Elliptic functions, Addition Formulae

Consider the ellipse

$\displaystyle \left(\frac{X}{a}\right)^2+ Y^2=1$

with eccentricity given by

${k =\frac{\sqrt{a^2-1}}{a}.}$

${k}$ is called the modulus. (We just assumed the length of the minor axis ${b=1.}$ )

Define ${u}$ to be the “angular” arc-length parameter given by

$du = rd\theta.$

Here ${r, \theta}$ are the polar coordinates.

We define elliptic functions by the formulae

$\displaystyle \mathrm{sn}(u,k) =\frac{X}{a}$

$\displaystyle \mathrm{cn}(u, k) =Y$

$\displaystyle \mathrm{dn}(u, k) =\frac{r}{a}.$

(These can be tough of deformations of ${\sin}$ and ${\cos}$ , in fact we get sines and cosines when we take the limit ${k \rightarrow 0}$ )

Further we can define other ellitptic functions as follows

$\displaystyle \mathrm{ns}(u)=\frac{1}{\mathrm{sn}(u)}, \quad \mathrm{nc}(u)=\frac{1}{\mathrm{cn}(u)}, \quad \mathrm{nd}(u)=\frac{1}{\mathrm{dn}(u)}$

$\displaystyle \mathrm{sc}(u)=\frac{\mathrm{sn}(u)}{\mathrm{cn}(u)}, \quad \mathrm{sd}(u)=\frac{\mathrm{sn}(u)}{\mathrm{dn}(u)}, \quad \mathrm{dc}(u)=\frac{\mathrm{dn}(u)}{\mathrm{cn}(u)}, \quad \mathrm{ds}(u)=\frac{\mathrm{dn}(u)}{\mathrm{sn}(u)},$ $\displaystyle \quad \mathrm{cs}(u)=\frac{\mathrm{cn}(u)}{\mathrm{sn}(u)}, \quad \mathrm{cd}(u)=\frac{\mathrm{cn}(u)}{\mathrm{dn}(u)}$

Inverse functions in terms of integrals:

The above defintions correspond to the following in terms of the elliptic integrals.

$\displaystyle u=\int_{0}^{\varphi} \frac{\mathrm{d} \theta}{\sqrt{1-k^2 \sin ^{2} \theta}}=\int_{0}^{\sin \varphi} \frac{\mathrm{d} t}{\sqrt{1-t^{2} (1-k^{2} t^{2})}}$

$\displaystyle \mathrm{sn} u=\sin \varphi,\quad \mathrm{cn} u=\cos \varphi, \quad \mathrm{dn} u=\sqrt{1-k^2 \sin ^{2} \varphi}$

$\displaystyle \begin{aligned} \mathrm{arcsn}(x, k) &=\int_{0}^{x} \frac{\mathrm{d} t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2} t^{2}\right)}} \\ \mathrm{arccn}(x, k) &=\int_{x}^{1} \frac{\mathrm{d} t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2}+k^{2} t^{2}\right)}} \\ \mathrm{arcdn}(x, k) &=\int_{x}^{1} \frac{\mathrm{d} t}{\sqrt{\left(1-t^{2}\right)\left(t^{2}+k^{2}-1\right)}} \end{aligned}$

$\displaystyle K(k) =\int_{0}^{\pi/2 } \frac{\mathrm{d} \theta}{\sqrt{1-k^2 \sin ^{2} \theta}}$

Identities:

$\displaystyle \begin{array}{l} \mathrm{cn}^{2}(u, k)+\mathrm{sn}^{2}(u, k)=1 \\ \mathrm{dn}^{2}(u, k)+k^{2} \mathrm{sn}^{2}(u, k)=1 \end{array}$

Derivatives:

$\displaystyle \begin{array}{l} \dfrac{{d}}{{d} u} \mathrm{sn}(u)=\mathrm{cn}(u) \mathrm{d} n(u) \\ \dfrac{{d}} {{d} u} \mathrm{cn}(u)=-\mathrm{sn}(u) \mathrm{dn}(u) \\ \dfrac{{d}}{{d} u} \mathrm{dn}(u)=-k^{2} \mathrm{sn}(u) \mathrm{cn}(u) \end{array} \ \ \ \ \ (1)$

Integrals:

$\displaystyle \begin{array}{l} \int ~\mathrm{sn}(x, k) \mathrm{d} x=k^{-1} \ln (~\mathrm{dn}(x, k)-k ~\mathrm{cn}(x, k)) \\ \int ~\mathrm{cn}(x, k) \mathrm{d} x=k^{-1} ~\mathrm{arccos}(~\mathrm{dn}(x, k)) \\ \int ~\mathrm{dn}(x, k) \mathrm{d} x=~\mathrm{arcsin}(~\mathrm{sn}(x, k))=~\mathrm{am}(x, k) \end{array}$

