Growth rate in the context of dynamical systems

Question

Consider the following question (I only need help with the last part)

The dynamics of a directly transmitted microparasite can be modelled by the system \begin{align*} \frac{d X}{d t} & =b N-\beta X Y-b X, \\ \frac{d Y}{d t} & =\beta X Y-(b+r) Y, \\ \frac{d Z}{d t} & =r Y-b Z, \end{align*} where $b, \beta$ and $r$ are positive constants and $X, Y$ and $Z$ are respectively the numbers of susceptible, infected and immune (i.e. infected by the parasite, but showing no further symptoms of infection) individuals in a population of size $N$, independent of $t$, where $N=X+Y+Z$. Consider the possible steady states of these equations. Show that there is a threshold population size $N_c$ such that if $N<N_c$ there is no steady state with the parasite maintained in the population. Show that in this case the number of infected and immune individuals decreases to zero for all possible initial conditions. Show that for $N>N_c$ there is a possible steady state with $X=X_s<N$ and $Y=Y_s>0$, and find expressions for $X_s$ and $Y_s$. By linearising the equations for $d X / d t$ and $d Y / d t$ about the steady state $X=X_s$ and $Y=Y_s$, derive a quadratic equation for the possible growth or decay rate in terms of $X_s$ and $Y_s$ and hence show that the steady state is stable.

I got to the last part and I am unsure what to do. I considered peturbating

$$ X= X_s+\epsilon \tilde X \quad Y=Y_s + \epsilon \tilde Y. $$ This got me \begin{align*} \frac{d \tilde{X} }{dt} = -\beta \tilde{Y} X_s - \beta \tilde{X} Y_s -b \tilde{X} \end{align*} and \begin{align*} \frac{d \tilde{Y} }{dt }= \beta (Y_s \tilde{X} + X_s \tilde{Y} ) - (b+ r) \tilde{Y}. \end{align*}

At this point I am stuck. What is the meaning of growth/decay rate here?

Answer 1

Hint We can rewrite the linearized system in matrix form as $$\pmatrix{\tilde X\\\tilde Y}' = A \pmatrix{\tilde X\\\tilde Y} ,$$ where $$A := \pmatrix{-\beta Y_s - b & -\beta X_s \\ \beta Y_s & \beta X_s - (b + r) } ,$$ which has general solution $$\pmatrix{\tilde X\\\tilde Y} = \pmatrix{\tilde X_0\\\tilde Y_0} e^{A t} .$$ In particular, to show that the steady state is stable, it suffices to show that the real parts of the eigenvalue(s) of $A$ are negative.

But the characteristic polynomial of $A$ (whose roots are the eigenvalues) can be written as $$\lambda^2 + \frac{b N}{X_s} \lambda + \beta (b + r) Y_s ,$$ and in particular its coefficients are positive, so the roots $\lambda$ have negative real part.