Hopf bifurcation analysis in a delayed Leslie–Gower predator–prey model incorporating additional food for predators, refuge and threshold harvesting of preys

Abstract

In this paper, we formulate and analyze a modified Leslie–Gower predator–prey model. Our model incorporates refuge of preys, additional fixed food for predators, harvesting of preys through a continuous threshold policy and a time delay as to account for predators maturity time. We first carry out a qualitative analysis of the model without time delay, showing existence of extinction, prey-free, predator-free and coexistence equilibria. We further study their stability conditions. Relying only on theoretical results of the model, we construct bifurcation diagrams involving refuge and harvest limit parameters. This led to summarize different scenarios for the model including elimination of one species or competition of both species that are proved possible. Furthermore, considering the time delay as bifurcation parameter, we analyze the stability of the coexistence equilibria and prove the system can undergoes a Hopf bifurcation. The direction of that Hopf bifurcation and the stability of the bifurcated periodic solution are determined by applying the normal form theory and the center manifold theorem. Numerical simulations are presented to illustrate our theoretical results.

Appendices

Proof of Theorem 1

We must show that each solution (x, y) of system (2)–(3), defined and continuous on $[-\tau ,A[$ where $A \in ]0,+\infty ]$, $x(t)>0$ and $y(t)>0$ for all $t\in [0,A[$. Suppose that it is not true. Then there exists a value of T in ]0, A[ such that for all $t\in [0,T[$, $x(t)>0$ and $y(t)>0$, and either $x(T)=0$ or $y(T)=0$.

For all $t\in [0,T[$ and from equations of (2), we have,

$$\begin{aligned}&x(t)>x(0)\exp \left( \displaystyle \int _{0}^{t}\left( \displaystyle r_{1}-b_{1}x(s)-q\lambda _{1}(1-m)y(s)\right. \right. \\&\quad \left. \left. -\frac{h}{h+x(s)-T_{1}}\right) \mathrm{d}s\right) \end{aligned}$$

and

$$\begin{aligned}&y(t)=y(0)exp\left( \displaystyle \int _{0}^{t}\left( \displaystyle r_{2}\right. \right. \\&\left. \left. \quad \left( 1 -\displaystyle \frac{y(s - \tau )}{q\alpha _{1}(1-m)x(s - \tau ) + (1 - q)\alpha _{A}K_{A}}\right) \right) \mathrm{d}s\right) . \end{aligned}$$

As x and y are defined and continuous on the compact $[-\tau ,T]$, there exists $M \ge 0$ such that:

$$\begin{aligned}&x(t)>x(0)exp\left( \displaystyle \int _{0}^{t}\left( \displaystyle r_{1}-b_{1}x(s)-a_{1}(1-m)y(s)\right. \right. \\&\quad \left. \left. -\frac{h}{h+x(s)-T_{1}}\right) \mathrm{d}s\right) \ge x(0)exp(-TM) \end{aligned}$$

and

$$\begin{aligned}&y(t)=y(0)exp\left( \displaystyle \int _{0}^{t}\left( \displaystyle r_{2}\right. \right. \\&\left. \left. \quad \left( 1 -\displaystyle \frac{y(s - \tau )}{q\alpha _{1}(1-m)x(s - \tau ) + (1 - q)\alpha _{A}K_{A}}\right) \right) \mathrm{d}s\right) \\&\quad \ge y(0)exp(-TM). \end{aligned}$$

Taking the limit as $t \rightarrow T$ gives

$$\begin{aligned} x(T)\ge x(0)exp(-TM)> 0 \end{aligned}$$

and

$$\begin{aligned} y(T)\ge y(0)exp(-TM)> 0, \end{aligned}$$

which contradicts the fact that either $x(T)= 0$ or $y(T)= 0$. Then, a solution of System (2)–(3) which starts in the positive quadrant ${\mathbb {R}}_{+}^2$ remains there.

Proof of Theorem 2

Using the first equation of System (2), we always have

$$\begin{aligned} \dot{x}(t)\le r_{1}x(t)\left( \displaystyle 1 -\displaystyle \frac{x(t)}{K_{1}}\right) . \end{aligned}$$

Applying a differential inequality [Hale, 1980] gives

$$\begin{aligned} {x}(t)\le \frac{1}{\displaystyle {\frac{1}{K_{1}} + \left( \displaystyle \frac{1}{x(0)}-\frac{1}{K_{1}}\right) e^{-r_{1}t}}}. \end{aligned}$$

Using assumption (5) gives $x(t)\le K_{1}$, for all positive value of t. It means that x(t) is bounded.

Let us use the second equation of System (3) to prove that y(t) is bounded. For all $t\ge 0$, we have

$$\begin{aligned} \dot{y}(t)\le r_{2}y(t). \end{aligned}$$

Integrating that differential inequality from $t-\tau$ to t gives

$$\begin{aligned} y(t-\tau )\ge y(t)e^{-r_{2}\tau }. \end{aligned}$$

Using the fact that $x(t)\le K_{1}$ gives the following differential inequality

$$\begin{aligned} \dot{y}(t)\le r_{2}y(t)\left( \displaystyle 1 -\displaystyle \frac{y(t)}{K}\right) \end{aligned}$$

where $K = \displaystyle \frac{(1-m)q\alpha _{1}K_{1}+(1-q)\alpha _{A}K_{A}}{e^{-r_{2}\tau }}$. Once more, applying a differential inequality [Hale, 1980] gives

$$\begin{aligned} {y}(t)\le \frac{1}{\displaystyle {\frac{1}{K} + \left( \displaystyle \frac{1}{y(0)}-\frac{1}{K}\right) e^{-r_{2}t}}}. \end{aligned}$$

Using assumption (6) gives $y(t)\le K$, for all positive value of t. Thus y(t) is bounded. One can easily verify that, using assumption (6), the inequalities $x(t)\le K_{1}$ and $e^{r_{2}\tau }\ge 1$ lead to $y(0)\le K$.

Proof of Theorem 3

A couple of variables (x, y) is an equilibrium of system (2)–(3) if it is a solution of the following systems on $[0,T_1]\times {\mathbb {R}}_{+}$ and on $[T_1,K_1]\times {\mathbb {R}}_{+}$, respectively.

$$\begin{aligned}&\left\{ \begin{array}{lll} 0 &{}=&{} r_{1}\left( 1 -\displaystyle \frac{x(t)}{K_{1}}\right) x(t) - q\lambda _{1}(1-m)x(t)y(t), \\ 0 &{}=&{} r_{2}\left( 1 -\displaystyle \frac{y(t)}{q\alpha _{1}(1-m)x(t) + (1 - q)\alpha _{A}K_{A}}\right) y(t), \end{array} \right. \nonumber \\ \end{aligned}$$

(20)

$$\begin{aligned}&\left\{ \begin{array}{lll} 0 &{}=&{} r_{1}\left( 1 -\displaystyle \frac{x(t)}{K_{1}}\right) x(t) - q\lambda _{1}(1-m)x(t)y(t) \\ &{}&{}\quad - \displaystyle \frac{h(x-T_{1})}{h+x-T_{1}}, \\ 0 &{}=&{} r_{2}\left( 1 -\displaystyle \frac{y(t)}{q\alpha _{1}(1-m)x(t) + (1 - q)\alpha _{A}K_{A}}\right) y(t). \end{array} \right. \nonumber \\ \end{aligned}$$

(21)

Firstly, we solve system (20) on $[0,T_1]\times {\mathbb {R}}_{+}$. Using the first equation of the system (20), we have $x=0$ or $x = \displaystyle \frac{r_{1}-q\lambda _{1}(1-m)y}{b_{1}}$.

