A two-grid discontinuous Galerkin method for a kind of nonlinear parabolic problems

Applied Mathematics and Computation 346 (2019) 96–108

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

A two-grid discontinuous Galerkin method for a kind of nonlinear parabolic problemsR Jiming Yang a,∗, Xiaoqing Xing b a b

College of Science, Hunan Institute of Engineering, Xiangtan, Hunan 411104, China School of Mathematical Sciences, South China Normal University, Guangzhou, Guangdong 510631, China

a r t i c l e

i n f o

MSC: 65M12 65M60 Keywords: Two-grid Nonlinear problems Discontinuous Galerkin method Convergence estimate

a b s t r a c t A discontinuous Galerkin approximation for the space variables with the backward Euler time discretisation for a kind of parabolic problems with the nonlinear diffusion, convection and source terms is investigated. To solve the strongly nonlinear algebra system, a two-grid method is proposed. With this algorithm, solving a nonlinear system on the fine discontinuous finite element space is reduced into solving a nonlinear problem on a coarse gird of size and solving a linear problem on a fine grid of size. Convergence estimates in H1 -norm are obtained. The numerical experiments are provided to confirm our theoretical analysis. © 2018 Elsevier Inc. All rights reserved.

1. Introduction The parabolic equations are often used to described the processes in science and technology, such as, fluid dynamics, hydrology and environmental protection. In this work, we consider the following nonlinear parabolic problem in R2 :



ut − ∇ · (a(u )∇ u ) + b(u )∇ u = f (u, ∇ u ), u(x, t ) = 0, (x, t ) ∈ ∂  × (0, T ], u(x, 0 ) = u0 (x ), x ∈ ,

(x, t ) ∈  × (0, T ], (1.1)

where b(u ) = (b1 (u ), b2 (u ))T is a vector function; ut = ∂∂ut ;  is a bounded, convex, polygonal domain with the smooth boundary ∂ . The equations describe the diffusion of a chemical species of concentration u in a porous medium with a source term f. It is assumed that a(u), b(u) are bounded smooth functions with positive upper and lower bounds, 0 < α ∗ ≤ a(u) ≤ α ∗ , b(u)∞ ≤ β ∗ , −∇ · (b(u )) > 0. Assume that a(u), bi (u )(i = 1, 2 ) are uniformly Lipschitz continuous with respect to u. f is twice continuously differentiable with derivatives up to second order bounded above by F. It is also assumed that u0 is smooth enough. Under the above assumptions, the model problem (1.1) has a unique solution in a certain Sobolev space [21].

R This work was supported by Project funded by National Natural Science Foundation of China (Grant No. 11671159), Hunan Provincial Natural Science Foundation of China, Guangdong Provincial Engineering Technology Research Center for Data Science, Guangdong Natural Science Foundation (Grant No. 2016A030313842). ∗ Corresponding author at: College of Science, Hunan Institute of Engineering, Xiangtan, Hunan 411104, China. E-mail addresses: [email protected] (J. Yang), [email protected] (X. Xing).

https://doi.org/10.1016/j.amc.2018.09.067 0 096-30 03/© 2018 Elsevier Inc. All rights reserved.

J. Yang, X. Xing / Applied Mathematics and Computation 346 (2019) 96–108

97

Recently, discontinuous Galerkin methods (DG) are very popular for solving the partial differential equations (see [14,17,19,20,24–29]) because of their attractive properties, such as the local mass conservation, the easiness for adaptivity, the flexibility in handling of complicated geometry, and high order accuracy. Two-grid method was firstly introduced as a discretization method by Xu [22,23] for nonsymmetric, indefinite and nonlinear partial differential equations. The two-grid method is effective for these problems. In [1], a two-grid approximation for the SIPG method for quasilinear elliptic PDE is presented. The hp-analysis for quasilinear elliptic PDEs and quasi-Newtonian fluid flows is given in [8,9]. Nonlinear parabolic equations are solved using the two-grid method with finite difference and mixed finite element methods by Dawson and Wheeler [10,11]. For the nonlinear parabolic problem (1.1), Chen and Liu [7] have construct the two-grid finite element method. In [2,3,6], two-grid finite volume element methods for semilinear and nonlinear parabolic problems are analysed. In [4,5], Chen et al. have studied two-grid methods with expanded mixed finite element solutions for nonlinear parabolic problems. Error estimates for discontinuous Galerkin method for the nonlinear parabolic problem (1.1) are given in [12,15,18]. But there is little literature about the error analysis of the twogrid discontinuous Galerkin method for the nonlinear parabolic problem (1.1). In [30], the error analysis of a semi-discrete two-grid discontinuous Galerkin method for nonlinear parabolic equations has been carried out. In this paper, we shall continue the study of [30] to obtain a error estimate in H1 norm of a full-discrete two-grid discontinuous Galerkin method for the proposed problem (1.1). The difference between [30] and this work lies in some cases: (i) the equations are different, the problem in [30] can be considered as a special case of our investigated problem; (ii)the difficulties which encountered are different, we handle with the nonlinear diffusion, convection and source terms; (iii)the discretisation forms are different, we discretize the time variable by implicit Euler. The paper is organized as follows. In Section 2, we introduce a full-discrete discontinuous Galerkin method. Error estimates of the discontinuous Galerkin approximation are presented in Section 3. Section 4 displays a two-grid discontinuous Galerkin method and its error analysis. The fifth section is the numerical experiments. The last part is the conclusions. 2. A full-discrete discontinuous Galerkin method 2.1. Notation Let Th = {E } be a quasi-uniform partitions of , with E being a triangle or quadrilateral with the diameter hE . Denote by  h the set of all interior edges for Th and by n the outward unit normal vector on each edge γ ∈  h ∪ ∂ . Let h = max hE be the maximal element diameter over all elements with the common edge γ ∈  h ∪ ∂ . For s ≥ 0, we define the following broken Sobolev space

