  7 Pages
English

# Universite de Nice Sophia Antipolis Master MathMods Finite Elements

- 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

Subjects

##### Weak formulation

Informations

Universit´edeNiceSophiaAntipolis Master MathMods  Finite Elements  2008/2009
Exercises  Chapter 1  Chapter 2 (Correction)
Exercise 1. (a) LetI=]0, l[,lR. Show that
C(l)>0,kuk¯C(, C(I)l)kukH(I) 0 1
¯ u∈ D(I).
¯ ¯ Letu∈ D(I). Letx , yIfondamental 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 22 |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 2222 l|u(x)| ≤l|u(y)|dy+l l|u(s)|ds , 0 0
or equivalently
|u(x)| ≤max 1/
l ,
  lkuk2+kuk2. L(I)L(I)
2 By using the the CauchySchwarz inequality inR, denotingC(l) = ¯ and taking the supremum onxI, one obtains the result.
1 0 ¯ Conclude thatH(I)C(I)in dimension1.
2 max 1/
l ,
l,
¯ 1 1 LetuH(I). SinceD(I) is dense inH(I) for the normk kH(I), there exists a sequence 1 1 ¯ (un)n0∈ D(I) which converges touinH(I) for the normk kH(I). Then the inequality (1) 1 holds for each fonctionun:
C( C(l)>0,kuk0¯1 n C(I)l)kunkH(I),n0.(2) 1 0 ¯ Since (un)n0is a Cauchy sequence inH(I), the sequence (un)n0is Cauchy inC(I) 0 ¯ for the uniform normk k0¯The spaceby the inequality (2). C(I) equipped with the norm C(I) 0 0 ¯ ¯ k k0¯there existsis complete: wC(I) such that (un)n0converges towinC(I) for C(I) the normk k0¯. C(I)
1