Stage-Structured Cannibalism in a Ratio-Dependent System with Constant Prey Refuge and Harvesting of Matured Predator

Abstract

A four dimensional stage-structured ratio-dependent predator-prey model with constant prey refuge is proposed and analyzed to study the effect of predation and cannibalism of the organisms at the highest trophic level with harvesting. Our analysis leads to different thresholds in terms of the model parameters acting as conditions under which the organisms associated with the system cannot thrive even in absence of predation. Local stability of the system is obtained in absence of one or more of the predators and in presence of all the predators. Moreover, it is shown that the system undergoes Hopf bifurcation when the carrying capacity of macro-algae crosses certain critical value. Computer simulations have been carried out to illustrate various analytical results.

Appendix

Proof of boundedness of the system (Theorem 4.1)

$\frac{dP}{dt}\le P(1-P)$ implies $P(t)\le 1$ as $t\rightarrow \infty$ and so corresponding to $\epsilon _1>0$, there exists $t_{\epsilon _1}>0$ such that $P(t)\le 1+\epsilon _1$, for all $t\ge t_{\epsilon _1}$. Let $\Sigma (t)= P(t)+x(t)+y(t)+z(t)$ Then $\frac{d}{dt}(\Sigma (t))\le 1-(\Sigma (t))D$, where $D=\min \{1,D_1,D_2,D_3\}$, for all $t\ge t_{\epsilon _1}$. Let u(t) be the solution of $\frac{du}{dt}+uD=1,$ satisfying $u(0)=\Sigma (0)$. Then $u(t)=\frac{1}{D}+\left( \Sigma (0)-\frac{1}{D}\right) e^{-tD}\rightarrow \frac{1}{D}$ as $t\rightarrow \infty .$ By comparison, it follows that $\lim _{t\rightarrow \infty }\sup [P(t)+x(t)+y(t)+z(t)]<\frac{1}{D}<\infty ,$ proving the theorem. $\square$

Proof of Theorem 4.2

(i) Since $P(t)\le 1$ as $t\rightarrow \infty$, corresponding to $\epsilon _1>0$, there exists $t_{\epsilon _1}>0$ such that $P(t)\le 1+\epsilon _1$, for all $t\ge t_{\epsilon _1}$. If $m_1\le D_1$, then $\frac{dx}{dt}\le x(m_1-D_1)$ and so $x(t)\le c_1e^{(m_1-D_1)t}\rightarrow 0$ as $t\rightarrow \infty$. Thus $lim_{t\rightarrow \infty }x(t)=0$ and consequently $lim_{t\rightarrow \infty }y(t)=0=lim_{t\rightarrow \infty }z(t)$. (ii) For all $t\ge t_{\epsilon _1}$, we have $\frac{dx}{dt}\le x(m_1-D_1)\left( 1-\frac{\lambda x}{1+\epsilon _1}\right)$. Also, $x(t)\le \frac{1}{D}$ as $t\rightarrow \infty$, where $D=\min \{1,D_1,D_2,D_3\}$. So, corresponding to $\epsilon _1>0$, there exists $t_{\epsilon _1}^{'}>0$ such that $x(t)\le \frac{1}{D}+\epsilon _1$, for all $t\ge t_{\epsilon _1}^{'}$. Let $T_\epsilon =\max \{t_{\epsilon _1},t_{\epsilon _1}^{'}\}$. Therefore, for all $t\ge T_\epsilon$, $\frac{dx}{dt}\le 0$ if $m_1>D_1$ and $\lambda >\frac{(1+\epsilon _1)}{\frac{1}{D}+\epsilon _1}$. Hence $lim_{t\rightarrow \infty }x(t)=0$ if $m_1>D_1$ and $\lambda >D$, and consequently, $lim_{t\rightarrow \infty }y(t)=0=lim_{t\rightarrow \infty }z(t)$.$\square$

Proof of Lemma 5.1.1

• (i) $(u,v,w,z)\rightarrow E_{00}$ as $t\rightarrow \infty$ implies that $u,v,w,z\rightarrow 0$ as $t\rightarrow \infty$. Since $v=\frac{x}{z}$ and $v,z\rightarrow 0$ as $t\rightarrow \infty$, it follows that x converges to zero faster than z as $t\rightarrow \infty$. Since $w=\frac{y}{z}$ and $w,z\rightarrow 0$ as $t\rightarrow \infty$, it follows that y converges to zero faster than z as $t\rightarrow \infty$. Also, $u=\frac{P}{x}$ and $u,x\rightarrow 0$ as $t\rightarrow \infty$, implies that P converges to zero faster than x as $t\rightarrow \infty$.

