Which of these 1-D representations of the Navier-Stokes equations is correct? The incompressible Navier Stokes equations can be written as 
A.
$$\frac{\partial (\rho \mathbf{v})}{\partial t} + \nabla \cdot (\rho \mathbf{v} \mathbf{v}) = S$$
or
B.
$$\frac{\partial \mathbf{v}}{\partial t} + (\mathbf{v} \cdot \nabla) \mathbf{v} = \frac{S}{\rho}$$
where $S$ is some source term. But it seems these not the same in 1-d...?
A.
$$\frac{\partial (\rho v)}{\partial t} + \frac{\partial (\rho v^2)}{\partial z}= S$$
$$\frac{\partial (\rho v)}{\partial t} + \rho\frac{\partial v^2}{\partial z} + v^2\frac{\partial \rho}{\partial z}= S$$
Let $\rho = 1$
$$\frac{\partial v}{\partial t} + \frac{\partial v^2}{\partial z} = S$$
$$\frac{\partial v}{\partial t} + 2v\frac{\partial v}{\partial z} = S$$
is not the same as
B.
$$\frac{\partial v}{\partial t} + v\frac{\partial v}{\partial z} = \frac{S}{\rho}$$
Let $\rho = 1$
$$\frac{\partial v}{\partial t} + v\frac{\partial v}{\partial z} = S$$
So what am I missing? Which version is correct?
 A: GENERAL DEVELOPMENT:
First we establish the equivalence of the two forms for the Navier-Stokes Equations given in the OP.
To do  this, we use straightforward product rule differentiation to show that
$$\begin{align}
\frac{\partial \rho \vec v}{\partial t}=\frac{\partial \rho }{\partial t}\vec v+\rho \frac{\partial \vec v }{\partial t} \tag 1
\end{align}$$
and 
$$\begin{align}
\nabla\cdot (\rho \vec v \vec v)&=(\rho \vec v \cdot \nabla)\vec v+\nabla \cdot (\rho \vec v)\vec v \tag 2\\\\
&=(\rho \vec v \cdot \nabla)\vec v-\frac{\partial \rho }{\partial t}\vec v \tag 3
\end{align}$$
In going from $(2)$ to $(3)$ we used the continuity relationship
$$\nabla \cdot (\rho \vec v)+\frac{\partial \rho}{\partial t} =0 \tag 4$$
which is an integral component of the physics of the problem.  Now, using $(1)$ and $(3)$ reveals that 
$$\begin{align}
\vec S&=\frac{\partial \rho \vec v}{\partial t}+\nabla\cdot (\rho \vec v \vec v)\\\\
&=\rho \frac{\partial \vec v }{\partial t} +(\rho \vec v \cdot \nabla)\vec v\tag 5
\end{align}$$
whereupon dividing $(4)$ by $\rho$ yields
$$ \frac{\vec S}{\rho}=\frac{\partial \vec v }{\partial t} +(\vec v \cdot \nabla)\vec v \tag 6$$
where we have established that the alternative forms are equivalent given $(4)$.

ONE-DIMENSIONAL CASE:
Note for the one-dimensional case, the reduced forms of $(4)$ and $(5)$ become
$$\frac{\partial \rho v}{\partial z} +\frac{\partial \rho}{\partial t}=0 \tag {4'}$$
and
$$\begin{align}
S&=\frac{\partial \rho v}{\partial t}+\frac{\partial (\rho v^2)}{\partial z}\\\\
&=\rho \frac{\partial  v }{\partial t} +\rho  v \frac{\partial v}{\partial z}
\end{align} \tag {5'}$$
If the density $\rho$ is constant, then we have further simplification of $(4')$ and $(5')$.  The continuity equation becomes
$$\frac{\partial  v}{\partial z} =0 \tag {4''}$$
while the Navier-Stokes equation becomes
$$\begin{align}
S&=\frac{\partial v}{\partial t}
\end{align} \tag {5''}$$
A: They are both correct.  Remember you also have the divergence free condition:
$$ \frac{\partial v}{\partial z} = 0 .$$
Note, this makes the 1D incompressible NS equation quite boring.
