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Let $d\in\left\{2,3\right\}$ and $\Omega\subseteq\mathbb R^d$ be a bounded domain. The evolution up to time $T>0$ of an incompressible Newtonian fluid with uniform density $\rho_0$ and viscosity $\nu$ is given by the instationary Navier-Stokes equations $$\left\{\begin{matrix}\displaystyle\left(\frac\partial{\partial t}+\boldsymbol u\cdot\nabla\right)\boldsymbol u&=&\displaystyle\nu\Delta\boldsymbol u-\frac 1\rho_0\nabla p+\boldsymbol f&&\text{in }\Omega\times (0,T)\\\nabla\cdot \boldsymbol u&=&0&&\text{in }\Omega\times (0,T)\end{matrix}\right.\;,\tag 1$$ where $\boldsymbol u:\Omega\times [0,T]\to\mathbb R^d$ and $p:\Omega\times [0,T]\to\mathbb R$ are the velocity field and pressure, respectively, and $\boldsymbol f:\Omega\times (0,T)\to\mathbb R^d$ is the sum of all external forces.

Now, I've read that the evolution of a density $\rho:\Omega\times[0,T]\to\mathbb R$ injected into the fluid is described by a PDE of the form $$\left(\frac\partial{\partial t}+\boldsymbol u\cdot\nabla\right)\rho=\kappa\Delta\rho+s\;\;\;\text{in }\Omega\times (0,T)\tag 2$$ where $\kappa\in\mathbb R$ is somehow a diffusion rate and $s:\Omega\times[0,T]$ is an (or the sum of many?) external source(s).

  1. How do we obtain $(2)$? I'm searching for a simple, but mathematically coherent derivation
  2. Obviously, $(2)$ looks very similar to $(1)$. Why did the term $-1/\rho_0\nabla p$ disappear? (Is $(2)$ a special case of a more general variant?)
  3. What do people mean when they say, that density is injected into a fluid?
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    $\begingroup$ Equation 2 is just a normal transport equation (or advection-diffusion), it's not modelling a fluid, but modelling something being carried by the fluid motion. $\endgroup$
    – David
    Dec 13 '15 at 23:02
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I am not a mathematician but an engineer with a specialty in fluid mechanics so forgive me if i skip some mathematical details :)...

What do people mean when they say, that density is injected into a fluid?

I think the problem here is the use of the word 'density' which in this context means 'species density' (aka concentration) as opposed to 'mass density'. The variable $\rho$ in equation (2) describes 'species density' as you say the incompressible fluid has uniform 'mass density' $\rho_0$. Then the injection of a mass of a certain species into the fluid at some point makes more sense (e.g. imagine injecting a tracer compound in the fluid).

How do we obtain (2)? I'm searching for a simple, but mathematically coherent derivation

In the same way the Navier-Stokes equations are a representation of the local conservation of momentum of an infinitessimal fluid volume, (2) is a representation of the local conservation of (species) mass. A species concentration can be transported by the velocity field (aka advection, i.e. the $\boldsymbol{u}\cdot\boldsymbol{\nabla}\rho$ term) and by a diffusion process (aka Fick's law, i.e. the $k\Delta\rho$ term). Furthermore, it may be produced or consumed ($s$ term) by e.g. a chemical reaction.

The derivation is fairly straightforward given that the accumulation of the 'species mass' in a infinitessimal fluid volume $dV$ is the result of the in- and outflux $\boldsymbol{f}$ of the 'species mass' at the closed surface $S$ of the volume $V$:

$$d_{t}\int_{V}\rho dV=-\int_{S}\boldsymbol{f}\cdot\boldsymbol{n}dS+\int_{V}sdV$$

Here, $\boldsymbol{n}$ indicates the unit normal to the surface and the convective flux $f$ has a advective and diffusive contribution, respectively:

$$\boldsymbol{f}=\rho\boldsymbol{u}+\boldsymbol{j} = \rho \boldsymbol{u} - k\boldsymbol{\nabla}\rho$$

Using Gauss' divergence theorem the integral equation is transformed:

$$\int_{V}\partial_{t}\rho dV=\int_{V}\left[-\boldsymbol{\nabla}\cdot\boldsymbol{f}+s\right]dV$$

and we obtain the differential form:

$$\partial_{t}\rho =-\boldsymbol{\nabla}\cdot\boldsymbol{f}+s$$

Substituting in the expression for the flux we retrieve the requested 'species density' equation:

$$\partial_{t}\rho+\boldsymbol{\nabla}\cdot\rho \boldsymbol{u}=k\Delta\rho+s$$

where we can subsequently simplify $\boldsymbol{\nabla}\cdot\rho \boldsymbol{u}=\rho\boldsymbol{\nabla}\cdot\boldsymbol{u}+\boldsymbol{u}\cdot\boldsymbol{\nabla}\rho=\boldsymbol{u}\cdot\boldsymbol{\nabla}\rho$ because of the continuity equation $\boldsymbol{\nabla}\cdot\boldsymbol{u}=0$.

