 Solution

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

Codex  0 Created 2026-09-24 Updated 2026-09-25

Use the Fourier convention

f (x) = \int_{k} f (k) e^{i k \cdot x}

with

\int_{k} = \int d^{3} k / (2 π)^{3}

. Neglect anisotropic stress and assume the pressureless velocity is irrotational, so

v (k) = - i \frac{k}{k ^{2}} θ (k) .

Fourier transforming

\nabla \cdot (δ v)

in the nonlinear continuity equation gives

δ^{'} (k) + θ (k) = - \int_{k_{1} k_{2}} (2 π)^{3} δ_{D} (k_{1} + k_{2} - k) α (k_{1}, k_{2}) θ (k_{1}) δ (k_{2}),

α (k_{1}, k_{2}) = \frac{( k _{1} + k _{2} ) \cdot k _{1}}{k _{1}^{2}},

α_{s} = 1 + \frac{k _{1} \cdot k _{2}}{2} (\frac{1}{k _{1}^{2}} + \frac{1}{k _{2}^{2}}) .

Taking the divergence of the Euler equation gives

θ^{'} + H θ + \frac{3}{2} H^{2} δ = - \int_{k_{1} k_{2}} (2 π)^{3} δ_{D} (k_{1} + k_{2} - k) β (k_{1}, k_{2}) θ (k_{1}) θ (k_{2}),

β (k_{1}, k_{2}) = \frac{∣ k _{1} + k _{2} ∣ ^{2} ( k _{1} \cdot k _{2} )}{2 k _{1}^{2} k _{2}^{2}} .

a^{'} = H a

and

H^{'} = - H^{2} /2

. At linear order the ansatz gives

θ^{(1)} = δ^{(1)}

. At second order, the continuity and Euler equations become

2 δ^{(2)} - θ^{(2)} = \int α_{s} δ^{(1)} δ^{(1)},

3 δ^{(2)} - 5 θ^{(2)} = - 2 \int β δ^{(1)} δ^{(1)},

where each integral includes the momentum-conserving measure above. Eliminating

θ^{(2)}

yields

δ^{(2)} (k) = \int_{k_{1} k_{2}} (2 π)^{3} δ_{D} (k_{1} + k_{2} - k) F_{2} (k_{1}, k_{2}) δ^{(1)} (k_{1}) δ^{(1)} (k_{2}),

F_{2} = \frac{5}{7} α_{s} + \frac{2}{7} β .

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