E∈Th

H s (Th ) = {v ∈ L2 () : v|E ∈ H s (E ), E ∈ Th }. For v ∈ H s (Th ), s > 1/2, { · } and [ · ] denote the trace operators for the average and the jump [13], respectively. Let Ei ∈ Th , E j ∈ Th and γ = ∂ Ei ∩ ∂ E j ∈ h with n oriented from Ei to Ej . Define

{v } =

 1 ( v|Ei )|γ + ( v|E j )|γ , 2

[v ] = ( v | E i ) | γ − ( v | E j ) | γ .

If γ is on the boundary ∂ , i. e. γ = ∂ Ei ∩ ∂ , then n is defined as the unit outward vector normal to γ , and the average and the jump on γ that belong to ∂  is modified as:

{v} = [v] = (v|Ei )|γ . Let Lp () be the standard Banach space defined on , with norm  · L p . We denote by Wm,p () the standard Sobolev  p space (m partial derivatives in Lp ) with norm  · m,p given by vm,p = Dα vLpp . When p = 2, we let H m () = W m,2 (), α ≤m

 · m =  · m,2 and  ·  =  · 0,2 . The usual L2 inner product ( · , · ) is used for the domain , the element E and the edge γ. We shall use the following discontinuous finite element space:

Vh = {v ∈ L2 () : v|E ∈ Pr (E ), E ∈ Th }, where Pr (E) denotes the space of polynomials of total degree less than or equal to r on E. Like [29], throughout the paper, we shall use K, Ki (i ∈ N) to denote generic positive constants that are independent of h, but might depend on r and the solution of PDEs with different values at different occurrences. Let ε denote a fixed positive constant that can be chosen arbitrarily small. 2.2. The full-discrete discontinuous Galerkin method To solve the problem (1.1), the following discontinuous Galerkin method is considered. First, we introduce the trilinear forms B(ω; u, v ) and Bλ (ω; u, v ):

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J. Yang, X. Xing / Applied Mathematics and Computation 346 (2019) 96–108

B(ω; u, v )

=

Bλ (ω; u, v )

=

  E∈Th

E

a(ω )∇ u · ∇vdx −

+J σ (u, v ), B(ω; u, v ) + λ(u, v ),

 where λ is a positive number, J σ (u, v ) =

γ ∈h

r 2 σγ hγ

  γ ∈h



γ {a (ω )∇ u · n}[v]ds −

  γ ∈h

γ {a (ω )∇v · n}[u]ds

γ [u][v]ds is the interior penalty term,

σ is a positive real function taking

constant value σ γ on the edge γ and is bounded, hγ is the size of γ . The weak form of discontinuous Galerkin approximating for the problem (1.1) reads:

(ut , v ) + B(u; u, v ) + (b(u )∇ u, v ) = ( f (u, ∇ u ), v ) v ∈ H01 ().

(2.1)

Let N be a positive integer. t = NT , t n = n t, n = 0, 1, . . . , N. Denote unh = uh (t n ). The full-discrete discontinuous Galerkin approximation to (1.1), i. e. the discontinuous Galerkin approximation for the space variables and the backward Euler method for the time discretisation, is to find unh ∈ Vh for n = 0, 1, . . . , N such that



unh −unh−1 t ,

 v + B(unh ; unh , v ) + (b(unh )∇ unh , v ) = ( f (unh , ∇ unh ), v ), v ∈ Vh ,

(2.2)

u0h = u0 , where u0 is an appropriate projection of u0 to be defined later. 3. Error estimates of the discontinuous Galerkin approximation In order to proceed the theoretical analysis, we need to state some important results. Lemma 3.1 (approximation properties [16]). Assume that u ∈ Hs () for s ≥ 2 and let r ≥ 2 and assume that a is a given positive constant. Then there exists an interpolant u ∈ Vh of u and a positive constant K satisfying



γ

{a∇ ( u − u ) · n}ds = 0, ∀γ ∈ h ,

 u − u∞,E ≤ K

hμ us,E , ∀E ∈ Th , r s−1

∇ ( u − u )E ≤ K

hμ−1 us,E , ∀E ∈ Th , r s−1

∇ 2 ( u − u )E ≤ K  u − uE ≤ K

hμ−2 us,E , ∀E ∈ Th , r s−2

hμ us,E , ∀E ∈ Th , r s−1

(3.1)

(3.2)

(3.3)

(3.4)

(3.5)

where μ = min(r + 1, s ). Moreover, for γ = ∂ Ei ∩ ∂ E j ,

∇ u∞,γ ≤ K ∇ u∞,Ei ∪E j .

(3.6)

Lemma 3.2 (Trace theorem). Suppose that  has a Lipschitz boundary, p is a real number with 1 ≤ p < ∞. Then there is a constant K, s. t. 1− 1

1

p vL p (∂ ) ≤ K vL p () vWp 1,p () , ∀v ∈ W 1,p ().

