Use the Fourier convention $f(\mathbf x)=\int_{\mathbf k}f(\mathbf k)e^{i\mathbf k\cdot\mathbf x}$ with $\int_{\mathbf k}=\int d^3k/(2\pi)^3$. Neglect <scalar anisotropic stress>[anisotropic stress] and assume the pressureless velocity is irrotational, so
$$
\mathbf v(\mathbf k)=-i\frac{\mathbf k}{k^2}\theta(\mathbf k).
$$
Fourier transforming $\nabla\mathbin\cdot(\delta\mathbf v)$ in the nonlinear continuity equation gives
$$
\delta'(\mathbf k)+\theta(\mathbf k)
=-\int_{\mathbf k_1\mathbf k_2}
(2\pi)^3\delta_D(\mathbf k_1+\mathbf k_2-\mathbf k)
\alpha(\mathbf k_1,\mathbf k_2)
\theta(\mathbf k_1)\delta(\mathbf k_2),
$$
where the <alpha mode-coupling kernel> and its symmetrization are
$$
\alpha(\mathbf k_1,\mathbf k_2)
=\frac{(\mathbf k_1+\mathbf k_2)\cdot\mathbf k_1}{k_1^2},
$$
$$
\boxed{\alpha_s
=1+\frac{\mathbf k_1\cdot\mathbf k_2}{2}
\left(\frac1{k_1^2}+\frac1{k_2^2}\right)}.
$$
Taking the divergence of the Euler equation gives
$$
\theta'+\mathcal H\theta+\frac32\mathcal H^2\delta
=-\int_{\mathbf k_1\mathbf k_2}
(2\pi)^3\delta_D(\mathbf k_1+\mathbf k_2-\mathbf k)
\beta(\mathbf k_1,\mathbf k_2)
\theta(\mathbf k_1)\theta(\mathbf k_2),
$$
with the symmetric <beta mode-coupling kernel>
$$
\boxed{\beta(\mathbf k_1,\mathbf k_2)
=\frac{|\mathbf k_1+\mathbf k_2|^2
(\mathbf k_1\cdot\mathbf k_2)}{2k_1^2k_2^2}}.
$$

In the <Einstein-de Sitter universe>, $a'=\mathcal Ha$ and $\mathcal H'=-\mathcal H^2/2$. At linear order the ansatz gives $\widetilde\theta^{(1)}=\widetilde\delta^{(1)}$. At second order, the continuity and Euler equations become
$$
2\widetilde\delta^{(2)}-\widetilde\theta^{(2)}
=\int\alpha_s\widetilde\delta^{(1)}\widetilde\delta^{(1)},
$$
$$
3\widetilde\delta^{(2)}-5\widetilde\theta^{(2)}
=-2\int\beta\widetilde\delta^{(1)}\widetilde\delta^{(1)},
$$
where each integral includes the momentum-conserving measure above. Eliminating $\widetilde\theta^{(2)}$ yields
$$
\widetilde\delta^{(2)}(\mathbf k)
=\int_{\mathbf k_1\mathbf k_2}(2\pi)^3\delta_D(\mathbf k_1+\mathbf k_2-\mathbf k)
F_2(\mathbf k_1,\mathbf k_2)
\widetilde\delta^{(1)}(\mathbf k_1)
\widetilde\delta^{(1)}(\mathbf k_2),
$$
with the <standard perturbation theory density kernel>
$$
\boxed{F_2=\frac57\alpha_s+\frac27\beta}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 312 / 2 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-312/2/i/solution