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Unformatted text preview: ACM 95b/100b Problem Set 3 Solutions Prepared by: Lei Zhang January 24, 2007 Total: 142 points Include grading section: 2 points Problem 1 (40 points) Consider the initial value problem y = f ( t, y ) , y (0) = y . and the trapezoidal rule y n +1 = y n + t 2 ( f n +1 + f n ) , where f n f ( t n , y n ) and t = t n +1 t n . a) (15 pts) Show that the method is second order with truncation error T n = 1 12 t 2 y 000 ( n ) , for some 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 ( t, y 1 ) f ( t, y 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 tL (  e n +1  +  e n  ) + 1 12 t 3 M. c) (15 pts) For initial condition y = y ( t ) and uniform step size t satisfying tL < 2 deduce the following bound on the solution error  e n  t 2 M 12 L " 1 + 1 2 tL 1 1 2 tL n 1 # . Solution 1 a) The truncation error, T n is defined by T n y ( t n +1 ) y ( t n ) t f n +1 + f n 2 . We can use Taylors theorem to evaluate each of the component pieces in this definition. We have y ( t n +1 ) = y ( t n ) + y ( t n ) t + y 00 ( t n ) 2 t 2 + y 000 ( t n ) 6 t 3 + O ( t 4 ) . We also have f n = y ( t n ) 1 and f n +1 = y ( t n +1 ) = y ( t n ) + y 00 ( t n ) t + y 000 ( t n ) 2 t 2 + O ( t 3 ) . Substituting the truncated Taylor series into the expression for T n , we obtain T n = y ( t n ) + y 00 ( t n ) 2 t + y 000 ( t n ) 6 t 2 + O ( t 3 ) y ( t n ) + y 00 ( t n ) t + y 000 ( t n ) 2 t 2 + O ( t 3 ) + y ( t n ) 2 = y 000 ( t n ) 6 t 2 y 000 ( t n ) 4 t 2 + O ( t 3 ) = y 000 ( n ) 12 t 2 . with n ( t n , t n +1 ). b) If  y 000 ( t )  M , then we have  T n  t 2 12 M. We can subtract the equation we have for the Trapezoidal iteration y n +1 y n t ( f n +1 + f n ) 2 = 0 from the equation T n = y ( t n +1 ) y ( t n ) t f ( t n +1 , y ( t n +1 )) + f ( t n , y ( t n )) 2 to obtain T n = e n +1 e n t f n +1 + f n 2 , or e n +1 = e n + t f n +1 + f n 2 + tT n , (1) where f n = f ( t n , y ( t n )) f ( t n , y n ) , and similarly with f n +1 . We can use the Lipschitz condition to simplify f n :  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 t 2 (  e n  +  e n +1  ) + t 3 M 12 , (2) as asserted in the problem statement....
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 Winter '09
 NilesA.Pierce

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