Lemma 3.3. For each E ∈ Th , let γ be an edge of E. Then there exists a positive constant K such that the following trace inequalities are valid: 2 2 1 v2γ ≤ K (h−1 E vE + hE |v|1,E ), ∀v ∈ H (E ), 2 2 2 ∇v · n2γ ≤ K (h−1 E |v|1,E + hE |v|2,E ), ∀v ∈ H (E ).

From [12,15], we have the following lemma. Lemma 3.4. Let u ∈ H r+1 () for r ≥ 1. Let u ∈ Vh be an interpolant of u satisfying B(u; u − u, v ) = 0, v ∈ Vh . There exists a constant K satisfying

u − u ≤ Khr+1 , u − u1 ≤ Khr ,

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99

ut − ut  ≤ Khr+1 , ut − ut 1 ≤ Khr . Theorem 3.1. Let un ∈ H r+1 () ∩ H01 (), r ≥ 1 and unh be the solutions of the problem (1.1) and the full-discrete discontinuous Galerkin approximation (2.2), respectively. If u0h = u0 , then for 1 ≤ n ≤ N,

un − unh  ≤ K (hr+1 + t ), where the positive constant K may depend on u and ∂∂ut . n n−1 Proof. Denote ∂t ξ n = ξ − ξt . For v ∈ Vh , we get the error equation from (2.1) and (2.2)

(utn − ∂t unh , v ) + Bλ (un ; un , v ) − Bλ (unh ; unh , v ) + (b(un )∇ un , v ) − (b(unh )∇ unh , v ) = ( f (un , ∇ un ) − f (unh , ∇ unh ), v ) + λ(un − unh , v ), v ∈ Vh .

(3.7)

Note that un − unh = (un − un ) + ( un − unh ) = ηn + ξ n . Taking v = ξ n , we have

(∂t ξ n , ξ n ) + Bλ (unh ; ξ n , ξ n ) = (∂t un − utn , ξ n ) − (∂t ηn , ξ n ) + Bλ (unh ; un , ξ n ) n n n n n − ((b(u ) − b(uh ))∇ u , ξ ) − (b(uh )∇ (un − unh ), ξ n ) − Bλ ( un ; un , ξ n ) − Bλ ( un ; η n , ξ n ) n + ( f (u , ∇ un ) − f (unh , ∇ unh ), ξ n ) + λ(un − unh , ξ n ).

(3.8)

Note that



ξ n − ξ n−1 n ,ξ t  1  n = (ξ − ξ n−1 , ξ n + ξ n−1 ) + (ξ n − ξ n−1 , ξ n − ξ n−1 ) 2 t

(∂t ξ n , ξ n ) =

1 (ξ n − ξ n−1 , ξ n + ξ n−1 ) 2 t 1 = (ξ n 2 − ξ n−1 2 ). 2 t ≥

(3.9)

Now, we need to bound the right items of (3.8). Applying the Cauchy-Schwarz inequality, we have

|(∂t un − utn , ξ n )| ≤ ∂t un − utn  · ξ n 

tn

uttn dt )2 + K ξ n 2  t n ≤K uttn 2 dt t + K ξ n 2 . ≤ K(

t n−1

(3.10)

t n−1

| − (∂t ηn , ξ n ) t | ≤ ∂t ηn  · ξ n  t ≤ K (∂t ηn 2 + ξ n 2 ) t

≤K

tn

t n−1

ηtn 2 dt + K ξ n 2 t.

(3.11)

Note that Bλ (un ; ηn , ξ n ) = Bλ (un ; un − un , ξ n ) = 0 by Lemma 3.4. And, using the definition of the trilinear form Bλ (ω; u, v ) and the interior penalty term J σ (u, v ), we have

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J. Yang, X. Xing / Applied Mathematics and Computation 346 (2019) 96–108

Bλ (unh ; un , ξ n ) − Bλ ( un ; un , ξ n ) − Bλ ( un ; η n , ξ n )



= (a(unh ) − a(un ))∇ un · ∇ξ n dx − E∈Th

−

γ ∈h γ

E



γ ∈h γ

{(a(unh ) − a(un ))∇ un · n}[ξ n ]ds

{(a(unh ) − a(un ))∇ξ n · n}[ un ]ds

≡ I1 + I2 + I3 . For I1 , I2 , I3 , we have

|I1 | ≤ ∇ u∞ a(unh ) − a(un )∇ξ n  ≤ K |∇ un |∞ unh − un  · ξ n 1 ≤ K (ηn 2 + ξ n 2 ) + εξ n 21 , |I2 | ≤

γ ∈h

≤K

K ∇ un ∞,γ {unh − un }γ [ξ n ]γ

({ηn }γ + {ξ n }γ )[ξ n ]γ

γ ∈h

≤K

(h−1/2 ηn E + h1/2 ∇ηn E + h−1/2 ξ n E ) · h−1/2 ξ n E

E∈Th



≤ K h−1 ηn  · hξ n 1 + ∇ηn  · hξ n 1 + h−1 ξ n  · hξ n 1

|I3 |



≤ K (ηn 2 + h2 ∇ηn 2 + ξ n 2 ) + εξ n 21 ,

=| {(a(unh ) − a(un ))∇ξ n · n}[ηn ]| γ ∈h γ

≤K

γ ∈h

≤K

∇ξ n ∞,γ {unh − un }γ [ηn ]γ h−1/2 ∇ξ n γ ({ηn }γ + {ξ n }γ )[ηn ]γ

γ ∈h

≤K

h−1 ∇ ξ n E (h−1/2 ηE + h1/2 ∇ ηn E + h−1/2 ξ n E )

