Arbitrary Lagrangian Eulerian approximation with

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key words: Arbitrary Lagrangian Eulerian, Finite Element, Navier-Stokes. 1. ... where ρ > 0 is the mass density, µ > 0 is the dynamic viscosity, f = (f1,f2) are the applied ... Let v be the velocity of the fluid in the Eulerian coordinates. ..... interaction problems, Analele Universitatii din Bucuresti Matematica 1998; 47(2):177–186. 6.
COMMUNICATIONS IN NUMERICAL METHODS IN ENGINEERING Commun. Numer. Meth. Engng 2000; 00:1–11 Prepared using cnmauth.cls [Version: 2002/09/18 v1.02]

Arbitrary Lagrangian Eulerian approximation with uniform size meshes for Navier-Stokes equations C.M. Murea∗ Laboratoire de Math´ ematiques, Informatique et Applications, Universit´ e de Haute-Alsace, 4, rue des Fr` eres Lumi` ere, 68093 MULHOUSE Cedex, France

SUMMARY An algorithm which permits to employ meshes with uniform size in the Arbitrary Lagrangian Eulerian (ALE) framework is introduced. A key is to keep inside the integral sign the approximation of the ALE time derivative. We give the conditions when two time advancing algorithms based on the backward Euler scheme provide identical approximations. Numerical results are presented for Navier-Stokes equations in a moving domain. Contrary to the classical ALE framework, for a domain which grows strikingly, when the time step decreases, so do the errors. In addition, the uniform size mesh technique c 2000 John Wiley & Sons, Ltd. provides more accurate results. Copyright key words:

Arbitrary Lagrangian Eulerian, Finite Element, Navier-Stokes

1. INTRODUCTION A successful method for solving numerically partial differential equations in moving domain is the Arbitrary Lagrangian Eulerian (ALE) framework [1], [2]. Most commonly, at each time step, a mesh of the moving domain is obtained as the image by mapping of a fixed mesh of a reference domain. Different strategies have been proposed to get a mesh of the moving domain by replacing the interior vertices of a fixed mesh in order to accommodate the displacement of the boundary. In [3], the displacement of an interior vertex is computed iteratively making a mean of the displacements of the neighboring vertices. The relocation of an interior vertex is given by making a weight mean of the displacements of the boundary vertices in [4, p. 90]. A method based on a relaxation technique to solve an unconstrained optimization problem is presented in [5]. In [6], the vertices are moved by solving an elasticity problem. We emphasize that all generated meshes will have the same number of nodes as the initial mesh and the inter-vertex connections stay unchanged. In [7], a method to relocate the nodes of the reference mesh is

∗ Correspondence

to: C.M. Murea, Laboratoire de Math´ematiques, Informatique et Applications, Universit´e de Haute-Alsace, 4, rue des Fr`eres Lumi`ere, 68093 MULHOUSE Cedex, France, e-mail: [email protected], http://www.edp.lmia.uha.fr/murea/, Phone: +33 3.89.33.60.34

c 2000 John Wiley & Sons, Ltd. Copyright

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C. M. MUREA

proposed. In addition, a strategy to optimize the triangulation by changing the inter-vertex connections is studied, but the number of vertices is still constant. In the case when the domain grows strikingly, the image by mapping of a fixed mesh will be distortioned and consequently the accuracy will be affected. The aim of this paper is to introduce an algorithm that permits to retriangulate the computing domain at each time step in order to get meshes of uniform size. We regenerate a mesh for the current time domain independently of the mesh used previously, so the meshes have not the same number of vertices during the numerical simulation. Contrary to the so called conservative scheme [8], we keep the ALE time derivative inside the integral sign. Numerical results presented in this paper for Navier-Stokes equations, show that, for domain which grows effectively, when the time step decreases, so do the errors, which is not the case for the classical ALE formulation. In addition, the uniform size mesh technique provides more accurate results.

