If lumping were used and eg m hi then cn1 would be

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Unformatted text preview: od, we evaluate (9.2.8) at tn+1 and use the approximation (9.2.9) to obtain n+1 n (V U ; U ) + A(V U n+1 ) = (V f n+1) t 8v 2 S0N : (9.2.11a) 9.2. Semi-Discrete Galerkin Problems and Mc n+1 11 ; cn + Kn+1cn+1 = ln+1: t (9.2.11b) The backward Euler method is implicit regardless of whether or not lumping is used. Computation of cn+1 requires inversion of 1 M + Kn+1: t The most popular of these simple schemes uses a weighted average of the forward and backward Euler methods with weights of 1 ; and , respectively. Thus, n+1 n (V U ; U ) + (1 ; )A(V U n ) + A(V U n+1 ) = (1 ; )(V f n) + (V f n+1) t 8V 2 S0N : (9.2.12a) and Mc n+1 ; cn + (1 ; )Kncn + Kn+1cn+1 = (1 ; )ln + ln+1: t (9.2.12b) The forward and backward Euler methods are recovered by setting = 0 and 1, respectively. Let us regroup terms involving cn and cn+1 in (9.2.12b) to obtain M + tKn+1 ]cn+1 = M ; (1 ; ) tKn]cn + t (1 ; )ln + ln+1]: Thus, determination of cn+1 requires inversion of M + tKn+1: (9.2.12c) In one dimension, this system would typically be tridiagonal as with Example 9.2.1. In higher dimensions it would be sparse. Thus, explicit inversion would never be performed. We would just solve the sparse system (9.2.12c) for cn+1. Taylor's series calculations reveal that the global discretization error is kc(tn) ; cnk = O( t) for almost all choices of 2 0 1] 6]. The special choice = 1=2 yields the Crank-Nicolson method which has a discretization error kc(tn) ; cnk = O( t2): The foregoing discussion involved one-step methods. Multistep methods are also used to solve time-dependent nite element problems and we'll describe them for an ODE in 12 Parabolic Problems the implicit form (9.2.7). The popular backward di erence formulas (BDFs) approximate y(t) in (9.2.7) by a k th degree polynomial Y(t) that interpolates y at the k + 1 times _ _ tn+1;i, i = 0 1 : : : k. The derivative y is approximated by Y. The Newton backward di erence form of the interpolating is most frequently used to represent Y 2, 3], but since we're more familiar with Lagrangian interpolation we'll write y(t) Y(t) = k X i=0 yn+1;iNi(t) where Ni(t) = t 2 (tn+1;k tn+1] k Y t ; tn+1;j : t ; tn+1;j j =0 j 6=i n+1;i (9.2.13a) (9.2.13b) The basis (9.2.13b) is represented by the usual Lagrangian shape functions (cf. Section 2.4), so Ni(tn+1;j ) = ij . Assuming yn+1;i, i = 1 2 : : : k, to be known, the unknown yn+1 is determined by collocation at tn+1. Thus, using (9.2.7) _ f (tn+1 Y(tn+1) Y(tn+1)) = 0: (9.2.14) Example 9.2.2. The simplest BDF formula is obtained by setting k = 1 in (9.2.13) to obtain Y(t) = yn+1N0(t) + ynN1 (t) Di erentiating Y(t) t N0(t) = t t ;;nt n+1 N1(t) = tt ; ttn+1 ; n n _ Y(t) = y n+1 ; yn tn+1 ; tn n+1 thus, the numerical method (9.2.13) is the backward Euler method n+1 n f (tn+1 yn+1 y ; y ) = 0: t ;t n+1 n Example 9.2.3. The second-order BDF follows by setting k = 2 in (9.2.13) to get Y(t) = yn+1N0(t) + ynN1 (t) + yn;1N2 (t) N0(t) = (t ; tn2)(tt; tn;1) N1(t) = (t ; tn+1)(t2; tn;1) 2 ;t N (t) = (t ; tn+1 )(t ; tn ) 2 where time steps are of duration t. 2 t2 9...
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