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Consider the initial value problem $(P_\lambda) \begin{cases}x' = \lambda \arctan(t) + \sin(x) \\ x(0) = 5\end{cases}$

I'm asked to find:

  • A value $\lambda_1 > 0$ such that if $|\lambda| < \lambda_1$ the maximal solution of $(P_\lambda)$ is bounded in the future.

  • A value $\lambda_2 > 0$ such that if $\lambda > \lambda_2$ then the maximal solution of $(P_\lambda)$ is not bounded in the future.

My approach was to find a coercive guiding function which in this situation amounts to finding $V$ such that:

  • $\lim_{t \to +\infty,-\infty} V(x) = +\infty$
  • $V'(x) (\lambda \arctan(t) + \sin(x)) \le 0$ on $]t_0,+\infty[ \times \mathbb{R}$

However, my attempts to bound this thing didn't succeed.

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As the sine is bounded to the interval $[-1,1]$, for $λ>\frac2\pi$ after $t>\tan(\frac1λ)$ the right side will always be positive and thus any solution from that time on monotonically increasing without bound.

Now consider closer the case $λ<\frac2\pi$. At $t=0$, the right side is negative and thus the solution falling towards the next lower null-cline at $$x_1(t)=\pi+\arcsin(λ\arctan(t)).$$ As the nullcline is growing in $t$, the solution will intersect it horizontally. Now the right side is positive, the solution will grow towards and follow the nullcline.

plots of some solutions around the critical parameter

solutions for $λ=\frac2\pi+\epsilon$. In gray are the nullclines for $λ=\frac2\pi$.

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  • $\begingroup$ I see that in the first case the solution has to be monotonically increasing but why without bound? couldn't it converge to a limit? $\endgroup$ – Rodrigo Jun 7 '19 at 9:35
  • $\begingroup$ $λ\arctan t$ converges to $λ\frac{\pi}2=1+ϵ$. Beginning at some $t_2>t_1=\tan(1/λ)$ you will get $λ\arctan t>1+ϵ/2$, so that the right side is always larger than $ϵ/2$, thus perpetual growth. $\endgroup$ – Lutz Lehmann Jun 7 '19 at 9:50
  • $\begingroup$ Nullcline wasn't a word that I used before. I understand that it is the solution for $x' = 0$. But it would be very helpful if you clarify what property of nullclines you are using here. The only related thing I see is that equilibrium points are "barriers" that solutions can't traverse. Is that what you are using for the second part? $\endgroup$ – Rodrigo Jun 7 '19 at 10:12
  • $\begingroup$ At nullclines the solution is horizonal. If the nullcline is a rising function as happens here, this means that a solution will pass from above to below. The further behavior depends on the sign of the right side below the nullcline. As it is positive, the solution will rise. But it can never again pass above the nullcline, as a positive slope on the nullcline is a contradiction in terms. As the nullcline is asymptotitically constant to $\pi+\arcsin(λ\frac{\pi}2)$, also the solutions will converge there. $\endgroup$ – Lutz Lehmann Jun 7 '19 at 10:23
  • $\begingroup$ Do you see an argument that the solution has to pass through the lower nullcline? This is the only bit I don't get still. It could potentially "scape" if the upper nullcline get down fast... $\endgroup$ – Rodrigo Jun 7 '19 at 16:38

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