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):


 A: 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.