Differential Equations:

$\displaystyle \mathrm{sn} u: \quad \quad \left(\frac{\mathrm{d} Z}{\mathrm{d} u}\right)^{2}=\left(1-Z^{2}\right)\left(1-k^{2} Z^{2}\right); \quad \quad \frac{\mathrm{d}^{2} Z}{\mathrm{d} u^{2}}+\left(1+k^{2}\right) Z-2 k^{2} Z^{3}=0$

$\displaystyle \mathrm{cn} u: \quad \quad \left(\frac{\mathrm{d} Z}{\mathrm{d} u}\right)^{2}=\left(1-Z^{2}\right)\left(1-k^{2}+k^{2} Z^{2}\right); \quad \quad \frac{\mathrm{d}^{2} Z}{\mathrm{d} u^{2}}+\left(1-2 k^{2}\right) Z+2 k^{2} Z^{3}=0$

$\displaystyle \mathrm{dn} u: \quad \quad \left(\frac{\mathrm{d} Z}{\mathrm{d} u}\right)^{2}=\left(Z^{2}-1\right)\left(1-k^{2}-Z^{2}\right); \quad \quad \frac{\mathrm{d}^{2} Z}{\mathrm{d} u^{2}}-\left(2-k^{2}\right) Z+2 Z^{3}=0$

Double Angle Formulae:

$\displaystyle \mathrm{sn}(2 u, k)=\frac{2 \mathrm{sn}(u, k) \mathrm{cn}(u, k) \mathrm{dn}(u, k)}{1-k^{2} \mathrm{sn}^{4}(u, k)}$

$\displaystyle \mathrm{cn}(2 u, k)=\frac{\mathrm{cn}^{2}(u, k)-\mathrm{sn}^{2}(u, k) \mathrm{dn}^{2}(u, k)}{1-k^{2} \mathrm{sn}^{4}(u, k)}=\frac{\mathrm{cn}^{4}(u, k)-k^{\prime 2} \mathrm{sn}^{4}(u, k)}{1-k^{2} \mathrm{sn}^{4}(u, k)}$

$\displaystyle \mathrm{dn}(2 u, k)=\frac{\mathrm{dn}^{2}(u, k)-k^{2} \sin ^{2}(u, k) \mathrm{cn}^{2}(u, k)}{1-k^{2} \mathrm{sn}^{4}(u, k)}=\frac{\mathrm{dn}^{4}(u, k)+k^{2} k^{\prime 2} \mathrm{sn}^{4}(u, k)}{1-k^{2} \sin ^{4}(u, k)}$

Change of modulus:

$\displaystyle \mathrm{sn}(u, 1 / k)=k \mathrm{sn}(u / k, k)$

$\displaystyle \mathrm{cn}(u, 1 / k)=\mathrm{dn}(u / k, k)$

$\displaystyle \mathrm{dn}(u, 1 / k)=\mathrm{cn}(u / k, k)$

Landen Transformations:

$\displaystyle k_{1}=\frac{1-k^{\prime}}{1+k^{\prime}}$

$\displaystyle k_{2}=\frac{2 \sqrt{k}}{1+k}$

$\displaystyle k_{2}^{\prime}=\frac{1-k}{1+k}$

$\displaystyle ~\mathrm{sn}(u, k)=\frac{\left(1+k_{1}\right) ~\mathrm{sn}\left(2 /\left(1+k_{1}\right), k_{1}\right)}{1+k_{1} ~\mathrm{sn}^{2}\left(z /\left(1+k_{1}\right), k_{1}\right)}$

$\displaystyle ~\mathrm{cn}(u, k)=\frac{~\mathrm{cn}\left(u /\left(1+k_{1}\right), k_{1}\right) ~\mathrm{dn}\left(z /\left(1+k_{1}\right), k_{1}\right)}{1+k_{1} ~\mathrm{sn}^{2}\left(u /\left(1+k_{1}\right), k_{1}\right)}$

$\displaystyle ~\mathrm{dn}(u, k)=\frac{~\mathrm{dn}^{2}\left(u /\left(1+k_{1}\right), k_{1}\right)-\left(1-k_{1}\right)}{1+k_{1}-~\mathrm{dn}^{2}\left(z /\left(1+k_{1}\right), k_{1}\right)}$

$\displaystyle ~\mathrm{sn}(u, k)=\frac{\left.\left(1+k_{2}^{\prime}\right) ~\mathrm{sn}\left(u /\left(1+k_{2}^{\prime}\right), k_{2}\right) ~\mathrm{cn}\left(u / 1+k_{2}^{\prime}\right), k_{2}\right)}{~\mathrm{dn}\left(u /\left(1+k_{2}^{\prime}\right), k_{2}\right)}$

