On the sector $0\leq\phi\leq1$, choose the <superpotential>
$$
W'(\phi)=\phi^2(1-\phi^2),
\qquad
\boxed{W(\phi)=\frac{\phi^3}{3}-\frac{\phi^5}{5}}.
$$
The static energy has the <Bogomolny bound> completion
$$
E=\frac12\int_{\mathbb R}(\phi'-W')^2dx
+W(1)-W(0)\geq W(1)-W(0).
$$
Equality holds for $\phi'=W'$, and therefore
$$
\boxed{E[\phi^{(0,1)}]=\frac13-\frac15=\frac2{15}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 313 / 1 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-313/1/ii/solution