A new approach to convergence analysis of linearized finite element method for nonlinear hyperbolic equation

R E S E A R C H

Open Access

A new approach to convergence analysis of

linearized finite element method for

nonlinear hyperbolic equation

Junjun Wang

1*

_{and Lijuan Guo}

1

*_{Correspondence:}

[email protected]

1_{School of Mathematics and} Statistics, Pingdingshan University, Pingdingshan, P.R. China

Abstract

We study a new way to convergence results for a nonlinear hyperbolic equation with bilinear element. Such equation is transformed into a parabolic system by setting the original solutionuasut=q. A linearized backward Euler finite element method (FEM)

is introduced, and the splitting skill is exploited to get rid of the restriction on the ratio betweenhand

τ

. The boundedness of the solutions about the time-discrete system inH2-norm is proved skillfully through temporal error. The spatial error is derived without the mesh-ratio, where some new techniques are utilized to deal with the problems caused by the new parabolic system. The final unconditional optimal error results ofuandqare deduced at the same time. Finally, a numerical example is provided to support the theoretical analysis. Herehis the subdivision parameter, and

τ

is the time step.

Keywords: Nonlinear hyperbolic equation; Parabolic system; Bilinear element; Linearized FEM; Optimal error results

1 Introduction

Consider the following nonlinear hyperbolic equation:

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

utt–∇ ·(a(u)∇u) =f(u), (X,t)∈Ω×(0,T],

u= 0, (X,t)∈∂Ω×(0,T], u(X, 0) =u0(X), ut(X, 0) =u1(X), X∈Ω,

(1)

whereΩ⊂R2is a rectangle with boundary∂Ωparallel to the coordinate axes, 0 <T<∞, X= (x,y), anda(u) andf(u) are known smooth functions onR, for which we assume that 0 <a0≤a(u)≤a1.

A nonlinear hyperbolic equation is a kind of important problems on nonlinear vibration, the permeation fluid mechanics, and so on. Indeed, such partial differential equations (PDEs) have attracted lots of attention to various methods, especially numerical meth-ods. For example, the two-grid method was studied for solving a type of nonlinear hyper-bolic equations, and the error estimate inH1-norm was deduced in [1]. Newton’s modified method was utilized to a nonlinear wave equation depending on different norms of the ini-tial conditions in [2], and optimal error results were given in theL2_{- andH}1_{-norms. The}

interpolation theory and integral identity skill were used to obtain a superclose result for the nonlinear hyperbolic equations with nonlinear boundary condition in [3]. Moreover, the global superconvergence was also obtained through the interpolated postprocessing technique. The Galerkin alternating-direction method was applied to a three-dimensional nonlinear hyperbolic equation in [4], and the error estimates in theH1- andL2-norms were deduced. A mixed FEM was discussed in [5] and [6], and optimal error estimates were derived.

The inverse inequality is usually employed to discuss the boundedness of numerical so-lution Un

h in a nonlinear evolution equation, and such an issue usually results in some

time-step restrictions, such asτ =O(h),hr=O(τ) (1≤r≤k+ 1,k≥0), andτ=O(h2_{) in} [4] and [6], respectively. To get rid of such a restriction, [7,8] took advantage of a special inequality for getting unconditional superclose results for nonlinear Sobolev equations. In [9] a corresponding time-discrete system to split the error into two parts, the temporal error and the spatial error, is introduced. Then the spatial error leads to the unconditional boundedness of a numerical solution in theL∞-norm. Subsequently, this so-called split-ting technique was also applied to the other nonlinear parabolic type equations in [10–18]. Later, in [19] and [20] a second-order scheme for the nonlinear hyperbolic equation and the unconditional superconvergence analysis by using the splitting skill were given. It can be seen that constructing a linearized form for a nonlinear hyperbolic equation is not an easy task in comparison with nonlinear parabolic equations. In fact, there are lots of lit-erature referring to parabolic equations [21–24]. In [24] a special technique to change sine-Gordon equation into a parabolic system throughut=qwas used, and optimal order

error estimates of the Crank–Nicolson fully discrete scheme were obtained.

Inspired by [24], in this paper, we consider the unconditional convergent estimates for (1), which is a much more general nonlinear model than that in [24], with a bilinear ele-ment. First of all, we change a nonlinear hyperbolic equation into a nonlinear parabolic system. Such a practice can be used to avoid the difficulty in constructing a linearized scheme for a nonlinear hyperbolic equation and also give the error analysis foruandq=ut

at the same time. Then we develop a linearized backward Euler FE scheme for the non-linear parabolic system and apply the idea of splitting technique in [10–20] to split the error into the temporal and spatial errors. We obtain a temporal error, which implies the regularities of the solutions about the time-discrete equations. The spatial error result is exploited to get rid of the restriction on the ratio between handτ. The unconditional optimal error results ofuandqare simultaneously deduced. Note that, differently from [17,18], we utilize some new tricks such as rewriting some error terms, the new mean-value technique, and some other skills to handle new difficulties brought by the special nonlinear parabolic system during the process. Further, the results in this paper also hold for linear conforming triangular elements but do not hold for some other particular ele-ments; for example, the biquadratic finite element forvh|k= 0 cannot be true, wherevh

belongs to the FE space. Some numerical results in the last section also show the validity of the theoretical analysis.

