1

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.

Messi Lio
  • 765
  • 2
  • 12

1 Answers1

2

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.

Martin R
  • 113,040
  • Funny, I had just started writing "The right hand side is locally Lipschitz" when your answer appeared. – copper.hat Sep 30 '22 at 05:58
  • Martin R Thanks a lot.... – Messi Lio Sep 30 '22 at 06:04
  • @Martin R Could you pls mention the 'extension procedure' from the local solution (for the given IVP) to $\Bbb R$ unless the question has existence hypothesis? – Messi Lio Sep 30 '22 at 06:28
  • 1
    @MessiLio: There is a theorem (sorry, I do not have a reference right now) that the “maximal solution” either exists everywhere, or “explodes” at the end of its maximal definition interval. The latter can not happen here because the solution is bounded. – Martin R Sep 30 '22 at 06:38
  • 1
    @MessiLio: The statement is something like “the maximal solution leaves every compact set” and you might find a reference with these keywords. Here the maximal solution can not leave the interval $[-1, 1]$ in $y$-direction, therefore it must leave every interval in $t$-direction. – Martin R Sep 30 '22 at 08:27