Spectral discretization of the time-dependent Navier-Stokes problem with mixed boundary conditions

Mohamed Abdelwahed; Nejmeddine Chorfi

doi:10.1515/anona-2022-0253

Open Access Published by De Gruyter May 11, 2022

Spectral discretization of the time-dependent Navier-Stokes problem with mixed boundary conditions

Mohamed Abdelwahed and Nejmeddine Chorfi

From the journal Advances in Nonlinear Analysis

https://doi.org/10.1515/anona-2022-0253

Abstract

In this work, we handle a time-dependent Navier-Stokes problem in dimension three with a mixed boundary conditions. The variational formulation is written considering three independent unknowns: vorticity, velocity, and pressure. We use the backward Euler scheme for time discretization and the spectral method for space discretization. We present a complete numerical analysis linked to this variational formulation, which leads us to a priori error estimate.

Keywords: time-dependent Navier-Stokes problem; mixed boundary conditions; implicit Euler scheme; spectral method

MSC 2010: 35Q30; 65M70

1 Introduction

Many physical cases can be modeled by Navier-Stokes equations with mixed boundary conditions, for instance, a reservoir of water covered by a membrane or a flow in a piping network. Different types of nonstandard boundary conditions are suggested in the pioneering papers [12,13,24]. In this paper, we consider non-stationary Navier-Stokes equations provided with a mixed boundary condition. On some parts of the boundary, we provide an homogeneous Dirichlet condition of the velocity. On the other part, the normal component of the velocity and the tangential components of the vorticity are given.

Let Ω be a bounded simply-connected open domain in R 3 , with a Lipschitz continuous connected boundary ∂ Ω and [ 0 ; T ] an interval in R , where T is a positive constant. We consider a partition without overlap of ∂ Ω into two connected parts Γ m and Γ ,

∂ Ω = Γ ∪ Γ m and Γ ∩ Γ m = ∅ ,

where the index “ m ” in Γ m stands for membrane. We denote by x = ( x , y , z ) and n the unit outward normal vector to Ω on its boundary ∂ Ω . We intend to work with the following time-dependent Navier-Stokes system:

(1) ∂ u ∂ t ( x , t ) − ν Δ u ( x , t ) + ( u ⋅ ∇ u ) ( x , t ) + ∇ P ( x , t ) = f ( x , t ) in Ω × [ 0 , T ] , div u ( x , t ) = 0 in Ω × [ 0 , T ] , u ( x , t ) = 0 on Γ × [ 0 , T ] , u ( x , t ) . n ( x ) = 0 on Γ m × [ 0 , T ] , curl u ( x , t ) × n ( x ) = 0 on Γ m × [ 0 , T ] , u ( x , 0 ) = u 0 in Ω ,

where f represents a density of body forces and the viscosity ν is a positive constant. The unknowns are the velocity u and the pressure P of the fluid.

Using the basic idea in [22,31], by introducing the vorticity ω = curl u as a new unknown (see also [4,6,23]), the convection term can be written as follows:

u . ∇ u = ω × u + 1 2 grad ∣ u ∣ 2 ,

and system (1) is fully equivalent as follows:

(2) ∂ u ∂ t + ν curl ω + ω × u + ∇ p = f in Ω × [ 0 , T ] , div u = 0 in Ω × [ 0 , T ] , ω = curl u in Ω × [ 0 , T ] , u . n = 0 on ∂ Ω × [ 0 , T ] , u × n = 0 on Γ × [ 0 , T ] , ω × n = 0 on Γ m × [ 0 , T ] , u = u 0 in Ω ,

where the dynamical pressure p is defined by:

p = P + 1 2 ∣ u ∣ 2 .

We assume that ∂ Γ m = ∂ Γ a Lipschitz-continuous sub-manifold of ∂ Ω .

A priori error analysis of the finite element discretization of the stationary Stokes and Navier-Stokes problem with mixed boundary conditions have first been performed in [5,14,31] and extended to a posteriori error analysis for the time-dependent Stokes and Navier-Stokes problem with mixed boundary conditions in [17,18]. The case of the discretization by spectral, and spectral element methods, relying on this type of boundary conditions are studied in several papers (see [1,2,7,8,10,11,19,21]).

We propose in this work the discretization of the considered problem using Euler’s implicit scheme with respect to the time variable combined with the spectral method with respect to the space variables. Unlike boundary conditions on normal component of velocity and tangential components of vorticity, which are defined on the whole boundary [4,6,30], mixed boundary conditions produce a lack of regularity of the solution especially for the vorticity. Consequently, we propose a new formulation inspired from [4,30] for the two-dimensional domain and [6] for the three-dimensional domain. We prove that the time semi-discrete problem has a solution under the conditions, where Γ m is of class C 1 , 1 or convex with sufficiently large viscosity.

For the spectral discretization, we assume that Ω is a cube, where Γ m is one of its faces. Since the vorticity is a potential-vector in three dimension, it is required to choose appropriate polynomial spaces, which are the spectral analogs of Nédélec’s finite elements spaces on cubic three-dimensional meshes (see [29]). The full spectral discrete problem is constructed using the Galerkin method with numerical integration [15,16]. We then prove its well-posedness. To deal with the nonlinear aspects of the problem, the theorem of Brezzi et al. [20] is used. We thus prove optimal error estimates for the vorticity and the velocity. The choice of the discrete space of the pressure gives us a non-optimal inf-sup condition. Hence, we have the luck of optimality on the approximation of the pressure.

This article is organized as follows: Section 2 presents the variational formulation of the continuous problem. Section 3 is devoted to the well-posedness of time semi-discrete problem by using Euler’s implicit scheme in time. The well-posedness of the full spectral discrete problem is detailed in Section 4. Finally, the analysis of the a priori error estimate for velocity, vorticity, and pressure is presented in Section 5.

2 The continuous variational formulation

Before writing the variational formulation of the problem (2), we begin by defining the following Sobolev spaces:

W m , p ( Ω ) = { v ∈ L p ( Ω ) ; ∂ α v ∈ L p ( Ω ) ; ∀ ∣ α ∣ ≤ m } ,

which is a Banach space equipped with the following norm and semi-norm:

‖ v ‖ m , p , Ω = ∑ ∣ α ∣ ≤ m ∫ Ω ∣ ∂ α v ( x ) ∣ p 1 p and ∣ v ∣ m , p , Ω = ∑ ∣ α ∣ = m ∫ Ω ∣ ∂ α v ( x ) ∣ p 1 p .

If p = 2 , W m , 2 ( Ω ) = H m ( Ω ) is an Hilbert space equipped with the scalar product,

( v , w ) m , Ω = ∑ ∣ α ∣ ≤ m ( ∂ α v , ∂ α w ) 2 1 2 .

We denote by ( ⋅ , ⋅ ) the L 2 ( Ω ) scalar product, L 0 2 ( Ω ) the space of functions in L 2 ( Ω ) which have a null integral on Ω , and D ( Ω ) the space of indefinitely differentiable functions with a compact support in Ω . We consider H ( div , Ω ) the domain of the div operator as follows:

H ( div , Ω ) = { v ∈ L 2 ( Ω ) 3 ; div v ∈ L 2 ( Ω ) }

provided with the norm:

‖ v ‖ H ( div , Ω ) = ( ‖ v ‖ L 2 ( Ω ) 3 2 + ‖ div v ‖ L 2 ( Ω ) 2 ) 1 2 .

We remind (see [25, Chap I, Sec 2]) that the normal trace operator is defined from H ( div , Ω ) into H − 1 2 ( ∂ Ω ) , such that for any scalar function w smooth enough, we have:

⟨ v ⋅ n , w ⟩ = ∫ Ω div v ( x ) w ( x ) d x + ∫ Ω v ( x ) ⋅ ∇ w ( x ) d x ,

where ⟨ . , . ⟩ is the duality product between H − 1 2 ( ∂ Ω ) and H 1 2 ( ∂ Ω ) . Then, we define the kernel of the normal trace operator in H ( div , Ω ) as follows:

H 0 ( div , Ω ) = { v ∈ H ( div , Ω ) ; v ⋅ n = 0 on ∂ Ω } .

We also consider H ( curl , Ω ) the domain of the curl operator:

H ( curl , Ω ) = { v ∈ L 2 ( Ω ) 3 ; curl v ∈ L 2 ( Ω ) 3 } ,

provided with the norm:

‖ v ‖ H ( curl , Ω ) = ( ‖ v ‖ L 2 ( Ω ) 3 2 + ‖ curl v ‖ L 2 ( Ω ) 3 2 ) 1 2 .

