Mathematical Physics - Volume II - Numerical Methods

2.4 Determining of finite element matrices and finite element system matrices 47

obtain full discretization, it is necessary to assume how U ( t ) behaves in time. Usual methods for this are implicit and explicit, as was the case with the finite difference method. In the case of the explicit method, the time domain 0 ≤ t ≤ T is divided into M equal intervals with length of ∆ t = T / M , and the following equation is used d U ( n ∆ t ) dt ≈ U n + 1 − U n ∆ t , U n = U ( n ∆ t ) (2.107) for the purpose of determining the first derivative of a state variable in the interval of t = n ∆ t to ( n + 1 ) ∆ t , and by replacing this into (2.105), we obtain: U ( n + 1 ) = ( I − ∆ t C − 1 K ) U n + ∆ t C − 1 f n . (2.108) According to the above, based on known values of U 0 , we obtain U 1 followed by U 2 , and so on. As with the finite difference method, the explicit time integration is conditionally stable, which requires limited, smaller values of ∆ t , proportional to the mesh step h . In the case of implicit methods, instead of (2.107), the following equations are used to determine the first derivative: d U [( n + 1 ) ∆ t ] dt ≈ U n + 1 − U n ∆ t , ili d U [( n + 1 / 2 ) ∆ t ] dt ≈ U n + 1 − U n ∆ t (2.109) which provides an unconditionally stable solution, when replaced into (2.105). Hyperbolic partial differential equation (2.100) requires more complex initial conditions, in cluding first derivatives, since this equation has a second derivative with respect to time: We will once again adopt a simple-form boundary condition, as in the case of (2.101). By multiplying (2.99) with v ( z ) and integrating along z , we obtain: Z l 0 ρ ( z ) ∂ 2 u ( z , t ) ∂ t 2 v ( z ) d z − ∂ ∂ z k ( z ) ∂ u ( z , t ) ∂ z v ( z ) − f ( z , t ) v ( z ) d z = 0 (2.111) from which follows that: Z l 0 ρ ( z ) ∂ 2 u ( z , t ) ∂ t 2 v ( z ) d z + Z l 0 k ( z ) ∂ u ( z , t ) ∂ z ∂ v ( z ) ∂ z d z − Z l 0 f ( z , t ) v ( z ) d z = − k ( l ) ∂ u ( l , t ) ∂ z v ( l )+ k ( 0 ) ∂ u ( 0 , t ) ∂ z v ( 0 ) za ∀ v ∈ H 1 0 ( 0 , l ) . (2.112) If we introduce u h ( z , t ) into equations (2.103) and (2.104), in the same way, we obtain: u ( z , 0 ) = ˜ u ( z ) i ∂ u ( z , 0 ) ∂ t = ˆ g ( z ) 0 < z < l . (2.110)

C + KU ( t ) = f ( t ) U ( 0 ) = ˜ U d U ( 0 ) / dt = ˜ g d 2 U ( t ) dt 2

(2.113)

where

C i j = Z

l

0 (2.114) is the mass matrix, and stiffness matrix K i j and force vector f i have the same form as in (2.105). Equation (2.113) is also solved using explicit or implicit methods. In the case of explicit methods, the second derivative with respect to time is represented as: d 2 U ( n ∆ t ) d t 2 ≈ U n + 1 − 2 U n + U n − 1 ∆ t 2 , (2.115) ρ ( z ) φ i ( z ) φ j ( z ) d z