Differentiating the potential gives
$$
U'(\phi)=2(1-\phi^2)(\epsilon-\phi).
$$
Thus $\phi=\pm1$ are stationary points, and
$$
U''(1)=4(1-\epsilon)>0,
\qquad
U''(-1)=4(1+\epsilon)>0
$$
because $|\epsilon|<1$. Both are local minima. Their energies are
$$
U(1)=\frac{4\epsilon}{3},
\qquad
U(-1)=-\frac{4\epsilon}{3}.
$$

For $\epsilon=0$, the static energy can be completed to a square:
$$
E=\int dx\left\{\frac12[\phi'-(1-\phi^2)]^2
+\phi'(1-\phi^2)\right\}.
$$
The increasing <scalar-field kink> therefore obeys the first-order <Bogomolny equation>
$$
\boxed{\phi'=1-\phi^2}.
$$
With center $X$, its solution and energy are
$$
\boxed{\phi_K(x)=\tanh(x-X)},
\qquad
\boxed{E_K=\int_{-1}^{1}(1-\phi^2)\,d\phi=\frac43}.
$$
The antikink uses the opposite sign.

For small positive $\epsilon$, the true vacuum $\phi=-1$ lies below the false vacuum $\phi=1$ by
$$
\Delta U=U(1)-U(-1)=\frac{8\epsilon}{3}.
$$
This pressure exerts force $\Delta U$ on a kink with $-1$ on its left and $+1$ on its right. Dividing by its leading mass $4/3$ gives acceleration toward the false-vacuum side:
$$
\boxed{\ddot X=2\epsilon+O(\epsilon^2)}.
$$

An antikink followed by a kink encloses a region of the lower vacuum while approaching $\phi=1$ at both infinities. Vacuum pressure pushes the pair apart, whereas their attraction pulls them together. At a static separation $s$,
$$
\frac{8\epsilon}{3}=32e^{-2s},
$$
so
$$
\boxed{s\simeq\frac12\log\frac{12}{\epsilon}}.
$$
This estimate is self-consistent for $\epsilon\ll1$, when the two soliton cores are well separated.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 308 / 1 / Solution

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
$$
\boxed{M=\int_0^1\phi(1-\phi^2)\,d\phi=\frac14}.
$$

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

A scalar-field kink is a one-dimensional finite-energy static solution approaching different vacua as $x\to-\infty$ and $x\to+\infty$. Its boundary values define a topological sector.

= Kink
{synonym}


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