Put $\phi=\phi_Bg(u)$, $u=(x-x_0)/\xi_0$, and $\xi_0^2=-2\kappa/a$. Since $b\phi_B^2=-a$, the interface equation reduces to
$$
g''=2g(g^2-1).
$$
Multiplication by $g'$ and use of $g(\pm\infty)=\pm1$, $g'(\pm\infty)=0$ gives the first integral
$$
(g')^2=(1-g^2)^2.
$$
For the increasing profile, $g'=1-g^2$, so $\operatorname{artanh}g=u$ after shifting $x_0$. Hence the <phi-four diffuse interface> is
$$
\boxed{\phi(x)=\pm\phi_B\tanh\left(\frac{x-x_0}{\xi_0}\right).}
$$
The sign chooses the orientation and the translation zero mode $x_0$ sets the interface position.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 344 / 1 / d / Solution

{title2=$\phi^4$ interface}

For $F=\int[a\phi^2/2+b\phi^4/4+\kappa|\nabla\phi|^2/2]$ with $a<0<b,\kappa$, the planar coexistence interface is
$$
\phi(x)=\phi_B\tanh\frac{x-x_0}{\xi_0},
\qquad
\phi_B=\sqrt{-a/b},
\qquad
\xi_0=\sqrt{-2\kappa/a}.
$$

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