2. NAVIER-STOKES EQUATIONS IN ARBITRARY LAGRANGIAN EULERIAN FRAMEWORK Let Ωt be a two-dimensional domain depending on time parameter t ∈ [0, T ]. We assume that the evolution in time of the domain is known. We denote by ΣD t a part of the boundary ∂Ωt D and we set ΣN = ∂Ω \ Σ . t t t The Navier-Stokes equations in moving domain are: find the velocity v (·, t) : Ωt → R2 and the pressure p (·, t) : Ωt → R, such that   ∂v ρ + (v · ∇)v − ∇ · σ = f , ∀t ∈ (0, T ), ∀x ∈ Ωt (1) ∂t ∇ · v = 0, ∀t ∈ (0, T ), ∀x ∈ Ωt (2) v = g, ∀t ∈ (0, T ), ∀x ∈ ΣD (3) t σn = h, ∀t ∈ (0, T ), ∀x ∈ ΣN t v (x, 0) = v0 (x) , in Ω0

(4) (5)

where ρ > 0 is the mass density, µ > 0 is the dynamic viscosity, f = (f1 , f2 ) are the applied volume forces, g is the prescribed boundary velocity, h is the prescribed boundary stress,  1 σ = −pI2 + 2µ (v) is the stress tensor,  (v) = 2 ∇v + (∇v)T is the strain rate tensor, v0 is the initial fluid velocity. b a reference fixed domain. Let At , t ∈ [0, T ] be a family of transformations We denote by  Ω  b = Ωt , where x b represent the ALE coordinates and b = (b such that At Ω x1 , x b2 ) ∈ Ω

x = (x1 , x2 ) = At (b x) the Eulerian coordinates. We assume that both At and its inverse 1 A−1 are of class C . In addition, we suppose that the derivative by rapport of time t exists t ∂At b In the sequel, J bAt stands for the jacobian matrix of the ALE map (b x ) for all x ∈ Ω. ∂t b → At (b x x), defined by ∂At (b x) bAt (b J x) = ∂b x   bAt (b and JbAt (b x) = det J x) denotes its determinant. c 2000 John Wiley & Sons, Ltd. Copyright Prepared using cnmauth.cls

Commun. Numer. Meth. Engng 2000; 00:1–11

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ALE WITH UNIFORM SIZE MESHES

Let v be the velocity of the fluid in the Eulerian coordinates. The corresponding function b × [0, T ] → R2 is defined by v b : Ω b (b in the ALE framework v x, t) = v (At (b x) , t) = v (x, t). (x, t) = ∂ v (b We denote the ALE time derivative by ∂v x , t) and the domain velocity by ∂t x ∂t t (b x ). ϑ(x, t) = ∂A ∂t The Navier-Stokes equations in the ALE framework (see [9]) give: find the fluid velocity v and the fluid pressure p such that   ∂v ρ + ((v − ϑ) · ∇) v − 2µ∇ ·  (v) + ∇p = f , ∀t ∈ (0, T ), ∀x ∈ Ωt , ∂t x

∇·v

= 0,

∀t ∈ (0, T ), ∀x ∈ Ωt .

Multiplying the above equations by w and q respectively and using the Green formula, we can write the weak form of the Navier-Stokes equations: find the velocity v = g on Σ D t and the pressure p such that Z Z ∂v ρ (((v − ϑ) · ∇) v) · w + a (v, w) + b (w, p) = ` (w) (6) ρ ·w+ ∂t x Ωt Ωt b (v, q) = 0 (7) for all w = 0 on ΣD t and for all q. We have used the notations Z 2µ  (v) :  (w) , a (v, w) = Ωt Z b (w, q) = − (∇ · w) q, Ωt Z Z ` (w) = f ·w+ h · w. Ωt

ΣN t

Remark 1. Following [10], the system (6)–(7) is called the non-conservative weak formulation. Proposition 1. If ∂w ∂t x = 0, then Z Z Z d ∂v · w = v · w − v · w (∇ · ϑ) . (8) dt Ωt Ωt ∂t x Ωt Proof. We recall the Euler expansion formula (see [9, Lemma 7.1, p. 23]), which gives

∂ JbAt = JbAt (∇ · ϑ) . ∂t Using the change of variables, the chain rule and the Euler expansion formula, we have Z Z Z Z Z b b ∂w ∂ JbAt ∂b v d d b·w b JbAt = b· b·w b b JbAt + v v v ·w JA t + v·w = dt Z Ωt dt Ω ∂t ∂t Ω Z∂t Ω Ω Z b ∂ w ∂b v b JbAt + b· b·w b (∇ · ϑ) JbAt = ·w v JbA + v ∂t t Z Ω ZΩ ∂t Z Ω ∂v ∂w v· v · w (∇ · ϑ) . ·w+ + = ∂t x Ωt ∂t x Ωt Ωt Using the hypothesis ∂w 2 ∂t x = 0, we get the desired conclusion.

