Can someone familiar with ODE's help me work through this problem?

Consider the system of differential equations in $\mathbb{R}^3$ given by $$x'=(\epsilon x+2y)(z+1)$$$$y'=(-x+\epsilon y)(z+1)$$$$z'=-z^3$$where $\epsilon$ is a parameter.

I want to describe the stability of is equilibria.

Question

Can someone familiar with ODE's help me work through this problem?

Consider the system of differential equations in $\mathbb{R}^3$ given by $$x'=(\epsilon x+2y)(z+1)$$$$y'=(-x+\epsilon y)(z+1)$$$$z'=-z^3$$where $\epsilon$ is a parameter.

I want to describe the stability of is equilibria.

richyw · Answer

so first I do $$x'=0$$$$y'=0$$$$z'=0$$and find that the only equilibrium point is $(0,0,0)$

richyw · Answer

Now I linearized the system, and got $$\left[\begin{matrix}x'\y'\z'\end{matrix}\right]=\left[\begin{matrix}\epsilon & 2 & 0 \ -1 &\epsilon &0 \ 0&0&0\end{matrix} \right] \left[\begin{matrix} x \ y \ z \end{matrix}\right]$$

richyw · Answer

eigenvalues are $\epsilon\pm\sqrt{2}i$ and 0

richyw · Answer

my book says that from linearization we can only conclude that the origin is unstable if $\epsilon>0$. Because when, $z=0$, the x,y-plane is invariant and the system is linear on this plane.

richyw · Answer

I was wondering if anyone could explain to me how I know this and what it means...

anonymous · Answer

been a while richy. let me refresh my memory

anonymous · Answer

oh yea..
the Jacobian. duh. :)

anonymous · Answer

so, what this tells us graphically is how the system changes with respect to each variable.
remembering from 1D cases, when the Jacobian becomes "0", the function has a critical value with repect to that variable.. in other words, invarient. z=0 is the equation for XY plane.

anonymous · Answer

Now, about roots for stability:
if the roots are positive, function is continuously increasing -> unstable
if negative, decreasing -> stable
if zero, saddle point.
COMPLEX -> goes beating round the bush
positive real part -> goes round with increasing radius -> unstable
negative real part -> goes round with decreasing radius and finally converges -> stable
purely imaginary -> the radius does not change but goes cyclic -> marginally stable

anonymous · Answer

this root criterion for stability of DEs has been interpreted and explained by many people in many different ways suiting the applications; but this is the central theme for all.

richyw · Answer

thanks!

anonymous · Answer

yw