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Part CCSIR NET December 2024weak-minimum-need-not-be-strong

Weak minimum need not be strong

Define S := {y∈C1[−1,1]:y(−1)=−1,y(1)=3y \in C^{1}[-1, 1] : y(-1) = -1, y(1) = 3}. Let φ\varphi be the extremal of the functional J:S→RJ : S \to \mathbb{R} given by J[y]=∫J[y] = \int₋11[(y′)3+(y′)2]_{1}^{1}[(y')^{3} + (y')^{2}]dx. Define ‖y‖(∞):=max⁡(x∈[−1,1])(\infty) := \max(x\in[-1,1])|y(x)| for every y ∈ S and let B0(φ,ε):=B_{0}(\varphi, \varepsilon) := {y ∈ S : ‖y−φy - \varphi‖(∞)<ε(\infty) < \varepsilon}, B1(φ,ε):=B_{1}(\varphi, \varepsilon) := {y ∈ S : ‖y−φy - \varphi‖(∞)+(\infty) + ‖y′−φ′y' - \varphi'‖(∞)<ε(\infty) < \varepsilon}. Then which of the following statements are true?

  1. A.φ(x)=2x+1\varphi(x) = 2x + 1 for every x ∈ [−1, 1]✓ Correct
  2. B.There exists ε>0\varepsilon > 0 such that J[y]≥J[φ]J[y] \ge J[\varphi] for every y∈B0(φ,ε)y \in B_{0}(\varphi, \varepsilon)
  3. C.There exists ε>0\varepsilon > 0 such that J[y]≥J[φ]J[y] \ge J[\varphi] for every y∈B1(φ,ε)y \in B_{1}(\varphi, \varepsilon)✓ Correct
  4. D.There exists ε>0\varepsilon > 0 such that J[y]≤J[φ]J[y] \le J[\varphi] for every y∈B1(φ,ε)y \in B_{1}(\varphi, \varepsilon)

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 pointwise vs uniform.

See pricing

50 are analysed free — try those first.

The trap it tests

Pointwise vs uniform

The mode of convergence — or of continuity — was the wrong one.

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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