Solution

ID: past-exam-of-the-calculus-of-variations-course-of-the-university-of-oxford/2010/3/a/i/solution

The function has weak derivative and the required endpoint values, so it belongs to the admissible Sobolev space. Its fourth power is , and hence
Since the integrand is nonnegative, zero is a global lower bound. Thus is a global minimizer. Its derivative does not belong to , since its power is proportional to ; positive regularization therefore needs different competitors.

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