# Spectral Galerkin Method in Space and Time for the 2Dg-Navier-Stokes Equations

Authors: Dao Trong Quyet
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We prove theH2-stability and L2-error analysis of the spectral
Galerkin method in space and time with the implicit/explicit Euler scheme for the 2D g-Navier-Stokes equations in bounded domains when the initial data belong to H1.


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## Body

## 1. Introduction

LetΩ be a bounded domain in ℝ2 with sufficiently smooth boundary Γ. In this paper, we study the spectral Galerkin method in space and time for the following 2D g-Navier-Stokes equations:
(1)∂u∂t-νΔu+(u·∇)u+∇p=f(x,t)in(0,+∞)×Ω,∇·(gu)=0in(0,+∞)×Ω,u=0on(0,+∞)×Γ,u(0,x)=u0(x),x∈Ω,
where u=u(x,t)=(u1,u2) is the unknown velocity vector, p=p(x,t) is the unknown pressure, ν>0 is the kinematic viscosity coefficient, and u0 is the initial velocity.Theg-Navier-Stokes equations are a variation of the standard Navier-Stokes equations. More precisely, when g≡const we get the usual Navier-Stokes equations. The 2D g-Navier-Stokes equations arise in a natural way when we study the standard 3D problem in thin domains. We refer the reader to [1] for a derivation of the 2D g-Navier-Stokes equations from the 3D Navier-Stokes equations and a relationship between them. As mentioned in [1], good properties of the 2D g-Navier-Stokes equations can lead to an initiate of the study of the Navier-Stokes equations on the thin three-dimensional domain Ωg=Ω×(0,g). In the last few years, the existence and long-time behavior of both weak and strong solutions to the 2D g-Navier-Stokes equations have been studied extensively (cf. [2–9]). In this paper, we aim to study numerical approximation of the strong solutions to problem (1). To do this, we assume that (G)
g∈W1,∞(Ω) such that
(2)0<m0≤g(x)≤M0000∀x=(x1,x2)∈Ω,0000|∇g|∞<m0λ11/2,whereλ1>0 is the first eigenvalue of the g-Stokes operator in Ω (i.e., the operator A defined in Section 2.1 below);(F)
f∈W1,∞(ℝ+;Hg); that is, f,ft∈L∞(ℝ+;Hg).In this paper, in order to study the numerical approximation of strong solutions to the 2Dg-Navier-Stokes equations we will use the spectral Galerkin method in space and time, which is based on the eigen-subspaces of the g-Stokes operator. As mentioned in [10] for the Navier-Stokes equations, this method enables us to avoid solving the fully nonlinear g-Navier-Stokes equations on the low-frequency subspace, whereas to obtain the low-frequency component of the numerical solution, the usual multilevel spectral methods and the postprocessing Galerkin methods need to solve the fully nonlinear g-Navier-Stokes equations on the low-frequency subspace. In what follows, we will explain the spectral Galerkin method used in the paper. For the related function spaces, we refer the reader to Section 2.1.Letw1,w2,… and λ1,λ2,… be the eigenvectors and eigenvalues of the g-Stokes operator. For a fixed integer m, let Pm be the orthogonal projection of Hg onto Hm=span{w1,…,wm}. Then, the spectral Galerkin method in space is defined as follows: find um(t)∈Hm such that
(3)umt+νAum+νCum+PmB(um,um)=Pmf,00000000000000m00t>0,um(0)=Pmu0.
In order to simplify the implementation of the scheme, we restrict ourselves to the semi-implicit Euler scheme applied to the spectral Galerkin method in space. We consider a spectral Galerkin method in space and time with the implicit/explicit Euler scheme: find umn+1 (n≥0) such that
(4)1Δt(umn+1-umn)+νAumn+1+νCumn+1+PmB(umn,umn)=Pmfn+1,000000000000000000000000000m00000t>0,um0=Pmu0,
where Δt>0 is the time step size and
(5)fn+1=1Δt∫tntn+1‍f(t)dt,tn=nΔt.Here, the linear term is treated implicitly to avoid serve time step limitations, whereas the nonlinear term is kept explicitly so that the corresponding discrete system is easily invertible. It is well known that this type of scheme is only stable under some restriction on the time step size. We will obtainH2-stability uniform in time stated in Theorem 13, provided that the following condition holds
(6)CΔtlnλmλ1≤1
for some positive constant C depending on the data (u0,ν,f,Ω). As mentioned in [11] for the case of 2D Navier-Stokes equations, the stability condition (6) is a significant improvement compared with the results provided by the nonlinear Galerkin method [12] and the multilevel method [13, 14].We also derive anL2-error estimate of the numerical solution umn under the stability condition (6):
(7)|u(tn)-umn|2≤𝒦τ-1(tn)e(16/ν3γ03λ1)c02Cf2tn(λm+1-2+Δt2)00000000000000000000000000000000000000∀n≥1,
where Cf=supt≥0|f(t)|, τ(t)=min{1,t} and 𝒦 denotes a general positive constant depending only on the data (ν,Ω,|∇g|∞,λ1,Cf,∥u0∥). Noting that τ-1(tn) is a singular factor near t=0.Compared to He's works [11] on the spectral method of the 2D Navier-Stokes equations, here we have to address some additional difficulties. Firstly, to treat the more general condition ∇·(gu)=0, instead of the usual function spaces used for the Navier-Stokes equations, we use the function spaces Hg,Vg which are defined suitably for the g-Navier-Stokes equations (see Section 2.1 for details). Secondly, we have to deal with the term Cu in the equation, which only appears for the g-Navier-Stokes equations. It is worthy noticing that when g≡1, we of course recover the results for the Navier-Stokes equations in [11].The paper is organized as follows. In the next section, we recall some results on function spaces and inequalities for the nonlinear terms related to theg-Navier-Stokes equations, and some discrete Gronwall inequalities are frequently used later. In Section 3, we prove several estimates for the strong solution and the Galerkin approximate solutions of problem (1). In Section 4, we study the error analysis of the spectral Galerkin method in space. Stability and error analysis of the spectral Galerkin method in space and time are discussed in the last section.


## 2. Preliminaries

### 2.1. Function Spaces and Operators

LetL2(Ω,g)=(L2(Ω))2 and H01(Ω,g)=(H01(Ω))2 be endowed, respectively, with the inner products
(8)(u,v)g=∫Ωu·vgdx,u,v∈L2(Ω,g),((u,v))g=∫Ω∑j=12‍∇uj·∇vjgdx,u=(u1,u2),0000000000000l0000v=(v1,v2)∈H01(Ω,g),
and norms |u|2=(u,u)g, ∥u∥2=((u,u))g. Thanks to assumption (G), the norms |·| and ∥·∥ are equivalent to the usual ones in (L2(Ω))2 and in (H01(Ω))2.Let(9)𝒱={u∈(C0∞(Ω))2:∇·(gu)=0}.
Denote by Hg the closure of 𝒱 in L2(Ω,g), and denote by Vg the closure of 𝒱 in H01(Ω,g). It follows that Vg⊂Hg≡Hg′⊂Vg′, where the injections are dense and continuous. We will use ∥·∥* for the norm in Vg′, and 〈·,·〉 for duality pairing between Vg and Vg′.We now define the trilinear formb by
(10)b(u,v,w)=∑i,j=12‍∫Ω‍ui∂vj∂xiwjgdx,
whenever the integrals make sense. It is easy to check that if u,v,w∈Vg, then
(11)b(u,v,w)=-b(u,w,v).
Hence
(12)b(u,v,v)=0,∀u,v∈Vg.SetA:Vg→Vg′ by 〈Au,v〉=((u,v))g, B:Vg×Vg→Vg′ by 〈B(u,v),w〉=b(u,v,w). Denote D(A)={u∈Vg:Au∈Hg}, then D(A)=H2(Ω,g)∩Vg and Au=-PgΔu, for all u∈D(A), where Pg is the ortho-projector from L2(Ω,g) onto Hg. Consequently, there exists an orthogonal basis of Hg consisting of the eigenvectors wj of A:
(13)Awj=λjwj,0<λ1≤λ2≤⋯,0000000000λj⟶∞asj⟶∞.
Furthermore, we can also define the sth power As of A for all s∈ℝ. The space D(As) is the Hilbert space when equipped with the scalar product (As,As) and norm |As.|, where (·,·) and |·| denote the scalar product and norm in Hg. In particular, D(A0)=Hg and D(A1/2)=Vg.LetHm=span{w1,…,wm}. Then, the following estimates hold:
(14)λ1|v|2≤∥v∥2∀v∈Vg,∥vm∥2≤λm|vm|2∀vm∈Hm,(15)λm+1|wm|2≤∥wm∥2∀wm∈Vg∖Hm.Using the Hölder inequality, the Ladyzhenskaya inequality (whenn=2):
(16)|u|L4≤c|u|1/2|∇u|1/2,∀u∈H01(Ω),
and the interpolation inequalities, as in [15, 16], one can prove the following.Lemma 1.
Ifn=2, then(17)|b(u,v,w)|≤{c0|u|1/2∥u∥1/2∥v∥|w|1/2∥w∥1/2,00000000000000∀u,v,w∈Vg,c0|u|1/2∥u∥1/2∥v∥|Aw|1/2|w|1/2,∀u∈Vg,v∈D(A),w∈Hg,c0|u|1/2|Au|1/2∥v∥|w|,∀u∈D(A),v∈Vg,w∈Hg,c0|u|∥v∥|w|1/2|Aw|1/2,∀u∈Hgv∈Vg,w∈D(A),(18)|b(u,v,w)|+|b(w,v,u)|≤c0|u|1/2∥u∥1/2∥v∥1/2|Av|1/2|w||b(u,v,w)|+|b(w,vu)|≤c0∀u∈Vg,v∈D(A),w∈Hg,(19)|b(u,v,w)|+|b(w,v,u)|≤c0(1+ln|Au|2λ1∥u∥2)1/2∥u∥∥v∥|w||b(uv)|2+∀u∈D(A),v∈Vg,w∈Hg,
where c0 are appropriate constants depending only on Ω.Lemma 2 (see [3]).
Letu∈L2(0,T;D(A))∩L∞(0,T;Vg), then the function Bu defined by
(20)(Bu(t),v)g=b(u(t),u(t),v),∀v∈Hg,a.e.t∈[0,T],
belongs to L4(0,T;Hg), and therefore also belongs to L2(0,T;Hg).Lemma 3 (see [4]).
Letu∈L2(0,T;Vg), then the function Cu defined by
(21)(Cu(t),v)g=((∇gg·∇)u,v)g=b(∇gg,u,v),∀v∈Vg,
belongs to L2(0,T;Hg), and hence also belongs to L2(0,T;Vg′). Moreover,
(22)|Cu(t)|≤|∇g|∞m0·∥u(t)∥,fora.e.t∈(0,T),∥Cu(t)∥*≤|∇g|∞m0λ11/2·∥u(t)∥,fora.e.t∈(0,T).Since(23)-1g(∇·g∇)u=-Δu-(∇gg·∇)u,
we have
(24)(-Δu,v)g=((u,v))g+((∇gg·∇)u,v)g=(Au,v)g+((∇gg·∇)u,v)g,∀u,v∈Vg.


### 2.2. Discrete Gronwall Inequalities

Hereafter, we will frequently use the following modified discrete Gronwall lemmas.Lemma 4 (see [12]).
Let0<Δt<1,ak,gk,hk for integers k≥0 be nonnegative numbers such that
(25)1Δt(ak+1-ak)≤gkak+hk∀0≤k≤J.
If
(26)Δt∑k=0J‍gk≤α1,Δt∑k=0J‍hk≤α2,
for J≤n0-1 and
(27)Δt∑k=k0k0+n0-1‍gk≤α1,Δt∑k=k0k0+n0-1‍hk≤α2,Δt∑k=k0k0+n0-1‍ak≤a0+α3,
for J≥n0, 0≤k0≤J+1-n0, then
(28)aJ+1≤eα1(a0+α2+α3).Lemma 5 (see [13]).
Letα and ak,bk,hk,k≥0 be nonnegative numbers such that
(29)(1+γΔt)ak+1-ak+bk+1Δt≤hkΔt∀0≤k≤J.
Then,
(30)aJ+1+Δt∑k=1J+1‍(1+γΔt)-(J+2-k)bk≤(1+γΔt)-(J+1)a0+Δt∑k=0J‍(1+γΔt)-(J+1-k)hk.Lemma 6 (see [11]).
Letβ and ak,bk,gk,hk for integers k≥0 be nonnegative numbers such that
(31)an+Δt∑k=0n‍bk≤Δt∑k=0n‍gkak+Δt∑k=0n‍hk+β∀n≥0.
Suppose that gkΔt<1, for all k≥0, and set σk=(1-gkΔt)-1. Then,
(32)an+Δt∑k=0n‍bk≤exp{Δt∑k=0n‍σkgk}(Δt∑k=0n‍hk+β)∀n≥0.
Moreover, if
(33)ar+Δt∑k=rn‍bk≤Δt∑k=rn‍gkak+Δt∑k=rn‍hk+β∀0≤r≤n,
and gkΔt<1 for all 0≤k≤n, then
(34)ar+Δt∑k=rn‍bk≤exp{Δt∑k=0n‍σkgk}(Δt∑k=0n‍hk+β);0000000000000000000000000000000∀0≤r≤n.


