Prove the solution $y(x)$ of $y'=(\sin(x)-y)^3,y(0)=0$ is bounded. Prove the solution $y(x)$ of $y'=(\sin(x)-y)^3,y(0)=0$ is bounded.
My solution:
$B$ is a rectangle , $B:=\{(x,y)||x|\leq a , |y|\leq b \}$.
Denote $f(x,y)=(\sin(x)-y)^3\leq(1-y)^3\leq (1-b)^3=M$
$h=\min\{ a,\frac{b}{M}\}=\min\{ a,\frac{b}{(1-b)^3}\}\leq \min\{ a,-\frac{1}{b^2}\}$
$-\frac{1}{b^2}$ has no extreme points, how can I show that $y(x)$ is bounded ?
Thanks !
 A: The question is misleading, it can only be shown to be forwards-bounded, for $x\ge0$.
The leading qualitative observation is that for large $y$ it behaves like $y'=-y^3$ which indeed converges towards the $x$ axis for increasing $x$.
More specifically, note that on the line $y=2$ the vector field points downwards while on $y=-2$ it points upwards. Thus the strip $|y|\le 2$ is a trapping region in forward time, thus any solution starting inside will also stay inside.
A: Let $y:[0,\omega_+) \to \mathbb{R}$ be the soution of this IVP (nonextendable to the right). Fix any $c > 1$ and assume that $y(\xi) > c$ for some $\xi \in  (0,\omega_+)$. Go from $\xi$ to the first $\eta$ to the left of $\xi$ with $y(\eta)=c$ (this exists since $y(0)=0$). Now $y(x) > c$ $(x \in (\eta,\xi])$, hence $y'(\eta) \ge 0$. On the other hand $y'(\eta)=(\sin(\eta)-c)^3<0$, a contradiction. Thus $y(x) \le c$ on $[0,\omega_+)$. The same way show that  $y(x) \ge -c$ on $[0,\omega_+)$. Thus $y$ is bounded and $\omega_+= \infty$, and $|y(x)| \le c$ on $[0, \infty)$.
Remark: $c \to 1$ leads to $|y(x)| \le 1$ on $[0, \infty)$.
A: Hint.
Making $z = \sin x- y$ we have
$$
z'= -z^3+\cos x\Rightarrow z z' = z^4+z\cos x
$$
then
$$
\frac 12z^2 = \int (z\cos x -z^4)dx\le \int(z-z^4)dx
$$
which leads to a contradiction if $z$ is unbounded.
A: Maybe the easiest way is to show that
$$
\dot{y} = (u - y)^3
$$
is input to state stable which works because $u=\sin(x)$ is bounded ($|u| \leq 1$). Now take
$$
V(y) = \frac{1}{2}y^2
$$
as Lyapunov function. You end up with
$$
\dot{V}(y) = -y (y - u)^3
$$
So if $|y| > 1$ then $\dot{V}$ is negative and so $y$ must be bounded for all $y(0) \in R$ not only $y(0) = 0$.
