# Navier-Stokes: Incompressible momentum equation + compressible continuity equation?

In a paper I have seen, somebody wrote down the Navier-Stokes as \begin{align} \rho\frac{\text{D}u}{\text{D}t} &= -\nabla p + \nabla \cdot \sigma + F\,,\\ \partial_t \rho + \nabla \cdot \left(\rho u\right) &= 0 \end{align} with the stress tensor $$\sigma_{ij} = \mu \partial_{(i} u_{j)}\,.$$ ($$\rho$$: mass density, $$u$$: velocity field, $$p$$: pressure, $$\sigma$$: viscous stress tensor, $$F$$: external body force, $$\mu$$: viscosity; brackets in the indices denote symmetrization, of course)

It looks like the momentum balance equation for an incompressible fluid together with the mass balance for a compressible fluid. Is this just an imprecision (which I think is the case) or some common approximation that I happen to not know about? As far as I am aware, this would not be the Boussinesq approximation.

The more general formula, $$\frac{\partial\rho}{\partial t}+\nabla\cdot\left(\rho\mathbf u\right)=0,\tag{1}$$ can be written in terms of the material derivative, $$\frac{D\rho}{Dt}+\rho\nabla\cdot\mathbf u=0,\tag{2}$$ simply by expanding the spatial derivative.
With an incompressible flow, the density is constant in time along the line of the flow, $$\frac{D\rho}{Dt}=0,$$ which requires $$\nabla\cdot\mathbf u=0$$ in Eq. (2). Hence, Eq. (1) can be used for incompressible flows, so long as the condition $$\nabla\cdot\mathbf u=0$$ is satisfied. So unless the authors neglected to mention incompressibility, I wouldn't call it "imprecise", though I guess your mileage may vary.