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Part BCSIR NET June 2024boundary-term-gives-a-natural-condition

Boundary term gives a natural condition

Let B(0,1) = {(x,y)∈R2(x, y) \in \mathbb{R}^{2} | x2+y2<1x^{2} + y^{2} < 1} be the open unit disc in R2,∂B(0,1)\mathbb{R}^{2}, \partial{}B(0,1) denote the boundary of B(0,1), and ν\nu denote the unit outward normal to ∂B(0,1)\partial{}B(0,1). Let f:R2→Rf : \mathbb{R}^{2} \to \mathbb{R} be a given continuous function. The Euler–Lagrange equation of the minimization problem min { ½∬_B(0,1) |∇u\nabla{}u|2^{2} dxdy + ½∬B(0,1)e(u2)_B(0,1) e^(u^{2}) dxdy +∫∂B(0,1)fu+ \int_\partial{}B(0,1) f u ds } subject to u∈C1(u \in C^{1}(closure of B(0,1)) is

  1. A.Δu=−\Delta{}u = -ue(u2)^(u^{2}) in B(0,1),∂u/∂ν=fB(0,1), \partial{}u/\partial\nu = f on ∂B(0,1)\partial{}B(0,1)
  2. B.Δu=\Delta{}u = ue(u2)+f^(u^{2}) + f in B(0,1), u = 0 on ∂B(0,1)\partial{}B(0,1)
  3. C.Δu=\Delta{}u = ue(u2)^(u^{2}) in B(0,1),∂u/∂ν=−fB(0,1), \partial{}u/\partial\nu = -f on ∂B(0,1)\partial{}B(0,1)✓ Correct
  4. D.Δu=\Delta{}u = ue(u2)^(u^{2}) in B(0,1),∂u/∂ν+u=fB(0,1), \partial{}u/\partial\nu + u = f on ∂B(0,1)\partial{}B(0,1)

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 boundary and endpoint.

See pricing

50 are analysed free — try those first.

The trap it tests

Boundary and endpoint

The statement turns at the edge of the interval, the domain, or the parameter range.

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

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