1. Introduction
In the present work, our main concern is with derivation of Cartesian vector solutions for the compressible Navier–Stokes equation with Coriolis force and density-dependent viscosity, which takes the following form [1]:
$\begin{eqnarray}\left\{\begin{array}{l}{\rho }_{t}+{\rm{div}}(\rho {\boldsymbol{u}})=0,\\ {(\rho {\boldsymbol{u}})}_{t}+{\rm{div}}(\rho {\boldsymbol{u}}\displaystyle \otimes {\boldsymbol{u}})+\rho {\boldsymbol{J}}{\boldsymbol{u}}-{\rm{div}}(h(\rho )D({\boldsymbol{u}}))\\ -{\rm{\nabla }}(g(\rho ){\rm{div}}\,{\boldsymbol{u}})+{\rm{\nabla }}p(\rho )={\bf{0}},\end{array}\right.\end{eqnarray}$
where ${\boldsymbol{x}}={({x}_{1},{x}_{2},\cdots ,\,{x}_{N})}^{{\rm{T}}}\in {R}^{N}$ is the spatial coordinate, J is an antisymmetric matrix arising from the Coriolis force, ρ(x, t), u(x, t) and p(x, t) denote the density, velocity and pressure of the fluid, respectively. While, $\begin{eqnarray}D({\boldsymbol{u}})=\frac{{\rm{\nabla }}{\boldsymbol{u}}+{({\rm{\nabla }}{\boldsymbol{u}})}^{{\rm{T}}}}{2},\end{eqnarray}$
is the strain tensor, h(ρ) and g(ρ) are the Lamé viscosity coefficients satisfying $\begin{eqnarray}h(\rho )\geqslant 0,\quad \quad \quad h(\rho )+Ng(\rho )\geqslant 0.\end{eqnarray}$
The Navier–Stokes equation has drawn extensive scientific interest being the cornerstone in modelling fluids, plasmas, astrophysics, oceanography and atmospheric dynamics [2–4]. In particular, the Navier–Stokes (NS) equation with Coriolis force and density-dependent viscosity has exerted a profound influence on the rotational fluid dynamics, which is fundamental to understand and analyze a wide array of phenomena, including the behavior of the Gulf stream, the dynamics of hurricanes, the operation of chemical reactors, the functionality of rotating machines and the intricacies of electrochemical reactors featuring rotating electrodes [5, 6]. Therefore, there have been extensive and intensive studies carried on the NS equation with Coriolis force and different density-dependent viscosity, which is manifested by a large number of related papers [7–15]. For examples, Olshanskii and Reusken investigated numerical solutions to the NS equation with Coriolis force by the robust multigrid method [7]. Codina studied the numerical solutions via the discretization method [8]. Lee and his co-authors derived some analytical solutions for the NS equation with Coriolis force in two dimensions [9]. Yang and his co-authors proved the global existence and uniqueness of weak solutions for the NS equation with the density-dependent viscosity chosen by h(ρ) = 0 and g(ρ) = ξ [10], where ξ is a constant. Mellet and Vasseur showed the existence of weak solutions for the NS equation with h(ρ) = μ and g(ρ) = ξ [11], where μ and ξ are constants. Yuen constructed radially symmetric solutions for the NS equation with h(ρ) = 0 and g(ρ) = κρθ [12]. An and Zhu derived elliptic vortex solutions and pulsrodon solutions for the NS equation with h(ρ) = 0 and g(ρ) = κρθ [13]. Guo and Xin obtained spherically symmetric solutions for the NS equation with h(ρ) = ρθ and g(ρ) = (θ − 1)ρθ [14]. An and Yuen derived drifting solutions for the NS equation with h(ρ) = k1ρθ and g(ρ) = k2ρθ[15].
It is known that in geophysical flows, there are many mathematical models corresponding to (1.1 ). Among them, the Euler equation descriptive of the motion of inviscid fluids is expressed exactly as (1.1 ) with h(ρ) = 0 and g(ρ) = 0 [16]. In [17], An and her co-authors have showed that the Euler equation admits the Cartesian vector solutions via the matrix theory and decomposition technique. Subsequently, Chow and his co-authors have proved that the damped Euler equation also admits the Cartesian vector solutions [18]. After that, Fan and Yuen have demonstrated that the Euler equation with Coriolis force admits the Cartesian solutions, too [19]. What needs to be pointed out is that in the Cartesian solutions, the velocity function u(x, t) takes a linear function form of spatial coordinates x. In fact, seeking this type of solution has a long history for fluid flows [20–27], especially for the Euler and NS equations. One of the important results in this direction is the work of Craik and Criminale [28], in which a comprehensive analysis of solutions to the incompressible NS equation was given. Therefore, it is natural to inquire whether the compressible NS equation with Coriolis force and density-dependent viscosity (1.1 ) admits the general Cartesian vector solutions of this form:
$\begin{eqnarray}\begin{array}{r}{\boldsymbol{u}}={\boldsymbol{A}}{\boldsymbol{x}}+{\boldsymbol{b}}(t)\end{array}\end{eqnarray}$
for an appropriate function p(x, y, t). In the above, A is an N × N time-dependent matrix and b(t) is an N-dimensional vector, which are defined by: $\begin{eqnarray}\begin{array}{rcl}{\boldsymbol{b}}(t) & = & {({b}_{1}(t),{b}_{2}(t),\cdots ,\,{b}_{N}(t))}^{{\rm{T}}},\\ {\boldsymbol{A}} & = & {({a}_{ij}(t))}_{N\times N}.\end{array}\end{eqnarray}$
The main goal of this paper is to provide a positive answer to the above question. Here, by using the matrix technique and curve integration technique, we show the existence of general Cartesian vector solutions u = Ax + b(t) for the N-dimensional compressible NS equation (1.1 ) with the density-dependent viscosity chosen by h(ρ) = k1ρθ and g(ρ) = k2ρθ. These solutions are shown to be global and explicitly expressed by appropriate formulae. In particular, we discuss two special cases on the matrix A wherein some interesting new solutions are obtained. Moreover, the structure and properties of Cartesian vector solutions are discussed in a special case. Based on the importance and applications of the NS equation, the Cartesian vector solutions obtained here may be helpful for experts in understanding how the real fluid will flow and can also provide benchmarks for testing and improving some numerical codes for computing more complicated flows.
The organization of the paper is arranged as follows: In section 2 , we show that when the viscosity coefficients are chosen by h(ρ) = k1ρθ and g(ρ) = k2ρθ with θ being a constant, the N-dimensional compressible NS equation (1.1 ) admits the Cartesian vector solutions wherein A satisfies appropriate matrix equations. To shed light on the solvable cases of the matrix equations, we consider two specials cases of A in sections 3 and 4 . In the former, we reveal that when A is an antisymmetric constant matrix and the product of A and J is commutative, the exact rotational solutions can be obtained. In the latter, we show that when A is a time-dependent matrix and J = 0, the exact irrotational solutions can be derived. Meanwhile, some illustrative examples are given for these two solvable cases, which generalize the results obtained in [12, 17, 19, 29]. In section 5 , we analyze the structure and properties of Cartesian vector solutions in a special case. Finally, a short conclusion is attached.
