Finite Element Analysis. Barna Szabó

Finite Element Analysis

alt="p plus 1"/> Lobatto points are used. Throughout this book we will be concerned with errors of approximation that can be controlled by the design of mesh and the assignment of polynomial degrees. We will assume that the errors of integration and errors in mapping are negligibly small in comparison with the errors of discretization.

Exercise 1.9 Assume that c left-parenthesis x right-parenthesis equals c Subscript k is constant on Ik. Using the Lagrange shape functions of degree p equals 3 , with the nodes located in the Lobatto points, compute m 33 Superscript left-parenthesis k right-parenthesis numerically using 4 Lobatto points. Determine the relative error of the numerically integrated term. Refer to Remark 1.6 and Appendix E.

Exercise 1.10 Assume that c left-parenthesis x right-parenthesis equals c Subscript k is constant on Ik. Using the Lagrange shape functions of degree p equals 2 , compute m 11 Superscript left-parenthesis k right-parenthesis and m 13 Superscript left-parenthesis k right-parenthesis in terms of ck and ℓk.

1.3.4 Computation of the right hand side vector

Computation of the right hand side vector involves evaluation of the functional upper F left-parenthesis v right-parenthesis , usually by numerical means. In particular, we write:

(1.73)

The element‐level integral is computed from the definition of vn on Ik:

(1.74)

where

(1.75)

which is computed from the given data and the shape functions.

Example 1.5 Let us assume that f left-parenthesis x right-parenthesis is a linear function on Ik. In this case can be written as

f left-parenthesis x right-parenthesis equals StartFraction 1 minus xi Over 2 EndFraction f left-parenthesis x Subscript k Baseline right-parenthesis plus StartFraction 1 plus xi Over 2 EndFraction f left-parenthesis x Subscript k plus 1 Baseline right-parenthesis equals f left-parenthesis x Subscript k Baseline right-parenthesis upper N 1 left-parenthesis xi right-parenthesis plus f left-parenthesis x Subscript k plus 1 Baseline right-parenthesis upper N 2 left-parenthesis xi right-parenthesis period

Using the Legendre shape functions we have:

StartLayout 1st Row 1st Column r 1 Superscript left-parenthesis k right-parenthesis Baseline equals 2nd Column f left-parenthesis x Subscript k Baseline right-parenthesis StartFraction script l Subscript k Baseline Over 2 EndFraction integral Subscript negative 1 Superscript plus 1 Baseline upper N 1 squared d xi plus f left-parenthesis x Subscript k plus 1 Baseline right-parenthesis StartFraction script l Subscript k Baseline Over 2 EndFraction integral Subscript negative 1 Superscript plus 1 Baseline upper N 1 upper N 2 d xi equals StartFraction script l Subscript k Baseline Over 6 EndFraction left-parenthesis 2 f left-parenthesis x Subscript k Baseline right-parenthesis plus f left-parenthesis x Subscript k plus 1 Baseline right-parenthesis right-parenthesis 2nd Row 1st Column r 2 Superscript left-parenthesis k right-parenthesis Baseline equals 2nd Column f left-parenthesis x Subscript k Baseline right-parenthesis StartFraction script l Subscript k Baseline Over 2 EndFraction integral Subscript negative 1 Superscript plus 1 Baseline upper N 1 upper N 2 d xi plus f left-parenthesis x Subscript k plus 1 Baseline right-parenthesis StartFraction script l Subscript k Baseline Over 2 EndFraction integral Subscript negative 1 Superscript plus 1 Baseline upper N 2 squared d xi equals StartFraction script l Subscript k Baseline Over 6 EndFraction left-parenthesis f left-parenthesis x Subscript k Baseline right-parenthesis plus 2 f left-parenthesis x Subscript k plus 1 Baseline right-parenthesis right-parenthesis 3rd Row 1st Column r 3 Superscript left-parenthesis k right-parenthesis Baseline equals 2nd Column f left-parenthesis x Subscript k Baseline right-parenthesis StartFraction script l Subscript k Baseline Over 2 EndFraction integral Subscript negative 1 Superscript plus 1 Baseline upper N 1 upper N 3 d xi plus f left-parenthesis x Subscript k plus 1 Baseline right-parenthesis StartFraction script l Subscript k Baseline Over 2 EndFraction integral Subscript negative 1 Superscript plus 1 Baseline upper N 2 upper N 3 d xi 4th Row 1st Column equals 2nd Column minus StartFraction script l Subscript k Baseline Over 6 EndFraction StartRoot three halves EndRoot left-parenthesis f left-parenthesis x Subscript k Baseline right-parenthesis plus f left-parenthesis x Subscript k plus 1 Baseline right-parenthesis right-parenthesis period EndLayout

Exercise 1.11 Assume that f left-parenthesis x right-parenthesis is a linear function on Ik. Using the Legendre shape functions compute r 4 Superscript left-parenthesis k right-parenthesis and show that r Subscript i Superscript left-parenthesis k right-parenthesis Baseline equals 0 for i greater-than 4 . Hint: Make use of eq.

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