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.
  • ...?
  • $\begingroup$ A good reason to get existence and unicity of solutions to $x'=v(x)$ with $\int_0^1x(t)dt= p$ for all $p\in\mathbb{R}^n$, would be the following (say for a field $v\in C^1(\mathbb{R}^n)$). Suppose you are able to build another equation $y'=V(t,y)$ such that all solution starting at time $0$ are still defined at time $1$, and $y(1)=\int_0^1 x(t) dt$, where $x$ solves $x'=v(x)$ with the same initial condition $x(0)=y(0)$. Then, since $ y(0) \mapsto y(1)$ is a homeomorphism, the result would be achieved. However I do not see clearly which $v$ allow such $V$. $\endgroup$ – Pietro Majer Jan 25 '13 at 17:08

For scalar equations, $$\dot x= v(t,x),\qquad x(t)\in\mathbb{R}$$ that is $n=1$, the problem is well-posed, provided (1) the solutions of Cauchy initial value problem (IVP) at $t=0$ are defined at least in the interval $[0,1]$, (which is ensured, for instance, assuming a linear growth: $|v(t,x)|\le a|x|+b\\ $), and (2) there are unique solutions to any IVP (this is the case, for instance, in the Lipschitz hypotheses, of course true for a smooth $v$). The reason is simply the fact that solution are totally ordered (a direct consequence of the uniqueness: they can't cross), that makes their mean on $[0,1]$ depend bi-continuously on the initial value $x(0)$.

If we drop the uniqueness of solutions of IVP's, and consider a scalar equation with only continuous $v$, like in the classical example $\dot x= |x|^{1/2}$, then there can be several solutions with the same mean value, e.g. on $[-1,1]$ (for instance, in the above example, there is a continuum of odd solutions, hence with zero mean value).

Incidentally, an interesting fact here is that for scalar equation with continuous nonlinearity, the mean on the interval can be used to parametrize an arc of solutions connecting two given solutions of the same IVB (a set which is called the "Peano brush", emanating from $x_0$ at $t=t_0$). In other words, the Peano brush is not only connected (which is true for any $n$), but also arc-wise connected, (which is in general false for $n > 1$). Also note that if we change the above example into the equation $\dot x= |x|^{1/2}\mathrm{sgn} x$, it still lacks uniqueness of IVP, but now it has a backward uniqueness, the reason being that $|x|^{1/2}\mathrm{sgn}$ is a monotone function. As a consequence, prescribing the mean value on $[0,1]$ does produce a well-posed problem for this equation.

For systems (i.e. $n > 1$) these problems (solutions of ODE with assigned value of a given functional $L$, like the average on $[0,1]$) have been widely studied I think in the $70's$. For special classes of continuous nonlinearities, such conditions may give well-posed problems where the usual uniqueness of IVP fails. For equations where the uniqueness of IVP is already ensured, like in your case, there can be good motivations for these problems (as I guess you already have). In both situations I suspect there is no general result, but interesting results may arise from particular equations. A first nice category for systems comes again from monotonicity. The corresponding theorem should be (I should check) :

Let $v:\mathbb{R}^n\rightarrow\mathbb{R}^n$ be continuous, monotone (i.e., $(v(x)-v(y))(x-y)\ge0$ for all $x$ and $y$, e.g. the gradient of a convex function), with linear growth. Then, for any $p\in \mathbb{R}^n$ there is a unique solution to $\dot x=v(x)$ with $\int_0^1 x(t)dt=p$.


$\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\ve}{\varepsilon}$ $\newcommand{\pa}{\partial}$ $\newcommand{\eL}{\mathscr{L}}$

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

$$\phi(t) dt \to \mu_0(dt) $$


$$ \delta L_\phi\to \delta L_{\mu_0}. $$

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.

  • 1
    $\begingroup$ A natural condition should be that for all $x$, $DV(x)$ is in a convex set of invertible linear maps, e.g. definite positive or monotonic, so integrating $DV(x)$ along $u_y$ one gets an invertible $DL_\phi$. $\endgroup$ – Pietro Majer Jan 25 '13 at 17:42

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