Evolutionary dynamics of body size subject to dispersal and advection

Abstract

The aim of this paper is to investigate the evolution of body size of population subject to dispersal and advection. For this purpose, we consider a reaction–diffusion system for competing species with hostile boundary conditions, which models population dynamics in advective environments. First, with the methods of adaptive dynamics, we discuss the stability of evolutionary singular strategy; specifically, evolutionary conditions for continuously stable strategy and evolutionary branching are obtained. Second, by theoretical analysis and numerical simulations, we investigate the influence of dispersal and advection on the evolution trend of body size and high-level polymorphism. Our results show that in environments without advection, the result of evolution mainly depends on competition coefficients, intrinsic growth rates and biological dispersal. In particular, the evolution of the body size will tend to the slow disperser with the increase in the dispersal rate if regardless of the influence of body size on competitiveness and growth rate. But there will be different results in advective environments, the increasing dispersal rate may lead to the evolution of the body size toward a intermediate level. A small or a moderate dispersal rate is likely to stimulate the emergence of evolutionary branching. In addition, our results predict that for hostile boundary conditions, advection can promote evolutionary stability, which prevent the emergence of evolutionary branching.

Appendices

Appendix A

The following differential system is constructed about eigenvalue

$$\begin{aligned} \left\{ \begin{array}{ll} d(s_{i})\frac{\partial ^{2} N_{i}}{\partial x^{2}}-q(s_{i})\frac{\partial N_{i}}{\partial x}=-\lambda (s_{i})N_{i},\\ N_{i}(0)=N_{i}(L)=0, \end{array} \right. \end{aligned}$$

$(i=1,2)$. It is not difficult to verify that if $q^{2}(s_{i})-4d(s_{i})\lambda (s_{i})\ge 0$, then $N_{i}=0$. If $q^{2}(s_{i})-4d(s_{i})\lambda (s_{i})<0$,

$$\begin{aligned} N_{i}(x)= & {} e^{\frac{q(s_{i})}{2d(s_{i})}x}\bigg (C_{1}\cos \frac{\sqrt{4d(s_{i})\lambda (s_{i})-q^{2}(s_{i})}}{2d(s_{i})}x\\{} & {} +C_{2}\sin \frac{\sqrt{4d(s_{i})\lambda (s_{i})-q^{2}(s_{i})}}{2d(s_{i})}x\bigg ). \end{aligned}$$

According to $N_{i}(0)=N_{i}(L)=0$, we can get $C_{1}=0$,

$$\begin{aligned} \frac{\sqrt{4d(s_{i})\lambda (s_{i})-q^{2}(s_{i})}}{2d(s_{i})}L=k\pi , \end{aligned}$$

where k is positive integer. Therefore, the dominant eigenvalue of the advection–diffusion operator satisfies

$$\begin{aligned} \lambda (s_{i})=\frac{q^{2}(s_{i})}{4d(s_{i})}+\frac{\pi ^{2}d(s_{i})}{L^{2}}. \end{aligned}$$

Appendix B: Proof of Proposition 2

Evolutionary singular strategy is the solution of the following equation

$$\begin{aligned} r'(s)-\lambda '(s)+\frac{\beta }{1+v}(r(s)-\lambda (s))=0. \end{aligned}$$

Let

$$\begin{aligned} \phi _{1}(s)= & {} r'(s)+\frac{\beta }{1+v}r(s)=b\Big (\sqrt{s^{2}+e}\!-\!s\Big )\\{} & {} \left( \frac{\beta }{1+v}\!-\!\frac{1}{\sqrt{s^{2}+e}}\right) , \\ \phi _{2}(s)= & {} \lambda '(s)\!+\!\frac{\beta }{1+v}\lambda (s)\\= & {} \frac{\pi ^{2}}{L^{2}}d(s,p_{d})\left( \frac{\sigma _{d}^{2}\beta +(1+v) s_{d}}{(1+v)\sigma _{d}^{2}}\!-\!\frac{s}{\sigma _{d}^{2}}\right) , \end{aligned}$$

then

$$\begin{aligned} \phi _{1}(0)= & {} \frac{b(\beta \sqrt{e}-1-v)}{1+v},\phi _{2}(0)\\ {}= & {} \frac{\pi ^{2}k_{1}p_{d}\big (\beta \sigma _{d}^{2}+(1+v) s_{d}\big )}{L^{2}(k_{2}+p_{d})(1+v)\sigma _{d}^{2}}\exp \left( \frac{-s_{d}^{2}}{2\sigma _{d}^{2}} \right) . \end{aligned}$$

The evolutionary singular strategy can be regarded as the solution of the following equation equivalently,

$$\begin{aligned} \phi _{1}(s)-\phi _{2}(s)=0. \end{aligned}$$

We discuss the following four situations, respectively.

• (i) $\kappa _{1}<1$, $\kappa _{2}<1$ is equivalent to $\phi _{1}(0)<0$, ${\tilde{s}}_{2}<{\tilde{s}}_{1}$. This implies $\phi _{1}({\tilde{s}}_{2})-\phi _{2}({\tilde{s}}_{2})<0$, $\phi _{1}({\tilde{s}}_{1})-\phi _{2}({\tilde{s}}_{1})>0$, due to zero point theorem, there is at least one singular point $s^{*}\in (\tilde{s_{2}},\tilde{s_{1}})$, and $\phi _{1}(s^{*})=\phi _{2}(s^{*})<0$. And it is not difficult to verify that there is no evolutionary singular strategy outside $(\tilde{s_{2}},\tilde{s_{1}})$.

