Stability and Hopf bifurcation for a delayed predator–prey model with stage structure for prey and Ivlev-type functional response

Abstract

In this paper, we mainly investigate a delayed predator–prey model with stage structure for prey and Ivlev-type functional response. Four assumptions about this model are made as follows: (1) there are a single predator and a single prey population in the model; (2) the prey is divided by the age into two stage structures, immature and mature; (3) the trophic transfer from prey to predator is incomplete; (4) there are two time delays due to the time of maturation of prey and the time of gestation of predator. Some properties of equilibria for the model without delays are provided. Further, the stability of equilibria and existence of Hopf bifurcation are studied by discussing the different cases of time delays for the model with delays. We observe that delays can cause a stable equilibrium to become unstable one, even occur Hopf bifurcation when delays pass through their corresponding critical values. Meanwhile, we derive explicit formulae to determine the properties of Hopf bifurcation such as the direction of Hopf bifurcation and the stability of periodic solutions. Numerical simulations of all theoretical analyses are given for verifying our theoretical results. In the present work, we demonstrate that the validity and universality of oscillations induced by the time delays both theoretically and numerically. These results of this paper may be helpful for us to further understand the role of the critical values of time delays in stabilizing the predator–prey model.

Appendices

Appendix A

In Appendix, we will provide some details of Theorems 2.1–2.3.

Proof of Theorem 2.1

Let

$$\begin{aligned} \begin{array}{l} L_{1}:\, y = g_{1}(x_{2})=\frac{(ab - k d_{2})x_{2}-b_{1}kx_{2}^{2}}{k(1-e^{-\alpha x_{2}})},\\ L_{2}:\, y = g_{2}(x_{2})=\frac{1-e^{-\beta x_{2}}-d}{\delta }. \end{array} \end{aligned}$$

Setting $z=\alpha x_{2},~m=\alpha \left( \frac{ab-kd_{2}}{kb_{1}}\right)$, we obtain

$$\begin{aligned} g_{1}(x_{2})= & {} g_{1}\left( \frac{z}{\alpha }\right) = \frac{(ab-d_{2}k)\frac{z}{\alpha }-b_{1}k(\frac{z}{\alpha })^{2}}{k(1-e^{-z})} \\= & {} \frac{b_{1}}{\alpha ^{2}}\left( \frac{mz-z^{2}}{1-e^{-z}}\right) , \end{aligned}$$

therefore,

$$\begin{aligned} \begin{aligned}&\frac{\mathrm{d}g_{1}(x_2)}{\mathrm{d}x_{2}} = \frac{\mathrm{d}g_{1}}{\mathrm{d}z}\frac{\mathrm{d}z}{\mathrm{d}x_{2}}\\&\quad =\alpha \frac{b_{1}}{\alpha ^{2}}\frac{(m-2z)(1-e^{-z}) -(mz-z^{2})e^{-z}}{(1-e^{-z})^{2}}\\&\quad = \frac{b_{1}}{\alpha (1-e^{-z})^{2}}\left[ m-2z+e^{-z}(2z-mz+z^{2}-m)\right] \\&\quad = \frac{b_{1}e^{-z}}{\alpha (1-e^{-z})^{2}}\left[ (m-2z)e^{z}-m+z^{2}+(2-m)z\right] . \end{aligned} \end{aligned}$$

Denoting

$$\begin{aligned} \Phi (z)=(m-2z)e^{z}-m+z^{2}+(2-m)z, \end{aligned}$$

we have

$$\begin{aligned} \lim _{z\rightarrow 0^{+}}\Phi (z) = 0,~\Phi (m) = m\left( 1 - e^{m}\right) < 0, \end{aligned}$$

and

$$\begin{aligned} \frac{\mathrm{d}\Phi }{\mathrm{d}z}=(m-2z-2)(e^{z}-1). \end{aligned}$$

(A.1)

Therefore, when $m\in (0,2]$, $\Phi (z) < 0$ for $z\in (0,+\infty )$. When $m\in (2,+\infty )$, then there exists a $z_0\in \left( \frac{1}{2}(m-2),m\right)$ such that $\Phi (z) < 0$ for $z \in (z_0,+\infty )$ and $\Phi (z) > 0$ for $z \in (0,z_0)$.

Hence, when $m\in (0,2]$,

$$\begin{aligned} \begin{array}{l} \dfrac{\mathrm{d}g_1(x_2)}{\mathrm{d}x_2} < 0,\quad \text {for}~x_2 \in \left( 0,+\infty \right) , \end{array} \end{aligned}$$

that is, $y=g_{1}(x_{2})$ is an decreasing function on $\left( 0, +\infty \right)$.

When $m\in (2,+\infty )$,

$$\begin{aligned} \begin{array}{l} \dfrac{\mathrm{d}g_1(x_2)}{\mathrm{d}x_2} > 0,\quad \text {for}~x_2\in (0,\frac{z_0}{\alpha }),\\ \dfrac{\mathrm{d}g_1(x_2)}{\mathrm{d}x_2} < 0,\quad \text {for}~x_2\in (\frac{z_0}{\alpha },+\infty ). \end{array} \end{aligned}$$

that is, $y=g_{1}(x_{2})$ is an increasing function on $(0,\frac{z_0}{\alpha })$ and is an decreasing function on $(\frac{z_0}{\alpha }, +\infty )$. From $\frac{\mathrm{d}g_{2}(x_{2})}{\mathrm{d}x_{2}}=\frac{\beta e^{-\beta x_{2}}}{\delta }>0$ we have that $y=g_{2}(x_{2})$ is an increasing function on $[0,+\infty )$. Furthermore,

$$\begin{aligned} g_{1}\left( \frac{ab - kd_{2}}{kb_{1}}\right) =0,\quad g_{2}\left( -\, \frac{1}{\beta }\ln (1-d)\right) =0. \end{aligned}$$

Notice that $\displaystyle \lim _{x_2\rightarrow 0^{+}}g_1(x_2) = \frac{ab - kd_2}{\alpha k}>0$ and $\displaystyle \lim _{x_2\rightarrow 0^{+}}g_2(x_2) = -\frac{d}{\delta } < 0$. If $(C_{3})$ hold, there exists a unique intersection $\left( x^{*}_{2},y^{*}\right)$ of line $L_{1}$ and line $L_{2}$. From the above discussion, we obtain a unique positive equilibrium $E_{2}(x^{*}_{1},x^{*}_{2},y^{*})$ with

$$\begin{aligned} x^{*}_{1}= & {} \frac{ax^{*}_{2}}{k},\,\,y^{*}=\frac{1-d-e^{-\beta x^{*}_{2}}}{\delta },\,\, 0 \\&<-\, \frac{1}{\beta }\ln (1-d)<x^{*}_{2}<\frac{ab-kd_{2}}{b_{1}k}. \end{aligned}$$

