Why does Cauchy's EOM not contain $\frac{ D (\rho u_i)}{Dt}$ but instead contains $\rho \frac{D u_i}{Dt}$? In the book of Elementary fluid dynamics by Acheson, it is given that the Cauch'y equation of motion is
$$
\rho \frac{D u_i}{Dt} = \sum_{j=x,y,z} \frac{\partial T_{ij}}{\partial x_i} + \rho g_i \quad \forall i=x,y,z
$$
The author, during the derivation of this equation, describes the LHS as the linear momentum and the RHS as the net force, just like a Newton's second law.
However, if that is the case, shouldn't LHS be $\frac{ D (\rho u_i)}{Dt}$ ? After all, the density can also change and the author do not make any assumption about incompressibility.
 A: It's essentially a consequence of continuity. From Evan's response, what you essentially want to establish is $$\frac{d}{dt} \int_V \rho \mathbf{u} \, d^3x = \int_V \rho \frac{D\mathbf{u}}{Dt} \, d^3x $$
which you can do in the following way. From Reynold's transport theorem, we know that
$$
\frac{\mathrm{d}}{\mathrm{d} t} \int_{V(t)} \rho \mathbf{u} \,d^3x =\int_{V(t)}\left(\frac{D}{D t}(\rho \mathbf{u})+(\rho \mathbf{u}) \nabla \cdot \mathbf{u}\right) d^3x \tag{1} 
$$
note that the continuity equation is
$$ \frac{D\rho}{Dt} + \rho \nabla \cdot \mathbf{u} = 0$$
so we can expand $D(\rho \mathbf{u})/Dt$ from $(1)$ as
$$ \frac{D(\rho \mathbf{u})}{Dt} = \rho \frac{D \mathbf{u}}{Dt} + \mathbf{u}\frac{D\rho}{Dt} \\ = \rho \frac{D \mathbf{u}}{Dt} -\mathbf{u}(\rho \nabla \cdot \mathbf{u})$$
putting this into $(1)$ yields
$$\frac{d}{dt} \int_V \rho \mathbf{u} \, d^3x = \int_V \rho \frac{D\mathbf{u}}{Dt} \, d^3x $$
as desired.
The intuition here is that you're trying to track the rate of change of some quantity $X$ multiplied by $\rho dV$ while simultaneously following $dV$. Well, $\rho dV$ represents a mass element, and mass is assumed to be conserved. So the if you expand this out in a product rule, the sum is only concerned about how $X$ is changing, hence the rate of change of $X\rho dV$ should just be $DX/DT$ multiplied by $\rho dV$.
A: Look at the integral version of the balance of linear momentum first. It is in the form $\dot{{\bf G}} = {\bf F}$, where ${\bf G}$ is the linear momentum of a finite blob of continuum and ${\bf F}$ is the net force acting on the blob. For a blob occupying a current region $\mathscr{R}$, the integral form is
$\frac{D}{Dt} \int_{\mathscr{R}} \rho {\bf v} dv = \int_{\partial \mathscr{R}} {\bf t} da + \int_{\mathscr{R}} \rho {\bf b} dv$,
where ${\bf t}$ are the supplied tractions, and ${\bf b}$ are body forces. You should be able to prove that
$\frac{D}{Dt} \int_{\mathscr{R}} \rho {\bf v} dv = \int_{\mathscr{R}} \rho \frac{D {\bf v}}{Dt} dv$
on your way to deriving the local, or differential version of the balance law.
A: Sorry for my poor english. My native language is french.
To put the previous answers into words: you reason on a material particle of fluid. It is a closed system of constant mass. Even though the fluid is compressible, its mass $dm$ does not change. You have to write the newton's law in the classical form $dm\frac{d\vec{V_P}}{dt}=dm\frac{D(\vec{v})}{Dt}=...$ with $\vec{V_P}$ the velocity of the material fluid particle.
But it would be a mistake to believe that for an open system of variable mass, it suffices to enter the mass in the derivative. For example, for a rocket, you cannot write $\frac{d(m(t)\vec{v)}}{dt}=m\vec{g}$
For an open system, the calculations given in the previous answers justify a very simple result, easy to admit:
the rate of change in momentum in an open system is equal to the force plus the rate of momentum entering through the surface.
I can try to get the result with a little less integrals. (But I need one !)
Consider the open system limited by a small fixed volume. Along $x$, its momentum is $dP_x=\rho v_xd\tau$ In projection on Ox, we must write:
$$\frac{\partial(dP_x)}{\partial t}=d\tau\frac{\partial(\rho v_x)}{\partial t}={\rm dF}_{x surface}+x\ rate\ of\ momentum\ inward$$
Note that the derivative here is an ordinary partial derivative with respect to time.
The hardest part is finding the amount of x momentum going into the small volume.  It is the flux on the surface of the small volume element :
$$x\ rate\ of\ momentum\ inward=-\iint{\rho v_x\vec{v}\vec{dS}}$$
Using Green's theorem : $\iint{\rho v_x\vec{v}\vec{dS}}=\vec{\nabla}(\rho v_x\vec{v})d\tau$
And : $\vec{\nabla}(\rho v_x\vec{v})=v_x\vec{\nabla}(\rho\vec{v})+\rho\vec{v}\vec{\nabla}(v_x)$
So it remains: $d\tau\frac{\partial(\rho v_x)}{\partial t}={\rm dF}_{x surface}-v_x\vec{\nabla}(\rho\vec{v})d\tau-\rho\vec{v}\vec{\nabla}(v_x)d\tau$
But : $\frac{\partial(\rho v_x)}{\partial t}=\rho\frac{\partial(v_x)}{\partial t}+v_x\frac{\partial(\rho)}{\partial t}$
Using the mass conservation : $v_x\frac{\partial(\rho)}{\partial t}=-v_x\vec{\nabla}(\rho\vec{v})$ it remains $\rho d\tau\frac{\partial(v_x)}{\partial t}={\rm dF}_{x surface}-\rho\vec{v}\vec{\nabla}(v_x)d\tau$
You see the material derivative appear $\frac{\partial(v_x)}{\partial t}+\vec{v}\vec{\nabla}(v_x)=\frac{D(v_x)}{Dt}$ and finally we find the expected result $\rho\frac{D(v_x)}{Dt}={\rm dF}_{x surface}$
Not sure it is the easy way !
