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[SOLVED] Finding Eigenvalues for Different r Values

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
\mathcal{J} = \begin{pmatrix}
-\sigma & \sigma & 0\\
1 & -1 & -\sqrt{b(r - 1)}\\
\sqrt{b(r - 1)} & \sqrt{b(r - 1)} & - b
\end{pmatrix}
$$
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
Feb 7, 2012
2,703
$$
\mathcal{J} = \begin{pmatrix}
-\sigma & \sigma & 0\\
1 & -1 & -\sqrt{b(r - 1)}\\
\sqrt{b(r - 1)} & \sqrt{b(r - 1)} & - b
\end{pmatrix}
$$
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. (Yawn) )
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
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
Feb 7, 2012
2,703
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 $$\lambda = \frac{18b\sigma(r-1) - b(b+\sigma+1)(r+\sigma)}{2(b+\sigma+1)^2 -6b(r+\sigma)}.$$