The set of stochastic Eqs. (1)–(3) for the area 2), referring the pair (N,M) = (1,1) is:

$${\left(\frac{d{\left(\rho \right)}_{col_{st}}}{d\tau}\right)}_{1,1}=-\frac{d{\left(\rho \right)}_{st}}{d\tau};$$

$$\left\{\begin{array}{l}{\left(\frac{d{\left(\rho \overrightarrow{U}\right)}_{col_{st}}}{d\tau}\right)}_{1,1}=-\left(\frac{d{\left(\rho \overrightarrow{U}\right)}_{st}}{d\tau}\right);\\ {}{div}{\left({\tau}_{i,j}\right)}_{col_{st1}}=\frac{d{\left(\rho \overrightarrow{U}\right)}_{st}}{d\tau}.\end{array}\right.$$

(11)

$$\left\{\begin{array}{l}{\left(\frac{d{(E)}_{{col}_{st}}}{d\tau}\right)}_{1,1}=-{\left(\frac{d{(E)}_{st}}{d\tau}\right)}_{1,1};\\ {}{div}{\left(\lambda \frac{\partial T}{\partial {x}_j}+{u}_i{\tau}_{i,j}\right)}_{col_{st1}}={\left(\frac{d{(E)}_{st}}{d\tau}\right)}_{1,1}.\end{array}\right.$$

Then, taking into account the definition of the velocity *u*_{1} of laminar motion in a plane jet [61, 64].

$${u}_1=0.4543{\left(\frac{{\left({U}_m^2h\right)}^2}{\nu \cdot {x}_1}\right)}^{1/3}\left(1-{\left( th\left(\xi \right)\right)}^2\right).$$

(12)

Taking into account that

$$\xi =0.2752{\left(\frac{\left({U}_m^2h\right)}{\nu^2}\right)}^{1/3}\frac{x_2}{x_1^{2/3}}$$

(13)

and introducing the relation

$$\frac{th\left(\xi \right)}{ch\left(\xi \right)}=\frac{\xi -\frac{\xi^3}{3}+\frac{2{\xi}^5}{15}.\dots }{1+\frac{\xi^2}{2!}+\frac{\xi^4}{4!}.\dots }=f\left(\xi \right),$$

(14)

the derivative \(\frac{du_1}{dx_1}\) is

$$\frac{{d u}_1}{{d x}_1}=-\frac{1}{3}0.4543{\left(\frac{{\left({U}_m^2h\right)}^2}{\nu \cdot {x}_1}\right)}^{1/3}\frac{1}{x_1}-0.4543{\left(\frac{{\left({U}_m^2h\right)}^2}{\nu \cdot {x}_1}\right)}^{1/3}\cdot 2f\left(\xi \right)\frac{d\xi}{{d x}_1},$$

(15)

$$\frac{d\xi}{{d x}_1}=\left[0.2752{\left(\frac{\left({U}_m^2h\right)}{\nu^2}\right)}^{1/3}\frac{x_2}{x_1^{2/3}}\right]\cdot \left(-\frac{2}{3}\frac{1}{x_1}\right)=-\frac{2}{3}\frac{1}{x_1}\cdot \xi,$$

(16)

$$\frac{du_1}{dx_1}=-\frac{1}{3}0.4543{\left(\frac{{\left({U}_m^2h\right)}^2}{\nu \cdot {x}_1}\right)}^{1/3}\frac{1}{x_1}-0.4543{\left(\frac{{\left({U}_m^2h\right)}^2}{\nu \cdot {x}_1}\right)}^{1/3}\cdot 2f\left(\xi \right)\left(-\frac{2}{3}\frac{1}{x_1}\cdot \xi \right)=0.4543{\left(\frac{{\left({U}_m^2h\right)}^2}{\nu \cdot {x}_1}\right)}^{1/3}\frac{1}{x_1}\left(-\frac{1}{3}+\frac{4}{3}\cdot {\xi}^2\right),$$

