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Des outils pour ladaptationde maillage anisotrope en

dimension 2 et 3F. Hecht

Laboratoire Jacques-Louis Lions

Universite Pierre et Marie Curie

Paris, France

http://www.ann.jussieu.fr/~hecht mailto:hecht@ann.jussieu.fr

Cermics, Novembre, 2003 1

http://www.ann.jussieu.fr/~hechtmailto:hecht@ann.jussieu.fr

PLAN

Introduction

un schema dadaptation

Metrique et Maillage unite

Outil lie aux metriques

intersection, interpolation, gradation

Metrique versu indicateur derreur

Metrique pour element fini P1

Metrique pour element fini P2

Probleme de convergence (pas le temps)

http://www.ann.jussieu.fr/~hecht/freefem++.htm

Exemples / Demo of FreeFem++

Conclusion / Future (pas le temps)

Cermics, Novembre, 2003 2

Introduction : Le vrai probleme

Nous voulons resoudre une EDP, dans un domain avec la methode des

elements finis,

Le plus vite possible avec une precision donnee .

The theorie of error indicator given, the level of the error, but how to

compute the local mesh size to get a equidistribuate error.

Cermics, Novembre, 2003 3

Scheme of mesh adaption

i=0 ;

Soit Thi initial mesh

loop

compute ui the solution on mesh Thi

evaluate the level of error

si < 0 break

compute the new local mesh size

construct a mesh according to prescribe the mesh size.

Remarque : comment transforme un indicateur derreur local K en carte de

taille. Moralement, il faut estimer :

Khi

Cermics, Novembre, 2003 4

A main IDEA

The difficulty is to find a tradeoff between the error estimate and the mesh

generation, because this two work are strongly different.

To do that, we propose way based on a metric M and unit mesh w.r.t M The metric is a way to control the mesh size.

remark : The class of the mesh which can be created by the metric, is very

large.

Cermics, Novembre, 2003 5

Example of mesh adapted for 2 functions

f1 = (10 x3 + y3) + atan2(0.001, (sin(5 y) 2 x))

f2 = (10 y3 + x3) + atan2(0.01, (sin(5 x) 2 y)).

Enter ? for help Enter ? for help Enter ? for help

Cermics, Novembre, 2003 6

Metric / unit Mesh

In Euclidean geometry the length || of a curve of IRd parametrized by (t)t=0..1is

|| = 10

< (t), (t) >dt

We introduce the metric M(x) as a field of d d symmetric positive definitematrices, and the length ` of w.r.t Mis :

` = 10

< (t),M((t))(t) >dt

The key-idea is to construct a mesh where the lengths of the edges are

close to 1 accordingly to M.

Cermics, Novembre, 2003 7

Remark on the Metric

Let S be a surface , parametrized by

F (u) R3 with(u) IR2, and let (t) = F ((t)), t [0,1]

be a curve on the surface. The length of the curve is

|| = 10

< (t),(t) >dt

|| = 10

< (t), tFF(t) >dt

and on a parameteric surface the metric is

M = tFF

Cermics, Novembre, 2003 8

the Metric versus mesh size

at a point P ,

M =(

a bb c

)

M = R(

1 00 2

)R1

where R = (v1, v2) is the matrix construct with the 2 unit eigenvectors vi and

1, 2 the 2 eigenvalues.

The mesh size hi in direction vi is given by 1/

i

i =1

h2i

Cermics, Novembre, 2003 9

Remark on metric :

If the metric is independant of position, then geometry is euclidian. But the

circle in metric become ellipse in classical space.

Infact, the unit ball (ellipse) in a metric given the mesh size in all the

direction, because the size of the edge of mesh is close to 1 the metric.

If the metric is dependant of the position, then you can speak about

Riemmanian geometry, and in this case the sides of triangle are geodesics,

but the case of mesh generation, you want linear edge.

Cermics, Novembre, 2003 10

Metrix Tools

The unit ball B(M) in a metrix M plot the maximum mesh size on all the direction.If you two unknows u and v, we just compute the metrix Mv and Mu , find a metrix Muvcall intersection such that :

B(Muv) B(Mu) B(Mv)

Near the wall we inforce the metric such that :

Mw = (n t)(

f(y+)/h2wall 00 f(y+)R

)(n t)t

Where R is the radius of curvature, n, t the normal and tangent vector and f(y+) is asmooth decreasing function from 1 to 0.

