# Limit of solution of Cauchy problem

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 $$\lim_{t\to+\infty}y(t)=+\infty$$.
• 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! – qcc101 Jan 10 at 19:11
• For $t>1$ $y'\ge\arctan(t^3(\alpha-1))\ge\arctan((\alpha-1))>0$. – Julián Aguirre Jan 10 at 19:19