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For my class in dynamical models in biology, we analyze the local stability of steady states of systems of differential equations by taking a linear approximation and finding the Jacobian matrix. After this, the eigenvalues of this Jacobian are evaluated to determine local asymptotic stability. After a lot of algebra, I computed the characteristic polynomial i.e. det($J - \lambda * I$) of this Jacobian to be:

$\lambda^3 + \lambda^2(BN - b - r) + \lambda(-3b^2-2br+2bBN) + (b^2BN-b^2r-b^3) = 0$

where $\lambda$ is the eigenvalue and everything else is a constant. Because this has no obvious factorization, I tried this technique:

Let x, y, z be arbitrary constants such that $(\lambda - x)(\lambda - y)(\lambda - z) = \lambda^3 + \lambda^2(BN - b - r) + \lambda(-3b^2-2br+2bBN) + (b^2BN-b^2r-b^3)$

and after some expansion and reduction, I have this system:

$x + y + z = b + r - BN$

$xy + yz + xz = -3b^2-2br+2bBN$

$xyz = b^2r+b^3-b^2BN$

I am sure there is a way to isolate the values of x, y, and z so that I can compute my eigenvalues, but I do not know how to proceed. Can anyone tell me how to solve this final system with x, y, and z?

note: the original system is:

$dS/dt = f = bN - BSI - bS$

$dI/dt = g = BSI - rI - bI$

$dR/dt = h = rI - bR$

and the Jacobian I computed was:

first row: $-BI - b, -BS, 0$

second row: $BI, BS - r - b, 0$

third row: $0, r, -b$

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    $\begingroup$ What’s the original matrix? $\endgroup$
    – amd
    Commented Oct 7, 2018 at 16:15
  • $\begingroup$ @amd I updated the question. You can see it now $\endgroup$ Commented Oct 7, 2018 at 16:23
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    $\begingroup$ Try Laplace, develop for the last column. One obvious factor will be $-b-\lambda$. $\endgroup$ Commented Oct 7, 2018 at 16:28
  • $\begingroup$ @MichaelHoppe I actually did do Laplace Determinant Expansion on the last column, and I substituted the steady state (S = N, I = 0, R = 0) after which I was left with the problem described above. Why do you say $-b-\lambda$ is obvious? $\endgroup$ Commented Oct 7, 2018 at 16:47

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$$\det\begin{pmatrix} -BI - b-\lambda& -BS& 0\\ BI &BS - r - b-\lambda& 0\\ 0&r&-b-\lambda \end{pmatrix}=$$ $$=(-b-\lambda)\det\begin{pmatrix} -BI - b-\lambda&-BS\\ BI &BS - r - b-\lambda \end{pmatrix}$$

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