= Square completion for a one-dimensional kink
{title2=$E\geq|W(\phi_+)-W(\phi_-)|$}
For the static energy $E=\int[\phi'^2+W'(\phi)^2]dx/2$, <completing the square> gives
$$
E=\frac12\int(\phi'\mp W')^2dx\pm[W(\phi_+)-W(\phi_-)].
$$
The <Bogomolny bound> is the absolute boundary difference. Equality requires the <Bogomolny equation> $\phi'=\pm W'(\phi)$ with the sign selected by that difference. Differentiating it gives the <Euler-Lagrange field equation> $\phi''=W'W''$. A stationary solution of that second-order equation need not saturate the bound in a general field theory.
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