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Navier–Stokes equations

From Wikipedia, the free encyclopedia
Equations describing the motion of viscous fluid substances
Part of a series on
Continuum mechanics
J=−Ddφdx{\displaystyle J=-D{\frac {d\varphi }{dx}}}

TheNavier–Stokes equations (/nævˈjeɪˈstoʊks/nav-YAYSTOHKS) arepartial differential equations which describe the motion ofviscous fluid substances. They were named after French engineer and physicistClaude-Louis Navier and the Irish physicist and mathematicianGeorge Gabriel Stokes. They were developed over several decades of progressively building the theories, from 1822 (Navier) to 1842–1850 (Stokes).

The Navier–Stokes equations mathematically expressmomentum balance forNewtonian fluids and make use ofconservation of mass. They are sometimes accompanied by anequation of state relatingpressure,temperature anddensity.[1] They arise from applyingIsaac Newton's second law tofluid motion, together with the assumption that thestress in the fluid is the sum of adiffusingviscous term (proportional to thegradient of velocity) and apressure term—hence describingviscous flow. The difference between them and the closely relatedEuler equations is that Navier–Stokes equations takeviscosity into account while the Euler equations model onlyinviscid flow. As a result, the Navier–Stokes are anelliptic equation and therefore have better analytic properties, at the expense of having less mathematical structure (e.g. they are nevercompletely integrable).

The Navier–Stokes equations are useful because they describe the physics of many phenomena ofscientific andengineering interest. They may be used tomodel the weather,ocean currents, waterflow in a pipe and air flow around awing. The Navier–Stokes equations, in their full and simplified forms, help with the design ofaircraft and cars, the study ofblood flow, the design ofpower stations, the analysis ofpollution, and many other problems. Coupled withMaxwell's equations, they can be used to model and studymagnetohydrodynamics.

The Navier–Stokes equations are also of great interest in a purely mathematical sense. Despite their wide range of practical uses, it has not yet been proven whether smooth solutions alwaysexist in three dimensions—i.e., whether they are infinitely differentiable (or even just bounded) at all points in thedomain. This is called theNavier–Stokes existence and smoothness problem. TheClay Mathematics Institute has called this one of theseven most important open problems in mathematics and has offered aUS$1 million prize for a solution or a counterexample.[2][3]

Flow velocity

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The solution of the equations is aflow velocity. It is avector field—to every point in a fluid, at any moment in a time interval, it gives a vector whose direction and magnitude are those of the velocity of the fluid at that point in space and at that moment in time. It is studied in three spatial dimensions and one time dimension, and higher-dimensional analogues are studied in both pure and applied mathematics. Once the velocity field is calculated, other quantities of interest such aspressure ortemperature may be found using dynamical equations and relations. This is different from what one normally sees inclassical mechanics, where solutions are typically trajectories of position of aparticle or deflection of acontinuum. Studying velocity instead of position makes more sense for a fluid, although for visualization purposes one can compute varioustrajectories. In particular, thestreamlines of a vector field, interpreted as flow velocity, are the paths along which a massless fluid particle would travel. These paths are theintegral curves whose derivative at each point is equal to the vector field, and they can represent visually the behavior of the vector field at a point in time.

General continuum equations

[edit]
Main article:Derivation of the Navier–Stokes equations
See also:Cauchy momentum equation § Conservation form

The Navier–Stokes momentum equation can be derived as a particular form of theCauchy momentum equation, whose general convective form is:DuDt=1ρ∇⋅σ+f.{\displaystyle {\frac {\mathrm {D} \mathbf {u} }{\mathrm {D} t}}={\frac {1}{\rho }}\nabla \cdot {\boldsymbol {\sigma }}+\mathbf {f} .}By setting theCauchy stress tensorσ{\textstyle {\boldsymbol {\sigma }}} to be the sum of a viscosity termτ{\textstyle {\boldsymbol {\tau }}} (thedeviatoric stress) and a pressure term−pI{\textstyle -p\mathbf {I} } (volumetric stress), we arrive at:

Cauchy momentum equation  (convective form)

ρDuDt=−∇p+∇⋅τ+ρa{\displaystyle \rho {\frac {\mathrm {D} \mathbf {u} }{\mathrm {D} t}}=-\nabla p+\nabla \cdot {\boldsymbol {\tau }}+\rho \,\mathbf {a} }

where

In this form, it is apparent that in the assumption of an inviscid fluid – no deviatoric stress – Cauchy equations reduce to theEuler equations.

Assumingconservation of mass, with the known properties ofdivergence andgradient we can use the masscontinuity equation, which represents the mass per unit volume of ahomogenous fluid with respect to space and time (i.e.,material derivativeDDt{\displaystyle {\frac {\mathbf {D} }{\mathbf {Dt} }}}) of any finite volume (V) to represent the change of velocity in fluid media:DmDt=∭V(DρDt+ρ(∇⋅u))dVDρDt+ρ(∇⋅u)=∂ρ∂t+(∇ρ)⋅u+ρ(∇⋅u)=∂ρ∂t+∇⋅(ρu)=0{\displaystyle {\begin{aligned}&{\frac {\mathbf {D} m}{\mathbf {Dt} }}=\iiint \limits _{V}\left({\frac {\mathbf {D} \rho }{\mathbf {Dt} }}+\rho (\nabla \cdot \mathbf {u} )\right)\,dV\\[5pt]&{\frac {\mathbf {D} \rho }{\mathbf {Dt} }}+\rho (\nabla \cdot \mathbf {u} )={\frac {\partial \rho }{\partial t}}+(\nabla \rho )\cdot \mathbf {u} +\rho (\nabla \cdot \mathbf {u} )={\frac {\partial \rho }{\partial t}}+\nabla \cdot (\rho \mathbf {u} )=0\end{aligned}}}where

Note 1 – Refer to the mathematical operator del represented by the nabla (∇{\displaystyle \nabla })symbol.

to arrive at the conservation form of the equations of motion. This is often written:[4]

Cauchy momentum equation(conservation form)

∂∂t(ρu)+∇⋅(ρu⊗u)=−∇p+∇⋅τ+ρa{\displaystyle {\frac {\partial }{\partial t}}(\rho \,\mathbf {u} )+\nabla \cdot (\rho \,\mathbf {u} \otimes \mathbf {u} )=-\nabla p+\nabla \cdot {\boldsymbol {\tau }}+\rho \,\mathbf {a} }

where⊗{\textstyle \otimes } is theouter product of the flow velocity (u{\displaystyle \mathbf {u} }):u⊗u=uuT{\displaystyle \mathbf {u} \otimes \mathbf {u} =\mathbf {u} \mathbf {u} ^{\mathrm {T} }}

The left side of the equation describes acceleration, and may be composed of time-dependent and convective components (also the effects of non-inertial coordinates if present). The right side of the equation is in effect a summation of hydrostatic effects, the divergence of deviatoric stress and body forces (such as gravity).

All non-relativistic balance equations, such as the Navier–Stokes equations, can be derived by beginning with the Cauchy equations and specifying the stress tensor through aconstitutive relation. By expressing the deviatoric (shear) stress tensor in terms ofviscosity and the fluidvelocity gradient, and assuming constant viscosity, the above Cauchy equations will lead to the Navier–Stokes equations below.

Convective acceleration

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See also:Cauchy momentum equation § Convective acceleration
An example of convection. Though the flow may be steady (time-independent), the fluid decelerates as it moves down the diverging duct (assuming incompressible or subsonic compressible flow), hence there is an acceleration happening over position.

A significant feature of the Cauchy equation and consequently all other continuum equations (including Euler and Navier–Stokes) is the presence of convective acceleration: the effect of acceleration of a flow with respect to space. While individual fluid particles indeed experience time-dependent acceleration, the convective acceleration of the flow field is a spatial effect, one example being fluid speeding up in a nozzle.

Compressible flow

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Remark: here, the deviatoric stress tensor is denotedτ{\textstyle {\boldsymbol {\tau }}} as it was in thegeneral continuum equations and in theincompressible flow section.

The compressible momentum Navier–Stokes equation results from the following assumptions on the Cauchy stress tensor:[5]

Since thetrace of the rate-of-strain tensor in three dimensions is thedivergence (i.e. rate of expansion) of the flow:tr⁡(ε)=∇⋅u.{\displaystyle \operatorname {tr} ({\boldsymbol {\varepsilon }})=\nabla \cdot \mathbf {u} .}

Given this relation, and since the trace of the identity tensor in three dimensions is three:tr⁡(I)=3.{\displaystyle \operatorname {tr} ({\boldsymbol {I}})=3.}

the trace of the stress tensor in three dimensions becomes:tr⁡(σ)=−3p+(3λ+2μ)∇⋅u.{\displaystyle \operatorname {tr} ({\boldsymbol {\sigma }})=-3p+(3\lambda +2\mu )\nabla \cdot \mathbf {u} .}

So by alternatively decomposing the stress tensor intoisotropic anddeviatoric parts, as usual in fluid dynamics:[6]σ=−[p−(λ+23μ)(∇⋅u)]I+μ(∇u+(∇u)T−23(∇⋅u)I){\displaystyle {\boldsymbol {\sigma }}=-\left[p-\left(\lambda +{\tfrac {2}{3}}\mu \right)\left(\nabla \cdot \mathbf {u} \right)\right]\mathbf {I} +\mu \left(\nabla \mathbf {u} +\left(\nabla \mathbf {u} \right)^{\mathrm {T} }-{\tfrac {2}{3}}\left(\nabla \cdot \mathbf {u} \right)\mathbf {I} \right)}

Introducing thebulk viscosityζ{\textstyle \zeta },ζ≡λ+23μ,{\displaystyle \zeta \equiv \lambda +{\tfrac {2}{3}}\mu ,}

we arrive to the linearconstitutive equation in the form usually employed inthermal hydraulics:[5]

Linear stress constitutive equation(expression used for fluids)

σ=−[p−ζ(∇⋅u)]I+μ[∇u+(∇u)T−23(∇⋅u)I]{\displaystyle {\boldsymbol {\sigma }}=-[p-\zeta (\nabla \cdot \mathbf {u} )]\mathbf {I} +\mu \left[\nabla \mathbf {u} +(\nabla \mathbf {u} )^{\mathrm {T} }-{\tfrac {2}{3}}(\nabla \cdot \mathbf {u} )\mathbf {I} \right]}

which can also be arranged in the other usual form:[7]σ=−pI+μ(∇u+(∇u)T)+(ζ−23μ)(∇⋅u)I.{\displaystyle {\boldsymbol {\sigma }}=-p\mathbf {I} +\mu \left(\nabla \mathbf {u} +(\nabla \mathbf {u} )^{\mathrm {T} }\right)+\left(\zeta -{\frac {2}{3}}\mu \right)(\nabla \cdot \mathbf {u} )\mathbf {I} .}

Note that in the compressible case the pressure is no more proportional to theisotropic stress term, since there is the additional bulk viscosity term:p=−13tr⁡(σ)+ζ(∇⋅u){\displaystyle p=-{\frac {1}{3}}\operatorname {tr} ({\boldsymbol {\sigma }})+\zeta (\nabla \cdot \mathbf {u} )}

and thedeviatoric stress tensorσ′{\displaystyle {\boldsymbol {\sigma }}'} is still coincident with the shear stress tensorτ{\displaystyle {\boldsymbol {\tau }}} (i.e. the deviatoric stress in a Newtonian fluid has no normal stress components), and it has a compressibility term in addition to the incompressible case, which is proportional to the shear viscosity:

σ′=τ=μ[∇u+(∇u)T−23(∇⋅u)I]{\displaystyle {\boldsymbol {\sigma }}'={\boldsymbol {\tau }}=\mu \left[\nabla \mathbf {u} +(\nabla \mathbf {u} )^{\mathrm {T} }-{\tfrac {2}{3}}(\nabla \cdot \mathbf {u} )\mathbf {I} \right]}

Both bulk viscosityζ{\textstyle \zeta } and dynamic viscosityμ{\textstyle \mu } need not be constant – in general, they depend on two thermodynamics variables if the fluid contains a single chemical species, say for example, pressure and temperature. Any equation that makes explicit one of thesetransport coefficient in theconservation variables is called anequation of state.[8]

The most general of the Navier–Stokes equations become

Navier–Stokes momentum equation (convective form)

ρDuDt=ρ(∂u∂t+(u⋅∇)u)=−∇p+∇⋅{μ[∇u+(∇u)T−23(∇⋅u)I]}+∇[ζ(∇⋅u)]+ρa.{\displaystyle \rho {\frac {\mathrm {D} \mathbf {u} }{\mathrm {D} t}}=\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+(\mathbf {u} \cdot \nabla )\mathbf {u} \right)=-\nabla p+\nabla \cdot \left\{\mu \left[\nabla \mathbf {u} +(\nabla \mathbf {u} )^{\mathrm {T} }-{\tfrac {2}{3}}(\nabla \cdot \mathbf {u} )\mathbf {I} \right]\right\}+\nabla [\zeta (\nabla \cdot \mathbf {u} )]+\rho \mathbf {a} .}

in index notation, the equation can be written as[9]

