[PDF] Séance n 4 Eléments finis en dimension 1 et 2 Corrigé

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Elementsnisendimension1et2

Corrige

6Decembre2005

0

0 (1) 8 d dx(pdudx)+qu=fx2I; p(b)du dx(b)+u(b)=0; p(a)du dx(a)+u(a)=0: eectueuneintegrationparparties Z I pdu dxdvdxdx+Z I quvdxp(b)dudx(b)v(b)+p(a)dudx(a)v(a)=Z I fv: 1

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associeeauprobleme(1): (2)8 :Trouveru2H1(I)telque: a(u;v)=l(v);8v2H1(I); avec a(u;v)=Z I pdudxdvdxdx+Z I quvdx+u(b)v(b)+u(a)v(a); l(v)=Z I fvdx: lacontinuitedel(:)eta(:;:): jl(v)j=jZ I fvdxjkfkL2kvkL2; kfkL2kvkH1: ja(u;v)jpZ I jdu dxjjdvdxjdx+qZ I jujjvjdx+ju(b)jjv(b)j+ju(a)jjv(a)j

Enutilisantl'inegalitedeCauchy-Schwartz:

p Z I jdu dxjjdvdxj+qZ I jujjvjdxpkdudxkL2kdvdxkL2+qkukL2kvkL2; ju(b)jjv(b)j+ju(a)jjv(a)j2C2IkukH1kvkH1; onobtient: ja(u;v)jCkukH1kvkH1;C=p

2max(p;q)+2C2I:

Lacoercivitedeaestimmediate:

a(u;u)Z I pjdu dxj2dx+Z I qu2dxmin(p;q)kuk2 H1: obtientalors d dx(pdudx)+qu=fausensdesdistributions p(b)du dx(b)+u(b)=0: 2

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1.3-Onal'expressionsuivantedewj:

w j(x)=8 :x j1x xj1xjsix2[xj1;xj]; x j+1x xj+1xjsix2[xj;xj+1];

0sinon

combinaisonlineairedecoecientsjtelleque: n X j=0 jwj(x)=0: baseestlibre. pointsdistincts.Onendeduit: h+v(xi+1)xxih

Enutilisantl'identite

(3)v(x)=v(xi)+(xxi)v0(xi)+Z x x i(xt)v00(t)dt pourx=xi+1,onobtient hvj[xi;xi+1]=v(xi)+(xxi)v0(xi)+(xxi) hZ xi+1 x i(xi+1t)v00(t)dt jvhvj2= Z x x i(xt)v00(t)dt+(xxi) hZ xi+1 x i(xi+1t)v00(t)dt 2 Z x x i(xt)2dtZ x x ijv00(t)j2+(xxi)2 h2Z xi+1 x i(xi+1t)2dtZ xi+1 x ijv00(t)j2dt; 2

3h3kv00k2

L2[xi;xi+1]

3

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Onendeduitalors:

kvhvk2

L2(I)=X

iZ xi+1 x ijvhvj2dx; 2 3h4X ikv00k2

L2[xi;xi+1];

2

3h4kvk2

H2(I);

d'ou kvhvkL2(I)Ch2kvkH2(I) dhv dxj[xi;xi+1]=v(xi+1)v(xi)h=v0(xi)+1hZ xi+1 x i(xit)v00(t)dt; v

0(x)=v0(xi)+Z

x x iv00(t)dt:

Onaalorspourtoutx2[xi;xi+1]:

jvhvj2= Z x x iv00(t)dt1 hZ xi+1 x i(xi+1t)v00(t)dt 2 (xxi)Z x x ijv00(t)j2+1h2Z xi+1 x i(xi+1t)2dtZ xi+1 x ijv00(t)j2dt; 4

3hkv00k2

L2[xi;xi+1];

d'ou kv0dhv dxkL2(I)ChkvkH2(I):

1.5-Lasolutiondiscreteestdeniepar:

a(uh;v)=(f;v)8v2Vh;

Lasolutionuestdeniepar

a(u;v)=(f;v)8v2H1(I):

CommeVhH1(I),onfaitladierencepourv2Vh:

a(uhu;v)=08v2Vh ceciimplique a(huu;v)=a(huuh;v)8v2Vh: 4

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Onchoisitlafonctiontestv=huuh.Onaalors

a(huu;huuh)=a(huuh;huuh): khuuhk2

H1ja(huu;huuh)jCkhuukH1khuuhkH1

d'ou khuuhkH1CkhuukH1:

Al'aidedel'inegalitetriangulaire

kuuhkH1kuhukH1+khuuhkH1(C+1)khuukH1: kuuhkH1Ch:

Exercice2.ElementsnisP1endimension2

prouverd'abordque:

1(S1)=1;2(S1)=0;3(S1)=0:

S

1=2(S1)S23(S1)S3

11(S1)(2(S1);3(S1)6=(0;0))

S

Supposonsmaintenantque2(S1)6=0,onaalors:

2(S1)(S2S3)=0

demontredelam^ememaniereque:

1(S2)=0;2(S2)=1;3(S2)=0;

1(S3)=0;2(S3)=0;3(S3)=1:

Oncherche1delaforme

1(x;y)=(xx2)+(yy2)+

5

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1(S2)=0donne

laforme:

1(x;y)=C[(y2y3)(xx2)+(x3x2)(yy2)]

1(x;y)=(y2y3)(xx2)+(x3x2)(yy2)

(y2y3)(x1x2)+(x3x2)(y1y2) basedeP1.

2.2-soitvh2Vhmontronsquevh2H1(

).PardenitiondeVhonavhjTl2P18l= .Parconsequentvh estdansH1( )(cfTD2). carlarestrictiondeWI prendendeuxpointsdistincts. 1 MI

Fig.1{FonctiondebaseWI

6

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2.4-Lafamille(WI)1INestlibrecar

X I

IWI=0)X

I

IWI(MJ)=0)J=08J=1;N:

(WI)1INestgeneratricecar:

8vh2Vh;vh(M)=X

Iv h(MI)WI(M)

Eneet,enposantuh=vhP

Ivh(MI)WI,ona:

u h(MJ)=081JN: d'ouuh=0.

Exercice3.ElementsnisP2endimension1

3.1-Onnotex1

i=1=2(xi+1+xi).Onposewj w j(x)=8 :(xx1 j1)(xj1x) (xjx1 j1)(xj1xj)six2[xj1;xj]; (xx1 j)(xj+1x) (xjx1 j)(xj+1xj)six2[xj;xj+1];

0sinon

etw1j w

1j(x)=8

:(xxj)(xj+1x) (x1 jxj)(xj+1x1 j)six2[xj;xj+1];

0sinon

i)=0,w1j(x1 i)=ijetw1j(xi)=0. 7

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xjxj+1w jw1j x 1 jx1 j1xj1

Fig.2{Lesfonctiondebasewjetw1j

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