The tangential trace operator is defined from H ( curl , Ω ) into H − 1 2 ( ∂ Ω ) 3 , for any smooth enough vector field w ,

⟨ v × n , w ⟩ = ∫ Ω v ( x ) . curl w ( x ) d x − ∫ Ω curl v ( x ) ⋅ w ( x ) d x .

Then, we introduce the kernel of the tangential operator in H ( curl , Ω )

H 0 ( curl , Ω ) = { v ∈ H ( curl , Ω ) ; v × n = 0 on ∂ Ω } .

The restriction of the tangential trace operator to Γ maps the space H ( curl , Ω ) into the dual space ( H 00 1 2 ( Γ ) ′ ) 3 of ( H 00 1 2 ( Γ ) ) 3 , see ([27], Chap. 1, Th. 11.7) for the definition of this space. Then, due to the condition five in the system (2), we define the following space:

(3) H ∗ ( curl , Ω ) = { v ∈ H ( curl , Ω ) ; v × n = 0 on Γ } .

Then, the trivial definition of the velocity’s space is

(4) X ( Ω ) = H 0 ( div , Ω ) ∩ H ∗ ( curl , Ω ) .

This space is provided with the semi-norm

(5) ( ‖ div v ‖ L 2 ( Ω ) 2 + ‖ curl v ‖ L 2 ( Ω ) 3 2 ) 1 2 .

However, since Ω is a simply connected domain, this semi-norm is a norm on the space X ( Ω ) equivalent to the graph norm of H ( div , Ω ) ∩ H ( curl , Ω ) (see ([9], Cor. 3.16)), and there exists a positive constant C ˘ only depending on the domain Ω , such that

(6) ∀ v ∈ X ( Ω ) , ‖ v ‖ L 2 ( Ω ) 3 ≤ C ˘ ‖ v ‖ X ( Ω ) .

We also consider the following spaces, which depend on time. Let S a separable Banach space. We consider C n ( 0 , T ; S ) the set of time C n classes functions with a value on S . C n ( 0 , T ; S ) is a Banach space related to the norm:

‖ v ‖ C n ( 0 , T ; S ) = sup 0 ≤ t ≤ T ∑ j = 0 n ‖ ∂ t j v ‖ S

such that ∂ t j v is the time partial derivative up the order j of the function v . Let also the spaces

L p ( 0 , T ; S ) = { v measurable on ] 0 , T [ such that ∫ 0 T ‖ v ( ⋅ , t ) ‖ S p d t < ∞ } ,

and

H s ( 0 , T ; S ) = { v ∈ L 2 ( 0 , T ; S ) ; ∂ m v ∈ L 2 ( 0 , T ; S ) ; m ≤ s } .

L p ( 0 , T ; S ) is a Banach space equipped with the norm:

‖ v ‖ L p ( 0 , T ; S ) = ∫ 0 T ‖ v ( ⋅ , t ) ‖ S p d t 1 p , for 1 ≤ p < + ∞ sup 0 ≤ t ≤ T ‖ v ( ⋅ , t ) ‖ S , for p = + ∞

and H s ( 0 , T ; S ) is an Hilbert space when it is equipped with the following scalar product:

( v , w ) H s ( 0 , T ; S ) = ∑ i = 0 s ( ∂ i v , ∂ i w ) L 2 ( 0 , T ; S ) 2 1 2 .

Finally, we define also L ( S ) the Banach space of the linear and continuous functions from S into R provided with the norm

∀ L ∈ L ( S ) , ‖ L ‖ L ( S ) = sup f ∈ S / { 0 } ∣ L ( f ) ∣ ‖ f ‖ S .

Now, we suppose that the data f belong to the space L 2 ( 0 , T ; X ( Ω ) ′ ) , where X ( Ω ) ′ is the dual space of X ( Ω ) . We denote by u ( t ) = u ( ⋅ , t ) , ω ( t ) = ω ( ⋅ , t ) , p ( t ) = p ( ⋅ , t ) and we consider the following variational formulation on the interval ] 0 , T [ .

Find ( ω ( t ) , u ( t ) , p ( t ) ) ∈ L 2 ( Ω ) 3 × X ( Ω ) × L 0 2 ( Ω ) , such that

(7) ∀ v ∈ X ( Ω ) , ( ∂ t u ( t ) , v ) + a ( ω ( t ) , u ( t ) ; v ) + K ( ω ( t ) , u ( t ) ; v ) + b ( v , p ( t ) ) = ⟨ f ( t ) , v ⟩ , ∀ q ∈ L 0 2 ( Ω ) , b ( u ( t ) , q ) = 0 , ∀ ϑ ∈ L 2 ( Ω ) 3 , c ( ω ( t ) , u ( t ) ; ϑ ) = 0 , u ( 0 ) = u 0 in Ω ,

where ⟨ . , . ⟩ is the duality product between X ( Ω ) ′ and X ( Ω ) . The forms a ( . , . ; . ) , b ( , . , ) and c ( . , . ; . ) are defined as follows:

a ( ω ( t ) , u ( t ) ; v ) = ν ∫ Ω curl ω ( x , t ) ⋅ v ( x ) d x , b ( u ( t ) , q ) = − ∫ Ω div u ( x , t ) q ( x ) d x , c ( ω ( t ) , u ( t ) ; ϑ ) = ∫ Ω ω ( x , t ) ⋅ ϑ ( x ) d x − ∫ Ω u ( x , t ) ⋅ curl ϑ ( x ) d x .

While the trilinear form K ( . , . ; . ) is defined by

(8) K ( ω ( t ) , u ( t ) ; v ) = ∫ Ω ( ω × u ) ( x , t ) ⋅ v ( x ) d x .

The proof of the equivalence between problems (2) and (7) is difficult because the density of D ( Ω ) in X ( Ω ) is unlikely when Γ m has a positive measure. Then, we only give the following partial result.

Proposition 1

If the data f belong to the space L 2 ( 0 , T ; X ( Ω ) ′ ) , any solution ( ω ( t ) , u ( t ) , p ( t ) ) in L 2 ( Ω ) 3 × X ( Ω ) × L 0 2 ( Ω ) of problem (7) such that ω ( t ) × u ( t ) belongs to L 2 ( Ω ) 3 is a solution of the problem (2) in the sense of distributions.

Proof 1

According to the definition of the space X ( Ω ) , each solution ( ω ( t ) , u ( t ) , p ( t ) ) of problem (7) satisfies equations (4) and (5) of problem (2). Equations (2) and (3) in problem (7) give us equations (2) and (3) in problem (2), respectively. Let v ∈ D ( Ω ) 3 in the first equation of problem (7). Then, we conclude the first equation of problem (2). Finally, we consider a regular function φ on ∂ Ω with compact support in Γ m such that φ ⋅ n = 0 on ∂ Ω . If v is equal to the lifting of φ in X ( Ω ) ∩ H ( Ω ) 3 yields equation (6) of problem (2).□

We note that a ( ⋅ , ⋅ ; ⋅ ) and c ( ⋅ , ⋅ ; ⋅ ) , are respectively, continuous on the spaces L 2 ( Ω ) 3 × X ( Ω ) × X ( Ω ) and L 2 ( Ω ) 3 × X ( Ω ) × L 2 ( Ω ) 3 . While b ( ⋅ , ⋅ ) is continuous on the space X ( Ω ) × L 0 2 ( Ω ) . Then, its kernel

V = { φ ∈ X ( Ω ) ; ∀ q ∈ L 0 2 ( Ω ) , b ( φ , q ) = 0 } ,

is a closed subspace of X ( Ω ) , which coincides with the space of divergence free functions in X ( Ω ) . We also consider

W = { ( ϑ , φ ) ∈ L 2 ( Ω ) 3 × V ; ∀ ψ ∈ L 2 ( Ω ) 3 , c ( ϑ , φ ; ψ ) = 0 } = { ( ϑ , φ ) ∈ L 2 ( Ω ) 3 × V ; φ = curl ϑ } ,

the kernel of the form c ( . , . ; . ) .

If ( ω ( t ) , u ( t ) , p ( t ) ) is solution of problem (7), then ( ω ( t ) , u ( t ) ) is solution of the following reduced problem:

Find ( ω ( t ) , u ( t ) ) in W such that,

(9) ∀ v ∈ V , ∂ u ∂ t ( t ) , v + a ( ω ( t ) , u ( t ) ; v ) + K ( ω ( t ) , u ( t ) ; v ) = ⟨ f ( t ) , v ⟩ .

The existence of a solution for the problem (9) requires the continuity of the nonlinear term K ( ⋅ , ⋅ ; ⋅ ) , which requires the following assumption.

Assumption 1

We assume that the domain Ω has a C 1 , 1 boundary or is a polyhedron with no re-entrant corners inside Γ m .

