Spectral Galerkin Method in Space and Time for the 2D 𝑔-Navier-Stokes Equations

Abstract

We prove the 𝐻²-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.

1. Introduction

Let Ω 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: 𝜕𝑢()()𝜕𝑡−𝜈Δ𝑢+𝑢⋅∇𝑢+∇𝑝=𝑓𝑥,𝑡in(0,+∞)×Ω,∇⋅𝑔𝑢=0in(0,+∞)×Ω,𝑢=0on()𝑢()0,+∞×Γ,0,𝑥=𝑢₀(𝑥),𝑥∈Ω, (1) 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.

The 𝑔-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 Ω_𝑔=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∈𝑊^1,∞(Ω) such that 0<𝑚₀≤𝑔(𝑥)≤𝑀₀∀𝑥=𝑥₁,𝑥₂∈Ω,∇𝑔_∞<𝑚₀𝜆₁^1/2, (2) where 𝜆₁>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∈𝑊^1,∞(ℝ⁺;𝐻_𝑔); that is, 𝑓,𝑓_𝑡∈𝐿^∞(ℝ⁺;𝐻_𝑔).

In this paper, in order to study the numerical approximation of strong solutions to the 2D 𝑔-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.

Let 𝑤₁,𝑤₂,… and 𝜆₁,𝜆₂,… be the eigenvectors and eigenvalues of the 𝑔-Stokes operator. For a fixed integer 𝑚, let 𝑃_𝑚 be the orthogonal projection of 𝐻_𝑔 onto 𝐻_𝑚=span{𝑤₁,…,𝑤_𝑚}. Then, the spectral Galerkin method in space is defined as follows: find 𝑢_𝑚(𝑡)∈𝐻_𝑚 such that 𝑢_𝑚𝑡+𝜈𝐴𝑢_𝑚+𝜈𝐶𝑢_𝑚+𝑃_𝑚𝐵𝑢_𝑚,𝑢_𝑚=𝑃_𝑚𝑓,𝑡>0,𝑢_𝑚(0)=𝑃_𝑚𝑢₀. (3) 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 𝑢_𝑚^𝑛+1 (𝑛≥0) such that 1𝑢Δ𝑡^𝑚𝑛+1−𝑢^𝑛_𝑚+𝜈𝐴𝑢_𝑚^𝑛+1+𝜈𝐶𝑢_𝑚^𝑛+1+𝑃_𝑚𝐵𝑢^𝑛_𝑚,𝑢^𝑛_𝑚=𝑃_𝑚𝑓^𝑛+1,𝑡>0,𝑢⁰_𝑚=𝑃_𝑚𝑢₀, (4) where Δ𝑡>0 is the time step size and 𝑓^𝑛+1=1Δ𝑡^𝑡𝑛+1_𝑡𝑛𝑓(𝑡)𝑑𝑡,𝑡_𝑛=𝑛Δ𝑡. (5)

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 obtain 𝐻²-stability uniform in time stated in Theorem 13, provided that the following condition holds 𝜆𝐶Δ𝑡ln^𝑚𝜆₁≤1 (6) 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 an 𝐿²-error estimate of the numerical solution 𝑢^𝑛_𝑚 under the stability condition (6): 𝑢𝑡_𝑛−𝑢^𝑛_𝑚²≤𝒦𝜏⁻¹𝑡_𝑛𝑒^{(16/𝜈3𝛾30𝜆1)𝑐20𝐶2𝑓𝑡𝑛}𝜆⁻²_𝑚+1+Δ𝑡²∀𝑛≥1, (7) where 𝐶_𝑓=sup_𝑡≥0|𝑓(𝑡)|, 𝜏(𝑡)=min{1,𝑡} and 𝒦 denotes a general positive constant depending only on the data (𝜈,Ω,|∇𝑔|_∞,𝜆₁,𝐶_𝑓,‖𝑢₀‖). Noting that 𝜏⁻¹(𝑡_𝑛) 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 𝐻_𝑔,𝑉_𝑔 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 the 𝑔-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

Let 𝐿²(Ω,𝑔)=(𝐿²(Ω))² and 𝐻¹₀(Ω,𝑔)=(𝐻¹₀(Ω))² be endowed, respectively, with the inner products (𝑢,𝑣)_𝑔=_Ω𝑢⋅𝑣𝑔𝑑𝑥,𝑢,𝑣∈𝐿²Ω,𝑔,((𝑢,𝑣))_𝑔=_Ω²_𝑗=1∇𝑢_𝑗⋅∇𝑣_𝑗𝑔𝑑𝑥,𝑢=𝑢₁,𝑢₂,𝑣=𝑣₁,𝑣₂∈𝐻¹₀Ω,𝑔, (8) 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 (𝐻¹₀(Ω))².

Let 𝒱=𝑢∈𝐶^∞₀(Ω)²∶∇⋅𝑔𝑢=0. (9) Denote by 𝐻_𝑔 the closure of 𝒱 in 𝐿²(Ω,𝑔), and denote by 𝑉_𝑔 the closure of 𝒱 in 𝐻¹₀(Ω,𝑔). It follows that 𝑉_𝑔⊂𝐻_𝑔≡𝐻^′_𝑔⊂𝑉^′_𝑔, where the injections are dense and continuous. We will use ‖⋅‖_∗ for the norm in 𝑉^′_𝑔, and ⟨⋅,⋅⟩ for duality pairing between 𝑉_𝑔 and 𝑉^′_𝑔.

We now define the trilinear form 𝑏 by 𝑏(𝑢,𝑣,𝑤)=²_{𝑖,𝑗=1Ω}𝑢_𝑖𝜕𝑣_𝑗𝜕𝑥_𝑖𝑤_𝑗𝑔𝑑𝑥, (10) whenever the integrals make sense. It is easy to check that if 𝑢,𝑣,𝑤∈𝑉_𝑔, then 𝑏(𝑢,𝑣,𝑤)=−𝑏(𝑢,𝑤,𝑣). (11) Hence 𝑏(𝑢,𝑣,𝑣)=0,∀𝑢,𝑣∈𝑉_𝑔. (12)

Set 𝐴∶𝑉_𝑔→𝑉^′_𝑔 by ⟨𝐴𝑢,𝑣⟩=((𝑢,𝑣))_𝑔, 𝐵∶𝑉_𝑔×𝑉_𝑔→𝑉^′_𝑔 by ⟨𝐵(𝑢,𝑣),𝑤⟩=𝑏(𝑢,𝑣,𝑤). Denote 𝐷(𝐴)={𝑢∈𝑉_𝑔∶𝐴𝑢∈𝐻_𝑔}, then 𝐷(𝐴)=𝐻²(Ω,𝑔)∩𝑉_𝑔 and 𝐴𝑢=−𝑃_𝑔Δ𝑢, for all 𝑢∈𝐷(𝐴), where 𝑃_𝑔 is the ortho-projector from 𝐿²(Ω,𝑔) onto 𝐻_𝑔. Consequently, there exists an orthogonal basis of 𝐻_𝑔 consisting of the eigenvectors 𝑤_𝑗 of 𝐴: 𝐴𝑤_𝑗=𝜆_𝑗𝑤_𝑗,0<𝜆₁≤𝜆₂𝜆≤⋯,_𝑗⟶∞as𝑗⟶∞. (13) Furthermore, we can also define the 𝑠th power 𝐴^𝑠 of 𝐴 for all 𝑠∈ℝ. The space 𝐷(𝐴^𝑠) is the Hilbert space when equipped with the scalar product (𝐴^𝑠,𝐴^𝑠) and norm |𝐴^𝑠.|, where (⋅,⋅) and |⋅| denote the scalar product and norm in 𝐻_𝑔. In particular, 𝐷(𝐴⁰)=𝐻_𝑔 and 𝐷(𝐴^1/2)=𝑉_𝑔.

Let 𝐻_𝑚=span{𝑤₁,…,𝑤_𝑚}. Then, the following estimates hold: 𝜆₁|𝑣|²≤‖𝑣‖²∀𝑣∈𝑉_𝑔,𝑣_𝑚²≤𝜆_𝑚𝑣_𝑚²∀𝑣_𝑚∈𝐻_𝑚,𝜆_𝑚+1𝑤_𝑚²≤𝑤_𝑚²∀𝑤_𝑚∈𝑉_𝑔⧵𝐻_𝑚. (14)(15)

Using the Hölder inequality, the Ladyzhenskaya inequality (when 𝑛=2): |𝑢|_𝐿⁴≤𝑐|𝑢|^1/2|∇𝑢|^1/2,∀𝑢∈𝐻¹₀(Ω), (16) and the interpolation inequalities, as in [15, 16], one can prove the following.

Lemma 1. If 𝑛=2, then|𝑏(≤𝑐𝑢,𝑣,𝑤)|₀|𝑢|^1/2‖𝑢‖^1/2‖𝑣‖|𝑤|^1/2‖𝑤‖^1/2,∀𝑢,𝑣,𝑤∈𝑉_𝑔,𝑐₀|𝑢|^1/2‖𝑢‖^1/2‖𝑣‖|𝐴𝑤|^1/2|𝑤|^1/2,∀𝑢∈𝑉_𝑔,𝑣∈𝐷(𝐴),𝑤∈𝐻_𝑔,𝑐₀|𝑢|^1/2|𝐴𝑢|^1/2‖𝑣𝑤|,(𝐴)‖|∀𝑢∈𝐷,𝑣∈𝑉_𝑔,𝑤∈𝐻_𝑔,𝑐₀|𝑢|‖𝑣‖|𝑤|^1/2|𝐴𝑤|^1/2,∀𝑢∈𝐻_𝑔𝑣∈𝑉_𝑔,𝑤∈𝐷(𝐴),|𝑏(𝑢,𝑣,𝑤)|+|𝑏(𝑤,𝑣,𝑢)|≤𝑐₀|𝑢|^1/2‖𝑢‖^1/2‖𝑣‖^1/2|𝐴𝑣|^1/2|𝑤|∀𝑢∈𝑉_𝑔,𝑣∈𝐷(𝐴),𝑤∈𝐻_g,|𝑏(𝑢,𝑣,𝑤)|+|𝑏(𝑤,𝑣,𝑢)|≤𝑐₀||1+ln𝐴𝑢²𝜆₁‖𝑢‖^21/2‖𝑢𝑣𝑤|(𝐴)‖‖‖|∀𝑢∈𝐷,𝑣∈𝑉_𝑔,𝑤∈𝐻_𝑔, (17)(18)(19) where 𝑐₀ are appropriate constants depending only on Ω.

Lemma 2 (see [3]). Let 𝑢∈𝐿²(0,𝑇;𝐷(𝐴))∩𝐿^∞(0,𝑇;𝑉_𝑔), then the function 𝐵𝑢 defined by (𝐵𝑢(𝑡),𝑣)_𝑔=𝑏(𝑢(𝑡),𝑢(𝑡),𝑣),∀𝑣∈𝐻_𝑔,𝑎.𝑒.𝑡∈[0,𝑇], (20) belongs to 𝐿⁴(0,𝑇;𝐻_𝑔), and therefore also belongs to 𝐿²(0,𝑇;𝐻_𝑔).

Lemma 3 (see [4]). Let 𝑢∈𝐿²(0,𝑇;𝑉_𝑔), then the function 𝐶𝑢 defined by (𝐶𝑢(𝑡),𝑣)_𝑔=∇𝑔𝑔⋅∇𝑢,𝑣_𝑔=𝑏∇𝑔𝑔,𝑢,𝑣,∀𝑣∈𝑉_𝑔, (21) belongs to 𝐿²(0,𝑇;𝐻_𝑔), and hence also belongs to 𝐿²(0,𝑇;𝑉^′_𝑔). Moreover, |𝐶𝑢(𝑡)|≤∇𝑔_∞𝑚₀⋅‖𝑢(𝑡(),‖‖)‖,𝑓𝑜𝑟𝑎.𝑒.𝑡∈0,𝑇𝐶𝑢(𝑡)_∗≤∇𝑔_∞𝑚₀𝜆₁^1/2⋅‖𝑢(𝑡)‖,𝑓𝑜𝑟𝑎.𝑒.𝑡∈(0,𝑇). (22)

Since −1𝑔∇⋅𝑔∇𝑢=−Δ𝑢−∇𝑔𝑔⋅∇𝑢, (23) we have (−Δ𝑢,𝑣)_𝑔=((𝑢,𝑣))_𝑔+∇𝑔𝑔⋅∇𝑢,𝑣_𝑔=(𝐴𝑢,𝑣)_𝑔+∇𝑔𝑔⋅∇𝑢,𝑣_𝑔,∀𝑢,𝑣∈𝑉_𝑔. (24)

2.2. Discrete Gronwall Inequalities

Hereafter, we will frequently use the following modified discrete Gronwall lemmas.

