Varying the quadratic action gives
$$
(z^2\mathcal R')'-z^2\nabla^2\mathcal R=0.
$$
For a Fourier mode,
$$
\boxed{\mathcal R_k''+2\frac{z'}z\mathcal R_k'
+k^2\mathcal R_k=0}.
$$
At late times $\epsilon\propto a^{-6}$, so
$$
z=a\sqrt{2\epsilon}\propto a^{-2}\propto\eta^2,
\qquad
\boxed{\frac{z'}z\simeq\frac2\eta}.
$$
The equation becomes $\mathcal R_k''+4\mathcal R_k'/\eta+k^2\mathcal R_k=0$. At $k=0$, its independent solutions are a constant and $\eta^{-3}$. The leading late-time solution in the form given in the question is therefore
$$
\boxed{\mathcal R_k(\eta)
=A_k\frac{1+ik\eta}{\eta^3}e^{-ik\eta}},
\qquad
\boxed{q=3}.
$$
This is the <growing curvature perturbation in ultra-slow-roll inflation>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 312 / 2 / iii / Solution

In ultra-slow-roll inflation, $z=a\sqrt{2\epsilon}\propto a^{-2}\propto\eta^2$. The comoving curvature mode equation is
$$
\mathcal R_k''+\frac4\eta\mathcal R_k'+k^2\mathcal R_k=0,
$$
and its Bunch-Davies mode is proportional to $(1+ik\eta)e^{-ik\eta}/\eta^3$. Unlike ordinary slow roll, the superhorizon curvature perturbation grows as $a^3$.


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