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In this paper, an isogeometric error estimate for transport equation is obtained in 2D to prove the convergence of isogeometric method. The result that we have obtained, generalizes Ern result, got in finite elements method. For the time discretization, the two stage Heun scheme is used to prove this result. For a polynomial of degree
*k*
≥1
, the order of convergence in space is 2 and in time is
_{}.

Some phenomena of the daily life such as particles transport in an electric field, the signal transport along a wire, evolution of cars on a road [

Isogeometric Analysis has been introduced by Thomas Hughes, Austin Cottrell and Yuri Bazilevs in 2005 [

The objectives of Isogeometric Analysis are to generalize and improve upon Finite Element Analysis (FEA) in the following ways:

1) To provide more accurate modeling of complex geometries and to exactly represent common engineering shapes such as circles, cylinders, spheres, ellipsoids, etc.

2) To fix exact geometries at the coarsest level of discretization and eliminate geometrical errors.

3) To vastly simplify mesh refinement of complex industrial geometries by eliminating the necessity to communicate with the CAD (Computer Aided Design) description of geometry.

4) To provide refinement procedures, including classical h- and p-refinements analogues, and to develop a new refinement procedure called k-refinement [

The idea of Isogeometric Analysis is to build a geometry model and, rather than develop a finite element model approximating the geometry, directly use the functions describing the geometry in analysis [

Isogeometric Analysis is approached, using continuous or discontinuous Galerkin method. In the context of space semidiscretization by discontinuous Galerkin methods, explicit RK schemes are used to approximate in time systems of ordinary differential equations. These schemes have been developed by Cockburn and Shu [

1) Error equation.

2) An energy identity obtained from error equations.

3) A stability estimate using Gronwall lemma, Young inequality and inverse and trace inequalities for finite elements method.

In the literature, there exist many numerical methods to solve transport equation [

1) Constructed a parametrization of the physical domain, indispensable to describe this domain.

2) Constructed a parametric mesh making a tensor product of knot vectors.

3) Introduced the discrete space on the physical domain, using our parametrization.

Moreover, instead of using inverse and trace inequalities for finite elements method, we will use isogeometric inverse and trace inequalities to obtain our convergence result. As far as the discretization in time is concerned, the explicit two stage Heun scheme is used. Now, we consider the following model:

{ ∂ t u ( x , t ) + ∇ ⋅ ( β ( x ) u ( x , t ) ) = 0 , x ∈ Ω ⊂ ℝ 2 , t ∈ [ 0 ; t f ] u ( . , t = 0 ) = u 0 u = 0 in ∂ Ω − × [ 0 ; t f ] (1)

where Ω is a bounded open set in ℝ 2 , u : Ω × [ 0 ; t f ] ↦ R 2 is a scalar-valued function representing the unknown, t f is a finite time, ∂ Ω − = { x ∈ ∂ Ω , β ( x ) ⋅ n ( x ) < 0 } , n is the unit outward normal to the domain boundary, β is the advective velocity, β ∈ [ L ∞ ( Ω ) ] 2 , ∇ ⋅ β ∈ L ∞ ( Ω ) and u 0 is the initial datum.

Let us introduce some notations and assumptions:

• Assume β is a Lipschitz continuous functions i.e.

∃ L β , ∀ ( x , y ) ∈ Ω 2 , ‖ β ( x ) − β ( y ) ‖ ≤ L β ‖ x − y ‖ .

where ‖ x − y ‖ denotes the Euclidean norm of ( x − y ) in R 2 .

• We set τ c : = 1 max { ‖ β ′ ‖ L ∞ ( Ω ) ; L β } and β c : = ‖ β ‖ [ L ∞ ( Ω ) ] 2 .

• We set τ ⋆ : = min ( t f , τ c ) .

• Assume h ≤ β c τ ⋆ .

• ∀ x ∈ ℝ , x ⊖ : = 1 2 ( | x | − x ) and x ⊕ : = 1 2 ( | x | + x ) .

• Let l ∈ ℕ , we consider the space

C l ( V ) : = C l ( [ 0, t f ] , V ) , (2)

where V is a Hilbert space and equipped with the scalar product defined by:

( u , s ) V = ( u , s ) L 2 ( Ω ) + ( ∇ ⋅ ( β u ) , ∇ ⋅ ( β s ) ) L 2 ( Ω ) , ∀ ( u , s ) ∈ V 2 . (3)

The associated norm is:

‖ u ‖ V 2 = ‖ u ‖ L 2 ( Ω ) 2 + ‖ ∇ ⋅ ( β u ) ‖ L 2 ( Ω ) 2 , ∀ u ∈ V . [

This paper is organized as follows. In the first section, we will describe univariate B-splines. In the second one, we will describe bivariate B-splines and geometry of the physical domain. In the third one, we present main results of this work. In the fourth one, we will state inverse and isogeometric inequalities. In the fifth one, we will talk about the functional setting and space semidiscretization. In the sixth one, we will look into the explicit two stage Heun scheme analysis.

