Allee effect can simplify the dynamics of a prey-predator model

Abstract

In this work, we investigate a prey-predator model which includes the Allee effect phenomena in prey growth function, density dependent death rate for predators and ratio dependent functional response. we fulfill a comprehensive bifurcation analysis, constructing the two-parametric bifurcation diagrams which describes the effect of density dependent death rate parameter, and also show possible phase portraits. We have also investigated the model in the absence of Allee effect and corresponding bifurcation diagram has been presented to show the dynamical changes in the system. Then we compare the properties of the ratio dependent prey-predator model with and without the Allee effect and show that Allee effect has a significant role in the dynamics. Allee effect can preserve local extinction of populations and suppress the stability of interior equilibrium point. Finally, all the analytical results are validated with the help of numerical simulations.

Appendices

A Appendix 1

Transversality conditions for saddle-node bifurcation: Let V and W be the eigenvectors of $J_{E^{*}_{SN}}$ and $\left[ J_{E^{*}_{SN}}\right] ^{T}$ corresponding to zero eigenvalue respectively.

$$\begin{aligned} V= \left[ \begin{array}{cc} v_{1} \\ v_{2} \end{array}\right] ,\ W= \left[ \begin{array}{cc} w_{1} \\ w_{2} \end{array}\right] . \end{aligned}$$

Then V should satisfy the following matrix equation

$$\begin{aligned} \left[ \begin{array}{cc} T_{11}(u_{sn*}, v_{sn*})-a_{22}(u_{sn*}, v_{sn*}) &{} -a_{12}(u_{sn*}, v_{sn*}) \\ a_{22}(u_{sn*}, v_{sn*}) &{} a_{12}(u_{sn*}, v_{sn*})-D-2E_{SN}v_{sn*} \end{array}\right] \left[ \begin{array}{cc} v_{1} \\ v_{2} \end{array}\right] =\left[ \begin{array}{cc} 0 \\ 0 \end{array}\right] , \end{aligned}$$

where $T_{11}(u_{sn*}, v_{sn*})=(1-u_{sn*})(\frac{u_{sn*}}{A}-1)-u_{sn*}(\frac{u_{sn*}}{A}-1)+\frac{u_{sn*}}{A}(1-u_{sn*})$, $a_{12}(u_{sn*}, v_{sn*}) =\frac{Bu_{sn*}^2(u_{sn*}^2-Cv_{sn*}^2)}{(u_{sn*}^2+Cv_{sn*}^2)^2}$, $a_{22}(u_{sn*}, v_{sn*})=\frac{2BCu_{sn*}v_{sn*}^3}{(u_{sn*}^2+Cv_{sn*}^2)^2}$. Multiplying the second row by $T_{11}-a_{22}$, first row by $a_{22}$ and substracting the first row from above matrix, we get

$$\begin{aligned} \left[ \begin{array}{cccccc} T_{11}(u_{sn*}, v_{sn*}){-}a_{22}(u_{sn*}, v_{sn*}) &{} -a_{12}(u_{sn*}, v_{sn*}) \\ 0&{} (T_{11}(u_{sn*}, v_{sn*})\\ &{} -a_{22}(u_{sn*}, v_{sn*}))(a_{12}(u_{sn*}, v_{sn*})\\ &{} -D-2E_{SN}v_{sn*})\\ &{}+a_{12}(u_{sn*}, v_{sn*})a_{22}(u_{sn*}, v_{sn*}) \end{array}\right] \left[ \begin{array}{cc} v_{1} \\ v_{2} \end{array}\right] =\left[ \begin{array}{cc} 0 \\ 0 \end{array}\right] . \end{aligned}$$

To get the eigenvector the term $(T_{11}(u_{sn*}, v_{sn*})-a_{22}(u_{sn*}, v_{sn*}))(a_{12}(u_{sn*}, v_{sn*})-D-2E_{SN}v_{sn*})+a_{12}(u_{sn*}, v_{sn*})a_{22}(u_{sn*}, v_{sn*})=0$, which implies

$$\begin{aligned} (a_{12}(u_{sn*}, v_{sn*})-D-2E_{SN}v_{sn*})=-\frac{a_{22(D-2E_{SN}v_{sn*})}}{T_{11}(u_{sn*}, v_{sn*})} \end{aligned}$$

(11)

and from above matrix equation, we have

$$\begin{aligned} a_{22}(u_{sn*}, v_{sn*})v_{1}+(a_{12}(u_{sn*}, v_{sn*})-D-2E_{SN}v_{sn*})v_{2}=0. \end{aligned}$$

(12)

Using Eqs. (11) and (12), we get

$$\begin{aligned} V=\Big [1,\frac{T_{11}(u_{sn*}, v_{sn*})}{D+2E_{SN}v_{sn*}} \Big ]^T. \end{aligned}$$

