 # Part VIII: Navier-Stokes, Existence & Uniqueness

At long last, we’ve finally managed to derive the famous Navier-Stokes equation; the cornerstone of most of fluid mechanics. Notice all the assumptions we had to make to get to it! To refresh your memory, this is what it looks like, along with descriptions of each term in it: The Navier-Stokes equation, which describes momentum density transport in a simple, Newtonian, isotropic, incompressible fluid. The terms on the left-hand side are mathematical consequences of the concept of transport, while the terms on the right-hand side (save for the external force density) are defined by intermolecular fluid interactions.

There are many different, equivalent ways to write the Navier-Stokes equation, but the most common is this one, which incorporates a couple of simplifications thanks to conservation of mass: $\boxed\rho\left(\frac{\partial \vec{v}}{\partial t}\ + \nabla \vec{v} \cdot \vec{v} \right) = \vec{f} - \nabla p\ +\ \mu \nabla^2 \vec{v}$

(Note that some people write the dot product in the flux term backwards based on their notation for the gradient, but it’s all just preference.) Another equivalent form I briefly mentioned when discussing transport theory, which is particularly useful when discussing the curvature of flows, is commonly known as the Lamb form: $\displaystyle \rho\left(\frac{\partial \vec{v}}{\partial t}\ + \frac{\nabla \left( \vec{v} \cdot \vec{v}\right)}{2} - \vec{v}\times\left(\nabla\times\vec{v}\right) \right) = \vec{f} - \nabla p\ -\ \mu \nabla \times \left(\nabla \times \vec{v}\right)$

In whatever form it takes, this equation (along conservation of mass and varying simplifying assumptions) will be the chief mathematical tool we’ll use to determine the shape and speeds of flows. That being said, it turns out that the Navier-Stokes equation is notoriously difficult to deal with except in the simplest of situations, and it’s all thanks to the flux term: $\rho \nabla \vec{v} \cdot \vec{v}$

This little mathematical pest has been the bane of fluid mechanicists for a solid hundred years, and it induces such a monumental headache on any mathematician attempting to solve the Navier-Stokes equation that there is actually a million-dollar bounty out for anyone who is even able to prove that the Navier-Stokes equation always has “sensible” solutions for a physical input—this is one of the most important mathematical problems ever conceived!

The reason this flux term is such a problem is because of a property called nonlinearity, which is best shown rather than told. To demonstrate, let’s consider a silly non-physical problem in 1-D, where we consider two imaginary fluids whose velocity is solely determined by the following equations:

Applying some calculus tricks, what you’ll wind up finding is that the expressions for the velocities in each of the fluids is markedly different:

Note that for Fluid 2, if I set the constants to $0$, the velocity is linearly proportional to the force. This means that if I double the force, the velocity everywhere is going to be doubled, and so on—a property unsurprisingly called linearity. For Fluid 1, however, this isn’t true; setting the constant to $0$ and doubling the force instead leads to an increase in the velocity by a factor of $\sqrt{2}$! As innocuous as it seems, nearly every single tool ever made by mathematicians and scientists to solve partial differential equations like the Navier-Stokes equation relies on linearity, meaning any attempts to solve the Navier-Stokes equation relies on conceptual machinery on the outside of most of mathematical history. But that’s not even the worst part; as it turns out, nonlinearity is often comorbid with a host of horrendous properties, and the velocity of Fluid 1 is no exception.

For example, consider what would happen if I attempt to find the flow velocity for Fluid 1 if I take the velocity at the origin to be zero, $v(0) = 0$; because of that $\pm$ sign in front of the velocity expression for Fluid 1, I actually have two possible fluid velocities! This is incredibly strange from a physical perspective; one could conjecture a situation where I have two exact copies of this system, where I’m applying the same force under the same conditions to each of them, and somehow get different flows. This goes straight against any rational understanding of physics, and is called non-uniqueness. Even worse, no matter what value of the constant $c_1$ I choose based on some physical information, or which expression for the velocity of Fluid 1 I pick, there is always some value of $x$ past which my velocity suddenly turns into an imaginary number! This means that in some regions, we don’t even have an expression for the flow velocity, which is even more mind-boggling from a physical perspective and is referred to by mathematicians as existence failure (this is a really cool name). And this is only in one dimensions, folks—just imagine how much worse it gets in 3. Two solutions for the velocity of Fluid 1, each solving with zero velocity at the origin, corresponding to a fluid with . Neither solution is defined in the region .

When you include time into the equations, things get even worse; you can start off with a flow that makes sense, and find it evolving into something seemingly nonsensical! Take, for example, what happens if you now take the ODE describing Fluid 1 and add a time rate-of-change term, turning it into a PDE: $\frac{\partial v}{\partial t} + \rho v\frac{\partial v}{\partial x} = f$

This equation represents momentum transport in a one-dimensional fluid with no intermolecular interactions, and is commonly called Burgers’ equation. Take a look at what happens if you take a perfectly nice initial flow field for such a fluid and let it evolve in time for a constant unit $f$ and $\rho$: The solution to the PDE for a smooth initial condition. The function describing the velocity eventually develops a discontinuity in space, causing the solution to become ill-defined. The numerical solver fails to handle this phenomenon, and glitches out once it develops.