Throughout this paper, we denote the natural inner production inL2_{(Ω) by (·,·) and} the norm by · 0, and letH01(Ω) ={u∈H1(Ω) :u|∂Ω= 0}. Further, we use the classical Sobolev spacesWm,p(Ω), 1≤p≤ ∞, denoted byWm,p, with norm·m,p. Whenp= 2, we

simply write·m,pas·m. Besides, we define the spaceLp(a,b;Y) with normfLp_(a,b;Y)=

2 Conforming FE approximation scheme

LetΩbe a rectangle in the (x,y) plane with edges parallel to the coordinate axes, and let

Γh be a regular rectangular subdivision. GivenK∈Γh, let the four vertices and edges be

ai,i= 1∼4, andli=aiai+1,i= 1∼4 (mod4), respectively. LetVh be the usual bilinear

FE space, and letVh0={vh∈Vh,vh|∂Ω= 0}. Also, it can be found in [25] that ifu∈H2(Ω), then

∇(u–Ihu),∇vh

= 0, vh∈Vh0, (2)

whereIhbe the so-called Ritz projection operator onVh0.

Set {tn:tn=nτ; 0≤n≤N} be a uniform partition of [0,T] with time step τ =T/N.

We denoteσn_=σ(X,t

n). For a sequence of functions{σn}Nn=0, we denote∂¯tσn=σ n_–σn–1

τ , n= 1, 2, . . . ,N. With these notations, settingut=q, the weak form of (1) is seekingu,q∈

H1

0(Ω) such that, for allv∈H01(Ω),

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

(∂¯tun,v) = (qn,v) + (Rn1,v), v∈H01(Ω), (∂¯tqn,v) + (a(un–1)τ ni=1∇qi,∇v) + (a(un–1)∇u0,∇v)

= (f(un–1_{),v) + (R}n

2+Rn3+Rn4,v), v∈H01(Ω),

(3)

where

Rn₁=∂¯tun–unt, R2n=∂¯tqn–qnt, Rn4= –

fun–1–fun, Rn₃= –∇ ·

aun–1τ n

i=1

∇qi–aun

tn

0 ∇q ds

–∇ ·∇u0aun–1–aun.

We develop the linearized Galerkin FEM to problem (3): seekUn

h,Qnh∈Vh0such that

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

(∂¯tU_hn,vh) = (Q_hn,vh), vh∈Vh0, (∂¯tQnh,vh) + (a(Uhn–1)τ

n

i=1∇Qih,∇vh)

+ (a(Un–1

h )∇Uh0,∇vh) = (f(Uhn–1),vh), vh∈Vh0,

(4)

whereU0

h=Ihu0andQh0=Ihu1. A well-known consequence is that the linear system (4)

may always be solved forU_hnandQn_h; see [26].

3 Error estimates for the time-discrete system

To get rid of the ratio restriction betweenhandτ, we introduce a time-discrete system as follows:

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

¯

∂tUn=Qn, (X,t)∈Ω,

¯

∂tQn–∇ ·(a(Un–1)τ ni=1∇Qi) –∇ ·(a(Un–1)∇U0)

=f(Un–1), (X,t)∈Ω,

Un_{= 0, Q}n_{= 0, (X,t)∈∂Ω,}

U0=u0(X), Q0=u1(X), X∈Ω.

The existence and uniqueness of solutions for this linear elliptic system (5) are obvious. To show the unconditional results, the regularities ofUn_andQn_{are inevitable, and we}

therefore need some estimates forun_–Un_andqn_–Qn_{. In what follows, we sete}n_un_–Un_, δnqn_–Qn_{(n= 1, 2, . . . ,N), analyze the temporal errors and give the regularity results}

forUnandQn. It is easy to see thate0=δ0= 0.

Theorem 1 Let um and Um (m= 0, 1, 2, . . . ,N)be solutions of(1)and(5), respectively, u,q∈L2_(0,T;H3_(Ω)),u

t,qt ∈L∞(0,T;H2(Ω)),and utt∈L2(0,T;L2(Ω)).Then for m=

1, . . . ,N,there existsτ0such that whenτ≤τ0,we have

em₂+τ

_m

i=2

∂¯tei

2 2

1 2

+δm₁+τ

m

i=1

δi

2 +τ

_m

i=2

δi2₂

1 2

≤C0τ, (6)

∂¯tUm₂+Qm₂≤C0, (7)

where C0is a positive constant independent of m,h,andτ.

Proof SettingK01 +max1≤m≤N(um0,∞+ √

τ( m_i=1¯∂tui20,∞)

1

2), we begin to prove (6)–

(7) by mathematical induction. Whenm= 1, by (1) and (5) we have the error equation

⎧ ⎨ ⎩

¯

∂te1=δ1+R11, ¯

∂tδ1–∇ ·(a(u0)τ∇δ1) =R12+R13.

(8)

Withδ0= 0, multiplying the second equation of (8) byδ1and integrating it overΩ, we get

1

τ∇δ

12 0+τa

1

2_u0_δ12

0= –

au

u0∇u0τ∇δ1,δ1–R1₂+R1₃,δ1

≤Cτ∇δ1₀δ1₀+Cτδ1₀. (9)

Further, sincee1_∈H2_(Ω)∩H1

0(Ω), using the first equation of (8), we get

_e1

2=τ∂¯te 1

2≤Cτδ 1

0+CτR 1

12. (10)

Thus there exist positive constantsτ1,τ2,C1,C2such that whenτ≤τ1, we have

e1₂+τ∂¯te1₂+δ1₁+τδ1₂≤C1τ, (11)

which implies

U1–_τU0

2

+Q1₂≤C2, (12)

U1_0,∞≤e1_0,∞+u1_0,∞≤CC1τ+u1_0,∞≤K0, (13)

By mathematical induction we assume that (6) and (7) hold form≤n– 1. Then there existsτ3such that

Um_0,∞+√τ

_m

i=1

∂¯tUi

2 0,∞

1 2

≤Cem₂+C√τ

_m

i=1

∂¯tei

2 2

1 2

+um_0,∞+√τ

_m

i=1

∂¯tui

2 0,∞

1 2

≤CC0τ+CC0 √

τ+um_0,∞+√τ

_m

i=1

∂¯tui

2 0,∞

1 2

≤K0, (14)

whereτ≤τ3=min{1/2CC0, 1/4C2C20}.