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Remark 2. If we replace in (6) the time derivative by the right hand side of (8), we get an = 0. In order to obtain the last equality, we equivalent weak form under the hypotheses ∂w ∂t x b (b b does not depend on t. can take w (At (b x) , t) = w x), where w 3. TIME ADVANCING SCHEME Let N ∈ N∗ be the number of time steps and ∆t = T /N the time step. We set tn = n∆t for n = 0, 1, . . . , N . We will indicate vn+1 (x), pn+1 (x) the approximations of v(x, tn+1 ), p(x, tn+1 ) b = A−1 for x ∈ ΩF x) and therefore x tn+1 . We set x = Atn+1 (b tn+1 (x). To approximate the time derivative, we can use the first order finite difference scheme )−v(x,tn ) ∂v x, tn+1 ) ≈ v(x,tn+1∆t ∂t (b (x),tn v(x,tn+1 )−v Atn ◦A−1 t

=

n+1

= ≈



∆t

We can observe that the derivative v

n+1

v(Atn+1 (x),tn+1 )−v(Atn (x),tn ) ∆t (x) vn+1 (x)−vn Atn ◦A−1 t n+1

∆t

∂v x, tn+1 ) which depends ∂t (b (x)−vn Atn ◦A−1 (x) t



.

b can be on the ALE coordinates x

n+1 approached by the expression written using the Eulerian coordinates ∆t x. The time-advancing scheme is: knowing the velocity v n : Ωtn → R2 at the previous time step, the current domain Ωtn+1 and the domain velocity ϑn+1 , find the velocity vn+1 : Ωtn+1 → R2 n+1 verifying vn+1 = gn+1 on ΣD : Ωtn+1 → R, such that tn+1 and the pressure p   n+1 Z Z    v − Vn ·w+ ρ Vn − ϑn+1 · ∇ vn+1 · w ρ ∆t Ωt Ωtn+1 Z Z n+1  + 2µ  vn+1 :  (w) − (∇ · w) pn+1 

Ωtn+1

= −

Z

Z

f n+1 · w + Ωtn+1

Ωtn+1

Z

Ωtn+1

hn+1 · w, ΣN t

∀w = 0 on ΣD tn+1 ,

(9)

n+1

 ∇ · vn+1 q = 0,

∀q,

(10)

  where Vn (x) = vn Atn ◦ A−1 tn+1 (x) .

Remark 3. This algorithm is based on the backward Euler scheme in order to approach the time derivative and on a semi-implicit treatment of the convective term. We will see that if we apply the backward Euler scheme to the right hand side of (8), we will obtain a different algorithm, in general. In practice, the domain velocity can be approached by the first order finite difference scheme x − Atn ◦ A−1 x) x) − Atn (b Atn+1 (b tn+1 (x) = . ∆t ∆t b → R2 be arbitrary fields. b :Ω Proposition 2. Let vn+1 : Ωtn+1 → R2 , vn : Ωtn → R2 and w 2 2 We set w(·, tn+1 ) : Ωtn+1 → R and w(·, tn ) : Ωtn → R as following  b (b w Atn+1 (b x) , tn+1 = w (Atn (b x) , tn ) = w x) . (11) ϑn+1 (x) =

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Commun. Numer. Meth. Engng 2000; 00:1–11

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Then Z 6=

  vn+1 (x) − vn Atn ◦ A−1 tn+1 (x) ∆t

Ωtn+1

1 ∆t Z −

Z

vn+1 (x) · w(x, tn+1 ) − Ωtn+1

Ωtn+1

1 ∆t

Z

5

· w(x, tn+1 ) vn (y) · w(y, tn )

Ω tn

 vn+1 (x) · w(x, tn+1 ) ∇ · ϑn+1 (x)

(12)

 excepting the case when JbAtn+1 = JbAtn and ∇ · ϑn+1 = 0.

