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Part CCSIR NET December 2025the-boundary-condition-at-x-equals-0-kills-the-log-singular-term-outright-leaving-the-trivial-y-equals-x

The boundary condition at x equals 0 kills the log singular term outright leaving the trivial y equals x

Suppose y(x) is the extremal of the variational problem J(y)=∫01((y′)2sin⁡x+(2cosx)y)J(y) = \int_{0}^{1} ((y')^{2}\sin x + (2cos x)y) dx subject to y(0)=0, y(1)=1. Then which of the following statements are true?

  1. A.y(1/2) = 1
  2. B.y′(0) = 1✓
  3. C.y(1/4) = 2
  4. D.y′(1/2) = 1✓

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

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50 are analysed free — try those first.

Related counterexample: A weak minimum of a functional is a strong minimum

More on this topic

The chapter behind this: Euler–Lagrange, first integrals and null Lagrangians — free to read

From Calculus of Variations › Euler–Lagrange equation and standard functionals

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