Let $f:\mathbb{R} \rightarrow \mathbb{R}$, $f \in C^{1}$,
and $f(0)=f(1)=0$, $f(x)>0$ for $x \in (0,1)$.

Prove that solution (with maximal domain) $u$ of problem:
$x(0)=x_{0} \in (0,1)$
$\lim_{t \rightarrow -\infty}$ $u(t)=0$
$\lim_{t \rightarrow +\infty}$ $u(t)=1$

Thanks a lot for your help.


There are a few bits of evidence that will give you the proof if you put them together in the right way.

Can you show that $x(t)\equiv 1$ is a solution of the ODE?

Can you show that $x(t)\equiv 0$ is a solution of the ODE?

What does the uniqueness part of this theorem tell you about a solution (considered as a curve in the $t-x$ plane) that passes through a point of the form $(t_0,1)$? Same question for $(t_0,0)$?

Can a solution passing through the point $(t,x)=(0,x_0)$, with $x_0\in(0,1)$, ever meet a point of the form $(t_0,1)$ for some $t_0\in\mathbb{R}$? ...of the form $(t_0,0)$?

What is the sign of $\frac{dx}{dt}$ for a solution $x(t)$?

What can you say about a function $t\to x(t)$ that is increasing and bounded above by 1?

Putting everything together you should be able to get this: the solution $u$ of the problem is increasing and bounded above by 1, and therefore has a limit $\ell\leq 1$ satisfying $$\lim_{t\to\infty}u(t)=\ell.$$

Now consider what happens if $\ell<1$ - in particular, what would this tell you about $f(\ell)$?

  • $\begingroup$ Thank you. I've completed this solution. Very nice problem. $\endgroup$ – John Dec 12 '12 at 12:05

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