Differential Equations Solutions 174

# Differential Equations Solutions 174 - + 1 = j h 2 sin kj n...

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184 Chapter 30. Solutions: Case Study: Fast Solvers and Sylvester Equations (b) The algorithm fails if L ( i, i )+ R ( j, j ) = 0 for some value of i and j . The main diagonal elements of triangular matrices are the eigenvalues of the matrix, so it is necessary and suﬃcient that L and R have no common eigenvalues. (c) If AU + UB = C , then WLW U + UYRY = C . Multiplying on the left by W and on the right by Y , we obtain L b U + b UR = b C . CHALLENGE 30.3. The algorithm of Challenge 2(a) reduces to U ( i, j )= F ( i, j ) / ( L ( i, i )+ R ( j, j )) for i, j =1 ,...,n , which requires n 2 additions and divi- sions. CHALLENGE 30.4. (a) Recall the identities sin( a ± b )=sin a cos b ± cos a sin b. If we form B x times the j th column of V , then the k th element is v k 1 ,j +2 v k,j v k +1 ,j h 2 = α j h 2 µ sin ( k 1) n +1 +2sin kjπ n +1 sin ( k +1) n
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Unformatted text preview: + 1 = j h 2 sin kj n + 1 cos j n + 1 + cos kj n + 1 sin j n + 1 + 2 sin kj n + 1 sin kj n + 1 cos j n + 1 cos kj n + 1 sin j n + 1 = j h 2 2 2 cos j n + 1 sin kj n + 1 = 1 h 2 2 2 cos j n + 1 v k,j = j v k,j . Stacking these elements we obtain B x v j = j v j . (b) This follows by writing the k th component of Vy . CHALLENGE 30.5. See the website for the programs. The results are shown in Figures 30.1 and 30.2. All of the algorithms give accurate results, but as n gets large, the eciency of the fast algorithm of Challenge 4 becomes more apparent....
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## This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.

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