Analysis of a delayed malaria transmission model including vaccination with waning immunity and reinfection

Abstract

To study the impact of infection delays in human and mosquito populations, vaccination with waning immunity and reinfection on the malaria transmission process, a malaria transmission model with these factors is developed and investigated. The local stability of disease-free and endemic equilibria have been discussed explicitly. By taking the delay as the bifurcation parameter, the existence of Hopf bifurcation is analyzed in four cases. Using normal form theory and center manifold theorem, direction and stability of Hopf bifurcation are discussed. Numerically, the bifurcation diagrams show that both delays can destabilize the endemic equilibrium and cause Hopf bifurcation and irregular oscillations, and that stability switches can occur mainly because of the delay in human. In addition, the malaria transmission case of Nigeria is studied. Numerical analysis reveals that ignoring the waning of immunity and reinfection may underestimate the infection risk and enlarge the critical value of Hopf bifurcation. Moreover, combined with sensitivity analysis, we can see that even though vaccination is not so effective in reducing the basic reproduction number, it is efficient for controlling the disease.

Appendices

Proof of Theorem 3.3

For any $\phi \in \mathbb {X}$, define a functional $g(\phi ):=(g_1(\phi ),g_2(\phi ),g_3(\phi ),g_4(\phi ),g_5(\phi ))^T:\mathbb {X}\rightarrow \mathbb {R}^5$ as

$$\begin{aligned} g(\phi )= \begin{pmatrix} \displaystyle b_h(1-p)-\beta _h\phi _1(-\tau _h)\phi _5(-\tau _h)-(\mu _h+\eta )\phi _1(0) \\ \displaystyle b_hp+\eta \phi _1(0)-\epsilon \beta _h\phi _2(-\tau _h)\phi _5(-\tau _h)-\mu _h \phi _2(0) \\ \displaystyle \beta _h( \phi _1(-\tau _h)+\epsilon \phi _2(-\tau _h)+\sigma \phi _4(-\tau _h))\phi _5(-\tau _h)-(\mu _h+\alpha +\gamma )\phi _3(0)\\ \displaystyle \gamma \phi _3(0)-\sigma \beta _h \phi _4(-\tau _h) \phi _5(-\tau _h)-\mu _h \phi _4(0)\\ \displaystyle \beta _m(\frac{b_m}{\mu _m}- \phi _5(-\tau _m))\phi _3(-\tau _m)- \mu _m \phi _5(0) \end{pmatrix}. \end{aligned}$$

Since $g(\phi )$ is continuous and Lipschitz in each compact set in $\mathbb {X}$, it follows from Theorems 2.2.1 and 2.2.3 in [22] that there exists a unique solution $u(t,\phi )=(S_h(t,\phi ),V(t,\phi ),I_h(t,\phi ),R_h(t,\phi ), I_m(t,\phi ))$ with respect to initial value $\phi$. So the system (1.2) has a unique solution $u(t,\phi )$ on its maximal existence interval $[0,\sigma _\phi )$. It is easy to see that $g_i(\phi )\ge 0$ if $\phi _i(0)\ge 0$, for $i=1,2,3,4,5$. Therefore, we can get the unique solution $u(t,\phi )$ on $\forall t\in [0,\sigma _\phi )$ which is non-negative.

Furthermore, let $N_h(t):=S_h(t)+V(t)+I_h(t)+R_h(t)$, it is easy to get from system (1.2) that $\frac{dN_h(t)}{dt}=b_h-\mu _hN_h-\alpha I_h$, which implies $\limsup _{t\rightarrow \infty }N_h(t)\le \frac{b_h}{\mu _h}$. Besides, $\limsup _{t\rightarrow \infty }N_m(t)=\frac{b_m}{\mu _m}$. Therefore, $u(t,\phi )$ is bounded. And then $\sigma _\phi =\infty$ by Theorem 2.3.1 in [22]. The proof is completed.

