When deriving the sound wave equation: $${1 \over c^2} {\partial^2 p' \over \partial t^2 }= \Delta^2 p' $$ by linearizing the Euler equation:
$$\rho {d v \over dt }= - \nabla p $$ and the continuity equation:
$$ {\partial \rho \over \partial t } + \nabla (\rho v)=0$$ using an approach of small deviations $\rho', v', p'$from an equilibrium $\rho_0, v_0, p_0$ with $v_0=0$. So $$p=p_0+p'\\ v=0+v' \\ \rho=\rho_0+\rho'$$
Why can we just neglect the convective term in the euler equation? Meaning why can we use: $$\rho_0 {\partial v' \over \partial t }= \nabla p'$$?
Why can we just assume $ (v'\nabla)v'\approx 0$ ?
Answer
The assumption when linearizing is that the deviations/perturbations are very small compared to the reference (averaged) values.
Typically the derivatives of the deviations are of the same order as the deviations themselves. Consider the deviations having this functional form in 1D: $$u'=\Delta u\sin kx\quad \partial_x u'=k\Delta u\cos kx$$ The deviation and its derivative are of order $O\left(\Delta u\right)\ll O\left(1\right)$. The convective terms then are: $$u'\partial_{x}u'=\frac{1}{2}\Delta u^{2}k\sin2kx$$ which is order $O\left(\Delta u^2\right)\ll O\left(\Delta u\right)$ and therefore negligible compared to other terms.
An analysis of the equations without evaluating the derivatives can be done if we start more general with the continuity and Cauchy momentum equation (neglecting viscous stresses): $$\partial_{t}\rho+\boldsymbol{\nabla}\cdot\rho\boldsymbol{v}=0$$
$$\partial_{t}\rho\boldsymbol{v}+\boldsymbol{\nabla}\cdot\left[\rho\boldsymbol{v}\otimes\boldsymbol{v}+p\boldsymbol{I}\right]=0$$
linearizing:
$$\partial_{t}\left(\rho_{0}+\rho'\right)+\boldsymbol{\nabla}\cdot\left(\rho_{0}+\rho'\right)\boldsymbol{v}'=0$$
$$\partial_{t}\left(\rho_{0}+\rho'\right)\boldsymbol{v}'+\boldsymbol{\nabla}\cdot\left[\left(\rho_{0}+\rho'\right)\boldsymbol{v}'\otimes\boldsymbol{v}'+\left(p_{0}+p'\right)\boldsymbol{I}\right]=0$$
i am sure you will agree that $\rho' \boldsymbol{v}' \ll \rho_0 \boldsymbol{v}'$ and $\rho'\boldsymbol{v}'\otimes\boldsymbol{v}'\ll\rho_0\boldsymbol{v}'\otimes\boldsymbol{v}'\ll p'\boldsymbol{I}$, which yields the linearized equations you are looking for:
$$\partial_{t}\rho'+\rho_{0}\boldsymbol{\nabla}\cdot\boldsymbol{v}'=0$$
$$\rho_{0}\partial_{t}\boldsymbol{v}'=-\boldsymbol{\nabla} p'$$
Appendix: Using the identity: $$\boldsymbol{\nabla}\cdot\left(\boldsymbol{A}\otimes\boldsymbol{B}\right)=\boldsymbol{B}\left(\boldsymbol{\nabla}\cdot\boldsymbol{A}\right)+\left(\boldsymbol{A}\cdot\boldsymbol{\nabla}\right)\boldsymbol{B}$$
we can rewrite the momentum equation in simplified form: $$\begin{align}\partial_{t}\rho\boldsymbol{v}+\boldsymbol{\nabla}\cdot\left(\rho\boldsymbol{v}\otimes\boldsymbol{v}\right) &=\boldsymbol{v}\partial_{t}\rho+\rho\partial_{t}\boldsymbol{v}+\boldsymbol{v}\boldsymbol{\nabla}\cdot\left(\rho\boldsymbol{v}\right)+\left(\rho\boldsymbol{v}\cdot\boldsymbol{\nabla}\right)\boldsymbol{v} \\ &=\boldsymbol{v}\left[\partial_{t}\rho+\boldsymbol{\nabla}\cdot\left(\rho\boldsymbol{v}\right)\right]+\rho\left[\partial_{t}\boldsymbol{v}+\left(\boldsymbol{v}\cdot\boldsymbol{\nabla}\right)\boldsymbol{v}\right]\end{align} $$
where the first term on the last line is identically zero due to the continuity equation.
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