Higher-order ODEs as first-order ODEs

Question

"A second order ODE that lives in $\mathbb{R}^{n+1}$ is a first-order ODE that lives in $\mathbb{R}^{2n+1}$".

I don't understand this point and I think I'm missing some basic ideas here. Can someone fill me in? I was going through this(section 3.3).

Answer 1

It's easiest to see how this works in the case of a one dimensional second order ODE, i.e. an equation of the form $$F(x, \dot x, \ddot x, t) = 0$$

where $x$ and $t$ are in $\mathbb{R}$, so the ODE "lives in" $\mathbb{R}^{1+1}$. Now, let's say you're able to solve for $\ddot x$. Then you have

$$\ddot x = G(x, \dot x, t)$$

Now let $\dot x = y$, so that $\ddot x = \dot y$ . We now have a differential equation for the vector $\begin{bmatrix}x \\y\end{bmatrix}$: $$\frac{d}{dt}\begin{bmatrix}x \\y\end{bmatrix} = \begin{bmatrix}y \\G(x,y,t)\end{bmatrix}$$.

This is a first order equation for a vector in $\mathbb{R}^2$, and you've still got $t\in\mathbb{R}$, so your equation went from being second order in $\mathbb{R}^{1+1}$ to first order in $\mathbb{R}^{2+1}$. This generalizes to higher dimensions straightforwardly. If you want some further reading in this regard, or just another source from which to learn this topic, I highly recommend V.I. Arnold's Ordinary Differential Equations.