Proceeding in a similar way the eigenvector of $[J_{E_{SN}^{*}}]^{T}$ is given by,

$$\begin{aligned} W=\Big [-1,\frac{a_{12}(u_{sn*}, v_{sn*})T_{11}(u_{sn*}, v_{sn*})}{a_{22}(u_{sn*}, v_{sn*})(D+2E_{SN}v_{sn*})} \Big ]^T \end{aligned}$$

Let $F(u,v)=(F_{1}(u,v),F_{2}(u,v))^{T}$. Then $F_{E}(u,v)=(F_{1E}(u,v),F_{2E}(u,v))^{T}$. We can find from system (3)–(4), $F_{1E} |_{(E_{SN}^{*};E_{SN})}=\frac{dF_{1}}{dE}|_{(E_{SN}^{*};E_{SN})}=0$ and $F_{2E} |_{(E_{SN}^{*};E_{SN})}=\frac{dF_{2}}{dE}|_{(E_{SN}^{*};E_{SN})}=-v_{sn*}^2$ and the first transversality condition for saddle node bifurcation becomes

$$\begin{aligned} W^TF_{E}(u, v)\Big |_{[E_{SN}^{*};E_{SN}]}= & {} \left[ \begin{array}{cc} -1 &{} \frac{a_{12}(u_{sn*}, v_{sn*})T_{11}(u_{sn*}, v_{sn*})}{a_{22}(u_{sn*}, v_{sn*})(D+2E_{SN}v_{sn*})}\\ \end{array}\right] \left[ \begin{array}{cc} 0\\ -v_{sn*}^2 \end{array}\right] \\= & {} -v_{sn*}^2\frac{a_{12}(u_{sn*}, v_{sn*})T_{11}(u_{sn*}, v_{sn*})}{a_{22}(u_{sn*}, v_{sn*})(D+2E_{SN}v_{sn*})}. \end{aligned}$$

Now for second transversality condition,

$D^2F(u,v)(V,V)=\displaystyle \sum _{i,j=1}^{2}\frac{\partial ^2F(u,v)}{\partial u_i \partial u_j}v_i v_j$, where $(u,v)=(u_1,u_2)$(say). Then,

$$\begin{aligned} D^2 \left( \begin{array}{cc} F_1(u,v) \\ F_2(u,v) \\ \end{array} \right) (V,V)= & {} \left( \begin{array}{cc} \displaystyle \sum _{i,j=1}^{2}\frac{\partial ^2F_1(u,v)}{\partial u_i \partial u_j}v_i v_j \\ \displaystyle \sum _{i,j=1}^{2}\frac{\partial ^2F_2(u,v)}{\partial u_i \partial u_j}v_i v_j \\ \end{array} \right) \\= & {} \left( \begin{array}{cc} F_{1u_{1}^2}v_{1}^2+2F_{1u_1 u_2}v_1 v_2+F_{1u_{2}^2}v_{2}^2 \\ F_{2u_{1}^2}v_{1}^2+2F_{2u_1 u_2}v_1 v_2+F_{2u_{2}^2}v_{2}^2 \\ \end{array} \right) , \end{aligned}$$

where $V=(v_{1},v_{2})^T$, $F_{1u_iu_j}=\frac{\partial ^2F_1}{\partial u_i\partial u_j}$ for $i,j=1,2$ and similarly for $F_2$.

Using the equilibrium relation from Eq. (6), we get

$$\begin{aligned}&D^{2}F(E_{SN}^{*};E_{SN})(V,V)\\&\quad {=}\left( \begin{array}{cc} I_{0}{-}v_{sn*}^3S{+}2Su_{sn*}v_{sn*}^2\left( \frac{T_{11}(u_{sn*}, v_{sn*})}{D+2E_{SN}v_{sn*}}\right) -Su^2_{sn*}v_{sn*}\left( \frac{T_{11}(u_{sn*}, v_{sn*})}{D+2E_{SN}v_{sn*}}\right) ^2 \\ v_{sn*}^3S{-}2Su_{sn*}v_{sn*}^2\left( \frac{T_{11}(u_{sn*}, v_{sn*})}{D{+}2E_{SN}v_{sn*}}\right) +(Su^2_{sn*}v_{sn*}-2E_{SN})\left( \frac{T_{11}(u_{sn*}, v_{sn*})}{D+2E_{SN}v_{sn*}}\right) ^2 \end{array}\right) , \end{aligned}$$

where $I_{0}=-\frac{6u_{sn*}}{A}+\frac{2(1+A)}{A}$, $S=\frac{2BC(Cv_{sn*}^2-3u_{sn*}^2)}{(u_{sn*}^2+Cv_{sn*}^2)^3}$. We find the expression

