7.1: Finding Equilibrium Points
( \newcommand{\kernel}{\mathrm{null}\,}\)
Finding equilibrium points of a continuous-time model dxdt=G(x) can be done in the same way as for a discrete-time model, i.e., by replacing all x’s with xeq’s (again, note that these could be vectors). This actually makes the left hand side zero, because xeq is no longer a dynamical variable but just a static constant. Therefore, things come down to just solving the following equation
0=G(xeq)
with regard to xeq. For example, consider the following logistic growth model:
dxdt=rx(1−xK)
Replacing all the x’s with xeq’s, we obtain
0=rxeq(1−xK)
xeq=0,K
It turns out that the result is the same as that of its discrete-time counterpart(see Eq.(5.1.6)).
Find the equilibrium points of the following model:
dxdt=x2−rx+1
Find the equilibrium points of the following model of a simple pendulum:
d2θdt2=−gLsinθ
The following model is called a Susceptible-Infected-Recovered (SIR) model, a mathematical model of epidemiological dynamics. S is the number of susceptible individuals, I is the number of infected ones, and R is the number of recovered ones. Find the equilibrium points of this model.
dSdt=−aSIdIdt=aSI−bIdRdt=bI