Replacing x by 0 in the second equation of system (20) gives $y=0$ or $y=(1-q)\alpha _{A}K_A$. Then we have the equilibria $E_{1}^{0}$ and $E_{2}^{0}$.

Replacing x by $\displaystyle \frac{r_{1}-q\lambda _{1}(1-m)y}{b_{1}}$ in the second equation of system (20) gives $y=0$ or $y=(1-m)q\alpha _{1}x + (1-q)\alpha _{A}K_A$. If $y=0$, we have $x=\displaystyle \frac{r_{1}}{b_{1}}$. If $y=(1-m)q\alpha _{1}x + (1-q)\alpha _{A}K_A$, we have $x=\displaystyle \frac{r_{1}-\lambda _{1}(1-m)q(1-q)\alpha _{A}K_{A}}{b_{1}+q^{2}\lambda _{1}\alpha _{1}(1-m)^{2}}$ and $y=\displaystyle \frac{r_{1}\alpha _{1}(1-m)q+(1-m)b_{1}\alpha _{A}K_{A}}{b_{1}+q^{2}\lambda _{1}\alpha _{1}(1-m)^{2}}$. Then we have the equilibria $E_{3}^{0}$ and $E_{4}^{0}$.

Secondly, we solve system (21) on $[T_1,K_1]\times {\mathbb {R}}_{+}$.

Using the second equation of system (21) gives $y=0$ or $y=(1-m)q\alpha _{1}x + (1-q)\alpha _{A}K_A$. If $y=0$, we find that x is a solution of equation (10). Moreover, if we consider the function f defined by $f(x) = -b_{1}x^{3}+(r_{1}-b_{1}(h-T_{1}))x^{2}+(r_{1}(h-T_{1})-h)x+hT_{1}$, then we have $f(T_1) = hT_{1}(r_{1} - b_{1}T_{1})> 0$ using assumption (4). We also have $f(K_1) = -h(K_{1} - T_{1}) < 0$ because $T_{1}<K_{1}$. Thus, by the intermediate value theorem [9, 46], there exists at least one solution of equation (10). So we have the equilibria $E_{1}^{\varphi }$. If $y=(1-m)q\alpha _{1}x + (1-q)\alpha _{A}K_A$, x is a solution of equation (11). Moreover, if we consider the function g defined by $g(x) = A_{1}^{\varphi }x^{3}+ A_{2}^{\varphi }x^{2}+A_{3}^{\varphi }x+hT_{1}$, we have $g(T_{1}) = hT_{1}(r_{1} - b_{1}T_{1} - q\lambda _{1}(1 - m)(1 - q)\alpha _{A}K_A + q\alpha _{1}(1 - m)T_{1})$. We have $g(K_1) = -(K_{1}(K_{1} - T_{1}) + hT_1)(q\lambda _{1}(1 - m)(1 - q)\alpha _{A}K_A + q^{2}\lambda _{1}\alpha _{1}K_{1}(1 - m)^{2})< 0$ because $T_{1}<K_{1}$. Thus, if $r_{1} - b_{1}T_{1} - q\lambda _{1}(1 - m)(1 - q)\alpha _{A}K_{A} + q(1 - m)\alpha _{1}T_{1} > 0$ then $g(T_1)>0$. By the intermediate value theorem [9, 46], (11) has at least one solution in $[K_{1}, T_{1}]$. Thus we have the equilibrium $E_{2}^{\varphi }$.

Proof of Theorem 4

• 1. The Jacobian matrix of system (2)–(3) at the equilibrium $E_{1}^{0}$ is:

  $$\begin{aligned} J_{E_{1}^{0}}= \left( \begin{array}{cc} r_{1} &{}\quad 0\\ 0 &{}\quad r_{2} \end{array}\right) . \end{aligned}$$

  $J_{E_{1}^{0}}$ has two positive eigenvalues $(r_{1} \quad and \quad r_{2})$. So the equilibrium $E_{1}^{0}$ is an unstable node.

• 2. The Jacobian matrix of system (2)–(3) at the equilibrium $E_{2}^{0}$ is:

  $$\begin{aligned} J_{E_{2}^{0}}= \left( \begin{array}{cc} r_{1} - q\lambda _{1}(1-m)(1-q)\alpha _{A}K_{A} &{}\quad 0\\ r_{2}(1-m)\alpha _{1}q &{}\quad -r_{2} \end{array}\right) . \end{aligned}$$

  $J_{E_{2}^{0}}$ has one negative eigenvalue $(-r_{2})$. The stability of the equilibrium $E_{2}^{0}$ depends on the sign of the second eigenvalue $r_{1} - q\lambda _{1}(1-m)(1-q)\alpha _{A}K_{A}$. Moreover, the discriminant of the characteristic equation is $\varDelta _{E_{2}^{0}} = (r_ + r_{1} - q\lambda _{1}(1-m)(1-q)\alpha _{A}K_{A})^2 > 0$. Thus if $r_{1} < q\lambda _{1}(1-m)(1-q)\alpha _{A}K_{A}$, then the equilibrium $E_{2}^{0}$ is stable node and the equilibrium $E_4^0$ does not exist. If $r_{1} > q\lambda _{1}(1-m)(1-q)\alpha _{A}K_{A}$, the equilibrium $E_{2}^{0}$ is a saddle (unstable).

• 3. The Jacobian matrix of system (2)–(3) at the equilibrium $E_{3}^{0}$ is:

  $$\begin{aligned} J_{E_{3}^{0}}= \left( \begin{array}{cc} -r_{1} &{}\quad -q\lambda _{1}(1-m)K_{1}\\ 0 &{}\quad r_{2} \end{array}\right) . \end{aligned}$$

  $J_{E_{3}^{0}}$ has one positive eigenvalue $(r_{2})$. The equilibrium $E_{3}^{0}$ is a saddle (unstable).

• 4. The Jacobian matrix of system (2)–(3) at the equilibrium $E_{4}^{0}$ is:

  $$\begin{aligned}&J_{E_{4}^{0}}= \left( \begin{array}{cc} -b_{1}x_{4}^{0} &{}\quad -q\lambda _{1}(1-m)x_{4}^{0}\\ (1-m)\alpha _{1}r_{2}q &{}\quad -r_{2} \end{array}\right) .\\&Tr(J_{E_{4}^{0}})=-(b_{1}x_{4}^{0}+r_{2})< 0,\\&|(J_{E_{4}^{0}})|= r_{2}b_{1}x_{4}^{0}+(1-m)^{2}\alpha _{1}\lambda _{1}r_{2}q^{2}x_{4}^{0}> 0. \end{aligned}$$

  Moreover, the discriminant of the characteristic equation is $\varDelta _{E_{4}^{0}} = (b_{1}x_{4}^{0} - r_2)^{2} - 4(1 - m)^{2}\alpha _{1}\lambda _{1}r_{2}q^{2}x_{4}^{0}.$ Therefore, if $\varDelta _{E_{4}^{0}}> 0$, then the equilibrium $E_{4}^{0}$ is a stable node. If $\varDelta _{E_{4}^{0}}< 0$, then the equilibrium $E_{4}^{0}$ is a stable spiral. If $\varDelta _{E_{4}^{0}}= 0$, then the equilibrium $E_{4}^{0}$ is a stable degenerate node.

