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Part CCSIR NET June 2023coupled-euler-lagrange

Coupled euler lagrange

Let y(x) and z(x) be the stationary functions (extremals) of dx subject to y(0) = 1, y(1) = 0, z(0) = −1, z(1) = 2. Which of the following statements are correct?

  1. A.z(x) + 3y(x) = 2 for x ∈ [0,1].✓
  2. B.3z(x) + y(x) = 2 for x ∈ [0,1].
  3. C.y(x) + z(x) = 2x for x ∈ [0,1].✓
  4. D.y(x) + z(x) = x for x ∈ [0,1].

Solution

Euler–Lagrange: 2y″ + z″ = 0 and 2z″ + y″ = 0 ⇒ y″ = z″ = 0. So y = 1 − x and z = −1 + 3x. Then y + z = 2x and z + 3y = 2.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: A weak minimum of a functional is a strong minimum

More on this topic

From Calculus of Variations › Euler–Lagrange equation and standard functionals

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