Then we begin to prove (6) and (7) form=n. Subtracting (5) from (1), we obtain

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

¯

∂ten=δn+Rn1, ¯

∂tδn–∇ ·(a(Un–1)τ ni=1∇δi) –∇ ·(τ

n

i=1∇qi(a(un–1) –a(Un–1))) –∇ ·(∇u0_(a(un–1_{) –a(U}n–1₎₎₎

=f(un–1_{) –f(U}n–1_{) +R}n

2+Rn3+Rn4.

(15)

Multiplying the second equation of (15) byδn_{and integrating, we get}

1 2τ∇δ

n2

0–∇δ

n–12 0

+

aU_hn–1τ n

i=1

δi,δn

= –

au

Un–1∇Un–1

τ n

i=1 ∇δi

,δn

–

∇ ·

τ n

i=1

∇qiaun–1–aUn–1

,δn

–∇ ·∇u0aun–1–aUn–1,δn

–fun–1–fUn–1,δn–Rn₂+Rn₃+Rn₄,δn. (16)

Observe that (a(Un–1_)τ n

i=1δi,δn) cannot be bounded directly; we rewrite it as

aUn–1τ n

i=1

δi,δn

=τ

Ω

aUn–1

n–1

i=1

δi·δn+τa21_Un–1_δn2

0

=1 2τ

Ω

aUn–1

_n

i=1

δi

2 –1

2τ

Ω

aUn–1

_n–1

i=1

δi

2

+1 2τa

1

2_Un–1_δn2

Then we have

1 2τ∇δ

n2

0–∇δ

n–12 0

+1 2τ

a

1 2_Un–1

n

i=1

δi

2

0 –1

2τ

a

1 2_Un–2

n–1

i=1

δi

2

0

+1 2τa

1

2_Un–1_δn2

0

≤Cτ2∂¯tUn–1₂

n–1

i=1

δi

2

0 –

au

Un–1∇Un–1

τ n

i=1 ∇δi

,δn

–

τ n

i=1

qiaun–1–aUn–1,δn

–

τ n

i=1 ∇qiau

Un–1∇en–1,δn

–

τ n

i=1

∇qi∇un–1au

un–1–au

Un–1,δn

–u0aun–1–aUn–1,δn–∇u0au

Un–1∇en–1,δn

–∇u0∇un–1au

un–1–au

Un–1,δn–fun–1–fUn–1,δn

–Rn₂+Rn₃+Rn₄,δn

10

i=1 Ai.

In what follows, we will boundAi,i= 2∼10, one by one. Note the particularity ofδn

on the left-hand side, so we have to use new ways to handleδn_{on the right-hand side}

instead of applying the Young inequality directly. In view of Green’s formula, it follows that

A9= –

fu

μn₁–1en–1,δn

=fuu

μn₁–1∇μn₁–1en–1,∇δn+fu

μn₁–1∇en–1,∇δn

≤C∇en–12₀+C∇δn2₀,

whereμn₁–1=Un–1_+λn–1

1 en–1and 0 <λn1–1< 1.

ForA2∼A8,A10, it is not so obvious to be dealt with. We choose to rewrite δnby

τ n_i=1∂¯tδiand then try to transferτ from one side in the inner product to the other;

more precisely,

A4= –

τ n

i=1 ∇qiau

Un–1∇en–1,τ n

i=1 ¯

∂tδi

=

au

Un–1∇en–1∇qn,τ n–1

i=1

δi

+

τ n–1

i=1 ∇qiau

Un–2∂¯t∇en–1,τ n–1

i=1

δi

+

τ n–1

i=1

∇qi∇en–1au(U

n–1_{) –a}

u(Un–2)

τ ,τ

n–1

i=1

δi

–∂¯t

τ n

i=1 ∇qiau

Un–1∇en–1,τ n

i=1

δi

≤C∂¯t∇en–1₀

τ

n–1

i=1

δi

0

+C∇en–1₀∂¯tUn–1_0,∞

τ

n–1

i=1

δi

+C∇en–1₀ τ n–1 i=1 δi 0 –∂¯t

τ n

i=1 ∇qiau

Un–1∇en–1,τ n–1

i=1

δi

≤C∂¯t∇en–1

2 0+C∇e

n–12

0+C∂¯tU

n–12 0,∞ τ n–1 i=1 δi 2 0 +C τ n–1 i=1 δi 2 0 –∂¯t

τ n

i=1 ∇qi_a

u

Un–1_∇en–1_,τ

n i=1 δi . Similarly,

A10=

¯

∂tRn2+∂¯tRn3+∂¯tRn4,τ

n–1

i=1

δi

–∂¯t

Rn₂+Rn₃+Rn₄,τ n

i=1

δi

≤Cτ2+C

τ n–1 i=1 δi 2 0 –∂¯t

Rn₂+Rn₃+Rn₄,τ n i=1 δi .

ForA2, we rewrite it as follows:

A2= –

au

Un–1∇Un–1

τ n

i=1 ∇δi

,τ n

i=1

∂¯tδi

=

au

Un–2∇Un–2∇δn,τ n–1 i=1 δi + τ n i=1 ∇δi

au

Un–2∂¯t∇Un–1,τ n–1 i=1 δi + τ n i=1 ∇δi

∇Un–1au(U

n–1_{) –a}

u(Un–2)

τ ,τ

n–1

i=1

δi

–∂¯t

au

Un–1∇Un–1

τ n

i=1 ∇δi

,τ n i=1 δi

A2i.

