Finite Element Analysis. Barna Szabó

Finite Element Analysis

left-parenthesis alpha x Superscript alpha minus 1 Baseline minus left-parenthesis alpha plus 1 right-parenthesis x Superscript alpha Baseline right-parenthesis v Superscript prime Baseline d x period"/>

On integrating by parts, we get the following expression which is better suited for numerical evaluation:

(1.123) upper F left-parenthesis v right-parenthesis equals minus integral Subscript 0 Superscript script l Baseline u Subscript upper E upper X Baseline v Superscript double-prime Baseline d x period

We address the following questions: (a) How does the error in energy norm depend on the parameter α, the mesh Δ and the p‐distribution p? and (b) How is this error distributed among the elements? Understanding these relationships is necessary for making sound choices of discretization based on a priori information concerning the regularity of the exact solution.

We compute the potential energy of the difference between the exact solution and its linear interpolant for the kth element:

pi overbar Subscript upper E upper X Superscript left-parenthesis k right-parenthesis Baseline equals one half integral Subscript x Subscript k Baseline Superscript x Subscript k plus 1 Baseline Baseline left-parenthesis u prime Subscript upper E upper X Baseline minus StartFraction u Subscript upper E upper X Baseline left-parenthesis x Subscript k plus 1 Baseline right-parenthesis minus u Subscript upper E upper X Baseline left-parenthesis x Subscript k Baseline right-parenthesis Over x Subscript k plus 1 Baseline minus x Subscript k Baseline EndFraction right-parenthesis squared d x period

The exact solution for alpha equals 0.75 and its linear interpolant for upper M left-parenthesis normal upper Delta right-parenthesis equals 5 , uniform mesh, are shown in Fig. 1.12.

To obtain the potential energy of the difference between the exact solution and its linear interpolant for the kth element, denoted by pi overbar Subscript upper F upper E Superscript left-parenthesis k right-parenthesis , we need to solve:

(1.124)

Graph depicts the exact solution for α=0.75 and its linear interpolant for M(Δ)=5, uniform mesh.

Figure 1.12 The exact solution for

and its linear interpolant for

, uniform mesh.

The solution is

(1.125) a Subscript i Superscript left-parenthesis k right-parenthesis Baseline equals StartFraction script l Subscript k Baseline Over 2 EndFraction r Subscript i Superscript left-parenthesis k right-parenthesis Baseline comma i equals 3 comma 4 comma ellipsis comma p Subscript k Baseline plus 1 period

Using eq. (1.106) we get

(1.126)

where u overtilde Subscript upper E upper X Superscript left-parenthesis k right-parenthesis is the difference between u Subscript upper E upper X and its linear interpolant:

(1.127)

and compute

pi overbar Subscript upper F upper E Superscript left-parenthesis k right-parenthesis Baseline equals minus one half sigma-summation Underscript i equals 3 Overscript p Subscript k Baseline plus 1 Endscripts a Subscript i Superscript left-parenthesis k right-parenthesis Baseline r Subscript i Superscript left-parenthesis k right-parenthesis Baseline period

Referring to Theorem 1.5, the error in energy norm associated with the kth element is

(1.128) double-vertical-bar e Subscript k Baseline double-vertical-bar Subscript upper E left-parenthesis upper I Sub Subscript k Subscript right-parenthesis Baseline equals StartRoot pi overbar Subscript upper F upper E Superscript left-parenthesis k right-parenthesis Baseline minus pi overbar Subscript upper E upper X Superscript left-parenthesis k right-parenthesis Baseline EndRoot

and the relative error in energy norm associated with the kth element is:

(1.129)Скачать книгу