Conversely, $(P,x,y,z)\rightarrow E_0$ with P converges to zero faster than x and x, y converge to zero faster than z as $t\rightarrow \infty$ implies $u,v,w\rightarrow 0$ as $t\rightarrow \infty$ and so $(u,v,w,z)\rightarrow E_{00}$ as $t\rightarrow \infty$.

• (ii) Again, $(u,v,w,z)\rightarrow E_{01}$ as $t\rightarrow \infty$ implies that $u\rightarrow u_{01}$ and $v,w,z\rightarrow 0$ as $t\rightarrow \infty$. Since $v=\frac{x}{z}$ and $v,z\rightarrow 0$ as $t\rightarrow \infty$, it follows that x converges to zero at a faster rate then $z\rightarrow 0$ for $t\rightarrow \infty$. Since $w=\frac{y}{z}$ and $w,z\rightarrow 0$ as $t\rightarrow \infty$, it follows that y converges to zero at a faster rate than z for $t\rightarrow \infty$. Also, $u=\frac{P}{x}$ with $u\rightarrow u_{01}>0$ and $x\rightarrow 0$ as $t\rightarrow \infty$, implies that P converges to zero at a finite rate as x for $t\rightarrow \infty$.

Conversely, $(P,x,y,z)\rightarrow E_0$ with P converges to zero at a finite rate as x and x, y both converge to zero faster than z as $t\rightarrow \infty$ implies u converges to some positive constant and $v,w\rightarrow 0$ as $t\rightarrow \infty$ and so $(u,v,w,z)\rightarrow E_{01}$ as $t\rightarrow \infty$.

• (iii) $(u,v,w,z)\rightarrow E_{02}$ as $t\rightarrow \infty$ implies that $u,z\rightarrow 0$ and $v\rightarrow \ v_{02},w\rightarrow \ w_{02}$ as $t\rightarrow \infty$. Since $v=\frac{x}{z}$ with $v\rightarrow v_{02}$ and $z\rightarrow 0$ as $t\rightarrow \infty$, it follows that x converges to zero at a finite rate as z for $t\rightarrow \infty$. Since $w=\frac{y}{z}$ with $w\rightarrow w_{02}$ and $z\rightarrow 0$ as $t\rightarrow \infty$, it follows that y converges to zero at a finite rate as z for $t\rightarrow \infty$. Also, $u=\frac{P}{x}$ and $u,x\rightarrow 0$ as $t\rightarrow \infty$, implies that P converges to zero faster than x as $t\rightarrow \infty$.

Conversely, $(P,x,y,z)\rightarrow E_0$ with P converges to zero faster than x and x, y converge to zero at a finite rate as z for $t\rightarrow \infty$ implies $u\rightarrow 0$ as $t\rightarrow \infty$ and u, w converge to some positive constant as $t\rightarrow \infty$ and so $(u,v,w,z)\rightarrow E_{02}$ as $t\rightarrow \infty$.

• (iv) $(u,v,w,z)\rightarrow E_{03}$ as $t\rightarrow \infty$ implies that $z\rightarrow 0$ and $u\rightarrow \ u_{03}, v\rightarrow \ v_{03},w\rightarrow \ w_{03}$ as $t\rightarrow \infty$. Since $v=\frac{x}{z}$ with $v\rightarrow v_{03}$ and $z\rightarrow 0$ as $t\rightarrow \infty$, it follows that x converges to zero at a finite rate as z for $t\rightarrow \infty$. Again, $w=\frac{y}{z}$ with $w\rightarrow w_{03}$ and $z\rightarrow 0$ as $t\rightarrow \infty$. Therefore, it follows that y converges to zero at a finite rate as z for $t\rightarrow \infty$. Also, $u=\frac{P}{x}$ and $u\rightarrow u_{03}$ and $x\rightarrow 0$ as $t\rightarrow \infty$, implies that P converges to zero at a finite rate as x for $t\rightarrow \infty$.

