The competitive exclusion principle in stochastic environments

Abstract

In its simplest form, the competitive exclusion principle states that a number of species competing for a smaller number of resources cannot coexist. However, it has been observed empirically that in some settings it is possible to have coexistence. One example is Hutchinson’s ‘paradox of the plankton’. This is an instance where a large number of phytoplankton species coexist while competing for a very limited number of resources. Both experimental and theoretical studies have shown that temporal fluctuations of the environment can facilitate coexistence for competing species. Hutchinson conjectured that one can get coexistence because nonequilibrium conditions would make it possible for different species to be favored by the environment at different times. In this paper we show in various settings how a variable (stochastic) environment enables a set of competing species limited by a smaller number of resources or other density dependent factors to coexist. If the environmental fluctuations are modeled by white noise, and the per-capita growth rates of the competitors depend linearly on the resources, we prove that there is competitive exclusion. However, if either the dependence between the growth rates and the resources is not linear or the white noise term is nonlinear we show that coexistence on fewer resources than species is possible. Even more surprisingly, if the temporal environmental variation comes from switching the environment at random times between a finite number of possible states, it is possible for all species to coexist even if the growth rates depend linearly on the resources. We show in an example (a variant of which first appeared in Benaim and Lobry ’16) that, contrary to Hutchinson’s explanation, one can switch between two environments in which the same species is favored and still get coexistence.

Appendices

Appendix A: Proof of Theorem 4.1

If the number of species is strictly greater than the number of resources, $n>m$, the system

$$\begin{aligned} \sum _{i=1}^nc_{i} b_{ij}=0, ~j=1,\dots ,m \end{aligned}$$

(A.1)

admits a nontrivial solution $(c_1, \dots , c_n)$.

Theorem A.1

Assume that $\lim _{\Vert \mathbf {x}\Vert \rightarrow \infty } R_j(\mathbf {x})=-\infty , j=1,\dots ,m$. Suppose further that n species interact according to (4.3), the number of species is greater than the number of resources $n>m$ and the resources depend on the species densities according to (2.2) so that they eventually get exhausted. Suppose further that $g_i(\mathbf {x})=1$ and

$$\begin{aligned} 0<r_m\le \liminf _{\Vert \mathbf {x}\Vert \rightarrow \infty } \frac{|R_{j}(\mathbf {x})|}{|R_1(\mathbf {x})|}\le r^M<\infty , \end{aligned}$$

for $j=1,\dots ,m$. Let $(c_1,\dots ,c_n)$ be a non-trivial solution to (A.1) and assume that $\sum _{i=1}^n c_i\left( \alpha _i+\frac{\sigma _{ii}}{2}\right) \ne 0$. Then, for any starting densities $\mathbf {x}(0)\in (0,\infty )^n$ with probability 1

$$\begin{aligned} \limsup _{t\rightarrow 0}\dfrac{\ln \min \{x_1(t),\dots , x_n(t)\}}{t}<0. \end{aligned}$$

Proof of Theorem 4.1

Suppose $g_i(\mathbf {x})=1$, $i=1,\dots , n$ and $\sum _jb_{ij}>b_m>0$ for any i and some $b_m>0$. Note that if $\sum _j b_{ij}=0$ then we can remove $R_j$ from the equation. Assume that

$$\begin{aligned} 0<r_m\le \liminf _{\Vert \mathbf {x}\Vert \rightarrow \infty } \frac{|R_{j}(\mathbf {x})|}{|R_1(\mathbf {x})|}\le r^M<\infty . \end{aligned}$$

Then, since $\lim _{\Vert \mathbf {x}\Vert \rightarrow \infty } R_j(\mathbf {x})=-\infty$, we have when $|\mathbf {x}|$ large that:

$$\begin{aligned} \sum _{i}\left( -x_i\alpha _i+x_i\sum _j b_{ij} R_j(\mathbf {x})\right)\le & {} -\sum _{i}\left( x_i\alpha _i +x_i\sum _j b_{ij} |R_{j}(\mathbf {x})|\right) \\\le & {} -\sum _{i}\left( x_i\alpha _i +x_i r_m \sum _j b_{ij} |R_{1}(\mathbf {x})|\right) \\\le & {} - r_mb_m\left( \sum _i x_i\right) |R_1(\mathbf {x})|\\\le & {} -\frac{r_m b_m}{mR_m}\left( \sum _i x_i\right) \sum _j |R_j(\mathbf {x})| \end{aligned}$$

which together with the linearity of the diffusion part implies that Assumption 1.1 from the work by HN16 holds with $\mathbf {c}=(1,\dots , 1)$. As a result, for any starting point $\mathbf {x}(0)\in (0,\infty )^n$ the SDE (4.3) has a unique positive solution and by Hening and Nguyen (2018a) (equation (5.22)) with probability 1

$$\begin{aligned} \limsup _{\Vert \mathbf {x}\Vert \rightarrow \infty }\frac{\ln \Vert \mathbf {x}\Vert }{t}\le 0. \end{aligned}$$

