Mathematical Physics - Volume II - Numerical Methods

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

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Flux balance in the strip which contains P b gives the following: σ n ( s ) ≡ σ ( s ) · n ( s ) = ˆ σ ( s ) , s ∈ ∂ Ω 2 .

(3.49)

Hence, the following holds:

σ n ( s ) ≡ p ( s )[ u ( s ) − ˆ u ( s )] . (3.50) If we now use the constitutive equation (3.43) to eliminate σ and σ n from (3.46) - (3.50), it is possible to mathematically formulate the boundary problem: 1. Boundaries ∂ Ω 1 , ∂ Ω 2 and the inter-boundary Γ are defined by parameric equations: x = x ( s ) , y = y ( s ) , s ∈ ∂ Ω or s ∈ Γ ; 2. Source distribution is defined as f = f ( x , y ) in Ω i , i = 1 , 2; 3. Material coefficients are defined as k i = k i ( x , y ) for ( x , y ) ∈ Ω i , i = 1 , 2; 4. State variable ˆ u ( s ) for s ∈ ∂ Ω i , is given; 5. The value of boundary coefficients is p ( s ) and ˆ u ( s ) on ∂ Ω 2 or ˆ σ ( s ) is given for s ∈ ∂ Ω 2 . For a problem formulated in this way, it is necessary to find a function u = u ( x , y ) which fulfills the following requirements: 1. The partial differential equation in internal points of smooth subdomains Ω 1 and Ω 2 − ∇ · k ( x , y ) ∇ u ( x , y ) − f ( x , y ) = 0 za ( x , y ) ∈ Ω i ; (3.51) 2. The jump condition at the inter-boundary Γ [ | k ∇ u · n | ] = 0 s ∈ Γ ; (3.52) 3. Essential boundary conditions at ∂ Ω 1 u ( s ) = ˆ u ( s ) s ∈ ∂ Ω 1 ; (3.53) 4. Natural boundary conditions at ∂ Ω 2 − k ( s ) ∂ u ( s ) ∂ n = ˆ σ ( s ) s ∈ ∂ Ω 2 . (3.54) 3.3 Variation formulation of the boundary problem As before, we will multiply equation (3.51) with a sufficiently smooth weight function v and determine the integral with respect to all domains with smooth output: Z Ω 1 [ − ∇ · ( k ∇ u ) − f ] v d x d y + Z Ω 2 [ − ∇ · ( k ∇ u ) − f ] v d x d y = 0 . (3.55) Two-dimensional partial integration is necessary in order to reduce the first terms in both integrals to their first derivatives. By applying the derivative product we obtain:

∇ · ( vk ∇ u ) = k ∇ u · ∇ v + v ∇ · ( k ∇ u ) v ∇ · ( k ∇ u ) = ∇ · vk ∇ u − k ∇ u · ∇ v ,

(3.56)

By replacing (3.56) into (3.55), we obtain: Z Ω 1

( k ∇ u · ∇ v + buv − f v ) d x d y + Z Ω 2

( k ∇ u · ∇ v + buv − f v ) d x d y −

(3.57)

− Z

∇ · ( vk ∇ u ) d x d y − Z Ω 2

∇ · ( vk ∇ u ) d x d y = 0 .

Ω 1