Navier–Stokes momentum equation (index notation)

ρ(∂ui∂t+uk∂ui∂xk)=−∂p∂xi+∂∂xk[μ(∂ui∂xk+∂uk∂xi−23δik∂ul∂xl)]+∂∂xi(ζ∂uℓ∂xℓ)+ρai.{\displaystyle \rho \left({\frac {\partial u_{i}}{\partial t}}+u_{k}{\frac {\partial u_{i}}{\partial x_{k}}}\right)=-{\frac {\partial p}{\partial x_{i}}}+{\frac {\partial }{\partial x_{k}}}\left[\mu \left({\frac {\partial u_{i}}{\partial x_{k}}}+{\frac {\partial u_{k}}{\partial x_{i}}}-{\frac {2}{3}}\delta _{ik}{\frac {\partial u_{l}}{\partial x_{l}}}\right)\right]+{\frac {\partial }{\partial x_{i}}}\left(\zeta {\frac {\partial u_{\ell }}{\partial x_{\ell }}}\right)+\rho a_{i}.}

The corresponding equation in conservation form can be obtained by considering that, given the masscontinuity equation, the left side is equivalent to:

ρDuDt=∂∂t(ρu)+∇⋅(ρu⊗u){\displaystyle \rho {\frac {\mathrm {D} \mathbf {u} }{\mathrm {D} t}}={\frac {\partial }{\partial t}}(\rho \mathbf {u} )+\nabla \cdot (\rho \mathbf {u} \otimes \mathbf {u} )}

to give finally:

Navier–Stokes momentum equation (conservative form)

∂∂t(ρu)+∇⋅(ρu⊗u+[p−ζ(∇⋅u)]I−μ[∇u+(∇u)T−23(∇⋅u)I])=ρa.{\displaystyle {\frac {\partial }{\partial t}}(\rho \mathbf {u} )+\nabla \cdot \left(\rho \mathbf {u} \otimes \mathbf {u} +[p-\zeta (\nabla \cdot \mathbf {u} )]\mathbf {I} -\mu \left[\nabla \mathbf {u} +(\nabla \mathbf {u} )^{\mathrm {T} }-{\tfrac {2}{3}}(\nabla \cdot \mathbf {u} )\mathbf {I} \right]\right)=\rho \mathbf {a} .}

Apart from its dependence of pressure and temperature, the second viscosity coefficient also depends on the process, that is to say, the second viscosity coefficient is not just a material property. Example: in the case of a sound wave with a definitive frequency that alternatively compresses and expands a fluid element, the second viscosity coefficient depends on the frequency of the wave. This dependence is called thedispersion. In some cases, thesecond viscosityζ{\textstyle \zeta } can be assumed to be constant in which case, the effect of the volume viscosityζ{\textstyle \zeta } is that the mechanical pressure is not equivalent to the thermodynamicpressure:[10] as demonstrated below.∇⋅(∇⋅u)I=∇(∇⋅u),{\displaystyle \nabla \cdot (\nabla \cdot \mathbf {u} )\mathbf {I} =\nabla (\nabla \cdot \mathbf {u} ),}p¯≡p−ζ∇⋅u,{\displaystyle {\bar {p}}\equiv p-\zeta \,\nabla \cdot \mathbf {u} ,}However, this difference is usually neglected most of the time (that is whenever we are not dealing with processes such as sound absorption and attenuation of shock waves,[11] where second viscosity coefficient becomes important) by explicitly assumingζ=0{\textstyle \zeta =0}. The assumption of settingζ=0{\textstyle \zeta =0} is called as theStokes hypothesis.[12] The validity of Stokes hypothesis can be demonstrated for monoatomic gas both experimentally and from the kinetic theory;[13] for other gases and liquids, Stokes hypothesis is generally incorrect. With the Stokes hypothesis, the Navier–Stokes equations become

Navier–Stokes momentum equation (convective form, Stokes hypothesis)

ρDuDt=ρ(∂u∂t+(u⋅∇)u)=−∇p+∇⋅{μ[∇u+(∇u)T−23(∇⋅u)I]}+ρa.{\displaystyle \rho {\frac {\mathrm {D} \mathbf {u} }{\mathrm {D} t}}=\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+(\mathbf {u} \cdot \nabla )\mathbf {u} \right)=-\nabla p+\nabla \cdot \left\{\mu \left[\nabla \mathbf {u} +(\nabla \mathbf {u} )^{\mathrm {T} }-{\tfrac {2}{3}}(\nabla \cdot \mathbf {u} )\mathbf {I} \right]\right\}+\rho \mathbf {a} .}

If the dynamicμ and bulkζ{\displaystyle \zeta } viscosities are assumed to be uniform in space, the equations in convective form can be simplified further. By computing the divergence of the stress tensor, since the divergence of tensor∇u{\textstyle \nabla \mathbf {u} } is∇2u{\textstyle \nabla ^{2}\mathbf {u} } and the divergence of tensor(∇u)T{\textstyle \left(\nabla \mathbf {u} \right)^{\mathrm {T} }} is∇(∇⋅u){\textstyle \nabla \left(\nabla \cdot \mathbf {u} \right)}, one finally arrives to the compressible Navier–Stokes momentum equation:[14]

Navier–Stokes momentum equation with uniform shear and bulk viscosities (convective form)

DuDt=−1ρ∇p+ν∇2u+(13ν+ξ)∇(∇⋅u)+a.{\displaystyle {\frac {D\mathbf {u} }{Dt}}=-{\frac {1}{\rho }}\nabla p+\nu \,\nabla ^{2}\mathbf {u} +({\tfrac {1}{3}}\nu +\xi )\,\nabla (\nabla \cdot \mathbf {u} )+\mathbf {a} .}

whereDDt{\textstyle {\frac {\mathrm {D} }{\mathrm {D} t}}} is thematerial derivative.ν=μρ{\displaystyle \nu ={\frac {\mu }{\rho }}} is the shearkinematic viscosity andξ=ζρ{\displaystyle \xi ={\frac {\zeta }{\rho }}} is the bulk kinematic viscosity. The left-hand side changes in the conservation form of the Navier–Stokes momentum equation.By bringing the operator on the flow velocity on the left side, one also has:

Navier–Stokes momentum equation with uniform shear and bulk viscosities (convective form)

(∂∂t+u⋅∇−ν∇2−(13ν+ξ)∇(∇⋅))u=−1ρ∇p+a.{\displaystyle \left({\frac {\partial }{\partial t}}+\mathbf {u} \cdot \nabla -\nu \,\nabla ^{2}-({\tfrac {1}{3}}\nu +\xi )\,\nabla (\nabla \cdot )\right)\mathbf {u} =-{\frac {1}{\rho }}\nabla p+\mathbf {a} .}

The convective acceleration term can also be written asu⋅∇u=(∇×u)×u+12∇u2,{\displaystyle \mathbf {u} \cdot \nabla \mathbf {u} =(\nabla \times \mathbf {u} )\times \mathbf {u} +{\tfrac {1}{2}}\nabla \mathbf {u} ^{2},}where the vector(∇×u)×u{\textstyle (\nabla \times \mathbf {u} )\times \mathbf {u} } is known as theLamb vector.

For the special case of anincompressible flow, the pressure constrains the flow so that the volume offluid elements is constant:isochoric flow resulting in asolenoidal velocity field with∇⋅u=0{\textstyle \nabla \cdot \mathbf {u} =0}.[15]

Incompressible flow

[edit]

The incompressible momentum Navier–Stokes equation results from the following assumptions on the Cauchy stress tensor:[5]

This is constitutive equation is also called theNewtonian law of viscosity.Dynamic viscosityμ need not be constant – in incompressible flows it can depend on density and on pressure. Any equation that makes explicit one of thesetransport coefficient in theconservative variables is called anequation of state.[8]

The divergence of the deviatoric stress in case of uniform viscosity is given by:∇⋅τ=2μ∇⋅ε=μ∇⋅(∇u+∇uT)=μ∇2u{\displaystyle \nabla \cdot {\boldsymbol {\tau }}=2\mu \nabla \cdot {\boldsymbol {\varepsilon }}=\mu \nabla \cdot \left(\nabla \mathbf {u} +\nabla \mathbf {u} ^{\mathrm {T} }\right)=\mu \,\nabla ^{2}\mathbf {u} }because∇⋅u=0{\textstyle \nabla \cdot \mathbf {u} =0} for an incompressible fluid.

Incompressibility rules out density and pressure waves like sound orshock waves, so this simplification is not useful if these phenomena are of interest. The incompressible flow assumption typically holds well with all fluids at lowMach numbers (say up to about Mach 0.3), such as for modelling air winds at normal temperatures.[16] the incompressible Navier–Stokes equations are best visualized by dividing for the density:[17]

Incompressible Navier–Stokes equations with uniform viscosity (convective form)

DuDt=∂u∂t+(u⋅∇)u=ν∇2u−1ρ∇p+1ρf{\displaystyle {\frac {D\mathbf {u} }{Dt}}={\frac {\partial \mathbf {u} }{\partial t}}+(\mathbf {u} \cdot \nabla )\mathbf {u} =\nu \,\nabla ^{2}\mathbf {u} -{\frac {1}{\rho }}\nabla p+{\frac {1}{\rho }}\mathbf {f} }

whereν=μρ{\textstyle \nu ={\frac {\mu }{\rho }}} is called thekinematic viscosity. By isolating the fluid velocity, one can also state:

Incompressible Navier–Stokes equations with constant viscosity (alternative convective form)

(∂∂t+u⋅∇−ν∇2)u=−1ρ∇p+1ρf.{\displaystyle \left({\frac {\partial }{\partial t}}+\mathbf {u} \cdot \nabla -\nu \,\nabla ^{2}\right)\mathbf {u} =-{\frac {1}{\rho }}\nabla p+{\frac {1}{\rho }}\mathbf {f} .}

If the density is constant throughout the fluid domain, or, in other words, if all fluid elements have the same density,ρ{\textstyle \rho }, then we have

Incompressible Navier–Stokes equations with constant density and viscosity (convective form)

DuDt=ν∇2u−∇pρ+1ρf,{\displaystyle {\frac {D\mathbf {u} }{Dt}}=\nu \,\nabla ^{2}\mathbf {u} -\nabla {\frac {p}{\rho }}+{\frac {1}{\rho }}\mathbf {f} ,}

wherep/ρ{\textstyle p/\rho } is called the unitpressure head.

In incompressible flows, the pressure field satisfies thePoisson equation,[9]

∇2p=−ρ∂ui∂xk∂uk∂xi=−ρ∂2uiuk∂xkxi,{\displaystyle \nabla ^{2}p=-\rho {\frac {\partial u_{i}}{\partial x_{k}}}{\frac {\partial u_{k}}{\partial x_{i}}}=-\rho {\frac {\partial ^{2}u_{i}u_{k}}{\partial x_{k}x_{i}}},}

which is obtained by taking the divergence of the momentum equations.

A laminar flow example

Velocity profile (laminar flow):ux=u(y),uy=0,uz=0{\displaystyle u_{x}=u(y),\quad u_{y}=0,\quad u_{z}=0}for thex-direction, simplify the Navier–Stokes equation:0=−dPdx+μ(d2udy2){\displaystyle 0=-{\frac {\mathrm {d} P}{\mathrm {d} x}}+\mu \left({\frac {\mathrm {d} ^{2}u}{\mathrm {d} y^{2}}}\right)}

Integrate twice to find the velocity profile with boundary conditionsy =h,u = 0,y = −h,u = 0:u=12μdPdxy2+Ay+B{\displaystyle u={\frac {1}{2\mu }}{\frac {\mathrm {d} P}{\mathrm {d} x}}y^{2}+Ay+B}

From this equation, substitute in the two boundary conditions to get two equations:0=12μdPdxh2+Ah+B0=12μdPdxh2−Ah+B{\displaystyle {\begin{aligned}0&={\frac {1}{2\mu }}{\frac {\mathrm {d} P}{\mathrm {d} x}}h^{2}+Ah+B\\0&={\frac {1}{2\mu }}{\frac {\mathrm {d} P}{\mathrm {d} x}}h^{2}-Ah+B\end{aligned}}}

Add and solve forB:B=−12μdPdxh2{\displaystyle B=-{\frac {1}{2\mu }}{\frac {\mathrm {d} P}{\mathrm {d} x}}h^{2}}

Substitute and solve forA:A=0{\displaystyle A=0}

Finally this gives the velocity profile:u=12μdPdx(y2−h2){\displaystyle u={\frac {1}{2\mu }}{\frac {\mathrm {d} P}{\mathrm {d} x}}\left(y^{2}-h^{2}\right)}

It is well worth observing the meaning of each term (compare to theCauchy momentum equation):

∂u∂t⏟Variation+(u⋅∇)u⏟Convectiveacceleration⏞Inertia (per volume)=∂∂−∇w⏟Internalsource+ν∇2u⏟Diffusion⏞Divergence of stress+g⏟Externalsource.{\displaystyle \overbrace {{\vphantom {\frac {}{}}}\underbrace {\frac {\partial \mathbf {u} }{\partial t}} _{\text{Variation}}+\underbrace {{\vphantom {\frac {}{}}}(\mathbf {u} \cdot \nabla )\mathbf {u} } _{\begin{smallmatrix}{\text{Convective}}\\{\text{acceleration}}\end{smallmatrix}}} ^{\text{Inertia (per volume)}}=\overbrace {{\vphantom {\frac {\partial }{\partial }}}\underbrace {{\vphantom {\frac {}{}}}-\nabla w} _{\begin{smallmatrix}{\text{Internal}}\\{\text{source}}\end{smallmatrix}}+\underbrace {{\vphantom {\frac {}{}}}\nu \nabla ^{2}\mathbf {u} } _{\text{Diffusion}}} ^{\text{Divergence of stress}}+\underbrace {{\vphantom {\frac {}{}}}\mathbf {g} } _{\begin{smallmatrix}{\text{External}}\\{\text{source}}\end{smallmatrix}}.}

The higher-order term, namely theshear stress divergence∇⋅τ{\textstyle \nabla \cdot {\boldsymbol {\tau }}}, has simply reduced to thevector Laplacian termμ∇2u{\textstyle \mu \nabla ^{2}\mathbf {u} }.[18] This Laplacian term can be interpreted as the difference between the velocity at a point and the mean velocity in a small surrounding volume. This implies that – for a Newtonian fluid – viscosity operates as adiffusion of momentum, in much the same way as theheat conduction. In fact neglecting the convection term, incompressible Navier–Stokes equations lead to a vectordiffusion equation (namelyStokes equations), but in general the convection term is present, so incompressible Navier–Stokes equations belong to the class ofconvection–diffusion equations.

