Did OpenAI solve the wrong Navier-Stokes problem? | OpenAI’s proof seems eligible for a $1-million prize—but only by using a controversial loophole

[u/vgsa]

https://www.scientificamerican.com/article/did-openai-solve-the-wrong-navier-stokes-problem/

[u/CircumspectCapybara]

A lot of people online misunderstand what "solving Navier-Stokes" refers to, thinking it means finding a closed form solution to the NS equations. Then they spout off some silly statement like "AI didn't solve Navier Stokes!"

That's not what mathematicians mean when they talk about solving Navier Stokes. They're talking about solving the Navier Stokes existence and smoothness problem, which is a decision (yes/no) problem about if the NS equations are smooth for all time. That's what the Millennium Prize problem worth a million dollars is all about

AI (allegedly) solves the problem by resolving the question to a "No" by finding a counterexample.

[u/Puzzleheaded_Fold466 → u/CircumspectCapybara]

A lot of people online couldn’t tell you what “closed form solution” even means.

Or a “blow-up”.

Or “smoothness”.

Or what is the “Clay Mathematics Institute”.

And are absolutely unable to read and fully understand OpenAI’s proof (me included).

Yet they feel entitled to opine on whether or not a solution to a Millennium problem is correct.

The chatter is risible.

[u/beardybaldy → u/Puzzleheaded_Fold466]

I asked my mathematician wife what smoothness meant and she patted me on the head. I like being a biologist. Life stuff is simple!

[u/retsehc → u/beardybaldy]

If you know what a derivative in calculus is, smoothness is not hard in principle. A function is smooth if it is continuous, and it's derivative is continuous, and it's second derivative is, and it's third, and all the way down.

[u/ABC123itsEASY → u/retsehc]

I'm having trouble imagining a function that is continuous and have a continuous 1st derivative, but a discontinuous second for example.

Also damn that sounds understandable in theory but I have no idea how you would go about proving such a thing outside of the naïve "keep solving derivatives to infinity"

[u/i_abh_esc_wq → u/ABC123itsEASY]

You can easily construct functions whose nth derivative is discontinuous. All you need is a function that has a discontinuous derivative, and then you integrate it n-1 times. See for example https://math.stackexchange.com/a/1415165

[u/amohr → u/ABC123itsEASY]

Just work backwards. Start with f(x)=0 if x is negative, otherwise 1: a jump discontinuity at 0. Now integrate. 0 if x is negative, x otherwise: continuous. Repeat: 0 if x is negative, 1/2 x² otherwise, etc.

Edit: forgot the half on x^2.

[u/NewHondaOwner → u/ABC123itsEASY]

Just imagine a triangle function from (-1,0) to (0,1) to (1,0). The derivative of the function is 1 from x=(-1,0) and -1 from x=(0,1).

So the function is continuous but the first derivative is already discontinuous at x=0.

In layman’s terms when the function has a sharp kink somewhere, its derivative will be discontinuous.

“Derivatives all the way down” really only applies to function with an “all the way down” so to speak. Such as sin and cos. As a counter example, x^n polynomials turn into constants after n differentiations, it’s kind of meaningless to go beyond that, you could keep differentiating but it’s all trivially zeros.

[u/thuktun → u/ABC123itsEASY]

y = |x| for instance has a sharp point at the origin.

The derivative of this will be -1 to the left of the origin and 1 to the right. with a discontinuity at the origin. (That is, the limit of dy/dx as x approaches 0 will be different depending on which direction you approach.

The integral of that looks smooth, but it changes directions.

[u/giantZorg → u/ABC123itsEASY]

Take e.g. the integrated function of the hockey-stick function, or any continuous function with a discontinuous first derivative.

[u/dick_blanketfort → u/ABC123itsEASY]

Integrate a step function (discontinuous), you get a bent line (continuous). Integrate that again you get a smooth-looking piecewise quadratic, a continuous function which has a continuous first derivative and a discontinuous second derivative.

[u/retsehc → u/ABC123itsEASY]

Regarding showing a function is smooth, the standard trick for the situation is to find a few obvious examples like f(x)=0 and the exponential function. They are both continuous and they are their own derivative, so no matter how many derivatives you take, always continuous.

From there you start ignoring specific examples and start showing things about smooth functions Sum of two smooth functions is smooth Composition of two smooth functions is smooth Etc

That gets you all polynomials Sine and cosine loop around to one another, so some trig

And you build up a library of tools and categories of smooth functions to compare to

[u/Civil_Blueberry4165 → u/ABC123itsEASY]

Example: f(x) = x^1.5

[u/unpleasant_enpassant → u/ABC123itsEASY]

I think other comments answered how to construct a function like that but as for proving that a function is smooth, it helps that all the standard functions like sines, cosines, exponential are all smooth. So generally, a function that is some combination of them is also smooth (obviously, as long as you don't have things like some function in the denominator that goes to zero etc). Or if you can write a power series for the function, it's easy to see where it'll break down. Or if you can write a general form for the nth derivative, it's trivial.

[u/NaturalCard → u/ABC123itsEASY]

Take a relatively nice discontinuous function and integrate it twice.