E∈Th

·(h−1/2 ηn E + h1/2 ∇ηn E )

≤K h−2 ∇ ξ E (h−1/2 ηn E + h1/2 ∇ ηn E + h−1/2 ξ n E )h2 un 2 E∈Th

≤ K (ηn 2 + h2 ∇ηn 2 + ξ n 2 ) + εξ n 21 , where Lemma 3.3, the boundedness of ∇ u in (3.6) of Lemma 3.1, and the Lipschitz continuity of a(u) with respect to u are used. By virtue of Lemma 3.2 and the boundedness of ∇ un and b( · ), we obtain

| − ((b(un ) − b(unh ))∇ un , ξ n )| ≤ b(un ) − b(unh )∇ un ∞ ξ n  ≤ K un − unh ∇ un ∞ ξ n  ≤ K ηn + ξ n ∇ un ∞ ξ n  ≤ K ξ n 2 + K ηn ξ n  ≤ K ξ n 2 + K η n 2 , n n n n | − (b(uh )∇ (u − uh ), ξ )| = |(b(unh )∇ (ξ n + ηn ), ξ n )| ≤ b(unh )∞ ∇ξ n  · ξ n  + b(unh )∞ (∇ηn , ξ n ) ≤

β ∗ ξ n 1 ξ n  + β ∗



ηn ξ n · nds

γ ∈∂ 

≤ K ξ

n

1 ξ  + K ηn L2 (∂ ) ξ n L2 (∂ ) n

1

1

≤ K ξ n 1 ξ n  + K ηn  2 ηn 12 ξ n  2 ξ n 12 1

1

1

≤ K ξ n 1 ξ n  + K ηn  2 ηn 12 h 2 ξ n 1 1

1

≤ K ξ n 2 + εξ n 21 + Khηn ηn 1

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101

≤ K ξ n 2 + εξ n 21 + K ηn 2 + Kh2 ηn 21 . Using the Taylor expansions and the boundedness assumptions on f(u, ∇ u), we get that

( f (un , ∇ un ) − f (unh , ∇ unh ), ξ n ) ≤ (F (un − unh ), ξ n ) + (F (∇ un − ∇ unh ), ξ n ) = (F ξ n , ξ n ) + (F ηn , ξ n ) + (F ∇ξ n , ξ n ) + (F ∇ηn , ξ n ) ≡

4

Ti .

(3.12)

i=1

To estimate T1 − T4 , we proceed as follows.

|T1 | ≤ K ξ n 2 , |T2 | ≤ K (ηn 2 + ξ n 2 ), |T3 | ≤ εξ n 21 + K ξ n 2 , |T4 | ≤ K ηn 2 + K h2 ηn 21 + K ξ n 2 . And,

|λ(un − unh , ξ n )|

≤

K η n 2 + K ξ n 2 .

For a sufficiently large σ , we have

Bλ (unh ; ξ n , ξ n )



= a(unh )∇ ξ n · ∇ ξ n dx − 2 E∈Th

≥ α∗

∇ ξ n 2E − δ

E∈Th

γ ∈h

E∈Th

≥ K1

γ ∈h γ

E

{a(unh )∇ ξ n · n}[ξ n ]ds + Jσ (ξ n , ξ n ) + λ(ξ n , ξ n )

{∇ ξ n }2γ − δ −1

γ ∈h

[ξ n ]2γ + Jσ (ξ n , ξ n ) + λ(ξ n , ξ n )

δ −1 σ n n ∇ξ n 2E + (1 − )J ( ξ , ξ ) + λ ( ξ n , ξ n ) σ

≥ K0 ξ n 1 ,

(3.13)

where δ is a positive number from the Cauchy’s inequality with δ . Multiplying by 2 t and summing Eq. (3.8) from n = 1 to n = l (1 ≤ l ≤ N ), collecting all the above estimates and using ξ (0 ) = 0, we have

ξ l 2 + 2K0

l

ξ n 21 t ≤ K ( t )2

0

n=1

+K

l

tl

ηn 2 t + Kh2

n=1

+K

l



uttn 2 dt + K

tl 0

l

ηtn 2 dt

ηn 21 t

n=1

ξ n 2 t + 5ε

n=1

l

ξ n 21 t.

(3.14)

n=1

Choosing ε small enough and applying the discrete Gronwall’s lemma, we get

ξ l 2 +

l

ξ n 21 t ≤ K ( t )2 + K

n=1



tl 0

ηtn 2 dt + K

l

ηn 2 t + Kh2

n=1

l

ηn 21 t.

(3.15)

n=1

Combining the above inequality with Lemma 3.4, we obtain

ξ l 2 ≤ K ( t )2 + K h2(r+1) . The triangle inequality and Lemma 3.4 yield the desired result.

