Editing Jacobi elliptic functions (section)

==Jacobi elliptic functions as solutions of nonlinear ordinary differential equations==
===Derivatives with respect to the first variable===
The [[derivative]]s of the three basic Jacobi elliptic functions (with respect to the first variable, with <math>m</math> fixed) are:
<math display=block>\frac{\mathrm{d}}{\mathrm{d}z} \operatorname{sn}(z) = \operatorname{cn}(z) \operatorname{dn}(z),</math>
<math display=block>\frac{\mathrm{d}}{\mathrm{d}z} \operatorname{cn}(z) = -\operatorname{sn}(z) \operatorname{dn}(z),</math>
<math display=block>\frac{\mathrm{d}}{\mathrm{d}z} \operatorname{dn}(z) = - m \operatorname{sn}(z) \operatorname{cn}(z).</math>

These can be used to derive the derivatives of all other functions as shown in the table below (arguments (u,m) suppressed):

{| class="wikitable" style="text-align:center"
|+ Derivatives <math>\frac{\mathrm d}{\mathrm du} \operatorname{pq}(u,m)</math>
!colspan="2" rowspan="2"|
!colspan="4"|q
|-
! c
! s
! n
! d
|-
!rowspan="6"|p
|-
! c
|0 ||−ds ns ||−dn sn ||  −m' nd sd
|-
! s
|dc nc || 0 ||cn dn || cd nd
|-
! n
|dc sc || −cs ds || 0 || ''m'' cd sd
|-
! d
|m' nc sc || −cs ns || −''m'' cn sn ||0
|}

Also
:<math>\frac{\mathrm d}{\mathrm dz}\mathcal{E}(z)=\operatorname{dn}(z)^2.</math>

With the [[#Addition theorems|addition theorems above]] and for a given ''m'' with 0&nbsp;<&nbsp;''m''&nbsp;<&nbsp;1 the major functions are  therefore solutions to the following nonlinear [[ordinary differential equation]]s:

* <math>\operatorname{am}(x)</math> solves the differential equations <math>\frac{\mathrm d^2y}{\mathrm dx^2}+m\sin (y)\cos (y)=0</math> and
:<math>\left(\frac{\mathrm dy}{\mathrm dx}\right)^2=1-m\sin(y)^2</math> (for <math>x</math> not on a branch cut)
* <math>\operatorname{sn}(x)</math> solves the differential equations <math>\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + (1+m) y - 2 m y^3 = 0</math> and <math> \left(\frac{\mathrm{d} y}{\mathrm{d}x}\right)^2 = (1-y^2) (1-m y^2)</math>
* <math>\operatorname{cn}(x)</math> solves the differential equations <math>\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + (1-2m) y + 2 m y^3 = 0</math> and <math> \left(\frac{\mathrm{d} y}{\mathrm{d}x}\right)^2 = (1-y^2) (1-m + my^2)</math>
* <math>\operatorname{dn}(x)</math> solves the differential equations <math>\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - (2 - m) y + 2 y^3 = 0</math> and <math> \left(\frac{\mathrm{d} y}{\mathrm{d}x}\right)^2 = (y^2 - 1) (1 - m - y^2)</math>

The function which exactly solves the [[Pendulum (mechanics)#Simple gravity pendulum|pendulum differential equation]],
:<math>\frac{\mathrm d^2 \theta}{\mathrm dt^2}+c\sin \theta=0,</math>
with initial angle <math>\theta_0</math> and zero initial angular velocity is

:<math>\begin{align}\theta&=2\arcsin (\sqrt{m}\operatorname{cd}(\sqrt{c}t,m))\\
&=2\operatorname{am}\left(\frac{1+\sqrt{m}}{2}(\sqrt{c}t+K),\frac{4\sqrt{m}}{(1+\sqrt{m})^2}\right)-2\operatorname{am}\left(\frac{1+\sqrt{m}}{2}(\sqrt{c}t-K),\frac{4\sqrt{m}}{(1+\sqrt{m})^2}\right)-\pi\end{align}</math>
where <math>m=\sin (\theta_0/2)^2</math>, <math>c>0</math> and <math>t\in\mathbb{R}</math>.

===Derivatives with respect to the second variable===

With the first argument <math>z</math> fixed, the derivatives with respect to the second variable <math>m</math> are as follows:

:<math>\begin{align}\frac{\mathrm d}{\mathrm dm}\operatorname{sn}(z)&=\frac{\operatorname{dn}(z)\operatorname{cn}(z)((1-m)z-\mathcal{E}(z)+m\operatorname{cd}(z)\operatorname{sn}(z))}{2m(1-m)},\\
\frac{\mathrm d}{\mathrm dm}\operatorname{cn}(z)&=\frac{\operatorname{sn}(z)\operatorname{dn}(z)((m-1)z+\mathcal{E}(z)-m\operatorname{sn}(z)\operatorname{cd}(z))}{2m(1-m)},\\
\frac{\mathrm d}{\mathrm dm}\operatorname{dn}(z)&=\frac{\operatorname{sn}(z)\operatorname{cn}(z)((m-1)z+\mathcal{E}(z)-\operatorname{dn}(z)\operatorname{sc}(z))}{2(1-m)},\\
\frac{\mathrm d}{\mathrm dm}\mathcal{E}(z)&=\frac{\operatorname{cn}(z)(\operatorname{sn}(z)\operatorname{dn}(z)-\operatorname{cn}(z)\mathcal{E}(z))}{2(1-m)}-\frac{z}{2}\operatorname{sn}(z)^2.\end{align}</math>