Conversely, $(P,x,y,z)\rightarrow E_0$ with P converges to zero at a finite rate as x and x, y converge to zero at a finite rate as z for $t\rightarrow \infty$ implies u, v, w converge to some positive constant as $t\rightarrow \infty$ and so $(u,v,w,z)\rightarrow E_{03}$ as $t\rightarrow \infty$.$\square$

Proof of Lemma 5.1.2

The variational matrix at $E_{00}$ is

$$\begin{aligned} V(E_{00})=\left[ \begin{array}{cccc} 1+D_1-\frac{m_1}{a_1}+\frac{m_2}{a_2} &{} 0 &{} 0 &{}0 \\ 0 &{}D_3-D_1+\frac{h}{c}+\frac{m_1}{a_1}-\frac{m_2}{a_2} &{}0 &{}0\\ 0 &{}\frac{m_2}{a_2} &{} -\mu -D_2+D_3+\frac{h}{c} &{}0\\ 0 &{}0 &{}0 &{}-D_3-\frac{h}{c} \end{array} \right] \end{aligned}$$

(6)

The eigenvalues are $\rho _{01}=-D_3-\frac{h}{c}<0, \rho _{02}=-\mu -D_2+D_3+\frac{h}{c}, \rho _{03}=D_3-D_1+\frac{h}{c}+\frac{m_1}{a_1}-\frac{m_2}{a_2}$ and $\rho _{04}=1+D_1-\frac{m_1}{a_1}+\frac{m_2}{a_2}$.

If $\frac{m_1}{a_1}-\frac{m_2}{a_2}-D_1>0$, then $\rho _{03}>0$.

If $\frac{m_1}{a_1}-\frac{m_2}{a_2}-D_1<0$, then $\rho _{04}>0$

and if $\frac{m_1}{a_1}-\frac{m_2}{a_2}-D_1=0$, then $\rho _{03}$ and $\rho _{04}>0$. Therefore, the system (3) is always unstable at $E_{00}$. $\square$

Proof of Lemma 5.1.3

$$\begin{aligned} V(E_{01})=\left[ \begin{array}{cccc} U^0_u|_{E_{01}}&{} -\frac{m_2u_{01}^2}{a_2^2} &{} 0 &{}0 \\ 0 &{}D_3-D_1+\frac{h}{c}+\frac{m_1u_{01}}{a_1+u_{01}}-\frac{m_2}{a_2} &{}0 &{}0\\ 0 &{}\frac{m_2}{a_2} &{} -\mu -D_2+D_3+\frac{h}{c} &{}0\\ 0 &{}0 &{}0 &{}-D_3-\frac{h}{c} \end{array} \right] \end{aligned}$$

(7)

where $U^0_u|_{E_{01}}=1+D_1+\frac{m_2}{a_2}-m_1\frac{(a_1+u_{01})(1+2u_{01})-u_{0}(1+u_{01})}{(a_1+u_{01})^2}$.

The eigenvalues are $\rho _{11}=-D_3-\frac{h}{c}<0, \rho _{12}=-\mu -D_2+D_3+\frac{h}{c}, \rho _{13}=D_3-D_1+\frac{h}{c}+\frac{m_1u_{01}}{a_1+u_{01}}-\frac{m_2}{a_2}$ and $\rho _{14}=1+D_1+\frac{m_2}{a_2}-m_1\frac{(a_1+u_{01})(1+2u_{01})-u_{0}(1+u_{01})}{(a_1+u_{01})^2}$.

