Quasi-periodic non-stationary solutions of 3D Euler equations for incompressible flow

Sergey V. Ershkov

doi:10.1016/j.jksus.2015.05.005

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Original article

27 (

); 369-374

doi:

10.1016/j.jksus.2015.05.005

Quasi-periodic non-stationary solutions of 3D Euler equations for incompressible flow

Sergey V. Ershkov

Institute for Time Nature Explorations, M.V. Lomonosov’s Moscow State University, Leninskie Gory, 1-12, Moscow 119991, Russia

Received: 2015-2-5, Accepted: 2015-5-17, Published: 2015-10-01

Disclaimer:
This article was originally published by Elsevier and was migrated to Scientific Scholar after the change of Publisher.

Peer review under responsibility of King Saud University.

Abstract

A novel derivation of non-stationary solutions of 3D Euler equations for incompressible inviscid flow is considered here. Such a solution is the product of 2 separated parts: one consisting of the spatial component and the other being related to the time dependent part.

Spatial part of a solution could be determined if we substitute such a solution to the equations of motion (equation of momentum) with the requirement of scale-similarity in regard to the proper component of spatial velocity. So, the time-dependent part of equations of momentum should depend on the time-parameter only.

The main result, which should be outlined, is that the governing (time-dependent) ODE-system consists of 2 Riccati-type equations in regard to each other, which has no solution in general case. But we obtain conditions when each component of time-dependent part is proved to be determined by the proper elliptical integral in regard to the time-parameter t, which is a generalization of the class of inverse periodic functions.

Keywords

Euler equations

Riccati equations

Non-stationary solutions

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1 Introduction: the Euler system of equations

In accordance with (Landau and Lifshitz, 1987; Ladyzhenskaya, 1969; Lighthill, 1986 ), the Euler system of equations for incompressible flow of inviscid fluid should be presented in the Cartesian coordinates as below (under the proper initial conditions):

(1.1)

\nabla \cdot \vec{u} = 0,

(1.2)

\frac{\partial \vec{u}}{\partial t} + (\vec{u} \cdot \nabla) \vec{u} = - \frac{\nabla p}{ρ} + \vec{F},

where u is the flow velocity, a vector field; ρ is the fluid density, p is the pressure, F represents external force (per unit of mass in a volume) acting on the fluid; besides, we assume external force F above to be the force, which has a potential ϕ represented by F = −∇ϕ.

2 The originating system of PDE for Euler equations

Using the identity (u · ∇)u = (1/2)∇(u²) − u × (∇ × u), we could present the Euler equations in the case of incompressible flow of inviscid fluid u = {u₁, u₂, u₃} as below (Saffman, 1995; Milne-Thomson, 1950):

(2.1)

\begin{matrix} \nabla \cdot \vec{u} = 0, \\ \frac{\partial \vec{u}}{\partial t} = \overset{⇀}{u} \times \vec{ω} - (\frac{1}{2} \nabla ({\vec{u}}^{2}) + \frac{\nabla p}{ρ} + \nabla ϕ) \end{matrix}

here we denote the curl field ω = ∇ × u, a pseudovector field (time-dependent) Kamke, 1971:

(2.2)

{ω_{x}, ω_{y}, ω_{z}} \equiv \{\begin{matrix} (\frac{\partial u_{3}}{\partial y} - \frac{\partial u_{2}}{\partial z}), & (\frac{\partial u_{1}}{\partial z} - \frac{\partial u_{3}}{\partial x}), & (\frac{\partial u_{2}}{\partial x} - \frac{\partial u_{1}}{\partial y}) \end{matrix}\}

also we denote ∇ϕ = {f_x, f_y, f_z} in (2.1); besides, let us choose ρ = 1 for simplicity.

3 Conditions for the space-part of exact solution

Let us search for solutions {u, p} of the system (2.1) in a form below:

(3.1)

u_{1} = U (t) \cdot u (x, y, z), u_{2} = V (t) \cdot v (x, y, z), u_{3} = W (t) \cdot w (x, y, z), p = P (t) \cdot p (x, y, z)

then we should obtain from (2.1) and expression (2.2) the proper system of PDE:

(3.2)

