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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+uu=pu=0

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

Definition

Consider perfect fluid (in-compressible and inviscid) in a domain ΩR3 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(xx)×ω(x)|xx|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=2γstγs=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 γψ=κ exp(i0sτds).
Then: ψ=κ exp(i0sτds), thus: ψψ=κ2, thus |ψ|=κψ|ψ|=exp(i0sτds).

We now explicitly derive Da Rios system:

κt+2τκs+τsκ=0τt+(τ212κ2κssκ)s=0

Since ψ=|ψ| exp(i0sτds) we get that iψt=iκt exp(i0sτds)

{Vortex filament eqn}mapHasimotoNLSEDe-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=γssγss
Thus:
2κκt=2γssγsst=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)Ncos(θ)B

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

(TNB)s=(0κ0κ0τ0τ0)(TNB)dds(TUV)=(0κ1κ2κ100κ200)(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×TssSt=S×Sss (Heisenberg spin chain equation)