Remark 1

Note that if Ω has a C 1 , 1 boundary or is convex, the space H 0 ( div , Ω ) ∩ H ( curl , Ω ) is included in H 1 ( Ω ) 3 (see [9], Thm 2.17 for the proof). When Ω is a polyhedron, we recall that the space of restrictions of functions of X ( Ω ) to Ω ⧹ V ¯ , where V is a neighborhood of the re-entrant corners of Ω inside Γ m , is embedded in L 4 ( Ω ⧹ V ¯ ) 3 (see ([14], Lem. 2.5) for the proof and more details). So, to give a sense to the nonlinear term K ( ⋅ , ⋅ ; ⋅ ) , we suppose Assumption 1 satisfied.

Spaces L 0 2 ( Ω ) and X ( Ω ) are compatible by the following inf-sup condition (see [14] or ([25] Chap. I, Cor. 2.4) for its proof):

There exists a constant β > 0 such that:

(10) ∀ q ∈ L 0 2 ( Ω ) , sup v ∈ X ( Ω ) b ( v , q ) ‖ v ‖ X ( Ω ) ≥ β ‖ q ‖ L 2 ( Ω ) .

When the assumption 1 is satisfied, the arguments for the proof of the existence of solution of problem (2) are exactly the same as in ([32], Chap. III, Theo. 1.1), see also ([26], Chap. V).

3 The time semi-discrete problem

The aim of this section is the time discretization of problem (2) by the implicit Euler scheme. We consider a partition of the interval [ 0 , T ] into sub-intervals [ t k − 1 , t k ] , for 1 ≤ k ≤ K , such that 0 = t 0 < t 1 < … < t K = T , where K is a positive integer. Let τ k = t k − t k − 1 , the time step, and the k -tuple τ = ( τ 1 , τ 2 , … , τ k ) such that ∣ τ ∣ = max 1 ≤ k ≤ K τ k .

The regularity parameter σ τ is defined as follows:

σ τ = max 2 ≤ k ≤ K τ k τ k − 1 .

For any data f ∈ C 0 ( 0 , T ; X ( Ω ) ′ ) , if v k = v ( . , t k ) , the time semi-discrete problem issued from Euler’s implicit method is written as follows:

(11) u k − u k − 1 τ k + ν curl ω k + ω k × u k + ∇ p k = f k in Ω , div u k = 0 in Ω , ω k = curl u k in Ω , u k ⋅ n = 0 on ∂ Ω , u k × n = 0 on Γ , ω k × n = 0 on Γ m , u 0 = u 0 in Ω .

Problem (11) is equivalent to the following variational formulation:

(12) u 0 = u 0 in Ω ,

and find ( ω k , u k ) 1 ≤ k ≤ K ∈ ( L 2 ( Ω ) 3 × X ( Ω ) ) K and p 1 ≤ k ≤ K k ∈ ( L 0 2 ( Ω ) ) K such that for all 1 ≤ k ≤ K ,

(13) ∀ v ∈ X ( Ω ) , a ˘ ( ω k , u k ; v ) + τ k K ( ω k , u k ; v ) + τ k b ( v , p k ) = F k ( v ) , ∀ q ∈ L 0 2 ( Ω ) , b ( u k , q ) = 0 , ∀ ϑ ∈ L 2 ( Ω ) 3 , c ( ω k , u k ; ϑ ) = 0 ,

where f k = f ( . , t k ) ,

a ˘ ( ω k , u k ; v ) = ( u k , v ) + τ k ν ( curl ω k , v )

and

F k ( v ) = ( u k − 1 , v ) + τ k ⟨ f k , v ⟩ .

Thus, we note that if ( ω k , u k , p k ) is solution of problems (12) and (13), we conclude that ( ω k , u k ) belongs to W and is solution of the following reduced problem:

(14) ∀ v ∈ V , a ˘ ( ω k , u k ; v ) + τ k K ( ω k , u k ; v ) = F ( v ) .

The main difficulty now is to prove that problem (12)–(14) admits a solution. We observe that the form a ˘ ( ⋅ , ⋅ ; ⋅ ) is continuous on the space L 2 ( Ω ) 3 × X ( Ω ) × X ( Ω ) and the functional F is linear and continuous on V . Now we show the two following properties of the form a ˘ ( . , . ; . ) .

Lemma 1

For all 1 ≤ k ≤ K , a ˘ ( . , . ; . ) satisfies

(15) ∀ v ∈ V \ { 0 } , sup ( ω k , u k ) ∈ W a ˘ ( ω k , u k ; v ) > 0 ,

and there exists α k > 0 such that

(16) ∀ ( ω k , u k ) ∈ W , sup v ∈ V a ˘ ( ω k , u k ; v ) ‖ v ‖ X ( Ω ) ≥ α k ( ‖ ω k ‖ L 2 ( Ω ) 3 + ‖ u k ‖ X ( Ω ) ) .

Proof 2

We prove the positivity of the form a ˘ ( . , . ; . ) . Assume that

∀ v ∈ V \ { 0 } , a ˘ ( ω k , u k ; v ) = 0 .

If ( ω k , u k ) = ( curl v , v ) in W , we obtain

‖ v ‖ L 2 ( Ω ) 3 2 + τ k ν ‖ curl v ‖ L 2 ( Ω ) 3 2 = 0 .

Then v = 0 , which is in contradiction with the fact that v ∈ V \ { 0 } .

Now we prove the inf-sup condition (16). Let v = u k in V , then:

a ˘ ( ω k , u k ; v ) = ‖ u k ‖ L 2 ( Ω ) 3 2 + τ k ν ∫ Ω curl ω k u k d x .

We integrate by part and since ω k = curl u k , div u k = 0 , we have:

(17) a ˘ ( ω k , u k ; v ) ≥ τ k ν 2 ‖ ω k ‖ L 2 ( Ω ) 3 2 + τ k ν 2 ‖ u k ‖ X ( Ω ) 2 .

Using (17) combined with the following inequality

‖ v ‖ X ( Ω ) ≤ ( ‖ u k ‖ X ( Ω ) 2 + ‖ ω k ‖ L 2 ( Ω ) 3 2 ) 1 2 ,

we conclude the desired result with α k = τ k ν 2 .□

To go further in the proof that problems (12)–(14) admit a solution, we investigate the properties of the trilinear form K ( . , . ; . ) .

Lemma 2

If Assumption 1 is satisfied, the trilinear form K ( ⋅ , ⋅ ; ⋅ ) is continuous on the space L 2 ( Ω ) 3 × X ( Ω ) × X ( Ω ) .

Proof 3

From Hölder inequality and the fact that the space X ( Ω ) is compactly imbedded in ( L 4 ( Ω ) ) 3 , we have

(18) K ( ω , u ; v ) ≤ C ‖ ω ‖ L 2 ( Ω ) 3 ‖ u ‖ L 4 ( Ω ) 3 ‖ v ‖ L 4 ( Ω ) 3 ≤ C ∗ ‖ ω ‖ L 2 ( Ω ) 3 ‖ u ‖ X ( Ω ) ‖ v ‖ X ( Ω ) ,

where C ∗ only depends on the domain Ω .□

By using the following equality proved by Green formula

∫ Ω ( curl u × u ) ⋅ v d x = − ∫ Ω ( curl u × v ) ⋅ u d x − ∫ Ω ( u ⋅ v ) div u d x ,

we conclude for u ∈ V , the antisymmetry property of the trilinear form K ( ⋅ , ⋅ ; ⋅ ) ,

(19) K ( ω , u ; u ) = 0 .

The existence of solution for problems (12)–(14) is proved for a large enough viscosity ν with respect to the norm of the functional F .

Proposition 2

For any data ( f , u 0 ) belongs to C 0 ( 0 , T ; X ( Ω ) ′ ) × L 2 ( Ω ) 3 . Knowing u k − 1 , if Assumption 1 holds, and there exists a constant C ◇ such that

(20) C ◇ ν − 2 ‖ F k ‖ L ( X ( Ω ) ) < 1 ,

then problems (12)–(14) has a solution ( ω k , u k ) ∈ W satisfying

(21) ‖ u k ‖ L 2 ( Ω ) 3 2 + ∑ i = 1 k τ i ‖ ω i ‖ L 2 ( Ω ) 3 2 ≤ ‖ u 0 ‖ L 2 ( Ω ) 3 2 + 1 ν ∑ i = 1 k τ i ‖ f i ‖ L 2 ( Ω ) 3 2 .

Proof 4

Let the iterative sequence ( ( ω j k , u j k ) ) j ≥ 0 , where ( ω 0 k , u 0 k ) = ( 0 , 0 ) and ( ω j k , u j k ) is solution in W of the following problem:

(22) ∀ v ∈ V , a ˘ ( ω j k , u j k ; v ) = F k ( v ) − τ k K ( ω j − 1 k , u j − 1 k ; v ) .

