Problem Statement
An investigation explores the discrete logistic map $x_{n+1} = r x_n(1 - x_n)$, where $r > 0$ and $x_0 \in (0, 1)$.
(a) Find the fixed points $x^*$ satisfying $f(x^*) = x^*$. [3]
(b) A fixed point is stable if $|f'(x^*)| < 1$. Determine the range of $r$ for which the non-zero fixed point is stable. [5]
(c) For $r = 3.2$, determine the period-2 cycle by solving $f(f(x)) = x$. [7]
Verified Solution & Marking Scheme
(a) Find fixed points
$r x^*(1 - x^*) = x^* \implies x^* [r(1 - x^*) - 1] = 0$
Case 1: $x^*_1 = 0$
Case 2: $r(1 - x^*) = 1 \implies 1 - x^* = \frac{1}{r} \implies x^*_2 = 1 - \frac{1}{r} = \frac{r - 1}{r}$ (for $r > 1$).
(b) Stability of the non-zero fixed point
$f(x) = rx - rx^2 \implies f'(x) = r - 2rx$
Evaluate at $x^* = \frac{r - 1}{r}$:
$f'(x^*) = r - 2r\left(\frac{r - 1}{r}\right) = r - 2(r - 1) = r - 2r + 2 = 2 - r$
Stability condition: $|f'(x^*)| < 1$:
$|2 - r| < 1 \implies -1 < 2 - r < 1$
Subtract 2: $-3 < -r < -1 \implies 1 < r < 3$
Thus, the non-zero population fixed point is asymptotically stable for $1 < r < 3$.
(c) Period-2 cycle for r = 3.2
For $r = 3.2 > 3$, the system bifurcates into a 2-cycle $(p, q)$ where $f(p) = q$ and $f(q) = p$.
The 2-cycle satisfies $f(f(x)) - x = 0$. Since fixed points $x = 0$ and $x = \frac{r-1}{r} = \frac{2.2}{3.2} = \frac{11}{16} = 0.6875$ are already roots, dividing out $x(x - x^*)$ gives the quadratic for the 2-cycle:
$r^2 x^2 - r(r + 1)x + (r + 1) = 0$
For $r = 3.2$:
$(3.2)^2 x^2 - 3.2(4.2)x + 4.2 = 0 \implies 10.24 x^2 - 13.44 x + 4.2 = 0$
Solving via quadratic formula:
$x = \frac{13.44 \pm \sqrt{(13.44)^2 - 4(10.24)(4.2)}}{2(10.24)} = \frac{13.44 \pm \sqrt{180.6336 - 172.032}}{20.48} = \frac{13.44 \pm \sqrt{8.6016}}{20.48}$
$\sqrt{8.6016} \approx 2.93285$
$p = \frac{13.44 - 2.93285}{20.48} \approx 0.513, \quad q = \frac{13.44 + 2.93285}{20.48} \approx 0.799$
Notice $f(0.513) = 3.2(0.513)(1 - 0.513) \approx 0.799$ and $f(0.799) \approx 0.513$.