(3.16) 

Remark 1. By Theorem 3.1, we can find that the numerical scheme is only first order in t. To balance the spatial error and the temporal error, one would take t = O(hr+1 ), which is a restriction to the backward Euler method. So, in the proof of the following theorems, we choose t = O(hr+1 ) is reasonable. Theorem 3.2. Let un ∈ H r+1 () ∩ H01 (), r ≥ 1 and unh be the solutions of the problem (1.1) and the full-discrete discontinuous Galerkin approximation (2.2), respectively. Let t = O(hr+1 ). If u0h = u0 , then for 1 ≤ n ≤ N,

un − unh 1 ≤ K (hr + t ),

(3.17)

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J. Yang, X. Xing / Applied Mathematics and Computation 346 (2019) 96–108

where the positive constant K may depend on u and ∂∂ut . Proof. Choosing v = ∂t ξ n in Eq. (3.7) and using the notations of ξ n and ηn , we get

(∂t ξ n , ∂t ξ n ) + Bλ (unh ; ξ n , ∂t ξ n ) = (∂t un − utn , ∂t ξ n ) − (∂t ηn , ∂t ξ n ) + Bλ (unh ; un , ∂t ξ n ) n n n n n n − Bλ (u ; u , ∂t ξ ) − Bλ (u ; η , ∂t ξ ) − ((b(un ) − b(unh ))∇ un , ∂t ξ n ) − (b(unh )∇ (un − unh ), ∂t ξ n ) + ( f (un , ∇ un ) − f (unh , ∇ unh ), ∂t ξ n ) + λ(un − unh , ∂t ξ n ). Note that



1

Bλ (unh ; ξ n , ∂t ξ n ) =

Bλ (unh ; ξ n + ξ n−1 , ξ n − ξ n−1 ) + Bλ (unh ; ξ n − ξ n−1 , ξ n − ξ n−1 )

(3.18)



2 t 1 ≥ B (un ; ξ n + ξ n−1 , ξ n − ξ n−1 ) 2 t λ h  1  = Bλ (unh ; ξ n , ξ n ) − Bλ (unh ; ξ n−1 , ξ n−1 ) 2 t  1 − B (un ; ∂t ξ n , ξ n ) − Bλ (unh ; ξ n , ∂t ξ n ) . 2 λ h

By the definition of the trilinear form Bλ ( · ; · , · ), we have

|Bλ (unh ; ∂t ξ n , ξ n ) − Bλ (unh ; ξ n , ∂t ξ n )|

=|−

γ

γ ∈h

+

γ ∈h

γ

{a(unh )∇∂t ξ n · n}[ξ n ]ds −

{a( )∇ξ · n}[∂t ξ ]ds + unh

n

n

γ ∈h

γ ∈h

γ

γ

{a(unh )∇ξ n · n}[∂t ξ n ]ds

{a(unh )∇∂t ξ n · n}[ξ n ]ds|

= 0. Then,





1 Bλ (unh ; ξ n , ∂t ξ n ) = 2 Bλ (unh ; ξ n , ξ n ) − Bλ (unh ; ξ n−1 , ξ n−1 ) . t

(3.19)

Similarly to (3.10), we get

|(∂t un − utn , ∂t ξ n )| ≤ ∂t un − utn  · ∂t ξ n   t n ≤K uttn 2 dt t + ε∂t ξ n 2 , t n−1

| − (∂t ηn , ∂t ξ n ) t | ≤ ∂t ηn  · ∂t ξ n  t

≤K

tn

t n−1

ηtn 2 dt + ε∂t ξ n 2 t.

It is easy to find that

Bλ (unh ; un , ∂t ξ n ) − Bλ (un ; un , ∂t ξ n ) − Bλ (un ; ηn , ∂t ξ n )



= (a(unh ) − a(un ))∇ un · ∇∂t ξ n dx − {(a(unh ) − a(un ))∇ un · n}[∂t ξ n ]ds E∈Th

−

γ ∈h γ

E



γ ∈h

γ

{(a(unh ) − a(un ))∇∂t ξ n · n}[ un ]ds

≡ J1 + J2 + J3 . For J1 , J2 , J3 , using Theorem 3.1, we have

|J1 | ≤ ∇ u∞ a(unh ) − a(un )∇∂t ξ n  ≤ K |∇ un |∞ unh − un  · ∂t ξ n 1 ≤ Kh−1 unh − un  · ∂t ξ n  ≤ Kh−2 unh − un 2 + ε∂t ξ n 2 ≤ K (hr + h−1 t )2 + ε∂t ξ n 2 ,

|J2 | ≤ K ∇ un ∞,γ {unh − un }γ [∂t ξ n ]γ γ ∈h

(3.20)

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103

≤ K {unh − un }γ [∂t ξ n ]γ ≤ Kh−1 unh − un  · ∂t ξ n 

|J3 |

≤ K (hr + h−1 t )2 + ε∂t ξ n 2 ,

=| {(a(unh ) − a(un ))∇∂t ξ n · n}[ηn ]| γ ∈h γ

≤K

γ ∈h

≤K

γ ∈h

≤K

∇∂t ξ n ∞,γ {unh − un }γ · [ηn ]γ h−1/2 ∇∂t ξ n γ {unh − un }γ · [ηn ]γ

h−1 ∂t ξ n 1,E h−1/2 {unh − un }E · (h−1/2 ηn E + h1/2 ∇ηn E )

