Mathematical Physics - Volume II - Numerical Methods

3.3 Variation formulation of the boundary problem

61

Two last integrals in (3.57) can be transformed into linear integrals by using the divergence theory, thus resulting in: − Z Ω 1 ∇ · ( vk ∇ u ) d x d y − Z Ω 2 ∇ · ( vk ∇ u ) d x d y = = Z ∂ ( Ω 1 ) k ∂ u ∂ n v d s − Z ∂ ( Ω 2 ) k ∂ u ∂ n v d s , (3.58) where ∂ ( Ω 1 ) and ∂ ( Ω 2 ) are the subdomain limits for domains Ω 1 and Ω 2 , the integration direction is a positive mathematical direction, and ∂ u / ∂ n = − ∇ u · n . δ(Ω )−Γ 2

Ω 2

k k =

2

S

Γ

n n =

2

n n =

1

Ω 1

1 k k =

δ(Ω )−Γ 1

Figure 3.8: Domain boundaries.

Figure 3.8 shows that the limit of each domain is divided into two parts - ∂ ( Ω i ) , which do not coincide with Γ are denoted ∂ ( Ω i ) − Γ , i = 1 , 2. In accordance with this, line integrals in (3.58) are separated in the following way: − Z ∂ ( Ω 1 ) − Γ k ∂ u ∂ n v d s − Z ∂ ( Ω 2 ) − Γ k ∂ u ∂ n v d s +

(3.59)

Γ

∂ u ∂ n 1

Γ

∂ u ∂ n 2

+ Z

v d s + Z

− k

− k

v d s ,

∂ u ∂ n

∂ u ∂ n need to be determined within the region Ω i . Taking into account

where ( − k

) i indicates that − k

the sign of the normal in the last two integrals in (3.59), we obtain: Z Γ " − k (+) ∂ u (+) ∂ n + k ( − ) ∂ u ( − ) ∂ n # v d s .

(3.60)

Subintegral function in (3.60) is equal to v [ | σ n ( s ) | ] , which is according to (3.46) equal to zero, hence integral in (3.60) is equal to zero.