(A.2)

By possibly replacing all $c_i$ by $-c_i$ we can assume that $\sum _{i=1}^n c_i\left( \alpha _i+\frac{\sigma _{ii}}{2}\right) >0$. Using this in conjunction with (4.3), (A.1) and Itô’s Lemma we see that

$$\begin{aligned} \dfrac{\sum _{i=1}^n c_i\ln x_i(t)}{t}=-\sum _{i=1}^nc_i\left( \alpha _i+\frac{\sigma _{ii}}{2}\right) +\frac{1}{t}\sum _{i=1}^n c_i\int _0^tE_i(s)\,ds \end{aligned}$$

Letting $t\rightarrow \infty$ and using that $\lim _{t\rightarrow \infty }\dfrac{\sum _{i=1}^n c_iE_i(t)}{t}=0$ with probability 1, we obtain that with probability 1

$$\begin{aligned} \lim _{t\rightarrow \infty }\dfrac{\sum _{i=1}^n c_i\ln x_i(t)}{t}=-\sum _{i=1}^nc_i\left( \alpha _i+\frac{\sigma _{ii}}{2}\right) < 0. \end{aligned}$$

In view of (A.2) this implies that with probability 1

$$\begin{aligned} \limsup _{t\rightarrow \infty } \dfrac{\ln \left( \min \{x_1(t),\dots , x_n(t)\}\right) }{t}<0. \end{aligned}$$

$\square$

Appendix B: Proof of Theorem 4.2

Theorem B.14

Assume two species interact according to

$$\begin{aligned} dx_i(t) = x_i(t)\left( -\alpha _i+b_iR(\mathbf {x}(t))\right) \,dt + x_i(t)\sqrt{\beta _i x_i(t)} \,dB_i(t), ~i=1,2, \end{aligned}$$

the resource R depends linearly on the species densities

$$\begin{aligned} R(\mathbf {x})={\overline{R}} - a_1 x_1(t) - a_2x_2(t) \end{aligned}$$

and $b_i{\overline{R}}>\alpha _i, i=1,2$. Then there exist $\beta _1, \beta _2>0$ such that the two species coexist.

Proof

Consider

$$\begin{aligned} dx_i(t) = x_i(t)\left( -\alpha _i+b_iR(\mathbf {x}(t))\right) \,dt + x_i(t)\sqrt{\beta _i x_i(t)} \,dB_i(t), ~i=1,2. \end{aligned}$$

If the species $x_2$ is absent species $x_1$ has the one-dimensional dynamics

$$\begin{aligned} dx(t) = x(t)\left( -\alpha _1 + b_1 ({\overline{R}}-a_1x(t))\right) dt+ x(t)\sqrt{\beta _1 x(t)} \,dB_1(t). \end{aligned}$$

Since $b_1{\overline{R}}>\alpha _1$, we can use Hening and Nguyen (2018a) to show that the process x(t) has a unique invariant measure on $(0,\infty )$, say $\mu _1$. Moreover, (Hening and Nguyen 2018a, Lemma 2.1 ) shows that

$$\begin{aligned} \int _0^\infty \left( -\alpha _1+b_1({\overline{R}}-a_1 x)-\beta _1x\right) \mu _1(dx)=0 \end{aligned}$$

or

$$\begin{aligned} \int _0^\infty x\mu _1(dx)=\dfrac{b_1{\overline{R}}-\alpha _1}{b_1a_1+\beta _1}. \end{aligned}$$

The invasion rate of $x_2$ with respect to $x_1$ can be computed by (4.2) as

$$\begin{aligned} \Lambda _{x_2}=\int _0^\infty \left( -\alpha _2+b_2({\overline{R}}-a_1 x)\right) \mu _1(dx) =(b_2{\overline{R}}-\alpha _2)- b_2a_1 \dfrac{b_1\overline{R}-\alpha _1}{b_1a_1+\beta _1}. \end{aligned}$$

