Are constant density and incompresibility equivalent?

It's easy to see from the mass conservation equation that $\rho=C^{te}$ $\Rightarrow$ $\vec{\nabla}.\vec{u}=0$ (incompresibility).

But I don't think it's evident that $\vec{\nabla}.\vec{u}=0$ $\Rightarrow$ $\rho=C^{te}$.

Because from the Mass conservation equation:

$$\frac{\partial \rho}{\partial t} + \vec{\nabla}(\rho \vec{u})=0$$ and using $\vec{\nabla}.\vec{u}=0$ gives :

$$\frac{\partial \rho}{\partial t}+\vec{u}.\vec{\nabla}\rho=\frac{D\rho}{Dt}=0$$ It's true that the langrangian derivative of $\rho$ is constant but I don't think that implies $\rho=C^{te}$.

As an example if $\vec{u}=u_{0}\vec{e_{x}}$ we get the advection equation:

$$\frac{\partial \rho}{\partial t} + u_{0}\frac{\partial \rho}{\partial x}=0$$

And the solution to this equation is $\rho=f(x-u_{0}t)$ with $f$ a function determined by the initial conditions.

So why do people use incompressibility and constant density as equivalents? Is it because they assume that the initial conditions of the density are constant (I don't know if this gives the $\rho=C^{te}$ solution)? so it can't evolve with time or space? Or I'm in the wrong and there's a problem with my reasoning?

• You already pointed it out: $\frac{D\rho}{Dt}=0$. You don't have to think about a fixed space, you're talking about a fluid; you are tracking that $dV$ along its movement: you check the variation in time and also along space. – FGSUZ Nov 1 '17 at 0:34

First, no actual material is actually incompressible, it is just an approximation. It is equivalent to an infinite speed of sound. In other words, incompressible really means that $\partial \rho / \partial p \approx 0$ -- no matter how much pressure you apply, the density doesn't change. For many gases, liquids, and solids, this might be an okay assumption.
When your flow of interest evolves at speeds much slower than the speed of sound, then approximating the speed of sound as infinite can help solve the equations. In other words, as the ratio of velocity to the speed of sound approaches 0, a flow behaves more and more "incompressible". Of course, this is just the Mach number $u/c$ and so we can say that low-Mach number flows can be approximated as "incompressible".