Why can't I have $\dot{x}(t) < 0$? I'm in the very beginning of "Ordinary Differential Equations and Dynamical Systems" by Gerald Teschl and on page 9 he starts with differential equations of the form:
$$\dot{x} = f(x), \quad x(0) = x_0, \quad f \in C(\mathbb{R})$$
To solve these he quickly goes to:
$$\int_{0}^t\frac{\dot{x}(s)ds}{f(x(s))} = t$$
Letting $F(x) =\int_{x_0}^x\frac{dy}{f(y)}$ he points out that any solution must satisfy that $F(x(t)) = t$. So, since $F(x)$ is is strictly monotone in some neighborhood of $x_0$, any solution $\phi$ will need to satisfy
$$\phi(t) = F^{-1}(t), \phi(0) = F^{-1}(0) = x_0$$
Assuming, WLOG, $f(x_0)>0$ we can find a maximal interval $(x_1, x_2) \ni x_0$ where $f$ is positive in this interval as well. So we can define:
$$T_+ = \lim_{x \uparrow x_2} F(x) \in (0, +\infty]$$
$$T_- = \lim_{x \downarrow x_2} F(x) \in [-\infty, 0)$$
So, $\phi \in C^1((T_-, T_+))$. So far, so good. But, where I get confused is the following statement. Teschl states that:

In particular, $\phi$ is defined for all $t > 0$ if and only if:
$$T_+ = \int_{x_0}^{x_2}\frac{dy}{f(y)} = +\infty$$

This is what I don't understand. I understand the "if" direction - if the integral evaluates to infinity, then the approaches $x_2$ but keeps slowing down so that it never actually reaches $x_2$. But, the "only if" direction confuses me.
$x_2$ is just the maximal point to the right of $x_0$ where the velocity stays positive (since it started out positive). To me, this indicates that if the velocity starts out positive, it cannot every become negative, but that doesn't seem right. Can anyone help me understand why this must be true?
 A: The function $f$ is strictly positive in the maximal interval $(x_1,x_2)$; thus, by the fundamental theorem of Calculus, $F$ is strictly monotone on $(x_1,x_2)$ which contains $x_0$. Moreover,
$F(x_0)=0$, $F(x)>0$ for $x_0<x<x_2$.
It follows that  $F$ maps $(x_0,x_2)$ to some open interval $(0,\beta)$  (possible infinite) and the inverse $F^{-1}:(\alpha,\beta)\rightarrow(x_1,x_2)$ is also differentiable and increasing. As the OP pointed out
$$x(t)=F^{-1}(t)$$
solves the initial value problem, and is defined on $(\alpha,\beta)$.
This implies that
$$x(-\beta)=\lim_{t\nearrow\beta}x(t)=\lim_{t\nearrow\beta}F^{-1}(t)=x_2$$
and
$$t=F(x(t))=\int^{x(t)}_{x(0)}\frac{dy}{f(y)}$$
Letting $t\nearrow\beta$ gives
$$\beta=\lim_{t\nearrow\beta}\int^{x(t)}_{x(0)}\frac{dy}{f(y)}$$
Therefore,  $x$ is defined over $(0,\infty)$ (i.e. $\beta=\infty$) iff
$$\int^{x(\beta-)}_{x(0)}\frac{dy}{f(y)}=\infty$$
A: Discussed this with a few friends at the pub last night. Thinking of $x(t)$ as the position of a particle, if it were ever to turn around, then $\dot{x}(t)$ would not be a function of $x$ since there would be some position where it would have two different velocities.