Coefficients of the characteristic equations

$$\begin{aligned} u_{04}&=- l_{11} - l_{22} - l_{33} - l_{44} - l_{55},\\ u_{03}&= l_{11} l_{22} + l_{11} l_{33} + l_{11} l_{44} + l_{22} l_{33} + l_{11} l_{55} + l_{22} l_{44} + l_{22} l_{55} + l_{33} l_{44} + l_{33} l_{55} + l_{44} l_{55},\\ u_{02}&=- l_{11} l_{22}( l_{33} + l_{44} + l_{55}) - l_{11} l_{33} ( l_{44} + l_{55} ) - l_{22} l_{55}(l_{33} + l_{44} )\\&\quad - l_{44} l_{55} (l_{33}+ l_{11})- l_{22} l_{33} l_{44},\\ u_{01}&= l_{11} l_{22} l_{33} l_{44} + l_{11} l_{22} l_{33} l_{55} + l_{11} l_{22} l_{44} l_{55} + l_{11} l_{33} l_{44} l_{55} + l_{22} l_{33} l_{44} l_{55},\\ u_{00}&=- l_{11} l_{22} l_{33} l_{44} l_{55},\; u_{14}=- m_{55},\; u_{13}= l_{11} m_{55} + l_{22} m_{55} + l_{33} m_{55} + l_{44} m_{55},\\ u_{12}&=-m_{55} ( l_{11} l_{22} + l_{11} l_{33} + l_{11} l_{44} + l_{22} l_{33} + l_{22} l_{44} + l_{33} l_{44}),\\ u_{11}&=m_{55} ( l_{11} l_{22} l_{33} + l_{11} l_{22} l_{44} + l_{11} l_{33} l_{44} + l_{22} l_{33} l_{44}),\\ u_{10}&=- l_{11} l_{22} l_{33} l_{44} m_{55},\; u_{24}=- h_{11} - h_{22} - h_{44},\\ u_{23}&=h_{11} ( l_{22} + l_{33}+ l_{44}+ l_{55} ) + h_{22} ( l_{11} + l_{33} + l_{44}+ l_{55})\\&\quad + h_{44} ( l_{11} + l_{22}+ l_{33} + l_{55}) - h_{34} l_{43},\\ u_{22}&= - h_{11} ( l_{22} ( l_{44}+ l_{33}+ l_{55})+ l_{33} ( l_{44}+ l_{55}) + l_{44} l_{55}) - h_{22} ( l_{33} (l_{44}+ l_{55})\\&\quad + l_{44} l_{55}+ l_{11} ( l_{33} + l_{44} + l_{55}))\\&\quad - h_{44} ( l_{11} ( l_{22} + l_{33}+ l_{55}) - l_{33} (l_{55} + l_{22} )- l_{22} l_{55}) + h_{34} l_{43} (l_{22} + l_{55} + l_{11}), \\ u_{21}&= h_{11} l_{22} ( l_{33} l_{44} + l_{33} l_{55}+ l_{44}l_{55} )+ l_{33} l_{44} l_{55}(h_{11}+ l_{33} l_{44} l_{55}) \\ {}&\quad + h_{22} l_{11} ( l_{33} l_{44} + l_{33} l_{55}+ l_{44} l_{55})\\&\quad - h_{34} l_{11}l_{43} ( l_{22} - l_{55}) - h_{34} l_{22} l_{43} l_{55}+ h_{44} l_{11} l_{22} ( l_{33} + l_{55} )\\ {}&\quad + h_{44} l_{33} l_{55}(l_{11}+ l_{22} ),\\ u_{20}&= h_{34} l_{11} l_{22} l_{43} l_{55} - h_{22} l_{11} l_{33} l_{44} l_{55} - h_{11} l_{22} l_{33} l_{44} l_{55}- h_{44} l_{11} l_{22} l_{33} l_{55},\\ u_{33}&= h_{11} m_{55} + h_{22} m_{55} - h_{35} m_{53} + h_{44} m_{55},\\ u_{32}&= -h_{11} m_{55}(l_{22} + l_{33} + l_{44} ) - h_{22} m_{55}(l_{11} + l_{33} + l_{44} )+ h_{34} l_{43} m_{55}\\&\quad - h_{44} m_{55}( l_{11}+ h_{44} m_{55}+ l_{33} )+ h_{35} m_{53}(l_{44} + l_{11} + l_{22} ),\\ u_{31}&= h_{11} m_{55} (l_{22} l_{33}+ l_{22} l_{44} + l_{33} l_{44} )+ h_{22}m_{55} (l_{11} l_{33} + l_{11} l_{44} + l_{33} l_{44} )\\&\quad - h_{35}m_{53} (l_{11} l_{22} + l_{11} l_{44} + l_{22} l_{44} )\\&\quad + h_{44} m_{55}(l_{11} l_{22}+ l_{11} l_{33} + l_{22} l_{33} ) - h_{34} l_{43} m_{55}(l_{11} + l_{22}),\\ u_{30}&=- ( l_{33} l_{44} m_{55}(h_{11} l_{22} + h_{22} l_{11} ) - h_{34} l_{11} l_{22} l_{43} m_{55} - h_{35} l_{11} l_{22} l_{44} m_{53}\\ {}&\quad + h_{44} l_{11} l_{22} l_{33} m_{55}),\\ u_{43}&= h_{11} h_{22} + h_{11} h_{44} + h_{22} h_{44},\\ u_{42}&= h_{11} (h_{34} l_{43} - h_{22} ( l_{44}+ l_{55} + l_{33})- h_{44} (l_{22} + l_{33}+ l_{55}))\\&\quad +h_{22}( h_{34} l_{43} - h_{44} ( l_{33}+ l_{11}+ l_{55})),\\ u_{41}&= h_{11} h_{22} l_{44} (l_{33}+ l_{55}) - h_{11} h_{34} l_{43}(l_{22} + l_{55}) + h_{11} l_{33} l_{55} (h_{44} + h_{22}) \\ {}&\quad + h_{11} h_{44} l_{22} ( l_{33}+ l_{55})\\&\quad - h_{22} h_{34} l_{43}(l_{11} + l_{55})+ h_{22} h_{44} l_{11} ( l_{33} + l_{55})+ h_{22} h_{44} l_{33} l_{55},\\ u_{40}&=h_{11} l_{55} ( h_{34} l_{22} l_{43} - h_{22} l_{33} l_{44} ) - h_{11} h_{44} l_{22} l_{33} l_{55} \\ {}&\quad + h_{22} h_{34} l_{11} l_{43} l_{55} - h_{22} h_{44} l_{11} l_{33} l_{55},\\ u_{52}&= -h_{11}(m_{55}(h_{22} + h_{44} )- h_{35} m_{53} ) - m_{53} ( h_{15} h_{31}- h_{35}( h_{22}+ h_{44}) \\&\quad + h_{25} h_{32} + h_{34} h_{45} )- h_{22} h_{44} m_{55},\\ u_{51}&= m_{53}(- h_{11} h_{35}( l_{22} + l_{44})+ h_{15} h_{31} ( l_{22} + l_{44} ) - h_{15} h_{32} l_{21} - h_{22} h_{35}( l_{11} + l_{44}),\\&\quad + h_{25} h_{32} ( l_{11} + l_{44}) + h_{34} h_{45} l_{11} - h_{35} h_{44} ( l_{11} + l_{22}) + h_{34} h_{45} l_{22} )+ m_{55}( h_{11} h_{22} ( l_{33} + l_{44} ), \\&\quad + h_{11} ( h_{44} l_{22} - h_{34} l_{43} + h_{44} l_{33})+ h_{22} (h_{44} l_{11}- h_{34} l_{43} + h_{44} l_{33} ) ),\\ u_{50}&= - h_{11} m_{55} (h_{22} l_{33} l_{44} - h_{34} l_{22} l_{43} + h_{44} l_{22} l_{33} )+ m_{53}(h_{11} h_{35} l_{22} l_{44}\\&\quad + h_{25} h_{32} l_{11} l_{44}- h_{35} h_{44} l_{11} l_{22} )\\&\quad - h_{15} l_{44} m_{53}( h_{31} l_{22}- h_{32} l_{21} ) + h_{22} l_{44} m_{53}(h_{35} l_{11}- h_{44} l_{33})\\&\quad - h_{34} ( h_{45} l_{11} l_{22} m_{53} - h_{22} l_{11} l_{43} m_{55}),\\ u_{62}&=- h_{11} h_{22} h_{44},\; u_{61}= h_{11} h_{22} h_{44} l_{33} - h_{11} h_{22} h_{34} l_{43} + h_{11} h_{22} h_{44} l_{55},\\ u_{60}&= h_{11} h_{22} h_{34} l_{43} l_{55} - h_{11} h_{22} h_{44} l_{33} l_{55},\\ u_{71}&= h_{11} h_{25} h_{32} m_{53} - h_{11} h_{22} h_{35} m_{53} + h_{15} h_{22} h_{31} m_{53} + h_{11} h_{22} h_{44} m_{55} \\ {}&\quad + h_{11} h_{34} h_{45} m_{53}\\&\quad - h_{11} h_{35} h_{44} m_{53} + h_{15} h_{31} h_{44} m_{53} + h_{22} h_{34} h_{45} m_{53} \\ {}&\quad - h_{22} h_{35} h_{44} m_{53} + h_{25} h_{32} h_{44} m_{53},\\ u_{70}&= - h_{11} h_{22} ( h_{44} l_{33} m_{55} - h_{35} l_{44} m_{53}- h_{34} l_{43} m_{55}) - h_{11} m_{53} (h_{25} h_{32} l_{44}\\&\quad + h_{34} h_{45} l_{22}- h_{35} h_{44} l_{22}) \\&\quad - h_{15} m_{53}( h_{22} h_{31} l_{44} + h_{31} h_{44} l_{22} - h_{32} h_{44} l_{21} )\\&\quad - h_{22} l_{11} m_{53}( h_{34} h_{45} - h_{35} h_{44} ) + h_{25} h_{32} h_{44} l_{11} m_{53},\\ u_{80}&=-m_{53} ( h_{11} h_{22} h_{34} h_{45} - h_{11} h_{22} h_{35} h_{44} + h_{11} h_{25} h_{32} h_{44} + h_{15} h_{22} h_{31} h_{44}).\\ p_{04}&=- l_{11} - 2 l_{22} - l_{33} - l_{55},\; p_{03}= 2 l_{11} l_{22} + l_{11} l_{33} + 2 l_{22} l_{33} + l_{11} l_{55} \\&\quad + 2 l_{22} l_{55} + l_{33} l_{55} + l_{22}^2,\\ p_{02}&=- l_{11} l_{22}^2 - l_{22}^2 l_{33} - l_{22}^2 l_{55} - l_{22}^2 m_{55} - 2 l_{11} l_{22} l_{33} - 2 l_{11} l_{22} l_{55} - l_{11} l_{33} l_{55} - 2 l_{22} l_{33} l_{55},\\ p_{01}&= l_{11} l_{22}^2 l_{33} + l_{11} l_{22}^2 l_{55} + l_{22}^2 l_{33} l_{55} + 2 l_{11} l_{22} l_{33} l_{55},\ p_{00}=- l_{11} l_{22}^2 l_{33} l_{55},\\ p_{14}&=- m_{55},\; p_{13}= l_{11} m_{55} + 2 l_{22} m_{55} + l_{33} m_{55},\; p_{12}=- 2 l_{11} l_{22} m_{55} \\&\quad - l_{11} l_{33} m_{55} - 2 l_{22} l_{33} m_{55},\\ p_{11}&= l_{11} l_{22}^2 m_{55} + l_{22}^2 l_{33} m_{55} + 2 l_{11} l_{22} l_{33} m_{55},\; p_{10}=- l_{11} l_{22}^2 l_{33} m_{55},\\ p_{24}&=-h_{11},\; p_{23}=2 h_{11} l_{22} + h_{11} l_{33} + h_{11} l_{55},\; p_{22}=- h_{11} l_{22}^2 - 2 h_{11} l_{22} l_{33} \\&\quad - 2 h_{11} l_{22} l_{55} - h_{11} l_{33} l_{55},\\ p_{21}&=h_{11} l_{22}^2 l_{33} + h_{11} l_{22}^2 l_{55} + 2 h_{11} l_{22} l_{33} l_{55},\; p_{20}=- h_{11} l_{22}^2 l_{33} l_{55},\\ p_{33}&= h_{11} m_{55} + h_{15} m_{53},\; p_{32}=- h_{15} l_{11} m_{53} - 2 h_{11} l_{22} m_{55} - 2 h_{15} l_{22} m_{53} - h_{11} l_{33} m_{55},\\ p_{31}&= h_{11} l_{22}^2 m_{55} + h_{15} l_{22}^2 m_{53} + 2 h_{15} l_{11} l_{22} m_{53} + 2 h_{11} l_{22} l_{33} m_{55},\\ p_{30}&= - h_{15} l_{11} l_{22}^2 m_{53} - h_{11} l_{22}^2 l_{33} m_{55}. \end{aligned}$$