Thanks to properties (15) and (16) of the form a ˘ ( ⋅ , ⋅ ; ⋅ ) and the continuity of the trilinear form K ( . , . ; . ) , problem (22) has a unique solution (see [11] for more details about the proof). We consider

(23) r k = ν τ k 2 C ∗ ,

where C ∗ is the continuity constant defined in (18). Let us now prove by induction on j ≥ 0 that the sequence ( ( ω j k , u j k ) ) j is bounded by r k in the norm of the space W ,

(24) ∀ j ≥ 0 , ( ‖ ω j k ‖ L 2 ( Ω ) 3 2 + ‖ u j k ‖ X ( Ω ) 2 ) 1 2 ≤ r k .

The estimation (24) is clearly true for j = 0 , and we assume that it holds for the iteration j − 1 . If v = u j k in the problem (22), we obtain using properties (16) and (18) Cauchy-Schwarz inequality and the following Young inequality:

(25) ∀ ε > 0 , 1 2 ε a 2 + ε 2 b 2 ,

that

ν τ k 2 ‖ ω j k ‖ L 2 ( Ω ) 3 2 + ν τ k 2 ‖ u j k ‖ X ( Ω ) 2 ≤ 1 2 ε 1 ‖ F k ‖ L ( X ( Ω ) ) 2 + ε 1 2 ‖ u j k ‖ L 2 ( Ω ) 3 2 + C ∗ 1 2 ε 2 ( ‖ ω j − 1 k ‖ L 2 ( Ω ) 3 2 + ‖ u j − 1 k ‖ X ( Ω ) 2 ) + ε 2 2 ‖ u j k ‖ X ( Ω ) 2 .

Then, from (6) and the fact that the estimation (24) holds for the iteration j − 1 , we have

ν τ k 2 ‖ ω j k ‖ L 2 ( Ω ) 3 2 + ν τ k 2 − ε 1 C ˘ 2 − C ∗ ε 2 2 ‖ u j k ‖ X ( Ω ) 2 ≤ 1 2 ε 1 ‖ F k ‖ L ( X ( Ω ) ) 2 + C ∗ 2 ε 2 r k 4 .

By choosing ε 1 = ν τ k 4 C ˘ , ε 2 = ν τ k 4 C ∗ , and combining (23) and (20), with C ◇ = 4 2 C ∗ , we obtain (24).

Furthermore, we show for any j ≥ 2 that

a ˘ ( ω j k − ω j − 1 k , u j k − u j − 1 k ; v ) = K ( ω j − 2 k , u j − 2 k ; v ) − K ( ω j − 1 k , u j − 1 k ; v ) = − K ( ω j − 1 k − ω j − 2 k , u j − 2 k ; v ) − K ( ω j − 1 k , u j − 1 k − u j − 2 k ; v ) .

Taking v = u j k − u j − 1 k and using the continuity of the trilinear form K ( ⋅ , ⋅ ; ⋅ ) combined with 25, we obtain

ν τ k 2 ( ‖ ω j k − ω j − 1 k ‖ L 2 ( Ω ) 3 2 + ‖ u j k − u j − 1 k ‖ X ( Ω ) 2 ) ≤ C ∗ 1 2 ε 1 ( ‖ ω j − 1 k − ω j − 2 k ‖ L 2 ( Ω ) 3 ‖ u j − 2 k ‖ X ( Ω ) ) 2 + ε 1 2 ‖ u j k − u j − 1 k ‖ X ( Ω ) 2 + 1 2 ε 2 ( ‖ ω j − 1 k ‖ L 2 ( Ω ) 3 ‖ u j − 1 k − u j − 2 k ‖ X ( Ω ) ) 2 + ε 2 2 ‖ u j k − u j − 1 k ‖ X ( Ω ) 2 .

Let ε 1 = ε 2 = ν τ k 4 C ∗ and by (24), we obtain

ν τ k 4 ( ‖ ω j k − ω j − 1 k ‖ L 2 ( Ω ) 3 2 + ‖ u j k − u j − 1 k ‖ X ( Ω ) 2 ) ≤ 2 C ∗ 2 r k 2 ν τ k ( ‖ ω j − 1 k − ω j − 2 k ‖ L 2 ( Ω ) 3 2 + ‖ u j − 1 k − u j − 2 k ‖ X ( Ω ) 2 ) .

Now thanks to (23), we conclude that

‖ ω j k − ω j − 1 k ‖ L 2 ( Ω ) 3 2 + ‖ u j k − u j − 1 k ‖ X ( Ω ) 2 ≤ 2 C ∗ ( ‖ ω j − 1 k − ω j − 2 k ‖ L 2 ( Ω ) 3 2 + ‖ u j − 1 k − u j − 2 k ‖ X ( Ω ) 2 ) .

Then ( ( ω j k , u j k ) ) j ≥ 0 is a Cauchy sequence in the closet space W , so that it converges to ( ω k , u k ) in W . By passing to the limit in (22), it is readily checked that ( ω k , u k ) is solution of problems (12)–(14).

In order to prove estimation (21), we take v = u k in the equality (14), since the trilinear form K ( ⋅ , ⋅ ; ⋅ ) verifies the antisymmetry property (19), we obtain

‖ u k ‖ L 2 ( Ω ) 3 2 + ν τ k ‖ ω k ‖ L 2 ( Ω ) 3 2 ≤ ( u k − 1 , u k ) + τ k ⟨ f k , u k ⟩ .

By using Cauchy-Schwarz inequality combined with Young inequality (25) and the fact that u k ∈ V , we obtain

‖ u k ‖ L 2 ( Ω ) 3 2 + ν τ k ‖ ω k ‖ L 2 ( Ω ) 3 2 ≤ τ k 2 ε 1 ‖ f k ‖ X ( Ω ) ′ 2 + ε 1 τ k 2 ‖ ω k ‖ L 2 ( Ω ) 3 2 + 1 2 ε 2 ‖ u k − 1 ‖ L 2 ( Ω ) 3 2 + ε 2 2 ‖ u k ‖ L 2 ( Ω ) 3 2 .

So, if ε 1 = ν and ε 2 = 1 , we have

(26) ‖ u k ‖ L 2 ( Ω ) 3 2 + ν τ k ‖ ω k ‖ L 2 ( Ω ) 3 2 ≤ τ k ν ‖ f k ‖ X ( Ω ) ′ 2 + ‖ u k − 1 ‖ L 2 ( Ω ) 3 2 .

Summing the inequality (26) over k , we conclude the estimate (21).□

Theorem 1

If Assumption 1 holds, for any data ( f , u 0 ) belongs to C 0 ( 0 , T ; X ( Ω ) ′ ) × L 2 ( Ω ) 3 , such that (20) is satisfied. At each time step, knowing u k − 1 problems (12) and (13) has at most a solution ( ω k , u k , p k ) in L 2 ( Ω ) 3 × X ( Ω ) × L 0 2 ( Ω ) .

Proof 5

The existence of ( ω k , u k ) is proved (see proposition (2)). To prove the existence of p k , we define for any v ∈ X ( Ω ) the following functional:

Σ k ( v ) = 1 τ k ( F k ( v ) − a ˘ ( ω k , u k ; v ) ) − K ( ω k , u k ; v ) .

The functional Σ k is linear, continuous on X ( Ω ) and vanishes on the space V . Then, using the inf-sup condition (10), there exists p k in L 0 2 ( Ω ) such that:

∀ v ∈ X ( Ω ) , Σ k ( v ) = − ( div v , p k )

and

‖ p k ‖ L 2 ( Ω ) ≤ 1 β sup v ∈ X ( Ω ) Σ k ( v ) ‖ v ‖ X ( Ω ) .

Let ( ω 1 k , u 1 k , p 1 k ) and ( ω 2 k , u 2 k , p 2 k ) be two solutions of problems (12) and (13) such that ω 1 i = ω 2 i and u 1 i = u 2 i for 0 ≤ i ≤ ( k − 1 ) . It follows from assumption (1) and condition (20) that the pair ( ω k , u k ) = ( ω 1 k − ω 2 k , u 1 k − u 2 k ) , belongs to W and satisfies

(27) ∀ v ∈ V a ˘ ( ω k , u k ; v ) = K ( ω 2 k , u 2 k ; v ) − K ( ω 1 k , u 1 k ; v ) = − K ( ω k , u 2 k ; v ) − K ( ω 1 k , u k ; v ) .

