[problem: Schwarzian Derivative Of Power Maps | 1] Let $n \ge 2$ be an integer and define the map \begin{align*} f: \mathbb{R} \setminus \{0\} &\to \mathbb{R} \\ x &\mapsto x^n. \end{align*} Compute the Schwarzian derivative $Sf(x)$ and verify that $Sf(x) < 0$ for all $x \neq 0$. [/problem]

[solution] **Step 1: Compute the first three derivatives.** The derivatives of $f(x) = x^n$ are \begin{align*} f'(x) &= n x^{n-1}, \\ f''(x) &= n(n-1) x^{n-2}, \\ f'''(x) &= n(n-1)(n-2) x^{n-3}. \end{align*} **Step 2: Compute the Schwarzian derivative.** By definition, $Sf(x) = \dfrac{f'''(x)}{f'(x)} - \dfrac{3}{2}\left(\dfrac{f''(x)}{f'(x)}\right)^2$. Substituting: \begin{align*} \frac{f'''(x)}{f'(x)} &= \frac{n(n-1)(n-2) x^{n-3}}{n x^{n-1}} = \frac{(n-1)(n-2)}{x^2}, \\[6pt] \frac{f''(x)}{f'(x)} &= \frac{n(n-1) x^{n-2}}{n x^{n-1}} = \frac{n-1}{x}. \end{align*} Therefore \begin{align*} Sf(x) &= \frac{(n-1)(n-2)}{x^2} - \frac{3}{2} \cdot \frac{(n-1)^2}{x^2} \\ &= \frac{n-1}{x^2}\left((n-2) - \frac{3(n-1)}{2}\right) \\ &= \frac{n-1}{x^2} \cdot \frac{2(n-2) - 3(n-1)}{2} \\ &= -\frac{(n-1)(n+1)}{2x^2} \\ &= -\frac{n^2 - 1}{2x^2}. \end{align*} **Step 3: Verify negativity.** For $n \ge 2$, the numerator $n^2 - 1 \ge 3 > 0$, and $x^2 > 0$ for $x \neq 0$. Therefore $Sf(x) < 0$ for all $x \in \mathbb{R} \setminus \{0\}$, confirming that every power map with $n \ge 2$ has strictly negative Schwarzian derivative away from the origin. [/solution]

---

[problem: Bifurcation Analysis Of A Cubic Map | 3] Consider the one-dimensional map defined on $[0, a]$ (with $a > 0$): \begin{align*} f: [0, a] &\to \mathbb{R} \\ x &\mapsto (a - x)^3. \end{align*} (i) Find the fixed points, compute the multiplier, and identify all bifurcation values. (ii) Find the period-2 orbit condition and determine the range of parameter values for which the period-2 orbit is stable. [/problem]

