On number of solutions of ODE $y’=-y^a, y(0)=0$.

Question

How many solutions the ODE of the type $y’=-y^a, y(0)=0, 0<a<1$ have? I am able to prove that the IVP $y’=y^a, y(0)=0$, where $0<a<1$, has infinite number of solutions by finding it’s two solutions. But I am confused how many solutions are there for ODE $y’=-y^a, y(0)=0$, where $0<a<1?$ I am having two related problems Does there exist a unique solution for $y' = -y^{1/3}, y(0) = 0$? and IVP $\frac{dy}{dx}=-\sqrt{y}, y(0)=0.$ Please give me conditions in general when the IVP $$y’=-y^a, y(0)=0$$ where $$0<a<1$$ has infinite many solutions? One can even tell in particular for $a=\frac{1}{n}$. According to me if we are able to find non zero solution then it has infinitely many solutions. One more question is If I restricted the domain for $x\geq 0$ , then I can say unique solution if $a=\frac{1}{2n}$?Thank you in advance.

Answer 1

Suppose there exists a differentiable solution $y: \mathbb{R}_+ \to \mathbb{R}$ and time $T > 0$ where $y(T) > 0.$ The solution $y$ is continuous by assumption and $y(0) = 0$ so there exists a maximal time $0 \leq c < T$ where $y(c) = 0.$ Because of this, $y((c, T]) \in \mathbb{R}_+.$ By Mean Value Theorem, there exists a time $c < b < T$ where

$$y'(b) = \frac{y(T) - y(c)}{T - c} = \frac{y(T)}{T - c} > 0.$$

But this contradicts the fact that,

$$y'(b) = - ( y(b) )^{a} < 0.$$

Therefore the solution $y(t)$ to the initial value problem cannot be positive for any $t > 0$. It remains to show the solution cannot be negative but there is some nuance here since it doesn't always make sense to take the $a$th power of a negative number. Assuming it makes sense and the result is negative real, a parallel Mean Value Theorem argument shows this too cannot be possible: mean value theorem demands the existence of a negative slope in an interval when in fact the ODE demands a positive one throughout the interval. Use this to conclude that in fact the solution is unique to this initial value problem.

This proof only works if you consider positive time $t.$ There is no contradiction if the solution extends to $t < 0$ and examples have been posted by other commenters as well as in linked questions.