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Let us consider an ODE $$\frac{dx_t^y}{dt}=g(x_t^y),$$ where y is the initial condition i.e. $x_0^y=y$.

Now, given a function $f$ (increasing and smooth) is it possible to find $g$ (i.e. an ODE) such that $$x_1^y = f(y).$$


The motivation comes from the study of particular flows on the bi-dimensional torus and their first return map. The specific question is quite long to explain in detail, but if you are interested I can try to state it better


Note added by B. Kloeckner: this is the improved version of the question, pasted from math.SE. I voted to reopen this version, as I think the question, while deserving this improvement by its authors, was unduly migrated, possibly because some of the voters did not realize that the ODE to be found is autonomous.

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Aded in edit: as pointed out by Christian Remling, my answer assumes implicitly that $f'$ never vanish.

The question is basically whether diffeomorphisms of the line isotopic to identity embed in flows; this is know to be very false in compact manifolds (see the work of Palis and this MO question), but I do not know the answer for the line. Here is a reduction of the problem.

Assume $g$ exists as wanted, let $y$ be a point not fixed by $f$ (if $y$ is fixed then we set $g(y)=0$), and let $x$ be the solution of $x'=g(x)$ such that $x(0)=y$. Then by separating variables, we get $$ \int_0^1 \frac{x'(t)}{g(x(t))} dt = 1$$ then by using the variable $u=x(t)$, we get $$ \int_y^{f(y)} \frac{du}{g(u)} =1$$ Denoting by $G$ an antiderivative of $1/g$ (which must be well defined away from the fixed points of $f$), we are reduced to solve the functional equation $$ G\circ f - G =1.$$ Now, given the structure of the line, it is not difficult to construct solutions of this equation by chosing them almost arbitrarily on a fundamental domain and then extend by the functional equation... but this works only on each interval between two fixed points of $f$. The issue of gluing these solutions at fixed points to get a decent $g$ seems not so easy to me (note that $G$ diverges at fixed points, which is pretty unsurprising given that $g$ should vanish there). However, in some cases (e.g. $f$ has no fixed point), it can certainly be done.

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In this generality, the answer is no, for the simple reason that a flow always has a positive derivative, so no $f$ with $f'=0$ somewhere can be realized in this way. (Such an $f$ could be fixed point free, of course.)

To spell this out more explicitly, if $\phi_t(y)$ denotes the flow of $\dot{x}=g(x)$, then $u(t;y)=d\phi_t(y)/dy$ obeys$$ \dot{u} = g'(u(t;y)) u , \quad\quad u(0)=1 . $$ This is solved by $u(t)=\exp(\int_0^t g'(u(s))\, ds)$; in particular, $u>0$ always, as claimed.

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