Mathematical Physics - Volume II - Numerical Methods

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

78

y

ξ

(0,1)

(3,1)

3

2

η

Ω

1

4

1

x

(0,0)

(3,0)

y

ξ

(0,1)

(3,1)

2

3

η

η

Ω

2

1

4

x

(0,0)

(3,0)

4

3

3

η

(2,2)

ξ

y

ξ

Ω ^

(0,1)

4

2

1

3 Ω

2

1

x

(0,0)

(1,0)

y

(0,2)

4

η

(1,1)

3

4 Ω

ξ

1

2

x

(3,0)

(0,0)

Figure 3.18: Finite element transformation.

By transforming into element Ω 3 , fig. 3.18.

( x i , y i ) = { ( 0 , 0 ) , ( 1 , 0 ) , ( 2 , 2 ) , ( 0 , 1 ) } ⇒ x = ψ 2 + 2 ψ 3 = 1 4 ( 3 + 3 ξ + η + ξη )

1 4 ( 3 + ξ + 3 η + ξη )

y = 2 ψ 3 + ψ 4 =

1 8

1 8

1 2

| J | = η . Considering that | J | > 0 for ξ , η ∈ ( − 1 , 1 ) , this transformation is invertible. Value of | J | is lowest at node 1 and highest at node 3, indicating relative elongation of different parts of ˆ Ω via transformation T 3 . By transforming element Ω 4 , fig. 3.18, we obtain: ( x i , y i ) = { ( 0 , 0 ) , ( 3 , 0 ) , ( 1 , 1 ) , ( 0 , 2 ) } ⇒ + ξ +

x = 3 ψ 2 + ψ 3 y = ψ 3 + 2 ψ 4 ( 5 − 3 ξ − 4 η )

(3.133)

1 8

| J | =