## 2.1. Function Spaces and Operators

LetL2(Ω,g)=(L2(Ω))2 and H01(Ω,g)=(H01(Ω))2 be endowed, respectively, with the inner products
(8)(u,v)g=∫Ωu·vgdx,u,v∈L2(Ω,g),((u,v))g=∫Ω∑j=12‍∇uj·∇vjgdx,u=(u1,u2),0000000000000l0000v=(v1,v2)∈H01(Ω,g),
and norms |u|2=(u,u)g, ∥u∥2=((u,u))g. Thanks to assumption (G), the norms |·| and ∥·∥ are equivalent to the usual ones in (L2(Ω))2 and in (H01(Ω))2.Let(9)𝒱={u∈(C0∞(Ω))2:∇·(gu)=0}.
Denote by Hg the closure of 𝒱 in L2(Ω,g), and denote by Vg the closure of 𝒱 in H01(Ω,g). It follows that Vg⊂Hg≡Hg′⊂Vg′, where the injections are dense and continuous. We will use ∥·∥* for the norm in Vg′, and 〈·,·〉 for duality pairing between Vg and Vg′.We now define the trilinear formb by
(10)b(u,v,w)=∑i,j=12‍∫Ω‍ui∂vj∂xiwjgdx,
whenever the integrals make sense. It is easy to check that if u,v,w∈Vg, then
(11)b(u,v,w)=-b(u,w,v).
Hence
(12)b(u,v,v)=0,∀u,v∈Vg.SetA:Vg→Vg′ by 〈Au,v〉=((u,v))g, B:Vg×Vg→Vg′ by 〈B(u,v),w〉=b(u,v,w). Denote D(A)={u∈Vg:Au∈Hg}, then D(A)=H2(Ω,g)∩Vg and Au=-PgΔu, for all u∈D(A), where Pg is the ortho-projector from L2(Ω,g) onto Hg. Consequently, there exists an orthogonal basis of Hg consisting of the eigenvectors wj of A:
(13)Awj=λjwj,0<λ1≤λ2≤⋯,0000000000λj⟶∞asj⟶∞.
Furthermore, we can also define the sth power As of A for all s∈ℝ. The space D(As) is the Hilbert space when equipped with the scalar product (As,As) and norm |As.|, where (·,·) and |·| denote the scalar product and norm in Hg. In particular, D(A0)=Hg and D(A1/2)=Vg.LetHm=span{w1,…,wm}. Then, the following estimates hold:
(14)λ1|v|2≤∥v∥2∀v∈Vg,∥vm∥2≤λm|vm|2∀vm∈Hm,(15)λm+1|wm|2≤∥wm∥2∀wm∈Vg∖Hm.Using the Hölder inequality, the Ladyzhenskaya inequality (whenn=2):
(16)|u|L4≤c|u|1/2|∇u|1/2,∀u∈H01(Ω),
and the interpolation inequalities, as in [15, 16], one can prove the following.Lemma 1.
Ifn=2, then(17)|b(u,v,w)|≤{c0|u|1/2∥u∥1/2∥v∥|w|1/2∥w∥1/2,00000000000000∀u,v,w∈Vg,c0|u|1/2∥u∥1/2∥v∥|Aw|1/2|w|1/2,∀u∈Vg,v∈D(A),w∈Hg,c0|u|1/2|Au|1/2∥v∥|w|,∀u∈D(A),v∈Vg,w∈Hg,c0|u|∥v∥|w|1/2|Aw|1/2,∀u∈Hgv∈Vg,w∈D(A),(18)|b(u,v,w)|+|b(w,v,u)|≤c0|u|1/2∥u∥1/2∥v∥1/2|Av|1/2|w||b(u,v,w)|+|b(w,vu)|≤c0∀u∈Vg,v∈D(A),w∈Hg,(19)|b(u,v,w)|+|b(w,v,u)|≤c0(1+ln|Au|2λ1∥u∥2)1/2∥u∥∥v∥|w||b(uv)|2+∀u∈D(A),v∈Vg,w∈Hg,
where c0 are appropriate constants depending only on Ω.Lemma 2 (see [3]).
Letu∈L2(0,T;D(A))∩L∞(0,T;Vg), then the function Bu defined by
(20)(Bu(t),v)g=b(u(t),u(t),v),∀v∈Hg,a.e.t∈[0,T],
belongs to L4(0,T;Hg), and therefore also belongs to L2(0,T;Hg).Lemma 3 (see [4]).
Letu∈L2(0,T;Vg), then the function Cu defined by
(21)(Cu(t),v)g=((∇gg·∇)u,v)g=b(∇gg,u,v),∀v∈Vg,
belongs to L2(0,T;Hg), and hence also belongs to L2(0,T;Vg′). Moreover,
(22)|Cu(t)|≤|∇g|∞m0·∥u(t)∥,fora.e.t∈(0,T),∥Cu(t)∥*≤|∇g|∞m0λ11/2·∥u(t)∥,fora.e.t∈(0,T).Since(23)-1g(∇·g∇)u=-Δu-(∇gg·∇)u,
we have
(24)(-Δu,v)g=((u,v))g+((∇gg·∇)u,v)g=(Au,v)g+((∇gg·∇)u,v)g,∀u,v∈Vg.


## 2.2. Discrete Gronwall Inequalities

Hereafter, we will frequently use the following modified discrete Gronwall lemmas.Lemma 4 (see [12]).
Let0<Δt<1,ak,gk,hk for integers k≥0 be nonnegative numbers such that
(25)1Δt(ak+1-ak)≤gkak+hk∀0≤k≤J.
If
(26)Δt∑k=0J‍gk≤α1,Δt∑k=0J‍hk≤α2,
for J≤n0-1 and
(27)Δt∑k=k0k0+n0-1‍gk≤α1,Δt∑k=k0k0+n0-1‍hk≤α2,Δt∑k=k0k0+n0-1‍ak≤a0+α3,
for J≥n0, 0≤k0≤J+1-n0, then
(28)aJ+1≤eα1(a0+α2+α3).Lemma 5 (see [13]).
Letα and ak,bk,hk,k≥0 be nonnegative numbers such that
(29)(1+γΔt)ak+1-ak+bk+1Δt≤hkΔt∀0≤k≤J.
Then,
(30)aJ+1+Δt∑k=1J+1‍(1+γΔt)-(J+2-k)bk≤(1+γΔt)-(J+1)a0+Δt∑k=0J‍(1+γΔt)-(J+1-k)hk.Lemma 6 (see [11]).
Letβ and ak,bk,gk,hk for integers k≥0 be nonnegative numbers such that
(31)an+Δt∑k=0n‍bk≤Δt∑k=0n‍gkak+Δt∑k=0n‍hk+β∀n≥0.
Suppose that gkΔt<1, for all k≥0, and set σk=(1-gkΔt)-1. Then,
(32)an+Δt∑k=0n‍bk≤exp{Δt∑k=0n‍σkgk}(Δt∑k=0n‍hk+β)∀n≥0.
Moreover, if
(33)ar+Δt∑k=rn‍bk≤Δt∑k=rn‍gkak+Δt∑k=rn‍hk+β∀0≤r≤n,
and gkΔt<1 for all 0≤k≤n, then
(34)ar+Δt∑k=rn‍bk≤exp{Δt∑k=0n‍σkgk}(Δt∑k=0n‍hk+β);0000000000000000000000000000000∀0≤r≤n.


## 3. Existence and Some Estimates of Strong Solutions

In this section, we will prove some estimates for the strong solutionu and the Galerkin approximate solutions um of problem (1). First, with the operators defined in Section 2.1, one can write this problem as follows:
(35)dudt+νAu+νCu+B(u,u)=f(t),t>0,u(0)=u0.Definition 7.
Foru0∈Vg given, a strong solution of problem (1) is a function u∈L2(0,T;D(A))∩C([0,T];Vg) for all T>0 such that u(0)=u0, and u satisfies (35) in Hg for a.e. t>0.Theorem 8.
Suppose thatf,ft∈L∞(ℝ+;Hg) and u0∈Vg. Then, problem (1) has a unique strong solution u satisfying the following estimates for all t>0,
(36)|u(t)|2≤e-νγ0λ1t|u0|2+1ν2γ02λ12Cf2,∫0t‍eνγ0λ1s∥u∥2ds≤2νγ0|u0|2+4ν3γ03λ12eνγ0λ1tCf2,(37)∫0t‍∥u(s)∥2ds≤1νγ0|u0|2+tCf2ν2γ02λ1,(38)∥u(t)∥2≤Cu0,f,(39)τ(t)(|ut(t)|2+|Au(t)|2)+τ2(t)∥ut(t)∥2≤𝒦,(40)∫0t‍e-νγ0λ1(t-s)(|Au|2+|ut|2+τ(s)∥ut∥2∫0t‍e-νγ0λ1(t-s)+τ2(s)|utt|2+τ2(s)|Aut|2)ds≤𝒦,
where γ0=1-(|∇g|∞/m0λ11/2)>0, τ(t)=min{1,t}, Cf=supt≥0|f|, Cu0,f=C(∥u0∥,Cf), and 𝒦 is a generic positive constant depending only on the data (ν,Ω,|∇g|∞,λ1,Cf,∥u0∥). Moreover, all above estimates are also valid for the Galerkin approximate solutions um of problem (35).Proof.
We refer to [3] for the proof of existence and uniqueness of the strong solution u and estimates (36)–(38). We now prove (39)–(40).
First, we take the scalar product of (35) with eνγ0λ1tAu and ν-1eνγ0λ1tut, respectively, and add the resulting relations to obtain
(41)ddt(eνγ0λ1t∥u∥2)+eνγ0λ1t(ν|Au|2+ν-1|ut|2)+eνγ0λ1tb(u,u,Au+ν-1ut)+νeνγ0λ1t(Cu,Au+ν-1ut)g=νγ0λ1eνλ1t∥u∥2+eνγ0λ1t(f,Au+ν-1ut).
Using Lemmas 1 and 3 and Cauchy's inequality, we have
(42)|b(u,u,Au+ν-1ut)|≤14(ν|Au|2+ν-1|ut|2)+c|u|2∥u∥2,|ν(Cu,Au+ν-1ut)g|≤18(ν|Au|2+ν-1|ut|2)+c∥u∥2,|(f,Au+ν-1ut)|≤18(ν|Au|2+ν-1|ut|2)+c|f|2.
By combining these inequalities with (41), we get
(43)ddt(eνγ0λ1t∥u∥2)+12eνγ0λ1t(ν|Au|2+ν-1|ut|2)≤eνγ0λ1t(νλ1+c|u|2∥u∥2+c)∥u∥2+ceνγ0λ1t|f|2.
Integrating (43) from 0 to t and using (36)–(38), we obtain, after multiplying by e-νγ0λ1t, that
(44)e-νγ0λ1t∫0t‍eνγ0λ1sτ(s)(ν|Au|2+ν-1|ut|2)ds≤𝒦∀t≥0.
In view of (44), there exists a sequence ϵk→0 such that
(45)τ2(ϵk)(ν|Au(ϵk)|2+ν-1|ut(ϵk)|2)⟶0.
Now, differentiating (35) with respect to t yields
(46)utt+νAut+νCut+B(ut,u)+B(u,ut)=ft.
We take the scalar product (46) with 2ut to obtain
(47)ddt|ut|2+2ν∥ut∥2+2ν(Cut,ut)g+2b(ut,u,ut)+2b(u,ut,ut)=2(ft,ut)g.
By Lemma 3, we have (48)ddt|ut|2+2ν∥ut∥2≤2ν|∇g|∞m0λ11/2∥ut∥2+2|b(ut,u,ut)|+2|b(u,ut,ut)|+2(ft,ut)g.
Using Lemma1 and Cauchy's inequality, we get
(49)ddt|ut|2+νγ0∥ut∥2≤c|ut|2∥u∥2+c|ft|2.
Multiplying the last inequality byτ(t)eνγ0λ1t, we have
(50)ddt(eνγ0λ1tτ(t)|ut|2)+νγ0eνγ0λ1tτ(t)∥ut∥2≤eνγ0λ1t(1+νγ0λ1+c∥u∥2)|ut|2+ceνγ0λ1t|ft|2.
Therefore, integrating (50) from ϵk to t, letting ϵk→0, and using (44) and (45), one finds, after multiplying by e-νγ0λ1t, that
(51)τ(t)|ut(t)|2+νe-νγ0λ1t∫0t‍eνγ0λ1sτ(s)∥ut∥2ds≤𝒦∀t≥0.
Moreover, in view of (35), (36)–(38), and (51), we see that
(52)τ(t)|Au(t)|2≤c|ft(t)|2+cτ(t)|ut(t)|2+c|u(t)|2∥u(t)∥4≤𝒦∀t≥0.
Also, in view of (51), there exists a sequence ϵk→0 such that
(53)τ2(ϵk)∥ut(ϵk)∥2⟶0.
We again take the scalar product (46) with 2Aut to obtain
(54)ddt∥ut∥2+2ν|Aut|2+2ν(Cut,Aut)g+2b(ut,u,Aut)+2b(u,ut,Aut)=2(ft,Aut)g.
By Lemma 3 and Cauchy's inequality, we have
(55)ddt∥ut∥2+2ν|Aut|2≤2ν|∇g|∞m0λ11/2|Aut|2+ν|∇g|∞λ11/22m0∥ut∥2+2b|(ut,u,Aut)|+2|b(u,ut,Aut)|+2(ft,Aut)g.
Using Lemma 1, Cauchy's inequality, and Young's inequality, we obtain
(56)ddt∥ut∥2+νγ0|Aut|2≤c∥u∥4∥ut∥2+c|ft|2+ν|∇g|∞λ11/22m0∥ut∥2.
Multiplying the last inequality by τ2(t)eνγ0λ1t, we have
(57)ddt(eνγ0λ1tτ2(t)∥ut∥2)+νγ0τ2(t)eνγ0λ1t|Aut|2≤ceνγ0λ1t|ft|2+eνγ0λ1t(2+νγ0λ1+c∥u∥4+ν|∇g|∞λ11/22m0)τ(t)∥ut∥2.
Integrating (57) from ϵk to t, letting ϵk→0, and using (36)–(38), (51), and (53), we obtain, after a final multiplication by e-νγ0λ1t,
(58)τ2(t)∥ut(t)∥2+νγ0e-νγ0λ1t∫0t‍eνγ0λ1sτ2(s)|Aut|2ds≤𝒦000000000000000000000000000000000000000∀t≥0.
Using Lemmas 1 and 3 and (46), we deduce that
(59)e-νγ0λ1t∫0t‍τ2(t)eνγ0λ1s|utt|2ds≤ce-νγ0λ1t×∫0t‍τ2(t)eνγ0λ1s(|Aut|2+∥u∥2|Aut|2+|ft|2)ds≤c∫0t‍τ2(t)eνγ0λ1s|Aut|2+∥u∥2|Aut|2ds+∀t≥0.
Combining (44), (51), (52), (58), and (59) yields (39) and (40).
Finally, we observe that the problem for the approximate solutionum is similar to problem (35), and
(60)|um(0)|≤|u0|,∥um(0)∥≤∥u0∥,
so um satisfies the same estimates as those for the strong solution u of problem (35).