\{\begin{matrix} \frac{\partial u_{1}}{\partial t} = (u_{2} \cdot ω_{z} - u_{3} \cdot ω_{y}) - \frac{1}{2} \frac{\partial}{\partial x} (u_{1}^{2} + u_{2}^{2} + u_{3}^{2}) - \frac{\partial}{\partial x} p - f_{x}, \\ \frac{\partial u_{2}}{\partial t} = (u_{3} \cdot ω_{x} - u_{1} \cdot ω_{z}) - \frac{1}{2} \frac{\partial}{\partial y} (u_{1}^{2} + u_{2}^{2} + u_{3}^{2}) - \frac{\partial}{\partial y} p - f_{y}, \\ \frac{\partial u_{3}}{\partial t} = (u_{1} \cdot ω_{y} - u_{2} \cdot ω_{x}) - \frac{1}{2} \frac{\partial}{\partial z} (u_{1}^{2} + u_{2}^{2} + u_{3}^{2}) - \frac{\partial}{\partial z} p - f_{z}, \end{matrix}

Besides, there exists the proper restriction from continuity equation (1.1) as below (∂u/∂x ≠ 0):

(3.3)

\begin{matrix} U (t) \frac{\partial u}{\partial x} + V (t) \frac{\partial v}{\partial y} + W (t) \frac{\partial w}{\partial z} = 0, \Rightarrow \\ \frac{\partial v}{\partial y} = χ \frac{\partial u}{\partial x}, \frac{\partial w}{\partial z} = λ \frac{\partial u}{\partial x}, {χ, λ} = const \end{matrix}

The system of equations (3.2) should be transformed under conditions (3.1) as below:

(3.4)

\{\begin{matrix} \frac{dU (t)}{dt} = \frac{V (t) \cdot v (x, y, z) \cdot (V (t) \cdot \frac{\partial v}{\partial x} - U (t) \cdot \frac{\partial u}{\partial y}) - W (t) \cdot w (x, y, z) \cdot (U (t) \cdot \frac{\partial u}{\partial z} - W (t) \cdot \frac{\partial w}{\partial x})}{u (x, y, z)} \\ - \frac{1}{2} \frac{\frac{\partial}{\partial x} (U {(t)}^{2} \cdot u^{2} (x, y, z) + V^{2} (t) \cdot v^{2} (x, y, z) + W^{2} (t) \cdot w^{2} (x, y, z))}{u (x, y, z)} - \frac{(P (t) \cdot \frac{\partial}{\partial x} p (x, y, z) + f_{x})}{u (x, y, z)}, \\ \frac{dV (t)}{dt} = \frac{W (t) \cdot w (x, y, z) \cdot (W (t) \cdot \frac{\partial w}{\partial y} - V (t) \cdot \frac{\partial v}{\partial z}) - U (t) \cdot u (x, y, z) \cdot (V (t) \cdot \frac{\partial v}{\partial x} - U (t) \cdot \frac{\partial u}{\partial y})}{v (x, y, z)} \\ - \frac{1}{2} \frac{\frac{\partial}{\partial y} (U {(t)}^{2} \cdot u^{2} (x, y, z) + V^{2} (t) \cdot v^{2} (x, y, z) + W^{2} (t) \cdot w^{2} (x, y, z))}{v (x, y, z)} - \frac{(P (t) \cdot \frac{\partial}{\partial y} p (x, y, z) + f_{y})}{v (x, y, z)}, \\ \frac{dW (t)}{dt} = \frac{U (t) \cdot u (x, y, z) \cdot (U (t) \cdot \frac{\partial u}{\partial z} - W (t) \cdot \frac{\partial w}{\partial x}) - V (t) \cdot v (x, y, z) \cdot (W (t) \cdot \frac{\partial w}{\partial y} - V (t) \cdot \frac{\partial v}{\partial z})}{w (x, y, z)} \\ - \frac{1}{2} \frac{\frac{\partial}{\partial z} (U {(t)}^{2} \cdot u^{2} (x, y, z) + V^{2} (t) \cdot v^{2} (x, y, z) + W^{2} (t) \cdot w^{2} (x, y, z))}{w (x, y, z)} - \frac{(P (t) \cdot \frac{\partial}{\partial z} p (x, y, z) + f_{z})}{w (x, y, z)}, \end{matrix}

thus, from the 1-st of Eq. (3.4) we should assume ({a_i} = const, i = 1, …, 9):

(3.5)

\begin{matrix} \frac{v (x, y, z) \cdot \frac{\partial v}{\partial x}}{u (x, y, z)} = a_{1}, - \frac{v (x, y, z) \cdot \frac{\partial u}{\partial y}}{u (x, y, z)} = a_{2}, - \frac{w (x, y, z) \cdot \frac{\partial u}{\partial z}}{u (x, y, z)} = a_{3}, \frac{w (x, y, z) \cdot \frac{\partial w}{\partial x}}{u (x, y, z)} = a_{4}, \\ - \frac{1}{2} \frac{\frac{\partial}{\partial x} (u^{2} (x, y, z))}{u (x, y, z)} = a_{5}, - \frac{1}{2} \frac{\frac{\partial}{\partial x} (v^{2} (x, y, z))}{u (x, y, z)} = a_{6}, - \frac{1}{2} \frac{\frac{\partial}{\partial x} (w^{2} (x, y, z))}{u (x, y, z)} = a_{7}, \\ - \frac{\frac{\partial}{\partial x} p (x, y, z)}{u (x, y, z)} = a_{8}, - \frac{f_{x}}{u (x, y, z)} = a_{9}, \end{matrix}

but the 2-nd of Eq. (3.4) yields as below ({b_i} = const, i = 1, …, 9):

