A doubt on the positively invariant sets for a system of ODEs

Question

Let $f:\mathbb{R}^2 \to (-\infty,0]$ be such that $f(0,0)=0$. Consider the following system of ODEs

$$ \begin{aligned} x'(t) &= f\big( x(t), y(t) \big)\\ y'(t) &= y^2(t) \end{aligned} $$

for $t > 0$. A set $S \subseteq \mathbb{R}^2$ is said to be positively invariant if the following holds.

Condition: Assume that $\big(x(0),y(0) \big) \in S$. For $t \ge 0,$ if the solution $\big(x(t),y(t) \big)$ exists, then $\big(x(t),y(t) \big) \in S.$

Then which of the following sets are positively invariant ?

1. $\{ (x,y) \in \mathbb{R}^2: x \le 0,y \le 0\}$
2. $\{ (x,y) \in \mathbb{R}^2: y \ge 0\}$
3. $\{ (x,y) \in \mathbb{R}^2: x \le 0, y\le -1\}$
4. $\{ (x,y) \in \mathbb{R}^2: x \le 0, y\ge 0\}$

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My attempt

Since, $x'(t)=f(x,y)$ and $f(\mathbb{R^2})$ cannot contain positive real numbers, we can say that $x(t)$ is monotonically decreasing. Also, solving for $y(t),$ we get $y(t)=\frac{-1}{t+C},$ for an arbitrary constant C.

I am not able to proceed any further. Is there any result or a theorem that helps me check whether the given subset of the plane is positively invariant.

Any help is highly appreciated.
Please do let me know of any references also.

Edit: Please assume the necessary smoothness conditions on $f.$

Answer 1

My attempt.

I guess, in general, the answers can depend on $f$. For instance, maybe $f$ can be so discontinuous that the system has only trivial solutions. So OP allowed us to impose the necessary smoothness conditions on $f$.

Moreover, since solutions of differential equations usually are provided on convex sets, for instance (half)intervals (finite or infinite), we interpret ``for $t \ge 0,$ if the solution $\big(x(t),y(t) \big)$ exists'' in Condition as if $\big(x(0),y(0) \big) \in S$, $T>0$ and $\big(x(t),y(t) \big)$ is a solution of the system for $t\in [0,T]$ then $\big(x(t),y(t) \big) \in S$ for each $t\in [0,T]$. Then we have $y(t)=-\frac 1{t+C}$ for some constant $C\in\mathbb R\setminus [-T,0]$. Moreover, if the function $x'(t)$ is continuous on $[0,T]$ (this happens, for instance, when $f(x,y)$ is continuous) then $x(t)=\int_0^t x'(s)ds=\int_0^t f(x(s),y(s))ds$ for each $t\in [0,T]$. Since $f(x(s),y(s))\le 0$ for each $s\in [0,t]$, we have $x(t)\le x(0)$.

1)) We have $y(0)\le 0$. Then $C>0$ and $y(t)<0$ for each $t\in [0,T]$. Moreover, if $f$ is continuous then since $x(t)\le x(0)\le 0$ for each $t\in [0,T]$, the set is positively invariant.

2)) We have $y(0)\ge 0$. Then $C<0$. Thus $C<-T$ and $y(t)>0$ for each $t\in [0,T]$. Thus the set is positively invariant.

3)) We have $y(0)\le -1$. This holds iff $0<C\le 1$. If $C>1-T$ then $y(T)>-1$ and then the set is not positively invariant. This can happen, for instance, when $f(x,y)=0$ for each $(x,y)\in\mathbb R^2$. Then $x(t)=x(0)$ for each $t\in [0,T]$.

But it remains to investigate whether the set can be positively invariant for another $f$. Indeed, it can happen that the system has no solutions with $C>1-T$. In this case if $f$ is continuous then since $x(t)\le x(0)\le 0$ for each $t\in [0,T]$, the set is positively invariant.

4)) For each $t\in [0,T]$ similarly to (2) we can show that $y(t)>0$ . Moreover, if $f$ is continuous then since $x(t)\le x(0)\le 0$ for each $t\in [0,T]$, the set is positively invariant.