hw0911ans - Homework#11 Due April 16 Read 27 and 28 in...

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Unformatted text preview: Homework #11 Due April 16. Read Sections 25, 26, 27, and 28 in Gelfand & Fomin. Since what I am doing in class differs from what is in the book, I will give a brief outline of the subject as I covered it. We are looking at the functional F ( u ) = integraldisplay b a F ( x, u ( x ) , u ′ ( x ) with (1) u ( a ) = c and u ( b ) = d. (2) where F is continuously differentiable. We made the following defini- tions. Definition 1. The function u ∈ C 1 is a weak mimimizer for F if there is an ǫ > 0 such that F ( v ) ≥ F ( u ) for all v ∈ C 1 which satisfy the boundary conditions (2) and || v- u || 1 < ǫ. Definition 2. The function u ∈ C 1 is a strong mimimizer for F if there is an ǫ > 0 such that F ( v ) ≥ F ( u ) for all v ∈ C 1 which satisfy the boundary conditions (2) and || v- u || < ǫ. We were able to prove the next result. Theorem 3. 1. If u is a weak minimizer for F , then δ F ( u, φ ) = 0 and δ 2 F ( u, φ ) ≥ for all φ ∈ C 1 ( a, b ) . 2. If u ∈ C 1 [ a, b ] is a weak extremal for F and there is k > suxh that δ 2 F ( u, φ ) ≥ k integraldisplay b a bracketleftbig φ ( x ) 2 + φ ′ ( x ) 2 bracketrightbig dx for all φ ∈ C 1 ( a, b ) then u is a strict weak mimimizer for F . A strong minimizer is always a weak minimizer, but this example shows that the opposite is not always true. Example 4. Consider the funtional F ( u ) = integraltext 1 [ u ′ ( x ) 2 + u ′ ( x ) 3 ] dx with u (0) = 0 and u (1) = 0. The function u ( x ) = 0 is a strict weak minimizer, but not a strong minimizer....
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This note was uploaded on 09/09/2009 for the course MATH 410 taught by Professor Staff during the Spring '08 term at Maryland.

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hw0911ans - Homework#11 Due April 16 Read 27 and 28 in...

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