Proof of Theorem 2.2

The Jacobian matrix of (3) at the equilibrium $(x_{1},x_{2},y)$ is

$$\begin{aligned}&J(x_1, x_2, y)\nonumber \\&\quad \!\! = \begin{pmatrix} -k &{}\quad a &{}\quad 0\\ b &{}\quad -2b_1x_2 - d_2 - \alpha y e^{-\alpha x_2} &{}\quad -(1 - e^{-\alpha x_2})\\ 0 &{}\quad \beta ye^{-\beta x_2} &{}\quad 1 - d - 2\delta y - e^{-\beta x_2} \end{pmatrix}.\nonumber \\ \end{aligned}$$

(A.2)

(i) The eigenvalues $\lambda _1$, $\lambda _2$ and $\lambda _3$ associated with the trivial equilibrium $E_0(0,0,0)$ are given by $\lambda _1 = -d < 0$, $\lambda _{2,3} = \frac{-(k + d_2)\pm \sqrt{(k+d_2)^2 - 4(kd_2-ab)}}{2} = \frac{-(k + d_2)\pm \sqrt{(k - d_2)^2 + 4ab}}{2}$. Hence, if $(C_1)$ holds, then $\lambda _2> 0$ and $\lambda _3<0$ imply that $E_0(0,0,0)$ is a saddle point. $E_0(0,0,0)$ is unstable. If $(C_2)$ holds, then $\lambda _{2} < 0$ and $\lambda _3 < 0$ imply that $E_0(0,0,0)$ is locally asymptotically stable.

(ii) If $(C_{3})$ holds, one of the eigenvalues associated with the equilibrium $E_1({\widehat{x}}_{1},{\widehat{x}}_{2},0)$ is given by $\lambda =1-d-e^{-\beta {\widehat{x}}_{2}}>0.$ Therefore, $E_{1}({\widehat{x}}_{1},{\widehat{x}}_{2},0)$ is unstable. $\square$

Proof of Theorem 2.3

From (A.2), we have the following characteristic equation of $J(x^{*}_{1},x^{*}_{2},y^{*})$

$$\begin{aligned}&\lambda ^{3} + \left( 2b_1x^{*}_2 + d_2 + \alpha e^{-\alpha x^{*}_2}y^{*} \right. \nonumber \\&\left. \quad -\, 1 + d + 2\delta y^{*} + e^{-\beta x^{*}_2} + k\right) \lambda ^{2}\nonumber \\&\quad +\, \left[ \left( 2b_1x^{*}_2 + d_2 + \alpha e^{-\alpha x^{*}_2}y^{*}\right) \left( d - 1 + 2\delta y^{*}+e^{-\beta x^{*}_2}\right) \right. \nonumber \\&\quad \left. +\, \beta e^{-\beta x^{*}_2}y^{*}\left( 1-e^{-\alpha x^{*}_2}\right) - ab\right. \nonumber \\&\quad +\, \left. 2b_1kx^{*}_2 + d_2 k + \alpha e^{-\alpha x^{*}_2}y^{*}k \right. \nonumber \\&\quad \left. -\, k + e^{-\beta x^{*}_2}k + 2\delta y^{*}k + dk\right] \lambda \nonumber \\&\quad +\, \left[ k\left( 2b_1x^{*}_2 + d_2 + \alpha e^{-\alpha x^{*}_2}y^{*}\right) \left( d - 1 + 2\delta y^{*} + e^{-\beta x^{*}_2}\right) \right. \nonumber \\&\quad +\, \left. k\beta \left( 1-e^{-\alpha x^{*}_2}\right) e^{-\beta x^{*}_2}y^{*} \right. \nonumber \\&\left. \quad -\, ab\left( d - 1 + 2\delta y^{*} + e^{-\beta x^{*}_2}\right) \right] =0. \end{aligned}$$

(A.3)

Since $m_{1} \triangleq d - 1 + 2\delta y^{*} + e^{-\beta x^{*}_{2}} = \delta y^{*}>0,$$m_{2} \triangleq 2b_{1}x^{*}_{2} + d_{2} + \alpha e^{-\alpha x^{*}_{2}}y^{*}>0,$$m_{3} \triangleq \beta e^{-\beta x^{*}_{2}}y^{*}(1-e^{-\alpha x^{*}_{2}})>0,$ then (A.3) becomes the following equation

$$\begin{aligned}&\lambda ^{3} + \left( m_1 + m_2 + k\right) \lambda ^{2} + \left[ m_1m_2 + m_3 - ab \right. \\&\left. \quad +\, k\left( m_1 + m_2\right) \right] \lambda + \left( km_1m_2 + km_3 - abm_1\right) =0. \end{aligned}$$

Let

$$\begin{aligned} A_{1}= & {} m_{1}+m_{2}+k,~~ \\ A_{2}= & {} m_{1}m_{2}+m_{3}-ab+k(m_{1}+m_{2}),~~ \\ A_{3}= & {} km_{1}m_{2}+km_{3}-abm_{1}, \end{aligned}$$

obviously, $A_{1}>0$ and

$$\begin{aligned}&A_{1}A_{2}-A_{3} =(m_{1}+m_{2}+k)\left[ m_{1}m_{2}+m_{3}-ab\right. \\&\left. \qquad +\, k(m_{1}+m_{2})\right] -(km_{1}m_{2}+km_{3}-abm_{1})\\&\quad =m^{2}_{1}m_{2}+m_{1}m^{2}_{2}+2km_{1}m_{2}\\&\qquad +\, km^{2}_{1}+(m_{1}+m_{2})m_{3}+k^{2}m_{1} \\&\qquad +\, (k+m_{2})(2b_{1}kx^{*}_{2}+kd_{2}+k\alpha e^{-\alpha x^{*}_{2}}y^{*}-ab). \end{aligned}$$

