= Solution
Differentiate the bound with respect to one $J(q)$, using $\partial S_0/\partial J(q)=-J(q)^{-2}$ and $\partial S_2/\partial J(q)=-q^2J(q)^{-2}$:
$$
0=\frac1{J(q)}-\frac{G(q)}{J(q)^2}-\frac{4B}{VJ(q)^2}(S_2+q^2S_0).
$$
Thus an interior minimizing <Gaussian variational kernel for a gradient quartic interaction> has
$$
\boxed{J(q)=\bar a+\bar\kappa q^2+\gamma q^4,\qquad
\bar a=a+\frac{4B}{V}S_2,\quad\bar\kappa=\kappa+\frac{4B}{V}S_0.}
$$
Use the half-space mode density given in the question. In the <thermodynamic limit>, the coupled <self-consistency equations> are
$$
\boxed{\begin{aligned}
\bar a&=a+2B\int_{|k|<\Lambda}\frac{k^2}{\bar a+\bar\kappa k^2+\gamma k^4}\frac{d^dk}{(2\pi)^d},\\
\bar\kappa&=\kappa+2B\int_{|k|<\Lambda}\frac1{\bar a+\bar\kappa k^2+\gamma k^4}\frac{d^dk}{(2\pi)^d}.
\end{aligned}}
$$
The admissible solution must have $J(q)>0$. The coefficient $\gamma$ is unchanged because the interaction's derivative contributes only a constant and a $q^2$ term.
The formal continuum integrals require the coarse-graining <ultraviolet cutoff> $\Lambda$ in three dimensions: the first radial integrand tends to a constant at large $k$, so taking $\Lambda=\infty$ would leave $\bar a$ divergent. The low-wavevector fluctuation argument is independent of this short-distance regularization.
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