Q5. Consider a discrete model given by N t+1 = rNt / (1+bN² t-1) = f(N₁), r>1. Investigate the linear stability about the positive steady state N* by setting N₁ = N* + n₁. Show that n satisfies the equation nt+1 - n + 2(r-1)r⁻¹nt-1 = 0. Hence show that r = 2 is a bifurcation value and that as r → 2 the steady state bifurcates to a periodic solution of period 6.
Key Points:
- Steady state N* for N t+1 = rN_t / (1+bN² t-1) is √((r-1)/b).
- Linearization uses Taylor expansion: n t+1 ≈ J₁n_t + J₂n t-1.
- Partial derivatives J₁ = 1 and J₂ = -2(r-1)r⁻¹ at N*.
- Linearized equation is n t+1 - n_t + 2(r-1)r⁻¹n t-1 = 0.
Answer: The discrete model under consideration is given by N t+1 = rN_t / (1+bN² t-1), where r > 1. To investigate the linear stability about the positive steady state N*, we first need to find N* and then linearize the system around it. **1. Finding the Positive Steady State N*** At a steady state, N t+1 = N_t = N t-1 = N*. Substituting this into the given equation: N* = rN* / (1+bN*²) Since N* must be a positive population, we can divide by N*: 1 = r / (1+bN*²) 1+bN*² = r bN*² = r-1 N*² = (r-1)/b Th...