(3.6)

\begin{matrix} \frac{w (x, y, z) \cdot \frac{\partial w}{\partial y}}{v (x, y, z)} = b_{1}, - \frac{w (x, y, z) \cdot \frac{\partial v}{\partial z}}{v (x, y, z)} = b_{2}, - \frac{u (x, y, z) \cdot \frac{\partial v}{\partial x}}{v (x, y, z)} = b_{3}, \frac{u (x, y, z) \cdot \frac{\partial u}{\partial y}}{v (x, y, z)} = b_{4}, \\ - \frac{1}{2} \frac{\frac{\partial}{\partial y} (u^{2} (x, y, z))}{v (x, y, z)} = b_{5}, - \frac{1}{2} \frac{\frac{\partial}{\partial y} (v^{2} (x, y, z))}{v (x, y, z)} = b_{6}, - \frac{1}{2} \frac{\frac{\partial}{\partial y} (w^{2} (x, y, z))}{v (x, y, z)} = b_{7}, \\ - \frac{\frac{\partial}{\partial y} p (x, y, z)}{v (x, y, z)} = b_{8}, - \frac{f_{y}}{v (x, y, z)} = b_{9}, \end{matrix}

besides, 3-rd of Eq. (3.4) yields ({c_i} = const, i = 1, …, 9):

(3.7)

\begin{matrix} \frac{u (x, y, z) \cdot \frac{\partial u}{\partial z}}{w (x, y, z)} = c_{1}, - \frac{u (x, y, z) \cdot \frac{\partial w}{\partial x}}{w (x, y, z)} = c_{2}, - \frac{v (x, y, z) \cdot \frac{\partial w}{\partial y}}{w (x, y, z)} = c_{3}, \frac{v (x, y, z) \cdot \frac{\partial v}{\partial z}}{w (x, y, z)} = c_{4}, \\ - \frac{1}{2} \frac{\frac{\partial}{\partial z} (u^{2} (x, y, z))}{w (x, y, z)} = c_{5}, - \frac{1}{2} \frac{\frac{\partial}{\partial z} (v^{2} (x, y, z))}{w (x, y, z)} = c_{6}, - \frac{1}{2} \frac{\frac{\partial}{\partial z} (w^{2} (x, y, z))}{w (x, y, z)} = c_{7}, \\ - \frac{\frac{\partial}{\partial z} p (x, y, z)}{w (x, y, z)} = c_{8}, - \frac{f_{z}}{w (x, y, z)} = c_{9} . \end{matrix}

4 The space-part of exact solution

As for the structure of space part of exact solution (3.1), the system of equations (3.5)–(3.7) could be solved by the proper analytical way as below:

Eq. (3.5) yields:

(4.1)

\begin{matrix} \frac{v (x, y, z) \cdot \frac{\partial v}{\partial x}}{u (x, y, z)} = a_{1}, \frac{\partial u}{\partial y} = - (\frac{a_{2}}{a_{1}}) \cdot \frac{\partial v}{\partial x}, \frac{w (x, y, z) \cdot \frac{\partial u}{\partial z}}{u (x, y, z)} = - a_{3}, \frac{\partial w}{\partial x} = - (\frac{a_{4}}{a_{3}}) \cdot \frac{\partial u}{\partial z}, \\ \frac{\partial}{\partial x} (u (x, y, z)) = - a_{5} a_{6} = - a_{1}, a_{7} = - a_{4}, \frac{\partial}{\partial x} p (x, y, z) = a_{8} \cdot u (x, y, z), f_{x} = - a_{9} \cdot u (x, y, z), \end{matrix}

Eq. (3.6) yields:

(4.2)

