This question is related to Can a non-constant solution of DE intersect an equilibrium solution. and Why can't solutions to Autonomous ODE intersect?, where I didn't find sufficient details to resolve my question.
Specifically, the question is: for an autonomous DE, $dy/dx=f(y)$, with a critical point $c$, why can't a non-constant solution $y(x)$ intersect the equilibrium solution, $y=c$? (Assume $f$ and $f'$ are continuous functions of $y$.)
I am a beginner at differential equations, and am reading Zill and Wright's book on this topic. While I find this result plausible in light of the existence and uniqueness theorem of solutions (see below), I'm having difficulty coming up with a complete & rigorous proof. I state my attempt below, and would greatly appreciate it if someone'd confirm or refute it.
My attempted proof: Suppose $y(x)$ is a solution to the autonomous DE, $dy/dx=f(y)$, and $y(\alpha)=c$ for some $\alpha \in \mathbb R.$ I show that we must have $y(x)=c$ as follows.
Since $f$ and $f'$ are continuous functions (of $y$), by the uniqueness theorem below, $y(x)=c$ is the unique solution to $dy/dx=f(y), y(\alpha)=c$ on interval $(\alpha-h, \alpha+h)$ for some $h>0.$ Let $\beta=\sup\{t: y(x)=c, x \in [\alpha, t)\}.$ Suppose $\beta < \infty$. Then $y(\beta)\ne c$, by the uniqueness theorem again. Hence $y(\beta)=c+\Delta, \Delta\ne 0.$ But this means $y(x)$ is discontinuous at $\beta$, contradicting the fact that $y(x)$ has to be continuous. So $\beta$ has to be $\infty$, and $y(x)=c$ on $[\alpha, \infty)$. The other half, $y(x)=c$ on $(-\infty, \alpha]$, is proved similarly.
Theorem Let $dy/dx=f(x,y), R=\{(x, y):a\le x\le b, c \le y \le d\}$, and $(x_0, y_0)\in R$. If $f$ and $\partial f/\partial y$ are continuous on $R$, then there exists a unique solution $y(x)$ on $(x_0-h, x_0+h)$ for some $h>0$ to the initial value problem, $dy/dx=f(x,y), y(x_0)=y_0.$
Is this proof correct? Is there a simpler proof? Thanks a lot!
It is fine to invoke the existence and uniqueness theorem, but it is not necessary to go into sup or $\epsilon/\delta$ discussions. It is sufficient to say "basta". At a hypothetical crossing point $(x_0,c)$ of the constant solution $x\mapsto y_0(x)\equiv c$ and some other solution $x\mapsto y_1(x)$ that satisfies $y_1(x_0)=c$, but is not $\equiv y_0(x)$ for $x_0-h< x<x_0+h$, we would have a violation of the uniqueness part of the theorem.