2 with operator splitting the local error is o t2

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Unformatted text preview: (5.2.3) is equivalent to the Peaceman-Rachford form of ADI (5.1.5) for the heat conduction equation. Unlike the Peaceman-Rachford implementation, however, a portion of the operator is neglected at each step. Let's apply (5.2.3) to the variable-coe 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 Multi-Dimensional 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 Crank-Nicolson 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 next-to-last 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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