[solution] **Step 1: Fixed points and multiplier.** Setting $f(x) = x$ gives $x = (a-x)^3$. The substitution $u = a - x$ (so $x = a - u$) transforms this to \begin{align*} a - u = u^3 \implies a = u^3 + u. \end{align*} Since $\phi(u) = u^3 + u$ is strictly increasing ($\phi'(u) = 3u^2 + 1 > 0$), there is a unique $u^*$ — and hence a unique fixed point $x^* = a - u^*$ — for each $a > 0$. The multiplier is \begin{align*} \lambda = f'(x^*) = -3(a - x^*)^2 = -3(u^*)^2. \end{align*} Since $\lambda \le 0$ for all $u^* \ge 0$, the multiplier never reaches $+1$ (no saddle-node bifurcation in this range). **Step 2: Period-doubling bifurcation.** The fixed point is stable when $|\lambda| < 1$, i.e., \begin{align*} 3(u^*)^2 < 1 \implies u^* < \frac{1}{\sqrt{3}}. \end{align*} At $u^* = 1/\sqrt{3}$, the multiplier is $\lambda = -1$, and the corresponding bifurcation parameter is \begin{align*} a_c = \left(\frac{1}{\sqrt{3}}\right)^3 + \frac{1}{\sqrt{3}} = \frac{1}{3\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{4}{3\sqrt{3}} = \frac{4\sqrt{3}}{9} \approx 0.770. \end{align*} For $a < a_c$ the fixed point is stable ($u^*$ is small, $|\lambda| < 1$). For $a > a_c$ it is unstable ($|\lambda| > 1$). The transition is a **period-doubling bifurcation**. **Step 3: Period-2 orbit condition.** A period-2 orbit $\{x_1, x_2\}$ with $x_1 \neq x_2$ satisfies $f(x_1) = x_2$ and $f(x_2) = x_1$. With $u = a - x_1$ and $v = a - x_2$: \begin{align*} a - u = v^3, \quad a - v = u^3. \end{align*} Subtracting the second from the first: $v - u = v^3 - u^3 = (v - u)(v^2 + uv + u^2)$. Since $u \neq v$, dividing by $v - u$ gives \begin{align*} u^2 + uv + v^2 = 1. \end{align*} This is an ellipse in the $(u, v)$ plane, which admits real solutions for a range of $a$. **Step 4: Multiplier of the period-2 orbit.** Using $f'(x_i) = -3(a - x_i)^2$, the orbit multiplier is \begin{align*} \Lambda = f'(x_1) f'(x_2) = (-3u^2)(-3v^2) = 9u^2 v^2 = 9P^2, \end{align*} where $P := uv$. We determine the range of $P$. From $u^2 + uv + v^2 = 1$: \begin{align*} (u + v)^2 = u^2 + 2uv + v^2 = 1 + P, \end{align*} which requires $P \ge -1$ for real solutions. The constraint $u^2 + v^2 = 1 - P \ge 0$ gives $P \le 1$, and the inequality $u^2 + v^2 \ge 2|uv|$ gives $1 - P \ge 2|P|$. For $P > 0$: $P \le 1/3$, with equality when $u = v = 1/\sqrt{3}$ (the fixed point, not a distinct orbit). So the admissible range for the period-2 orbit is $P \in [-1, 1/3)$. **Step 5: Stability range.** The orbit is stable when $|\Lambda| < 1$: \begin{align*} 9P^2 < 1 \iff |P| < \frac{1}{3}. \end{align*} At the period-doubling bifurcation ($u = v = 1/\sqrt{3}$, $P = 1/3$), the orbit is born with $\Lambda = 9 \cdot (1/3)^2 = 1$. As $a$ increases past $a_c$, the orbit opens up and $P$ decreases from $1/3$. The orbit is stable for $P \in (-1/3, 1/3)$ and becomes unstable when $P$ passes through $-1/3$, at which point $\Lambda = 1$ again. This secondary bifurcation ($\Lambda = 1$ for the period-2 orbit, i.e., a period-doubling of the second iterate $f^2$) initiates a period-4 orbit — the next step in the period-doubling cascade. [/solution]

---

[problem: Saddle-Node And Period-Doubling Bifurcations Of A Quartic Map | 2] Consider the one-dimensional map \begin{align*} f: \mathbb{R} &\to \mathbb{R} \\ x &\mapsto a - x^4, \end{align*} where $a \in \mathbb{R}$ is a parameter. (i) Find all bifurcation values of the fixed points. Classify each as saddle-node or period-doubling. (ii) Compute the Schwarzian derivative at the period-doubling bifurcation point and determine whether the bifurcation is supercritical or subcritical. [/problem]

[solution] **Step 1: Fixed points and multiplier.** The fixed point equation $a - x^4 = x$ gives $a = x^4 + x$. Define $\phi(x) = x^4 + x$. Then $\phi'(x) = 4x^3 + 1$, which vanishes at $x_0 = -4^{-1/3}$. The function $\phi$ has a global minimum at $x_0$: \begin{align*} a_{\min} = \phi(x_0) = 4^{-4/3} - 4^{-1/3} = 4^{-1/3}(4^{-1} - 1) = -\frac{3}{4^{4/3}}. \end{align*} For $a < a_{\min}$: no fixed points. For $a = a_{\min}$: one degenerate fixed point. For $a > a_{\min}$: two fixed points. The multiplier at a fixed point $x^*$ is $\lambda = f'(x^*) = -4(x^*)^3$. **Step 2: Saddle-node bifurcation ($\lambda = +1$).** Setting $-4(x^*)^3 = 1$ gives $x^* = -4^{-1/3} = x_0$. This is exactly the critical point of $\phi$, so the saddle-node occurs at \begin{align*} a_{\text{sn}} = -\frac{3}{4^{4/3}}. \end{align*} At this parameter, $f''(x_0) = -12 x_0^2 \neq 0$ (since $x_0 \neq 0$), confirming the non-degeneracy condition for a **generic saddle-node bifurcation**. Two fixed points are born as $a$ increases through $a_{\text{sn}}$. **Step 3: Period-doubling bifurcation ($\lambda = -1$).** Setting $-4(x^*)^3 = -1$ gives $(x^*)^3 = 1/4$, so $x^* = 4^{-1/3}$. The period-doubling occurs at \begin{align*} a_{\text{pd}} = 4^{-4/3} + 4^{-1/3} = \frac{5}{4^{4/3}}. \end{align*} At this parameter, the stable fixed point (on the branch with $x^* > 0$) loses stability. This is a **period-doubling bifurcation**. **Step 4: Schwarzian derivative and bifurcation type.** The higher derivatives of $f(x) = a - x^4$ are $f''(x) = -12x^2$ and $f'''(x) = -24x$. At $x^* = 4^{-1/3}$ where $f'(x^*) = -1$: \begin{align*} \frac{f'''(x^*)}{f'(x^*)} &= \frac{-24 x^*}{-4(x^*)^3} = \frac{6}{(x^*)^2}, \\[6pt] \frac{f''(x^*)}{f'(x^*)} &= \frac{-12(x^*)^2}{-4(x^*)^3} = \frac{3}{x^*}. \end{align*} Therefore \begin{align*} Sf(x^*) &= \frac{6}{(x^*)^2} - \frac{3}{2} \cdot \frac{9}{(x^*)^2} = \frac{1}{(x^*)^2}\left(6 - \frac{27}{2}\right) = -\frac{15}{2(x^*)^2} < 0. \end{align*} Since $Sf(x^*) < 0$, the period-doubling bifurcation is **supercritical**: for $a$ slightly above $a_{\text{pd}}$, a stable period-2 orbit exists. [/solution]

