# Intutition on local uniqueness theorem for separable variables

I'm looking for the intuition (and ideally a sketch of proof or reference) of the following theorem, which comes from and standard course in differential equations:

Let $a \in C(J_1),g \in C(J_2),t_0 \in J_1,x_0 \in J_2$ and consider the IVP $\begin{cases} x' = a(t)g(x) \\ x(t_0) = x_0 \\ \end{cases}$ then the IVP

1. Has solutions.
2. If $g(x_0) \neq 0$ then verifies the property of local uniqueness.
3. If $g(x_0) = 0$, $G \in C(J_2)$ such that $G(x_0) = 0$,$\exists \delta > 0.\forall x \in ]x_0,x_0+\delta[.G'(x) = \frac{1}{g(x)}$ and $a(t_0) \neq 0$ then the IVP doesn't verify the property of local uniqueness.

I've already used it in several occassions (see [1],[2]) but now that I know more from differential equations and I successfully interpreted Cauchy-Peano's or Picard-Lindelof's theorem I would like to see where did this theorem come from.

I'm especially interested in knowing what is the intuition for point 3. since for point 2. basic algebraic manipulation tells you that one can divide by $g(x)$ in a neighborhood of $t_0$, but what is the intuition behind 3.? I would like that this intuition helped me to better remember the theorem...

Other references

I don't think that this properly addresses my question since apparently they are considering that $a \neq 0,g \neq 0$ in all the interval which is not the case here. Also, this one doesn't work since they're applying Picard-Lindelof because they have enough regularity. For the search, I remember that there exist other theorems for exacts equations...

When I say intuition I mean I can understand the sitution geometrically. One diagram that is helping me a lot is vector fields which I draw when $(t,x) \in \mathbb{R}^2$.
I do not know of any geometrical intuition in 3. The basic intuition is that, if $\varphi(\cdot)$ is a nontrivial (that is, not equal constantly to zero on its whole interval of existence) solution of the IVP $$\tag{*} \begin{cases} x' = g(x) \\ x(0) = 0, \end{cases}$$ where $g \colon [0, \infty) \to [0, \infty)$ is continuous and $g(x) = 0$ only when $x = 0$, then, for any two $0 \le t_1 < t_2$ such that $\varphi(t_1) > 0$ and $\varphi(t_2) > 0$ there holds $$t_2 - t_1 = \int\limits_{\varphi(t_1)}^{\varphi(t_2)} \frac{d\xi}{g(\xi)}.$$ So, intuition is that the integral on the RHS equals the time needed for the described system to pass from state $x_1 := \varphi(t_1)$ to state $x_2 := \varphi(t_2)$. If $$\int\limits_{0}^{1} \frac{d\xi}{g(\xi)} < \infty$$ then the time needed to pass from state $0$ to state $1$ is finite, and we can suspect that there is a nontrivial solution to $(*)$. Indeed, it is quite easy to prove that that the function $\psi \colon (-\infty, T) \to [0, \infty)$ equal to $0$ for $t \le 0$ and defined by $$t = \int\limits_{0}^{\psi(t)} \frac{d\xi}{g(\xi)}, \qquad t \in(0, T)$$ is a nontrivial solution of $(*)$. If, on the other hand, $$\int\limits_{0}^{1} \frac{d\xi}{g(\xi)} = \infty,$$ then there is no possibility to pass from state $0$ to any positive state in finite time, therefore the trivial solution is a unique solution of $(*)$.
• Le's start form the RHS: $\int\limits_{\varphi(t_1)}^{\varphi(t_2)}\frac{d\xi}{g(\xi)}$ gives, after substitution $\xi=\varphi(\tau)$, $\int\limits_{t_1}^{t_2}\frac{\varphi'(\tau)\,d\tau}{g(\varphi(\tau))}=\int\limits_{t_1}^{t_2}\frac{g(\varphi(\tau))\,d\tau}{g(\varphi(\tau))}=\int\limits_{t_1}^{t_2}d\tau=t_2-t_1$. – user539887 Apr 20 at 18:10