= Solution
Use <Minkowski spacetime> signature $(+,-)$ and put $\kappa=|\lambda|>0$. The <Euler-Lagrange field equation> for the real <scalar field> is
$$
\boxed{\phi_{tt}-\phi_{xx}+2\kappa^2\phi(\phi^2-1)=0.}
$$
Indeed, differentiating the potential $U(\phi)=\kappa^2(\phi^2-1)^2/2$ gives $U'=2\kappa^2\phi(\phi^2-1)$, while the kinetic term contributes $\partial_\mu\partial^\mu\phi$.
Translation invariance gives the symmetric <stress-energy tensor>
$$
T^{\mu\nu}=\partial^\mu\phi\,\partial^\nu\phi
-\eta^{\mu\nu}\mathcal L.
$$
Its divergence is $(\partial_\mu\partial^\mu\phi+U')\partial^\nu\phi$, which vanishes by the <Euler-Lagrange field equation>. Consequently the conserved <energy> and physical spatial <momentum> are
$$
\boxed{E=\int_{\mathbb R}\left[\frac12\phi_t^2+\frac12\phi_x^2+
\frac{\kappa^2}{2}(\phi^2-1)^2\right]dx,\qquad
P=-\int_{\mathbb R}\phi_t\phi_x\,dx.}
$$
The minus sign in $P=\int T^{01}dx$ follows from $\partial^1=-\partial_x$. Conservation assumes the associated stress-energy fluxes vanish at spatial infinity, as they do for the localized fields below.
A static <finite-energy field configuration> approaches vacua $\phi=\pm1$. Multiplying $\phi''=U'(\phi)$ by $\phi'$ gives
$$
\frac12\phi'^2-U(\phi)=\text{constant}=0,
$$
where the vacuum limits fix the constant. For a <kink> increasing from $-1$ to $1$, $\phi'=\kappa(1-\phi^2)$. Integration gives $\operatorname{artanh}\phi=\kappa(x-X)$ and hence
$$
\boxed{\phi_{\mathrm K}(x)=\tanh[\kappa(x-X)].}
$$
Its negative is the <antikink>. The arbitrary center $X$ is a translational <collective coordinate>. The static <kink> mass is its rest <energy>; using $\phi'^2=2U$ and $y=\kappa(x-X)$ gives
$$
M=\int_{\mathbb R}\phi'^2dx
=\kappa\int_{-\infty}^{\infty}\operatorname{sech}^4y\,dy
=\kappa\int_{-1}^{1}(1-z^2)\,dz
=\boxed{\frac{4\kappa}{3}}.
$$
Equivalently, completing the static <energy> into a square gives $E=\frac12\int[\phi'-\kappa(1-\phi^2)]^2dx+\kappa[\phi-\phi^3/3]_{-\infty}^{\infty}$, so this <kink> saturates its <Bogomolny bound>.
A <Lorentz boost> with velocity $v$ gives
$$
\boxed{\phi(t,x)=\tanh[\kappa\gamma(x-X-vt)],
\qquad \gamma=(1-v^2)^{-1/2}.}
$$
Writing $z=\gamma(x-X-vt)$, the derivatives are $\phi_t=-\gamma v\phi_{\mathrm K}'(z)$ and $\phi_x=\gamma\phi_{\mathrm K}'(z)$. The <momentum> integral therefore becomes
$$
P=\gamma^2v\int\phi_{\mathrm K}'(z)^2\,\frac{dz}{\gamma}
=\boxed{\gamma Mv}.
$$
The boosted <energy> is
$$
E=\frac12\left[\gamma(1+v^2)+\gamma^{-1}\right]
\int\phi_{\mathrm K}'(z)^2dz
=\boxed{\gamma M},
$$
because $\gamma^{-1}=\gamma(1-v^2)$. Thus $E^2-P^2=M^2$ and $P/E=v$, exactly the <energy–momentum relation> of a relativistic particle of rest mass $4\kappa/3$. This is the <relativistic energy and momentum of a phi-four kink>. If $\lambda=0$, the double-well potential disappears and there is no localized static kink connecting these vacua.
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