Corrigé de l`examen (mars 2006) - CMAP
Transcription
Corrigé de l`examen (mars 2006) - CMAP
` !#"$% &')(*",+.-0/1&'2&3+)-0/5476981":)"<;=->@?A(B-?)>0",;CEDGFH8I8169/1>@"KJ &3>@>@/1L3MN:."G8POQ"SR%6K47",'2M,(*>T/U-:)?WV3X7Y6K>@;ZV9[3[\2C]V_^)",?.>@",;@J acbd.egf e=h.ijd.elknm biopif qrd%oHe=stvu w<x bGkegmyd)h z*{}|~~S] ,KKQ#K~j= 9B=Kg#BK=,KT V = {φ ∈ H 1(Ω) ]=%, φ = K {S.1* nKT *~TB T]*~K~K=KK BK=r=B]g~Sg~KQ 0 ]KB~SΓ.D¡}¢g$9T *~9B)K~K£1B~K= *~]g¤ φ ∈ V ¥ =~~S g¦* B~SH9TA9T]g ¥ =§=~¨] B~Sr=rgB~KK *~©B¨ª g={¨.n1*KT]*~TBT *~K~g g¤*~K«K]**= u ∈ V ]=%, Z h∇u · ∇φ dx = Z f φ dx ∀ φ ∈ V. ¬ B§Kl9~KQ]B~)T¦*]B~¦*=~gy*Kj1*~=]*~v*,®==]Qpl §1* T *~jTB T]*~K~K=¯r¡¢g$9T *~#¢=°@¯g*~ g =rgB±K~ g*~S ]B~$ B{3²³B~K ¥ *K©]B (h, v, q) ∈ U × V × V ¥ *~ ad Ω 1 L(h, v, q) = 2 Z 2 Ω |h∇v − σ0 | dx + Z h∇v · ∇q dx − Z f q dx. . 1B KT]*~NTBT *~K~gI=$9T]*~´B@®B~S]y=9BKl9~KQ]B~ K*~K~g9B Ω h gj$K.g¤=$KTBg~S}µ Z Ω Ω ∂L (h, u, p), φi = 0 ∀ φ ∈ V, ∂v (σ − σ0 ) · h∇φdx + ¬ B§=*~K =$Kg~SIl°T}T@®*~S Z h∇p · ∇φ dx = 0 ∀ φ ∈ V. =} *]*~K 9B~ Ω, −div (h∇p) = div (h(σ − σ0 )) K Γ ∂p = −h(σ − σ0 ) · n h ∂n N K p=0 ΓD . ¶ * ∂u = 0 K Γ *~<gKQ9g§Zg*~K *~NB¨ªN g§ N ,%K∂n=,=~S ΓN ∂p K Γ . = hσ · n h Ω Ω p∈V 0 ∂n z N ¨{§.§g Q*=£1*~Kl]*~*,®==]Q 9B 0 ¶ * hJ (h), ki = =)K*~K~g ¥ *K# * J 0 (h)k dx = h ¬ B§=*~K =$Kg~S ¥ *~ Z k ∈ L∞ (Ω) ¥ ∂L (h, u, p), ki. ∂h Ω y K = Q * g j 9T]g³KB§]BK*¤}µ σ = h∇u ∂L h (h, u, p), ki = ∂h Z J(h) (σ − σ0 ) · k∇u dx Ω Z h g k∇u · ∇p dx. Ω J 0 (h) = (σ − σ0 ) · ∇u + ∇u · ∇p { acbd.egf e=h.ijd.elknm q©knf qrd%oAe=stu w7` bGkegmyd)h z*{¯~ ~@®gl°B~S p(x) ≡ p(x ) l ,=¤ $] Tg}*~v*]g~S 3 ~u(x) ≡ u3 (x1 , x2 )~e3 9B~K¡¢g$9@]B~ p0 (x3 ) = ∆u3 (x1 , x2 ), g,$Ky$Kg=nK=¨ª l ª¨Kg *~ *~Sg*~°T~$ g={³¨] ==~S K ZKg *~ g¤ ~Kg~ x l ¥ $KQ yµZ~K*ZTg}]K= Q*g ¥ 3 ,]=¥ n@ª¨B³Bgp9 9B~K Ω, K ∂Ω. ¨{§% *Kg#¢*]]T *~K=,g~S}K*~K −∆u3 = 1 u3 = 0 Z 2 inf J(Ω) = |∇u3 | dx . Ω∈Uad Ω .1B KT]*~vBT @]B~K~KgBKZ,Q]g @ª¨Tn=« *¨*g ]g%$K H 1 (Ω) u3 ∈ 0 Z ∇u3 · ∇φ dx = Ω ¯~ 9T]gK= ¥ * ~K*B B=*T~Kgφ = u3 Z