$$\begin{aligned} W^{T}D^{2}&F&(E_{SN}^{*};E_{SN})(V,V)\\ =&I_{0}&+\frac{T_{11}Sv_{sn*}}{a_{22}(u_{sn*}, v_{sn*})}.\left[ \left( \frac{T_{11}(u_{sn*}, v_{sn*})}{D+2E_{SN}v_{sn*}}\right) u_{sn*}-v_{sn*}\right] ^2\nonumber \\- & {} \frac{2E_{SN}a_{12}(u_{sn*}, v_{sn*})}{a_{22}(u_{sn*}, v_{sn*})}\left( \frac{T_{11}(u_{sn*}, v_{sn*})}{D+2E_{SN}v_{sn*}}\right) ^3. \end{aligned}$$

B Appendix 2

We calculate the first lyapunov number for stability of Hopf-bifurcation.

We translate the equilibrium point $E_{1*}(u_{1*}, v_{1*})$ to origin with new coordinate By setting $x=u-u_{1*} , y=v-v_{1*}$. Then new system in coordinate (x, y) has power series expansion as given below

$$\begin{aligned} {\dot{x}}= & {} ax+by+(a_{20}x^2+a_{11}xy+a_{02}y^2)+(a_{30}x^3+a_{12}x^2y+a_{21}xy^2)\nonumber \\&+Q_{1}(|x, y|^4),\nonumber \\ {\dot{y}}= & {} cx+dy+(b_{20}x^2+b_{11}xy+b_{02}y^2)+(b_{30}x^3+b_{21}x^2y+b_{12}xy^2)\nonumber \\&+ Q_{2}(|x, y|^4). \end{aligned}$$

