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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 3x3 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

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  • $\begingroup$ $\dot{x} = y$ does not imply $x = y t + x_0$, because $y$ is not constant. $\endgroup$ – Robert Israel Aug 28 '18 at 15:20
  • $\begingroup$ @RobertIsrael I knew I was doing something very silly and obviously wrong, in which case I'm not sure on how to proceed as i've never encountered a problem such as this. $\endgroup$ – A. Bharj Aug 28 '18 at 15:24
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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.

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  • $\begingroup$ I see now, invariant manifolds exist spanning the $x-y$ lines and the $z$ axis. I think that at (0,0,z), $\dot y = \dot x = 0$ but $\dot z = -z$, thus the stable manifold is the z plane or the span of the vector $[0,0,1]$. This leaves the $x-y$ lines for the center manifold. So thus $x = 0$ and $y = 0$ is the center manifold? How do you know that the rest of the x-y plane is not invariant? $\endgroup$ – A. Bharj Aug 28 '18 at 15:44
  • $\begingroup$ If $\dot{z} = -z$ you have $z \to 0$ as $t \to +\infty$, so the $z$ axis would be in the stable manifold. The $xy$ plane should be the center manifold. $\endgroup$ – Robert Israel Aug 28 '18 at 17:39
  • $\begingroup$ Thanks for the clarification on this. Are you able to help at all with the Center manifold part? I've had an attempt but I'm having trouble since there are three equations due to the 3 dimensional nature of the problem. I understand that I have to restrict it to the $xy$ plane as this is where the center manifold lies but I'm having trouble doing so due the $z$ component in the the equation for $\dot y$. Even after omitting it, I don't appear to get anywhere. $\endgroup$ – A. Bharj Aug 28 '18 at 19:19
  • $\begingroup$ The $xy$ plane is the center manifold. On the $xy$ plane, $z = 0$ and $\dot{z} = 0$. On the $xy$ plane, the system is reduced to $2$-dimensional: $\dot{x} = y,\; \dot{y} = -x - x^2 y$. $\endgroup$ – Robert Israel Aug 28 '18 at 23:09

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