Let $M^n$ be a (compact) $n$-dimensional manifold. And we are given a set of $m$ basis functions $$B=\{\psi_i(x): \mathbb{R}^n \rightarrow \mathbb{R}| i=1,...,m \}$$ such that all of them are differentiable. Let $F \in \mathcal{X}(M)$ be a vector field, linearly built from the basis functions, i.e. $\exists A \in \mathbb{R}^{n \times m}$ $$F_j(x)= \sum_{i=1}^m A_{i,j}\psi_i(x) \quad, j=1,...,n$$
We define a DS by $$\dot{x}(t)=F(x(t))$$ and set $\Phi: M \times \mathbb{R} \rightarrow M, \quad (x_0, t) \mapsto \Phi(x_0,t)$ as the flow of the DS.
Now, we assume to be given an open subset $U \subset M$ such that $$ \forall x_0 \in U \ \exists \theta^{x_0} \in \mathbb{R}^m, \ \theta^{x_0} \neq 0 : \quad \Lambda(x(t))=\sum_{i=0}^m \theta_i^{x_0} \psi_i(\Phi(x_0,t))=0 \quad \forall t $$
My question is: Can we proof that the $\theta^{x_0}$ depend differentiable on $x_0$. Or in ther words, given the information above, can we proof that $\Lambda$ is a differentiable map on $U$, $\Lambda: U \rightarrow \mathbb{R}$.
We know that along each trajectories the $\theta^{x_0}$ are constant, but we don't know how they change fro mtrajectory to trajectory. Maybe Whitney's extension theorem comes intop play in the proof.