From a mathematical standpoint, what is the most general form of the Navier-Stokes tensor equation? I'm looking at this from an abstract perspective, so the equations need not be limited to the standard domain of $\mathbb{R}\times\mathbb{R}^3$. I've also seen references to a form for compressible flow as well as one for incompressible flow. Is one more general than the other, or are they separate entities? If there is a single over-arching equation that describes all possible special cases of the Navier-Stokes equation, what is it?
[Math] the most general form of the Navier-Stokes equation
fluid dynamicstensors
Related Solutions
Okay, I'm no expert but somewhat recently I read, while an undergraduate, Leray's paper on the existence of weak solution in the whole space $\mathbb{R}^3$. So maybe I can be of some help.
First I would say to get a good real analysis, graduate level, course. Mostly the theory of the $L^p$ spaces, the basic inequalities in measure theory and Arzela-Ascoli is what's needed. Some knowledge of Hausdorff measure is requiered for the more sophisticated estimate on the size of the set of singular times (which is good to know before going to more modern partial regularity results).
A course in functional analysis is advisable since you're going to need some weak-* convergence familiarity and several arguments with weak compactness in $L^2$ and $H^1$ are used.
A course in PDE is obvious, the arguments are based first on a linearization of the equation which of course assumes some knowledge of the theory for these equations (for instance the heat equation). Another thing from this area that you'll need is the theory of Sobolev spaces, and the course on which this'll be seen varies from program to program (I learned the basics from a real analysis course, for instance) so you should look for a course that'll cover this.
The rest of the background is covered by topics courses (if at all), like "singular" integral equations and differential inequalities (think variants of Gronwall's inequalities) and you'll have to look for courses that cover them.
I'm sure, as you approach the modern results on the regularity of N-S the prerrequisites will grow, but I think these are a good place to start, since it's background for any serious study of PDE.
As you suspect, the mechanical energy equation you derive is missing terms that account for viscous effects. The other terms you derive are correct.
The RHS should be $\nabla \cdot \mathbf{\sigma}$, the divergence of the stress tensor, and you have edited accordingly.
This clearly is the correct term to include because the rate of increase of momentum in a fixed volume $V$ of fluid should be related to the viscous stress force of the surrounding fluid acting on the bounding surface of the volume $\partial V$,
$$\text{rate of momentum change } = \int_{\partial V}\mathbf{\sigma} \cdot \mathbf{n} \, dS = \int_V \nabla \cdot \mathbf{\sigma} \, dV.$$
Your error stems from incorrectly forming the dot product of the velocity and the viscous term $\eta\,\nabla \cdot [\nabla \mathbf{v} + \nabla \mathbf{v}^T] = \eta \, \nabla^2 \mathbf{v} + \eta \, \nabla (\nabla \cdot \mathbf{v})$.
In particular, you are missing $ \mathbf{v} \cdot [\eta\,\nabla \cdot \nabla \mathbf{v}]$ which should give rise to the viscous dissipation term, representing the action of viscous forces in converting useful mechanical energy into heat.
The correct treatment of that term is
$$\tag{*} \eta\,\mathbf{v} \cdot\,\nabla \cdot [\nabla \mathbf{v} + \nabla \mathbf{v}^T] = \underbrace{-\eta \nabla \mathbf{v} \,\mathbf{:}\, \nabla \mathbf{v}}_{\text{viscous dissipation}} + \frac{\eta}{2}\nabla^2(|\mathbf{v}|^2) + \eta \, \mathbf{v} \cdot \nabla (\nabla \cdot \mathbf{v}), $$
where the double dot product represents $\sum_{i=1}^3 \sum_{j=1}^3 |\frac{\partial v_i}{\partial x_j}|^2$, a norm of the velocity gradient, and $|\mathbf{v}|$ is the usual Euclidean norm of the velocity vector.
