Search CTRL + K

Homework 1

Problem statement

Show that γt=γs×γss+α[γsss+32γss×(γs×γss)] for αR is equivalent to:

ψt=iψss+i2|ψ|2ψ+α(ψsss+32|ψ|2ψs)

Solution:

Recall the Bishop frame:

T=TU=cos(θ)Nsin(θ)BV=sin(θ)N+cos(θ)B

With Frenet Serret Equations:

dds(TUV)=(0κ1κ2κ100κ200)(TUV)

In the Bishop frame γt=γs×γss+α[γsss+32γss×(γs×γss)] for αR is rewritten to:

γt=κ2U+κ1V+α[(κ1U+κ2V)s+32(κ1U+κ2V)×(κ1Vκ2U)]=κ2U+κ1V+α[(κ1U+κ2V)s+32κ2T]=κ2U+κ1V+α[ka1sU+κ2sV+12κ2T].

with γss=κ1U+κ2V.

We note that γ has to satisfy the compatibility conditions (γt)ss=(γss)t.

We begin by computing Tt=(γs)t=(γt)s:

Tt=s(κ2U+κ1V+α[ka1sU+κ2sV+12κ2T])=κ2sU+κ1sV+α[ka1ssU+κ2ssV+12κ2(κ1U+κ2V))].

Since VT=0 (as we are working in an orthonormal frame) we obtain that t(VT)=0 thus:

VtT=TVt

Similarly, we get that UVt=UtV.
We are interested in computing VtU and in order to do so we compute (VtU)s instead:

(VtU)s=VtsU+VtUs=κ2(T)tUκ1VtT=κ2(T)tU+κ1VTt=κ2κ2sκ2α(κ1ss+12κ2κ2)+κ1κ1s+κ1α(κ2ss+12κ2κ1)=(κ1κ1s+κ2κ2s)+α(κ1κ2ssκ2κ1ss)=s(12(k12+κ22)+α(κ1κ2sκ2κ1s)).

Thus VtU=12κ2+α(κ1κ2sκ2κ1s)+C(t), where C is a real valued function in t.

We now note that γ has to satisfy the compatibility conditions (γt)ss=(γss)t and we use the fact that since UU=1 we get that UtU=0. Thus:

(γt)ssU=(γss)tUκ2ss+α(κ1sss+κκsκ1+12κ2κ1s)=κ1t+12κ2κ2+ακ2(κ1κ2sκ2κ1s)

Hence:

κ1t=κ2ss12κ2κ2+α(κ1sss+12κ2κ1s+κκsκ1κ2(κ1κ2sκ2κ1s)).

We observe that κκs=κ1κ1s+κ2κ2s, κ1s=kscos(θ)κ2τ and κ2s=kssin(θ)+κ1τ, thus as a consequence we compute the following identities:

κ2(κ1κ2sκ2κ1s)=κ2κ2τ,κ1sin(θ)=κ2cos(θ),κκsκ1=κ12κ1s+κ1κ2κ2s=κ12κ1s+κ1κ2κssin(θ)+κ12κ2τ=κ12κ1s+κ22κs+(κ2κ22))κ2τ=κ12κ1s+κ1κ2κssin(θ)+κ12κ2τ=κ12κ1s+κ22κscos(θ)+(κ2κ22))κ2τ=κ12κ1s+κ22κ1s+κ2κ2τ.=κ2κ1s+κ2κ2τ

Thus we finally obtain:

(1)κ1t=κ2ss12κ2κ2+α(κ1sss+32κ2κ1s).

Similarly one can also show that:

(2)κ2t=κ1ss+12κ1κ2+α(κ2sss+32κ2κ2s).

Now using the Hasimoto map ψ=κesτ=κ1+iκ2 with (1) and (2) we thus get afforemention equation that we wanted to show:

ψt=iψss+i2|ψ|2ψ+α(ψsss+32|ψ|2ψs).