23 the second order bdf follows by setting k 2 in 9213

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Unformatted text preview: .3. Finite Element Methods in Time 13 Di erentiating and setting t = tn+1 _ _ _ N0 (tn+1) = 2 3 t N1 (tn+1) = ; 2t N2(tn+1 ) = 2 1 t : Thus, n+1 n n;1 _ Y(tn+1) = 3y ;24yt + y and the second-order BDF is n+1 n n;1 f (tn+1 yn+1 3y ;24yt + y ) = 0: Applying this method to (9.2.4a) yields 3cn+1 ; 4cn + cn;1 + Kn+1cn+1 = ln+1: M 2t Thus, computation of cn+1 requires inversion of M + K: 2t Backward di erence formulas through order six are available 2, 3, 6, 7, 8]. 9.3 Finite Element Methods in Time It is, of course, possible to use the nite element method in time. This can be done on space-time triangular or quadrilateral elements for problems in one space dimension on hexahedra, tetrahedra, and prisms in two space dimensions and on four-dimensional parallelepipeds and prisms in three space dimensions. However, for simplicity, we'll focus on the time aspects of the space-time nite element method by assuming that the spatial discretization has already been performed. Thus, we'll consider an ODE system in the form (9.2.4a) and construct a Galerkin problem in time by multiplying it by a test function w 2 L2 and integrating on (tn tn+1] to obtain _ (w Mc)n + (w Kc)n = (w l)n where the L2 inner product in time is (w c)n = Zt n+1 tn 8w 2 L2 (tn tn+1] wT cdt: (9.3.1a) (9.3.1b) Only rst derivatives are involved in (9.2.4a) thus, neither the trial space for c nor the test space for w have to be continuous. For our initial method, let us assume that c(t) is continuous at tn. By assumption, c(tn) is known in this case and, hence, w(tn) = 0. 14 Parabolic Problems Example 9.3.1. Let us examine the method that results when c(t) and w(t) are linear on (tn tn+1]. We represent c(t) in the manner used for a spatial basis as c( ) cnNn( ) + cn+1Nn+1( ) where Nn( ) = 1 ; 2 are hat functions in time and Nn+1( ) = 1 + 2 = 2t ; tn ; tn+1 (9.3.2a) (9.3.2b) (9.3.2c) t de nes the canonical element in time. The test function w = Nn+1 ( ) 1 1 : : : 1]T (9.3.2d) vanishes at tn ( = ;1) and is linear on (tn tn+1). Transforming the integrals in (9.3.1a) to (;1 1) using (9.3.2c) and using (9.3.2a,b,d) yields t Z 1 1 + M cn+1 ; cn d _ (w Mc)n = 2 2 t tZ ;1 1 + K cn 1 ; + cn+1 1 + ]d : 2 ;1 2 2 2 (Again, we have written equality instead of for simplicity.) Assuming that M and K are independent of time, we have (w Kc)n = 1 _ (w Mc)n = M c 2; c n+1 n (w Kc)n = 6t K(cn + 2cn+1): Substituting these into (9.3.1a) cn+1 ; cn + t K(cn + 2cn+1) = t Z 1 1 + l( )d M2 6 2 2 or, if l is approximated like c, ;1 (9.3.3a) n+1 n M c 2; c + 6t K(cn + 2cn+1) = 6t (ln + 2ln+1): (9.3.3b) 1 M + 2 tK]cn+1 = M ; 3 tK]cn + 1 t ln + 2ln+1] 3 3 (9.3.3c) Regrouping terms 9.3. Finite Element Methods in Time 15 we see that the piecewise-linear Galerkin method in time is a weighted average scheme (9.2.12c) with = 2=3. Thus, at least to this low order, there is not much di erence between nite di erence and nite element methods. Other similarities appear in Problem 1 at the end of this section. Low-order schemes such as (9.2.12) are popular in nite element packages. Our preference is for BD...
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This document was uploaded on 03/16/2014 for the course CSCI 6860 at Rensselaer Polytechnic Institute.

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