In view of the embedding theorem, this yields

A22≤C

τ n i=1 δi 0

∂¯tUn–1₀

τ n–1 i=1 δi 0 ≤C τ n i=1 δi 2 0

+C∂¯tUn–1

2 0 τ n–1 i=1 δi 2 0 .

To get round the need ofUi∈H3(Ω),i= 1, 2, . . . ,n– 1, we splitUi,i= 1, 2, . . . ,n– 1, into two parts; with inductive assumption (14), it reduces to

A21= –

au

Un–2∇en–2∇δn,τ n–1 i=1 δi + au

Un–2∇un–2∇δn,τ n–1

i=1

δi

≤Cen–2₀δn

0 τ n–1 i=1 δi 0

+C∇δn

≤C

τ

n–1

i=1

δi

2

0 +a0

4e

n–22 0δ

n2

0+C∇δ

n2 0

≤C

τ

n–1

i=1

δi

2

0 +a0

4τδ

n2

0+C∇δ

n2 0.

Similarly, we have

A23≤CUn–1₀∂¯tUn–1₀

τ

n

i=1

δi

0

τ

n–1

i=1

δi

0

≤C

τ

n

i=1

δi

2

0

+C∂¯tUn–12₀

τ

n–1

i=1

δi

2

0 .

We splitA3as

A3= –

τ n

i=1

qiaun–1–aUn–1,τ n

i=1 ¯

∂tδi

=

τ n–1

i=1

qi(a(u

n–1_{) –a(U}n–1_{)) – (a(u}n–2_{) –a(U}n–2₎₎

τ ,τ

n–1

i=1

δi

+

aun–1–aUn–1qn,τ n–1

i=1

δi

–∂¯t

τ n

i=1

qiaun–1–aUn–1,τ n

i=1

δi

3

i=1 A3i.

We can see that

A32=

aun–1–aUn–1qn,τ n–1

i=1

δi

≤C∇en–12₀+C

τ

n–1

i=1

δi

2

0 .

Since

(a(un–1_{) –a(U}n–1_{)) – (a(u}n–2_{) –a(U}n–2₎₎

τ

=aμn₂–1∂¯ten–1+∂¯tun–1

aμn₃–1–aμn₂–1, (17)

where

μ₃n–1=un–2+τ λ₃n–1∂¯tun–1, μn2–1=Un–2+τ λ2n–1∂¯tUn–1,

0 <λn₂–1< 1, 0 <λn₃–1< 1,

and

μ₃n–1–μn₂–1=en–2+τ λ₂n–1∂¯ten–1+∂¯tun–1τ

we see that

A31≤

τ

n–1

i=1

qi

0,4

(a(un–1) –a(Un–1)) – (a(u_τ n–2) –a(Un–2))

0,4

τ

n–1

i=1

δi

0

≤C∂¯t∇en–1₀

τ

n–1

i=1

δi

0

+C∇en–2₀

τ

n–1

i=1

δi

0 +Cτ

τ

n–1

i=1

δi

0

≤Cτ2+C∂¯t∇en–12₀+C∇en–22₀+C

τ

n–1

i=1

δi

2

0 ,

whence

A3≤Cτ2+C∂¯t∇en–1

2

0+C∇e

n–22

0+C∇e

n–12 0+C

τ

n–1

i=1

δi

2

0

–∂¯t

τ n

i=1

qiaun–1–aUn–1,τ n

i=1

δi

.

RewritingA5,A6,A8, with (17), we obtain

A6=

u0(a(un–1) –a(Un–1)) – (a(un–2) –a(Un–2))

τ ,τ

n–1

i=1

δi

–∂¯t

u0aun–1–aUn–1,τ n

i=1

δi

≤u0_0,4(a(u

n–1_{) –a(U}n–1_{)) – (a(u}n–2_{) –a(U}n–2₎₎

τ

0,4

τ

n–1

i=1

δi

0

–∂¯t

u0aun–1_–aUn–1_,τ

n

i=1

δi

≤C∂¯t∇en–1₀

τ

n–1

i=1

δi

0

+C∇en–2₀

τ

n–1

i=1

δi

0

+Cτ

τ

n–1

i=1

δi

0 –∂¯t

u0aun–1–aUn–1,τ n

i=1

δi

≤Cτ2+C∂¯t∇en–12₀+C∇en–22₀+C

τ

n–1

i=1

δi

2

0

–∂¯t

u0aun–1–aUn–1,τ n

i=1

δi

,

A5= –

τ n

i=1

∇qi∇un–1au

un–1–au

Un–1,τ n

i=1 ¯

∂tδi

=

τ n–1

i=1

∇qi∇un–2(au(u

n–1_{) –a}

u(Un–1)) – (au(un–2) –au(Un–2))

τ ,τ

n–1

i=1

δi

+

au

un–1–au

Un–1τ n–1

i=1

∇qi∂¯t∇un–1,τ n–1 i=1 δi + au

un–1–au

Un–1∇un–1∇qn,τ n–1

i=1

δi

–∂¯t

τ n

i=1

∇qi∇un–1au

un–1–au

Un–1,τ n

i=1

δi

≤C(au(u

n–1_{) –a}

u(Un–1)) – (au(un–2) –au(Un–2)) τ 0 τ n–1 i=1 δi 0

+Cen–1₀

τ n–1 i=1 δi 0 –∂¯t

τ n

i=1

∇qi∇un–1au

un–1–au

Un–1,τ n

i=1

δi

≤Cτ2+C∂¯t∇en–1

2

0+C∇e

n–12

0+C∇e

n–22 0+C

τ n–1 i=1 δi 2 0

–∂¯t

τ n

i=1

∇qi∇un–1au

un–1–au

Un–1,τ n i=1 δi , and

A8=

∇u0∇un–2(au(u

n–1_{) –a}

u(Un–1)) – (au(un–2) –au(Un–2))