$\displaystyle ~\mathrm{cn}(u, k)=\frac{\left(1+k_{2}^{\prime}\right)\left(~\mathrm{dn}^{2}\left(u /\left(1+k_{2}^{\prime}\right), k_{2}\right)-k_{2}^{\prime}\right)}{k_{2}^{2} ~\mathrm{dn}\left(u /\left(1+k_{2}^{\prime}\right), k_{2}\right)}$

$\displaystyle ~\mathrm{dn}(u, k)=\frac{\left(1-k_{2}^{\prime}\right)\left(~\mathrm{dn}^{2}\left(u / 1+k_{2}^{\prime}\right), k_{2}\right)+k_{2}^{\prime}}{k_{2}^{2} ~\mathrm{dn}\left(u /\left(1+k_{2}^{\prime}\right), k_{2}\right)}$

Addition Theorems:

$\displaystyle ~\mathrm{sn}(u+v)=\dfrac{~\mathrm{sn} u ~\mathrm{cn} v ~\mathrm{dn} v+~\mathrm{sn} v ~\mathrm{cn} u ~\mathrm{dn} u}{1-k^{2} \sin ^{2} u \sin ^{2} v}$

$\displaystyle ~\mathrm{cn}(u+v)=\dfrac{~\mathrm{cn} u ~\mathrm{cn} v-~\mathrm{sn} u ~\mathrm{dn} u \sin v ~\mathrm{dn} v}{1-k^{2} \sin ^{2} u ~\mathrm{sn}^{2} v}$

$\displaystyle ~\mathrm{dn}(u+v)=\dfrac{~\mathrm{dn} u ~\mathrm{dn} v-k^{2} \sin u ~\mathrm{cn} u ~\mathrm{sn} v ~\mathrm{cn} v}{1-k^{2} ~\mathrm{sn}^{2} u ~\mathrm{sn}^{2} v}$

Periodicity: Elliptic functions are doubly periodic.

$\displaystyle k' :=\sqrt{1-k^2}$

$\displaystyle K'(k) :=K(k')=K(\sqrt{1-k^2})$

$\displaystyle K(k) =\int_{0}^{\pi/2 } \frac{\mathrm{d} \theta}{\sqrt{1-k^2 \sin ^{2} \theta}}$

$\displaystyle K'(k) =\int_{0}^{\pi/2 } \frac{\mathrm{d} \theta}{\sqrt{1-(1-k^2) \sin ^{2} \theta}}$

$\displaystyle \text{Periods of } ~~\mathrm{sn} u: \quad \quad 4 K, 2 \mathrm{i} K^{\prime}$

$\displaystyle \text{Periods of } ~~\mathrm{cn} u: \quad \quad 4 K_{1} 2 K+2 \mathrm{i} K^{\prime}$

$\displaystyle \text{Periods of } ~~ \mathrm{dn} u: \quad \quad 2 K, 4 \mathrm{i} K^{\prime}$

Nome and Theta functions:

$\displaystyle \vartheta_{1}(z, q) \equiv \sum_{n=-\infty}^{\infty}(-1)^{n-1 / 2} q^{(n+1 / 2)^{2}} e^{(2 n+1) i z}$

$\displaystyle \vartheta_{2}(z, q) \equiv \sum_{n=-\infty}^{\infty} q^{(n+1 / 2)^{2}} e^{(2 n+1) i z}$

$\displaystyle \vartheta_{3}(z, q) \equiv \sum_{n=-\infty}^{\infty} q^{n^{2}} e^{2 n i z}$

$\displaystyle \vartheta_{4}(z, q) \equiv \sum_{n=-\infty}^{\infty}(-1)^{n} q^{n^{2}} e^{2 n i z}\\$

$\displaystyle \vartheta_i(q):=\vartheta_i(0, q)$

$\displaystyle q=\exp \left(-\pi K^{\prime}(k) / K(k)\right)$

$\displaystyle k=\frac{\vartheta_{2}^{2}(q)}{\vartheta_{3}^{2}(q)}$

$\displaystyle k'=\frac{\vartheta_{4}^{2}(q)}{\vartheta_{3}^{2}(q)}$

$\displaystyle ~\mathrm{sn}(u, k)=\frac{\vartheta_{3}(q)}{\vartheta_{2}(q)} \frac{\vartheta_{1}\left(u \vartheta_{3}^{-2}; q\right)}{\vartheta_{4}\left(u \vartheta_{3}^{-2}; q\right)} \\$

$\displaystyle ~\mathrm{cn}(u, k)=\frac{\vartheta_{4}(q)}{\vartheta_{2}(q)} \frac{\vartheta_{2}\left(u \vartheta_{3}^{-2}; q\right)}{\vartheta_{4}\left(u \vartheta_{3}^{-2}; q\right)} \\$

$\displaystyle ~\mathrm{dn}(u, k)=\frac{\vartheta_{4}(q)}{\vartheta_{3}(q)} \frac{\vartheta_{3}\left(u \vartheta_{3}^{-2}; q\right)}{\vartheta_{4}\left(u \vartheta_{3}^{-2}; q\right)}$

Below we give a proof of one of the addition formulae, others can be derived similarly.

