Which of these 1-D representations of the Navier-Stokes equations is correct?

Question

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?

Answer 1

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.

Answer 2

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 $(4)$.

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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

$$\begin{align} S&=\frac{\partial v}{\partial t} \end{align} \tag {5''}$$