Therefore, the system (3) is stable at $E_{01}$ if $h>c(D_3-D_2-\mu ), A_{01}>1+D_1+\frac{m_2}{a_2}$ and $u_{01}<\frac{a_1\{a_2c(D_1-D_3)+m_2c-ha_2\}}{a_2c(m_1+D_3-D_1)+ha_2-m_2c}$, where $A_{01}=m_1\frac{(a_1+u_{01})(1+2u_{01})-u_{01}(1+u_{01})}{(a_1+u_{01})^2}$. $\square$

Proof of Lemma 5.1.4

$$\begin{aligned} V(E_{02})=\left[ \begin{array}{cccc} 1+D_1-\frac{m_1}{a_1}+\frac{m_2}{a_2+v_{02}}&{} 0 &{} 0 &{}0 \\ \frac{m_1v_{02}}{a_1} &{}V^0_v|_{E_{02}} &{}-\mu v_{02} &{}-\frac{v_{02}h}{c^2}\\ 0 &{}\frac{m_2a_2}{(a_2+v_{02})^2} &{} W^0_w|_{E_{02}} &{}-\frac{w_{02}h}{c^2}\\ 0 &{}0 &{}0 &{}\mu w_{02}-D_3-\frac{h}{c} \end{array} \right] \end{aligned}$$

(8)

where $V^0_v|_{E_{02}}=D_3-D_1-\mu w_{02}+\frac{h}{c}-\frac{m_2a_2}{(a_2+v_{02})^2}$ and $W^0_w|_{E_{02}}=-\mu (1+2w_{02})+D_3-D_2+\frac{h}{c}-\frac{m_3w_{02}(2a_3+w_{02})}{(a_3+w_{02})^2}$.

Two eigenvalues are $\rho _{21}=w_{02}-D_3-\frac{h}{c}$ and $\rho _{22}=1+D_1-\frac{m_1}{a_1}+\frac{m_2}{a_2+v_{02}}$.

The other two eigenvalues are given by $\rho ^2-\rho (V^0_v|_{E_{02}}+W^0_w|_{E_{02}})+V^0_v|_{E_{02}}W^0_w|_{E_{02}}+\frac{\mu v_{02}m_2a_2}{(a_2+v_{02})^2}=0$.

Therefore, the system (3) is locally asymptotically stable if $h>c(\mu w_{02}-D_3), v_{02}>\frac{a_1m_2-a_2\{m_1-a_1(1+D_1)\}}{m_1-a_1(1+D_1)}, V^0_v|_{E_{02}}+W^0_w|_{E_{02}}<0$ and $V^0_v|_{E_{02}}W^0_w|_{E_{02}}>0$. $\square$

Proof of Lemma 5.1.5

$$\begin{aligned} V(E_{03})=\left[ \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} U^0_u|_{E_{03}}&{} U^0_v|_{E_{03}} &{} 0 &{}U^0_z|_{E_{03}} \\ V^0_u|_{E_{03}} &{}V^0_v|_{E_{03}} &{}V^0_w|_{E_{03}} &{}V^0_z|_{E_{03}}\\ 0 &{}W^0_v|_{E_{03}} &{} W^0_w|_{E_{03}} &{}W^0_z|_{E_{03}}\\ 0 &{}0 &{}0 &{}Z^0_z|_{E_{03}} \end{array} \right] \end{aligned}$$

(9)

where $U^0_u|_{E_{03}}=1+D_1+\frac{m_2}{a_2+v_{03}}-m_1\frac{(a_1+u_{03})(1+2u_{03})-u_{03}(1+u_{03})}{(a_1+u_{03})^2}$,

$U^0_v|_{E_{03}}=-\frac{m_2u_{03}}{(a_2+v_{03})^2},$

$U^0_z|_{E_{03}}=-u^2_{03}v_{03},$

$V^0_u|_{E_{03}}=\frac{m_1a_1v_{03}}{(a_1+u_{03})^2},$

$V^0_v|_{E_{03}}=D_3-D_1-\mu w_{03}+\frac{h}{c}+\frac{m_1u_{03}}{a_1+u_{03}}-\frac{m_2a_2}{(a_2+v_{03})^2},$

$V^0_w|_{E_{03}}=-\mu v_{03},$

$V^0_z|_{E_{03}}=-\frac{hv_{03}}{c^2},$

$W^0_v|_{E_{03}}=\frac{m_2a_2}{(a_2+v_{03})^2},$

$W^0_w|_{E_{03}}=D_3-D_2-\mu -2\mu w_{03}+\frac{h}{c}-\frac{m_3a_3}{(a_3+w_{03})^2},$

