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

The Euler-Lagrange equation isTime-translation invariance gives the conserved energyIndeed, after integrating the spatial term by parts and imposing finite-energy boundary conditions,

The potential is even, so field reflection has $\alpha =-1$; spatial reflection has $\beta =-1$. If $K(x)={\varphi }^{(0,1)}(x)$, the four oriented interpolating solutions, allowing a common translation, areThe first and third are kinks under this orientation convention, and spatial reflection gives their antikinks.

On the sector $0\le \varphi \le 1$, choose the superpotentialThe static energy has the Bogomolny bound completionEquality holds for ${\varphi }^{\prime }=W^{\prime }$, and therefore

The first-order Bogomolny equation isPartial fractions and one integration giveThus

As $x\to -\infty$, the term $-1/\varphi$ dominates the implicit equation, soAs $x\to +\infty$, put $\delta =1-\varphi$. Thenand henceThe profile rises monotonically from $0$ to $1$, with an algebraic left tail and an exponential right tail.