How to use Lemniscate sine and Lemniscate cosine elliptic integrals?

Question

I’ve been reading up on lemniscate sine and cosine functions and found that they are essentially the lemniscate analogues of circular sine and cosine functions. The following wiki page elaborates a bit more on this

I’ve highlighted the parts of the article I didn’t quite understand. For example, what do variables s and c represent? What does t represent?

Below I’ve attached a visualization of the mental boundary I’m trying to overcome.

Is it that using the lemniscate sine and lemniscate cosine function give the lengths of the y and x components of the hypotenuse drawn from the origin (0,0) to endpoint of the lemniscate arc length?

The Wikipedia page: https://en.wikipedia.org/wiki/Lemniscatic_elliptic_function

Thank you in advance!

Answer 1

Just take a point $P$ in first quadrant which lies on lemniscate. Traverse the curve from origin to $P$ in first quadrant. Let the length of this traversed part of curve be $l$ then line segment $OP$ has a length equal to $\operatorname {sl} (l) $.

On the other hand if $l'$ is the length of curve traversed from point $(1,0)$ to $P$ (see yellow part in your image) in first quadrant then we have $OP=\operatorname{cl} (l') $.

Further note that neither $x$ nor $y$ of your image but the length $OP =\sqrt{x^2+y^2}$ is given by these lemniscatic functions.

Do you now understand what Wikipedia says? Let me know if some more clarification is needed.

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The polar equation of lemniscate is $r^2=\cos 2\theta$ and hence the arc-length $l'$ from $(1,0)$ to $P=(\rho\cos\phi, \rho\sin\phi) $ is given by $$l'=\int_{0}^{\phi}\sqrt{r^2+\left(\frac{dr}{d\theta}\right) ^2}\,d\theta =\int_{0}^{\phi}\frac{d\theta}{\sqrt{\cos 2\theta}} $$ Putting $t^2=\cos 2\theta $ and noting that $\rho=\sqrt{\cos 2\phi}$ we get $$l'=\int_{\rho} ^{1}\frac {dt} {\sqrt{1-t^4}}$$ Note that $\rho=OP=\operatorname{cl} (l') $ so we have $$l'=\int_{\operatorname {cl} (l')} ^{1}\frac{dt}{\sqrt{1-t^4}}$$ and similarly $$l=\int_{0}^{\operatorname {sl} (l)} \frac{dt} {\sqrt{1-t^4}}$$ and you get the formulas in Wikipedia (Wikipedia article replaces both $l, l'$ by $r$ and writes $s=\operatorname {sl} (r), c=\operatorname {cl} (r) $).

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The functions defined above can be expressed in terms of Jacobian elliptic functions of modulus $k=1/\sqrt {2}$ (and corresponding nome $q=e^{-\pi}$) as $$\operatorname {sl} (u) =\frac{1}{\sqrt{2}}\operatorname {sd} (\sqrt{2}u,k),\operatorname {cl} (u) =\operatorname{cn} (\sqrt{2}u,k)$$ These elliptic functions can then be evaluated using their Fourier series given below: \begin{align} \operatorname {sd} (u, k) &=\frac{2\pi}{Kkk'}\sum_{n=0}^{\infty} (-1)^n\cdot\frac{q^{n+(1/2)}\sin((2n+1)\pi u/(2K))}{1+q^{2n+1}}\notag\\ \operatorname {cn} (u, k) &=\frac{2\pi}{Kk}\sum_{n=0}^{\infty} \frac{q^{n+(1/2)}\cos((2n+1)\pi u/(2K))} {1+q^{2n+1}} \notag \end{align}

Answer 2

These integrals occur when integrating the quartic potential well:

$T = m \dot{x}^2 / 2$, $V = m k^2 x^4 / 2$

thus $L = - m \ddot{x} + 2 m k^2 x^3$

which gives an equation of motion $\ddot{x} = 2 k^2 x^3$

The equation $\ddot{x} = F(x)$ can be solved by multiplying both sides by $\dot{x}$ and integrating by parts: $$ \frac{d}{dt} \left( \frac{\dot{x}^2}{2} \right) = \ddot{x} \dot{x} = 2 k^2 \dot{x} x^3 = 2 k^2 \frac{d}{dt} \left( \frac{x^4}{4} \right)$$

After some manipulation you obtain Gauss's elliptic integral:

$k(t - t_0 ) = \int_0^x \frac{ds}{\sqrt{a^4 - s^4}}$

whose inverse gives the solution $x(t) = a\text{sl}(ak(t-t_0 ))$.

The integration constant $a$ arises from the first integral and $t_0$ from the second.