Prove that the functional subject to the mixed boundary conditions u(0) = 0, u'(l) = I has

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Prove that the functional

subject to the mixed boundary conditions u(0) = 0, u'(l) = I has no minimizer! Thus, omitting the extra boundary term in (11.106) is a fatal mistake.

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Applied Linear Algebra

ISBN: 978-0131473829

1st edition

Authors: Peter J. Olver, Cheri Shakiban

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Question Posted: Jun 18, 2016 09:06 AM

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Prove that the functional subject to the mixed boundary conditions u(0) = 0, u'(l) = I has

Question:

(u')² dx Plu] =

Step by Step Answer:

For any 0 the function ux boundary conditions but Ju 13 and hence Ju c...View the full answer

Applied Linear Algebra

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