Mathematical Physics - Volume II - Numerical Methods

Chapter 2. Finite element method

32

fluks kontinualno rasporedenih izvora -

f z z ( - ) δ 2 = fluks tackastog izvora ^ ^

f z ( )

z

k k = 1

k k = 2

diskontinuitet modula

Ω 3

Ω 4

Ω 1

Ω 2

z z = = 0 0

z 1

z l 4 =

z 2 (a)

z 3

f

σ ( ) b

σ ( ) a

z

a

z

b

(b)

f

σ ( ) a

σ 0

z

a

z z = = 0 0

(c)

Figure 2.1: One-dimensional problem with 5 nodes and 4 elements.

(3) Equation describing this problem is a second-order ordinary differential equation. Since f ( z ) is continuous, it follows in accordance with the mean integral values theorem that: b R a f ( z ) d z = ( b − a ) f ( ξ ) , where a < ξ < b and f ( ξ ) is the mean value of f ( z ) at the ( a , b ) interval. It follows that: P ( b ) − P ( a ) = ( b − a ) f ( ξ ) , (2.6) i.e. lim f ( ξ ) .

P ( b ) − P ( a ) b − a

= lim a → ¯ z − b → ¯ z +

a → ¯ z − b → ¯ z +

As f ( z ) is continuous, we have:

f ( ξ ) = f ( ¯ z ) .

lim a → ¯ z − b → ¯ z +

In addition:

P ( b ) − P ( a ) b − a

d P ( ¯ z ) d z

lim a → ¯ z − b → ¯ z +

=

,