This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 EulerLagrange 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 a and b be curves in R 2 defined by a ( t ) = ( a, t ) and 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 a to the image of b , where f ( x, y ) 1. We know by solving the EulerLagrange 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 a and b . Thus the shortest curves that we seek are horizontal lines from a to b ....
View Full
Document
 Spring '08
 staff

Click to edit the document details