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Lecture 07 - Elastica 1

We are going to be studying 'elastic' curves.

Definition

Bending energy of a curve (which will be a conserved quantity) is defined to be:

El=120Lκ2ds

If we also want to twist the curve then:

E^l=120Lκ2ds+0Lτ2ds

We want to obtain the equations of motion from the bending energy of a curve.


Consider a regular curve in Euclidean space R3, defined on a fixed interval [a1,a2]. (One can do the same on a Riemannian manifold instead by replacing derivatives with covariant derivatives instead).

Let Ω={γ : γ(ai)=αi,γ(ai)=αi}, now consider the subspace of Ω such that:

Ωu={γΩ : ||γ||=1}

Elastica minimizes the bending energy that is:

F=12γκ2ds=12γ||γ||2ds

We are interested in varying F subject to the constraint of the defined subspace, thus we define Fλ(γ)=12||γ||2+Λ(s)(||γ||21)ds, where Λ is the langrange multiplier. Lagrange multiplier says that the minimum of F on Ωu is a stationary point for Fλ for some values of Λ(s).

Derivation of equations of motion

If w be a vector along γ:

Fλ(w)=ϵFλ(γ+ϵw)|ϵ=0

We first compute the Fλ on the right:

Fλ(γ+ϵw)=12a1a2||(γ+ϵw)||2+Λ(s)(||(γ+ϵw)||21)ds

Thus:

ϵFλ(γ+ϵw)|ϵ=0=a1a2(γw+Λ(s)γw)ds=0

Integrating by parts gives us:

a1a2[γ(Λγ)]wds+(γw(Λγγ)w)|a1a2

Thus we obtain our Equations of motion:

E(γ)=γ(Λγ)=0(*)γΛγ=J

Da Rios Equation from the Elastic equations

γ=Tγ=κNγ=κ2T+κN+κτB

Thus () transforms to:

(κ2Λ(s))T+κN+κτB=J

Differentiating gives us:

(3κκΛ)T(κκ3Λκκτ2)N+(κτ+2κτ)B=0

Thus we obtain Λ(s) by projecting the above equation on T.

Λ(s)=32κ2+λ2

where λ is some constant. Thus substituting back we obtain:

J=κ2λ2T+κN+κτB

Thus once again differentiating we obtain our Da Rios equation:

κ+12κ3κτ2λκ2=0,κτ+2κτ=0.

We obtain from the second equation that (κ2τ)=0 implies κ2τ=c1, thus reducing our system to:

κ+12κ3c12κ3λκ2=0.

Notice that 2J=κ2λT+2κN+2κτB, thus:

(**)4||J||2=(κ2λ)2+4κ2+4κ2τ2=a2

Taking u=κ2 then () reduces to:

(uλ)2+u2u+4τ2u=a2

Let u2=P(u) and multiply u both sides of the equation then:

P(u)=u3+2λu2+(a2λ2)u4τ2

Notice the following:

Thus P(u) has 3 real roots thus:

P(u)=(u+α1)(u+α2)(u+α3)

Thus:

4λ=α3+α2α1a2λ2=α1α3+α1α2α2α34τ2=α1α2α3

After some laborious calculation:

u(s)=α3(1q2sn2(rs,p))

where:

p2=α3α2α3+α2q2=α3α2α3α3=k02r=12α3+α1ω2=p2q2κ2=κ02(1p2ω2sn2(k02ωs,p)