Similarly, one can compute the invasion rate of $x_1$ with respect to $x_2$ as

$$\begin{aligned} \Lambda _{x_1}=(b_1{\overline{R}}-\alpha _1)- b_1a_2 \dfrac{b_2\overline{R}-\alpha _2}{b_2a_2+\beta _2}. \end{aligned}$$

Since $b_i{\overline{R}}-\alpha _i>0$, one can easily see that $\Lambda _{x_2}>0$, and $\Lambda _{x_1}>0$ if $\beta _1>\dfrac{b_2a_1(b_1{\overline{R}}-\alpha _1)}{b_2\overline{R}-\alpha _2}-b_1a_1.$ and $\beta _2>\dfrac{b_1a_2(b_2\overline{R}-\alpha _2)}{b_1{\overline{R}}-\alpha _1}-b_2a_2.$ If both invasion rates are positive we get by Hening and Nguyen (2018a) that the species coexist. $\square$

Appendix C: Proof of Theorem 4.3

We construct an SDE example of two species competing for one abiotic resource and coexisting. We remark that this happens solely because of the random temporal environmental variation term.

Theorem C.1

Suppose the dynamics of the two species is given by

$$\begin{aligned} \begin{aligned} dx_1(t)&=x_1(t)(-\alpha _1+f({\overline{R}}-a_1x_1(t)-a_2x_2(t)))\,dt +\sigma _1x_1dB_1(t)\\ dx_2(t)&=x_2(t)(-\alpha _2+({\overline{R}}-a_1x_1(t)-a_2x_2(t)))\,dt \end{aligned} \end{aligned}$$

(C.1)

where f is a continuously differentiable Lipschitz function satisfying $\lim _{x\rightarrow -\infty }f(x)=-\infty$, $\dfrac{df(x)}{dx}>0, \dfrac{d^2f(x)}{dx^2}\le 0$ for all $x\in \mathbb {R}$ and $\dfrac{d^2f(x)}{dx^2}<0$ for x in some subinterval of $\left( -\infty ,\frac{{\overline{R}}}{a_1}\right)$. Let $a_1, a_2, \sigma _1, \alpha _1,\sigma _1, {\overline{R}}$ be any fixed positive constants satisfying $f({\overline{R}})>\alpha _1+\dfrac{\sigma _1^2}{2}$. Then there exists an interval $(c_0,c_1)\subset (0,\infty )$ such that the two species coexist for all $\alpha _2\in (c_0,c_1)$.

Proof

The dynamics of species $x_1$ in the absence of species $x_2$ is given by the one-dimensional SDE

$$\begin{aligned} dx(t)=x(t)(-\alpha _1+f({\overline{R}}-a_1x(t)))\,dt +\sigma _1xdB_1(t).\\ \end{aligned}$$

Since $\lim _{x\rightarrow \infty }f({\overline{R}}-a_1x)=-\infty$, and $f(\overline{R})>\alpha _1+\dfrac{\sigma _1^2}{2}$, this diffusion has a unique invariant probability measure $\mu$ on $(0,\infty )$ whose density is strictly positive on $(0,\infty )$ (see Borodin and Salminen (2016) or Mao (1997)). Moreover, by noting that $\lim _{t\rightarrow \infty }\frac{\ln x(t)}{t}=0$ with probability 1 (using Lemma 5.1 of Hening and Nguyen (2018a)) and using Itô’s formula one sees that

$$\begin{aligned} \int _0^\infty f({\overline{R}}-a_1x)\mu (dx)=\alpha _1+\dfrac{\sigma _1^2}{2}. \end{aligned}$$

(C.2)

Since f is a concave function and $\dfrac{d^2f(x)}{dx^2}>0$ for all x in some subinterval of $\left( -\infty ,\frac{\overline{R}}{a_1}\right)$ we must have by Jensen’s inequality that

$$\begin{aligned} \int _0^\infty f({\overline{R}}-a_1x)\mu (dx)< f\left( \int _0^\infty (\overline{R}-a_1x)\mu (dx)\right) . \end{aligned}$$

(C.3)

The fact that the function f is strictly increasing together with (C.2) and (C.3) forces

$$\begin{aligned} \varepsilon _0:=\int _0^\infty ({\overline{R}}-a_1x)\mu (dx)- f^{-1}\left( \alpha _1+\dfrac{\sigma _1^2}{2}\right) >0, \end{aligned}$$

