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# 1 11 00 11 00 is n1 11 00 j 1 r 11 00 j n j1 figure

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Unformatted text preview: e manner. Example 2.3.1. Let us use the forward time-centered space scheme (2.3.3a) to obtain an approximate solution of (2.3.1) with = 1 and the initial data U j ::: J j n ( )= x j U 2 if 0 2(1 ; ) if 1 2 x x = x< <x U 12 1 = : In Section 1.2, we saw that the exact Fourier series solution of this problem is 1 X8 ( )= ( )2 k =1 uxt k ;k2 2 t sin k e x: We'll solve this problem with = 0 1 ( = 10) and either = 0 001 or 0.01. Again, the selection of these parameters is rather arbitrary. Using (2.3.3b) with = 0 001 and = 0 1 gives = 0 1. Similarly, with = 0 01 we nd = 1. The two solutions are x : J t : t x : r : t : r : 2.2. A Heat Equation 19 = 0 0.1 0.2 0.3 0.4 0.5 0.6 =0 1 2 3 4 5 6 0.0 0 0.0 0.2 0.4 0.6 0.8 1.0 0.8 0.001 1 0.0 0.2 0.4 0.6 0.8 0.96 0.8 0.002 2 0.0 0.2 0.4 0.6 0.796 0.928 0.796 0.003 3 0.0 0.2 0.4 0.600 0.790 0.901 0.790 0.004 4 0.0 0.2 0.4 0.599 0.782 0.879 0.782 0.005 5 0.0 0.2 0.400 0.597 0.773 0.860 0.773 0.006 6 0.0 0.2 0.400 0.595 0.764 0.842 0.764 0.007 7 0.0 0.200 0.399 0.592 0.755 0.827 0.755 0.008 8 0.0 0.200 0.399 0.589 0.746 0.813 0.746 0.009 9 0.0 0.200 0.398 0.586 0.737...
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