2. Existence of the Cartesian vector solutions
For the N-dimensional compressible NS equation (1.1 ), from the γ-law aspect, we consider the case wherein the density ρ and pressure p satisfy a relation of:
$\begin{eqnarray}p(\rho )=k{\rho }^{\gamma },\end{eqnarray}$
with k > 0 and γ = cp/cu, where cp, cu are the specific heats capacities under constant volume and constant pressure, respectively. In particular, when γ = 1, the fluid is called isothermal. It can be applied to construct models with nondegenerate isothermal cores, which have a role in connection with the so-called Schonberg-Chandrasekhar [30]. When γ = 2, the fluid corresponds to the viscous Saint-Venant system for shallow water [16] and investigated in [14]. When γ = 5/3, 7/5, the equation can describe the dynamics of the monatomic and diatomic gas [5], respectively.It is noticed from equation (1.1 ) that we can take k = 1 without loss of generality by using a simple transformation ρ → k−1/γρ, k1 → k1k−1/γ and k2 → k2k−1/γ. Let 1.1 ) can be readily rewritten into the following form:
$\begin{eqnarray}\begin{array}{rcl}\bar{p} & = & \left\{\begin{array}{ccc}{\mathrm{ln}}\,\rho ,\,\, & {\rm{for}}\,\, & \gamma =1\,{,}\\ \frac{\gamma }{\gamma -1}{\rho }^{\gamma -1},\,\, & {\rm{for}}\,\, & \gamma \ne 1\,{,}\end{array}\right.\\ \iff \,\,\rho & = & \left\{\begin{array}{ccc}\exp (\bar{p}),\,\, & {\rm{for}}\,\, & \gamma =1\,{,}\\ \mu {\bar{p}}^{\frac{1}{\gamma -1}},\,\, & {\rm{for}}\,\, & \gamma \ne 1\,{,}\end{array}\right.\end{array}\end{eqnarray}$
with $\mu ={\left(\frac{\gamma -1}{\gamma }\right)}^{\frac{1}{\gamma -1}}$, then the compressible NS equation ( $\begin{eqnarray}\begin{array}{r}\left\{\begin{array}{l}{\rho }_{t}+{\rm{div}}(\rho \,{\boldsymbol{u}})=0,\\ {{\boldsymbol{u}}}_{t}+({\boldsymbol{u}}\cdot {\rm{\nabla }}){\boldsymbol{u}}+{\boldsymbol{J}}{\boldsymbol{u}}-{k}_{1}({\rm{\nabla }}\bar{p}\cdot {\rm{\nabla }}){\boldsymbol{u}}-{k}_{2}{\rm{div}}{\boldsymbol{u}}{\rm{\nabla }}\bar{p}-\frac{{k}_{1}(\gamma -1)}{\gamma }\bar{p}\bigtriangleup {\boldsymbol{u}}\\ \quad \quad -\frac{{k}_{2}(\gamma -1)}{\gamma }\bar{p}{\rm{\nabla }}\times {\rm{div}}\,{\boldsymbol{u}}+{\rm{\nabla }}\bar{p}={\boldsymbol{0}},\quad \quad {\rm{for}}\quad \gamma \ne 1,\\ {{\boldsymbol{u}}}_{t}+({\boldsymbol{u}}\cdot {\rm{\nabla }}){\boldsymbol{u}}+{\boldsymbol{J}}{\boldsymbol{u}}-{k}_{1}({\rm{\nabla }}\bar{p}\cdot {\rm{\nabla }}){\boldsymbol{u}}-{k}_{2}{\rm{div}}\,{\boldsymbol{u}}{\rm{\nabla }}\bar{p}-{k}_{1}\bigtriangleup {\boldsymbol{u}}\\ \quad \quad -{k}_{2}{\rm{\nabla }}\times {\rm{div}}\,{\boldsymbol{u}}+{\rm{\nabla }}\bar{p}={\boldsymbol{0}},\quad \quad {\rm{for}}\quad \gamma =1.\end{array}\right.\end{array}\end{eqnarray}$
Here, for an appropriate function $\bar{p}({\boldsymbol{x}},t)$, our aim is to find a sufficient condition on the existence of the Cartesian vector solutions for the N-dimensional compressible NS equation (2.3 ), which takes the following form: 2.2 ) between $\bar{p}$ and ρ, we only need to deal with $\bar{p}$ when solving the equation (2.3 ). The main results obtained are given in the following theorem:
$\begin{eqnarray}\begin{array}{r}{\boldsymbol{u}}={\boldsymbol{b}}(t)+{\boldsymbol{A}}{\boldsymbol{x}},\end{array}\end{eqnarray}$
where A is an N × N matrix function and b(t) is an N-dimensional vector function defined by: $\begin{eqnarray}\begin{array}{r}{\boldsymbol{b}}(t)={({b}_{1}(t),{b}_{2}(t),\cdots \,{,}{b}_{N}(t))}^{{\rm{T}}},\,\,\,{\boldsymbol{A}}={({a}_{ij}(t))}_{N\times N}.\end{array}\end{eqnarray}$
In the above, the elements bi(t) and aij(t) (i, j = 1, 2, ⋯ , N) are time-dependent functions. Due to the equivalent relation (We define the matrices B and C via:
$\begin{eqnarray}\begin{array}{r}{\boldsymbol{B}}={\boldsymbol{I}}-{k}_{2}{\rm{tr}}({\boldsymbol{A}}){\boldsymbol{I}}-{k}_{1}{\boldsymbol{A}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{r}{\boldsymbol{C}}={{\boldsymbol{B}}}^{-1}({{\boldsymbol{A}}}_{t}+J{\boldsymbol{A}}+{{\boldsymbol{A}}}^{2})/2,\end{array}\end{eqnarray}$
where I denotes an identity matrix, B−1 denotes the inverse matrix of b. If A and C satisfy the following matrix differential equations $\begin{eqnarray}\begin{array}{r}{{\boldsymbol{C}}}^{{\rm{T}}}={\boldsymbol{C}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{r}{{\boldsymbol{C}}}_{t}+(\gamma -1){\rm{tr}}({\boldsymbol{A}}){\boldsymbol{C}}+{\boldsymbol{C}}{\boldsymbol{A}}+{{\boldsymbol{A}}}^{{\rm{T}}}{\boldsymbol{C}}={\boldsymbol{0}},\end{array}\end{eqnarray}$
then the compressible NS equation ( $\begin{eqnarray}\begin{array}{r}{\boldsymbol{u}}={\boldsymbol{b}}(t)+{\boldsymbol{A}}{\boldsymbol{x}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{r}\bar{p}=-{{\boldsymbol{x}}}^{{\rm{T}}}({{\boldsymbol{B}}}^{-1}{{\boldsymbol{b}}}_{t}+{{\boldsymbol{B}}}^{-1}{\boldsymbol{A}}{\boldsymbol{b}}+{{\boldsymbol{B}}}^{-1}{\boldsymbol{J}}{\boldsymbol{b}})-{{\boldsymbol{x}}}^{{\rm{T}}}{\boldsymbol{C}}{\boldsymbol{x}}+c(t).\end{array}\end{eqnarray}$
