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# 5.1: Finding Equilibrium Points

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When you analyze an autonomous, ﬁrst-order discrete-time dynamical systems (a.k.a. iterative map)

$x_{t}=F(x_{t-1}).\label{(5.1)}$

one of the ﬁrst things you should do is to ﬁnd its equilibrium points (also called ﬁxed points or steady states), i.e., states where the system can stay unchanged over time. Equilibrium points are important for both theoretical and practical reasons. Theoretically, they are key points in the system’s phase space, which serve as meaningful references when we understand the structure of the phase space. And practically, there are many situations where we want to sustain the system at a certain state that is desirable for us. In such cases, it is quite important to know whether the desired state is an equilibrium point, and if it is, whether it is stable or unstable. To ﬁnd equilibrium points of a system, you can substitute all the $$x$$’s in the equation with a constant $$x_{eq}$$ (either scalar or vector) to obtain

$x_{eq} =F(x_{eq}).\label{(5.2)}$

and then solve this equation with regard to $$x_{eq}$$. If you have more than one state variable, you should do the same for all of them.

Example $$\PageIndex{1}$$:

For example, here is how you can ﬁnd the equilibrium points of the logistic growth model:

$x_{t} =x_{t-1} +rx_{t-1}\left(1-\frac{x_{t-1}}{K}\right)\label{(5.3)}$

Solution

Replacing all the $$x$$’s with $$x_{eq}$$ in Equation \ref{(5.3)}, we obtain the following:

$x_{eq} &=x_{eq} + rx_{eq} (1-\frac{x_{eq}}{K} ) \label{(5.4)}$

$0 &=rx_{eq}(1-\frac{x_{eq}}{K} \label{(5.5)}$

$x_{eq} &=0, \qquad{ K }\label{(5.6)}$

The result shows that the population will not change if there are no organisms $$(x_{eq} = 0)$$ or if the population size reaches the carrying capacity of the environment $$(x_{eq} = K)$$. Both make perfect sense.

Exercise $$\PageIndex{1}$$

Obtain the equilibrium point(s) of the following difference equation:

$x_{t} = 2x_{t-1} -x^{2}_{t-1}\label{(5.7)} \nonumber$

Exercise $$\PageIndex{2}$$

Obtain the equilibrium point(s) of the following two-dimensional difference equation model:

$x_{t}=x_{t-1}y_{t-1}\label{(5.8)} \nonumber$

$y_{t} =y_{t-1}(x_{t-1}-1)\label{(5.9)} \nonumber$

Exercise $$\PageIndex{3}$$

Obtain the equilibrium point(s) of the following difference equation:

$x_{t} =x_{t-1} -x^{2}_{t-2} +1\label{(5.10)} \nonumber$

Note that this is a second-order difference equation,so you will need to ﬁrst convert it into a ﬁrst-order form and then ﬁnd the equilibrium point(s).