In the usual case of an external field being aconservative field:g=−∇φ{\displaystyle \mathbf {g} =-\nabla \varphi }by defining thehydraulic head:h≡w+φ{\displaystyle h\equiv w+\varphi }

one can finally condense the whole source in one term, arriving to the incompressible Navier–Stokes equation with conservative external field:∂u∂t+(u⋅∇)u−ν∇2u=−∇h.{\displaystyle {\frac {\partial \mathbf {u} }{\partial t}}+(\mathbf {u} \cdot \nabla )\mathbf {u} -\nu \,\nabla ^{2}\mathbf {u} =-\nabla h.}

The incompressible Navier–Stokes equations with uniform density and viscosity and conservative external field is thefundamental equation ofhydraulics. The domain for these equations is commonly a 3 or fewer dimensionalEuclidean space, for which anorthogonal coordinate reference frame is usually set to explicit the system of scalar partial differential equations to be solved. In 3-dimensional orthogonal coordinate systems are 3:Cartesian,cylindrical, andspherical. Expressing the Navier–Stokes vector equation in Cartesian coordinates is quite straightforward and not much influenced by the number of dimensions of the euclidean space employed, and this is the case also for the first-order terms (like the variation and convection ones) also in non-cartesian orthogonal coordinate systems. But for the higher order terms (the two coming from the divergence of the deviatoric stress that distinguish Navier–Stokes equations from Euler equations) sometensor calculus is required for deducing an expression in non-cartesian orthogonal coordinate systems.A special case of the fundamental equation of hydraulics is theBernoulli's equation.

The incompressible Navier–Stokes equation is composite, the sum of two orthogonal equations,∂u∂t=ΠS(−(u⋅∇)u+ν∇2u)+fSρ−1∇p=ΠI(−(u⋅∇)u+ν∇2u)+fI{\displaystyle {\begin{aligned}{\frac {\partial \mathbf {u} }{\partial t}}&=\Pi ^{S}\left(-(\mathbf {u} \cdot \nabla )\mathbf {u} +\nu \,\nabla ^{2}\mathbf {u} \right)+\mathbf {f} ^{S}\\\rho ^{-1}\,\nabla p&=\Pi ^{I}\left(-(\mathbf {u} \cdot \nabla )\mathbf {u} +\nu \,\nabla ^{2}\mathbf {u} \right)+\mathbf {f} ^{I}\end{aligned}}}whereΠS{\textstyle \Pi ^{S}} andΠI{\textstyle \Pi ^{I}} are solenoidal andirrotational projection operators satisfyingΠS+ΠI=1{\textstyle \Pi ^{S}+\Pi ^{I}=1}, andfS{\textstyle \mathbf {f} ^{S}} andfI{\textstyle \mathbf {f} ^{I}} are the non-conservative and conservative parts of the body force. This result follows from theHelmholtz theorem (also known as the fundamental theorem of vector calculus). The first equation is a pressureless governing equation for the velocity, while the second equation for the pressure is a functional of the velocity and is related to the pressure Poisson equation.

The explicit functional form of the projection operator in 3D is found from the Helmholtz Theorem:ΠSF(r)=14π∇×∫∇′×F(r′)|r−r′|dV′,ΠI=1−ΠS{\displaystyle \Pi ^{S}\,\mathbf {F} (\mathbf {r} )={\frac {1}{4\pi }}\nabla \times \int {\frac {\nabla ^{\prime }\times \mathbf {F} (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\,\mathrm {d} V',\quad \Pi ^{I}=1-\Pi ^{S}}with a similar structure in 2D. Thus the governing equation is anintegro-differential equation similar toCoulomb's andBiot–Savart's law, not convenient for numerical computation.

An equivalent weak or variational form of the equation, proved to produce the same velocity solution as the Navier–Stokes equation,[19] is given by,(w,∂u∂t)=−(w,(u⋅∇)u)−ν(∇w:∇u)+(w,fS){\displaystyle \left(\mathbf {w} ,{\frac {\partial \mathbf {u} }{\partial t}}\right)=-{\bigl (}\mathbf {w} ,\left(\mathbf {u} \cdot \nabla \right)\mathbf {u} {\bigr )}-\nu \left(\nabla \mathbf {w} :\nabla \mathbf {u} \right)+\left(\mathbf {w} ,\mathbf {f} ^{S}\right)}

for divergence-free test functionsw{\textstyle \mathbf {w} } satisfying appropriate boundary conditions. Here, the projections are accomplished by the orthogonality of the solenoidal and irrotational function spaces. The discrete form of this is eminently suited to finite element computation of divergence-free flow, as we shall see in the next section. There, one will be able to address the question, "How does one specify pressure-driven (Poiseuille) problems with a pressureless governing equation?".

The absence of pressure forces from the governing velocity equation demonstrates that the equation is not a dynamic one, but rather a kinematic equation where the divergence-free condition serves the role of a conservation equation. This would seem to refute the frequent statements that the incompressible pressure enforces the divergence-free condition.

Weak form of the incompressible Navier–Stokes equations

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Strong form

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Consider the incompressible Navier–Stokes equations for aNewtonian fluid of constant densityρ{\textstyle \rho } in a domainΩ⊂Rd(d=2,3){\displaystyle \Omega \subset \mathbb {R} ^{d}\quad (d=2,3)}with boundary∂Ω=ΓD∪ΓN,{\displaystyle \partial \Omega =\Gamma _{D}\cup \Gamma _{N},}beingΓD{\textstyle \Gamma _{D}} andΓN{\textstyle \Gamma _{N}} portions of the boundary where respectively aDirichlet and aNeumann boundary condition is applied (ΓD∩ΓN=∅{\textstyle \Gamma _{D}\cap \Gamma _{N}=\emptyset }):[20]{ρ∂u∂t+ρ(u⋅∇)u−∇⋅σ(u,p)=f in Ω×(0,T)∇⋅u=0 in Ω×(0,T)u=g on ΓD×(0,T)σ(u,p)n^=h on ΓN×(0,T)u(0)=u0 in Ω×{0}{\displaystyle {\begin{cases}\rho {\dfrac {\partial \mathbf {u} }{\partial t}}+\rho (\mathbf {u} \cdot \nabla )\mathbf {u} -\nabla \cdot {\boldsymbol {\sigma }}(\mathbf {u} ,p)=\mathbf {f} &{\text{ in }}\Omega \times (0,T)\\\nabla \cdot \mathbf {u} =0&{\text{ in }}\Omega \times (0,T)\\\mathbf {u} =\mathbf {g} &{\text{ on }}\Gamma _{D}\times (0,T)\\{\boldsymbol {\sigma }}(\mathbf {u} ,p){\hat {\mathbf {n} }}=\mathbf {h} &{\text{ on }}\Gamma _{N}\times (0,T)\\\mathbf {u} (0)=\mathbf {u} _{0}&{\text{ in }}\Omega \times \{0\}\end{cases}}}u{\textstyle \mathbf {u} } is the fluid velocity,p{\textstyle p} the fluid pressure,f{\textstyle \mathbf {f} } a given forcing term,n^{\displaystyle {\hat {\mathbf {n} }}} the outward directed unit normal vector toΓN{\textstyle \Gamma _{N}}, andσ(u,p){\textstyle {\boldsymbol {\sigma }}(\mathbf {u} ,p)} theviscous stress tensor defined as:[20]σ(u,p)=−pI+2με(u).{\displaystyle {\boldsymbol {\sigma }}(\mathbf {u} ,p)=-p\mathbf {I} +2\mu {\boldsymbol {\varepsilon }}(\mathbf {u} ).}Letμ{\textstyle \mu } be the dynamic viscosity of the fluid,I{\textstyle \mathbf {I} } the second-orderidentity tensor andε(u){\textstyle {\boldsymbol {\varepsilon }}(\mathbf {u} )} thestrain-rate tensor defined as:[20]ε(u)=12((∇u)+(∇u)T).{\displaystyle {\boldsymbol {\varepsilon }}(\mathbf {u} )={\frac {1}{2}}\left(\left(\nabla \mathbf {u} \right)+\left(\nabla \mathbf {u} \right)^{\mathrm {T} }\right).}The functionsg{\textstyle \mathbf {g} } andh{\textstyle \mathbf {h} } are given Dirichlet and Neumann boundary data, whileu0{\textstyle \mathbf {u} _{0}} is theinitial condition. The first equation is the momentum balance equation, while the second represents themass conservation, namely thecontinuity equation. Assuming constant dynamic viscosity, using the vectorial identity∇⋅(∇f)T=∇(∇⋅f){\displaystyle \nabla \cdot \left(\nabla \mathbf {f} \right)^{\mathrm {T} }=\nabla (\nabla \cdot \mathbf {f} )}and exploiting mass conservation, the divergence of the total stress tensor in the momentum equation can also be expressed as:[20]∇⋅σ(u,p)=∇⋅(−pI+2με(u))=−∇p+2μ∇⋅ε(u)=−∇p+2μ∇⋅[12((∇u)+(∇u)T)]=−∇p+μ(Δu+∇⋅(∇u)T)=−∇p+μ(Δu+∇(∇⋅u)⏟=0)=−∇p+μΔu.{\displaystyle {\begin{aligned}\nabla \cdot {\boldsymbol {\sigma }}(\mathbf {u} ,p)&=\nabla \cdot \left(-p\mathbf {I} +2\mu {\boldsymbol {\varepsilon }}(\mathbf {u} )\right)\\&=-\nabla p+2\mu \nabla \cdot {\boldsymbol {\varepsilon }}(\mathbf {u} )\\&=-\nabla p+2\mu \nabla \cdot \left[{\tfrac {1}{2}}\left(\left(\nabla \mathbf {u} \right)+\left(\nabla \mathbf {u} \right)^{\mathrm {T} }\right)\right]\\&=-\nabla p+\mu \left(\Delta \mathbf {u} +\nabla \cdot \left(\nabla \mathbf {u} \right)^{\mathrm {T} }\right)\\&=-\nabla p+\mu {\bigl (}\Delta \mathbf {u} +\nabla \underbrace {(\nabla \cdot \mathbf {u} )} _{=0}{\bigr )}=-\nabla p+\mu \,\Delta \mathbf {u} .\end{aligned}}}Moreover, note that the Neumann boundary conditions can be rearranged as:[20]σ(u,p)n^=(−pI+2με(u))n^=−pn^+μ∂u∂n^.{\displaystyle {\boldsymbol {\sigma }}(\mathbf {u} ,p){\hat {\mathbf {n} }}=\left(-p\mathbf {I} +2\mu {\boldsymbol {\varepsilon }}(\mathbf {u} )\right){\hat {\mathbf {n} }}=-p{\hat {\mathbf {n} }}+\mu {\frac {\partial {\boldsymbol {u}}}{\partial {\hat {\mathbf {n} }}}}.}

