How can we rescale the time in the differential equations?

Question

$$Ax'(t)+ Bx(t)=Du(t)$$ where $t\in[0,T_f]$.

I want to rescale the time $t\in[0,T_f]$ to $[0,1]$.

How can we rescale it?

Answer 1

I'm willing to explain this in full detail since just writing $x(\tau)$ and $u(\tau)$ feels wrong (abuse of notation) and non-intuitive to me.

Let $\tau := t/T_f$, then $t \in [0, T_f]$ maps to $\tau \in [0,1]$. Note that $t=T_f \ \tau$ and $\dfrac{{\rm d}t}{{\rm d}\tau} =T_f$.

Let $y(\tau) := x(t)=x(T_f \ \tau)$ and $v(\tau) := u(t) =u(T_f \ \tau)$. Then (chain rule) $$\frac{{\rm d}y}{{\rm d}\tau} =\frac{{\rm d}x}{{\rm d}t} \frac{{\rm d}t}{{\rm d}\tau} =T_f \ \dot x(t), \qquad\text{so that }\dot x(t) =\frac{1}{T_f} \frac{{\rm d}y}{{\rm d}\tau}.$$ Then rewrite the ODE in terms of $\tau$ and $y$: $$\frac{A}{T_f} \frac{{\rm d}y}{{\rm d}\tau} +B \ y(\tau) = D \ v(\tau).$$

Answer 2

$$ \hbox{With} \ \ \ \ \ t = \lambda \tau\\ \frac{d}{dt}x(t) = \frac{d}{d\tau}x(\tau)\frac{d\tau}{dt} = \frac{1}{\lambda}\dot x(\tau) $$