# Rescaling time in differential equations

differential equationsdimensional analysishomework-and-exercisesoscillatorsunits

On a scientific paper, I found the following equations about a compass gait (one leg behaves like an inverted pendulum, the other one as a simple pendulum; $$\theta$$ and $$\phi$$ are time-dependent):

$$\ddot{\theta} – \sin\left( \theta – \gamma \right) = 0$$

$$\ddot{\theta} – \ddot{\phi} + \dot{\theta}^2 \, \sin(\phi) – \cos(\theta – \gamma) \, \sin(\phi) = 0$$

I got the following ones, and also, in the paper, there are these before the sentence (*):

$$\ddot{\theta} – \frac{g}{l} \, \sin\left( \theta – \gamma \right) = 0$$

$$\ddot{\theta} – \ddot{\phi} + \dot{\theta}^2 \, \sin(\phi) – \frac{g}{l} \, \cos(\theta – \gamma) \, \sin(\phi) = 0$$

(*) The explanation of the paper to pass from the last two equations to the first two equations is "we have rescaled time by $$\sqrt{l/g}$$".

Could you tell me what it means?

It's an usual procedure in deriving non-dimensional equations, from the dimensional ones: angles have no physical dimensions, they wanted a "scaled" (non-dimensional) time as well.

You just need to define a "scaled-time" independent variable

$$\tau = \sqrt{\frac{g}{\ell}} t$$

perform derivatives using the rules for composite functions,

$$\phi'(\tau) := \frac{d}{d\tau} \phi(t(\tau)) = \frac{dt}{d\tau} \frac{d}{dt} \phi(t) = \sqrt{\frac{\ell}{g}} \dfrac{d \phi}{d t} = \sqrt{\frac{\ell}{g}} \dot{\phi}(t) \ ,$$

$$\rightarrow \qquad \phi'' = \frac{\ell}{g} \ddot{\phi} \quad , \quad \phi'^2 = \frac{\ell}{g} \dot{\phi}^2$$

and multiply all the equations by the factor $$\frac{g}{\ell}.$$