From the proof of Theorem 2.1, if $m\in (0,2]$, $x_2^{*}\in (0,+\infty )$ or $m\in (2,+\infty )$, $x_2^{*}\in \left( \frac{z_0}{\alpha },+\infty \right)$, then we have

$$\begin{aligned} \begin{array}{l} y^{*} = g_1\left( x^{*}_2\right) = \dfrac{(ab-d_2k)x^{*}_2 - b_1k(x^{*}_2)^2}{k(1-e^{-\alpha x^{*}_2})}>0,\\ \end{array} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \dfrac{\mathrm{d}g_1(x^{*}_2)}{\mathrm{d}x_2}&= \dfrac{k\left( ab-d_{2}k-2b_{1}kx^{*}_2\right) \left( 1-e^{-\alpha x^{*}_2}\right) - \alpha ke^{-\alpha x^{*}_2}x^{*}_2\left( ab - d_2k - b_1kx^{*}_2\right) }{k^2\left( 1-e^{-\alpha x^{*}_2}\right) ^2}\\&= \dfrac{\left( ab-d_2k-2b_1kx^{*}_2\right) \left( 1 - e^{-\alpha x^{*}_2}\right) - \alpha e^{-\alpha x^{*}_2}x^{*}_2\left( ab - d_2k - b_1kx^{*}_2\right) }{k\left( 1 - e^{-\alpha x^{*}_2}\right) ^2} < 0. \end{aligned} \end{aligned}$$

Hence, we have

$$\begin{aligned}&\alpha e^{-\alpha x^{*}_{2}}x^{*}_{2}(ab-d_{2}k-b_{1}kx^{*}_{2})\\&\quad >(ab-d_{2}k-2b_{1}kx^{*}_{2})(1-e^{-\alpha x^{*}_{2}}), \end{aligned}$$

then

$$\begin{aligned}&2b_1kx^{*}_2 + kd_2 + k\alpha e^{-\alpha x^{*}_2}y^{*} - ab= 2b_1kx^{*}_2 + kd_2 -ab + k\alpha e^{-\alpha x^{*}_2}\frac{\left( ab - k d_2\right) x^{*}_2 - kb_1\left( x^{*}_2\right) ^2}{k\left( 1 - e^{-\alpha x^{*}_2}\right) }\\&\quad = \frac{(1-e^{-\alpha x^{*}_{2}})(2b_{1}kx^{*}_{2}+kd_{2}-ab)+\alpha e^{-\alpha x^{*}_2}x^{*}_2\left[ \left( ab - kd_2\right) - kb_1x^{*}_2\right] }{\left( 1 - e^{-\alpha x^{*}_2}\right) }\\&\quad = \frac{\alpha e^{-\alpha x^{*}_2}x^{*}_2\left[ \left( ab - kd_2\right) - kb_1x^{*}_2\right] - \left( 1 - e^{-\alpha x^{*}_2}\right) \left( ab - kd_2 - 2kb_1x^{*}_{2}\right) }{\left( 1-e^{-\alpha x^{*}_2}\right) } >0, \end{aligned}$$

therefore

$$\begin{aligned} A_1A_2 - A_3 > 0. \end{aligned}$$

From

$$\begin{aligned} \begin{aligned} A_2&= m_1m_2 + m_3 - ab + k\left( m_1 + m_2\right) \\&= \left( km_2 - ab\right) + m_1m_2 + m_3 + km_1 \\&= (2b_{1}kx^{*}_{2}+kd_{2}+k\alpha e^{-\alpha x^{*}_{2}}y^{*}-ab)\\&\quad +m_{1}m_{2}+m_{3}+km_{1},\\ A_3&= km_1m_2 + km_3 - abm_1 = \left( km_2 - ab\right) m_1 + km_3\\&= \left( 2kb_1x^{*}_2 + kd_2 + k\alpha e^{-\alpha x^{*}_2}y^{*} - ab\right) m_1 + km_3, \end{aligned} \end{aligned}$$

and the above discussion, we get $A_1>0,~A_2>0,~A_3>0$ and $A_1A_2 - A_3>0$. By the Routh–Hurwitz criterion, we see that the positive equilibrium $E_2(x^{*}_1,x^{*}_2,y^{*})$ is locally asymptotically stable. $\square$

Appendix B

$$\begin{aligned}&\langle q^*(s),q(\theta )\rangle \\&\quad = {\overline{q}}^*(0)\cdot q(0) - \int _{\theta = -1}^0\int _{\xi =0}^\theta {\overline{q}}^{*T}(\xi -\theta )\mathrm{d}\eta (\theta )q(\xi )\mathrm{d}\xi \\&\quad = {\overline{P}}\left( {\overline{v}}_1^*,{\overline{v}}_2^*,{\overline{v}}_3^*\right) (v_1,v_2,v_3)^T \\&\quad - \int _{\theta = -1}^0\int _{\xi =0}^\theta {\overline{P}}({\overline{v}}_1^*,{\overline{v}}_2^*,{\overline{v}}_3^*)e^{-\mathrm{i}\nu _{40}\tau _{40}(\xi -\theta )}\mathrm{d}\eta (\theta )\\&\qquad (v_1,v_2,v_3)^Te^{\mathrm{i}\nu _{40}\tau _{40}\xi }\mathrm{d}\xi \\&= {\overline{P}}\left( {\overline{v}}_1^*,{\overline{v}}_2^*,{\overline{v}}_3^*\right) (v_1,v_2,v_3)^T \\&\quad - {\overline{q}}^{*T}(0)\int _{\theta = -1}^0\int _{\xi =0}^\theta e^{\mathrm{i}\nu _{40}\tau _{40}\theta }\mathrm{d}\xi \mathrm{d}\eta (\theta )q(0)\\&= {\overline{P}}\left( {\overline{v}}_1^*,{\overline{v}}_2^*,{\overline{v}}_3^*\right) (v_1,v_2,v_3)^T \\&\quad - {\overline{q}}^{*T}(0)\int _{\theta = -1}^0 \xi e^{\mathrm{i}\nu _{40}\tau _{40}\theta }\bigg |_{\xi = 0}^{\theta }\mathrm{d}\eta (\theta )q(0)\\&= {\overline{P}}\left( {\overline{v}}_1^*,{\overline{v}}_2^*,{\overline{v}}_3^*\right) (v_1,v_2,v_3)^T \\&\quad - {\overline{q}}^{*T}(0)\int _{\theta = -1}^0 \theta e^{\mathrm{i}\nu _{40}\tau _{40}\theta }\mathrm{d}\eta (\theta )q(0)\\&= {\overline{P}}\left( v_1{\overline{v}}_1^* + v_2{\overline{v}}_2^* + v_3{\overline{v}}_3^*\right) \\&\quad + {\overline{q}}^{*T}(0)\tau _2^* \begin{pmatrix} 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad \beta y^*e^{-\beta x_2^*} &{}\quad 1 - e^{-\beta x_2^*} \end{pmatrix} e^{-\mathrm{i}\nu _{40}\tau _2^*}q(0) \\&\quad + {\overline{q}}^{*T}(0)\tau _{40} \begin{pmatrix} -b &{}\quad 0 &{}\quad 0\\ b &{}\quad 0 &{}\quad 0\\ 0 &{}\quad 0 &{}v 0 \end{pmatrix} e^{-\mathrm{i}\nu _{40}\tau _{40}}q(0)\\&= {\overline{P}}\left\{ v_1{\overline{v}}_1^* + v_2{\overline{v}}_2^* + v_3{\overline{v}}_3^* + \tau _2^*e^{-\mathrm{i}\nu _{40}\tau _2^*}\right. \\&\left. \quad \times \left[ \beta y^*e^{-\beta x_2^*}v_2{\overline{v}}_3^* +(1 - e^{-\beta x_2^*})v_3{\overline{v}}_3^*\right] \right. \\&\left. \quad - b\tau _{40} e^{-\mathrm{i}\nu _{40} \tau _{40}}v_1{\overline{v}}_1^* \right\} . \end{aligned}$$

