This is an attempt to revisit Leray's self-similar solutions of the 3-dimensional incompressible Navier-Stokes equation (NSE) and to show, that they are an isolated case of the general set of self-similar solutions. I'm seeking constructive feedback about this proof. It would be really helpful to see a detailed answer addressing the results.
This question is part of a study which I published on arXiv. You can read it in full detail here: Polihronov, "Charles Bouton and the Navier Stokes Global Regularity Conjecture" arXiv:1902.01985 [math.GM].
Update: This article lists the definitions and properties of differential invariants in a concise form. For full details see C. L. Bouton, Am. J. Math. vol21, p25 (1899). The reader is encouraged to familiarize themselves with the following differential invariants:
absolute invariant
absolute covariant
relative invariant
relative covariant.
and how they behave under ${\it finite}$ (not infinitesimal) transformation. Without this knowledge, it would be very difficult to grasp how Lie groups apply to the NSE.
This question is built mostly on the scaling properties of the NSE. Why is scaling so important? Because it determines the criticality of the NSE. Detailed explanation on this was written by T. Tao and can be found here.
Why are criticality and scaling important? Because the NSE was written originally for a continuum. This concept allows infinite rescalings and may lead to blow up in the solutions. More detailed explanation on this is again found in T. Tao's article referenced above.
The NSE is scale-invariant, and the scaling transform that preserves it has been discovered only recently by A. Ercan and M. L. Kavvas, Chaos vol25, p123126 (2015):
$$\vec{u}^\prime=k^{\alpha_x-\alpha_t}\vec{u}$$ $$(x,y,z)^\prime=k^{\alpha_x} (x,y,z)$$ $$t^\prime=k^{\alpha_t} t$$ $$p^\prime = k^{2(\alpha_x-\alpha_t)} p$$ $$\nu^\prime = k^{2\alpha_x-\alpha_t} \nu$$
where "$p$" stands for "$p/\rho$".
Regarding the viscosity $\nu$: it is a constant with a fixed value, or a constant parameter. In the dimensionalized NSE, $\nu$ has dimension, and is thus a scalable constant because its dimensions are scalable. Note that $\nu$ is a fixed constant at a fixed scale, however this does not mean that $\nu$ is non-scalable.
In the nondimensionalized NSE, all variables $x,y,z,t$ are dimensionless. Written in nondimensionalized form, the NSE retains its form and is still considered to have scaling invariance. It is for this reason that Leray's solutions (see below) were written for the nondimensionalized NSE as "self-similar" solutions, due to their scalability. Is then the fixed constant $\nu$ (or, the Reynolds number) non-scalable within the nondimensionalized NSE? Is scalability a property of $x,y,z,t$ only? Why should we set the viscosity to be non-scalable? Hard to answer, there is no unified viscosity theory, since the molecular origins of viscosity are unclear. Note that scalability is not to be confused with fixed constant values at a fixed scale. Scalability can be the property of a fixed constant to take fixed, but different values at different scales.
The viscosity is scalable in the dimensionalized NSE, it is an undeniable fact. We cannot hand pick $\nu$ as non-scalable in the nondimensionalized NSE because after all, it is still the same equation. Any nondimensionalized solution can be dimensionalized and must be a solution of both dimensionalized and nondimensionalized NSE.
Scalability of the Reynolds number/viscosity in the nondimensionalized NSE is known (discussions on this are found even in this forum). For this reason, in this work the fixed constant $\nu$ is considered scalable within the nondimensionalized NSE (when initially posted, this was considered obvious, or well-known).
We see at once that the classic NSE scaling trasform. $$\vec{u}^\prime = \frac{1}{k} \vec{u}$$ $$p^\prime = \frac{1}{k^2} p$$ $$(x,y,z)^\prime=k(x,y,z)$$ $$t^\prime=k^2 t$$ is an isolated case of the general trasform. when $\alpha_x=1$, $\alpha_t=2$.
Let's define isobaric weights as follows: $W(x^a) =W(y^a) =W(z^a) =a \alpha_x$;
$W(t^b) =b\alpha_t$;
when variables are multiplied, their weights are added.
