Critical thresholds for eventual extinction in randomly disturbed population growth models

Abstract

This paper considers several single species growth models featuring a carrying capacity, which are subject to random disturbances that lead to instantaneous population reduction at the disturbance times. This is motivated in part by growing concerns about the impacts of climate change. Our main goal is to understand whether or not the species can persist in the long run. We consider the discrete-time stochastic process obtained by sampling the system immediately after the disturbances, and find various thresholds for several modes of convergence of this discrete process, including thresholds for the absence or existence of a positively supported invariant distribution. These thresholds are given explicitly in terms of the intensity and frequency of the disturbances on the one hand, and the population’s growth characteristics on the other. We also perform a similar threshold analysis for the original continuous-time stochastic process, and obtain a formula that allows us to express the invariant distribution for this continuous-time process in terms of the invariant distribution of the discrete-time process, and vice versa. Examples illustrate that these distributions can differ, and this sends a cautionary message to practitioners who wish to parameterize these and related models using field data. Our analysis relies heavily on a particular feature shared by all the deterministic growth models considered here, namely that their solutions exhibit an exponentially weighted averaging property between a function of the initial condition, and the same function applied to the carrying capacity. This property is due to the fact that these systems can be transformed into affine systems.

Appendix A: Proof of Theorem 3.1

The continuous time evolution can be expressed in terms of the semigroup of linear contraction operators defined by

$$\begin{aligned} T(t)f(x) = E_xf(N(t)), \quad t\ge 0, x > 0, \end{aligned}$$

via its infinitesimal generator given by

$$\begin{aligned} Lf(x) = T^\prime (0)f(x) = {d\over dt}f(g(t,x))|_{t=0} + \lambda \{Ef(\mathcal{D}_1 x)-f(x)\}. \end{aligned}$$

To derive this simply observe that up to o(t) error as $t\downarrow 0$, either one or no disturbance will occur in the time interval [0, t). Thus

$$\begin{aligned} {T(t)f(x) - f(x)\over t}= & {} {f(g(t,x))e^{-\lambda t} - f(x)\over t} \\&+\,{1\over t}\int _0^tE\left( f(\mathcal{D}_1 g(s,x))\right) \lambda e^{-\lambda s}ds + o(t). \end{aligned}$$

The first term is, by the product differentiation rule,

$$\begin{aligned}&{f(g(t,x))e^{-\lambda t} - f(g(0,x))e^{-\lambda 0}\over t} \rightarrow {d\over dt}f(g(t,x))e^{-\lambda t}|_{t=0}\\&\quad = {d\over dt}f(g(t,x))|_{t=0} - \lambda f(x). \end{aligned}$$

The second term is $\lambda Ef(\mathcal{D}_1 x)$ in the limit as $t\downarrow 0$.

If $\mu$ is an invariant probability distribution for this continuous time evolution then one has essentially from the Fokker–Planck equation $L^*\mu = {d\over dt}\mu = 0$ for the adjoint operator, e.g., see Bhattacharya and Waymire (1990). In particular, for f belonging to the domain of L as an (unbounded) operator on $L^2(\mu )$,

$$\begin{aligned} 0 =<f,L^*\mu> = <Lf, \mu > = \int _0^\infty Lf(x)\mu (dx), f\in L^2(\mu ). \end{aligned}$$

In the case of the discrete time evolution, the one-step transition operator is defined by

$$\begin{aligned} Mf(x) = Ef(\mathcal{D}_1 g(T_1,x)), \quad x > 0. \end{aligned}$$

The condition for $\pi$ to be an invariant probability distribution for the discrete time evolution is that for integrable functions f,

$$\begin{aligned} \int _0^\infty Mf(x)\pi (dx) = \int _0^\infty f(x)\pi (dx). \end{aligned}$$

In particular, it suffices to consider indicator functions $f = 1_C, C\subset (0,\infty ),$ in which case one has

$$\begin{aligned} \int _0^\infty P(\mathcal{D}_1 g(T,x)\in C) \, \pi (dx) = \pi (C). \end{aligned}$$

These are the essential calculations required for the proof.

Let’s begin with part (i). First note from the definition of $\mu$ that

$$\begin{aligned} \int _0^\infty Lf(x)\mu (dx) = \int _0^\infty \int _0^\infty Lf(g(t,y))\lambda e^{-\lambda t} \, dt \, \pi (dy). \end{aligned}$$

Now, in view of the above calculation of L, one has

$$\begin{aligned}&\int _0^\infty Lf(g(t,y))\lambda e^{-\lambda t} \, dt \\&\quad = \int _0^\infty \left( {\partial f(g(t,x))\over \partial t} + \lambda \left[ Ef(\mathcal{D}_1 g(t,x))-f(g(t,x)) \right] \right) \lambda e^{-\lambda t} dt. \end{aligned}$$

After an integration by parts this yields

$$\begin{aligned} \int _0^\infty Lf(g(t,y))\lambda e^{-\lambda t}dt = \lambda \{Ef(\mathcal{D}_1 g(T,x))-f(x)\} \end{aligned}$$

Thus, using this and the invariance of $\pi$ for the discrete process, one has

$$\begin{aligned} \int _0^\infty Lf(x)\mu (dx) = \lambda \int _0^\infty \{Ef(\mathcal{D}_1 g(T,x))-f(x)\}\pi (dx) = 0. \end{aligned}$$

This proves part (i).

To prove part (ii), first apply L to the function $x\rightarrow P(\mathcal{D}_1 g(T,x)\in C)$. First note from the composition property and an indicated change of variable,

$$\begin{aligned} P(\mathcal{D}_1 g(T,x)\in C) = P(\mathcal{D}_1 g(T+t,x)\in C) = e^{\lambda t}\int _t^\infty P(\mathcal{D}_1 g(s,x)\in C)\lambda e^{-\lambda s}ds. \end{aligned}$$

In particular the first term of $LP(\mathcal{D}_1 g(T,x)\in C)$ is

$$\begin{aligned} {d\over dt}P(\mathcal{D}_1 g(T,x)\in C)|_{t=0} = \lambda \{P(\mathcal{D}_1 g(T+t,x)\in C) -P(\mathcal{D}_1 x\in C)\}. \end{aligned}$$

Adding this to the second term yields,

$$\begin{aligned} LP(\mathcal{D}_1 g(T,x)\in C) = \lambda \left\{ \int _0^\infty P(\mathcal{D}_1 g(T,y)\in C)P(\mathcal{D}_1 x\in dy) -P(\mathcal{D}_1 x\in C)\right\} . \end{aligned}$$

Integrating with respect to the continuous time invariant distribution $\mu$ yields

$$\begin{aligned} 0 = \lambda \int _0^\infty \left\{ \int _0^\infty P(\mathcal{D}_1 g(T_1,y)\in C)P(\mathcal{D}_1 x\in dy) -P(\mathcal{D}_1 x\in C)\right\} \mu (dx), \end{aligned}$$

or equivalently,

$$\begin{aligned} \int _0^\infty \int _0^\infty P(\mathcal{D}_1 g(T,y)\in C)P(\mathcal{D}_1 x\in dy)\mu (dx) = \int _0^\infty P(\mathcal{D}_1 x\in C)\mu (dx). \end{aligned}$$

But since by definition $\pi (dy) = \int _0^\infty P(\mathcal{D}_1 x\in dy)\mu (dx)$, this is precisely the condition

$$\begin{aligned} \int _0^\infty P(\mathcal{D}_1 g(T,y)\in C)\pi (dy) = \pi (C), \end{aligned}$$

i.e., that $\pi$ is an invariant probability for the discrete time distribution. $\square$