From (17), and taking v = u k in (27), we obtain

τ k ν 2 ( ‖ ω k ‖ L 2 ( Ω ) 3 2 + ‖ u k ‖ X ( Ω ) 2 ) ≤ ( ∣ K ( ω k , u 2 ; u k ) ∣ + ∣ K ( ω 1 , u k ; u k ) ∣ ) .

The antisymmetry property of the trilinear form K ( . , . ; . ) gives that K ( ω 1 , u k ; u k ) = 0 . Then by applying the continuity of K ( . , . ; . ) together with (21), we have

τ k ν 2 ( ‖ ω k ‖ L 2 ( Ω ) 3 2 + ‖ u k ‖ X ( Ω ) 2 ) ≤ C ν − 1 ‖ F k ‖ L ( X ( Ω ) ) ( ‖ ω k ‖ L 2 ( Ω ) 3 2 + ‖ u k ‖ X ( Ω ) 2 ) .

Thus, if condition (20) holds with C ◇ > 2 C , we obtain that ω k = 0 and u k = 0 .

Finally, for any v ∈ X ( Ω ) , we have

b ( v , p 1 k − p 2 k ) = 0 ,

then p 1 k − p 2 k = 0 according to the inf-sup condition (10).□

4 The full spectral discrete problem

In this section, we are interested to the spectral discretization of problems (12) and (13). Henceforth, we assume that

Ω = ] − 1 , 1 [ 3 and Γ m = ] − 1 , 1 [ 2 × { 1 } .

The spectral discretization is done by the same way of the Nédélec’s finite element method (see [29, Sec 2]). We introduce P n , m , s ( Ω ) the space of the restriction on Ω of the polynomials of degree n in the x direction, m in the y direction, and s in the z direction.

Let N ≥ 2 be an integer, we define the space

(28) X N = X ( Ω ) ∩ P N , N − 1 , N − 1 ( Ω ) × P N − 1 , N , N − 1 ( Ω ) × P N − 1 , N − 1 , N ( Ω ) ,

which approximates the space of the velocity X ( Ω ) and

(29) Y N = P N − 1 , N , N ( Ω ) × P N , N − 1 , N ( Ω ) × P N , N , N − 1 ( Ω ) ,

the space of the discrete vorticity. For the approximation of pressure in the space L 0 2 ( Ω ) , we consider the sub-space M N of L 0 2 ( Ω ) ∩ P N − 1 ( Ω ) , which we will define explicitly later.

For a better accuracy of the approximation of the trilinear form K ( . , . ; . ) , we will do a over numeric integration (see [28] and [8]). We choose the integer M ≥ N equal to the integer part of ( 1 + α ) N , where α is a fixed real number in the interval [ 0 , 1 ] . Setting ξ 0 = − 1 and ξ M = 1 , we introduce the M − 1 nodes ξ i , 1 ≤ i ≤ M − 1 and the M + 1 weights ρ i , 0 ≤ i ≤ M , of the Gauss-Lobatto quadrature formula. Denoting by P n ( − 1 , 1 ) the space of restriction to [ − 1 , 1 ] of polynomials with degree ≤ n , then we have:

(30) ∀ φ ∈ P 2 M − 1 ( − 1 , 1 ) , ∫ − 1 1 φ ( ζ ) d ζ = ∑ i = 0 M φ ( ξ i ) ρ i .

We also recall the following important inequality (see [15]):

(31) ∀ ϕ N ∈ P M ( − 1 , 1 ) , ‖ ϕ N ‖ L 2 ( − 1 , 1 ) 2 ≤ ∑ i = 0 M ϕ N 2 ( ξ i ) ρ i ≤ 3 ‖ ϕ N ‖ L 2 ( − 1 , 1 ) 2 .

Relying on this formula, we introduce the discrete scalar product on P M ( Ω ) defined by: for continuous functions φ and ψ on Ω ¯

(32) ( φ , ψ ) N = ∑ i = 0 M ∑ j = 0 M ∑ k = 0 M φ ( ξ i , ξ j , ξ k ) ψ ( ξ i , ξ j , ξ k ) ρ i ρ j ρ k ,

and we consider I N the Lagrange interpolating operator at the nodes ( ξ i , ξ j , ξ k ) , 0 ≤ i , j , k ≤ N , with values in P M ( Ω ) .

Assume that the data f and u 0 are, respectively, continuous on Ω ¯ × [ 0 , T ] and Ω ¯ , we construct the discrete problem from (12) and (13) using the Galerkin method combined with numerical integration.

(33) If u N 0 = I N ( u 0 ) ,

knowing u k − 1 , find ( ω N k , u N k , p N k ) in X N × Y N × M N such that for 1 ≤ k ≤ K ,

(34) ∀ v N ∈ X N , a ˘ N ( ω N k , u N k ; v N ) + τ k K N ( ω N k , u N k ; v N ) + τ k b N ( v N , p N k ) = F N k ( v N ) , ∀ q N ∈ M N , b N ( u N k , q N ) = 0 , ∀ ϑ N ∈ Y N , c N ( ω N k , u N k ; ϑ N ) = 0 ,

where the bilinear forms a ˘ N ( ⋅ , ⋅ ; ⋅ ) , b N ( ⋅ , ⋅ ) and c N ( ⋅ , ⋅ ; ⋅ ) are defined by

(35) a ˘ N ( ω N k , u N k ; v N ) = ( u N k , v N ) N + τ k ν ( curl ω N k , v N ) N , b N ( v N , q N ) = − ( div v N , q N ) N , and c N ( ω N k , u N k ; φ N ) = ( ω N k , φ N ) N − ( u N k , curl φ N ) N .

By using Cauchy-Schwarz inequality and (31), it follows that the bilinear forms a ˘ N ( ⋅ , ⋅ ; ⋅ ) , b N ( ⋅ , ⋅ ) and c N ( ⋅ , ⋅ ; ⋅ ) are, respectively, continuous on ( Y N × X N ) × X N , X N × M N and ( Y N × X N ) × Y N with norms bounded independently of N . The function F N k ( v N ) = ( u N k − 1 , v N ) N + τ k ( I N ( f k ) , v N ) N is linear and continuous on X N . Moreover, as a consequence of the exactness of property (30), the bilinear forms b ( ⋅ , ⋅ ) and b N ( ⋅ , ⋅ ) coincide on X N × M N . While the trilinear form K N ( . , . ; ) is defined as follows:

(36) K N ( ω N k , u N k ; v N ) = ( ω N k × u N k , v N ) N .

Now in order to define the discrete space M N of the pressure, we start by introducing the set of the spurious modes SP N of the discrete bilinear form b N ( ⋅ , ⋅ ) ,

SP N = { q N ∈ L 0 2 ( Ω ) ∩ P N − 1 ( Ω ) ; ∀ v N ∈ X N , b N ( v N , q N ) = 0 } .

For a fixed real number λ ∈ ] 0 , 1 [ , we denote by π [ λ N ] the orthogonal projection operator from L 2 ( ] − 1 , 1 [ ) onto P [ λ N ] , where [ λ N ] stands for the integer part of λ N . We set

Λ N − 1 = L N ′ − π [ λ N ] L N ′ , Λ N − 2 = L N − 1 ′ − π [ λ N ] L N − 1 ′ , Ξ N = χ N − π [ λ N ] χ N ,

where the polynomial χ N is defined by:

χ N ( ζ ) = L N ′ ( ζ ) N + 1 − L N − 1 ′ ( ζ ) N − 1 ,

and L N is the Lagrange polynomial defined on the interval ] − 1 , 1 [ . The proof of the next lemma is given in ([7], Lem. a.1).

Lemma 3

The space SP N is spanned by the polynomials

( Λ N − 1 ± Λ N − 2 ) ( x ) ( Λ N − 1 ± Λ N − 2 ) ( y ) ψ N ( z ) , ( Λ N − 1 ± Λ N − 2 ) ( x ) ψ N ( y ) Ξ N ( z ) , end ψ N ( x ) ( Λ N − 1 ± Λ N − 2 ) ( y ) Ξ N ( z ) ,

where ψ N runs through P N − 1 ( ] − 1 , 1 [ ) .

Now since SP N it has been characterized, the space M N is chosen as the subspace of L 0 2 ( Ω ) ∩ P N − 1 ( Ω ) equal to the orthogonal of the space SP N

(37) M N = SP N ⊥ .

We intend now to show the well-posedness of problems (33) and (34). We begin by introducing the discrete kernel V N of the bilinear form b N ( ⋅ , ⋅ )

(38) V N = { v N ∈ X N ; ∀ q N ∈ M N , b N ( v N , q N ) = 0 } .

This kernel coincides with X N ∩ V , i.e., the space of divergence-free polynomials in X N (see [7], Lem. 3.2., for the proof).

