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Part CCSIR NET June 2025meeting-the-boundary-conditions-does-not-make-a-function-an-extremal

Meeting the boundary conditions does not make a function an extremal

For α≥0\alpha \ge 0, consider the functional Jα[y]=∫12(y′)2/xαJ_\alpha[y] = \int_{1}^{2} (y')^{2}/x^\alpha dx defined for all continuously differentiable functions defined on the interval [1, 2] satisfying the conditions y(1) = 1, y(2) = 2. Then which of the following statements are true?

  1. A.y(x)=(1/15)(x4+14)y(x) = (1/15)(x^{4} + 14) is an extremal for J3J_{3}✓ Correct
  2. B.y(x)=(1/3)(x2+2)y(x) = (1/3)(x^{2} + 2) is an extremal for J1J_{1}✓ Correct
  3. C.y(x) = x is an extremal for J0J_{0}✓ Correct
  4. D.y(x)=(1/2)(x2−x+2)y(x) = (1/2)(x^{2} - x + 2) is an extremal for J1J_{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, and why this one tests converse assumed.

See pricing

50 are analysed free — try those first.

The trap it tests

Converse assumed

The theorem runs one way. You used it in the other.

Drill statements like this

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

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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