\frac{w (x, y, z) \cdot \frac{\partial w}{\partial y}}{v (x, y, z)} = b_{1}, \frac{\partial w}{\partial y} = - (\frac{b_{1}}{b_{2}}) \cdot \frac{\partial v}{\partial z}, (\frac{u (x, y, z) \cdot \frac{\partial v}{\partial x}}{v (x, y, z)}) \cdot (\frac{v (x, y, z) \cdot \frac{\partial v}{\partial x}}{u (x, y, z)}) = - b_{3} \cdot a_{1}, \Rightarrow \frac{\partial v}{\partial x} = \sqrt{(- b_{3} \cdot a_{1})}, \Rightarrow u (x, y, z) = - (\frac{b_{3}}{\sqrt{(- b_{3} \cdot a_{1})}}) \cdot v (x, y, z), \Rightarrow b_{3} = a_{5}, \frac{\partial u}{\partial y} = - (\frac{b_{4}}{b_{3}}) \cdot \frac{\partial v}{\partial x}, \Rightarrow (\frac{a_{2}}{a_{1}}) = (\frac{b_{4}}{b_{3}}), b_{5} = - b_{4}, \frac{\partial}{\partial y} (v (x, y, z)) = - b_{6}, b_{7} = - b_{1}, \frac{\partial}{\partial y} p (x, y, z) = - b_{8} \cdot v (x, y, z), \frac{\partial}{\partial x} p (x, y, z) = - a_{8} \cdot (\frac{b_{3}}{\sqrt{(- b_{3} \cdot a_{1})}}) \cdot v (x, y, z), f_{y} = - b_{9} \cdot v (x, y, z),

and Eq. (3.7) yields:

(4.3)

\begin{matrix} (\frac{u (x, y, z) \cdot \frac{\partial u}{\partial z}}{w (x, y, z)}) \cdot (\frac{w (x, y, z) \cdot \frac{\partial u}{\partial z}}{u (x, y, z)}) = - a_{3} \cdot c_{1}, \Rightarrow \frac{\partial u}{\partial z} = \sqrt{(- a_{3} \cdot c_{1})}, \Rightarrow \frac{\partial v}{\partial z} = - \frac{\sqrt{(a_{3} \cdot c_{1} \cdot b_{3} \cdot a_{1})}}{b_{3}}, \\ w (x, y, z) = - (\frac{a_{3}}{\sqrt{(- a_{3} \cdot c_{1})}}) \cdot u (x, y, z) = (\frac{a_{3}}{\sqrt{(- a_{3} \cdot c_{1})}}) \cdot (\frac{b_{3}}{\sqrt{(- b_{3} \cdot a_{1})}}) \cdot v (x, y, z), \frac{\partial w}{\partial x} = - (\frac{c_{2}}{c_{1}}) \cdot \frac{\partial u}{\partial z}, \Rightarrow \\ (\frac{a_{4}}{a_{3}}) = (\frac{c_{2}}{c_{1}}), - \frac{v (x, y, z) \cdot \frac{\partial w}{\partial y}}{w (x, y, z)} = c_{3}, \Rightarrow \frac{\partial w}{\partial y} = \sqrt{(- b_{1} \cdot c_{3})}, \frac{\partial w}{\partial y} = - (\frac{c_{3}}{c_{4}}) \cdot \frac{\partial v}{\partial z}, \Rightarrow (\frac{c_{3}}{c_{4}}) = (\frac{b_{1}}{b_{2}}) c_{5} = - c_{1}, \\ c_{6} = - c_{4}, \frac{\partial}{\partial z} (w (x, y, z)) = - c_{7}, \frac{\partial}{\partial z} p (x, y, z) = - c_{8} \cdot w (x, y, z) = - c_{8} \cdot (\sqrt{\frac{a_{3} \cdot b_{3}}{a_{1} \cdot c_{1}}}) \cdot v (x, y, z), f_{z} = - c_{9} \cdot w (x, y, z) . \end{matrix}

So, the space part of the solution should be presented as below:

(4.4)

u = - (\frac{b_{3}}{\sqrt{(- b_{3} \cdot a_{1})}}) \cdot v (x, y, z) = - b_{3} \cdot x + b_{3} \cdot b_{6} \cdot y + \sqrt{(- a_{3} \cdot c_{1})} \cdot z, v = x - b_{6} \cdot y - \frac{\sqrt{(- a_{3} \cdot c_{1})}}{b_{3}} \cdot z, w = (\sqrt{\frac{a_{3} \cdot b_{3}}{a_{1} \cdot c_{1}}}) \cdot v (x, y, z) = \sqrt{(- \frac{a_{3}}{c_{1}})} \cdot | b_{3} | \cdot x - b_{6} \cdot (\sqrt{\frac{a_{3} \cdot b_{3}}{a_{1} \cdot c_{1}}}) \cdot y - a_{3} \cdot z, (a_{1} \cdot b_{3}) = - 1, a_{2} = b_{6}, | a_{3} | \cdot | b_{3} | = a_{3} \cdot b_{3}, a_{3} = c_{7}, a_{5} = b_{3}, a_{8} \cdot b_{3} \cdot b_{6} = b_{8}, a_{8} = \frac{c_{8}}{| a_{1} | \cdot | c_{1} |}, p (x, y, z) = a_{8} \cdot (- b_{3} \cdot b_{6} \cdot x \cdot y - \sqrt{(- a_{3} \cdot c_{1})} \cdot x \cdot z + b_{6} \sqrt{(- a_{3} \cdot c_{1})} \cdot y \cdot z + \frac{b_{3}}{2} x^{2} + (\frac{b_{3} \cdot {(b_{6})}^{2}}{2}) \cdot y^{2} + \frac{| c_{1} | \cdot | a_{3} |}{2 b_{3}} \cdot z^{2})

