The lattice $\displaystyle \Lambda $ associated to the pendulum equation is rectangular

Question

The pendulum equation (for a suitable choice of parameters) is $\dot{\theta}^2 − \cos \theta = E.$

Taking $u = \cos \theta$, we get $$\dot{u}^{2} =( u+E)\left( 1-u^{2}\right)$$

It is well known that the solution to the above is $u = a \cdot \wp_{\Lambda}(t-t_0) +b$, where $\wp_{\Lambda}$ is the Weierstrass function associated with the lattice $\Lambda$. The values of $a,b,\Lambda$ depends on $E$, while the value of $t_0$ depends on the initial conditions.

The question is:

Show that the lattice $\displaystyle \Lambda $ is rectangular, i.e. generated by 1 and $\displaystyle iq$ for some $\displaystyle q\ >\ 0$, and that for physically meaningful solutions the imaginary part of the constant of integration $\displaystyle t_{0}$ must be $\displaystyle q/2$.

[From Riemann Surfaces by S. K. Donaldson, page 96]

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1. What is special about the pendulum equation that makes its lattice rectangular?
2. I presume by constant of integration they mean $t_0$ (mentioned above). Why does $\text{Im}(t_0)$ have to $q/2$ and why is $t_0$ called "the constant of integration".

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When trying to find the answer to 1, I came across a book which mentioned that the lattice is rectangular for any elliptic curve $y^2 = f(x)$, where $f$ is real and has 3 distinct roots. If one can prove this, we are done.

(Ref: Elliptic Curves: Number Theory and Cryptography, Second Edition. Page 286)

Answer 1

For the first part, let $$v = -u-E/3$$ so that $$v'^2 =v^3-v(1+E^2/3)+2E/3-2 E^3/27 $$

which corresponds to the elliptic curve

$$C:y^2=x^3-Ax-B, \qquad A = 1+\frac{E^2}3,\quad B=\frac{2E}3- \frac{2E^3}{27} $$

• $A > 0$ since $E$ is real

• $\Delta=4A^3-27B^2 = \prod_{i<j} (\gamma_i-\gamma_j)^2$ is $\in (-\infty,4A^3]$ since $A,B$ are real, and $\Delta> 0$ because the RHS cubic polynomial has 3 distinct real roots $\gamma_1,\gamma_2,\gamma_3$.

• The $j$-invariant is $$j(C) = 1728 \frac{4a^3}{4a^3-27b^2}\ge 1728$$

As $j(ir)$ is $(0,\infty)\to \Bbb{R}$, continuous, and $j(i)=1728,j(i\infty)=+\infty$, we get that $$j(C)=j(\tau)$$ for some $\tau$ purely imaginary.

The lattice $\Bbb{Z}+\tau \Bbb{Z}$ is rectangular and its Weierstrass function satisfies $$\wp_\tau'^2 = \wp_\tau^3-4 g_2 \wp_\tau/4-16 g_3$$ with $4 g_2=c^4 A,16 g_3=c^6 B$ for some $c$.

I don't see why $1$ should be in $\Lambda$, I think we need to set $v(z)=c^2 \wp_\tau(z/c+z_0)$.