For the <kink in a phi-six model>, choose the sector from $\phi=0$ to $\phi=2$ and the increasing sign of the <Bogomolny equation>:
$$
\phi'=\sqrt{2U(\phi)}
=\sqrt2\,\phi(4-\phi^2).
$$
For $y=\phi^2$ this becomes the <logistic differential equation>
$$
y'=2\sqrt2\,y(4-y).
$$
After translating the centre to $x_0$, its solution is
$$
\boxed{
\phi(x)=\frac2{\sqrt{1+e^{-8\sqrt2(x-x_0)}}}}.
$$
It tends to $0$ as $x\to-\infty$ and to $2$ as $x\to+\infty$. Spatial reflection and $\phi\mapsto-\phi$ generate the other kink and antikink sectors.

Because the first-order equation saturates the <Bogomolny bound>, its mass is
$$
\begin{aligned}
M&=\int_{-\infty}^{\infty}
\left[\frac12(\phi')^2+U(\phi)\right]dx
=\int_0^2\sqrt{2U(\phi)}\,d\phi\\
&=\sqrt2\int_0^2\phi(4-\phi^2)d\phi
=\boxed{4\sqrt2}.
\end{aligned}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 313 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-313/1/c/solution