ps3sol - ACM 95b/100b Problem Set 3 Solutions Prepared by:...

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ACM 95b/100b Problem Set 3 Solutions Prepared by: Jon Othmer February 3, 2006 Total: 142 points Include grading section: 2 points Problem 1 (40 points) Consider the initial value problem y 0 = f ( t,y ) , y (0) = y 0 . and the trapezoidal rule y n +1 = y n + Δ t 2 ( f n +1 + f n ) , where f n f ( t n ,y n ) and Δ t = t n +1 - t n . a) (15 pts) Show that the method is second order with truncation error T n = - 1 12 Δ t 2 y 000 ( ξ n ) , for some ξ n ( t n ,t n +1 ) , where y is the unknown analytical solution to the IVP. b) (10 pts) Suppose that | y 000 ( t ) | ≤ M for some positive constant independent of t and that f satisfies the Lipschitz condition | f ( 1 ) - f ( 2 ) | ≤ L | y 1 - y 2 | for all real t , y 1 , y 2 , where L is a positive constant independent of t . Show that the solution error e n y ( t n ) - y n satisfies | e n +1 | ≤ | e n | + 1 2 Δ tL ( | e n +1 | + | e n | ) + 1 12 Δ t 3 M. c) (15 pts) For initial condition y 0 = y ( t 0 ) and uniform step size Δ t satisfying Δ tL < 2 deduce the following bound on the solution error | e n | ≤ Δ t 2 M 12 L " ± 1 + 1 2 Δ tL 1 - 1 2 Δ tL ² n - 1 # . Solution 1 a) The truncation error, T n is defined by T n y ( t n +1 ) - y ( t n ) Δ t - f n +1 + f n 2 . We can use Taylor’s theorem to evaluate each of the component pieces in this definition. We have y ( t n +1 ) = y ( t n ) + y 0 ( t n t + y 00 ( t n ) 2 Δ t 2 + y 000 ( t n ) 6 Δ t 3 + O t 4 ) . We also have f n = y 0 ( t n ) 1
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and f n +1 = y 0 ( t n +1 ) = y 0 ( t n ) + y 00 ( t n t + y 000 ( t n ) 2 Δ t 2 + O t 3 ) . Substituting the truncated Taylor series into the expression for T n , we obtain T n = y 0 ( t n ) + y 00 ( t n ) 2 Δ t + y 000 ( t n ) 6 Δ t 2 + O t 3 ) - y 0 ( t n ) + y 00 ( t n t + y 000 ( t n ) 2 Δ t 2 + O t 3 ) + y 0 ( t n ) 2 = y 000 ( t n ) 6 Δ t 2 - y 000 ( t n ) 4 Δ t 2 + O t 3 ) = - y 000 ( ξ n ) 12 Δ t 2 . with ξ n ( t n ,t n +1 ). b) If | y 000 ( t ) | ≤ M , then we have | T n | ≤ Δ t 2 12 M. We can subtract the equation we have for the Trapezoidal iteration y n +1 - y n Δ t - ( f n +1 + f n ) 2 = 0 from the equation T n = y ( t n +1 ) - y ( t n ) Δ t - f ( t n +1 ,y ( t n +1 )) + f ( t n ( t n )) 2 to obtain T n = e n +1 - e n Δ t - Δ f n +1 + Δ f n 2 , or e n +1 = e n + Δ t ± Δ f n +1 + Δ f n 2 ² + Δ tT n , (1) where Δ f n = f ( t n ( t n )) - f ( t n n ) , and similarly with Δ f n +1 . We can use the Lipschitz condition to simplify Δ f n : | Δ f n | ≤ L | y ( t n ) - y n | = L | e n | . Applying the triangle inequality to Equation 1, we find | e n +1 | ≤ | e n | + L Δ t 2 ( | e n | + | e n +1 | ) + Δ t 3 M 12 , (2) as asserted in the problem statement. c) If we re-arrange the terms in Equation 2, we find ± 1 - 1 2 Δ tL ² | e n +1 | ≤ | e n | ± 1 + 1 2 Δ tL ² + 1 12 Δ t 3 Because we are given the value of y ( t 0 ), we can define e 0 = 0, and examine a few terms in the series that results for bounds on e n : | e 1 | ≤ 1 12 Δ t 3 M 1 - 1 2 Δ tL | e 2 | ≤ 1 12 Δ t 3 M 1 - 1 2 Δ tL ³ 1 + 1 2 Δ tL 1 - 1 2 Δ tL + 1 ´ | e 3 | ≤ 1 12 Δ t 3 M 1 - 1 2 Δ tL " ± 1 + 1 2 Δ tL 1 - 1 2 Δ tL ² 2 + 1 + 1 2 Δ tL 1 - 1 2 Δ tL + 1 # .
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This note was uploaded on 01/08/2011 for the course ACM 95b taught by Professor Nilesa.pierce during the Winter '09 term at Caltech.

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ps3sol - ACM 95b/100b Problem Set 3 Solutions Prepared by:...

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