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Unformatted text preview: 1 Homework #5 1. Find the extremals of F ( u ) = Z 1 ( a + u 00 ( x ) 2 ) dx with end conditions u (0) = 0 , u (0) = 1 , u (1) = 1 , and u (1) = 1. Solution Using the formula on page 42, the Euler-Lagrange equation is F y- d dx F y + d 2 dx 2 F y 00 = 0 , which yields 0 = 0- 0 + d 2 dx 2 (2 u 00 ( x )) = u (4) ( x ) . Integrating yields u ( x ) = c 3 x 3 + c 2 x 2 + c 1 x + c for some constants. Eval- uating for the given boundary conditions gives u ( x ) = x . 2. Find the shortest curve(s) y = u ( x ) for a ≤ x ≤ b with the endpoints u ( a ) = c and u ( b ) = d unspecified. Solve the problem without doing any computations. Solution Let γ a and γ b be curves in R 2 defined by γ a ( t ) = ( a, t ) and γ b ( t ) = ( b, t ) for all t ∈ R . Then the problem is to find an extremal curve of R x 2 x 1 f ( x, y ) p 1 + y 2 dx going from the image of γ a to the image of γ b , where f ( x, y ) ≡ 1. We know by solving the Euler-Lagrange equation that extremals will be straight lines. Also, for functionals of the above form, the tranvsersality conditions reduce to orthogonality. That is, extremal curves need to be perpendicular to both γ a and γ b . Thus the shortest curves that we seek are horizontal lines from γ a to γ b ....
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- Spring '08
- 3 g, Line segment