I have tackled many 2D systems, but not any with 3D. I'm convinced that the principles and concepts still hold, however the problem is the lengthy computation of the eigenvalues as we have a 3×3 matrix, and the fact that the equations are quite complex. I was curious if there are any easy tricks or something obvious that I am not seeing with regards to this question. So far I have the equations for the flows (I think) and the eigenvalues, but I am unsure of how to see what eigenvalue corresponds to what equation when I have more than two, thus I find it hard to see which equation would correspond to a stable or central manifold.
Consider the following autonomous vector field on $\mathbb{R}^3$:
$$\dot x = y$$
$$\dot y = −x − x^2y + zxy$$
$$\dot z = −z + xz^2$$
$\bullet$ Show that $(x, y, z) = (0, 0, 0)$ is a nonhyperbolic fixed point.
$\bullet$ Compute the stable manifold of $(x, y, z) = (0, 0, 0)$.
$\bullet$ Compute the centre manifold of $(x, y, z) = (0, 0, 0)$.
$\bullet$ Use the centre manifold reduction and the LaSalle invariance principle restricted to the centre manifold to show that $(x, y, z) = (0, 0, 0)$ is asymptotically
stable.
ANSWER ATTEMPT
$\bullet$Taking the Jacobian I have:
$ J = \begin{pmatrix}0 & 1 & 0 \\ -1-2xy+2y & -x^2 + zx & xy \\ z^2 & 0 & -1 +2xz \end{pmatrix}$
So $J(0,0,0) = \begin{pmatrix}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}$
Eigenvalues: $\begin{vmatrix} -\lambda & 1 & 0 \\ -1 & -\lambda & 0 \\ 0 & 0 & -1 – \lambda \end{vmatrix} = -\lambda(-\lambda)(-1-\lambda) – 1(-1)(-1-\lambda) = -\lambda^3 – \lambda^2 – \lambda – 1 = 0$
I can quickly see that $\lambda = – 1$ is a factor so we now have:
$(a\lambda^2 + b\lambda + c)(\lambda + 1) = 0$
$\Rightarrow a\lambda^3+ (a+b)\lambda^2 + (b+c)\lambda + c =-\lambda^3 – \lambda^2 – \lambda – 1$
$\Rightarrow a = -1, a+b = -1, b+c = -1, c = -1$
$\Rightarrow b = 0$
So we have the equation $(-\lambda^2 + -1)(\lambda + 1) = -(\lambda^2 + 1)(\lambda + 1)= (\lambda^2 + 1)(\lambda + 1) = (\lambda +i)(\lambda-i)(\lambda + 1) = 0 $
Thus the eigenvalues are : $\lambda = -1, \pm i$ which means the point $(0,0,0)$ is nonhyperbolic.
$\bullet$ Incorrect integration removed
This question is at the level I should be able to answer in a future exam, however it seems extremely calculation heavy for what it is. I feel I am missing something obvious.
EDIT: Centre manifold reduction
As we have $\dot x = \dot y = 0$ at $x = 0, y = 0$ at the centre manifold, I will restricting these equations to the centre manifold, leaving out the $z$ component.
$$\dot x = y$$
$$\dot y = −x − x^2y + zxy$$
The vector field is of the form
$$\dot x = Ax + f(x, y)$$
$$\dot y = By + g(x, y)$$
where
$A = 0, B = 0, f (x, y) = y, g(x, y) = -x^2y$
Assuming a centre manifold of the form $y = h(x) = ax^2 + bx^3 + O(x^4)$
Substituting this into the expression $Bh(x) + g(x, h(x)) = Dh(x) (Ax + f(x, h(x)))$ we have:
$-x^2y = (2ax+ 3bx^2 + O(x^3))y$
Equating coefficients of $x$:
$a = 0$
$3b = -1$
$\Rightarrow y = \frac{-x^2}{3}$ which is stable for $x$ sufficiently small
Best Answer
Hint: Note that the line $x=0,y=0$ is invariant since $\dot{x}=\dot{y}=0$ there, and the plane $z=0$ is invariant since $\dot{z}=0$ there.