In the above, the scalar function c(t) and vector function b(t) satisfy the following ordinary differential equations: $\begin{eqnarray}\begin{array}{l}{c}_{t}+(\gamma -1){\rm{tr}}({\boldsymbol{A}})c-{{\boldsymbol{B}}}^{{\rm{T}}}({{\boldsymbol{B}}}^{-1}{{\boldsymbol{b}}}_{t}+{{\boldsymbol{B}}}^{-1}{\boldsymbol{A}}{\boldsymbol{b}}\\ \quad +\,{{\boldsymbol{B}}}^{-1}{\boldsymbol{Jb}})=0,\quad \quad {\rm{for}}\quad \gamma \ne 1,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{c}_{t}+{\rm{tr}}({\boldsymbol{A}})c-{{\boldsymbol{b}}}^{{\rm{T}}}({{\boldsymbol{B}}}^{-1}{{\boldsymbol{b}}}_{t}+{{\boldsymbol{B}}}^{-1}{\boldsymbol{A}}{\boldsymbol{b}}\\ \quad +\,{{\boldsymbol{B}}}^{-1}{\boldsymbol{Jb}})=0,\quad \quad {\rm{for}}\quad \gamma =1\,{,}\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{({{\boldsymbol{B}}}^{-1}{{\boldsymbol{b}}}_{t}+{{\boldsymbol{B}}}^{-1}{\boldsymbol{A}}{\boldsymbol{b}}+{{\boldsymbol{B}}}^{-1}{\boldsymbol{Jb}})}_{t}\\ \quad +\,[(\gamma -1){\rm{tr}}({\boldsymbol{A}}){\boldsymbol{I}}+{{\boldsymbol{A}}}^{{\rm{T}}}]\\ \quad \times \,({{\boldsymbol{B}}}^{-1}{{\boldsymbol{b}}}_{t}+{{\boldsymbol{B}}}^{-1}{\boldsymbol{A}}{\boldsymbol{b}}+{{\boldsymbol{B}}}^{-1}{\boldsymbol{Jb}})+2{\boldsymbol{C}}{\boldsymbol{b}}={\bf{0}}\,.\end{array}\end{eqnarray}$
Here, we shall first prove how to obtain the analytical solutions (
$\begin{eqnarray}\begin{array}{l}{{\boldsymbol{u}}}_{t}+({\boldsymbol{u}}\cdot {\rm{\nabla }}){\boldsymbol{u}}+{\boldsymbol{Ju}}-{k}_{1}({\rm{\nabla }}\bar{p}\cdot {\rm{\nabla }}){\boldsymbol{u}}\\ -\,{k}_{2}{\rm{div}}\,{\boldsymbol{u}}{\rm{\nabla }}\bar{p}-\frac{(\gamma -1)}{\gamma }\bar{p}({k}_{1}\bigtriangleup {\boldsymbol{u}}\\ +\,{k}_{2}{\rm{\nabla }}\times {\rm{div}}\,{\boldsymbol{u}})+{\rm{\nabla }}\bar{p}\\ =\,{{\boldsymbol{b}}}_{t}+{{\boldsymbol{A}}}_{t}{\boldsymbol{x}}+[({\boldsymbol{b}}+{\boldsymbol{A}}{\boldsymbol{x}})\cdot \unicode{x025BF}]({\boldsymbol{b}}+{\boldsymbol{A}}{\boldsymbol{x}})\\ +\,{\boldsymbol{J}}({\boldsymbol{b}}+{\boldsymbol{A}}{\boldsymbol{x}})-{k}_{1}({\rm{\nabla }}\bar{p}\cdot {\rm{\nabla }})({\boldsymbol{b}}+{\boldsymbol{A}}{\boldsymbol{x}})\\ -\,{k}_{2}{\rm{div}}({\boldsymbol{b}}+{\boldsymbol{A}}{\boldsymbol{x}})\unicode{x025BF}\bar{p}-\frac{{k}_{1}(\gamma -1)}{\gamma }\bar{p}\bigtriangleup ({\boldsymbol{b}}+{\boldsymbol{A}}{\boldsymbol{x}})\\ -\,\frac{{k}_{2}(\gamma -1)}{\gamma }\bar{p}{\rm{\nabla }}\times {\rm{div}}({\boldsymbol{b}}+{\boldsymbol{A}}{\boldsymbol{x}})+{\rm{\nabla }}\bar{p}\quad \quad \quad \\ =\,{{\boldsymbol{b}}}_{t}+{{\boldsymbol{A}}}_{t}{\boldsymbol{x}}+{\boldsymbol{A}}({\boldsymbol{b}}+{\boldsymbol{A}}{\boldsymbol{x}})+{\boldsymbol{JA}}{\boldsymbol{x}}+{\boldsymbol{J}}{\boldsymbol{b}}\\ -\,{k}_{1}{\boldsymbol{A}}{\rm{\nabla }}\bar{p}-{k}_{2}{\rm{tr}}({\boldsymbol{A}}){\rm{\nabla }}\bar{p}+{\rm{\nabla }}\bar{p}\\ =\,{{\boldsymbol{b}}}_{t}+{\boldsymbol{A}}{\boldsymbol{b}}+{\boldsymbol{Jb}}+({{\boldsymbol{A}}}_{t}+{\boldsymbol{JA}}+{{\boldsymbol{A}}}^{2}){\boldsymbol{x}}\\ +\,({\boldsymbol{I}}-{k}_{2}{\rm{tr}}({\boldsymbol{A}}){\boldsymbol{I}}-{k}_{1}{\boldsymbol{A}})\unicode{x025BF}\bar{p}\\ =\,{\bf{0}}.\end{array}\end{eqnarray}$
For the convenience of subsequent computations, we introduce the following auxiliary matrices:
$\begin{eqnarray}\begin{array}{rcl}{{\boldsymbol{B}}}^{-1} & = & {({b}_{ij})}_{N\times N},\quad {\boldsymbol{J}}={({g}_{ij})}_{N\times N},\\ {\boldsymbol{C}} & = & \frac{1}{2}{{\boldsymbol{B}}}^{-1}({{\boldsymbol{A}}}_{t}+{\boldsymbol{JA}}+{{\boldsymbol{A}}}^{2})={({c}_{ij})}_{N\times N},\end{array}\end{eqnarray}$
where the elements cij are given by: $\begin{eqnarray}\begin{array}{rcl}{c}_{ij} & = & \frac{1}{2}\left(\displaystyle \sum _{k=1}^{N}{b}_{ik}{a}_{kj,t}+\displaystyle \sum _{k=1}^{N}\displaystyle \sum _{l=1}^{N}{b}_{ik}{a}_{kl}{a}_{lj}\right.\\ & & \left.+\displaystyle \sum _{k=1}^{N}\displaystyle \sum _{l=1}^{N}{b}_{ik}{g}_{kl}{a}_{lj}\right).\end{array}\end{eqnarray}$
So that the equation ( $\begin{eqnarray}\begin{array}{l}{Q}_{i}({x}_{1},\cdots ,{x}_{N})\equiv -\displaystyle \sum _{k=1}^{N}{b}_{ik}{b}_{kt}-\displaystyle \sum _{k=1}^{N}\displaystyle \sum _{j=1}^{N}{b}_{ij}({a}_{jk}+{g}_{jk}){b}_{k}-2\displaystyle \sum _{k=1}^{N}{c}_{ik}{x}_{k}\\ \,=\,\frac{\partial \bar{{p}}}{\partial {x}_{i}},\,i=1,2,\cdots ,\,N.\end{array}\end{eqnarray}$
In order to solve $\bar{p}({\boldsymbol{x}},t)$ through ( $\begin{eqnarray}\begin{array}{rcl}\frac{\partial {Q}_{i}({x}_{1},\cdots ,\,{x}_{N})}{\partial {x}_{j}} & = & \frac{\partial {Q}_{j}({x}_{1},\cdots ,\,{x}_{N})}{\partial {x}_{i}},\\ i,j & = & 1,2,\cdots ,\,{N},\end{array}\end{eqnarray}$
