= Solution
Work in <conformal Newtonian gauge>, where $\mathcal R=-\phi+\mathcal Hv$. Set
$$
A=4\pi Ga^2(\bar\rho+\bar P)=\mathcal H^2-\mathcal H',
\qquad c_a^2=\frac{\bar P'}{\bar\rho'}.
$$
The momentum equation gives $v=-(\phi'+\mathcal H\phi)/A$, hence
$$
\mathcal R=-\phi-\frac{\mathcal H}{A}(\phi'+\mathcal H\phi).
$$
This assumes nonzero background enthalpy, as required for the matter comoving slicing. Background energy conservation gives $\bar\rho'=-3\mathcal H(\bar\rho+\bar P)$ and thus
$$
\frac{A'}A=2\mathcal H+\frac{\bar\rho'+\bar P'}{\bar\rho+\bar P}
=-\mathcal H(1+3c_a^2).
$$
The density constraint gives $4\pi Ga^2\delta\rho=\nabla^2\phi-3\mathcal H(\phi'+\mathcal H\phi)$. Insert this into the pressure equation to obtain
$$
\phi''+3\mathcal H(1+c_a^2)\phi'
+\bigl[2\mathcal H'+(1+3c_a^2)\mathcal H^2\bigr]\phi
=c_a^2\nabla^2\phi.
$$
Differentiating the expression for $\mathcal R$, using the formula for $A'/A$ and then $A=\mathcal H^2-\mathcal H'$, groups the terms as
$$
-A\mathcal R'=\mathcal H\left\{
\phi''+3\mathcal H(1+c_a^2)\phi'
+\bigl[2\mathcal H'+(1+3c_a^2)\mathcal H^2\bigr]\phi
\right\}.
$$
The preceding equation proves the <gradient evolution identity for comoving curvature>:
$$
\boxed{-4\pi Ga^2(\bar\rho+\bar P)\mathcal R'
=\mathcal H\,\frac{\bar P'}{\bar\rho'}\,\nabla^2\phi}.
$$
For a Fourier mode, $\nabla^2\phi=-k^2\phi$. On <super-Hubble scales>, $k/\mathcal H\ll1$, its evolution is gradient-suppressed. With the stated regular adiabatic relation between $\mathcal R$ and $\phi$, \b[$\mathcal R$ is constant at leading long-wavelength order]. At finite wavelength there are corrections; the conclusion is not an exact claim that the Laplacian vanishes.
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