• 5. The Jacobian matrix of system (2)–(3) at the equilibrium $E_{1}^{\varphi }$ is:

  $$\begin{aligned} J_{E_{1}^{\varphi }}= \left( \begin{array}{cc} \displaystyle -b_{1}x_{1}^{\varphi }+ \frac{\varphi (x_{1}^{\varphi })}{x_{1}^{\varphi }}- \varphi '(x_{1}^{\varphi }) &{}\quad -q\lambda _{1}(1-m)x_{2}^{\varphi }\\ 0 &{}\quad r_{2} \end{array}\right) . \end{aligned}$$

  $J_{E_{1}^{\varphi }}$ has one positive eigenvalue $(r_{2})$. Thus, the equilibrium $E_{1}^{\varphi }$ is unstable. Moreover, if $-b_{1}x_{1\varphi } + \displaystyle \frac{h(x_{1}^{\varphi } - T_1)^2 - h^{2}T_1}{x_{1}^{\varphi }(x_{1}^{\varphi } + h - T_1)^2} < 0$, then the equilibrium $E_{1}^{\varphi }$ is a saddle. If $-b_{1}x_{1}^{\varphi } + \displaystyle \frac{h(x_{1}^{\varphi } - T_1)^2 - h^{2}T_1}{x_{1}^{\varphi }(x_{1}^{\varphi } + h - T_1)^2} > 0$, then the equilibrium $E_{1}^{\varphi }$ is an unstable node. If $-b_{1}x_{1}^{\varphi } + \displaystyle \frac{h(x_{1}^{\varphi } - T_1)^2 - h^{2}T_1}{x_{1}^{\varphi }(x_{1}^{\varphi } + h - T_1)^2} = 0$, then the equilibrium $E_{1}^{\varphi }$ is an unstable nonhyperbolic point.

• 6. The Jacobian matrix of system (2)–(3) at the equilibrium $E_{2}^{\varphi }$ is:

  $$\begin{aligned}&J_{E_{2}^{\varphi }}= \left( \begin{array}{cc} \displaystyle -b_{1}x_{2}^{\varphi }+ \frac{\varphi (x_{2}^{\varphi })}{x_{2}^{\varphi }}- \varphi '(x_{2}^{\varphi }) &{}\quad -q\lambda _{1}(1-m)x_{2}^{\varphi }\\ (1-m)\alpha _{1}r_{2}q &{}\quad -r_{2} \\ \end{array}\right) .\\&Tr(J_{E_{2}^{\varphi }})= \displaystyle -b_{1}x_{2}^{\varphi } - r_{2}+ \displaystyle \frac{h(x_{2}^{\varphi } - T_1)^2 - h^{2}T_1}{x_{2}^{\varphi }(x_{2}^{\varphi } + h - T_1)^2},\\&|J_{E_{2}^{\varphi }}|= -r_{2}\left( \displaystyle -b_{1}x_{2}^{\varphi }+ \displaystyle \frac{h(x_{2}^{\varphi } - T_1)^2 - h^{2}T_1}{x_{2}^{\varphi }(x_{2}^{\varphi } + h - T_1)^2}\right) \\&\quad + (1-m)^{2}q^{2}\alpha _{1}\lambda _{1}r_{2} x_{2}^{\varphi }. \end{aligned}$$

  The discriminant of the characteristic equation is:

  $$\begin{aligned} \varDelta _{E_{2}^{\varphi }}= & {} \left( \displaystyle r_{2} + b_{1}x_{2}^{\varphi } - \displaystyle \frac{h(x_{2}^{\varphi } - T_1)^2 - h^{2}T_1}{x_{2}^{\varphi }(x_{2}^{\varphi } + h - T_1)^2} \right) ^{2} \\&- 4(1-m)^{2}q^{2}\alpha _{1}\lambda _{1}r_{2} x_{2}^{\varphi }. \end{aligned}$$

  Then using the signs of $Tr(J_{E_{2}^{\varphi }})$, $|J_{E_{2}^{\varphi }}|$ and $\varDelta _{E_{2}^{\varphi }}$, and the table given in Jordan et al. [29], we have the type and the stability of the equilibrium $E_{2}^{\varphi }$ as in theorem (4). (For stability of equilibria in the plane, one can also refer to Auger et al. [3] and Hirsch et al. [27].)

Proof of Theorem 5

We proved that it is possible to have purely imaginary roots for the characteristic equation (13) and the transversality condition is satisfied. Thus, we have the third item of theorem 5. In what follows, we prove the first and the second items of theorem 5. By Rouche’s theorem [19] and the continuity in $\tau$, the characteristic equation (13) has roots with positive real parts if and only if it has purely imaginary roots. Let $\lambda (\tau )= \mu (\tau )+i\omega (\tau )$ where $\mu$ and $\omega$ are real depending on $\tau$. For $\tau =0$, the equilibrium E is stable. Thus, we have $\mu (0)<0$. By continuity, if $\tau$ is sufficiently small, we still have $\mu (\tau )<0$ and E is still stable. The change of stability will occur for some values of $\tau$ for which $\mu (\tau )=0$ and $\omega (\tau )\ne 0$; it means that $\lambda$ will be purely imaginary. Let $\tau _{e}$ be such that $\mu (\tau _{e})=0$ and $\omega (\tau _{e})=\omega _{e}\ne 0$ with $\lambda =i\omega (\tau _{e})$. In this case, the steady state loses stability and eventually becomes unstable when $\mu (\tau )$ becomes positive. In other words, if such a value $\omega _{e}$ does not exist, then the steady state E will remain stable for all $\tau$.

Proof of Theorem 6

For convenience, let $t=s\tau$, $x(s\tau )=x_{1}(s)$, $y(s\tau )=x_{2}(s)$ and $\tau =\tau _{0}+\mu$, $\mu \in {\mathbb {R}}$ so that $\mu =0$ is the Hopf bifurcating value for system (2)–(3). Then system (2)–(3) becomes equivalent to the system:

$$\begin{aligned} \dot{u}(t)=L_{\mu }(u_{t})+f(\mu ,u_{t}), \end{aligned}$$

(22)

where $u_{t}=(x_{1}(t),x_{2}(t))^{T} \in \mathcal {C}$ and $u_{t}(\theta )=u(t+\theta )=(x_{1}(t+\theta ),x_{1}(t+\theta ))^{T}\in \mathcal {C}$.

$L_{\mu }: \mathcal {C}\rightarrow {\mathbb {R}}^{2}$ is defined as follows:

$$\begin{aligned} L_{\mu }\phi= & {} (\tau _{0}+\mu )\left( \begin{array}{cc} A_{A\varphi } &{}\quad B_{A\varphi } \\ 0 &{}\quad 0 \end{array} \right) \left( \begin{array}{c} \phi _{1}(0) \\ \phi _{2}(0) \end{array} \right) \nonumber \\&+ (\tau _{0}+\mu )\left( \begin{array}{cc} 0 &{}\quad 0 \\ C_{A\varphi } &{}\quad D_{A\varphi } \end{array} \right) \left( \begin{array}{c} \phi _{1}(-1) \\ \phi _{2}(-1) \\ \end{array} \right) , \end{aligned}$$

(23)

where $A_{A\varphi }=-b_{1}x_{e}+ \displaystyle \frac{\varphi (x_{e})}{x_{e}} - \varphi '(x_{e})$, $B_{A\varphi }=-q\lambda _{1}(1-m)x_{e}$, $C_{A\varphi }= qr_{2}\alpha _{1}(1-m)$, $D_{A\varphi }=-r_{2}$.