• (ii) $\kappa _{1}<1$, $\kappa _{2}>1$ is equivalent to $\phi _{1}(0)<0$, ${\tilde{s}}_{2}>{\tilde{s}}_{1}$. This means $\phi _{1}({\tilde{s}}_{2})-\phi _{2}({\tilde{s}}_{2})>0$, $\phi _{1}({\tilde{s}}_{1})-\phi _{2}({\tilde{s}}_{1})<0$, in this case there is at least one singular point $s^{*}\in (\tilde{s_{1}},\tilde{s_{2}})$, and $\phi _{1}(s^{*})=\phi _{2}(s^{*})>0$. And there is no evolutionary singular strategy outside $(\tilde{s_{1}},\tilde{s_{2}})$.

• (iii) $\kappa _{1}>1$ implies $\phi _{1}(0)>0$, $0<\kappa _{3}<1$ means $0<\phi _{1}(0)<\phi _{2}(0)$. Considering $\phi _{1}({\tilde{s}}_{2})-\phi _{2}({\tilde{s}}_{2})>0$, there is at least one singular point $s^{*}\in (0,\tilde{s_{2}})$, and $\phi _{1}(s^{*})=\phi _{2}(s^{*})>0$. If $s>\tilde{s_{2}}$, $\varPhi _{1}(s)\varPhi _{1}(s)<0$, thus, there is no evolutionary singular strategy in $(\tilde{s_{2}},\infty )$.

• (iv) If $\kappa _{1}>1$ and $\kappa _{3}>1$, the singular points may not exist; if it exists, we can get $s^{*}\in (0,\tilde{s_{2}})$, and $\phi _{1}(s^{*})=\phi _{2}(s^{*})>0$.

The proof of the proposition is completed.

Appendix C: Proof of Proposition 3

Evolutionary singular strategies are the points that satisfy the following equation

$$\begin{aligned} D(s^{*})=\phi _{1}(s^{*})-\phi _{2}(s^{*})=0. \end{aligned}$$

Although we do not have the explicit expression of evolutionary singular strategies, we can calculate the derivative of $s^{*}$ with respect to $p_{d}$ indirectly by implicit differential equation. It is given by

$$\begin{aligned} \frac{\textrm{d}s^{*}}{\textrm{d}p_{d}}=-\frac{\frac{\partial D(s^{*})}{\partial p_{d}}}{\frac{\partial D(s^{*})}{\partial s^{*}}}=\frac{\frac{\partial \phi _{2}(s^{*})}{\partial p_{d}}}{\frac{\partial D(s^{*})}{\partial s^{*}}}=\frac{k_{2} \phi _{2}(s^{*})}{\big (p_{d}(k_{2}+p_{d})\big )\frac{\partial D(s^{*})}{\partial s^{*}}}.\nonumber \\ \end{aligned}$$

(32)

For the convergence stable strategy $s^{*}$, $\frac{\partial D(s^{*})}{\partial s^{*}}<0$, according to the proof of Proposition 2, if $\kappa _{i}<1\ (i=1,2)$, then the numerator of Eq. (32) is negative. Thus, the derivative of $s^{*}$ with respect to $p_{d}$ is positive. Otherwise, the derivative of $s^{*}$ with respect to $p_{d}$ is negative. The proof is complete.

Appendix D: Proof of Proposition 4

By calculation, we can obtain

$$\begin{aligned} \frac{\partial D(s^{*})}{\partial p_{d}}= & {} \frac{2q(s^{*},p_{q})d(s^{*},p_{d})q'_{s}(s^{*},p_{q})d'_{p_{d}}(s^{*},p_{d})}{4d^{3}(s^{*},p_{d})}\\{} & {} +\frac{q^{2}(s^{*},p_{q})d(s^{*},p_{d})d''_{s,p_{d}}(s^{*},p_{d})}{4d^{3}(s^{*},p_{d})}\nonumber \\{} & {} -\frac{2q^{2}(s^{*},p_{q})d'_{p_{d}}(s^{*},p_{d})d'_{s}(s^{*},p_{d})}{4d^{3}(s^{*},p_{d})}\\{} & {} +\frac{\beta q^{2}(s^{*},p_{q})d'_{p_{d}}(s^{*},p_{d})}{4(1+v)d^{2}(s^{*},p_{d})}\nonumber \\{} & {} -\frac{\pi ^{2}d''_{s,p_{d}}(s^{*},p_{d})}{L^{2}}-\frac{\beta \pi ^{2}d'_{p_{d}}(s^{*},p_{d})}{(1+v)L^{2}} \nonumber \\= & {} \frac{k_{2}q^{2}(s^{*},p_{q})}{4(k_{2}+p_{d})p_{d}d(s^{*},p_{d})}\\{} & {} \Big (2\frac{s_{q}-s^{*}}{\sigma ^{2}_{q}} -\frac{s_{d}-s^{*}}{\sigma ^{2}_{d}}+\frac{1+v}{\beta }\Big )\nonumber \\{} & {} +\frac{\pi ^{2}k_{2}d(s^{*},p_{d})}{L^{2}(k_{2}+p_{d})p_{d}}\Big (\frac{s^{*}-s_{d}}{\sigma ^{2}_{d}}-\frac{1+v}{\beta }\Big ). \end{aligned}$$

If condition (i) or condition (ii) of Proposition 4 holds, then $\frac{\textrm{d} s^{*}}{\textrm{d} p_{d}}>0$, and if condition (iii) or condition (iv) of Proposition 4 holds, then $\frac{d s^{*}}{d p_{d}}<0$. The proof is completed.