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Finding the stability of the origin fixed point in a Lorentz system.

The Lorentz equation:

$\dot{x}=\sigma\left ( y-x \right )$

$\dot{y}=rx-y-xz$

$\dot{z}=-bz+xy$

The Jacobian evaluated at the fixed point $\left ( 0,0,0 \right )$:

$J\mid _{\left ( 0,0,0 \right )}=\begin{bmatrix} -\sigma &\sigma &0 \\ r &-1 &0 \\ 0&0 &-b \end{bmatrix}$

We seek $det\left ( J \mid _{\left ( 0,0,0 \right )}-\lambda I \right )=0$

We compute the $3\times 3$ matrix:

This gives: to save you helpful souls from working it out

enter image description here

Setting the above result to 0, and rearranging the term to $\lambda^{3}$, $\lambda^{2}$, $\lambda$ and the constants, there is no integer valued solutions to be found.

How should I get around this conundrum?

Any explanation to expand my understanding is greatly appreciated.

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Hint: Don't expand the brackets. The first cofactor gives you one of the eigenvalues precisely, and the second cofactor is a simple quadratic equation which roots could be analyzed by Vieta's formula.

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  • $\begingroup$ So the first eigenvalue is -b. The other two is solved via the quadratic equation. I see it now. Thank you! $\endgroup$ Jul 29 '17 at 6:03
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    $\begingroup$ Yeah, but take into account that usually you don't need to solve quadratic equations to say something about its roots. For example, if you have equation $\lambda^2 +a \lambda + b = 0$ and $b < 0$ the roots are always real and have different signs. If $b > 0$ the roots always have the same real part the sign of which depends on $a$. It's easier to analyze the roots this way than by analyzing roots from discriminant formula. $\endgroup$
    – Evgeny
    Jul 29 '17 at 6:08
  • $\begingroup$ The above was computed with Mathematica. With handworking, it's can be quite a strenuous affair to get it to the form above produced by Mathematica. $\endgroup$ Jul 29 '17 at 6:16
  • $\begingroup$ Using your second comment: The origin is a saddle node. $\endgroup$ Jul 29 '17 at 6:35
  • $\begingroup$ Saddle-node is a degenerate equilibrium with 1 zero eigenvalue (+some additional conditions which are typically hold). For what values of parameters do you want to understand the type of equilibrium? $\endgroup$
    – Evgeny
    Jul 29 '17 at 6:43

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