Fix a smooth vector field $\vec v$ on $\mathbb R^n$. It is well-known that any trajectory $\vec x : [0,1]\to \mathbb R^n$ solving the ODE $\dot x(t) = \vec v(\vec x(t))$ is determined by its evaluation $\vec x(t)$ for any $t\in [0,1]$. My question is about when a solution is determined by its average value.

More precisely, let $\phi$ be a smooth function on $\mathbb R$ vanishing identically outside $[0,1]$ and satisfying $1 = \int \phi(t)\ \mathrm{d}t$. For any $f : [0,1] \to \mathbb R$, let $\langle f\rangle = \int f(t)\ \phi(t)\ \mathrm{d}t$. My question is:

Under what conditions does the data of the vector field $\vec v$, the averaging function $\phi$, and the value $\langle \vec x\rangle \in \mathbb R^n$ determine the classical trajectory $\vec x$?

Here are two examples. Fix $\vec a \in \mathbb R^n$ and consider the constant vector field $\vec v(x) = \vec a$. Then $$ \vec x(t) = \vec a t + \langle \vec x \rangle - \vec a \int t\ \phi(t)\ \mathrm{d}t$$ and so the answer is that any $\phi$ works.

On the other hand, on $\mathbb R^2$ consider vector field $\vec v \bigl( \begin{smallmatrix} x_1 \\ x_2 \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} 4\pi\ x_2 \\ -4\pi\ x_1 \end{smallmatrix} \bigr) $. Then classical trajectories are of the form $$ x_1(t) = r\cos (4\pi(t - \theta)), \quad x_2(t) = r\sin(4\pi(t-\theta))$$ for fixed $r,\theta$. Choose a smooth function $\varphi$ which is identically $0$ on $(-\infty,0]$ and identically $1$ on $[1,\infty)$, and set $\phi(t) = 2\varphi(2t) - 2\varphi(2t - 1)$. If I haven't made an arithmetic error, then for any $r,\theta$ we have $\langle \vec x \rangle = \bigl( \begin{smallmatrix} 0 \\ 0 \end{smallmatrix} \bigr)$.

I could imagine answers of the following forms:

- For any vector field $\vec v$, there exists an averaging function $\phi$ that works: take a sufficiently good approximation of a delta-function.
- If $\vec v$ is bounded in some appropriate norm (or if the trajectory $\vec x$ is known a priori to stay in a region in which $\vec v$ is bounded), then there is some $\epsilon$ depending on the bound such that any $\phi$ with domain $(0,\epsilon)$ works. Or perhaps what works is any $\phi$ which is "within $\epsilon$ of a delta-function" in the appropriate sense.
- For any $\phi$, the only vector fields that fail to have their solutions determined in the above way have such-and-such property, and hence are few and far between.
- ...?