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Exact solutions of the Navier–Stokes equations

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—are Poiseuille flow, Couette flow and the oscillatory Stokes boundary layer. But also, more interesting examples, solutions to the full non-linear equations, exist, such as Jeffery–Hamel flow, Von Kármán swirling flow, stagnation point flow, Landau–Squire jet, and Taylor–Green vortex. 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.

For example, in the case of an unbounded planar domain with two-dimensional — incompressible and stationary — flow in polar coordinates (r, φ), the velocity components (u, u) and pressure p are: u r = A r, u φ = B (1 r − r A ν + 1), p = − A 2 + B 2 2 r 2 − 2 B 2 ν r A ν A + B 2 r (2 A ν + 2) 2 A ν + 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}}}

where A and B are arbitrary constants. This solution is valid in the domain r ≥ 1 and for A < −2 ν.

In Cartesian coordinates, when the viscosity is zero (ν = 0), this is: v (x, y) = 1 x 2 + y 2 (A x + B y A y − B x), p (x, y) = − A 2 + B 2 2 (x 2 + y 2) {\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}}}

For example, in the case of an unbounded Euclidean domain with three-dimensional — incompressible, stationary and with zero viscosity (ν = 0) — radial flow in Cartesian coordinates (x, y, z), the velocity vector v and pressure p are: v (x, y, z) = A x 2 + y 2 + z 2 (x y z), p (x, y, z) = − A 2 2 (x 2 + y 2 + z 2). {\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 at x = y = z = 0.