Mathematical Physics - Volume II - Numerical Methods

Chapter 3. Comparison of finite element method and finite difference method

72

An illustration of finite element method application to solving of two-dimensional problems Let’s define the following problem: − ∆ u ( x , y ) = f ( x , y ) na Ω u = 0 na Γ 41 ,

∂ u ∂ n

(3.113)

na Γ 12 , Γ 25 , Γ 67 i Γ 74

= 0

∂ u ∂ n

+ β u = γ

na Γ 56 ,

where Ω is the polygonal domain, figure 3.15, and Γ 41 , Γ 12 · · · Γ 74 are boundary segments. In this case ∂ Ω 1 = Γ 41 and ∂ Ω 2 = Γ 12 ∪ Γ 25 ∪ Γ 56 ∪ Γ 67 ∪ Γ 74 . The analysis of this problem consists of:

y

Γ 25

2

5

Γ 12

Γ 56

1

Ω

6

Γ 41

Γ 67

Γ 74

4

7

x

(a)

2

5

(3)

(6)

(1)

3

6

1

(5)

(2)

(4)

4

7

(b)

Figure 3.15: Two-dimensional problem (a) and division into finite element (b).

1. We divide the domain into six triangular elements with seven nodes, fig. 3.15, within which we define a linear approximation u h for solution u of equation (3.112). Here ∂ Ω 1 = ∂ Ω 1 h and ∂ Ω 2 = ∂ Ω 2 h , since the domain boundary is linear and thus identical to the finite element boundary. 2. Next, by using (3.95) and (3.96), we calculate K e and f e , e = 1 , 2 , · · · , 6:

    K 1 11 K 1 21 K 1 31

    F

    f 1 1 f 1 2 f 1 3 0 0 0 0

   

1 12 1 22 1 32

1 13 1 23 1 33

K K K

K K K

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

K 1 =

1 =