Upper bounds for the spatial differential of the inverse of a flux

Question

It is well known that given a regular velocity field $b: \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}^n$ (say, continuous in time and uniformly Lipshitz in space), the flux $X$ associated to $b$ is a diffeomorphism for every fixed time $t$, and it is easy to deduce from the relation $\partial_t X (t,x)= b (t, X(t,x))$ and using the Gronwall inequality that

$$ \exp \left( -\int_0 ^t |\nabla b (s,X(s,x)| ds \right) \leq ||\nabla_x X(t,x)||\leq \exp \left( \int_0 ^t |\nabla b (s,X(s,x)| ds \right) ,$$

where I denote by $||\cdot ||$ the operator norm. Now, the question is very simple, yet I found little looking for the answer online, and it is: are there similar estimates for $X^{-1} (t,x)$ ? I mean, about $X(t,x)$ we know precisely who $\partial_t X$ is and we have an upper and lower bound on the operator norm of $\nabla _x X(t,x)$, what can we say about $\nabla _x X^{-1}$ ? I managed to get a lower bound on the norm, practically using simply the fact that $||A^{-1}|| \geq \frac{1}{||A||}$, but I don't see how to get an upper bound. The only one I can think of only works if $||A-I||<1$, which is not the case here.

I would bet that there exists an upper bound, because if I think about it, applying $X^{-1} (t,\cdot)$ to a point $x$ simply means going back in time for a time $t$ along the integral line passing from $x$. If there's a a bound going forward, there should be one going backwards.

Answer 1

I post an answer to expand on the comment. Considering $\nabla _x X (t, X(t,x)= [\nabla _x X(t,x)]^{-1}$, we can bound from above its operator norm. Assume for simplicity that the velocity $b$ is divergence free, so volumes are preserved. The operator norm of $A=\nabla_x X(t,x)$ is equal to its biggest singular value $\sigma_1$, while the norm of the inverse is equal to the reciprocal of the smallest singular value, i.e. $1/\sigma_n$. Now, since by the divergence assumption volumes are preserved, the jacobian of $\nabla_x X$ is identically 1, and by SVD decomposition it is equal to the product of the singular values. But then

$$ 1= \sigma_1 \sigma_2 \dots \sigma_n \leq (\sigma_1 )^{n-1} \sigma_n\implies \frac{1}{\sigma_n} \leq \sigma_1 ^{n-1} .$$