E∈Th

≤ Kh−2 ∂t ξ n 1 (hr+1 + t )h2 un 2 ≤ Kh−1 ∂t ξ n (hr+1 + t ) ≤ K (hr + h−1 t )2 + ε∂t ξ n 2 , where the inverse inequalities, Lemma 3.3, (3.6) of Lemma 3.1 and the boundedness of ∇ u are used. By virtue of Lemma 3.2, Theorem 3.1, and the boundedness of ∇ un and b( · ), we obtain

| − ((b(un ) − b(unh ))∇ un , ∂t ξ n )| ≤ b(un ) − b(unh )∇ un ∞ ∂t ξ n  ≤ K un − unh ∇ un ∞ ∂t ξ n  ≤ K un − unh 2 + ε∂t ξ n 2 ≤ K (hr+1 + t )2 + ε∂t ξ n 2 , | − (b(unh )∇ (un − unh ), ∂t ξ n )| = |(b(unh )∇ (ξ n + ηn ), ∂t ξ n )| ≤ b(unh )∞ ∇ξ n  · ∂t ξ n  + b(unh )∞ ∇ηn  · ∂t ξ n  ≤ β ∗ ξ n 1 ∂t ξ n  + β ∗ ∇ηn  · ∂t ξ n  ≤

K2 n 2 ξ 1 + ε∂t ξ n 2 + K ηn 21 . 2

Similarly to (3.12), by using Theorem 3.1 and the Young’s inequality, we have

( f (un , ∇ un ) − f (unh , ∇ unh ), ∂t ξ n ) ≤ (F (un − unh ), ∂t ξ n ) + (F (∇ un − ∇ unh ), ∂t ξ n ) ≤ K un − unh 2 + ε∂t ξ n 2 + (F ∇ξ n , ∂t ξ n ) + (F ∇ηn , ∂t ξ n ) ≤ K (h2(r+1) + ( t )2 ) + ε∂t ξ n 2 + K ηn 21 +

K2 n 2 ξ 1 . 2

(3.21)

And,

|λ(un − unh , ∂t ξ n )| ≤ K un − unh 2 + ε∂t ξ n 2 ≤ K (h2(r+1) + ( t )2 ) + ε∂t ξ n 2 . Multiplying by t and summing Eq. (3.18) from n = 1 to n = l (3.13), we have

ξ l 21 + 2K0

l

∂t ξ n 2 t ≤ K ( t )2

n=1



tl 0

uttn 2 dt + K

+K (hr + h−1 t )2 + 9ε



l

0

(1 ≤ l ≤ N ), collecting all the above estimates and using tl

ηtn 2 dt + K

l

ηn 21 t

n=1

∂t ξ n 2 t + K (h2(r+1) + ( t )2 )

n=1

+K2

l

ξ n 21 t.

n=1

Choosing ε small enough, using Lemma 3.4 and the discrete Gronwall lemma, we obtain

ξ l 21 +

l  n=1

∂t ξ n 2 t

≤

K ( t )2 + K h2(r+1) + K (hr + h−1 t )2 .

Let t = O(hr+1 ), combining the triangle inequality with Lemma 3.4, we finish the proof.

(3.22) 

104

J. Yang, X. Xing / Applied Mathematics and Computation 346 (2019) 96–108

Remark 2. If 2 < q < ∞, from (3.17) and similar to the proof of Theorem 3.4 in [22], we have

un − unh 1,q ≤ K (hr + t ).

(3.23)

Furthermore,

un − unh 0,q ≤ Khun − unh 1,q ≤ K (hr+1 + t ).

(3.24)

4. A two-grid discontinuous Galerkin method and its error analysis Here, we shall present a two-grid algorithm of the discontinuous Galerkin approximation (2.2) to solve the problem (1.1). We involve a nonlinear solver on the coarse-grid space and a linear solver on the fine grid. The two-grid discontinuous Galerkin method is given as follows. Algorithm 1. Step 1. On the coarse grid, find unH , such that



unH − unH−1 , vH t



+ B(unH ; unH , vH ) + (b(unH )∇ unH , vH ) = ( f (unH , ∇ unH ), vH ),

vH ∈ VH ,

u0H = u0 . Step 2. On the fine grid, find unh , such that



unh − unh−1

t



, vh

+ B(unH ; unh , vh ) + (b(unH )∇ unh , vh ) = ( f (unH , ∇ unH ) + f1 (unH , ∇ unH )(unh − unH )

+ f2 (unH , ∇ unH )(∇ unh − ∇ unH ), vh ), u0h

vh ∈ Vh ,

= u0 ,

where fi (·, · ) denotes the partial derivative with respect to the ith variable. Theorem 4.1. Let un ∈ H r+1 () ∩ H01 (), r ≥ 1 and unh be the solutions of the problem (1.1) at t = t n and Algorithm 1, respectively. If u0h = u0 , then

un − unh 1 ≤ K (hr + H r+1 + t ),

(4.25)

where the positive constant K may depend on u and ∂∂ut . Proof. For v ∈ Vh , we get the error equation from (1.1) and Algorithm 1

(utn − ∂t unh , v ) + Bλ (un ; un , v ) − Bλ (unH ; unh , v ) + (b(un )∇ un , v ) − (b(unH )∇ unh , v )



= ( f (un , ∇ un ) − ( f (unH , ∇ unH ) + f1 (unH , ∇ unH )(unh − unH ) + f2 (unH , ∇ unH )(∇ unh − ∇ unH )), v ) + λ(un − unh , v ).