Proof. We denote by E the difference between the left and right hand sides of the expression (12). We obtain that Z Z   1 1 (x) · w(x, t ) + vn Atn ◦ A−1 vn (y) · w(y, tn ) E = − n+1 tn+1 ∆t Ωtn+1 ∆t Ωtn Z  + vn+1 (x) · w(x, tn+1 ) ∇ · ϑn+1 (x). Ωtn+1

Using the change of variable for the first two integrals, it follows Z 1 vn (Atn (b x)) · w(Atn+1 (b x) , tn+1 )JbAtn+1 E = − ∆t Ω Z 1 vn (Atn (b x)) · w(Atn (b x)) , tn )JbAtn + ∆t Ω Z  + vn+1 (x) · w(x, tn+1 ) ∇ · ϑn+1 (x). Ωtn+1

From the relation (11), we get E

=

1 ∆t Z +

Z



  b x) −JbAtn+1 + JbAtn x)) · w(b vn (Atn (b

Ωtn+1

 vn+1 (x) · w(x, tn+1 ) ∇ · ϑn+1 (x).

b are arbitrary, Since vn+1 , vn and w we conclude that E vanishes only in the case when  2 JbAtn+1 = JbAtn and ∇ · ϑn+1 = 0.

Remark 4. Left and right hand parts of relation (12) represent backward Euler scheme applied to the left and right hand parts of equality (8), respectively. As a consequence, the algorithm presented in this paper and that one obtained discretising the right hand part of equality (8) produce two different approximations of the Navier-Stokes equations, in general. But, if the volume of the domain Ωt is constant in time, then JbAt is constant also and according to the Euler expansion formula, we obtain that ∇·ϑ = 0. In this case, the both algorithms produce the same approximations. If an incompressible, homogeneous fluid is limited by an impermeable boundary, then the volume of the domain is constant in time. c 2000 John Wiley & Sons, Ltd. Copyright Prepared using cnmauth.cls

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4. FINITE ELEMENT APPROXIMATION WITH UNIFORM SIZE MESHES In the commonly used ALE framework, the triangulation T n+1 of the computational domain b by the ALE Ωtn+1 is obtained as the image of a fixed mesh Tb of the reference domain Ω n+1 mapping. But, if the domain grows strikingly, the size of the mesh T increase drastically and consequently, the approximation looses in accuracy. This inconvenient disappears when we employs the algorithm (9)–(10). Firstly, let us remark that the time derivative approximation appears inside the integral sign in the first term of equation (9). In the classical approach, the time derivative is out of the integral sign and the shape functions have to verify the constraint ∂w ∂t x = 0. If all the meshes are obtained from a fixed mesh of the reference domain through the ALE map, this condition holds. Secondly, the algorithm (9)–(10) is defined independently of the finite element triangulation of Ωtn+1 . Let T n+1 and T n be independent triangulations of Ωtn+1 and Ωtn , respectively. We emphasize the fact that the two triangulations have not necessary the same number of vertices or triangles. The problem is how to define the discrete ALE  map and how  to −1 n n A ◦ A (x) and compute Atn ◦ A−1 (x) which appears in the definition of V (x) = v t n tn+1 tn+1 x−Atn ◦A−1 t

(x)

n+1 ϑn+1 (x) = . ∆t b be a reference domain chosen by the user. We denote by dn the P1 finite element Let Ω hn b verifying approximation of dn : Ωtn → Ω

∆dn d

n

= 0, in Ωtn

= gn , on ∂Ωtn

b The function dn is continuous on Ωtn where gn is the displacement which moves ∂Ωtn to ∂ Ω. hn and linear on each triangle of T n . In a similar way, we can construct dn+1 hn+1 piecewise linear n+1 b approximation of d : Ωtn+1 → Ω. bi the image of Ai by the map dn . Let Tb n be We denote by Ai the vertices of T n and by A hn bi of the domain Ω. b Analogous, from T n+1 using dn+1 , we get the triangulation of vertices A hn+1

b We emphasize that the two triangulations Tb n and Tb n+1 of Tb n+1 another triangulation of Ω. b have not the same number of vertices. Ω bi of Tb n , we define For each vertex A −1   bi − dnh (Ai ) , bi = A A dnh n

n

then we can define dnhn

dn+1 hn+1 (x)

x ∈ Ωtn+1 , then defined. Finally, we set

−1

: Tb n → T n by linear interpolation on each triangle of Tb n . If  b and as a consequence, the expression dn −1 ◦ dn+1 (x) is well ∈Ω hn hn+1 n Atn ◦ A−1 tn+1 (x) = dhn

−1

◦ dn+1 hn+1 (x).

 n −1 bn In general, dn+1 , we hn+1 (x) is not a vertex of T , so in order to compute its image by dhn n b need to interpolate on the mesh T . In practice, in order to simplify the computation, we can b = Ωtn and we have Tb n = T n . take Ω c 2000 John Wiley & Sons, Ltd. Copyright Prepared using cnmauth.cls

