Then Navier Stokes equations become $\nu =\alpha$ the viscosity constant:

$\frac{\partial v}{\partial t}+\left(N-10\right)\underset{n=1}{\overset{+\infty }{{\displaystyle \sum }}}\frac{v^{500n+2}}{n!}-{10}^{-2}\underset{p=0}{\overset{N-10}{{\displaystyle \sum }}}pv=f+\alpha {\nabla }^{2}v-\frac{1}{\rho }\nabla P$ (1)

$\frac{\partial v}{\partial t}+\left(N-10\right)\underset{n=1}{\overset{+\infty }{{\displaystyle \sum }}}\frac{v^{500n+2}}{n!}-\frac{\left(N-10\right)\left(N-9\right)}{2\times {10}^2}v=f+\alpha {\nabla }^{2}v-\frac{1}{\rho }\nabla P$ (2)

Let $v=h\left(t\right)g\left(r\right)$ with $r\in \left\{x,y,z\right\}$ a coordinate:

$\frac{1}{h}\frac{\partial h}{\partial t}+\left(N-10\right)\underset{n=1}{\overset{+\infty }{{\displaystyle \sum }}}\frac{v^{500n+1}}{n!}-\frac{\left(N-10\right)\left(N-9\right)}{2\times {10}^2}=\frac{f}{v}+\alpha \frac{1}{g}{\nabla }^{2}g-\frac{1}{v\rho }\nabla P$ (3)

with $\frac{1}{h}\frac{\partial h}{\partial t}=-K^2$; $K^2=\frac{1}{{\tau }_c}$ the temporal decay, we obtain:

$h=h_0\exp \left(-\frac{t}{{\tau }_c}\right)$ (4)

Let’s take ${\omega }^2=\frac{1}{{\tau }_c}+\frac{\left(N-10\right)\left(N-9\right)}{2\times {10}^2}$ we get:

${\displaystyle \iint f\text{d}{r}^2}+\alpha {\displaystyle \iint {\text{d}}^{2}v}-\frac{1}{\rho }{\displaystyle \iint \nabla P\text{d}{r}^2}-\left(N-10\right)\underset{n=1}{\overset{+\infty }{{\displaystyle \sum }}}{\displaystyle \iint \frac{v^{500n+2}}{n!}\text{d}{r}^2}+{\omega }^2{\displaystyle \iint v\text{d}{r}^2}=0$ (5)

$\frac{f{r}^2}{2}+\alpha \frac{v^2}{2}-\frac{Pr}{\rho }-\left(N-10\right)\underset{n=1}{\overset{+\infty }{{\displaystyle \sum }}}\frac{1}{n!}{\displaystyle \iint v^{500n+2}\text{d}{r}^2}+{\omega }^2{\displaystyle \iint v\text{d}{r}^2}=0$ (6)

with $g=\exp \left(ar\right)$ we have $\text{d}v=av\text{d}r$ that implies $\text{d}r=\frac{1}{a}\frac{\text{d}v}{v}$, then we get:

$a^2\frac{f{r}^2}{2}+a^2\alpha \frac{v^2}{2}-a^2\frac{Pr}{\rho }-\left(N-10\right)\underset{n=1}{\overset{+\infty }{{\displaystyle \sum }}}\frac{1}{n!}{\displaystyle \iint v^{500n}\text{d}{v}^2}+{\omega }^2{\displaystyle \iint \frac{1}{v}\text{d}{v}^2}=0$ (7)

$\begin{array}{l}{a}^2\frac{f{r}^2}{2}+a^2\alpha \frac{v^2}{2}-a^2\frac{Pr}{\rho }\\ -\left(N-10\right)\underset{n=1}{\overset{+\infty }{{\displaystyle \sum }}}\frac{v^{500n+2}}{n!\left(500n+1\right)\left(500n+2\right)}+{\omega }^2\left[\ln \left(v\right)-1\right]v=0\end{array}$ (8)

$\begin{array}{l}{a}^2\frac{f{r}^2}{2}+a^2\alpha \frac{v^2}{2}-a^2\frac{Pr}{\rho }-\left(N-10\right)\underset{n=1}{\overset{+\infty }{{\displaystyle \sum }}}\frac{v^{500n+2}}{n!\left(500n+1\right)\left(500n+2\right)}\\ ={\omega }^2\left[1-\ln \left(v\right)\right]v\end{array}$ (9)

$a$ and ${\tau }_c$ must numerically verify for $v$:

$\begin{array}{c}v\left(r\in \left\{x,y,z\right\}\right)=\frac{a^2}{{\omega }^2\left(1+\frac{t}{{\tau }_c}-ar\right)}\{\frac{f{r}^2}{2}+\alpha \frac{\exp \left[2\left(ar-\frac{t}{{\tau }_c}\right)\right]}{2}-\frac{Pr}{\rho }\\ -\left(N-10\right)\underset{n=1}{\overset{+\infty }{{\displaystyle \sum }}}\frac{\exp \left[\left(500n+2\right)\left(ar-\frac{t}{{\tau }_c}\right)\right]}{n!\left(500n+1\right)\left(500n+2\right)a^2}\}\end{array}$ (10)

by continuity equation, $a$ must also be verified:

$\frac{D\rho }{Dt}=-av$ (11)

$\left(\frac{\partial }{\partial t}+v\cdot \nabla \right)\rho =-av$ (12)

$\left(\frac{\partial }{\partial t}+v\cdot \nabla \right)\frac{\rho }{v}=-a$ (13)

$-\left(\frac{\partial }{\partial t}+v\cdot \nabla \right)\frac{\rho }{v}=a$ (14)

with $\lambda =a$, we get:

$\{\begin{array}{l}{\lambda }_{liquid}=-\left(\frac{\partial }{\partial t}+v\cdot \nabla \right)\rho \simeq -\frac{\partial \rho }{\partial t}\text{for a liquid}{\rho }_{liquid}\gg v_{fluid}\\ {\lambda }_{gaz}\simeq -\nabla \rho \text{for a gaz}{\rho }_{gaz}\le v_{gaz}\end{array}$

$\begin{array}{c}v\left(r\in \left\{x,y,z\right\}\right)=\frac{{\lambda }^2}{{\omega }^2\left(1+\frac{t}{{\tau }_c}-\lambda r\right)}\{\frac{f{r}^2}{2}+\alpha \frac{\exp \left[2\left(\lambda r-\frac{t}{{\tau }_c}\right)\right]}{2}-\frac{Pr}{\rho }\\ -\left(N-10\right)\underset{n=1}{\overset{+\infty }{{\displaystyle \sum }}}\frac{\exp \left[\left(500n+2\right)\left(\lambda r-\frac{t}{{\tau }_c}\right)\right]}{n!\left(500n+1\right)\left(500n+2\right){\lambda }^2}\}\end{array}$ (15)