= Solution
Let $\theta=\nabla\mathbin\cdot\mathbf v$. Vanishing <vorticity> implies in Fourier space
$$
\mathbf v(\mathbf q)=-i\frac{\mathbf q}{q^2}\theta(\mathbf q).
$$
Fourier transforming $\nabla\cdot(\delta\mathbf v)$ in the continuity equation gives
$$
\delta'(\mathbf k)+\theta(\mathbf k)
=-\int_{\mathbf q_1,\mathbf q_2}
\delta_D(\mathbf k-\mathbf q_1-\mathbf q_2)
\alpha(\mathbf q_1,\mathbf q_2)
\theta(\mathbf q_1)\delta(\mathbf q_2),
$$
with the <alpha mode-coupling kernel>
$$
\boxed{\alpha(\mathbf q_1,\mathbf q_2)
=\frac{(\mathbf q_1+\mathbf q_2)\cdot\mathbf q_1}{q_1^2}}.
$$
Taking the divergence of $(\mathbf v\cdot\nabla)\mathbf v$ and symmetrizing its two velocity arguments gives
$$
\theta'(\mathbf k)+\mathcal H\theta(\mathbf k)
+\frac32\mathcal H^2\Omega_m\delta(\mathbf k)
=-\int_{\mathbf q_1,\mathbf q_2}
\delta_D(\mathbf k-\mathbf q_1-\mathbf q_2)
\beta(\mathbf q_1,\mathbf q_2)
\theta(\mathbf q_1)\theta(\mathbf q_2),
$$
where the <beta mode-coupling kernel> is
$$
\boxed{\beta(\mathbf q_1,\mathbf q_2)
=\frac{|\mathbf q_1+\mathbf q_2|^2
(\mathbf q_1\cdot\mathbf q_2)}{2q_1^2q_2^2}}.
$$
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