Where

$$\begin{aligned} a= & {} \Big [\frac{\partial F_{1}(u, v)}{\partial u}\Big ] \Big |_{(E_{1*};B_{H})}=\frac{-3u_{1*}^2}{A}+2u_{1*}\left( \frac{1}{A}+1\right) -1-\frac{2B_{H}Cu_{1*}v_{1*}^3}{(u_{1*}^2+Cv_{1*}^2)^2},\\ b= & {} \Big [\frac{\partial F_{1}(u, v)}{\partial v}\Big ] \Big |_{(E_{1*};B_{H})}=-\frac{B_{H}u_{1*}^2(u_{1*}^2-Cv_{1*}^2)}{(u_{1*}^2+Cv_{1*}^2)^2},\\ c= & {} \Big [\frac{\partial F_{2}(u, v)}{\partial u}\Big ] \Big |_{(E_{1*};B_{H})}= \frac{2B_{H}Cu_{1*}v_{1*}^3}{(u_{1*}^2+Cv_{1*}^2)^2},\\ d= & {} \Big [\frac{\partial F_{2}(u, v)}{\partial v}\Big ] \Big |_{(E_{1*};B_{H})}=\frac{B_{H}u_{1*}^2(u_{1*}^2-Cv_{1*}^2)}{(u_{1*}^2+Cv_{1*}^2)^2}-D-2Ev_{1*},\\ a_{20}= & {} \frac{1}{2}\Big [\frac{\partial ^2 F_{1}(u, v)}{\partial u^2} \Big ] \Big |_{(E_{1*};B_{H})}=\frac{-3u_{1*}}{A}+\left( \frac{1}{A}+1\right) -\frac{B_{H}Cv_{1*}^3(Cv_{1*}^2-3u_{1*}^2)}{(u_{1*}^2+Cv_{1*}^2)^3}, \\ a_{11}= & {} \Big [\frac{\partial ^2 F_{1}(u, v)}{\partial u \partial v} \Big ]\Big |_{((E_{1*};B_{H})}=\frac{2B_{H}Cu_{1*}v_{1*}^2(Cv_{1*}^2-3u_{1*}^2)}{(u_{1*}^2+Cv_{1*}^2)^3},\\ a_{02}= & {} \frac{1}{2}\Big [\frac{\partial ^2 F_{1}(u, v)}{\partial v^2}\Big ] \Big |_{(E_{1*};B_{H})}=\frac{-B_{H}Cu_{1*}^2v_{1*}(Cv_{1*}^2-3u_{1*}^2)}{(u_{1*}^2+Cv_{1*}^2)^3},\\ a_{30}= & {} \frac{1}{6}\Big [\frac{\partial ^3 F_{1}(u, v)}{\partial u^3} \Big ] \Big |_{(E_{1*};B_{H})}=-\frac{1}{A}-\frac{B_{H}Cu_{1*}v_{1*}^3(u_{1*}^2-Cv_{1*}^2)}{3(u_{1*}^2+Cv_{1*}^2)^4},\\ a_{21}= & {} \frac{1}{2}\Big [\frac{\partial ^3 F_{1}(u, v)}{\partial u^2 \partial v} \Big ] \Big |_{(E_{1*};B_{H})}=-\frac{B_{H}Cv_{1*}^2(9u_{1*}^4-14Cu_{1*}^2v_{1*}^2+C^2v_{1*}^4)}{(u_{1*}^2+Cv_{1*}^2)^4},\\ a_{12}= & {} \frac{1}{2}\Big [\frac{\partial ^3 F_{1}(u, v)}{\partial u\partial v^2} \Big ] \Big |_{(E_{1*};B_{H})}=\frac{B_{H}Cv_{1*}(6u_{1*}^5-16Cu_{1*}^3v_{1*}^2+2C^2u_{1*}v_{1*}^4)}{(u_{1*}^2+Cv_{1*}^2)^4},\\ a_{03}= & {} \frac{1}{6}\Big [\frac{\partial ^3 F_{1}(u, v)}{\partial v^3} \Big ] \Big |_{(E_{1*};B_{H})}=\frac{-B_{H}Cu_{1*}^2(6Cu_{1*}^2v_{1*}^2-C^2v_{1*}^4-u_{1*}^4)}{(u_{1*}^2+Cv_{1*}^2)^4},\\ b_{20}= & {} \frac{1}{2}\Big [\frac{\partial ^2 F_{2}(u, v)}{\partial u^2} \Big ] \Big |_{(E_{1*};B_{H})}=\frac{B_{H}Cv_{1*}^3(Cv_{1*}^2-3u_{1*}^2)}{(u_{1*}^2+Cv_{1*}^2)^3},\\ b_{11}= & {} \Big [\frac{\partial ^2 F_{2}(u, v)}{\partial u \partial v} \Big ] \Big |_{(E_{1*};B_{H})}=\frac{-2B_{H}Cu_{1*}v_{1*}^2(Cv_{1*}^2-3u_{1*}^2)}{(u_{1*}^2+Cv_{1*}^2)^3},\\ b_{02}= & {} \frac{1}{2}\Big [\frac{\partial ^2 F_{2}(u, v)}{\partial v^2} \Big ] \Big |_{(E_{1*};B_{H})}=\frac{B_{H}Cu_{1*}^2v_{1*}(Cv_{1*}^2-3u_{1*}^2)}{(u_{1*}^2+Cv_{1*}^2)^3}-E, \\ b_{30}= & {} \frac{1}{6}\Big [\frac{\partial ^3 F_{2}(u, v)}{\partial u^3} \Big ] \Big |_{(E_{1*};B_{H})}=-\frac{1}{A}+\frac{4B_{H}Cu_{1*}v_{1*}^3(Cv_{1*}^2-u_{1*}^2)}{(u_{1*}^2+Cv_{1*}^2)^4},\\ b_{21}= & {} \frac{1}{2}\Big [\frac{\partial ^3 F_{2}(u, v)}{\partial u^2 \partial v} \Big ] \Big |_{(E_{1*};B_{H})}=-\frac{B_{H}Cv_{1*}^2(9u_{1*}^4-14Cu_{1*}^2v_{1*}^2+C^2v_{1*}^4)}{(u_{1*}^2+Cv_{1*}^2)^4}, \\ b_{12}= & {} \frac{1}{2}\Big [\frac{\partial ^3 F_{2}(u, v)}{\partial u \partial v^2} \Big ] \Big |_{(E_{1*};B_{H})}=\frac{B_{H}Cv_{1*}(6u_{1*}^5-16Cu_{1*}^3v_{1*}^2+2C^2u_{1*}v_{1*}^4)}{(u_{1*}^2+Cv_{1*}^2)^4},\\ b_{03}= & {} \frac{1}{6}\Big [\frac{\partial ^3 F_{2}(u, v)}{\partial v^3} \Big ] \Big |_{(E_{1*};B_{H})}=\frac{B_{H}Cu_{1*}^2(6Cu_{1*}^2v_{1*}-C^2v_{1*}^4-u_{1*}^4)}{(u_{1*}^2+Cv_{1*}^2)^4}. \end{aligned}$$