Proofs of Theorems 3.2–3.7

Proof of Theorem 3.2

When $R_0\le 1,$ and $\tau _h=\tau _m=0$, the disease-free equilibrium $P^0$ is locally asymptotically stable. When $\tau _h>0,\tau _m>0$, it is obvious that Eq. (3.1) has roots with positive real part if and only if equation

$$\begin{aligned} \lambda ^2+a_1\lambda +a_2(1-R_0^2e^{-\lambda \tau _h}e^{-\lambda \tau _m})=0 \end{aligned}$$

(5.1)

with $a_1=-(l_{33}+l_{55}),a_2=l_{55}l_{33}$ has roots with positive real part. By substituting $\lambda =i\kappa$ into Eq. (5.1) and separating the real and imaginary parts, we have

$$\begin{aligned} \begin{array}{l} a_1\kappa =-a_2R_0^2\sin \kappa (\tau _h+\tau _m),\ -\kappa ^2+a_2=a_2R_0^2\cos \kappa (\tau _h+\tau _m). \end{array} \end{aligned}$$

(5.2)

Squaring and taking the sum of Eq. (5.2) yields $\kappa ^4+(a_1^2-2a_2)\kappa ^2+a_2^2(1-R_0^4)=0,$ with $a_1^2-2a_2= (l_{33})^2+(l_{55})^2>0$ and $1-R_0^4>0$ since $R_0<1$. Hence, all roots of Eq. (3.1) have negative real parts. Here completes the proof. $\square$

When $R_0>1,$ for $\tau _m=\tau _h=0$, Eq. (3.2) reduced to the following equation

$$\begin{aligned} \lambda ^5+ n_4\lambda ^4+n_3\lambda ^3+n_2\lambda ^2+n_1\lambda +n_0=0 \end{aligned}$$

with $n_4=\sum _{j=0}^2u_{j4},\ n_3=\sum _{j=0}^4u_{j3},\ n_2=\sum _{j=0}^6u_{j2},\ n_1=\sum _{j=0}^7u_{j1},\ n_0=\sum _{j=0}^8u_{j0}.$ According to the Routh-Hurwitz criteria gives $Re(\lambda )<0$ if and only if

$$\begin{aligned} \small D_1=n_1>0,\ D_2=\begin{vmatrix} n_1&n_0 \\ n_3&n_2 \end{vmatrix}>0,\ D_3=\begin{vmatrix} n_1&n_0&0 \\ n_3&n_2&n_1\\ 1&n_4&n_3 \end{vmatrix}>0,\ D_4=\begin{vmatrix} n_1&n_0&0&0\\ n_3&n_2&n_1&n_0\\ 1&n_4&n_3&n_2\\ 0&0&1&n_4 \end{vmatrix}>0. \end{aligned}$$

