Ode periodic solutions

Question

I have a question concerning periodic odes: $y'=y^2f (t)$, where $f $ is a real continous and periodic function with period $T $, with initial condition $y(0)=y_0\gt0$. It is asked to show that there are always $y_0$ such that there is not a global solution and to find those conditions on $f $ such that there are global non-zero periodic solutions. I solved the equation with separation af variables getting $$y=\frac{1} {\frac{1}{y_0}-\int_{0}^t f (s)ds}.$$ And I have no othqer ideas on the first question. As far as the second is concerned I substituted $s+T$ obtaining the same solution, so I guess no more conditions are needed, but I feel I'm wrong. May you help me?

Answer 1

We introduce the antiderivative $F(t) = \int_0^t f(s)\, ds$ and the following decomposition of the integral: \begin{equation} F(t) = \int_0^{T\left\lfloor\frac{t}{T}\right\rfloor} f(s)\, ds + \int_{T\left\lfloor\frac{t}{T}\right\rfloor}^t f(s)\, ds = \left\lfloor t/T\right\rfloor\! F(T) + F(t-T\left\lfloor t/T\right\rfloor)) \, . \end{equation} If $F(T)$ is nonzero, then the choice $y_0 = 1/F(T)$ makes the denominator of $y$ vanish at the time $t=T$. Else, $F(T)=0$ and we can find $t^*$ in $[0,T[$ such that $F(t^*)$ is nonzero. The choice $y_0 = 1/F(t^*)$ makes the denominator vanish at the time $t=t^*$. Thus, we can always find $y_0$ such that no global solution exists.

Now, let us consider a $T$-periodic global solution $y$. For $t$ such that $y(t)$ is nonzero, one has \begin{equation} y(t+T)^{-1} - y(t)^{-1} = -\int_t^{t+T} f(s)\, ds = -F(T) = 0 \, . \end{equation} Furthermore, the denominator in $y$ cannot vanish, i.e. $1/y_0$ cannot belong to the image of $[0,T[$ through $F$. The inverse $1/y_0$ of the initial value is allowed to be strictly larger than $\max_{t\in[0,T[} F(t)$, or strictly smaller than $\min_{t\in[0,T[} F(t)$.