Weak form

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In order to find the weak form of the Navier–Stokes equations, firstly, consider the momentum equation[20]ρ∂u∂t−μΔu+ρ(u⋅∇)u+∇p=f{\displaystyle \rho {\frac {\partial \mathbf {u} }{\partial t}}-\mu \Delta \mathbf {u} +\rho (\mathbf {u} \cdot \nabla )\mathbf {u} +\nabla p=\mathbf {f} }multiply it for a test functionv{\textstyle \mathbf {v} }, defined in a suitable spaceV{\textstyle V}, and integrate both members with respect to the domainΩ{\textstyle \Omega }:[20]∫Ωρ∂u∂t⋅v−∫ΩμΔu⋅v+∫Ωρ(u⋅∇)u⋅v+∫Ω∇p⋅v=∫Ωf⋅v{\displaystyle \int \limits _{\Omega }\rho {\frac {\partial \mathbf {u} }{\partial t}}\cdot \mathbf {v} -\int \limits _{\Omega }\mu \Delta \mathbf {u} \cdot \mathbf {v} +\int \limits _{\Omega }\rho (\mathbf {u} \cdot \nabla )\mathbf {u} \cdot \mathbf {v} +\int \limits _{\Omega }\nabla p\cdot \mathbf {v} =\int \limits _{\Omega }\mathbf {f} \cdot \mathbf {v} }Counter-integrating by parts the diffusive and the pressure terms and by using the Gauss' theorem:[20]−∫ΩμΔu⋅v=∫Ωμ∇u⋅∇v−∫∂Ωμ∂u∂n^⋅v∫Ω∇p⋅v=−∫Ωp∇⋅v+∫∂Ωpv⋅n^{\displaystyle {\begin{aligned}-\int \limits _{\Omega }\mu \Delta \mathbf {u} \cdot \mathbf {v} &=\int _{\Omega }\mu \nabla \mathbf {u} \cdot \nabla \mathbf {v} -\int \limits _{\partial \Omega }\mu {\frac {\partial \mathbf {u} }{\partial {\hat {\mathbf {n} }}}}\cdot \mathbf {v} \\\int \limits _{\Omega }\nabla p\cdot \mathbf {v} &=-\int \limits _{\Omega }p\nabla \cdot \mathbf {v} +\int \limits _{\partial \Omega }p\mathbf {v} \cdot {\hat {\mathbf {n} }}\end{aligned}}}

Using these relations, one gets:[20]∫Ωρ∂u∂t⋅v+∫Ωμ∇u⋅∇v+∫Ωρ(u⋅∇)u⋅v−∫Ωp∇⋅v=∫Ωf⋅v+∫∂Ω(μ∂u∂n^−pn^)⋅v∀v∈V.{\displaystyle \int \limits _{\Omega }\rho {\dfrac {\partial \mathbf {u} }{\partial t}}\cdot \mathbf {v} +\int \limits _{\Omega }\mu \nabla \mathbf {u} \cdot \nabla \mathbf {v} +\int \limits _{\Omega }\rho (\mathbf {u} \cdot \nabla )\mathbf {u} \cdot \mathbf {v} -\int \limits _{\Omega }p\nabla \cdot \mathbf {v} =\int \limits _{\Omega }\mathbf {f} \cdot \mathbf {v} +\int \limits _{\partial \Omega }\left(\mu {\frac {\partial \mathbf {u} }{\partial {\hat {\mathbf {n} }}}}-p{\hat {\mathbf {n} }}\right)\cdot \mathbf {v} \quad \forall \mathbf {v} \in V.}In the same fashion, the continuity equation is multiplied for a test functionq belonging to a spaceQ{\textstyle Q} and integrated in the domainΩ{\textstyle \Omega }:[20]∫Ωq∇⋅u=0.∀q∈Q.{\displaystyle \int \limits _{\Omega }q\nabla \cdot \mathbf {u} =0.\quad \forall q\in Q.}The space functions are chosen as follows:V=[H01(Ω)]d={v∈[H1(Ω)]d:v=0 on ΓD},Q=L2(Ω){\displaystyle {\begin{aligned}V=\left[H_{0}^{1}(\Omega )\right]^{d}&=\left\{\mathbf {v} \in \left[H^{1}(\Omega )\right]^{d}:\quad \mathbf {v} =\mathbf {0} {\text{ on }}\Gamma _{D}\right\},\\Q&=L^{2}(\Omega )\end{aligned}}}Considering that the test functionv vanishes on the Dirichlet boundary and considering the Neumann condition, the integral on the boundary can be rearranged as:[20]∫∂Ω(μ∂u∂n^−pn^)⋅v=∫ΓD(μ∂u∂n^−pn^)⋅v⏟v=0 on ΓD +∫ΓN∫ΓN(μ∂u∂n^−pn^)⏟=h on ΓN⋅v=∫ΓNh⋅v.{\displaystyle \int \limits _{\partial \Omega }\left(\mu {\frac {\partial \mathbf {u} }{\partial {\hat {\mathbf {n} }}}}-p{\hat {\mathbf {n} }}\right)\cdot \mathbf {v} =\underbrace {\int \limits _{\Gamma _{D}}\left(\mu {\frac {\partial \mathbf {u} }{\partial {\hat {\mathbf {n} }}}}-p{\hat {\mathbf {n} }}\right)\cdot \mathbf {v} } _{\mathbf {v} =\mathbf {0} {\text{ on }}\Gamma _{D}\ }+\int \limits _{\Gamma _{N}}\underbrace {{\vphantom {\int \limits _{\Gamma _{N}}}}\left(\mu {\frac {\partial \mathbf {u} }{\partial {\hat {\mathbf {n} }}}}-p{\hat {\mathbf {n} }}\right)} _{=\mathbf {h} {\text{ on }}\Gamma _{N}}\cdot \mathbf {v} =\int \limits _{\Gamma _{N}}\mathbf {h} \cdot \mathbf {v} .}Having this in mind, the weak formulation of the Navier–Stokes equations is expressed as:[20]find u∈L2(R+[H1(Ω)]d)∩C0(R+[L2(Ω)]d) such that: {∫Ωρ∂u∂t⋅v+∫Ωμ∇u⋅∇v+∫Ωρ(u⋅∇)u⋅v−∫Ωp∇⋅v=∫Ωf⋅v+∫ΓNh⋅v∀v∈V,∫Ωq∇⋅u=0∀q∈Q.{\displaystyle {\begin{aligned}&{\text{find }}\mathbf {u} \in L^{2}\left(\mathbb {R} ^{+}\;\left[H^{1}(\Omega )\right]^{d}\right)\cap C^{0}\left(\mathbb {R} ^{+}\;\left[L^{2}(\Omega )\right]^{d}\right){\text{ such that: }}\\[5pt]&\quad {\begin{cases}\displaystyle \int \limits _{\Omega }\rho {\dfrac {\partial \mathbf {u} }{\partial t}}\cdot \mathbf {v} +\int \limits _{\Omega }\mu \nabla \mathbf {u} \cdot \nabla \mathbf {v} +\int \limits _{\Omega }\rho (\mathbf {u} \cdot \nabla )\mathbf {u} \cdot \mathbf {v} -\int \limits _{\Omega }p\nabla \cdot \mathbf {v} =\int \limits _{\Omega }\mathbf {f} \cdot \mathbf {v} +\int \limits _{\Gamma _{N}}\mathbf {h} \cdot \mathbf {v} \quad \forall \mathbf {v} \in V,\\\displaystyle \int \limits _{\Omega }q\nabla \cdot \mathbf {u} =0\quad \forall q\in Q.\end{cases}}\end{aligned}}}

Discrete velocity

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With partitioning of the problem domain and definingbasis functions on the partitioned domain, the discrete form of the governing equation is(wi,∂uj∂t)=−(wi,(u⋅∇)uj)−ν(∇wi:∇uj)+(wi,fS).{\displaystyle \left(\mathbf {w} _{i},{\frac {\partial \mathbf {u} _{j}}{\partial t}}\right)=-{\bigl (}\mathbf {w} _{i},\left(\mathbf {u} \cdot \nabla \right)\mathbf {u} _{j}{\bigr )}-\nu \left(\nabla \mathbf {w} _{i}:\nabla \mathbf {u} _{j}\right)+\left(\mathbf {w} _{i},\mathbf {f} ^{S}\right).}

It is desirable to choose basis functions that reflect the essential feature of incompressible flow – the elements must be divergence-free. While the velocity is the variable of interest, the existence of the stream function or vector potential is necessary by the Helmholtz theorem. Further, to determine fluid flow in the absence of a pressure gradient, one can specify the difference of stream function values across a 2D channel, or the line integral of the tangential component of the vector potential around the channel in 3D, the flow being given byStokes' theorem. Discussion will be restricted to 2D in the following.

We further restrict discussion to continuous Hermite finite elements which have at least first-derivative degrees-of-freedom. With this, one can draw a large number of candidate triangular and rectangular elements from theplate-bending literature. These elements have derivatives as components of the gradient. In 2D, the gradient and curl of a scalar are clearly orthogonal, given by the expressions,∇φ=(∂φ∂x,∂φ∂y)T,∇×φ=(∂φ∂y,−∂φ∂x)T.{\displaystyle {\begin{aligned}\nabla \varphi &=\left({\frac {\partial \varphi }{\partial x}},\,{\frac {\partial \varphi }{\partial y}}\right)^{\mathrm {T} },\\[5pt]\nabla \times \varphi &=\left({\frac {\partial \varphi }{\partial y}},\,-{\frac {\partial \varphi }{\partial x}}\right)^{\mathrm {T} }.\end{aligned}}}

Adopting continuous plate-bending elements, interchanging the derivative degrees-of-freedom and changing the sign of the appropriate one gives many families of stream function elements.

Taking the curl of the scalar stream function elements gives divergence-free velocity elements.[21][22] The requirement that the stream function elements be continuous assures that the normal component of the velocity is continuous across element interfaces, all that is necessary for vanishing divergence on these interfaces.

Boundary conditions are simple to apply. The stream function is constant on no-flow surfaces, with no-slip velocity conditions on surfaces.Stream function differences across open channels determine the flow. No boundary conditions are necessary on open boundaries, though consistent values may be used with some problems. These are all Dirichlet conditions.

The algebraic equations to be solved are simple to set up, but of course arenon-linear, requiring iteration of the linearized equations.

Similar considerations apply to three-dimensions, but extension from 2D is not immediate because of the vector nature of the potential, and there exists no simple relation between the gradient and the curl as was the case in 2D.

Pressure recovery

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Recovering pressure from the velocity field is easy. The discrete weak equation for the pressure gradient is,(gi,∇p)=−(gi,(u⋅∇)uj)−ν(∇gi:∇uj)+(gi,fI){\displaystyle (\mathbf {g} _{i},\nabla p)=-\left(\mathbf {g} _{i},\left(\mathbf {u} \cdot \nabla \right)\mathbf {u} _{j}\right)-\nu \left(\nabla \mathbf {g} _{i}:\nabla \mathbf {u} _{j}\right)+\left(\mathbf {g} _{i},\mathbf {f} ^{I}\right)}

where the test/weight functions are irrotational. Any conforming scalar finite element may be used. However, the pressure gradient field may also be of interest. In this case, one can use scalar Hermite elements for the pressure. For the test/weight functionsgi{\textstyle \mathbf {g} _{i}} one would choose the irrotational vector elements obtained from the gradient of the pressure element.

Non-inertial frame of reference

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The rotating frame of reference introduces some interesting pseudo-forces into the equations through thematerial derivative term. Consider a stationary inertial frame of referenceK{\textstyle K} , and a non-inertial frame of referenceK′{\textstyle K'}, which is translating with velocityU(t){\textstyle \mathbf {U} (t)} and rotating with angular velocityΩ(t){\textstyle \Omega (t)} with respect to the stationary frame. The Navier–Stokes equation observed from the non-inertial frame then becomes

Navier–Stokes momentum equation in non-inertial frame

ρ(∂u∂t+(u⋅∇)u)=−∇p+∇⋅{μ[∇u+(∇u)T−23(∇⋅u)I]}+∇[ζ(∇⋅u)]+ρf−ρ[2Ω×u+Ω×(Ω×x)+dUdt+dΩdt×x].{\displaystyle \rho \left({\frac {\partial \mathbf {u} }{\partial t}}+(\mathbf {u} \cdot \nabla )\mathbf {u} \right)=-\nabla p+\nabla \cdot \left\{\mu \left[\nabla \mathbf {u} +(\nabla \mathbf {u} )^{\mathrm {T} }-{\tfrac {2}{3}}(\nabla \cdot \mathbf {u} )\mathbf {I} \right]\right\}+\nabla [\zeta (\nabla \cdot \mathbf {u} )]+\rho \mathbf {f} -\rho \left[2\mathbf {\Omega } \times \mathbf {u} +\mathbf {\Omega } \times (\mathbf {\Omega } \times \mathbf {x} )+{\frac {\mathrm {d} \mathbf {U} }{\mathrm {d} t}}+{\frac {\mathrm {d} \mathbf {\Omega } }{\mathrm {d} t}}\times \mathbf {x} \right].}

Herex{\textstyle \mathbf {x} } andu{\textstyle \mathbf {u} } are measured in the non-inertial frame. The first term in the parenthesis representsCoriolis acceleration, the second term is due tocentrifugal acceleration, the third is due to the linear acceleration ofK′{\textstyle K'} with respect toK{\textstyle K} and the fourth term is due to the angular acceleration ofK′{\textstyle K'} with respect toK{\textstyle K}.