Definition 1. Let x 1 ≤ x 2 ≤ ⋯ ≤ x m be an increasing sequence of reals, B-splines functions of degree k are defined by Cox-de Boor-Mansfield recursion formula [

{ For 1 ≤ i ≤ m − 1 N i , 0 ( t ) = 1 if t ∈ [ x i , x i + 1 [ N i , 0 ( t ) = 0 otherwise (5)

{ For k ≥ 1 and 1 ≤ i ≤ m − k − 1 N i , k ( t ) = t − x i x i + k − x i N i , k − 1 ( t ) + x i + k + 1 − t x i + k + 1 − x i + 1 N i + 1 , k − 1 ( t ) , (6)

with the convention x 0 : = 0 for all real number x.

The set ( x i ) i = 1 m ( 1 ≤ i ≤ m ) is called knots vector.

Now, we want to look into bivariate B-splines, obtained from univariate B-splines.

The definition of bivariate B-splines follows easily through a tensor-product construction. Let us focus on the two-dimensional case. Notably, let us consider the unit square Ω ^ = [ 0 ; 1 ] 2 ⊂ ℝ 2 . Mimicking the one-dimensional case, given integers p l and n l for l = 1 , 2 . Let us introduce open knot vectors:

E l = { ξ 1 , l , ⋯ , ξ n l + p l + 2 , l }

and the associated vectors without repetitions for each direction l

ζ l = { ζ 1 , l , ⋯ , ζ m l , l }

There is a parametric cartesian mesh Q h associated with these knot vectors partitioning the parametric domain Ω ^ into a rectangular grid. So, we have:

Q h = { Q = ⊗ l = 1 , 2 ( ζ i l , l , ζ i l + 1 , d ) , 1 ≤ i l ≤ m l − 1 } [

For each element Q ∈ Q h , we associate a parametric mesh size h Q = h Q , max where h Q , max denotes the length of the largest edge of Q. Also, for each element, we define a shape regularity constant as in [

λ Q = h Q h Q , min (8)

where h Q , min denotes the length of the smallest edge of Q.

We associate with each knot vector E l , ( l = 1 , 2 ) univariate B-spline basis functions N i , p l of degree p l for i = 1 , ⋯ , n l .

On the mesh Q h , we define the tensor-product B-spline basis functions as in [

N ( i , j ) , ( p 1 , p 2 ) = N i , p 1 ⊗ N j , p 2 , i = 1 , ⋯ , n 1 , j = 1 , ⋯ , n 2 . (9)

N ( i , j ) , ( p 1 , p 2 ) = N i , p 1 N j , p 2 , i = 1 , ⋯ , n 1 , j = 1 , ⋯ , n 2 . (10)

The span of these functions form the space of two-dimensional splines over Ω ^ , denoted by:

S h = s p a n { N ( i , j ) , ( p 1 , p 2 ) } i = 1 , j = 1 n 1 , n 2

The physical domain Ω is defined through a geometrical mapping:

F = ∑ i = 1 n 1 ∑ j = 1 n 2 P i j N i , n 1 N j , n 2 [

where P i j ∈ ℝ 2 are the so-called control points. F is a parametrization of the physical domain Ω , that is,

F : [ 0,1 ] 2 → Ω

For each element Q in the parametric domain [ 0,1 ] 2 , there is a corresponding physical element K = F ( Q ) , as shown in

We assume throughout that F is invertible, with smooth inverse F − 1 , on each element Q ∈ Q h .

We define the physical mesh to be:

τ h = { K : K = F ( Q ) , Q ∈ Q h } = F ( Q h ) . (11)

We assume ( τ h ) is quasi-uniform:

∃ C > 0 , h ≤ C h K . (12)

with h K the diameter of K and h : = max K ∈ τ h h K .

We introduce V h , the space spanned by B-splines basis functions in Ω as the push-forward of the B-splines space S h .

V h : = s p a n { N ( i , j ) , ( p 1 , p 2 ) ∘ F − 1 } i = 1 , j = 1 n 1 , n 2 .

Given a function v ^ ∈ L 2 ( Ω ^ ) , we define a projective operator over the B-splines space S h as:

π S h : L 2 ( Ω ^ ) → S h , π S h v ^ : = ∑ i = 1 , j = 1 n 1 , n 2 φ ( v ^ ) N ( i , j ) , ( p 1 , p 2 ) ,

where the linear functionals φ ∈ L 2 ( Ω ^ ) ′ determine the dual basis for the set of B-splines.

The projective operator over the B-splines space V h , is defined as the push-forward of the operator π S h .

π h : L 2 ( Ω ) → V h , π h v : = ( π S h ( v ^ ) ) ∘ F − 1

This section is devoted to our convergence results obtained for respectively a polynomial of degree k ≥ 2 and a polynomial of degree k = 1 . We present our main results whose proofs are given in the subsection 6.6.

