ODE flow of the Chafee-Infante problem I'm currently listening to a lecture on dynamical systems. Unfortunately I am lacking some of the requirements for that course and thus ran into some problems with the latest problem set.

Show that the ODE flow 
  $$ \frac{d^2}{dx^2}v + f(v) = 0 $$
  for the Chafee-Infante nonlinearity $f(v)=v(1-v^2)$ is not global. 
  [...]

So far I thought the ODE flow is a function mapping a point $p$ in the phase-space to the point $\Phi^t(p)$ where the solution starting at $p$ will be after time $t$. Is this not correct? Because the problem as stated above even has a well defined potential and will easily give you full solutions $v(x)$, $x\in(-\infty,\infty)$ for any starting point.
Or does my misconception lie with the word "global"? If $\Phi$ is defined for all points and all times, what else could "global" ask for?
 A: Maybe there is some misconception here because you use the words "time" and "points" for your particular problem. For your equation the point is the pair $(v(x),v'(x))$ and "time" is the variable $x$. To prove that the flow is not global one usually needs to show that solutions blow up at a finite time (a prototypical example is $\dot v=1+v^2$), i.e., that the solutions, starting at a particular time $x_0$ cannot be extended to any time $x\to\pm \infty$. Unfortunately, I cannot right now prove this fact for your equation, but if you consider the problem $\ddot v=v^3$ then there is an explicit solution in terms of the Weierstrass elliptic function, which has vertical asymptotes, and which can help understand why the flow is not global in this case. 
Update:
I decided undelete my solution because it seems that the following argument helps to prove that the flow is not global. Since you can write down solution to your problem in implicit way:
$$
\pm \frac{1} {\sqrt 2}\int_{v_0}^v\frac{d\tau}{\sqrt{E-\tau^2/2+\tau^4/4}}=x-x_0,
$$
where $E$ is a constant, then it follows that for $v\to\infty$ (which also has to be proven) the improper integral above converges in this particular case, and then there is a finite $x$ for which $v\to\infty$. 
Though I have to admit that my argument is still vague. 
