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