Lemma 4 (see [12]). Let 0<Δ𝑡<1,𝑎_𝑘,𝑔_𝑘,ℎ_𝑘 for integers 𝑘≥0 be nonnegative numbers such that 1𝑎Δ𝑡^𝑘+1−𝑎_𝑘≤𝑔_𝑘𝑎_𝑘+ℎ_𝑘∀0≤𝑘≤𝐽. (25) If Δ𝑡^𝐽_𝑘=0𝑔_𝑘≤𝛼₁,Δ𝑡^𝐽_𝑘=0ℎ_𝑘≤𝛼₂, (26) for 𝐽≤𝑛₀−1 and Δ𝑡^{𝑘0+𝑛0−1}_𝑘=𝑘0𝑔_𝑘≤𝛼₁,Δ𝑡^{𝑘0+𝑛0−1}_𝑘=𝑘0ℎ_𝑘≤𝛼₂,Δ𝑡^{𝑘0+𝑛0−1}_𝑘=𝑘0𝑎_𝑘≤𝑎₀+𝛼₃, (27) for 𝐽≥𝑛₀, 0≤𝑘₀≤𝐽+1−𝑛₀, then 𝑎_𝐽+1≤𝑒^𝛼1𝑎₀+𝛼₂+𝛼₃. (28)

Lemma 5 (see [13]). Let 𝛼 and 𝑎_𝑘,𝑏_𝑘,ℎ_𝑘,𝑘≥0 be nonnegative numbers such that 1+𝛾Δ𝑡𝑎_𝑘+1−𝑎_𝑘+𝑏_𝑘+1Δ𝑡≤ℎ_𝑘Δ𝑡∀0≤𝑘≤𝐽. (29) Then, 𝑎_𝐽+1+Δ𝑡^𝐽+1_𝑘=11+𝛾Δ𝑡^{−(𝐽+2−𝑘)}𝑏_𝑘≤1+𝛾Δ𝑡^−(𝐽+1)𝑎₀+Δ𝑡^𝐽_𝑘=01+𝛾Δ𝑡^{−(𝐽+1−𝑘)}ℎ_𝑘. (30)

Lemma 6 (see [11]). Let 𝛽 and 𝑎_𝑘,𝑏_𝑘,𝑔_𝑘,ℎ_𝑘 for integers 𝑘≥0 be nonnegative numbers such that 𝑎_𝑛+Δ𝑡^𝑛_𝑘=0𝑏_𝑘≤Δ𝑡^𝑛_𝑘=0𝑔_𝑘𝑎_𝑘+Δ𝑡^𝑛_𝑘=0ℎ_𝑘+𝛽∀𝑛≥0. (31) Suppose that 𝑔_𝑘Δ𝑡<1, for all 𝑘≥0, and set 𝜎_𝑘=(1−𝑔_𝑘Δ𝑡)⁻¹. Then, 𝑎_𝑛+Δ𝑡^𝑛_𝑘=0𝑏_𝑘≤expΔ𝑡^𝑛_𝑘=0𝜎_𝑘𝑔_𝑘Δ𝑡^𝑛_𝑘=0ℎ_𝑘+𝛽∀𝑛≥0. (32) Moreover, if 𝑎_𝑟+Δ𝑡^𝑛_𝑘=𝑟𝑏_𝑘≤Δ𝑡^𝑛_𝑘=𝑟𝑔_𝑘𝑎_𝑘+Δ𝑡^𝑛_𝑘=𝑟ℎ_𝑘+𝛽∀0≤𝑟≤𝑛, (33) and 𝑔_𝑘Δ𝑡<1 for all 0≤𝑘≤𝑛, then 𝑎_𝑟+Δ𝑡^𝑛_𝑘=𝑟𝑏_𝑘≤expΔ𝑡^𝑛_𝑘=0𝜎_𝑘𝑔_𝑘Δ𝑡^𝑛_𝑘=0ℎ_𝑘+𝛽;∀0≤𝑟≤𝑛. (34)

3. Existence and Some Estimates of Strong Solutions

In this section, we will prove some estimates for the strong solution 𝑢 and the Galerkin approximate solutions 𝑢_𝑚 of problem (1). First, with the operators defined in Section 2.1, one can write this problem as follows: 𝑑𝑢()(𝑡)(0)𝑑𝑡+𝜈𝐴𝑢+𝜈𝐶𝑢+𝐵𝑢,𝑢=𝑓,𝑡>0,𝑢=𝑢₀. (35)

Definition 7. For 𝑢₀∈𝑉_𝑔 given, a strong solution of problem (1) is a function 𝑢∈𝐿²(0,𝑇;𝐷(𝐴))∩𝐶([0,𝑇];𝑉_𝑔) for all 𝑇>0 such that 𝑢(0)=𝑢₀, and 𝑢 satisfies (35) in 𝐻_𝑔 for a.e. 𝑡>0.

Theorem 8. Suppose that 𝑓,𝑓_𝑡∈𝐿^∞(ℝ⁺;𝐻_𝑔) and 𝑢₀∈𝑉_𝑔. Then, problem (1) has a unique strong solution 𝑢 satisfying the following estimates for all 𝑡>0, |𝑢(𝑡)|²≤𝑒^{−𝜈𝛾0𝜆1𝑡}𝑢₀²+1𝜈²𝛾²₀𝜆²₁𝐶²_𝑓,_𝑡0𝑒^{𝜈𝛾0𝜆1𝑠}‖𝑢‖²2𝑑𝑠≤𝜈𝛾₀𝑢₀²+4𝜈³𝛾³₀𝜆²₁𝑒^{𝜈𝛾0𝜆1𝑡}𝐶²_𝑓,_𝑡0‖𝑢(𝑠)‖²1𝑑𝑠≤𝜈𝛾₀𝑢₀²+𝑡𝐶²_𝑓𝜈²𝛾²₀𝜆₁,‖𝑢(𝑡)‖²≤𝐶_𝑢0,𝑓,𝜏(𝑡)𝑢_𝑡(𝑡)²+|𝐴𝑢(𝑡)|²+𝜏²(𝑡)𝑢_𝑡(𝑡)²≤𝒦,_𝑡0𝑒^{−𝜈𝛾0𝜆1(𝑡−𝑠)}|𝐴𝑢|²+𝑢_𝑡²+𝜏(𝑠)𝑢_𝑡²+𝜏²(𝑠)𝑢_𝑡𝑡²+𝜏²(𝑠)𝐴𝑢_𝑡²𝑑𝑠≤𝒦, (36)(37)(38)(39)(40) where 𝛾₀=1−(|∇𝑔|_∞/𝑚₀𝜆₁^1/2)>0, 𝜏(𝑡)=min{1,𝑡}, 𝐶_𝑓=sup_𝑡≥0|𝑓|, 𝐶_𝑢0,𝑓=𝐶(‖𝑢₀‖,𝐶_𝑓), and 𝒦 is a generic positive constant depending only on the data (𝜈,Ω,|∇𝑔|_∞,𝜆₁,𝐶_𝑓,‖𝑢₀‖). Moreover, all above estimates are also valid for the Galerkin approximate solutions 𝑢_𝑚 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 𝑒^{𝜈𝛾0𝜆1𝑡}𝐴𝑢 and 𝜈⁻¹𝑒^{𝜈𝛾0𝜆1𝑡}𝑢_𝑡, respectively, and add the resulting relations to obtain 𝑑𝑒𝑑𝑡^{𝜈𝛾0𝜆1𝑡}‖𝑢‖²+𝑒^{𝜈𝛾0𝜆1𝑡}𝜈|𝐴𝑢|²+𝜈⁻¹𝑢_𝑡²+𝑒^{𝜈𝛾0𝜆1𝑡}𝑏𝑢,𝑢,𝐴𝑢+𝜈⁻¹𝑢_𝑡+𝜈𝑒^{𝜈𝛾0𝜆1𝑡}𝐶𝑢,𝐴𝑢+𝜈⁻¹𝑢_𝑡𝑔=𝜈𝛾₀𝜆₁𝑒^{𝜈𝜆1𝑡}‖𝑢‖²+𝑒^{𝜈𝛾0𝜆1𝑡}𝑓,𝐴𝑢+𝜈⁻¹𝑢_𝑡. (41) Using Lemmas 1 and 3 and Cauchy's inequality, we have 𝑏𝑢,𝑢,𝐴𝑢+𝜈⁻¹𝑢_𝑡≤14𝜈|𝐴𝑢|²+𝜈⁻¹𝑢_𝑡²+𝑐|𝑢|²‖𝑢‖²,𝜈𝐶𝑢,𝐴𝑢+𝜈⁻¹𝑢_𝑡𝑔≤18𝜈|𝐴𝑢|²+𝜈⁻¹𝑢_𝑡²+𝑐‖𝑢‖²,𝑓,𝐴𝑢+𝜈⁻¹𝑢_𝑡≤18𝜈|𝐴𝑢|²+𝜈⁻¹𝑢_𝑡²+𝑐𝑓². (42) By combining these inequalities with (41), we get 𝑑𝑒𝑑𝑡^{𝜈𝛾0𝜆1𝑡}‖𝑢‖²+12𝑒^{𝜈𝛾0𝜆1𝑡}𝜈|𝐴𝑢|²+𝜈⁻¹𝑢_𝑡²≤𝑒^{𝜈𝛾0𝜆1𝑡}𝜈𝜆₁+𝑐|𝑢|²‖𝑢‖²+𝑐‖𝑢‖²+𝑐𝑒^{𝜈𝛾0𝜆1𝑡}𝑓². (43) Integrating (43) from 0 to 𝑡 and using (36)–(38), we obtain, after multiplying by 𝑒^{−𝜈𝛾0𝜆1𝑡}, that 𝑒^{−𝜈𝛾0𝜆1𝑡𝑡}₀𝑒^{𝜈𝛾0𝜆1𝑠}𝜏(𝑠)𝜈|𝐴𝑢|²+𝜈⁻¹𝑢_𝑡²𝑑𝑠≤𝒦∀𝑡≥0. (44) In view of (44), there exists a sequence 𝜖_𝑘→0 such that 𝜏²𝜖_𝑘𝜈𝐴𝑢𝜖_𝑘²+𝜈⁻¹𝑢_𝑡𝜖_𝑘²⟶0. (45) Now, differentiating (35) with respect to 𝑡 yields 𝑢_𝑡𝑡+𝜈𝐴𝑢_𝑡+𝜈𝐶𝑢_𝑡+𝐵𝑢_𝑡,𝑢+𝐵𝑢,𝑢_𝑡=𝑓_𝑡. (46) We take the scalar product (46) with 2𝑢_𝑡 to obtain 𝑑𝑢𝑑𝑡^𝑡2+2𝜈𝑢_𝑡²+2𝜈𝐶𝑢_𝑡,𝑢_𝑡𝑔+2𝑏𝑢_𝑡,𝑢,𝑢_𝑡+2𝑏𝑢,𝑢_𝑡,𝑢_𝑡=2𝑓_𝑡,𝑢_𝑡𝑔. (47) By Lemma 3, we have 𝑑𝑢𝑑𝑡^𝑡2+2𝜈𝑢_𝑡²≤2𝜈∇𝑔_∞𝑚₀𝜆₁^1/2𝑢_𝑡²+2𝑏𝑢_𝑡,𝑢,𝑢_𝑡+2𝑏𝑢,𝑢_𝑡,𝑢_𝑡+2𝑓_𝑡,𝑢_𝑡𝑔. (48)
Using Lemma 1 and Cauchy's inequality, we get 𝑑𝑢𝑑𝑡^𝑡2+𝜈𝛾₀𝑢_𝑡²≤𝑐𝑢_𝑡²‖𝑢‖²+𝑐𝑓_𝑡². (49)
Multiplying the last inequality by 𝜏(𝑡)𝑒^{𝜈𝛾0𝜆1𝑡}, we have 𝑑𝑒𝑑𝑡^{𝜈𝛾0𝜆1𝑡}𝜏(𝑡)𝑢_𝑡²+𝜈𝛾₀𝑒^{𝜈𝛾0𝜆1𝑡}𝜏(𝑡)𝑢_𝑡²≤𝑒^{𝜈𝛾0𝜆1𝑡}1+𝜈𝛾₀𝜆₁+𝑐‖𝑢‖²𝑢_𝑡²+𝑐𝑒^{𝜈𝛾0𝜆1𝑡}𝑓_𝑡². (50) Therefore, integrating (50) from 𝜖_𝑘 to 𝑡, letting 𝜖_𝑘→0, and using (44) and (45), one finds, after multiplying by 𝑒^{−𝜈𝛾0𝜆1𝑡}, that 𝜏(𝑡)𝑢_𝑡(𝑡)²+𝜈𝑒^{−𝜈𝛾0𝜆1𝑡𝑡}₀𝑒^{𝜈𝛾0𝜆1𝑠}𝜏(𝑠)𝑢_𝑡²𝑑𝑠≤𝒦∀𝑡≥0. (51) Moreover, in view of (35), (36)–(38), and (51), we see that 𝜏(𝑡)|𝐴𝑢(𝑡)|²≤𝑐𝑓_𝑡(𝑡)²+𝑐𝜏(𝑡)𝑢_𝑡(𝑡)²+𝑐|𝑢(𝑡)|²‖𝑢(𝑡)‖⁴≤𝒦∀𝑡≥0. (52) Also, in view of (51), there exists a sequence 𝜖_𝑘→0 such that 𝜏²𝜖_𝑘𝑢_𝑡(𝜖_𝑘)²⟶0. (53) We again take the scalar product (46) with 2𝐴𝑢_𝑡 to obtain 𝑑𝑢𝑑𝑡^𝑡2+2𝜈𝐴𝑢_𝑡²+2𝜈𝐶𝑢_𝑡,𝐴𝑢_𝑡𝑔+2𝑏𝑢_𝑡,𝑢,𝐴𝑢_𝑡+2𝑏𝑢,𝑢_𝑡,𝐴𝑢_𝑡=2𝑓_𝑡,𝐴𝑢_𝑡𝑔. (54) By Lemma 3 and Cauchy's inequality, we have 𝑑𝑢𝑑𝑡^𝑡2+2𝜈|𝐴𝑢_𝑡|²≤2𝜈∇𝑔_∞𝑚₀𝜆₁^1/2𝐴𝑢_𝑡²+𝜈∇𝑔_∞𝜆₁^1/22𝑚₀𝑢_𝑡²+2𝑏𝑢_𝑡,𝑢,𝐴𝑢_𝑡+2𝑏𝑢,𝑢_𝑡,𝐴𝑢_𝑡+2𝑓_𝑡,𝐴𝑢_𝑡𝑔. (55) Using Lemma 1, Cauchy's inequality, and Young's inequality, we obtain 𝑑𝑢𝑑𝑡^𝑡2+𝜈𝛾₀𝐴𝑢_𝑡²≤𝑐‖𝑢‖⁴𝑢_𝑡²+𝑐𝑓_𝑡²+𝜈∇𝑔_∞𝜆₁^1/22𝑚₀𝑢_𝑡². (56) Multiplying the last inequality by 𝜏²(𝑡)𝑒^{𝜈𝛾0𝜆1𝑡}, we have 𝑑𝑒𝑑𝑡^{𝜈𝛾0𝜆1𝑡}𝜏²(𝑡)𝑢_𝑡²+𝜈𝛾₀𝜏²(𝑡)𝑒^{𝜈𝛾0𝜆1𝑡}𝐴𝑢_𝑡²≤𝑐𝑒^{𝜈𝛾0𝜆1𝑡}𝑓_𝑡²+𝑒^{𝜈𝛾0𝜆1𝑡}2+𝜈𝛾₀𝜆₁+𝑐‖𝑢‖⁴+𝜈∇𝑔_∞𝜆₁^1/22𝑚₀𝜏(𝑡)𝑢_𝑡². (57) Integrating (57) from 𝜖_𝑘 to 𝑡, letting 𝜖_𝑘→0, and using (36)–(38), (51), and (53), we obtain, after a final multiplication by 𝑒^{−𝜈𝛾0𝜆1𝑡}, 𝜏²(𝑡)𝑢_𝑡(𝑡)²+𝜈𝛾₀𝑒^{−𝜈𝛾0𝜆1𝑡𝑡}₀𝑒^{𝜈𝛾0𝜆1𝑠}𝜏²(𝑠)𝐴𝑢_𝑡²𝑑𝑠≤𝒦∀𝑡≥0. (58) Using Lemmas 1 and 3 and (46), we deduce that 𝑒^{−𝜈𝛾0𝜆1𝑡𝑡}₀𝜏²(𝑡)𝑒^{𝜈𝛾0𝜆1𝑠}𝑢_𝑡𝑡²𝑑𝑠≤𝑐𝑒^{−𝜈𝛾0𝜆1𝑡}×^𝑡₀𝜏²(𝑡)𝑒^{𝜈𝛾0𝜆1𝑠}𝐴𝑢_𝑡²+‖𝑢‖²𝐴𝑢_𝑡²+𝑓_𝑡²𝑑𝑠∀𝑡≥0. (59) Combining (44), (51), (52), (58), and (59) yields (39) and (40).
Finally, we observe that the problem for the approximate solution 𝑢_𝑚 is similar to problem (35), and 𝑢_𝑚(0)≤𝑢₀,𝑢_𝑚(0)≤𝑢₀, (60) so 𝑢_𝑚 satisfies the same estimates as those for the strong solution 𝑢 of problem (35).

