0
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ Commented Jul 18, 2017 at 7:21
  • $\begingroup$ Close to the line $y=0$, the equation virtually simplifies to $\dot y=y$, which has an exponential solution thus a constant sign. $\endgroup$
    – user65203
    Commented Jul 18, 2017 at 7:56

1 Answer 1

4
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .