Yikes. You're very confused.

Your first equation, which is incorrect, is trying to calculate the viscous stress vector (not shear stress vector) at the wall. In general, the viscous stress vector will have a component normal to the wall and a component tangent to the wall. The component tangent to the wall is called the shear stress. The correct equation for calculating the viscous stress vector exerted by the wall on the fluid is: $$\vec{\tau}=\mu \left[{\vec{\nabla} \vec{u}+(\vec{\nabla} \vec{u})^T}\right]\cdot \vec{n}$$where $\vec{n}$ is the unit normal vector drawn from the fluid through the wall. (This uses the sign convention that tensile stresses are positive and compressive stresses are negative)

The normal component of the viscous stress vector at the wall is obtained by dotting the stress vector $\vec{\tau}$ with the unit normal $\vec{n}$:$$\tau_n=\vec{\tau}\cdot\vec{n}$$
The shear component of the viscous stress vector at the wall is obtained by dotting the stress vector $\vec{\tau}$ with the unit tangent to the wall $\vec{t}$:$$\tau_t=\vec{\tau}\cdot\vec{t}$$So you can also write that:$$\vec{\tau}=\tau_n\vec{n}+\tau_t\vec{t}$$
The unit tangent vector is oriented parallel to the fluid velocity in close proximity to the wall. More generally, the shear stress vector at the wall, including direction, can be obtained by subtracting the normal component of the viscous stress vector from the total viscous stress vector:

shear stress vector $=\vec{\tau}-\tau_n\vec{n}=\vec{\tau}-(\vec{\tau}\cdot\vec{n})\vec{n} $

Hope this helps.

**ADDENDUM**
For the case of $\vec{u}=u_x\vec{i}_x+u_y\vec{i}_y+u_z\vec{i}_z$ and $\vec{n}=\vec{i}_y$, I get the following at the wall:
$$(\vec{\nabla}\vec{u})^T\cdot \vec{n}=\frac{\partial v_y}{\partial y}\vec{i_y}$$
$$(\vec{\nabla}\vec{u})\cdot \vec{n}=\frac{\partial v_x}{\partial y}\vec{i}_x+\frac{\partial v_z}{\partial y}\vec{i}_z+\frac{\partial v_y}{\partial y}\vec{i}_y$$

$$\vec{\tau}=\mu\left(\frac{\partial v_x}{\partial y}\vec{i}_x+\frac{\partial v_z}{\partial y}\vec{i}_z+2\frac{\partial v_y}{\partial y}\vec{i}_y\right)$$

I can see now what you're saying about the shear stress terms in $\vec{\tau}$, but there is also a normal stress component too. But you are correct that the shear stress component is parallel to the velocity vector near the wall.

You have the right starting point with energy basically, but I'm finding your homework hint more useful than where you go from energy of a differential unit. It says "The energy density should be easy to identify." The energy density is:

$$ \frac{\text{energy}}{\text{volume}} = \frac{\text{mass}}{\text{volume}} \frac{\text{energy}}{\text{mass}} = \rho \frac{v^2}{2} + \rho e $$

For your problem, this is practical because we're already seeing this form in the answer.

Let's look at the components mentioned in the hint and see if those help us.

- internal energy
- kinetic energy
- energy lost to heat
- energy lost to work against pressure
- energy lost to work against viscous stress

Looking at these, and looking at the equation, we have a vivid picture painted of what all the terms mean. A fast deconstruction makes that look like the following if I denote the fluid energy (internal plus kinetic) as $w$ (I would rather call it $e$, but they took that).

$$ \frac{d}{dt} \rho w = \frac{d}{dx} \left( -v \rho w + \text{conduction - expansion + kinetic} \right) $$

These are all in the exact same order as what was given in the hint. It sounds like you're job is to walk backward from this view to some fundamental principle, specifically, conservation of energy. So start with a completely theoretical statement of that (say, equation 26) and start matching things up. Starting from the equation in that linked pdf, the first term already matches. That's easy, because you know what direction the fluid is flowing in. Next term, you have a Del dot-product with the same term. You should be able to work that out.

