Past exam of the topics in applied mathematics course of the University of Oxford / 2009 / A / 1 / c

For $u>0$, the second-iterate equation $H^2(u)=u$ is equivalent to $u+H(u)=2\log r$. Write $L=\log r$ and let paired points be $L-d,L+d$. The condition $H(L-d)=L+d$ becomes $\tanh (d/2)=d/L$. At $r=9$, $L=2\log 3>2$, so strict concavity of the positive hyperbolic tangent gives one nonzero positive $d$, as well as $d=0$. Here it is exactly $d=\log 3$.

Thus the four nonnegative fixed points of $H^2$ areZero has periodic-orbit multiplier $81$ and the central point has periodic-orbit multiplier $(1-2\log 3{)}^2>1$, so both are unstable. The outer two points are exchanged by $H$, since $H(\log 3)=3\log 3$ and $H(3\log 3)=\log 3$. Their common second-iterate periodic-orbit multiplier isIts magnitude is below one, so both outer fixed points of $H^2$ are stable. This verifies the crossings and stability in the supplied graph rather than reading stability from crossing direction alone.

The Ricker population map two-cycle consists of the two stable points of the second iterate, exchanged by the original map. Thus generic positive orbits approach alternating valuesThere are exceptions: zero stays extinct, while an orbit starting at $\log 9$ or at one of its finite preimages eventually stays at that unstable fixed point.

To justify the large-time conclusion, put $b=9/e$ and $a=H(b)\simeq 1.087>1$. Every first image is at most $b$. A positive orbit below $a$ grows until it enters $[a,b]$; if it never entered, its monotone limit below $a<\log 9$ would contradict the fixed-point equation. The interval $[a,b]$ is invariant. On it $H$ is decreasing, so $H^2$ is increasing. Its only fixed points there are the three positive points already found; its graph signs make iterations below the unstable middle point converge to the lower stable point, and those above converge to the upper stable point. Interchanging even and odd subsequences under $H$ then gives the stated two-cycle. This is regular period-two behaviour, not chaos at this parameter.

Let $L=\log r$. The given expansion implies $u_{\ast }=2L+O(L/r)$ and thereforeUsing $H^{\prime }(u)=r{e}^{-u}(1-u)$ gives preciselyThe paired low value is $v_{\ast }=H(u_{\ast })=(2L/r)[1+O(L/r)]$. It is small, so $e^{-v_{\ast }}(1-v_{\ast })=1+O(L/r)$ andBy the chain rule the second-iterate periodic-orbit multiplier is their product:Since $L^2/r\to 0$ while $L\to \infty$, this tends to $-\infty$ and eventually has magnitude above one. Thus the large second-iterate fixed point, and hence its associated two-cycle, is unstable for sufficiently large $r$. The high state's small derivative alone does not imply stability: the low state's derivative supplies the compensating factor of order $r$.