Fix $f: \mathbb{R}\times \mathbb{R}^n \mapsto \mathbb{R}^n$ and $v_0:\mathbb{R}^n \mapsto \mathbb{R}^n$. Let $X_t$ be the solution to the second-order ODE

$$\frac{d^2}{dt^2}X_t = f(t,X_t), \quad X_0 = id, \frac{d}{dt}X_t|_{t = 0}=v_0.$$

Under what conditions on $f$ and $v_0$ is $X_t$ a diffeomorphism?

  • $\begingroup$ Do you want it to be a diffeo for all time, or is just for a finite amount of time ok? (The finite time question is easier) $\endgroup$ – Kevin Casto Jun 15 at 15:46
  • $\begingroup$ @KevinCasto Preferably, for all time. $\endgroup$ – Ben Jun 15 at 16:03

First, I just want to consider the requirement that $X_t$ be a local diffeomorphism, which by the inverse function theorem is equivalent to requiring the Jacobian of $X_t$ to never vanish. We have $$ X_t = id + \int_0^t V_s\,ds = id + \int_0^t \left(v_0 + \int_0^s f_r(X_r)\,dr\right)\, ds \\= id + v_0 t + \int_0^t \int_0^s f_r(X_r)\,dr\,ds$$ and so, taking Jacobians, $$D X_t = I + (D v_0)t + \int_0^t \int_0^s D f_r(X_r) \bullet D X_r\,dr\,ds$$ where $\bullet$ denotes matrix multiplication. Mirroring the proof of Picard-Lindelöf, we can define a sequence converging to $DX_T$ by "repeatedly plugging the whole RHS into the term inside the integral", which should converge by the Banach fixed point theorem, assuming $f$ is Lipschitz. So we get an infinite sum that looks like $$DX_t = I + (Dv_0)t + \int_0^t \int_0^s D f_r(X_r) + r\,D f_r(X_r)\bullet D v_0\,dr\,ds \\ + \int_0^t\int_0^s\int_0^r\int_0^u Df_v(X_v)+v\,Df_v(X_v)\bullet D v_0\,dv\,du\,dr\,ds + \cdots $$

Now, it seems to me that a very natural condition to impose is that $Dv_0$ be nilpotent. After all, consider the case where $f = 0$, so $DX_t = I + D v_0\,t$. This matrix will be non-invertible as soon as $t$ hits $-1/\lambda$ for any nonzero eigenvalue $\lambda$ of $Dv_0$. So if $Dv_0$ has any nonzero eigenvalues, $DX_t$ will be singular in some finite time.*

Once we start thinking this way, it's natural to continue and require $D f_t$ to be nilpotent at each point, and to commute with $Dv_0$, so that their product is also nilpotent. We will then have $DX_t = I + nilpotent$, so that it's unipotent, and in particular a local diffeo.

Going from that to a global diffeo is a much much harder problem. Based on preliminary googling (search for example 'global univalence'), this is an active area of research that a lot of work has been done in, but which still has a lot of open conjectures -- it's quite close to the famous Jacobian conjecture for polynomial maps. One conjecture I encountered here on page 5 is the following:

Conjecture 3 ($C^1$ Unipotent Jacobian Univalence Conjecture). If $f: \mathbb R^n \to \mathbb R^n$ is $C^1$ and $Df$ is unipotent, then $f$ is injective.

which would apply in this case. This seems to have been proven for $n = 2$, and there are other results about so-called "P-matrices" and "N-matrices" where it is also known. Surjectivity also seems to be an easier problem than injectivity. In any case, I think the moral is that you'll probably have to do a dig through the literature to see if you can find something that applies in your case.

*I'm going off your question that we want $X_t$ to be defined for all time, positive and negative. If you only care about $t>0$, you could make the weaker assumption that the eigenvalues of $Dv_0$ and $Df_t(X_t)$ are always positive (and still that they commute), which would related to Conjecture 2 from the same paper, since the eigenvalues of $DX_t$ would be bounded away from 0.


All ODE can be considered as first-order by increasing the dimension: in your case, consider $$ \frac{d}{dt}\begin{pmatrix}y\\x\end{pmatrix}=\begin{pmatrix}f(t,x)\\y\end{pmatrix}. $$ Then assuming for instance $f$ locally Lipschitz in $x$ and say $L^1$ in time, you get a flow by the standard Cauchy-Lipshitz Theorem.

  • $\begingroup$ This doesn't answer my question. Although you get a flow by increasing the dimension, the flow isn't the same as the function introduced. $\endgroup$ – Ben Jun 15 at 19:01
  • $\begingroup$ @Ben You get a flow $Φ(𝑡,𝑥_0,𝑦_0)$ by the standard Cauchy-Lipschitz Theorem. Don't you have with your notations $𝑋(𝑡,𝑥)=Φ_2(𝑡,𝑥,𝑣_0(𝑥))$? – Bazin 3 hours ago $\endgroup$ – Bazin Jun 16 at 20:33
  • $\begingroup$ We do indeed have that equality. However, $\phi_2(t,x,v_0(x))$ is not necessarily a diffeomorphism. $\endgroup$ – Ben Jun 16 at 22:24

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