Cermics, Novembre, 2003 11

Interpolation de Metric

Generalement, la Metrique, nest connue quaux sommets du maillage, il

faut donc interpole.

linterpolation direct des matrices (revient a dire que la fonction est P3 (on

genere trop de points).

literpolation en h revient a interpole avec

M(t) =((t)M

12

A + (1 t)M12B

)2car 1/h2 (trop lent)

une derniere solution est :

M(t) =((t)M1A + (1 t)M

1B

)1

Cermics, Novembre, 2003 12

Remark

The quality of the generated triangles, depend on the variation of the M inspace

The idea is to bound the variation of the M

H- variation

In the case of isotropic metric

M(P ) =1

h2PId

the length of a edge PQ of T

`(PQ) = |PQ| 10

1

h(t)dt

where h(t) = (1 t) hP + t hQ, linear interpolation

`(PQ) = |PQ|ln(hQ) ln(hP)

hQ hP the H-variation is

v(PQ) =hQ hP|PQ|

Cermics, Novembre, 2003 13

Definition

let be Max(v(PQ)) an all mesh edges PQ where v(PQ) =hQhP|PQ| )

let be the maximal ratio of the Euclidean length of all the two adjacent

and opposite edges in a new unit mesh

Result :

' ln()

Cermics, Novembre, 2003 14

Size correction w/r H-variation

In the isotropic case, the problem is to bound v(PQ) by a threshold value

thus simply modify hP and hQ accordingly :

hP = min(hP , hQ + |PQ|)

hQ = min(hQ, hP + |PQ|)

in anisotropic case

M(P ) = M(P ) M(Q)(1 + `P (PQ))2

M(Q) = M(Q) M(P )(1 + `Q(PQ))2 ,

where `P (PQ) (resp. `Q(PQ)) is the length of PQ w/r to M(P ) (resp.M(Q))

Cermics, Novembre, 2003 15

Metric computation with Hessien for P1 Lagrange finite element

The le Cea lemma say : the error is bound with the interpolation error. In a

classical Adaption way, the metric tensor M is constructed in order toequilibrate the error of interpolation.

Find a metric tensor M, such that, the adapted mesh constructed from Mminimizes the interpolation error :

||U h(U)||X

where h is the finite element interpolation operator considered

Cermics, Novembre, 2003 16

En dimension un et la solution u est connu

notons : u = inf[a,b] u et u = sup[a,b] u

.

1

2u(x a)(x b) (u1u)

1

2u(x a)(x b)

dou

(b a)2

8u u1u

(b a)2

8u

donc les erreurs dinterpolation sur [a, b] sont :

u1u||L (b a)2

8|u|

u1u||L1 (b a)3

12|u|

u1u||L2 (b a)

52

2

30|u|

Cermics, Novembre, 2003 17

En dimension Un Ladaptation revient a decouper le segment [x0, xN ] telle

que

Li = xi+1xi

|u| =

LN

soit independant de i.

La question naturelle est ba

|u| u1u|| Cp

ou p est un exposant a determiner en fonction de la norme et C une

contante independante de la longueur b a.

Cermics, Novembre, 2003 18

En dimension 2

With the P1 finite element the error interpolation is :

||uhu||T 1

2sup

x,yT( t ~ax |H(y)| ~ax )

where a is a vertex of triangle T

where |H| have the same eigen-vectors and the eigen-value of |H| is the absof the eigen-value of H,

We take M = 10 |H| and where 0 is the expected error.

Cermics, Novembre, 2003 19

Suite

||uhu||T 1

2sup

x,yT( t ~ax |H(y)| ~ax )

il y a 2 cas, le maximum est atteint en c a linterieur de T ou sur une arete

de T . Dans le second cas nous sommes ramene au cas 1D. Donc c considere

comme interne et on a donc : (uhu)xi (c) = 0, La formule de Taylor MacLaurin dit qil existe un point d ]a, c[ tel que

0 = (uhu)(a) = u(c) +1

2

(t ~ac

2(uhu)xixj

(d)

)~ac

mais

t ~ad

(2(uhu)

xixj(d)

)~ad sup

x,yT( t ~ax |H(y)| ~ax)

Cermics, Novembre, 2003 20

La solution uh est discrete

Nous voulons calculer H dune fonction P1 par morceaux. Approximation duHessien Hkh a chaque sommet k

Methode 1

Hkhij = 1

3|k|

k

iuhjwk

Remarques : cette approximation ne converge par si h 0, et le stencilest juste forme des triangles contenant k.

Methode 2 faire une double projection :

H = L2

(L2

(u

))

ou L2 est la projection L2 sur lespace des fonctions elements finis P1

Lagrange.

Remarques : cette approximation converge telle si h 0 ?. Le stencil estforme des triangles voisins des triangles contenant k.

Cermics, Novembre, 2003 21

Probleme ouvert

Pour que la methode adaptation proposer converge, juste pour le probleme

dinterpolation, (on calcul le Hessien a partir de linterpolation)

Question : construire un schema pour calculer une approximation du Hessien

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