Using the notations of ηn and ξ n in Section 3, choosing v = ∂t ξ n in the above equation, we get

(∂t ξ n , ∂t ξ n ) + Bλ (unH ; ξ n , ∂t ξ n ) = (∂t un − utn , ∂t ξ n ) − (∂t ηn , ∂t ξ n ) + Bλ (unH ; un , ∂t ξ n ) − Bλ ( un ; un , ∂t ξ n ) − Bλ (un ; ηn , ∂t ξ n ) n − ((b(u ) − b(unH ))∇ un , ∂t ξ n ) − (b(unH )∇ (un − unh ), ∂t , ξ n )

+ ( f (un , ∇ un ) − f (unH , ∇ unH ) − f1 (unH , ∇ unH )(unh − unH )

− f2 (unH , ∇ unH )(∇ unh − ∇ unH ), ∂t ξ n ) + +λ(un − unh , ∂t ξ n ).

(4.26)

Similarly to (3.20) in Section 3, we have

Bλ (unH ; un , ∂t ξ n ) − Bλ (un ; un , ∂t ξ n ) − Bλ (un ; ηn , ∂t ξ n )



= (a(unH ) − a(un ))∇ un · ∇∂t ξ n dx − {(a(unH ) − a(un ))∇ un · n}[∂t ξ n ]ds E∈Th

−

γ ∈h γ

E



γ ∈h γ

{(a(unH ) − a(un ))∇∂t ξ n · n}[ un ]ds

≡ L1 + L2 + L3 . For L1 , L2 , by using the inverse inequalities, the boundedness of ∇ u, (3.6) of Lemma 3.1, and the Lipschitz continuity of a(u) with respect to u, with Theorem 3.1, we have

J. Yang, X. Xing / Applied Mathematics and Computation 346 (2019) 96–108

105

|L1 | ≤ ∇ u∞ a(unH ) − a(un )∇∂t ξ n  ≤ K |∇ un |∞ unH − un  · ∂t ξ n 1 −1 ≤ Kh unH − un  · ∂t ξ n  ≤ Kh−2 unH − un 2 + ε∂t ξ n 2 ≤ K (h−1 H r+1 + h−1 t )2 + ε∂t ξ n 2 ,

|L2 | ≤ K ∇ un ∞,γ {unH − un }γ [∂t ξ n ]γ γ ∈h

≤ K {unH − un }γ [∂t ξ n ]γ ≤ Kh−1 unH − un  · ∂t ξ n  ≤ K (h−1 H r+1 + h−1 t )2 + ε∂t ξ n 2 , Note that ηn = un − un and [un ] = 0. For L3 , combining Theorem 3.1 with the inverse inequalities, Lemma 3.1, Lemma 3.3, and the Lipschitz continuity of a(u) with respect to u, we get

|L3 | = |



γ ∈h γ

≤K

{(a(unH ) − a(un ))∇∂t ξ n · n}[ηn ]|

∇∂t ξ n ∞,γ {unH − un }γ · [ηn ]γ

γ ∈h

≤K

h−1/2 ∇∂t ξ n γ {unH − un }γ · [ηn ]γ

γ ∈h

≤K

h−1 ∂t ξ n 1,E h−1/2 {unH − un }E · (h−1/2 ηn E + h1/2 ∇ηn E )

E∈Th

≤ Kh−2 ∂t ξ n 1 (H r+1 + t )h2 un 2 ≤ Kh−1 ∂t ξ n (H r+1 + t ) ≤ K (h−1 H r+1 + h−1 t )2 + ε∂t ξ n 2 , Similarly to the estimates in Section 3, with Theorem 3.1 and the boundedness of ∇ un and b( · ), we have

| − ((b(un ) − b(unH ))∇ un , ∂t ξ n )| ≤ b(un ) − b(unH )∇ un ∞ ∂t ξ n  ≤ K un − unH ∇ un ∞ ∂t ξ n  ≤ K un − unH 2 + ε∂t ξ n 2 ≤ K (H r+1 + t )2 + ε∂t ξ n 2 , n n n n | − (b(uH )∇ (u − uh ), ∂t ξ )| = |(b(unH )∇ (ξ n + ηn ), ∂t ξ n )| ≤ K ξ n 21 + ε∂t ξ n 2 + K ηn 21 . Note that a Taylor expansion yields



f (un , ∇ un ) − f (unH , ∇ unH ) − f1 (unH , ∇ unH )(unh − unH ) − f2 (unH , ∇ unH )(∇ unh − ∇ unH )





= f (unH , ∇ unH ) + f1 (unH , ∇ unH )(un − unH ) + f2 (unH , ∇ unH )(∇ un − ∇ unH )



+ f11 (un − unH )2 + 2 f12 (un − unH )(∇ un − ∇ unH ) + f22 (∇ un − ∇ unH )T (∇ un − ∇ unH )



− f (unH , ∇ unH ) − f1 (unH , ∇ unH )(unh − unH ) − f2 (unH , ∇ unH )(∇ unh − ∇ unH )



= f1 (unH , ∇ unH )(un − unh ) + f2 (unH , ∇ unH )(∇ un − ∇ unh )



+ f11 (un − unH )2 + 2 f12 (un − unH )(∇ un − ∇ unH ) + f22 (∇ un − ∇ unH )T (∇ un − ∇ unH ), where

f11 =

f12 =

f22 =



1 0 1

0





(1 − s ) f11 (uns , ∇ uns )ds,

1 0



(1 − s ) f12 (uns , ∇ uns )ds,

(1 − s ) f22 (uns , ∇ uns )ds,

with

uns = unH + s(un − unH ), ∇ uns = ∇ unH + s(∇ un − ∇ unH ).