(17)

$${\left(\frac{du_1}{dx_1}\right)}^2={\left[0.4543{\left(\frac{{\left({U}_m^2h\right)}^2}{\nu \cdot {x}_1}\right)}^{1/3}\frac{1}{x_1}\left(-\frac{1}{3}+\frac{4}{3}\cdot {\xi}^2\right)\right]}^2.$$

(18)

In case of small values of *ξ* for laminar flow we have next formulas:

$${\left(\frac{du_1}{dx_1}\right)}^2={\left[0.4543{\left(\frac{{\left({U}_m^2h\right)}^2}{\nu \cdot {x}_1}\right)}^{1/3}\frac{1}{x_1}\left(-\frac{1}{3}\right)\right]}^2=\frac{1}{9}{\left[0.4543{\left(\frac{{\left({U}_m^2h\right)}^2}{\nu \cdot {x}_1}\right)}^{1/3}\frac{1}{x_1}\right]}^2,$$

(19)

$${\left(\frac{du_1}{dx_1}\right)}^2=\left[\frac{1}{9}{0.4543}^2{\left(\frac{{\left({U}_m^2h\right)}^2}{\nu \cdot {x}_1}\right)}^{1/3}{\left(\frac{{\left({U}_m^2h\right)}^2}{\nu \cdot {x}_1}\right)}^{1/3}\frac{1}{x_1^2}\right]=\left[\frac{1}{9}{0.4543}^2{\left(\frac{\left({U}_m^8{h}^4\right)}{x_1^2{\nu}^2}\right)}^{1/3}\frac{1}{x_1^2}\right],$$

(20)

$${\left(\frac{du_1}{dx_1}\right)}^2=\left[\frac{1}{9}{0.4543}^2{\left(\frac{\left({U}_m^8{h}^4\right)}{x_1^2{\nu}^2}\right)}^{1/3}\frac{1}{x_1^2}\right]=\left[\frac{1}{9}{0.4543}^2{U}_m^2{\left(\frac{h^2}{x_1^2}\right)}^{1/3}{\left(\frac{\left({U}_m^2{h}^2\right)}{\nu^2}\right)}^{1/3}\frac{1}{x_1^2}\right].$$

(21)

So finally we have

$$\mu {\left(\frac{du_1}{dx_1}\right)}^2=\rho \nu \left[\frac{1}{9}{0.4543}^2{U}_m^2{\left(\frac{h}{x_1}\right)}^{2/3}{\left({Re}\right)}^{2/3}\frac{1}{x_1^2}\right].$$

(22)

For the flow region of the beginning of turbulence initiation, we assume the same dependence, but with a different degree “*n*”. As is known, if the velocity *u*_{1} in laminar motion is ~ (*x*_{1})^{-1/3}, then for a developed turbulent region *u*_{1} ~ (*x*_{1})^{-1/n} and “*n* = 2”. Thus, for the transition region the values of degree “*n*” are in the interval 2 ≤ *n* ≤ 3. Next, we write down the relation determined earlier from the equivalence of measures for the region (1, 1) - the generation of turbulence [33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58]:

$$\frac{{\left[\rho \nu {\left(\frac{\partial {u}_1}{\partial {x}_2}\right)}^2\Delta \tau \right]}_{st och}}{{\left[\rho \nu {\left(\frac{\partial {u}_1}{\partial {x}_2}\right)}^2\Delta \tau \right]}_{lamin}}=\left|{{Re}_{st}}-\frac{1}{{Re}_{st}}\right|.$$

(23)

