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

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.

• Just an idea: a way to look at it is that the operator norm of $A^{-1}$ is equal to the reciprocal of the smallest singular value of $A$. Since the flow changes volumes in a controlled way, the smallest singular value of $A=\nabla _x X$ cannot be "too small", or in the other directions you couldn't stretch enough to have an admissible volume. Commented Jun 30, 2023 at 18:20

## 1 Answer

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

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