$W^0_z|_{E_{03}}=-\frac{hw_{03}}{c^2},$

$Z^0_z|_{E_{03}}=\mu w_{03}-\frac{h}{c}.$

One eigenvalue is $\mu w_{03}-\frac{h}{c},$ the three other eigenvalues are obtained form the equation

$\eta ^3+A_{03}\eta ^2+B_{03}\eta +C_{03}=0$, where

$A_{03}=-(U^0_u|_{E_{03}}+V^0_v|_{E_{03}}+W^0_w|_{E_{03}}), B_{03}=U^0_u|_{E_{03}}(V^0_v|_{E_{03}}+W^0_w|_{E_{03}})+V^0_v|_{E_{03}}W^0_w|_{E_{03}}-V^0_w|_{E_{03}}W^0_v|_{E_{03}}, C_{03}=U^0_v|_{E_{03}}V^0_u|_{E_{03}}W^0_v|_{E_{03}}-U^0_u|_{E_{03}}(V^0_v|_{E_{03}}W^0_w|_{E_{03}}-V^0_w|_{E_{03}}W^0_v|_{E_{03}})$.

Therefore, the system (3) is stable at $E_{03}$ if $w_{03}>\frac{h}{c\mu }, A_{01}>0$ and $A_{03}B_{03}>C_{03}$. $\square$

Proof of Lemma 5.2.1

(i) $(P,u,v,z)\rightarrow E_{10}$ as $t\rightarrow \infty$ implies that $P\rightarrow 1$ and $u,v,z\rightarrow 0$ as $t\rightarrow \infty$. Since $u=\frac{x}{z}$ and $u,z\rightarrow 0$ as $t\rightarrow \infty$, it follows that x converges to zero faster than z as $t\rightarrow \infty$. Since $v=\frac{y}{z}$ and $v,z\rightarrow 0$ as $t\rightarrow \infty$, it follows that y converges to zero faster than z as $t\rightarrow \infty$.

Conversely, $(P,x,y,z)\rightarrow E_1$ with $P\rightarrow 1$ as $t\rightarrow \infty$ and x, y converge to zero faster than z as $t\rightarrow \infty$ implies $u,v\rightarrow 0$ as $t\rightarrow \infty$ and so $(P,u,v,z)\rightarrow E_{10}$ as $t\rightarrow \infty$.

(ii) Again, $(P,u,v,z)\rightarrow E_{11}$ as $t\rightarrow \infty$ implies that $P\rightarrow 1$ and $u\rightarrow u_1,v\rightarrow v_1, z\rightarrow 0$ as $t\rightarrow \infty$. Since $u=\frac{x}{z}$ with $u\rightarrow u_1>0,z\rightarrow 0$ as $t\rightarrow \infty$, it follows that x converges to zero at a finite rate as $z\rightarrow 0$ for $t\rightarrow \infty$. Since $v=\frac{y}{z}$ and $v\rightarrow v_1>0,z\rightarrow 0$ as $t\rightarrow \infty$, it follows that y converges to zero at a fine rate as z for $t\rightarrow \infty$.

Conversely, $(P,x,y,z)\rightarrow E_1$ with $P\rightarrow 1$ as $t\rightarrow \infty$ and x, y both converge to zero at a finite rate as z for $t\rightarrow \infty$ implies u, v converge to some positive constants as $t\rightarrow \infty$ and so $(P,u,v,z)\rightarrow E_{11}$ as $t\rightarrow \infty$. $\square$

Proof of Lemma 5.2.2

The variational matrix at $E_{10}$ is

$$\begin{aligned} V(E_{10})=\left[ \begin{array}{cccc} -1 &{} 0 &{} 0 &{}0 \\ 0 &{}D_3-D_1-\frac{m_2}{a_2}+\frac{h}{c} &{}0 &{}0\\ 0 &{}\frac{m_2}{a_2} &{} D_3-D_2-\mu -\frac{m_3}{a_3}+\frac{h}{c} &{}0\\ 0 &{}0 &{}0 &{}-D_3-\frac{h}{c} \end{array} \right] \end{aligned}$$

(10)

At $E_{10}$ the eigenvalues of the variational matrix are $-1, D_3-D_1+\frac{h}{c}-\frac{m_2}{a_2},D_3-D_2-\mu +\frac{h}{c}-\frac{m_3}{a_3}$ and $-D_3-\frac{h}{c}$.

