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Part CCSIR NET December 2023weak-vs-strong-extremum

Weak vs strong extremum

Define S = {y∈C1[0,π]:y(0)=y(π)=0y \in C^{1}[0, \pi] : y(0) = y(\pi) = 0}, ‖f‖∞=max⁡_\infty = \max |f|, B0(0,ε)=B_{0}(0, \varepsilon) = {f ∈ S : ‖f‖∞<ε_\infty < \varepsilon} and B1(0,ε)=B_{1}(0, \varepsilon) = {f ∈ S : ‖f‖∞+_\infty + ‖f′‖∞<ε_\infty < \varepsilon}. For the functional J[y]=∫0π(1−(y′)2)y2J[y] = \int_{0}^\pi (1 - (y')^{2}) y^{2} dx, there exists ε>0\varepsilon > 0 such that

  1. A.J[y] ≤ J[0] for all y∈B0(0,ε)y \in B_{0}(0, \varepsilon)
  2. B.J[y] ≤ J[0] for all y∈B1(0,ε)y \in B_{1}(0, \varepsilon)
  3. C.J[y] ≥ J[0] for all y∈B0(0,ε)y \in B_{0}(0, \varepsilon)
  4. D.J[y] ≥ J[0] for all y∈B1(0,ε)y \in B_{1}(0, \varepsilon)✓ Correct

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 existence vs uniqueness.

See pricing

50 are analysed free — try those first.

The trap it tests

Existence vs uniqueness

A theorem giving one was read as giving both.

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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