---

[problem: Bifurcation Curves In A Two Parameter Family | 3] Consider the two-parameter family of maps \begin{align*} f_{a,b}: \mathbb{R} &\to \mathbb{R} \\ x &\mapsto a - bx - x^3. \end{align*} (i) Derive parametric equations for the saddle-node and period-doubling bifurcation curves in the $(a, b)$ parameter plane. Express the saddle-node curve as an implicit equation. (ii) On the period-doubling curve, determine the multiplier of the period-2 orbit at the moment of birth, and explain the result in terms of the second iterate. [/problem]

[solution] **Step 1: Fixed point equation and multiplier.** The fixed point condition $a - bx - x^3 = x$ gives \begin{align*} a = (b+1)x + x^3. \end{align*} The multiplier is $\lambda = f'(x) = -b - 3x^2$. **Step 2: Saddle-node bifurcation curve ($\lambda = +1$).** Setting $\lambda = +1$: $-b - 3x^2 = 1$, so $b = -1 - 3x^2$. Substituting into the fixed point equation: \begin{align*} a = (-1 - 3x^2 + 1)x + x^3 = -3x^3 + x^3 = -2x^3. \end{align*} The parametric equations for the saddle-node curve are $a = -2x^3$, $b = -1 - 3x^2$ with $x \in \mathbb{R}$. **Step 3: Implicit equation for the saddle-node curve.** From $a = -2x^3$: $x = -(a/2)^{1/3}$, so $x^2 = (a/2)^{2/3}$. Substituting into $b = -1 - 3x^2$: \begin{align*} b + 1 = -3(a/2)^{2/3}. \end{align*} Cubing: $(b+1)^3 = -27(a/2)^2 = -27a^2/4$. Rearranging: \begin{align*} \frac{a^2}{4} + \frac{(b+1)^3}{27} = 0. \end{align*} This is a semicubical parabola (cusp) in the $(a, b)$ plane with its cusp at $(a, b) = (0, -1)$. **Step 4: Period-doubling bifurcation curve ($\lambda = -1$).** Setting $\lambda = -1$: $-b - 3x^2 = -1$, so $b = 1 - 3x^2$. Substituting into the fixed point equation: \begin{align*} a = (1 - 3x^2 + 1)x + x^3 = 2x - 2x^3. \end{align*} The parametric equations for the period-doubling curve are \begin{align*} a = 2x - 2x^3, \quad b = 1 - 3x^2, \quad x \in \mathbb{R}. \end{align*} At $x = 0$: $(a, b) = (0, 1)$. As $|x|$ increases, $b$ decreases below $1$ and $|a|$ initially increases before decreasing. **Step 5: Multiplier of the period-2 orbit at birth.** At any point on the period-doubling curve, the fixed point $x^*$ has $f'(x^*) = -1$. A period-2 orbit of $f$ is a fixed point of $g = f \circ f$. At the moment of birth, the period-2 orbit is degenerate: $x_1 = x_2 = x^*$. By the chain rule, its multiplier is \begin{align*} \Lambda = g'(x^*) = f'(f(x^*)) \cdot f'(x^*) = f'(x^*)^2 = (-1)^2 = 1. \end{align*} This holds at **every** point on the period-doubling curve. The multiplier $\Lambda = +1$ for $g = f^2$ means that the birth of the period-2 orbit is a **saddle-node bifurcation of the second iterate**: the two period-2 points appear (or disappear) as a pair of fixed points of $g$ that collide at $x^*$ with multiplier $+1$, exactly as in a tangent bifurcation. This is a universal feature of period-doubling bifurcations — the second iterate always undergoes a saddle-node at the period-doubling parameter. [/solution]