# Confusion with periodic orbits, classical dynamical systems

I have a question related to two theorems in the book Differential Equations, Dynamical Systems, and Linear Algebra [Hirsch & Smale,1974]. First let me describe the framework.

Let us consider a $C^1$-vector field $f:\mathbb R^n\to\mathbb R^n$ and an ODE $$\dot x = f(x)\quad(1)$$ with initial condition $x(0) = x_0$. The correspondent flow we denote by $\phi(t,x)$, i.e. $$\begin{cases} \frac{\partial \phi}{\partial t}(t,x) &= f(x) \\ \\ \phi(0,x) &= x. \end{cases}$$ I.e. if $x(t)$ solves the problem $(1)$ with initial conditions $x(0) = x_0$ then $x(t) = \phi(t,x_0)$.

Let us consider a $n-1$-dimensional hyperplane $$H = \{x\in \mathbb R^n:h(x) = 0\}$$ where $h$ is a linear function: $h(0) = 0$. An open subset $S$ of $H$ is called a local section at $0$ if $0\in S$ and $h(f(x))\neq 0$ for any $x\in S$. In other words, $S$ is an $n-1$-dimensional section at $0$ which is transversal to the flow.

Proposition 1. (p. 243) Let $S$ be a local section at $0$ as above, and suppose $\phi(t_0,x_0) = 0$. There is an open set $U$ containing $x_0$ and a unique $C^1$ function $\tau:U\to\mathbb R$ such that $\tau(x_0) = t_0$ and $\phi(\tau(x),x) \in S$ for all $x\in U$.

The meaning of the proposition is clear: if from the point $x_0$ we hit the section $S$ in $t_0$ time units, then for all $x$ in a small neighborhood of $x_0$ we still hit $S$ in the time period which varies smoothly.

Note, that if $x_0\in S$ then $\tau(x)$ is not necessary a first hitting time of $S$ starting from $x$. I think that $$D \tau(x_0)\cdot f(x_0)<0$$ since we move on the same orbit, in a small neighborhood $U$ of $x_0$ and so $\tau$ have to decrease. In particular, $D \tau(x_0)\neq 0$.

Nevertheless, in Proposition 3 of the book (page 283-284) it is proved that if $x_0 = 0$ belongs to the periodic orbit, then $D\tau(0) = 0$. My guess is that the derivative here is $n-1$-dim vector of the restricted function $\tau|_S$, just the notation is used the same for the derivatives of $\tau$ and $\tau|_S$.

I hope you can help me to resolve this confusion, it particular - to justify formally that $D\tau\cdot f<0$ (for the unrestricted $\tau$). Applying the implicit function theorem we have $$D\tau = -\frac{Dh \;D_x\phi}{Dh\cdot f}.$$

If some points are left unclear, I will be happy to fix them.

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