Mathematical Physics - Volume II - Numerical Methods

2.1 Finite element application to solving of one-dimensional problems

37

for arbitrary β i . This expression can also be represented in a simplified form N ∑ i = 1 β i N ∑ j = 1 K i j α j − F i ! = 0 ,

(2.36)

where

K i j = Z

l

k φ ′ i φ ′ j d z

(2.37)

0

l 0 + Z

l

F i = P φ i

f φ i d z .

(2.38)

0

K i , j is the stiffness matrix of the problem for base functions φ i , of N × N -th order, and F i is the load vector , for the same base functions, of the N -th order. Since β i values are arbitrary, the condition below needs to be fulfilled in order for equation (2.36) to hold: N ∑ j = 1 K i j α j = F i i = 1 , 2 , . . . , N . (2.39) Since functions φ i were selected as independent, equations (2.39) are also independent, hence matrix K i j is invertible, and coefficients α j now become:

N ∑ i = 1

1 F

( K ) i j −

α j =

(2.40)

i .

Once these coefficients are obtained, the approximate solution u N are also known, based on equation (2.32). Thus, it can be seen that Galerkin’s method represents a very powerful tool, but only if a systematic technique for constructing of base functions is present. Hence, this method only became widely applied once finite element method was introduced, as will be shown in the following paragraphs.

2.1.3 Finite element base functions

There are significant difficulties in selecting of base functions, since apart from the fact that the members of space H 1 0 are independent, nothing else is known about them. The situation is further complicated with the increase of the number of dimensions of the boundary problem, since this makes the boundary conditions considerably more complex. Additionally, taking into account the dependence of the solution and the finite element on the selected functions φ i , it is clear that the selection of base functions is of crucial importance in practical applications of the Galerkin’s method. In order to demonstrate how these base functions are determined, we will start by considering a homogeneous differential equation: ( ku ′ ) ′ = 0 z ∈ Ω i Ω i = [ − h / 2 , h / 2 ] , (2.41) The exact solution of this equation is given as: ku + C 1 z + C 0 = 0 . (2.42) If we introduce the following boundary conditions: u = u i for z = − h / (2.43) u = u i + 1 for z = h / 2 , (2.44)