Navier-Stokes is hard?

Navier-Stokes is hard?
Author: James Lowman

I was recently asked the question; “How can the Navier-Stokes equations both describe our observable world and not be known to always have solutions in 3D?

For those that don’t know, the Navier-Stokes (NS) equations represent a mathematical model that describes fluid. It is derived from quantities in physics that are conserved. Those are:

  • Conservation of mass
  • Conservation of momentum
  • Conservation of energy

The Navier-Stokes equations are yet to be considered solved. There is a prize association called “The Millennium Prize Problems” which offers one million dollars to anyone that can solve one of the seven problems they have listed, and the existence and smoothness of a solution to the Navier-Stokes equations is on that list.

So the question I was asked seems completely reasonable. How can these equations describe physical phenomena with knowledge of a solution? The answer touches on an interesting, and modern, intersection in science.

We use a number of assumption to simplify a NS system in order to approximate a solution. These approximations are usually handled by a computational fluid dynamics software package. They are high powered suites that allow for customized physics simulations. The assumptions most often used to simplify the NS equations are things like the fluids being in-compressible, the turbulence affects are averaged, or that the fluid is Newtonian. Suffice to say, even with a high powered computational engine, more often then not a scaled down and simplified version of the NS equations are all that we can easily approximate.

So if the solutions to the fluid problems are all approximate, how can we know if they are describing real physics? To compare a NS system solution to an actual physical fluid flow, we would need to perform an experiment and collect data on such a flow. This results in flawed data, because the act of introducing measurement tools into fluid domains involves disrupting the fluid. The best measurements of fluid dynamics themselves are approximations.

So what we have ended up with is approximate mathematical approximations of fluids approximating approximate experiments. Its not hard to see why the student’s question needed clarification.

The answer is: Because it works.

That’s a pretty hand-waved explanation, that no student should accept. But for anyone interested, the derivation of Navier-Stokes is a fun derivation that follows from those conservation laws that were listed, and experiments can be found in open source publications that show how much effort was placed on collecting data with minimal interference. The interested party might be able discern that experimental data matches model solutions (reasonably well), and we can start to believe that Navier-Stokes delivers on the promise of predicting real world physics.

The Quadratic Equation

The Quadratic Equation

The Quadratic Equation

Author: James Lowman


The quadratic equation is always the answer.

Time and time again,  while tutoring students,  I encounter a resistance to using  an  unfailing  tool  called  the  quadratic  equation.   This  simple  algebraic mathematical statement allows a student to find the roots of any clumsy second order polynomial with ease. I can only assume that, while the quadratic equation is drilled into memory, its useful nature is under-reported in high school mathematics.

For those who aren’t familiar:

\( ax^2 + bx + c = 0 \)                (1)

\( x=\frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)                    (2)

If you can get an equation into the form of equation??, then you can use the Coefficients a, b, and c in order to solve for x in equation??. The ± in equation?? implies that there are two solutions for x (this being the result of solving a second order polynomial equation).  These two solutions represent the roots of equation??. The roots are the values of x that allow equation?? to equal 0 on the left hand side. Roots  are  of  interest  to  us,  because  they  are  a  quick  and easy  way  to determine factors of a quadratic (not to mention all the mathematical reasons to identify when a function is equal to zero).  Students are drilled on factoring. I rarely see a student that is unable to factor a simple quadratic, but I often encounter  those  that  hav problems  when  the  factors  fail  to  be  immediately obvious.  The quadratic equation is always the answer. I often despair that education relies entirely on problem singularity, a belief that there is only one correct way to approach a question.  When I see a student struggle  for minutes  to  try  and  factor  a  difficult  quadratic,  I  can’t  help  but wonder why they shy away from the quadratic equation.  Sometimes it is hyper- focus that keeps them from finding an alternate path forward,  but more and more I encounter fear.  Factoring is supposed to be an easy shortcut, while the quadratic equation has the confusing ± and the scary √ .  But a shortcut fails to be short when it takes time to muddle through possible variations.  Take, for example, the following quadratic:

 \( 10x^2 + 13x – 30 = 0 \)        (3)

Some  people  may  be  able  to  see  immediately  that  this  equation  can  be factored.  Others  might have enough experience with factoring to only fumble through one or two permutations before generating the factored result quickly. The  rest  of  us,  myself  included,  might  struggle  with  5  or  more  permutations before coming close to a factored result.  Yet the quadratic yields the answer in 4 quick lines:

\( x=\frac{-13 \pm \sqrt{169 + 1200}}{20} \)              (4)

\( x=\frac{-13 \pm \sqrt{1369}}{20} \)                     (5)

\( x=\frac{-13 \pm 37}{20} \)                           (6)

so \( x= \frac{6}{5} \) and \( x=\frac{-5}{2} \)         (7)


The quadratic equation is always the answer.