which holds if and only if $\begin{eqnarray*}\begin{array}{r}{c}_{ij}={c}_{ji},\,\,i,j=1,2,\cdots ,\,N.\end{array}\end{eqnarray*}$
This condition means that ${\boldsymbol{C}}=\frac{1}{2}{{\boldsymbol{B}}}^{-1}({{\boldsymbol{A}}}_{t}+{\boldsymbol{JA}}+{{\boldsymbol{A}}}^{2})$ is a symmetric matrix, that is, the condition (From the condition (
$\begin{eqnarray}\begin{array}{rcl}{\rm{d}}\bar{p}({\boldsymbol{x}},t) & = & \displaystyle \sum _{i=1}^{N}\frac{\partial \bar{p}({\boldsymbol{x}})}{\partial {x}_{i}}{\rm{d}}{x}_{i}\\ & = & \displaystyle \sum _{i=1}^{N}{Q}_{i}({x}_{1},\cdots ,{\,x}_{N}){\rm{d}}{x}_{i}.\end{array}\end{eqnarray}$
Therefore, the second kind of curvilinear integral of $\bar{p}({\boldsymbol{x}},t)$ is independent of its integration route according to the curve integration technique (see references in [31, 32]). This allows us to select a special integration route to get the value of $\bar{p}$: $\begin{eqnarray}\begin{array}{rcl}\bar{p}({\boldsymbol{x}},t) & = & \displaystyle \sum _{i=1}^{N}{\displaystyle \int }_{(0,0,\cdots ,\,0)}^{({x}_{1},{x}_{2},\cdots ,\,{x}_{N})}{Q}_{i}({x}_{1},{x}_{2},\cdots ,\,{x}_{N}){\rm{d}}{x}_{i}\\ & = & {\displaystyle \int }_{0}^{{x}_{1}}{Q}_{1}({x}_{1},0,\cdots ,\,0){\rm{d}}{x}_{1}\\ & & +\,{\displaystyle \int }_{0}^{{x}_{2}}{Q}_{2}({x}_{1},{x}_{2},0,\cdots ,\,0){\rm{d}}{x}_{2}\\ & & +\,\cdots +{\displaystyle \int }_{0}^{{x}_{N}}{Q}_{N}({x}_{1},{x}_{2},\cdots ,{\,x}_{N}){\rm{d}}{x}_{N}\\ & = & -\displaystyle \sum _{i=1}^{N}\left[\displaystyle \sum _{k=1}^{N}{b}_{ik}{b}_{kt}+\displaystyle \sum _{k=1}^{N}\displaystyle \sum _{j=1}^{N}{b}_{ij}({a}_{jk}+{g}_{jk}){b}_{k}\right]{x}_{i}\\ & & -\,\displaystyle \sum _{i=1}^{N}{c}_{ii}{x}_{i}^{2}-2\displaystyle \sum _{i,k=1,\,i\lt k}^{N}{c}_{ik}{x}_{i}{x}_{k}+c(t)\\ & = & -{{\boldsymbol{x}}}^{{\rm{T}}}({{\boldsymbol{B}}}^{-1}{{\boldsymbol{b}}}_{t}+{{\boldsymbol{B}}}^{-1}{\boldsymbol{Ab}}+{{\boldsymbol{B}}}^{-1}{\boldsymbol{Jb}})-{{\boldsymbol{x}}}^{{\rm{T}}}{\boldsymbol{Cx}}+c(t).\end{array}\end{eqnarray}$
Next, we shall prove that the functions (
$\begin{eqnarray}\begin{array}{l}{\rho }_{t}+{\rm{div}}(\rho {\boldsymbol{u}})\\ =\,{\rho }_{t}+\rho {\rm{div}}\,{\boldsymbol{u}}+{\boldsymbol{u}}\cdot {\rm{\nabla }}\rho ={\rho }_{t}+\rho {\rm{tr}}({\boldsymbol{A}})+{\boldsymbol{u}}\cdot {\rm{\nabla }}\rho \\ =\,\frac{\mu }{\gamma -1}{\bar{p}}^{\frac{1}{\gamma -1}-1}\{{\bar{p}}_{t}+(\gamma -1)\bar{p}\,{\rm{tr}}({\boldsymbol{A}})+{\boldsymbol{u}}\cdot {\rm{\nabla }}\bar{p}\}\\ =\,-\frac{\mu }{\gamma -1}{\bar{p}}^{\frac{1}{\gamma -1}-1}\{{{\boldsymbol{x}}}^{{\rm{T}}}[{{\boldsymbol{C}}}_{t}+(\gamma -1){\rm{tr}}({\boldsymbol{A}}){\boldsymbol{C}}+2{{\boldsymbol{A}}}^{{\rm{T}}}{\boldsymbol{C}}]{\boldsymbol{x}}\\ +\,{{\boldsymbol{x}}}^{{\rm{T}}}{[{{\boldsymbol{B}}}^{-1}({{\boldsymbol{b}}}_{t}+{\boldsymbol{A}}{\boldsymbol{b}}+{\boldsymbol{Jb}})]}_{t}\\ +\,{{\boldsymbol{x}}}^{{\rm{T}}}[(\gamma -1){\rm{tr}}({\boldsymbol{A}}){\boldsymbol{I}}+{{\boldsymbol{A}}}^{{\rm{T}}}]{{\boldsymbol{B}}}^{-1}({{\boldsymbol{b}}}_{t}+{\boldsymbol{A}}{\boldsymbol{b}}+{\boldsymbol{J}}{\boldsymbol{b}})\\ +\,2{{\boldsymbol{x}}}^{{\rm{T}}}{\boldsymbol{Cb}}-[{c}_{t}+(\gamma -1){\rm{tr}}({\boldsymbol{A}})c\\ -\,{{\boldsymbol{b}}}^{{\rm{T}}}({{\boldsymbol{B}}}^{-1}{{\boldsymbol{b}}}_{t}+{{\boldsymbol{B}}}^{-1}{\boldsymbol{A}}{\boldsymbol{b}}+{{\boldsymbol{B}}}^{-1}{\boldsymbol{Jb}})]\}\\ =\,0.\end{array}\end{eqnarray}$
In the above, we have used this condition: $\begin{eqnarray*}\begin{array}{r}{{\boldsymbol{x}}}^{{\rm{T}}}[{{\boldsymbol{C}}}_{t}+(\gamma -1){\rm{tr}}({\boldsymbol{A}}){\boldsymbol{C}}+2{{\boldsymbol{A}}}^{{\rm{T}}}{\boldsymbol{C}}]{\boldsymbol{x}}=0,\end{array}\end{eqnarray*}$
which is equivalent to $\begin{eqnarray*}\begin{array}{l}{[{{\boldsymbol{C}}}_{t}+(\gamma -1){\rm{tr}}({\boldsymbol{A}}){\boldsymbol{C}}+2{{\boldsymbol{A}}}^{{\rm{T}}}{\boldsymbol{C}}]}^{{\rm{T}}}\\ \quad =-[{{\boldsymbol{C}}}_{t}+(\gamma -1){\rm{tr}}({\boldsymbol{A}}){\boldsymbol{C}}+2{{\boldsymbol{A}}}^{{\rm{T}}}{\boldsymbol{C}}],\end{array}\end{eqnarray*}$
that is, $\begin{eqnarray*}\begin{array}{r}{{\boldsymbol{C}}}_{t}+(\gamma -1){\rm{tr}}({\boldsymbol{A}}){\boldsymbol{C}}+{\boldsymbol{CA}}+{{\boldsymbol{A}}}^{{\rm{T}}}{\boldsymbol{C}}={\bf{0}}.\end{array}\end{eqnarray*}$
As for the case of γ = 1, it can be proved in a similar way, in which one only needs to replace $\mu {\bar{p}}^{\frac{1}{\gamma -1}}$ with $\exp (\bar{p})$ in the proof procedure of equation (
We would like to point out that although the condition (2.8 )–(2.9 ) looks simple, it is actually a complicated matrix differential system containing N2 scalar equations. Therefore, it is very difficult to derive its general solutions. In the following, some special techniques are devised to make its analytical solutions obtainable.