As it was shown in the papers [33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58], the Eq. (23) determines the relative increase of the energy transferred from the deterministic state to the random one. Subscript “lamin” refers to the laminar flow, and subscript “stoch” refers to the non-laminar flow. Taking into account that for transition regime formula (22) may be written as \(\mu {\left(\frac{du_1}{dx_1}\right)}^2=\rho \nu \left[\frac{1}{n^2}{0.4543}^2{\left(\frac{U_m}{x_1}\right)}^2{\left(\frac{h}{x_1}\right)}^{2/n}{\left({Re}\right)}^{2/n}\right]\). Then in accordance with [44,45,46,47,48,49, 60], the Eq. (23) is

$$\frac{\left[\rho \nu {{\left(\frac{\partial {u}_1}{\partial {x}_2}\right)}^2}_{stoch}\right]\Delta {\tau}_{stoch}}{\left[\rho \nu {{\left(\frac{\partial {u}_1}{\partial {x}_2}\right)}^2}_{lamin}\right]\Delta {\tau}_l}=\frac{9}{n^2}{\left[\left({Re}\right)\left(\frac{h}{L_2}\right)\left(\frac{L_2}{x_1}\right)\right]}^{\left(2/n\right)-2/3}\frac{\Delta {\tau}_{stoch}}{\Delta {\tau}_l}.$$

(24)

Easy to see that for laminar flow *n* = 3, the right side of Eq. (24) is equal to 1. Let us determine the value of the left side of the Eq. (23) (or the right side of Eq. 24) for various values of the indicator “*n*” and then compare with the calculated value of the right side of the Eq. (23). For the left side of the equation, if *n* = 2.5, we have \({K}_{\tau }=\frac{\Delta {\tau}_{stoch}}{\Delta {\tau}_l}\approx 0.8\div 1.2\) [56,57,58] and in accordance with [61,62,63,64,65,66], \(\left(\frac{h}{L_2}\right)\cdot \left(\frac{L_2}{x_1}\right)=\left(2.5\div 14.5\right)\cdot \left(0.01\div 0.0157\right)=0.025\div 0.22765\). As can be seen, the experimental spread of values refers to the magnitude of the turbulence scale \(\left(\frac{h}{L_2}\right)\). Therefore, the main attention will be focused on the correspondence of the calculated values of the right side of Eq. (24) to the left side of this equation. Then it can be written that

$$\frac{\rho \nu {{\left(\frac{\partial {u}_1}{\partial {x}_2}\right)}^2}_{stoch}}{\rho \nu {{\left(\frac{\partial {u}_1}{\partial {x}_2}\right)}^2}_{lamin}}{K}_{\tau }=\frac{9}{6.25}\cdot {\left[\left(\frac{h}{L_2}\right)\cdot \left(\frac{L_2}{x_1}\right)\right]}^{0.1333}\cdot {\left({Re}\right)}^{0.1333}\cdot {K}_{\tau }=1.44\cdot {\left[0.025\div 0.22765\right]}^{0.1333}\cdot {\left(7.25\div 25\right)}^{0.1333}\cdot \left(0.8\div 1.2\right),$$

(25)

$$\frac{\rho \nu {{\left(\frac{\partial {u}_1}{\partial {x}_2}\right)}^2}_{stoch}}{\rho \nu {{\left(\frac{\partial {u}_1}{\partial {x}_2}\right)}^2}_{lamin }}{K}_{\tau }=(1.44)\cdot \left(0.61\div 0.821\right)\cdot {{Re}}^{0.133}\cdot \left(0.8\div 1.2\right)=\left(0.878\div 1.18\right)\cdot \left(1.3\div 1.53\right)\cdot \left(0.8\div 1.2\right).$$

(26)

So for *n* = 2.5, the left side of the equation is

$$\frac{\rho \nu {{\left(\frac{\partial {u}_1}{\partial {x}_2}\right)}^2}_{stoch}}{\rho \nu {{\left(\frac{\partial {u}_1}{\partial {x}_2}\right)}^2}_{lamin }}{K}_{\tau }=\left(0.913\div 1.62\right)\div \left(1.235\div 2.18\right)=1.5.$$

(27)

