Setup
Let $d\in\mathbb N$. Let $f\colon\mathbb R^d\to\mathbb R^d$ be $L$-Lipschitz continuous and bounded by $C$ for some fixed constants $L,C\in(0,\infty)$, i.e., $\Vert f(x)\Vert \leq C$ and $\Vert f(x) - f(y)\Vert \leq L \Vert x - y \Vert$ for all $x,y\in\mathbb R^d$. Let $\varphi\colon\mathbb R^d \times [0, 1] \to \mathbb R^d$ be the flow induced by $f$, i.e., if $u \colon [0, 1] \to \mathbb R^d$ is the solution to the initial value problem $u(0) = u_0$, $\frac{d}{dt} u(t) = f(u(t))$, then $\varphi(u_0, t) = u(t)$ for $t\in[0,1]$, $u_0 \in \mathbb R^d$.
Questions
Assume we do not know the vector field $f$, but have access to the flow $\varphi$. Can we reconstruct $f$, i.e.:
- (i) is $f$ unique given $\varphi$?
- (ii) is there a closed form description of $f$ given $\varphi$?
This question comes in different flavors depending on what we know about the flow $\varphi$ and what we want to know about $f$ (the main question is 4.):
- We know $\varphi\colon\mathbb R^d \times [0, 1] \to \mathbb R^d$ on its whole domain and want to know $f\colon\mathbb R^d\to\mathbb R^d$ on its whole domain.
- We know $\varphi\colon[0,1]^d \times [0, 1] \to \mathbb R^d$ on a hypercube and want to know $f$ on the same hypercube $f\colon[0,1]^d\to\mathbb R^d$.
- We know $\varphi(\cdot, 1)\colon[0,1]^d \to \mathbb R^d$ on a hypercube at time $1$ and want to know $f$ on the same hypercube $f\colon[0,1]^d\to\mathbb R^d$.
- We know $\varphi\colon(\cdot, \delta) \to \mathbb R^d$ on a hypercube at time $\delta$ and want to know $f$ on the same hypercube $f\colon[0,1]^d\to\mathbb R^d$. Here $\delta>0$ can be chosen arbitrarily small. To be more precise, the question is whether following statement is true: For every $L$-Lipschitz, $C$-bounded vector field $f$, there is $\delta>0$, such that any other $L$-Lipschitz, $C$-bounded vector field $\tilde f$ with associated flow $\tilde\varphi$ and $\varphi(x, \delta) = \tilde\varphi(x, \delta)$ for all $x \in [0,1]^d$ agrees with $f$ on the hypercube, i.e., $\tilde f(x) = f(x)$ for all $x \in [0,1]^d$.
Some Answers
For 1. and 2. we can use $\frac{\partial}{\partial t} \varphi(x, 0) = f(x)$ to answer (i) and (ii) affirmative.
For 3., take $d=2$, define the vector fileds $f_0(x) = 0$ and $f_1(x_1, x_2) = (-x_2, x_1)$ with associated flows $\varphi_0$ and $\varphi_1$. Then $\varphi_0(x, 2 \pi) = x = \varphi_1(x, 2 \pi)$ and both $f_0$ and $f_1$ fulfill the boundedness and Lipschitz-conditions for $L, C$ large enough, but $f_0 \neq f_1$ on $[0,1]^d$.
For 4. we neither can take the derivative as in 1 or 2, nor can we make a construction as in 3. as for small $\delta$, we would need large values of the vector field, which is required to be bounded.
Reference Request
Does this problem (or something similar) come up in the literature?
Unhappily, such a $\delta$ doesn't exist, even in dimension $1$ (if we build a counterexample in dimension $1$, it is easy to build one in any finite dimension by setting the other coordinates of the vector fields at $0$). For this, consider the following lemma,
Lemma : For all $C > 0$, $L > 0$, $\delta > 0$, there exists a diffeomorphism $\phi : \mathbb{R} \rightarrow \mathbb{R}$ that verifies,
$*$ For all $x$, $\phi\!\left(x + \frac{\delta C}{2}\right) = \phi(x) + \frac{\delta C}{2}$ i.e. $\phi - \mathbb{id}$ is $\frac{\delta C}{2}$-periodic.
$*$ $\frac{1}{2} \leqslant \phi' \leqslant 2$.
$*$ $|\phi''| \leqslant \frac{L}{C}$.
$*$ $\phi \neq \mathrm{id}$.
And in this case, we have for all $x$, $\phi^{-1}\!\left(x + \frac{\delta C}{2}\right) = \phi^{-1}(x) + \delta C$ i.e. $\phi^{-1} - \mathrm{id}$ is $\frac{\delta C}{2}$-periodic and $\left|\frac{C\phi''}{2\phi'}\right| \leqslant L$.
