Differentiating by a function

Question

I'm trying to learn some physics, and so it has come to pass that I came across an example in Arnol'd's book, which reads $$ x''(t) = \frac{dU}{dx}, $$ where $U(x) := gx$ is a function of $x$ (this is supposed to be a fancy way to write down the equation that describes a stone falling down onto the Earth). Now one might assume that the RHS is some sort of Fréchet derivative of $U$, but then at least one should evaluate it at $x$, so that the equation would read $$ x''(t) = \frac{dU}{dx}(x). $$ But OK, I see how that might just be an abbreviated notation. Still, it is mysterious to me, and I feel like as though I'm missing some deep mathematical point that is supposed to prepare me for the subsequent chapters. An entirely different matter is the solution of the second order autonomous ODE given at this Wikipedia page, because the derivative by $t$ is not even continuous, as the standard example $$ t \mapsto \cos(nt) ~~~~ (n \in \mathbb N) $$ shows. My questions are the following two:

• How does one make sense of these differential expressions, ie. how is one to put them onto a rigorous footing?
• How does one make rigorous (in the sense of justifying every step, not just showing that what one arrives at is a solution) the mnemonic given for the first order ODE in the same article one heading earlier, which is also required? Is there a natural notion of infinitesimals or an infinitesimal calculus in which all these steps may be performed?

Answer 1

OK, it turns out that a few pages later, Arnol'd provides an explanation. It goes like this: Suppose one is given a point particle $x$ which moves about in a 3D space $\mathbb R^3$, and suppose then that a function $U: \mathbb R^3 \to \mathbb R$ is given; physicists seem to call such a thing a "scalar field". One can show that the direction of steepest descent of $U$ at a point $y \in \mathbb R^3$ is given by the additive inverse of the gradient $$ -\nabla U (y) = -\begin{pmatrix} \partial_1 U(y) \\ \partial_2 U(y) \\ \partial_3 U(y) \end{pmatrix}. $$ This is, in fact, not only a direction, but also an acceleration, because the vector is not normed. That is to say, if the particle moves according to the law $$ x''(t) = -\nabla U(x(t)), $$ then it moves in the direction of the steepest descent, and the acceleration is proportional (in fact equal) to the slope in that direction.

EDIT: Regarding the 2nd order ODE: I've found that the mystery disappears if one assumes the solution $x$ exists and is invertible. Then the derived equation is not, in fact, to be solved for $v$, but for the function $v \circ t$, where $t = t(x)$ is supposed to be a function of $x$. And the strange fraction has nothing to do with any chain rule, but may instead be understood as a limit. The equation for $v \circ t$ is then a sufficient condition for $x$ to be a solution.

Answer 2

You're really over-complicating things.

What you have is nothing but a real-valued function of a real variable, $U \colon \mathbf{R} \to \mathbf{R}$, and $dU/dx$ simply means its ordinary derivative $U' \colon \mathbf{R} \to \mathbf{R}$, well known from single-variable calculus. And the ODE asks you to find a function $f \colon \mathbf{R} \to \mathbf{R}$ such that $f'' = U' \circ f$, or in other words $$ f''(t) = U'(f(t)) . $$ Here I have used a separate letter $f$ for the sought function, i.e., “$x=f(t)$”, in order to avoid the abuse of notation “$x=x(t)$”, which seems to be what confused you.