Theorem 1. (Convergence for RK2, k ≥ 2 )

Assume the 4 3 CFL Condition:

δ t ≤ ϱ ′ τ ⋆ − 1 3 ( h β c ) 4 3 for some positive real number ϱ ′ (13)

and d t s u ∈ C 0 ( H k + 1 − s ( Ω ) ) for s ∈ { 0,1 } . Then,

‖ u N − u h N ‖ L 2 ( Ω ) + ( ∑ m = 0 N − 1 δ t | u m − u h m | β 2 ) 1 2 ≲ exp ( 4 C r 8 τ ⋆ t f ) ( χ 1 δ t 2 + χ 2 h k + 1 2 ) (14)

with

χ 1 = r 4 t f 1 2 τ ⋆ 1 2 ‖ d t 3 u ‖ C 0 ( L 2 ( Ω ) ) (15)

χ 2 = t f 1 2 r 5 β c 1 2 ‖ u ‖ C 0 ( H k + 1 ( Ω ) ) + t f 1 2 r 5 β c − 1 2 ‖ d t u ‖ C 0 ( H k ( Ω ) ) (16)

| v | β 2 = ∫ ∂ Ω 1 2 | β ⋅ n | v 2 + ∑ F ∈ F h i ∫ F 1 2 | β ⋅ n F | [ v ] 2 , (17)

where

r = max ( 1 , max K ∈ τ h ( λ Q λ K ) ) (18)

and δ t is the time step.

Theorem 2. (Convergence for RK2, k = 1 )

Assume the 4 3 CFL Condition, assume d t s u ∈ C 0 ( H k + 1 − s ( Ω ) ) for s ∈ { 0,1 } and 4 − 3 r 2 > 0 . Then,

‖ u N − u h N ‖ L 2 ( Ω ) + ( ∑ m = 0 N − 1 δ t | u m − u h m | β 2 ) 1 2 ≲ 1 4 − 3 r 2 exp ( 8 C r 8 τ ⋆ t f ) ( χ 1 δ t 2 + χ 2 h k + 1 2 ) (19)

with

χ 1 = r 4 t f 1 2 τ ⋆ 1 2 ‖ d t 3 u ‖ C 0 ( L 2 ( Ω ) ) (20)

and

χ 2 = t f 1 2 r 5 β c 1 2 ‖ u ‖ C 0 ( H k + 1 ( Ω ) ) + t f 1 2 r 5 β c − 1 2 ‖ d t u ‖ C 0 ( H k ( Ω ) ) (21)

In this section, we present isogeometric inverse and trace inequalities, useful tools to analyze partial differential equations.

Let K ∈ τ h and Q = F − 1 ( K ) .

Theorem 3. (see [

∀ h > 0 , ∀ K ∈ τ h and ∀ v h ∈ ℙ 2 k ( τ h ) , ‖ ∇ v h ‖ [ L 2 ( K ) ] 2 ≤ C ⋆ h K − 1 ‖ v h ‖ L 2 ( K ) (22)

where C ⋆ depends on k and on the parametrization F.

Theorem 4. (see [

∀ v h ∈ ℙ 2 k ( τ h ) , ‖ v h ‖ L 2 ( ∂ K ) 2 ≤ C λ Q λ K h K − 1 ‖ v h ‖ L 2 ( K ) 2 (23)

where C depends only on p 1 and p 2 , λ Q is the local shape regularity constant of Q, and λ K is the shape regularity constant of K.

We set p : = min { p 1 , p 2 } (see. [

Theorem 5. (see [

∑ K ∈ τ h | u − π h u | H l ( K ) 2 ≤ C h 2 ( s − l ) ‖ u ‖ H s ( Ω ) 2 , (24)

where C is independent on h.

Theorem 6. Given the integer s such that 0 ≤ s ≤ p + 1 and a function u ∈ H s ( Ω ) , then:

∑ K ∈ τ h ‖ u − π h u ‖ L 2 ( ∂ K ) 2 ≤ C max K ∈ τ h ( λ Q λ K ) h ( 2 s − 1 ) ‖ u ‖ H s ( Ω ) 2 , (25)

where C is independent on h.

Proof 1. Let u ∈ H s ( Ω ) . Using the inequality (23), we have:

∑ K ∈ τ h ‖ u − π h u ‖ L 2 ( ∂ K ) 2 ≤ C ∑ K ∈ τ h λ Q λ K h K − 1 ‖ u − π h u ‖ L 2 ( K ) 2 , (26)

( τ h ) h being quasi-uniform, h ≤ C h K .

h ≤ C h K ⇒ C − 1 h K − 1 ≤ h − 1 ⇒ h K − 1 ≤ C h − 1 ⇒ C h K − 1 ≤ C ′ h − 1 (27)

So

∑ K ∈ τ h ‖ u − π h u ‖ L 2 ( ∂ K ) 2 ≤ C max K ∈ τ h ( λ Q λ K ) h − 1 ∑ K ∈ τ h ‖ u − π h u ‖ L 2 ( K ) 2 (28)

Using the inequality (24), we have:

∑ K ∈ τ h ‖ u − π h u ‖ L 2 ( K ) 2 ≤ C h 2 s ‖ u ‖ H s ( Ω ) 2 (29)

Thus, we get:

∑ K ∈ τ h ‖ u − π h u ‖ L 2 ( ∂ K ) 2 ≤ C max K ∈ τ h ( λ Q λ K ) h ( 2 s − 1 ) ‖ u ‖ H s ( Ω ) 2 (30)

In this part, we introduce some basic notations for space-time functions and important theorems.