$$\begin{aligned} \begin{array}{l} {\overline{P}} = \dfrac{1}{v_1{\overline{v}}_1^* + v_2{\overline{v}}_2^* + v_3{\overline{v}}_3^* + \tau _2^*e^{-\mathrm{i}\nu _{40}\tau _2^*}\left[ \beta y^*e^{-\beta x_2^*}v_2{\overline{v}}_3^* +(1 - e^{-\beta x_2^*})v_3{\overline{v}}_3^*\right] -\, b\tau _{40} e^{-\mathrm{i}\nu _{40} \tau _{40}}v_1{\overline{v}}_1^*}. \end{array} \end{aligned}$$

Appendix C

Using the same notations as in Hassard et al. [47], we first compete the coordinates to describe the center manifold $C_0$ at $\mu = 0$. Let $X_t$ be the solution of (30) when $\mu = 0$. Define

$$\begin{aligned} z(t)= & {} \langle q^*,X_t\rangle , ~~W(t,\theta ) = X_t(\theta ) - z(t)q(\theta ) \nonumber \\&-\, {\overline{z}}(t){\overline{q}}(\theta ) = X_t(\theta ) - 2\mathrm{Re}\{z(t)q(\theta )\}. \end{aligned}$$

(C.1)

On the center manifold $C_0$ we have $W(t,\theta )=W(z(t),{\overline{z}}(t),\theta )$, where

$$\begin{aligned} W(z(t),{\overline{z}}(t),\theta )= & {} W_{20}(\theta )\dfrac{z^2(t)}{2} + W_{11}(\theta )z(t){\overline{z}}(t) \nonumber \\&+\, W_{02}(\theta )\dfrac{{\overline{z}}^2(t)}{2} \nonumber \\&+\, W_{30}(\theta )\dfrac{z^3(t)}{6} + \cdots , \end{aligned}$$

(C.2)

z(t) and ${\overline{z}}(t)$ are local coordinates for center manifold $C_0$ in the directions of $q^*$ and ${\overline{q}}^*$. Note that W is real if $X_t$ is, we shall deal with real solutions only. For the solution $X_t \in C_0$ of (30), since $\mu = 0$, we have

$$\begin{aligned} {\dot{z}}(t)&= \langle q^*,{\dot{X}}_t\rangle = \langle q^*,A(0)X_t + R(0)X_t\rangle \\&= \langle q^*,A(0)X_t\rangle + \langle q^*,R(0)X_t\rangle \\&= \langle A^*(0)q^*,X_t\rangle + {\overline{q}}^*(0)\cdot F(0,X_t)\\&=\mathrm{i}\nu _{40} \tau _{10}'z(t) + {\overline{q}}^*(0)\cdot F\left( 0,W(z(t),{\overline{z}}(t),0) \right. \\&\left. \quad +\, 2{{\mathrm{Re}}}\{z(t)q(0)\}\right) \\&\triangleq \mathrm{i}\nu _{40} \tau _{10}'z(t) + {\overline{q}}^{*}(0)\cdot F_0(z(t),{\overline{z}}(t)). \end{aligned}$$

We rewrite this equation as

$$\begin{aligned} {\dot{z}}(t) = \mathrm{i}\nu _{40} \tau _{40}z(t) + g(z(t),{\overline{z}}(t)), \end{aligned}$$

(C.3)

where $g(z(t),{\overline{z}}(t)) = {\overline{q}}^*(0)\cdot F_0(z(t),{\overline{z}}(t))$, and expand $g(z(t),{\overline{z}}(t))$ in powers of z(t) and ${\overline{z}}(t)$, that is

$$\begin{aligned} g\left( z(t),{\overline{z}}(t)\right)= & {} g_{20}\dfrac{z^2(t)}{2} + g_{11}z(t){\overline{z}}(t) + g_{02}\dfrac{{\overline{z}}^2(t)}{2} \nonumber \\&+\, g_{21}\dfrac{z^2(t){\overline{z}}(t)}{2} + \cdots . \end{aligned}$$

(C.4)

then it follows from (C.1) that

$$\begin{aligned} X_t(\theta )= & {} W(t,\theta ) + 2\mathrm{{Re}}\{z(t)q(t)\} \nonumber \\= & {} W_{20}(\theta )\dfrac{z^2(t)}{2} + W_{11}(\theta )z(t){\overline{z}}(t)\nonumber \\&+ W_{02}(\theta )\dfrac{{\overline{z}}^2(t)}{2} + (v_1,v_2,v_3)^{T}e^{\mathrm{i}\nu _{40} \tau _{40}\theta }z(t)\nonumber \\&+ ({\overline{v}}_1,{\overline{v}}_2,{\overline{v}}_3)^{T}e^{-\mathrm{i}\nu _{40} \tau _{40}\theta }{\overline{z}}(t) + \cdots . \end{aligned}$$

(C.5)