Similarly, we introduce the discrete kernel W N of the form c N ( ⋅ , ⋅ ; ⋅ ) as follows:

(39) W N = { ( ϑ N , v N ) ∈ Y N × V N ; ∀ φ N ∈ Y N , c N ( ϑ N , v N ; φ N ) = 0 } .

We remark that for any solution ( ω N k , u N k ; p N ) of problems (33) and (34), the pair ( ω N k , u N k ) is solution of the following reduced discrete problem:

Knowing u k − 1 , find ( ω N k , u N k ) ∈ W N , such that for 1 ≤ k ≤ K ,

(40) ∀ v N ∈ V N , a ˘ N ( ω N k , u N k ; v N ) + τ k K N ( ω N k , u N k ; v N ) = F N k ( v N ) .

The well-posedness of the discrete reduced problems (33)–(40) is proved by the same arguments as for the continuous reduced problems (12)–(14). We show the following two properties of the form a ˘ N ( ⋅ , ⋅ , ⋅ ) .

Lemma 4

For 1 ≤ k ≤ K , the positivity property of the form a ˘ N ( ⋅ , ⋅ ; ⋅ ) holds:

(41) ∀ v N ∈ V N ⧹ { 0 } , sup ( ω N k , u N k ) ∈ W N a ˘ N ( ω N k , u N k ; v N ) > 0 .

Proof 6

For any v N ∈ V N , ( curl v N , v N ) belongs to W N . Taking ( ω N k , u N k ) = ( curl v N , v N ) , we have

(42) a ˘ N ( ω N k , u N k ; v N ) = ( v N , v N ) N + τ k ν ( curl v N , curl v N ) N ≥ τ k ν ( curl v N , curl v N ) N .

From the property (31), we obtain

a ˘ N ( ω N k , u N k ; v N ) ≥ τ k ν ‖ curl v N ‖ L 2 ( Ω ) 3 2 .

Then if a ˘ N ( ω N k , u N k ; v N ) = 0 , we conclude that v N = 0 , using property (6) and the fact that v N is divergence free.□

The form a ˘ N ( ⋅ , ⋅ ; ⋅ ) also verifies the following inf-sup condition.

Lemma 5

For any 1 ≤ k ≤ K , there exists α k > 0 independent of N such that, for all ( ω N k , u N k ) in W N ,

(43) sup v N ∈ V N a ˘ N ( ω N k , u N k ; v N ) ‖ v N ‖ X ( Ω ) ≥ α k ( ‖ ω N k ‖ L 2 ( Ω ) 3 + ‖ u N k ‖ X ( Ω ) ) .

Proof 7

For any ( ω N k , u N k ) in W N , by taking v N = u N k and using (31) yields

(44) a ˘ N ( ω N k , u N k ; v N ) = ( u N k , u N k ) N + τ k ν ( ω N k , ω N k ) N ≥ τ k ν ‖ ω N k ‖ L 2 ( Ω ) 3 2 .

On the other hand, setting ω N k = curl u N k in the definition of W N , we obtain

( curl u N k , curl u N k ) N = ( ω N k , curl u N k ) N .

By Cauchy-Schwarz inequality combined with (31),

‖ curl u N k ‖ L 2 ( Ω ) 3 ≤ 3 ‖ ω N k ‖ L 2 ( Ω ) 3 .

Since u N k is divergence free, we conclude that

a ˘ N ( ω N k , u N k ; v N ) ≥ τ k ν 2 ‖ ω N k ‖ L 2 ( Ω ) 3 2 + τ k ν 18 ‖ u N k ‖ X ( Ω ) 2 .

We therefore prove the desired inf-sup condition (43).□

Proposition 3

For any data ( f , u 0 ) continuous on ( [ 0 , T ] × Ω ¯ ) × Ω ¯ and knowing u N k − 1 at each time step k , problems (33)–(40) admits a solution ( ω N k , u N k ) in W N , which satisfies, for any 1 ≤ k ≤ K ,

(45) ‖ u N k ‖ L 2 ( Ω ) 3 2 + ν ∑ i = 1 k ‖ ω N i ‖ L 2 ( Ω ) 3 2 ≤ ‖ u N 0 ‖ L 2 ( Ω ) 3 2 + 2 7 2 2 ν ∑ i = 1 k τ i ‖ I N ( f i ) ‖ L 2 ( Ω ) 3 2 .

Proof 8

We define the mapping ϕ N from W N into its dual space by:

(46) ∀ ( ω N k , u N k ) ∈ W N ∀ ( θ N , v N ) ∈ W N , ⟨ ϕ N ( ω N k , u N k ) , ( θ N , v N ) ⟩ = a ˘ N ( ω N k , u N k ; v N ) + τ k K N ( ω N k , u N k ; v N ) − F N k ( v N ) .

We are equipped W N with the following norm:

( ‖ ω N k ‖ L 2 ( Ω ) 3 2 + ‖ u N k ‖ X ( Ω ) 2 ) 1 2 .

We check that ϕ N is continuous since the space W N is finite dimensional, and from the antisymmetry property (19) of the trilinear form K N ( ⋅ , ⋅ ; ⋅ ) , we obtain

⟨ ϕ N ( ω N k , u N k ) , ( ω N k , u N k ) ⟩ = a ˘ N ( ω N k , u N k ; u N k ) − F N k ( u N k ) .

Now, using Lemma 5 with α k = τ k ν 18 leads to

⟨ ϕ N ( ω N k , u N k ) , ( ω N k , u N k ) ⟩ ≥ τ k ν 18 ( ‖ ω N k ‖ L 2 ( Ω ) 3 2 + ‖ u N k ‖ X ( Ω ) 2 ) − F N k ( u N k ) .

Then, by Young’s inequality (25), we have

⟨ ϕ N ( ω N k , u N k ) , ( ω N k , u N k ) ⟩ ≥ 5 τ k ν 9 ( ‖ ω N k ‖ L 2 ( Ω ) 3 2 + ‖ u N k ‖ X ( Ω ) 2 ) − 1 τ k ν ‖ F N k ‖ L ( X ( Ω ) ) 2 .

So, setting

ϖ N = 3 5 τ k ν ‖ F N k ‖ L ( X ( Ω ) ) ,

we check that ⟨ ϕ N ( ω N k , u N k ) , ( ω N k , u N k ) ⟩ is nonnegative on the sphere of W N with radius ϖ N . Then, by using Brouwer’s fixed point theorem (see ([25], Chap. IV, Cor. 1.1)), we conclude that problems (33)–(40) have a solution ( ω N k , u N k ) in W N .

Replacing v N = u N k in the first equality of (34), and considering the antisymmetry property (19) of the trilinear form K N ( ⋅ , ⋅ ; ⋅ ) , we have:

( u N k − u N k − 1 , u N k ) N + ν τ k ( curl ω N k , u N k ) N = τ k ( I N ( f k ) , u N k ) N .

By using property (31), Cauchy-Schwarz inequality and integrating by part, we obtain:

( u N k − u N k − 1 , u N k ) + ν τ k ( ω N k , ω N k ) ≤ 27 τ k ‖ I N ( f k ) ‖ L 2 ( Ω ) 3 ‖ u N k ‖ L 2 ( Ω ) 3 .

Thanks to the inequality

a ( a − b ) = 1 2 a 2 − 1 2 b 2 + 1 2 ( a − b ) 2 ,

we obtain

1 2 ‖ u N k ‖ L 2 ( Ω ) 3 2 − 1 2 ‖ u N k − 1 ‖ L 2 ( Ω ) 3 2 + 1 2 ‖ u N k − u N k − 1 ‖ L 2 ( Ω ) 3 2 + ν τ k ‖ ω N k ‖ L 2 ( Ω ) 3 2 ≤ 27 τ k ‖ I N ( f k ) ‖ L 2 ( Ω ) 3 ‖ u N k ‖ L 2 ( Ω ) 3 .

By using Young’s inequality (25) leads to,

1 2 ‖ u N k ‖ L 2 ( Ω ) 3 2 − 1 2 ‖ u N k − 1 ‖ L 2 ( Ω ) 3 2 + ν h n ‖ ω N k ‖ L 2 ( Ω ) 3 2 ≤ 27 τ k ε ‖ I N ( f k ) ‖ L 2 ( Ω ) 3 2 2 + ‖ u N k ‖ L 2 ( Ω ) 3 2 2 ε .

Finally, by choosing ε = 27 2 ν and by summing over k , we obtain the desired result (45).□

Due to the definition of the space M N in (37), we consider the following inf-sup condition proved in ([7], lem. a.6):

There exists a positive constant γ independent of N such that

(47) ∀ q N ∈ M N , sup v N ∈ X ( Ω ) b N ( v N , q N ) ‖ v N ‖ X ( Ω ) ≥ γ N − 1 ‖ q N ‖ L 2 ( Ω ) .

