Let $M^n$ be a (compact) $n$ susbet of $\mathbb{R}^n$. And we are given a set of $m$ basis functions $$B=\{\psi_i(x): M^n \rightarrow \mathbb{R}| i=1,...,m \}$$ such that all of them are differentiable and non-constant. We define a DS by $$\dot{x}(t)=F(x(t))$$ and set $\Phi: M \times \mathbb{R} \rightarrow M, \quad (x_0, t) \mapsto \Phi(x_0,t)$ as the flow of the DS.
Now, we assume to be given an open subset $U \subset M$ such that $$ \forall x_0 \in U \ \exists \theta(x_0) \in \mathbb{R}^m, \ \theta^{x_0} \neq 0 : \quad \Lambda(x(t))=\sum_{i=1}^m \theta_i(x_0) \psi_i(\Phi(x_0,t))=-\theta_0(x_0) \quad \forall t $$ I want to show that $\Lambda$ is a (local) first integral. 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, sis it true that $$ \exists x_0^1,x_0^2 \in U: \quad \Lambda(x^1(t))=-\theta_0(x_0^1) \neq -\theta_0(x_0^2)=\Lambda(x^2(t))\;? $$
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.)