= Solution
The background <Einstein-de Sitter universe> satisfies $a'=\mathcal H a$ and $\mathcal H'=-\mathcal H^2/2$. 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 gives
$$
(\delta^{(n)})'=n\mathcal H a^n\mathcal F_n,\qquad
(\theta^{(n)})'=-\left(n-\frac12\right)\mathcal H^2a^n\mathcal G_n.
$$
At 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 gives
$$
2F_2-G_2=\alpha(q_1,q_2),\qquad
\frac52G_2-\frac32F_2=\beta(q_1,q_2).
$$
Hence
$$
F_2=\frac57\alpha+\frac27\beta,\qquad
G_2=\frac37\alpha+\frac47\beta.
$$
With the paper's unsymmetrized convention $\alpha=\beta=(q_1+q_2)/q_1$, one possible unsymmetrized representation is
$$
\boxed{F_2(q_1,q_2)=G_2(q_1,q_2)=\frac{q_1+q_2}{q_1}}.
$$
Only the symmetric part enters a <convolution> of two identical linear fields. Symmetrizing gives
$$
F_{2,s}=G_{2,s}=\frac12\left[\frac{q_1+q_2}{q_1}+\frac{q_1+q_2}{q_2}\right]
=\boxed{\frac{(q_1+q_2)^2}{2q_1q_2}}.
$$
This is the <one-dimensional cosmological density kernel>. The physical symmetric one-dimensional Euler kernel is also $(q_1+q_2)^2/(2q_1q_2)$: 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.
Back to article page