Moving on, Del dot q is also exactly what you're looking for (to get the conduction term). You'll need some finesse arguing why the signs should be the way they are. Then, volumetric heat contribution can be equated to frictional heating.

## Best Answer

First, the difficulty of the Clay problem is not because of any limitation on our understanding of where the Navier Stokes equation comes from, or how to solve it. In practical situations, you just solve it on a computer. The Clay problem is asking whether the numerical solution converges to a differentiable continuum limit at small lattice spacings, and the reason this is difficult is because it might not be true. There might special configurations of small vortices which produce even smaller vortices in a scale invariant way, which shrink to zero size with negligible energy in finite time. This is heuristically unlikely based on our current turbulence models, but it might be possible in special initial conditions. The existence (or nonexistence) of shrinking size blowing-up velocity solutions would have little bearing on practical simulations of fluid flow, where the difficulties come from the turbulent scales, not the microscopic blow ups, if they even exist.

The linear approximation for momentum flow in the Navier Stokes equation is only for regions where the velocity field is locally linear, for small enough regions where the velocity is described by its value and its first derivative. The reason to believe this is both experimental and theoretical.

If you place a sheet of water between two plates to make a thin lubrication layer, and slide the two layers against each other, the flow in the middle region stays laminar as long as the layer is thin. The profile of the velocity is linear, and the momentum flow from one plate to the other is the friction force you measure between the plates. This friction force is linear in the plate velocity for all realistic velocity gradients acheivable in an every-day fluid.

Since the plate friction is proportional to the velocity, the flow of momentum (the internal stress) is to counter the local velocity gradient, and it is proportional to the velocity gradient. This is just saying that velocity moves down a velocity gradient, or that velocity diffuses (component by component). Diffusion is a smoothing process, it takes functions with sharp corners and local blowups to smoother functions. In k-space, it shrinks high-k modes by a factor of $exp(-Dk^2)$, so that unless the k-modes are exponentially growing to begin with, they end up smooth.

If there were blow-ups in Navier Stokes, then inside the very high velocity microscopic blow-up region you would expect that the next order term would become important, so that some higher derivative operator would kick in to smooth out the blow up. It is likely that the ordinary viscosity term is enough to smooth out the blow ups by itself.

The approximate velocity-squared forces for fast moving objects are caused by complicated turbulent flow in the bulk fluid. The reason the force goes as the velocity squared is that if you move twice as fast you are displacing twice as much fluid to twice as fast a speed. In the linear Stokes regime, you are dragging a big clump of fluid with you as you move, with a power-law decay of the velocity away from your position, and this, plus the fundamental linear momentum flow, make a friction proportional to velocity. But at high speeds, the moving body leaves behind a turbulent disturbed wake which is comparable to its geometric cross-section.

You can view the mismatch between the growth rate of the linear-in-velocity Stokes friction and the quadratic-in-velocity inertial friction as the fundamental reason there is turbulence in the first place. The linear flow is not very efficient at getting momentum from one place to another in the fluid, so that some small advection speeds up the momentum flow from the body to the bulk fluid, and leads to more advection, which amplifies until the steady state turbulent profile is set up. In the turbulent profile, the velocity gradients are much larger than in the laminar flow, so that the momentum removed from the body is much larger, and grows as the velocity squared times the cross section.

But this phenomenological law does not change the fact that the momentum flow locally in the fluid is still proportional to the velocity gradient.

The theoretical reason to expect a force linear in the velocity is a Taylor expansion. The friction is zero at zero velocity, and grows from there. You can in principle compute the linear viscosity term in statistical mechanics using a Kubo-type formula, and the length scale is the mean-free path, which is the atomic scale for a fluid like water. This means that each successive gradient term is suppressed by the ratio of the scale to the atomic scale, so the next order Taylor coefficient, the quadratic in velocity force, will kick in when the velocity gradient is comparable to 1/mean-free-time (the mean free path over the mean thermal velocity), which is smaller than typical macroscopic scales by a lot, so this term is theoretically negligible unless you have appreciable velocity gradients at the mean-free-path scale. For air, the mean free path is many dozens of time larger than the atomic scale, and the continuum approximation breaks down sooner.