 Solution

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 $r>4$.