Is this function an alternative solution to the nonlinear pendulum?
Introduction
I am working with the differential equation of the frictionless nonlinear pendulum: $$\ddot{\theta}(t) + b\,\sin(\theta(t)) = 0 \tag{Eq. 1}$$
Which it already have known exact solution: $$\theta(t)=2\sin^{-1}\left(\sin\left(\frac{\theta_0}{2}\right)\text{sn}\left[K\left(\textstyle{\sin^2\left(\frac{\theta_0}{2}\right)}\right)-\sqrt{b}\,t\,;\,\sin^2\left(\frac{\theta_0}{2}\right)\right] \right)\tag{Eq. 2}$$ with $\text{sn}[z;m]$ one of the Jacobi elliptic functions with mode $m$, $K(z)$ the Complete elliptic integral of the first kind, and $\theta_0 = \theta(t=0)$.
Also, on Wolfram-alpha the solution to Eq. 1 is given by: $$\theta(t)=\pm\,\text{am}\left[\frac{1}{2}\sqrt{(2b+c_1)(t+c_2)^2}\,;\,\frac{4b}{2b+c_1}\right] \tag{Eq. 3}$$ with $\text{am}[z\,;\,m]$ the Jacobi amplitude function which fulfills $\text{sn}[z\,;\,m] = \sin\left(\text{am}[z\,;\,m]\right)$, and $c_1,\,c_2$ are integration constants.
If the initial conditions $\theta_0 = \frac{\pi}{2}$ and $\theta'(0)=0$ are introduced in Wolfram-Alpha the solution takes the form: $$\theta(t) = 2\,\sin^{-1}\left(\frac{\text{cd}\left[\sqrt{b}\,t\,;\,\frac{1}{2}\right]}{\sqrt{2}}\right)\tag{Eq. 4}$$ where $\text{cd}[z\,;\,m]$ is another secondary Jacobi elliptic function.
Main text
I was trying to apply the Weierstrass substitution $x(t)=\tan\left(\frac{\theta(t)}{2}\right)$ to Eq. 1 since it made possible to change the sine function into polynomials: $$ \begin{array}{c} (i)\,\,\sin(\theta) = \frac{2x}{1+x^2}\,; & (ii)\,\,\dot{\theta} = \frac{2\dot{x}}{1+x^2}\,; & (iii)\,\,\ddot{\theta} = \frac{\ddot{x}(1+x^2)-2x(\dot{x})^2}{(1+x^2)^2} \end{array}$$ and since $1+x^2 > 0,\,\,\forall x \in \mathbb{R}$ with this change of variable Eq. 1 becomes: $$\ddot{x}+b\,x = \frac{2x(\dot{x})^2}{(1+x^2)} \tag{Eq. 5}$$ where it can be seen that Wolfram Alpha find the solution to Eq. 5 given by complex numbers: $$ x(t) = \pm i\,\text{sn}\left[i\sqrt{b+c_1}\,(t+c_2)\,;\,\frac{c_1}{b+c_1}\right] \tag{Eq. 6}$$
By a "happy accident" trying to solve Eq. 5, I found about the existence of the Lemniscate elliptic functions: lemniscate sine function $sl(x)$ and lemniscate cosine function $cl(x)$, which haves only one variable and fulfills the following properties:
- $(iv)\quad sl' = cl\,(1+sl^2)$
- $(v)\quad cl' = -sl\,(1+cl^2)$
- $(vi)\quad (sl\,cl)' = cl^2-sl^2$
- $(vii)\quad sl^2+sl\,cl+cl^2 = 1$
- $(viii)\quad (sl')^2+sl^4 = 1$
- $(ix)\quad sl^4+sl^4cl^2+sl^2cl^2 = sl^2$
- $(x)\quad (1+sl^2)(1+cl^2)=2$
- $(xi)\quad cl(t) = \text{cn}\left[\sqrt{2}\,t\,;\,\frac{1}{\sqrt{2}}\right]$
- $(xii)\quad sl(t) = \displaystyle{\frac{\text{sd}\left[\sqrt{2}\,t\,;\,\frac{1}{\sqrt{2}}\right]}{\sqrt{2}}}$
Here is where the "issue" begins, since I found the following by accident, I don´t know if it is completely right, and maybe there is an "initial constants issues" that could made it an incomplete solution, as Eq. 4 is different from Eq. 3... but I believe that the following function is a solution to Eq. 5: $$ x(t) = cl\left(\sqrt{\frac{b}{2}}\,t-c_0\right) \tag{Eq. 7}$$
Differentiating Eq. 7 through property $(v)$ which lead to (I will be omitting the argument): $$ x'(t) = -sl\,(1+cl^2)\sqrt{\frac{b}{2}} \tag{Eq. 8}$$ And differentiating Eq. 8 through properties $(iv),\,(v),\,(x)$ which lead to: $$ \begin{array}{r c l} x''(t) & = & -\sqrt{\frac{b}{2}}\left\{cl(1+sl^2)(1+cl^2)\sqrt{\frac{b}{2}}+sl\,(2\,cl)(-sl(1+cl^2))\sqrt{\frac{b}{2}} \right\} \\ & = & -\frac{b}{2}\left\{2\,cl-2\,cl\,sl^2(1+cl^2)) \right\} \\ & = & -b\left\{cl-cl\,sl^2(1+cl^2)) \right\} \tag{Eq. 9} \end{array}$$
Now, solving the Left-Hand-Side (LHS) and Right-Hand-Side (LHS) of Eq. 5 separately will lead to: $$\text{LHS} = x'' + b\,x = -b\,cl + b\,cl\,sl^2(1+cl^2)+b\,cl = b\,cl\,sl^2(1+cl^2) \tag{Eq. 10}$$ $$\text{RHS} = \frac{2x(\dot{x})^2}{(1+x^2)} = \frac{2\,cl\left(-sl(1+cl^2)\sqrt{\frac{b}{2}}\right)^2}{1+cl^2} = \frac{b\,cl\,sl^2(1+cl^2)^2}{1+cl^2} = b\,cl\,sl^2(1+cl^2) \tag{Eq. 11}$$
If I didn´t make any mistakes, since Eq. 10 = Eq. 11, I think that Eq. 7 is indeed a solution of Eq. 5, so a solution to Eq. 1 will be: $$ \theta(t) = 2\tan^{-1}\left(cl\left(\sqrt{\frac{b}{2}}\,t-c_0\right)\right) \tag{Eq. 12}$$ Where the integration constant will be given by (using their own inverse functions): $$c_0 = \frac{1}{\sqrt{2}}F\left[\cos^{-1}\left(\tan\left(\frac{\theta_0}{2}\right)\right)\,;\,\frac{1}{\sqrt{2}}\right] \tag{Eq. 13}$$ with $F[z,\,;\,m]$ the Incomplete elliptic integral of the first kind. Here since the constant is dependent of having well obtained Eq. 12, I am not so interested in it, being Eq. 12 the main topic from the questions.
