I consider the following differential equation
$$ x’(t)= rx(t)(1-\frac{x(t)}{K}) $$
Where $r$ and $K$ are constant.
I consider an initial condition $x(0)=x_0\in (0,K)$.
In my lecture notes, it is written that given this initial condition and by the uniqueness of the Cauchy-Lipschitz theorem we must have $x(t)\in (0,K)$ (and $x(t)$ constant equals to $K$ for $t\geq K$).
I don’t see how to make the link between this initial condition and the codomain of $x$ neither why we are restricted to $(0,K)$ ?
I have just noticed that when the derivative is $0$ (hence constant) we have $x(t)=K$, but I do not see how it prevents $x(t)$ to vary for $t\geq K$.
If you have any hint or explanations to give to me I will be glad.
Thank you a lot !
When $t\le K$ derivative is positive and $x'(t)$ is increasing with respect to $x(t)$.
and $x(t)$ like below
When $t \ge K$ derivative is negative and $x'(t)$ is decreasing with respect to $x(t)$.
So as a sample solution, we can plot $x'(t)$as below
Hope that helps you.
Anf finally about $x=k$ it's the maximum point of $x',x$ graph like below