We can calculate Lypunov first coefficient by using (13) as given below

$$\begin{aligned} l_{1}= & {} \frac{-3\pi }{2bD_{1}^{\frac{3}{2}}}\big [ac(a^2_{11}+a_{11}b_{02}+a_{02}b_{11})+ab(b_{11}^2+a_{20}b_{11}+a_{11}b_{02})\\&+\,c^2(a_{11}a_{02}+2a_{02}b_{02})\\&-\,2ac(b_{02}^2-a_{20}a_{02})-2ab(a_{20}^2-b_{20}b_{02})-b^2(2a_{20}b_{20}+b_{11}b_{20})+(bc-2a^2)\\&(b_{11}b_{02}-a_{11}a_{20})-(a^2+bc)\lbrace 3(cb_{03-}ba_{30})+2a(a_{21}+b_{12)}+(ca_{12}-bb_{21})\rbrace \big ], \end{aligned}$$

where $D_1=ad-bc$. we get the expression for Lypunov first coefficient as:

$$\begin{aligned} l_{1}=\frac{3\pi }{2bD_1^\frac{3}{2}}L. \end{aligned}$$

where

$$\begin{aligned} L= & {} \big [ac(a^2_{11}+a_{11}b_{02}+a_{02}b_{11})+ab(b_{11}^2+a_{20}b_{11}+a_{11}b_{02})+c^2(a_{11}a_{02}+2a_{02}b_{02})\\&-2ac(b_{02}^2-a_{20}a_{02})-2ab(a_{20}^2-b_{20}b_{02})-b^2(2a_{20}b_{20}+b_{11}b_{20})+(bc-2a^2)\\&(b_{11}b_{02}-a_{11}a_{20})-(a^2+bc)\lbrace 3(cb_{03-}ba_{30})+2a(a_{21}+b_{12)}+(ca_{12}-bb_{21})\rbrace \big ]. \end{aligned}$$

The limit cycle is unstable for $l_{1} >0$ and stable if $l_{1} <0$. Therefore, the Hopf-bifurcation is subcritical if $l_{1} >0$ and supercritical if $l_{1} <0$.

C Appendix 3

Transversality conditions for Bogdanov-Takens bifurcation : We transform the equilibrium point $E_{SN}^{*}$ to origin by $x=u-u_{sn*}, y=v-v_{sn*}$ and we get

$$\begin{aligned} \dot{x_{1}}= & {} a'x_{1}+b'x_{2}-\frac{2\lambda _{1}Cu_{sn*}v_{sn*}^3x_{1}}{(u_{sn*}^2+Cv_{sn*}^2)^2}-\frac{\lambda _{1}u_{sn*}^2(u_{sn*}^2-Cv_{sn*}^2)x_{2}}{(u_{sn*}^2+Cv_{sn*}^2)^2}\\&+\,\frac{p_{11}}{2}x_{1}^2+p_{12}x_{1}x_{2} + \frac{p_{22}}{2}x_{2}^2+...,\\ \dot{x_{2}}= & {} c'x_{1}+d'x_{2}+\frac{2\lambda _{1}Cu_{sn*}v_{sn*}^3x_{1}}{(u_{sn*}^2+Cv_{sn*}^2)^2}+\left( \frac{\lambda _{1}u_{sn*}^2(u_{sn*}^2-Cv_{sn*}^2)}{(u_{sn*}^2+Cv_{sn*}^2)^2}-2\lambda _{2}v_{sn*} \right) x_{2} \\&+\,\frac{q_{11}}{2}x_{1}^2+q_{12}x_{1}x_{2}+\frac{q_{22}}{2}x_{2}^2+...., \end{aligned}$$

where $a'$, $b'$, $c'$ ,$d'$ are the element of the Jacobian matrix evaluated at an equilibrium point $E_{SN}^{*}$ and $p_{11}$, $p_{12}$, $p_{22}$, $q_{11}$, $q_{12}$, $q_{22}$ are as follow

