tailieunhanh - Handbook of mathematics for engineers and scienteists part 148

Handbook of mathematics for engineers and scienteists part 148. Tài liệu toán học quốc tế để phục vụ cho các bạn tham khảo, tài liệu bằng tiếng anh rất hữu ích cho mọi người. | . Calculus of Variations and Optimal Control 997 Example 3 Hilbert s example . In this example a unique admissible extremal exists that provides a global extremum but is not a differentiable function. Let J .r f t2 3 x t 2 dt min x x t x 0 0 x 1 1. Jo The Euler equation becomes dt 2t2 3x t 0. The extremals are given by the equation x t C1t1 3 C2. The unique extremal satisfying the endpoint conditions is the function x t t1 3 which does not belong to the class C1 0 1 . Nevertheless it provides the global minimum in this problem on the set of all absolutely continuous functions x t that satisfy the boundary conditions and for which the integral J is finite. Indeed J t J x ft f1 1 .A2 . 2 r . r1 . I t .3 2 3 h t dt J X 3 J h t dt J t ht dt - J Example 4 Weierstrass s example . In this example the problem has no solutions and no admissible extremals even in the class of absolutely continuous functions. Let J x i t2 x t 2 dt min x x t x 0 0 x 1 1. J0 The Euler equation becomes dt 2t2xt 0. The extremals are given by the equation x t C1 t 1 C2 none of them satisfies the boundary condition x 0 0. Furthermore J x 0 and J x 0 for any absolutely continuous function x t . Obviously the lower bound in this problem is zero. To prove this it suffices to consider the sequence of admissible functions xn t arctan nt arctann. One has J X z1 t nÙta dt 1 n J0 1 n2t2 2 arctan2 n J0 dt t1 dt arctan2 n J 1 n n2t2 arctan2 n 0. Example 5. In this example there exists a unique admissible extremal that does not provide an extremum. Let 3n 2 J x x t 2 - x2 dt min x x t x 0 x 3n 2 0. Jo The Euler equation becomes x t x 0. The extremals are given by the equation x t C1 sin t C2 cos t. The only extremal satisfying the endpoint conditions is x t 0. The admissible extremal x t 0 does not provide a local minimum. Consider the sequence of admissible functions xn t n 1 sin 2t 3 . Obviously xn t 0 in C 0 3n 2 but 5n J xn 0 JM- 12n2 This example shows that the Euler equation is a necessary but not .