To identify equilibria of the exponential growth model in discrete time, we replace n(t + 1) and n(t) with h in the recursion equation, n(t + 1) = Rn(t), to get the equilibrium condition, h = Rh (Step 1). We then solve for h (Step 2). You might be tempted to divide both sides by h to simplify this equation, leaving 1 = R. At this point you might conclude that there is no equilibrium of the model unless R happens to equal one. This is known as a special case of the parameters, because it is extremely unlikely that R will equal exactly one for any real population. In general, it is a great idea to simplify an equilibrium condition by canceling out a term that multiplies both sides of the condition, but when we do so, we must check whether the canceled term could itself equal zero. For the exponential model, when we divide both sides of the equilibrium condition h = Rn by the term h, the canceled term could be zero if h = 0. Therefore, n = 0 satisfies the equilibrium condition and is a valid equilibrium of the exponential model.

We can check to make sure that we have correctly identified an equilibrium by setting n(t) to the potential equilibrium value (zero in the present case) and confirming that n(t + 1) remains at this value (Step 3). Indeed, plugging n(t) = 0 into the recursion equation for the exponential model gives n(t + 1) = R x 0 = 0. Thus, an equilibrium of the exponential model occurs when there are no individuals within the population.

An alternative way to solve the equilibrium condition h = Rh is to subtract h from both sides, giving 0 = Rh - h, which can be factored into 0 = (R - 1 )h. Solving for h, we can again see that h = 0 is an equilibrium. Indeed, it is the only equilibrium of the model. The only other way to satisfy this equilibrium condition is to set R to a specific value, R = 1. Remember, however, that we are seeking values of the variable n at which no change is predicted, and therefore it does not make sense to say that R = 1 is an equilibrium. Rather, when R = 1, the equilibrium condition is always satisfied, and all initial values of n represent equilibria, because the population exactly replaces itself regardless of the population size.

In the logistic model, applying Recipe 5.1 to the recursion equation,

n(t + 1) = n{t) + r n(t)y 1--— J (recursion equation from 3.5a)

gives the equilibrium condition h = h + r 1 ~ (equilibrium condition for 3.5a).

Subtracting h from both sides demonstrates that rh(l - n/K) must equal zero at an equilibrium. Besides the special case of zero growth (r = 0, in which case all populations remain constant in size), the equilibrium condition has two possible solutions; either h = 0 or (1 - h/K) = 0. The second solution is satisfied when h = K. Thus, we conclude that there are two states in the logistic model at which the population size will remain constant: when there are no individuals present (n = 0) or when the population is at the carrying capacity in = K).

If you are well practiced in solving equations, it might be obvious that the solution to (1 - h/K) = 0 is h = K. Until solving such equations becomes second nature, however, there is a general recipe to find values of h that satisfy any linear equation.

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