489e−03

1.2258e−02 9.5836e−03 7.7773e−03 6.9271e−03

and Fig. 4.1(c) are the figures of numerical solution (U˜ h , Q˜ h ), while Fig. 4.1(b) and Fig. 4.1(d) are the figures of exact solution (u , q). We compare the CPU cost of the TGM with the conventional Galerkin mixed FEM in Table 4.6 on the same space mesh size h = 1/36 but at different time levels. It can be seen clearly that the computing time required for the TGM is less than that of the mixed FEM. Therefore, the TGM of our paper is indeed a very effective algorithm for solving the nonlinear problem (1.1).

Table 4.6 Errors and CPU cost of FEM and TGM. tn

0.2 0.4 0.6 0.8 1

FEM

TGM

2 ˜n un − I 2h U h h , 2 ˜n qn − I 2h Q h h

CPU (s)

2 ˜n un − I 2h U h h , 2 ˜n qn − I 2h Q h h

CPU (s)

4.6132e−03, 3.7772e−03, 3.0938e−03, 2.5370e−03, 2.0860e−03,

83.2652 157.8789 222.6819 297.3213 370.3174

4.6132e−03, 3.7772e−03, 3.0941e−03, 2.5379e−03, 2.0880e−03,

31.4646 58.9715 84.4391 110.5724 135.8005

4.6179e−03 3.7922e−03 3.1192e−03 2.5680e−03 2.1138e−03

4.6186e−03 3.7941e−03 3.1220e−03 2.5712e−03 2.1169e−03

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Fig. 4.1. The picture of numerical solution and exact solution.

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