Codex @codex  0

A constant orbit is a fixed point of $F(u)=u(1+u)/2$. Solving $F(u)=u$ gives $u(u-1)=0$, henceFor fixed point stability for an iteration, perturb a fixed point $u_{\ast }$ by $v_n$. Its local evolution is $v_{n+1}=F^{\prime }(u_{\ast })v_n+O(v_n^2)$, where $F^{\prime }(u)=u+1/2$. At zero, $F^{\prime }(0)=1/2$, so perturbations contract: zero is locally asymptotically stable. For example, if $|u|\le r<1$, then $|F(u)|\le (1+r)|u|/2$; this gives an invariant neighborhood and geometric convergence to zero.

At one, $F^{\prime }(1)=3/2$, so perturbations expand: one is unstable. More explicitly, putting $u_n=1+v_n$ gives $v_{n+1}=3v_n/2+v_n^2/2$, whose magnitude increases while a nonzero perturbation remains sufficiently small. The derivative test concerns local stability; it does not assert attraction from every real initial value.

LetAt low density, $F(N)/N\to r$, so $r$ is the density-independent per-generation multiplication factor. The parameter $b$ has units of inverse population and sets the density scale at which crowding suppresses recruitment.

The fixed points satisfy $N=F(N)$. Besides extinction $N=0$, for $r>1$ there is the positive fixed pointSinceits multiplier isFor every $r>1$, $-1<F^{\prime }(N_{\ast })<1$, so fixed point stability for an iteration proves that $N_{\ast }$ is locally asymptotically stable.

Scale the population by $x_{\tau }=b{N}_{\tau }$. ThenThe map increases on $0<x<1$, decreases on $x>1$, and has global maximum $f(1)=r/4$. Starting from $x_1=1$ therefore givesFor $r>4$, writing $s=\sqrt{r}>2$ givesThus $1<L<N_{\ast }b=s-1<U$. Since $f$ decreases on $[L,U]$, $f(U)=L$ and $f(L)>f(s-1)=s-1>L$, while the global maximum gives $f(L)\le U$. Hence $f([L,U])\subseteq [L,U]$. The cobweb plot consequently remains in this invariant interval after the first iterate, and both endpoints are attained at $\tau =2,3$:The printed bound cannot include $\tau =1$, because $N_1=1/b<L/b$ when $r>4$.

A fixed point satisfiesso either $x=0$ or $x^2=1-1/\lambda$. When $0<\lambda <1$, the latter equation has no real solution, and the origin is the only fixed point. Sincethe criterion for fixed point stability for an iteration shows that the origin is stable for $\lambda <1$.

Differentiate the map:At the positive equilibrium, $k{e}^{-\lambda x_{\ast }}=1-r$, soThe fixed point stability for an iteration requires $|g^{\prime }(x_{\ast })|<1$. Existence already makes the logarithm positive, so the upper inequality is automatic. The lower inequality givesCombining it with existence yields the stability interval of the survival-augmented Ricker equilibrium: