On the uniqueness of solution for the IVP $y'=1-y^2,~y(0)=0$ to analyse the solution behavior throughout $\Bbb R$

Question

I was solving a multiple select question (CSIR (India)-JRF Mathematical Sciences, December 2013):

Q. Let $y: \Bbb R \rightarrow \mathbb{R}$ sitisfies the initial value problem, $$y^{\prime}(t)=1-y^2(t),~ y(0)=0, ~t \in \mathbb{R}.$$ Then,which of the following is/are true,

(1) $y\left(t_1\right)=1$ for some $t_1 \in \Bbb R$.

(2) $y(t)>-1 \quad \forall t \in \mathbb{R}$.

(3) $y$ is strictly increasing in $\mathbb{R}$.

(4) $y$ is increasing in $(0,1)$ and decreasing in $(1, \infty)$.

Options (1) and (4) can be eliminated by an exact solution $$y=\frac{e^{2t}-1}{e^{2t}+1},$$ using the integration $$\int \frac{1}{1-y^2}dy=\int dt+C.$$

But, without the uniqueness of solution (throughout $\Bbb R$), how can we establish options (2) and (3)?

I couldn't apply the Picard's theorem due to unsatisfied hypothesis (Lipschitz condition w.r.t $y$ in $f(t,y)=1-y^2$ on a vertical strip $(a,b) \times \Bbb R$ containing $(0,0)$). However, I can say that the solution is unique in some neighbourhood of $0$ since $\frac{\partial f}{\partial y}(t,y)$ is continuous in any closed rectangle $R=[a,b] \times [c,d]$ which is not sufficient for option (3) and (4).

I also tried to apply the theorem developped here in the answer, but failed to absorb the condition $f(t,0)=0$.

Any analytical justfication (like, 'If we are able to prove $|y|<1$ throughout $\Bbb R$, then I can take the option (3) as a true one') is appreciable.

Thank you in advance.

Answer 1

The right-hand side is locally Lipschitz continuous with respect to $y$, so that any solution to $y' = 1-y^2$ is locally unique.

Since $y_1(y) = 1$ and $y_2(t) = -1$ are (global) solutions to $y' = 1-y^2$, it follows that any solution is either identically $1$, identically $-1$, or does not take the values $1, -1$ at all.

This is sufficient to conclude that the solution $y$ to the initial value problem $y' = 1-y^2$, $y(0) = 0$, satisfies $-1 < y(t) < 1$ for all $t$ in its domain. (It also follows that any local solution to the IVP can be extended to $\Bbb R$.)

So (2) and (3) are true, and (1) and (4) are false.