4. Spectral Galerkin Method in Space

For a fixed integer 𝑚, let 𝑃_𝑚 be the orthogonal projection of 𝐻_𝑔 onto 𝐻_𝑚=span{𝑤₁,…,𝑤_𝑚} and 𝑄_𝑚=𝐼−𝑃_𝑚. Then, every solution 𝑢 of problem (1) can be decomposed uniquely into 𝑢=𝑝+𝑞,where𝑝=𝑃_𝑚𝑢,𝑞=𝑄_𝑚𝑢. (61) Now, we apply 𝑃_𝑚 and 𝑄_𝑚 to (35) to obtain 𝑝_𝑡+𝜈𝐴𝑝+𝜈𝐶𝑝+𝑃_𝑚𝐵(𝑢,𝑢)=𝑃_𝑚𝑓∀𝑡>0,𝑞_𝑡+𝜈𝐴𝑞+𝜈𝐶𝑞+𝑄_𝑚𝐵(𝑢,𝑢)=𝑄_𝑚𝑓∀𝑡>0, (62)(63) and the initial conditions 𝑝(0)=𝑃_𝑚𝑢₀, 𝑞(0)=𝑄_𝑚𝑢₀.

Using Theorem 8 and the property of 𝑃_𝑚, we arrive at the following estimates of 𝑞(𝑡)=𝑄_𝑚𝑢(𝑡): 𝑞(𝑡)²≤𝒦𝜆⁻¹_𝑚+1,𝜏(𝑡)𝑞(𝑡)²+𝜆⁻¹_𝑚+1𝑞(𝑡)²≤2𝒦𝜆⁻²_𝑚+1∀𝑡≥0,𝑒^{−𝜈𝛾0𝜆1𝑡𝑡}₀𝑒^{𝜈𝛾0𝜆1𝑠}𝑞²+𝜆⁻¹_𝑚+1𝑞²𝑑𝑠≤2𝒦𝜆⁻²_𝑚+1∀𝑡≥0. (64)(65) We now define the spectral Galerkin method as follows: find 𝑢_𝑚(𝑡)∈𝐻_𝑚 such that 𝑢_𝑚𝑡+𝜈𝐴𝑢_𝑚+𝜈𝐶𝑢_𝑚+𝑃_𝑚𝐵𝑢_𝑚,𝑢_𝑚=𝑃_𝑚𝑓∀𝑡>0, (66) with the initial condition 𝑢_𝑚(0)=𝑃_𝑚𝑢₀.

In order to give an analysis of the error 𝑢−𝑢_𝑚 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;𝐻_𝑔), the dual problem: find Φ(𝑠)∈𝐻_𝑚 such that 𝑣,Φ_𝑠𝑔−𝑎(𝑣,Φ)−𝜈(𝐶𝑣,Φ)_𝑔−𝑏𝑣,𝑢_𝑚,Φ−𝑏𝑢_𝑚=,𝑣,Φ𝑣,𝑒^{𝜈𝛾0𝜆1𝑠}𝜉_𝑔,0≤𝑠<𝑡, (67) for all 𝑣∈𝐻_𝑚 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;𝑉_𝑔)∩𝐿²(0,𝑡;𝐷(𝐴)).

Next, we prove a regularity result of problem (67).

Lemma 9. If 𝑢₀∈𝑉_𝑔, then the solution Φ(𝑠) of problem (67) satisfies 𝑒^{−𝜈𝛾0𝜆1𝑠}‖Φ(𝑠)‖²+^𝑡_𝑠𝑒^{−𝜈𝛾0𝜆1𝑟}|𝐴Φ|²+Φ_𝑠²2𝑑𝑟≤𝒦exp𝜈𝛾₀𝑐²₀^𝑡₀𝑢_𝑚²𝑑𝑟^𝑡₀𝑒^{𝜈𝛾0𝜆1𝑟}𝜉²𝑑𝑟. (68)

Proof. Taking 𝑣=−2Φ in (67), we obtain −𝑑|Φ|𝑑𝑠²+2𝜈‖Φ‖²+2𝜈(𝐶Φ,Φ)_𝑔+2𝑏Φ,𝑢_𝑚,Φ=−2Φ,𝑒^{𝜈𝛾0𝜆1𝑠}𝜉_𝑔∀0≤𝑠<𝑡. (69) Using Lemma 3, we have −𝑑|Φ|𝑑𝑠²+2𝜈𝛾₀‖Φ‖²≤2Φ,𝑒^{𝜈𝛾0𝜆1𝑠}𝜉_𝑔+2𝑏Φ,𝑢_𝑚,Φ∀0≤𝑠<𝑡. (70) Using Lemma 1 and Cauchy's inequality, we get 2𝑏Φ,𝑢_𝑚,Φ≤𝜈𝛾₀2‖Φ‖²+2𝜈𝛾₀𝑐²₀𝑢_𝑚²|Φ|²,2Φ,𝑒^{𝜈𝛾0𝜆1𝑠}𝜉_𝑔≤𝜈𝛾₀4‖Φ‖²+4𝜈𝛾₀𝜆₁𝑒^{2𝜈𝛾0𝜆1𝑠}𝜉². (71) Then, we have −𝑑|Φ|𝑑𝑠²+54𝜈𝛾₀‖Φ‖²≤2𝜈𝛾₀𝑐²₀𝑢_𝑚²|Φ|²+4𝜈𝛾₀𝜆₁𝑒^{2𝜈𝛾0𝜆1𝑠}𝜉²∀0≤𝑠<𝑡. (72) Multiplying this inequality by 𝑒^{−𝜈𝛾0𝜆1𝑠} and using (14), we obtain −𝑑𝑒𝑑𝑠^{−𝜈𝛾0𝜆1𝑠}|Φ|²+𝜈𝛾₀4𝑒^{−𝜈𝛾0𝜆1𝑠}‖Φ‖²≤2𝜈𝛾₀𝑐²₀𝑒^{−𝜈𝛾0𝜆1𝑠}𝑢_𝑚²|Φ|²+4𝜈𝛾₀𝜆₁𝑒^{𝜈𝛾0𝜆1𝑠}𝜉². (73) Multiplying this inequality by 𝑒^{(2/𝜈𝛾0)𝑐20∫𝑠0‖𝑢𝑚‖2𝑑𝑟} yields −𝑑𝑒𝑑𝑠^{(2/𝜈𝛾0)𝑐20∫𝑠0‖𝑢𝑚‖2𝑑𝑟}𝑒^{−𝜈𝛾0𝜆1𝑠}|Φ|²+𝜈𝛾₀4𝑒^{(2/𝜈𝛾0)𝑐20∫𝑠0‖𝑢𝑚‖2𝑑𝑟}𝑒^{−𝜈𝛾0𝜆1𝑠}‖Φ‖²≤4𝜈𝛾₀𝜆₁𝑒^{(2/𝜈𝛾0)𝑐20∫𝑠0‖𝑢𝑚‖2𝑑𝑟}𝑒^{𝜈𝛾0𝜆1𝑠}𝜉². (74) Integrating (74) from 𝑠 to 𝑡 and noting that Φ(𝑡)=0, we obtain, after multiplying by 𝑒^{−(2/𝜈𝛾0)𝑐20∫𝑠0‖𝑢𝑚‖2𝑑𝑟}, that 𝑒^{−𝜈𝛾0𝜆1𝑠}|Φ(𝑠)|²+𝜈𝛾₀4^𝑡_𝑠𝑒^{−𝜈𝛾0𝜆1𝑟}‖Φ‖²≤4𝑑𝑟𝜈𝛾₀𝜆₁𝑒^{(2/𝜈𝛾0)𝑐20∫𝑡0‖𝑢𝑚‖2𝑑𝑟𝑡}₀𝑒^{𝜈𝛾0𝜆1𝑟}𝜉²𝑑𝑟∀0≤𝑠≤𝑡. (75) Moreover, inserting 𝑣=−2𝐴Φ into (67), we get −𝑑‖Φ‖𝑑𝑡²+2𝜈|𝐴Φ|²+2𝜈(𝐶𝐴Φ,Φ)_𝑔−2𝑏𝐴Φ,Φ,𝑢_𝑚+𝑏𝑢_𝑚,Φ,𝐴Φ=−2𝐴Φ,𝑒^{𝜈𝛾0𝜆1𝑠}𝜉^𝑘_𝑔. (76) Using Lemma 3, (14), and Cauchy's inequality, we have 2𝜈(𝐶𝐴Φ,Φ)_𝑔≤2𝜈∇𝑔_∞𝑚₀𝜆₁^1/2|𝐴Φ|²+𝜈∇𝑔_∞𝜆_𝑚2𝑚₀𝜆₁^1/2‖Φ‖². (77) Therefore, −𝑑‖Φ‖𝑑𝑡²+2𝜈𝛾₀|𝐴Φ|²≤2𝐴Φ,𝑒^{𝜈𝛾0𝜆1𝑠}𝜉^𝑘_𝑔+2𝑏𝐴Φ,Φ,𝑢_𝑚+𝑏𝑢_𝑚+𝜈,Φ,𝐴Φ∇𝑔_∞𝜆_𝑚2𝑚₀𝜆₁^1/2‖Φ‖². (78) Multiplying this inequality, by 𝑒^{−𝜈𝛾0𝜆1𝑠} and using (14), we obtain −𝑑𝑒𝑑𝑡^{−𝜈𝛾0𝜆1𝑠}‖Φ‖²+𝜈𝛾₀𝑒^{−𝜈𝛾0𝜆1𝑠}|𝐴Φ|²≤2𝐴Φ,𝜉^𝑘_𝑔+2𝑒^{−𝜈𝛾0𝜆1𝑠}𝑏𝐴Φ,Φ,𝑢_𝑚+𝑏𝑢_𝑚+𝜈,Φ,𝐴Φ∇𝑔_∞𝜆_𝑚2𝑚₀𝜆₁^1/2𝑒^{−𝜈𝛾0𝜆1𝑠}‖Φ‖². (79) Using Lemma 1, Cauchy's inequality and Young's inequality, we have 2𝑏𝐴Φ,Φ,𝑢_𝑚+2𝑏𝑢_𝑚,Φ,𝐴Φ≤𝑐₀|𝐴Φ|^3/2‖Φ‖^1/2𝑢_𝑚^1/2𝑢_𝑚^1/2≤𝜈𝛾₀2|𝐴Φ|²+𝑐‖Φ‖²𝑢_𝑚²𝑢_𝑚²,2𝐴Φ,𝜉^𝑘_𝑔≤𝜈𝛾₀4𝑒^{−𝜈𝛾0𝜆1𝑠}|𝐴Φ|²+4𝜈𝛾₀𝑒^{𝜈𝛾0𝜆1𝑠}𝜉^𝑘2. (80) Combining the above estimates with (79) yields −𝑑𝑒𝑑𝑡^{−𝜈𝛾0𝜆1𝑠}‖Φ‖²+𝜈𝛾₀4𝑒^{−𝜈𝛾0𝜆1𝑠}|𝐴Φ|²≤𝑐𝑒^{−𝜈𝛾0𝜆1𝑠}‖Φ‖²𝑢_𝑚²𝑢_𝑚²+𝜈∇𝑔_∞𝜆_𝑚2𝑚₀𝜆₁^1/2𝑒^{−𝜈𝛾0𝜆1𝑠}‖Φ‖²+4𝜈𝛾₀𝑒^{𝜈𝛾0𝜆1𝑠}𝜉². (81) Integrating this inequality from 𝑠 to 𝑡 and using (75), we have 𝑒^{−𝜈𝛾0𝜆1𝑠}‖Φ‖²+𝜈𝛾₀4^𝑡_𝑠𝑒^{−𝜈𝛾0𝜆1𝑟}|𝐴Φ|²𝜈𝑑𝑟≤𝑐1+∇𝑔_∞𝜆_𝑚2𝑚₀𝜆₁^1/2+sup_𝑡≥0𝑢_𝑚(𝑡)²𝑢_𝑚(𝑡)²×𝑒^{(2/𝜈𝛾0)𝑐20∫𝑡0‖𝑢𝑚‖2𝑑𝑟𝑡}₀𝑒^{𝜈𝛾0𝜆1𝑟}𝜉²𝑑𝑟. (82) Taking 𝑣=Φ_𝑠 in (67), then using (14), Lemmas 1 and 3, we obtain Φ_𝑠≤𝜈|𝐴Φ|+𝜈∇𝑔_∞𝜆_𝑚𝑚₀𝜆₁^1/2‖Φ‖+𝑐𝑢_𝑚|𝐴Φ|+𝑒^{𝜈𝛾0𝜆1𝑠}𝜉. (83) Hence, 𝑒^{−𝜈𝛾0𝜆1𝑠}Φ_𝑠²≤𝑐1+𝑢_𝑚²𝑒^{−𝜈𝛾0𝜆1𝑠}|𝐴Φ|²+𝑐𝑒^{−𝜈𝛾0𝜆1𝑠}‖Φ‖²+𝑐𝑒^{𝜈𝛾0𝜆1𝑠}𝜉². (84) Integrating (84) from 𝑠 to 𝑡 and using (82), (75), and Theorem 8, we complete the proof.

