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

Boundary term gives a natural condition

Let B(0,1) = { $(x, y) \in R^{2}$ | $x^{2} + y^{2} < 1$ } be the open unit disc in $R^{2}, \partial B (0, 1)$ denote the boundary of B(0,1), and $ν$ denote the unit outward normal to $\partial B (0, 1)$ . Let $f : R^{2} \to R$ be a given continuous function. The Euler–Lagrange equation of the minimization problem min { ½∬_B(0,1) | $\nabla u$ | $^{2}$ dxdy + ½∬ $_{B} (0, 1) e^{(} u^{2})$ dxdy $+ \int_{\partial} B (0, 1) f u$ ds } subject to $u \in C^{1} ($ closure of B(0,1)) is

A. $Δ u = -$ ue $^{(} u^{2})$ in $B (0, 1), \partial u / \partial ν = f$ on $\partial B (0, 1)$
B. $Δ u =$ ue $^{(} u^{2}) + f$ in B(0,1), u = 0 on $\partial B (0, 1)$
C. $Δ u =$ ue $^{(} u^{2})$ in $B (0, 1), \partial u / \partial ν = - f$ on $\partial B (0, 1)$ ✓
D. $Δ u =$ ue $^{(} u^{2})$ in $B (0, 1), \partial u / \partial ν + u = f$ on $\partial B (0, 1)$

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.

See pricing

50 are analysed free — try those first.

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

Boundary term gives a natural condition

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