Thus, if we choose for simplicity the proper constants as:

(4.5)

c_{1} = - 1, a_{8} = 1 (\Rightarrow b_{8} = b_{3} \cdot b_{6}, c_{8} = 1 / | b_{3} |)

the space part of the solution should be presented as below (|a₃| · |b₃| = a₃ · b₃):

(4.6)

\begin{matrix} u = - b_{3} \cdot x + b_{3} \cdot b_{6} \cdot y + \sqrt{a_{3}} \cdot z \\ v = x - b_{6} \cdot y - \frac{\sqrt{a_{3}}}{b_{3}} \cdot z, \\ w = \sqrt{a_{3}} \cdot | b_{3} | \cdot x - b_{6} \cdot | b_{3} | \cdot \sqrt{a_{3}} \cdot y - a_{3} \cdot z, \\ p (x, y, z) = - b_{3} \cdot b_{6} \cdot x \cdot y - \sqrt{a_{3}} \cdot x \cdot z + b_{6} \sqrt{a_{3}} \cdot y \cdot z + \frac{b_{3}}{2} x^{2} + (\frac{b_{3} \cdot {(b_{6})}^{2}}{2}) \cdot y^{2} + \frac{| a_{3} |}{2 b_{3}} \cdot z^{2} \end{matrix}

5 Time-dependent part of exact solution

As for the structure of time-dependent part of exact solution (3.1) with space part (4.6), it could be obtained from system of Eq. (3.4) which should be transformed as below:

(5.1)

\{\begin{matrix} \frac{dU (t)}{dt} = V (t) \cdot U (t) \cdot a_{2} + W (t) \cdot U (t) \cdot a_{3} + U {(t)}^{2} \cdot a_{5} + P (t) \cdot a_{8} + a_{9}, \\ \frac{dV (t)}{dt} = W (t) \cdot V (t) \cdot b_{2} + U (t) \cdot V (t) \cdot b_{3} + V^{2} (t) \cdot b_{6} + P (t) \cdot b_{8} + b_{9}, \\ \frac{dW (t)}{dt} = U (t) \cdot W (t) \cdot c_{2} + V (t) \cdot W (t) \cdot c_{3} + W^{2} (t) \cdot c_{7} + P (t) \cdot c_{8} + c_{9}, \end{matrix}

where a₂ = b₆, |a₃| · |b₃| = a₃ · b₃, a₅ = b₃, a₈ = 1, b₈ = b₃ · b₆, c₇ = a₃, c₈ = 1/|b₃|; besides, it should be accomplished along with the continuity equation (3.3):

U (t) \frac{\partial u}{\partial x} + V (t) \frac{\partial v}{\partial y} + W (t) \frac{\partial w}{\partial z} = 0, \Rightarrow U (t) \cdot b_{3} + V (t) \cdot b_{6} + W (t) \cdot a_{3} = 0,

so, we have 4 equations for the obtaining of 4 functions U(t), V(t), W(t), P(t).

Along with the invariant from the continuity equation (besides, a₉ · b₃ + b₉ · b₆ + c₉ · a₃ = 0): $\frac{d (U (t) \cdot b_{3} + V (t) \cdot b_{6} + W (t) \cdot a_{3})}{dt} = 0,$ a system of Eq. (5.1) immediately yields the invariant for function P(t) as below:

(5.2)