Proof of Theorem 3.3

For $\tau _h=0$, Eq. (3.2) reduced to the following equation

$$\begin{aligned}{} & {} \lambda ^5+ E_4\lambda ^4+E_3\lambda ^3+E_2\lambda ^2+E_1\lambda +E_0+\left( W_4\lambda ^4+W_3\lambda ^3+W_2\lambda ^2+W_1\lambda +W_0\right) \nonumber \\{} & {} \quad e^{-\lambda \tau _m}=0 \end{aligned}$$

(5.3)

with $E_4=u_{04}+u_{24},\ E_3=u_{03}+u_{23}+u_{43},\ E_j=u_{0j}+u_{2j}+u_{4j}+u_{6j}\;( j=0,1,2),\ W_4=u_{14},\ W_3=u_{13}+u_{33},\ W_2=u_{12}+u_{32}+u_{52}, W_1=u_{11}+u_{31}+u_{51}+u_{71},\ W_0=u_{10}+u_{30}+u_{50}+u_{70}+u_{80}.$ Suppose that $\lambda =i\kappa$ is a root of Eq. (5.3), then we have

$$\begin{aligned} \begin{array}{l} b_{11}(\kappa )\sin \kappa \tau _m+b_{12}(\kappa )\cos \kappa \tau _m=b_{13}(\kappa ),\ b_{12}(\kappa )\sin \kappa \tau _m-b_{11}(\kappa )\cos \kappa \tau _m\\ \qquad =b_{23}(\kappa ), \end{array} \end{aligned}$$

(5.4)

where $b_{11}(\kappa )=W_4\kappa ^4-W_2\kappa ^2+W_0,\ b_{12}(\kappa )=W_3\kappa ^3-W_1\kappa ,\ b_{13}(\kappa )=\kappa ^5-E_3\kappa ^3+E_1\kappa ,\ b_{23}(\kappa )=E_4\kappa ^4-E_2\kappa ^2+E_0,$ which implies

$$\begin{aligned} \kappa ^{10}+c_4\kappa ^8+c_3\kappa ^6+c_2\kappa ^4+c_1\kappa ^2+c_0=0 \end{aligned}$$

(5.5)

with $c_4=-2E_3+E_4^2-W_4^2,\ c_3=2E_1+E_3^2-2E_4E_2+2W_4W_2-W_3^2,\ c_2=E_2^2-W_2^2-2E_3E_1+2E_4E_0-2W_4W_0+2W_3W_1,\ c_1=E_1^2-2E_0E_2+2W_2W_0-W_1^2,\ c_0=E_0^2-W_0^2.$ For simplicity denote $\nu =\kappa ^2$, then Eq. (5.5) turns into

$$\begin{aligned} \mathcal {L}(\nu ):=\nu ^{5}+c_4\nu ^4+c_3\nu ^3+c_2\nu ^2+c_1\nu +c_0=0. \end{aligned}$$

(5.6)

If the assumption: $({\textbf {H}}_1):$ Eq. (5.6) has a positive root $\nu _0$ is satisfied, then, Eq. (5.5) has a positive root $\kappa _0=\sqrt{\nu _0}$. Eliminating $\sin \kappa \tau _m$ in Eq. (5.4) and letting $\kappa =\kappa _0$, we can obtain that

$$\begin{aligned} \tau _{m}^*=\frac{1}{\kappa _0}\arccos \left( \frac{b_{13}(\kappa _0)b_{12}(\kappa _0)-b_{23}(\kappa _0)b_{11}(\kappa _0)}{b_{12}^2(\kappa _0)+b_{11}^2(\kappa _0)}\right) . \end{aligned}$$

Substituting $\lambda (\tau _m)$ into Eq. (5.3), taking derivative with respect to $\tau _m$, we obtain

$$\begin{aligned}{} & {} \left( 5\lambda ^5+\sum _{j=4}^1jE_j\lambda ^{j-1} +(\sum _{j=4}^1jW_j\lambda ^{j-1}-\tau _m\sum _{j=4}^0W_j\lambda ^j)e^{-\lambda \tau _m}\right) \frac{d\lambda }{dt} \\{} & {} \quad =\lambda \sum _{j=4}^0W_j\lambda ^je^{-\lambda \tau _m} \end{aligned}$$

Therefore,

$$\begin{aligned} \left( \frac{d\lambda }{d\tau _m}\right) ^{-1}=\frac{(5\lambda ^4+\sum _{j=4}^1jE_j\lambda ^{j-1})e^{\lambda \tau _m}+\sum _{j=4}^1jW_j\lambda ^{j-1}}{\lambda \sum _{j=4}^0W_j\lambda ^j} -\frac{\tau _m}{\lambda }. \end{aligned}$$