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In this way, we can triangulate the domain Ωtn+1 independently of the mesh used at the previous time step T n . We can use that in order to have meshes with similar size. Some triangulation generator provide mesh with size controlled by the length of the segments on the boundary of the domain. If we want a mesh size of about h, we will approach the boundary of length of ∂Ω the domain by a polygon of about segments of equal lengths. h 5. NUMERICAL BENCHMARK 5.1. All the meshes are obtained from a fixed mesh of the reference domain through the ALE map In order to test the time accuracy of the algorithm presented before, we will perform the b = [0, 1] × [0, L], where benchmark proposed in [10, pp. 92-94]. The reference domain is Ω N D b b b b N the boundaries of the L = 6 for instant. We denote by Σ = [0, 1] × {L} and Σ = ∂ Ω \ Σ reference domain. The ALE map is given by      2πt At (b x1 , x b2 ) = x b1 , 1 − 0.4 sin (b x2 − 0.5) + 0.5 . 10  Remark 5. The jacobian of this particular ALE map is JbAt = 1 − 0.4 sin 2πt 10 , therefore the algorithm presented in this paper and the conservative one introduced in [10] produce different approximations. For the physical parameters ρ = 1, µ = 1, f = (f1 , f2 ) = (0, 0), the exact solution is   2  2V (x1 − L) 2V (x1 − L) 2V (x2 − 0.5) , , p(x1 , x2 , t) = − v(x1 , x2 , t) = − (1 + 2V t) (1 + 2V t) (1 + 2V t) where V = 0.2. On the top, left and bottom boundary, we have imposed the velocity profile, while on the right boundary, we have imposed external forces   4V h = (h1 , h2 ) = − ,0 . (1 + 2V t) We have performed the simulation in the time interval [0, T = 10], for the time steps ∆t = 1/2, 1/4, 1/8, 1/16, 1/32 as in [10], where the time accuracy of a conservative algorithm was tested. The reference mesh (see Figure 1) is of 696 vertices and 1250 triangles, it is unstructured and its size is h = 0.152882. The P1 +bubble finite element was employed for the approximation of the fluid velocity, while the fluid pressure was approached by the P1 finite element. The computations have been produced employing the software FreeFem++ [11]. All the meshes are images of the mesh plotted in Figure 1 through the ALE map. Consequently, all the meshes have the same number of vertices, triangles and the same connections. During the simulation, the mesh size varies from hmin = 0.15 to hmax = 0.212897. Videoanimations from the numerical simulation results can be found on the web site of the author http://www.edp.lmia.uha.fr/murea/simulation.html qP  N n 2 The norm of L2 0, T ; L2 (Ωt ) was approached by n=1 ∆t kp kL2 (Ωtn ) . Figures 2 and 3 show good concordance between the exact and the computed solution. c 2000 John Wiley & Sons, Ltd. Copyright Prepared using cnmauth.cls

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Figure 1. The reference mesh 7e-06

5.55e-10

6.9e-06

5.5e-10

6.8e-06 5.45e-10 6.7e-06 L^2 error in velocity

5.4e-10 6.6e-06 5.35e-10 6.5e-06 5.3e-10

6.4e-06

5.25e-10

6.3e-06 6.2e-06

5.2e-10

6.1e-06

5.15e-10

6e-06 0

0.1

0.2

0.3

0.4

0.5

5.1e-10 0

time step

0.1

0.2

0.3

0.4

0.5

Figure 2. The errors in the norm L2 (Ωt ) between the computed and the exact velocity in function of the time step at the time instant t=2 (left). The global errors in the norm L2 0, T ; L2 (Ωt ) in function of the time step (right) 

0.2

0.6

0.18 0.5

0.16

L^2 error in pression

0.14 0.4 0.12 0.1

0.3

0.08 0.2

0.06 0.04

0.1 0.02 0 0

0.1

0.2

0.3

0.4

0.5

time step

0 0

0.1

0.2

0.3

0.4

0.5

Figure 3. The errors in the norm L2 (Ωt ) between the computed and the exact pressure in function of the time step at the time instant t=2 (left). The global errors in the norm L2 0, T ; L2 (Ωt ) in function of the time step (right) 