P (t) \cdot (a_{8} \cdot b_{3} + b_{8} \cdot b_{6} + c_{8} \cdot a_{3}) = - [V (t) \cdot U (t) \cdot a_{2} + W (t) \cdot U (t) \cdot a_{3} + U {(t)}^{2} \cdot a_{5}] \cdot b_{3} - [W (t) \cdot V (t) \cdot b_{2} + U (t) \cdot V (t) \cdot b_{3} + V^{2} (t) \cdot b_{6}] \cdot b_{6} - [U (t) \cdot W (t) \cdot c_{2} + V (t) \cdot W (t) \cdot c_{3} + W^{2} (t) \cdot c_{7}] \cdot a_{3},

thus, we should exclude expression (5.2) for function P(t) from the analysis of equations of system (5.1) and also we should exclude the continuity equation, which means that one of 3 functions U(t), V(t), W(t) is the linear combination of two others:

(5.3)

W (t) = - U (t) \cdot \frac{b_{3}}{a_{3}} - V (t) \cdot \frac{b_{6}}{a_{3}}

So, analyzing the system (5.1), we finally should obtain the system of 2 ordinary differential equations of the 1-st order for any 2 of 3 functions U(t), V(t), W(t) (the last 3-rd function could be obtained by expressing it from the continuity equation above).

These governing ODE-equations form together a system of 2 Riccati-type equations in regard to each other, which is the system of 2 ordinary differential equations of the 1-st kind with the right parts, consisting of polynomials of the 2-nd extent in regard to the functions U(t), V(t), W(t).

Riccati type of equations has no analytical solution in general case (Kamke, 1971). We should note also that modern methods exist for obtaining the solution of Riccati equations with a good approximation (Bender and Orszag, 1999; Rosu et al., 2012; Christianto and Smarandache, 2008 ). But if we choose proper constants for the system (5.1):

(5.4)

\begin{matrix} a_{9} = b_{9} = c_{9} = 0, (a_{8} \cdot b_{3} + b_{8} \cdot b_{6} + c_{8} \cdot a_{3}) = 1, \Rightarrow \\ (b_{3} + b_{3} \cdot {(b_{6})}^{2} + a_{3} / | b_{3} |) = 1, \Rightarrow a_{3} = | b_{3} | - b_{3} \cdot | b_{3} | - b_{3} \cdot | b_{3} | \cdot {(b_{6})}^{2}, \end{matrix}

the 1-st and 2-nd equations of system (5.1) could be transformed as presented below:

(5.5)

\{\begin{matrix} \frac{dU (t)}{dt} = V (t) \cdot U (t) \cdot (b_{3} \cdot (\frac{b_{2} \cdot b_{6}}{a_{3}} + c_{3} - 2 b_{6}) + b_{6} \cdot (c_{2} - b_{3})) \\ + U {(t)}^{2} \cdot b_{3} \cdot (c_{2} - b_{3}) + V^{2} (t) \cdot b_{6} \cdot (\frac{b_{2} \cdot b_{6}}{a_{3}} + c_{3} - 2 b_{6}), \\ \frac{dV (t)}{dt} = U (t) \cdot V (t) \cdot (b_{3} \cdot (1 - \frac{b_{2}}{a_{3}}) + b_{6} \cdot b_{8} \cdot (c_{2} - b_{3}) + b_{3} \cdot b_{8} \cdot (\frac{b_{2} \cdot b_{6}}{a_{3}} + c_{3} - 2 b_{6})) \\ + V^{2} (t) \cdot (b_{6} \cdot (1 - \frac{b_{2}}{a_{3}}) + b_{6} \cdot b_{8} \cdot (\frac{b_{2} \cdot b_{6}}{a_{3}} + c_{3} - 2 b_{6})) + U^{2} (t) \cdot b_{3} \cdot b_{8} \cdot (c_{2} - b_{3}), \end{matrix}

where a₂ = b₆, see (4.4). Besides, if we additionally choose b₂ = a₃, system (5.5) above could be reduced to the simplified regular form below:

(5.6)

\{\begin{matrix} \frac{dU (t)}{dt} = (b_{8} \cdot U (t) + C) \cdot U (t) \cdot (b_{3} \cdot (c_{3} - b_{6}) + b_{6} \cdot (c_{2} - b_{3})) + U {(t)}^{2} \cdot b_{3} \cdot (c_{2} - b_{3}) + {(b_{8} \cdot U (t) + C)}^{2} \cdot b_{6} \cdot (c_{3} - b_{6}), \\ \frac{dV (t)}{dt} = b_{8} \cdot \frac{dU (t)}{dt} \Rightarrow V (t) = b_{8} \cdot U (t) + C, C = const \end{matrix}

where the 1-st equation of system (5.6) has a proper solution below (b₈ = b₃ · b₆):

(5.7)

