Local dynamics of a predator–prey community in a moderate period of time

Abstract

In this work, we have introduced an ecological model of a prey–predator system. It is assumed that the prey species grows logistically, but the total number of predator is constant in the time interval. Positivity and boundedness of the solution ensure that the proposed model is well-posed. Local stability conditions of the equilibrium points have been analysed by the Routh–Hurwitz criterion. The persistence of the system has also been shown under a parametric restriction. Numerical analysis has indicated that both axial and interior steady states can exist only for moderate consumption rate (searching efficiency). But if this rate becomes high (or low), then only the prey-free equilibrium (or one of the interior equilibriums) exists as a steady state. Further, the equilibrium points can change their stability through transcritical and saddle-node bifurcations by varying the consumption rate of the predator. Analytical results provide an interesting phenomenon about this model: the system can never show any oscillating behaviour for any parametric values, i.e. no limit cycle can occur through Hopf bifurcation around an equilibrium point. The axial equilibrium becomes stable from an unstable situation when the consumption rate becomes high and the interior state which is stable remains stable as time goes by.

Appendix

Proof of Theorem 1

Proof

As the right-hand side of system (2) is continuous and locally Lipschitzian on space of continuous functions C, the solution (x(t), y(t)) of (2) with positive initial conditions $(x_{0},y_{0})$ exists and is unique on $[0,\tau ),$ where $0<\tau \le +\infty$ (Hale 1977). From the first equation of (2), we have

$$\begin{aligned} \frac{{\mathrm{{d}}}x}{{\mathrm{{d}}}t}& = {} x(1-x)-\frac{\xi Pxy}{1+ax}, \\ {{\text {i.e.}}}\ x(t)& = {} x_{0}\exp \left[ \int ^t_0 \left\{ (1-x(s))-\frac{P\xi y(s)}{1+a x(s)}\right\} \,{{\mathrm{d}}}s\right] \\&> 0, \ {{\text {for}}} \ x_{0}> 0. \end{aligned}$$

Next we show that $y(t)>0, \ \forall \, t\in [0,\tau )$. If it does not hold, then $\exists t_{1} \in [0,\tau )$ such that $y(t_{1})=0,\ {\dot{y}}(t_{1})\le 0$ and $y(t)>0, \ \forall \, t\in [0,t_{1})$. From the $2^{nd}$ equation of (2), we have

$$\begin{aligned} \frac{{{\mathrm{d}}}y(t_{1})}{{\mathrm{{d}}}t}=\sigma \{1-y(t_{1})\}-\frac{\xi Kx(t_{1})y(t_{1})}{1+ax(t_{1})}-\zeta y(t_{1})=\sigma > 0, \end{aligned}$$

which is a contradiction to ${\dot{y}}(t_{1})\le 0$. So, $y(t)> 0, \ \forall \, t\in [0,\tau )$. Hence, the theorem is proved. $\square$

Proof of Theorem 2

Proof

From the first equation of (2):

$$\begin{aligned} \begin{aligned} \frac{{\mathrm{{d}}}x}{{\mathrm{{d}}}t}&= x(1-x)-\frac{\xi Pxy}{1+ax}, \\&<x(1-x) \end{aligned} \\ \Rightarrow \limsup _{t\rightarrow \infty }x(t)\le 1. \end{aligned}$$

Let $W= x(t)+ y(t)$.

$$\begin{aligned} \begin{aligned} {{\text {So,}}}\ \frac{{{\mathrm{d}}}W}{{\mathrm{{d}}}t}&= \frac{{\mathrm{{d}}}x}{{\mathrm{{d}}}t}+ \frac{{{\mathrm{d}}}y}{{\mathrm{{d}}}t} \\&=\left[ x(1-x)+\sigma -\frac{\xi (K+P)xy}{1+ax}-\sigma y-\zeta y\right] \\&\le 2x+\sigma -x-\sigma y \\&\le (2+\sigma )-\kappa W,\ {{\text {where}}}\ \kappa =\min \{1,\sigma \} \\ {{\text {Hence,}}}\ W(t)&\le \frac{(2+\sigma )}{\kappa }(1-\exp (-\kappa t))+ W(x_{0}, y_{0})\\&\quad \exp (-\kappa t)\Rightarrow \lim _{t\rightarrow \infty } W(t)&\le \frac{(2+\sigma )}{\kappa }. \end{aligned} \end{aligned}$$

