Universite de Nice Sophia Antipolis Master MathMods Finite Elements V Dolean

Sophia - Antipolis Master

Description

Niveau: Supérieur, Master
Universite de Nice Sophia-Antipolis Master MathMods - Finite Elements - 2008/2009 V. Dolean Exercises - Chapter 0 (correction) Exercise 1. Find the ”stiffness” matrix K for linear basis functions. If the right hand side f is piecewise linear i.e. f(x) = n∑ j=1 fj?j(x) determine the matrix M called ”mass” matrix such that : KU = MF. Answer. The linear basis functions are given by : ?i(x) = ? ???? ???? x? xi?1 hi , x ? [xi?1, xi], xi+1 ? x hi+1 , x ? [xi, xi+1], 0, x /? [xi?1, xi+1]. According to the expression of the ”stiffness” matrix we can write : Kii = ∫ 1 0 (??i) 2(x)dx = ∫ xi xi?1 (??i) 2(x)dx+ ∫ xi+1 xi (??i) 2(x)dx = 1 hi + 1 hi+1 . Ki,i+1 = Ki+1,i = ∫ 1 0 ??i(x)? ? i+1(x)dx = ∫ xi+1 xi ??i(x)? ? i+1(x)dx = ? 1 hi+1 .

• let wi ?


• nodal basis consisting


• ?f ???v??


• stiffness matrix


• x? xi?1 ?


• mathmods - finite elements


• defined only up


• basis functions


• dx ≤

Sujets

• Master
• Stiffness matrix
• Basis function

Informations

• Publié par: profil-infoe-2012
• Nombre de lectures: 30
• Langue: English

Extrait

Universit´edeNiceSophia-Antipolis Master MathMods - Finite Elements - 2008/2009 V. Dolean

Exercises - Chapter 0 (correction) Exercise 1. Find the ”stiffness” matrixKfor linear basis functions. If the right hand sidefis piecewise linear i.e. n X f(x) =f j φ j (x) j=1 determine the matrixMcalled ”mass” matrix such that :KU=MF. Answer. The linear basis functions are given by :  x−x i−1 , x∈[x i−1 , x i ],   h i x i+1 −x φ i (x) = , x∈[x i , x i+1 ], h i+1   0/, x∈[x i−1 , x i+1 ]. According to the expression of the ”stiffness” matrix we can write : Z ZZ 1x i x i [phone] ) (x)dx+. K ii = (φ) (x)dx= (φ i (φ i ) (x)dx= + i h i h i+1 0x i−1 x i Z Z 1x i +1 1 0 00 0 (x−. K i,i+1 =K i+1,i =φ i )φ i (x)dx=φ(x)φ(x)dx= +1i i+1 h i+1 0x i all the other elements being null since in all the other cases the basis functionsφ i andφ j cannot be simultaneously non-zero. The right hand side can be written as : Z n Z n Z 1 11 X X b i =f(x)φ i (x)dx=f j φ i (x)φ j (x)dx=M ij f j , M ij =φ i (x)φ j (x)dx. 0 00 j=1j=1 Thus, the ”mass” matrix is formed by the elementsM ij which can be computed as follows (by performing a variable changex=x i−1 +thin the integral on [x i−1 , x i ] andx=x i +th in the integral on [x i , x i+1 ]) : Z ZZ 1x i x i +1 2 22 M ii =φ(x)dx=φ(x)dx+φ(x)dx i ii 0x i−1 x i Z Z 1 1 2 h i +h i+1 2 =h i t dt+h i+1 (1−t)dt= 3 0 0 Z ZZ 1x i +1 1 h i+1 M i,i+1 =M i+1,i =φ i (x)φ i+1 (x)dx=φ i (x)φ i+1 (x)dx=h i+1 (1−t)tdt=. 6 0x i 0 all the other elements being null since in all the other cases the basis functionsφ i andφ j cannot be simultaneously non-zero. Exercise 2. Give the weak formulation for the two-point boundary value problem :  00 −u+u=f, x∈(0,1), u(0) =u(1) = 0.

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