Other equations

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The Navier–Stokes equations are strictly a statement of the balance of momentum. To fully describe fluid flow, more information is needed, how much depending on the assumptions made. This additional information may include boundary data (no-slip,capillary surface, etc.), conservation of mass,balance of energy, and/or anequation of state.

Continuity equation for incompressible fluid

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Main article:Continuity equation

Regardless of the flow assumptions, a statement of theconservation of mass is generally necessary. This is achieved through the masscontinuity equation, as discussed above in the "General continuum equations" within this article, as follows:DmDt=∭V(DρDt+ρ(∇⋅u))dVDρDt+ρ(∇⋅u)=∂ρ∂t+(∇ρ)⋅u+ρ(∇⋅u)=∂ρ∂t+∇⋅(ρu)=0{\displaystyle {\begin{aligned}{\frac {\mathbf {D} m}{\mathbf {Dt} }}&={\iiint \limits _{V}}({{\frac {\mathbf {D} \rho }{\mathbf {Dt} }}+\rho (\nabla \cdot \mathbf {u} )})dV\\{\frac {\mathbf {D} \rho }{\mathbf {Dt} }}+\rho (\nabla \cdot {\mathbf {u} })&={\frac {\partial \rho }{\partial t}}+({\nabla \rho })\cdot {\mathbf {u} }+{\rho }(\nabla \cdot \mathbf {u} )={\frac {\partial \rho }{\partial t}}+\nabla \cdot ({\rho \mathbf {u} })=0\end{aligned}}}A fluid media for which thedensity (ρ{\displaystyle \rho }) is constant is calledincompressible. Therefore, the rate of change ofdensity (ρ{\displaystyle \rho }) with respect to time(∂ρ∂t){\displaystyle ({\frac {\partial \rho }{\partial t}})} and thegradient of density(∇ρ){\displaystyle (\nabla \rho )} are equal to zero(0){\displaystyle (0)}. In this case the general equation of continuity,∂ρ∂t+∇⋅(ρu)=0{\displaystyle {\frac {\partial \rho }{\partial t}}+\nabla \cdot ({\rho \mathbf {u} })=0}, reduces to:ρ(∇⋅u)=0{\displaystyle \rho (\nabla {\cdot }{\mathbf {u} })=0}. Furthermore, assuming thatdensity (ρ{\displaystyle \rho }) is a non-zero constant(ρ≠0){\displaystyle (\rho \neq 0)} means that the right-hand side of the equation(0){\displaystyle (0)} is divisible bydensity (ρ{\displaystyle \rho }). Therefore, the continuity equation for anincompressible fluid reduces further to:(∇⋅u)=0{\displaystyle (\nabla {\cdot {\mathbf {u} }})=0}This relationship,(∇⋅u)=0{\textstyle (\nabla {\cdot {\mathbf {u} }})=0}, identifies that thedivergence of the flow velocityvector (u{\displaystyle \mathbf {u} }) is equal to zero(0){\displaystyle (0)}, which means that for anincompressible fluid theflow velocity field is asolenoidal vector field or adivergence-free vector field. Note that this relationship can be expanded upon due to its uniqueness with thevector Laplace operator(∇2u=∇(∇⋅u)−∇×(∇×u)){\displaystyle (\nabla ^{2}\mathbf {u} =\nabla (\nabla \cdot \mathbf {u} )-\nabla \times (\nabla \times \mathbf {u} ))}, andvorticity(ω→=∇×u){\displaystyle ({\vec {\omega }}=\nabla \times \mathbf {u} )} which is now expressed like so, for anincompressible fluid:∇2u=−(∇×(∇×u))=−(∇×ω→){\displaystyle \nabla ^{2}\mathbf {u} =-(\nabla \times (\nabla \times \mathbf {u} ))=-(\nabla \times {\vec {\omega }})}

Stream function for incompressible 2D fluid

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Taking thecurl of the incompressible Navier–Stokes equation results in the elimination of pressure. This is especially easy to see if 2D Cartesian flow is assumed (like in the degenerate 3D case withuz=0{\textstyle u_{z}=0} and no dependence of anything onz{\textstyle z}), where the equations reduce to:ρ(∂ux∂t+ux∂ux∂x+uy∂ux∂y)=−∂p∂x+μ(∂2ux∂x2+∂2ux∂y2)+ρgxρ(∂uy∂t+ux∂uy∂x+uy∂uy∂y)=−∂p∂y+μ(∂2uy∂x2+∂2uy∂y2)+ρgy.{\displaystyle {\begin{aligned}\rho \left({\frac {\partial u_{x}}{\partial t}}+u_{x}{\frac {\partial u_{x}}{\partial x}}+u_{y}{\frac {\partial u_{x}}{\partial y}}\right)&=-{\frac {\partial p}{\partial x}}+\mu \left({\frac {\partial ^{2}u_{x}}{\partial x^{2}}}+{\frac {\partial ^{2}u_{x}}{\partial y^{2}}}\right)+\rho g_{x}\\\rho \left({\frac {\partial u_{y}}{\partial t}}+u_{x}{\frac {\partial u_{y}}{\partial x}}+u_{y}{\frac {\partial u_{y}}{\partial y}}\right)&=-{\frac {\partial p}{\partial y}}+\mu \left({\frac {\partial ^{2}u_{y}}{\partial x^{2}}}+{\frac {\partial ^{2}u_{y}}{\partial y^{2}}}\right)+\rho g_{y}.\end{aligned}}}

Differentiating the first with respect toy{\textstyle y}, the second with respect tox{\textstyle x} and subtracting the resulting equations will eliminate pressure and anyconservative force. For incompressible flow, defining thestream functionψ{\textstyle \psi } throughux=∂ψ∂y;uy=−∂ψ∂x{\displaystyle u_{x}={\frac {\partial \psi }{\partial y}};\quad u_{y}=-{\frac {\partial \psi }{\partial x}}}results in mass continuity being unconditionally satisfied (given the stream function is continuous), and then incompressible Newtonian 2D momentum and mass conservation condense into one equation:∂∂t(∇2ψ)+∂ψ∂y∂∂x(∇2ψ)−∂ψ∂x∂∂y(∇2ψ)=ν∇4ψ{\displaystyle {\frac {\partial }{\partial t}}\left(\nabla ^{2}\psi \right)+{\frac {\partial \psi }{\partial y}}{\frac {\partial }{\partial x}}\left(\nabla ^{2}\psi \right)-{\frac {\partial \psi }{\partial x}}{\frac {\partial }{\partial y}}\left(\nabla ^{2}\psi \right)=\nu \nabla ^{4}\psi }

where∇4{\textstyle \nabla ^{4}} is the 2Dbiharmonic operator andν{\textstyle \nu } is thekinematic viscosity,ν=μρ{\textstyle \nu ={\frac {\mu }{\rho }}}. We can also express this compactly using theJacobian determinant:∂∂t(∇2ψ)+∂(ψ,∇2ψ)∂(y,x)=ν∇4ψ.{\displaystyle {\frac {\partial }{\partial t}}\left(\nabla ^{2}\psi \right)+{\frac {\partial \left(\psi ,\nabla ^{2}\psi \right)}{\partial (y,x)}}=\nu \nabla ^{4}\psi .}

This single equation together with appropriate boundary conditions describes 2D fluid flow, taking only kinematic viscosity as a parameter. Note that the equation forcreeping flow results when the left side is assumed zero.

Inaxisymmetric flow another stream function formulation, called theStokes stream function, can be used to describe the velocity components of an incompressible flow with onescalar function.

The incompressible Navier–Stokes equation is adifferential algebraic equation, having the inconvenient feature that there is no explicit mechanism for advancing the pressure in time. Consequently, much effort has been expended to eliminate the pressure from all or part of the computational process. The stream function formulation eliminates the pressure but only in two dimensions and at the expense of introducing higher derivatives and elimination of the velocity, which is the primary variable of interest.

Properties

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Nonlinearity

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The Navier–Stokes equations arenonlinearpartial differential equations in the general case and so remain in almost every real situation.[23][24] In some cases, such as one-dimensional flow andStokes flow (or creeping flow), the equations can be simplified to linear equations. The nonlinearity makes most problems difficult or impossible to solve and is the main contributor to theturbulence that the equations model.

The nonlinearity is due toconvective acceleration, which is an acceleration associated with the change in velocity over position. Hence, any convective flow, whether turbulent or not, will involve nonlinearity. An example of convective butlaminar (nonturbulent) flow would be the passage of a viscous fluid (for example, oil) through a small convergingnozzle. Such flows, whether exactly solvable or not, can often be thoroughly studied and understood.[25]

Turbulence

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Turbulence is the time-dependentchaotic behaviour seen in many fluid flows. It is generally believed that it is due to theinertia of the fluid as a whole: the culmination of time-dependent and convective acceleration; hence flows where inertial effects are small tend to be laminar (theReynolds number quantifies how much the flow is affected by inertia). It is believed, though not known with certainty, that the Navier–Stokes equations describe turbulence properly.[26]

The numerical solution of the Navier–Stokes equations for turbulent flow is extremely difficult, and due to the significantly different mixing-length scales that are involved in turbulent flow, the stable solution of this requires such a fine mesh resolution that the computational time becomes significantly infeasible for calculation ordirect numerical simulation. Attempts to solve turbulent flow using a laminar solver typically result in a time-unsteady solution, which fails to converge appropriately. To counter this, time-averaged equations such as theReynolds-averaged Navier–Stokes equations (RANS), supplemented with turbulence models, are used in practicalcomputational fluid dynamics (CFD) applications when modeling turbulent flows. Some models include theSpalart–Allmaras,k–ω,k–ε, andSST models, which add a variety of additional equations to bring closure to the RANS equations.Large eddy simulation (LES) can also be used to solve these equations numerically. This approach is computationally more expensive—in time and in computer memory—than RANS, but produces better results because it explicitly resolves the larger turbulent scales.

Applicability

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Further information:Discretization of Navier–Stokes equations

Together with supplemental equations (for example, conservation of mass) and well-formulated boundary conditions, the Navier–Stokes equations seem to model fluid motion accurately; even turbulent flows seem (on average) to agree with real world observations.

The Navier–Stokes equations assume that the fluid being studied is acontinuum (it is infinitely divisible and not composed of particles such as atoms or molecules), and is not moving atrelativistic velocities. At very small scales or under extreme conditions, real fluids made out of discrete molecules will produce results different from the continuous fluids modeled by the Navier–Stokes equations. For example,capillarity of internal layers in fluids appears for flow with high gradients.[27] For largeKnudsen number of the problem, theBoltzmann equation may be a suitable replacement.[28] Failing that, one may have to resort tomolecular dynamics or various hybrid methods.[29]

Another limitation is simply the complicated nature of the equations. Time-tested formulations exist for common fluid families, but the application of the Navier–Stokes equations to less common families tends to result in very complicated formulations and often to open research problems. For this reason, these equations are usually written forNewtonian fluids where the viscosity model islinear; truly general models for the flow of other kinds of fluids (such as blood) do not exist.[30]

Application to specific problems

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The Navier–Stokes equations, even when written explicitly for specific fluids, are rather generic in nature and their proper application to specific problems can be very diverse. This is partly because there is an enormous variety of problems that may be modeled, ranging from as simple as the distribution of static pressure to as complicated asmultiphase flow driven bysurface tension.

Generally, application to specific problems begins with some flow assumptions and initial/boundary condition formulation, this may be followed byscale analysis to further simplify the problem.

Visualization of(a) parallel flow and(b) radial flow

Parallel flow

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Assume steady, parallel, one-dimensional, non-convective pressure-driven flow between parallel plates, the resulting scaled (dimensionless)boundary value problem is:d2udy2=−1;u(0)=u(1)=0.{\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} y^{2}}}=-1;\quad u(0)=u(1)=0.}

The boundary condition is theno slip condition. This problem is easily solved for the flow field:u(y)=y−y22.{\displaystyle u(y)={\frac {y-y^{2}}{2}}.}

From this point onward, more quantities of interest can be easily obtained, such as viscous drag force or net flow rate.

Radial flow

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Difficulties may arise when the problem becomes slightly more complicated. A seemingly modest twist on the parallel flow above would be theradial flow between parallel plates; this involves convection and thus non-linearity. The velocity field may be represented by a functionf(z) that must satisfy:d2fdz2+Rf2=−1;f(−1)=f(1)=0.{\displaystyle {\frac {\mathrm {d} ^{2}f}{\mathrm {d} z^{2}}}+Rf^{2}=-1;\quad f(-1)=f(1)=0.}

Thisordinary differential equation is what is obtained when the Navier–Stokes equations are written and the flow assumptions applied (additionally, the pressure gradient is solved for). Thenonlinear term makes this a very difficult problem to solve analytically (a lengthyimplicit solution may be found which involveselliptic integrals androots of cubic polynomials). Issues with the actual existence of solutions arise forR>1.41{\textstyle R>1.41} (approximately; this is not√2), the parameterR{\textstyle R} being the Reynolds number with appropriately chosen scales.[31] This is an example of flow assumptions losing their applicability, and an example of the difficulty in "high" Reynolds number flows.[31]

Convection

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A type of natural convection that can be described by the Navier–Stokes equation is theRayleigh–Bénard convection. It is one of the most commonly studied convection phenomena because of its analytical and experimental accessibility.