Therefore, the system is stable at $E_{10}$ if $h<\min \left\{ D_1-D_3+\frac{m_2}{a_2},\mu +D_2-D_3+\frac{m_3}{a_3}\right\} .$ $\square$

Proof of Lemma 5.2.3

The variational matrix at $E_{11}$ is

$$\begin{aligned} V(E_{11})=\left[ \begin{array}{cccc} -1 &{} 0 &{} 0 &{}-m_1u_1\\ 0 &{}V^1_u|_{E_{11}} &{}-\mu u_1 &{}-m_1a_1u_1^2+\frac{hu_1}{c^2}\\ 0 &{}\frac{m_2a_2}{(a_2+u_1)^2} &{}W^1|_{E_{11}} &{}-\frac{v_1h}{c^2}\\ 0 &{}0 &{}0 &{} \mu v_1-D_3-\frac{h}{c} \end{array} \right] \end{aligned}$$

(11)

where $V^1_u|_{E_{11}}=m_1u_1-D_1+D_3+\frac{h}{c}-v_1\mu -\frac{m_2a_2}{(a_2+u_1)^2}$

and $W^1|_{E_{11}}=-\mu -D_2+D_3+\frac{h}{c}-2v_1\mu -\frac{m_3a_3}{(a_3+v_1)^2}$.

The two eigenvalues are $-1$ and $v_1\mu -D_3-\frac{h}{c}$. Other two eigenvalues are given by the equation $\eta ^2+A_{11}\eta +B_{11}=0$,

where $A_{11}=\mu (1+3v_1)+D_2+D_1-2D_3-m_1u_1-\frac{2h}{c}+\frac{m_2a_2}{(a_1+u_1)^2}+\frac{m_3a_3}{(a_3+v_1)^2}$ and $B_{11}=\left\{ D_3-D_1-\mu v_1+m_1u_1+\frac{h}{c}-\frac{m_2a_2}{(a_3+u_1)^2}\right\} \left\{ D_3-D_2-\mu (1+2v_1)+\frac{h}{c}-\frac{m_3a_3}{(a_3+v_1)^2}\right\} +\frac{\mu m_2a_2u_1}{(a_2+u_1)^2}.$

Therefore, the system (4) is stable at $E_{11}$ if $h>c(\mu v_1-D_3)$ and $A_{11},B_{11}>0$. $\square$

Proof of Lemma 5.3.1

$(P,x,w,z)\rightarrow E_{20}$ as $t\rightarrow \infty$ implies that $P\rightarrow \frac{\lambda -D_1}{\lambda }, x\rightarrow \frac{\lambda -D_1}{\lambda ^2}, w\rightarrow w_1$ and $z\rightarrow 0$ as $t\rightarrow \infty$. Since $w=\frac{y}{z}$ and $z\rightarrow 0$ as $t\rightarrow \infty$, it follows that y converges to zero at a finite rate as z for $t\rightarrow \infty$.

Conversely, $(P,x,y,z)\rightarrow E_2$ with $P\rightarrow \frac{\lambda -D_1}{\lambda }, x\rightarrow \frac{\lambda -D_1}{\lambda ^2}, w\rightarrow w_1$ and y converge to zero at a finite rate as z for $t\rightarrow \infty$ implies w converges to some positive constant as $t\rightarrow \infty$ and so $(P,x,w,z)\rightarrow E_{20}$ as $t\rightarrow \infty$. $\square$

Proof of Lemma 5.3.2

The variational matrix at $E_{20}$ is

$$\begin{aligned} V(E_{20})=\left[ \begin{array}{cccc} 1-\frac{2(1-\lambda )}{\lambda }-\frac{a_1D_1^2}{m_1\lambda ^2} &{} -\frac{D_1^2}{m_1} &{} 0 &{}0\\ \frac{a_1D_1^2}{m_1} &{}\frac{-D_1(m_1-D_1)}{m_1} &{}0 &{}-\frac{m_2a_2\lambda .^2}{\lambda -D_1}\\ 0 &{}0 &{}W^2_w|_{E_{20}} &{}m_2+\frac{hw_1}{c^2}\\ 0 &{}0 &{}0 &{} \mu w_1-D_3-\frac{h}{c} \end{array} \right] \end{aligned}$$

(12)

where $W^2_w|_{E_{20}}=D_3-D_2-\mu (1+2w_1)+\frac{h}{c}-\frac{m_3a_3}{(a_3+w_1)^2}$.

