me180_hw02

# me180_hw02 - N for each p along with the exact solution 1...

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HOMEWORK 2: HIGHER ORDER ELEMENTS Consider the followng boundary value problem, with domain Ω = (0 ,L ): d dx ± A 1 du dx ² = k 2 sin ( 2 πkx L ) A 1 = 0 . 1 k = 16 L = 1 u (0) = Δ 1 = given constant = 0 u ( L ) = Δ 2 = given constant = 1 (1) Solve this with the ﬁnite element method using order p equal-sized elements. In order to achieve e N def = || u - u N || A 1 (Ω) || u || A 1 (Ω) TOL = 0 . 05 , || u || A 1 (Ω) def = s Z Ω du dx A 1 du dx dx (2) how many ﬁnite elements ( N ) are needed for p = 1 N =? p = 2 N =? p = 3 N =? (3) Plot the numerical solutions for several values of

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Unformatted text preview: N , for each p , along with the exact solution 1 • Plot e N as a function of the element size h for each p • Plot e N as a function of the number of degrees of freedom for each p • Determine the relationship between the error and the element size for each p • Note: Please be careful with the quadrature order. ..you will need higher order Gauss rules for quadratic and cubic elements. 2...
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me180_hw02 - N for each p along with the exact solution 1...

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