# Question on dynamic of $y'(t)=\sin(y(t))$ and on nature of equilibrium point of $y'=|\sin(y)|$.

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 $\dot{x}=f(x)$: $\lim_{t\rightarrow\infty}x(t)=x^*\Rightarrow f(x^*)=0$).

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.

• 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... – user659895 May 26 at 21:04
• @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. – Martin R May 27 at 3:40