τ ,τ

n–1 i=1 δi +

∇u0au

un–1–au

Un–1∂¯t∇un–1,τ n–1

i=1

δi

–∂¯t

∇u0∇un–1au

un–1–au

Un–1,τ n

i=1

δi

≤Cτ2+C∂¯t∇en–1

2

0+C∇e

n–12

0+C∇e

n–22 0 +C τ n–1 i=1 δi 2 0 –∂¯t

∇u0∇un–1au

un–1–au

Un–1,τ n i=1 δi .

Finally,A7can be bounded as

A7=

∇u0au

Un–2∂¯t∇en–1,τ n–1 i=1 δi +

∇u0∇en–1au(U

n–1_{) –a}

u(Un–2)

τ ,τ

n–1

i=1

δi

–∂¯t

∇u0au

Un–1∇en–1,τ n

i=1

δi

≤C∂¯t∇en–1₀

τ n–1 i=1 δi 0

+C∇en–1₀∂¯tUn–1_0,∞

–∂¯t

∇u0au

Un–1∇en–1,τ n

i=1

δi

≤C∂¯t∇en–1

2 0+C∇e

n–12 0+C

τ n–1 i=1 δi 2 0

+C∂¯tUn–1

2 0,∞ τ n–1 i=1 δi 2 0 –∂¯t

∇u0au

Un–1∇en–1,τ n i=1 δi . Moreover, because

_{∇ ¯}∂ten₀=∇

δn+Rn₁₀≤C∇δn₀+C∇Rn₁₀≤C∇δn₀+Cτ,

∂¯ten₀=

δn+Rn₁₀≤Cδn₀+CRn₁₀≤Cδn₀+Cτ, (18)

_∇en₀≤C√τ

_n

i=1

_{∇ ¯}∂tei2₀

1 2

≤C√τ

_n

i=1

_∇δi2

0

1 2

+Cτ,

we have

1

τ∇δ n2

0–∇δ

n–12 0 +τ a 1 2_Un–1

n i=1 δi 2 0 –τ a 1 2_Un–2

n–1 i=1 δi 2 0

+τδn2₀

≤Cτ2+C∇δn2

0+C∇δ

n–12 0+Cτ

n

i=1

_∇δi2

0 +C τ n i=1 δi 2 0

+C∂¯tUn–1

2 2 τ n–1 i=1 δi 2 0 +C τ n–1 i=1 δi 2 0 –∂¯t

au

Un–1∇Un–1

τ n

i=1 ∇δi

,τ n i=1 δi

–∂¯t

τ n

i=1 ∇qiau

Un–1∇en–1,τ n

i=1

δi

–∂¯t

τ n

i=1

qiaun–1–aUn–1,τ n

i=1

δi

–∂¯t

u0aun–1–aUn–1,τ n

i=1

δi

–∂¯t

∇u0au

Un–1∇en–1,τ n

i=1

δi

–∂¯t

Rn₂+Rn₃+Rn₄,τ n

i=1

δi

–∂¯t

τ n

i=1

∇qi∇un–1au

un–1–au

Un–1,τ n

i=1

δi

–∂¯t

∇u0∇un–1au

un–1–au

Summing this inequality from 2 ton, we get

_∇δn2

0+ τ n i=1 δi 2 0 + n i=2

τ δi2

0

≤∇δ12

0+τ

2_δ12 0+Cτ

2_+Cτ

n

i=1

_∇δi2

0+Cτ

n i=1 τ i–1 j=1 δj 2 0

+Cτ2 n i=2 i–1 j=1

_∇δj2

0+Cτ

n

i=1

∂¯tUi–12₀

τ i–1 j=1 δj 2 0 – au

Un–1∇Un–1

τ n

i=1 ∇δi

,τ n i=1 δi

+au

U0∇U0τ∇δ1,τ δ1–

τ n

i=1

qiaun–1–aUn–1,τ n i=1 δi – τ n i=1 ∇qiau

Un–1∇en–1,τ n i=1 δi –

u0aun–1–aUn–1,τ n i=1 δi –

∇u0au

Un–1∇en–1,τ n i=1 δi –

Rn₂+Rn₃+Rn₄,τ n

i=1

δi

+R1₂+R1₃+R1₄,τ δ1–

τ n

i=1

∇qi∇un–1au

un–1–au

Un–1,τ n i=1 δi –

∇u0∇un–1au

un–1–au

Un–1,τ n i=1 δi . (19) Due to au

Un–1∇Un–1

τ n

i=1 ∇δi

,τ n i=1 δi = au

Un–1∇un–1

τ n

i=1 ∇δi

,τ n i=1 δi – au

Un–1∇en–1

τ n

i=1 ∇δi

,τ n i=1 δi

≤Cτ14

τ n i=1 δi 2 0 +C τ n i=1 ∇δi

2 0 +1 4 τ n i=1 δi 2 0

≤Cτ12

τ n i=1 δi 2 0 +1 2 τ n i=1 δi 2 0 +Cτ

n

i=1

after obvious estimates and a kickback ofτ n_i=1∇δi2

0, together with our earlier estimate forn= 1, we obtain

_∇δn2

0+τ 2

n

i=1

δi

2

0 +τ2

n

i=2

δi2

0≤Cτ

2_{. (21)}

Here by (18) we have

τ

n

i=1 ¯

∂tei

2 +τ

_n

i=2

∂¯tei

2 2

1 2

≤Cτ (22)

and, further,

en

2=τ

n

i=1 ¯

∂tei

2

≤Cτ. (23)

