Skip to content
Part CCSIR NET December 2024small-values-permit-large-slopes

Small values permit large slopes

Define S := {y∈C1[0,1]:y(0)=y(1)=0y \in C^{1}[0, 1] : y(0) = y(1) = 0}, ‖y‖(∞):=max⁡(x∈[0,1])(\infty) := \max(x\in[0,1])|y(x)| for every y∈S,B0(0,ε):=y \in S, B_{0}(0, \varepsilon) := {y ∈ S : ‖y‖(∞)<ε(\infty) < \varepsilon}, B1(0,ε):=B_{1}(0, \varepsilon) := {y ∈ S : ‖y‖(∞)+(\infty) + ‖y′‖(∞)<ε(\infty) < \varepsilon}. Consider the functional J:S→RJ : S \to \mathbb{R} given by J[y]=∫01[(y′)2−2x(y′)4]J[y] = \int_{0}^{1}[(y')^{2} - 2x(y')^{4}]dx, then there exists an ε>0\varepsilon > 0 such that

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

ShareWhatsAppTelegram