= Solution
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.
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