Consider the following autonomous vector field:
$$\dot x = −x$$
$$\dot y = \sin y$$
where $x \in \mathbb{R}^2, -\pi ≤ y ≤ \pi$
$\bullet$ Find all fixed points.
$\bullet$ Determine the linearized stability properties of each fixed point.
$\bullet$ Determine the global stable and unstable manifolds of the origin.
$\bullet$ Determine whether any nonhyperbolic fixed points are stable or unstable? (You must justify your answer.)
ATTEMPT:
$\bullet$ Fixed points: $(0, -\pi), (0,0), (0,\pi)$
$\bullet$ Linearisation by the Jacobian:
$J = \begin{pmatrix}-1 & 0 \\ 0 & \cos y \end{pmatrix}$
So we have:
$J(0,-\pi) = \begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}$, $J(0,0) = \begin{pmatrix}-1 & 0 \\ 0 & 1 \end{pmatrix}$, $J(0,\pi) = \begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}$
Which represent a sink (stable), a saddle (unstable), and a sink (stable) respectively,
$\bullet$ For the global stable manifold we can see by simply solving for $x$:
$\dot x = -x \Rightarrow \int \frac{\dot x}{x} dt = \int \frac{dx}{x} = -\int dt$
$\Rightarrow x = e^{-t+c}$
Using $x(0) = x_0$ an arbitrary initial condition $\Rightarrow x_0 = e^c$
$x = x_0 e^{-t}$ is the global stable manifold as $x \rightarrow 0$ as $t \rightarrow \infty$ for all initial conditions
My problem here is with the unstable manifold due to the $\sin$ function and that there doesn't seem to be any nonhyperbolic fixed points, as I have two sinks and a saddle? I feel like that the unstable manifold should simply be the region $-\pi \le y \le \pi$ but I just wanted clarification.
As far as I am taught, the saddle points is hyperbolic so I'm not sure where the nonhyperbolic fixed points come from, unless I am missing something obvious or have done something wrong.