It follows together with (32) that

$$\begin{aligned}&g(z(t),{\bar{z}}(t)) = {\bar{q}}^*(0)F_0(z(t),{\overline{z}}(t))={\overline{q}}^*(0)F_0(0,X_t)\nonumber \\&= {\overline{q}}^*(0)\tau _{40} \begin{pmatrix} 0\\ -y^*\displaystyle {\sum _{i=2}^{\infty }}{\frac{1}{i!}f_1^i(x^*)\left( W^{(2)}(0)+zv_2+{\bar{z}}{\bar{v}}_2\right) ^i} +\displaystyle {\sum _{i=1}^{\infty }}f_1^i(x^*)\left( W^{(2)}(0)+zv_2+{\bar{z}}{\bar{v}}_2 \right) ^i\\ \times \left( W^{(3)}(0) +zv_3+{\bar{z}}{\bar{v}}_3 \right) -b_1\left( W^{(2)}(0)+zv_2+{\bar{z}}{\bar{v}}_2\right) ^2 -\delta \left( W^{(3)}(0)+zv_3+{\bar{z}}{\bar{v}}_3\right) ^2\\ -y^*\displaystyle {\sum _{i=2}^{\infty }}{\frac{1}{i!}f_2^i(x^*)\left( W^{(2)}\left( \frac{-\tau _2^*}{\tau _{40}}\right) +zv_2e^{-\mathrm{i}\nu _{40}\tau _2^*}+{\bar{z}}{\bar{v}}_2e^{\mathrm{i}\nu _{40}\tau _2^*} \right) ^i} -\displaystyle {\sum _{i=1}^{\infty }}f_2^i(x^*)\left( W^{(2)}+zv_2e^{-\mathrm{i}\nu _{40}\tau _2^*} \right. \\ \displaystyle \left. +{\bar{z}}{\bar{v}}_2e^{\mathrm{i}\nu _{40}\tau _2^*}\right) ^i \left( W^{(3)}\left( \dfrac{-\tau _2^*}{\tau _{40}}\right) +zv_3e^{-\mathrm{i}\nu _{40}\tau _2^*}+{\bar{z}}{\bar{v}}_3e^{\mathrm{i}\nu _{40}\tau _2^*} \right) \end{pmatrix} \nonumber \\&={\overline{P}}\tau _{50}\left\{ v_2^*\left[ y^*\displaystyle {\sum _{i=1}^{\infty }}f_1^i(x^*)\left( W_{20}^{(2)}(0)\frac{z^2}{2}+W_{11}^{(2)}(0)z{\bar{z}} +\,W_{02}^{(2)}(0)\frac{{\bar{z}}^2}{2} +zv_2+{\bar{z}}{\bar{v}}_2 + \cdots \right) ^i\left( W_{20}^{(3)}(0)\frac{z^2}{2}+W_{11}^{(3)}(0)z{\bar{z}} \right. \right. \right. \nonumber \\&\quad \left. \left. +\,W_{02}^{(3)}(0)\frac{{\bar{z}}^2}{2}+zv_3+{\bar{z}}{\bar{v}}_3 + \cdots \right) -b_1\left( W_{20}^{(2)}(0)\frac{z^2}{2}+\, W_{11}^{(2)}(0)z{\bar{z}} +W_{02}^{(2)}(0)\frac{{\bar{z}}^2}{2}+zv_2+{\bar{z}}{\bar{v}}_2 + \cdots \right) ^2\right] \nonumber \\&\quad +\, v_3^*\left[ -\delta \left( W_{20}^{(3)}(0)\frac{z^2}{2} +W_{11}^{(3)}(0)z{\bar{z}}\right. \right. \nonumber \\&\left. \left. \quad +\, W_{02}^{(3)}(0)\frac{{\bar{z}}^2}{2}+zv_3+{\bar{z}}{\bar{v}}_3 + \cdots \right) ^2 \right. -\,y^*\displaystyle {\sum _{i=2}^{\infty }}\frac{1}{i!}f_2^i(x^*)\left( W_{20}^{(2)}\left( \frac{-\tau _2^*}{\tau _{40}}\right) \frac{z^2}{2}+W_{11}^{(2)}\left( \frac{-\tau _2^*}{\tau _{40}}\right) z{\bar{z}} \right. \nonumber \\&\left. \quad +\,W_{02}^{(2)}\left( \frac{-\tau _2^*}{\tau _{40}}\right) \frac{{\bar{z}}^2}{2}+zv_2e^{-\mathrm{i}\nu _{40}\tau _2^*}+{\bar{z}}{\bar{v}}_2e^{\mathrm{i}\nu _{40}\tau _2^*} + \cdots \right) ^i\nonumber \\&\quad -\,\displaystyle {\sum _{i=1}^{\infty }}f_2^i(x^*)\left( W_{20}^{(2)}\left( \frac{-\tau _2^*}{\tau _{40}}\right) \frac{z^2}{2}+W_{11}^{(2)}\left( \frac{-\tau _2^*}{\tau _{40}}\right) z{\bar{z}} \right. \nonumber \\&\left. \quad +\,W_{02}^{(2)}\left( \frac{-\tau _2^*}{\tau _{40}}\right) \frac{{\bar{z}}^2}{2}+zv_2e^{-\mathrm{i}\nu _{40}\tau _2^*}+{\bar{z}}{\bar{v}}_2e^{\mathrm{i}\nu _{40}\tau _2^*} + \cdots \right) ^i \nonumber \\&\quad \left. \left. \times \left( W_{20}^{(3)}\left( \frac{-\tau _2^*}{\tau _{40}}\right) \frac{z^2}{2}+W_{11}^{(3)}\left( \frac{-\tau _2^*}{\tau _{40}}\right) z{\bar{z}} \right. \right. \right. \nonumber \\&\left. \left. \left. \quad +\,W_{02}^{(3)}\left( \frac{-\tau _2^*}{\tau _{40}}\right) \frac{{\bar{z}}^2}{2}+zv_3e^{-\mathrm{i}\nu _{40}\tau _2^*}+{\bar{z}}{\bar{v}}_3e^{\mathrm{i}\nu _{40}\tau _2^*} + \cdots \right) \right] \right\} . \end{aligned}$$

(C.6)

