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Let consider a simple pendulum of angular displacement $\theta(t)$ that starts at $\theta(0) = \alpha \in (-\pi,\pi)$ with no initial velocity.

$\theta$ is given by the ODE $$ \left\{ \begin{aligned} & \theta^{\prime\prime}+\omega^2 \sin \theta = 0 \\ & \theta(0) = \alpha,\ \theta^\prime(0) = 0 \end{aligned} \right. $$ where $\omega := \sqrt{g/\ell}$, with $g$ the acceleration due to the gravity and $\ell$ is the length of the pendulum.

Physics textbooks tells us that under the small-angle approximation $\theta \ll 1$, since $\sin \theta \simeq \theta$, one gets the linearised equation $$ \left\{ \begin{aligned} & \theta_1^{\prime\prime}+\omega^2 \theta_1 = 0 \\ & \theta_1(0) = \alpha,\ \theta_1^\prime(0) = 0 \end{aligned} \right. $$ and one has $\theta \simeq \theta_1$.

Of course this is not very convincing... A lot of ODEs have chaotic behaviour, so it is not clear at all that replacing $\sin \theta$ by $\theta_1$ will not lead to huge changes to the solution.

I am looking for a rigorous mathematical analysis of this problem. I am particularly interested for a result that looks like an error estimate of the form $$\forall t \geq 0,\quad \vert \theta(t) - \theta_1(t) \vert \leq F(t,\alpha)$$ where $F$ is as explicit as possible, with of course $F(t,\alpha) \to 0$ when $\alpha \to 0$.

I would also be interested for the high-order approximations $\theta_n$, where we replace $\sin \theta$ by the $n$ first terms of its Taylor series. For instance, $$ \left\{ \begin{aligned} & \theta_3^{\prime\prime}+\omega^2 \left( \theta_3 - \frac{1}{6} \theta_3^3+\frac{1}{150}\theta_3^5 \right) = 0 \\ & \theta_3(0) = \alpha,\ \theta_3^\prime(0) = 0 \end{aligned} \right. $$

Thanks!

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  • $\begingroup$ The rigorous argument here starts by showing that the exact solution is periodic. This basically follows from the energy conservation equation. You can then estimate the period and amplitude of the underlying oscillator. Given a discrepancy between the periods, the two oscillators will slowly drift out of phase. As for the case of adding higher order terms, these turn out to spoil the approximation due to the introduction of "spurious resonance", causing growing amplitude. These terms are called "secular" in multi scale analysis and there are particular multi scale methods for handling them. $\endgroup$ – Ian Oct 20 '19 at 0:02
  • $\begingroup$ Here the idea is to write the solution as a sinusoid with slowly varying frequency and phase shift and then separate scales to resolve the asymptotic dynamics of the frequency and phase shift. This form is still approximate but it avoids introducing any resonances. $\endgroup$ – Ian Oct 20 '19 at 0:03
  • $\begingroup$ Thanks ! Do you have any reference for (1) the rigorous argument using that the exact solution is periodic and (2) for the higher order terms ? $\endgroup$ – Héhéhé Oct 20 '19 at 0:32
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You won't get an error bound for all $t$ like you ask because the frequency of the approximation and the frequency of the real pendulum differ, so $\theta$ will drift away from the approximation. The fact that the real solution is periodic comes from conservation of energy and the form of the equation. The maximum amplitude has to repeat, then you are in the same state as the last maximum, so the swing will repeat, and the position is periodic.

I don't believe you can get a closed form solution for the exact equation. What we usually do is expand the solution in a series. You write $$\theta''(t)+\omega^2 \theta(t)=\omega^2(\theta(t)-\sin(\theta(t))\approx \omega^2\left(\frac {\theta(t)^3}6-\frac{\theta(t)^5}{120}+\ldots\right)$$ The leading term cancels on the left, so you can solve the right ignoring the left, then plug each solution into the right and solve again. You get a series expansion in harmonics of the original $\omega$

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constant energy curves

As told in the comments, using the energy function simplifies the situation somewhat. Multiply with $2θ'(t)$ and integrate to get $$ θ'(t)^2+2ω^2(1-\cosθ(t))=θ'(t)^2+\left(2ω\sin\frac{θ(t)}2\right)^2=ω^2R^2=ω^2\left(2\sin\frac{α}2\right)^2 $$ It follows that the points $(\frac{θ'(t)}ω, 2\sin\frac{θ(t)}2)$ lie on a circle of the small radius $R$.