$$\begin{aligned} a'= & {} \Big [\frac{\partial F^{1}_{0}(u, v)}{\partial u}\Big ] \Big |_{(E_{SN}^{*};B_{BT};E_{BT})}=\frac{-3u_{sn*}^2}{A}+2u_{sn*}\left( \frac{1}{A}+1\right) \\&-1-\frac{2B_{BT}Cu_{sn*}v_{sn*}^3}{(u_{sn*}^2+Cv_{sn*}^2)^2},\\ b'= & {} \Big [\frac{\partial F^{1}_{0}(u, v)}{\partial v}\Big ] \Big |_{(E_{SN}^{*};B_{BT};E_{BT})}=-\frac{B_{BT}u_{sn*}^2(u_{sn*}^2-Cv_{sn*}^2)}{(u_{sn*}^2+Cv_{sn*}^2)^2},\\ c'= & {} \Big [\frac{\partial F^{2}_{0}(u, v)}{\partial u}\Big ] \Big |_{(E_{SN}^{*};B_{BT};E_{BT})}= \frac{2B_{BT}Cu_{sn*}v_{sn*}^3}{(u_{sn*}^2+Cv_{sn*}^2)^2},\\ d'= & {} \Big [\frac{\partial F^{2}_{0}(u, v)}{\partial v}\Big ] \Big |_{(E_{SN}^{*};B_{BT};E_{BT})}=\frac{B_{BT}u_{sn*}^2(u_{sn*}^2-Cv_{sn*}^2)}{(u_{1*}^2+Cv_{sn*}^2)^2}-D-2E_{BT}v_{sn*},\\ p_{11}= & {} \Big [\frac{\partial ^2 F^{1}_{0}(u, v)}{\partial u^2} \Big ] \Big |_{(E_{SN}^{*};B_{BT};E_{BT})}=\frac{-6u_{sn*}}{A}+2\left( \frac{1}{A}+1\right) \\&-\frac{2B_{BT}Cv_{sn*}^3(Cv_{sn*}^2-3u_{sn*}^2)}{(u_{sn*}^2+Cv_{sn*}^2)^3},\\ p_{12}= & {} \Big [\frac{\partial ^2 F^{1}_{0}(u, v)}{\partial u \partial v} \Big ]\Big |_{(E_{SN}^{*};B_{BT};E_{BT})}=\frac{2B_{BT}Cu_{sn*}v_{sn*}^2(Cv_{sn*}^2-3u_{sn*}^2)}{(u_{sn*}^2+Cv_{sn*}^2)^3},\\ p_{22}= & {} \Big [\frac{\partial ^2 F^{1}_{0}(u, v)}{\partial v^2}\Big ] \Big |_{(E_{SN}^{*};B_{BT};E_{BT})}=\frac{-2B_{BT}Cu_{sn*}^2v_{sn*}(Cv_{sn*}^2-3u_{sn*}^2)}{(u_{sn*}^2+Cv_{sn*}^2)^3}, \\ q_{11}= & {} \Big [\frac{\partial ^2 F^{2}_{0}(u, v)}{\partial u^2} \Big ] \Big |_{(E_{SN}^{*};B_{BT};E_{BT})}=\frac{2B_{BT}Cv_{sn*}^3(Cv_{sn*}^2-3u_{sn*}^2)}{(u_{sn*}^2+Cv_{sn*}^2)^3},\\ q_{12}= & {} \Big [\frac{\partial ^2 F^{2}_{0}(u, v)}{\partial u \partial v} \Big ] \Big |_{(E_{SN}^{*};B_{BT};E_{BT})}=\frac{-2B_{BT}Cu_{sn*}v_{sn*}^2(Cv_{sn*}^2-3u_{sn*}^2)}{(u_{sn*}^2+Cv_{sn*}^2)^3},\\ q_{22}= & {} \Big [\frac{\partial ^2 F^{2}_{0}(u, v)}{\partial v^2} \Big ] \Big |_{(E_{SN}^{*};B_{BT};E_{BT})}=\frac{2B_{BT}Cu_{sn*}^2v_{sn*}(Cv_{sn*}^2-3u_{sn*}^2)}{(u_{sn*}^2+Cv_{sn*}^2)^3}-2E_{BT}. \end{aligned}$$

Now we use affine transformation $y_{1}=x$, $y_{2}=a'x+b'y$ in (13) to get the new transformed system in $(y_{1}, y_{2})$ as:

$$\begin{aligned} \dot{y_{1}}= & {} y_{2}+\frac{u_{sn*}\lambda _{1}(a'u_{sn*}(u_{sn*}^2-Cv_{sn*}^2)-2b'Cv_{sn*}^3)}{b'(u_{sn*}^2+Cv_{sn*}^2)}y_{1}\\&-\frac{u_{sn*}^2\lambda _{1}(u_{sn*}^2-Cv_{sn*}^2)}{b'(u_{sn*}^2+Cv_{sn*}^2)}y_{2} +\Big (\frac{p_{11}}{2}-\frac{a'p_{12}}{b'} \\&+\frac{a'^2p_{22}}{2b'^2}\Big )y_{1}^2+\Big (\frac{p_{12}}{b'}-\frac{a'p_{22}}{b'^2}\Big )y_{1}y_{2} +\frac{p_{22}}{2b'^2}y_{2}^2. \\ \dot{y_{2}}= & {} (b'c'-a'd')y_{1}+(a'+d')y_{2}\\&+\Big (\frac{(a'-b')(a'\lambda _{1}u_{sn*}^2(u_{sn*}^2-Cv_{sn*}^2)-2\lambda _{1}b'Cu_{sn*}v_{sn*}^3 ) }{b'(u_{sn*}^2+Cv_{sn*}^2)} \\&+2a'\lambda _{2}v_{sn*}\Big ) y_{1} -\Big (2\lambda _{2}v_{sn*}+\frac{(a'-b')\lambda _{1}u_{sn*}^2(u_{sn*}^2-Cv_{sn*}^2)}{b'(u_{sn*}^2+Cv_{sn*}^2)}\Big )y_{2}\\&+\Big (\frac{1}{b'}(a'p_{12}+b'q_{12}) \\&-\frac{a'}{b'^2}(a'p_{22}+b'q_{22})\Big )y_{1}y_{2}+\Big (\frac{1}{2}(a'p_{11}+b'q_{11})\\&-\frac{a'}{b'}(a'p_{12}+b'q_{12})+\frac{a'^2}{2b'^2}(a'p_{22}\\&+b'q_{22})\Big )y_{1}^2+\Big (\frac{a'p_{22}+b'q_{22}}{2b'^2}\Big )y_{2}^2. \end{aligned}$$