Lemma 10. If 𝑢₀∈𝑉_𝑔, then the error 𝑃_𝑚𝑢(𝑡)−𝑢_𝑚(𝑡) satisfies 𝑃_𝑚𝑢(𝑡)−𝑢_𝑚(𝑡)²+𝜈𝛾₀4𝑒^{−𝜈𝛾0𝜆1𝑡𝑡}₀𝑒^{𝜈𝛾0𝜆1𝑠}𝑃_𝑚𝑢−𝑢_𝑚²𝑑𝑠≤𝒦𝑒^{(2/𝜈𝛾0)𝑐20∫𝑡0‖𝑢(𝑟)‖2𝑑𝑟}𝜆⁻¹_𝑚+1∀𝑡≥0. (85)

Proof. We set 𝑒(𝑡)=𝑃_𝑚𝑢(𝑡)−𝑢_𝑚(𝑡) and subtract (66) from (62) to obtain 𝑒_𝑡+𝜈𝐴𝑒+𝜈𝐶𝑒+𝑃_𝑚𝐵𝑄_𝑚𝑢+𝑒,𝑢+𝑃_𝑚𝐵𝑢_𝑚,𝑄𝑢_𝑚+𝑒=0, (86) with 𝑒(0)=0. Taking the scalar product of (86) with 2𝑒, we obtain 𝑑|𝑒|𝑑𝑡²+2𝜈‖𝑒‖²+2(𝐶𝑒,𝑒)_𝑔+2𝑏(𝑒,𝑢,𝑒)+𝑏𝑢_𝑚,𝑄_𝑚𝑢,𝑒+𝑏𝑄_𝑚𝑢,𝑢,𝑒=0. (87) Using Lemma 3, we get 𝑑|𝑒|𝑑𝑡²+2𝜈𝛾₀‖𝑒‖²≤2𝑏(𝑒,𝑢,𝑒)+𝑏𝑢_𝑚,𝑄_𝑚𝑢,𝑒+𝑏𝑄_𝑚𝑢,𝑢,𝑒. (88) Multiplying the last inequality by 𝑒^{𝜈𝛾0𝜆1𝑡} and using (14), we obtain 𝑑𝑒𝑑𝑡^{𝜈𝛾0𝜆1𝑡}|𝑒|²+𝜈𝛾₀𝑒^{𝜈𝛾0𝜆1𝑡}‖𝑒‖²≤2𝑒^{𝜈𝛾0𝜆1𝑡}𝑏(𝑒,𝑢,𝑒)+𝑏𝑢_𝑚,𝑄_𝑚𝑢,𝑒+𝑏𝑄_𝑚𝑢,𝑢,𝑒. (89) Due to Lemma 1 and Cauchy's inequality, we have 2|𝑏(𝑒,𝑢,𝑒)|≤2𝑐₀|𝑒|‖𝑒‖‖𝑢‖≤𝜈𝛾₀2‖𝑒‖²+2𝜈𝛾₀𝑐²₀‖𝑢‖²|𝑒|²,2𝑏𝑢_𝑚,𝑄_𝑚𝑢,𝑒+2𝑏𝑄_𝑚𝑢,𝑢,𝑒≤2𝑐₀𝜆₁^−1/2𝑄_𝑚𝑢|𝐴𝑢|+𝐴𝑢_𝑚‖𝑒‖≤𝜈𝛾₀4‖𝑒‖²+𝑐|𝐴𝑢|²+𝐴𝑢_𝑚²𝑄_𝑚𝑢². (90) Combining (89) with the above estimate yields 𝑑𝑒𝑑𝑡^{𝜈𝛾0𝜆1𝑡}|𝑒|²+𝜈𝛾₀4𝑒^{𝜈𝛾0𝜆1𝑡}‖𝑒‖²≤2𝜈𝛾₀𝑐²₀𝑒^{𝜈𝛾0𝜆1𝑡}‖𝑢‖²|𝑒|²+𝑐𝑒^{𝜈𝛾0𝜆1𝑡}|𝐴𝑢|²+𝐴𝑢_𝑚²𝑄_𝑚𝑢². (91) Multiplying (91) by 𝑒^{−(2/𝜈𝛾0)𝑐20∫𝑠0‖𝑢‖2𝑑𝑟} yields 𝑑𝑒𝑑𝑡^{−(2/𝜈𝛾0)𝑐20∫𝑠0‖𝑢‖2𝑑𝑟}𝑒^{𝜈𝛾0𝜆1𝑡}|𝑒|²+𝜈𝛾₀4𝑒^{−(2/𝜈𝛾0)𝑐20∫𝑠0‖𝑢‖2𝑑𝑟}𝑒^{𝜈𝛾0𝜆1𝑡}‖𝑒‖²≤𝑐𝑒^{−(2/𝜈𝛾0)𝑐20∫𝑠0‖𝑢‖2𝑑𝑟}𝑒^{𝜈𝛾0𝜆1𝑡}|𝐴𝑢|²+𝐴𝑢_𝑚²𝑄_𝑚𝑢²≤𝑐𝑒^{𝜈𝛾0𝜆1𝑡}|𝐴𝑢|²+𝐴𝑢_𝑚²𝑄_𝑚𝑢². (92) Integrating this inequality from 0 to 𝑡 and using (64), Theorem 8, we obtain, after a final multiplication by 𝑒^{(2/𝜈𝛾0)𝑐20∫𝑠0‖𝑢‖2𝑑𝑟}𝑒^{−𝜈𝛾0𝜆1𝑡}, that |𝑒|²+𝜈𝛾₀4𝑒^{−𝜈𝛾0𝜆1𝑡𝑡}₀𝑒^{𝜈𝛾0𝜆1𝑠}‖𝑒‖²𝑑𝑠≤𝒦𝑒^{(2/𝜈𝛾0)𝑐20∫𝑡0‖𝑢‖2𝑑𝑟}𝜆⁻¹_𝑚+1∀𝑡≥0, (93) which is (85).

Lemma 11. If 𝑢₀∈𝑉_𝑔, then the error 𝑃_𝑚𝑢(𝑡)−𝑢_𝑚(𝑡) satisfies the following bound: 𝜏(𝑡)𝑃_𝑚𝑢(𝑡)−𝑢_𝑚(𝑡)²≤𝒦𝑒^{(4/𝜈𝛾0)𝑐20∫𝑡0(‖𝑢‖2+‖𝑢𝑚‖2)𝑑𝑟}𝜆⁻²_𝑚+1∀𝑡≥0. (94)

Proof. Take 𝑣=𝑒(𝑠) and 𝜉=𝑒(𝑠) in (67) to obtain 𝑒,Φ_𝑠𝑔−𝑎(𝑒,Φ)−𝜈(𝐶𝑒,Φ)_𝑔−𝑏𝑒,𝑢_𝑚,Φ−𝑏𝑢_𝑚,𝑒,𝜙=𝑒^{𝜈𝛾0𝜆1𝑠}|𝑒|². (95) Multiplying (86) by Φ, we have 𝑒_𝑠,Φ_𝑔+𝑎(𝑒,Φ)+𝜈(𝐶𝑒,Φ)_𝑔+𝑏𝑢_𝑚,𝑒+𝑄_𝑚𝑢,Φ+𝑏𝑒+𝑄_𝑚𝑢,𝑢_𝑚,Φ+𝑏𝑢−𝑢_𝑚,𝑢−𝑢_𝑚,Φ=0. (96) Adding (96) and (95), we get 𝑒^{𝜈𝛾0𝜆1𝑠}|𝑒|²=𝑑()𝑑𝑠𝑒,Φ_𝑔+𝑏𝑢_𝑚,𝑄_𝑚𝑄𝑢,Φ+𝑏_𝑚𝑢,𝑢_𝑚,Φ+𝑏𝑢−𝑢_𝑚,𝑢−𝑢_𝑚,Φ. (97) Using Lemma 1, we have 𝑏𝑢_𝑚,𝑄_𝑚𝑢,Φ+𝑏𝑄_𝑚𝑢,𝑢_𝑚,Φ≤𝑐𝑢_𝑚𝑄_𝑚𝑢|,𝑏|𝐴Φ𝑢−𝑢_𝑚,𝑢−𝑢_𝑚,Φ≤𝑐𝑢−𝑢_𝑚𝑢−𝑢_𝑚‖𝐴Φ‖. (98) Hence, we deduce from (97) that 𝑒^{𝜈𝛾0𝜆1𝑠}|𝑒|²≤𝑑()𝑑𝑠𝑒,Φ_𝑔+𝑐𝑢_𝑚𝑄_𝑚𝑢+𝑢−𝑢_𝑚𝑢−𝑢_𝑚|𝐴Φ|. (99) Integrating (99) from 0 to 𝑡 and noting that 𝑒(0)=Φ(𝑡)=0, we obtain ^𝑡₀𝑒^{𝜈𝛾0𝜆1𝑠}|𝑒|²𝑑𝑠≤𝑐^𝑡₀𝑒^{𝜈𝛾0𝜆1𝑠}𝑢_𝑚²𝑄_𝑚𝑢²+𝑢−𝑢_𝑚²𝑢−𝑢_𝑚²𝑑𝑠^1/2×^𝑡₀𝑒^{−𝜈𝛾0𝜆1𝑠}|𝐴Φ|²𝑑𝑠^1/2. (100) Using (64), (65), Theorem 8, and Lemmas 9 and 10, we deduce from (100) that𝑒^{−𝜈𝛾0𝜆1𝑡𝑡}₀𝑒^{𝜈𝛾0𝜆1𝑠}|𝑒|²𝑑𝑠≤𝒦𝑒^{(4/𝜈𝛾0)𝑐20∫𝑡0(‖𝑢‖2+‖𝑢𝑚‖2)𝑑𝑟}𝜆⁻²_𝑚+1. (101) Now, multiplying (91) by 𝜏(𝑡), we have 𝑑𝜏(𝑡)𝑒𝑑𝑡^{𝜈𝛾0𝜆1𝑡}|𝑒|²+𝜈𝛾₀4𝜏(𝑡)𝑒^{𝜈𝛾0𝜆1𝑡}‖𝑒‖²≤21+𝜈𝛾₀𝑐²₀‖𝑢‖²𝑒^{𝜈𝛾0𝜆1𝑡}|𝑒|²+𝑐𝑒^{𝜈𝛾0𝜆1𝑡}𝜏(𝑡)|𝐴𝑢|²+𝐴𝑢_𝑚²𝑄_𝑚𝑢². (102) Integrating (102) from 0 to 𝑡 and using Theorem 8, we obtain, after a final multiplication by 𝑒^{−𝜈𝛾0𝜆1𝑡}, 𝜏(𝑡)|𝑒(𝑡)|²+𝜈𝛾₀4𝑒^{−𝜈𝛾0𝜆1𝑡𝑡}₀𝜏(𝑠)𝑒^{𝜈𝛾0𝜆1𝑠}‖𝑒‖²𝑑𝑠≤𝒦𝑒^{−𝜈𝛾0𝜆1𝑡𝑡}₀𝑒^{𝜈𝛾0𝜆1𝑡}|𝑒|²+𝑄_𝑚𝑢²𝑑𝑠. (103) Using (65) and (101) in (103) yields 𝜏(𝑡)|𝑒(𝑡)|²+𝜈𝛾₀4𝑒^{−𝜈𝛾0𝜆1𝑡𝑡}₀𝜏(𝑠)𝑒^{𝜈𝛾0𝜆1𝑠}‖𝑒‖²𝑑𝑠≤𝒦𝑒^{(4/𝜈𝛾0)𝑐20∫𝑡0(‖𝑢‖2+‖𝑢𝑚‖2)𝑑𝑟}𝜆⁻²_𝑚+1, (104) which is (94).

Finally, by combining Lemma 11 with (64) and using Theorem 8, we get the following error estimate.

Theorem 12. If 𝑢∈𝑉_𝑔, then the error 𝑢(𝑡)−𝑢_𝑚(𝑡) satisfies the following bound: 𝜏(𝑡)𝑢(𝑡)−𝑢_𝑚(𝑡)²≤𝒦𝑒^{(8/𝜈3𝛾30𝜆1)𝑐20𝐶2𝑓𝑡}𝜆⁻²_𝑚+1∀𝑡≥0. (105)

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 {𝑢^𝑛_𝑚}⊂𝐻_𝑚 such that 𝑑_𝑡𝑢_𝑚^𝑛+1+𝜈𝐴𝑢_𝑚^𝑛+1+𝜈𝐶𝑢_𝑚^𝑛+1+𝑃_𝑚𝐵𝑢^𝑛_𝑚,𝑢^𝑛_𝑚=𝑃_𝑚𝑓^𝑛+1, (106) for 𝑛≥0 with the initial condition 𝑢⁰_𝑚=𝑃_𝑚𝑢₀, where Δ𝑡>0 is a time step such that 𝑛₀Δ𝑡=1 for some integer 𝑛₀ and 𝑑_𝑡𝑢_𝑚^𝑛+1=1𝑢Δ𝑡^𝑚𝑛+1−𝑢^𝑛_𝑚,𝑓^𝑛+1=1Δ𝑡^𝑡𝑛+1_𝑡𝑛𝑓(𝑡)𝑑𝑡,𝑡_𝑛=𝑛Δ𝑡. (107)(108) In order to derive the 𝐿²-bound on the error 𝑢_𝑚(𝑡_𝑛)−𝑢^𝑛_𝑚, 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 𝑢_𝑚^𝑘−1,𝜉^𝑘∈𝐻_𝑚, 1≤𝑘≤𝑛, find Φ^𝑘−1∈𝐻_𝑚, 1≤𝑘≤𝑛 such that 𝑣,𝑑_𝑡Φ^𝑘_𝑔−𝑎𝑣,Φ^𝑘−1−𝜈𝐶𝑣,Φ^𝑘−1_𝑔−𝑏𝑣,𝑢_𝑚^𝑘−1,Φ^𝑘−1−𝑏𝑢_𝑚^𝑘−1,𝑣,Φ^𝑘−1=1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘𝑣,𝜉^𝑘_𝑔∀𝑣∈𝐻_𝑚, (109) with an initial condition Φ^𝑛=0.

The following theorem provides the 𝐻²-stability of scheme (106).

