# Diffeomorphisms as solutions to second-order ODEs

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?

• 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) – Kevin Casto Jun 15 at 15:46
• @KevinCasto Preferably, for all time. – 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.

• 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. – Ben Jun 15 at 19:01
• @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 – Bazin Jun 16 at 20:33
• We do indeed have that equality. However, $\phi_2(t,x,v_0(x))$ is not necessarily a diffeomorphism. – Ben Jun 16 at 22:24