Thus, when $\lambda =i\kappa _0$, we can get $\text {Re}\left( \frac{d\lambda }{d\tau _m}\right) ^{-1}_{\lambda =i\kappa _0}=\frac{g^{'}(\nu _0)}{b_{12}^2(\nu _0^2)+b_{11}^2(\nu _0^2)}.$ It can be seen that $\text {Re}\left( \frac{d\lambda }{d\tau _m}\right) ^{-1}_{\lambda =i\kappa _0}\ne 0$ if the assumption: $({\textbf {H}}_2): \mathcal {L}^{'}(\nu _0)=\frac{d\mathcal {L}(\nu )}{d\nu }|_{\nu =\nu _0}\ne 0$ is satisfied. Therefore, by the Hopf bifurcation theorem [25], Theorem 3.2 can be obtained if $({\textbf {H}}_1)$ and $({\textbf {H}}_2)$ hold. $\square$

Proof of Theorem 3.4

For $\tau _m=0$, Eq. (3.3) reduced to the following equation

$$\begin{aligned}{} & {} \lambda ^5+ F_4\lambda ^4+F_3\lambda ^3+F_2\lambda ^2+F_1\lambda +F_0+\left( X_4\lambda ^4+X_3\lambda ^3+X_2\lambda ^2+X_1\lambda +X_0\right) \nonumber \\{} & {} \quad e^{-\lambda \tau _h}=0 \end{aligned}$$

(5.7)

with $F_j=p_{0j}+p_{1j}\;(j=0,1,2,3,4),\; X_4=p_{24},\; X_j=p_{2j}+p_{3j}\;(j=0,1,2,3).$

Suppose that $\lambda =i\kappa$ is a root of Eq. (5.7), then we have

$$\begin{aligned} \begin{array}{l} d_{11}(\kappa )\sin \kappa \tau _h+d_{12}(\kappa )\cos \kappa \tau _h=d_{13}(\kappa ),\ d_{12}(\kappa )\sin \kappa \tau _h-d_{11}(\kappa )\cos \kappa \tau _h=d_{23}(\kappa ), \end{array}\nonumber \\ \end{aligned}$$

(5.8)

where $d_{11}(\kappa )=X_4\kappa ^4-X_2\kappa ^2+X_0,\ d_{12}(\kappa )=X_3\kappa ^3-X_1\kappa ,\ d_{13}(\kappa )=\kappa ^5-F_3\kappa ^3+F_1\kappa ,\ d_{23}(\kappa )=F_4\kappa ^4-F_2\kappa ^2+F_0,$ which implies

$$\begin{aligned} \kappa ^{10}+r_4\kappa ^8+r_3\kappa ^6+r_2\kappa ^4+r_1\kappa ^2+r_0=0 \end{aligned}$$

(5.9)

with $r_4=-2F_3+F_4^2-X_4^2,\ r_3=2F_1+F_3^2-2F_4F_2+2X_4X_2-X_3^2,\ r_2=F_2^2-X_2^2-2F_3F_1+2F_4F_0-2X_4X_0+2X_3X_1,\ r_1=F_1^2-2F_0F_2+2X_2X_0-X_1^2,\ r_0=F_0^2-X_0^2.$ For simplicity denote $\nu =\kappa ^2$, then Eq. (5.9) turns into

$$\begin{aligned} \mathcal {L}_1(\nu ):=\nu ^{5}+r_4\nu ^4+r_3\nu ^3+r_2\nu ^2+r_1\nu +r_0=0. \end{aligned}$$

(5.10)

If the assumption: $({\textbf {H}}_3):$ Eq. (5.10) has a positive root $\nu _1$ is satisfied, then, Eq. (5.9) has a positive root $\kappa _1=\sqrt{\nu _1}$. Eliminating $\sin \kappa \tau _h$ in Eq. (5.8) and letting $\kappa =\kappa _1$, we can obtain that $\tau _h^*=\frac{1}{\kappa _1}\arccos \left( \frac{d_{13}(\kappa _1)d_{12}(\kappa _1)-d_{23}(\kappa _1)d_{11}(\kappa _1)}{d_{12}^2(\kappa _1)+d_{11}^2(\kappa _1)}\right) .$

Similar to the proof of Theorem 3.3, differentiating Eq. (5.7) with respect $\tau _h$ and substituting $\lambda =i\kappa _1$, we can get $\text {Re}\left( \frac{d\lambda }{d\tau _h}\right) ^{-1}_{\lambda =i\kappa _1}=\frac{\mathcal {L}_1^{'}(\nu _1)}{d_{12}^2(\nu _1^2)+d_{11}^2(\nu _1^2)}.$ Thus, $\text {Re}\left( \frac{d\lambda }{d\tau _h}\right) ^{-1}_{\lambda =i\kappa _1}\ne 0$ if the assumption: $({\textbf {H}}_4): \mathcal {L}_1^{'}(\nu _1)=\frac{d\mathcal {L}_1(\nu )}{d\nu }|_{\nu =\nu _1}\ne 0$ is satisfied. Therefore, by the Hopf bifurcation theorem [25], Theorem 3.4 can be obtained if $({\textbf {H}}_3)$ and $({\textbf {H}}_4)$ hold. $\square$