After time passes, the velocity becomes discontinuous, and jumps immediately from one value to another; a phenomenon commonly referred to as a shock or shock wave. Mathematically, this is a problem not just because $\frac{dv}{dx} = \infty$ at the discontinuity, but because the velocity has effectively two values there, causing it to be ill-defined! What happens here, from a philosophical perspective, is astounding; we’ve gone and defined an object mathematically using the rules of calculus, and the object has then evolved over time into something that violates the very mathematical principles we used to define it. This is exactly the problem (or at least, one of them) people run into when trying to solve the Navier-Stokes equation; the objects one uses to define flows for fluids can very rapidly outgrow the conceptual framework we used to define them.

Consequently, nearly every analytical attempt at using the Navier-Stokes equations to describe a flow necessitates ensuring that the momentum flux term in the equations is either negligible or handled indirectly. One way to verify this negligiblity is by taking the ratio of the magnitudes of each component of the flux term with the magnitudes of the components of the only other term that involves spatial gradients of the velocity—the viscosity term: $\mathbf{Re} = \rho \nabla \vec{v} \cdot \vec{v} \oslash \mu \nabla^2 \vec{v}$

The $\oslash$ symbol represents that we are dividing every term of a tensor by every other term of another tensor, effectively acting like the division counterpart of the outer product $\otimes$. Since each of these quantities is a 3-D vector, this ratio actually contains 9 distinct ratios, and I refer to it as the Reynolds tensor. The Reynolds tensor is useful in the sense that it is telling us whether or not, for a given flow, we should expect to observe the same kind of mind-boggling behavior we saw above in the Burgers’ equation, where it would be located, and it what directions we expect it to manifest. Because calculations involving the Reynolds tensor are pretty unwieldy, fluid mechanicists almost always single out a specific component of the Reynolds tensor, guess “characteristic” values for the density/viscosity/velocity/velocity gradients/lengths that they assume are valid everywhere in the flow, and come up with a single scalar ratio called the Reynolds number which they use instead. Sometimes this leads to spurious assumptions and misdrawn conclusions, but alas. The components of the Reynolds tensor in Cartesian coordinates, along with a typical Reynolds number-type approximation. Calculating the Reynolds tensor is often unwieldy, so most scientists simply estimate a single number based on quantities in the problem they expect to see and call it the Reynolds number, shown below.

Nearly all of the time, we use the Reynolds tensor/number to do sanity checks on any simplifying assumptions we make to the Navier-Stokes equation—either by using a conjectured Reynolds tensor/number to justify removing some term from the Navier-Stokes equations, or by calculating the Reynolds tensor/number for a solution to a simplified Navier-Stokes equation to validate whether or not the simplification was sensible in the first place. And one setting where we’ll be able to get rid of that flux term and, as a result, easily construct specific conclusions about flow shapes and speeds is a setting we’ve looked at before—pipe flow.

1. Some people think adding a term representing viscosity to equations like the ones described above help get rid of the shocks & other strange behaviors we saw just now. Try solving the ODE $\rho v(x)\frac{dv}{dx} - \frac{d^2 v}{dx^2} = f$ for a bunch of different $\rho/f/\mu$ using some math computer programs to see whether or not that line of thinking is correct.

2. How would neglecting the flux term in a problem change the physics of the fluid you would observe, and would it make sense? Is there a fluid that can possess a zero flux term?

2. In what situations do you think simply using the Reynolds number instead of the Reynolds tensor would be sufficient? In what situations would it fail?

3. Try calculating the Reynolds tensor for simple flows; maybe a constant flow, or a flow that changes linearly in one direction in space. What components are zero, and which aren’t? What do they correspond to?

4. Shock waves appear to make no sense in a differential equations sense, but seem to make perfect sense physically; is there some mathematical tool we were using before that would be able to make mathematical sense of shock waves?

5. In Part II, we talked about how pressure ensures liquid doesn’t just accumulate into a hyperdense thin film at the bottom of a cup. How is that connected to the shock wave phenomenon we saw above?

6. If you want to get a glimpse of how linearity is useful for solving differential equations, solve $\frac{dv}{dt} = f$ for $f=1$ and $2$ for arbitrary constants. Can the sum of both of these solutions represent a solution for $\frac{dv}{dt} = 3$?

## 4 thoughts on “Part VIII: Navier-Stokes, Existence & Uniqueness”

1. Where did you learn of the operation \oslash, the operator used to define your Reynolds tensor? (Is there a reference or particular textbook that defines this operator?). Does this operator have a name, e.g., the outer quotient or dyadic quotient? How would you write this operation in tensor (indicial) notation? E.g., if the outer product is c_ij = a_i b_j then would the \oslash be written as c_ij = a_i (b_j)^-1 ?

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1. The \oslash operation is just a Hadamard or element-wise division; I don’t think it has specific widespread use beyond computer science circles, but you can find it discussed in (https://en.wikipedia.org/wiki/Hadamard_product_(matrices)#Analogous_operations) and the cited textbooks therein. The indicial notation for this Hadamard division (or whatever you’d like to call it) in this case is c_ij = a_i *(1/b_j), which makes it clear that you can define it by virtue of the following (traditional) outer product: c = a \otimes (1/b), where c is a tensor, a and b are vectors, and the (1/b) is intended to be applied to all elements of b. Hope this helps!

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1. It can’t be Hadamard since in the Hadamard takes two matrices of the same dimensions and produces another matrix of the same dimension. So if you were to perform a Hadamard product (or division) of the convection flux and viscous vectors as you did above, you’d end up with a vector and not the 2nd rank tensor.

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2. Oh, you’re right! In that case, just consider it as c_ij = a_i/b_j, which you can interpret as the outer product of a_i and 1/b_j; it just didn’t seem very elegant to write that way!

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