I have the following Cauchy problem:

\begin{align} y'(t) = \arctan(t^3(y-1)) \\ y(0) = \alpha \end{align}

I want to study the limit of the solution on the boundary.

This is what I have done so far:

I know that the function is $C^\infty$ so it is Lipshitz and then I have uniqueness and existence of global solution for each $\alpha$.

The constant solution is y = 1. By uniqueness, other solutions cannot interest this line.

If $\alpha < 1$, then I have that the function increases monotonically up until $\alpha$ and then decreases monotonically to infinity.

If $\alpha > 1$, then I have that the function decreases monotonically down to $\alpha$ and then increases monotonically to infinity.

This means that I always have the limit both at $-\infty$ and at $\infty$ for whatever alpha. However my problem now is actually finding the limit.

Any suggestion?


I will do the case $\alpha>1$ and $t\to+\infty$. By uniqueness, $y(t)>1$ for all $t>0$. This implies that $y'(t)\ge0$ and that $y$ is increasing. In particular, $y(t)\ge\alpha$ for all $t>0$. Then $y'(t)\ge\arctan(t^3(\alpha-1))$ for all $t>0$. It follows now easily that $\lim_{t\to+\infty}y(t)=+\infty$.

  • $\begingroup$ I do not understand how to go from the fact that $y'(t) \geq arctan(t^3(\alpha-1))$ to the fact that $y(t) \to \infty$ EDIT: Nevermind by contradiction follows easily. Thank you! $\endgroup$ – qcc101 Jan 10 at 19:11
  • $\begingroup$ For $t>1$ $y'\ge\arctan(t^3(\alpha-1))\ge\arctan((\alpha-1))>0$. $\endgroup$ – Julián Aguirre Jan 10 at 19:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.