So, all solutions of system (2) will enter into the region:

$$\begin{aligned} \Omega = \left\{ (x,y): 0\le x(t)\le 1; 0\le W(t)\le \frac{(2+\sigma )}{\kappa }+\epsilon , \epsilon >0\right\} . \end{aligned}$$

$\square$

Proof of Theorem 3

Proof

Choose $\epsilon >0$, then $\exists \ T_{1}>0,$ such that $x(t)< {\overline{x}}+ \epsilon ,\ \forall \ t>T_{1}$.

Also, for $0<\epsilon _{1}<{\underline{y}}-\frac{(1+2a)}{P\xi }, \ \exists \, \, T_{2}>0,$ such that $y(t)> {\underline{y}}- \epsilon _{1}$, $\forall \, t>T_{2}$.

Therefore, for all $t> \max \{T_{1},T_{2}\}$, we have

$$\begin{aligned} \begin{aligned} \frac{{\mathrm{{d}}}x}{{\mathrm{{d}}}t}&= x(1-x)-\frac{\xi Pxy}{1+ax} \\&\le \left\{ 1-\frac{P\xi ({\underline{y}}- \epsilon _{1})}{1+a({\overline{x}}+ \epsilon )}\right\} x \\&\le \left\{ 1-\frac{P\xi ({\underline{y}}- \epsilon _{1})}{1+2a}\right\} x \\&=-\mu x \left( {{\text {where}}}\ \mu =\left\{ \frac{P\xi ({\underline{y}}- \epsilon _{1})}{1+2a}-1\right\} >0\right). \end{aligned} \end{aligned}$$

Hence, $\lim _{t\rightarrow \infty }x=0$. $\square$

Proof of Theorem 4

Proof

Let us consider the average Lyapunov function is $V(x, y)=x^{\theta _{1}}y^{\theta _{2}}$ where each $\theta _{i}$ for $i=1,2$ is assumed to be positive. In the interior of ${\mathbb {R}}_{+}^{2},$ we have

$$\begin{aligned} \frac{{\dot{V}}}{V}& = {} \phi (x, y)=\theta _{1}\left[ 1-x-\frac{P\xi y}{1+ax}\right] \\&+\frac{\theta _{2}}{y}\left[ \sigma (1-y)-\frac{K\xi xy}{1+ax}-\zeta y\right]. \end{aligned}$$

To prove the permanence, we need to show $\phi (x, y)>0$ at $E_{1}\left( 0,\frac{\sigma }{\sigma +\zeta }\right)$. The value of $\phi (x, y)$ at the boundary equilibria $E_{1}$ is as follows:

$E_{1}:\phi \left( 0,\frac{\sigma }{\sigma +\zeta }\right) = \frac{\theta _{1}}{\sigma +\zeta }[\sigma +\zeta -P\xi \sigma ]$.