Proof of Theorem 3.5

For $\tau _m=\tau _h=\tau >0$, Eq. (3.3) reduced to the following equation

$$\begin{aligned} \lambda ^5+ \sum _{j=4}^0g_j\lambda ^j+\sum _{j=4}^0y_{1j}\lambda ^je^{-\lambda \tau }+\sum _{j=3}^0y_{2j}\lambda ^je^{-2\lambda \tau }=0 \end{aligned}$$

(5.11)

with $g_j=p_{0j},\ y_{1j}=p_{1j}+p_{2j},j=0,\cdots ,4,\ y_{2j}=p_{3j},\ j=0,\cdots ,3.$

Multiplying $e^{\lambda \tau }$ on both sides of Eq. (5.11), we can get

$$\begin{aligned} \left( \lambda ^5+ \sum _{j=4}^0g_j\lambda ^j\right) e^{\lambda \tau }+\sum _{j=4}^0y_{1j}\lambda ^j+\sum _{j=3}^0y_{2j}\lambda ^je^{-\lambda \tau }=0 \end{aligned}$$

(5.12)

Let $\lambda =i\kappa$ be a root of Eq. (5.12), then we have $e_{11}(\kappa )\sin \kappa \tau +e_{12}(\kappa )\cos \kappa \tau =e_{13}(\kappa ),\ e_{21}(\kappa )\sin \kappa \tau +e_{22}(\kappa )\cos \kappa \tau =e_{23}(\kappa ),$ with $e_{11}(\kappa )=g_4\kappa ^4+(-g_2+y_{22})\kappa ^2+g_0-y_{20},\ e_{12}(\kappa )=\kappa ^5-(g_3+y_{23})\kappa ^3+(g_1+y_{21})\kappa ,\ e_{21}=-\kappa ^5+(g_3-y_{23})\kappa ^3-(g_1-y_{21})\kappa ,\ e_{22}=g_4\kappa ^4-(g_2+y_{22})\kappa ^2+g_0+y_{20},\ e_{13}(\kappa )=y_{13}\kappa ^3-y_{11}\kappa ,\ e_{23}(\kappa )=-y_{14}\kappa ^4+y_{12}\kappa ^2-y_{10},$ which implies

$$\begin{aligned} \sin \kappa \tau =\frac{e_{13}e_{22}-e_{12}e_{23}}{e_{11}e_{22}-e_{12}e_{21}},\ \cos \kappa \tau =\frac{e_{11}e_{23}-e_{13}e_{21}}{e_{11}e_{22}-e_{12}e_{21}}. \end{aligned}$$

(5.13)

Consequently, the following equation with respect to $\kappa$ is obtained

$$\begin{aligned} \kappa ^{20}+\sum _{j=9}^0s_j\kappa ^{2j}=0 \end{aligned}$$

(5.14)

