# Logistic Differential equation $\dot{y}= y(1-y)$ and its stationary solutions.

Given the logistic differential eqation $$\dot{y}=y(1-y)$$ with someinital value $y(0)$. The stationary solutions of the differential equation would be $y(t)=0$ or $y(t)=1$ for all $t>0$. Why is it then that the solutions with inital value $$y(0)<0, 0<y(0)<1,y(0)>1$$ will always stay in the regions $$(-\infty,0), (0,1),(1,\infty)$$ respectively?

• Why should there be regions intuitively? Should a solution ever be able to "cross" from one region to another? Assume the solution's continuous for this, and think about the IVT and the stationary solutions. Commented Jul 18, 2017 at 7:21
• Close to the line $y=0$, the equation virtually simplifies to $\dot y=y$, which has an exponential solution thus a constant sign.
– user65203
Commented Jul 18, 2017 at 7:56

Suppose that you have a solution $y_{1}(t)$ satisfying some initial value $y_{1}(0)=y_{0}<0$, such that for some $t>0$ you have $y_{1}(t)>0$. Then by the intermediate value theorem you have some $t_{0}>0$ such that $y_{1}(t{0})=0$.
Now consider the IVP with initial value $y(t_{0})=0$. You have two solutions - the stationary solution $y=0$ and the solution $y_{1}$. Which is a contradiction to the uniqueness of solution of this IVP.
Similarily for $t<0$ and the other cases.