Thanks to this inf-sup condition (47), the existence of the pressure follows the same proof as given in Theorem 1, then from Proposition 3, and we have the next full result.

Theorem 4.1

For any data ( f , u 0 ) continuous on ( [ 0 , T ] × Ω ¯ ) × Ω ¯ and knowing u N k − 1 at each time step k , problems (33) and (34) has a solution ( ω N k , u N k , p N k ) in Y N × X N × M N . Moreover, the part ( ω N k , u N k ) of this solution satisfies (45).

5 Error estimates

This section is devoted to the a priori error estimate based on Brezzi-Rappaz-Raviart theorem [20]. We begin by writing both problems (12)–(13) and (33)–(40) in another form.

For any data f ∈ L 2 ( 0 , T ; X ( Ω ) ′ ) , we define the operator S such that S F k is the solution ( ω k , u k ) of the following reduced problem:

u 0 = u 0 in Ω ,

for all 1 ≤ k ≤ K , knowing u k − 1 , find ( ω k , u k ) ∈ W such that

∀ v ∈ V , a ˘ ( ω k , u k ; v ) = F k ( v ) .

Let the mapping G defined from the space X = L 2 ( Ω ) 3 × V into the dual space of X ( Ω ) by:

∀ ( ω k , u k ) ∈ X , 1 ≤ k ≤ K , ∀ v ∈ X ( Ω ) , ⟨ G ( ω k , u k ) , v ⟩ = τ k K ( ω k , u k ; v ) − F k ( v ) .

Then, problem (14) is equivalently written as follows:

(48) ( ω k , u k ) + S G ( ω k , u k ) = 0 .

Similarly for the discrete problem, let X N = Y N × V N , and we denote by S N the operator such that S N F k is the solution ( ω N k , u N k ) of the problem:

u N 0 = I N ( u 0 ) , knowing u N k − 1 , find ( ω N k , u N k ) ∈ W N , such that for any 1 ≤ k ≤ K ,

(49) ∀ v N ∈ V N , a ˘ N ( ω N k , u N k ; v N ) = F N k ( v N ) .

We recall the following properties, and see [11] for the proof.

The stability property of the operator S N
(50) ‖ S N F k ‖ X ≤ C sup v N ∈ V N ∣ F k ( v N ) ∣ ‖ v N ‖ X ( Ω ) .
If S F k belongs to L 2 ( Ω ) 3 × H s ( Ω ) 3 for any s ≥ 1 , we hold the following error:
(51) ‖ ( S − S N ) F k ‖ ≤ C N − s ‖ S F k ‖ L 2 ( Ω ) 3 × H s ( Ω ) 3 .

We consider also the mapping G N defined from X N into the dual space of X N by:

∀ ( ω N k , u N k ) ∈ X N , 1 ≤ k ≤ K , ∀ v N ∈ X N , ⟨ G N ( ω N k , u N k ) , v N ⟩ = τ k K N ( ω N k , u N k ; v N ) − F N k ( v N ) .

Then the problem (40) is equivalent to the problem:

(52) ( ω N k , u N k ) + S N G N ( ω N k , u N k ) = 0 .

We recall the following result (see [15], 13.28) needed to prove the continuity of K N ( ⋅ , ⋅ ; ⋅ ) ,

(53) ∀ φ m ∈ P m ( Ω ) , ‖ I N ( φ m ) ‖ L 2 ( Ω ) ≤ C 1 + m M 2 ‖ φ m ‖ L 2 ( Ω ) .

Lemma 6

If Assumption 1 is satisfied, for any ( ω N , u N , v N ) belongs to Y N × X N × X N , we have

(54) ∣ K N ( ω N , u N ; v N ) ∣ ≤ C ‖ ω N ‖ L 2 ( Ω ) 3 ‖ u N ‖ X ( Ω ) ‖ v N ‖ X ( Ω ) ,

where C is a positive constant.

Proof 9

Let

K N ( ω N , u N ; v N ) = ( ω N × u N , v N ) N = ( ω N , I N ( u N × v N ) ) N .

Then, if we combine Cauchy-Schwarz inequality, (31), (53), and Hölder’s inequality, we obtain the following:

∣ K N ( ω N , u N ; v N ) ∣ ≤ C ‖ ω N ‖ L 2 ( Ω ) 3 ‖ u N ‖ L 4 ( Ω ) 3 ‖ v N ‖ L 4 ( Ω ) 3 .

So by assumption (1), we conclude the result (54).□

Let D the Fréchet derivative. We make the following assumption.

Assumption 2

The solution ( ω k , u k , p k ) of problems (12) and (13) belongs to H s − 3 4 ( Ω ) × H s ( Ω ) 3 × H s − 1 ( Ω ) for all s > 3 2 .
The operator Id + S D G ( ω k , u k ) is an isomorphism of X .

In the first part of this assumption, we give a regularity result for the solution ( ω k , u k , p k ) of our problem. While the regularity results for the Navier-Stokes problem with mixed boundary conditions in a three-dimensional domain are so far unknown, this result is for a convex polyhedron domain.

Now we are ready to prove the following preliminary lemmas, which need to use Brezzi-Rappaz-Raviart theorem [20]. An approximation ( ω N k ♯ , u N k ♯ ) of ( ω k , u k ) in Y N ♯ × V N ♯ is required and satisfies:

(55) ‖ ( ω k − ω N k ♯ , u k − u N k ♯ ) ‖ X ≤ C N ♯ 1 − s ‖ u k ‖ H s ( Ω ) 3 , for s ≥ 1 ,

where N ♯ is equal to the integer part of 2 μ N − 1 for a fixed μ ∈ ] 0 , 1 ] (see [7], chap. 4).

We introduce L ( X N ) the space of endomorphisms on X N . To state the following lemma, we also make the following additional assumption.

Assumption 3

The operator S transforms the space L 2 ( Ω ) 3 into the space H s − 3 4 ( Ω ) 3 × H s ( Ω ) 3 , for all s > 1 .

Lemma 7

If we consider the Assumptions 1–3, there exists an integer N ◇ such that, for all N ≥ N ◇ , the operator Id + S N D G N ( ω N k ♯ , u N k ♯ ) is an isomorphism of X N , where the norm of its inverse is bounded independently of N.

Proof 10

We start by doing the following expansion:

(56) Id + S N D G N ( ω N k ♯ , u N k ♯ ) = Id + S D G ( ω k , u k ) − ( S − S N ) D G ( ω k , u k ) − S N ( D G ( ω k , u k ) − D G ( ω N k ♯ , u N k ♯ ) ) − S N ( D G ( ω N k ♯ , u N k ♯ ) − D G N ( ω N k ♯ , u N k ♯ ) ) .

As Id + S D G ( ω k , u k ) does not depend on N , and according to the second part of the assumption 3, we need to prove that the last three terms of the right-hand side in equality (56) tend to 0 when N tends to + ∞ using the norm of L ( X N ) .

We remark that, for any ( θ , ϑ ) in X , we have

D G ( ω k , u k ) ⋅ ( θ , ϑ ) = ω k × ϑ + θ × u k .

Using Assumption 1, the first part of Assumption 2 and the fact that the embedding of H s ( Ω ) , respectively, H s − 3 4 ( Ω ) , into L ∞ ( Ω ) , respectively L 4 ( Ω ) is compact, we obtain that the mapping ( ω k , u k ) ↦ ω k × ϑ + θ × u k transforms the unit sphere of X into a compact subset of L 2 ( Ω ) 3 . Due to the assumption (3), the operator S transform this compact into a compact subset of H s − 3 4 ( Ω ) × H s ( Ω ) 3 .

From property (51), we obtain the next convergence

(57) lim N ↦ + ∞ ‖ ( S − S N ) D G ( ω k , u k ) ‖ L ( X ) = 0 .
Since for all ( θ N , ϑ N ) ∈ X N
( D G ( ω k , u k ) − D G ( ω N k ♯ , u N k ♯ ) ) ⋅ ( θ N , ϑ N ) = G ( ω k − ω N k ♯ , ϑ N ) + G ( θ N , u k − u N k ♯ ) ,
and then, we obtain from (50) and Lemma 2 that
‖ S N ( D G ( ω k , u k ) − D G ( ω N k ♯ , u N k ♯ ) ) ‖ L ( X ) ≤ sup v N ∈ V N ⟨ G ( ω k − ω N k ♯ , ϑ N ) + G ( θ N , u k − u N k ♯ ) , v N ⟩ ‖ v N ‖ X ( Ω ) ≤ C ∗ ( ‖ ω k − ω N k ♯ ‖ L 2 ( Ω ) 3 ‖ ϑ N ‖ X ( Ω ) + ‖ θ N ‖ L 2 ( Ω ) 3 ‖ u k − u N k ♯ ‖ X ( Ω ) ) .

