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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?

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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 $\lim_{t\to+\infty}y(t)=+\infty$.

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  • $\begingroup$ 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

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