$f:\mathcal {R}\times \mathcal {C}\rightarrow {\mathbb {R}}^{2}$ is defined as follows:

$$\begin{aligned} f(\mu ,\phi )= (\tau _{0}+\mu )(f_{1},f_{2})^{T}, \end{aligned}$$

(24)

where $\phi (\theta )=(\phi _{1}(\theta ),\phi _{2}(\theta ))^{T} \in \mathcal {C}$

$$\begin{aligned} f_{1}=a_{11}\phi _{1}^{2}(0)+a_{12}\phi _{1}(0)\phi _{2}(0)+a_{13}\phi _{1}^{3}(0), \end{aligned}$$

and

$$\begin{aligned} f_{2}= & {} a_{21}\phi _{1}^{2}(-1)+a_{22}\phi _{1}(-1)\phi _{2}(0)\\&+a_{23}\phi _{1}(-1)\phi _{2}(-1)+a_{24}\phi _{2}(-1)\phi _{2}(0)\\&+a_{25}\phi _{1}^{3}(-1) +a_{26}\phi _{1}^{2}(-1)\phi _{2}(0)\\&+a_{27}\phi _{1}^{2}(-1)\phi _{2}(-1)\\&+a_{28}\phi _{1}(-1)\phi _{2}(0)\phi _{2}(-1), \end{aligned}$$

with

$$\begin{aligned} a_{11}= & {} -b_{1}+ \displaystyle \frac{h^{2}}{(x_{e}+h-T_1)^{2}} , a_{12}=-q\lambda _{1}(1-m), \\ a_{13}= & {} \displaystyle \frac{h^{2}}{(x_{e}+h-T_1)^{4}}\\ a_{21}= & {} - \displaystyle \frac{r_{2}(1-m)^{2}q^{2}\alpha _{1}^{2}}{y_{e}}, a_{22}= \displaystyle \frac{r_{2}(1-m)q\alpha _{1}}{y_{e}}, \\ a_{23}= & {} \displaystyle \frac{r_{2}(1-m)q\alpha _{1}}{y_{e}}, a_{24}=- \displaystyle \frac{r_{2}}{y_{e}}, \\ a_{25}= & {} -\displaystyle \frac{r_{2}(1-m)^{3}q^{3}\alpha _{1}^{3}}{y_{e}^{2}}, a_{26}=- \displaystyle \frac{r_{2}(1-m)^{2}q^{2}\alpha _{1}^{2}}{y_{e}^{2}},\\ a_{27}= & {} - \displaystyle \frac{r_{2}(1-m)^{2}q^{2}\alpha _{1}^{2}}{y_{e}^{2}}, a_{28}= \displaystyle \frac{r_{2}(1-m)q\alpha _{1}}{y_{e}^{2}}. \end{aligned}$$

$L_{\mu }$ is a one parameter family of bounded linear operators in $\mathcal {C}[-1,0]\rightarrow {\mathbb {R}}^{2}$. Then by the Riesz representation theorem, there exists a matrix whose components are bounded variation functions $\eta (\theta ,\mu )$ in $[-1,0]\rightarrow {\mathbb {R}}^{2}$ such that

$$\begin{aligned} L_{\mu }=\int _{-1}^{0}\mathrm{d}\eta (\theta ,\mu )\phi (\theta ). \end{aligned}$$

(25)

In fact, we can choose

$$\begin{aligned}&\eta (\theta ,\mu )= (\tau _{0}+\mu )\left( \begin{array}{cc} A_{A\varphi } &{}\quad B_{A\varphi } \\ 0 &{}\quad 0 \\ \end{array} \right) \left( \begin{array}{c} \phi _{1}(0) \\ \phi _{2}(0) \\ \end{array} \right) \delta (\theta )\nonumber \\&\quad + (\tau _{0}+\mu )\left( \begin{array}{cc} 0 &{}\quad 0 \\ C_{A\varphi } &{}\quad D_{A\varphi } \end{array} \right) \left( \begin{array}{c} \phi _{1}(-1) \\ \phi _{2}(-1) \end{array} \right) \delta (\theta +1),\nonumber \\ \end{aligned}$$

(26)

where $\delta$ is the Dirac function. Then equation (22) is satisfied.

For $\phi \in \mathcal {C}^{1}[-1,0]$, let us define

$$\begin{aligned} A(\mu )\phi =\left\{ \begin{array}{ll} \displaystyle \frac{\mathrm{d}\phi (\theta )}{\mathrm{d}\theta } , &{}\quad \text{ if }\quad -1\le \theta <0 \\ \int _{-1}^{0}\mathrm{d}\eta (\theta ,\mu )\phi (\theta ), &{}\quad \text{ if }\quad \theta =0 \end{array}\right. \nonumber \\ \end{aligned}$$

(27)

and

$$\begin{aligned} R(\mu )\phi =\left\{ \begin{array}{ll} \left( \begin{array}{c} 0\ \\ 0\ \\ \end{array} \right) , &{}\quad \text{ if }\quad -1\le \theta <0, \\ f(\mu , \phi ), &{}\quad \text{ if }\quad \theta =0. \end{array}\right. \end{aligned}$$

(28)

System (2)–(3) is then transformed into the operator equation of the form (29) as follows, in order to study the Hopf bifurcation problem

$$\begin{aligned} \dot{u}(t)=A(\mu )u_{t}+Ru_{t}. \end{aligned}$$

(29)

Define the adjoint operator for $\psi \in \mathcal {C}^{1}([0,1], ({\mathbb {R}}^{2})^{\star })$,

$$\begin{aligned} A^{\star }(\mu )\psi (s)=\left\{ \begin{array}{ll} \displaystyle -\frac{\mathrm{d}\phi (s)}{\mathrm{d}s} &{}\quad \text{ if }\quad 0 < s \le 1, \\ \int _{-1}^{0}\mathrm{d}\eta ^{T}(s,\mu )\psi (-s) &{}\quad \text{ if }\quad s = 0. \end{array}\right. \nonumber \\ \end{aligned}$$

(30)

In order to normalize the eigenvectors of the operator A and the adjoint operator $A^{\star }$, we need to introduce the following bilinear form:

$$\begin{aligned}&\langle \psi , \phi \rangle = \overline{\psi }(0).\phi (0)\nonumber \\&\quad - \int _{\theta =-1}^{0}\int _{\xi =0}^{\theta }\overline{\psi }^{T}(\xi -\theta )\mathrm{d}\eta (\theta )\phi (\xi )\mathrm{d}\xi , \end{aligned}$$

(31)

where $\eta (\theta )=\eta (\theta ,0)$.

By the discussion and the transformation $t=s\tau$, we know that $i\tau _{0}\omega _{0}$ and $-i\tau _{0}\omega _{0}$ are the eigenvalues of A(0) and other eigenvalues have strictly negative real parts. Hence, they are also eigenvalues of $A^{\star }$. Now we are going to compute the eigenvectors of A(0) and $A^{\star }$ corresponding to their respective eigenvalues $i\tau _{0}\omega _{0}$ and $-i\tau _{0}\omega _{0}$. If we suppose that $q(\theta )= (q^{(1)}(\theta ), q^{(2)}(\theta ))^T = (1,q_{1})^{T}e^{i\tau _{0}\omega _{0}\theta }$ is the eigenvector of A(0) corresponding to the eigenvalue $i\tau _{0}\omega _{0}$, then by the definition we have $A(0)q(0)=i\tau _{0}\omega _{0}q(0)$. Then using the definition of A(0) and the expressions given by (22), (24) and (25) gives:

$$\begin{aligned}&\tau _{0}\left( \begin{array}{cc} A_{A\varphi } &{}\quad B_{A\varphi } \\ 0 &{}\quad 0 \\ \end{array}\right) q(0)+ \tau _{0}\left( \begin{array}{cc} 0 &{}\quad 0 \\ C_{A\varphi } &{}\quad \quad D_{A\varphi } \\ \end{array}\right) q(-1)\\&\quad = i\tau _{0}\omega _{0}q(0), \end{aligned}$$

or equivalently $\tau _{0}\left( \begin{array}{cc} A_{A\varphi } &{}\quad B_{A\varphi } \\ 0 &{}\quad 0 \\ \end{array} \right) \left( \begin{array}{c} 1\ \\ q_{1} \end{array} \right) + \tau _{0}\left( \begin{array}{cc} 0 &{} \quad 0 \\ C_{A\varphi } &{}\quad D_{A\varphi } \\ \end{array} \right) \left( \begin{array}{c} 1\ \\ q_{1} \end{array} \right) e^{-i\tau _{0}\omega _{0}}= i\tau _{0}\omega _{0}\left( \begin{array}{c} 1\ \\ q_{1} \end{array} \right)$

