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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....
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- Spring '08
- Derivative, Mathematical analysis, Continuous function, Boundary conditions, Hilbert, FP