Theorem 13. Under the assumptions of Theorem 8, if Δ𝑡 and 𝑚 satisfy the following condition: 8𝜈𝛾₀𝑐²₀𝜆ln^𝑚𝜆₁Δ𝑡𝐶_𝑢0,𝑓≤1, (110) then the semi-implicit Euler scheme is the 𝐻²-stability; that is, Δ𝑡^𝑛_𝑘=1𝑢^𝑘_𝑚²≤1𝜈𝛾₀𝑢₀²+1𝜈²𝛾²₀𝜆₁𝐶²_𝑓𝑡_𝑛∀𝑛≥0,Δ𝑡^𝑛_𝑘=𝑘0+1𝑢^𝑘_𝑚²≤1𝜈𝛾₀𝑢₀²+2𝜈³𝛾³₀𝜆²₁𝐶²_𝑓+2𝜈²𝛾²₀𝜆₁𝐶²_𝑓𝑡_𝑛−𝑡_𝑘0∀𝑛≥𝑘₀≥0,𝑢^𝑛_𝑚²≤1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑛𝑢₀²+1𝜈²𝛾²₀𝜆²₁𝐶²_𝑓∀𝑛≥0,𝑢^𝑛_m²≤𝐶_𝑢0,𝑓∀𝑛≥0,Δ𝑡^𝑛_𝑘=11+𝜈𝛾₀𝜆₁Δ𝑡^{−(𝑛−1−𝑘)}×𝐴𝑢^𝑘_𝑚²+𝑑_𝑡𝑢^𝑘_𝑚²+𝜏𝑡_𝑘𝑑_𝑡𝑢^𝑘_𝑚²+𝜏𝑡_𝑛𝐴𝑢^𝑛_𝑚²+𝜏𝑡_𝑛𝑑_𝑡𝑢^𝑛_𝑚²≤𝒦∀𝑛≥0. (111)(112)(113)(114)(115)

Proof. Clearly, scheme (106) defines a unique sequence {𝑢^𝑛_𝑚}⊂𝐻_𝑚. Now, we will prove (111)–(115).
Taking the scalar product of (106) with 2𝑢_𝑚^𝑛+1Δ𝑡, we obtain 𝑢_𝑚^𝑛+12−𝑢^𝑛_𝑚²+𝑢_𝑚^𝑛+1−𝑢^𝑛_𝑚²+2𝜈𝑢_𝑚^𝑛+12Δ𝑡+2𝜈Δ𝑡𝐶𝑢_𝑚^𝑛+1,𝑢_𝑚^𝑛+1_𝑔+2𝑏𝑢^𝑛_𝑚,𝑢^𝑛_𝑚−𝑢_𝑚^𝑛+1,𝑢_𝑚^𝑛+1𝑓Δ𝑡=2^𝑛+1,𝑢_𝑚^𝑛+1_𝑔Δ𝑡. (116) Using Lemma 3 and Cauchy's inequality, we have 𝑢_𝑚^𝑛+12−𝑢^𝑛_𝑚²+𝑢_𝑚^𝑛+1−𝑢^𝑛_𝑚²+2𝜈𝑢_𝑚^𝑛+12Δ𝑡≤2𝜈∇𝑔_∞𝑚₀𝜆₁^1/2𝑢_𝑚^𝑛+12Δ𝑡+2𝑓^𝑛+1,𝑢_𝑚^𝑛+1_𝑔𝑢Δ𝑡−2𝑏^𝑛_𝑚,𝑢^𝑛_𝑚−𝑢_𝑚^𝑛+1,𝑢_𝑚^𝑛+1Δ𝑡. (117) Hence, 𝑢_𝑚^𝑛+12−𝑢^𝑛_𝑚²+𝑢_𝑚^𝑛+1−𝑢^𝑛_𝑚²+2𝜈𝛾₀𝑢_𝑚^𝑛+12𝑓Δ𝑡≤2^𝑛+1,𝑢_𝑚^𝑛+1_𝑔Δ𝑡−2𝑏𝑢^𝑛_𝑚,𝑢^𝑛_𝑚−𝑢_𝑚^𝑛+1,𝑢_𝑚^𝑛+1Δ𝑡. (118) By (19), we have 2𝑓^𝑛+1,𝑢_𝑚^𝑛+1_𝑔≤𝜈𝛾₀2𝑢_𝑚^𝑛+12+2𝜈𝛾₀𝜆₁𝑓^𝑛+122𝑏𝑢Δ𝑡,_𝑛𝑚,𝑢^𝑛_𝑚−𝑢_𝑚^𝑛+1,𝑢_𝑚^𝑛+1≤𝜈𝛾₀2𝑢_𝑚^𝑛+12+𝐿𝑢^𝑛_𝑚𝑢^𝑛_𝑚−𝑢_𝑚^𝑛+12, (119) where 𝐿𝑢^𝑛_𝑚=8𝜈𝛾₀𝑐²₀1+ln𝐴𝑢^𝑛_𝑚²𝜆₁𝑢^𝑛_𝑚²𝑢^𝑛_𝑚². (120) Substituting those into (118), we get 𝑢_𝑚^𝑛+12−𝑢^𝑛_𝑚²+1−𝐿𝑢^𝑛_𝑚Δ𝑡𝑢_𝑚^𝑛+1−𝑢^𝑛_𝑚²+𝜈𝛾₀𝑢_𝑚^𝑛+122Δ𝑡≤𝜈𝛾₀𝜆₁𝑓^𝑛+12Δ𝑡. (121) Next, by taking the scalar product of (106) with 2𝐴𝑢_𝑚^𝑛+1Δ𝑡, we obtain 𝑢_𝑚^𝑛+12−𝑢^𝑛_𝑚²+𝑢_𝑚^𝑛+1−𝑢^𝑛_𝑚²+2𝜈𝐴𝑢_𝑚^𝑛+12Δ𝑡+2𝜈Δ𝑡𝐶𝑢_𝑚^𝑛+1,𝐴𝑢_𝑚^𝑛+1_𝑔+2𝑏𝑢^𝑛_𝑚,𝑢^𝑛_𝑚,𝐴𝑢_𝑚^𝑛+1𝑓Δ𝑡=2^𝑛+1,𝐴𝑢_𝑚^𝑛+1_𝑔Δ𝑡. (122) Using Lemma 3 and Cauchy's inequality, we have 2𝜈𝐶𝑢_𝑚^𝑛+1,𝐴𝑢_𝑚^𝑛+1_𝑔≤2𝜈∇𝑔_∞𝑚₀𝜆₁^1/2𝐴𝑢_𝑚^𝑛+12+𝜈∇𝑔_∞𝜆₁^1/22𝑚₀𝑢_𝑚^𝑛+12. (123) Substituting into (122), we get 𝑢_𝑚^𝑛+12−𝑢^𝑛_𝑚²+𝑢_𝑚^𝑛+1−𝑢^𝑛_𝑚²+2𝜈𝛾₀𝐴𝑢_𝑚^𝑛+12𝑓Δ𝑡≤2^𝑛+1,𝐴𝑢_𝑚^𝑛+1_𝑔Δ𝑡−2𝑏𝑢^𝑛_𝑚,𝑢^𝑛_𝑚,𝐴𝑢_𝑚^𝑛+1+𝜈Δ𝑡∇𝑔_∞𝜆₁^1/22𝑚₀𝑢_𝑚^𝑛+12Δ𝑡. (124) Due to (17)–(19), Cauchy's inequality, and Young's inequality, we have 2𝑏𝑢^𝑛_𝑚,𝑢_𝑚^𝑛+1−𝑢^𝑛_𝑚,𝐴𝑢_𝑚^𝑛+1≤𝜈𝛾₀4𝐴𝑢_𝑚^𝑛+12+𝐿𝑢^𝑛_𝑚𝑢_𝑚^𝑛+1−𝑢^𝑛_𝑚²,2𝑏𝑢^𝑛_𝑚,𝑢_𝑚^𝑛+1,𝐴𝑢_𝑚^𝑛+1≤𝜈𝛾₀𝐴𝑢_𝑚^𝑛+12+2𝜈𝛾₀⁻³𝑐⁴₀𝑢^𝑛_𝑚²𝑢^𝑛_𝑚²𝑢_𝑚^𝑛+12,2𝑓^𝑛+1,𝐴𝑢_𝑚^𝑛+1_𝑔≤𝜈𝛾₀2𝐴𝑢_𝑚^𝑛+12+2𝜈𝛾₀𝐶²_𝑓. (125) By combining (124) with the above estimates, we obtain 𝑢_𝑚^𝑛+12−𝑢^𝑛_𝑚²+1−𝐿𝑢^𝑛_𝑚Δ𝑡𝑢_𝑚^𝑛+1−𝑢^𝑛_𝑚²≤2𝜈𝛾₀⁻³𝑐⁴₀𝑢^𝑛_𝑚²𝑢^𝑛_𝑚²𝑢_𝑚^𝑛+122Δ𝑡+𝜈𝛾₀𝐶²_𝑓+𝜈Δ𝑡∇𝑔_∞𝜆₁^1/22𝑚₀𝑢_𝑚^𝑛+12Δ𝑡. (126) On the other hand, from (17)–(19), Cauchy's inequality, and Young's inequality, we have 2𝑏𝑢^𝑛_𝑚,𝑢_𝑚^𝑛+1−𝑢^𝑛_𝑚,𝐴𝑢_𝑚^𝑛+1≤𝜈𝛾₀4𝐴𝑢_𝑚^𝑛+12+𝐿𝑢^𝑛_𝑚𝑢_𝑚^𝑛+1−𝑢^𝑛_𝑚²,2𝑏𝑢^𝑛_𝑚,𝑢_𝑚^𝑛+1,𝐴𝑢_𝑚^𝑛+1≤𝜈𝛾₀8𝐴𝑢_𝑚^𝑛+12+𝑐𝑢^𝑛_𝑚²𝑢^𝑛_𝑚²𝑢_𝑚^𝑛+12,2𝑓^𝑛+1,𝐴𝑢_𝑚^𝑛+1_𝑔≤𝜈𝛾₀8𝐴𝑢_𝑚^𝑛+12+𝑐𝐶²_𝑓. (127) Combining (124) with those estimates, we obtain 1+𝜈𝛾₀𝜆₁Δ𝑡𝑢_𝑚^𝑛+12−𝑢^𝑛_𝑚²+1−𝐿𝑢^𝑛_𝑚Δ𝑡𝑢_𝑚^𝑛+1−𝑢^𝑛_𝑚²+𝜈𝛾₀2𝐴𝑢_𝑚^𝑛+12𝑢Δ𝑡≤𝑐^𝑛_𝑚²𝑢^𝑛_𝑚²𝑢_𝑚^𝑛+12Δ𝑡+𝑐𝐶²_𝑓Δ𝑡+𝑐𝑢_𝑚^𝑛+12Δ𝑡. (128) Next, from (106) we have 𝑑_𝑡𝑢_𝑚^𝑛+12+𝜈𝐴𝑢_𝑚^𝑛+1,𝑑_𝑡𝑢_𝑚^𝑛+1+𝜈𝐶𝑢_𝑚^𝑛+1,𝑑_𝑡𝑢_𝑚^𝑛+1_𝑔+𝑏𝑢^𝑛,𝑢^𝑛,𝑑_𝑡𝑢_𝑚^𝑛+1=𝑓^𝑛+1,𝑑_𝑡𝑢_𝑚^𝑛+1_𝑔. (129) Using (11), Lemmas 1 and 3, and Cauchy's inequality, we get 𝑏𝑢^𝑛,𝑢^𝑛,𝑑_𝑡𝑢_𝑚^𝑛+1≤𝑐𝑢^𝑛_𝑚²𝑢^𝑛_𝑚𝑢_𝑚^𝑛+12,𝜈𝐴𝑢_𝑚^𝑛+1,𝑑_𝑡𝑢_𝑚^𝑛+1_𝑔≤14𝑑_𝑡𝑢_𝑚^𝑛+12+𝜈²𝐴𝑢_𝑚^𝑛+12,𝜈𝐶𝑢_𝑚^𝑛+1,𝑑_𝑡𝑢_𝑚^𝑛+1_𝑔≤14𝑑_𝑡𝑢_𝑚^𝑛+12+𝑐𝑢_𝑚^𝑛+12,𝑓^𝑛+1,𝑑_𝑡𝑢_𝑚^𝑛+1_𝑔≤14𝑑_𝑡𝑢_𝑚^𝑛+12+𝑐𝐶²_𝑓. (130) Therefore, we have 𝛾₀𝑑16𝜈^𝑡𝑢_𝑚^𝑛+12−𝜈𝛾₀4𝐴𝑢_𝑚^𝑛+12≤𝑐𝑢^𝑛_𝑚²𝑢^𝑛_𝑚²𝑢_𝑚^𝑛+12+𝑐𝐶²_𝑓+𝑐𝑢_𝑚^𝑛+12. (131) Combining this inequality with (128) yields 