Finite Element Analysis - Barna Szabó

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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

      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)StartFraction 2 Over script l Subscript k Baseline EndFraction Start 4 By 4 Matrix 1st Row 1st Column 1 2nd Column 0 3rd Column midline-horizontal-ellipsis 4th Column 0 2nd Row 1st Column 0 2nd Column 1 3rd Column midline-horizontal-ellipsis 4th Column 0 3rd Row 1st Column Blank 2nd Column Blank 3rd Column down-right-diagonal-ellipsis 4th Column Blank 4th Row 1st Column 0 2nd Column 0 3rd Column midline-horizontal-ellipsis 4th Column 1 EndMatrix Start 4 By 1 Matrix 1st Row a 3 Superscript left-parenthesis k right-parenthesis Baseline 2nd Row a 4 Superscript left-parenthesis k right-parenthesis Baseline 3rd Row vertical-ellipsis 4th Row a Subscript p Sub Subscript k Subscript plus 1 Superscript left-parenthesis k right-parenthesis Baseline EndMatrix equals Start 4 By 1 Matrix 1st Row r 3 Superscript left-parenthesis k right-parenthesis Baseline 2nd Row r 4 Superscript left-parenthesis k right-parenthesis Baseline 3rd Row vertical-ellipsis 4th Row r Subscript p Sub Subscript k Subscript plus 1 Superscript left-parenthesis k right-parenthesis EndMatrix dot

Graph depicts the exact solution for α=0.75 and its linear interpolant for M(Δ)=5, uniform mesh.
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)r Subscript i Superscript left-parenthesis k right-parenthesis Baseline equals minus StartRoot StartFraction 2 i minus 3 Over 2 EndFraction EndRoot StartFraction 2 Over script l Subscript k Baseline EndFraction integral Subscript negative 1 Superscript 1 Baseline left-parenthesis kappa u overtilde Subscript upper E upper X Superscript left-parenthesis k right-parenthesis Baseline right-parenthesis Subscript x equals upper Q Sub Subscript k Subscript left-parenthesis xi right-parenthesis Baseline StartFraction d upper P Subscript i minus 2 Baseline Over d xi EndFraction d xi comma i equals 3 comma 4 comma ellipsis comma p Subscript k Baseline plus 1

      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)u overtilde Subscript upper E upper X Superscript left-parenthesis k right-parenthesis Baseline equals left-parenthesis u Subscript upper E upper X Baseline right-parenthesis Subscript x equals upper Q Sub Subscript k Subscript left-parenthesis xi right-parenthesis Baseline minus left-parenthesis StartFraction 1 minus xi Over 2 EndFraction u Subscript upper E upper X Baseline left-parenthesis x Subscript k Baseline right-parenthesis plus StartFraction 1 plus xi Over 2 EndFraction u Subscript upper E upper X Baseline left-parenthesis x Subscript k plus 1 Baseline right-parenthesis right-parenthesis

      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

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