(C.4)

where $f^{-1}$ is the inverse of f—it exists because f is strictly increasing. As a result, the invasion rate of species $x_2$ with respect to $x_1$, of the invariant probability measure $\mu$, can be computed using (4.2) as

$$\begin{aligned} \Lambda _{x_2}=-\alpha _2+\int _0^\infty ({\overline{R}}-a_1x)\mu (dx)=\varepsilon _0+ f^{-1}\left( \alpha _1+\dfrac{\sigma _1^2}{2}\right) - \alpha _2. \end{aligned}$$

This implies that $\Lambda _{x_2}>0$ if and only if

$$\begin{aligned} \alpha _2<\varepsilon _0+ f^{-1}\left( \alpha _1+\dfrac{\sigma _1^2}{2}\right) . \end{aligned}$$

(C.5)

The dynamics of species $x_2$ in the absence of species $x_1$ is

$$\begin{aligned} dy(t)=y(t)({\overline{R}}-\alpha _2-a_2y(t))dt. \end{aligned}$$

The positive solutions of this equation converge to the point $y^*=\dfrac{{\overline{R}}-\alpha _2}{a_2}$ if and only if

$$\begin{aligned} {\overline{R}}>\alpha _2. \end{aligned}$$

(C.6)

The invasion rate of $x_1$ with respect to $x_2$ will be

$$\begin{aligned} \Lambda _{x_1}=-\alpha _1-\frac{\sigma _1^2}{2}+f(\alpha _2). \end{aligned}$$

Note that since the function f is increasing we get $\Lambda _{x_1}>0$ if and only if

$$\begin{aligned} \alpha _2>f^{-1}\left( \alpha _1+\frac{\sigma _1^2}{2}\right) . \end{aligned}$$

(C.7)

Note that $f^{-1}\left( \alpha _1+\dfrac{\sigma _1^2}{2}\right) <{\overline{R}}$ since by assumption $f({\overline{R}})>\alpha _1+\dfrac{\sigma _1^2}{2}$. As a result, making use of the inequalities (C.5), (C.6) and (C.7) we get that $\Lambda _{x_2}>0, \Lambda _{x_1}>0$ if any only if

$$\begin{aligned} \alpha _2\in \left( f^{-1}\left( \alpha _1+\dfrac{\sigma _1^2}{2}\right) , \overline{R}\wedge \left( f^{-1}\left( \alpha _1+\dfrac{\sigma _1^2}{2}\right) +\varepsilon _0\right) \right) . \end{aligned}$$

This implies by Theorem 3.1 or by Benaim (2018) [Theorem 4.4 and Definition 4.3] that the two species coexist. $\square$

Appendix D: Proof of Theorem 5.1

Theorem D.2

Assume that

$$\begin{aligned} \lim _{\Vert \mathbf {x}\Vert \rightarrow \infty } \left( -\alpha _i(u)+\sum _{j=1}^mb_{ij}(u)R_j(\mathbf {x},u) \right) <0, i=1,\dots , n, u=1,\dots ,N.\quad \quad \quad \end{aligned}$$

(D.1)

Suppose further that there exists a vector $(c_1,\dots ,c_n)$ that is simultaneously a solution to the systems (5.5) for all $u\in \{1,\dots ,N\}$. Then, with probability 1,

$$\begin{aligned} \limsup _{t\rightarrow 0}\dfrac{\ln \min \{x_1(t),\dots , x_n(t)\}}{t}<0 \end{aligned}$$

except possibly for the critical case when

$$\begin{aligned} \sum _{i=1}^n c_i\sum _{k=1}^N\alpha _i(k)\nu _k=0, \end{aligned}$$

(D.2)

where $(\nu _k)_{k\in \mathcal {N}}$ is the invariant probability measure of the Markov chain (r(t)).

Proof

Under the condition (D.1), there exists an $M>0$ such that the set $K_M:=\{\mathbf {x}\in \mathbb {R}^n: \Vert \mathbf {x}\Vert \le M\}$ is a global attractor of (5.4). As a result, the solution to (5.4) eventually enters and never leaves the compact set $K_M$. In particular, this shows that the process $\mathbf {x}(t)$ is bounded. Next, note that we can assume that

$$\begin{aligned} \sum _{i=1}^n c_i\sum _{k=1}^N\alpha _i(k)\pi _k>0. \end{aligned}$$

(D.3)