which can be written as

$$\begin{aligned} \dot{y_{1}}= & {} y_{2}+\xi _{00}(\lambda )+\xi _{10}(\lambda )y_{1}+\xi _{01}(\lambda )y_{2}+\frac{1}{2}\xi _{20}(\lambda )y_{1}^2+\xi _{11}(\lambda )y_{1}y_{2}\\&+\frac{1}{2}\xi _{02}(\lambda )y_{2}^2+O_{1}(y_{1}y_{2}).\\ \dot{y_{2}}= & {} \eta _{00}(\lambda )+\eta _{10}(\lambda )y_{1}+\eta _{01}(\lambda )y_{2}+\frac{1}{2}\eta _{20}(\lambda )y_{1}^2+\eta _{11}(\lambda )y_{1}y_{2}\\&+\frac{1}{2}\eta _{02}(\lambda )y_{2}^2+O_{2}(y_{1}y_{2}).\\ \end{aligned}$$

where $\lambda =(\lambda _{1}, \lambda _{2})$

$$\begin{aligned} \xi _{00}(\lambda )= & {} 0, \xi _{10}(\lambda )=\frac{u_{sn*}\lambda _{1}(a'u_{sn*}(u_{sn*}^2-Cv_{sn*}^2)-2b'Cv_{sn*}^3)}{b'(u_{sn*}^2+Cv_{sn*}^2)},\\ \xi _{01}(\lambda )= & {} \frac{-u_{sn*}^2\lambda _{1}(u_{sn*}^2-Cv_{sn*}^2)}{b'(u_{sn*}^2+Cv_{sn*}^2)}\\ \xi _{20}(\lambda )= & {} 2\Big (\frac{p_{11}}{2}-\frac{a'p_{12}}{b'}+\frac{a'^2p_{22}}{2b'^2} \Big ), \xi _{11}(\lambda )=\Big (\frac{p_{12}}{b'}-\frac{a'p_{22}}{b'^2}\Big ),\\ \xi _{02}(\lambda )= & {} \frac{p_{22}}{b'^2}, \eta _{00}(\lambda )=0,\\ \eta _{10}(\lambda )= & {} (b'c'-a'd')\\&+\left( \frac{(a'-b')\lbrace a'\lambda _{1}u_{sn*}^2(u_{sn*}^2-Cv_{sn*}^2)-2\lambda _{1}b'Cu_{sn*}v_{sn*}^3 \rbrace }{b'(u_{sn*}^2+Cv_{sn*}^2)}+2a'\lambda _{2}v_{sn*}\right) \\ \eta _{01}(\lambda )= & {} (a'+d')-\Big (2\lambda _{2}v_{sn*}+\frac{(a'-b')\lambda _{1}u_{sn*}^2(u_{sn*}^2-Cv_{sn*}^2)}{b'(u_{sn*}^2+Cv_{sn*}^2)}\Big ) \\ \eta _{20}(\lambda )= & {} 2\Big (\frac{1}{2}(a'p_{11}+b'q_{11})-\frac{a'}{b'}(a'p_{12}+b'q_{12})+\frac{a'^2}{2b'^2}(a'p_{22}+b'q_{22})\Big ) \\ \eta _{11}(\lambda )= & {} \Big (\frac{1}{b'}(a'p_{12}+b'q_{12})-\frac{a'}{b'^2}(a'p_{22}+b'q_{22})\Big ), \eta _{02}(\lambda )=2\Big (\frac{a'p_{22}+b'q_{22}}{2b'^2}\Big ) \end{aligned}$$

Now

$$\begin{aligned} \xi _{20}(0)&=2\Big (\frac{p_{11}}{2}-\frac{a'p_{12}}{b'}+\frac{a'^2p_{22}}{2b'^2}\Big )_{\lambda =0},\nonumber \\ \eta _{11}(0)&=\Big (\frac{1}{b'}(a'p_{12}+b'q_{12})-\frac{a'}{b'^2}(a'p_{22}+b'q_{22}) \Big )_{\lambda =0},\nonumber \\ \eta _{20}(0)&=2\Big (\frac{1}{2}(a'p_{11}+b'q_{11})-\frac{a'}{b'}(a'p_{12}+b'q_{12})+\frac{a'^2}{2b'^2}(a'p_{22}+b'q_{22})\Big )_{\lambda =0}. \end{aligned}$$

