Let $U$ be an (open) set in $\mathbb{R}^n$. And we are given a set of $m$ basis functions $$B=\{\psi_i(x): U \rightarrow \mathbb{R}\mid i=1,\ldots,m \}$$ such that all of them are differentiable and non-constant. We define a DS by $$\dot{x}(t)=F(x(t))$$ and set $\Phi: U \times \mathbb{R} \rightarrow U, \quad (x_0, t) \mapsto \Phi(x_0,t)$ as the flow of the DS.
Now we look at a function which is a linear combination in the basis functions ($\theta \in \mathbb{R}^{m+1}$) $$\Lambda: V \rightarrow \mathbb{R}, \ x \mapsto \sum_{i=0}^m \theta_i \psi_i(x)$$ where $V \subseteq U$ is some open possibly smaller subset of $U$.
Now we know (by some different result) that for each point/ inital condition $x_0=x(0)$ and the according trajectory $x(t)$ there is a non-zero vector $\theta(x_0) \in \mathbb{R}^{m+1}$ (with components $\theta_i(x_0), \quad i=0,...,m$) such that
$$ \Lambda(x_0)=\sum_{i=0}^m \theta_i(x_0) \psi_i(\Phi(x_0,t))=0$$
So, $\Lambda$ is a function which maps $V$ to zero. The special porperty is that along a trajectory $\theta_i(x_0) \ \forall i $ is constant, hence its value is defined by the value at the initial condition. But for different trajectories/initial conditions $x_0' \neq x_0$ the value of $\theta_i(x'_0) \neq \theta_i(x_0)$ is allowed to be different from $\theta(x_0)$. (We actually want to prove that this has to be the case)
Now we set the function $$\Lambda'(x(t))=\sum_{i=1}^m \theta_i(x_0) \psi_i(\Phi(x_0,t))=- \theta_0(x_0)$$ for which we want to prove that it is a local (i.e. on $V$) 1. Integral of $F$. I know that the Lie derivative $L_F(\Lambda)=0$ is zero, which is the first thing I need to prove. But I am struggling to prove that the function is not locally constant.
My question is the following: Can we prove that $\Lambda' $ is not locally constant? More formally, $$\exists x_0,x_0' \in V: \quad \Lambda'(x_0)=-\theta_0(x_0) \neq -\theta_0(x_0')=\Lambda'(x_0')$$ I.e. are there two initial conditions $x_0 \neq x_0'$ for which the the first (or zeroth depending on the convention) entry of the $\theta$-vector needs to be different $\theta_0(x_0) \neq \theta_0(x_0')$
I think this has to be true, since for two different inital conditions $\Phi(x_0^1,t) \neq \Phi(x_0^2,t) \ \forall t$. And since all functions are non-constant the parameters can't be the same but I am lacking a formal proof. Especially how I can conclude from the fact that the parameters are not the same that the value of the function is not the same. (If it helps , the vector field $F$ itself is a linear combination of the basis functions $\psi_i$ in each dimension.)
