The question I will ask makes sense in much more generality, but I will leave the translation to the experts, since I'm only looking for a special case (and it would not surprise me if the answer does not generalize). I will give some background, and then ask my question as a conjecture, set apart from the main text.
Let $mathbb R^n$ have its usual metric, and pick a differential one-form (= vector field) $B$ (the "magnetic potential") and a differential zero-form (= function) $C$ (the "electric potential"). Then consider the following second-order ODE for parameterized paths $gamma: [0,T] to mathbb R^n$:
$$ 0 = ddot gamma + dB cdot dotgamma + dC quadquad text{(EOM)} $$
I'll let you pick the signs for how the two-form $dB$ eats the vector $dotgamma$; just be consistent.
Then (EOM) is nondegenerate, and so a solution is determined by its initial conditions $(dotgamma(0),gamma(0))$. For each $T in mathbb R$, let $phi_T: mathbb R^{2n} to mathbb R^n$ be the "flow by time $T$" (actually, it is defined only on an open subset of $mathbb R^{2n}$, given by $phi_T(v,q) = gamma(T)$ for the solution $gamma\,$ to (EOM) with initial conditions $(dotgamma(0),gamma(0)) = (v,q)$. Then $phi_T$ is smooth; in fact, it is smooth in the $T$ variable as well. This follows from some standard fundamental result in ODEs, for which I don't have a good reference.
A path $gamma: [0,T] to mathbb R^n$ is classical if it satisfies (EOM); its duration is the number $T$. We can also consider paths with negative duration by flowing backwards, although we will not need to do so.
Definition: A point $(v,q) in mathbb R^{2n}$ is focal for duration $T$ iff ($phi_T(v,q)$ is defined and) $det(partial phi_T(v,q)/partial v) = 0$; i.e. fix the $q$, think of $phi_T(-,q)$ as a function of $v$ only, and ask that its differential is degenerate. By identifying $(v,q)$ with its classical path, we will talk about "focal (classical) paths" for given durations.
It is a standard results (see e.g. Milnor's Morse Theory) that for a given point $(v,q) in mathbb R^{2n}$, the durations $Tin mathbb R$ for which it is focal are discretely separated. Note that every $(v,q)$ is focal for duration $T=0$.
Proposition: Let $gamma$ be a classical path of duration $T$. Then it is non-focal if and only if it extends to a family of classical paths smoothly parametrized by the boundary positions $(gamma(0),gamma(T))$.
Sketch of Proof: Being focal for duration $T$ is a closed condition on $mathbb R^{2n}$, so we can vary $gamma(0) = q$ while remaining non-focal. But for non-focal paths we can vary $gamma(T)$ via the inverse function theorem.
Anyway, pick $q in mathbb R^n$, and $v = B(q)$ (or $-B(q)$ depending on your sign convention: for experts, I want the momentum to vanish). Then for some $epsilon>0$, for all $Tin (0,epsilon)$, $(v,q)$ is non-focal for duration $T$. Thus, for each $T in (0,epsilon)$, I can find an open neighborhood $q in mathcal O_0 subseteq mathbb R^n$ and another open neighborhood $mathcal O_1 subseteq mathbb R^n$ so that for $(q_0,q_1) in mathcal O_0 times mathcal O_1$, there is a non-focal classical path $gamma$ of duration $T$ with $gamma(0) = q_0$, $gamma(T) = q_1$, depending smoothly on the boundary conditions, and such that the classical path of duration $T$ and initial conditions $(dotgamma(0),gamma(0)) = (v,q)$ is contained within this family.
Note that as $T to 0$, the classical path with initial conditions $(dotgamma(0),gamma(0)) = (v,q)$ ends at a point very close to $q$. I don't know if I can take $mathcal O_1$ to actually contain $q$.
I would like to reverse the direction of choices: I'd like to pick $mathcal O_0,mathcal O_1$ first.
Question/Conjecture: Let $q in mathbb R^n$. Then there exist open neighborhood $mathcal O_0,mathcal O_1 subseteq mathbb R^n$, with $q in mathcal O_0,mathcal O_1$, and $epsilon>0$ such that:
- There exists a family of classical paths $gamma$ with boundary values varying in $mathcal O_0,mathcal O_1$ and with duration varying in $(0,epsilon)$. I.e. let $Delta = { (T,t) in mathbb R^2 : T in (0,epsilon), tin [0,T] }$; then there is a smooth function $gamma: mathcal O_0 times mathcal O_1 times Delta to mathbb R^n$ with: (a) $gamma(q_0,q_1,T,-)$ is classical for each $(q_0,q_1,T) in mathcal O_0 times mathcal O_1 times (0,epsilon)$, and (b) $gamma(q_0,q_1,T,0) = q_0$ and $gamma(q_0,q_1,T,T) = q_1$.
- For each $T in (0,epsilon)$, the classical path of duration $T$ with initial conditions $(B(q),q)$ appears as some $gamma(q,q_1,T,-)$.
For comparison, the corresponding theorem about geodesics on a Riemannian manifold is standard: around any point you can find a small neighborhood such that any two points in the neighborhood can be connected by a unique geodesic that does not leave the neighborhood. In fact, it follows from the proposition and the observation that changing the duration of a geodesic for fixed boundary conditions amounts just to a linear reparameterization.
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