Exact solutions of the Navier–Stokes equations

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Some exact solutions to the Navier–Stokes equations exist. Examples of degenerate cases—with the non-linear terms in the Navier–Stokes equations equal to zero—arePoiseuille flow,Couette flow and the oscillatoryStokes boundary layer. But also, more interesting examples, solutions to the full non-linear equations, exist, such asJeffery–Hamel flow,Von Kármán swirling flow,stagnation point flow,Landau–Squire jet, andTaylor–Green vortex.[32][33][34] Time-dependentself-similar solutions of the three-dimensional non-compressible Navier–Stokes equations in Cartesian coordinate can be given with the help of theKummer's functions with quadratic arguments.[35] For the compressible Navier–Stokes equations the time-dependent self-similar solutions are however theWhittaker functions again with quadratic arguments when thepolytropicequation of state is used as a closing condition.[36] Note that the existence of these exact solutions does not imply they are stable: turbulence may develop at higher Reynolds numbers.

Under additional assumptions, the component parts can be separated.[37]

A two-dimensional example

For example, in the case of an unbounded planar domain withtwo-dimensional — incompressible and stationary — flow inpolar coordinates(r,φ), the velocity components(ur,uφ) and pressurep are:[38]ur=Ar,uφ=B(1r−rAν+1),p=−A2+B22r2−2B2νrAνA+B2r(2Aν+2)2Aν+2{\displaystyle {\begin{aligned}u_{r}&={\frac {A}{r}},\\u_{\varphi }&=B\left({\frac {1}{r}}-r^{{\frac {A}{\nu }}+1}\right),\\p&=-{\frac {A^{2}+B^{2}}{2r^{2}}}-{\frac {2B^{2}\nu r^{\frac {A}{\nu }}}{A}}+{\frac {B^{2}r^{\left({\frac {2A}{\nu }}+2\right)}}{{\frac {2A}{\nu }}+2}}\end{aligned}}}

whereA andB are arbitrary constants. This solution is valid in the domainr ≥ 1 and forA < −2ν.

In Cartesian coordinates, when the viscosity is zero (ν = 0), this is:v(x,y)=1x2+y2(Ax+ByAy−Bx),p(x,y)=−A2+B22(x2+y2){\displaystyle {\begin{aligned}\mathbf {v} (x,y)&={\frac {1}{x^{2}+y^{2}}}{\begin{pmatrix}Ax+By\\Ay-Bx\end{pmatrix}},\\p(x,y)&=-{\frac {A^{2}+B^{2}}{2\left(x^{2}+y^{2}\right)}}\end{aligned}}}

A three-dimensional example

For example, in the case of an unbounded Euclidean domain withthree-dimensional — incompressible, stationary and with zero viscosity (ν = 0) — radial flow inCartesian coordinates(x,y,z), the velocity vectorv and pressurep are:[citation needed]v(x,y,z)=Ax2+y2+z2(xyz),p(x,y,z)=−A22(x2+y2+z2).{\displaystyle {\begin{aligned}\mathbf {v} (x,y,z)&={\frac {A}{x^{2}+y^{2}+z^{2}}}{\begin{pmatrix}x\\y\\z\end{pmatrix}},\\p(x,y,z)&=-{\frac {A^{2}}{2\left(x^{2}+y^{2}+z^{2}\right)}}.\end{aligned}}}

There is a singularity atx =y =z = 0.

A three-dimensional steady-state vortex solution

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Wire model of flow lines along aHopf fibration

A steady-state example with no singularities comes from considering the flow along the lines of aHopf fibration. Letr{\textstyle r} be a constant radius of the inner coil. One set of solutions is given by:[39]ρ(x,y,z)=3Br2+x2+y2+z2p(x,y,z)=−A2B(r2+x2+y2+z2)3u(x,y,z)=A(r2+x2+y2+z2)2(2(−ry+xz)2(rx+yz)r2−x2−y2+z2)g=0μ=0{\displaystyle {\begin{aligned}\rho (x,y,z)&={\frac {3B}{r^{2}+x^{2}+y^{2}+z^{2}}}\\p(x,y,z)&={\frac {-A^{2}B}{\left(r^{2}+x^{2}+y^{2}+z^{2}\right)^{3}}}\\\mathbf {u} (x,y,z)&={\frac {A}{\left(r^{2}+x^{2}+y^{2}+z^{2}\right)^{2}}}{\begin{pmatrix}2(-ry+xz)\\2(rx+yz)\\r^{2}-x^{2}-y^{2}+z^{2}\end{pmatrix}}\\g&=0\\\mu &=0\end{aligned}}}

for arbitrary constantsA{\textstyle A} andB{\textstyle B}. This is a solution in a non-viscous gas (compressible fluid) whose density, velocities and pressure goes to zero far from the origin. (Note this is not a solution to the Clay Millennium problem because that refers to incompressible fluids whereρ{\textstyle \rho } is a constant, and neither does it deal with the uniqueness of the Navier–Stokes equations with respect to anyturbulence properties.) It is also worth pointing out that the components of the velocity vector are exactly those from thePythagorean quadruple parametrization. Other choices of density and pressure are possible with the same velocity field:

Other choices of density and pressure

Another choice of pressure and density with the same velocity vector above is one where the pressure and density fall to zero at the origin and are highest in the central loop atz = 0,x2 +y2 =r2:ρ(x,y,z)=20B(x2+y2)(r2+x2+y2+z2)3p(x,y,z)=−A2B(r2+x2+y2+z2)4+−4A2B(x2+y2)(r2+x2+y2+z2)5.{\displaystyle {\begin{aligned}\rho (x,y,z)&={\frac {20B\left(x^{2}+y^{2}\right)}{\left(r^{2}+x^{2}+y^{2}+z^{2}\right)^{3}}}\\p(x,y,z)&={\frac {-A^{2}B}{\left(r^{2}+x^{2}+y^{2}+z^{2}\right)^{4}}}+{\frac {-4A^{2}B\left(x^{2}+y^{2}\right)}{\left(r^{2}+x^{2}+y^{2}+z^{2}\right)^{5}}}.\end{aligned}}}

In fact in general there are simple solutions for any polynomial functionf where the density is:ρ(x,y,z)=1r2+x2+y2+z2f(x2+y2(r2+x2+y2+z2)2).{\displaystyle \rho (x,y,z)={\frac {1}{r^{2}+x^{2}+y^{2}+z^{2}}}f\left({\frac {x^{2}+y^{2}}{\left(r^{2}+x^{2}+y^{2}+z^{2}\right)^{2}}}\right).}

Viscous three-dimensional periodic solutions

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Two examples of periodic fully-three-dimensional viscous solutions are described in.[40]These solutions are defined on a three-dimensionaltorusT3=[0,L]3{\displaystyle \mathbb {T} ^{3}=[0,L]^{3}} and are characterized by positive and negativehelicity respectively.The solution with positive helicity is given by:ux=4233U0[sin⁡(kx−π/3)cos⁡(ky+π/3)sin⁡(kz+π/2)−cos⁡(kz−π/3)sin⁡(kx+π/3)sin⁡(ky+π/2)]e−3νk2tuy=4233U0[sin⁡(ky−π/3)cos⁡(kz+π/3)sin⁡(kx+π/2)−cos⁡(kx−π/3)sin⁡(ky+π/3)sin⁡(kz+π/2)]e−3νk2tuz=4233U0[sin⁡(kz−π/3)cos⁡(kx+π/3)sin⁡(ky+π/2)−cos⁡(ky−π/3)sin⁡(kz+π/3)sin⁡(kx+π/2)]e−3νk2t{\displaystyle {\begin{aligned}u_{x}&={\frac {4{\sqrt {2}}}{3{\sqrt {3}}}}\,U_{0}\left[\,\sin(kx-\pi /3)\cos(ky+\pi /3)\sin(kz+\pi /2)-\cos(kz-\pi /3)\sin(kx+\pi /3)\sin(ky+\pi /2)\,\right]e^{-3\nu k^{2}t}\\u_{y}&={\frac {4{\sqrt {2}}}{3{\sqrt {3}}}}\,U_{0}\left[\,\sin(ky-\pi /3)\cos(kz+\pi /3)\sin(kx+\pi /2)-\cos(kx-\pi /3)\sin(ky+\pi /3)\sin(kz+\pi /2)\,\right]e^{-3\nu k^{2}t}\\u_{z}&={\frac {4{\sqrt {2}}}{3{\sqrt {3}}}}\,U_{0}\left[\,\sin(kz-\pi /3)\cos(kx+\pi /3)\sin(ky+\pi /2)-\cos(ky-\pi /3)\sin(kz+\pi /3)\sin(kx+\pi /2)\,\right]e^{-3\nu k^{2}t}\end{aligned}}}wherek=2π/L{\displaystyle k=2\pi /L} is the wave number and the velocity components are normalized so that the average kinetic energy per unit of mass isU02/2{\displaystyle U_{0}^{2}/2} att=0{\displaystyle t=0}.The pressure field is obtained from the velocity field asp=p0−ρ0‖u‖2/2{\displaystyle p=p_{0}-\rho _{0}\|{\boldsymbol {u}}\|^{2}/2} (wherep0{\displaystyle p_{0}} andρ0{\displaystyle \rho _{0}} are reference values for the pressure and density fields respectively).Since both the solutions belong to the class ofBeltrami flow, the vorticity field is parallel to the velocity and, for the case with positive helicity, is given byω=3ku{\displaystyle \omega ={\sqrt {3}}\,k\,{\boldsymbol {u}}}. These solutions can be regarded as a generalization in three dimensions of the classic two-dimensional Taylor-GreenTaylor–Green vortex.

Wyld diagrams

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Wyld diagrams are bookkeepinggraphs that correspond to the Navier–Stokes equations via aperturbation expansion of the fundamentalcontinuum mechanics. Similar to theFeynman diagrams inquantum field theory, these diagrams are an extension ofMstislav Keldysh's technique for nonequilibrium processes in fluid dynamics.[citation needed] In other words, these diagrams assigngraphs to the (often)turbulent phenomena in turbulent fluids by allowingcorrelated and interacting fluid particles to obeystochastic processes associated topseudo-randomfunctions inprobability distributions.[41]

Representations in 3D

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Note that the formulas in this section make use of the single-line notation for partial derivatives, where, e.g.∂xu{\textstyle \partial _{x}u} means the partial derivative ofu{\textstyle u} with respect tox{\textstyle x}, and∂y2fθ{\textstyle \partial _{y}^{2}f_{\theta }} means the second-order partial derivative offθ{\textstyle f_{\theta }} with respect toy{\textstyle y}.

A 2022 paper provides a less costly, dynamical and recurrent solution of the Navier-Stokes equation for 3D turbulent fluid flows. On suitably short time scales, the dynamics of turbulence is deterministic.[42]

Cartesian coordinates

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From the general form of the Navier–Stokes, with the velocity vector expanded asu=(ux,uy,uz){\textstyle \mathbf {u} =(u_{x},u_{y},u_{z})}, sometimes respectively namedu{\textstyle u},v{\textstyle v},w{\textstyle w}, we may write the vector equation explicitly,x: ρ(∂tux+ux∂xux+uy∂yux+uz∂zux)=−∂xp+μ(∂x2ux+∂y2ux+∂z2ux)+13μ ∂x(∂xux+∂yuy+∂zuz)+ρgx{\displaystyle {\begin{aligned}x:\ &\rho \left({\partial _{t}u_{x}}+u_{x}\,{\partial _{x}u_{x}}+u_{y}\,{\partial _{y}u_{x}}+u_{z}\,{\partial _{z}u_{x}}\right)\\&\quad =-\partial _{x}p+\mu \left({\partial _{x}^{2}u_{x}}+{\partial _{y}^{2}u_{x}}+{\partial _{z}^{2}u_{x}}\right)+{\frac {1}{3}}\mu \ \partial _{x}\left({\partial _{x}u_{x}}+{\partial _{y}u_{y}}+{\partial _{z}u_{z}}\right)+\rho g_{x}\\\end{aligned}}}y: ρ(∂tuy+ux∂xuy+uy∂yuy+uz∂zuy)=−∂yp+μ(∂x2uy+∂y2uy+∂z2uy)+13μ ∂y(∂xux+∂yuy+∂zuz)+ρgy{\displaystyle {\begin{aligned}y:\ &\rho \left({\partial _{t}u_{y}}+u_{x}{\partial _{x}u_{y}}+u_{y}{\partial _{y}u_{y}}+u_{z}{\partial _{z}u_{y}}\right)\\&\quad =-{\partial _{y}p}+\mu \left({\partial _{x}^{2}u_{y}}+{\partial _{y}^{2}u_{y}}+{\partial _{z}^{2}u_{y}}\right)+{\frac {1}{3}}\mu \ \partial _{y}\left({\partial _{x}u_{x}}+{\partial _{y}u_{y}}+{\partial _{z}u_{z}}\right)+\rho g_{y}\\\end{aligned}}}z: ρ(∂tuz+ux∂xuz+uy∂yuz+uz∂zuz)=−∂zp+μ(∂x2uz+∂y2uz+∂z2uz)+13μ ∂z(∂xux+∂yuy+∂zuz)+ρgz.{\displaystyle {\begin{aligned}z:\ &\rho \left({\partial _{t}u_{z}}+u_{x}{\partial _{x}u_{z}}+u_{y}{\partial _{y}u_{z}}+u_{z}{\partial _{z}u_{z}}\right)\\&\quad =-{\partial _{z}p}+\mu \left({\partial _{x}^{2}u_{z}}+{\partial _{y}^{2}u_{z}}+{\partial _{z}^{2}u_{z}}\right)+{\frac {1}{3}}\mu \ \partial _{z}\left({\partial _{x}u_{x}}+{\partial _{y}u_{y}}+{\partial _{z}u_{z}}\right)+\rho g_{z}.\end{aligned}}}

Note that gravity has been accounted for as a body force, and the values ofgx{\textstyle g_{x}},gy{\textstyle g_{y}},gz{\textstyle g_{z}} will depend on the orientation of gravity with respect to the chosen set of coordinates.

