Universite de Nice Sophia Antipolis Master MathMods Finite Elements V Dolean
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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 .
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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 .
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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) =fjφj(x) j=1 determine the matrixMcalled ”mass” matrix such that :KU=MF. Answer. The linear basis functions are given by : x−xi−1 , x∈[xi−1, xi], hi xi+1−x φi(x) = , x∈[xi, xi+1], hi+1 0/, x∈[xi−1, xi+1]. According to the expression of the ”stiffness” matrix we can write : Z ZZ 1xixi+1 1 1 020202 ) (x)dx+. Kii= (φ) (x)dx= (φi(φi) (x)dx= + i hihi+1 0xi−1xi Z Z 1xi+1 1 0 00 0 (x−. Ki,i+1=Ki+1,i=φi)φi(x)dx=φ(x)φ(x)dx= +1i i+1 hi+1 0xi all the other elements being null since in all the other cases the basis functionsφiandφj cannot be simultaneously non-zero. The right hand side can be written as : ZnZnZ 1 11 X X bi=f(x)φi(x)dx=fjφi(x)φj(x)dx=Mijfj, Mij=φi(x)φj(x)dx. 0 00 j=1j=1 Thus, the ”mass” matrix is formed by the elementsMijwhich can be computed as follows (by performing a variable changex=xi−1+thin the integral on [xi−1, xi] andx=xi+th in the integral on [xi, xi+1]) : Z ZZ 1xixi+1 2 22 Mii=φ(x)dx=φ(x)dx+φ(x)dx i ii 0xi−1xi Z Z 1 1 2hi+hi+1 2 =hit dt+hi+1(1−t)dt= 3 0 0 Z ZZ 1xi+1 1 hi+1 Mi,i+1=Mi+1,i=φi(x)φi+1(x)dx=φi(x)φi+1(x)dx=hi+1(1−t)tdt=. 6 0xi0 all the other elements being null since in all the other cases the basis functionsφiandφj 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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