## 4. Spectral Galerkin Method in Space

For a fixed integerm, let Pm be the orthogonal projection of Hg onto Hm=span{w1,…,wm} and Qm=I-Pm. Then, every solution u of problem (1) can be decomposed uniquely into (61)u=p+q,wherep=Pmu,q=Qmu.
Now, we apply Pm and Qm to (35) to obtain
(62)pt+νAp+νCp+PmB(u,u)=Pmf∀t>0,(63)qt+νAq+νCq+QmB(u,u)=Qmf∀t>0,
and the initial conditions p(0)=Pmu0, q(0)=Qmu0.Using Theorem8 and the property of Pm, we arrive at the following estimates of q(t)=Qmu(t):
(64)|q(t)|2≤𝒦λm+1-1,τ(t)(|q(t)|2+λm+1-1∥q(t)∥2)≤2𝒦λm+1-2∀t≥0,(65)e-νγ0λ1t∫0t‍eνγ0λ1s(|q|2+λm+1-1∥q∥2)ds≤2𝒦λm+1-2∀t≥0.
We now define the spectral Galerkin method as follows: find um(t)∈Hm such that
(66)umt+νAum+νCum+PmB(um,um)=Pmf∀t>0,
with the initial condition um(0)=Pmu0.In order to give an analysis of the erroru-um in the L2-norm, we begin with a technical result concerning a dual linearized g-Navier-Stokes problem which is a similar problem to that used in [17]. We consider, for any given t>0 and ξ∈L2(0,t;Hg), the dual problem: find Φ(s)∈Hm such that
(67)(v,Φs)g-a(v,Φ)-ν(Cv,Φ)g-b(v,um,Φ)-b(um,v,Φ)=(v,eνγ0λ1sξ)g,0≤s<t,
for all v∈Hm with Φ(t)=0. It is easy to see that (67) is a well-posed problem and has a unique solution Φ∈L∞(0,t;Vg)∩L2(0,t;D(A)).Next, we prove a regularity result of problem (67).Lemma 9.
Ifu0∈Vg, then the solution Φ(s) of problem (67) satisfies
(68)e-νγ0λ1s∥Φ(s)∥2+∫st‍e-νγ0λ1r(|AΦ|2+|Φs|2)dr≤𝒦exp{2νγ0c02∫0t‍∥um∥2dr}∫0t‍eνγ0λ1r|ξ|2dr.Proof.
Takingv=-2Φ in (67), we obtain
(69)-dds|Φ|2+2ν∥Φ∥2+2ν(CΦ,Φ)g+2b(Φ,um,Φ)=-2(Φ,eνγ0λ1sξ)g∀0≤s<t.
Using Lemma 3, we have
(70)-dds|Φ|2+2νγ0∥Φ∥2≤2|(Φ,eνγ0λ1sξ)g|+2|b(Φ,um,Φ)|00000000000000000000000000000000000000∀0≤s<t.
Using Lemma 1 and Cauchy's inequality, we get
(71)2|b(Φ,um,Φ)|≤νγ02∥Φ∥2+2νγ0c02∥um∥2|Φ|2,2|(Φ,eνγ0λ1sξ)g|≤νγ04∥Φ∥2+4νγ0λ1e2νγ0λ1s|ξ|2.
Then, we have
(72)-dds|Φ|2+54νγ0∥Φ∥2≤2νγ0c02∥um∥2|Φ|2+4νγ0λ1e2νγ0λ1s|ξ|2∀0≤s<t.
Multiplying this inequality by e-νγ0λ1s and using (14), we obtain
(73)-dds(e-νγ0λ1s|Φ|2)+νγ04e-νγ0λ1s∥Φ∥2≤2νγ0c02e-νγ0λ1s∥um∥2|Φ|2+4νγ0λ1eνγ0λ1s|ξ|2.
Multiplying this inequality by e(2/νγ0)c02∫0s‍∥um∥2dr yields
(74)-dds(e(2/νγ0)c02∫0s‍∥um∥2dre-νγ0λ1s|Φ|2)+νγ04e(2/νγ0)c02∫0s‍∥um∥2dre-νγ0λ1s∥Φ∥2≤4νγ0λ1e(2/νγ0)c02∫0s‍∥um∥2dreνγ0λ1s|ξ|2.
Integrating (74) from s to t and noting that Φ(t)=0, we obtain, after multiplying by e-(2/νγ0)c02∫0s‍∥um∥2dr, that
(75)e-νγ0λ1s|Φ(s)|2+νγ04∫st‍e-νγ0λ1r∥Φ∥2dr≤4νγ0λ1e(2/νγ0)c02∫0t‍∥um∥2dr∫0t‍eνγ0λ1r|ξ|2dr∀0≤s≤t.
Moreover, inserting v=-2AΦ into (67), we get
(76)-ddt∥Φ∥2+2ν|AΦ|2+2ν(CAΦ,Φ)g-2{b(AΦ,Φ,um)+b(um,Φ,AΦ)}=-2(AΦ,eνγ0λ1sξk)g.
Using Lemma 3, (14), and Cauchy's inequality, we have
(77)2ν(CAΦ,Φ)g≤2ν|∇g|∞m0λ11/2|AΦ|2+ν|∇g|∞λm2m0λ11/2∥Φ∥2.
Therefore,
(78)-ddt∥Φ∥2+2νγ0|AΦ|2≤2|(AΦ,eνγ0λ1sξk)g|+2|b(AΦ,Φ,um)+b(um,Φ,AΦ)|+ν|∇g|∞λm2m0λ11/2∥Φ∥2.
Multiplying this inequality, by e-νγ0λ1s and using (14), we obtain
(79)-ddt(e-νγ0λ1s∥Φ∥2)+νγ0e-νγ0λ1s|AΦ|2≤2|(AΦ,ξk)g|+2e-νγ0λ1s|b(AΦ,Φ,um)+b(um,Φ,AΦ)|+ν|∇g|∞λm2m0λ11/2e-νγ0λ1s∥Φ∥2.
Using Lemma 1, Cauchy's inequality and Young's inequality, we have
(80)2|b(AΦ,Φ,um)|+2|b(um,Φ,AΦ)|≤c0|AΦ|3/2∥Φ∥1/2∥um∥1/2|um|1/2≤νγ02|AΦ|2+c∥Φ∥2∥um∥2|um|2,2|(AΦ,ξk)g|≤νγ04e-νγ0λ1s|AΦ|2+4νγ0eνγ0λ1s|ξk|2.
Combining the above estimates with (79) yields
(81)-ddt(e-νγ0λ1s∥Φ∥2)+νγ04e-νγ0λ1s|AΦ|2≤ce-νγ0λ1s∥Φ∥2∥um∥2|um|2+ν|∇g|∞λm2m0λ11/2e-νγ0λ1s∥Φ∥2+4νγ0eνγ0λ1s|ξ|2.
Integrating this inequality from s to t and using (75), we have
(82)e-νγ0λ1s∥Φ∥2+νγ04∫st‍e-νγ0λ1r|AΦ|2dr≤c(1+ν|∇g|∞λm2m0λ11/2+supt≥0|um(t)|2∥um(t)∥2)×e(2/νγ0)c02∫0t‍∥um∥2dr∫0t‍eνγ0λ1r|ξ|2dr.
Taking v=Φs in (67), then using (14), Lemmas 1 and 3, we obtain
(83)|Φs|≤ν|AΦ|+ν|∇g|∞λmm0λ11/2∥Φ∥+c∥um∥|AΦ|+eνγ0λ1s|ξ|.
Hence,
(84)e-νγ0λ1s|Φs|2≤c(1+∥um∥2)e-νγ0λ1s|AΦ|2+ce-νγ0λ1s∥Φ∥2+ceνγ0λ1s|ξ|2.
Integrating (84) from s to t and using (82), (75), and Theorem 8, we complete the proof.Lemma 10.
Ifu0∈Vg, then the error Pmu(t)-um(t) satisfies
(85)|Pmu(t)-um(t)|2+νγ04e-νγ0λ1t∫0t‍eνγ0λ1s∥Pmu-um∥2ds≤𝒦e(2/νγ0)c02∫0t‍∥u(r)∥2drλm+1-1∀t≥0.Proof.
We sete(t)=Pmu(t)-um(t) and subtract (66) from (62) to obtain
(86)et+νAe+νCe+PmB(Qmu+e,u)+PmB(um,Qum+e)=0,
with e(0)=0. Taking the scalar product of (86) with 2e, we obtain
(87)ddt|e|2+2ν∥e∥2+2(Ce,e)g+2(b(e,u,e)+b(um,Qmu,e)+b(Qmu,u,e))=0.
Using Lemma 3, we get
(88)ddt|e|2+2νγ0∥e∥2≤2|b(e,u,e)+b(um,Qmu,e)+b(Qmu,u,e)|.
Multiplying the last inequality by eνγ0λ1t and using (14), we obtain
(89)ddt(eνγ0λ1t|e|2)+νγ0eνγ0λ1t∥e∥2≤2eνγ0λ1t|b(e,u,e)+b(um,Qmu,e)+b(Qmu,u,e)|.
Due to Lemma 1 and Cauchy's inequality, we have
(90)2|b(e,u,e)|≤2c0|e|∥e∥∥u∥≤νγ02∥e∥2+2νγ0c02∥u∥2|e|2,2|b(um,Qmu,e)|+2|b(Qmu,u,e)|≤2c0λ1-1/2|Qmu|(|Au|+|Aum|)∥e∥≤νγ04∥e∥2+c(|Au|2+|Aum|2)|Qmu|2.
Combining (89) with the above estimate yields
(91)ddt(eνγ0λ1t|e|2)+νγ04eνγ0λ1t∥e∥2≤2νγ0c02eνγ0λ1t∥u∥2|e|2+ceνγ0λ1t(|Au|2+|Aum|2)|Qmu|2.
Multiplying (91) by e-(2/νγ0)c02∫0s‍∥u∥2dr yields
(92)ddt(e-(2/νγ0)c02∫0s‍∥u∥2dreνγ0λ1t|e|2)+νγ04e-(2/νγ0)c02∫0s‍∥u∥2dreνγ0λ1t∥e∥2≤ce-(2/νγ0)c02∫0s‍∥u∥2dreνγ0λ1t(|Au|2+|Aum|2)|Qmu|2≤ceνγ0λ1t(|Au|2+|Aum|2)|Qmu|2.
Integrating this inequality from 0 to t and using (64), Theorem 8, we obtain, after a final multiplication by e(2/νγ0)c02∫0s‍∥u∥2dre-νγ0λ1t, that
(93)|e|2+νγ04e-νγ0λ1t∫0t‍eνγ0λ1s∥e∥2ds≤𝒦e(2/νγ0)c02∫0t‍∥u∥2drλm+1-1∀t≥0,
which is (85).Lemma 11.
Ifu0∈Vg, then the error Pmu(t)-um(t) satisfies the following bound:
(94)τ(t)|Pmu(t)-um(t)|2≤𝒦e(4/νγ0)c02∫0t‍(∥u∥2+∥um∥2)drλm+1-2∀t≥0.Proof.
Takev=e(s) and ξ=e(s) in (67) to obtain
(95)(e,Φs)g-a(e,Φ)-ν(Ce,Φ)g-b(e,um,Φ)-b(um,e,ϕ)=eνγ0λ1s|e|2.
Multiplying (86) by Φ, we have
(96)(es,Φ)g+a(e,Φ)+ν(Ce,Φ)g+b(um,e+Qmu,Φ)+b(e+Qmu,um,Φ)+b(u-um,u-um,Φ)=0.
Adding (96) and (95), we get
(97)eνγ0λ1s|e|2=dds(e,Φ)g+b(um,Qmu,Φ)+b(Qmu,um,Φ)+b(u-um,u-um,Φ).
Using Lemma 1, we have
(98)|b(um,Qmu,Φ)|+|b(Qmu,um,Φ)|≤c∥um∥|Qmu||AΦ|,|b(u-um,u-um,Φ)|≤c|u-um||u-um|∥AΦ∥.
Hence, we deduce from (97) that
(99)eνγ0λ1s|e|2≤dds(e,Φ)g+c(∥um∥|Qmu|+|u-um|∥u-um∥)|AΦ|.
Integrating (99) from 0 to t and noting that e(0)=Φ(t)=0, we obtain
(100)∫0t‍eνγ0λ1s|e|2ds≤c(∫0t‍eνγ0λ1s(∥um∥2|Qmu|2+|u-um|2∥u-um∥2)ds)1/2×(∫0t‍e-νγ0λ1s|AΦ|2ds)1/2.
Using (64), (65), Theorem 8, and Lemmas 9 and 10, we deduce from (100) that(101)e-νγ0λ1t∫0t‍eνγ0λ1s|e|2ds≤𝒦e(4/νγ0)c02∫0t‍(∥u∥2+∥um∥2)drλm+1-2.
Now, multiplying (91) by τ(t), we have
(102)ddt(τ(t)eνγ0λ1t|e|2)+νγ04τ(t)eνγ0λ1t∥e∥2≤(1+2νγ0c02∥u∥2)eνγ0λ1t|e|2+ceνγ0λ1tτ(t)(|Au|2+|Aum|2)|Qmu|2.
Integrating (102) from 0 to t and using Theorem 8, we obtain, after a final multiplication by e-νγ0λ1t,
(103)τ(t)|e(t)|2+νγ04e-νγ0λ1t∫0t‍τ(s)eνγ0λ1s∥e∥2ds≤𝒦e-νγ0λ1t∫0t‍eνγ0λ1t(|e|2+|Qmu|2)ds.
Using (65) and (101) in (103) yields
(104)τ(t)|e(t)|2+νγ04e-νγ0λ1t∫0t‍τ(s)eνγ0λ1s∥e∥2ds≤𝒦e(4/νγ0)c02∫0t‍(∥u∥2+∥um∥2)drλm+1-2,
which is (94).Finally, by combining Lemma11 with (64) and using Theorem 8, we get the following error estimate.Theorem 12.
Ifu∈Vg, then the error u(t)-um(t) satisfies the following bound:
(105)τ(t)|u(t)-um(t)|2≤𝒦e(8/ν3γ03λ1)c02Cf2tλm+1-2∀t≥0.