The two eigenvalues are $\mu w_1-D_3-\frac{h}{c}$ and $D_3-D_2-\mu (1+2w_1)+\frac{h}{c}-\frac{m_3a_3}{(a_3+w_1)^2}$.

Other two eigenvalues are given by the equation $\eta ^2+A_{21}\eta +B_{21}=0$, where

$A_{21}=1+D_1-\frac{2D_1}{\lambda }-\frac{D_1^2(\lambda ^2-a_1^2)}{m_1\lambda ^2}$ and $B_{21}=\frac{a_1D_1^4}{m_1^2}+D_1(m_1-D_1)\left( \frac{1}{m_1}+\frac{a_1D_1^2}{m_1^2\lambda ^2}-\frac{2D_1}{m_1\lambda }\right)$.

Therefore, the system (5) is locally asymptotically stable at $E_{20}$ if $\mu w_1-D_3<\frac{h}{c}<\mu (1+2w_1)+D_2-D_3+\frac{m_3a_3}{(a_3+w_1)^2}$ and $A_{21},B_{21}>0$. $\square$

Proof of Lemma 5.4.1

Since $\lim _{t\rightarrow \infty }\sup [P(t)+x(t)+y(t)+z(t)]\le \frac{1}{D}$, where $D=\min \{D_1,D_2,D_3\}$, it follows that there exists $T_1>0$ and $M_i<\frac{1}{D}, (i=1,2,3)$, such that for all $t\ge T_1$, we have $x(t)\le M_1, y(t)\le M_2$ and $z(t)\le M_3$.

Now if $m_1<a_1$, then $\frac{dP}{dt}|_{P=P_1}\ge 0$ implies that there exists $T_2>0$ such that $P(t)\ge P_1=1-\frac{m_1}{a_1}>0$ for all $t\ge T_2$.

Again, $\frac{dP}{dt}|_{P=1}\le 0$ implies that there exists $T_3>0$ such that $P(t)\le 1$, for all $t\ge T_3$.

Therefore, $P_1\le P(t)\le 1$, for all $t\ge \max \{T_2,T_3\}$.

For $t>\max \{T_1,T_2,T_3\}$ we have $\frac{dx}{dt}\ge x\left( \frac{m_1P_1}{a_1x+P_1}-D_1-\frac{m_2}{a_2}\right)$

This implies $\frac{dx}{dt}\mid _{x=x_1}\ge 0$ for $t>\max \{T_1,T_2,T_3\}$ where $x_1=\frac{(a_1-m_1)(m_1-D_1)\left( a_2-\frac{m_2\lambda }{a_1D_1}\right) }{a_2(a_1-m_1)+a_1m_2}$.

If $\lambda <\min \left\{ D,\frac{a_1a_2D_1}{m_2}\right\}$, then $x_1>0,$ and so in this case there exists $T_4>0$ such that $x_1\le x(t)\le M_1$ for all $t>T_4$.

Again for $t>T_4$ we have

$\frac{dy}{dt}\ge \frac{m_2x_1z}{a_2M_3+x_1}-(\mu +D_2)M_2-\frac{m_3M_2}{a_3}$.

Therefore, $\frac{dy}{dt}>0$ if $z(t)>\frac{M_2(a_2M_3+x_1)\{a_3(\mu +D_2)+m_3\}}{a_3m_2x_1}>0$, for all $t>T_4$.

Let there exists $z_1$ such that $\frac{M_2(a_2M_3+x_1)\{a_3(\mu +D_2)+m_3\}}{a_3m_2x_1}<z_1<M_3.$

Therefore, $\frac{dy}{dt}>0$ for $z(t)\ge z_1>0 \forall t \ge T_4$ and so in this case $\exists T_5>0$ and $0<y_1<M_2$ such that $y(t)\ge y_1, \forall t\ge T_5.$

Therefore, $\forall t\ge T_5$, if $z(t)\ge z_1$ then $y_1\le y(t)\le M_2$ and $z_1\le z(t)\le M_3$.

