{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

410HW6 - 1 Homework#6 1 Hilbert’s theorem says Suppose...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 Homework #6 1. Hilbert’s theorem says: Suppose that F ( x,u,p ) is continuously differen- tiable for a ≤ x ≤ b and for all ( u,p ) ∈ R 2 . Suppose that F p is con- tinuously differentiable and that F pp ( x,u,p ) > 0 for a ≤ x ≤ b and for all ( u,p ) ∈ R 2 . If u is a weak piecewise continuously differentiable local extremal for F ( u ) = R b a F ( x,u ( x ) ,u ( x )) dx , then u ∈ C 2 [ a,b ]. (a) Show that the piecewise continuously differentiable function u ( x ) = x for 0 ≤ x ≤ 1 / 2 1- x for 1 / 2 ≤ x ≤ 1 is a global minimum for the functional F ( u ) = Z 1 ( u ( x ) 2- 1 ) 2 dx. Solution Note that F ( x,u,p ) = ( p 2- 1) 2 is non-negative, so if there is a function u satisfying F ( x,u,u ) = 0 for all but a finite number of x values in [0 , 1], then such a u is obviously a global minimum. Using the given u , it is clear that F ( x,u,u ) = 0 for all x ∈ [0 , 1] except for x = 1 2 , where the derivative of u is not defined. Immedi- ately, we have F ( u ) = 0 for the given u , so that u must be a global minimum. (b) Why is this example not a counterexample to Hilbert’s theorem? Solution It is quickly calculated that F pp = 12 p 2- 4 , which may not always be positive, as the theorem would require....
View Full Document

{[ snackBarMessage ]}

Page1 / 4

410HW6 - 1 Homework#6 1 Hilbert’s theorem says Suppose...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online