parametrization by modified angle

Now introduce $\phi$ as the angle on that circle, $\phi(0)=\frac\pi2$. Then $$ θ'(t)=ωR\cosϕ(t), ~~2\sin\frac{θ(t)}2=R\sinϕ(t). $$ Then also $$ \cos\left(\frac{θ(t)}2\right)θ'(t)=R\cos(ϕ(t))ϕ'(t) $$

connecting the pendulum angle and the modified phase space angle

Assuming that the solution is not constant with $θ'(t)=0$, which could only happen for $α=0$, then this means that $$ ω\cos\left(\frac{θ(t)}2\right)=ϕ'(t) \\ \implies ω=\left(1-\frac{R^2}{4}\sin^2ϕ(t)\right)^{-\frac12}ϕ'(t) =ϕ'(t)+\frac{R^2}{8}\sin^2(ϕ(t))ϕ'(t)-\frac{R^4}{32}\sin^4(ϕ(t))ϕ'(t)\pm... $$ Using $\sin^2ϕ=\frac{1-\cos(2ϕ)}2$, $\sin^4ϕ=\frac{\cos(4ϕ)-4\cos(2ϕ)+6}8$ etc., the integration of the right side will give trend terms and oscillation terms. With the two terms here this results in \begin{align} ωt&=\left[ϕ(t)-\frac\pi2\right]+\frac{R^2}{16}\left[ϕ(t)-\frac\pi2\right]-\frac{R^2}{32}\sin(2ϕ(t))-\frac{3R^4}{128}\left[ϕ(t)-\frac\pi2\right]+\frac{R^4}{128}\sin(2ϕ(t))-\frac{R^4}{1024}\sin(4ϕ(t)) \\ &=\left(1+\frac{R^2}{16}-\frac{3R^4}{128}+O(R^6)\right)\left[ϕ(t)-\frac\pi2\right]-\left(\frac{R^2}{32}-\frac{R^4}{128}+O(R^6)\right)\sin(2ϕ(t))-\frac{R^4}{1024}\sin(4ϕ(t))+O(R^6) \end{align}

deriving approximation formulas

The linear approximation corresponds to setting $R=0$ in the last formula, so that $\phi(t)=\frac\pi2+ωt$. In the next order approximation one gets $$ ϕ(t)=\frac\pi2+\left(1-\frac{R^2}{16}\right)ωt-\frac{R^2}{32}\sin(2ωt) $$ so that one gets a correction to the frequency and at the same time small oscillations in the phase. The first implies a continuously growing phase difference in the small angle approximation, leading to a long period oscillation of the error of period $T=\frac{32\pi}{ωR^2}$, containing about $16R^{-2}$ oscillations of the pendulum.

For the rather large value of $θ(0)=α=0.4$ I get then the following plot of numerical solution of the pendulum, the small angles approximation and the second order approximation, and the error of the approximations against the numerical solution.

plot of numerical solution and approximations

As can be seen, the second order approximation gives a very good impression of what the error of the small angle approximation looks like, for a range of about $t< 40R^{-4}$

Using the second order approximation to replace the more exact numerical solution gives in the next plot a small but increasing phase shift in the first error plot in the next picture

enter image description here

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Just a note on this subject.

Starting from $$ \theta '' + \omega ^{\,2} \sin \theta = 0\quad \left| {\; - \pi < \theta \le \pi } \right. $$ it is interesting to see what happens if we take half of the angle $$ 0 = 2\alpha '' + \omega ^{\,2} \sin 2\alpha = 2\left( {\alpha '' + \left( {\omega ^{\,2} \cos \alpha } \right)\sin \alpha } \right)\quad \left| {\; - \pi /2 < \alpha \le \pi /2} \right. $$ So we have halved the oscillation, in change of introducing a variable frequency ($\cos \alpha$ is non negative).

Continuing we have $$ 0 = n\alpha '' + \omega ^{\,2} \sin n\alpha = n\left( {\alpha '' + \left( {{1 \over n}\omega ^{\,2} \,U_{\,n - 1} (\cos \alpha )} \right)\sin \alpha } \right)\quad \left| {\; - \pi /n < \alpha \le \pi /n} \right. $$ where $U_{\,n - 1}$ is the Chebyshev polynomial of 2nd kind.

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