Derive the use of the Jacobi Amplitude function with the nonlinear pendulum diffequation

Question

I've been playing around with the pendulum differential equation ${\theta}''+\frac{g}{L}\sin{\theta}=0$, and have found many general solutions using the Jacobi Amplitude function $\text{am}(u,m)$ often multiplied with a trig function using the integral that evaluates period time. I'm somewhat of a beginner to the topic only being fluent in the first three courses of calculus as I am working on this for a physics project. All of the derivation videos I've watched make sense, but always arrive at the idea that these equations are 'really hard to solve' and only reveal how to derive the period time. I've tried applying generalizing $\theta$ to a power series and trying to get a solution for the coefficiens of $\theta$, $a_n$, but I couldn't solve this because of the nonlinear term. What I have so far is this: $$ \text{let } \theta(t)=\sum_{n\ge0} a_nt^n \\\text{if }\theta(0)=\theta_0=a_0 \text{ and }\theta'(0)=0=a_1 \\\text{so }\theta(t)=\theta_0+a_2t^2+a_3t^3+...+a_nt^n=\theta_0+\sum_{n\ge2}a_nt^n \\\text{and }\theta''(t)=2a_2+6a_3t+12a_4t^2+20a_5t^3+...+n(n-1)a_nt^{n-1} \\\text{then substituting}\sum_{n\ge0}n(n-1)a_nt^{n-2}+\frac{g}{L}\sin({\theta_0+\sum_{n\ge2}a_nt^n})=0 $$ how can I go from here to find that the coefficients of $a_n$ match the coefficients of some elliptic function. Also let me know if the concept for 'finding' an elliptic function in a diffeq is misconstrued or if a better approach without generalizing to a power series exists. Let me know if this question isn't clear so I can help you help me ;). PS, this is my first stack exchange post, so please let me know if there are problems with my latex or other things, thanks

Answer 1

Solving the ODE with a power series will work. What you are missing is the expansion of the non-linear term in a power series. After combing terms, you then obtain an equation $$\sum_n f_n t^n = 0\tag{1}$$ with $f_n$ some combinations of $a_n$. In order that the (1) is fulfilled at all times, we need $f_n$. From this you obtain a recurrence relation and you can find $a_{n\geq 2}$ as a function of $\theta_0$. Note that this procedure can turn out to be rather time-consuming.

Note that Jacobi elliptic functions are nothing more than defining (a more general version of) the functions that you obtain with this procedure and giving it a name. This allows to study properties, table them, and implement them in computer algebra systems, so that future generation (i.e. you) do not have to go through all the trouble again.

Edit: As Claude Leibovici has pointed out the procedure does not lead to a nice solution. Still one can learn the solution term by term. In order, to have nice expressions one needs to think a bit more about the problem. In particular, it is important to choose other initial condition in order to make the solution compact.

In general, the solution has two initial condition. One initial condition can be absorbed by using the fact that the ODE does not explicitly depend on time such that with $f(t)$ also $f(t+t_0)$ such that $t_0$ is a constant of integration.

This motivate the initial condition $$\theta(0)= 0, \theta'(0) = 2 \omega,\text{ with }\omega \in \mathbb{R},$$ for the equation $$\theta'' + k \sin \theta = 0.$$ Calling the solution to this equation $A(t,\omega)$ the general solution has the form $A(t+ t_0, \omega)$. Following your procedure, you will find $$A(t,\omega) =2 \omega t -\frac{1}{3} k \omega t^3+\frac{1}{60} k \omega \left(k+4 \omega ^2\right) t^5-\frac{1}{2520} k \omega \left(k^2+44 k \omega ^2+16 \omega ^4\right) t^7+O\left(t^9\right) $$

Performing the expansion to higher orders, you will rediscover the series representation of the function $2\mathop{\rm am}(\omega t; \sqrt{k}/\omega)$, with `am' the Jacobi amplitude function, see 22.16.6.

Answer 2

Let $$k=\frac g L \qquad s=\sin(a_0) \qquad c=\cos(a_0)$$ and the series expansion will be $$\color{blue}{\large\theta(t)=a_0+s\sum_{n=1}^\infty (-1)^n\, \frac{\alpha_n}{(2n)! }\, k^n\,t^{2n}}$$ the very first coefficients being $$\left( \begin{array}{cc} n & \alpha_n \\ 1 & 1 \\ 2 & c \\ 3 & c^2-3 s^2 \\ 4 & c^3-33 c s^2 \\ 5 & c^4-306 c^2 s^2+189 s^4 \\ 6 & c^5-2766 c^3 s^2+8289 c s^4 \\ 7 & c^6-24909 c^4 s^2+255987 c^2 s^4-68607 s^6 \\ 8 & c^7-224199 c^5 s^2+6988167 c^3 s^4-766073 \\ 9 & c^8-2017812 c^6 s^2+181346094 c^4 s^4-553514148 c^2 s^6+82908441 s^8 \\ \end{array} \right)$$

which, for sure, could simplify using $c^2+s^2=1$ and/or multiple angle formulae.

I am afraid that it would be quite tedious (not to say more) to find the general $\alpha_n$.

Answer 3

The question is about power series solutions to the equation of motion of the nonlinear simple gravity pendulum and its relation to the Jacobi amplitude function. This is a well studied problem with two centuries of history.

It usually begins with the derivation from Newtonion physics of an idealized model that leads to the second order autonomous ordinary differential equation (where $m$ denotes a dimensionless parameter)

$$ {\theta}''(t) + m\sin{\theta}(t)=0. \tag1 $$

One solution method begins with the initial Ansatz

$$ \theta_1(t) := \theta(t) = a_0 + a_1t + O(t^2). \tag2 $$

Now use the ODE $(1)$ to derive a second Ansatz

$$ \theta_2(t) := a_0 + \left( \int_0^t a_1 + \left( \int_0^t -m\, \theta_1(t)\,dt\right)\,dt \right)\!. \tag3 $$

Expand this expression (with $s := \sin a_0 ,\, c := \cos a_0 $) to get

$$ \theta_2(t) = a_0 + a_1 t - m\, s \frac{t^2}{2!} - a_1 m\,c \frac{t^3}{3!} + O(t^4). \tag4 $$

Similarly, the third Ansatz expands to

$$ \theta_3(t) = a_0 + a_1 t - m\, s \frac{t^2}{2!} - a_1 m\,c \frac{t^3}{3!} + \\ m\, s(a_1^2 + m\,c)\frac{t^4}{4!} + a_1 m(a_1^2 c + m\left(c^2-3s^2)\right)\frac{t^5}{5!} + O(t^6). \tag5 $$

There seems to be no closed form for the successive coefficients of these power series. They can be simplified somewhat in the special cases where $\,a_0=0\,$ or else $\,a_1=0\,$ to give odd or even solutions.

How does this relate t0 the Jacobi amplitude? One common definition of this function is that it is the solution to the ODE

$$ \text{am}'(t) = \sqrt{1-m\,\sin^2(\text{am}(t))}, \quad \text{am}(0)=0. \tag6 $$

This definition easily implies that

$$ 2\,\text{am}(t) = 2t - 2m\frac{t^3}{3!} + 2m(4 + m)\frac{t^5}{5!} - 2m(16 + 44m + m^2)\frac{t^5}{5!} + O(t^9) \tag7 $$

is an odd solution to the original ODE $(1)$.