Prove that if the initial value of $\dot{x} = X(x)$ belongs to $Z$ and the vector field is tangent to curve Z, then the solution also belongs to $Z$.

Question

Let $Z = \{x\in \mathbb{R}^{2}: F(x) = 0\}$ be a simple disjoint closed curve, and the vector field $X$ is perpendicular to the gradient of $F$ on $Z$. Prove that if the initial value of $\dot{x} = X(x)$ belongs to $Z$, then the solution also belongs to $Z$.

We can get that the vector field is tangent to curve Z. So, in pictures, it's like true by using the existence and uniqueness theorem. How to do complete proof?

Answer 1

Let $\alpha(t)$ be a solution with initial value in $Z$, that is, $F(\alpha(0))=0$. Then $$ \frac{d}{dt}F(\alpha(t)) = \langle \text{grad} F(\alpha(t)),\alpha(t)' \rangle = \langle \text{grad} F(\alpha(t)),X(\alpha(t)) \rangle = 0 $$ therefore, $t \mapsto F(\alpha(t))$ is constant and we must have $F(\alpha(t)) = 0$ for all $t$, that is, $\alpha(t) \in Z$ for all $t$.