Trying a change of variable like a reciprocal $z=\dfrac{v}{\|v\|^2}$ on the Navier-Stokes Equation

Question

Trying a change of variable like a reciprocal $z=\dfrac{v}{\|v\|^2}$ on the Navier-Stokes Equation

I would like to know how the Navier-Stokes equation would look like if is applied the following change of variable and develops all the new derivatives associated with the new denominator:

$$z=\dfrac{v}{\|v\|^2}=\left[\begin{array}{c} \frac{v_x}{v_x^2+v_y^2+v_z^2}\\ \frac{v_y}{v_x^2+v_y^2+v_z^2}\\ \frac{v_z}{v_x^2+v_y^2+v_z^2}\end{array}\right],\quad v_k(\vec{x},t)\ \forall k=\{x,\ y,\ z\}$$

$$\Rightarrow \frac{\partial}{\partial t}\left(\dfrac{v}{\|v\|^2}\right)+\left(\dfrac{v}{\|v\|^2}\cdot \nabla\right)\dfrac{v}{\|v\|^2}=-\frac{1}{\rho}\nabla p+\nu\ \Delta\!\!\left(\dfrac{v}{\|v\|^2}\right)+f(\vec{x},t)$$

It is possible to have after developing the derivatives any term as a fractional power of the variables (like a square root)?

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Motivation

Following what I tried on this question 1 I want to try the change of variable similar to $z=\frac{1}{v}$ on the classic Navier-Stokes Equation for incompresible fluids with uniform viscosity:

$$\frac{\partial v}{\partial t}+(v\cdot \nabla)v=-\frac{1}{\rho}\nabla p+\nu \Delta v+f(\vec{x},t)$$

to see if similar terms as the mentioned on Eq. 11 of this another question 2 could be used to identify if there are points where the fluid stops moving in finite time.

After seen this video 1 where are studied blown-up situations on the NS equation, I suspicious that maybe it could be possible to use a reciprocal transform to identify clues about finite duration solutions on the NS equation.

@md2perpe recommended me to try the change of variable $z=\dfrac{v}{\|v\|^2}$ since it fulfill $z\cdot v=1$ so it could be interpreted as an inverse of a vector.

I tried in wolfram-alpha to change it by replacing term by term but I got troubles with the vectorial calculus (I am very rusty on it, so I hope you could explain the procedures too).

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Update

After some really interesting comments by @CalvinKhor and @Funktorality I would like to comment the following:

@Funktorality shared this paper by Buckmaster and Vicol where are shown the existence and non-uniqueness of weak solutions to the NS equation, some of them showing finite duration, which tells mi idea is not a complete non-sense.

@CalvinKhor shared this paper by Bonforte and Figalli showing the existence of solutions of finite duration for the fast diffusion equation $u_t=\Delta u^m,\,0<m<1$, so at least there are other PDEs able to show the requested behavior.

Both mentioned papers are a bit advanced for my current knowledge, but maybe could guide more experienced users on imaging an answer.

Also, @whpowell96 have shown in the comments some worries about my question implying solving the NS Clay's Millenium prize dilemma, but I don't think it is the case since that claims are for smooth solutions, and it don't says nothing about continuous piecewise solutions to Non-Lipschitz differential equations like the one I asked before for the following ODE: $y'=-\text{sgn}(y)\sqrt{|y|}$ that admit the piecewise solution $y(t)=\frac{\text{sgn}(y(0))}{4}(T-t)^2\theta(T-t)$ with a finite extinction time $T=2\sqrt{|y(0)|}\ll\infty$ and $\theta(t)$ the Heaviside step function, which coincidentally is related to a physical example about fluids (thanks @CalvinKhor for sharing this example).

If I am mistaken and it is impossible for the NS to hold these kind of continuous piecewise solutions of finite duration, please share it as an answer explaining why.

As an example of some PDE with solutions of finite duration, I believe that the equation $\nabla |u|^{\frac12}=0$ accept these kind of solutions, as I am studying with an Ansatz in this another question.

Answer 1

Suggestion.

If you mean to make the change of variables $\vec z= \frac{\vec v}{||\vec v||^2}$ then note that inversely $\vec v=\vec z/||\vec z||^2$. Plug this directly into the original Navier-Stokes equation satisfied by $\vec v$ and expand out.

The main tool you will need next is the product rule for expanding the vector Laplacian of an expression which is the product of a scalar function $f$ and a vector field $\vec z$: $$\triangle (f \vec z) = f\triangle \vec z+ 2 (\nabla f\cdot \nabla) \vec z+ (\triangle f) \vec z$$

In your case, $f=||\vec z||^{-2}$

Also you will need

(i) $v\cdot \nabla= \nabla_{f \vec z} \cdot \nabla = f (\vec z\cdot \nabla) $ and

(ii) $(\nabla f\cdot \nabla) \vec z$ is just what you get by applying $(\nabla f\cdot \nabla)=\sum_k \frac{\partial f}{\partial x_k} \frac{\partial}{\partial x_k}$ to each component of $\vec z$.

The remaining details are messy but straightforward.

P.S. As a practice exercise, you might first consider a simpler special sub-case in which the velocity field has the form $\vec v= grad F$ for some potential function $F$.