Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 4 / 16A / c

Table of contents
- Solution c

Solution

For

U (y) = c^{2} - y^{2}

, completing the square gives the Bogomolny bound

I [y] = \frac{1}{2} \int_{- \infty}^{\infty} [(y^{'} - U (y))^{2} + 2 U (y) y^{'}] d x = \frac{1}{2} \int_{- \infty}^{\infty} (y^{'} - U (y))^{2} d x + \int_{- c}^{c} (c^{2} - y^{2}) d y = \frac{1}{2} \int_{- \infty}^{\infty} (y^{'} - U (y))^{2} d x + \frac{4 c ^{3}}{3} .

Consequently

I [y] \geq \frac{4 c ^{3}}{3},

with equality exactly when the Bogomolny equation

y^{'} = c^{2} - y^{2}

holds everywhere. Separating variables, or differentiating the proposed form directly, gives all solutions with the required limits:

y (x) = c tanh {c (x - x_{0})}, x_{0} \in R .

These are the translated phi-four kinks.

Differentiating the first-order equation yields

y^{''} = - 2 y y^{'} = (- 2 y) (c^{2} - y^{2}) = U^{'} (y) U (y),

which is precisely the Euler-Lagrange equation from part (a). Thus every configuration saturating the first-order bound is automatically a stationary solution of the second-order variational equation.

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