The continuity equation reads:∂tρ+∂x(ρux)+∂y(ρuy)+∂z(ρuz)=0.{\displaystyle \partial _{t}\rho +\partial _{x}(\rho u_{x})+\partial _{y}(\rho u_{y})+\partial _{z}(\rho u_{z})=0.}

When the flow is incompressible,ρ{\textstyle \rho } does not change for any fluid particle, and itsmaterial derivative vanishes:DρDt=0{\textstyle {\frac {\mathrm {D} \rho }{\mathrm {D} t}}=0}. The continuity equation is reduced to:∂xux+∂yuy+∂zuz=0.{\displaystyle \partial _{x}u_{x}+\partial _{y}u_{y}+\partial _{z}u_{z}=0.}

Thus, for the incompressible version of the Navier–Stokes equation the second part of the viscous terms fall away (seeIncompressible flow).

This system of four equations comprises the most commonly used and studied form. Though comparatively more compact than other representations, this is still anonlinear system ofpartial differential equations for which solutions are difficult to obtain.

Cylindrical coordinates

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A change of variables on the Cartesian equations will yield[16] the following momentum equations forr{\textstyle r},ϕ{\textstyle \phi }, andz{\textstyle z}[43]r: ρ(∂tur+ur∂rur+uφr∂φur+uz∂zur−uφ2r)=−∂rp+μ(1r∂r(r∂rur)+1r2∂φ2ur+∂z2ur−urr2−2r2∂φuφ)+13μ∂r(1r∂r(rur)+1r∂φuφ+∂zuz)+ρgr{\displaystyle {\begin{aligned}r:\ &\rho \left({\partial _{t}u_{r}}+u_{r}{\partial _{r}u_{r}}+{\frac {u_{\varphi }}{r}}{\partial _{\varphi }u_{r}}+u_{z}{\partial _{z}u_{r}}-{\frac {u_{\varphi }^{2}}{r}}\right)\\&\quad =-{\partial _{r}p}\\&\qquad +\mu \left({\frac {1}{r}}\partial _{r}\left(r{\partial _{r}u_{r}}\right)+{\frac {1}{r^{2}}}{\partial _{\varphi }^{2}u_{r}}+{\partial _{z}^{2}u_{r}}-{\frac {u_{r}}{r^{2}}}-{\frac {2}{r^{2}}}{\partial _{\varphi }u_{\varphi }}\right)\\&\qquad +{\frac {1}{3}}\mu \partial _{r}\left({\frac {1}{r}}{\partial _{r}\left(ru_{r}\right)}+{\frac {1}{r}}{\partial _{\varphi }u_{\varphi }}+{\partial _{z}u_{z}}\right)\\&\qquad +\rho g_{r}\\[8px]\end{aligned}}}φ: ρ(∂tuφ+ur∂ruφ+uφr∂φuφ+uz∂zuφ+uruφr)=−1r∂φp+μ(1r ∂r(r∂ruφ)+1r2∂φ2uφ+∂z2uφ−uφr2+2r2∂φur)+13μ1r∂φ(1r∂r(rur)+1r∂φuφ+∂zuz)+ρgφ{\displaystyle {\begin{aligned}\varphi :\ &\rho \left({\partial _{t}u_{\varphi }}+u_{r}{\partial _{r}u_{\varphi }}+{\frac {u_{\varphi }}{r}}{\partial _{\varphi }u_{\varphi }}+u_{z}{\partial _{z}u_{\varphi }}+{\frac {u_{r}u_{\varphi }}{r}}\right)\\&\quad =-{\frac {1}{r}}{\partial _{\varphi }p}\\&\qquad +\mu \left({\frac {1}{r}}\ \partial _{r}\left(r{\partial _{r}u_{\varphi }}\right)+{\frac {1}{r^{2}}}{\partial _{\varphi }^{2}u_{\varphi }}+{\partial _{z}^{2}u_{\varphi }}-{\frac {u_{\varphi }}{r^{2}}}+{\frac {2}{r^{2}}}{\partial _{\varphi }u_{r}}\right)\\&\qquad +{\frac {1}{3}}\mu {\frac {1}{r}}\partial _{\varphi }\left({\frac {1}{r}}{\partial _{r}\left(ru_{r}\right)}+{\frac {1}{r}}{\partial _{\varphi }u_{\varphi }}+{\partial _{z}u_{z}}\right)\\&\qquad +\rho g_{\varphi }\\[8px]\end{aligned}}}z: ρ(∂tuz+ur∂ruz+uφr∂φuz+uz∂zuz)=−∂zp+μ(1r∂r(r∂ruz)+1r2∂φ2uz+∂z2uz)+13μ∂z(1r∂r(rur)+1r∂φuφ+∂zuz)+ρgz.{\displaystyle {\begin{aligned}z:\ &\rho \left({\partial _{t}u_{z}}+u_{r}{\partial _{r}u_{z}}+{\frac {u_{\varphi }}{r}}{\partial _{\varphi }u_{z}}+u_{z}{\partial _{z}u_{z}}\right)\\&\quad =-{\partial _{z}p}\\&\qquad +\mu \left({\frac {1}{r}}\partial _{r}\left(r{\partial _{r}u_{z}}\right)+{\frac {1}{r^{2}}}{\partial _{\varphi }^{2}u_{z}}+{\partial _{z}^{2}u_{z}}\right)\\&\qquad +{\frac {1}{3}}\mu \partial _{z}\left({\frac {1}{r}}{\partial _{r}\left(ru_{r}\right)}+{\frac {1}{r}}{\partial _{\varphi }u_{\varphi }}+{\partial _{z}u_{z}}\right)\\&\qquad +\rho g_{z}.\end{aligned}}}

The gravity components will generally not be constants, however for most applications either the coordinates are chosen so that the gravity components are constant or else it is assumed that gravity is counteracted by a pressure field (for example, flow in horizontal pipe is treated normally without gravity and without a vertical pressure gradient). The continuity equation is:∂tρ+1r∂r(ρrur)+1r∂φ(ρuφ)+∂z(ρuz)=0.{\displaystyle {\partial _{t}\rho }+{\frac {1}{r}}\partial _{r}\left(\rho ru_{r}\right)+{\frac {1}{r}}{\partial _{\varphi }\left(\rho u_{\varphi }\right)}+{\partial _{z}\left(\rho u_{z}\right)}=0.}

This cylindrical representation of the incompressible Navier–Stokes equations is the second most commonly seen (the first being Cartesian above). Cylindrical coordinates are chosen to take advantage of symmetry, so that a velocity component can disappear. A very common case is axisymmetric flow with the assumption of no tangential velocity (uϕ=0{\textstyle u_{\phi }=0}), and the remaining quantities are independent ofϕ{\textstyle \phi }:ρ(∂tur+ur∂rur+uz∂zur)=−∂rp+μ(1r∂r(r∂rur)+∂z2ur−urr2)+ρgrρ(∂tuz+ur∂ruz+uz∂zuz)=−∂zp+μ(1r∂r(r∂ruz)+∂z2uz)+ρgz1r∂r(rur)+∂zuz=0.{\displaystyle {\begin{aligned}\rho \left({\partial _{t}u_{r}}+u_{r}{\partial _{r}u_{r}}+u_{z}{\partial _{z}u_{r}}\right)&=-{\partial _{r}p}+\mu \left({\frac {1}{r}}\partial _{r}\left(r{\partial _{r}u_{r}}\right)+{\partial _{z}^{2}u_{r}}-{\frac {u_{r}}{r^{2}}}\right)+\rho g_{r}\\\rho \left({\partial _{t}u_{z}}+u_{r}{\partial _{r}u_{z}}+u_{z}{\partial _{z}u_{z}}\right)&=-{\partial _{z}p}+\mu \left({\frac {1}{r}}\partial _{r}\left(r{\partial _{r}u_{z}}\right)+{\partial _{z}^{2}u_{z}}\right)+\rho g_{z}\\{\frac {1}{r}}\partial _{r}\left(ru_{r}\right)+{\partial _{z}u_{z}}&=0.\end{aligned}}}