## 5. Spectral Galerkin Method in Space and Time

### 5.1. Stability Analysis

In this subsection, we consider the semi-implicit Euler scheme applied to the spatially discrete spectral Galerkin approximation, show the stability of this scheme, and establish some preliminaries related to the error analysis uniform in time.We consider the semi-implicit Euler scheme and define recursively a solution{umn}⊂Hm such that
(106)dtumn+1+νAumn+1+νCumn+1+PmB(umn,umn)=Pmfn+1,
for n≥0 with the initial condition um0=Pmu0, where Δt>0 is a time step such that n0Δt=1 for some integer n0 and
(107)dtumn+1=1Δt(umn+1-umn),(108)fn+1=1Δt∫tntn+1‍f(t)dt,tn=nΔt.
In order to derive the L2-bound on the error um(tn)-umn, we will begin with a time discrete duality argument which is similar to the one used in [11, 17]. We consider the dual scheme correponding to scheme (106): for any fixed n≥1 and umk-1,ξk∈Hm, 1≤k≤n, find Φk-1∈Hm, 1≤k≤n such that
(109)(v,dtΦk)g-a(v,Φk-1)-ν(Cv,Φk-1)g-b(v,umk-1,Φk-1)-b(umk-1,v,Φk-1)=(1+νγ0λ1Δt)k(v,ξk)g-b(umk-1,v,Φk-1)==(1+νγ0λ1Δt)k∀v∈Hm,
with an initial condition Φn=0.The following theorem provides theH2-stability of scheme (106).Theorem 13.
Under the assumptions of Theorem8, if Δt and m satisfy the following condition:
(110)8νγ0c02lnλmλ1ΔtCu0,f≤1,
then the semi-implicit Euler scheme is the H2-stability; that is,
(111)Δt∑k=1n‍∥umk∥2≤1νγ0|u0|2+1ν2γ02λ1Cf2tn∀n≥0,(112)Δt∑k=k0+1n‍∥umk∥2≤1νγ0|u0|2+2ν3γ03λ12Cf2+2ν2γ02λ1Cf2(tn-tk0)0000000000000000000000000000∀n≥k0≥0,(113)|umn|2≤(1+νγ0λ1Δt)-n|u0|2+1ν2γ02λ12Cf2∀n≥0,(114)∥umn∥2≤Cu0,f∀n≥0,(115)Δt∑k=1n‍(1+νγ0λ1Δt)-(n-1-k)×(|Aumk|2+|dtumk|2+τ(tk)∥dtumk∥2)+τ(tn)|Aumn|2+τ(tn)|dtumn|2≤𝒦∀n≥0.Proof.
Clearly, scheme (106) defines a unique sequence {umn}⊂Hm. Now, we will prove (111)–(115).
Taking the scalar product of (106) with 2umn+1Δt, we obtain
(116)|umn+1|2-|umn|2+|umn+1-umn|2+2ν∥umn+1∥2Δt+2νΔt(Cumn+1,umn+1)g+2b(umn,umn-umn+1,umn+1)Δt=2(fn+1,umn+1)gΔt.
Using Lemma 3 and Cauchy's inequality, we have
(117)|umn+1|2-|umn|2+|umn+1-umn|2+2ν∥umn+1∥2Δt≤2ν|∇g|∞m0λ11/2∥umn+1∥2Δt+2(fn+1,umn+1)gΔt-2b(umn,umn-umn+1,umn+1)Δt.
Hence,
(118)|umn+1|2-|umn|2+|umn+1-umn|2+2νγ0∥umn+1∥2Δt≤2(fn+1,umn+1)gΔt-2b(umn,umn-umn+1,umn+1)Δt.
By (19), we have
(119)2(fn+1,umn+1)g≤νγ02∥umn+1∥2+2νγ0λ1|fn+1|2Δt,2|b(umn,umn-umn+1,umn+1)|≤νγ02∥umn+1∥2+L(umn)|umn-umn+1|2,
where
(120)L(umn)=8νγ0c02(1+ln|Aumn|2λ1∥umn∥2)∥umn∥2.
Substituting those into (118), we get
(121)|umn+1|2-|umn|2+(1-L(umn)Δt)|umn+1-umn|2+νγ0∥umn+1∥2Δt≤2νγ0λ1|fn+1|2Δt.
Next, by taking the scalar product of (106) with 2Aumn+1Δt, we obtain
(122)∥umn+1∥2-∥umn∥2+∥umn+1-umn∥2+2ν|Aumn+1|2Δt+2νΔt(Cumn+1,Aumn+1)g+2b(umn,umn,Aumn+1)Δt=2(fn+1,Aumn+1)gΔt.
Using Lemma 3 and Cauchy's inequality, we have
(123)2ν(Cumn+1,Aumn+1)g≤2ν|∇g|∞m0λ11/2|Aumn+1|2+ν|∇g|∞λ11/22m0∥umn+1∥2.
Substituting into (122), we get
(124)∥umn+1∥2-∥umn∥2+∥umn+1-umn∥2+2νγ0|Aumn+1|2Δt≤2(fn+1,Aumn+1)gΔt-2b(umn,umn,Aumn+1)Δt+ν|∇g|∞λ11/22m0∥umn+1∥2Δt.
Due to (17)–(19), Cauchy's inequality, and Young's inequality, we have
(125)2|b(umn,umn+1-umn,Aumn+1)|≤νγ04|Aumn+1|2+L(umn)∥umn+1-umn∥2,2|b(umn,umn+1,Aumn+1)|≤νγ0|Aumn+1|2+2(νγ0)-3c04|umn|2∥umn∥2∥umn+1∥2,2(fn+1,Aumn+1)g≤νγ02|Aumn+1|2+2νγ0Cf2.
By combining (124) with the above estimates, we obtain
(126)∥umn+1∥2-∥umn∥2+(1-L(umn)Δt)∥umn+1-umn∥2≤2(νγ0)-3c04|umn|2∥umn∥2∥umn+1∥2Δt+2νγ0Cf2Δt+ν|∇g|∞λ11/22m0∥umn+1∥2Δt.
On the other hand, from (17)–(19), Cauchy's inequality, and Young's inequality, we have
(127)2|b(umn,umn+1-umn,Aumn+1)|≤νγ04|Aumn+1|2+L(umn)∥umn+1-umn∥2,2|b(umn,umn+1,Aumn+1)|≤νγ08|Aumn+1|2+c|umn|2∥umn∥2∥umn+1∥2,2(fn+1,Aumn+1)g≤νγ08|Aumn+1|2+cCf2.
Combining (124) with those estimates, we obtain
(128)(1+νγ0λ1Δt)∥umn+1∥2-∥umn∥2+(1-L(umn)Δt)∥umn+1-umn∥2+νγ02|Aumn+1|2Δt≤c|umn|2∥umn∥2∥umn+1∥2Δt+cCf2Δt+c∥umn+1∥2Δt.
Next, from (106) we have
(129)|dtumn+1|2+ν(Aumn+1,dtumn+1)+ν(Cumn+1,dtumn+1)g+b(un,un,dtumn+1)=(fn+1,dtumn+1)g.
Using (11), Lemmas 1 and 3, and Cauchy's inequality, we get
(130)|b(un,un,dtumn+1)|≤c|umn|2∥umn∥∥umn+1∥2,ν|(Aumn+1,dtumn+1)g|≤14|dtumn+1|2+ν2|Aumn+1|2,ν|(Cumn+1,dtumn+1)g|≤14|dtumn+1|2+c∥umn+1∥2,|(fn+1,dtumn+1)g|≤14|dtumn+1|2+cCf2.
Therefore, we have
(131)γ016ν|dtumn+1|2-νγ04|Aumn+1|2≤c|umn|2∥umn∥2∥umn+1∥2+cCf2+c∥umn+1∥2.
Combining this inequality with (128) yields
(132)(1+νγ0λ1Δt)∥umn+1∥2-∥umn∥2+(1-L(umn)Δt)∥umn+1-umn∥2+γ016ν|dtumn+1|2Δt+νγ04|Aumn+1|2Δt≤c|umn|2∥umn∥2∥umn+1∥2Δt+cCf2Δt+c∥umn+1∥2Δt.
Moreover, we deduce from (106) that
(133)dttumn+1+νAdtumn+1+νCdtumn+1+PmB(dtumn,umn)+PmB(umn,dtumn)=Pmdtfn+1,
where dtum0 is defined by
(134)dtum0+νAum0+νCum0+PmB(um0,um0)=Pmf(0).
Taking the scalar product of (133) with 2dtumn+1Δt, we obtain
(135)|dtumn+1|2-|dtumn|2+|dtumn+1-dtumn|2+2ν∥dtumn+1∥2Δt+2ν(Cdtumn+1,dtumn+1)Δt+2b(dtumn-dtumn+1,umn,dtumn+1)Δt+2b(dtumn+1,umn,dtumn+1)Δt+2b(dtumn+1,dtumn-dtumn+1,dtumn+1)Δt=2(dtfn+1,dtumn+1)Δt.
By Lemma 3, we get
(136)|dtumn+1|2-|dtumn|2+|dtumn+1-dtumn|2+2νγ0∥dtumn+1∥2Δt≤2(dtfn+1,dtumn+1)Δt-2b(dtumn-dtumn+1,umn,dtumn+1)Δt-2b(dtumn+1,umn,dtumn+1)Δt-2b(dtumn+1,dtumn-dtumn+1,dtumn+1)Δt.
Using Lemma 1 and (19), we have
(137)2|b(dtumn-dtumn+1,umn,dtumn+1)|+2|b(dtumn+1,dtumn-dtumn+1,dtumn+1)|≤νγ02∥dtumn+1∥2+12(L(umn)+L(umn+1))|dtumn+1-dtumn|2,2|b(dtumn+1,umn,dtumn+1)|≤νγ08∥dtumn+1∥2+8νγ0c02∥umn∥2|dtumn+1|2,2(dtfn+1,dtumn+1)≤νγ08∥dtumn+1∥2+8νγ0λ1supt≥0|ft(t)|2.
By combining (136) with the above estimates, we arrive at
(138)(1+νγ0λ1Δt)|dtumn+1|2-|dtumn|2+(1-12(L(umn)+L(umn+1))Δt)|dtumn+1-dtumn|2+νγ04∥dtumn+1∥2Δt≤8νγ0c02∥umn∥2|dtumn+1|2Δt+νγ04∥dtumn+1∥2Δt≤+8νγ0λ1supt≥0|ft(t)|2.
