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hw4_sol - ME254/ECE271C Solutions to HW 4 Winter 09 1 Note...

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ME254/ECE271C Winter ’09 Solutions to HW 4 1. Note: These solutions were written several years back, so the terminology may not exactly match that being used in class this quarter, but you should be able to figure things out from context. The technique also uses a short cut that’s somewhat special to this problem. Letting ˙ x 1 = x 2 , ˙ x 2 = u , we have the calculus of variations problem: min x 1 ,x 2 Z 1= t f 0 ( ˙ x 2 ) 2 dt 3 ˙ x 1 = x 2 x 1 ( t f ) 2 + x 2 ( t f ) 2 = 1 x 1 (0) = - 2 2 , x 2 (0) = 5 2 Let the Lagrange multiplier function associated with the differential constraint be λ . Then, we seek an extremal for: L := ( ˙ x 2 ) 2 + λ ( t ) [ ˙ x 1 - x 2 ] . The E-L equations are: x 2 = - λ ˙ λ = 0 λ ( t ) = c 1 Hence, x 2 = - c 1 ˙ x 2 = - 1 2 c 1 t + c 2 x 2 ( t ) = - 1 4 c 1 t 2 + c 2 t + 5 2 . (1) Since ˙ x 1 = x 2 , we have x 1 ( t ) = - 1 12 c 1 t 3 + 1 2 c 2 t 2 + 5 2 t - 2 2 . (2) From the target set we have x 1 (1) 2 + x 2 (1) 2 = 1 - 1 4 c 1 + c 2 + 5 2 2 + - 1 12 c 1 + 1 2 c 2 + 3 2 2 = 1 (3) which is one of the equations to be solved to determine c 1 and
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