To write the self-similar solutions of the NSE, one can use Bouton's theorem (see C. L. Bouton, Am. J. Math. vol21, p25 (1899)).
About Bouton's work: While it is true that his article is more than 100 years old and may use older notation, his first theorem is very important. In essence, it states that the self-similar solutions of the NSE must be isobaric functions. This is the main reason why his name is the title of my article, because through his theorem I was able to construct the general form of all NSE's self-similar solutions. Bouton's work on differential invariants deserves much credit and recognition alongside its modern counterparts.
According to Bouton's theorem, scaling invariance of the self-similar solutions means that they are isobaric functions (polynomials or rational functions or representations thereof) of $x,y,z,t$. More precisely, $\vec{u}$ must be isobaric in $x,y,z,t$ of weight $W=\alpha_x-\alpha_t$, while $p$ must be isobaric in $x,y,z,t$ of weight $2(\alpha_x-\alpha_t)$.
In the literature, Leray is credited with proposing the form of NSE self-similar solutions as
$$\vec{u} = \frac{1}{\sqrt{t}} \mathbf{F} \left( \frac{x}{\sqrt{t}}, \frac{y}{\sqrt{t}}, \frac{z}{\sqrt{t}} \right)$$
(analogously for $p$)
but these are an isolated case $\alpha_x=1$, $\alpha_t=2$ of the general self-similar solution of the NSE. To find it, write Bouton's theorem above as
$$c\left( x\frac{\partial \vec{u}}{\partial x} + y\frac{\partial \vec{u}}{\partial y} + z\frac{\partial \vec{u}}{\partial z} \right) +at \frac{\partial \vec{u}}{\partial t} = b\vec{u}$$
and find (I used Maple)
$$\vec{u} = t^{\frac{b}{a}} \mathbf{F} \left( \frac{x}{t^\frac{c}{a}}, \frac{y}{t^\frac{c}{a}}, \frac{z}{t^\frac{c}{a}} \right)$$
then seek
$$\frac{b}{a} = \frac{\alpha_x-\alpha_t}{\alpha_t}$$ $$\frac{c}{a} = \frac{\alpha_x}{\alpha_t}$$
which gives Bouton's self-similar solutions
$$\vec{u} = t^{\frac{\alpha_x-\alpha_t}{\alpha_t}} \mathbf{F} \left( \frac{x}{t^{\frac{\alpha_x}{\alpha_t}}} , \frac{y}{t^{\frac{\alpha_x}{\alpha_t}}}, \frac{z}{t^{\frac{\alpha_x}{\alpha_t}}}\right)$$
These are identical with Leray's solutions as long as $\alpha_t=2\alpha_x$.
The expression for $p$ is analogous. It is a scalar function with isobaric weight of zero multiplied by $t^{2(\alpha_x-\alpha_t)/\alpha_t}$.
Could you kindly provide constructive feedback about this proof. It would be really helpful to see a detailed answer addressing the results, rather than argumentative comments.
So far, one opinion was given (I have not voted on it yet), however we would like to see constructive reasoning on the mathematics of viscosity scaling. The facts that viscosity traditionally was considered non-scalable; or that Charles Bouton's article was published more than 100 years ago do not contribute to the constructive discussion that is needed.
Thank-you
It seems that scaling is an important topic for you, c.f. your prior questions 1 2 3 4.
In view of your persistence, I am confused as to why you ignored me asking in link 1 above if you could demonstrate the scaling from Kavvas and Ercan is indeed as you said. The computation would have revealed the below error.
In fact, I see no evidence in the paper that K&E claim to have a new scaling for Navier--Stokes that was just missed by everyone else until 2015.
Instead it seems that (among other things) they talk about relaxing the scalings of other things like the density and forcing term, which allows you to compare different physical scenarios. This is not the fixed viscosity scenario considered by Leray / Tao etc. Obviously if the viscosity is fixed then from your expression for $\nu'$, one immediately gets $\alpha_t=2\alpha_x$ which you seem to accept, makes it collapse to the same one used by other people.
Being a refutation of the premises, I see no need to address the rest of the post. Which is fortunate for me, because as in link 3 above, it seems we will not agree on the relevant definitions, nor do I want to check a paper for you that's over 100 years old...