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Solutions of Exercises on Addition Theorem of Elliptic Functions and Theta Functions
Problem 1. (Darboux’s Derivation of the Addition Theorem for Jacobian Elliptic Sine Function by Using the Motion of Two Pendulums). The Jaco- bian elliptic sine function with modulus k is denote by sn w which satisfies the differential equation
( d dw sn w
1 − x^2
1 − k^2 x^2
and is odd with the initial conditions sn (0) = 0 and sn ′(0) = 1, where sn ′(w) denotes the derivative of sn w. (The initial value sn ′(0) = 1 specifies which branch of the square root is being chosen.)
Derive the addition theorem for the Jacobian ellitpic sine function
sn (w 1 + w 2 ) = sn w 1 sn ′^ (w 2 ) + sn w 2 sn ′^ (w 1 ) 1 − k^2 sn^2 w 1 sn^2 w 2
by following the steps outlined below. Recall that the formula of motion for a pendulum of length a and initial angle α is given by
sin θ 2
= sin α 2
sn(
g a
t),
where θ is the angle made with the vertical line. We normalize the constants so that g = a and we have sin θ 2 sin α 2 = sn t.
We consider the motion of two pendulums, the first one with angle made with the vertical line denoted by θ and the time t denoted by u and the second one with angle made with the vertical line denoted by ϕ and the time t denoted by v. We run the second pendulum backward in time and with an offset in time so that u + v = c for some constant c. Introduce the quantities
ξ =
sin θ 2 sin α 2 , η =
sin ϕ 2 sin α 2
so that ξ = sn u and η = sn v. Darboux’s derivation of using the motion of two pendulums is motivated by the special case of k = 0 when sn w = sin w and the addition formula with u + v equal to constant c becomes
sin c = sin v
d du sin u
− sin u
d du sin v
(because (^) dud sin v is equal to sin′^ v when u + v equal to constant c) which sug- gests the argument of imitating the conservation of the angular momentum of the point (sin u, sin v) in R^2 (with respect to the time variable u), which for the general case becomes (ξ, η) = (sn u, sn v).
(a) Use the differential equations dξ du
(1 − ξ^2 ) (1 − k^2 ξ^2 ) , dη dv
(1 − η^2 ) (1 − k^2 η^2 )
to derive η d^2 ξ du^2 − ξ d^2 η du^2 = 2 k^2 ξη(ξ^2 − η^2 ) and η^2
dξ du
− ξ^2
dη du
= η^2 − ξ^2 + k^2 ξ^2 η^2
ξ^2 − η^2
(b) By taking the quotients of the two equations in (a), verify that ( η d^2 ξ du^2 − ξ d^2 η du^2
η
dξ du
− ξ
dη du
= 2k^2 ξη
η
dξ du
dη du
k^2 ξ^2 η^2 − 1
(c) By integrating the equation in (b), verify that ( η
dξ du
− ξ
dη du
1 − k^2 ξ^2 η^2
= C
for some constant C. (d) Use the initial value u = 0 (with v = c) to verify that C = sn c.
By squaring () and (\) we get
()^2
dξ du
1 − ξ^2
1 − k^2 ξ^2
and
(\)^2
dη dv
1 − η^2
1 − k^2 η^2
Since u + v = c is a constant, it follows that ( dη du
dη dv
and
(^ ˜\)^2.
dη du
1 − η^2
1 − k^2 η^2
From η ()^2 − ξ (˜)^2 we obtain
η^2
dξ du
− ξ^2
dη du
= η^2 − ξ^2 + k^2 ξ^2 η^2
ξ^2 − η^2
(b) By taking the dividing the following two equations from (a) (the first equation by the second equations)
η d^2 ξ du^2 − ξ d^2 η du^2 = 2 k^2 ξη(ξ^2 − η^2 ),
η^2
( (^) dξ du
− ξ^2
( (^) dη du
= η^2 − ξ^2 + k^2 ξ^2 η^2 (ξ^2 − η^2 ) ,
we get
η d (^2) ξ du^2 −^ ξ^
d^2 η du^2 η^2
( (^) dξ du
− ξ^2
( (^) dη du
) 2 =^
2 k^2 ξη(ξ^2 − η^2 ) η^2 − ξ^2 + k^2 ξ^2 η^2 (ξ^2 − η^2 )
2 k^2 ξη −1 + k^2 ξ^2 η^2
which is the same as ( η d^2 ξ du^2 − ξ d^2 η du^2
η
dξ du
− ξ
dη du
= 2k^2 ξη
η
dξ du
dη du
k^2 ξ^2 η^2 − 1
(c) By taking the logarithmic derivative of ( η
dξ du
− ξ
dη du
1 − k^2 ξ^2 η^2
with respect to u which means
d du log
η
( (^) dξ du
− ξ
( (^) dη du
1 − k^2 ξ^2 η^2
we get η
d^2 ξ du^2
− ξ
d^2 η du^2
η
( (^) dξ du
− ξ
( (^) dη du
2 ξηk^2
η dξdu + ξ dηdu
1 − k^2 ξ^2 η^2
which is identically equal to 0 by the last equation of (a). Thus
log
η
( (^) dξ du
− ξ
( (^) dη du
1 − k^2 ξ^2 η^2 = constant,
which means that ( η
dξ du
− ξ
dη du
1 − k^2 ξ^2 η^2
= C
for some constant C.