Theorem 7. (see. [

‖ ϕ ‖ C l ( V ) = max 0 ≤ m ≤ l ‖ d t m ϕ ‖ C 0 ( V ) (31)

with

‖ ϕ ‖ C 0 ( V ) = max t 0 ≤ t ≤ t f ‖ ϕ ( t ) ‖ V [

We want to specify mathematically the meaning of the boundary condition 1. Our aim is to give a meaning to such traces in the space. Thus, we need to investigate the trace on ∂ Ω of functions in the space defined by:

L 2 ( | β ⋅ n | , ∂ Ω ) = { v is defined on ∂ Ω , ∫ ∂ Ω | β ⋅ n | v 2 < ∞ } [

Considering ( τ h ), we present following notations:

• Interfaces are collected in the set F h i and boundary faces are collected in the set F h b . We set F h : = F h b ∪ F h i . ∀ T ∈ τ h , F T : = { F ∈ F h , F ⊂ ∂ T } .

• ∀ F ∈ F h i , the mean of v is denoted by {{v}}.

• The jump of v is denoted by [v].

• Assume P Ω = { Ω i } 1 ≤ i ≤ N Ω is a partition of Ω such that, for the exact solution u,

u ∈ V ⋆ = V ∩ H 1 2 + ε ( P Ω ) , ε > 0 [

where

V = { u ∈ L 2 ( Ω ) , ∇ ⋅ ( β u ) ∈ L 2 ( Ω ) } (34)

We set

V ⋆ h : = V ⋆ + V h with V h = ℙ 2 k ( τ h ) = { v ∈ L 2 ( Ω ) ; ∀ T ∈ τ h , v / T ∈ ℙ 2 k ( T ) } (35)

We define the discrete operator A h : V ⋆ h → V h such as ∀ ( v , w h ) ∈ V ⋆ h × V h ,

( A h v , w h ) L 2 ( Ω ) = ∫ Ω ∇ ⋅ ( β v ) w h + ∫ ∂ Ω ( β ⋅ n ) ⊖ v w h − ∑ F ∈ F h i ∫ F ( β n F ) [ v ] { { w h } } + ∑ F ∈ F h i 1 2 ∫ F | β ⋅ n F | [ v ] [ w h ] [

For all v ∈ V ⋆ h , set:

‖ v ‖ ⋆ ⋆ 2 = ‖ v ‖ u w b , ⋆ 2 + β c h ‖ ∇ v ‖ L 2 ( Ω ) 2 (37)

‖ v ‖ u w b , ⋆ 2 = ‖ v ‖ u w b 2 + ∑ T ∈ τ h β c ‖ v ‖ L 2 ( ∂ T ) 2 (38)

‖ v ‖ u w b 2 = 1 τ c ‖ v ‖ L 2 ( Ω ) 2 + | v | β 2 (39)

E h n = ‖ ε π n ‖ ⋆ ⋆ + ‖ ζ π n ‖ ⋆ ⋆ + τ ⋆ 1 2 ‖ d t 3 u ‖ C 0 ( L 2 ( Ω ) ) δ t 2 + τ ⋆ − 1 2 ‖ ε h n ‖ L 2 ( Ω ) (40)

We abbreviate as a ≲ b the inequality a ≤ C b with positive C independent of β , h , δ t . The value of C can change at each occurrence [

We now state some assumptions on the discrete operator A h . The first one (41) is important to introduce the notion of numerical fluxes:

1) For all ( v , w h ) ∈ V ⋆ h × V h ,

( A h v , w h ) = − ∫ Ω ( β ⋅ ∇ w h ) v + ∫ ∂ Ω ( β ⋅ n ) ⊕ v w h + ∑ F ∈ F h i ∫ F ( β ⋅ n F ) { { v } } [ w h ] + ∑ F ∈ F h i 1 2 ∫ F | β ⋅ n F | [ v ] [ w h ] [

2) From equality (41), Cauchy-Schwarz inequality and inverse inequality (22), we can infer:

For all ( v , w h ) ∈ H 1 ( Ω ) × V h ,

( A h ( v − π h v ) , w h ) L 2 ( Ω ) ≲ ‖ v − π h v ‖ u w b , ⋆ ‖ w h ‖ u w b (42)

3) The three next assumptions are useful to bound the operator A h .

For all

v ∈ V ⋆ h , ‖ A h v ‖ L 2 ( Ω ) ≲ r β c 1 2 h − 1 2 ‖ v ‖ ⋆ ⋆ with r = max ( 1, max K ∈ τ h ( λ Q λ K ) ) (43)

For all

v h ∈ V h , ‖ v h ‖ ⋆ ⋆ ≲ r β c 1 2 h − 1 2 ‖ v h ‖ L 2 ( Ω ) (44)

For all

v h ∈ V h , ‖ A h v ‖ L 2 ( Ω ) ≲ r 2 β c h − 1 ‖ v h ‖ L 2 ( Ω ) (45)

4) The two next inequalities are bounds of δ t ( α h n , ε h n ) L 2 ( Ω ) − δ t 2 | ε h n | β 2 and δ t ( β h n , ζ h n ) L 2 ( Ω ) − δ t 2 | ζ h n | β 2 .

