[SOLVED]Finding Eigenvalues for Different r Values

dwsmith

Well-known member
$$\mathcal{J} = \begin{pmatrix} -\sigma & \sigma & 0\\ 1 & -1 & -\sqrt{b(r - 1)}\\ \sqrt{b(r - 1)} & \sqrt{b(r - 1)} & - b \end{pmatrix}$$
From a quick try and error, I was able to find that when $r = 1.3456171$ we will have 3 negative eigenvalues.
But when $r = 1.3456172$, there will be a complex-conjugate pair of eigenvalues.
Is there a mathematically more elegant way to determine this r value?
$b = \frac{8}{3}$ and $\sigma = 10$

Opalg

MHB Oldtimer
Staff member
$$\mathcal{J} = \begin{pmatrix} -\sigma & \sigma & 0\\ 1 & -1 & -\sqrt{b(r - 1)}\\ \sqrt{b(r - 1)} & \sqrt{b(r - 1)} & - b \end{pmatrix}$$
From a quick try and error, I was able to find that when $r = 1.3456171$ we will have 3 negative eigenvalues.
But when $r = 1.3456172$, there will be a complex-conjugate pair of eigenvalues.
Is there a mathematically more elegant way to determine this r value?
$b = \frac{8}{3}$ and $\sigma = 10$
This is an interesting problem. The eigenvalue equation is $\det(\mathcal{J} - \lambda I) = 0$, which simplifies to $$\lambda^3 + (b+\sigma+1)\lambda^2 + b(r+\sigma)\lambda + 2b\sigma(r-1) = 0.\qquad(1)$$ At the critical value of $r$ (the one where the bifurcation occurs, from three real roots to a complex conjugate pair), the eigenvalue equation will have a repeated root. That root will also be a root of the derived equation. So differentiate the eigenvalue equation to get $$3\lambda^2 +2(b+\sigma+1)\lambda + b(r+\sigma) = 0.\qquad(2)$$ Solve those equations together, to find that $$\lambda = \frac{18b\sigma(r-1) - b(b+\sigma+1)(r+\sigma)}{2(b+\sigma+1)^2 -6b(r+\sigma)}.$$ The next stage is to substitute that value of $\lambda$ into equation (2), to get a cubic equation for $r$. If you have stamina and patience enough to do that, you should find that, for the given values of $b$ and $\sigma$, one of the solutions is $r = 1.345617...$. (Rather you than me, though. )

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dwsmith

Well-known member
Solve those equations together, to find that $$\lambda = \frac{18b\sigma(r-1) - b(b+\sigma+1)(r+\sigma)}{2(b+\sigma+1)^2 -6b(r+\sigma)}.$$
How did you get this lambda? I solved for the quadratic and got something different. I also set the quadratic and cubic equal and then solved for lambda but didn't arrive at this lambda either.

Opalg

MHB Oldtimer
Staff member
How did you get this lambda? I solved for the quadratic and got something different. I also set the quadratic and cubic equal and then solved for lambda but didn't arrive at this lambda either.
Starting from $$\lambda^3 + (b+\sigma+1)\lambda^2 + b(r+\sigma)\lambda + 2b\sigma(r-1) = 0\qquad(1)$$ and $$3\lambda^2 +2(b+\sigma+1)\lambda + b(r+\sigma) = 0,\qquad(2)$$ I multiplied (1) by 3, and (2) by $\lambda$, and subtracted, getting $$(b+\sigma+1)\lambda^2 + b(r+\sigma)\lambda + 6b\sigma(r-1) = 0.\qquad(3)$$ I then multiplied (2) by $(b+\sigma+1)$, and (3) by 3, and again subtracted, to get the equation $$\lambda = \frac{18b\sigma(r-1) - b(b+\sigma+1)(r+\sigma)}{2(b+\sigma+1)^2 -6b(r+\sigma)}.$$