Universite de Nice Sophia Antipolis Master MathMods Finite Elements

Sophia - Antipolis Master

Description

Niveau: Supérieur, Master
Universite de Nice Sophia-Antipolis Master MathMods - Finite Elements - 2008/2009 Exercises - Chapter 1 - Chapter 2 (Correction) Exercise 1. (a) Let I =]0, l[, l ? R. Show that ?C(l) > 0 , ?u?C0(I¯) ≤ C(l) ?u?H1(I) , ?u ? D(I¯) . (1) Let u ? D(I¯). Let x , y ? I¯. The fondamental theorem of analysis gives u(x) = u(y) + ∫ x y u?(s) ds , which leads to |u(x)| ≤ |u(y)| + ∫ l 0 |u?(s)| ds , By means of Cauchy-Schwarz inequality, one gets |u(x)| ≤ |u(y)|+ √ l ( ∫ l 0 |u?(s)|2 ds ) 1 2 . Integrating the above inequality over [0, l] in the y variable, and applying once again the Cauchy-Schwarz inequality, give l |u(x)| ≤ √ l ( ∫ l 0 |u(y)|2 dy ) 1 2 + l √ l ( ∫ l 0 |u?(s)|2 ds ) 1 2 , or equivalently |u(x)| ≤ max ( 1/ √ l , √ l

• let


• then v?


• all ? ?


• variational problem


• lax-milgram theorem


• mapping ?


• norm ?


• cauchy- schwarz inequality


• pde

Sujets

• Master
• Weak formulation

Informations

• Publié par: profil-mupheg-2012
• Nombre de lectures: 18
• Langue: English

Extrait

Universit´edeNiceSophiaAntipolis Master MathMods  Finite Elements  2008/2009

Exercises  Chapter 1  Chapter 2 (Correction)

Exercise 1. (a) LetI=]0, l[,l∈R. Show that

∃C(l)>0,kuk ¯ ≤C(, C(I) l)kuk H(I) 0 1

¯ ∀u∈ D(I).

¯ ¯ Letu∈ D(I). Letx , y∈Ifondamental theorem of analysis gives. The Z x ′ u(x) =u(y) +u(s)ds , y

which leads to Z l ′ |u(x)| ≤ |u(y)|+|u(s)|ds , 0 By means of CauchySchwarz inequality, one gets 1 Z  l 2 2 ′ |u(x)| ≤ |u(y)|+l|u(s)|ds . 0

(1)

Integrating the above inequality over [0, l] in theyvariable, and applying once again the CauchySchwarz inequality, give 1 1 Z  Z  l l 2 2 ′ 2 2 l|u(x)| ≤l|u(y)|dy+l l|u(s)|ds , 0 0

or equivalently

 |u(x)| ≤max 1/

l ,

   ′ lkuk 2 +kuk 2 . L(I)L(I)

2 By using the the CauchySchwarz inequality inR, denotingC(l) = ¯ and taking the supremum onx∈I, one obtains the result.

1 0 ¯ Conclude thatH(I)⊂C(I)in dimension1.

 2 max 1/

l ,

 l,

¯ 1 1 Letu∈H(I). SinceD(I) is dense inH(I) for the normk k H(I) , there exists a sequence 1 1 ¯ (u n ) n≥0 ∈ D(I) which converges touinH(I) for the normk k H(I) . Then the inequality (1) 1 holds for each fonctionu n :

C( ∃C(l)>0,kuk 0 ¯ 1 n C (I) ≤l)ku n k H(I) ,∀n≥0.(2) 1 0 ¯ Since (u n ) n≥0 is a Cauchy sequence inH(I), the sequence (u n ) n≥0 is Cauchy inC(I) 0 ¯ for the uniform normk k 0 ¯ The spaceby the inequality (2). C(I) equipped with the norm C(I) 0 0 ¯ ¯ k k 0 ¯ there existsis complete: w∈C(I) such that (u n ) n≥0 converges towinC(I) for C(I) the normk k 0 ¯ . C(I)

1