1+𝜈𝛾₀𝜆₁Δ𝑡𝑢_𝑚^𝑛+12−𝑢^𝑛_𝑚²+1−𝐿𝑢^𝑛_𝑚Δ𝑡𝑢_𝑚^𝑛+1−𝑢^𝑛_𝑚²+𝛾₀𝑑16𝜈^𝑡𝑢_𝑚^𝑛+12Δ𝑡+𝜈𝛾₀4𝐴𝑢_𝑚^𝑛+12𝑢Δ𝑡≤𝑐^𝑛_𝑚²𝑢^𝑛_𝑚²𝑢_𝑚^𝑛+12Δ𝑡+𝑐𝐶²_𝑓Δ𝑡+𝑐𝑢_𝑚^𝑛+12Δ𝑡. (132) Moreover, we deduce from (106) that 𝑑_𝑡𝑡𝑢_𝑚^𝑛+1+𝜈𝐴𝑑_𝑡𝑢_𝑚^𝑛+1+𝜈𝐶𝑑_𝑡𝑢_𝑚^𝑛+1+𝑃_𝑚𝐵𝑑_𝑡𝑢^𝑛_𝑚,𝑢^𝑛_𝑚+𝑃_𝑚𝐵𝑢^𝑛_𝑚,𝑑_𝑡𝑢^𝑛_𝑚=𝑃_𝑚𝑑_𝑡𝑓^𝑛+1, (133) where 𝑑_𝑡𝑢⁰_𝑚 is defined by 𝑑_𝑡𝑢⁰_𝑚+𝜈𝐴𝑢⁰_𝑚+𝜈𝐶𝑢⁰_𝑚+𝑃_𝑚𝐵𝑢⁰_𝑚,𝑢⁰_𝑚=𝑃_𝑚𝑓(0). (134) Taking the scalar product of (133) with 2𝑑_𝑡𝑢_𝑚^𝑛+1Δ𝑡, we obtain 𝑑_𝑡𝑢_𝑚^𝑛+12−𝑑_𝑡𝑢^𝑛_𝑚²+𝑑_𝑡𝑢_𝑚^𝑛+1−𝑑_𝑡𝑢^𝑛_𝑚²+2𝜈𝑑_𝑡𝑢_𝑚^𝑛+12Δ𝑡+2𝜈𝐶𝑑_𝑡𝑢_𝑚^𝑛+1,𝑑_𝑡𝑢_𝑚^𝑛+1𝑑Δ𝑡+2𝑏_𝑡𝑢^𝑛_𝑚−𝑑_𝑡𝑢_𝑚^𝑛+1,𝑢^𝑛_𝑚,𝑑_𝑡𝑢_𝑚^𝑛+1𝑑Δ𝑡+2𝑏_𝑡𝑢_𝑚^𝑛+1,𝑢^𝑛_𝑚,𝑑_𝑡𝑢_𝑚^𝑛+1𝑑Δ𝑡+2𝑏_𝑡𝑢_𝑚^𝑛+1,𝑑_𝑡𝑢^𝑛_𝑚−𝑑_𝑡𝑢_𝑚^𝑛+1,𝑑_𝑡𝑢_𝑚^𝑛+1𝑑Δ𝑡=2_𝑡𝑓^𝑛+1,𝑑_𝑡𝑢_𝑚^𝑛+1Δ𝑡. (135) By Lemma 3, we get 𝑑_𝑡𝑢_𝑚^𝑛+12−𝑑_𝑡𝑢^𝑛_𝑚²+𝑑_𝑡𝑢_𝑚^𝑛+1−𝑑_𝑡𝑢^𝑛_𝑚²+2𝜈𝛾₀𝑑_𝑡𝑢_𝑚^𝑛+12𝑑Δ𝑡≤2_𝑡𝑓^𝑛+1,𝑑_𝑡𝑢_𝑚^𝑛+1𝑑Δ𝑡−2𝑏_𝑡𝑢^𝑛_𝑚−𝑑_𝑡𝑢_𝑚^𝑛+1,𝑢^𝑛_𝑚,𝑑_𝑡𝑢_𝑚^𝑛+1𝑑Δ𝑡−2𝑏_𝑡𝑢_𝑚^𝑛+1,𝑢^𝑛_𝑚,𝑑_𝑡𝑢_𝑚^𝑛+1𝑑Δ𝑡−2𝑏_𝑡𝑢_𝑚^𝑛+1,𝑑_𝑡𝑢^𝑛_𝑚−𝑑_𝑡𝑢_𝑚^𝑛+1,𝑑_𝑡𝑢_𝑚^𝑛+1Δ𝑡. (136) Using Lemma 1 and (19), we have 2𝑏𝑑_𝑡𝑢^𝑛_𝑚−𝑑_𝑡𝑢_𝑚^𝑛+1,𝑢^𝑛_𝑚,𝑑_𝑡𝑢_𝑚^𝑛+1+2𝑏𝑑_𝑡𝑢_𝑚^𝑛+1,𝑑_𝑡𝑢^𝑛_𝑚−𝑑_𝑡𝑢_𝑚^𝑛+1,𝑑_𝑡𝑢_𝑚^𝑛+1≤𝜈𝛾₀2𝑑_𝑡𝑢_𝑚^𝑛+12+12𝐿𝑢^𝑛_𝑚+𝐿𝑢_𝑚^𝑛+1𝑑_𝑡𝑢_𝑚^𝑛+1−𝑑_𝑡𝑢^𝑛_𝑚²,2𝑏𝑑_𝑡𝑢_𝑚^𝑛+1,𝑢^𝑛_𝑚,𝑑_𝑡𝑢_𝑚^𝑛+1≤𝜈𝛾₀8𝑑_𝑡𝑢_𝑚^𝑛+12+8𝜈𝛾₀𝑐²₀𝑢^𝑛_𝑚²𝑑_𝑡𝑢_𝑚^𝑛+12,2𝑑_𝑡𝑓^𝑛+1,𝑑_𝑡𝑢_𝑚^𝑛+1≤𝜈𝛾₀8𝑑_𝑡𝑢_𝑚^𝑛+12+8𝜈𝛾₀𝜆₁sup_𝑡≥0𝑓_𝑡(𝑡)². (137) By combining (136) with the above estimates, we arrive at 1+𝜈𝛾₀𝜆₁Δ𝑡𝑑_𝑡𝑢_𝑚^𝑛+12−𝑑_𝑡𝑢^𝑛_𝑚²+11−2𝐿𝑢^𝑛_𝑚+𝐿𝑢_𝑚^𝑛+1Δ𝑡𝑑_𝑡𝑢_𝑚^𝑛+1−𝑑_𝑡𝑢^𝑛_𝑚²+𝜈𝛾₀4𝑑_𝑡𝑢_𝑚^𝑛+128Δ𝑡≤𝜈𝛾₀𝑐²₀𝑢^𝑛_𝑚²𝑑_𝑡𝑢_𝑚^𝑛+12+8Δ𝑡𝜈𝛾₀𝜆₁sup_𝑡≥0𝑓_𝑡(𝑡)². (138) Multiplying (138) by 𝜏(𝑡_𝑛+1) and noting that 𝜏(𝑡_𝑛+1)≤𝜏(𝑡_𝑛)+Δ𝑡, we obtain 1+𝜈𝛾₀𝜆₁Δ𝑡𝜏𝑡_𝑛+1𝑑_𝑡𝑢_𝑚^𝑛+12−𝜏𝑡_𝑛𝑑_𝑡𝑢^𝑛_𝑚²+11−2𝐿𝑢^𝑛_𝑚+𝐿𝑢_𝑚^𝑛+1𝑡Δ𝑡×𝜏_𝑛+1𝑑_𝑡𝑢_𝑚^𝑛+1−𝑑_𝑡𝑢^𝑛_𝑚²+𝜈𝛾₀4𝜏𝑡_𝑛+1𝑑_𝑡𝑢_𝑚^𝑛+12≤𝑑Δ𝑡_𝑡𝑢^𝑛_𝑚²8Δ𝑡+𝜈𝛾₀𝑐²₀𝑢^𝑛_𝑚²𝑑_𝑡𝑢_𝑚^𝑛+12+8Δ𝑡𝜈𝛾₀𝜆₁sup_𝑡≥0𝑓_𝑡(𝑡)²Δ𝑡. (139) 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 1−𝐿𝑢^𝑛_𝑚Δ𝑡≥0∀0≤𝑛≤𝐽. (140) 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 𝑎_𝑘=𝑢^𝑘_𝑚²,𝑏_𝑘ℎ=0,_𝑘=2𝜈𝛾₀𝜆₁𝐶²_𝑓,𝛾=𝜈𝛾₀𝜆₁, (141) we obtain (113) for 𝑛=𝐽+1.
Furthermore, setting 𝑎_𝑘=𝑢^𝑘_𝑚²,𝑔_𝑘=2𝜈𝛾₀⁻³𝑐⁴₀𝑢^𝑘_𝑚²𝑢_𝑚^𝑘+12,ℎ_𝑘=2𝜈𝛾₀𝐶²_𝑓+𝜈∇𝑔_∞𝜆₁^1/22𝑚₀𝑢_𝑚^𝑘+12, (142) in (126), using (111)–(113) and (140), we obtain 1𝑎Δ𝑡^𝑘+1−𝑎_𝑘≤𝑔_𝑘𝑎_𝑘+ℎ_𝑘∀0≤𝑘≤𝐽, (143) with Δ𝑡^𝐽_𝑘=0𝑔_𝑘≤𝛼₁,Δ𝑡^𝐽_𝑘=0ℎ_𝑘≤𝛼₁,𝑎₀≤𝑢₀², (144) for all 𝐽+1≤𝑛₀−1 and Δ𝑡^{𝑘0+𝑛0−1}_𝑘=𝑘0𝑔_𝑘≤𝛼₁,Δ𝑡^{𝑘0+𝑛0−1}_𝑘=𝑘0ℎ_𝑘≤𝛼₂,Δ𝑡^{𝑘0+𝑛0−1}_𝑘=𝑘0𝑎_𝑘≤𝑎₀+𝛼₃,0≤𝑘₀≤𝐽+1−𝑛₀, (145) for all 𝐽+1≥𝑛₀. Applying Lemma 4 to (143), we obtain (114) for 𝑛=𝐽+1.
Applying again Lemma 4 to (132) with 𝑎_𝑘=𝑢^𝑘_𝑚²,𝑏_𝑘=116𝜈𝛾₀𝑑_𝑡𝑢^𝑘_𝑚²+𝜈𝛾₀4𝐴𝑢^𝑘_𝑚²,𝛾=𝜈𝛾₀𝜆₁,ℎ_𝑘=𝑐𝑢^𝑘_𝑚²𝑢^𝑘_𝑚²𝑢_𝑚^𝑘+12+𝑐𝐶²_𝑓+𝑐𝑢_𝑚^𝑘+12, (146) and using (111)–(114) and (140), we obtain Δ𝑡^𝐽+1_𝑘=11+𝜈𝛾₀𝜆₁Δ𝑡^{−(𝐽+2−𝑘)}𝐴𝑢^𝑘_𝑚+𝑑_𝑡𝑢^𝑘_𝑚²≤𝒦. (147) Applying Lemma 5 to (139) with 𝑎_𝑘=𝜏𝑡_𝑘𝑑_𝑡𝑢^𝑘_𝑚²,𝑏_𝑘=𝜈𝛾₀4𝜏𝑡_𝑘𝑑_𝑡𝑢^𝑘_𝑚²,𝛾=𝜈𝛾₀𝜆₁,ℎ_𝑘=𝑑_𝑡𝑢^𝑘_𝑚²8Δ𝑡+𝜈𝛾₀𝑐²₀𝑢^𝑘_𝑚²𝑑_𝑡𝑢_𝑚^𝑘+12+8𝜈𝛾₀𝜆₁sup_𝑡≥0𝑓_𝑡(𝑡)², (148) and using (114), (140), and (147) yields 𝜏𝑡_𝐽+1𝑑_𝑡𝑢_𝑚^𝐽+12+𝜈𝛾₀4Δ𝑡^𝐽+1_𝑘=11+𝜈𝛾₀𝜆₁Δ𝑡^{−(𝐽+2−𝑘)}𝜏𝑡_𝑘𝑑_𝑡𝑢^𝑘_𝑚²≤𝒦. (149) Finally, due to (17)–(19), we have 𝑏𝑢^𝐽_𝑚,𝑢^𝐽−𝑢_𝑚^𝐽+1,𝐴𝑢_𝑚^𝐽+1≤𝜈𝛾₀8𝐴𝑢_𝑚^𝐽+12+𝑐𝐿𝑢^𝐽_𝑚𝑑_𝑡𝑢_𝑚^𝐽+12Δ𝑡²,𝑏𝑢^𝐽_𝑚,𝑢_𝑚^𝐽+1,𝐴𝑢_𝑚^𝐽+1≤𝜈𝛾₀8𝐴𝑢_𝑚^𝐽+12+𝑐𝑢^𝐽_𝑚²𝑢^𝐽_𝑚²𝑢_𝑚^𝐽+12. (150) Combining these inequalities with (106) and (140) yields 𝜏𝑡_𝐽+1𝐴𝑢_𝑚^𝐽+12≤𝑐𝜏𝑡_𝐽+1𝑑_𝑡𝑢_𝑚^𝐽+12+𝑐𝑢^𝐽_𝑚²𝑢^𝐽_𝑚²𝑢_𝑚^𝐽+12+𝑐𝐶²_𝑓+𝑐𝜏𝑡_𝐽+1𝑑_𝑡𝑢_𝑚^𝐽+12Δ𝑡. (151) Using (113), (114) and (149) in this inequality yields 𝜏𝑡_𝐽+1𝐴𝑢_𝑚^𝐽+12≤𝒦. (152) Combining this inequality with (149) and (147) implies (115) for 𝑛=𝐽+1.

