= Solution
Take $\lambda>0$, $m>0$, use metric $(+,-)$ and set $\kappa=\sqrt{2\lambda}\,m$. The two <scalar-field vacua> are $\phi=\pm m$; the positive <kink> joins $-m$ at the left end to $m$ at the right. The <square completion for a one-dimensional kink> gives
$$
E=\frac12\int_{\mathbb R}\left[\phi_x-\sqrt{2\lambda}(m^2-\phi^2)\right]^2dx+\sqrt{2\lambda}\left[m^2\phi-\frac{\phi^3}{3}\right]_{-\infty}^{+\infty}.
$$
Thus its <Bogomolny bound> is $E\geq4\sqrt{2\lambda}m^3/3$. Equality requires $\phi_x=\sqrt{2\lambda}(m^2-\phi^2)$; separating variables, or differentiating the resulting <hyperbolic tangent>, gives
$$
\boxed{\phi_K(x)=m\tanh\!\bigl[\kappa(x-X)\bigr],\qquad M=E_K=\frac43\sqrt{2\lambda}\,m^3.}
$$
$X$ is the translational <collective coordinate>. The <antikink> is $-\phi_K$ for the centered odd profile. Differentiating the first-order equation gives the static <Euler-Lagrange field equation> $\phi_{xx}=4\lambda\phi(\phi^2-m^2)$. Hence the <Bogomolny equation> really does solve the second-order theory. The two distinct vacua and the above <kink> require the positive parameters; the degenerate case $m=0$, or the unstable negative-$\lambda$ potential, does not have this topological <kink> sector.
For a slowly moving <kink>, put $\phi(x,t)=\phi_K(x-X(t))$. The saturated first-order equation implies $V(\phi_K)=\phi_K'^2/2$, so
$$
\int_{\mathbb R}\phi_K'^2dx=E_K=M.
$$
The <stress-energy tensor> has $T^{00}=\phi_t^2/2+\phi_x^2/2+V$ and $T^{01}=-\phi_t\phi_x$. Using $\phi_t=-v\phi_K'$ in the approximate translate therefore gives $E=M+Mv^2/2$ and $P=Mv$. Equivalently, substitution into the action gives the <collective-coordinate effective Lagrangian> $L_{\rm eff}=-M+M\dot X^2/2$. These are the <translational dynamics of a phi-four kink>, with
$$
\boxed{E=M+\frac12Mv^2+O(v^4),\qquad P=Mv+O(v^3).}
$$
The displayed remainder orders refer to the exact constant-velocity solution. The uncontracted translated profile is only an approximation: its error as a field starts at $O(v^2)$, but the static profile is an <energy> stationary point, so the corresponding static-energy error starts at $O(v^4)$ and does not change the displayed $v^2$ coefficient.
Since the relativistic action is invariant under <Lorentz transformations>, a <Lorentz boost> gives the exact moving <kink>, for $|v|<1$,
$$
\boxed{\phi(x,t)=m\tanh\!\left[\kappa\gamma_v(x-X-vt)\right],\qquad \gamma_v=(1-v^2)^{-1/2}.}
$$
Its conserved <energy> and <momentum> are $E=\gamma_v M$ and $P=\gamma_v Mv$, whose expansions agree with the preceding calculation. The width contraction is essential to solving the exact time-dependent equation.
Use the centered profile $K(x)=m\tanh(\kappa x)$. An appropriate approximate separated-lump initial condition is
$$
\boxed{\phi(x,0)=K(x+a)-K(x)+K(x-a),\qquad \phi_t(x,0)=0.}
$$
It tends to $-m$ as $x\to-\infty$ and $m$ as $x\to+\infty$. Near $-a$ and $a$ it is a positive <kink>, and near zero it is an <antikink>; the other two tails cancel to exponentially small accuracy. The condition $\kappa a\gg1$ makes these interpretations accurate. This sum is legitimate smooth initial data, not an exact static multi-kink solution. Its <topological charge> is
$$
Q=\frac{\phi(+\infty)-\phi(-\infty)}{2m}=1,
$$
and its <energy> is close to $3M$, with exponentially small interactions.
Adjacent <kink>–<antikink> pairs attract: the <kink–antikink attraction from the stress tensor> gives a negative midpoint pressure and hence an inward force on each outside <kink>. Direct overlap of the two outer <kink> tails is exponentially smaller than each adjacent <kink>–<antikink> overlap. The initial field is odd and its velocity zero, so the equation's reflection-plus-sign symmetry preserves $\phi(-x,t)=-\phi(x,t)$. The central zero remains at $x=0$ and the outer cores move in symmetrically. They collide with the central <antikink>. Excess <energy> can excite localized <kink> oscillations and outgoing radiation; the usual relaxational outcome is a remaining centered <kink> plus radiation after transient collisions or oscillations. On an infinite line radiation can carry <energy> away from the core even though total <energy> is conserved. Conservation of $Q=1$ prevents complete disappearance into vacuum, but it does not determine every bounce or exclude a temporarily re-emerging <kink>–<antikink> pair. The nonintegrable collision should not be described as exact elastic passage of all three solitons.
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