 Solution

Mass conservation in the Zeldovich approximation gives $\rho d^{3}r=\bar{\rho }{a}^3{d}^{3}q$. Expanding the trajectory Jacobian determinant to first order therefore yields $\boxed{\delta =-{\nabla }_q\cdot \Psi }$. At the same order, ${\nabla }_r=a^{-1}{\nabla }_q$ andThe last equality uses ${\nabla }_q\times \Psi =0$ and the vector calculus $\nabla \mathrm{div}={\nabla }^2+\nabla \times \nabla \times$.

Split the gravitational potential into its homogeneous-background and perturbed parts. The background has ${\nabla }_r{\Phi }_0=(4\pi G\bar{\rho }/3)r$. For the perturbation, Poisson equation and irrotationality givewith a harmonic or uniform-acceleration contribution set to zero by the perturbation boundary conditions. Its divergence is $-4\pi G\bar{\rho }{\nabla }_q\cdot \Psi =4\pi G\bar{\rho }\delta$, as required. Insert this and the pressure gradient into $\ddot{r}=\ddot{a}(q+\Psi )+2\dot{a}\dot{\Psi }+a\ddot{\Psi }$. The background acceleration cancels using $\ddot{a}/a=-4\pi G\bar{\rho }/3$, leavingTake minus the comoving divergence, and expand $\delta (q,t)={\sum }_k{\delta }_k(t)e^{ik\cdot q}$. Since ${\nabla }_q^2$ acts on a Fourier mode by $-k^2$, each mode satisfiesHere the expansion coordinate is comoving; confusing it with the physical coordinate would lose the factor $a^{-2}$.