Z φ dx ∀ φ ∈ H01 (Ω). Ω *~ ]*B$KvN1*~K= *~ B,®gl]g¤µ 2 |∇u3 | dx = Ω Z u3 dx. Ω { g BKZKvg*~K *~_B¨ª_ gK²³ =jp5B~S]¨KyK~ KQ]gT]=KrK.B¦* B~K¦*j KK==~S°B 1B*§j=*K l{%n.B¦B]B~¨ ¦*=~=£ B§K§1*~K= *~*$®g= =Kg£Kg,ªg*~S] B~S g 5g$90 ]*~v#l°T§l}g*~K]*~vB¨ª g L(Ω, v, q, λ) = Z v dx + Ω Z (∆v + 1)q dx + Ω Z vλ dx, ∂Ω v, q, λ *~SK=1*~K= *~nK H 1(R2 ) 5=gTB BKg *~SK*~K=~ ~Kgg~K9T~$ g}K Ωl{ .y1B KT]*~ZTBT *~K~gI=$9T]*~B@®*~S]= h ∂L (Ω, u, p, µ), φi = 0 ∀ φ ∈ H 1 (R2 ), ∂v gj$K.g¤=$KTBg~S}µ Z φ dx + Ω Z ∆φp dx + Ω ¬ B§~$ g¦* T]*~9B©KB g© ,=~$ Z φ(1 + ∆p) dx + Ω Z ∂Ω ¬ B§=*~K =$Kg~SIl°T}T@®*~S ∂φ ∂n Z φµ dx = 0 ∀ φ. ∂Ω p + φ(µ − p ∈ H 1 (R2 ) ∂p ) ds = 0 ∀ φ. ∂n g}* *~ 9B~K Ω, K ∂Ω, g§$KKK$K£, p = u 5rK*K=±g£B¨]T B@®*~$l{ ¬ BTgKH KQ]gT]=KK)T¦*]B~¦** B*©gB~KK *~B¨ªv]=TB¨ −∆p = 1 p=0 µ= {§.g Q*=K1*= J 0 (Ω)(θ) = |~*¨]=~S§K*~K 0 J (Ω)(θ) = Z ∂Ω ∂p . ∂n ∂L (Ω, u, p, µ)(θ). ∂Ω ∂uµ u + (∆u + 1)p + Huµ + ∂n θ · n ds. B K==*~KKQ]*~KT¨ª ]=v²³K=*K TB=K©K µ *~=~gKKQ 0 J (Ω)(θ) = Z ∂Ω u l p =yK Z 2 ∂u ∂u µ θ · n ds = θ · n ds. ∂n ∂n ∂Ω ¯~~B°B~S rKQ]K T gK%K§.B¦B]B~K¦BrBKpg*~S] B~S r#B BK Ω ¥ `gB∈~KRK *~ *¨]B]}g¤}K*~K Z `+ ∂Ω 2 ! θ · n ds = 0 ∀θ ∈ W 1,∞ (R2 ; R2 ). 2 ∂u ∂n ¬ B§=*~K =$Kg~S ¥ y=*~KKQ]B~#B]ZB K$K, ∂u ∂n = −` K ∂Ω. ¨{ ¨ Ω g¤¯K~yK $K©K±]**~ R *~n=¯ TggAlª¨KgQ]==~$H³*¨]B~ Kj¡¢g$9T *~v#¢=]T§g~g,*K*~K~K=g}]BBg 1 u(r) = (R2 − r 2 ). 4 |~ Bg Q9G$K ∂u (R) = − R $Kg¤g~ =*~K]B~S K ∂Ω ¥ K*~KG ∂nBQ]³=}2= Kgn*K§yTBgK§KKQ]gT]=K± g*~K *~v#¢* .B¦*]T~K¦* ` = − R { { ³g*~S] B ¥ *K4.K~K£ ==]*~ gl°B~¦*KT Ω §g*~K]*~j#¢*]ZT ~K=9B=]y*g9=B{ ¯~7 l ¥ gBy B *~ u =g*~°T~$ 5~$K § K³n* ]Zg Q*=°B~K¦Bg~S]g=jTK p~$KB{ ¨H*~N ∂Ω ¥ KTg9B~K~ =*~ K Ω ¥ 9B=*~S]~$KQ]Kv<Kg*=vK u 5$#¢*~ K*gg¦*K= r*~g~<gKKQ³$KnKg*=n~KB BBK )g¤~$KB{ ¬ B±g*~Kg$K=~$ ¥ =*~KKQ]*~Z#¢* BQ]©l°BQ£Bg Q9g³*~vBK B ` = =jK*~yK~KygB~KK *~NB¨ª< g}K gKB~K~ ∂u = 0 KB *K 0¥ ¶ { K= ~Kg©*~S§=*~KKKQ}µyK~Kjg*~S ]BK= *~v T∂n©*~~g¨³ B* 2 ∂Ω Z dx = − Ω Z ∆u dx = − Ω Z ∂Ω ∂u ds = 0 ! ∂n
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