The following lemma provides the stability of scheme (109).

Lemma 14. If 𝑢₀∈𝑉_𝑔,Δ𝑡 and 𝑚 satisfy (110), then problem (109) admits a unique solution {Φ^𝑘}^𝑛₀⊂𝐻_𝑚 satisfying 1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑟Φ^𝑟2+Δ𝑡^𝑛_𝑘=𝑟1+𝜈𝛾₀𝜆₁Δ𝑡^{−(𝑘−1)}𝐴Φ^𝑘24≤𝒦exp𝜈𝛾₀𝑐²₀Δ𝑡^𝑛_𝑘=1𝑢^𝑘_𝑚²Δ𝑡^𝑛_𝑘=11+𝜈𝛾₀𝜆₁Δ𝑡^𝑘𝜉^𝑘2. (153)

Proof. In view of (17), (110), and (114), we can prove that the following bilinear form 1()()()Δ𝑡𝑣,Φ+𝑎𝑣,Φ+𝜈𝐶𝑣,Φ+𝑏𝑣,𝑢_𝑚^𝑘−1𝑢,Φ+𝑏_𝑚^𝑘−1,𝑣,Φ∀𝑣,Φ∈𝐻_𝑚 (154) is elliptic. Hence, problem (109) has a unique solution Φ^𝑘−1 for 1≤𝑘≤𝑛. Next, by taking 𝑣=−2(1+𝜈𝛾₀𝜆₁Δ𝑡)^−𝑘Φ^𝑘−1Δ𝑡 in (109) and using (11), we have −21+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘Φ^𝑘−1,Φ^𝑘_𝑔+21+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘Φ^𝑘−12+2𝜈1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘Φ^𝑘−12Δ𝑡+2𝜈1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘𝐶Φ^𝑘−1,Φ^𝑘−1_𝑔Δ𝑡+21+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘𝑏Φ^𝑘−1,𝑢_𝑚^𝑘−1,Φ^𝑘−1ΦΔ𝑡=−2^𝑘−1,𝜉^𝑘_𝑔Δ𝑡. (155) Using Lemma 3, we get 21+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘Φ^𝑘−12+2𝜈𝛾₀1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘Φ^𝑘−12ΦΔ𝑡≤2^𝑘−1,𝜉^𝑘_𝑔Δ𝑡+21+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘Φ^𝑘−1,Φ^𝑘_𝑔+21+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘𝑏Φ^𝑘−1,𝑢_𝑚^𝑘−1,Φ^𝑘−1Δ𝑡. (156) Using (17) and Cauchy's inequality, we have 2Φ^𝑘−1,Φ^𝑘_𝑔≤Φ^𝑘−12+Φ^𝑘2,2Φ^𝑘−1,𝜉^𝑘_𝑔≤𝜈𝛾₀41+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘Φ^𝑘−12+4𝜈𝛾₀𝜆₁1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘𝜉^𝑘2,2𝑏Φ^𝑘−1,𝑢_𝑚^𝑘−1,Φ^𝑘−1≤𝜈𝛾₀2Φ^𝑘−12+2𝜈𝛾₀𝑐²₀Φ^𝑘−12𝑢^𝑘−12. (157) Combining above inequalities and using (14), we obtain 1+𝜈𝛾₀𝜆₁Δ𝑡^{−(𝑘−1)}Φ^𝑘−12−1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘Φ^𝑘2+𝜈𝛾₀41+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘Φ^𝑘−12≤2Δ𝑡𝜈𝛾₀𝑐²₀1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘𝑢_𝑚^𝑘−12Φ^𝑘−12+4Δ𝑡𝜈𝛾₀𝜆₁1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘𝜉^𝑘2Δ𝑡. (158) Summing (158) from 𝑟+1 to 𝑛, noting that Φ^𝑛=0, and using Theorem 13, we arrive at 1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑟Φ^𝑟2+𝜈𝛾₀4Δ𝑡^𝑛_𝑘=𝑟1+𝜈𝛾₀𝜆₁Δ𝑡^−(𝑘+1)Φ^𝑘2≤2𝜈𝛾₀𝑐²₀Δ𝑡^𝑛_𝑘=𝑟1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘𝑢^𝑘_𝑚²Φ^𝑘2+4𝜈𝛾₀𝜆₁Δ𝑡^𝑛_𝑘=𝑟1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘𝜉^𝑘2 (159) for all 0≤𝑟≤𝑛. Let 𝑎_𝑟=1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑟Φ^𝑟2,𝑏_𝑟=𝜈𝛾₀41+𝜈𝛾₀𝜆₁Δ𝑡^−(𝑟+1)Φ^𝑟2,ℎ_𝑟=4𝜈𝛾₀𝜆₁1+𝜈𝛾₀𝜆₁Δ𝑡^𝑟𝜉^𝑘2,𝑔_𝑟=2𝜈𝛾₀𝑐²₀𝑢^𝑘_𝑚²,𝛽=0, (160) in (159) to obtain 𝑎_𝑟+Δ𝑡^𝑛_𝑘=𝑟𝑏_𝑘≤Δ𝑡^𝑛_𝑘=𝑟𝑔_𝑘𝑎_𝑘+Δ𝑡^𝑛_𝑘=𝑟ℎ_𝑘∀0≤𝑟≤𝑛, (161) with 𝜎_𝑟=(1−𝑔_𝑟Δ𝑡)⁻¹≤2. Applying Lemma 6 to this inequality yields 1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑟Φ^𝑟2+𝜈𝛾₀2Δ𝑡^𝑛_𝑘=𝑟1+𝜈𝛾₀𝜆₁Δ𝑡^−(𝑘+1)Φ^𝑘24≤𝒦exp𝜈𝛾₀𝑐²₀Δ𝑡^𝑛_𝑘=1𝑢^𝑘_𝑚²Δ𝑡^𝑛_𝑘=𝑟1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘𝜉^𝑘2. (162) Moreover, by taking 𝑣=−2(1+𝜈𝛾₀𝜆₁Δ𝑡)^−𝑘𝐴Φ^𝑘−1Δ𝑡 in (109), we have −21+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘𝐴Φ^𝑘−1,Φ^𝑘_𝑔+21+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘Φ^𝑘−12+2𝜈1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘𝐴Φ^𝑘−12Δ𝑡+2𝜈1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘𝐶𝐴Φ^𝑘−1,Φ^𝑘−1_𝑔Δ𝑡+21+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘𝑏𝐴Φ^𝑘−1,Φ^𝑘−1,𝑢_𝑚^𝑘−1Δ𝑡+21+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘𝑏𝑢_𝑚^𝑘−1,Φ^𝑘−1,𝐴Φ^𝑘−1Δ𝑡=2𝐴Φ^𝑘−1,𝜉^𝑘_𝑔Δ𝑡. (163) Using Lemma 3 and Cauchy's inequality, we deduce that 2𝜈1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘𝐶𝐴Φ^𝑘−1,Φ^𝑘−1_𝑔≤2𝜈1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘∇𝑔_∞𝑚₀𝜆₁^1/2𝐴Φ^𝑘−12+𝑐1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘Φ^𝑘−12. (164) Therefore, 21+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘Φ^𝑘−12+2𝜈𝛾₀1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘𝐴Φ^𝑘−12Δ𝑡≤2𝐴Φ^𝑘−1,𝜉^𝑘_𝑔Δ𝑡+21+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘𝐴Φ^𝑘−1,Φ^𝑘_𝑔+21+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘𝑏𝐴Φ^𝑘−1,Φ^𝑘−1,𝑢_𝑚^𝑘−1Δ𝑡+21+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘𝑏𝑢_𝑚^𝑘−1,Φ^𝑘−1,𝐴Φ^𝑘−1Δ𝑡+𝑐1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘Φ^𝑘−12Δ𝑡. (165) Using (17)–(19) and Cauchy’s inequality, we have 2𝑏𝐴Φ^𝑘−1,Φ^𝑘−1,𝑢_𝑚^𝑘−1+2𝑏𝑢_𝑚^𝑘−1,Φ^𝑘−1,𝐴Φ^𝑘−1≤𝜈𝛾₀4𝐴Φ^𝑘−12+𝑐𝑢_𝑚^𝑘−12𝑢_𝑚^𝑘−12Φ^𝑘−12,2𝐴Φ^𝑘−1,Φ^𝑘_𝑔=2Φ^𝑘−1,Φ^𝑘_𝑔≤Φ^𝑘−12+Φ^𝑘2,2𝐴Φ^𝑘−1,𝜉^𝑘_𝑔≤𝜈𝛾₀41+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘𝐴Φ^𝑘−12+4𝜈𝛾₀1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘𝜉^𝑘2. (166) Combining above inequality, noting that 𝜆₁‖𝑣‖²≤|𝐴𝑣|², we obtain 1+𝜈𝛾₀𝜆₁Δ𝑡^{−(𝑘−1)}Φ^𝑘−12−1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘Φ^𝑘2+𝜈𝛾₀21+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘𝐴Φ^𝑘−12Δ𝑡≤𝑐1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑘𝑢_𝑚^𝑘−12𝑢_𝑚^𝑘−12+1Φ^𝑘−12Δ𝑡+𝑐1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘𝜉^𝑘2Δ𝑡 (167) for all 0≤𝑘≤𝑛. Summing (167) from 𝑟+1 to 𝑛, we obtain 1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑟Φ^𝑟2+𝜈𝛾₀2Δ𝑡^𝑛_𝑘=𝑟1+𝜈𝛾₀𝜆₁Δ𝑡^−(𝑘+1)𝐴Φ^𝑘2≤𝑐sup_𝑘≥0𝑢^𝑘_𝑚²𝑢^𝑘_𝑚²+1Δ𝑡^𝑛_𝑘=𝑟1+𝜈𝛾₀𝜆₁Δ𝑡^−(𝑘+1)Φ^𝑘2+𝑐Δ𝑡^𝑛_𝑘=𝑟+11+𝜈𝛾₀𝜆₁Δ𝑡^𝑘𝜉^𝑘2∀0≤𝑟≤𝑛. (168) Combining (168) with (162) and using Theorem 13, we complete the proof.

5.2. Error Analysis

In this subsection, we will establish the 𝐻¹- 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 𝑡_𝑛 to 𝑡_𝑛+1, we obtain 𝑑_𝑡𝑢_𝑚𝑡_𝑛+1+𝜈Δ𝑡^𝑡𝑛+1_𝑡𝑛𝐴𝑢_𝑚(𝑡)𝜈𝑑𝑡+Δ𝑡^𝑡𝑛+1_𝑡𝑛𝐶𝑢_𝑚(𝑡)+1𝑑𝑡Δ𝑡^𝑡𝑛+1_𝑡𝑛𝑃_𝑚𝐵𝑢_𝑚(𝑡),𝑢_𝑚(𝑡)𝑑𝑡=𝑓^𝑛+1. (169) Subtracting (106) from (153) and setting 𝑒^𝑛=𝑢_𝑚(𝑡_𝑛)−𝑢^𝑛_𝑚, we have 𝑑_𝑡𝑒^𝑛+1+𝜈𝐴𝑒^𝑛+1+𝜈𝐶𝑒^𝑛+1+𝑃_𝑚𝐵𝑒^𝑛,𝑢^𝑛_𝑚+𝑃_𝑚𝐵𝑢_𝑚𝑡_𝑛,𝑒^𝑛=𝑃_𝑚𝐸_𝑛+1, (170) with 𝑒⁰=0 and 𝐸_𝑛+1=𝜈Δ𝑡^𝑡𝑛+1_𝑡𝑛𝐴𝑢_𝑚𝑡_𝑛+1−𝑢_𝑚(𝑡)+𝜈𝑑𝑡Δ𝑡^𝑡𝑛+1_𝑡𝑛𝐶𝑢_𝑚𝑡_𝑛+1−𝑢_𝑚(𝑡)+1𝑑𝑡Δ𝑡^𝑡𝑛+1_𝑡𝑛𝐵𝑢_𝑚𝑡_𝑛,𝑢_𝑚𝑡_𝑛−𝐵𝑢_𝑚(𝑡),𝑢_𝑚(𝑡)=𝜈𝑑𝑡Δ𝑡^𝑡𝑛+1_𝑡𝑛𝑡−𝑡_𝑛𝐴𝑢_𝑚𝑡𝜈𝑑𝑡+Δ𝑡^𝑡𝑛+1_𝑡𝑛𝑡−𝑡_𝑛𝐶𝑢_𝑚𝑡−1𝑑𝑡Δ𝑡^𝑡𝑛+1_𝑡𝑛𝑡_𝑛+1−𝑡𝐵𝑢_𝑚𝑡,𝑢_𝑚+𝐵𝑢_𝑚,𝑢_𝑚𝑡𝑑𝑡. (171) To derive a bound on 𝑒^𝑛, we need to provide the following estimates of 𝐸_𝑛.

Lemma 15. Under the assumptions of Theorem 13, the error 𝐸_𝑛 satisfies the following bounds: Δ𝑡^𝑛_𝑘=11+𝜈𝛾₀𝜆₁^𝑘−1𝐴⁻¹𝑃_𝑚𝐸_𝑘²≤𝒦1+𝜈𝛾₀𝜆₁^𝑛Δ𝑡²∀𝑛≥1,Δ𝑡^𝑛_𝑘=11+𝜈𝛾₀𝜆₁^𝑘−1𝐴^−1/2𝑃_𝑚𝐸_𝑘²≤𝒦1+𝜈𝛾₀𝜆₁^𝑛Δ𝑡∀𝑛≥1,Δ𝑡^𝑛_𝑘=11+𝜈𝛾₀𝜆₁^𝑘−1𝜏𝑡_𝑘𝐴^−1/2𝑃_𝑚𝐸_𝑘²≤𝒦1+𝜈𝛾₀𝜆₁^𝑛Δ𝑡²∀𝑛≥1. (172)(173)(174)

Proof. In view of (17)–(19), Lemma 3, and Theorem 8, we deduce from (171) that 𝐴⁻¹𝑃_𝑚𝐸_𝑛=sup_{𝑣∈𝐻𝑚}𝐸_𝑛,𝑣|𝐴𝑣|≤Δ𝑡^{−1/2𝑡𝑛}_𝑡𝑛−1𝑐₁𝑡−𝑡_𝑛−1²+∇𝑔_∞𝑚₀𝑡−𝑡_𝑛−1²𝑢_𝑚𝑡²+𝑐₂𝑡_𝑛−𝑡²𝑢_𝑚²𝑢_𝑚𝑡²𝑑𝑡^1/2≤𝑐Δ𝑡^{−1/2𝑡𝑛}_𝑡𝑛−1𝑡−𝑡_𝑛−1²+𝑡_𝑛−𝑡²𝑢_𝑚²𝑢_𝑚𝑡²𝑑𝑡^1/2,𝐴^−1/2𝑃_𝑚𝐸_𝑛=sup_{𝑣∈𝐻𝑚}𝐸_𝑛,𝑣‖𝑣‖≤𝑐Δ𝑡^−1/2×^𝑡𝑛_𝑡𝑛−1𝑡−𝑡_𝑛−1²+𝑡_𝑛−𝑡²𝑢_𝑚²×𝑢_𝑚𝑡²𝑑𝑡^1/2. (175) By Theorem 8, (175), and the fact that (1+𝜈𝛾₀𝜆₁)^𝑘−1≤𝑒^{𝜈𝛾0𝜆1𝑡𝑘−1}, we have 1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘−1𝐴⁻¹𝑃_𝑚𝐸_𝑘²Δ𝑡≤𝒦Δ𝑡³1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘−1∀𝑘≥𝑛₀+1,1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘−1𝐴^−1/2𝑃_𝑚𝐸_𝑘²Δ𝑡≤𝒦Δ𝑡³1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘−1∀𝑘≥𝑛₀+1,1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘−1𝐴⁻¹𝑃_𝑚𝐸_𝑘²Δ𝑡≤𝒦Δ𝑡^2𝑡𝑘_𝑡𝑘−1𝑒^{𝜈𝛾0𝜆1𝑡}𝑢_𝑚𝑡²𝑑𝑡∀1≤𝑘≤𝑛₀,1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘−1𝐴^−1/2𝑃_𝑚𝐸_𝑘²Δ𝑡≤𝒦Δ𝑡^2𝑡𝑘_𝑡𝑘−1𝑒^{𝜈𝛾0𝜆1𝑡}𝑢_𝑚𝑡²𝑑𝑡∀1≤𝑘≤𝑛₀. (176)(177)(178) For 𝑛≥𝑛₀+1, summing (176) from 𝑛₀+1 to 𝑛, we have Δ𝑡^𝑛_𝑘=𝑛0+11+𝜈𝛾₀𝜆₁Δ𝑡^𝑘−1𝐴⁻¹𝑃_𝑚𝐸_𝑘²+𝐴^−1/2𝑃_𝑚𝐸_𝑘²≤𝒦Δ𝑡^3𝑛_𝑘=𝑛0+11+𝜈𝛾₀𝜆₁Δ𝑡^𝑘−1≤𝒦Δ𝑡³1+𝜈𝛾₀𝜆₁Δ𝑡^𝑛01−1+𝜈𝛾₀𝜆₁Δ𝑡^𝑛−𝑛0−𝜈𝛾₀𝜆₁Δ𝑡≤𝒦Δ𝑡²1+𝜈𝛾₀𝜆₁Δ𝑡^𝑛. (179) For 𝑛≤𝑛₀, we sum (177) from 1 to 𝑛 and use Theorem 8 to obtain Δ𝑡^𝑛_𝑘=11+𝜈𝛾₀𝜆₁Δ𝑡^𝑘−1𝐴⁻¹𝑃_𝑚𝐸_𝑘²≤𝒦Δ𝑡²𝑒^{𝜈𝛾0𝜆1}≤𝒦Δ𝑡²1+𝜈𝛾₀𝜆₁Δ𝑡^𝑛∀𝑛≤𝑛₀, (180) which, together with (179), gives (172).
Finally, multiplying (178) by 𝜏^𝑗(𝑡_𝑘), 𝑗=0,1; 2≤𝑘≤𝑛₀, and noting that Δ𝑡≤𝜏(𝑡_𝑘−1), 𝜏(𝑡_𝑘)≤2𝜏(𝑡_𝑘−1), we obtain 1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘−1𝜏^𝑗𝑡_𝑘𝐴^−1/2𝑃_𝑚𝐸_𝑘²Δ𝑡≤𝒦Δ𝑡^{1−𝑗𝑡𝑘}_𝑡𝑘−1𝑒^{𝜈𝛾0𝜆1𝑡}𝜏^𝑗(𝑡)𝑢_𝑚𝑡²𝑑𝑡∀2≤𝑘≤𝑛₀. (181) Observing again (170) with 𝑛=1 and using Theorem 8 yields 𝜏^𝑗𝑡₁𝐴^−1/2𝑃_𝑚𝐸_𝑘²Δ𝑡≤𝒦sup_{0≤𝑡≤𝑡1}𝑢_𝑚(𝑡)−𝑢_𝑚(0)²Δ𝑡^1−𝑗. (182) For 2≤𝑛≤𝑛₀, summing (181) from 2 to 𝑛 and using Theorem 12, we obtain Δ𝑡^𝑛_𝑘=21+𝜈𝛾₀𝜆₁Δ𝑡^𝑘−1𝜏^𝑗𝑡_𝑘𝐴^−1/2𝑃_𝑚𝐸_𝑘²≤𝒦Δ𝑡^1−𝑗𝑒^{𝜈𝛾0𝜆1}. (183) Combining (182) and (183) with (179) yields (173) and (174).