Obviously, (2) looks very similar to (1). Why did the term $−1/ρ_0∇p$ disappear? (Is (2) a special case of a more general variant?)

All conservation equations look similar because the starting point, i.e. conservation of mass, momentum, energy, entropy, etc. are similar. For the Navier-Stokes equations, we are interested in the accumulation of momentum in an infinitessimal fluid volume:

$$d_{t}\int_{V}\rho\boldsymbol{u} dV=-\int_{S}\boldsymbol{\sigma}\cdot\boldsymbol{n}dS+\int_{V}fdV$$

I have changed notation slightly where $\sigma$ is now the in- and outflux through the surface $S$ and $f$ is a source or sink of momentum acting on the volume $V$ (aka known as a body force, see Newton's second law), but the structure of the equations are the same. The change of notation was to make the point that the convective flux $\sigma$ is different from the convective flux $f$ in that $\sigma$ is now a rank 2 tensor, however still contains a advective and diffusive contribution: $$\boldsymbol{\sigma}=\rho\boldsymbol{u}\otimes\boldsymbol{u}+\boldsymbol{\tau}=\rho\boldsymbol{u}\otimes\boldsymbol{u}+p\boldsymbol{I}-\mu\left[\boldsymbol{\nabla}\boldsymbol{u}+\boldsymbol{\nabla}\boldsymbol{u}^{T}\right]$$

Here $\rho\boldsymbol{u}\otimes\boldsymbol{u}$ denotes the advective flux of momentum, whereas $\tau=p\boldsymbol{I}-\mu\left[\boldsymbol{\nabla}\boldsymbol{u}+\boldsymbol{\nabla}\boldsymbol{u}^{T}\right]$ represents the diffusive flux of momentum, most often refered to as the stress tensor. This stress tensor contains two types of stress; normal and shear stress. Normal stresses are caused by mainly the pressure whereas shear stresses are caused by viscosity due to velocity gradients across laminae of fluid (see Newtonian fluids).

The reason why the pressure term is in the Navier-Stokes equations is because a pressure gradient directly leads to changes in momentum, whereas it has no direct effect on the transport of 'species mass'. It instead would be reflected in the convective term of the 'species mass' equation.

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  • $\begingroup$ What do you mean by $$\textit{it's a pertubation of }\rho\textit{ on top of }ρ_0$$\;? And how can we derive your first centered equation? $\endgroup$
    – 0xbadf00d
    Dec 14 '15 at 22:08
  • $\begingroup$ @0xbadf00d: I should not have said that as it is not true, the sum of the species densities is by definition the mass density $\rho_0$. please forget it. The first equation has no derivation (i know of) but is a sort of 'law' often used in engineering that states intuitively that: accumulation quantity = quantity in - quantity out + production - consumption. I guess you could derive it using infinitessimals but it doesn't provide new insights as far as i am concerned. $\endgroup$
    – nluigi
    Dec 15 '15 at 9:44
  • $\begingroup$ I think I've almost got it, but it's still unclear to me why $\Delta\rho$ appears in your formula and why your $f$ needs to have the form you've given. I've tried to derive (a part) of $(2)$ using Reynolds' transport theorem and asked another question in order to clarify how I need to proceed: math.stackexchange.com/questions/1582114/… $\endgroup$
    – 0xbadf00d
    Dec 19 '15 at 13:17
  • $\begingroup$ @0xbadf00d: the laplacian of $\rho$ is simply the result of $\nabla\cdot\nabla\rho=\nabla^2\rho=\Delta\rho$. $f$ is just the most used form of advection and diffusion, there could however also be other terms that need to be included. An example is the thermophoresis where a mass flux is induced by a temperature gradient. I have a feeling you aren't satisfied with my answers because you would like 'pure' mathematical proofs, unfortunately i am not qualified to give them as i usually approach these problem purely from an engineering perspective. $\endgroup$
    – nluigi
    Dec 19 '15 at 14:05
  • $\begingroup$ @nluigi Maybe my confusion is due to an elementary misunderstanding. I've translated my thoughts into a question at PhysicsSE: physics.stackexchange.com/questions/225143/… $\endgroup$
    – 0xbadf00d
    Dec 20 '15 at 17:53

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