Mathematical Physics - Volume II - Numerical Methods

2.1 Finite element application to solving of one-dimensional problems

41

Since z i + 1 − z i = h , it is clear that | z − ¯ z | ≤ h / 2, hence we have the following:

h 2

E ′′ ( ζ ) | .

| E ( ¯ z ) | ≤

(2.72)

8 |

Since a linear interpolation of the second derivative is considered, we obtain:

E ′′ = g ′′ − g ′′ h = g ′′

(2.73)

and by replacing into equation (2.72) and determining the maximum for all elements, we obtain:

h 2 8

max | g ′′ ( z ) | .

max | E ( z ) | ≤

(2.74)

Since the second derivative of g is limited, g ′′ ≤ c ≤ ∞ , we finally get that: ∥ E ∥ ∞ = max | E ( z ) | ≤ Ch 2 ,

(2.75)

where C is a constant, independent of h . Using an analogue method for a Lagrange interpolation of the k -th order, the following is obtained:

k + 1 .

∥ E ∥ ∞ ≤ Ch

(2.76)

Since in the general case, Taylor series contains terms of every order up to k , it is necessary for the interpolant, as well as the interpolation functions for every element, to be able to represent each of these terms. For example, if interpolation functions contain independent terms proportional to z 0 (const), z 2 , z 3 , . . . , z k , but do not contain a term proportional to z 1 , the error will be proportional to h instead of h k + 1 . If the constants are missing from the interpolation functions, the convergence is non-existent. Hence, it is of great importance for the interpolation function set to contain complete polynomials. During the determining of the error, we assumed that function g has (continuous) derivatives whose order is ≤ k . However, if g only has derivatives up to order s , where 0 < s < k , only the first s terms in the interpolation polynomial will participate in the approximate representation of g , regardless of k , hence instead of using expression (2.73) to define the error, we use the following one:

max | g ( z ) − g h ( z ) | ≤ Ch s .

(2.77)

It is clear that the error cannot be decrease by increasing the interpolation function order, but can still be decreased by reducing the step of mesh h . Finally, we will give an example of finite element interpolation: for function g ( h ) = sin ( π x ) at the interval 0 ≤ z ≤ 1, using two quadratic elements, Fig. 2.5. Nodes are given by z = 0; 0 , 25; 0 , 5; 0 , 75; 1, whereas function values in these nodes are g ( z ) = 0; 0 , 707; 1; 0 , 707; 0, such that the interpolant g h ( z ) = 0 , 707 φ 2 ( z )+ φ 3 ( z )+ 0 , 707 φ 4 ( z ) .