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Model functions solve the constant curvature jacobi equation

Statement

For every real k∈R the comparison functions sn⁡k,cs⁡k of Comparison sine, cosine and cotangent functions satisfy, on all of R, sn⁡k′′+k sn⁡k=0,cs⁡k′′+k cs⁡k=0, with initial data sn⁡k(0)=0,sn⁡k′(0)=cs⁡k(0)=1,cs⁡k′(0)=0, and the Wronskian identity cs⁡k(t)2+k sn⁡k(t)2=1for every t∈R. In particular, for k>0 the pair (sn⁡k,cs⁡k) is the solution of the scalar Jacobi equation with the spherical initial data, for k=0 of the Euclidean initial-value problem, and for k<0 of the hyperbolic initial-value problem. The identity shows that sn⁡k and cs⁡k never vanish simultaneously; for k≤0, cs⁡k is everywhere positive. The assertions are purely about the explicitly displayed model functions and use no geometric or choice hypothesis.

Facts & Assumptions

Given: A real number k∈R and the piecewise formulas for sn⁡k,cs⁡k=sn⁡k′ (Comparison sine, cosine and cotangent functions).

[F1]

The comparison functions are sn⁡k(t)=sin⁡(k t)/k and cs⁡k(t)=cos⁡(k t) for k>0; sn⁡k(t)=t and cs⁡k(t)=1 for k=0; and sn⁡k(t)=sinh⁡(−k t)/−k, cs⁡k(t)=cosh⁡(−k t) for k<0 (Comparison sine, cosine and cotangent functions); cs⁡k=sn⁡k′ is part of that definition.

[F2]

Sine and cosine are differentiable with sin⁡′=cos⁡, cos⁡′=−sin⁡, and the chain rule gives (sin⁡(ct))′=ccos⁡(ct), (cos⁡(ct))′=−csin⁡(ct) (The derivatives of sine and cosine are cosine and minus sine, The chain rule, in one line from Carathéodory: if g is differentiable at c and f is differentiable at g(c), then f∘g is differentiable at c with (f∘g)′(c)=f′(g(c)) g′(c)); moreover sin⁡2+cos⁡2=1 (Parity and the Pythagorean identity for sine and cosine).

[F3]

The hyperbolic functions satisfy sinh⁡′=cosh⁡, cosh⁡′=sinh⁡ and cosh⁡2−sinh⁡2=1 (The six hyperbolic functions and their natural domains, Addition formulas, identities, parity, and derivatives of the hyperbolic functions); with the chain rule, (sinh⁡(ct))′=ccosh⁡(ct) and (cosh⁡(ct))′=csinh⁡(ct) (The chain rule, in one line from Carathéodory: if g is differentiable at c and f is differentiable at g(c), then f∘g is differentiable at c with (f∘g)′(c)=f′(g(c)) g′(c)).

Proof

1.1F1F2

The positive-curvature case. [F1, F2] Let k>0 and c:=k. By [F1], sn⁡k(t)=sin⁡(ct)/c and cs⁡k(t)=cos⁡(ct). Differentiating with [F2], sn⁡k′(t)=cos⁡(ct)=cs⁡k(t), sn⁡k′′(t)=−csin⁡(ct)=−c2sn⁡k(t)=−ksn⁡k(t), and cs⁡k′(t)=−csin⁡(ct), cs⁡k′′(t)=−c2cos⁡(ct)=−kcs⁡k(t). At t=0: sn⁡k(0)=sin⁡0/c=0, sn⁡k′(0)=cos⁡0=1=cs⁡k(0) and cs⁡k′(0)=−csin⁡0=0.

1.2F1

The flat case. [F1] Let k=0. By [F1], sn⁡0(t)=t and cs⁡0(t)=1. Hence sn⁡0′′=0=−0⋅sn⁡0, cs⁡0′′=0=−0⋅cs⁡0, and the initial data are sn⁡0(0)=0, sn⁡0′(0)=1=cs⁡0(0), cs⁡0′(0)=0.

1.3F1F3

The negative-curvature case. [F1, F3] Let k<0 and c:=−k>0. By [F1], sn⁡k(t)=sinh⁡(ct)/c and cs⁡k(t)=cosh⁡(ct). Differentiating with [F3], sn⁡k′=cosh⁡(ct)=cs⁡k, sn⁡k′′=csinh⁡(ct)=−ksn⁡k, and cs⁡k′′=c2cosh⁡(ct)=−kcs⁡k. At t=0: sn⁡k(0)=0, sn⁡k′(0)=cosh⁡0=1, cs⁡k(0)=1 and cs⁡k′(0)=sinh⁡0=0.

2.1F1F4step 1.1step 1.2step 1.3∎

Assembly and the Wronskian identity. [F1, F4, step 1.1, step 1.2, step 1.3] Steps 1.1, 1.2 and 1.3 cover the three cases k>0, k=0 and k<0, which exhaust R: in every case sn⁡k′′+ksn⁡k=0 and cs⁡k′′+kcs⁡k=0 on all of R, with sn⁡k(0)=0, cs⁡k(0)=1, cs⁡k′(0)=0 and cs⁡k=sn⁡k′, so sn⁡k′(0)=1. For the remaining identity define w(t):=cs⁡k(t)2+ksn⁡k(t)2. By the product rule and the two differential equations just established, w′=2cs⁡kcs⁡k′+2ksn⁡ksn⁡k′=2cs⁡k(−ksn⁡k)+2ksn⁡kcs⁡k=0 on all of R. Hence [F4] makes w constant, and w(0)=cs⁡k(0)2+ksn⁡k(0)2=1. Therefore cs⁡k2+ksn⁡k2=1 everywhere, so sn⁡k and cs⁡k never vanish simultaneously; the explicit case analysis shows that the same formulas continue to hold at every real t, including t=0 and the reflected negative times. No choice or completeness hypothesis is used.

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