Prove or disprove: if $y'=y-y^2$ and $y(0)=a$ where $0<a<1$ then $a<y(x)\leq 1$ for any $x>0$

Prove or disprove: if $y'=y-y^2$ and $y(0)=a$ where $0<a<1$ then $a<y(x)\leq 1$ for any $x>0$.

I thought about using PL and Grownwall, but this led me to nothing.

Solution 1:

This is probably to be solved with the principles of one-dimensional dynamics. In $y'=f(y)$, with $f$ locally Lipschitz or even differentiable:

The roots of $f$ are the only stationary points, that is, give rise to constant solutions.
By uniqueness, no other solution can touch or cross the constant solutions. In the contraposition, solutions starting between roots of $f$ remain between these roots.
All non-constant solutions are strictly monotonous. In the case of a growing solution that is bounded above by a root, the solution exists for the whole positive time axis and converges towards this root. Falling and the behavior in direction of falling arguments give similar claims.

Now observe that $f(y)=y-y^2=y(1-y)$ has roots $0$ and $1$ and apply these principles to directly get the claim without computing the solutions.

$$y' = y - y^2 \iff \dfrac{dy}{y(1-y)} = dx \iff y = \dfrac{e^x}{c + e^x}$$

Hence we have that $y(0) = a = \dfrac{1}{1+c} \iff c= \dfrac{1-a}{a} \ge 0$

Now we may deduce that $a < y(x) \le 1$