Unformatted text preview: (5.2.3) is equivalent to the PeacemanRachford form of ADI (5.1.5) for the heat conduction equation. Unlike the PeacemanRachford implementation, however, a portion of the
operator is neglected at each step.
Let's apply (5.2.3) to the variablecoe cient heat conduction equation
n n jk jk u =( u ) +( u )
t x x y y where = (x y). Suppose that we select
1u = ( L u)
x x 2u = ( L u)
y y 5.2. Operator Splitting 13 We discretize each operator using (4.4.4) and introduce the shorthand notation
^2 U L1 U = x2
x n n x jk ( jk jk where
x ( U )=
n jk x j jk +1 2
= (U +1
n k j k ; U)
n jk ; x U)
n jk x2
j ;1 2
= k (U n
jk ; U ;1 ):
n j k A similar formula may be written for L2 . Upon using (5.2.3) r0 ^2 ^
(I 2 )U
y ; y 0
+1 = (I + r ^2 )U
y n 2 jk y (5.2.4a) n
jk 0
0
^
(I ; r2 ^2 )U +1 = (I + r2 ^2 )U +1 (5.2.4b) r0 = t= x2 (5.2.4c) x x n x n x jk jk where r 0 = t= y 2 : x y The combined method (5.2.4a, 5.2.4b) is solved in alternating directions, like the ADI
method (5.1.5). The way that we have split the operator, (5.2.4a) would be solved by
columns and (5.2.4b) would be solved by rows. Proceeding in the opposite manner is
acceptable.
Example 5.2.2. Consider the Taylor's series expansions about time level n + 1=2 t + t2 2 + O( t3)]u +1 2
u
I + 2 4 2!
2
u = I 2t + 4 t2! 2 + O( t3)]u +1 2:
+1 = n L jk n ; jk n L L = jk n L = jk Adding and subtracting u +1 u = t + O( t3)]u +1 2
n jk ; n jk n n L jk = jk 2
u +1 + u = 2 I + 4 t2! 2 + O( t3)]u +1 2 :
n Eliminating u +1 2 n jk n L = jk = jk u +1 u = 2t (u +1 + u ) + O( t3):
n jk ; n jk L n n jk jk 14 MultiDimensional Parabolic Problemss
Splitting the operator into its component parts I 2t
; 1; L t ]u +1 = I + t + t ]u + O( t3):
22
21 22
L n L jk n L jk This is just the trapezoidal rule integration of (5.2.1) for a time step. Indeed, were we to
discretize the spatial operators L1 and L2 using centered di erences, we would obtain the
same CrankNicolson scheme (5.1.4a) that we rejected. Now, instead, let's factor each
operator as I 2t 1 L t t )(I
21 2 2 = (I
L L t)
22
L t2 4 L1L2: ; Thus, we have
(I ; 2t L1)(I ; 2t L2)u +1 = (I + 2t L1)(I + 2t L2)u +
n n jk jk t2 +1
3
4 L1 L2(u ; u ) + O( t ):
We have already shown that the nexttolast term on the right is O( t3) thus, it may
be combined with the discretization error to obtain
n n jk jk (I ; 2t L1)(I ; 2t L2)u +1 = (I + 2t L1)(I + 2t L2)u + O( t3 ):
If we neglect the local discretization error and discretize the spatial operators, we obtain
n n jk jk (I ; 2t L1 )(I ; 2t L2 )U +1 = (I + 2t L1 )(I + 2t L2 )U
Peaceman and Rachford 5] solved (5.2...
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 Spring '14
 JosephE.Flaherty
 Numerical Analysis

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