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Lecture 03 - Dynamics

Let us consider unbounded, inviscid, incompressible fluid flows. In the absence of external force, the motion of such fluid with unit density is described by the Euler equations:

(1)∂tu+u⋅∇u=−∇p∇⋅u=0

where u(x,t) is the velocity and p is the pressure.

Definition

Consider perfect fluid (in-compressible and inviscid) in a domain Ω⊂R3 then vorticity is defined to be:

ω=∇×V

If the curl vanishes then we call it an irrotational fluid.

The velocity u(x) can be determined from the vorticity ω(x) through the Biot-Savart Law:

u(x)=−14π∫(x−x′)×ω(x′)|x−x′|3dx′
Definition

A vortex tube is a tubular region of the fluid that has a much higher vorticity than the surrounding fluid, eg: Smoke rings, whirlpools.

One can avoid the singularity in equation (1) by simply ignoring the nonlocal contribution of the filament and replace the Biot-Savart’s law by a velocity expression that depends only
on the local curvature of the vortex filament. If a vortex filament described by γ(s), where s is an arc length parameter measured along the filament. The equation of a filament is:

γt=γs×γss

Recall that γs=T and γss=Ts=κN
Thus: γt=T×κN=κB, which is known as a binormal flow.

We now contemplate on what we call 'moving curves', γ(s,t), which the curve is not only parameterized by arc-length by also by time. That is the the curves evolve over time.
Consider arc length |∂sγ|:

∂t|γs|2=2γst⋅γs=2γs⋅(γs×γss)s=0.

As a consequence we get the following conserved quantities
Energy: ∫|∂ssγ|2ds=∫κ2ds
Linear Momentum: ∫γ×∂sγ ds
Angular Momentum: ∫γ×(γ×∂sγ) ds
Helicity: ∫κ2τ ds
Total Torsion: ∫τ ds


Let us define a complex curvature quantity, we associate to a curve γ→ψ=κ exp(i∫0sτds′).
Then: ψ∗=κ exp(−i∫0sτds′), thus: ψψ∗=κ2, thus |ψ|=κ⟹ψ|ψ|=exp(i∫0sτds′).

We now explicitly derive Da Rios system:

κt+2τκs+τsκ=0τt+(τ2−12κ2−κssκ)s=0

Since ψ=|ψ| exp(i∫0sτds′) we get that iψt=iκt exp(i∫0sτds′)

{Vortex filament eqn}→mapHasimotoNLSE⟷De-Rios System

2D-Case

γ is a solution for the planar filament equation.

γt=κ22T+κsN

iff κ is a solution to the modified KDV equation:

κt=32κ2κs+κsss
Proof

Consider the relation
κ2=|Ts|2=γss⋅γss
Thus:
2κκt=2γss⋅γsst=2(κN)⋅(κ22T+κsN)ss
=2(κN)⋅(κκsT+κ22Ts+κssN+κsNs)s
⟹κt=32κ2κs+κsss


Consider θ=∫0sτds′
Bishop frame:
(TNB)→(TUV)
Where:

T=TU=Ncos(θ)−sin(θ)BV=sin(θ)N−cos(θ)B

Let us define κ1=κcos(θ) and κ2=κsin(θ) then our Frenet Serret equations change to:

(TNB)s=(0κ0−κ0τ0−τ0)(TNB)⟷dds(TUV)=(0κ1κ2−κ100−κ200)(TUV)

Notice that κB=−κ2U+κ1V, if we are considering a vortex filament then: γt=κB as noted here thus for a filament in the Bishop frame:

γt=−κ2U+κ1V

and since κN=κ1U+κ2V we get: γss=κ1U+κ2V.

End of class after-notes

γt=γs×γssγst=γss×γss+γs×γsssγs=TThus we get that:Tt=T×Tss⟷St=S×Sss (Heisenberg spin chain equation)