Multiplying (138) by τ(tn+1) and noting that τ(tn+1)≤τ(tn)+Δt, we obtain
(139)(1+νγ0λ1Δt)τ(tn+1)|dtumn+1|2-τ(tn)|dtumn|2+(1-12(L(umn)+L(umn+1))Δt)×τ(tn+1)|dtumn+1-dtumn|2+νγ04τ(tn+1)∥dtumn+1∥2Δt≤|dtumn|2Δt+8νγ0c02∥umn∥2|dtumn+1|2Δt+8νγ0λ1supt≥0|ft(t)|2Δt.
Now, we will prove (111)–(115) by induction. Obviously, (111)–(115) are true for n=0. Assuming that (111)–(115) hold for n=0,1,…,J, we need to prove (111)–(115) for n=J+1. In view of (110) and the inductive assumption, we obtain
(140)1-L(umn)Δt≥0∀0≤n≤J.
Summing (121) from 0 to J and k0+1 to J, respectively, and using (140), we obtain (111) and (112) with n=J+1 after a final multiplication by (νγ0)-1. Noting that λ1|v|2≤∥v∥2, using (140) in (121), and then applying Lemma 5 with
(141)ak=|umk|2,bk=0,hk=2νγ0λ1Cf2,γ=νγ0λ1,
we obtain (113) for n=J+1.
Furthermore, setting(142)ak=∥umk∥2,gk=2(νγ0)-3c04|umk|2∥umk+1∥2,hk=2νγ0Cf2+ν|∇g|∞λ11/22m0∥umk+1∥2,
in (126), using (111)–(113) and (140), we obtain
(143)1Δt(ak+1-ak)≤gkak+hk∀0≤k≤J,
with
(144)Δt∑k=0J‍gk≤α1,Δt∑k=0J‍hk≤α1,a0≤∥u0∥2,
for all J+1≤n0-1 and
(145)Δt∑k=k0k0+n0-1‍gk≤α1,Δt∑k=k0k0+n0-1‍hk≤α2,Δt∑k=k0k0+n0-1‍ak≤a0+α3,0≤k0≤J+1-n0,
for all J+1≥n0. Applying Lemma 4 to (143), we obtain (114) for n=J+1.
Applying again Lemma4 to (132) with
(146)ak=∥umk∥2,bk=116νγ0|dtumk|2+νγ04|Aumk|2,γ=νγ0λ1,hk=c|umk|2∥umk∥2∥umk+1∥2+cCf2+c∥umk+1∥2,
and using (111)–(114) and (140), we obtain
(147)Δt∑k=1J+1‍(1+νγ0λ1Δt)-(J+2-k)(|Aumk|+|dtumk|2)≤𝒦.
Applying Lemma 5 to (139) with
(148)ak=τ(tk)|dtumk|2,bk=νγ04τ(tk)∥dtumk∥2,γ=νγ0λ1,hk=|dtumk|2Δt+8νγ0c02∥umk∥2|dtumk+1|2+8νγ0λ1supt≥0|ft(t)|2,
and using (114), (140), and (147) yields
(149)τ(tJ+1)|dtumJ+1|2+νγ04Δt∑k=1J+1‍(1+νγ0λ1Δt)-(J+2-k)τ(tk)∥dtumk∥2≤𝒦.
Finally, due to (17)–(19), we have
(150)|b(umJ,uJ-umJ+1,AumJ+1)|≤νγ08|AumJ+1|2+cL(umJ)∥dtumJ+1∥2Δt2,|b(umJ,umJ+1,AumJ+1)|≤νγ08|AumJ+1|2+c|umJ|2∥umJ∥2∥umJ+1∥2.
Combining these inequalities with (106) and (140) yields
(151)τtJ+1|AumJ+1|2≤cτ(tJ+1)|dtumJ+1|2+c|umJ|2∥umJ∥2∥umJ+1∥2+cCf2+cτ(tJ+1)∥dtumJ+1∥2Δt.
Using (113), (114) and (149) in this inequality yields
(152)τ(tJ+1)|AumJ+1|2≤𝒦.
Combining this inequality with (149) and (147) implies (115) for n=J+1.The following lemma provides the stability of scheme (109).Lemma 14.
Ifu0∈Vg,Δt and m satisfy (110), then problem (109) admits a unique solution {Φk}0n⊂Hm satisfying
(153)(1+νγ0λ1Δt)-r∥Φr∥2+Δt∑k=rn‍(1+νγ0λ1Δt)-(k-1)|AΦk|2≤𝒦exp(4νγ0c02Δt∑k=1n‍∥umk∥2)Δt∑k=1n‍(1+νγ0λ1Δt)k|ξk|2.Proof.
In view of (17), (110), and (114), we can prove that the following bilinear form
(154)1Δt(v,Φ)+a(v,Φ)+ν(Cv,Φ)+b(v,umk-1,Φ)+b(umk-1,v,Φ)∀v,Φ∈Hm
is elliptic. Hence, problem (109) has a unique solution Φk-1 for 1≤k≤n. Next, by taking v=-2(1+νγ0λ1Δt)-kΦk-1Δt in (109) and using (11), we have
(155)-2(1+νγ0λ1Δt)-k(Φk-1,Φk)g+2(1+νγ0λ1Δt)-k|Φk-1|2+2ν(1+νγ0λ1Δt)-k∥Φk-1∥2Δt+2ν(1+νγ0λ1Δt)-k(CΦk-1,Φk-1)gΔt+2(1+νγ0λ1Δt)-kb(Φk-1,umk-1,Φk-1)Δt=-2(Φk-1,ξk)gΔt.
Using Lemma 3, we get
(156)2(1+νγ0λ1Δt)-k|Φk-1|2+2νγ0(1+νγ0λ1Δt)-k∥Φk-1∥2Δt≤2|(Φk-1,ξk)g|Δt+2(1+νγ0λ1Δt)-k|(Φk-1,Φk)g|+2(1+νγ0λ1Δt)-k|b(Φk-1,umk-1,Φk-1)|Δt.
Using (17) and Cauchy's inequality, we have
(157)2|(Φk-1,Φk)g|≤|Φk-1|2+|Φk|2,2|(Φk-1,ξk)g|≤νγ04(1+νγ0λ1Δt)-k∥Φk-1∥2+4νγ0λ1(1+νγ0λ1Δt)k|ξk|2,2|b(Φk-1,umk-1,Φk-1)|≤νγ02∥Φk-1∥2+2νγ0c02|Φk-1|2∥uk-1∥2.
Combining above inequalities and using (14), we obtain
(158)(1+νγ0λ1Δt)-(k-1)|Φk-1|2-(1+νγ0λ1Δt)-k|Φk|2+νγ04(1+νγ0λ1Δt)-k∥Φk-1∥2Δt≤2νγ0c02(1+νγ0λ1Δt)-k∥umk-1∥2|Φk-1|2Δt+4νγ0λ1(1+νγ0λ1Δt)k|ξk|2Δt.
Summing (158) from r+1 to n, noting that Φn=0, and using Theorem 13, we arrive at
(159)(1+νγ0λ1Δt)-r|Φr|2+νγ04Δt∑k=rn‍(1+νγ0λ1Δt)-(k+1)∥Φk∥2≤2νγ0c02Δt∑k=rn‍(1+νγ0λ1Δt)-k∥umk∥2|Φk|2+4νγ0λ1Δt∑k=rn‍(1+νγ0λ1Δt)k|ξk|2
for all 0≤r≤n. Let
(160)ar=(1+νγ0λ1Δt)-r|Φr|2,br=νγ04(1+νγ0λ1Δt)-(r+1)∥Φr∥2,hr=4νγ0λ1(1+νγ0λ1Δt)r|ξk|2,gr=2νγ0c02∥umk∥2,β=0,
in (159) to obtain
(161)ar+Δt∑k=rn‍bk≤Δt∑k=rn‍gkak+Δt∑k=rn‍hk∀0≤r≤n,
with σr=(1-grΔt)-1≤2. Applying Lemma 6 to this inequality yields
(162)(1+νγ0λ1Δt)-r|Φr|2+νγ02Δt∑k=rn‍(1+νγ0λ1Δt)-(k+1)∥Φk∥2≤𝒦exp(4νγ0c02Δt∑k=1n‍∥umk∥2)Δt∑k=rn‍(1+νγ0λ1Δt)k|ξk|2.
Moreover, by taking v=-2(1+νγ0λ1Δt)-kAΦk-1Δt in (109), we have
(163)-2(1+νγ0λ1Δt)-k(AΦk-1,Φk)g+2(1+νγ0λ1Δt)-k∥Φk-1∥2+2ν(1+νγ0λ1Δt)-k|AΦk-1|2Δt+2ν(1+νγ0λ1Δt)-k(CAΦk-1,Φk-1)gΔt+2(1+νγ0λ1Δt)-kb(AΦk-1,Φk-1,umk-1)Δt+2(1+νγ0λ1Δt)-kb(umk-1,Φk-1,AΦk-1)Δt=2(AΦk-1,ξk)gΔt.
Using Lemma 3 and Cauchy's inequality, we deduce that
(164)2ν(1+νγ0λ1Δt)-k|(CAΦk-1,Φk-1)g|≤2ν(1+νγ0λ1Δt)-k|∇g|∞m0λ11/2|AΦk-1|2+c(1+νγ0λ1Δt)-k∥Φk-1∥2.
Therefore,
(165)2(1+νγ0λ1Δt)-k∥Φk-1∥2+2νγ0(1+νγ0λ1Δt)-k|AΦk-1|2Δt≤2(AΦk-1,ξk)gΔt+2(1+νγ0λ1Δt)-k(AΦk-1,Φk)g+2(1+νγ0λ1Δt)-k|b(AΦk-1,Φk-1,umk-1)|Δt+2(1+νγ0λ1Δt)-k|b(umk-1,Φk-1,AΦk-1)|Δt+c(1+νγ0λ1Δt)-k∥Φk-1∥2Δt.
Using (17)–(19) and Cauchy’s inequality, we have
(166)2|b(AΦk-1,Φk-1,umk-1)|+2|b(umk-1,Φk-1,AΦk-1)|≤νγ04|AΦk-1|2+c|umk-1|2∥umk-1∥2∥Φk-1∥2,2(AΦk-1,Φk)g=2((Φk-1,Φk))g≤∥Φk-1∥2+∥Φk∥2,2(AΦk-1,ξk)g≤νγ04(1+νγ0λ1Δt)-k|AΦk-1|2+4νγ0(1+νγ0λ1Δt)k|ξk|2.
Combining above inequality, noting that λ1∥v∥2≤|Av|2, we obtain
(167)(1+νγ0λ1Δt)-(k-1)∥Φk-1∥2-(1+νγ0λ1Δt)-k∥Φk∥2+νγ02(1+νγ0λ1Δt)-k|AΦk-1|2Δt≤c(1+νγ0λ1Δt)-k(|umk-1|2∥umk-1∥2+1)∥Φk-1∥2Δt+c(1+νγ0λ1Δt)k|ξk|2Δt
for all 0≤k≤n. Summing (167) from r+1 to n, we obtain
(168)(1+νγ0λ1Δt)-r∥Φr∥2+νγ02Δt∑k=rn‍(1+νγ0λ1Δt)-(k+1)|AΦk|2≤csupk≥0{(|umk|2∥umk∥2+1)}Δt∑k=rn‍(1+νγ0λ1Δt)-(k+1)∥Φk∥2+cΔt∑k=r+1n‍(1+νγ0λ1Δt)k|ξk|2∀0≤r≤n.
Combining (168) with (162) and using Theorem 13, we complete the proof.


