Mathematical Physics - Volume II - Numerical Methods

Chapter 2. Finite element method

40

From which it follows that functions ψ i are linearly independent. This total of k + 1 functions define the base for the set of k -th order polynomials, as well as for those of lesser order, hence we can say that this base is complete, which means that any and all polynomials of the k -th order (or lower) can be uniquely represented using the Lagrange interpolation functions. In other words, Lagrange interpolation functions can be used as the base functions for the finite element.

Example 2.2 For k = 1 ( ξ 1 = − 1 , ξ 2 = 1 )

ξ − ξ 2 ξ 1 − ξ 2 ξ − ξ 1 ξ 2 − ξ 1

1 2 1 2

ψ 1 =

( 1 − ξ )

(2.61)

=

ψ 2 =

( 1 + ξ ) ,

(2.62)

=

(2.63)

For k = 2 ( ξ 1 = − 1 , ξ 2 = 0 , ξ 3 = 1 ) ψ 1 = ( ξ − ξ 2 )( ξ − ξ 3 ) ( ξ 1 − ξ 2 )( ξ 1 − ξ 3 )

ξ ( ξ − 1 ) − 1 ( − 2 )

1 2

ξ ( ξ − 1 ) ,

(2.64)

=

=

( ξ − ξ 1 )( ξ − ξ 3 ) ( ξ 2 − ξ 1 )( ξ 2 − ξ 3 ) ( ξ − ξ 1 )( ξ − ξ 2 ) ( ξ 3 − ξ 1 )( ξ 3 − ξ 2 )

( ξ + 1 )( ξ − 1 ) 1 ( − 1 )

= ( 1 − ξ 2 ) ,

ψ 2 =

(2.65)

=

( ξ + 1 ) ξ 2 · 1

1 2

ψ 3 =

ξ ( 1 + ξ ) .

(2.66)

=

=

2.2.1 Interpolation error

Lagrange linear interpolation error can be determined as the difference in the values of the interpolated function g and the Lagrange interpolation function g h : E = g − g h . (2.67) Let us observe an arbitrary element Ω e ( z ) defined within a mesh as z i < z < z i + 1 and derive a Taylor series from error E , within an element in the vicinity of an arbitrary point ¯ z in the following way: E ( z ) = E ( ¯ z )+ E ′ ( ¯ z )( z − ¯ z )+ 1 2 E ′′ ( ζ )( z − ¯ z ) 2 , (2.68) where ζ ∈ [ z , ¯ z ] , and E ( z i ) = E ( z i + 1 ) = 0. Let u now select ¯ z as a point where the error reaches its maximum value, such that E ′ ( ¯ z ) = 0

1 2

E ′′ ( ζ )( z − ¯ z ) 2 .

E ( z ) = E ( ¯ z )+

(2.69)

Now, let us consider the boundary point z , which is closest to ¯ z , e.g. z i :

1 2

2 = 0 ,

E ′′ ( ζ )( z i − ¯ z )

E ( z i ) = E ( ¯ z )+

(2.70)

It follows from the above expression that the absolute value of the error is equal to:

1 2 |

2 .

E ′′ ( ζ ) | ( z i − ¯ z )

| E ( ¯ z ) | =

(2.71)