What is the domain of the action in classical mechanics?

Question

1. In classical mechanics we define the action as $$S[x(t)] = \int L(x(t),\dot{x}(t),t)\,\mathrm{d}t$$ and continue to minimize the action and derive the Euler-Lagrange equations. Given $S$ is a definite integral, it is a functional and hence maps to $\mathbb{R}$. I am wondering what space we are pulling the $x(t)$'s from? I would think it should be something like $\mathrm{C}^2(\mathbb{R})$ since we want $F = m\ddot{x}$ to make sense.

2. However, I know when we start talking about the Feynman path-integral the paths that come out are actually non-differentiable so clearly the domain has been enlarged in some sense. What is the domain in this setting then?

Answer 1

I think there should be some reason, why people are always so reluctant to specify this domain. Maybe this is because if the domain is specified, one ends up going outside that domain in applications anyway. All I can say about this is how the most basic choice of domain works:

Let $X\subseteq\mathbb{R}^n$ be open and $-\infty<\alpha<\beta<\infty$. Let $L:[\alpha,\beta]\times X\times \mathbb{R}^n\to \mathbb{R}$ be continuously differentiable. Then the integral $$S[x]:=\int_{\alpha}^{\beta}L(t,x(t),\dot{x}(t))\ dt$$ exists for any $x$ in the Banach space $C^1([\alpha,\beta],X)$.

One may ask what the stationary points of $S$ are in this space. Usually one wants specific boundary conditions for $x$. In this case one can restrict the domain to something like $\{x\in C^1([\alpha,\beta],X): x(\alpha)=a, x(\beta)=b\}$, which is still a Banach space. (edit: As pointed out by Philippe Malot, it is not a vector space, unless $a=b=0$. If that condition is fulfilled, it is a Banach space.) Here one later considers "variations that vanish on the boundary".

To get the Euler-Lagrange equations in this context one has:

Theorem If $x\in C^1([\alpha,\beta],X)$ (or in the subspace with fixed bounary values) is a local extremum of $S$ (defined as above) and further the function $$[\alpha,\beta]\ni t\mapsto \partial_3 L(t,x(t), \dot{x}(t))$$ is continuously differentiable, then the Euler-Lagrange equation holds: $$\partial_2 L(t,x(t),\dot{x}(t))=\frac{d}{dt}\{\partial_3 L(t,x(t),\dot{x}(t))\}.$$

Certainly this is way to restrictive for most applications. But at least it's a true, mathematically precise result. None of the above is even remotely useful when working towards path integrals. One can generalize the above to include piecewise differentiable functions, which allows them to have corners (and in non-novex variational problems these are often extremizers), but that is still fairly restrictive.

In a more geometric view of Classical Mechanics one works with smooth manifolds, and requires just about everything to be a smooth function. There is a lot of materials on Variational Calculus on smooth manifolds (I don't know any introductory source that I could recomend, in fact I'm looking for good references for this!). In a mathematical treatment of Classical Field Theory one would define the Lagrangian on a jet bundle.

Reference for quoted results:

Amann and Escher. Analysis 2. Birkhäuser, Basel, 2006.