Let $y'=\sin(y)$ an ODE. I just try to imagine the dynamic behind.

Q1) First, on $\mathbb R$ we only have $\mathcal C^1$ piecwise solution, no ? Because, when for example $y(t_0)\in (0,\pi/2)$, then, $y(t)\to \pi/2$, and when the particle will arrive in $\pi/2$, then it will sotp and never go again. And so, I can't get a smooth solution on $\mathbb R$. Am I right ?

Q2) Now, if $y'=|\sin(y)|$, what will be the nature of $y=\pi$ ? Because it's an equilibrium point, but it will be stable at left and unstable at right. Can I call it a saddle ?


For each $(t_0, y_0) \in \Bbb R^2$ the initial value problem $$ \tag{*} y'(t) = \sin(y(t)) \, , \quad y(t_0) = y_0 $$ has a unique solution on $\Bbb R$ because the right-hand side is a Lipschitz-continuous function of $y$.

If $y_0 = k \pi$ with $k \in \Bbb Z$ then the constant function $y(t) \equiv t_0$ is the unique solution.

If $y_0 \in (k \pi, (k+1) \pi) $ for some $k \in \Bbb Z$ then the solution $y$ cannot take any of the values $l \pi$ with $l \in \Bbb Z$, and since it is continuous, $$ k \pi < y(t) < (k+1) \pi \quad\text{ for all } t \in \Bbb R \, . $$ It follows that $y$ is strictly increasing if $k$ is even, and strictly decreasing if $k$ is odd. Consequently, both limits $\lim_{t\to \pm \infty}y(t)$ exits and are either $k \pi$ or $(k+1)\pi$ (compare Autonomous ODE $\dot{x}=f(x)$: $\lim_{t\rightarrow\infty}x(t)=x^*\Rightarrow f(x^*)=0$).

Regarding your questions:

First, on $\mathbb R$ we only have $\mathcal C^1$ piecewise solution.

No. Repeated differentiation of $(*)$ shows that every solution is $\mathcal C^\infty$ on $\Bbb R$.

When the particle will arrive in $\pi/2$, then it will stop and never go again.

No. The only constant solutions are $y(t) \equiv k \pi$. All other solutions are strictly increasing or strictly decreasing.

The same conclusions hold for the initial value problem $$ \tag{**} y'(t) = |\sin(y(t))| \, , \quad y(t_0) = y_0 $$ with the only difference that the non-constant solutions are all strictly increasing. As above, every non-constant solution lies in a strip $$ k \pi < y(t) < (k+1) \pi \quad\text{ for all } t \in \Bbb R \, . $$ In particular, either $y'(t) = \sin(y(t))$ for all $t$ or $y'(t) = -\sin(y(t))$ for all $t$, so that the solutions of $(**)$ are $\mathcal C^\infty$ functions as well.

  • $\begingroup$ Sorry, but why if $y_0=k\pi$, then when $y_0\in (k\pi, (k+1)\pi)$, then $y$ cannot take any value of $l\pi$ ? And for $y'(t)=|\sin(y(t))|$ what is the nature of the equilibrium points ? are they saddle ? Because they are stable at left, but unstable at right... $\endgroup$ – user659895 May 26 at 21:04
  • $\begingroup$ @user659895: If $y(t_1) = l\pi$ for some $t_1$ then $y(t) \equiv l \pi$ because of the uniqueness of solutions. Therefore solutions with an initial value $y_0\in (k\pi, (k+1)\pi)$ cannot take the values $l \pi$. – I cannot answer your other questions because I am not familiar with the terms “saddle” or “stable at left/right” in this context. $\endgroup$ – Martin R May 27 at 3:40

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