Now, $\phi \left( 0,\frac{\sigma }{\sigma +\zeta }\right)$ is positive only when $\sigma +\zeta -P\xi \sigma >0$ holds. So, system (2) is permanent (Freedman and Ruan 1995; Mondal and Samanta 2019) if the stated condition is fulfilled. $\square$

Proof of Theorem 5

Proof

For $E_{1}=\left( 0,\frac{\sigma }{\sigma +\zeta }\right)$:

$$\begin{aligned} J|_{E_{1}}=\left( \begin{array}{ccc} 1-\frac{P\xi \sigma }{\sigma +\zeta } &{} 0 \\ -\frac{K\xi \sigma }{\sigma +\zeta } &{} -(\sigma +\zeta ) \end{array} \right) . \end{aligned}$$

The eigenvalues are the roots of the equation: $\lambda ^{2}+P_{1}\lambda +P_{2}=0,$ where $P_{1}=\sigma +\zeta +\frac{P\xi \sigma }{\sigma +\zeta }-1$ and $P_{2}=(\sigma +\zeta )\left( \frac{P\xi \sigma }{\sigma +\zeta }-1\right)$.

By Routh–Hurwitz criterion, the roots have negative real parts if $P_{1},P_{2}>0$. Now, $P_{2}>0$ implies $P\xi \sigma >\sigma +\zeta$ and this condition gives $P_{1}>0$ also. $\square$

Proof of Theorem 6

Proof

For $E^{*}_{L}=(x^{*}_{L},y^{*}_{L})$:

$$\begin{aligned} J|_{E^{*}_{L}}=\left( \begin{array}{ccc} b_{11} &{}\quad b_{12} \\ b_{21} &{}\quad b_{22} \end{array} \right) , \end{aligned}$$

where $b_{11}=-x^{*}_{L}+\frac{aP\xi x^{*}_{L}y^{*}_{L}}{(1+ax^{*}_{L})^{2}}; \ b_{12}=-\frac{P\xi x^{*}_{L}}{1+ax^{*}_{L}}; \ b_{21}=-\frac{K\xi y^{*}_{L}}{(1+ax^{*}_{L})^{2}}; \ b_{22}=-\frac{\sigma }{y^{*}_{L}}.$

The eigenvalues are the roots of the equation: $\lambda ^{2}+Q_{1}\lambda +Q_{2}=0,$ where $Q_{1}\equiv Q_{1}(x^{*}_{L}, y^{*}_{L})=x^{*}_{L}+\frac{\sigma }{y^{*}_{L}}-\frac{aP\xi x^{*}_{L}y^{*}_{L}}{(1+ax^{*}_{L})^{2}}=\frac{\sigma }{y^{*}_{L}}+\frac{(1-a)x^{*}_{L}+2ax^{*2}_{L}}{1+ax^{*}_{L}}$

and $Q_{2}\equiv Q_{2}(x^{*}_{L}, y^{*}_{L})=\frac{\sigma }{y^{*}_{L}}\left\{ x^{*}_{L}-\frac{aP\xi x^{*}_{L}y^{*}_{L}}{(1+ax^{*}_{L})^{2}}\right\} -\frac{PK\xi ^{2}x^{*}_{L}y^{*}_{L}}{(1+ax^{*}_{L})^{3}}=\frac{\sigma x^{*}_{L}(1-a+2ax^{*}_{L})}{y^{*}_{L}(1+ax^{*}_{L})}-\frac{K\xi x^{*}_{L}(1-x^{*}_{L})}{(1+ax^{*}_{L})^{2}}$.

By Routh–Hurwitz criterion, the roots have negative real parts if $Q_{1},Q_{2}>0$.

Now $x^{*}_{L}<1$. So, $Q_{2}>0$ implies $1-a+2ax^{*}_{L}>0$ and this implies $Q_{1}> 0$ too. $\square$

Proof of Theorem 8

Proof

For $E^{*}_{L}(x^{*}_{L},y^{*}_{L}):$ let $H(x,y)=\frac{1}{xy};\ h_{1}(x,y)=x(1-x)-\frac{P\xi xy}{1+ax}\ {{\text {and}}}\ h_{2}(x,y)=\sigma (1-y)-\frac{K\xi xy}{1+ax}-\zeta y$.