Then we conclude that there existτ4,τ5,C3,C4such that whenτ≤τ4, we have

en₂+τ

_n

i=2

∂¯tei

2 2

1 2

+δn₁+τ

n

i=1

δi

2 +τ

_n

i=2

δi2₂

1 2

≤C3τ, (24)

which leads to

en₂≤τ41_{, ∂}¯_tUn

2≤C4, (25)

Un_0,∞+√τ

_n

i=1

∂¯tUi

2 0,∞

1 2

≤Cen₂+C√τ

_n

i=1

∂¯tei

2 2

1 2

+un_0,∞+√τ

_n

i=1

∂¯tui

2 0,∞

1 2

≤CC3τ+CC3 √

τ+un_0,∞+√τ

_n

i=1

∂¯tui

2 0,∞

1 2

≤K0, (26)

where τ ≤τ5=min{1/2CC3, 1/4C2C32}. Clearly, C3,C4have nothing to do withC0, and thus (6) and (7) hold form=nif we take C0≥ 4i=1Ci andτ0≤min1≤τ≤5τi. Then the

induction is closed. The proof is completed.

Remark1 The special method used to tackle the left-hand side of (16) is important to deduce the regularities ofUn_andQn_{in theH}2_{-norm. Further, the terms includingδ}n

on the right-hand side needs innovative technologies to treat.

4 Error estimates for spatial-discrete system and optimal error results

In this section, we will establishτ-independent optimal error results forun_andqn_through

the spatial results. We decompose the errors as follows:

Qi–Qi_h=Qi–IhQi+IhQi–Qihri+θi, i= 1, 2, . . . ,n,

and we are now ready for the unconditional spatial results.

Theorem 2 Let um _{and U}m

h be solutions of(3)and(4),respectively,for m= 1, 2, . . . ,N.

Under the conditions of Theorem1,there existτ₀,h₀such that,forτ ≤τ₀and h≤h₀,we have

um–U_hm₀+qm–Qm_h0=Oh2+τ (27)

and

_∇

um_–Um

h0+∇

qm_–Qm

h0=O(h+τ). (28) Proof Before discussing (27) and (28), we shall pause to give the results

ξm₀+θm₀+τ

_m

i=1

_∇θi2₀

1 2

≤C₀hh+τ12 ₍₂₉₎

by mathematical induction, whereC₀is a positive constant independent ofm,τ, andh. Since IhUm0,∞ + √τ( _im₌₂¯∂tIhUi20,∞)

1

2 ≤ C, let K

0 1 + IhUm0,∞ + √τ ×

( m_i=2¯∂tIhUi20,∞)

1

2_{. We begin withm= 1:}

_¯

∂tθ1,vh

+aU0τ∇θ1,∇vh

= –∂¯tr1,vh

–aU0τ∇r1,∇vh

–aU0–aU_h0τ∇Q1_h,∇vh

–aU_h0∇η0,∇v –∇U0aU0–aU_h0,∇v+fU0–fU_h0,vh

. (30)

Takingvh=θ1in (30), we get

1

τθ

12 0+τa

1

2_U0∇_θ12

0

= –∂¯tr1,θ1

–aU0τ∇r1,∇θ1

–aU0–aU_h0τ∇Q1_h,∇θ1–aU_h0∇η0,∇θ1

–∇U0aU0–aU_h0,∇θ1+fU0–fU_h0,θ1. (31)

It is easy to see that

_¯

∂tr1,θ1

≤Ch2∂¯tU1₂θ1₀≤Ch4+Cθ1

2 0,

∇U0aU0–aU_h0,∇θ1≤Cr0₀∇θ1₀≤Ch2τ+ 1 8τθ

12 0,

Denotingγ(X)|K=_|K1_|

Kγ(X)dXand then using the mean-value technique, we obtain

aU0τ∇r1,∇θ1

=

K

aU0–aU0_τ∇r1_,∇θ1

K

–

K

aU0_|

K

τ∇e1–∇Ihe1

,∇θ1_K+

K

aU0_|

K

τ∇u1–∇Ihu1

,∇θ1_K

≤Ch2τU1₂∇θ1₀+Chτe1₂∇θ1₀≤Ch4+Ch2τ2+Cτ2∇θ12₀,

aU_h0∇η0,∇θ1

=

K

aU_h0–aU0

h

∇η0,∇θ1_K+

K

aU0

h

|K

∇η0,∇θ1_K

≤Ch2u0₂∇θ1₀≤Ch√τ√1 τθ

1 0≤Ch

2_τ+ 1 8τθ

12 0.

By Theorem1we have

aU0–aU_h0τ∇Q1_h,∇θ1

= –aU0–aU_h0τ∇θ1,∇θ1

–aU0–aU_h0τ∇r1,∇θ1–aU0–aU_h0τ∇δ1,∇θ1

+aU0–aU_h0τ∇q1,∇θ1

≤Ch2τU0₂∇θ1_0,∞∇θ1₀

+Ch3τU0₂U1₂∇θ1

0,∞+Ch

2_τ∇δ1 0∇θ

1 0,∞

+Ch2τ∇q1_0,∞∇θ1

0

≤Ch4+Ch2τ2+Chτ∇θ12₀+Cτ2∇θ12₀.