Spherical coordinates

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Inspherical coordinates, ther{\textstyle r},ϕ{\textstyle \phi }, andθ{\textstyle \theta } momentum equations are[16] (note the convention used:θ{\textstyle \theta } is polar angle, orcolatitude,[44]0≤θ≤π{\textstyle 0\leq \theta \leq \pi }):r: ρ(∂tur+ur∂rur+uφrsin⁡θ∂φur+uθr∂θur−uφ2+uθ2r)=−∂rp+μ(1r2∂r(r2∂rur)+1r2sin2⁡θ∂φ2ur+1r2sin⁡θ∂θ(sin⁡θ∂θur)−2ur+∂θuθ+uθcot⁡θr2−2r2sin⁡θ∂φuφ)+13μ∂r(1r2∂r(r2ur)+1rsin⁡θ∂θ(uθsin⁡θ)+1rsin⁡θ∂φuφ)+ρgr{\displaystyle {\begin{aligned}r:\ &\rho \left({\partial _{t}u_{r}}+u_{r}{\partial _{r}u_{r}}+{\frac {u_{\varphi }}{r\sin \theta }}{\partial _{\varphi }u_{r}}+{\frac {u_{\theta }}{r}}{\partial _{\theta }u_{r}}-{\frac {u_{\varphi }^{2}+u_{\theta }^{2}}{r}}\right)\\&\quad =-{\partial _{r}p}\\&\qquad +\mu \left({\frac {1}{r^{2}}}\partial _{r}\left(r^{2}{\partial _{r}u_{r}}\right)+{\frac {1}{r^{2}\sin ^{2}\theta }}{\partial _{\varphi }^{2}u_{r}}+{\frac {1}{r^{2}\sin \theta }}\partial _{\theta }\left(\sin \theta {\partial _{\theta }u_{r}}\right)-2{\frac {u_{r}+{\partial _{\theta }u_{\theta }}+u_{\theta }\cot \theta }{r^{2}}}-{\frac {2}{r^{2}\sin \theta }}{\partial _{\varphi }u_{\varphi }}\right)\\&\qquad +{\frac {1}{3}}\mu \partial _{r}\left({\frac {1}{r^{2}}}\partial _{r}\left(r^{2}u_{r}\right)+{\frac {1}{r\sin \theta }}\partial _{\theta }\left(u_{\theta }\sin \theta \right)+{\frac {1}{r\sin \theta }}{\partial _{\varphi }u_{\varphi }}\right)\\&\qquad +\rho g_{r}\\[8px]\end{aligned}}}φ: ρ(∂tuφ+ur∂ruφ+uφrsin⁡θ∂φuφ+uθr∂θuφ+uruφ+uφuθcot⁡θr)=−1rsin⁡θ∂φp+μ(1r2∂r(r2∂ruφ)+1r2sin2⁡θ∂φ2uφ+1r2sin⁡θ∂θ(sin⁡θ∂θuφ)+2sin⁡θ∂φur+2cos⁡θ∂φuθ−uφr2sin2⁡θ)+13μ1rsin⁡θ∂φ(1r2∂r(r2ur)+1rsin⁡θ∂θ(uθsin⁡θ)+1rsin⁡θ∂φuφ)+ρgφ{\displaystyle {\begin{aligned}\varphi :\ &\rho \left({\partial _{t}u_{\varphi }}+u_{r}{\partial _{r}u_{\varphi }}+{\frac {u_{\varphi }}{r\sin \theta }}{\partial _{\varphi }u_{\varphi }}+{\frac {u_{\theta }}{r}}{\partial _{\theta }u_{\varphi }}+{\frac {u_{r}u_{\varphi }+u_{\varphi }u_{\theta }\cot \theta }{r}}\right)\\&\quad =-{\frac {1}{r\sin \theta }}{\partial _{\varphi }p}\\&\qquad +\mu \left({\frac {1}{r^{2}}}\partial _{r}\left(r^{2}{\partial _{r}u_{\varphi }}\right)+{\frac {1}{r^{2}\sin ^{2}\theta }}{\partial _{\varphi }^{2}u_{\varphi }}+{\frac {1}{r^{2}\sin \theta }}\partial _{\theta }\left(\sin \theta {\partial _{\theta }u_{\varphi }}\right)+{\frac {2\sin \theta {\partial _{\varphi }u_{r}}+2\cos \theta {\partial _{\varphi }u_{\theta }}-u_{\varphi }}{r^{2}\sin ^{2}\theta }}\right)\\&\qquad +{\frac {1}{3}}\mu {\frac {1}{r\sin \theta }}\partial _{\varphi }\left({\frac {1}{r^{2}}}\partial _{r}\left(r^{2}u_{r}\right)+{\frac {1}{r\sin \theta }}\partial _{\theta }\left(u_{\theta }\sin \theta \right)+{\frac {1}{r\sin \theta }}{\partial _{\varphi }u_{\varphi }}\right)\\&\qquad +\rho g_{\varphi }\\[8px]\end{aligned}}}θ: ρ(∂tuθ+ur∂ruθ+uφrsin⁡θ∂φuθ+uθr∂θuθ+uruθ−uφ2cot⁡θr)=−1r∂θp+μ(1r2∂r(r2∂ruθ)+1r2sin2⁡θ∂φ2uθ+1r2sin⁡θ∂θ(sin⁡θ∂θuθ)+2r2∂θur−uθ+2cos⁡θ∂φuφr2sin2⁡θ)+13μ1r∂θ(1r2∂r(r2ur)+1rsin⁡θ∂θ(uθsin⁡θ)+1rsin⁡θ∂φuφ)+ρgθ.{\displaystyle {\begin{aligned}\theta :\ &\rho \left({\partial _{t}u_{\theta }}+u_{r}{\partial _{r}u_{\theta }}+{\frac {u_{\varphi }}{r\sin \theta }}{\partial _{\varphi }u_{\theta }}+{\frac {u_{\theta }}{r}}{\partial _{\theta }u_{\theta }}+{\frac {u_{r}u_{\theta }-u_{\varphi }^{2}\cot \theta }{r}}\right)\\&\quad =-{\frac {1}{r}}{\partial _{\theta }p}\\&\qquad +\mu \left({\frac {1}{r^{2}}}\partial _{r}\left(r^{2}{\partial _{r}u_{\theta }}\right)+{\frac {1}{r^{2}\sin ^{2}\theta }}{\partial _{\varphi }^{2}u_{\theta }}+{\frac {1}{r^{2}\sin \theta }}\partial _{\theta }\left(\sin \theta {\partial _{\theta }u_{\theta }}\right)+{\frac {2}{r^{2}}}{\partial _{\theta }u_{r}}-{\frac {u_{\theta }+2\cos \theta {\partial _{\varphi }u_{\varphi }}}{r^{2}\sin ^{2}\theta }}\right)\\&\qquad +{\frac {1}{3}}\mu {\frac {1}{r}}\partial _{\theta }\left({\frac {1}{r^{2}}}\partial _{r}\left(r^{2}u_{r}\right)+{\frac {1}{r\sin \theta }}\partial _{\theta }\left(u_{\theta }\sin \theta \right)+{\frac {1}{r\sin \theta }}{\partial _{\varphi }u_{\varphi }}\right)\\&\qquad +\rho g_{\theta }.\end{aligned}}}

Mass continuity will read:∂tρ+1r2∂r(ρr2ur)+1rsin⁡θ∂φ(ρuφ)+1rsin⁡θ∂θ(sin⁡θρuθ)=0.{\displaystyle {\partial _{t}\rho }+{\frac {1}{r^{2}}}\partial _{r}\left(\rho r^{2}u_{r}\right)+{\frac {1}{r\sin \theta }}{\partial _{\varphi }(\rho u_{\varphi })}+{\frac {1}{r\sin \theta }}\partial _{\theta }\left(\sin \theta \rho u_{\theta }\right)=0.}

These equations could be (slightly) compacted by, for example, factoring1r2{\textstyle {\frac {1}{r^{2}}}} from the viscous terms. However, doing so would undesirably alter the structure of the Laplacian and other quantities.

See also

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Citations

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  1. ^McLean, Doug (2012)."Continuum Fluid Mechanics and the Navier-Stokes Equations".Understanding Aerodynamics: Arguing from the Real Physics. John Wiley & Sons. pp. 13–78.ISBN 978-1-119-96751-4.The main relationships comprising the NS equations are the basic conservation laws for mass, momentum, and energy. To have a complete equation set we also need an equation of state relating temperature, pressure, and density...
  2. ^"Millennium Prize Problems—Navier–Stokes Equation",claymath.org, Clay Mathematics Institute, March 27, 2017, archived fromthe original on 2015-12-22, retrieved2017-04-02
  3. ^Fefferman, Charles L."Existence and smoothness of the Navier–Stokes equation"(PDF).claymath.org. Clay Mathematics Institute. Archived fromthe original(PDF) on 2015-04-15. Retrieved2017-04-02.
  4. ^Batchelor (1967), pp. 137 & 142
  5. ^abcdBatchelor (1967), pp. 142–148
  6. ^Chorin, Alexandre E.; Marsden, Jerrold E. (1993).A Mathematical Introduction to Fluid Mechanics. p. 33.
  7. ^Bird, Stewart, Lightfoot, Transport Phenomena, 1st ed., 1960, eq. (3.2-11a)
  8. ^abBatchelor (1967), p. 165
  9. ^abLandau, Lev Davidovich, and Evgenii Mikhailovich Lifshitz. Fluid mechanics: Landau And Lifshitz: course of theoretical physics, Volume 6. Vol. 6. Elsevier, 2013.
  10. ^Landau & Lifshitz (1987), pp. 44–45, 196
  11. ^White (2006), p. 67
  12. ^Stokes, G. G. (1845). On the theories of the internal friction of fluids in motion, and of the equilibrium and motion of elastic solids.
  13. ^Vincenti, W. G., Kruger Jr., C. H. (1975). Introduction to physical gas dynamic. Introduction to physical gas dynamics/Huntington.
  14. ^Batchelor (1967), pp. 147 & 154
  15. ^Batchelor (1967), p. 75
  16. ^abcAcheson (1990)
  17. ^Abdulkadirov, Ruslan; Lyakhov, Pavel (2022-02-22)."Estimates of Mild Solutions of Navier–Stokes Equations in Weak Herz-Type Besov–Morrey Spaces".Mathematics.10 (5): 680.doi:10.3390/math10050680.ISSN 2227-7390.
  18. ^Batchelor (1967), pp. 21 & 147
  19. ^Temam, Roger (2001),Navier–Stokes Equations, Theory and Numerical Analysis, AMS Chelsea, pp. 107–112
  20. ^abcdefghijklQuarteroni, Alfio (2014-04-25).Numerical models for differential problems (Second ed.). Springer.ISBN 978-88-470-5522-3.
  21. ^Holdeman, J. T. (2010), "A Hermite finite element method for incompressible fluid flow",Int. J. Numer. Methods Fluids,64 (4):376–408,Bibcode:2010IJNMF..64..376H,doi:10.1002/fld.2154,S2CID 119882803
  22. ^Holdeman, J. T.; Kim, J. W. (2010), "Computation of incompressible thermal flows using Hermite finite elements",Comput. Meth. Appl. Mech. Eng.,199 (49–52):3297–3304,Bibcode:2010CMAME.199.3297H,doi:10.1016/j.cma.2010.06.036
  23. ^Potter, M.; Wiggert, D. C. (2008).Fluid Mechanics. Schaum's Outlines. McGraw-Hill.ISBN 978-0-07-148781-8.
  24. ^Aris, R. (1989).Vectors, Tensors, and the basic Equations of Fluid Mechanics. Dover Publications.ISBN 0-486-66110-5.
  25. ^Parker, C. B. (1994).McGraw Hill Encyclopaedia of Physics (2nd ed.). McGraw-Hill.ISBN 0-07-051400-3.
  26. ^Encyclopaedia of Physics (2nd Edition),R.G. Lerner, G.L. Trigg, VHC publishers, 1991,ISBN 3-527-26954-1 (Verlagsgesellschaft),ISBN 0-89573-752-3 (VHC Inc.)
  27. ^Gorban, A.N.; Karlin, I. V. (2016),"Beyond Navier–Stokes equations: capillarity of ideal gas",Contemporary Physics (Review article),58 (1):70–90,arXiv:1702.00831,Bibcode:2017ConPh..58...70G,doi:10.1080/00107514.2016.1256123,S2CID 55317543
  28. ^Cercignani, C. (2002), "The Boltzmann equation and fluid dynamics", in Friedlander, S.; Serre, D. (eds.),Handbook of mathematical fluid dynamics, vol. 1, Amsterdam: North-Holland, pp. 1–70,ISBN 978-0-444-50330-5
  29. ^Nie, X.B.; Chen, S.Y.; Robbins, M.O. (2004),"A continuum and molecular dynamics hybrid method for micro-and nano-fluid flow",Journal of Fluid Mechanics (Research article),500:55–64,Bibcode:2004JFM...500...55N,doi:10.1017/S0022112003007225,S2CID 122867563
  30. ^Öttinger, H.C. (2012),Stochastic processes in polymeric fluids, Berlin, Heidelberg: Springer Science & Business Media,doi:10.1007/978-3-642-58290-5,ISBN 978-3-540-58353-0
  31. ^abShah, Tasneem Mohammad (1972). "Analysis of the multigrid method".NASA Sti/Recon Technical Report N.91: 23418.Bibcode:1989STIN...9123418S.
  32. ^Wang, C. Y. (1991), "Exact solutions of the steady-state Navier–Stokes equations",Annual Review of Fluid Mechanics,23:159–177,Bibcode:1991AnRFM..23..159W,doi:10.1146/annurev.fl.23.010191.001111
  33. ^Landau & Lifshitz (1987), pp. 75–88
  34. ^Ethier, C. R.; Steinman, D. A. (1994), "Exact fully 3D Navier–Stokes solutions for benchmarking",International Journal for Numerical Methods in Fluids,19 (5):369–375,Bibcode:1994IJNMF..19..369E,doi:10.1002/fld.1650190502
  35. ^Barna, I.F. (2011)."Self-Similar Solutions of Three-Dimensional Navier–Stokes Equation".Communications in Theoretical Physics.56 (4):745–750.arXiv:1102.5504.Bibcode:2011CoTPh..56..745I.doi:10.1088/0253-6102/56/4/25.
  36. ^Barna, I.F.; Mátyás, L. (2014)."Analytic solutions for the three-dimensional compressible Navier-Stokes equation".Fluid Dynamics Research.46 (5) 055508.arXiv:1309.0703.Bibcode:2014FlDyR..46e5508B.doi:10.1088/0169-5983/46/5/055508.
  37. ^"Navier Stokes Equations".www.claudino.webs.com. Archived fromthe original on 2015-06-19. Retrieved2023-03-11.
  38. ^Ladyzhenskaya, O. A. (1969),The Mathematical Theory of viscous Incompressible Flow (2nd ed.), p. preface, xi
  39. ^Kamchatno, A. M. (1982),"Topological solitons in magnetohydrodynamics"(PDF),Soviet Journal of Experimental and Theoretical Physics,55 (1): 69,Bibcode:1982JETP...55...69K,archived(PDF) from the original on 2016-01-28
  40. ^Antuono, M. (2020), "Tri-periodic fully three-dimensional analytic solutions for the Navier–Stokes equations",Journal of Fluid Mechanics,890 A23,Bibcode:2020JFM...890A..23A,doi:10.1017/jfm.2020.126,S2CID 216463266
  41. ^McComb, W. D. (2008),Renormalization methods: A guide for beginners, Oxford University Press, pp. 121–128,ISBN 978-0-19-923652-7
  42. ^Georgia Institute of Technology (August 29, 2022)."Physicists uncover new dynamical framework for turbulence".Proceedings of the National Academy of Sciences of the United States of America.119 (34) e2120665119.Phys.org.doi:10.1073/pnas.2120665119.PMC 9407532.PMID 35984901.S2CID 251693676.
  43. ^de' Michieli Vitturi, Mattia,Navier–Stokes equations in cylindrical coordinates, retrieved2016-12-26
  44. ^Eric W. Weisstein (2005-10-26),Spherical Coordinates,MathWorld, retrieved2008-01-22

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