calculus_of_variations.20100309.4b9712b0b07e45.42100094

# calculus_of_variations.20100309.4b9712b0b07e45.42100094 -...

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Study sheet { Calculus of Variations Harry Dankowicz Mechanical Science and Engineering University of Illinois at Urbana-Champaign danko@uiuc.edu Objective: To investigate the conditions for a local extremum of a functional. Observation 1 Consider the functional J ( q ) = Z 1 0 μ _ q 2 2 ¡ q dt (1) and let f 1 ( s ) def = J ( st ( t ¡ 1)) = 1 6 s (1 + s ) : (2) This is a quadratic function of s which attains its minimum at s = s ¤ def = ¡ 1 = 2, where f 1 ( s ¤ ) = ¡ 1 = 24. On the other hand, let f 2 ( s ) = J (sin( st ) ¡ t sin s ) = cos s ¡ 1 s + sin s 2 ¡ sin 2 s 2 + s 4 ( s + cos s sin s ) : (3) This function attains its mininum at s = s ¤¤ def ¼ 1 : 28875, where f 2 ( s ¤¤ ) ¼ ¡ 0 : 0396599 > f 1 ( s ¤ ). Finally, let f 3 ( s ) = J ¡ e st ¡ 1 ¡ t ( e s ¡ 1) ¢ = (2 + s ( e s ¡ 1))(2 + s + e s ( s ¡ 2)) 4 s : (4) This function attains its minimum at s = s ¤¤¤ def = 0, where f 3 ( s ¤¤¤ ) = 0 > f 1 ( s ¤ ). Now suppose that Q ( t ) is an arbitrary function for which Q (0) = Q (1) = 0 and let f ( s ) = J μ ¡ 1 2 t ( t ¡ 1) + sQ = ¡ 1 24 + s Z 1 0 μ _ Q μ 1 2 ¡ t ¡ Q dt + s 2 2 Z 1 0 _ Q 2 dt = ¡ 1 24 + s 2 2 Z 1 0 _ Q 2 dt (5) where we used integration by parts to evaluate the ¯rst integral. It follows that nonzero values of s necessarily result in values of f larger than f 1 ( s ¤ ). 1

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HHHHH Observation 2 Suppose that q ¤ ( t ) is the special function among all those which equal zero when t = 0 and t = 1, which results in the smallest value of the functional J ( q ) = Z 1 0 μ _ q 2 2 ¡ q dt (6) and suppose that Q ( t ) is an arbitrary function for which Q (0) = Q (1) = 0. Let f ( s ) = J ( q ¤ + sQ ) = Z 1 0 μ _ q ¤ 2 2 ¡ q ¤ + s Z 1 0 ³ _ Q _ q ¤ ¡ Q ´ dt + s 2 2 Z 1 0 _ Q 2 dt = Z 1 0 μ _ q ¤ 2 2 ¡ q ¤ ¡ s Z 1 0 Q q ¤ + 1) dt + s 2 2 Z 1 0 _ Q 2 dt (7) where the middle integral has been simpli¯ed using integration by parts. Clearly the property assumed for q ¤ holds only if the middle integral equals zero for all Q , i.e., if Ä q ¤ = ¡ 1 ) q ¤ = ¡ t 2 2 + C 1 t + C 2 ; (8) where q ¤ (0) = q ¤ (1) = 0 (9) imply that C 1 = 1 2 ;C 2 = 0 ; (10) i.e., q ¤ ( t ) = ¡ 1 2 t ( t ¡ 1) : (11) Theory: Let Q be a collection of functions. Let F
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## calculus_of_variations.20100309.4b9712b0b07e45.42100094 -...

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