δ t ( α h n , ε h n ) L 2 ( Ω ) − δ t 2 | ε h n | β 2 ≲ δ t ( E h n ) 2 (46)

δ t ( β h n , ζ h n ) L 2 ( Ω ) − δ t 2 | ζ h n | β 2 ≲ r 4 δ t ( E h n ) 2 (47)

5) The two last inequalities are obtained thanks to CFL condition and isogeometric inverse and trace inequalities:

‖ ε π m ‖ ⋆ ⋆ 2 ≲ r 2 β c h 2 k + 1 ‖ u m ‖ H k + 1 ( Ω ) 2 (48)

‖ ζ π m ‖ ⋆ ⋆ 2 ≲ r 2 β c h 2 k + 1 ( ‖ u m ‖ H k + 1 ( Ω ) 2 + β c − 2 ‖ d t u m ‖ H k ( Ω ) 2 ) (49)

For the time discretization, we are interested in an explicit scheme: the two stage Heun scheme.

In this section, we want to tackle the convergence analysis of the two stage Heun scheme.

Let δ t be the time step such as t f = N δ t where N is an integer. For n ∈ { 0, ⋯ , N } , we define the discrete times t n : = n δ t and u n = u ( t n ) . Assume δ t ≤ τ ⋆ with τ ⋆ = min ( t f , τ c ) .

We consider the following explicit scheme:

{ u h n , 1 = u h n − δ t A h u h n u h n + 1 = 1 2 ( u h n + u h n , 1 ) − 1 2 δ t A h u h n , 1 with u h 0 = π h u 0 [

This step is to identify the error equation governing the time evolution of ε h n and ζ h n .

We set

ε h n = u h n − π h u n (50)

ε π n = u n − π h u n (51)

ζ h n = w h n − π h w n (52)

ζ π n = w n − π h w n (53)

with

w = u + δ t d t u [

From (50) and (51), we have u n − u h n = ε π n − ε h n .

From (52) and (53), we have w n − w h n = ζ π n − ζ h n .

We get:

ζ h n = ε h n − δ t A h ε h n + δ t α h n [

ε h n + 1 = 1 2 ( ε h n + ζ h n ) − 1 2 δ t A h ζ h n + 1 2 δ t β h n [

where

α h n = A h ε π n (57)

where

β h n = A h ζ π n − π h θ n (58)

and

θ n = 1 δ t ∫ t n t n + 1 ( t n + 1 − t ) 2 u t 3 d t (59)

This step is to derive an energy identity for our scheme (6.1).

‖ ε h n + 1 ‖ L 2 ( Ω ) 2 − ‖ ε h n ‖ L 2 ( Ω ) 2 + δ t | ε h n | β 2 + δ t | ζ h n | β 2 = ‖ ε h n + 1 − ζ h n ‖ L 2 ( Ω ) 2 + δ t ( α h n , ε h n ) L 2 ( Ω ) + δ t ( β h n , ζ h n ) L 2 ( Ω ) − Λ h n [

with

Λ h n = δ t ( Λ ε h n , ε h n ) L 2 ( Ω ) + δ t ( Λ ζ h n , ζ h n ) L 2 ( Ω ) (61)

Our aim is to bound the right terms in the energy identity (60).

We want now to establish a stability lemma for a polynomial of degree k = 1 (68). To get it, we need the next lemma.

Lemma 1. Let π h 0 denote the L^{2}-orthogonal projection onto ℙ 2 0 ( τ h ) . ℙ 2 0 ( τ h ) is spanned by piecewise constant functions on τ h .

Then, ∀ ( v h , w h ) ∈ V h × V h ,

( A h v h , w h − π h 0 w h ) ≤ C ⋆ ⋆ r β c 1 2 h − 1 2 ‖ v h ‖ u w b ‖ w h − π h 0 w h ‖ L 2 ( Ω ) (62)

where C ⋆ ⋆ is independent of h , δ t and of β .

Proof 2. This result is obtained using Cauchy-Schwarz inequality and equality (41).

This lemma is a preliminary stability bound.

Lemma 2. Assume u ∈ C 3 ( L 2 ( Ω ) ) ∩ C 0 ( H 1 ( Ω ) ) .

Assume the CFL condition:

δ t ≤ ϱ h β c for some positive real number ϱ . (63)

Thus,

‖ ε h n + 1 ‖ L 2 ( Ω ) 2 − ‖ ε h n ‖ L 2 ( Ω ) 2 + δ t 2 | ε h n | β 2 + δ t 2 | ζ h n | β 2 ≤ ‖ ε h n + 1 − ζ h n ‖ L 2 ( Ω ) 2 + C r 4 δ t ( E h n ) 2 (64)

where C is independent of h , δ t and β .