It is noticed from (
In theorem
3. The first reduction: constant matrix A
Observation on the form of the matrices B and C in (2.6 )–(2.7 ) inspires us to choose A to be an antisymmetric constant and k1 = 0, which can lead to the rotational Cartesian solutions. The result is built in the following theorem.
If the parameter k1 = 0 and A is a constant matrix satisfying:
$\begin{eqnarray}\begin{array}{r}{{\boldsymbol{A}}}^{{\rm{T}}}=-{\boldsymbol{A}},\quad \quad {\boldsymbol{AJ}}={\boldsymbol{JA}},\end{array}\end{eqnarray}$
then the compressible Navier–Stokes equation ( $\begin{eqnarray}\begin{array}{r}{\boldsymbol{u}}={\boldsymbol{b}}(t)+{\boldsymbol{A}}{\boldsymbol{x}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{r}\bar{p}=-{{\boldsymbol{x}}}^{{\rm{T}}}({{\boldsymbol{b}}}_{t}+{\boldsymbol{A}}{\boldsymbol{b}}+{\boldsymbol{Jb}})-{{\boldsymbol{x}}}^{{\rm{T}}}{\boldsymbol{C}}{\boldsymbol{x}}+c(t),\end{array}\end{eqnarray}$
where b(t) and c(t) are described by $\begin{eqnarray}\begin{array}{r}{{\boldsymbol{b}}}_{tt}+{{\boldsymbol{Jb}}}_{t}={\bf{0}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{r}{c}_{t}-{{\boldsymbol{b}}}^{{\rm{T}}}{{\boldsymbol{b}}}_{t}-{{\boldsymbol{b}}}^{{\rm{T}}}{\boldsymbol{A}}{\boldsymbol{b}}-{{\boldsymbol{b}}}^{{\rm{T}}}{\boldsymbol{J}}{\boldsymbol{b}}=0.\end{array}\end{eqnarray}$
We just need to validate that conditions (
It is seen that if A is a constant anti-symmetry matrix, then the following relations hold:
$\begin{eqnarray}\begin{array}{r}{\boldsymbol{A}}+{{\boldsymbol{A}}}^{{\rm{T}}}=0,\quad {\rm{tr}}({\boldsymbol{A}})=0,\end{array}\end{eqnarray}$
Moreover, substituting k1 = 0, AJ = JA $\begin{eqnarray}\begin{array}{r}{\boldsymbol{B}}={\boldsymbol{I}},\quad \quad {\boldsymbol{C}}=\frac{1}{2}({\boldsymbol{JA}}+{{\boldsymbol{A}}}^{2}),\end{array}\end{eqnarray}$
so that we can have $\begin{eqnarray}\begin{array}{rcl}{{\boldsymbol{C}}}^{{\rm{T}}} & = & \frac{1}{2}{({\boldsymbol{JA}}+{{\boldsymbol{A}}}^{2})}^{{\rm{T}}}=\frac{1}{2}({{\boldsymbol{A}}}^{T}{{\boldsymbol{J}}}^{{\rm{T}}}+{{\boldsymbol{A}}}^{{\rm{T}}}{{\boldsymbol{A}}}^{{\rm{T}}})\\ & = & \frac{1}{2}({\boldsymbol{AJ}}+{{\boldsymbol{A}}}^{2})=\frac{1}{2}({\boldsymbol{JA}}+{{\boldsymbol{A}}}^{2})={\boldsymbol{C}},\end{array}\end{eqnarray}$
which indicates that C is a symmetry matrix and the condition ( $\begin{eqnarray}\begin{array}{r}{{\boldsymbol{C}}}_{t}+(\gamma -1){\rm{tr}}({\boldsymbol{A}}){\boldsymbol{C}}+{\boldsymbol{CA}}+{{\boldsymbol{A}}}^{{\rm{T}}}{\boldsymbol{C}}={\bf{0}},\end{array}\end{eqnarray}$
which implies that (With the aid of (
It's remarked that by using theorem 2 , various special solutions can be obtained for the compressible NS equation when the reduced system is solvable. Here, we give two examples.
For the 2D compressible Navier–Stokes equation with k1 = 0, setting A and J to be the following form:
$\begin{eqnarray*}\begin{array}{r}{\boldsymbol{A}}={a}_{0}\left(\begin{array}{cc}0 & 1\\ -1 & 0\end{array}\right),\quad {\boldsymbol{J}}={g}_{0}\left(\begin{array}{cc}0 & -1\\ 1 & 0\end{array}\right),\end{array}\end{eqnarray*}$
where a0 and g0 are arbitrary constants. Substituting A and J into ( $\begin{eqnarray*}\begin{array}{rcl}{\boldsymbol{B}} & = & {\boldsymbol{I}},\quad {\boldsymbol{C}}=\frac{1}{2}({\boldsymbol{JA}}+{{\boldsymbol{A}}}^{2})=\frac{{a}_{0}({g}_{0}-{a}_{0})}{2}\left(\begin{array}{cc}1 & 0\\ 0 & 1\end{array}\right),\\ {\boldsymbol{b}} & = & {g}_{0}\left(\begin{array}{c}\cos ({a}_{0}t)\\ \sin ({a}_{0}t)\end{array}\right),\quad c(t)={c}_{0}.\end{array}\end{eqnarray*}$
Therefore, according to ( $\begin{eqnarray}\begin{array}{rcl}{u}_{1} & = & {a}_{0}{x}_{2}+{g}_{0}\cos ({a}_{0}t),\\ {u}_{2} & = & -{a}_{0}{x}_{1}+{g}_{0}\sin ({a}_{0}t),\\ \bar{p} & = & -{{\boldsymbol{x}}}^{{\rm{T}}}({{\boldsymbol{b}}}_{t}+{\boldsymbol{A}}{\boldsymbol{b}}+{\boldsymbol{J}}{\boldsymbol{b}})-{{\boldsymbol{x}}}^{{\rm{T}}}{\boldsymbol{C}}{\boldsymbol{x}}+c(t),\\ & = & \frac{{a}_{0}({a}_{0}-{g}_{0})}{2}({x}_{1}^{2}+{x}_{2}^{2})\\ & & +{a}_{0}{g}_{0}[\cos ({a}_{0}t){x}_{2}-\sin ({a}_{0}t){x}_{1}]+{c}_{0}.\end{array}\end{eqnarray}$
Interestingly, we find that the solution (3.10 ) takes a similar structure to what was given by An and Zhu in [13]. It is noticed that the solution obtained here is rotational unless a0 = 0 since rot u = −2a0 and meanwhile it has no additional Ermakov structure, which are two differences to the results in [13]. The structure of the solution (3.10 ) is depicted in figure 1. From this figure, we can clearly see that steady flow is rotational and the velocities of all particles of the fluid move from inwards to the origin, which is different to the behaviors of the radially symmetric solution given by Yuen et al in [12, 29].