Proof of Addition formula (Darboux’s Proof- System of Two Pendulums)

Consider two coupled systems with ${u, v}$ related by

$\displaystyle u+v = A$

$\displaystyle \begin{array}{l} x :=~\mathrm{sn}(u, k) \\ y :=~\mathrm{sn}(v, k) \end{array}$

$\displaystyle \left(\frac{d x}{d u}\right)^{2}=\left(1-x^{2}\right)\left(1-k^{2} x^{2}\right)$

$\displaystyle \left(\frac{d y}{d v}\right)^{2}=\left(1-y^{2}\right)\left(1-k^{2} y^{2}\right)$

$\displaystyle \frac{d^{2} x}{d u^{2}}=-\left(1+k^{2}\right) x+2 k^{2} x^{3}$

$\displaystyle \frac{d^{2} y}{d v^{2}}=-\left(1+k^{2}\right) y+2 k^{2} y^{3}$

$\displaystyle \begin{aligned} \frac{d}{d u}\left(y \frac{d x}{d u}-x \frac{d y}{d u}\right)=y \frac{d^{2} x}{d u^{2}}-x\frac{d^{2} y}{d u^{2}} &=y\left(-\left(1+k^{2}\right) x+2 k^{2} x^{2}\right)-x\left(-\left(1+k^{2}\right) y+2 k^{2} y^{2}\right) \\ &=2 k^{2} x y\left(x^{2}-y^{2}\right) \end{aligned}$

$\displaystyle \begin{aligned} y^{2}\left(\frac{d x}{d u}\right)^{2}-x^{2}\left(\frac{d y}{d u}\right)^{2} &=y^{2}\left(1-x^{2}-k^{2} x^{2}+k^{2} x^{4}\right)-x^{2}\left(1-y^{2}-k^{2} y^{2}+k^{2} y^{4}\right)\\ &=y^{2}-x^{2}+k^{2} x^{2} y^{2}\left(x^{2}-y^{2}\right) \end{aligned}$

$\displaystyle \dfrac{\frac{d}{d u}\left(y \frac{d x}{d u}-x \frac{d y}{d u}\right)}{y^{2}\left(\frac{d x}{d u}\right)^{2}-x^{2}\left(\frac{d y}{d u}\right)^{2} } =\frac{-2 k^{2} x y}{1-k^{2} x^{2} y^{2}}$

$\displaystyle \dfrac{\frac{d}{d u}\left(y \frac{d x}{d u}-x \frac{d y}{d u}\right)}{ \left(y \frac{d x}{d u}-x \frac{d y}{d u}\right) \left(y \frac{d x}{d u}+x \frac{d y}{d u}\right)} =\frac{-2 k^{2} x y}{1-k^{2} x^{2} y^{2}}$

$\displaystyle \dfrac{\frac{d}{d u}\left(y \frac{d x}{d u}-x \frac{d y}{d u}\right)}{ \left(y \frac{d x}{d u}-x \frac{d y}{d u}\right) } =\frac{-2 k^{2} x y \left(y \frac{d x}{d u}+x \frac{d y}{d u}\right)}{1-k^{2} x^{2} y^{2}}$

$\displaystyle \implies \frac{d}{d u} \ln \left(y \frac{d x}{d u}-x \frac{d y}{d u}\right)=\frac{d}{d u} \ln \left(1-k^{2} x^{2} y^{2}\right)$

$\displaystyle \implies \frac{y \frac{d x}{d u}-x \frac{d y}{d u}}{1-k^{2} x^{2} y^{2}}= C$

$\displaystyle \implies \frac{~\mathrm{sn}(v) ~\mathrm{cn}(u) ~\mathrm{dn}(u)+~\mathrm{sn}(u) ~\mathrm{cn}(v) ~\mathrm{dn}(v)}{1-k^{2} ~\mathrm{sn}^{2}(u) ~\mathrm{sn}^{2}(v)} = C$

$\displaystyle C= F(A) =F(u+v)$ . Taking ${v=0}$ gives

$\displaystyle F(u) =~\mathrm{sn}(u)$

$\displaystyle \implies F(u+v)=~\mathrm{sn}(u+v)=C=\frac{~\mathrm{sn}(v) ~\mathrm{cn}(u) ~\mathrm{dn}(u)+~\mathrm{sn}(u) ~\mathrm{cn}(v) ~\mathrm{dn}(v)}{1-k^{2} ~\mathrm{sn}^{2}(u) ~\mathrm{sn}^{2}(v)}$

References:
1. Wikipedia
2. DLMF https://dlmf.nist.gov/22
3. Mathworld

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