### 5.2. Error Analysis

In this subsection, we will establish theH1- and L2-error estimates uniform in time for the fully discrete spectral Galerkin method with the explicit time discretization for the nonlinear term. To do this, by integrating (35) from tn to tn+1, we obtain
(169)dtum(tn+1)+νΔt∫tntn+1‍Aum(t)dt+νΔt∫tntn+1‍Cum(t)dt+1Δt∫tntn+1‍PmB(um(t),um(t))dt=fn+1.
Subtracting (106) from (153) and setting en=um(tn)-umn, we have
(170)dten+1+νAen+1+νCen+1+PmB(en,umn)+PmB(um(tn),en)=PmEn+1,
with e0=0 and
(171)En+1=νΔt∫tntn+1‍A(um(tn+1)-um(t))dt+νΔt∫tntn+1‍C(um(tn+1)-um(t))dt+1Δt∫tntn+1‍(B(um(tn),um(tn))+1Δt∫tntn+1‍-B(um(t),um(t)))dt=νΔt∫tntn+1‍(t-tn)Aumtdt+νΔt∫tntn+1‍(t-tn)Cumtdt-1Δt∫tntn+1‍(tn+1-t)(B(umt,um)+B(um,umt))dt.
To derive a bound on en, we need to provide the following estimates of En.Lemma 15.
Under the assumptions of Theorem13, the error En satisfies the following bounds:
(172)Δt∑k=1n‍(1+νγ0λ1)k-1|A-1PmEk|2≤𝒦(1+νγ0λ1)nΔt2∀n≥1,(173)Δt∑k=1n‍(1+νγ0λ1)k-1|A-1/2PmEk|2≤𝒦(1+νγ0λ1)nΔt∀n≥1,(174)Δt∑k=1n‍(1+νγ0λ1)k-1τ(tk)|A-1/2PmEk|2≤𝒦(1+νγ0λ1)nΔt2∀n≥1.Proof.
In view of (17)–(19), Lemma 3, and Theorem 8, we deduce from (171) that
(175)|A-1PmEn|=supv∈Hm|(En,v)||Av|≤Δt-1/2(∫tn-1tn‍(c1((t-tn-1)2+|∇g|∞m0(t-tn-1)2)|umt|2Δ∫tn-1tn‍t-1/2+c2(tn-t)2∥um∥2|umt|2{(c1((t-tn-1)2+|∇g|∞m0(t-tn-1)2)|umt|2})dt{∫tn-1tn‍})1/2≤cΔt-1/2(∫tn-1tn‍((t-tn-1)2+(tn-t)2∥um∥2)|umt|2dt)1/2,|A-1/2PmEn|=supv∈Hm|(En,v)|∥v∥|A-1/2PmEn|≤cΔt-1/2|A-1/2PmEn|×(∫tn-1tn‍((t-tn-1)2+(tn-t)2∥um∥2)|A-1/2PmEn|∫tn-1tn‍×∥umt∥2dt∫tn-1tn‍)1/2.
By Theorem 8, (175), and the fact that (1+νγ0λ1)k-1≤eνγ0λ1tk-1, we have
(176)(1+νγ0λ1Δt)k-1|A-1PmEk|2Δt≤𝒦Δt3(1+νγ0λ1Δt)k-1∀k≥n0+1,(1+νγ0λ1Δt)k-1|A-1/2PmEk|2Δt≤𝒦Δt3(1+νγ0λ1Δt)k-1∀k≥n0+1,(177)(1+νγ0λ1Δt)k-1|A-1PmEk|2Δt≤𝒦Δt2∫tk-1tk‍eνγ0λ1t|umt|2dt∀1≤k≤n0,(178)(1+νγ0λ1Δt)k-1|A-1/2PmEk|2Δt≤𝒦Δt2∫tk-1tk‍eνγ0λ1t∥umt∥2dt∀1≤k≤n0.
For n≥n0+1, summing (176) from n0+1 to n, we have
(179)Δt∑k=n0+1n‍(1+νγ0λ1Δt)k-1(|A-1PmEk|2+|A-1/2PmEk|2)≤𝒦Δt3∑k=n0+1n‍(1+νγ0λ1Δt)k-1≤𝒦Δt3(1+νγ0λ1Δt)n01-(1+νγ0λ1Δt)n-n0-νγ0λ1Δt≤𝒦Δt2(1+νγ0λ1Δt)n.
For n≤n0, we sum (177) from 1 to n and use Theorem 8 to obtain
(180)Δt∑k=1n‍(1+νγ0λ1Δt)k-1|A-1PmEk|2≤𝒦Δt2eνγ0λ1≤𝒦Δt2(1+νγ0λ1Δt)n∀n≤n0,
which, together with (179), gives (172).
Finally, multiplying (178) by τj(tk), j=0,1; 2≤k≤n0, and noting that Δt≤τ(tk-1), τ(tk)≤2τ(tk-1), we obtain
(181)(1+νγ0λ1Δt)k-1τj(tk)|A-1/2PmEk|2Δt≤𝒦Δt1-j∫tk-1tk‍eνγ0λ1tτj(t)∥umt∥2dt∀2≤k≤n0.
Observing again (170) with n=1 and using Theorem 8 yields
(182)τj(t1)|A-1/2PmEk|2Δt≤𝒦sup0≤t≤t1∥um(t)-um(0)∥2Δt1-j.
For 2≤n≤n0, summing (181) from 2 to n and using Theorem 12, we obtain
(183)Δt∑k=2n‍(1+νγ0λ1Δt)k-1τj(tk)|A-1/2PmEk|2≤𝒦Δt1-jeνγ0λ1.
Combining (182) and (183) with (179) yields (173) and (174).Now, we prove the following error estimate.Lemma 16.
Under the assumptions of Theorem13, one has
(184)|en|2+Δt∑k=1n‍(1+νγ0λ1Δt)-(n-1-k)×(νγ08∥ek∥2Δt+|ek-ek-1|2)≤𝒦exp(8νγ0c02Δt∑k=1n‍∥umk∥2)Δt∀n≥1.Proof.
Taking the scalar product of (170) with 2en+1Δt, using (11), Lemma 3 and noting that λ1|v|2≤∥v∥2, we have
(185)(1+νγ0λ1Δt)|en+1|2-|en|2+|en+1-en|2+νγ0∥en+1∥2Δt+2(b(en+1,umn,en+1)+b(en-en+1,umn,en+1)+b(um(tn),en-en+1,en+1))Δt≤2(En+1,en+1)Δt.
Using (17)–(19), we get
(186)2|b(en+1,umn,en+1)|≤νγ04∥en+1∥2+4νγ0c02∥umn∥2|en+1|2,2|b(en-en+1,umn,en+1)|+2|b(um(tn),en-en+1,en+1)|≤9νγ016∥en+1∥2+49(L(umn)+L(um(tn)))|en+1-en|2,2|(En+1,en+1)|≤νγ016∥en+1∥2+16νγ0|A-1/2PmEn+1|2.
Hence, by combining the above inequalities with (185) and using Theorem 8, (110) and (114), we obtain
(187)(1+νγ0λ1Δt)n+1|en+1|2-(1+νγ0λ1Δt)n|en|2+(1+νγ0λ1Δt)n(νγ08∥en+1∥2Δt+19|en+1-en|2)≤4νγ0c02∥umn∥2(1+νγ0λ1Δt)n|en+1|2Δt+16νγ0(1+νγ0λ1Δt)n|A-1/2PmEn+1|2Δt,
for all n≥1. Summing (187) from 1 to n-1 and using Lemma 15, we obtain
(188)(1+νγ0λ1Δt)n|en|2+∑k=1n‍(1+νγ0λ1Δt)k-1(νγ08∥ek∥2Δt+19|ek-ek-1|2)≤4νγ0c02Δt∑k=1n‍(1+νγ0λ1Δt)k-1∥umk-1∥2|ek|2+16νγ0Δt∑k=1n‍(1+νγ0λ1Δt)k-1|A-1/2PmEk|2.
We set
(189)ak=(1+νγ0λ1Δt)k|ek|2,bk=(1+νγ0λ1Δt)k-1(νγ08∥ek∥2+19|ek-ek-1|2Δt-1),gk=4νγ0c02∥umk-1∥2,hk=16νγ0Δt(1+νγ0λ1Δt)k-1|A-1/2PmEk|2,β=0
in (188) to obtain
(190)an+Δt∑k=1n‍bk≤Δt∑k=1n‍gkak+Δt∑k=1n‍hk,
with σk=(1-gkΔt)-1≤2. Applying Lemma 6 to (190) and using Lemma 15, we deduce that
(191)(1+νγ0λ1Δt)n|en|2+∑k=1n‍(1+νγ0λ1Δt)k-1(νγ08∥ek∥2Δt+19|ek-ek-1|2)≤(1+νγ0λ1Δt)n𝒦exp(8νγ0c02Δt∑k=0n-1‍∥umk∥2)Δt.
Multiplying (191) by (1+νγ0λ1Δt)-n, we get (184).Lemma 17.
Under the assumptions of Theorem13, one has
(192)τ(tn)|um(tn)-umn|2≤𝒦exp(16νγ0c02Δt∑k=0n-1‍∥umk∥2)Δt2∀n≥1.Proof.
Replacingn+1 by k in (170) and taking the scalar product of (170) with Φk-1, we obtain
(193)(dtek,Φk-1)g+a(ek,Φk-1)+ν(Cek,Φk-1)g+b(ek-1,umk-1,Φk-1)+b(umk-1,ek-1,Φk-1)+b(ek-1,ek-1,Φk-1)=(Ek,Φk-1)g.
Next, setting v=ek, ξk=ek, 1≤k≤n in (109), we get
(194)(ek,dtΦk)g-a(ek,Φk-1)-ν(Cek,Φk-1)g-b(ek,umk-1,Φk-1)-b(umk-1,ek,Φk)=(1+νγ0λ1Δt)k|ξk|2.
Adding (193) to (194), we arrive at
(195)(1+νγ0λ1Δt)k|ξk|2=1Δt((ek,Φk)g-(ek-1,Φk-1)g)+b(ek-1-ek,umk-1,Φk-1)+b(umk-1,ek-1-ek,Φk-1)+b(ek-1,ek-1,Φk-1)-(Ek,Φk-1)g.
From (17) and (18), noting that λ1|v|2≤∥v∥2, one finds that
(196)|b(ek-1-ek,umk-1,Φk-1)+b(umk-1,ek-1-ek,Φk-1)|≤c|ek-ek-1|∥umk-1∥|AΦk-1|,|b(ek-1,ek-1,Φk-1)|≤c∥ek-1∥|ek-1||AΦk-1|,|(Ek,Φk-1)g|≤|A-1PmEk||AΦk-1|.
Combining (195) with the above estimates yields
(197)(1+νγ0λ1Δt)k|ξk|2Δt≤((ek,Φk)g-(ek-1,Φk-1)g)+|A-1PmEk||AΦk-1|Δt+c(|ek-ek-1|∥umk-1∥+∥ek-1∥|ek-1|)|AΦk-1|Δt,
with e0=Φn=0. Summing (197) from 1 to n and using Lemma 15, we obtain
(198)Δt∑k=1n‍(1+νγ0λ1Δt)k|ξk|2≤𝒦Δt∑k=1n‍(1+νγ0λ1Δt)k×(|ek-ek-1|2∥umk-1∥2+∥ek-1∥2|ek-1|2)+Δt∑k=1n‍(1+νγ0λ1Δt)k|A-1PmEk|2.
Applying Lemmas 15 and 16 and Theorem 8 in (198), we get
(199)(1+νγ0λ1Δt)-nΔt∑k=1n‍(1+νγ0λ1Δt)k|ξk|2≤𝒦exp(16νγ0c02Δt∑n=0n-1‍∥umk∥2)Δt2∀n≥1.
Now, multiplying (187) by τ(tn+1) and noting that
(200)τ(tn+1)≤Δt+τ(tn),Δt≤τ(tn)∀1≤n≤N,e0=0,
we arrive at
(201)(1+νγ0λ1Δt)n+1τ(tn+1)|en+1|2-(1+νγ0λ1Δt)nτ(tn)|en|2≤𝒦(1+νγ0λ1Δt)n|en|2Δt+k(1+νγ0λ1Δt)nτ(tn+1)|A-1/2PmEn+1|2Δt.
Summing (201) from 0 to n-1 and using (199) and (174), we obtain, after a final multiplication by (1+νγ0λ1Δt)-n, the estimate (192).Finally, combining Theorems12, 8, and 13 with Lemma 17, we obtain the error estimate of the numerical solution {umn}.Theorem 18.
Under the assumptions of Theorem13, the following error estimates holds:
(202)|u(tn)-umn|2≤𝒦τ-1(tn)e(16/ν3γ03λ1)c02Cf2tn(λm+1-2+Δt2)00000000000000000000000000000000000000∀n≥1.