Proof 3. Using CFL condition, energy identity (60), inequalities (46) and (47), we get (64).

Lemma 3. (Stability lemma, k ≥ 2 ) Assume u ∈ C 3 ( L 2 ( Ω ) ) ∩ C 0 ( H 1 ( Ω ) ) .

Assume the ( 4 3 CFL Condition)

δ t ≤ ϱ ′ τ ⋆ − 1 3 ( h β c ) 4 3 for some positive real number ϱ ′ (65)

Then, we infer:

‖ ε h n + 1 ‖ L 2 ( Ω ) 2 − ‖ ε h n ‖ L 2 ( Ω ) 2 + δ t 2 | ε h n | β 2 + δ t 2 | ζ h n | β 2 ≤ C r 8 δ t ( E h n ) 2 (66)

Proof 4. The stability lemma (66) for a polynomial of degree k ≥ 2 is obtained by bounding the term ‖ ε h n + 1 − ζ h n ‖ L 2 ( Ω ) 2 in the energy identity (60).

Lemma 4. (Stability lemma, k = 1 ) Assume u ∈ C 3 ( L 2 ( Ω ) ) ∩ C 0 ( H 1 ( Ω ) ) .

Assume the CFL condition:

δ t ≤ ϱ h β c for some positive real number ϱ with ϱ ≤ min { 1 8 ( C ′ ) − 2 ; 1 2 ( C ⋆ C ⋆ ⋆ ) − 2 3 } (67)

Thus,

‖ ε h n + 1 ‖ L 2 ( Ω ) 2 − ‖ ε h n ‖ L 2 ( Ω ) 2 + δ t ( 1 2 − 3 8 r 2 ) | ε h n | β 2 + δ t ( 1 2 − 3 8 r 2 ) | ζ h n | β 2 ≤ C r 4 δ t ( E h n ) 2 (68)

where C is independent of h , δ t and β .

Proof 5. This lemma is proven as in [

Proof of theorem 1

‖ u N − u h N ‖ L 2 ( Ω ) = ‖ u N − π h u n + π h u n − u h N ‖ L 2 ( Ω ) ≤ ‖ u N − π h u n ‖ L 2 ( Ω ) + ‖ π h u n − u h N ‖ L 2 (Ω)

‖ u N − u h N ‖ L 2 ( Ω ) ≤ ‖ ε π N ‖ L 2 ( Ω ) + ‖ ε h N ‖ L 2 (Ω)

Using the triangle and Young inequalities, we deduce:

( ∑ m = 0 N − 1 δ t | u m − u h m | β 2 ) 1 2 ≲ ∑ m = 0 N − 1 δ t 1 2 ‖ ε π m ‖ ⋆ ⋆ + ∑ m = 0 N − 1 δ t 1 2 | ε h m | β

Thus, we obtain:

‖ u N − u h N ‖ L 2 ( Ω ) + ( ∑ m = 0 N − 1 δ t | u m − u h m | β 2 ) 1 2 ≲ ‖ ε π N ‖ L 2 ( Ω ) + ‖ ε h N ‖ L 2 ( Ω ) + ∑ m = 0 N − 1 δ t 1 2 ‖ ε π m ‖ ⋆ ⋆ + ∑ m = 0 N − 1 δ t 1 2 | ε h m | β (69)

Let n ∈ { 0, ⋯ , N }

Set b n = δ t 2 | ε h n | β 2 + δ t 2 | ζ h n | β 2

Given ( a + b + c + d ) 2 ≤ 4 a 2 + 4 b 2 + 4 c 2 + 4 d 2

From the relation (66), we deduce that:

‖ ε h n + 1 ‖ L 2 ( Ω ) 2 + b n ≤ ‖ ε h n ‖ L 2 ( Ω ) 2 + C r 8 δ t ( 4 ‖ ε π n ‖ ⋆ ⋆ 2 + 4 ‖ ζ π n ‖ ⋆ ⋆ 2 + 4 τ ⋆ ‖ d t 3 u ‖ C 0 ( L 2 ( Ω ) ) 2 δ t 4 + 4 δ t τ ⋆ − 1 ‖ ε h n ‖ L 2 ( Ω ) 2 )

Set d n = 4 C r 8 δ t ( ‖ ε π n ‖ ⋆ ⋆ 2 + ‖ ζ π n ‖ ⋆ ⋆ 2 + τ ⋆ ‖ d t 3 u ‖ C 0 ( L 2 ( Ω ) ) 2 δ t 4 ) (70)

whence, we have:

‖ ε h n + 1 ‖ L 2 ( Ω ) 2 ≤ ( 1 + 4 C r 8 δ t τ ⋆ − 1 ) ‖ ε h n ‖ L 2 ( Ω ) 2 + d n − b n (71)

Applying the Gronwall lemma, we get for n = N − 1 :

‖ ε h N ‖ L 2 ( Ω ) 2 ≤ exp ( 4 C r 8 τ ⋆ ( t f − t 0 ) ) ‖ ε h 0 ‖ L 2 ( Ω ) 2 + ∑ i = 0 N − 1 exp ( 4 C r 8 τ ⋆ ( t f − t i + 1 ) ) ( d i − b i ) (72)

