= Solution
The preceding summer supplies $\gamma M_n$ new seeds. After one winter, $\alpha\gamma M_n$ survive and the fraction $\mu$ germinates. Seeds from summer $n-1$ must survive two winters and fail to germinate after the first, so their germinating contribution is $\nu\alpha^2(1-\mu)\gamma M_{n-1}$. Older seeds cannot contribute. Applying the density-dependent maturation function gives
$$
\boxed{M_{n+1}=N(aM_n+bM_{n-1}),\qquad a=\alpha\mu\gamma,\quad b=\nu\gamma\alpha^2(1-\mu).}
$$
This is a <seed-bank population recurrence>. Put $A=a+b$. Its <fixed points> satisfy $M_*=N(AM_*)$. The zero solution always exists. If $A>1$, the graph of $N(AM)$ leaves the origin with slope $A>1$, is concave because $N'$ decreases, and is bounded above by $N_{\max}$; it therefore crosses the straight line $M$ again at a positive value. <Concavity> makes this positive crossing unique: the ratio $N(S)/S$ is nonincreasing, starts at one, tends to zero, and is strictly decreasing at any level below one that it attains after the initial slope has dropped. Thus \b[there are exactly two nonnegative steady states when $A>1$].
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ii/paper-4-seed-bank.png]
{title=Concave seed-bank recruitment crossing the identity at extinction and a positive population}
Linearization at extinction uses $N'(0)=1$:
$$
\delta M_{n+1}=a\delta M_n+b\delta M_{n-1},\qquad
r^2-ar-b=0.
$$
The positive root is $r_+=(a+\sqrt{a^2+4b})/2$. Since the polynomial at $r=1$ is $1-A<0$, and tends to positive infinity, $r_+>1$. Its associated two-component perturbation can be chosen nonnegative, so the growing mode is physically admissible. \b[Extinction is unstable whenever $\boxed{a+b>1}$].
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