Now, $H(x,y)>0$ in the interior of positive quadrant of x–y plane. At $E^{*}_{L}$:

$$\begin{aligned} \begin{aligned} \Delta (x,y)&=\frac{\partial }{\partial x}(Hh_{1})+\frac{\partial }{\partial y}(Hh_{2})\\&=-\frac{1}{y}+\frac{aP\xi }{(1+ax)^{2}}-\frac{\sigma }{xy^{2}}\\&=-\frac{1}{xy}\left[ x+\frac{\sigma }{y}-\frac{aP\xi xy}{(1+ax)^{2}}\right]. \end{aligned} \end{aligned}$$

Hence, $\Delta (x,y)$ does not change sign and is not identically zero in $\Omega$. So, by Bendixson–Dulac criterion, there exists no limit cycle in $\Omega$. Therefore, if $E^{*}_{L}(x^{*}_{L},y^{*}_{L})$ is locally asymptotically stable, then it will be globally asymptotically stable in $\Omega$ in the positive quadrant of x–y plane. $\square$

Proof of Theorem 10

Proof

$$\begin{aligned} J|_{E_{1}}=\left( \begin{array}{ccc} 1-\frac{P\xi \sigma }{\sigma +\zeta } &{} 0 \\ -\frac{K\xi \sigma }{\sigma +\zeta } &{} -(\sigma +\zeta ) \end{array} \right). \end{aligned}$$

Let $\xi _{[TC]}$ be the value of $\xi$ such that $J|_{E_{1}}$ has a simple zero eigenvalue at $\xi =\xi _{[TC]}$.

So, at $\xi =\xi _{[TC]}:$

$$\begin{aligned} J|_{E_{1}}=\left( \begin{array}{ccc} 0 &{} 0 \\ -\frac{K\xi \sigma }{\sigma +\zeta } &{} -(\sigma +\zeta ) \end{array} \right) . \end{aligned}$$

Here, $\lambda _{1}=-(\sigma +\zeta )<0$.

After some calculations: $V=\left( -(\sigma +\zeta ), \frac{K\xi \sigma }{\sigma +\zeta }\right) ^{{{\mathrm{T}}}}=(v_{1},v_{2})\ {{\text {and}}} \ W=(1,0)^{{{\mathrm{T}}}}$.

Therefore,

$$\begin{aligned} \Omega _{1}& = {} W^{{{\mathrm{T}}}} \cdot f_{\xi }(E_{1}, \xi _{[TC]})\\& = {} -\frac{Pxy}{1+ax}\bigg |_{E_{1}} =0,\\ \Omega _{2}& = {} W^{{{\mathrm{T}}}}\left[ Df_{\xi }(E_{1}, \xi _{[TC]})V\right] \\& = {} -\frac{P\sigma }{(\sigma +\zeta )} v_{1} =P\sigma \ne 0\\ {{\text {and}}}\ \Omega _{3}& = {} W^{{{\mathrm{T}}}}\left[ D^{2}f(E_{1}, \xi _{[TC]})(V,V)\right] \\& = {} \left\{ -2+\frac{aP\xi \sigma }{\sigma +\zeta }\right\} (\sigma +\zeta )^{2}+2PK\sigma \xi ^{2} \ne 0. \end{aligned}$$

By Sotomayor’s theorem, system (2) undergoes a transcritical bifurcation around $E_{1}$ at $\xi =\xi _{[TC]}$. $\square$

Proof of Theorem 11

Proof

$$\begin{aligned} J|_{{\overline{E}}^{*}}=\left( \begin{array}{ccc} b_{11} &{}\quad b_{12} \\ b_{21} &{}\quad b_{22} \end{array} \right) , \end{aligned}$$

where $b_{11}=-{\overline{x}}^{*}+\frac{aP\xi {\overline{x}}^{*}{\overline{y}}^{*}}{(1+a{\overline{x}}^{*})^{2}}; \ b_{12}=-\frac{P\xi {\overline{x}}^{*}}{1+a{\overline{x}}^{*}}; \ b_{21}=-\frac{K\xi {\overline{y}}^{*}}{(1+a{\overline{x}}^{*})^{2}}; \ b_{22}=-\frac{\sigma }{{\overline{y}}^{*}}$.

