 Solution

There are $k{x}_n$ seeds and their germinating proportion is $e^{-\gamma x_n}$, giving the survival-augmented Ricker map with zero survival. A positive equilibrium point satisfies $e^{-\gamma x_{\ast }}=1/k$, soIt exists when $k>1$ and is strictly linearly stable when $|1-\log k|<1$. Thus the usual strict linear-stability interval is $\boxed{1<k<e^2}$.

With winter survival, surviving adults and new plants add:Writing $L=\log [k/(1-s)]$, the linearization of a dynamical system has multiplier $1-(1-s)L$. Hence the strict linear-stability interval becomesIts lower endpoint is less than $1$. Its upper endpoint exceeds $e^2$: for $t=1-s\in (0,1)$, $\log t+2/t$ is decreasing in $t$ and equals $2$ at $t=1$.

The nonlinear stability at the Ricker flip threshold needs a separate check. The upper endpoints have multiplier $-1$, so the strict linear test alone does not decide stability there. They are in fact locally asymptotically stable, with algebraic rather than geometric convergence. To see this, put $u=\gamma (x-x_{\ast })$ at $L=2/(1-s)$. The local map isand its second iterate is $u-2(s^2-s/2+1/6)u^3+O(u^4)$. The cubic coefficient in parentheses is positive for every $s\in [0,1)$. Thus, if “stable” includes nonlinear endpoint stability, the upper inequalities can be replaced by $\le$; the lower endpoints have no positive equilibrium.