# 5 substituting into 942a yields m tk n1r m 1

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Unformatted text preview: ation n = N X k=1 0 k ( k )nrk (9.4.7a) r: (9.4.7b) when the initial perturbation has the form 0 = N X k=1 0 kk Using (9.4.7a), we see that (9.4.2) will be absolutely stable when j kj 1 k = 1 2 : : : N: (9.4.8) 9.4. Convergence and Stability 21 The eigenvalues and eigenvectors of many tridiagonal matrices are known. Thus, the analysis is often straight forward for one-dimensional problems. Analyses of two- and three-dimensional problems are more di cult however, eigenvalue-eigenvector pairs are known for simple problems on simple regions. Example 9.4.1. Consider the eigenvalue problem (9.4.6) and rearrange terms to get M + tK] k rk = M ; (1 ; ) tK]rk or ( k ; 1)Mrk = ; or k + (1 ; )] tKrk ;Krk = k Mrk where ;1 k + (1 ; )] t Thus, k is an eigenvalue and rk is an eigenvector of ;M;1 K. Let us suppose that M and K correspond to the mass and sti ness matrices of the k = k one-dimensional heat conduction problem of Example 9.2.1. Then, using (9.2.4b,c), we have 2 2 ;1 32 r 3 24 1 32 r 3 k1 6 ;1 2 ;1 7 6 rk2 7 k h 6 1 4 1 7 6 rk12 7 p6 76 7 6 76 k 7 ;h 6 . . . 7 6 ... 7 = 6 6 . . . 7 6 ... 7 : 4 54 5 4 54 5 ;1 2 rk N ; 1 14 rk N ;1 The di usivity p and mesh spacing h have been assumed constant. Also, with Dirichlet boundary conditions, the dimension of this system is N ; 1 rather than N . It is di cult to see in the above form, but writing this eigenvalue-eigenvector problem in component form p (r ; 2r + r ) = k h (r + 4r + r ) j j +1 h j;1 j j+1 6 j;1 j = 1 2 ::: N ;1 we may infer that the components of the eigenvector are rkj = sin kNj j = 1 2 : : : N ; 1: This guess of rk may be justi ed by the similarity of the discrete eigenvalue problem to a continuous one however, we will not attempt to do this. Assuming it to be correct, we substitute rkj into the eigenvalue problem to nd p (sin k (j ; 1) ; 2 sin k j + sin k (j + 1) ) h N N N 22 But and Hence, Parabolic Problems k = 6h (sin k (j ; 1) + 4 sin kNj + sin k (j + 1) ) N N j = 1 2 : : : N ; 1: sin k (j ; 1) + sin k (j + 1) = 2 sin kNj cos k N N N p (cos k ; 1) sin k j = k h (cos k + 2) sin k j : hN N 6 N N cos k =N ; 1 : cos k =N + 2 With cos k =N ranging on ;1 1], we see that ;12p=h2 0. Determining k in k terms of k 1 + k (1 ; ) t = 1 + k t : k= 1; k t 1; k t We would like j k j 1 for absolute stability. With k 0, we see that the requirement that k 1 is automatically satis ed. Demanding the k ;1 yields k = 6p h2 j k j t(1 ; 2 ) 2: 1=2 then 1 ; 2 0 and the above inequality is satis ed for all choices of k and t. Methods of this class are unconditionally absolutely stable. When < 1=2, we have to satisfy the condition 1: pt 2 h 6(1 ; 2 ) If we view this last relation as a restriction of the time step t, we see that the forward Euler method ( = 0) has the smallest time step. Since all other methods listed in Table 9.3.1 are unconditionally stable, there would be little value in using the forward Euler method without lumping the mass matrix. With lumping, the stability restriction of the forward Euler method ac...
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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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