For this, consider a non zero smooth compactly supported function $\theta : ]0,1[ \rightarrow \mathbb{R}$ and set $L_1 = \|\theta'\|_\infty$ and $L_2 = \|\theta''\|_\infty$. Let, $$ \phi : \left\{\begin{array}{rcl} \mathbb{R} & \rightarrow & \mathbb{R} \\ \frac{\delta C}{2}n \leqslant x < \frac{\delta C}{2}(n + 1) & \mapsto & x + K\theta\!\left(\frac{x - \delta Cn/2}{\delta C/2}\right) \end{array}\right. $$ where $K = \min\left\{\frac{\delta C}{4L_1},\frac{L\delta^2C}{4L_2}\right\} > 0$. The fact that $\theta$ is smooth and $\theta = 0$ in a neighborhood of $0$ and $1$ implies that $\phi$ is smooth and the facts that $\theta \neq 0$ and $K > 0$ imply that $\phi \neq \mathrm{id}$.
We have for all $x$, $$ \phi'(x) = 1 + \frac{2K}{\delta C}\theta'\!\left(\frac{x - \delta Cn/2}{\delta C/2}\right) \in \left[1 - \frac{2KL_1}{\delta C},1 + \frac{2KL_1}{\delta C}\right] \subset \left[\frac{1}{2},2\right], $$ because $K \leqslant \frac{\delta C}{4L_1}$. And, $$ |\phi''(x)| = \left|\frac{4K}{\delta^2C^2}\theta''\!\left(\frac{x - \delta Cn/2}{\delta C/2}\right)\right| \leqslant \frac{4KL_2}{\delta^2C^2} \leqslant \frac{L}{C}, $$ because $K \leqslant \frac{L\delta^2C}{4L_2}$.
$\phi' > 0$ hence $\phi$ is a diffeomorphism. $\phi - \mathrm{id}$ is $\frac{\delta C}{2}$-periodic hence for all $x$, \begin{align*} \phi^{-1}\!\left(x + \frac{\delta C}{2}\right) & = \phi^{-1}\!\left(\phi(\phi^{-1}(x)) + \frac{\delta C}{2}\right)\\ & = \phi^{-1}\!\left(\phi\!\left(\phi^{-1}(x) + \frac{\delta C}{2}\right)\right)\\ & = \phi^{-1}(x) + \frac{\delta C}{2}, \end{align*} so $\phi^{-1}$ also verifies this property. The last inequality $\left|\frac{C\phi''}{2\phi'}\right| \leqslant L$ immediately comes from $\phi' \geqslant \frac{1}{2}$ and $|\phi''| \leqslant \frac{L}{C}$.
Now, let $f_1 = \frac{C}{2}$ be a constant vector field on $\mathbb{R}$ (which is clearly bounded by $C$ and $L$-Lipschitzian) whose flow is $\varphi_1 : (t,x) \mapsto x + \frac{Ct}{2}$. Let $$ \varphi_2 : (t,x) \mapsto \phi(\varphi_1(t,\phi^{-1}(x))). $$ We have for all $x$, $\varphi_2(0,x) = x$ and for all $(t,s)$, \begin{align*} \varphi_2(t + s,x) & = \phi(\varphi_1(t + s,\phi^{-1}(x)))\\ & = \phi(\varphi_1(t,\varphi_1(s,\phi^{-1}(x))))\\ & = \phi(\varphi_1(t,\phi^{-1}(\phi(\varphi_1(s,\phi^{-1}(x))))))\\ & = \varphi_2(t,\varphi_2(s,x)). \end{align*} It means that $\varphi_2$ is a semi-group. Therefore, it is the flow of the vector field $f_2 : x \mapsto \partial_t\varphi_2(0,x)$. And for all $x$, \begin{align*} f_2(x) & = \partial_t(\phi(\varphi_1(t,\phi^{-1}(x))))|_{t = 0}\\ & = \phi'(\varphi_1(t,\phi^{-1}(x)))\partial_t\varphi_1(t,\phi^{-1}(x))|_{t = 0}\\ & = \frac{C}{2}\phi'(\phi^{-1}(x)). \end{align*} In particular, $|f(x)| \leqslant C$ for all $x$ because $|\phi'| \leqslant 2$ and, $$ f'(x) = \frac{C}{2}\phi''(\phi^{-1}(x))(\phi^{-1})'(x) = \frac{C\phi''(\phi^{-1}(x))}{2\phi'(\phi^{-1}(x))}. $$ In particular, $|f'(x)| \leqslant L$ so $f$ is $L$-Lipschitzian, because $\left|\frac{C\phi''}{2\phi'}\right| \leqslant L$. Now, by $\frac{\delta C}{2}$-periodicity of $\phi - \mathrm{id}$ and $\phi^{-1} - \mathrm{id}$, we have for all $x$, \begin{align*} \varphi_2(\delta,x) & = \phi(\varphi_1(\delta,\phi^{-1}(x)))\\ & = \phi\left(\phi^{-1}(x) + \frac{\delta C}{2}\right)\\ & = \phi(\phi^{-1}(x)) + \frac{\delta C}{2}\\ & = \varphi_1(\delta,x). \end{align*} It proves that $\varphi_1(\delta,\cdot) = \varphi_2(\delta,\cdot)$ and we can build such a vector field for any $\delta > 0$ arbitrarily small. Notice that $f_1$ doesn't depend on $\delta$, which is what you wanted, and all what is done in $\mathbb{R}$ can be restricted to $[0,1]$.