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Part BCSIR NET December 2025the-beltrami-identity-gives-k-squared-not-k-check-the-boundary-condition-arithmetic

The beltrami identity gives k squared not k check the boundary condition arithmetic

Suppose y(x) is the extremal of the variational problem J(y)=∫04(y′)2/y2J(y) = \int_{0}^{4} (y')^{2}/y^{2} dx subject to y(0)=1,y(4)=e8y(0) = 1, y(4) = e^{8}. Then which of the following statements is true?

  1. A.y(loge2)=2y(log_{e} 2) = 2.
  2. B.y(loge3)=9y(log_{e} 3) = 9.✓
  3. C.y(loge4)=4y(log_{e} 4) = 4.
  4. D.y(loge5)=5y(log_{e} 5) = 5.

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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