The eigenvalues are the roots of the equation: $\lambda ^{2}+Q_{1}\lambda +Q_{2}=0,$ where $Q_{1}=-(b_{11}+b_{22})={\overline{x}}^{*}+\frac{\sigma }{{\overline{y}}^{*}}-\frac{aP\xi {\overline{x}}^{*}{\overline{y}}^{*}}{(1+a{\overline{x}}^{*})^{2}}=\frac{\sigma }{{\overline{y}}^{*}}+\frac{(1-a){\overline{x}}^{*}+2a{\overline{x}}^{*2}}{1+a{\overline{x}}^{*}}$

and $Q_{2}=(b_{11}b_{22}-b_{12}b_{21})=\frac{\sigma }{{\overline{y}}^{*}}\left\{ {\overline{x}}^{*}-\frac{aP\xi {\overline{x}}^{*}{\overline{y}}^{*}}{(1+a{\overline{x}}^{*})^{2}}\right\} -\frac{PK\xi ^{2}{\overline{x}}^{*}{\overline{y}}^{*}}{(1+a{\overline{x}}^{*})^{3}}=\frac{\sigma {\overline{x}}^{*}(1-a+2a{\overline{x}}^{*})}{{\overline{y}}^{*}(1+a{\overline{x}}^{*})}-\frac{K\xi {\overline{x}}^{*}(1-{\overline{x}}^{*})}{(1+a{\overline{x}}^{*})^{2}}$.

At $\zeta =\zeta _{\left[ {{\mathrm{Sn}}}\right] },\ Q_{2}=Det\left( J|_{{\overline{E}}^{*}}\right) =0$. Then the characteristic equation for ${\overline{E}}^{*}$ reduces to $\lambda ^{2}-Tr\left( J|_{{\overline{E}}^{*}}\right) \lambda = 0$. So, $J|_{E^{*}}$ has a simple zero eigenvalue and a eigenvalue with negative real part at $\zeta =\zeta _{\left[ {{\mathrm{Sn}}}\right] }$.

At $\zeta =\zeta _{\left[ {{\mathrm{Sn}}}\right] }$:

$$\begin{aligned} J|_{{\overline{E}}^{*}}=A\left( \begin{array}{ccc} 1 &{}\quad 0 \\ 1 &{}\quad 0 \end{array} \right) , \end{aligned}$$

where $A=K\xi {\overline{x}}^{*}\frac{(1-{\overline{x}}^{*})}{(1+a{\overline{x}}^{*})^{2}}$. Here, $V=(1,0)^{{{\mathrm{T}}}}\ {{\text {and}}}\ W=(1,-1)^{{{\mathrm{T}}}}$.

$$\begin{aligned} {{\text {So,}}}\ \Omega _{1}& = {} W^{{{\mathrm{T}}}}.f_{\zeta }({\overline{E}}^{*}, \zeta _{\left[ {{\mathrm{Sn}}}\right] }) \\&=y|_{{\overline{E}}^{*}} \ne 0 \\ {{\text {and}}}\ \Omega _{2}& = {} W^{{{\mathrm{T}}}}\left[ D^{2}f({\overline{E}}^{*}, \zeta _{\left[ {{\mathrm{Sn}}}\right] })(V,V)\right] \\&=-2+\frac{(P-2K)a\xi {\overline{y}}^{*}}{(1+a{\overline{x}}^{*})^{3}} \ \ne 0. \end{aligned}$$

Thus, using Sotomayor’s theorem, system (2) undergoes a saddle-node bifurcation around ${\overline{E}}^{*}$ at $\zeta =\zeta _{\left[ {{\mathrm{Sn}}}\right] }$. $\square$