Vtyjkyuk

Consider the following autonomous vector field:

$$dot x = −x$$

$$dot y = sin y$$

where $x in mathbbR^2, -pi ≤ y ≤ pi$

$bullet$ 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:

$J = beginpmatrix-1 & 0 \ 0 & cos y endpmatrix$

So we have:

$J(0,-pi) = beginpmatrix-1 & 0 \ 0 & -1endpmatrix$,
$J(0,0) = beginpmatrix-1 & 0 \ 0 & 1 endpmatrix$,
$J(0,pi) = beginpmatrix-1 & 0 \ 0 & -1endpmatrix$

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 $x$:

$dot x = -x Rightarrow int fracdot xx dt = int fracdxx = -int dt$

$Rightarrow x = e^-t+c$

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.