## 5.1. Stability Analysis

In this subsection, we consider the semi-implicit Euler scheme applied to the spatially discrete spectral Galerkin approximation, show the stability of this scheme, and establish some preliminaries related to the error analysis uniform in time.We consider the semi-implicit Euler scheme and define recursively a solution{umn}⊂Hm such that
(106)dtumn+1+νAumn+1+νCumn+1+PmB(umn,umn)=Pmfn+1,
for n≥0 with the initial condition um0=Pmu0, where Δt>0 is a time step such that n0Δt=1 for some integer n0 and
(107)dtumn+1=1Δt(umn+1-umn),(108)fn+1=1Δt∫tntn+1‍f(t)dt,tn=nΔt.
In order to derive the L2-bound on the error um(tn)-umn, we will begin with a time discrete duality argument which is similar to the one used in [11, 17]. We consider the dual scheme correponding to scheme (106): for any fixed n≥1 and umk-1,ξk∈Hm, 1≤k≤n, find Φk-1∈Hm, 1≤k≤n such that
(109)(v,dtΦk)g-a(v,Φk-1)-ν(Cv,Φk-1)g-b(v,umk-1,Φk-1)-b(umk-1,v,Φk-1)=(1+νγ0λ1Δt)k(v,ξk)g-b(umk-1,v,Φk-1)==(1+νγ0λ1Δt)k∀v∈Hm,
with an initial condition Φn=0.The following theorem provides theH2-stability of scheme (106).Theorem 13.
Under the assumptions of Theorem8, if Δt and m satisfy the following condition:
(110)8νγ0c02lnλmλ1ΔtCu0,f≤1,
then the semi-implicit Euler scheme is the H2-stability; that is,
(111)Δt∑k=1n‍∥umk∥2≤1νγ0|u0|2+1ν2γ02λ1Cf2tn∀n≥0,(112)Δt∑k=k0+1n‍∥umk∥2≤1νγ0|u0|2+2ν3γ03λ12Cf2+2ν2γ02λ1Cf2(tn-tk0)0000000000000000000000000000∀n≥k0≥0,(113)|umn|2≤(1+νγ0λ1Δt)-n|u0|2+1ν2γ02λ12Cf2∀n≥0,(114)∥umn∥2≤Cu0,f∀n≥0,(115)Δt∑k=1n‍(1+νγ0λ1Δt)-(n-1-k)×(|Aumk|2+|dtumk|2+τ(tk)∥dtumk∥2)+τ(tn)|Aumn|2+τ(tn)|dtumn|2≤𝒦∀n≥0.Proof.
Clearly, scheme (106) defines a unique sequence {umn}⊂Hm. Now, we will prove (111)–(115).
Taking the scalar product of (106) with 2umn+1Δt, we obtain
(116)|umn+1|2-|umn|2+|umn+1-umn|2+2ν∥umn+1∥2Δt+2νΔt(Cumn+1,umn+1)g+2b(umn,umn-umn+1,umn+1)Δt=2(fn+1,umn+1)gΔt.
Using Lemma 3 and Cauchy's inequality, we have
(117)|umn+1|2-|umn|2+|umn+1-umn|2+2ν∥umn+1∥2Δt≤2ν|∇g|∞m0λ11/2∥umn+1∥2Δt+2(fn+1,umn+1)gΔt-2b(umn,umn-umn+1,umn+1)Δt.
Hence,
(118)|umn+1|2-|umn|2+|umn+1-umn|2+2νγ0∥umn+1∥2Δt≤2(fn+1,umn+1)gΔt-2b(umn,umn-umn+1,umn+1)Δt.
By (19), we have
(119)2(fn+1,umn+1)g≤νγ02∥umn+1∥2+2νγ0λ1|fn+1|2Δt,2|b(umn,umn-umn+1,umn+1)|≤νγ02∥umn+1∥2+L(umn)|umn-umn+1|2,
where
(120)L(umn)=8νγ0c02(1+ln|Aumn|2λ1∥umn∥2)∥umn∥2.
Substituting those into (118), we get
(121)|umn+1|2-|umn|2+(1-L(umn)Δt)|umn+1-umn|2+νγ0∥umn+1∥2Δt≤2νγ0λ1|fn+1|2Δt.
Next, by taking the scalar product of (106) with 2Aumn+1Δt, we obtain
(122)∥umn+1∥2-∥umn∥2+∥umn+1-umn∥2+2ν|Aumn+1|2Δt+2νΔt(Cumn+1,Aumn+1)g+2b(umn,umn,Aumn+1)Δt=2(fn+1,Aumn+1)gΔt.
Using Lemma 3 and Cauchy's inequality, we have
(123)2ν(Cumn+1,Aumn+1)g≤2ν|∇g|∞m0λ11/2|Aumn+1|2+ν|∇g|∞λ11/22m0∥umn+1∥2.
Substituting into (122), we get
(124)∥umn+1∥2-∥umn∥2+∥umn+1-umn∥2+2νγ0|Aumn+1|2Δt≤2(fn+1,Aumn+1)gΔt-2b(umn,umn,Aumn+1)Δt+ν|∇g|∞λ11/22m0∥umn+1∥2Δt.
Due to (17)–(19), Cauchy's inequality, and Young's inequality, we have
(125)2|b(umn,umn+1-umn,Aumn+1)|≤νγ04|Aumn+1|2+L(umn)∥umn+1-umn∥2,2|b(umn,umn+1,Aumn+1)|≤νγ0|Aumn+1|2+2(νγ0)-3c04|umn|2∥umn∥2∥umn+1∥2,2(fn+1,Aumn+1)g≤νγ02|Aumn+1|2+2νγ0Cf2.
By combining (124) with the above estimates, we obtain
(126)∥umn+1∥2-∥umn∥2+(1-L(umn)Δt)∥umn+1-umn∥2≤2(νγ0)-3c04|umn|2∥umn∥2∥umn+1∥2Δt+2νγ0Cf2Δt+ν|∇g|∞λ11/22m0∥umn+1∥2Δt.
On the other hand, from (17)–(19), Cauchy's inequality, and Young's inequality, we have
(127)2|b(umn,umn+1-umn,Aumn+1)|≤νγ04|Aumn+1|2+L(umn)∥umn+1-umn∥2,2|b(umn,umn+1,Aumn+1)|≤νγ08|Aumn+1|2+c|umn|2∥umn∥2∥umn+1∥2,2(fn+1,Aumn+1)g≤νγ08|Aumn+1|2+cCf2.
Combining (124) with those estimates, we obtain
(128)(1+νγ0λ1Δt)∥umn+1∥2-∥umn∥2+(1-L(umn)Δt)∥umn+1-umn∥2+νγ02|Aumn+1|2Δt≤c|umn|2∥umn∥2∥umn+1∥2Δt+cCf2Δt+c∥umn+1∥2Δt.
Next, from (106) we have
(129)|dtumn+1|2+ν(Aumn+1,dtumn+1)+ν(Cumn+1,dtumn+1)g+b(un,un,dtumn+1)=(fn+1,dtumn+1)g.
Using (11), Lemmas 1 and 3, and Cauchy's inequality, we get
(130)|b(un,un,dtumn+1)|≤c|umn|2∥umn∥∥umn+1∥2,ν|(Aumn+1,dtumn+1)g|≤14|dtumn+1|2+ν2|Aumn+1|2,ν|(Cumn+1,dtumn+1)g|≤14|dtumn+1|2+c∥umn+1∥2,|(fn+1,dtumn+1)g|≤14|dtumn+1|2+cCf2.
Therefore, we have
(131)γ016ν|dtumn+1|2-νγ04|Aumn+1|2≤c|umn|2∥umn∥2∥umn+1∥2+cCf2+c∥umn+1∥2.
Combining this inequality with (128) yields
(132)(1+νγ0λ1Δt)∥umn+1∥2-∥umn∥2+(1-L(umn)Δt)∥umn+1-umn∥2+γ016ν|dtumn+1|2Δt+νγ04|Aumn+1|2Δt≤c|umn|2∥umn∥2∥umn+1∥2Δt+cCf2Δt+c∥umn+1∥2Δt.
Moreover, we deduce from (106) that
(133)dttumn+1+νAdtumn+1+νCdtumn+1+PmB(dtumn,umn)+PmB(umn,dtumn)=Pmdtfn+1,
where dtum0 is defined by
(134)dtum0+νAum0+νCum0+PmB(um0,um0)=Pmf(0).
Taking the scalar product of (133) with 2dtumn+1Δt, we obtain
(135)|dtumn+1|2-|dtumn|2+|dtumn+1-dtumn|2+2ν∥dtumn+1∥2Δt+2ν(Cdtumn+1,dtumn+1)Δt+2b(dtumn-dtumn+1,umn,dtumn+1)Δt+2b(dtumn+1,umn,dtumn+1)Δt+2b(dtumn+1,dtumn-dtumn+1,dtumn+1)Δt=2(dtfn+1,dtumn+1)Δt.
By Lemma 3, we get
(136)|dtumn+1|2-|dtumn|2+|dtumn+1-dtumn|2+2νγ0∥dtumn+1∥2Δt≤2(dtfn+1,dtumn+1)Δt-2b(dtumn-dtumn+1,umn,dtumn+1)Δt-2b(dtumn+1,umn,dtumn+1)Δt-2b(dtumn+1,dtumn-dtumn+1,dtumn+1)Δt.
Using Lemma 1 and (19), we have
(137)2|b(dtumn-dtumn+1,umn,dtumn+1)|+2|b(dtumn+1,dtumn-dtumn+1,dtumn+1)|≤νγ02∥dtumn+1∥2+12(L(umn)+L(umn+1))|dtumn+1-dtumn|2,2|b(dtumn+1,umn,dtumn+1)|≤νγ08∥dtumn+1∥2+8νγ0c02∥umn∥2|dtumn+1|2,2(dtfn+1,dtumn+1)≤νγ08∥dtumn+1∥2+8νγ0λ1supt≥0|ft(t)|2.
By combining (136) with the above estimates, we arrive at
(138)(1+νγ0λ1Δt)|dtumn+1|2-|dtumn|2+(1-12(L(umn)+L(umn+1))Δt)|dtumn+1-dtumn|2+νγ04∥dtumn+1∥2Δt≤8νγ0c02∥umn∥2|dtumn+1|2Δt+νγ04∥dtumn+1∥2Δt≤+8νγ0λ1supt≥0|ft(t)|2.
Multiplying (138) by τ(tn+1) and noting that τ(tn+1)≤τ(tn)+Δt, we obtain
(139)(1+νγ0λ1Δt)τ(tn+1)|dtumn+1|2-τ(tn)|dtumn|2+(1-12(L(umn)+L(umn+1))Δt)×τ(tn+1)|dtumn+1-dtumn|2+νγ04τ(tn+1)∥dtumn+1∥2Δt≤|dtumn|2Δt+8νγ0c02∥umn∥2|dtumn+1|2Δt+8νγ0λ1supt≥0|ft(t)|2Δt.
Now, we will prove (111)–(115) by induction. Obviously, (111)–(115) are true for n=0. Assuming that (111)–(115) hold for n=0,1,…,J, we need to prove (111)–(115) for n=J+1. In view of (110) and the inductive assumption, we obtain
(140)1-L(umn)Δt≥0∀0≤n≤J.
Summing (121) from 0 to J and k0+1 to J, respectively, and using (140), we obtain (111) and (112) with n=J+1 after a final multiplication by (νγ0)-1. Noting that λ1|v|2≤∥v∥2, using (140) in (121), and then applying Lemma 5 with
(141)ak=|umk|2,bk=0,hk=2νγ0λ1Cf2,γ=νγ0λ1,
we obtain (113) for n=J+1.
Furthermore, setting(142)ak=∥umk∥2,gk=2(νγ0)-3c04|umk|2∥umk+1∥2,hk=2νγ0Cf2+ν|∇g|∞λ11/22m0∥umk+1∥2,
in (126), using (111)–(113) and (140), we obtain
(143)1Δt(ak+1-ak)≤gkak+hk∀0≤k≤J,
with
(144)Δt∑k=0J‍gk≤α1,Δt∑k=0J‍hk≤α1,a0≤∥u0∥2,
for all J+1≤n0-1 and
(145)Δt∑k=k0k0+n0-1‍gk≤α1,Δt∑k=k0k0+n0-1‍hk≤α2,Δt∑k=k0k0+n0-1‍ak≤a0+α3,0≤k0≤J+1-n0,
for all J+1≥n0. Applying Lemma 4 to (143), we obtain (114) for n=J+1.
Applying again Lemma4 to (132) with
(146)ak=∥umk∥2,bk=116νγ0|dtumk|2+νγ04|Aumk|2,γ=νγ0λ1,hk=c|umk|2∥umk∥2∥umk+1∥2+cCf2+c∥umk+1∥2,
and using (111)–(114) and (140), we obtain
(147)Δt∑k=1J+1‍(1+νγ0λ1Δt)-(J+2-k)(|Aumk|+|dtumk|2)≤𝒦.
Applying Lemma 5 to (139) with
(148)ak=τ(tk)|dtumk|2,bk=νγ04τ(tk)∥dtumk∥2,γ=νγ0λ1,hk=|dtumk|2Δt+8νγ0c02∥umk∥2|dtumk+1|2+8νγ0λ1supt≥0|ft(t)|2,
and using (114), (140), and (147) yields
(149)τ(tJ+1)|dtumJ+1|2+νγ04Δt∑k=1J+1‍(1+νγ0λ1Δt)-(J+2-k)τ(tk)∥dtumk∥2≤𝒦.
Finally, due to (17)–(19), we have
(150)|b(umJ,uJ-umJ+1,AumJ+1)|≤νγ08|AumJ+1|2+cL(umJ)∥dtumJ+1∥2Δt2,|b(umJ,umJ+1,AumJ+1)|≤νγ08|AumJ+1|2+c|umJ|2∥umJ∥2∥umJ+1∥2.
Combining these inequalities with (106) and (140) yields
(151)τtJ+1|AumJ+1|2≤cτ(tJ+1)|dtumJ+1|2+c|umJ|2∥umJ∥2∥umJ+1∥2+cCf2+cτ(tJ+1)∥dtumJ+1∥2Δt.
Using (113), (114) and (149) in this inequality yields
(152)τ(tJ+1)|AumJ+1|2≤𝒦.
Combining this inequality with (149) and (147) implies (115) for n=J+1.The following lemma provides the stability of scheme (109).Lemma 14.
Ifu0∈Vg,Δt and m satisfy (110), then problem (109) admits a unique solution {Φk}0n⊂Hm satisfying
(153)(1+νγ0λ1Δt)-r∥Φr∥2+Δt∑k=rn‍(1+νγ0λ1Δt)-(k-1)|AΦk|2≤𝒦exp(4νγ0c02Δt∑k=1n‍∥umk∥2)Δt∑k=1n‍(1+νγ0λ1Δt)k|ξk|2.Proof.
In view of (17), (110), and (114), we can prove that the following bilinear form
(154)1Δt(v,Φ)+a(v,Φ)+ν(Cv,Φ)+b(v,umk-1,Φ)+b(umk-1,v,Φ)∀v,Φ∈Hm
is elliptic. Hence, problem (109) has a unique solution Φk-1 for 1≤k≤n. Next, by taking v=-2(1+νγ0λ1Δt)-kΦk-1Δt in (109) and using (11), we have
(155)-2(1+νγ0λ1Δt)-k(Φk-1,Φk)g+2(1+νγ0λ1Δt)-k|Φk-1|2+2ν(1+νγ0λ1Δt)-k∥Φk-1∥2Δt+2ν(1+νγ0λ1Δt)-k(CΦk-1,Φk-1)gΔt+2(1+νγ0λ1Δt)-kb(Φk-1,umk-1,Φk-1)Δt=-2(Φk-1,ξk)gΔt.
Using Lemma 3, we get
(156)2(1+νγ0λ1Δt)-k|Φk-1|2+2νγ0(1+νγ0λ1Δt)-k∥Φk-1∥2Δt≤2|(Φk-1,ξk)g|Δt+2(1+νγ0λ1Δt)-k|(Φk-1,Φk)g|+2(1+νγ0λ1Δt)-k|b(Φk-1,umk-1,Φk-1)|Δt.
Using (17) and Cauchy's inequality, we have
(157)2|(Φk-1,Φk)g|≤|Φk-1|2+|Φk|2,2|(Φk-1,ξk)g|≤νγ04(1+νγ0λ1Δt)-k∥Φk-1∥2+4νγ0λ1(1+νγ0λ1Δt)k|ξk|2,2|b(Φk-1,umk-1,Φk-1)|≤νγ02∥Φk-1∥2+2νγ0c02|Φk-1|2∥uk-1∥2.
Combining above inequalities and using (14), we obtain
(158)(1+νγ0λ1Δt)-(k-1)|Φk-1|2-(1+νγ0λ1Δt)-k|Φk|2+νγ04(1+νγ0λ1Δt)-k∥Φk-1∥2Δt≤2νγ0c02(1+νγ0λ1Δt)-k∥umk-1∥2|Φk-1|2Δt+4νγ0λ1(1+νγ0λ1Δt)k|ξk|2Δt.
Summing (158) from r+1 to n, noting that Φn=0, and using Theorem 13, we arrive at
(159)(1+νγ0λ1Δt)-r|Φr|2+νγ04Δt∑k=rn‍(1+νγ0λ1Δt)-(k+1)∥Φk∥2≤2νγ0c02Δt∑k=rn‍(1+νγ0λ1Δt)-k∥umk∥2|Φk|2+4νγ0λ1Δt∑k=rn‍(1+νγ0λ1Δt)k|ξk|2
for all 0≤r≤n. Let
(160)ar=(1+νγ0λ1Δt)-r|Φr|2,br=νγ04(1+νγ0λ1Δt)-(r+1)∥Φr∥2,hr=4νγ0λ1(1+νγ0λ1Δt)r|ξk|2,gr=2νγ0c02∥umk∥2,β=0,
in (159) to obtain
(161)ar+Δt∑k=rn‍bk≤Δt∑k=rn‍gkak+Δt∑k=rn‍hk∀0≤r≤n,
with σr=(1-grΔt)-1≤2. Applying Lemma 6 to this inequality yields
(162)(1+νγ0λ1Δt)-r|Φr|2+νγ02Δt∑k=rn‍(1+νγ0λ1Δt)-(k+1)∥Φk∥2≤𝒦exp(4νγ0c02Δt∑k=1n‍∥umk∥2)Δt∑k=rn‍(1+νγ0λ1Δt)k|ξk|2.
Moreover, by taking v=-2(1+νγ0λ1Δt)-kAΦk-1Δt in (109), we have
(163)-2(1+νγ0λ1Δt)-k(AΦk-1,Φk)g+2(1+νγ0λ1Δt)-k∥Φk-1∥2+2ν(1+νγ0λ1Δt)-k|AΦk-1|2Δt+2ν(1+νγ0λ1Δt)-k(CAΦk-1,Φk-1)gΔt+2(1+νγ0λ1Δt)-kb(AΦk-1,Φk-1,umk-1)Δt+2(1+νγ0λ1Δt)-kb(umk-1,Φk-1,AΦk-1)Δt=2(AΦk-1,ξk)gΔt.
Using Lemma 3 and Cauchy's inequality, we deduce that
(164)2ν(1+νγ0λ1Δt)-k|(CAΦk-1,Φk-1)g|≤2ν(1+νγ0λ1Δt)-k|∇g|∞m0λ11/2|AΦk-1|2+c(1+νγ0λ1Δt)-k∥Φk-1∥2.
Therefore,
(165)2(1+νγ0λ1Δt)-k∥Φk-1∥2+2νγ0(1+νγ0λ1Δt)-k|AΦk-1|2Δt≤2(AΦk-1,ξk)gΔt+2(1+νγ0λ1Δt)-k(AΦk-1,Φk)g+2(1+νγ0λ1Δt)-k|b(AΦk-1,Φk-1,umk-1)|Δt+2(1+νγ0λ1Δt)-k|b(umk-1,Φk-1,AΦk-1)|Δt+c(1+νγ0λ1Δt)-k∥Φk-1∥2Δt.
Using (17)–(19) and Cauchy’s inequality, we have
(166)2|b(AΦk-1,Φk-1,umk-1)|+2|b(umk-1,Φk-1,AΦk-1)|≤νγ04|AΦk-1|2+c|umk-1|2∥umk-1∥2∥Φk-1∥2,2(AΦk-1,Φk)g=2((Φk-1,Φk))g≤∥Φk-1∥2+∥Φk∥2,2(AΦk-1,ξk)g≤νγ04(1+νγ0λ1Δt)-k|AΦk-1|2+4νγ0(1+νγ0λ1Δt)k|ξk|2.
Combining above inequality, noting that λ1∥v∥2≤|Av|2, we obtain
(167)(1+νγ0λ1Δt)-(k-1)∥Φk-1∥2-(1+νγ0λ1Δt)-k∥Φk∥2+νγ02(1+νγ0λ1Δt)-k|AΦk-1|2Δt≤c(1+νγ0λ1Δt)-k(|umk-1|2∥umk-1∥2+1)∥Φk-1∥2Δt+c(1+νγ0λ1Δt)k|ξk|2Δt
for all 0≤k≤n. Summing (167) from r+1 to n, we obtain
(168)(1+νγ0λ1Δt)-r∥Φr∥2+νγ02Δt∑k=rn‍(1+νγ0λ1Δt)-(k+1)|AΦk|2≤csupk≥0{(|umk|2∥umk∥2+1)}Δt∑k=rn‍(1+νγ0λ1Δt)-(k+1)∥Φk∥2+cΔt∑k=r+1n‍(1+νγ0λ1Δt)k|ξk|2∀0≤r≤n.
Combining (168) with (162) and using Theorem 13, we complete the proof.