(13)

To check the non degeneracy conditions of Bogdanov-Takens bifurcation we have to check the following quantities:

$$\begin{aligned} (i)\left[ \begin{array}{cc} a' &{} b'\\ c' &{} d' \end{array}\right] \ne 0_{2\times 2},\quad (ii) \xi _{20}(0)+ \eta _{11}(0)\ne 0, \quad (ii)\eta _{20}(0)\ne 0. \end{aligned}$$

The first condition is clearly satisfied. Also

$$\begin{aligned} \xi _{20}(0)+\eta _{11}(0)= & {} 2\Big (\frac{p_{11}}{2}-\frac{a'p_{12}}{b'}+\frac{a'^2p_{22}}{2b'^2}\Big )\\&+ \Big (\frac{1}{b'}(a'p_{12}+b'q_{12})-\frac{a'}{b'^2}(a'p_{22}+b'q_{22}) \Big ) \end{aligned}$$

D Appendix 4

Here we discuss the stability of the axial equilibrium point $E_{0}(0,0)$ of the model (9)–(10). Now we transform the variables to new variables by replacing $p=\frac{v}{u}$ and we get the following system

$$\begin{aligned} \frac{du}{dT}= & {} u(1-u)-\frac{B pu}{1+Cp^2} , \end{aligned}$$

(14)

$$\begin{aligned} \frac{dp}{dT}= & {} \frac{B p}{1+Cp^2}+\frac{B p^2}{1+Cp^2} -Dp-E p^2u-(1-u)p . \end{aligned}$$

(15)

We are interested here of the axial equilibrium points on the p-axis only. Axial equilibrium points on the p-axis are (0, 0) and $(0,\mu _{1,2})$, where $\mu _{1,2}$ are the two roots of the quadratic equation

$$\begin{aligned}&\frac{B}{1+Cp^2}+\frac{B p}{1+Cp^2} -D-1=0 \\&\quad \Rightarrow p=\frac{B\pm \sqrt{B^2+4C(1+D)(B-D-1)}}{2C(1+D)} \end{aligned}$$

So $\mu _{1}=\frac{B+ \sqrt{B^2+4C(1+D)(B-D-1)}}{2C(1+D)}$ and $\mu _{2}=\frac{B- \sqrt{B^2+4C(1+D)(B-D-1)}}{2C(1+D)}$. Eigenvalues of the Jacobian matrix evaluated at (0, 0) is 1 and $B-(1+D)$. Eigenvalues evaluated at $(0,\mu _{1,2})$ are $-\frac{B \mu _{1,2}}{1+C\mu _{1,2}^2}$ and $-D-1+\frac{B (1-C\mu _{1,2}^2+2 \mu _{1,2})}{(1+C\mu _{1,2}^2)^2}$. Now we discuss the different cases

• (i) If $(B-D-1)<0$ and $B^2+4C(1+D)(B-D-1)<0$, then $(0,\mu _{1,2})$ do not exist and (0, 0) is saddle point.

• (ii) If $(B-D-1)<0$ and $B^2+4C(1+D)(B-D-1)>0$, then $(0,\mu _{1,2})$ exist. (0, 0) is saddle point and $(0,\mu _{1,2})$ are stable points.

• (iii) If $(B-D-1)>0$ then $B^2+4C(1+D)(B-D-1)>B^2>0$. (0, 0) is unstable and only $(0,\mu _{1})$ exists.

We construct a bifurcation diagram on $B-E$ plane (see Fig. 10), which describes the above results and also the stability nature of origin. Two curves $(B-D-1)=0$ and $B^2+4C(1+D)(B-D-1)=0$ divide the bifurcation diagram into three subregions and corresponding phase portraits are also shown in Fig. 11.

Hence, by blowing-down back (14)–(15), the line $u=0$ is collapsed to the origin of the system (9)–(10). When $(0,\mu _{1,2})$ are stable, then invariant attracting curve is also mapped to an curve in the first quadrant which passes through the origin. So, the axial equilibrium point $E_{0}(0,0)$ of the model (9)–(10) is stable at that region.

Fig. 10

Bifurcation diagram in $B-E$ plane of the axial equilibrium point $E_{0}(0,0)$

Fig. 11

Stability of $E_{0}(0,0)$ corresponding to Fig.10. a for region A1, b for region A2, (c) for region A3