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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. SemiDiscrete 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 CrankNicolson
method which has a discretization error kc(tn) ; cnk = O( t2):
The foregoing discussion involved onestep methods. Multistep methods are also used
to solve timedependent 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 secondorder 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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 Spring '14
 JosephE.Flaherty
 The Land

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