Date modified: Wed 2025-06-25 . 05:04 PM
Date created: Wed 2025-06-25 . 04:51 PM
# Unique global solution IVP without satisfying Picard's. Consider for $y=y(x)$, the IVP $y'= f(x,y) = \begin{cases} 10 & \text{if } y > 0 \\ 0 & \text{if } y = 0 \\ -10 & \text{if }y < 0\end{cases}$ with $y(0) = 0$. Note that here $f(x,y)$ is not continuous on any open rectangle about $(0,0)$. We claim $y(x) = 0$, the constant solution, is the unique global solution. Suppose to the contrary that $w(x)$ is a non-constant solution to this IVP. Note $w$ is differentiable. Suppose it is the case that $w(a) > 0$ for some $a > 0$. Then let $T = \inf \{x > 0 : w(x) > 0\}$. Note $T$ exists because the set there is nonempty and bounded below by $0$. So $T \ge 0$. By continuity, $w(T) = 0$. So $w'(T) = 0$. Also, on some small neighborhood $(T, T + \epsilon)$, we have $w(x) > 0$. So for all sufficiently large $n$ , $w(T+ \frac{1}{n}) > 0$. Let us write $w(T+\frac{1}{n}) = b_{n}$ Now by mean value theorem, there exists some $c \in (T,T+\frac{1}{n})$ where $w'(c) = \frac{b_{n}-0}{1 / n} = n b_{n}$. But $w(c) > 0$, so we must have $w'(c) = 10$. Hence $nb_{n} = 10$, or $b_{n} = \frac{10}{n}$. Now, as $w$ is differentiable at $x = T$, we must have this limit of difference quotient $$ 0 = w'(T) = \lim_{n \to\infty} \frac{w(T + 1 / n) - w(T)}{ 1 / n} = 10. $$This gives a contradiction.