Comparing the coefficients with (C.4), we obtain

$$\begin{aligned} g_{20}= & {} 2{\overline{P}}\tau _{40}\left\{ {\bar{v}}_2^*\left( -b_1v_2^2-\alpha e^{-\alpha x_2^*}v_2v_3 +\dfrac{1}{2}\alpha ^2y^*e^{-\alpha x_2^*}v_2^2\right) \right. \nonumber \\&\left. +\,{\bar{v}}_3^*\left[ -\delta v_3^2+\beta e^{-2\mathrm{i}\nu _{40}\tau _2^*}v_2\left( -\dfrac{1}{2}\beta y^*e^{-\alpha x_2^*}v_2 \right. \right. \right. \nonumber \\&\left. \left. \left. +\, e^{-\beta x_2^*}v_3\right) \right] \right\} ,\nonumber \\ g_{11}= & {} {\overline{P}}\tau _{40}\left\{ -2\delta v_3{\bar{v}}_3{\bar{v}}_3^*-2b_1v_2{\bar{v}}_2{\bar{v}}_2^* + e^{-\alpha x_2^*}{\bar{v}}_2^* \right. \nonumber \\&\left. \quad \left( -\alpha v_2{\bar{v}}_3 - \alpha {\bar{v}}_2v_3 + \alpha ^2y^*v_2{\bar{v}}_2\right) \right. \nonumber \\&\left. +\,{\bar{v}}_3^*\left[ -\beta ^2y^*e^{-\alpha x_2^*}v_2{\bar{v}}_2 \right. \right. \nonumber \\&\left. \left. +\,e^{-\beta x_2^*}\left( \beta v_2{\bar{v}}_3+\beta {\bar{v}}_2v_3\right) \right] \right\} ,\nonumber \\ g_{02}= & {} 2{\overline{P}}\tau _{40}\left[ -\delta {\bar{v}}_3^2{\bar{v}}_3^*-b_1{\bar{v}}_2^2{\bar{v}}_2^* + \beta e^{-2\mathrm{i}\nu _{40}\tau _2^*}{\bar{v}}_2{\bar{v}}_3^*\right. \nonumber \\&\left. \left( -\,\dfrac{1}{2}\beta y^*e^{-\alpha x_2^*}{\bar{v}}_2+e^{-\beta x_2^*}{\bar{v}}_3\right) \right. \nonumber \\&\left. +\,\alpha e^{-\alpha x_2^*} {\bar{v}}_2 {\bar{v}}_2^*\left( \frac{1}{2}\alpha y^*{\bar{v}}_2 - {\bar{v}}_3\right) \right] ,\nonumber \\ g_{21}= & {} 2{\overline{P}}\tau _{40}\left\{ b_1{\bar{v}}_2^*\left( -W_{20}^{(2)}(0){\bar{v}}_2- 2W_{11}^{(2)}(0)v_2\right) \right. \nonumber \\&\left. +\,\delta {\bar{v}}_3^*\left( -W_{20}^{(3)}(0){\bar{v}}_3 - 2W_{11}^{(3)}(0)v_3\right) \right. \nonumber \\&+\,\dfrac{1}{2}\beta e^{\mathrm{i}\nu _{40}\tau _2^*}{\bar{v}}_3^*\left( -\beta y^*e^{-\alpha x_2^*}W_{20}^{(2)}\left( -\dfrac{\tau _2^*}{\nu _{40}}\right) {\bar{v}}_2\right. \nonumber \\&\left. +\, e^{-\beta x_2^*}\left( W_{20}^{(2)}\left( -\dfrac{\tau _2^*}{\nu _{40}}\right) {\bar{v}}_3+W_{20}^{(3)}\left( -\dfrac{\tau _2^*}{\nu _{40}}\right) {\bar{v}}_2\right) \right) \nonumber \\&+\, \beta e^{-\mathrm{i}\nu _{40}\tau _2^*}{\bar{v}}_3^*\left( e^{-\alpha x_2^*}\left( -\beta y^*W_{11}^{(2)}(0)v_2 \right. \right. \nonumber \\&\left. \left. +\,\dfrac{1}{2}\beta ^2y^*v_2^2{\bar{v}}_2\right) \right. \nonumber \\&\left. +e^{-\beta x_2^*}\left( -\beta v_2{\bar{v}}_2v_3+W_{11}^{(3)}(0)v_2\right. \right. \nonumber \\&\left. \left. +\,W_{11}^{(2)}(0)v_3-\dfrac{1}{2}\beta v_2^2{\bar{v}}_3\right) \right) \nonumber \\&+\,e^{-\alpha x_2^*}{\bar{v}}_2^*\left[ \alpha ^2\left( -\dfrac{1}{2}\alpha y^*v_2^2{\bar{v}}_2+ \dfrac{1}{2} y^*W_{20}^{(2)}(0){\bar{v}}_2\right. \right. \nonumber \\&\left. \left. +\, v_2{\bar{v}}_2v_3 +y^*W_{11}^{(2)}(0)v_2 +\dfrac{1}{2} v_2^2{\bar{v}}_3\right) \right. \nonumber \\&\left. \left. +\, \alpha \left( -\dfrac{1}{2}W_{20}^{(3)}(0){\bar{v}}_2-\dfrac{1}{2}W_{20}^{(2)}(0){\bar{v}}_3 \right. \right. \right. \nonumber \\&\left. \left. \left. -\,W_{11}^{(2)}(0)v_3-W_{11}^{(3)}(0)v_2\right) \right] \right\} . \end{aligned}$$

(C.7)

Since there are $W_{20}(\theta )$ and $W_{11}(\theta )$ in $g_{21}$, we still need to compute them. From (30), (C.1) and (C.3), we have

$$\begin{aligned} \begin{aligned} {\dot{W}}&= {\dot{X}}_t - {\dot{z}}q - \dot{{\bar{z}}}{\bar{q}}\\&= {\left\{ \begin{array}{ll} AW - 2\mathrm{{Re}}\{{\overline{q}}^*(0)\cdot F_0(z,{\overline{z}})q(\theta )\}, &{} \theta \in [-1,0)\\ AW - 2\mathrm{{Re}}\{{\overline{q}}^*(0)\cdot F_0(z,{\overline{z}})q(0)\} + F_0(z,{\overline{z}}), &{} \theta = 0 \end{array}\right. }\\&\triangleq AW + H(z,{\overline{z}},\theta ), \end{aligned} \end{aligned}$$

(C.8)

where

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

(C.9)

On the other hand, on $C_0$ near to the origin, we have ${\dot{W}} = W_{z}{\dot{z}} + W_{{\overline{z}}}\dot{{\overline{z}}}$. Using (C.2) and (C.3) to replace $W_{z}$ and ${\dot{z}}$ and their conjugates by their power series expansions, comparing the coefficients with the right-hand side of (C.8), we obtain

$$\begin{aligned} \left( 2\mathrm{i}\nu _{40} \tau _{40}I - A\right) W_{20}(\theta )= & {} H_{20}(\theta ),~~ \nonumber \\ -AW_{11}(\theta )= & {} H_{11}(\theta ), \end{aligned}$$

(C.10)

where I denotes the $3\times 3$ identity matrix.