This implies $\left\{ \begin{array}{ll} A_{A\varphi }+B_{A\varphi }q_{1} &{} = i\omega _{0},\\ (C_{A\varphi }+D_{A\varphi }q_{1})e^{-i\tau _{0}\omega _{0}} &{} = q_{1}i\omega _{0}, \end{array}\right.$

and    $q_{1}= \displaystyle \frac{C_{A\varphi }}{-D_{A\varphi } + i\omega _{0}e^{i\tau _{0}\omega _{0}}}.$

Thus, $q^{(1)}(\theta )= e^{i\tau _{0}\omega _{0}\theta }$ and $q^{(2)}(\theta )= \displaystyle \frac{C_{A\varphi }e^{i\tau _{0}\omega _{0}}}{-D_{A\varphi } + i\omega _{0}e^{i\tau _{0}\omega _{0}}}.$

Now let us compute the eigenvector $q^{\star }$ of $A^{\star }$. Suppose that we have $q^{\star }(s) = G_{A\varphi }(1,q_{1}^{\star })^{T}e^{i\tau _{0}\omega _{0}s} , 0\le s \le 1$. Then we have the following relation

$$\begin{aligned}&\tau _{0}\left( \begin{array}{cc} A_{A\varphi } &{} \quad 0 \\ B_{A\varphi } &{} \quad 0 \\ \end{array}\right) q^{\star }(0)+ \tau _{0}\left( \begin{array}{cc} 0 &{} \quad C_{A\varphi } \\ 0 &{} \quad D_{A\varphi } \end{array}\right) q^{\star }(-1)\\&\quad = -i\tau _{0}\omega _{0}q^{\star }(0), \end{aligned}$$

which is equivalent to $\tau _{0}\left( \begin{array}{cc} A_{A\varphi } &{} \quad 0 \\ B_{A\varphi } &{} \quad 0 \end{array} \right) \left( \begin{array}{c} 1\ \\ q^{\star }_{1} \end{array} \right) + \tau _{0}\left( \begin{array}{cc} 0 &{} \quad C_{A\varphi } \\ 0 &{} \quad D_{A\varphi } \end{array} \right) \left( \begin{array}{c} 1\ \\ q^{\star }_{1} \end{array} \right) e^{-i\tau _{0}\omega _{0}}= -i\tau _{0}\omega _{0}\left( \begin{array}{c} 1\ \\ q^{\star }_{1} \end{array} \right) .$

This implies $\left\{ \begin{array}{ll} A_{A\varphi }+C_{A\varphi }q^{\star }_{1}e^{-i\tau _{0}\omega _{0}} &{} = -i\omega _{0},\\ B_{A\varphi }+D_{A\varphi }q^{\star }_{1})e^{-i\tau _{0}\omega _{0}} &{} = -q_{1}^{\star }i\omega _{0}, \end{array}\right.$ and    $q^{\star }_{1}= \displaystyle -\frac{B_{A\varphi }}{D_{A\varphi } + i\omega _{0}e^{i\tau _{0}\omega _{0}}}.$

Now let’s compute $G_{A\varphi }$ by using the orthogonality condition $\langle q^{\star }(s), q(\theta ) \rangle = 1$. By using (30), we have:

$$\begin{aligned}&\langle q^{\star }(s), q(\theta )\rangle \\&\quad = \overline{q^{\star }}(0).q(0)- \displaystyle \int _{\theta =-1}^{0}\displaystyle \int _{\xi =0}^{\theta }\overline{q^{\star }}^{T}\\&\qquad (\xi -\theta )\mathrm{d}\eta (\theta )q(\xi )\mathrm{d}\xi \\&\quad = \overline{q^{\star }}(0).q(0)- \displaystyle \int _{\theta =-1}^{0}\displaystyle \int _{\xi =0}^{\theta }\overline{G_{A\varphi }}(1,\overline{q_{1}^{\star }}) e^{-i\omega _{0}\tau _{0}(\xi -\theta )}\\&\qquad \mathrm{d}\eta (\theta )(1,q_{1})^{T} e^{i\omega _{0}\tau _{0}\xi } \\&\quad = \overline{G_{A\varphi }}\left\{ \displaystyle 1+q_{1}\overline{q_{1}^{\star }}-(1,\overline{q_{1}^{\star }})\right. \\&\qquad \left. \int _{-1}^{0}\theta e^{i\omega _{0}\tau _{0}\theta } \mathrm{d}\eta (\theta )(1,q_{1})^{T}\right\} \\&\quad = \overline{G_{A\varphi }}\left\{ \displaystyle 1+q_{1}\overline{q_{1}^{\star }}+ \tau _{0}e^{i\omega _{0}\tau _{0}}(C_{A\varphi }+D_{A\varphi }q_{1})\overline{q_{1}}^{\star } \right\} . \end{aligned}$$

Then,

$$\begin{aligned} G_{A\varphi } = [1+\overline{q_{1}}q_{1}^{\star }+ \tau _{0}e^{-i\omega _{0}\tau _{0}}(C_{A\varphi } + D_{A\varphi } \overline{q_{1}})q_{1}^{\star }]^{-1}. \end{aligned}$$

Now we are going to compute the coordinates to describe the center manifold $\mathcal {C}_0$ at $\mu =0$. Let us define

$$\begin{aligned} z(t) = \langle q^{*},u_{t} \rangle \end{aligned}$$

(32)

and

$$\begin{aligned} W(t, \theta )= u_{t}(\theta )-2Re\{z(t)q(\theta )\}, \end{aligned}$$

(33)

where $u_{t}$ is a solution of (28) when $\mu =0$.

On the center manifold $\mathcal {C}_{0}$, we have:

$$\begin{aligned} W(t, \theta )= W(z(t), \overline{z}(t), \theta ), \end{aligned}$$

where

$$\begin{aligned} W(t, \theta )= & {} W(z(t), \overline{z}(t), \theta )= W_{20}(\theta )\frac{z^{2}}{2}+W_{11}(\theta )z\overline{z}\nonumber \\&+W_{02}(\theta )\frac{\overline{z}^{2}}{2}+\cdots \end{aligned}$$

(34)

In (33), z and $\overline{z}$ represent the local coordinates of the center manifold $\mathcal {C}_{0}$ in the direction of q and $q^{*}$, respectively.

Now let us reduce equation (28) to an ordinary differential equation using a single complex variable on the center manifold. Since $\mu =0$ and for $u_{t}$ a solution of (28) belonging to the center manifold $\mathcal {C}_{0}$, we have:

$$\begin{aligned} \dot{z}(t)= & {} \langle q^{*},\dot{u}(t) \rangle =\langle q^{*},Au_{t}+Ru_{t} \rangle \\= & {} \langle q^{*},Au_{t}\rangle +\langle q^{*},Ru_{t}\rangle \\= & {} \langle A^{*} q^{*},u_{t}\rangle +\langle q^{*},Ru_{t}\rangle \\= & {} i\tau _{0}\omega _{0}z+\overline{q}^{*}f_{0}(z,\overline{z}). \end{aligned}$$

The following equation:

$$\begin{aligned} \dot{z}(t)= i\tau _{0}\omega _{0}z+\overline{q}^{*}f_{0}(z,\overline{z}) \end{aligned}$$

(35)

can be rewritten as:

$$\begin{aligned} \dot{z}(t)=i\tau _{0}\omega _{0}z+g(z,\overline{z}), \end{aligned}$$

(36)

where

$$\begin{aligned} g(z,\overline{z})= g_{20}\frac{z^{2}}{2}+ g_{11}z\overline{z}+g_{02}\frac{\overline{z}^{2}}{2}+g_{21}\frac{z^{2}\overline{z}}{2}+\cdots \end{aligned}$$

(37)

In what follows, we are going to expand g in powers of z and $\overline{z}$ in order to obtain from the first three coefficients of this expansion, the value of $\mu _2$ which indicates the direction of the Hopf bifurcation (that is to say if the Hopf bifurcation is supercritical or subcritical) and the value of $\beta _2$ which determines the stability. To do so, we use the algorithm presented by Hassard et al. [26].