106

J. Yang, X. Xing / Applied Mathematics and Computation 346 (2019) 96–108 Table 1 Convergence rates of different mesh size H with r = 1. h

H

H1 error

rate

1/4 1/16 1/64 1/256

1/2 1/4 1/8 1/16

1.51e − 3 3.83e − 4 9.68e − 5 2.41e − 5

1.98 1.98 2.01

Table 2 Convergence rates of different mesh size H with r = 2. h

H

H1 error

rate

1/4 1/16 1/64 1/256

1/2 1/4 1/8 1/16

2.87e − 4 3.64e − 5 4.63e − 6 5.61e − 7

2.98 2.97 2.99

Then, by using the boundedness assumption on f, (3.23), (3.24) and the Holder’s inequality, we get



( f (un , ∇ un ) − f (unH , ∇ unH ) − f1 (unH , ∇ unH )(unh − unH ) − f2 (unH , ∇ unH )(∇ unh − ∇ unH ), ∂t ξ n ) ≤ K un − unh 21 + ε∂t ξ n 2 + K (un − unH 2 + ∇ (un − unH )40,4 + un − unH 40,4 ) ≤ K (ξ n 21 + ηn 21 ) + K (H 2(r+1) + ( t )2 + H 4r + H 4(r+1) ) + ε∂t ξ n 2 . Note that

|λ(un − unh , ∂t ξ n )| ≤ K un − unh 2 + ε∂t ξ n 2 ≤ K (ξ n 2 + ηn 2 ) + ε∂t ξ n 2 . Multiplying by t and summing Eq. (4.26) from n = 1 to n = l (3.13) and (3.19), we obtain

ξ l 21 + 2K0

l

∂t ξ n 2 t ≤ K ( t )2

n=1



tl 0

uttn 2 dt + K



(1 ≤ l ≤ N ), collecting all the above estimates and using tl

0

+ K (h−1 H r+1 + h−1 t )2 + K

ηtn 2 dt + 6ε

l

∂t ξ n 2 t

n=1 l

(ηn 21 + ξ n 21 ) t

n=1

+ K (H 2(r+1) + ( t )2 + H 4r + H 4(r+1) ). Choosing ε small enough, applying Lemma 3.4 and the Gronwall’s lemma, we get that if r ≥ 1, then

ξ l 1 ≤ K (hr + H r+1 + t ). Combining the triangle inequality with Lemma 3.4, we obtain (4.25), which completes the proof.

(4.27) 

Remark 3. Theorem 4.1 indicates that the optimal convergence rate in the full-discrete two-grid discontinuous Galerkin r+1 approximation using the r-order discontinuous finite element space can be achieved if h = O(H r ). 5. Numerical experiments Here, we shall solve the problem (1.1) by using the proposed two-grid discontinuous Galerkin method. We take a(u ) = u, b1 (u ) = b2 (u ) = u. The initial value, boundary value, and f(u, ∇ u) are decided so that the exact solution of (1.1) is u(x, t ) = e−t x(1 − x )y(1 − y ) for (x, y ) ∈  = (0, 1 ) × (0, 1 ). The simulation time is T = 1.0. To solve this problem, we use the discontinuous Galerkin method with σγ = 10.0 and with the piecewise linear and quadratic discontinuous finite element space (r = 1 and r = 2). Let H and h be the space steps of the coarse grid and the fine grid, respectively. Let the time step be t = H r+1 . Tables 1 and 2 show convergence rates of our two-grid discontinuous Galerkin method with different mesh size H. We also give the errors and CPU time of the two-grid discontinuous Galerkin method and the standard discontinuous Galerkin method in Tables 3 and 4. It can be seen that the two-grid discontinuous Galerkin method spends less time than the standard discontinuous Galerkin method to achieve a good accuracy from the above data. Our proposed two-grid discontinuous Galerkin algorithm is effective. The numerical results coincide with the theoretical analysis.

J. Yang, X. Xing / Applied Mathematics and Computation 346 (2019) 96–108

107

Table 3 H1 errors and CPU time of the standard discontinuous Galerkin method (SDG) and the two-grid discontinuous Galerkin method (Two-grid DG) with r = 1. SDG h 1/9 1/16

H 1/3 1/4

Two-grid DG H1 error 5.69e − 3 3.78e − 4

CPU time (s) 22.43 74.42

H1 error 5.79e − 3 3.83e − 4

CPU time (s) 1.08 2.01

Table 4 H1 errors and CPU time of the standard discontinuous Galerkin method (SDG) and the two-grid discontinuous Galerkin method (Two-grid DG) with r = 2. SDG h 1/8 1/64

H 1/4 1/16

H1 error 3.57e − 5 5.51e − 7

Two-grid DG CPU time (s) 48.26 2992.35

H1 error 3.64e − 5 5.61e − 7

CPU time (s) 2.17 92.04

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