From (C.8), we know that for $\theta \in [-1,0)$,

$$\begin{aligned}&H(z,{\overline{z}},\theta ) = -{\overline{q}}^*(0)\cdot f_0(z,{\overline{z}})q(\theta ) - q^*(0)\cdot \nonumber \\&\quad {\overline{f}}_0(z,{\overline{z}}){\overline{q}}(\theta ) = -g(z,{\overline{z}})q(\theta ) - {\overline{g}}(z,{\overline{z}}){\overline{q}}(\theta ).\nonumber \\ \end{aligned}$$

(C.11)

Substituting (C.4) into (C.9) gives

$$\begin{aligned} H(z,{\overline{z}},\theta )= & {} \left[ -g_{20}q(\theta ) - {\overline{g}}_{02}{\overline{q}}(\theta )\right] \frac{z^2}{2} \nonumber \\&+ \left[ -g_{11}q(\theta ) - {\overline{g}}_{11}{\overline{q}}(\theta )\right] z{\overline{z}} \nonumber \\&+ \left[ -g_{02}q(\theta ) - {\overline{g}}_{20}{\overline{q}}(\theta )\right] \frac{{\overline{z}}^2}{2} + \cdots .\nonumber \\ \end{aligned}$$

(C.12)

Comparing the coefficients in (C.12) with those in (C.9) gives that

$$\begin{aligned} H_{20}(\theta )= & {} -g_{20}q(\theta ) - {\overline{g}}_{02}{\overline{q}}(\theta ), \nonumber \\ H_{11}(\theta )= & {} -g_{11}q(\theta ) - {\overline{g}}_{11}{\overline{q}}(\theta ). \end{aligned}$$

(C.13)

From (C.10), (C.13) and the definition of A, we have

$$\begin{aligned} {\dot{W}}_{20}(\theta ) = 2\mathrm{i}\nu _{40} \tau _{40}W_{20}(\theta ) + g_{20}q(\theta ) + {\overline{g}}_{02}{\overline{q}}(\theta ). \end{aligned}$$

(C.14)

Notice that $q(\theta ) = (v_1,v_2,v_3)^Te^{\mathrm{i}\nu _{40} \tau _{40}\theta } = q(0)e^{\mathrm{i}\nu _{40} \tau _{40}\theta }$, hence, using the method of variation of constants, the solution of (C.14) is given by

$$\begin{aligned} W_{20}(\theta )= & {} \dfrac{\mathrm{i}g_{20}}{\nu _{40} \tau _{40}}q(0)e^{\mathrm{i}\nu _{40} \tau _{40}\theta } + \dfrac{\mathrm{i}{\overline{g}}_{02}}{3\nu _{40} \tau _{40}}{\overline{q}}(0)e^{-\mathrm{i}\nu _{40} \tau _{40}\theta } \nonumber \\&+\, E_1e^{2\mathrm{i}\nu _{40} \tau _{40}\theta }, \end{aligned}$$

(C.15)

where $E_1 = \left( E_1^{(1)},E_1^{(2)},E_1^{(3)}\right) ^T \in R^3$ is a constant vector.

Similarly, from (C.10), (C.13) and the definition of A, we have

$$\begin{aligned} {\dot{W}}_{11}(\theta ) = g_{11}q(\theta ) + {\overline{g}}_{11}{\overline{q}}(\theta ), \end{aligned}$$

(C.16)

and

$$\begin{aligned} W_{11}(\theta )= & {} -\dfrac{\mathrm{i}g_{11}}{\nu _{40} \tau _{40}}q(0)e^{\mathrm{i}\nu _{40} \tau _{40}\theta } \nonumber \\&+\, \dfrac{\mathrm{i}{\overline{g}}_{11}}{\nu _{40} \tau _{40}}{\overline{q}}(0)e^{-\mathrm{i}\nu _{40} \tau _{40}\theta } + E_2, \end{aligned}$$

(C.17)

where $E_2 = \left( E_2^{(1)},E_2^{(2)},E_2^{(3)}\right) ^T \in {\mathbb {R}}^3$ is also a constant vector.

In what follows, we shall seek appropriate constant vectors $E_1$ and $E_2$ in (C.15) and (C.17), respectively. From (C.10) and the definition of A, we know that when $\theta = 0$,

$$\begin{aligned} A(0)W_{20}(\theta ) = \int _{-1}^0\mathrm{d}\eta (\theta )W_{20}(\theta ) = 2\mathrm{i}\nu _{40} \tau _{40}W_{20}(0) - H_{20}(0) \end{aligned}$$

(C.18)

and

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

(C.19)

where $\eta (\theta ) = \eta (\theta ,0)$. And from (C.8), we can obtain when $\theta = 0$,

$$\begin{aligned}&H_{20}(0) = - g_{20}q(0) - {\overline{g}}_{02}{\overline{q}}(0) + 2\nu _{40} \nonumber \\&\quad \begin{pmatrix} 0 \\ -b_1v_2^2-\alpha e^{-\alpha x^*}v_2v_3+\frac{1}{2}\alpha ^2y^*e^{-\alpha x^*}v_2^2 \\ -\delta v_3^2-\frac{1}{2}\beta ^2y^*e^{-\alpha x^*-2\mathrm{i}\nu _{40} \tau _{40}}v_2^2+\beta e^{-\alpha x^*-2\mathrm{i}\nu _{40} \tau _{40}}v_2v_3 \end{pmatrix}, \end{aligned}$$

(C.20)

$$\begin{aligned}&H_{11}(0) = - g_{11}q(0) - {\overline{g}}_{11}{\overline{q}}(0) + \tau _{50} \nonumber \\&\quad \begin{pmatrix} 0 \\ -2b_1v_2{\bar{v}}_2- \alpha e^{-\alpha x^*}v_2{\bar{v}}_3-\alpha e^{-\alpha x^*}{\bar{v}}_2v_3+\alpha ^2y^*e^{-\alpha x^*}v_2{\bar{v}}_2 \\ -2\delta v_3{\bar{v}}_3-\beta ^2y^*e^{-\alpha x^*}v_2{\bar{v}}_2+\beta e^{-\alpha x^*}v_2{\bar{v}}_3+\beta e^{-\alpha x^*}{\bar{v}}_2v_3 \end{pmatrix}, \end{aligned}$$

(C.21)

$$\begin{aligned}&\left( 2\mathrm{i}\omega \tau _{50}I - \int _{-1}^0e^{2\mathrm{i}\omega \tau _{50}\theta }\mathrm{d}\eta (\theta )\right) E_1 = 2\tau _{50}\nonumber \\&\quad \begin{pmatrix} 0 \\ -b_1v_2^2-\alpha e^{-\alpha x^*}v_2v_3+\frac{1}{2}\alpha ^2y^*e^{-\alpha x^*}v_2^2 \\ -\delta v_3^2-\frac{1}{2}\beta ^2y^*e^{-\alpha x^*-2\mathrm{i}\nu _{40} \tau _{40}}v_2^2+\beta e^{-\alpha x^*-2\mathrm{i}\nu _{40} \tau _{40}}v_2v_3 \end{pmatrix}.\nonumber \\ \end{aligned}$$

(C.22)