Allocating all the estimates obtained, we have

1

τθ

12 0+τ∇θ

12 0

≤Ch4+Ch2τ+Chτ∇θ12₀+Cθ12₀+Cτ2∇θ12₀. (32)

Thus there existτ₁,τ₂,h₁,h₂,C₁ such that, forτ≤τ₁andh≤h₁, we have

θ1₀+τ∇θ1₀≤C₁h(h+√τ), (33)

which implies

U_h1_0,∞≤Ch–1ξ1₀+IhU1_0,∞

≤CC₁h+CC₁√τ+IhU1_0,∞≤K0, (34)

By mathematical induction we assume that (29) holds form≤n– 1. Then there existτ₃

andh₃such that

U_hm_0,∞+√τ

_m

i=2

∂¯tUhi

2 0,∞

1 2

≤Ch–1

ξm₀+√τ

_m

i=2

∂¯tξi

2 0

1 2

+

IhUm_0,∞+ √

τ

_m

i=2

∂¯tIhUi

2 0,∞

1 2

≤2CC₀h+ 2CC₀√τ+

IhUm_0,∞+ √

τ

_m

i=2

∂¯tIhUi

2 0,∞

1 2

≤K₀, (35)

whereh≤h3≤1/4CC0andτ ≤τ3≤1/6(CC0)2.

Then we prove that (29) also holds form=n. By (4) and (5) we derive the error equations

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

(∂¯tξn,vh) = –(∂¯tηn,vh) + (θn,vh) + (rn,vh),

(∂¯tθn,vh) + (a(Uhn–1)τ n

i=1∇θi,∇vh)

= –(∂¯trn,vh) – (a(Un–1)τ ni=1∇ri,∇vh)

– ((a(Un–1_{) –a(U}n–1

h ))τ n

i=1∇IhQi,∇vh) – (a(Uhn–1)∇η0,∇v)

– (∇U0_(a(Un–1_{) –a(U}n–1

h )),∇v) + (f(Un–1) –f(Uhn–1),vh).

(36)

Forvh=θnin the second equation of (36), we have

τ

aU_hn–1

n

i=1

∇θi,∇θn

=τ

Ω

aU_hn–1

n–1

i=1

∇θi· ∇θn+τa12_Un–1 h

∇θn2₀

=1 2τ

Ω

aU_hn–1

_n

i=1 ∇θi

2 –1

2τ

Ω

aU_hn–1

_n–1

i=1 ∇θi

2

+1 2τa

1 2_Un–1

h

∇θn2₀,

and hence we find

1 2τθ

n2

0–θ

n–12 0

+1 2τa

1 2_Un–1

h

∇θn2

0

+1 2τ

a

1 2_Un–1

h

n i=1

∇θi

2

0 –1

2τ

a

1 2_Un–2

h

n–1

i=1 ∇θi

2

0

≤Cτ2∂¯tUhn–10,∞

n–1

i=1 ∇θi

2

0

–∂¯trn,θn

–

aUn–1τ n

i=1

∇ri,∇θn

–

aUn–1–aU_hn–1τ n

i=1

∇IhQi,∇θn

–aU_hn–1∇η0,∇θn

–∇U0aUn–1–aU_hn–1,∇θn–fUn–1–fU_hn–1,θn

7

i=1

Bi. (37)

Obviously,

B1≤Cτ2

n–1

i=1 ∇θi

2

0 ,

B2≤Ch2∂¯tUhn2θ

n

0≤Ch

4_+Cθn2

0, (38)

B7≤Ch4+Cξn–1 2 0+Cθ

n2

0.

Similarly to the proof ofA2∼A8 andA10, we rewriteθnby τ ni=1∂¯tθi and then try to

transferτfrom one side to the other in the inner product. For simplicity and concreteness, with the help of (2), we show that

B5=

au

μn₄–1∂¯tUhn–1∇η0,τ n–1

i=1 ∇θi

–1

τ

Ω

∇η0

aU_hn–1τ n

i=1

∇θi–aU_hn–2τ n–1

i=1 ∇θi

=

K

au

μn₄–1∂¯tξn–1–au

μn₄–1∂¯tξn–1

∇η0,τ n–1

i=1 ∇θi

K

+

K

au

μn₄–1∂¯tξn–1|K

∇η0,τ n–1

i=1 ∇θi

–

au

μn₄–1∂¯tηn–1∇η0,τ n–1

i=1 ∇θi

+

K

au

μn₄–1∂¯tUn–1–au

μn₄–1∂¯tUn–1

∇η0,τ n–1

i=1 ∇θi

K

+

K

au

μn₄–1∂¯tUn–1|K

∇η0,τ n–1

i=1 ∇θi

–1

τ

Ω

∇η0

aU_hn–1τ n

i=1

∇θi–aU_hn–2τ n–1

i=1 ∇θi

≤Ch2∂¯tξn–1₁u0_2,4

τ

n–1

i=1 ∇θi

0,4 +Ch2

τ

n–1

i=1 ∇θi

0

–1

τ

Ω

∇η0

aU_hn–1τ n

i=1

∇θi–aU_hn–2τ n–1

i=1 ∇θi

≤Ch4+C∂¯tξn–1

2 0+C

τ

n–1

i=1 ∇θi

2

0

–1

τ

Ω

∇η0

aU_hn–1τ n

i=1

∇θi–aU_hn–2τ n–1

i=1 ∇θi

whereμn₄–1=U_hn–2+τ λn₄–1∂¯tUhn–1and 0 <λn4–1< 1. Again transferringτ from one part of the inner product to the other, we have

B3= –

aUn–1τ n

i=1 ∇ri,τ

n

i=1 ∇ ¯∂tθi

=

aUn–2∇rn,τ n–1

i=1 ∇θi

+

a(Un–1) –a(Un–2)

τ τ

n

i=1 ∇ri,τ

n–1

i=1 ∇θi

–∂¯t

aUn–1τ n

i=1 ∇ri,τ

n

i=1 ∇θi

3

i=1 B3i.