A substitution of (28) in (36) leads to:

$$\begin{aligned} \dot{W}(t)= & {} \dot{u_{t}}-\dot{z}q-\dot{\overline{z}}\overline{q}\\= & {} A(\mu )u_{t}+Ru_{t}-[i\tau _{0}\omega _{0}z+\overline{q}^{*}(0)f_{0}(z,\overline{z})]q\\&- [-i\tau _{0}\omega _{0}\overline{z}+{q}^{*}(0)\overline{f}_{0}(z,\overline{z})]\overline{q}\\= & {} AW+2ARe(zq)+Ru_{t}-2Re[\overline{q}^{*}(0)f_{0}(z,\overline{z})q(\theta )]\\&-2Re[i\tau _{0}\omega _{0}zq(\theta )]\\= & {} AW-2Re[\overline{q}^{*}(0)f_{0}(z,\overline{z})q(\theta )]+Ru_{t}. \end{aligned}$$

Then we obtain the following equations

$$\begin{aligned} \dot{W}=\left\{ \begin{array}{ll} AW-2Re[\overline{q}^{*}(0)f_{0}(z,\overline{z})q(\theta )]&{}\quad \text{ if }\quad -1\le \theta <0, \\ AW-2Re[\overline{q}^{*}(0)f_{0}(z,\overline{z})q(\theta )] + f &{}\quad \text{ if }\quad \theta =0. \end{array}\right. \nonumber \\ \end{aligned}$$

(38)

Equations (37) can be rewritten as:

$$\begin{aligned} \dot{W}=AW+H(z,\overline{z},\theta ), \end{aligned}$$

(39)

where

$$\begin{aligned} H(z,\overline{z},\theta )= & {} H_{20}(\theta )\frac{z^{2}}{2}+H_{11}(\theta )z\overline{z}\nonumber \\&+H_{02}(\theta )\frac{\overline{z}^{2}}{2}+\cdots \end{aligned}$$

(40)

Derivating W in (33) with respect to t, we have:

$$\begin{aligned} \dot{W}=W_{z}\dot{z}+W_{\overline{z}}\dot{\overline{z}}. \end{aligned}$$

(41)

Then using (33), (38) and (39) gives:

$$\begin{aligned} (A-2i\tau _{0}\omega _{0})W_{20}(\theta )=-H_{20}(\theta ), \end{aligned}$$

(42)

and

$$\begin{aligned} AW_{11}(\theta )=-H_{11}(\theta ). \end{aligned}$$

(43)

Using equations (34) and (35) gives:

$$\begin{aligned} g(z,\overline{z})= & {} \overline{q}^{*}(0)f_{0}(z,\overline{z})= \tau _{0}\overline{G_{A\varphi }}\left( \begin{array}{c} 1\ \\ q^{\star }_{1}\ \\ \end{array} \right) \left( \begin{array}{c} f_{1}\ \\ f_{2} \end{array} \right) \nonumber \\= & {} \tau _{0}\overline{G_{A\varphi }}(f_{1}+q^{\star }_{1}f_{2}), \end{aligned}$$

(44)

where

$$\begin{aligned} f_{1}= & {} a_{11}x_{1}^{2}(t)+a_{12}x_{1}(t)x_{2}(t)+a_{13}x_{1}^{3}(t), \\ f_{2}= & {} a_{21}x_{1}^{2}(t-1)+a_{22}x_{1}(t-1)x_{2}(t)\\&+ a_{23}x_{1}(t-1)x_{2}(t-1)+a_{24}x_{2}(t-1)x_{2}(t)\\&+ a_{25}x_{1}^{3}(t-1) \\&+ a_{26}x_{1}^{2}(t-1)x_{2}(t)+a_{27}x_{1}^{2}(t-1)x_{2}(t-1)\\&+ a_{28}x_{1}(t-1)x_{2}(t)x_{2}(t-1), \end{aligned}$$

and

$$\begin{aligned} x_{1t}(0)= & {} q^{(1)}(0)z+\overline{q}^{(1)}(0)\overline{z}+W_{20}^{(1)}(0)\frac{z^{2}}{2}\\&+W_{11}^{(1)}(0)z\overline{z}+ W_{02}^{(1)}(0)\frac{\overline{z}^{2}}{2}+ \cdots \\ x_{2t}(0)= & {} q^{(2)}(0)z+\overline{q}^{(2)}(0)\overline{z}+W_{20}^{(2)}(0)\frac{z^{2}}{2}\\&+W_{11}^{(2)}(0)z\overline{z}+ W_{02}^{(2)}(0)\frac{\overline{z}^{2}}{2}+\cdots \\ x_{1t}(-1)= & {} q^{(1)}(-1)z+\overline{q}^{(1)}(-1)\overline{z}+W_{20}^{(1)}(-1)\frac{z^{2}}{2}\\&+W_{11}^{(1)}(-1)z\overline{z}+ W_{02}^{(1)}(-1)\frac{\overline{z}^{2}}{2}+\cdots \\ x_{2t}(-1)= & {} q^{(2)}(-1)z+\overline{q}^{(2)}(-1)\overline{z}+W_{20}^{(2)}(-1)\frac{z^{2}}{2}\\&+W_{11}^{(2)}(-1)z\overline{z}+ W_{02}^{(2)}(-1)\frac{\overline{z}^{2}}{2}+ \cdots \end{aligned}$$

An identification by using (34) and (43) gives us the following coefficients of g:

