TITLE: When is a solution to an ODE determined by its average value?
QUESTION [5 upvotes]: 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.
...?

REPLY [2 votes]: $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\ve}{\varepsilon}$ $\newcommand{\pa}{\partial}$ $\newcommand{\eL}{\mathscr{L}}$ 
Fix a point $p_0\in \bR^n$ and a neighborhood $U$ of $p_0$ such that  for any  $p\in U$ the solution $x_p(t)$ of the IVP
$$ \dot{x}(t)= V(x(t)),\;\;x(0)=p $$
exists for all $t\in [0,1]$.   Define
$$L_\phi: U\to \bR^n,\;\; L_\phi(p)=\int_0^1 x_p(t) \phi(t) dt. $$
Clearly $L$ is smooth, and if its differential at $p_0$ is injective, then your question has, locally, a positive answer.     If $y\in T_{p_0}\bR^n$ and $ \delta L_\phi$ denotes the differential of $L_\phi$ at $p_0$,  then
$$ \delta L_\phi(y)= \int_0^1 u_y(t) \phi(t) dt, $$
where $u_y(t)$ is the solution of the  linear Cauchy problem
$$ u_y(0)=y,\;\;\frac{d}{dt} u_y(t)= A(t) u_y(t),  $$
and $A(t)$ denotes the differential of $V$ at $x_{p_0}(t)$.   
Thus, if the linear, nonautonomous  version of your question has a  positive  answer, then the nonlinear version  has  a positive answer, locally. 
The linear  case seems quite approachable. I might  have something more to say about this a bit later.
Addendum.  (I'm back from my class.) As Pietro indicated, positivity (monotonicity) conditions on $V$ guarantee a positive answer for any nonnegative $\phi$.    Here is a condition on $\phi$.   
Observe that the above arguments work without change if we replace $\phi(t) dt$ with an arbitrary measure $\mu(dt)$ on $[0,1]$.  Denote by $\mu_0$ the Dirac measure at $0$. Note that if 
$$\phi(t) dt \to \mu_0(dt) $$
then
$$ \delta L_\phi\to \delta L_{\mu_0}. $$
Clearly $\delta L_{\mu_0}$ is  the identity operator thus invertible. In particular if $\phi(t) dt$ is sufficiently close to the  Dirac $\mu_0$ the operator $\delta L_\phi$ will also be invertible.
Thus, for any  smooth vector field $V$ the problem has a positive local answer  if $\phi(t) dt$ is sufficiently close to the Dirac measure.