Continuity equation for a perturbed system

Question

I'm reading J. Frank's book, "Accretion Power in Astrophysics", and on chapter 2, section 2.4, he studies the perturbation of a system initially in equilibrium. The perturbation is given by,

$$ P \rightarrow P_{0} + P^{'}, \hspace{0.5cm} \rho \rightarrow \rho_{0} + \rho^{'}, \hspace{0.5cm} \boldsymbol{v} \rightarrow \boldsymbol{v}^{'}, $$ where $P$, $\rho$ and $v$ are the perturbed pressure, volumetric density and velocity, respectively. The analogous symbols with $_{0}$ subscript are the initial values for these quantities (where $v_{0} = 0$), and $'$ stands for the perturbed variables.

With this perturbation, he obtains the following expression for the continuity equation,

$$ \frac{\partial \rho^{'}}{\partial t} + \rho_{0}\boldsymbol{\nabla}\cdot \boldsymbol{v^{'}} = 0. $$

The problem is I can't get this expression. I tried the following: starting with the continuity equation,

$$ \frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \cdot (\rho \boldsymbol{v}) = 0, $$ and opening the divergence operator, writing the quantities explicitly showing its initial and perturbed terms, we have,

$$ \frac{\partial (\rho_{0} + \rho^{'})}{\partial t} + \boldsymbol{v}^{'} \cdot \boldsymbol{\nabla} (\rho_{0} + \rho^{'}) + (\rho_{0} + \rho^{'}) \boldsymbol{\nabla}\cdot \boldsymbol{v^{'}} = 0. $$

I know the term $\partial \rho_{0}/\partial t = 0$, because the system was in equilibrium, and we can rearrange the last two terms as,

$$ \boldsymbol{\nabla}\cdot (\rho_{0}\boldsymbol{v^{'}}) + \boldsymbol{\nabla}\cdot (\rho^{'}\boldsymbol{v^{'}}), $$

but I don't know how these terms can turn out to be $\rho_{0}\boldsymbol{\nabla}\cdot \boldsymbol{v^{'}}$.

Update for people who have the same doubt

In this case, we are applying perturbation theory to variables which corresponds to a system which is initially in equilibrium, that is, we are varying slightly its values. So, if the ' quantities are small, the multiplication of two of these quantities is even smaller, and, in a first approximation, can be neglected. So the term,

$$ \boldsymbol{\nabla} \cdot (\rho^{'}\boldsymbol{v^{'}}) \approx 0, $$ and then, we get, as a final result for the continuity equation, the expression,

$$ \frac{\partial \rho^{'}}{\partial t} + \rho_{0}\boldsymbol{\nabla} \cdot \boldsymbol{v}^{'} = 0. $$

Answer 1

You seem to have done everything right, so it's difficult to see what the problem is. However, if you're asking why

$$ \boldsymbol{\nabla} \cdot (\rho_o \textbf{v}')$$

can be written as

$$\rho_0 \boldsymbol{\nabla} \cdot \textbf{v}',$$

it's simply because the $\rho_0$ is a constant, and can be taken in front of the derivative.