# If $T_t$ is a diffeomorphism and $t\mapsto T_t(x)$ is differentiable, can we find a map $v$ with $v(t,T_t(x))=\frac{\partial T}{\partial t}(t,x)$?

Let $$d\in\mathbb N$$, $$\tau>0$$, $$U\subseteq\mathbb R^d$$ be open and $$T_t$$ be a $$C^1$$-diffeomorphism from $$U$$ onto an open subset of $$\mathbb R^d$$ for $$t\in[0,\tau)$$ with $$T_0=\operatorname{id}_U$$. Note that $$V:=\bigcup_{t\in[0,\:\tau)}T_t(U)$$ is open.

Assume $$[0,\tau)\ni t\mapsto T_t(x)$$ is differentiable for all $$x\in U$$. Can we show (under suitable additional assumptions, if necessary) that there is a $$v:[0,\tau)\times V\to\mathbb R^d$$ with $$v\left(t,T_t(x)\right)=\frac{\partial T}{\partial t}(t,x)\tag1$$ for all $$t\in[0,\tau)\times V$$?

If $$U=\mathbb R^d$$ (and hence $$V=\mathbb R^d$$), we may simply set $$v(t,x):=\frac{\partial T}{\partial t}\left(t,T_t^{-1}(x)\right)\tag2.$$

EDIT 1: I want to choose $$v$$ such that it is (jointly) continuous. By assumption, $$[0,\tau)\times U\ni(t,x)\mapsto T_t(x)\tag3$$ is partially differentiable in both the first and the second variable. So, it should be differentiable and hence (jointly) continuous.

EDIT 2: I wonder whether any differentiability properties of $$v$$ with respect to the second variable carry over to $$v$$. I've found the following excerpt in a book, which seems to indicate this, but I actually don't understand how they conclude (2.76):  We are looking for a function $$v:[0,\tau)\times V\to\mathbb R^d$$ satisfying:

$$v\left(t,T_t(x)\right)=\dfrac{\partial T}{\partial t}(t,x) \qquad \forall (t,x)\in[0,\tau)\times U\tag1$$

Fix $$\bar{t}\in [0,\tau)$$. The equation: $$v\left(\bar{t},T_\bar{t}(x)\right)=\frac{\partial T}{\partial t}(\bar{t},x) \qquad \forall x \in U$$ is a condition on the value of $$v(\bar{t}, \cdot)$$ on the set $$T_\bar{t}(U)$$. It can be restated as follows: $$v\left(\bar{t},y\right)=\frac{\partial T}{\partial t}(\bar{t},T_\bar{t}^{-1}y) \qquad \forall y\in T_\bar{t}(U)$$

Consider now the set $$A := \bigcup_{t \in [0,\tau)} \big(\{t\}\times T_t(U)\big) \subseteq [0,\tau)\times V$$

and define $$f: A \to \mathbb{R}^d$$ as follows: $$f(t,y):= \dfrac{\partial T}{\partial t}(t,T_t^{-1}y) \qquad \forall (t,y)\in A$$ Then $$(1)$$ is equivalent to the following: $$v(t,y)=f(t,y) \qquad \forall (t,y) \in A$$

If you don't mind continuity of $$v$$, you can define $$v$$ on $$A^c:=[0,\tau)\times V - A$$ arbitrarily and to be equal to $$f$$ on $$A$$.

## LOOKING FOR CONTINUITY

If you need $$v$$ continuous (or even more regular), then the problem is whether $$f$$ admits a continuous (or even more regular) extension to the set $$[0,\tau)\times V$$.

In what follows, we will consider only the simpler case in which $$T_t(U)=U$$ for all $$t \in [0,\tau)$$.

In this case, $$A=[0,\tau)\times V = [0,\tau)\times U$$ and thus $$v$$ and $$f$$ must be the same function. Therefore, $$v$$ is necessarily defined as follows: $$v(t,y):= \dfrac{\partial T}{\partial t}(t,T_t^{-1}y) \qquad \forall (t,y)\in A$$

If we assume

• $$(t,x)\mapsto\dfrac{\partial T}{\partial t}(t,x)$$ is jointly continuous in $$t,x$$.
• $$(t,y)\mapsto T_t^{-1}(y)$$ is jointly continuous in $$t,y$$.

then $$v(t,y)$$ is jointly continuous in $$t,y$$ since it is the composition of the continuous mapping $$(t,y)\mapsto (t,T_t^{-1}(y))$$ with the continuous mapping $$\dfrac{\partial T}{\partial t}$$.

Be aware that if $$(t,x)\mapsto\dfrac{\partial T}{\partial t}(t,x)$$ or $$(t,y)\mapsto T_t^{-1}(y)$$ lack of jointly continuity then it is in general unlikely for $$v$$ to be jointly continuous.

• Thank you for your answer. The reason why I thought this definition of $v$ is not suitable, is that I surely want $v$ to be jointly continuous. However, the assumptions should imply that $(3)$ is differentiable (on the product space) and hence (jointly) continuous. So, your map $v$ should be jointly continuous as well, right? Jul 28, 2020 at 15:02
• @0xbadf00d Well, this is a different question :). I'll think about it and modify the answer accordingly. Jul 28, 2020 at 15:40
• I think the easiest solution is to assume that $T_t(U)=U$. This is actually what the reference I've added to the question seems to do. Do you see how they conclude (2.76)? Jul 28, 2020 at 16:56
• Thank you for your edit. Do you have any thoughts on (2.76)? Jul 31, 2020 at 4:15
• @0xbadf00d I suspect it is a matter of function spaces definition. Can you please tell me how is $C^0(0,\epsilon; C^k(\bar{D},\mathbb{R}^N))$ is defined in the book? Moreover, nowhere it is stated that $T$ can be derived with respect to $t$, nontheless $\partia T / \partial t$ appears... Jul 31, 2020 at 8:18