The following theorem characterizes a critical point of the energy of the velocity vector field of a curve in Rn.
Proof.
Let α:I→Rn be a unit speed curve in Rn and [a,b]⊂I, α(a)=p, α(b)=q. There exists a real-valued function λ on [a,b], λ(s)=(s-a)(b-s), λ(a)=λ(b)=0, and λ(s)≠0 for all s∈(a,b). Let {V1,V2,…,Vr} be the Frenet frame field on α and(8)λsV1s=v1s,v2s,…,vns, vi:a,b⟶R.Let the collection of curves be(9)αks=α1s+kv1s,α2s+kv2s,…,αns+kvnsfor sufficiently small k.
For k=0, α0(s)=α(s), and λ(a)=λ(b)=0, we have vi(a)=vi(b)=0 1≤i≤n and αka=p, αkb=q.
These results show that αk is the curve segment from p to q. Assume this collection αks=α(s,k) for all curves. The expression for the energy of the vector field V1k of αk from p to q becomes E(V1k).
Now, let TCk be the tangent bundle. So we have V1k:Ck→TCk, where TCk=∪t∈ITαk(t)Ck, Ck=αkI, and Tαk(t)Ck denotes straight line generated V1k. Let π:TCk→Ck be the bundle projection. By using (5) we calculate the energy of V1k as (10)EV1k=12∫abgSdV1kV1kαs,k,dV1kV1kαs,kds,where ds is the differential arc length. From (4) we have(11)gSdV1kV1k,dV1kV1k=dπdV1kV1k,dπdV1kV1k+KdV1kV1k,KdV1kV1k.Since V1k is a section, we have d(π)∘d(V1k)=dπ∘V1k=didCk=idTCk. By Proposition 4, we also have that(12)KdV1kV1k=∇V1kV1k=V1′k=∂V1k∂s,giving(13)gSdV1kV1k,dV1kV1k=V1k,V1k+V1′k,V1′k.Using these results in (10) we get (14)EV1k=12∫abV1k,V1k+V1′k,V1′kds,where V1k=1/ws,kdα/∂s(s,k); ws,k=dα/∂s(s,k),dα/∂s(s,k).
By Definition 7, if V1k is a harmonic, then k=0 should be a critical point of EV1k. Suppose that ∂EV1k/∂kk=0=0.
From (14) we obtain(15)∂EV1k∂k=∂∂k12∫abV1k,V1k+V1′k,V1′kds=12∫ab∂∂kV1k,V1k+∂V1k∂s,∂V1k∂sds.Since V1k,V1k=1 we have ∂/∂kV1k,V1k=0 and we get(16)∂EV1k∂k=12∫ab∂∂k∂V1k∂s,∂V1k∂sds=∫ab∂2V1k∂s∂k,∂V1k∂sds.We can write(17)∂∂s∂V1k∂k,∂V1k∂s=∂2V1k∂s∂k,∂V1k∂s+∂V1k∂k,∂2V1k∂s2.Thus, we can deduce(18)∂2V1k∂s∂k,∂V1k∂s=∂∂s∂V1k∂k,∂V1k∂s-∂V1k∂k,∂2V1k∂s2.Substituting (18) in (16), for k=0,(19)∂EV1k∂kk=0=∫ab∂∂s∂V1k∂ks,0,∂V1k∂ss,0-∂V1k∂ks,0,∂2V1k∂s2s,0ds,(20)∂EV1k∂kk=0=∂V1k∂ks,0,∂V1k∂ss,0ab-∫ab∂V1k∂ks,0,∂2V1k∂s2s,0ds.From (8) and (9), we obtain(21)dα∂ks,k=λsV1s,(22)dα∂ss,0=α′s=V1s,0.Now we calculate the partial derivatives of (22) with respect to s and k; using Frenet formulas, we get(23)∂V1k∂ss,0=∂2α∂s2s,0=α′′s=V1′k=k1sV2s,(24)∂V1k∂ks,k=∂2α∂s∂ks,k=∂2α∂k∂ss,k.From (21), we have(25)∂V1k∂ks,kk=0=∂V1k∂ks,0=λ′sV1s+λsk1sV2s.It follows from (23) and (25) that(26)∂V1k∂ks,0,∂V1k∂ss,0=λsk12s.Considering the candidate function λa=λb=0, we get(27)∂V1k∂ks,0,∂V1k∂ss,0ab=λbk12b-λak12a=0.From (23), we get(28)∂2V1k∂s2s,0=-k12sV1s+k1′sV2s+k1sk2sV3s.Therefore, (25) and (28) give (29)∂V1k∂ks,0,∂2V1k∂s2s,0=-λsk12s′+3λsk1sk′1s.Substituting (27) and (29) in (20) yields(30)∂EV1k∂kk=0=-∫ab-λsk12s′+3λsk1sk′1sds,∂EV1k∂kk=0=-λsk12sab-3∫abλsk1sk′1sds=0.Since λa=λb=0, it gives -λsk12sab=0 and(31)∂EV1k∂kk=0=-3∫abλsk1sk′1sds=0.This completes the proof of the theorem. Also, it is trivial that geodesics and curves with constant curvature satisfy the theorem.