Let $T=\max \{T_1,T_2,\ldots ,T_5\}$. Then for $t>T$ if $D_1<m_1<a_1$ and $\lambda <\min \left\{ D,\frac{a_1a_2D_1}{m_2}\right\}$ hold, then there exists finite positive real numbers $P_1, x_1, y_1, z_1, M_1, M_2, M_3$ with $M_1+M_2+M_3<\frac{1}{D}, P_1=1-\frac{m_1}{a_1}, x_1=\frac{(a_1-m_1)(m_1-D_1)\left( a_2-\frac{m_2\lambda }{a_1D_1}\right) }{a_2(a_1-m_1)+a_1m_2}$ and $z_1>\frac{M_2(a_2M_3+x_1)\{a_3(\mu +D_2)+m_3\}}{a_3m_2x_1}$ such that $P_1\le P(t)\le 1, x_1 \le x(t) \le M_1, y_1 \le y(t) \le M_2, z_1 \le z(t) \le M_3$. $\square$

Proof of Lemma 5.4.2

The characteristic equation of the variational matrix at $E^*$ is

$u^4+Q_1u^3+Q_2u^2+Q_3u+Q_4=0$, where

$Q_1=-F^1_P|_{E^*}-F^2_x|_{E^*}-F^3_y|_{E^*}-F^4_z|_{E^*}, Q_2=(F^1_P|_{E^*}+F^2_x|_{E^*})(F^3_y|_{E^*}+F^4_z|_{E^*})+F^1_P|_{E^*}F^2_x|_{E^*}-F^1_x|_{E^*}F^2_P|_{E^*}+F^3_y|_{E^*}F^4_z|_{E^*}-\mu F^3_z|_{E^*}, Q_3=(\mu F^3_z|_{E^*}-F^3_y|_{E^*}F^4_z|_{E^*})(F^1_P|_{E^*}+F^2_x|_{E^*})+(F^1_x|_{E^*}F^2_P|_{E^*}-F^1_P|_{E^*}F^2_x|_{E^*})(F^3_y|_{E^*}+F^4_z|_{E^*})-\mu F^2_z|_{E^*}F^3_x|_{E^*}$

and $Q_4=\mu F^1_P|_{E^*}F^2_z|_{E^*}F^3_x|_{E^*}+(F^1_P|_{E^*}F^2_x|_{E^*}-F^1_x|_{E^*}F^2_P|_{E^*})(F^3_y|_{E^*}F^4_z|_{E^*}-\mu F^3_z|_{E^*})$, with

$F^1_P|_{E^*}=1-2P^*-\frac{m_1a_1x^{*2}}{(a_1x^*+P^*)^2}; F^1_x|_{E^*}=-\frac{m_1P^{*2}}{(a_1x^*+P^*)^2}; F^2_P|_{E^*}=\frac{m_1a_1x^{*2}}{(a_1x^*+P^*)^2}; F^2_x|_{E^*}=\frac{m_1P^{*2}}{(a_1x^*+P^*)^2}-D_1-\frac{m_2a_2z^{*2}}{(a_2z^*+x^*)^2}; F^2_z|_{E^*}=-\frac{m_2x^{*2}}{(a_2z^*+x^*)^2}; F^3_x|_{E^*}=\frac{m_2a_2z^{*2}}{(a_2z^*+x^*)^2}; F^3_y|_{E^*}=-\mu -D_2-\frac{m_3a_3z^{*2}}{(a_3z^*+y^*)^2}; F^3_z|_{E^*}=\frac{m_2x^{*2}}{(a_2z^*+x^*)^2}-\frac{m_3y^{*2}}{(a_3z^*+y^*)^2}; F^4_y|_{E^*}=\mu ; F^4_z|_{E^*}=-D_3-\frac{hc}{(c+z^*)^2}$.

The roots of this equation will have negative real parts if $Q_1>0$ and $Q_1Q_2>Q_3$ and $Q_1Q_2Q_3=Q_3^2+Q_1^2Q_4$.$\square$