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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?
Cloudscape
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  • You can interpret this via the chain rule; $\frac{dU}{dx} = \frac{dU}{dt}/(\frac{dx}{dt})$ – Andrea B. Dec 09 '21 at 16:24
  • @AmejiB. I honestly don't know the chain rule that you're applying, but I'd rather like to get to know it! Enlighten us! – Cloudscape Dec 09 '21 at 17:02
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    In physics, chain rule operates almost like treating derivatives as fractions. You have to be more careful in multivariate calculations, but since we just have everything as a function of time, namely x(t) and U(x) = U(x(t)), you can compute dU/dt = U'(x(t)) x'(t) = dU/dx dx/dt. Sorry for lack of LaTeX here – Andrea B. Dec 09 '21 at 17:23
  • @user264745 right. I guess to answer OP's last question, a real analysis class would treat this rigorously. But to get the point across, it's enough to abuse notation – Andrea B. Dec 09 '21 at 17:39
  • The first question is asking how to make sense of "differentiating by a function" (chain rule). The second is how differentials can be made rigorous (real analysis) – Andrea B. Dec 09 '21 at 17:57
  • @AmejiB. What's about discontinuity function. He gave as an example. I guess while dealing with $x(t), x^{\prime}, $ etc we don't have to worry about that because it is physically impossible. – user264745 Dec 09 '21 at 18:12
  • Exactly, that's why physicists usually don't deal with such details :P – Andrea B. Dec 09 '21 at 18:17
  • Regarding your very last question, see https://math.stackexchange.com/questions/27425/what-am-i-doing-when-i-separate-the-variables-of-a-differential-equation (and the many similar questions linking to it). – Hans Lundmark Dec 09 '21 at 18:37
  • @AmejiB. The equation that you give implies that $U'(x(t)) = dU/dx$. And of course I am aware of the rigorous calculus as first developed by Bolzano. My question was about manipulating with infinitesimals the right way. Perhaps I've got to find a source myself. – Cloudscape Dec 09 '21 at 21:57
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    Because $U$ is a function of $x$; $U(x) = gx$. Therefore, the natural setting is to take $dU/dx$. We can take a $t$ derivative of $U(x(t))$, but $U'$ should be coming from $x$, since we still plug in $x$ or $x(t)$. – Andrea B. Dec 09 '21 at 22:06
  • @AmejiB. OK I might accept that. Still, I feel uneasy writing things down like this. – Cloudscape Dec 09 '21 at 22:07

2 Answers2

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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.

Hans Lundmark
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  • You are ignoring the constraint on $x''(t) = \frac{dU}{dx}$. You can’t use generic $U$, potential energy function. By definition of potential energy function, $\nabla U$ exist. In particular, $U^{\prime}$ exist by definition. If one take generic $U$, then $U$ might not be differentiable. So talking about $x''(t) = \frac{dU}{dx}$ without any context(from where it came from) makes no sense. – user264745 Dec 09 '21 at 21:19
  • @HansLundmark Of course I did see the interpretation that you describe, but at the very least the notation is overloaded when one differentiates by a function. Therefore, I naturally thought I was missing something, and that's what my question is about. – Cloudscape Dec 09 '21 at 22:01
  • @user264745: It goes without saying that $U$ has to be differentiable if we are to talk about $dU/dx$... My point was only that $U'$ is the ordinary single-variable derivative, not “some sort of Fréchet derivative”. – Hans Lundmark Dec 10 '21 at 05:19
  • @AlgebraicsAnonymous: Why do you keep saying that we are differentiating with respect to a function? In the expressions $U(x)=gx$ and $dU/dx$, $x$ is just a variable. It's only later that the function $x(t)$ enters. – Hans Lundmark Dec 10 '21 at 05:21
  • @HansLundmark $x$ was defined to be a function by my book. That's why I assumed that $dU/dx$ means differentiating by a function. But see my answer below. – Cloudscape Dec 11 '21 at 12:18
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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.

Cloudscape
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