For an easy derivation, we can expand $\mathbf{v} \cdot\,\nabla \cdot [\nabla \mathbf{v}+ \nabla \mathbf{v}^T] $ using Cartesian coordinates and the Einstein convention (repeated index implies summation over that index). First we obtain
$$\nabla \cdot [\nabla \mathbf{v}+ \nabla \mathbf{v}^T]_{i} = \partial_j(\partial_j v_i + \partial_iv _j) = \partial_j(\partial_j v_i) + \partial_i(\partial_jv_j),$$
and forming the dot product we get
$$ \mathbf{v} \cdot\,\nabla \cdot [\nabla \mathbf{v} + \nabla \mathbf{v}^T] = v_i\,\partial_j(\partial_j v_i) + v_i \,\partial_i(\partial_jv_j) \\ = -\partial_jv_i \, \partial_j v_i + \partial_j[v_i(\partial_j v_i)] + v_i \,\partial_i(\partial_jv_j) \\ = -\partial_jv_i \, \partial_j v_i + \frac{1}{2}\partial_j\,\partial_j (v_iv_i) + v_i \,\partial_i(\partial_jv_j), $$
which is the component form of the coordinate-free term on the RHS of (*).
Best Answer
First of all, some assumption: the first one is that a fluid is treated as a continuum, that is not made unto discrete particles but rather a continuous substance.
Another assumption: all the fields of interest like pressure, velocity, density, temperature and so on are differentiable, weakly at least.
The equations are derived from the basic principles of conservation of mass, momentum, and energy. For that matter, sometimes it is necessary to consider a finite arbitrary volume, called a control volume, over which these principles can be applied.
We start with the so called material derivative
Changes in properties of a moving fluid can be measured in two different ways. One can measure a given property by either carrying out the measurement on a fixed point in space as particles of the fluid pass by, or by following a parcel of fluid along its streamline. The derivative of a field with respect to a fixed position in space is called the Eulerian derivative while the derivative following a moving parcel is called the convective or material derivative.
It's defined as
$$\frac{D}{Dt} =! \frac{\partial}{\partial t} + v\cdot \nabla$$
where $v$ is the velocity of the fluid.
The first term on the right-hand side of the equation is the ordinary Eulerian derivative (i.e. the derivative on a fixed reference frame, representing changes at a point with respect to time) whereas the second term represents changes of a quantity with respect to position. This "special" derivative is in fact the ordinary derivative of a function of many variables along a path following the fluid motion; it may be derived through application of the chain rule.
For example, the measurement of changes in wind velocity in the atmosphere can be obtained with the help of an anemometer in a weather station or by mounting it on a weather balloon. The anemometer in the first case is measuring the velocity of all the moving particles passing through a fixed point in space, whereas in the second case the instrument is measuring changes in velocity as it moves with the fluid.
Now we need some conservation laws in particular for mass, momentum and energy.
This is done via the Reynolds transport theorem, an integral relation stating that the sum of the changes of some extensive property defined over a control volume $\Omega$ must be equal to what is lost (or gained) through the boundaries of the volume plus what is created/consumed by sources and sinks inside the control volume. This is expressed by the following integral equation:
$$\frac{d}{dt}\int_{\Omega} L\ \text{d}V = -\int_{\partial\Omega} Lv\cdot n\ \text{d}A - \int_{\Omega} Q\ \text{d}V$$
Where $L$ is a set of extensive properties as written above, $Q$ represents the sources and sinks in the fluid.
A straightforward application of the divergence theorem and Leibniz rule lead us to this conclusion:
$$\frac{\partial L}{\partial t} + \nabla\cdot (Lv) + Q = 0$$
From this valuable relation (a very generic continuity equation), three important concepts may be concisely written: conservation of mass, conservation of momentum, and conservation of energy. Validity is retained if is a vector, in which case the vector-vector product in the second term will be a dyad.