We splitB31andB32and estimate them as follows:

B31=

K

aUn–2–aUn–2∇_rn_,τ n–1

i=1 ∇θi

K

–

K

aUn–2_|

K

∇ei–∇Ihei,τ n–1

i=1 ∇θi

K

+

K

aUn–2|

K

∇ui–∇Ihui,τ n–1

i=1 ∇θi

K

≤Ch2

τ

n–1

i=1 ∇θi

0

+Ch√τ

τ

n–1

i=1 ∇θi

0

≤Ch4+Ch2τ+C

τ

n–1

i=1 ∇θi

2

0 ,

B32=

a(Un–1_{) –a(U}n–2₎

τ τ

n

i=1

∇ei–∇Ihei

,τ n–1

i=1 ∇θi

+

K

a(Un–1_{) –a(U}n–2₎

τ –

a(Un–1_{) –a(U}n–2₎

τ

×τ n

i=1

∇ui–∇Ihui

,τ n–1

i=1 ∇θi

K

+

K

a(Un–1_{) –a(U}n–2₎

τ

K

τ n

i=1

∇ui–∇Ihui

,τ n–1

i=1 ∇θi

K

≤Ch√τ

τ

n–1

i=1 ∇θi

0 +Ch2

τ

n–1

i=1 ∇θi

0

≤Ch4+Ch2τ+C

τ

n–1

i=1 ∇θi

2

Then we have

B3≤Ch4+Ch2τ+C

τ

n–1

i=1 ∇θi

2

0 –∂¯t

aUn–1τ n

i=1 ∇ri,τ

n

i=1 ∇θi

.

Note that

B6=

∇U0au

μn₅–2∂¯tξn–1+∂¯tηn–1

,τ n–1

i=1 ∇θi

+

∇U0ξn–1+ηn–1au(μ n–1

5 ) –au(μn5–2)

τ ,τ

n–1

i=1 ∇θi

–∂¯t

∇U0au

μn₅–1ξn–1+ηn–1,τ n

i=1 ∇θi

≤Ch4+Cξn–12

0+C∂¯tξ

n–12 0+C

τ

n–1

i=1 ∇θi

2

0

–∂¯t

∇U0au

μn₅–1ξn–1+ηn–1,τ n

i=1 ∇θi

,

where

μn₅–1=Un–1+λn₅–1ξn–1+ηn–1, 0 <λn₅–1< 1, and

au(μn5–1) –au(μn5–2)

τ

≤∂¯tUn–1+λn5–1∂¯tξn–1+∂¯tηn–1.

RewritingB4and splitting it into several parts, we obtain

B4=

aUn–1–aU_hn–1τ n

i=1 ∇ri,τ

n

i=1 ∇ ¯∂tθi

–

aUn–1–aU_hn–1τ n

i=1

∇Qi,∇θn

= –

aUn–2–aU_hn–2∇rn,τ n–1

i=1 ∇θi

–

τ n

i=1

∇ri(a(U

n–1_{) –a(U}n–1

h )) – (a(Un–2) –a(Uhn–2))

τ ,τ

n–1

i=1 ∇θi

+∂¯t

aUn–1–aU_hn–1τ n

i=1 ∇ri,τ

n

i=1 ∇θi

–

aUn–1–aU_hn–1τ n

i=1

∇Qi,∇θn

It is obvious that

–

aUn–2–aU_hn–2∇rn,τ n–1

i=1 ∇θi

≤ChUn₂Ch2Un–2₂+ξn–2

0

τ

n–1

i=1 ∇θi

0,∞

≤Ch4+Cξn–22₀+C

τ

n–1

i=1 ∇θi

2

0 .

Because

(a(Un–1) –a(U_hn–1)) – (a(Un–2) –a(U_hn–2))

τ

=au

μn₆–1∂¯tUn–1–au

μn₇–1∂¯tUhn–1

=au

μn₇–1∂¯tξn–1+∂¯tηn–1

+∂¯tUn–1

au

μn₆–1–au

μn₇–1, (39)

where

μ₆n–1=Un–2+τ λ₆n–1∂¯tUn–1, μ7n–1=Uhn–2+τ λ7n–1∂¯tUhn–1,

0 <μn₆–1,μn₇–1< 1,

and

μn₆–1–μn₇–1=ξn–2+ηn–2+τ λn₇–1∂¯tξn–1+∂¯tηn–1

+τ∂¯tUn–1

λn₆–1–λn₇–1,

it follows that

τ n

i=1

∇ri(a(U

n–1_{) –a(U}n–1

h )) – (a(Un–2) –a(Uhn–2))

τ ,τ

n–1

i=1 ∇θi

≤Ch2

τ

n

i=1 Qi

2

(a(Un–1) –a(Uhn–1)) – (a(Un–2) –a(Uhn–2)) τ

0

τ

n–1

i=1 ∇θi

0,∞

≤∂¯tξn–1₀+∂¯tηn–1₀+ξn–2₀+ηn–2₀+τ

Ch

τ

n–1

i=1 ∇θi

0

≤Ch4+Ch2τ+∂¯tξn–1

2 0+ξ

n–22 0+C

τ

n–1

i=1 ∇θi

2