I'm reading Arnold's book on classical mechanics. This is kind of a dumb question, but I'm having problems understanding his explanation for existence and uniqueness of Newton's laws. On page $8$ he discusses Newton's law $\textbf{x}''=\textbf{F}(\textbf{x},\textbf{x}',t)$ and how it follows from a basic existence and uniqueness theorem that a unique solution exists $\textbf{x}$ for some time. But I'm having some trouble seeing this:

I know the theorem he's referring to for non-autonomous systems: If $J\subseteq\mathbb{R}$ is open and $U\subseteq\mathbb{R}^n$ is open and $V:J\times U\to\mathbb{R}^n$ is a (smooth) vector-valued function, then for any $\textbf{y}_0\in\mathbb{R}^n$ and small enough time there exists a unique (smooth) solution $\textbf y$ with $\textbf{y}(0)=\textbf{y}_0$.

Now at first I thought he meant just replace $\textbf{y}\mapsto\textbf{x}'$ and use the above theorem twice (first solve for $\textbf{x}'$ then for $\textbf x$): Then you'd have $(\textbf{x}')'=\textbf{F}(\textbf{x},\textbf{x}',t)$. But this doesn't work because $\textbf{F}$ is also a function of $\textbf x$. Can someone please explain?

  • $\begingroup$ Isn't this just an application of Picard's theorem for each instant in time? It is a second order ODE. $\endgroup$ Commented Mar 8, 2016 at 3:24
  • $\begingroup$ Basically. Picard's theorem is the one I mentioned in the question. I know that if $x:\mathbb{R}\to\mathbb{R}$ and you have a second order ODE, then you can use a standard trick setting $(y=x',y'=x'')$ to turn it into a vector equation. The problem is $\textbf x$ is vector valued, so for say the first component of $\textbf{x}''$, it would be a function of all components of $\textbf{x}$ and $\textbf{x}'$, not just their first components functions. $\endgroup$
    – user153582
    Commented Mar 8, 2016 at 4:00
  • $\begingroup$ But it's all vectors; equivalent to three equations, all at the same time. Maybe you want to pose this in math, as it's a bit formal for physics. $\endgroup$ Commented Mar 8, 2016 at 4:06

1 Answer 1


You have a vector equation

$$ \vec{x}'' = \vec{F}(\vec{x},\vec{x}',t) $$

Let's define a new vector of length $2n$, $\vec{X} = (\vec{x}',\vec{x})$. Then in terms of this new vector, our differential equation becomes $$ \vec{X}' = (\vec{F}(\vec{X},t), \vec{G}(\vec{X},t)) $$ where $\vec{G}((\vec{a},\vec{b}),t)=\vec{a}$ just projects onto the first $n$ components of a length $2n$ vector. If you like, we can define a new function $\vec{Q}$ such that $\vec{Q}(\vec{X},t)=(\vec{F},\vec{G})$. Then

$$ \vec{X}' = \vec{Q}(\vec{X},t) $$

You now have a first order equation that, I believe, is of the form you want.

In general, you can always convert a high-order equation to a first order equation in a higher-dimensional vector space.


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