{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online