## 5.2. Error Analysis

In this subsection, we will establish theH1- and L2-error estimates uniform in time for the fully discrete spectral Galerkin method with the explicit time discretization for the nonlinear term. To do this, by integrating (35) from tn to tn+1, we obtain
(169)dtum(tn+1)+νΔt∫tntn+1‍Aum(t)dt+νΔt∫tntn+1‍Cum(t)dt+1Δt∫tntn+1‍PmB(um(t),um(t))dt=fn+1.
Subtracting (106) from (153) and setting en=um(tn)-umn, we have
(170)dten+1+νAen+1+νCen+1+PmB(en,umn)+PmB(um(tn),en)=PmEn+1,
with e0=0 and
(171)En+1=νΔt∫tntn+1‍A(um(tn+1)-um(t))dt+νΔt∫tntn+1‍C(um(tn+1)-um(t))dt+1Δt∫tntn+1‍(B(um(tn),um(tn))+1Δt∫tntn+1‍-B(um(t),um(t)))dt=νΔt∫tntn+1‍(t-tn)Aumtdt+νΔt∫tntn+1‍(t-tn)Cumtdt-1Δt∫tntn+1‍(tn+1-t)(B(umt,um)+B(um,umt))dt.
To derive a bound on en, we need to provide the following estimates of En.Lemma 15.
Under the assumptions of Theorem13, the error En satisfies the following bounds:
(172)Δt∑k=1n‍(1+νγ0λ1)k-1|A-1PmEk|2≤𝒦(1+νγ0λ1)nΔt2∀n≥1,(173)Δt∑k=1n‍(1+νγ0λ1)k-1|A-1/2PmEk|2≤𝒦(1+νγ0λ1)nΔt∀n≥1,(174)Δt∑k=1n‍(1+νγ0λ1)k-1τ(tk)|A-1/2PmEk|2≤𝒦(1+νγ0λ1)nΔt2∀n≥1.Proof.
In view of (17)–(19), Lemma 3, and Theorem 8, we deduce from (171) that
(175)|A-1PmEn|=supv∈Hm|(En,v)||Av|≤Δt-1/2(∫tn-1tn‍(c1((t-tn-1)2+|∇g|∞m0(t-tn-1)2)|umt|2Δ∫tn-1tn‍t-1/2+c2(tn-t)2∥um∥2|umt|2{(c1((t-tn-1)2+|∇g|∞m0(t-tn-1)2)|umt|2})dt{∫tn-1tn‍})1/2≤cΔt-1/2(∫tn-1tn‍((t-tn-1)2+(tn-t)2∥um∥2)|umt|2dt)1/2,|A-1/2PmEn|=supv∈Hm|(En,v)|∥v∥|A-1/2PmEn|≤cΔt-1/2|A-1/2PmEn|×(∫tn-1tn‍((t-tn-1)2+(tn-t)2∥um∥2)|A-1/2PmEn|∫tn-1tn‍×∥umt∥2dt∫tn-1tn‍)1/2.
By Theorem 8, (175), and the fact that (1+νγ0λ1)k-1≤eνγ0λ1tk-1, we have
(176)(1+νγ0λ1Δt)k-1|A-1PmEk|2Δt≤𝒦Δt3(1+νγ0λ1Δt)k-1∀k≥n0+1,(1+νγ0λ1Δt)k-1|A-1/2PmEk|2Δt≤𝒦Δt3(1+νγ0λ1Δt)k-1∀k≥n0+1,(177)(1+νγ0λ1Δt)k-1|A-1PmEk|2Δt≤𝒦Δt2∫tk-1tk‍eνγ0λ1t|umt|2dt∀1≤k≤n0,(178)(1+νγ0λ1Δt)k-1|A-1/2PmEk|2Δt≤𝒦Δt2∫tk-1tk‍eνγ0λ1t∥umt∥2dt∀1≤k≤n0.
For n≥n0+1, summing (176) from n0+1 to n, we have
(179)Δt∑k=n0+1n‍(1+νγ0λ1Δt)k-1(|A-1PmEk|2+|A-1/2PmEk|2)≤𝒦Δt3∑k=n0+1n‍(1+νγ0λ1Δt)k-1≤𝒦Δt3(1+νγ0λ1Δt)n01-(1+νγ0λ1Δt)n-n0-νγ0λ1Δt≤𝒦Δt2(1+νγ0λ1Δt)n.
For n≤n0, we sum (177) from 1 to n and use Theorem 8 to obtain
(180)Δt∑k=1n‍(1+νγ0λ1Δt)k-1|A-1PmEk|2≤𝒦Δt2eνγ0λ1≤𝒦Δt2(1+νγ0λ1Δt)n∀n≤n0,
which, together with (179), gives (172).
Finally, multiplying (178) by τj(tk), j=0,1; 2≤k≤n0, and noting that Δt≤τ(tk-1), τ(tk)≤2τ(tk-1), we obtain
(181)(1+νγ0λ1Δt)k-1τj(tk)|A-1/2PmEk|2Δt≤𝒦Δt1-j∫tk-1tk‍eνγ0λ1tτj(t)∥umt∥2dt∀2≤k≤n0.
Observing again (170) with n=1 and using Theorem 8 yields
(182)τj(t1)|A-1/2PmEk|2Δt≤𝒦sup0≤t≤t1∥um(t)-um(0)∥2Δt1-j.
For 2≤n≤n0, summing (181) from 2 to n and using Theorem 12, we obtain
(183)Δt∑k=2n‍(1+νγ0λ1Δt)k-1τj(tk)|A-1/2PmEk|2≤𝒦Δt1-jeνγ0λ1.
Combining (182) and (183) with (179) yields (173) and (174).Now, we prove the following error estimate.Lemma 16.
Under the assumptions of Theorem13, one has
(184)|en|2+Δt∑k=1n‍(1+νγ0λ1Δt)-(n-1-k)×(νγ08∥ek∥2Δt+|ek-ek-1|2)≤𝒦exp(8νγ0c02Δt∑k=1n‍∥umk∥2)Δt∀n≥1.Proof.
Taking the scalar product of (170) with 2en+1Δt, using (11), Lemma 3 and noting that λ1|v|2≤∥v∥2, we have
(185)(1+νγ0λ1Δt)|en+1|2-|en|2+|en+1-en|2+νγ0∥en+1∥2Δt+2(b(en+1,umn,en+1)+b(en-en+1,umn,en+1)+b(um(tn),en-en+1,en+1))Δt≤2(En+1,en+1)Δt.
Using (17)–(19), we get
(186)2|b(en+1,umn,en+1)|≤νγ04∥en+1∥2+4νγ0c02∥umn∥2|en+1|2,2|b(en-en+1,umn,en+1)|+2|b(um(tn),en-en+1,en+1)|≤9νγ016∥en+1∥2+49(L(umn)+L(um(tn)))|en+1-en|2,2|(En+1,en+1)|≤νγ016∥en+1∥2+16νγ0|A-1/2PmEn+1|2.
Hence, by combining the above inequalities with (185) and using Theorem 8, (110) and (114), we obtain
(187)(1+νγ0λ1Δt)n+1|en+1|2-(1+νγ0λ1Δt)n|en|2+(1+νγ0λ1Δt)n(νγ08∥en+1∥2Δt+19|en+1-en|2)≤4νγ0c02∥umn∥2(1+νγ0λ1Δt)n|en+1|2Δt+16νγ0(1+νγ0λ1Δt)n|A-1/2PmEn+1|2Δt,
for all n≥1. Summing (187) from 1 to n-1 and using Lemma 15, we obtain
(188)(1+νγ0λ1Δt)n|en|2+∑k=1n‍(1+νγ0λ1Δt)k-1(νγ08∥ek∥2Δt+19|ek-ek-1|2)≤4νγ0c02Δt∑k=1n‍(1+νγ0λ1Δt)k-1∥umk-1∥2|ek|2+16νγ0Δt∑k=1n‍(1+νγ0λ1Δt)k-1|A-1/2PmEk|2.
We set
(189)ak=(1+νγ0λ1Δt)k|ek|2,bk=(1+νγ0λ1Δt)k-1(νγ08∥ek∥2+19|ek-ek-1|2Δt-1),gk=4νγ0c02∥umk-1∥2,hk=16νγ0Δt(1+νγ0λ1Δt)k-1|A-1/2PmEk|2,β=0
in (188) to obtain
(190)an+Δt∑k=1n‍bk≤Δt∑k=1n‍gkak+Δt∑k=1n‍hk,
with σk=(1-gkΔt)-1≤2. Applying Lemma 6 to (190) and using Lemma 15, we deduce that
(191)(1+νγ0λ1Δt)n|en|2+∑k=1n‍(1+νγ0λ1Δt)k-1(νγ08∥ek∥2Δt+19|ek-ek-1|2)≤(1+νγ0λ1Δt)n𝒦exp(8νγ0c02Δt∑k=0n-1‍∥umk∥2)Δt.
Multiplying (191) by (1+νγ0λ1Δt)-n, we get (184).Lemma 17.
Under the assumptions of Theorem13, one has
(192)τ(tn)|um(tn)-umn|2≤𝒦exp(16νγ0c02Δt∑k=0n-1‍∥umk∥2)Δt2∀n≥1.Proof.
Replacingn+1 by k in (170) and taking the scalar product of (170) with Φk-1, we obtain
(193)(dtek,Φk-1)g+a(ek,Φk-1)+ν(Cek,Φk-1)g+b(ek-1,umk-1,Φk-1)+b(umk-1,ek-1,Φk-1)+b(ek-1,ek-1,Φk-1)=(Ek,Φk-1)g.
Next, setting v=ek, ξk=ek, 1≤k≤n in (109), we get
(194)(ek,dtΦk)g-a(ek,Φk-1)-ν(Cek,Φk-1)g-b(ek,umk-1,Φk-1)-b(umk-1,ek,Φk)=(1+νγ0λ1Δt)k|ξk|2.
Adding (193) to (194), we arrive at
(195)(1+νγ0λ1Δt)k|ξk|2=1Δt((ek,Φk)g-(ek-1,Φk-1)g)+b(ek-1-ek,umk-1,Φk-1)+b(umk-1,ek-1-ek,Φk-1)+b(ek-1,ek-1,Φk-1)-(Ek,Φk-1)g.
From (17) and (18), noting that λ1|v|2≤∥v∥2, one finds that
(196)|b(ek-1-ek,umk-1,Φk-1)+b(umk-1,ek-1-ek,Φk-1)|≤c|ek-ek-1|∥umk-1∥|AΦk-1|,|b(ek-1,ek-1,Φk-1)|≤c∥ek-1∥|ek-1||AΦk-1|,|(Ek,Φk-1)g|≤|A-1PmEk||AΦk-1|.
Combining (195) with the above estimates yields
(197)(1+νγ0λ1Δt)k|ξk|2Δt≤((ek,Φk)g-(ek-1,Φk-1)g)+|A-1PmEk||AΦk-1|Δt+c(|ek-ek-1|∥umk-1∥+∥ek-1∥|ek-1|)|AΦk-1|Δt,
with e0=Φn=0. Summing (197) from 1 to n and using Lemma 15, we obtain
(198)Δt∑k=1n‍(1+νγ0λ1Δt)k|ξk|2≤𝒦Δt∑k=1n‍(1+νγ0λ1Δt)k×(|ek-ek-1|2∥umk-1∥2+∥ek-1∥2|ek-1|2)+Δt∑k=1n‍(1+νγ0λ1Δt)k|A-1PmEk|2.
Applying Lemmas 15 and 16 and Theorem 8 in (198), we get
(199)(1+νγ0λ1Δt)-nΔt∑k=1n‍(1+νγ0λ1Δt)k|ξk|2≤𝒦exp(16νγ0c02Δt∑n=0n-1‍∥umk∥2)Δt2∀n≥1.
Now, multiplying (187) by τ(tn+1) and noting that
(200)τ(tn+1)≤Δt+τ(tn),Δt≤τ(tn)∀1≤n≤N,e0=0,
we arrive at
(201)(1+νγ0λ1Δt)n+1τ(tn+1)|en+1|2-(1+νγ0λ1Δt)nτ(tn)|en|2≤𝒦(1+νγ0λ1Δt)n|en|2Δt+k(1+νγ0λ1Δt)nτ(tn+1)|A-1/2PmEn+1|2Δt.
Summing (201) from 0 to n-1 and using (199) and (174), we obtain, after a final multiplication by (1+νγ0λ1Δt)-n, the estimate (192).Finally, combining Theorems12, 8, and 13 with Lemma 17, we obtain the error estimate of the numerical solution {umn}.Theorem 18.
Under the assumptions of Theorem13, the following error estimates holds:
(202)|u(tn)-umn|2≤𝒦τ-1(tn)e(16/ν3γ03λ1)c02Cf2tn(λm+1-2+Δt2)00000000000000000000000000000000000000∀n≥1.



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*Source: 805685-2013-07-25.xml*