Conservation of Momentum
The most elemental form of the Navier–Stokes equations is obtained when the conservation relation is applied to momentum. Writing momentum as $\rho v$ gives:
$$\frac{\partial \rho v}{\partial t} + \nabla\cdot (\rho v v) + Q = 0$$
Where $v v$ is what we called before a dyad: a special case of tensor product, which results in a second rank tensor; the divergence of a second rank tensor is again a vector (also called: a first rank tensor). Noting that a body force (notated $b$) is a source or sink of momentum (per volume) and expanding the derivatives completely:
$$v\frac{\partial\rho}{\partial t} + \rho \frac{\partial v}{\partial t} + v v \cdot \nabla\rho + \rho v\cdot \nabla v = b$$
Note that the gradient of a vector is a special case of the covariant derivative, the operation results in second rank tensors; except in Cartesian coordinates, it's important to understand that this isn't simply an element by element gradient. Rearranging and recognizing that:
$$v \cdot \nabla\rho + \rho v\cdot \nabla v = \nabla\cdot(\rho v)$$
With an easy calculation we can show that we can rewrite the general formulation above as
$$v\left(\frac{\partial\rho}{\partial t} + \nabla\cdot(\rho v)\right) + \rho\left(\frac{\partial v}{\partial t} + v\cdot\nabla v\right) = b$$
The leftmost expression enclosed in parentheses is, by mass continuity (shown in a moment), equal to zero. Noting that what remains on the left side of the equation is the convective derivative:
$$\rho\left(\frac{\partial v}{\partial t} + v\cdot\nabla v\right) = b ~~~~~ \to ~~~~~ \rho\frac{Dv}{Dt} = b$$
Conservation of mass
Mass may be considered also. Taking $Q = 0$ (no sources or sinks of mass) and putting in density:
$$\frac{\partial\rho}{\partial t} + \nabla\cdot(\rho v) = 0$$
where $\rho$ is the mass density (per volume).This equation is called the mass continuity equation, or simply "the" continuity equation. This equation generally accompanies the Navier–Stokes equation.
In case of an incompressible fluid, $\rho = $ constant and so we gain
$$\nabla\cdot v = 0$$
GENERAL FORM
The generic body force $b$ seen previously is made specific first by breaking it up into two new terms, one to describe forces resulting from stresses and one for "other" forces such as gravity. By examining the forces acting on a small cube in a fluid, it may be shown that
$$\rho \frac{Dv}{Dt} = \nabla\cdot\sigma + f$$
Where $\sigma$ is the stress tensor and $f$ accounts the body force present.
This equation is called the Cauchy momentum equation and describes the non-relativistic momentum conservation of any continuum that conserves mass. $\sigma$ is a rank two symmetric tensor given by its covariant components:
$$\sigma_{ik} = \begin{pmatrix} \sigma_{xx} & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \sigma_{yy} & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \tau_{zz} \end{pmatrix} $$
$\sigma$ are the normal stresses and $\tau$ are the shear stresses.
The tensor can be split up to two terms like
$$\sigma_{ij} = -p\mathbb{1} + \mathbb{T}$$
where $T$ is the deviator stress components, $1$ is the identity matrix and
$$p = -\frac{1}{3}(\sigma_{xx} + \sigma_{yy} + \sigma_{zz})$$
The motivation for doing this is that pressure is typically a variable of interest, and also this simplifies application to specific fluid families later on since the rightmost tensor $\mathbb{T}$ in the equation above must be zero for a fluid at rest. Note that $\mathbb{T}$ is traceless. The Navier–Stokes equation may now be written in the most general form:
$$\boxed{\rho\frac{Dv}{Dt} = -\nabla p + \nabla\cdot\mathbb{T} + f}$$
This equation is still incomplete. For completion, one must make hypotheses on the form of $\mathbb{T}$ , that is, one needs a constitutive law for the stress tensor which can be obtained for specific fluid families; additionally, if the flow is assumed compressible an equation of state will be required, which will likely further require a conservation of energy formulation.