Otherwise, if $\sum _{i=1}^n c_i\sum _{k=1}^N\alpha _i(k)\pi _k<0$, we can replace $c_i$ by $-c_i$, $i=1,\dots , n$ and then get (D.3). Using (5.1) and the fact that that $c_i$’s solve (5.5) simultaneously we get

$$\begin{aligned} \dfrac{\sum _{i=1}^n c_i\ln x_i(t)}{t}=-\dfrac{1}{t}\int _0^t\sum _{i=1}^n c_i\alpha _i(r(s))ds \end{aligned}$$

Letting $t\rightarrow \infty$ and using the ergodicity of the Markov chain (r(t)) we obtain that with probability 1

$$\begin{aligned} \limsup _{t\rightarrow \infty }\dfrac{\sum _{i=1}^n c_i\ln x_i(t)}{t}= & {} -\liminf _{t\rightarrow \infty }\dfrac{1}{t}\int _0^t\sum _{i=1}^n c_i\alpha _i(r(s))ds\\= & {} -\sum _{i=1}^n c_i\sum _{k=1}^N\alpha _i(k)\pi _k< 0 \text { a.s.} \end{aligned}$$

Since $\mathbf {x}(t)$ is bounded, this implies that with probability 1

$$\begin{aligned} \limsup _{t\rightarrow \infty } \dfrac{\ln \min \{x_1(t),\dots , x_n(t)\}}{t}<0. \end{aligned}$$

$\square$

Appendix E: Proof of Theorem 5.2

According to Benaïm and Lobry (2016), Malrieu and Zitt (2017), and Malrieu and Phu (2016) it is enough to find an example for which the invasion rates $\Lambda _{x_1}, \Lambda _{x_2}$ are positive. We will follow Benaïm and Lobry (2016) in order to compute the invasion rates of the two species. Set for $u=1,2$$\mu _u=-\alpha _1(u)+b_1(u){\overline{R}}, \nu _u=-\alpha _2(u)+b_2(u){\overline{R}}$${\overline{a}}_u=\dfrac{b_1(u)a_1(u)}{\mu _u}, \overline{b}_u=\dfrac{b_1(u)a_2(u)}{\mu _u}, \overline{c}_u=\dfrac{b_2(u)a_1(u)}{\nu _u}, \overline{d}(u)=\dfrac{b_2(u)a_2(u)}{\nu _u}$, $p_u=\dfrac{1}{{\overline{a}}_u}, q_u=\dfrac{1}{{\overline{d}}_u}, \gamma _1=\dfrac{q_{12}}{\mu _u}, \gamma _2=\dfrac{q_{21}}{\nu _u}$. If $p_1\ne p_2$, suppose without loss of generality that $p_1<p_2$. Define the functions

$$\begin{aligned} \theta (x)=\dfrac{|x-p_1|^{\gamma _1-1}|p_2-x|^{\gamma _2-1}}{x^{1+\gamma _1+\gamma _2}} \end{aligned}$$

and

$$\begin{aligned} P(x)=\dfrac{{\overline{a}}_2-{\overline{a}}_1}{|{\overline{a}}_2-\overline{a}_1|}\left[ \dfrac{\nu _2}{\mu _2}(1-{\overline{c}}_2 x)(1-\overline{a}_1x)-\dfrac{\nu _1}{\mu _1}(1-{\overline{c}}_1 x)(1-{\overline{a}}_2x)\right] . \end{aligned}$$

By Benaïm and Lobry (2016) we have

$$\begin{aligned} \Lambda _{x_2}={\left\{ \begin{array}{ll} \dfrac{1}{q_{12}+q_{21}}\left( q_{21}\nu _1(1-{\overline{c}}_1 p)+q_{12}\nu _2(1-{\overline{c}}_2p)\right) &{}\text { if } p_1=p_2=p\\ p_1p_2\dfrac{\int _{p_1}^{p_2}\theta (x)P(x)dx}{\int _{p_1}^{p_2}\theta (x)dx}&{}\text { if } p_1<p_2 \end{array}\right. } \end{aligned}$$

(E.1)

The expression for $\Lambda _{x_1}$ can be obtained by swapping $\mu _i$ and $\nu _i$, $({\overline{a}}_i,{\overline{c}}_i)$ with $({\overline{d}}_i,\overline{b}_i)$, and $p_i$ with $q_i$.

For the example from Figs. 3 and 4 we have used the integral equation from (E.1) together with the numerical integration package of Mathematica in order to find $\Lambda _{x_1}>0$ and $\Lambda _{x_2}>0$. This implies by Theorem 3.1, Benaïm and Lobry (2016) or by Benaim (2018) [Theorem 4.4 and Definition 4.3] that the two species coexist.