So, by using (55), we conclude that

(58) lim N ↦ + ∞ ‖ S N ( D G ( ω k , u k ) − D G ( ω N k ♯ , u N k ♯ ) ) ‖ L ( X ) = 0 .

(3) From the definition of G and G N , we have for all ( θ N , ϑ N ) ∈ X N and v N ∈ V N

⟨ G ( ω N k ♯ , u N k ♯ ) ⋅ ( θ N , ϑ N ) , v N ⟩ = K ( ω N k ♯ , ϑ N ; v N ) + K ( θ N , u N k ♯ ; v N ) , ⟨ G N ( ω N k ♯ , u N k ♯ ) ⋅ ( θ N , ϑ N ) , v N ⟩ = K N ( ω N k ♯ , ϑ N ; v N ) + K N ( θ N , u N k ♯ ; v N ) .

So, according to the exactness of the quadrature formula (30) on the polynomial space P 2 M − 1 ( Ω ) , the last term in the right-hand side of the equality (56) vanishes.

Following the second part of Assumption (2), we consider η = ‖ ( Id + S D G ( ω k , u k ) ) − 1 ‖ L ( X ) , when N is large enough. Then, terms (57) and (58) become bounded by 1 2 η . So, we obtain that the norm of ( I d + S D G ( ω k , u k ) ) − 1 is bounded by 2 η .□

Now, we prove the next lemma, which deals with the Lipschitz property of the operator S N .

Lemma 8

There exists a real number L > 0 , such that for any ( ω ˆ N , u ˆ N ) ∈ X N

‖ S N ( D G N ( ω N k ♯ , u N k ♯ ) − D G N ( ω ˆ N , u ˆ N ) ) ‖ L ( X ) ≤ L ( ‖ ω N k ♯ − ω ˆ N ‖ L 2 ( Ω ) 3 + ‖ u N k ♯ − u ˆ N ‖ X ( Ω ) ) .

Proof 11

We have, for any ( θ N , ϑ N ) ∈ X N and v N ∈ V N ,

⟨ ( D G N ( ω N k ♯ , u N k ♯ ) − D G N ( ω ˆ N , u ˆ N ) ) ⋅ ( θ N , ϑ N ) , v N ⟩ = K N ( ω N k ♯ − ω ˆ N , ϑ N ; v N ) + K N ( θ N , u N k ♯ − u ˆ N ; v N ) .

Then, if we combine Lemma 6 and (50) with the aforementioned equality, it leads to the desired Lipschitz property□

Lemma 9

If the part one of Assumption (2) is satisfied, for any data f in the space L 2 ( 0 , T ; H μ ( Ω ) ) ; μ > 3 2 , we have the following estimate:

(59) ‖ ( ω k , u k ) + S N G N ( ω k , u k ) ‖ X ≤ C ( ∣ τ ∣ + N 1 − s ) ( ‖ ω k ‖ H s − 3 4 ( Ω ) 3 + ‖ u k ‖ H s ( Ω ) 3 ) + ( ∣ τ ∣ + N − μ ) ‖ f ‖ L 2 ( 0 , T ; H μ ( Ω ) ) ,

where C is a constant only depending on the data f and u j , 0 ≤ j ≤ k − 1 .

Proof 12

By using equality (48), we obtain

(60) ‖ I d + S N G N ( ω k , u k ) ‖ X ≤ ‖ ( S − S N ) G N ( ω k , v k ) ‖ X + ‖ S N ( G N ( ω k , u k ) − G ( ω k , u k ) ) ‖ X .

The first term on the right-hand side of inequality (60) is bounded thanks to (51). We obtain from the definition of G and G N that

⟨ ( G N ( ω k , u k ) − G ( ω k , u k ) ) , v N ⟩ = F N k ( v N ) − F k ( v N ) = ( u N k − 1 , v N ) N − ( u k − 1 , v N ) + τ k ( ( I N ( f k ) , v N ) N − ( f k , v N ) ) .

Then, the bound of the second term on the right-hand side of inequality (60) is deduced from the approximation properties of the operators Π N − 1 and I N (see [15]).□

Now we state the main result of the a priori error estimate.

Theorem 2

Assume that the data function f belongs to the space L 2 ( 0 , T ; H μ ( Ω ) ) ; μ > 3 2 . Under Assumptions 1–3, there exist an integer N ∗ and a positive real number τ ∗ , such that for all N ≥ N ∗ and ∣ τ ∣ ≤ τ ∗ , problems (33)and (34) has a unique solution. Moreover, this solution satisfies for all 1 ≤ k ≤ K the following error estimate:

(61) ‖ ω k − ω N k ‖ L 2 ( Ω ) 3 + ‖ u k − u N k ‖ X ( Ω ) + N − 1 ‖ p k − p N k ‖ L 2 ( Ω ) ≤ C ( ∣ τ ∣ + N 1 − s ) ( ‖ ω k ‖ H s − 3 4 ( Ω ) 3 + ‖ u k ‖ H s ( Ω ) 3 + ‖ p k ‖ H s ( Ω ) ) + ( ∣ τ ∣ + N − μ ) ‖ f ‖ L 2 ( 0 , T ; H μ ( Ω ) ) ,

where C is a constant only depending on the data f and u j , 0 ≤ j ≤ k − 1 .

Proof 13

Combining Lemmas 7–9 together with the Brezzi-Rappaz-Raviart theorem (see [20]) give for N large enough, the existence and a local uniqueness of the solution ( ω N k , u N k ) of problem (52). This solution satisfies

‖ ω k − ω N k ‖ L 2 ( Ω ) 3 + ‖ u k − u N k ‖ X ( Ω ) ≤ C ( ∣ τ ∣ + N 1 − s ) ( ‖ ω k ‖ H s − 3 4 ( Ω ) 3 + ‖ u k ‖ H s ( Ω ) 3 ) + ( ∣ τ ∣ + N − μ ) ‖ f ‖ L 2 ( 0 , T ; H μ ( Ω ) ) .

Moreover, using inf-sup condition (47), there exists for all 1 ≤ k ≤ K , a unique pressure p N k in M N , such that:

∀ v N ∈ V N , b N ( v N , p N k ) = 1 τ k ( F N k ( v N ) − a ˘ N ( ω N k , u N k ; v N ) ) − K ( ω N k , u N k ; v N ) .

Furthermore, having for any q N ∈ M N ,

b N ( v N , p N k − q N ) = b ( v N , p k − q N ) − F k ( v N ) + F N k ( v N ) + a ˘ ( ω k − ω N k , u k − u N k ; v N ) + ( a ˘ − a ˘ N ) ( ω N k , u N k ; v N ) + K ( ω k , u k ; v N ) − K ( ω N k , u N k ; v N ) ,

then the bound of ‖ p k − p N k ‖ L 2 ( Ω ) is deduced from both a triangular inequality and (47).□

6 Conclusion and future work

This work deals with the numerical analysis of the implicit Euler scheme in time and the spectral discretization in space of the Navier-Stokes equations with mixed boundary conditions in a three-dimensional domain. We prove using Brouwer fixed point theorem that the new discrete formulation has at most one solution. We show that the error is optimal for the vorticity and the velocity and a quasi-optimal for the pressure. The algorithm of resolution, and the numerical implementation of these results in a cube and a more complex domain using the spectral elements method will be the subject of our forthcoming work.

Funding information: Researchers Supporting Project number (RSP-2021/153), King Saud University, Riyadh, Saudi Arabia.
Conflict of interest: The authors state no conflict of interest.
Data availability statement: We declare there are none.

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Received: 2022-03-05

Revised: 2022-04-10

Accepted: 2022-04-14

Published Online: 2022-05-11

This work is licensed under the Creative Commons Attribution 4.0 International License.

Spectral discretization of the time-dependent Navier-Stokes problem with mixed boundary conditions

Abstract

1 Introduction

2 The continuous variational formulation

Proposition 1

Proof 1

Assumption 1

Remark 1

3 The time semi-discrete problem

Lemma 1

Proof 2

Lemma 2

Proof 3

Proposition 2

Proof 4

Theorem 1

Proof 5

4 The full spectral discrete problem

Lemma 3

Lemma 4

Proof 6

Lemma 5

Proof 7

Proposition 3

Proof 8

Theorem 4.1

5 Error estimates

Lemma 6

Proof 9

Assumption 2

Assumption 3

Lemma 7

Proof 10

Lemma 8

Proof 11

Lemma 9

Proof 12

Theorem 2

Proof 13

6 Conclusion and future work

References

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