Figure 1. The vortex structure of the solution ${\boldsymbol{u}}={({u}_{1},{u}_{2})}^{{\rm{T}}}$ given by ( |
For the 3D compressible Navier–Stokes equation with k1 = 0, let A and J be the antisymmetric matrices of this form:
$\begin{eqnarray*}\begin{array}{rcl}{\boldsymbol{A}} & = & {a}_{0}\left(\begin{array}{ccc}0 & 1 & -1\\ -1 & 0 & 1\\ 1 & -1 & 0\end{array}\right),\\ {\boldsymbol{J}} & = & {g}_{0}\left(\begin{array}{ccc}0 & -1 & 1\\ 1 & 0 & -1\\ -1 & 1 & 0\end{array}\right).\end{array}\end{eqnarray*}$
Insertion them into ( $\begin{eqnarray*}\begin{array}{r}\begin{array}{l}{\boldsymbol{B}}={\boldsymbol{I}},\quad {\boldsymbol{C}}=\frac{{a}_{0}({a}_{0}-{g}_{0})}{2}\left(\begin{array}{ccc}-2 & 1 & 1\\ 1 & -2 & 1\\ 1 & 1 & -2\end{array}\right),\\ {\boldsymbol{b}}={b}_{0}\left(\begin{array}{c}t\\ t\\ t\end{array}\right),\quad c(t)={c}_{0}+\frac{3}{2}{b}_{0}^{2}{t}^{2},\end{array}\end{array}\end{eqnarray*}$
where a0, b0 and c0 are arbitrary constants. Then from ( $\begin{eqnarray}\begin{array}{rcl}{u}_{1} & = & {a}_{0}({x}_{2}-{x}_{3})+{b}_{0}t,\\ {u}_{2} & = & {a}_{0}({x}_{3}-{x}_{1})+{b}_{0}t,\\ {u}_{3} & = & {a}_{0}({x}_{1}-{x}_{2})+{b}_{0}t,\\ \bar{p} & = & -{{\boldsymbol{x}}}^{{\rm{T}}}({{\boldsymbol{b}}}_{t}+{\boldsymbol{A}}{\boldsymbol{b}}+{\boldsymbol{J}}{\boldsymbol{b}})-{{\boldsymbol{x}}}^{{\rm{T}}}{\boldsymbol{C}}{\boldsymbol{x}}+c(t),\\ & = & ({a}_{0}^{2}-{a}_{0}{g}_{0})({x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2}-{x}_{1}{x}_{2}-{x}_{1}{x}_{3}-{x}_{2}{x}_{3})\\ & & -{b}_{0}({x}_{1}+{x}_{2}+{x}_{3})+\frac{3}{2}{b}_{0}^{2}{t}^{2}+{c}_{0}.\end{array}\end{eqnarray}$
It is of interest to notice that the solution (3.11 ) to the Navier–Stokes equation takes the same form as that given by Fan and Yuen in [19] for the Euler equation, In fact, if we require the parameter k2 to be zero and g0 = −a0, then the equation (2.3 ) becomes the Euler equation and the solution (3.11 ) is just the solution for the Euler equation in [19].
4. The second reduction: time-dependent matrix A
Observation on the form of the matrix B and C in (2.6 )–(2.7 ) motives us to decompose A into A = f(t)I when J = 0, which can lead to the explicit irrotational solutions for the Navier–Stokes equation (2.3 ). We would like to point out that different to the above section, here no restrictions are imposed. The main result is described in the following theorem.
Suppose that the matrix A can be decomposed into the form of
$\begin{eqnarray}{\boldsymbol{A}}=f(t){\boldsymbol{I}},\end{eqnarray}$
and J = 0, then C defined by ( $\begin{eqnarray}\begin{array}{l}[1-({k}_{1}+N{k}_{2})f]{f}_{tt}+2f{f}_{t}+({k}_{1}+N{k}_{2})({f}_{t}^{2}-{f}^{2}{f}_{t})\\ +\,[N(\gamma -1)+2][1-({k}_{1}+N{k}_{2})f]({f}_{t}+{f}^{2})f=0.\end{array}\end{eqnarray}$
Under this condition, we can obtain an explicit general solution of the Navier–Stokes equation, which is given by: $\begin{eqnarray}{\boldsymbol{u}}={\boldsymbol{b}}(t)+f{\boldsymbol{x}},\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}\bar{p}=-\frac{{f}_{t}+{f}^{2}}{2(1-{k}_{1}f-N{k}_{2}f)}({x}_{1}^{2}+\cdots +{x}_{N}^{2})\\ -\,\frac{{{\boldsymbol{x}}}^{{\rm{T}}}({{\boldsymbol{b}}}_{t}+f{\boldsymbol{b}})}{1-{k}_{1}f-N{k}_{2}f}+c(t),\end{array}\end{eqnarray}$
where b(t) and c(t) satisfy $\begin{eqnarray}\begin{array}{r}{c}_{t}+N(\gamma -1)fc-\frac{{{\boldsymbol{b}}}^{{\rm{T}}}({{\boldsymbol{b}}}_{t}+f{\boldsymbol{b}})}{1-{k}_{1}f-N{k}_{2}f}=0,{\rm{for}}\ \gamma \ne 1,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{r}{c}_{t}+Nfc-\frac{{{\boldsymbol{b}}}^{{\rm{T}}}({{\boldsymbol{b}}}_{t}+f{\boldsymbol{b}})}{1-{k}_{1}f-N{k}_{2}f}=0,\,{\rm{for}}\ \gamma =1,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{\left(\frac{{{\boldsymbol{b}}}_{t}+f{\boldsymbol{b}}}{1-{k}_{1}f-N{k}_{2}f}\right)}_{t}+f[N(\gamma -1)+1]\\ \times \,\frac{{{\boldsymbol{b}}}_{t}+f{\boldsymbol{b}}}{1-{k}_{1}f-N{k}_{2}f}+\frac{({f}_{t}+{f}^{2}){\boldsymbol{b}}}{1-{k}_{1}f-N{k}_{2}f}={\boldsymbol{0}},\end{array}\end{eqnarray}$
with f satisfying the equation (It is seen that when the matrix A is decomposed as
$\begin{eqnarray*}{\boldsymbol{A}}=f(t){\boldsymbol{I}},\end{eqnarray*}$
substituting it together with J = 0 into ( $\begin{eqnarray}\begin{array}{rcl}{\boldsymbol{B}} & = & (1-N{k}_{2}f-{k}_{1}f){\boldsymbol{I}},\\ {\boldsymbol{C}} & = & \frac{1}{2}{{\boldsymbol{B}}}^{-1}({{\boldsymbol{A}}}_{t}+{{\boldsymbol{A}}}^{2}+{\boldsymbol{JA}})=\frac{{f}_{t}+{f}^{2}}{2(1-{k}_{1}f-N{k}_{2}f)}{\boldsymbol{I}},\end{array}\end{eqnarray}$
which shows that C is a symmetric matrix. When inserting the expressions of A and C into (It is noticed that the intrusion of the function f makes the solution of equation (4.2 ) difficultly accessible. In the sequel, we restrict to the solvable case, in which the parameters are required to satisfy the following condition: 1.3 ). And under this condition, equation (4.2 ) is readily reducible to: 4.11 ) is discussed into two cases:
$\begin{eqnarray}{k}_{1}+N{k}_{2}=0.\end{eqnarray}$
It is of interest to find that the above condition is compatible with the condition given in ( $\begin{eqnarray}{f}_{tt}+[N(\gamma -1)+4)]\,f{f}_{t}+[N(\gamma -1)+2]\,{f}^{3}=0,\end{eqnarray}$
which can be written in the form of: $\begin{eqnarray}\begin{array}{r}{({f}_{t}+{f}^{2})}_{t}+[N(\gamma -1)+2]\,f({f}_{t}+{f}^{2})=0.\end{array}\end{eqnarray}$
The solution of (Case 1: If ft + f2 = 0, then we can obtain a solution:
$\begin{eqnarray}\begin{array}{rc} & f=\frac{1}{t+\alpha },\end{array}\end{eqnarray}$
where α is a constant of integration. It is noticed that the solution has no constraint on the numbers N and γ, which is an extension result given in [17].Case 2: If ft + f2 ≠ 0, then we reformulate (4.11 ) into this form: 1.1 ) is dependent on the gas parameter γ and dimensional parameter N. Therefore, the following two cases needs to be considered:
$\begin{eqnarray}\begin{array}{r}\frac{{({f}_{t}+{f}^{2})}_{t}}{{f}_{t}+{f}^{2}}=-[N(\gamma -1)+2]f,\end{array}\end{eqnarray}$
which shows that the solution f(t) for the NS equation ((i). When the parameters γ and N satisfy $\gamma =1-\frac{2}{N}$, then the system (4.13 ) reduces to 4.12 ). It is worth pointing out that the solution (4.15 ) is analogues to what was given by Chow et al for the damped compressible Euler equation [18]. Indeed, when setting the Lamé viscosity coefficients h(ρ) and g(ρ) in (2.1 ) to be zero, it is just the solution derived by Chow and his co-authors in [18]. Here, the intrusion of the viscosity coefficients h(ρ) and g(ρ) renders it to constitute a general solution of the NS equation.
