Number Theory: An Introduction to Mathematics

3 Elliptic Functions 513

then

(dz/dt)^2 =( 1 +

√

λ)^2 { 4 λ 0 z^3 − 4 ( 1 +λ 0 )z^2 + 4 z},

where

λ 0 = 4

√

λ/( 1 +

√

λ)^2. (19)

Sincez( 0 )=z′( 0 )=0andz′′( 0 )=0, it follows thatz(t)=S(( 1 +

√

λ)t,λ 0 ). Thus

S(( 1 +

√

λ)t,λ 0 )=( 1 +

√

λ)^2 S(t,λ)/[1+

√

λS(t,λ)]^2. (20)

The inequality 0<λ<1 impliesλ<λ 0 <1. Hence, by regarding (19) as a quadratic
equation for

√

λ, we obtain

√

λ=[1−( 1 −λ 0 )^1 /^2 ]^2 /λ 0. (21)

If we write

√

λ 0 =c 0 /a 0 ,wherec 0 =(a 02 −b^20 )^1 /^2 and 0<b 0 <a 0 ,then

√

λ=(a 0 −b 0 )/(a 0 +b 0 )=c 1 /a 1 ,

where

a 1 =(a 0 +b 0 )/ 2 , b 1 =(a 0 b 0 )^1 /^2 , c 1 =(a^21 −b^21 )^1 /^2.

Since 1+

√

λ=a 0 /a 1 , we can rewrite (20) in the form

S(a 0 t,λ 0 )=( 1 +c 1 /a 1 )^2 S(a 1 t,λ 1 )/[1+(c 1 /a 1 )S(a 1 t,λ 1 )]^2 ,

whereλ 1 =λ=(c 1 /a 1 )^2. Repeating the process, we obtain

S(an− 1 t,λn− 1 )=( 1 +cn/an)^2 S(ant,λn)/[1+(cn/an)S(ant,λn)]^2 ,

whereλn=(cn/an)^2 .Asn→∞,

an→μ:=M(a,b), cn→ 0 ,λn→ 0.

SinceS(t, 0 )=sin^2 t, for some (not very large)n=Nwe haveS(aNt,λN)≈sin^2 μt,
which may be considered as known. Then, by taking successivelyn=N,N− 1 ,..., 1
we can calculateS(a 0 t,λ 0 ). Moreover, we can start the process by takinga 0 =1,
b 0 =( 1 −λ 0 )^1 /^2.
We now consider periodicity properties. Ifλ=1andS(h)=1 for some nonzero
h∈Cthen, by (13),S( 2 h)=0. FurthermoreS′( 2 h)=0, by (9). It follows that
S(t)has period 2h,sinceS(t+ 2 h)is a solution of the differential equation (7) which
satisfies the same initial conditions (8) asS(t). It remains to show that there exists such
anh.
Suppose first thatλ∈Rand 0<λ<1. SinceS′′( 0 )=2, we haveS′(t)>0for
smallt>0. IfS′(t)>0for0<t<T,thenS(t)is a positive increasing function for
0 <t<T.Sincegλ[S(t)]>0, we must also haveS(t)<1for0<t<T.Fromthe
relation

t=

∫S(t)

0

dx/gλ(x)^1 /^2 ,