Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 85 / 2

The associated even map is simply $g(x)=\mu -x^2$. For nonzero $x$, the odd quadratic Lorenz map has the two branches $f_L(x)=x^2-\mu$ on the positive side and $f_L(x)=\mu -x^2$ on the negative side. Their limiting values at zero are $-\mu$ and $+\mu$, while the prescribed point value is $f_L(0)=-\mu$. Thus the branches join continuously only at $\mu =0$, where the origin is a fixed point and $f_L(x)=x|x|$.

The small fixed point of $g$ isFor small negative $\mu$, $r<0$ and the odd quadratic Lorenz map has two stable fixed points $r,-r$. Their periodic-orbit multiplier is $2|r|<1$. As $\mu \uparrow 0$ these fixed points collide at zero. For small positive $\mu$, $r>0$ andso a stable symmetric two-cycle is present, with periodic-orbit multiplier $4r^2<1$. Hence the global transition isThe large fixed points $\pm (1+r)$ remain unstable near this transition and are not created or destroyed there. The parabola $g$ itself is smooth as $\mu$ passes zero; its fixed point crosses the critical point and changes the sign of its periodic-orbit multiplier. The change of orbit type occurs in the sign-lifted discrete dynamical system, whose discontinuity switches orientation.

Write $g(x)=\mu -x^2$ and $f_L(x)=-\mathrm{sgn}(x)g(x)$. Suppose first that $g^k(x)\ne 0$ for $0\le k<n$, so no critical hit occurs before the final iterate. Set $A_n=(-1{)}^n{\prod }_{k=0}^{n-1}\mathrm{sgn}(g^k(x))\in \{1,-1\}$. The case $n=1$ is the defining identity. If $f_L^n(x)=A_n{g}^n(x)$ and $g^n(x)\ne 0$, evenness of $g$ and oddness of $f_L$ away from zero giveInduction proves the sign lift of an even map formulaon noncritical orbit segments. A final zero value is harmless; using oddness at zero in a subsequent step is not.

The PDF's assertion that this holds for every $x$ is false with its stated $\mathrm{sgn}(0)=1$ convention. For $\mu >0$, choose $x=\sqrt{\mu }$. Then $g(x)=f_L(x)=0$, sobut the proposed right side for $n=2$ is $g^2(x)\mathrm{sgn}(x)\mathrm{sgn}(g(x))=+\mu$. This explicit counterexample identifies the missing noncritical qualification. An exact recurrence valid even at critical hits can instead be written with $y_k=g^k(x)$ and $f_L^k(x)=A_k{y}_k$: start with $A_0=1$, and updateThe second rule resets the sign because $f_L(0)=-g(0)$ independently of the previous sign. This proves the intended identity wherever it is valid and accounts for the exceptional critical itineraries without changing the map's definition.

For a period-$n$ orbit of $g(x)=\mu -x^2$ that avoids zero, the chain rule gives its multiplier of a periodic orbit of an iteration asThe sign of $g^{\prime }$ is positive on the negative half-axis and negative on the positive half-axis, exactly as the parabola sketch shows. ConsequentlyThis product is independent of the chosen starting point because changing $\hat{x}$ only cyclically reorders the factors. If the orbit contains zero, the derivative product is zero and its orientation is undefined. One must then use the actual zero periodic-orbit multiplier, rather than infer a positive or negative orientation from the special assignment $\mathrm{sgn}(0)=1$. The nonzero-periodic-orbit multiplier cases needed in part (d) avoid this exception automatically.

Assume the $g$ cycle has least period $n$ and nonzero periodic-orbit multiplier $P=(g^n{)}^{\prime }(\hat{x})$. It then avoids zero. Combining the iterate identity with the periodic-orbit multiplier-sign formula givesThus $P>0$ makes $\hat{x}$ return after $n$ iterates, whereas $P<0$ gives $f_L^n(\hat{x})=-\hat{x}$. In the latter case oddness on the noncritical orbit impliesTo prove the least periods, note that a $g$ cycle cannot contain two distinct points $x$ and $-x$: evenness would give them the same image, but $g$ permutes its distinct cycle points bijectively. Therefore the $n$ absolute values in the cycle are distinct. Since $|f_L^k(\hat{x})|=|g^k(\hat{x})|$, any return under $f_L$ must occur at a multiple of $n$. For $P<0$ the first such return fails because $\hat{x}\ne 0$, so $2n$ is minimal. Hence the period transfer from a quadratic map to its Lorenz sign lift isThere are two symmetry-related $f_L$ cycles in the positive case and one cycle containing both signs in the negative case. Away from zero $f_L^{\prime }(x)=2|x|$, so their periodic-orbit multipliers are $|P|$ and $|P{|}^2$, respectively. A stable $g$ cycle therefore gives stable lifted cycles in both cases.

The positive fixed point $r=(-1+\sqrt{1+4\mu })/2$ of $g$ has periodic-orbit multiplier $-2r$. At $\mu =3/4$, $r=1/2$ and the periodic-orbit multiplier crosses $-1$. Factoring the second-iterate fixed-point equation givesThe second factor yields a two-cycleIt exists for $\mu >3/4$ and has periodic-orbit multiplier $4pq=4(1-\mu )$. Its stability range is $3/4<\mu <5/4$, so $g$ undergoes a supercritical period-doubling bifurcation at $\mu =3/4$. At $\mu =1$ the two-cycle is $\{0,1\}$ and its periodic-orbit multiplier is zero, making it superstable.

For positive $\mu$ and $0<|x|<\sqrt{\mu }$, direct composition gives the local second iterateThe one-sided limits at zero are opposite, and agree with zero exactly when $\mu =1$ in this neighborhood. There $F_1(x)=\mathrm{sgn}(x)(2x^2-x^4)$ is continuous at zero and fixes it. Thus the local return branch has the same discontinuity-closing transition as the map in part (a), with effective parameter ${\mu }^2-\mu$ crossing zero.

For $3/4<\mu <1$, both $p,q$ are positive and the period transfer from a quadratic map to its Lorenz sign lift gives the two stable cyclesFor $1<\mu <5/4$, $p>0$ and $q<0$, so the periodic-orbit multiplier of the $g$ cycle is negative and the lifted orbit isIt is one stable four-cycle. Hence the gluing of cycles in an iterated Lorenz map givesEquivalently, $F_{\mu }$ changes from two small stable fixed points $\pm q$ to a stable small two-cycle $\pm (-q)$. At the critical value itself, the actual convention gives $f_L(0)=-1$, $f_L(-1)=0$, $f_L(1)=0$: the surviving critical two-cycle is $\{0,-1\}$, and $+1$ is preperiodic. This critical itinerary must be evaluated directly, not with part (b)'s noncritical sign product.

There is also a precise qualification to the printed use of “continuous”. The whole second iterate is not continuous everywhere at $\mu =1$: $F_1(1)=-1$, while ${\lim }_{x\to 1^{-}}{F}_1(x)=1$ and ${\lim }_{x\to 1^{+}}{F}_1(x)=-1$. Jumps persist at the nonzero preimages of the original discontinuity. Therefore the stated global-bifurcation criterion holds for the return map restricted to a neighborhood of zero, as in the sketches; if continuity on all of $\mathbb{R}$ is demanded literally, the claim about $f_L^2$ needs that restriction. The orbit creation and destruction above is unaffected.