\frac{dU (t)}{(A \cdot U^{2} (t) + B \cdot U (t) + D)} = dt,

\begin{matrix} A = (b_{8} \cdot b_{3} \cdot (c_{3} - b_{6}) + b_{8} \cdot b_{6} \cdot (c_{2} - b_{3}) + b_{3} \cdot (c_{2} - b_{3}) + {(b_{8})}^{2} \cdot b_{6} \cdot (c_{3} - b_{6})) \\ B = C \cdot (b_{3} \cdot (c_{3} - b_{6}) + b_{6} \cdot (c_{2} - b_{3}) + 2 b_{8} \cdot b_{6} \cdot (c_{3} - b_{6})), D = C^{2} \cdot b_{6} \cdot (c_{3} - b_{6}) \end{matrix}

The left side of expression (5.7) could be transformed to the proper elliptical integral (Lawden, 1989) in regard to function U(t):

(5.8)

\int \frac{dU (t)}{(A \cdot U^{2} (t) + B \cdot U (t) + D)} = \{\begin{matrix} \frac{2}{\sqrt{Δ}} \arctan (\frac{2 A \cdot U (t) + B}{\sqrt{Δ}}), Δ > 0 \\ - \frac{2}{\sqrt{- Δ}} Arth (\frac{2 A \cdot U (t) + B}{\sqrt{- Δ}}), Δ < 0 \end{matrix} Δ = (4 A \cdot D - B^{2})

6 Discussion

In fluid mechanics, a lot of authors have been executing their researches to obtain the analytical solutions of Euler and Navier–Stokes equations (Drazin and Riley, 2006), even for 3D case of compressible gas flow (Ershkov and Schennikov, 2001). But there is an essential deficiency of non-stationary solutions indeed.

Our presentation (3.1) of the non-stationary solutions of 3D Euler equations (1.1) and (1.2) for incompressible flow is considered here. The spatial part of such a solution is determined by equalities (4.4), under the given initial conditions; but the time-dependent part is determined by Eqs. (5.1)–(5.3).

Besides, the real example of exact solution is obtained. The spatial part of such a solution is presented by the equalities (4.5) and (4.6), but the time-dependent part is presented by equalities (5.2)–(5.4) and (5.8).

Also, we should especially note that the components of flow velocity (4.6) of the solution (3.1) will be uniformly increasing when (x, y, z) → ∞. So, such a solution should be defined within the limited domain of the meanings of variables (x, y, z), it should be given by the initial conditions.

The explicit solutions for U(t) can easily be obtained from Eqs. (5.7) and (5.6), therefore, the explicit form of the special solutions (3.1) should be provided: $u_{1} = U (t) \cdot u (x, y, z), u_{2} = V (t) \cdot v (x, y, z), u_{3} = W (t) \cdot w (x, y, z), p = P (t) \cdot p (x, y, z),$ where: $\begin{matrix} u = - b_{3} \cdot x + b_{3} \cdot b_{6} \cdot y + \sqrt{a_{3}} \cdot z \\ v = x - b_{6} \cdot y - \frac{\sqrt{a_{3}}}{b_{3}} \cdot z, \\ w = \sqrt{a_{3}} \cdot | b_{3} | \cdot x - b_{6} \cdot | b_{3} | \cdot \sqrt{a_{3}} \cdot y - a_{3} \cdot z, \\ p (x, y, z) = - b_{3} \cdot b_{6} \cdot x \cdot y - \sqrt{a_{3}} \cdot x \cdot z + b_{6} \sqrt{a_{3}} \cdot y \cdot z + \frac{b_{3}}{2} x^{2} + (\frac{b_{3} \cdot {(b_{6})}^{2}}{2}) \cdot y^{2} + \frac{| a_{3} |}{2 b_{3}} \cdot z^{2}, \end{matrix}$ $\begin{matrix} P (t) = - [V (t) \cdot U (t) \cdot b_{6} + W (t) \cdot U (t) \cdot a_{3} + U {(t)}^{2} \cdot b_{3}] \cdot b_{3} \\ - [W (t) \cdot V (t) \cdot a_{3} + U (t) \cdot V (t) \cdot b_{3} + V^{2} (t) \cdot b_{6}] \cdot b_{6} - [U (t) \cdot W (t) \cdot c_{2} + V (t) \cdot W (t) \cdot c_{3} + W^{2} (t) \cdot a_{3}] \cdot a_{3}, \end{matrix}$ $W (t) = - U (t) \cdot \frac{b_{3}}{a_{3}} - V (t) \cdot \frac{b_{6}}{a_{3}}, V (t) = (b_{3} \cdot b_{6}) \cdot U (t) + C, C = const,$ here we should choose a₃ = {|b₃| − b₃ · |b₃| − b₃ · |b₃| · (b₆)²} ≠ 0, →b₃ · (1 + (b₆)²) ≠ 1, but the key function U(t) should be given as below (b₈ = b₃ · b₆): $\begin{matrix} \{\begin{matrix} U (t) = \frac{\sqrt{Δ} \cdot \tan ((\frac{\sqrt{Δ}}{2}) \cdot t) - B}{2 A}, Δ > 0 \\ U (t) = \frac{\sqrt{- Δ} \cdot \tanh (- (\frac{\sqrt{- Δ}}{2}) \cdot t) - B}{2 A}, Δ < 0 \end{matrix} Δ = (4 A \cdot D - B^{2}) \\ A = (b_{8} \cdot b_{3} \cdot (c_{3} - b_{6}) + b_{8} \cdot b_{6} \cdot (c_{2} - b_{3}) + b_{3} \cdot (c_{2} - b_{3}) + {(b_{8})}^{2} \cdot b_{6} \cdot (c_{3} - b_{6})) \\ B = C \cdot (b_{3} \cdot (c_{3} - b_{6}) + b_{6} \cdot (c_{2} - b_{3}) + 2 b_{8} \cdot b_{6} \cdot (c_{3} - b_{6})), D = C^{2} \cdot b_{6} \cdot (c_{3} - b_{6}) \end{matrix}$