Then for *n* = 2.0 and the same initial data for the critical Reynolds number (*Re*)_{cr} ~ 7.25 ÷ 25, \({K}_{\tau }=\frac{\Delta {\tau}_{stoch}}{\Delta {\tau}_l}\approx 0.8\div 1.2\) [54,55,56,57,58] and \(\left(\frac{h}{L_2}\right)\cdot \left(\frac{L_2}{x_1}\right)=\left(2.5\div 14.5\right)\cdot \left(0.01\div 0.0157\right)=0.025\div 0.22765\) [61,62,63,64,65,66,67], the next equation can be written as

$$\frac{\rho \nu {{\left(\frac{\partial {u}_1}{\partial {x}_2}\right)}^2}_{st och}}{\rho \nu {{\left(\frac{\partial {u}_1}{\partial {x}_2}\right)}^2}_{lamin }}{K}_{\tau }=\frac{9}{4}\cdot {\left[\left(\frac{h}{L_2}\right)\cdot \left(\frac{L_2}{x_1}\right)\right]}^{1/3}\cdot {\left({Re}\right)}^{1/3}\cdot {K}_{\tau }=\left|{{Re}}_{st}-\left(1/{{Re}}_{st}\right)\right|.$$

(28)

Substituting the numerical values, it is possible to find that

$$\frac{\rho \nu {{\left(\frac{\partial {u}_1}{\partial {x}_2}\right)}^2}_{stoch}}{\rho \nu {{\left(\frac{\partial {u}_1}{\partial {x}_2}\right)}^2}_{lamin }}\cdot {K}_{\tau }=\frac{9}{4}\cdot {\left[\left(\frac{h}{L_2}\right)\cdot \left(\frac{L_2}{x_1}\right)\right]}^{1/3}\cdot {\left({Re}\right)}^{1/3}\cdot {K}_{\tau }=2.25\cdot {\left[0.025\div 0.22765\right]}^{1/3}\cdot {\left(7.25\div 25\right)}^{1/3}\cdot \left(0.8\div 1.2\right).$$

(29)

So for *n* = 2.0, the left side of the equation is

$$\frac{\rho \nu {{\left(\frac{\partial {u}_1}{\partial {x}_2}\right)}^2}_{stoch}}{\rho \nu {{\left(\frac{\partial {u}_1}{\partial {x}_2}\right)}^2}_{lamin }}{K}_{\tau }=2.25\cdot \left(0.292\div 0.608\right)\cdot \left(0.8\div 1.2\right)\cdot \left(1.94\div 2.924\right)=\left(1.02\div 2.3\right)\div \left(2.12\div 4.81\right)\approx 2.5.$$

(30)

Then determine the right side of Eq. (23). Within the framework of the available data in the case of the origin of turbulence, experimental values of turbulent Reynolds number are *Re*_{st} ~ 0.4÷0.5 [61,62,63,64,65,66,67]. In the case of the developed turbulence, according to [2], experimental values of turbulent Reynolds number are *Re*_{st} ~ 2÷4. So the right side of Eq. (23) for *n* = 3.0 ÷ 2.5 has value

$$\left|{{Re}_{st}}\hbox{-} 1/{{Re}_{st}}\right|=\left|\left(0.5\div 0.4\right)\hbox{-} 1/\left(0.5\div 0.4\right)\right|\approx \left(1.5\div 2.1\right)\approx 1.7,$$

and for *n* = 2.0 the value is

$$\left|{{Re}_{st}}\hbox{-} 1/{{Re}_{st}}\right|=\left|\left(2\div 4\right)\hbox{-} 1/\left(2\div 4\right)\right|\approx \left(1.5\div 3.8\right)\approx 2.7.$$

(31)

Thus, the estimate of the profile index is satisfactorily determined by the obtained relation *n* = 2.0, corresponding to the turbulent flow value of index ‘*n*’ in empirical formula *u*_{1} ~ (*x*_{1})^{-1/2} [61, 64]. It should be noted that the main last studies [61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119] do not contain information on analytical solutions for the jet profile in the laminar-turbulent transition regime.