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The differential equation for the pendulum is $$\ddot{\theta}=-\frac{g}{L}\sin\theta$$ But physics professors (on youtube at least) turn this equation into $\ddot{\theta}=-\frac{g}{L}\theta$ and $\theta$ is assumed to be small. So what if the angle could be as big as you want? What is the real solution to $\ddot{\theta}=-\frac{g}{L}\sin\theta$? Is it unsolvable (and if it is, please show why. I don't understand Differential Algebra yet but I just want to familiarize myself with it)? I will accept any solution, non-elementary or not, but I don't want an analytic solution.

Edit: @mr_e_man said that a change of variables could transform the differential equation into $$\frac{d^2\theta}{d\tau^2}-\sin\theta=0$$

Kamal Saleh
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  • I think it's solvable with elliptic functions (but I haven't checked). – mr_e_man Dec 15 '22 at 02:12
  • The simple pendulum problem is discussed in Chapter 1 of Armitage and Eberlein's Elliptic Functions (https://books.google.com/books?id=zyxAb4ro-oMC&pg=PA1&source=gbs_toc_r&cad=3#v=onepage&q&f=false ) , which you can view at the link (the book is still in print, so a complete file is not available online). –  Dec 15 '22 at 02:16
  • Note that the constants can be eliminated by a change of variables: Given $d^2\theta/dt^2=-g/L\sin\theta$, if we define $\tau=t\sqrt{g/L}$, then this becomes $d^2\theta/d\tau^2=-\sin\theta$. – mr_e_man Dec 15 '22 at 02:19
  • Some discussions from a computational view can be found here: http://ed.quantum-bg.org/elliptic-integrals.pdf , https://www.cfm.brown.edu/people/dobrush/am33/Mathematica/ch4/solution.html . The solution to the pendulum motion is found numerically in general. –  Dec 15 '22 at 02:27
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    Your question is vague. Do you mean "Is there a $C^2$ solution, an analytic solution?a solution expressible in terms of the 'elementary 'functions,?etc. " As the first comment says, there is a solution in terms of elliptic fuctions but you should also study general existence theorems for differential equations, series solutions and the Galois theory of differential equations. – P. Lawrence Dec 15 '22 at 03:07
  • What do you think an "analytic solution" is? – David K Dec 15 '22 at 05:58
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    3blue1brown made a series about differential equations, one of the first topics he touches is the pendulum, it will explain it perfectly for you: https://youtu.be/p_di4Zn4wz4 – knight5478 Dec 15 '22 at 06:24
  • Thanks @tau_knight ! – Kamal Saleh Dec 15 '22 at 15:59

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With $\theta=\theta(x)$ $$\theta''=-\frac{g}{L}\sin(\theta)$$ Switch variables $$\frac {x''}{[x']^3}=\frac{g}{L}\sin(\theta)$$ Reduction of order $p=x'$ gives $$\frac {p'}{p^3}=\frac{g}{L}\sin(\theta)\quad \implies \quad \frac 1{p^2}+c_1=\frac{g}{L}\cos(\theta)\quad \implies \quad p=x'=\pm \sqrt { \frac{L}{c_1+g \cos (\theta )}}$$ One more integration gives $$x+c_2=\pm \sqrt L\int \frac{d \theta}{\sqrt{c_1+g \cos (\theta )}}$$ Using the tangent half-angle substitution $$\int \frac{d \theta}{\sqrt{c_1+g \cos (\theta )}}=\frac{2}{\sqrt{c_1+g}}\,F\left(\frac{\theta }{2}|\frac{2 g}{c_1+g}\right)$$ where appears the elliptic integral of the first kind.

The problem is now to inverse the result to obtain $\theta(x)$.

But, if $\theta$ is small $$F\left(\frac{\theta }{2}|\frac{2 g}{c_1+g}\right)=\frac{\theta }{2}\left(1+\frac{g }{12 (c_1+g)}\theta ^2+\frac{g(7 g-2 c_1)}{480 (c_1+g)^2} \theta ^4 +O\left(\theta ^6\right) \right)$$ and we could use power series reversion.

Let $$y=\pm\frac{(x+c_2) \sqrt{c_1+g}}{2 \sqrt{L}}$$ and an approximation will be $$\theta=2 y-\frac{2 g }{3 (c_1+g)}y^3+\frac{g (2 c_1+3 g)}{15 (c_1+g)^2}y^5+O\left(y^7\right)$$

All of the above would seriously simply after inreoducing the boudary conditions.

Claude Leibovici
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