$\begin{eqnarray}{f}_{tt}+2\,f{f}_{t}=0,\end{eqnarray}$
which admits the solution of $\begin{eqnarray}\begin{array}{r}f=\alpha \tanh (\alpha t+\beta ),\end{array}\end{eqnarray}$
where α and β are constants. This solution is bounded and its form is different from ((ii). When the parameters γ and N satisfy $\gamma \ne 1-\frac{2}{N}$, the form of (4.13 ) suggests us to introduce an auxiliary function h via: 4.13 ) is transformed into the following form: 4.17 ), we can get: 4.16 ), we can obtain the solution of f, which is given by: 4.19 ) not only takes a similar form to what was obtained by An et al for the compressible Euler equation [17] wherein no restrictions were imposed on the parameters γ and N, but also to what was given by An and Zhu for the compressible Navier–Stokes equation [13], wherein it admits the integrable Ermakov structure. Moreover, one can easily validate that when $\gamma =1-\frac{2}{N}$, the solution (4.19 ) becomes to (4.12 ) by taking c1 = 0 and c2 = 1. Thus, (4.19 ) is a more general solution of equation (4.11 ) in the case of $\gamma =1-\frac{2}{N}$.
$\begin{eqnarray}\begin{array}{r}h={{\rm{e}}}^{\displaystyle \int \,f{\rm{d}}t}.\end{array}\end{eqnarray}$
Thus, the equation ( $\begin{eqnarray*}\begin{array}{r}\frac{{\rm{d}}}{{\rm{d}}t}\,{\mathrm{ln}}({f}_{t}+{f}^{2})=\frac{{\rm{d}}}{{\rm{d}}t}\,{\mathrm{ln}}\,{h}_{tt}{h}^{-1}=\frac{{\rm{d}}}{{\rm{d}}t}\,{\mathrm{ln}}\,{h}^{-[N(\gamma -1)+2]}.\end{array}\end{eqnarray*}$
Integrating the above expression with respect to t yields $\begin{eqnarray}{h}_{tt}{h}^{N(\gamma -1)+1}={\rm{const}}.\end{eqnarray}$
By solving ( $\begin{eqnarray}h=\left\{\begin{array}{lll}{c}_{1}{t}^{2}+{c}_{2}t+{c}_{3}, & {\rm{if}} & \gamma =1-\frac{1}{N},\\ {({c}_{1}t+{c}_{2})}^{\frac{2}{N(\gamma -1)+2}}, & {\rm{if}} & \gamma \ne 1-\frac{1}{N},2-\frac{2}{N},\end{array}\right.\end{eqnarray}$
where c1, c2 and c3 are integration constants. Substituting the expression of h into ( $\begin{eqnarray}f=\left\{\begin{array}{lll}\frac{2{c}_{1}t+{c}_{2}}{{c}_{1}{t}^{2}+{c}_{2}t+{c}_{3}}, & {\rm{if}} & \gamma =1-\frac{1}{N},\\ \frac{2{c}_{1}}{[N(\gamma -1)+2]({c}_{1}t+{c}_{2})}, & {\rm{if}} & \gamma \ne 1-\frac{1}{N},1-\frac{2}{N}.\end{array}\right.\end{eqnarray}$
Interestingly, we find that the solution (To sum up, we have shown in the above that some explicit solutions of the NS equation (2.3 ) can be obtained via theorem 3 when requiring the parameters satisfying the condition (4.9 ). To shed light on this aspect, we shall give some illustrative examples.
For the 2D compressible NS equation, we take k1 + 2k2 = 0, $\gamma =\frac{1}{2}$, b(t) = 0, c(t) = 0 and
$\begin{eqnarray*}{\boldsymbol{A}}=f(t){\boldsymbol{I}}=\frac{2{c}_{1}t+{c}_{2}}{{c}_{1}{t}^{2}+{c}_{2}t+{c}_{3}}\left(\begin{array}{cc}1 & 0\\ 0 & 1\end{array}\right),\end{eqnarray*}$
which satisfy ( $\begin{eqnarray}\begin{array}{rcl}{u}_{1} & = & \frac{2{c}_{1}t+{c}_{2}}{{c}_{1}{t}^{2}+{c}_{2}t+{c}_{3}}{x}_{1},\\ {u}_{2} & = & \frac{2{c}_{1}t+{c}_{2}}{{c}_{1}{t}^{2}+{c}_{2}t+{c}_{3}}{x}_{2},\\ \bar{p} & = & -\frac{{c}_{1}}{{c}_{1}{t}^{2}+{c}_{2}t+{c}_{3}}({x}_{1}^{2}+{x}_{2}^{2}).\end{array}\end{eqnarray}$
It is obvious that the solution (4.20 ) takes the similar structure to the solution given in [17]. Here, the solution has an irrotational structure, which can be seen in figure 2. When comparing it with the rotational solution given in (3.11 ), one can find that they are quite different to each other. In order to study the dynamical behaviors, we present the numerical figure in figure 3.
Figure 2. The irrotational structure of the velocity solution ${\boldsymbol{u}}={({u}_{1},{u}_{2})}^{{\rm{T}}}$ given by ( |
Figure 3. Time evolutions of the velocity solution u1 given by ( |
For the 3D compressible NS equation, we take k1 + 3k2 = 0, $\gamma =\frac{1}{3}$, b(t) = 0, c(t) = 0 and
$\begin{eqnarray*}{\boldsymbol{A}}=f(t){\boldsymbol{I}}=\alpha \tanh (\alpha t+\beta )\left(\begin{array}{ccc}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right),\end{eqnarray*}$
which satisfy ( $\begin{eqnarray}\begin{array}{rcl}{u}_{1} & = & \alpha \tanh (\alpha t+\beta ){x}_{1},\\ {u}_{2} & = & \alpha \tanh (\alpha t+\beta ){x}_{2},\\ {u}_{3} & = & \alpha \tanh (\alpha t+\beta ){x}_{3},\\ \bar{p} & = & \frac{1}{2}({x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2})[{\alpha }^{2}{\tanh }^{2}(\alpha t+\beta )\\ & & +{{\rm{{\rm{sech}} }}}^{2}(\alpha t+\beta )].\end{array}\end{eqnarray}$
It is seen that the velocity of the fluid is bounded with respect to time, whose dynamical behavior is displayed in figure 4. Such a kind of bounded solution was previously obtained for the damped Euler equation in [18]. However, for the Navier–Stokes equation, it seems to be the first time to obtain this type of solution.