From the definition of A, we have

$$\begin{aligned} \int _{-1}^0e^{2\mathrm{i}\nu _{40} \tau _{40}\theta }\mathrm{d}\eta (\theta ) = A(\mu )e^{2\mathrm{i}\nu _{40} \tau _{40}\theta } = L_{\mu }\left( e^{2\mathrm{i}\nu _{40} \tau _{40}\theta }\right) . \end{aligned}$$

therefore, when $\mu = 0$, we have

$$\begin{aligned}&\int _{-1}^0e^{2\mathrm{i}\nu _{40} \tau _{40}\theta }\mathrm{d}\eta (\theta ) \\&\quad =\tau _{40} \begin{pmatrix} -c &{}\quad a &{}\quad 0\\ 0 &{}\quad -2b_1x_2^* - d_2 - \alpha y^* e^{-\alpha x_2^*} &{}\quad -\left( 1 - e^{-\alpha x_2^*}\right) \\ 0 &{}\quad 0 &{}\quad - 2\delta y^* - d \end{pmatrix} \\&\qquad +\, \tau _{40} \begin{pmatrix} 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad \beta y^*e^{-\beta x_2^*} &{}\quad 1 - e^{-\beta x_2^*} \end{pmatrix} e^{-2\mathrm{i}\nu _{40}\tau _{2}^*} \\&\qquad +\, \tau _{40} \begin{pmatrix} -b &{}\quad 0 &{}\quad 0\\ b &{}\quad 0 &{}\quad 0\\ 0 &{}\quad 0 &{}\quad 0 \end{pmatrix} e^{-2\mathrm{i}\nu _{40} \tau _{40}}, \end{aligned}$$

thus, we have

$$\begin{aligned}&\begin{pmatrix} 2\mathrm{i}\nu _{40}+c + be^{-2\mathrm{i}\nu _{40}\tau _{40}} &{}\quad a &{}\quad 0\\ -be^{-2\mathrm{i}\nu _{40}\tau _{40}} &{}\quad 2\mathrm{i}\nu _{40}+2b_1x_2^* + d_2 + \alpha y^* e^{-\alpha x_2^*} &{}\quad \left( 1 - e^{-\alpha x_2^*}\right) \\ 0 &{}\quad -\beta y^*e^{-\beta x_2^*} e^{-2\mathrm{i}\nu _{40}\tau _2^*} &{}\quad 2\mathrm{i}\nu _{40}+ 2\delta y^* + d - (1 - e^{-\beta x_2^*})e^{-2\mathrm{i}\nu _{40}\tau _2^*} \end{pmatrix}E_1 \nonumber \\&\quad = 2 \begin{pmatrix} 0 \\ -b_1v_2^2-\alpha e^{-\alpha x^*}v_2v_3+\frac{1}{2}\alpha ^2y^*e^{-\alpha x^*}v_2^2 \\ -\delta v_3^2-\frac{1}{2}\beta ^2y^*e^{-\alpha x^*-2\mathrm{i}\nu _{40} \tau _{40}}v_2^2+\beta e^{-\alpha x^*-2\mathrm{i}\nu _{40} \tau _{40}}v_2v_3 \end{pmatrix}, \end{aligned}$$

(C.23)

hence,

$$\begin{aligned} \begin{array}{l} E_1=2 \begin{pmatrix} 2\mathrm{i}\nu _{40}+c + be^{-2\mathrm{i}\nu _{40}\tau _{40}} &{}\quad a &{}\quad 0\\ -be^{-2\mathrm{i}\nu _{40}\tau _{40}} &{}\quad 2\mathrm{i}\nu _{40}+2b_1x_2^* + d_2 + \alpha y^* e^{-\alpha x_2^*} &{}\quad \left( 1 - e^{-\alpha x_2^*}\right) \\ 0 &{}\quad -\beta y^*e^{-\beta x_2^*} e^{-2\mathrm{i}\nu _{40}\tau _2^*} &{}\quad 2\mathrm{i}\nu _{40}+ 2\delta y^* + d - (1 - e^{-\beta x_2^*})e^{-2\mathrm{i}\nu _{40}\tau _2^*} \end{pmatrix}^{-1}\\ ~~~~~~\times \begin{pmatrix} 0 \\ -b_1v_2^2-\alpha e^{-\alpha x^*}v_2v_3+\frac{1}{2}\alpha ^2y^*e^{-\alpha x^*}v_2^2 \\ -\delta v_3^2-\frac{1}{2}\beta ^2y^*e^{-\alpha x^*-2\mathrm{i}\nu _{40} \tau _{40}}v_2^2+\beta e^{-\alpha x^*-2\mathrm{i}\nu _{40} \tau _{40}}v_2v_3 \end{pmatrix}. \end{array} \end{aligned}$$

(C.24)

Similarly, substituting (C.17) and (C.15) into (C.19), we obtain

$$\begin{aligned}&\int _{-1}^0\mathrm{d}\eta (\theta )E_2 = - \begin{pmatrix} 0 \\ -2b_1v_2{\bar{v}}_2- \alpha e^{-\alpha x^*}v_2{\bar{v}}_3-\alpha e^{-\alpha x^*}{\bar{v}}_2v_3+\alpha ^2y^*e^{-\alpha x^*}v_2{\bar{v}}_2 \\ -2\delta v_3{\bar{v}}_3-\beta ^2y^*e^{-\alpha x^*}v_2{\bar{v}}_2+\beta e^{-\alpha x^*}v_2{\bar{v}}_3+\beta e^{-\alpha x^*}{\bar{v}}_2v_3 \end{pmatrix}, \end{aligned}$$

(C.25)

hence,

$$\begin{aligned} \begin{array}{l} E_2= \begin{pmatrix} -c - b &{}\quad a &{}\quad 0\\ b &{}\quad -2b_1x_2^* - d_2 - \alpha y^* e^{-\alpha x_2^*} &{}\quad -\left( 1 - e^{-\alpha x_2^*}\right) \\ 0 &{}\quad \beta y^*e^{-\beta x_2^*} &{}\quad - 2\delta y^* - d + (1 - e^{-\beta x_2^*}) \end{pmatrix}^{-1}\\ ~~~~~~\times \begin{pmatrix} 0 \\ -2b_1v_2{\bar{v}}_2- \alpha e^{-\alpha x^*}v_2{\bar{v}}_3-\alpha e^{-\alpha x^*}{\bar{v}}_2v_3+\alpha ^2y^*e^{-\alpha x^*}v_2{\bar{v}}_2 \\ -2\delta v_3{\bar{v}}_3-\beta ^2y^*e^{-\alpha x^*}v_2{\bar{v}}_2+\beta e^{-\alpha x^*}v_2{\bar{v}}_3+\beta e^{-\alpha x^*}{\bar{v}}_2v_3 \end{pmatrix}. \end{array} \end{aligned}$$

(C.26)