For example, if we choose c₃ = b₆ (C ≠ 0) it should simplify the expression for U(t): $\begin{matrix} U (t) = \frac{\sqrt{- Δ} \cdot \tanh (- (\frac{\sqrt{- Δ}}{2}) \cdot t) - B}{2 A}, Δ = - B^{2} \\ A = b_{3} \cdot (c_{2} - b_{3}) \cdot (1 + {(b_{6})}^{2}), B = C \cdot b_{6} \cdot (c_{2} - b_{3}), D = 0 \end{matrix}$ and, if we additionally choose c₂ = 2b₃, b₃ = 2/C (C ≠ 0), b₆ = 1, we should obtain (see Figs. 1–5 ): $U (t) = \frac{C^{2}}{8} \cdot (\tanh (- t) - 1), Δ = - 4, A = {(b_{3})}^{2} \cdot 2, B = 2, D = 0$

A schematic plot of the function ∼ (x − y + 1) * {tanh(−t) − 1}, here we designate: x ∈ (−50, 50), t = y ∈ (0, 25).

A schematic plot of the function ∼ (x − y + 1) * {tanh(−t) − 1}, here we designate: x ∈ (−1, 1), t = y ∈ (0, 1).

A schematic plot of the function ∼ (x − y − 10) * {tanh(−t) − 1}, here we designate: x ∈ (−1, 1), t = y ∈ (0, 1).

A schematic plot of the function ∼ (x − y + 30) * {tanh(−t) − 1}, here we designate: x ∈ (−1, 1), t = y ∈ (0, 1).

We assume at Figs. 1–5 that in the expressions (x − y − 10), (x − y + 1), (x − y + 10), (x − y + 30) set of meanings {−10, 1, 10, 30} is varying according to the varying of the range of variable z; besides, the factor {tanh(−t) − 1} could be schematically presented (for imagination of the plots of solutions) by the changing of parameter t to variable y, for example.

At Figs. 6–8 we schematically imagined solutions for the case Δ > 0.

A schematic plot of the function ∼ (x − y − 10) * {tan(t) − 1}, here we designate: x ∈ (−1, 1), t = y ∈ (0, 1).

A schematic plot of the function ∼ (x − y + 1) * {tan(t) − 1}, here we designate: x ∈ (−1, 1), t = y ∈ (0, 1).

Also, we should note that since some solutions are unbounded (see for instance Eq. (5.8) for Δ > 0), such a solution should be defined within the limited range of the meanings of time-parameter t (it should be given by the initial conditions).

Besides, we should additionally note that the only periodic (and unbounded) solutions are the ones given by U(t) for Δ > 0 (Figs. 6–8 ), since the hyperbolic tangent is a non periodic but bounded function in this case (Figs. 1–5 ).

7 Conclusion

A new presentation of non-stationary solutions of 3D Euler equations for incompressible inviscid flow is considered here. Such a solution is the product of 2 separated parts: - spatial and the time-dependent parts.

Spatial part of a solution could be determined if we substitute such a solution to the equations of motion (equation of momentum) under the demand of scale-similarity in regard to the proper component of spatial velocity. So, the time-dependent part of equations of momentum should depend on the time-parameter only.

Acknowledgements

I am thankful to unknown esteemed Reviewers for valuable advices in preparing of this manuscript. Especially I appreciate Dr. Rizwan Irshad for his kind help and assistance during navigating in submission process of my articles to JKSUS.

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