Now, we prove the following error estimate.

Lemma 16. Under the assumptions of Theorem 13, one has 𝑒^𝑛2+Δ𝑡^𝑛_𝑘=11+𝜈𝛾₀𝜆₁Δ𝑡^{−(𝑛−1−𝑘)}×𝜈𝛾₀8𝑒^𝑘2Δ𝑡+𝑒^𝑘−𝑒^𝑘−128≤𝒦exp𝜈𝛾₀𝑐²₀Δ𝑡^𝑛_𝑘=1𝑢^𝑘_𝑚²Δ𝑡∀𝑛≥1. (184)

Proof. Taking the scalar product of (170) with 2𝑒^𝑛+1Δ𝑡, using (11), Lemma 3 and noting that 𝜆₁|𝑣|²≤‖𝑣‖², we have 1+𝜈𝛾₀𝜆₁Δ𝑡𝑒^𝑛+12−𝑒^𝑛2+𝑒^𝑛+1−𝑒^𝑛2+𝜈𝛾₀𝑒^𝑛+12𝑏𝑒Δ𝑡+2^𝑛+1,𝑢^𝑛_𝑚,𝑒^𝑛+1+𝑏𝑒^𝑛−𝑒^𝑛+1,𝑢^𝑛_𝑚,𝑒^𝑛+1+𝑏𝑢_𝑚𝑡_𝑛,𝑒^𝑛−𝑒^𝑛+1,𝑒^𝑛+1𝐸Δ𝑡≤2_𝑛+1,𝑒^𝑛+1Δ𝑡. (185)
Using (17)–(19), we get 2𝑏𝑒^𝑛+1,𝑢^𝑛_𝑚,𝑒^𝑛+1≤𝜈𝛾₀4𝑒^𝑛+12+4𝜈𝛾₀𝑐²₀𝑢^𝑛_𝑚²𝑒^𝑛+12,2𝑏𝑒^𝑛−𝑒^𝑛+1,𝑢^𝑛_𝑚,𝑒^𝑛+1+2𝑏𝑢_𝑚𝑡_𝑛,𝑒^𝑛−𝑒^𝑛+1,𝑒^𝑛+1≤9𝜈𝛾₀𝑒16^𝑛+12+49𝐿𝑢^𝑛_𝑚+𝐿𝑢_𝑚𝑡_𝑛𝑒^𝑛+1−𝑒^𝑛2,2𝐸_𝑛+1,𝑒^𝑛+1≤𝜈𝛾₀𝑒16^𝑛+12+16𝜈𝛾₀𝐴^−1/2𝑃_𝑚𝐸_𝑛+1². (186) Hence, by combining the above inequalities with (185) and using Theorem 8, (110) and (114), we obtain 1+𝜈𝛾₀𝜆₁Δ𝑡^𝑛+1𝑒^𝑛+12−1+𝜈𝛾₀𝜆₁Δ𝑡^𝑛𝑒^𝑛2+1+𝜈𝛾₀𝜆₁Δ𝑡^𝑛𝜈𝛾₀8𝑒^𝑛+121Δ𝑡+9𝑒^𝑛+1−𝑒^𝑛2≤4𝜈𝛾₀𝑐²₀𝑢^𝑛_𝑚²1+𝜈𝛾₀𝜆₁Δ𝑡^𝑛𝑒^𝑛+12+Δ𝑡16𝜈𝛾₀1+𝜈𝛾₀𝜆₁Δ𝑡^𝑛𝐴^−1/2𝑃_𝑚𝐸_𝑛+1²Δ𝑡, (187) for all 𝑛≥1. Summing (187) from 1 to 𝑛−1 and using Lemma 15, we obtain 1+𝜈𝛾₀𝜆₁Δ𝑡^𝑛𝑒^𝑛2+^𝑛_𝑘=11+𝜈𝛾₀𝜆₁Δ𝑡^𝑘−1𝜈𝛾₀8𝑒^𝑘21Δ𝑡+9𝑒^𝑘−𝑒^𝑘−12≤4𝜈𝛾₀𝑐²₀Δ𝑡^𝑛_𝑘=11+𝜈𝛾₀𝜆₁Δ𝑡^𝑘−1𝑢_𝑚^𝑘−12𝑒^𝑘2+16𝜈𝛾₀Δ𝑡^𝑛_𝑘=11+𝜈𝛾₀𝜆₁Δ𝑡^𝑘−1𝐴^−1/2𝑃_𝑚𝐸_𝑘². (188) We set 𝑎_𝑘=1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘𝑒^𝑘2,𝑏_𝑘=1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘−1𝜈𝛾₀8𝑒^𝑘2+19𝑒^𝑘−𝑒^𝑘−12Δ𝑡⁻¹,𝑔_𝑘=4𝜈𝛾₀𝑐²₀𝑢_𝑚^𝑘−12,ℎ_𝑘=16𝜈𝛾₀Δ𝑡1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘−1𝐴^−1/2𝑃_𝑚𝐸_𝑘²,𝛽=0 (189) in (188) to obtain 𝑎_𝑛+Δ𝑡^𝑛_𝑘=1𝑏_𝑘≤Δ𝑡^𝑛_𝑘=1𝑔_𝑘𝑎_𝑘+Δ𝑡^𝑛_𝑘=1ℎ_𝑘, (190) with 𝜎_𝑘=(1−𝑔_𝑘Δ𝑡)⁻¹≤2. Applying Lemma 6 to (190) and using Lemma 15, we deduce that 1+𝜈𝛾₀𝜆₁Δ𝑡^𝑛𝑒^𝑛2+^𝑛_𝑘=11+𝜈𝛾₀𝜆₁Δ𝑡^𝑘−1𝜈𝛾₀8𝑒^𝑘21Δ𝑡+9𝑒^𝑘−𝑒^𝑘−12≤1+𝜈𝛾₀𝜆₁Δ𝑡^𝑛8𝒦exp𝜈𝛾₀𝑐²₀Δ𝑡^𝑛−1_𝑘=0𝑢^𝑘_𝑚²Δ𝑡. (191) Multiplying (191) by (1+𝜈𝛾₀𝜆₁Δ𝑡)^−𝑛, we get (184).

Lemma 17. Under the assumptions of Theorem 13, one has 𝜏𝑡_𝑛𝑢_𝑚𝑡_𝑛−𝑢^𝑛_𝑚²≤𝒦exp16𝜈𝛾₀𝑐²₀Δ𝑡^𝑛−1_𝑘=0𝑢^𝑘_𝑚²Δ𝑡²∀𝑛≥1. (192)

Proof. Replacing 𝑛+1 by 𝑘 in (170) and taking the scalar product of (170) with Φ^𝑘−1, we obtain 𝑑_𝑡𝑒^𝑘,Φ^𝑘−1_𝑔+𝑎𝑒^𝑘,Φ^𝑘−1+𝜈𝐶𝑒^𝑘,Φ^𝑘−1_𝑔+𝑏𝑒^𝑘−1,𝑢_𝑚^𝑘−1,Φ^𝑘−1+𝑏𝑢_𝑚^𝑘−1,𝑒^𝑘−1,Φ^𝑘−1+𝑏𝑒^𝑘−1,𝑒^𝑘−1,Φ^𝑘−1=𝐸_𝑘,Φ^𝑘−1_𝑔. (193) Next, setting 𝑣=𝑒^𝑘, 𝜉^𝑘=𝑒^𝑘, 1≤𝑘≤𝑛 in (109), we get 𝑒^𝑘,𝑑_𝑡Φ^𝑘_𝑔−𝑎𝑒^𝑘,Φ^𝑘−1−𝜈𝐶𝑒^𝑘,Φ^𝑘−1_𝑔−𝑏𝑒^𝑘,𝑢_𝑚^𝑘−1,Φ^𝑘−1−𝑏𝑢_𝑚^𝑘−1,𝑒^𝑘,Φ^𝑘=1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘𝜉^𝑘2. (194) Adding (193) to (194), we arrive at 1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘𝜉^𝑘2=1𝑒Δ𝑡^𝑘,Φ^𝑘_𝑔−𝑒^𝑘−1,Φ^𝑘−1_𝑔+𝑏𝑒^𝑘−1−𝑒^𝑘,𝑢_𝑚^𝑘−1,Φ^𝑘−1+𝑏𝑢_𝑚^𝑘−1,𝑒^𝑘−1−𝑒^𝑘,Φ^𝑘−1+𝑏𝑒^𝑘−1,𝑒^𝑘−1,Φ^𝑘−1−𝐸_𝑘,Φ^𝑘−1_𝑔. (195) From (17) and (18), noting that 𝜆₁|𝑣|²≤‖𝑣‖², one finds that 𝑏𝑒^𝑘−1−𝑒^𝑘,𝑢_𝑚^𝑘−1,Φ^𝑘−1+𝑏𝑢_𝑚^𝑘−1,𝑒^𝑘−1−𝑒^𝑘,Φ^𝑘−1≤𝑐𝑒^𝑘−𝑒^𝑘−1𝑢_𝑚^𝑘−1𝐴Φ^𝑘−1,𝑏𝑒^𝑘−1,𝑒^𝑘−1,Φ^𝑘−1≤𝑐𝑒^𝑘−1𝑒^𝑘−1𝐴Φ^𝑘−1,𝐸_𝑘,Φ^𝑘−1_𝑔≤𝐴⁻¹𝑃_𝑚𝐸_𝑘𝐴Φ^𝑘−1. (196) Combining (195) with the above estimates yields 1+𝜈𝛾₀𝜆₁Δ𝑡^𝑘𝜉^𝑘2≤𝑒Δ𝑡_𝑘,Φ^𝑘_𝑔−𝑒^𝑘−1,Φ^𝑘−1_𝑔+𝐴⁻¹𝑃_𝑚𝐸_𝑘𝐴Φ^𝑘−1𝑒Δ𝑡+𝑐^𝑘−𝑒^𝑘−1𝑢_𝑚^𝑘−1+𝑒^𝑘−1𝑒^𝑘−1𝐴Φ^𝑘−1Δ𝑡, (197) with 𝑒⁰=Φ^𝑛=0. Summing (197) from 1 to 𝑛 and using Lemma 15, we obtain Δ𝑡^𝑛_𝑘=11+𝜈𝛾₀𝜆₁Δ𝑡^𝑘𝜉^𝑘2≤𝒦Δ𝑡^𝑛_𝑘=11+𝜈𝛾₀𝜆₁Δ𝑡^𝑘×𝑒^𝑘−𝑒^𝑘−12𝑢_𝑚^𝑘−12+𝑒^𝑘−12𝑒^𝑘−12+Δ𝑡^𝑛_𝑘=11+𝜈𝛾₀𝜆₁Δ𝑡^𝑘𝐴⁻¹𝑃_𝑚𝐸_𝑘². (198) Applying Lemmas 15 and 16 and Theorem 8 in (198), we get 1+𝜈𝛾₀𝜆₁Δ𝑡^−𝑛Δ𝑡^𝑛_𝑘=11+𝜈𝛾₀𝜆₁Δ𝑡^𝑘𝜉^𝑘2≤𝒦exp16𝜈𝛾₀𝑐²₀Δ𝑡^𝑛−1_𝑛=0𝑢^𝑘_𝑚²Δ𝑡²∀𝑛≥1. (199) Now, multiplying (187) by 𝜏(𝑡_𝑛+1) and noting that 𝜏𝑡_𝑛+1≤Δ𝑡+𝜏𝑡_𝑛,Δ𝑡≤𝜏𝑡_𝑛∀1≤𝑛≤𝑁,𝑒⁰=0, (200) we arrive at 1+𝜈𝛾₀𝜆₁Δ𝑡^𝑛+1𝜏𝑡_𝑛+1𝑒^𝑛+12−1+𝜈𝛾₀𝜆₁Δ𝑡^𝑛𝜏𝑡_𝑛𝑒^𝑛2≤𝒦1+𝜈𝛾₀𝜆₁Δ𝑡^𝑛𝑒^𝑛2Δ𝑡+𝑘1+𝜈𝛾₀𝜆₁Δ𝑡^𝑛𝜏𝑡_𝑛+1𝐴^−1/2𝑃_𝑚𝐸_𝑛+1²Δ𝑡. (201) 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 Theorems 12, 8, and 13 with Lemma 17, we obtain the error estimate of the numerical solution {𝑢^𝑛_𝑚}.

Theorem 18. Under the assumptions of Theorem 13, the following error estimates holds: 𝑢𝑡_𝑛−𝑢^𝑛_𝑚²≤𝒦𝜏⁻¹𝑡_𝑛𝑒^{(16/𝜈3𝛾30𝜆1)𝑐20𝐶2𝑓𝑡𝑛}𝜆⁻²_𝑚+1+Δ𝑡²∀𝑛≥1. (202)

Acknowledgment

This work is supported by a grant from the Le Quy Don Technical University.