 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-4/16a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Write the admissible variation as

y_{ε} = y + ε η

, where

η (x) \to 0

at both ends. Differentiating the functional under the integral sign gives the first variation

δ I [y; η] = \int_{- \infty}^{\infty} (y^{'} η^{'} + U (y) U^{'} (y) η) d x .

δ I [y; η] = \int_{- \infty}^{\infty} [- y^{''} + U (y) U^{'} (y)] η d x .

y^{''} = U (y) U^{'} (y) .

A second differentiation gives the second variation

δ^{2} I [y; η] = \int_{- \infty}^{\infty} {(η^{'})^{2} + [U^{'} (y)^{2} + U (y) U^{''} (y)] η^{2}} d x .

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