quiz and ans - smaller than necessary Answer We know that...

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AMSC/CMSC 660 Quiz 9 , Fall 2006 1. (10) Recall that a Hamiltonian system is a system of ODEs for which there exists a scalar Hamiltonian function H ( y ) so that y 0 = D y H ( y ) where D is a block-diagonal matrix with blocks equal to 0 1 - 1 0 . Derive the Hamiltonian system for H ( y ) = 1 2 y 2 1 + 1 2 y 2 2 + 1 2 y 2 3 + 1 2 y 2 4 + 1 2 y 2 1 y 2 2 y 2 3 y 2 4 . Answer: y 0 = D y H ( y ) = 0 1 0 0 - 1 0 0 0 0 0 0 1 0 0 - 1 0 y 1 + y 1 y 2 2 y 2 3 y 2 4 y 2 + y 2 1 y 2 y 2 3 y 2 4 y 3 + y 2 1 y 2 2 y 3 y 2 4 y 4 + y 2 1 y 2 2 y 2 3 y 4 = y 2 + y 2 1 y 2 y 2 3 y 2 4 - ( y 1 + y 1 y 2 2 y 2 3 y 2 4 ) y 4 + y 2 1 y 2 2 y 2 3 y 4 - ( y 3 + y 2 1 y 2 2 y 3 y 2 4 ) 1
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2. (10) Suppose we have used the Adams-Bashforth and Adams-Moulton methods of order 3 to form two estimates of y ( t n +1 ), the solution to a differ- ential equation. These formulas are: y ab n +1 = y n + h 12 (23 f n - 16 f n - 1 + 5 f n - 2 ) e rror : 3 h 4 8 y (4) ( η ) . y am n +1 = y n + h 12 (5 f n +1 + 8 f n - f n - 1 ) e rror : - h 4 24 y (4) ( η ) . How would you estimate the local error in the Adams-Moulton formula? How would you use that estimate to change h in order to keep the estimated local error less than a user-supplied local error tolerance τ without taking steps
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Unformatted text preview: smaller than necessary? Answer: We know that if our old values are correct, y ab n +1-y ( t n +1 ) = 3 h 4 8 y (4) ( η ) . y am n +1-y ( t n +1 ) =-h 4 24 y (4) ( ν ) . Subtracting, we obtain y ab n +1-y am n +1 = 3 h 4 8 y (4) ( η )-(-h 4 24 y (4) ( ν )) where η, ν are in the interval containing y ab n +1 , y am n +1 , and the true value. Since 3 / 8+1 / 24 = 10 / 24, the error in AM can be estimated as ² = | y ab n +1-y am n +1 | / 10. Now, if ² > τ , we might reduce h by a factor of 2 and retake the step. If ² << τ , we might double h in preparation for the next step (expecting that the local error might increase by a factor of 2 4 ). 2...
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