Maximal Interval Solution Proof for ODE

ordinary differential equations

I have the following problem: let $f$ locally Lipschitz and consider the equation:
$$\frac{dx}{dt} = f(x)$$
Suppose $x_1 < x_2$ and that $f(x_1) = f(x_2)= 0$, prove that the PVI solution:
$$\left\{\begin{matrix}
\frac{dx}{dt} = f(x)\\
x(t_0) = x_0
\end{matrix}\right.$$
where $x_0 \in (x_1, x_2)$ has a maximum interval of $\mathbb{R}$.

The truth is I'm not sure how to show that. I understand that I would have to show that the solution is well defined in all of $\Bbb R$, but only that it is locally Lipschitz I know that it is locally continuous after the problem data and continuity it would have to reach a maximum and a minimum but I don't know how to use said result. Any suggestions or ideas on how to show that the maximum interval is all of $\mathbb{R}$?

Best Answer

Let $(a, b)$ be the maximal open interval of existence of $x(t)$, with $-\infty\le a< t_0<b\le \infty$. The following alternative holds

  1. $f(x_0) = 0$. In that case, $x(t) = x_0$ is the (only) solution and $(a, b) = {\mathbb R}$
  2. $f(x_0) \not= 0$. In that case, $\text{sgn}(f(x(t))) = \text{sgn}(f(x_0))$ for all $t\in (a,b)$ and $x(t)$ is strictly monotonic in $(a,b)$.

Indeed, in case 2, if $f(x(\tau)) = 0$ for some $\tau\in(a, b)$, then it means that $x(t)$ is constant equal to $x(\tau)$ on all of $(a,b)$ but then $f(x(t_0)) = 0$, a contradiction. As $f(x(t))$ is continuous, if it is never $0$, its sign never changes, hence $x'(t)$ has constant sign.

This plus the fact that $x_1< x(t)< x_2$ imply that $m_a = \lim_{x\to a} x(t)$ and $m_b = \lim_{x\to b} x(t)$ exist and satisfy $x_1\le m_a, m_b\le x_2$.

If $-\infty < a$, the solution can be extended to an interval $(a-\epsilon, b)$ by considering the unique solution satisfying $x(a) = m_a$. This is a consequence of the proof of the Picard-Lindelöf (or Cauchy-Lipschitz) theorem. Similarly, $b$ cannot be $< \infty$. Finally, $(a, b) = {\mathbb R}$.

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