Figure 4. The bounded structure of the velocity solution u1 given by ( |
5. The property of solutions to the NS equation
In this section, we shall investigate the structure and properties of solutions to the NS equation. For simplicity, we shall only consider the special case wherein k1 = 0, b(t) = 0 and c(t) = 0. Thus, under this case the solutions of the NS equations (2.10 ) and (2.11 ) can be written into the following form:
$\begin{eqnarray}\begin{array}{r}{\boldsymbol{u}}({\boldsymbol{x}},t)={\boldsymbol{A}}{\boldsymbol{x}},\,\,\,\bar{p}({\boldsymbol{x}},t)=-{{\boldsymbol{x}}}^{{\rm{T}}}{\boldsymbol{C}}{\boldsymbol{x}},\end{array}\end{eqnarray}$
where the matrix C is given by $\begin{eqnarray}\begin{array}{r}{\boldsymbol{C}}=\frac{1}{2(1-{k}_{2}{\rm{tr}}({\boldsymbol{A}}))}({{\boldsymbol{A}}}^{2}+{\boldsymbol{JA}}).\end{array}\end{eqnarray}$
Then, the solution u(x, t) can be viewed as a linear transformation on x ∈ RN via: $\begin{eqnarray*}{\boldsymbol{A}}:{\boldsymbol{x}}\longrightarrow A{\boldsymbol{x}}={\boldsymbol{u}},\end{eqnarray*}$
while $\bar{p}({\boldsymbol{x}},t)$ is a quadratic form associated with the matrix C via: $\begin{eqnarray*}{\boldsymbol{C}}:{\boldsymbol{x}}\longrightarrow -{{\boldsymbol{x}}}^{{\rm{T}}}{\boldsymbol{C}}{\boldsymbol{x}}=\bar{p}({\boldsymbol{x}},t).\end{eqnarray*}$
Here, we would like to show that there exists an almost linear superposition principle for the solutions (5.1 ) of the NS equation, which is established in the following theorem:
Assuming that A and C satisfy conditions of theorem
$\begin{eqnarray*}\begin{array}{rcl}{\boldsymbol{u}}({\boldsymbol{x}}+{\boldsymbol{y}},t) & = & {\boldsymbol{u}}({\boldsymbol{x}},t)+{\boldsymbol{u}}({\boldsymbol{y}},t),\\ \bar{p}({\boldsymbol{x}}+{\boldsymbol{y}},t) & = & \bar{p}({\boldsymbol{x}},t)+\bar{p}({\boldsymbol{y}},t)-2{{\boldsymbol{x}}}^{{\rm{T}}}{\boldsymbol{C}}{\boldsymbol{y}},\end{array}\end{eqnarray*}$
is a solution of the NS equation (In theorem
$\begin{eqnarray}\rho =\left\{\begin{array}{l}\exp (| | {\boldsymbol{A}}{\boldsymbol{x}}| {| }^{2}),\quad {\rm{for}}\,\gamma =1,\\ \mu | | {\boldsymbol{A}}{\boldsymbol{x}}| {| }^{\frac{2}{\gamma -1}},\quad \,\,{\rm{for}}\,\gamma \ne 1,\end{array}\right.\end{eqnarray}$
is well-defined and positive.If A = J is an antisymmetric matrix, then the following relations hold:
$\begin{eqnarray*}{{\boldsymbol{A}}}^{{\rm{T}}}=-{\boldsymbol{A}},\quad \quad {\rm{tr}}({\boldsymbol{A}})=0.\end{eqnarray*}$
Thus, the matrix C in ( $\begin{eqnarray*}{\boldsymbol{C}}=\frac{1}{2(1-{k}_{2}{\rm{tr}}({\boldsymbol{A}}))}({{\boldsymbol{A}}}^{2}+{\boldsymbol{JA}})={{\boldsymbol{A}}}^{2}.\end{eqnarray*}$
For any arbitrary x ≠ 0, we can have: $\begin{eqnarray}\begin{array}{rcl}\bar{p}({\boldsymbol{x}},t) & = & -{{\boldsymbol{x}}}^{{\rm{T}}}{\boldsymbol{C}}{\boldsymbol{x}}=-{{\boldsymbol{x}}}^{{\rm{T}}}{{\boldsymbol{A}}}^{2}{\boldsymbol{x}}={{\boldsymbol{x}}}^{{\rm{T}}}{{\boldsymbol{A}}}^{{\rm{T}}}{\boldsymbol{Ax}}\\ & = & | | {\boldsymbol{Ax}}| {| }^{2}\gt 0,\end{array}\end{eqnarray}$
which reveals that −C is positive definite. Therefore, by using the relation of (For the 2D compressible NS equation, we take b(t) = 0, c(t) = 0, k1 = 0 and A = J is an antisymmetric constant matrix
$\begin{eqnarray*}{\boldsymbol{A}}=\left(\begin{array}{cc}0 & {a}_{12}\\ -{a}_{12} & 0\end{array}\right).\end{eqnarray*}$
Then according to theorem $\begin{eqnarray*}\begin{array}{rcl}\rho & = & \exp [{a}_{12}^{2}({x}_{1}^{2}+{x}_{2}^{2})],\,\,{\rm{for}}\quad \gamma =1;\\ \rho & = & \mu {[{a}_{12}^{2}({x}_{1}^{2}+{x}_{2}^{2})]}^{\frac{1}{\gamma -1}},\,\,{\rm{for}}\quad \gamma \gt 1,\end{array}\end{eqnarray*}$
which is an elliptic cylinder.6. Conclusion
The Navier–Stokes equation is an important physical model, which has been widely used in modelling fluids, plasmas, astrophysics, oceanography and atmospheric dynamics. In this paper, based on the matrix and curve integration techniques, we have built a sufficient condition for the existence of Cartesian vector solutions for the N-dimensional NS equation with Coriolis force and density-dependent viscosity. Among these solutions, some constitute the generalization of the results previously obtained by other authors and the bounded solution obtained in section 4 seems to be new. Based on the importance and applications of the NS equation, the results obtained here may provide a new perspective for expert to understand how the real fluid will flow. And they also provide benchmarks for testing and improving some numerical codes for computing more complicated flows.
However, there still remain many challenging problems that deserve deep considerations:
1). It is seen that the key to find solutions for the NS equation (2.3 ) is to solve the matrix differential equation (2.9 ), which contains N2 scalar equations. Due to its complexity, here we only handle two special solvable cases. Therefore, it is worth considering whether some other special cases that can lead to solutions for the NS equation exist.
2). It is noticed that the velocity function in the Cartesian vector solutions obtained here takes a linear function form of the spatial coordinates. Therefore, it is natural to inquire whether the solutions expressed by quadratic, cubic, or other nonlinear functions of the Cartesian coordinates exist in the NS equation (2.3 ). If they do exist, how to find the corresponding pressure function seems to be an intriguing research.
3). We know that in geophysical flows, there are many physical models which are closely linked to the NS equation, such as the Euler-Poisson and Navier–Stokes-Poisson equations. Therefore, it would be of interest to study whether the method can be applied to derive Cartesian vector solutions of these models.