= Solution
For a compactly supported variation $\phi\mapsto\phi+\varepsilon\eta$, the first variation of the <energy functional> is
$$
\delta E=\int_{\mathbb R}\{\phi'\eta'+U'(\phi)\eta\}\,dx=\int_{\mathbb R}\{-\phi''+U'(\phi)\}\eta\,dx.
$$
Integration by parts has no endpoint term. Since $\eta$ is arbitrary, the static <Euler-Lagrange field equation> is
$$
\boxed{\phi''=U'(\phi).}
$$
Integrating the specified polynomial derivative gives $U(\phi)=\phi^2/2-\phi^4+\phi^6/2+C$. The polynomial factors as a nonnegative square times $\phi^2$; its global minimum is zero when $C=0$. Therefore
$$
\boxed{U(\phi)=\tfrac12\phi^2(1-\phi^2)^2,\qquad \phi_{\rm vac}=-1,0,1.}
$$
These are the three constant <scalar-field vacua>. For a localized finite-energy <kink>, the two ends approach vacua and the derivative tends to zero. Multiplying the field equation by $\phi'$ gives the first integral
$$
\tfrac12(\phi')^2-U(\phi)=0.
$$
A nonconstant kink must stay within one interval between neighboring vacua. A proposed direct connection from $-1$ to $1$ would reach $\phi=0$ at some finite point, where the first integral forces $\phi'=0$. Uniqueness for the smooth second-order differential equation then forces the identically zero solution, a contradiction. This is the <intermediate-vacuum obstruction to a kink>; two elementary kinks can span the outer vacua only in an infinite-separation limit, not as one finite-width static kink.
There are consequently \b[four oriented elementary sectors: $-1\to0$, $0\to-1$, $0\to1$ and $1\to0$]. If opposite orientations are grouped together, there are two kink families and their antikinks. Spatial reflection reverses orientation, and $\phi\mapsto-\phi$ exchanges the two adjacent intervals; both preserve the energy. All four therefore have the same rest <mass>, even though the central and outer vacua have different fluctuation curvatures.
For the sector $0\to1$, define $W(\phi)=\phi^2/2-\phi^4/4$, so $U=(W')^2/2$ and $W'=\phi(1-\phi^2)$. The <Bogomolny bound> follows by <completing the square>:
$$
E=\frac12\int_{\mathbb R}[\phi'-W'(\phi)]^2dx+W(1)-W(0)\ge\frac14.
$$
It is saturated by the <Bogomolny equation>
$$
\boxed{\phi'=\phi(1-\phi^2),\qquad M_{\rm kink}=\frac14.}
$$
Let $y=\phi^2$ on this positive branch. Then $y'=2y(1-y)$, so $\log[y/(1-y)]=2(x-x_0)$. Thus the explicit <kink in a phi-six model> is
$$
\boxed{\phi_{0\to1}(x)=\frac1{\sqrt{1+e^{-2(x-x_0)}}}.}
$$
It has the required limits and satisfies the first integral. The translated center $x_0$ is arbitrary. All oriented elementary solutions are
$$
\phi_{\sigma,\eta}(x)=\sigma[1+e^{-2\eta(x-x_0)}]^{-1/2},\qquad\sigma,\eta\in\{1,-1\}.
$$
For $\eta=1$ the endpoints are $0\to\sigma$, and for $\eta=-1$ they are $\sigma\to0$. Direct evaluation $\int_0^1\sqrt{2U(\phi)}\,d\phi=\int_0^1\phi(1-\phi^2)d\phi=1/4$ independently gives their common <mass>.
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