(d) We are going to use the initial value u = 0 (with v = c) to verify that C = sn c. We set ξ(u) = sn u and ξ(v) = sn v = sn(c − u). When u = 0, we have ξ(u) = 0 and (^) duξ (u) = 1 and η = sn c. Hence evaluation at u = 0 yields
C =
η
( (^) dξ du
− ξ
( (^) dη du
1 − k^2 ξ^2 η^2 = sn c.
(e) Finally we make the substitution ξ = sn u and η = sn v = sn (c − u) into the equation η
( (^) dξ du
− ξ
( (^) dη du
1 − k^2 ξ^2 η^2 = sn c.
Solution. (a) From
P(w 1 + w 2 ) = −P(w 1 ) − P(w 2 ) +
P′(w 1 ) − P′(w 2 ) P(w 1 ) − P(w 2 )
it follows from the evenness of P(w) and the oddity of P′(w) (with the sub- stitution w 1 = z, w 2 = 1w and the substitution w 1 = z, w 2 = z respectively) that
(∗) P(z − w) = −P(z) − P(w) +
P′(z) + P′(w) P(z) − P(w)
and
(∗∗) P(z − w) = −P(z) − P(w) +
P′(z) − P′(w) P(z) − P(w)
Subtracting the equation (∗∗) from the equation (∗), we get
P(z − w) − P(z + w)
=
P′(z) + P′(w) P(z) − P(w)
P′(z) − P′(w) P(z) − P(w)
(P′(z))^2 + 2P′(z)P′(w) + (P′(w))^2
(P′(z))^2 − 2 P′(z)P′(w) + (P′(w))^2
4 (P(z) − P(w))^2
P′(z)P′(w) [P(z) − P(w)]^2
(b) By applying l’Hˆopital’s rule (or power series expansion) to
P(w 1 + w 2 ) = −P(w 1 ) − P(w 2 ) +
P′(w 1 ) − P′(w 2 ) P(w 1 ) − P(w 2 )
with w 1 = w → z and w 2 = z when z is not a period, we obtain
P(2z) = − (^) wlim→z P(w) − P(z) + lim w→z
P′(w) − P′(z) P(w) − P(z)
= − 2 P(z) +
P′′(z) P′(z)
The differentiation of both sides of
(P′(w))^2 = 4 (P(w))^3 − g 2 P(w) − g 3
with respect to w yields
2 P′(w)P′′(w) = 12 (P(w))^2 P′(w) − g 2 P′(w)
which can be rewritten as
2 P′′(w) = 12 (P(w))^2 − g 2.
Thus (P′′(w))^2 =
6 (P(w))^2 − g 2 2
Hence
P(2z) == − 2 P(z) +
P′′(z) P′(z)
= − 2 P(z) +
1 4
6 (P(w))^2 − g 22
4 (P(z))^3 − g 2 P(z) − g 3
= P(z)^4 + 12 g 2 P(z)^2 + 2g 3 P(z) + 161 g^22 4 P(z)^3 − g 2 P(z) − g 3
Problem 3. (Application of Argument Principle to Theta Functions). Let ω 1 , ω 2 ∈ C be R-linearly independent. Let η 1 , η 2 be nonzero complex num- bers and ξ 1 , ξ 2 ∈ C. Suppose f (w) is an entire function on C such that
f (w + ων ) = eην^ w+ξν^ f (w) for ν = 1, 2 and w ∈ C.
Let c ∈ C such that f is nowhere zero on the four sides of the parallelogram Ω with vertices a, a + ω 1 , a + ω 2 , a + ω 1 + ω 2. Use the argument principle to show that the number of zeroes of f in the interior of the parallelogram Ω is equal to 1 2 πi (ω 2 η 1 − ω 1 η 2 ).