$$\begin{aligned} g_{20}= & {} 2\tau _{0}\overline{G_{A\varphi }}\{ a_{11}(q^{(1)}(0))^{2}+a_{12}q^{1}(0)q^{(2)}(0)\\&+ q_{1}^{*}(a_{21}(q^{(1)}(-1))^{2} +a_{22}q^{(1)}(-1)q^{(2)}(0)\\&+ a_{23}q^{(1)}(-1)q^{(2)}(-1)+a_{24}q^{(2)}(0)q^{(2)}(-1)) \}, \\ g_{11}= & {} \tau _{0}\overline{G_{A\varphi }}\{2a_{11}q^{(1)}(0)\overline{q}^{(1)}(0)\\&+ a_{12}(q^{(1)}(0)q^{(2)}(0)+\overline{q}^{(1)}(0)q^{(2)}(0))\\&+ q_{1}^{*}(2a_{21}q^{(1)}(-1)\overline{q}^{(1)}(-1)\\&+ a_{22}(q^{(1)}(-1)\overline{q}^{(2)}(0)+q^{(2)}(0)\overline{q}^{(1)}(-1))\\&+ a_{23}(q^{(1)}(-1)\overline{q}^{(2)}(-1)+q^{(2)}(-1)\overline{q}^{(1)}(-1))\\&+ a_{24}(q^{(2)}(-1)\overline{q}^{(2)}(0)+q^{(2)}(0)\overline{q}^{(2)}(-1)))\},\\ g_{02}= & {} 2\tau _{0}\overline{G_{A\varphi }}\{a_{11}(\overline{q}^{(1)}(0))^{2}\\&+ a_{21}\overline{q}^{1}(0)\overline{q}^{(2)}(0)+ q_{1}^{*}(a_{21}(\overline{q}^{(2)}(-1))^{2}\\&+ a_{22}\overline{q}^{(1)}(-1)\overline{q}^{(2)}(0)\\&+ a_{23}\overline{q}^{(1)}(-1)\overline{q}^{(2)}(-1)\\&+ a_{24}\overline{q}^{(2)}(0)\overline{q}^{(2)}(-1))\}, \\ g_{21}= & {} 2\tau _{0}\overline{G_{A\varphi }}\{a_{11}(\overline{q}^{(1)}(0)W_{20}^{(1)}(0)\\&+ 2q^{(1)}(0)W_{11}^{(1)}(0))+ a_{12}(q^{(1)}(0)W_{11}^{(2)}(0)\\&+ \frac{1}{2}\overline{q}^{(1)}(0)W_{20}^{(2)}(0)\\&+ \frac{1}{2}\overline{q}^{(2)}(0)W_{20}^{(1)}(0)\\&+ \overline{q}^{(2)}(0)W_{11}^{(1)}(0))+3a_{13}(q^{(1)}(0))^{2}\overline{q}^{(1)}(0)\\&+ q_{1}^{*}(a_{21}(\overline{q}^{(1)}(-1)W_{20}^{(1)}(-1)\\&+ 2q^{(1)}(-1)W_{11}^{(1)}(-1))+a_{22}(q^{(1)}(-1)W_{11}^{(2)}(0)\\&+ \frac{1}{2}\overline{q}^{(1)}(-1)W_{20}^{(2)}(0)\\&+ \frac{1}{2}\overline{q}^{(2)}(0)W_{20}^{(1)}(-1)\\&+ q^{(2)}(0)W_{11}^{(1)}(-1))+a_{23}(q^{(1)}(-1)W_{11}^{(2)}(-1)\\&+ \frac{1}{2} \overline{q}^{(1)}(-1)W_{20}^{(2)}(-1)+q^{(2)}(-1)W_{11}^{(1)}(-1)\\&+ \frac{1}{2}\overline{q}^{(2)}(-1)W_{20}^{(1)}(-1))+a_{24}(q^{(2)}(0)W_{11}^{(2)}(-1)\\&+ \frac{1}{2}\overline{q}^{(2)}(0)W_{20}^{(2)}(-1)+q^{(2)}(-1)W_{11}^{(2)}(0)\\&+\frac{1}{2}\overline{q}^{(2)}(-1)W_{20}^{(2)}(0))\\&+ 3a_{25}(q^{(1)}(-1))^{2}\overline{q}^{(1)}(-1)\\&+ a_{26}((q^{(1)}(-1))^{2}\overline{q}^{(2)}(0)\\&+ 2q^{(1)}(-1)\overline{q}^{(1)}(-1)q^{(2)}(0))\\&+ a_{27}((q^{(1)}(-1))^{2}\overline{q}^{(2)}(-1)\\&+ 2q^{(1)}(-1)\overline{q}^{(1)}(-1)q^{(2)}(-1))\\&+ a_{28}(q^{(1)}(-1)\overline{q}^{(2)}(0)q^{(2)}(-1))\\&+ q^{(1)}(-1)\overline{q}^{(2)}(-1)q^{(1)}(0)\\&+ q^{(2)}(-1)\overline{q}^{(1)}(-1)q^{(2)}(0))\}. \end{aligned}$$

Now we reach to the step of computation of $W_{20}(\theta )$. Using (27) and (38) gives:

$$\begin{aligned} H(z,\overline{z}, \theta )= & {} -2Re[\overline{q}^{*}(0)f_{0}(z,\overline{z})q(\theta )]+Ru_{t}\\= & {} -gq(\theta )-\overline{g}\overline{q}(\theta )+Ru_{t} \end{aligned}$$

$$\begin{aligned}= & {} -\left( \displaystyle g_{20}\frac{z^{2}}{2}+g_{11}z\overline{z}+g_{02}\frac{\overline{z}^{2}}{2}\right) q(\theta )\nonumber \\&-\left( \displaystyle \overline{g}_{20}\frac{\overline{z}^{2}}{2}+\overline{g}_{11}z\overline{z}+\overline{g}_{02}\frac{{z}^{2}}{2}\right) \overline{q}(\theta )+Ru_{t}, \end{aligned}$$

(45)

where

$$\begin{aligned} Ru_{t}= \tau _{0} \left( \begin{array}{c} f_{1}\ \\ f_{2} \end{array} \right) =2\tau _{0}A^{A\varphi }\frac{z^{2}}{2}+\tau _{0}B^{A\varphi }z\overline{z}, \end{aligned}$$

with

$$\begin{aligned} A^{A\varphi }= & {} \left( \begin{array}{c} A_{11}^{A\varphi }\ \\ A_{21}^{A\varphi } \end{array} \right) , \\ B^{A\varphi }= & {} \left( \begin{array}{c} B_{11}^{A\varphi }\ \\ B_{21}^{A\varphi } \end{array} \right) , \end{aligned}$$

where

$$\begin{aligned} A_{11}^{A\varphi }= & {} a_{11}(q^{(1)}(0))^{2}+a_{12}q^{(1)}(0)q^{(2)}(0), \\ A_{21}^{A\varphi }= & {} a_{21}(q^{(1)}(-1))^{2}+a_{22}q^{(1)}(-1)q^{(2)}(0)\\&+ a_{23}q^{(1)}(-1)q^{(2)}(-1)\\&+ a_{24}q^{(2)}(-1)q^{(2)}(0), \\ B_{11}^{A\varphi }= & {} 2a_{11}q^{(1)}(0)\overline{q}^{(1)}(0)\\&+ a_{12}(q^{(1)}(0)\overline{q}^{(2)}(0)+\overline{q}^{(1)}(0){q}^{(2)}(0)), \\ B_{21}^{A\varphi }= & {} 2a_{21}q^{(1)}(-1)\overline{q}^{(1)}(-1)\\&+ a_{22}(q^{(1)}(-1)\overline{q}^{(2)}(0)+\overline{q}^{(1)}(-1){q}^{(2)}(0)) \\&+ a_{23}(q^{(1)}(-1)\overline{q}^{(2)}(-1)+\overline{q}^{(1)}(-1){q}^{(2)}(-1))\\&+ a_{24}(q^{(2)}(-1)\overline{q}^{(2)}(0)+\overline{q}^{(2)}(-1){q}^{(2)}(0)). \end{aligned}$$

A comparison of the coefficients of equations (39) and (44) gives the following equalities:

$$\begin{aligned} H_{20}(\theta )= & {} \left\{ \begin{array}{ll} -gq(\theta )-\overline{g}\overline{q}(\theta )&{}\quad \text{ if }\quad -1\le \theta <0, \\ -gq(\theta )-\overline{g}\overline{q}(\theta )+2\tau _{0}A &{}\quad \text{ if }\quad \theta =0. \end{array}\right. \nonumber \\ \end{aligned}$$

(46)

$$\begin{aligned} H_{11}(\theta )= & {} \left\{ \begin{array}{ll} -gq(\theta )-\overline{g}\overline{q}(\theta )&{}\quad \text{ if }\quad -1\le \theta <0, \\ -gq(\theta )-\overline{g}\overline{q}(\theta )+\tau _{0}B &{}\quad \text{ if }\quad \theta =0. \end{array}\right. \nonumber \\ \end{aligned}$$

(47)

