Question

1. Periodic solutions and their loss of stability The logistic difference equation (Nn+1 - Nn)/ Δt=r. Nn. (1 - N/K) possesses chaotic solutions for r Δt > 2.569946 .... The above equation can be made dimensionless and rewritten in the simpler form: Xn+1 = f(X.n), where f(x) = ax(1 - x), a= 1+r Δt, a > 1 and Xn = Nn . r. Δt/((1+r Δt)K). We want to investigate the equilibrium solutions to this equation and their loss of stability as a increases. a. Find the equilibria of Eq. (7.7) and determine their stability. b. For a > 3, the solution to Eq. (7.7) includes a 2-cycle. That is, it obeys, along with the equilibria you just found in (a), this new equation: Xn+2 = X = X2.

Answer 1

The logistic difference equation can be rewritten as Xn+1 = f(Xn), where f(x) = ax(1 - x) and a = 1+r Δt. To find the equilibria, set Xn+1 = Xn and solve. Determine stability by calculating the derivative of f(x) at the equilibrium points. For a > 3, the solution includes a 2-cycle.

Step-by-step explanation:

The logistic difference equation (Nn+1 - Nn)/ Δt=r. Nn. (1 - N/K) possesses chaotic solutions for r Δt > 2.569946. The equation can be made dimensionless and rewritten as Xn+1 = f(Xn), where f(x) = ax(1 - x), a = 1+r Δt, and Xn = Nn . r. Δt/((1+r Δt)K).

a. To find the equilibria of Eq. (7.7), we set Xn+1 = Xn and solve for Xn. This gives us the equilibria. To determine the stability, we calculate the derivative of f(x) at the equilibrium points. If the derivative is less than 1 in absolute value, the equilibrium is stable. If it is greater than 1 in absolute value, the equilibrium is unstable.

b. For a > 3, the solution to Eq. (7.7) includes a 2-cycle. This means that it obeys the equation Xn+2 = Xn in addition to the equilibria found in part (a).

Answer 2

The equation Xn+1 = f(X.n) is used to study periodic solutions and their stability. To find the equilibria of the equation, we solve X = f(X). The stability of the equilibria can be determined using the derivative of f(X). For a > 3, the equation also includes a 2-cycle.

Step-by-step explanation:

The given equation, Xn+1 = f(X.n), where f(x) = ax(1 - x), a = 1 + r Δt and a > 1, is used to study periodic solutions and their stability.

1. To find the equilibria of the equation, we set Xn+1 = Xn = X, resulting in the equation X = f(X) which can be rearranged to get a quadratic equation: X^2 - (a+1)X + a = 0. The solutions to this equation are the equilibria.
2. The stability of the equilibria can be determined by analyzing the behavior of f'(X), the derivative of f(X). If |f'(X)| > 1, the equilibrium is unstable. If |f'(X)| < 1, the equilibrium is stable.

For a > 3, the equation Xn+2 = Xn shows the presence of a 2-cycle in addition to the equilibria found in (a).