So

‖ ε h N ‖ L 2 ( Ω ) 2 ≤ ∑ i = 0 N − 1 exp ( 4 C r 8 τ ⋆ ( t f − t i + 1 ) ) ( d i − b i ) for u h 0 = π h u 0 ⇒ ε h 0 = 0 (73)

So

‖ ε h N ‖ L 2 ( Ω ) + ( ∑ i = 0 N − 1 δ t 1 2 | ε h i | β ) ≲ ( ∑ i = 0 N − 1 exp ( 4 C r 8 τ ⋆ t f ) d i ) 1 2 (74)

where

d i = 4 C r 8 δ t ( ‖ ε π i ‖ ⋆ ⋆ 2 + ‖ ζ π i ‖ ⋆ ⋆ 2 + τ ⋆ ‖ d t 3 u ‖ C 0 ( L 2 ( Ω ) ) 2 δ t 4 )

Therefore, we have:

( ∑ i = 0 N − 1 δ t 1 2 ‖ ε π i ‖ ⋆ ⋆ ) ≲ ( ∑ i = 0 N − 1 exp ( 4 C r 8 τ ⋆ t f ) d i ) 1 2 (75)

From inequalities (69), (74) and (75), we get:

‖ u N − u h N ‖ L 2 ( Ω ) + ( ∑ m = 0 N − 1 δ t | u m − u h m | β 2 ) 1 2 ≲ ‖ ε π N ‖ L 2 ( Ω ) + exp ( 4 C r 8 τ ⋆ t f ) ( ∑ i = 0 N − 1 d i ) 1 2 (76)

From inequalities (48), (49) and CFL condition, we obtain:

( ∑ i = 0 N − 1 d i ) 1 2 ≲ t f 1 2 r 5 β c 1 2 h k + 1 2 ‖ u m ‖ H k + 1 ( Ω ) + t f 1 2 r 5 β c − 1 2 h k + 1 2 ‖ d t u m ‖ H k ( Ω ) + r 4 t f 1 2 τ ⋆ 1 2 ‖ d t 3 u ‖ C 0 ( L 2 ( Ω ) ) δ t 2

Using inverse inequality (24),

‖ ε π N ‖ L 2 ( Ω ) 2 ≲ h 2 k + 2 ‖ u m ‖ H k + 1 ( Ω ) 2

So

‖ ε π N ‖ L 2 ( Ω ) ≲ h k + 1 ‖ u ‖ C 0 ( H k + 1 (Ω) )

Therefore, inequality (76) becomes:

‖ u N − u h N ‖ L 2 ( Ω ) + ( ∑ m = 0 N − 1 δ t | u m − u h m | β 2 ) 1 2 ≲ h k + 1 ‖ u ‖ C 0 ( H k + 1 ( Ω ) ) + exp ( 4 C r 8 τ ⋆ t f ) ( t f 1 2 r 5 β c 1 2 h k + 1 2 ‖ u ‖ C 0 ( H k + 1 ( Ω ) ) + t f 1 2 r 5 β c − 1 2 h k + 1 2 ‖ d t u ‖ C 0 ( H k ( Ω ) ) ) + exp ( 4 C r 8 τ ⋆ t f ) r 4 t f 1 2 τ ⋆ 1 2 ‖ d t 3 u ‖ C 0 ( L 2 ( Ω ) ) δ t 2 (77)

Set

χ 1 = r 4 t f 1 2 τ ⋆ 1 2 ‖ d t 3 u ‖ C 0 ( L 2 (Ω))

and

χ 2 = t f 1 2 r 5 β c 1 2 ‖ u ‖ C 0 ( H k + 1 ( Ω ) ) + t f 1 2 r 5 β c − 1 2 ‖ d t u ‖ C 0 ( H k ( Ω ) ) (78)

h k + 1 ‖ u ‖ C 0 ( H k + 1 ( Ω ) ) = h h k + 1 2 ‖ u ‖ C 0 ( H k + 1 ( Ω ) ) ≲ t f 1 2 β c 1 2 h k + 1 2 ‖ u ‖ C 0 ( H k + 1 ( Ω ) ) for h ≲ t f β c

h k + 1 ‖ u ‖ C 0 ( H k + 1 ( Ω ) ) ≲ χ 2 h k + 1 2

We obtain thus:

‖ u N − u h N ‖ L 2 ( Ω ) + ( ∑ m = 0 N − 1 δ t | u m − u h m | β 2 ) 1 2 ≲ exp ( 4 C r 8 τ ⋆ t f ) ( χ 1 δ t 2 + χ 2 h k + 1 2 )

Proof of theorem 2

Using inequality (68), we get inequality:

