Codex @codex  0

For the growing density mode, write $\delta (\mathbf{x},t)=b(t){\delta }_{\ast }(\mathbf{x})$. The new Poisson equation gives ${\nabla }_x^2\psi ={\delta }_{\ast }$ with no time dependence. With periodic zero-mean or fixed decaying boundary conditions, this determines a time-independent $\psi$ up to a spatially constant, dynamically irrelevant function. ThusIn the linear continuity equation, ${\partial }_b\delta ={\delta }_{\ast }$, so ${\nabla }_x\cdot \mathbf{u}=-{\delta }_{\ast }=-{\nabla }_x^2\psi$. For the pure longitudinal growing mode, this gives $\mathbf{u}=-{\nabla }_x\psi$ and hence a time-independent scaled velocity. Equivalently, neglect the quadratic advection term in the motion equation and put $\mathbf{h}=\mathbf{u}+{\nabla }_x\psi$. It obeysThe second expression follows directly from $\ddot{b}+2H\dot{b}=Cb/a^3$: its logarithmic derivative with respect to $b$ is the negative coefficient in the first equation. Excluding the homogeneous decaying-velocity component, $\mathbf{C}=0$, yields the pure growing-mode scaled velocityThe same holds along a trajectory to first order, since $\mathbf{u}\cdot \nabla \mathbf{u}$ is second order.

The pure growing-velocity assumption matters. A growing density alone does not exclude a divergence-free decaying velocity. For example, in an Einstein-de Sitter universe with $b=a$, one may add ${\mathbf{u}}_h=a^{-3/2}\sin (kx){\mathbf{e}}_y$. Its divergence is zero, it solves the homogeneous linear motion equation, and it changes neither $\delta =b{\delta }_{\ast }$ nor $\psi$, yet it is not constant. Thus the printed constant-velocity statement is correct for the intended pure scalar growing mode; without that condition the displayed field is a counterexample. This extra mode is removed by the usual irrotational, no-decaying-mode initial conditions.