In applications like fluid-structure interaction, it is required to know the forces from the fluid acting on the wall, given by −σn, where σ = −pI2 + 2µ (v). Figure 4 shows that the differences are small between the computed and the exact applied stresses on the left boundary. Theoretically, the algorithm is of first order accurate in time. General works on NavierStokes approximations are [12], [13], [14]. In the case of a convex and fixed domain Ω, for the P1 + bubble/P1 finite elements, we have (see [12], for example) kv(n∆t) − vhn kL2 (Ω) ≤ O(∆t) + O(h2 ), c 2000 John Wiley & Sons, Ltd. Copyright Prepared using cnmauth.cls

kp(n∆t) − pnh kL2 (Ω) ≤ O(∆t) + O(h). Commun. Numer. Meth. Engng 2000; 00:1–11

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ALE WITH UNIFORM SIZE MESHES

0.00026

6e-07

0.00024 L^2 error in forces on the left boundary

5.5e-07

0.00022 5e-07 0.0002

4.5e-07 0.00018

4e-07

0.00016

0.00014 0

0.1

0.2

0.3

0.4

0.5

3.5e-07 0

time step

0.1

0.2

0.3

0.4

0.5

Figure 4. The errors in the norm L2 (Ωt ) between the computed and the exact forces exerced on the left boundary in function of the time step at the time instant t=2 (left). The global errors in the norm L2 0, T ; L2 (Ωt ) in function of the time step (right) 

Consequently, when the time step decreases and the mesh size is constant, the errors diminish. In the right plot of Figure 2, we observe that the error in velocity is not a monotone function. The reason is that the size of the meshes of the computational domain Ωn is not constant in time. We will see in the next section that the loose in accuracy is more important when the domain grows strikingly. 5.2. Approximation using uniform size meshes In order to prove the importance of using uniform size meshes, we will take a case when the deformation of the domain is more important. We use the same benchmark as in the precedent section, but for domain length L = 0.6 and the time interval [2.5, 7.5]. In the first case, all the meshes are obtained through the ALE map from a fixed mesh of 57 vertices and 88 triangles. During the simulation, we obtain the mesh size: h = 0.159099 at t = 2.5, h = 0.25 at t = 5, h = 0.35 at t = 7.5 (see Figure 5).

Figure 5. Meshes obtained through the ALE map from a fixed mesh at time instants: t = 2.5 (left), t = 5 (middle), t = 7.5 (left)

In the second case, we retriangulate at each time step in order to get meshes with similar size c 2000 John Wiley & Sons, Ltd. Copyright Prepared using cnmauth.cls

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C. M. MUREA

h. We obtain at time instant t = 2.5 a mesh of 57 vertices, 88 triangles and size h = 0.159099, at t = 5 a mesh of 84 vertices, 134 triangles and size h = 0.15298, at t = 7.5 a mesh of 114 vertices, 186 triangles and size h = 0.159099 (see Figure 6). In the both cases, the meshes at the time instant t = 2.5 are identical.

Figure 6. Meshes obtained by retriangulation at time instants: t = 2.5 (left), t = 5 (middle), t = 7.5 (left)

-8

-2

-8.5 -3 -9 log10(L2 error in pressure)

log10(L2 error in velocity)

-4 -9.5

-10

-10.5

-11

-5

-6

-7 -11.5 -8 -12

-12.5 -1.6

-1.4

-1.2

-1 -0.8 log10(time step)

-0.6

-0.4

-0.2

-9 -1.6

-1.4

-1.2

-1 -0.8 log10(time step)

-0.6

-0.4

-0.2

-3.5

log10(L2 error in forces on the left boundary)

-4 -4.5 -5 -5.5 -6 -6.5 -7 -7.5 -8 -8.5 -1.6

-1.4

-1.2

-1 -0.8 log10(time step)

-0.6

-0.4

-0.2

Figure 7. The errors of velocity (top left), pressure (top right), forces exerced on the left boundary (bottom) in the norm L2 0, T ; L2 (Ωt ) in function of the time step using the log 10 scale. 

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Now, observe Figure 7. When uniform size mesh technique is used, the slope of the graphics is about 2 for the velocity and about 1.6 for the pressure. In the case when all the meshes are obtained through the ALE map from a fixed mesh, the errors increase when the time step decreases. This is a undesirable phenomena. Contrary, when we employ meshes with similar size, when the time step decreases, so do the errors. Moreover, the uniform size mesh technique provides more accurate results.

CONCLUSIONS We have presented an algorithm which permits to employ meshes with uniform size in a ALE framework. Though the computational time will be longer if we retriangulate the domain at each time step, the advantage is that the numerical results are more accurate.

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Commun. Numer. Meth. Engng 2000; 00:1–11