Question

Consider a discrete model given by

N_{t+1}=\frac{rN_{t}}{1+bN^{2}_{t-1}}=f(N_{t})r> 1.

Investigate the linear stability about the positive steady state N^{*} by setting

N_{t=}N^{*}+n_{t.} Show that n_{t}  satisfies the equation

n_{t+1-n_{t+}}2(r-1)r^{-}n_{t-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.

08 Mar 2024
Answer :
Word Count : 370
We will numerically analyze the given discrete model: \[ N_{t+1} = \frac{rN_t}{1 + bN_{t-1}^2} \] and investigate the linear stability around the positive steady state \(N^*\). Then, we will analyze how the bifurcation occurs as \( r \to 2 \), leading to a period-6 solution. --- ### Step 1: Find the Positive Steady State \(N^*\) The steady state satisfies: \[ N^* = \frac{rN^*}{1 + b(N^*)^2} \] Rearrange: \[ 1 + b(N^*)^2 = r \] \[ b(N^*)^2 = r - 1 \] \[ N^* = \sqrt{\frac{r-1}{b}} \] This is the positive steady state. ____ _______ ________ ___ ____.
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