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Navier–Stokes equations

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These are partial differential equations that describe the motion of fluid.

PDEs

Navier–Stokes equations for incompressible flow describe the development of its velocity $v$: $$\rho \left(\frac{\partial v}{\partial t} + v \cdot \nabla v\right) = -\nabla p + \mu \nabla^2 v + f.$$ Here

  • $\rho$ is the density,
  • $p$ is the pressure,
  • $\nabla$ is the gradient,
  • $\nabla^2$ is the (vector) Laplacian,
  • $\mu$ is the (constant) dynamic viscosity,
  • $f$ is the external forces (per unit volume), such as gravity or centrifugal force.

There is also the incompressability condition: $$\nabla \cdot v =0.$$

From Wikipedia: $$\overbrace{\rho \Big( \underbrace{\frac{\partial \mathbf{v}}{\partial t}}_{ \begin{smallmatrix} \text{Unsteady}\\ \text{acceleration} \end{smallmatrix}} + \underbrace{\mathbf{v} \cdot \nabla \mathbf{v}}_{ \begin{smallmatrix} \text{Convective} \\ \text{acceleration} \end{smallmatrix}}\Big)}^{\text{Inertia (per volume)}} = \overbrace{\underbrace{-\nabla p}_{ \begin{smallmatrix} \text{Pressure} \\ \text{gradient} \end{smallmatrix}} + \underbrace{\mu \nabla^2 \mathbf{v}}_{\text{Viscosity}}}^{\text{Divergence of stress}} + \underbrace{\mathbf{f}}_{ \begin{smallmatrix} \text{Other} \\ \text{body} \\ \text{forces} \end{smallmatrix}}. $$

Switch to differential forms

Now in terms of differential forms, the velocity, $v$ is, as always, a $1$-form.

According to Differential Forms, Fluids, And Finite Models by Scott O. Wilson, the Navier-Stokes equations on the Riemannian manifold $R^3$ take form: $$\frac{\partial v}{\partial t} = − \star (v ∧ \star dv) + \frac{1}{2}d||v||^2 − dp + \mu d^∗dv. $$ And, of course, $$d^∗v = 0.$$

We are supposed to use the fundamental correspondence here. Compare the terms with the PDE: $$\frac{\partial v}{\partial t} + v \cdot \nabla v = -\nabla p + \mu \nabla^2 v.$$ $$\frac{\partial v}{\partial t} + \star (v ∧ \star dv) - \frac{1}{2}d||v||^2 = − dp + \mu d^∗dv. $$ Just $\rho$ and $f$ are dropped...

Simplification

Now the second equation implies that also $$d^∗\frac{\partial v}{\partial t} = 0.$$ The left hand side is in the kernel of $d^∗$. Hence so is the-right hand side: $$d^∗RHS = 0.$$

Using the Hodge decomposition of differential forms we conclude that the $d$-exact part of the right-hand side must be zero, as follows. The decomposition of RHS: $$RHS = A + B + C,$$ where $A$ is $d$-exact: $A = da$ and $B$ is $d^*$-exact (co-exact): $B=d^*b$, and $C=d^*c$ is both (harmonic). The three are pair-wise orthogonal.

Om particular, in the RHS:

  • $− \star (v ∧ \star dv)$ has parts of both kinds,
  • $\frac{1}{2}d||v||^2 − dp$ is $d$-exact, and
  • $− \mu d^∗dv$ is $d^*$-exact.

Apply the second equation: $$0 = d^∗RHS = d^∗(A + B + C) = d^∗(A + d^*b + C) = d^∗A + d^∗d^*b + d^∗C = d^∗(A + C).$$ Then, $A + C$ is co-closed. But it's also exact. Now we use the fact that exact forms are orthogonal to co-closed forms. Hence $A + C = 0$. Or, the $d$-exact part of the right-hand side is zero.

In particular, the two middle terms in the RHS are $0$: $$RHS = − \star (v ∧ \star dv) + \mu d^∗dv.$$ And the exact part of the first term is zero. The last term has none.

Then what's left in the RHS is what's orthogonal, the co-closed part. Let $\pi$ be the (orthogonal) projection on the kernel of $d^∗$, $\it{ker}d^∗=$ the set of all co-closed forms. Then the equation takes the form $$\frac{\partial v}{\partial t} = \pi\Big( − \star (v ∧ \star dv)\Big) + \mu d^∗dv.$$

Observations

First, suppose in addition that $v$ is $d$-closed (so harmonic too). Then the right hand side of the last equation is zero, and therefore $v$ is a steady state solution.

Second, observe about the first term: $$< \pi\Big( − \star (v ∧ \star dv)\Big), v> = <\Big( − \star (v ∧ \star dv)\Big), v>$$ $$=<dv, v ∧ v> $$ $$ = 0,$$ since $d$ is the adjoint of $d^*$. Then, if we multiply our equation by $v$, we have $$<\frac{\partial v}{\partial t},v> = <\pi\Big( − \star (v ∧ \star dv)\Big) + \mu d^∗dv,v>$$ and $$\frac{\partial}{\partial t}<v,v> = <\mu d^∗dv,v>,$$ hence $$\frac{\partial}{\partial t}||v||^2 = \mu <d^∗dv,v> = - \mu <dv,dv> = -\mu ||dv||^2 .$$ Therefore the norm is decreasing.

In case of variable viscosity, the last equation is $$\frac{\partial}{\partial t}||v||^2 = <d^∗d\mu v,v> = - <d\mu v,dv> .$$

Note

DARPA. Mathematical Challenge 4: 21st Century Fluids

Classical fluid dynamics and the Navier-Stokes Equation were extraordinarily successful in obtaining quantitative understanding of shock waves, turbulence and solitons, but new methods are needed to tackle complex fluids such as foams, suspensions, gels and liquid crystals.