When we substitute (45) in (41) and (46) in (42), respectively, we obtain the following differential equations:

$$\begin{aligned} \left\{ \begin{array}{l} \dot{W}_{20}(\theta )= 2i\tau _{0}\omega _{0}W_{20}(\theta )+g_{20}q(\theta )+\overline{g}_{20}\overline{q}(\theta ), \\ \dot{W}_{11}(\theta )=g_{11}q(\theta )+\overline{g}_{11}\overline{q}(\theta ), \end{array}\right. \nonumber \\ \end{aligned}$$

(48)

which have the following solutions:

$$\begin{aligned} \left\{ \begin{array}{l} {W}_{20}(\theta )= \displaystyle \frac{ig_{20}}{\tau _{0}\omega _{0}}q(0)e^{i\tau _{0}\omega _{0}\theta }+ \frac{\overline{g}_{02}}{3i\tau _{0}\omega _{0}}\overline{q}(0)e^{-i\tau _{0}\omega _{0}\theta }\\ \qquad \quad +E_{1A\varphi }e^{2i\tau _{0} \omega _{0}\theta } , \\ {W}_{11}(\theta )= \displaystyle \frac{g_{11}}{i\tau _{0}\omega _{0}}q(0)e^{i\tau _{0}\omega _{0}\theta }+ \frac{\overline{g}_{11}}{i\tau _{0}\omega _{0}}\overline{q}(0)e^{-i\tau _{0}\omega _{0}\theta }\\ \qquad +E_{2A\varphi }, \end{array}\right. \nonumber \\ \end{aligned}$$

(49)

where $E_{1A\varphi }=(E_{1A\varphi }^{(1)},E_{1A\varphi }^{(2)})^{T}$ and $E_{2A\varphi }=(E_{2A\varphi }^{(1)},E_{2A\varphi }^{(2)})^{T}$ are constant vectors belonging to ${\mathbb {R}}^{2}$.

Now, let us compute the constant vectors $E_{1A\varphi }$ and $E_{2A\varphi }$ by using (41) and (42). We have:

$$\begin{aligned} \int _{-1}^{0}\mathrm{d}\eta (\theta )W_{20}(\theta )=2i\tau _{0}\omega _{0}W_{20}(0)-H_{20}(0), \end{aligned}$$

(50)

and

$$\begin{aligned} \int _{-1}^{0}\mathrm{d}\eta (\theta )W_{11}(\theta )=-H_{11}(0). \end{aligned}$$

(51)

Then, using the fact that,

$$\begin{aligned}&\left( \displaystyle i\tau _{0}\omega _{0}I-\int _{-1}^{0}e^{i\tau _{0}\omega _{0}\theta }\mathrm{d}\eta (\theta )\right) q(0)=0, \\&\left( \displaystyle -i\tau _{0}\omega _{0}I-\int _{-1}^{0}e^{-i\tau _{0}\omega _{0}\theta }\mathrm{d}\eta (\theta )\right) \overline{q}(0)=0, \end{aligned}$$

and using the first equation of (48), equations (44) and (49), we have:

$$\begin{aligned} \left( \displaystyle 2i\tau _{0}\omega _{0}I-\int _{-1}^{0}e^{2i\tau _{0}\omega _{0}\theta }\mathrm{d}\eta (\theta )\right) E_{1}=2\tau _{0}\left( \begin{array}{c} A_{11}^{A\varphi }\ \\ A_{21}^{A\varphi } \end{array} \right) \end{aligned}$$

which implies $\left( \begin{array}{cc} 2i\omega _{0}-A_{A\varphi } &{}\quad -B_{A\varphi }\\ -C_{A\varphi }e^{-2i\tau _{0}\omega _{0}} &{}\quad 2i\omega _{0}-D_{A\varphi }e^{-2i\tau _{0}\omega _{0}} \\ \end{array}\right) \left( \begin{array}{c} E_{1A\varphi }^{(1)}\ \\ E_{1A\varphi }^{(2)} \end{array} \right) =2\left( \begin{array}{c} A_{11}^{A\varphi }\ \\ A_{21}^{A\varphi } \end{array} \right)$ and finally

$$\begin{aligned} \left\{ \begin{array}{l} E_{1A\varphi }^{(1)}= \displaystyle \frac{2B_{A\varphi }A_{21}^{A\varphi }+2A_{11}^{A\varphi }(2i\omega _{0}-D_{A\varphi }e^{-2i\tau _{0}\omega _{0}})}{(A_{A\varphi }D_{A\varphi }-B_{A\varphi }C_{A\varphi })e^{-2i\tau _{0}\omega _{0}}-2i\omega _{0}D_{A\varphi } e^{-2i\tau _{0}\omega _{0}}-2i\omega _{0}A_{A\varphi }-4\omega _{0}^{2}} , \\ E_{1A\varphi }^{(2)}= \displaystyle \frac{2A_{11}^{A\varphi }C_{A\varphi }e^{-2i\tau _{0}\omega _{0}}+2A_{21}^{A\varphi }(2i\omega _{0}-A_{A\varphi })}{(A_{A\varphi }D_{A\varphi }-B_{A\varphi }C_{A\varphi })e^{-2i\tau _{0}\omega _{0}}-2i\omega _{0}D_{A\varphi } e^{-2i\tau _{0}\omega _{0}}-2i\omega _{0}A_{A\varphi }-4\omega _{0}^{2}}. \end{array}\right. \end{aligned}$$

(52)

Similarly, using the second equation of (48), (46) and (50) gives:

$$\begin{aligned}&\left( \displaystyle \int _{-1}^{0}e^{-i\tau _{0}\omega _{0}}\mathrm{d}\eta (\theta )\right) E_{2A\varphi }=\tau _{0}\left( \begin{array}{c} B_{11}^{A\varphi }\ \\ B_{21}^{^{A\varphi }} \end{array} \right) \\&\quad \Rightarrow \left( \begin{array}{cc} A_{A\varphi } &{} B_{A\varphi } \\ C_{A\varphi } &{} D_{A\varphi } \end{array}\right) \left( \begin{array}{c} E_{2A\varphi }^{(1)}\ \\ E_{2A\varphi }^{(2)} \end{array} \right) =-\left( \begin{array}{c} B_{11}^{A\varphi }\ \\ B_{21}^{A\varphi } \end{array} \right) , \end{aligned}$$

and finally

$$\begin{aligned} \left\{ \begin{array}{l} E_{2A\varphi }^{(1)}= \displaystyle \frac{B_{A\varphi }B_{21}^{A\varphi }-D_{A\varphi }B_{11}^{A\varphi }}{A_{A\varphi }D_{A\varphi } - B_{A\varphi }C_{A\varphi }}, \\ E_{2A\varphi }^{(2)}= \displaystyle \frac{C_{A\varphi }B_{11}^{A\varphi }-A_{A\varphi }B_{21}^{A\varphi }}{A_{A\varphi }D_{A\varphi } - B_{A\varphi }C_{A\varphi }}. \end{array}\right. \end{aligned}$$

(53)

Finally, from (48), (50) and (52) we can now calculate values which are useful for the determination of the period of the bifurcating solution and its stability. We have the following values:

$$\begin{aligned} C_{1}(0)= & {} \frac{i}{2\tau _{0}\omega _{0}}\left( \displaystyle g_{20}g_{11}-2|g_{11}|^{2}-\frac{1}{3}|g_{02}|^{2}\right) +\frac{1}{2}g_{21}, \\ \mu _{2}= & {} -\frac{ReC_{1}(0)}{Re(\frac{\mathrm{d}\lambda }{\mathrm{d}\tau }(\tau _{0}))}, \\ \beta _{2}= & {} 2ReC_{1}(0), \\ T_{2}= & {} -\frac{Im(C_{1}(0))+\mu _{2}Im(\frac{\mathrm{d}\lambda }{\mathrm{d}\tau }(\tau _{0}))}{\tau _{0}\omega _{0}}. \end{aligned}$$

This ends the proof.