‖ ε h N ‖ L 2 ( Ω ) 2 + ∑ i = 0 N − 1 b i ≤ ∑ i = 0 N − 1 exp ( 8 C r 8 τ ⋆ t f ) d i (79)

with

b i = δ t ( 1 2 − 3 8 r 2 ) | ε h i | β 2 + δ t ( 1 2 − 3 8 r 2 ) | ζ h i | β 2

b i = δ t ( 4 − 3 r 2 2 ) | ε h i | β 2 + δ t ( 4 − 3 r 2 8 ) | ζ h i | β 2

( ‖ ε h N ‖ L 2 ( Ω ) + ∑ i = 0 N − 1 4 − 3 r 2 2 2 δ t 1 2 | ε h i | β + ∑ i = 0 N − 1 4 − 3 r 2 2 2 δ t 1 2 | ζ h i | β ) 2 ≲ ‖ ε h N ‖ L 2 ( Ω ) 2 + ∑ i = 0 N − 1 δ t 4 − 3 r 2 8 | ε h i | β 2 + ∑ i = 0 N − 1 δ t 4 − 3 r 2 8 | ζ h i | β 2

Thus

‖ ε h N ‖ L 2 ( Ω ) + ∑ i = 0 N − 1 4 − 3 r 2 2 2 δ t 1 2 | ε h i | β + ∑ i = 0 N − 1 4 − 3 r 2 2 2 δ t 1 2 | ζ h i | β ≲ ( ∑ i = 0 N − 1 exp ( 8 C r 8 τ ⋆ t f ) d i ) 1 2

So

‖ ε h N ‖ L 2 ( Ω ) + ∑ i = 0 N − 1 4 − 3 r 2 δ t 1 2 | ε h i | β ≲ ( ∑ i = 0 N − 1 exp ( 8 C r 8 τ ⋆ t f ) d i ) 1 2 (80)

whence

‖ ε h N ‖ L 2 ( Ω ) + ∑ i = 0 N − 1 δ t 1 2 | ε h i | β ≲ 1 4 − 3 r 2 ( ∑ i = 0 N − 1 exp ( 8 C r 8 τ ⋆ t f ) d i ) 1 2 for 4 − 3 r 2 ≤ 1 (81)

Therefore,

‖ u N − u h N ‖ L 2 ( Ω ) + ( ∑ m = 0 N − 1 δ t | u m − u h m | β 2 ) 1 2 ≲ h k + 1 ‖ u ‖ C 0 ( H k + 1 ( Ω ) ) + 1 4 − 3 r 2 exp ( 8 C r 8 τ ⋆ t f ) × ( t f 1 2 r 5 β c 1 2 h k + 1 2 ‖ u ‖ C 0 ( H k + 1 ( Ω ) ) + t f 1 2 r 5 β c − 1 2 h k + 1 2 ‖ d t u ‖ C 0 ( H k ( Ω ) ) ) + 1 4 − 3 r 2 exp ( 8 C r 8 τ ⋆ t f ) r 4 t f 1 2 τ ⋆ 1 2 ‖ d t 3 u ‖ C 0 ( L 2 ( Ω ) ) δ t 2 (82)

‖ u N − u h N ‖ L 2 ( Ω ) + ( ∑ m = 0 N − 1 δ t | u m − u h m | β 2 ) 1 2 ≲ 1 4 − 3 r 2 h k + 1 ‖ u ‖ C 0 ( H k + 1 ( Ω ) ) + 1 4 − 3 r 2 exp ( 8 C r 8 τ ⋆ t f ) × ( t f 1 2 r 5 β c 1 2 h k + 1 2 ‖ u ‖ C 0 ( H k + 1 ( Ω ) ) + t f 1 2 r 5 β c − 1 2 h k + 1 2 ‖ d t u ‖ C 0 ( H k ( Ω ) ) ) + 1 4 − 3 r 2 exp ( 8 C r 8 τ ⋆ t f ) r 4 t f 1 2 τ ⋆ 1 2 ‖ d t 3 u ‖ C 0 ( L 2 ( Ω ) ) δ t 2 for 1 ≤ 1 4 − 3 r 2 (83)

‖ u N − u h N ‖ L 2 ( Ω ) + ( ∑ m = 0 N − 1 δ t | u m − u h m | β 2 ) 1 2 ≲ 1 4 − 3 r 2 exp ( 8 C r 8 τ ⋆ t f ) ( χ 1 δ t 2 + χ 2 h k + 1 2 ) (84)

Remark 1. When F is the identity mapping, K = Q so λ K = λ Q = 1 [

The isogeometric method has been used to establish an error estimate for transport equation in 2D using the explicit two stage Heun scheme, for smooth solutions, in the energy norm comprising the L^{2}-norm and the jumps. These results generalize Ern results. An extension of this present paper is to tackle Burgers equation to get an isogeometric error estimate.

This article has been written in the framework of my Phd works, supported by CEA-SMA (Centre d’Excellence Africain en Sciences Mathématiques et Applications), funded by World Bank.

The authors declare that there is no conflict of interest as far as the publication of this paper is concerned.

Goudjo, A. and Aguemon, U. (2019) An Isogeometric Error Estimate for Transport Equation in 2D. Advances in Pure Mathematics, 9, 777-793. https://doi.org/10.4236/apm.2019.99037