The Big Questions
Is Eq. 12 a "true" solution for the frictionless nonlinear pendulum?... Hope if there are calculation mistakes or conceptual ones, please show on your answer how to properly find the solution (if there is any by using the Weierstrass Substitution).
Are all the presented solutions equivalent?... I am not familiar with nonlinear differential equations, and I have read that sometimes don´t fulfill uniqueness of solutions, so I would like to know if they are the same solution or not (myself I tried to plot them but "softwares" have a bad time working with elliptic functions). Also, I have no formal training on elliptic functions (I discovered them on Wikipedia), so maybe equivalence among solutions could be "easily" proved (I hope) through Jacobi Theta Functions or Neville Theta Functions, or other elliptic functions, but unfortunately for me are too complicated and I don´t understand how to work with them from the Wiki web pages.
There is a general solution in terms of lemniscatic sine/cosine functions? This for avoiding the second parameter of the Elliptic Functions. Here note that solution of Eq. 7 and 12 only depends of one integration constant!! which must be wrong for a second order ODE!!... I believe the solution I found works only when starting from rest $\dot{\theta}(t)=0$, but I am not really sure - Hope you can give the general form of this solution in terms of the lemniscatic function (I tried to work with Eq. 6 on Wolfram Alpha but a complex elliptic function was too heavy for the free version I think).
Motivation
The nonlinear pendulum with friction is known for being the simple and classical example of a mechanical system with friction included, by so far there is not known exact solution to its differential equation: $$\ddot{\theta}(t) + a\,\dot{\theta}+b\,\sin(\theta(t)) = 0 \tag{Eq. 14}$$
By using the Weierstrass Substitution $x(t)=\tan\left(\frac{\theta(t)}{2}\right)$ to Eq. 14 it will becomes: $$\ddot{x}+a\,\dot{x}+b\,x = \frac{2x(\dot{x})^2}{(1+x^2)} \tag{Eq. 15}$$ which is highly similar to Eq. 5, so I am trying to find if something of the form $x(t)=q(t)\,cl(\sqrt{b/2}\,t-c)$ could be a solution for a function $q(t)$ to be determined (but it rapidly becomes in a mess).
Added Later
I was trying to develop Eq. 6 in something similar to Eq. 7, but things are more confusing now... using what is said on Wikipedia I can use the following property: $$ (xiii)\quad \text{sn}[ix;\,m] = i\,\text{sc}[x;\,1-m]$$ where $\text{sc}[x;\,m]=\frac{\text{sn}[x;\,m]}{\text{cn}[x;\,m]}$ another Jacobi Elliptic function. But reading the reference of Wikipedia, it takes it from Page 504 of Wiki-Reference where the property is display as $\text{sn}[ix;\,k] = i\,\text{sc}[x;\,k']$ where in Page 493 is defined the complementary modulus as $k^2+k'^2=1$, which is different from property $(xiii)$ - I don´t know if there is a mistake or both are right because of other properties of the Jacobi elliptic functions, but since in Wolfram-Alpha is implemented the Wikipedia version, I will use it as if it is correct.
With this, Eq. 6 becomes: $$ x(t) = \pm \text{sc}\left[\sqrt{b+c_1}\,(t+c_2)\,;\,\frac{b}{b+c_1}\right] \tag{Eq. 16}$$
Which will imply that: $$\theta(t) = 2 \tan^{-1}\left(\text{sc}\left[\sqrt{b+c_1}\,(t+c_2)\,;\,\frac{b}{b+c_1}\right] \right) \tag{Eq. 17}$$ which should be the same solution as Eq. 2 by Uniqueness (as I get convinced from here - but for me is not trivial at all, I don´t know if after matching the integration constants will lead to some new properties, but is of reach of my current skills).
At least, using property $(xi)$ in Eq. 7, one can see the proposed solution as: $$x(t) = \text{cn}\left[\sqrt{b}t-c_3;\,\frac{1}{\sqrt{2}}\right] \tag{Eq. 18}$$ but I have not been able to match the constants so Eq. 18 and Eq. 16 becomes equivalent, which should be the case at some pair of integrating constant due Uniqueness of solutions (I don't believe I accidentally found something that drop Uniqueness theorem of Lipschitz Diff. Eq... so If you found the corresponding constants, will be highly appreciated).
