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