 Solution

The background Einstein-de Sitter universe satisfies $a^{\prime }=\mathcal{H}a$ and ${\mathcal{H}}^{\prime }=-{\mathcal{H}}^2/2$. Introduce ${\mathcal{F}}_n$ and ${\mathcal{G}}_n$ for the spatial convolutions in the Fourier transform representation with kernels $F_n$ and $G_n$, respectively, so ${\delta }^{(n)}=a^n{\mathcal{F}}_n$ and ${\theta }^{(n)}=-\mathcal{H}{a}^n{\mathcal{G}}_n$. Differentiation givesAt first order the cosmological continuity equation yields $F_1-G_1=0$, while the cosmological Euler equation gives $3F_1/2-3G_1/2=0$. The growing-mode normalization therefore gives $\boxed{G_1=F_1=1}$.

At second order, take the unsymmetrized order of the inputs printed in the paper. Matching the alpha mode-coupling kernel and beta mode-coupling kernel terms givesHenceWith the paper's unsymmetrized convention $\alpha =\beta =(q_1+q_2)/q_1$, one possible unsymmetrized representation isOnly the symmetric part enters a convolution of two identical linear fields. Symmetrizing givesThis is the one-dimensional cosmological density kernel. The physical symmetric one-dimensional Euler kernel is also $(q_1+q_2{)}^2/(2q_1{q}_2)$: symmetrizing the printed $\beta$ reproduces it, so its apparently different unsymmetrized form is not a contradiction. Unsymmetrized kernels themselves are not unique, since antisymmetric parts integrate to zero. Zero input momenta require the usual separate treatment of the homogeneous background.