Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 4 / 14E (Dynamical Systems)

The fixed-point equation is $x[a{x}^2+bx+\mu -1]=0$. Thuswhen the discriminant is nonnegative. At zero the multiplier is $F^{\prime }(0)=\mu$, so it is linearly stable for $-1<\mu <1$ and unstable for $|\mu |>1$. At a nonzero fixed point, eliminate $\mu$ to obtain the branch and its multiplier:The branch is stable precisely when $-2<x(b+2ax)<0$. These formulas specify stability throughout the diagram, including any extra flip points outside the region requested.

If $b=0$ and $a>0$, the two nonzero branches $x=\pm \sqrt{(1-\mu )/a}$ exist for $\mu <1$ and have multiplier $3-2\mu >1$. Thus $\mu =1$ is a subcritical pitchfork bifurcation. If $b\ne 0$, the branch through zero has $x\sim (1-\mu )/b$ and $M\sim 2-\mu$; its stability exchanges with zero at $\mu =1$, a transcritical bifurcation. The two nonzero roots meet atwhere $M=1$. The graph of $\mu (x)$ has nonzero second derivative $-2a$, proving a saddle-node bifurcation. At zero and $\mu =-1$, the multiplier is $-1$, indicating a period-doubling bifurcation; generically its nondegeneracy coefficient is $a+b^2$. If $a+b^2=0$, necessarily $b\ne 0$ here, composition at $\mu =-1$ instead gives $F^2(x)-x=4b^4{x}^5+O(x^6)$. For $\delta =\mu +1>0$ small, the leading return equation is $-2\delta x+4b^4{x}^5=0$, producing a two-cycle with leading amplitudes $x\sim \pm [\delta /(2b^4){]}^{1/4}$. Its return multiplier is $1+8\delta +o(\delta )>1$. Thus this special case still has a flip bifurcation, but it is degenerate and subcritical rather than the generic cubic normal form.

For $a,b>0$, the fold lies above $\mu =1$. Between it and the transcritical point, the branch $-b/(2a)<x<0$ is stable provided its multiplier remains above $-1$, while the other branch is unstable. For $a,b<0$, the fold lies below $\mu =1$: immediately above it the branch $x<-b/(2a)$ is stable and the branch between $-b/(2a)$ and zero is unstable; the branch crossing zero becomes stable for $\mu >1$. All these statements are subject to the displayed exact multiplier criterion if the diagram is extended. The sketch uses representative parameter values and stops before additional flips, as stipulated.

Separate fixed points from roots of $F^2(x)=x$. Fixed points are zero, and $\pm \sqrt{\mu -1}$ for $\mu \ge 1$. The remaining factors in the given factorization yield first $x^2=\mu +1$. For $\mu >-1$ these form the symmetric two-cycle $\{\sqrt{\mu +1},-\sqrt{\mu +1}\}$ because $F(x)=-x$. Its multiplier is $(-2\mu -3{)}^2>1$, so this cycle is always unstable.

The quartic factor gives $s^2-\mu s+1=0$ for $s=x^2$, hence $s_{\pm }=(\mu \pm \sqrt{{\mu }^2-4})/2$. Positive roots require $\mu \ge 2$. At $\mu =2$ they give the fixed points $\pm 1$, not genuine cycles. For $\mu >2$, $s_{+}{s}_{-}=1$ and $F(\sqrt{s_{+}})=\sqrt{s_{-}}$, giving exactly two additional cycles:There are therefore at most three two-cycles. For either of the latter two, its multiplier isThus they are stable for $2<\mu <\sqrt{5}$ and unstable for $\mu >\sqrt{5}$, with a flip at $\sqrt{5}$. The nonzero fixed points have multiplier $3-2\mu$, hence are stable for $1<\mu <2$. Zero is stable for $-1<\mu <1$. At $\mu =1$ there is a supercritical pitchfork; at $\mu =2$ both nonzero fixed points lose stability to these two-cycles. The symmetric unstable cycle terminates at zero at the subcritical flip $\mu =-1$. These are the cubic map period-two branches.

Beyond $\sqrt{5}$ one expects period-four branches and further period doublings, potentially a cascade toward chaotic behaviour. This local argument does not claim that every larger parameter is chaotic; periodic windows and escape may also occur.