 Solution

Put $\varphi ={\varphi }_{B}g(u)$, $u=(x-x_0)/{\xi }_0$, and ${\xi }_0^2=-2\kappa /a$. Since $b{\varphi }_B^2=-a$, the interface equation reduces toMultiplication by $g^{\prime }$ and use of $g(\pm \infty )=\pm 1$, $g^{\prime }(\pm \infty )=0$ gives the first integralFor the increasing profile, $g^{\prime }=1-g^2$, so $\mathrm{artanh}g=u$ after shifting $x_0$. Hence the phi-four diffuse interface isThe sign chooses the orientation and the translation zero mode $x_0$ sets the interface position.