Write the <scalar field> energy as $E=T+V$, where
$$
T=\frac12\int_{-\infty}^{\infty}(\phi')^2\,dx,
\qquad
V=\int_{-\infty}^{\infty}U(\phi)\,dx.
$$
Under the <Derrick scaling> $\phi_\lambda(x)=\phi(\lambda x)$, a <change of variables> gives
$$
E(\lambda)=\lambda T+\lambda^{-1}V.
$$
A finite-energy solution of the <Euler-Lagrange equation> is stationary under this admissible variation. The <Derrick virial identity> is therefore
$$
0=E'(1)=T-V,
\qquad
\boxed{\frac12\int_{-\infty}^{\infty}(\phi')^2\,dx
=\int_{-\infty}^{\infty}U(\phi)\,dx}.
$$

The static field equation is $\phi''=U'(\phi)$, so integration gives
$$
U(\phi)=\frac12\phi^6-\phi^4+\frac12\phi^2+C
=\frac12\phi^2(1-\phi^2)^2+C.
$$
The polynomial before $C$ is nonnegative and vanishes, so requiring the minimum to be zero fixes $C=0$. Hence the <vacuum manifold> is
$$
\boxed{U^{-1}(0)=\{-1,0,1\}},
$$
which has three elements. A finite-energy <scalar-field kink> can join only adjacent vacua: a solution cannot cross the intermediate vacuum at finite $x$ because its first integral would have $\phi=\phi'=0$ there and the <Picard-Lindelof theorem> would make it constant. There are therefore four oriented topological sectors,
$$
-1\to0,\qquad0\to-1,\qquad0\to1,\qquad1\to0,
$$
comprising two increasing kinks and their two antikinks. Symmetry under $\phi\mapsto-\phi$ and spatial reflection generates all four from one profile.

For the $0\to1$ sector, completing the square gives the <Bogomolny bound>
$$
\begin{aligned}
E
&=\frac12\int_{-\infty}^{\infty}
\left[\phi'-\phi(1-\phi^2)\right]^2dx
+\int_{-\infty}^{\infty}\phi'\phi(1-\phi^2)\,dx\\
&\geq\int_0^1\phi(1-\phi^2)\,d\phi.
\end{aligned}
$$
Equality holds for the <Bogomolny equation>
$$
\boxed{\phi'=\phi(1-\phi^2)}.
$$
With $y=\phi^2$, this becomes the <logistic differential equation> $y'=2y(1-y)$. Translation invariance supplies an arbitrary center $x_0$, and the explicit <kink in a phi-six model> is
$$
\boxed{\phi(x)=\frac1{\sqrt{1+e^{-2(x-x_0)}}}}.
$$
It tends to $0$ and $1$ at the two spatial ends and saturates the bound. Its mass is consequently
$$
\boxed{M=\int_0^1\phi(1-\phi^2)\,d\phi=\frac14}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 313 / 1 / Solution

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
$$
\begin{aligned}
M&=\int_{-\infty}^{\infty}
\left[\frac12(\phi')^2+U(\phi)\right]dx
=\int_0^2\sqrt{2U(\phi)}\,d\phi\\
&=\sqrt2\int_0^2\phi(4-\phi^2)d\phi
=\boxed{4\sqrt2}.
\end{aligned}
$$

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

{title2=$\phi^6$ kink}

Kinks in a phi-six model commonly arise from a nonnegative degree-six potential with three degenerate vacua. For the unit-vacuum normalization
$$
U(\phi)=\frac12\phi^2(1-\phi^2)^2,
$$
the kink joining $0$ to $1$ is
$$
\phi(x)=\frac1{\sqrt{1+e^{-2(x-x_0)}}}
$$
and has mass $1/4$. A rescaled model with $U(\phi)=\phi^2(\phi^2-4)^2$ has the elementary kink
$$
\phi(x)=\frac2{\sqrt{1+e^{-8\sqrt2(x-x_0)}}},
$$
and its mass is $4\sqrt2$ in the normalization $E=\int[(\phi')^2/2+U]dx$.


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