with

$$\begin{aligned} s_9&= - 2g_4^2 - y_{14}^2 - 4g_3 ,\ s_8= 4g_1 + 2g_4(g_2 - y_{22}) + (g_4^2 + 2g_3)^2 - (y_{13} \\&\quad - g_4y_{14})^2 + 2y_{14}(y_{12} + g_4y_{13}\\&\quad + y_{14}(g_3 + y_{23})),+ 2g_4(g_2 + y_{22}) + 2(g_3 + y_{23})(g_3 - y_{23}),\\ s_7&= 2(y_{13} - g_4y_{14})(y_{11} - g_4y_{12} - y_{14}(g_2 - y_{22}) + y_{13}(g_3 - y_{23})) - 2g_4(g_0 - y_{20})\\&\quad - 2(g_4^2 + 2g_3)(2g_1 + g_4(g_2 - y_{22}) + g_4(g_2 + y_{22}) + (g_3 + y_{23})(g_3 - y_{23}))\\&\quad - 2y_{14}(y_{10} + g_4y_{11} + y_{14}(g_1 + y_{21}) + y_{13}(g_2 + y_{22}) + y_{12}(g_3 + y_{23})) \\&\quad - 2g_4(g_0 + y_{20}) - 2(g_1 + y_{21})(g_3 - y_{23}) - 2(g_2 + y_{22})(g_2 - y_{22})\\&\quad - 2(g_3 + y_{23})(g_1 - y_{21}) - (y_{12} + g_4y_{13} + y_{14}(g_3 + y_{23}))^2 ,\\ s_6&= 2(g_4^2 + 2g_3)(g_4(g_0 - y_{20}) + g_4(g_0 + y_{20}) + (g_1 + y_{21})(g_3 - y_{23})+ (g_2 + y_{22})(g_2 - y_{22}) ,\\&\quad + (g_3 + y_{23})(g_1 - y_{21})) - (y_{11} - g_4y_{12} - y_{14}(g_2 - y_{22})+ y_{13}(g_3 - y_{23}))^2\\&\quad + 2(y_{12} + g_4y_{13} + y_{14}(g_3 + y_{23}))(y_{10} + g_4y_{11} + y_{14}(g_1 + y_{21})\\&\quad + y_{13}(g_2 + y_{22}) + y_{12}(g_3 + y_{23})) + 2(g_0 + y_{20})(g_2 - y_{22}) + 2(g_1 + y_{21})(g_1 - y_{21})\\&\quad + 2(g_2 + y_{22})(g_0 - y_{20}) + (2g_1 + g_4(g_2 - y_{22}) + g_4(g_2 + y_{22}) + (g_3 + y_{23})(g_3 - y_{23}))^2 \\&\quad + 2(y_{13} - g_4y_{14})(g_4y_{10} + y_{14}(g_0 - y_{20}) - y_{13}(g_1 - y_{21}) + y_{12}(g_2 - y_{22}) - y_{11}(g_3 - y_{23}))\\&\quad + 2y_{14}(y_{13}(g_0 + y_{20}) + y_{12}(g_1 + y_{21}) + y_{11}(g_2 + y_{22}) + y_{10}(g_3 + y_{23})),\\ s_5&=- 2(y_{11} - g_4y_{12} - y_{14}(g_2 - y_{22}) + y_{13}(g_3 - y_{23}))(g_4y_{10} + y_{14}(g_0 - y_{20})\\&\quad - y_{13}(g_1 - y_{21}) + y_{12}(g_2 - y_{22}) - y_{11}(g_3 - y_{23})) \\&\quad - 2(y_{12} + g_4y_{13} + y_{14}(g_3 + y_{23}))(y_{13}(g_0 + y_{20}) \\&\quad + y_{12}(g_1 + y_{21}) + y_{11}(g_2 + y_{22}) + y_{10}(g_3 + y_{23})) \\&\quad - 2(2g_1 + g_4(g_2 - y_{22}) + g_4(g_2 + y_{22})\\&\quad + (g_3 + y_{23})(g_3 - y_{23}))(g_4(g_0 - y_{20}) + g_4(g_0 + y_{20}) + (g_1 + y_{21})(g_3 - y_{23})\\&\quad + (g_2 + y_{22})(g_2 - y_{22}) + (g_3 + y_{23})(g_1 - y_{21})) - (y_{10} + g_4y_{11} \\&\quad + y_{14}(g_1 + y_{21}) + y_{13}(g_2 + y_{22})\\&\quad + y_{12}(g_3 + y_{23}))^2- 2y_{14}(y_{11}(g_0 + y_{20}) + y_{10}(g_1 + y_{21})) \\&\quad - 2(g_4^2 + 2g_3)((g_0 + y_{20})(g_2 - y_{22}) \\&\quad + (g_1 + y_{21})(g_1 - y_{21})+ (g_2 + y_{22})(g_0 - y_{20})) - 2(g_0 + y_{20})(g_0 - y_{20}) \\&\quad - 2(y_{13} - g_4y_{14})(y_{12}(g_0 - y_{20}) - y_{11}(g_1 - y_{21}) + y_{10}(g_2 - y_{22})), \end{aligned}$$

$$\begin{aligned} s_4&=2((g_0 + y_{20})(g_2 - y_{22}) + (g_1 + y_{21})(g_1 - y_{21}) \\&\quad + (g_2 + y_{22})(g_0 - y_{20}))(2g_1 + g_4(g_2 - y_{22})\\&\quad + g_4(g_2 + y_{22}) + (g_3 + y_{23})(g_3 - y_{23})) - (g_4y_{10} + y_{14}(g_0 - y_{20}) \\&\quad - y_{13}(g_1 - y_{21}) + y_{12}(g_2 - y_{22}) ,\\&\quad - y_{11}(g_3 - y_{23}))^2 + 2(y_{13}(g_0 + y_{20}) + y_{12}(g_1 + y_{21}) \\&\quad + y_{11}(g_2 + y_{22}) + y_{10}(g_3 + y_{23}))(y_{10} \\&\quad + g_4y_{11} + y_{14}(g_1 + y_{21})+ y_{13}(g_2 + y_{22}) + y_{12}(g_3 + y_{23}))\\&\quad + 2(y_{11}(g_0 + y_{20}) + y_{10}(g_1 + y_{21}))(y_{12} \\&\quad + g_4y_{13} + y_{14}(g_3 + y_{23}))+ (g_4(g_0 - y_{20}) + g_4(g_0 + y_{20}) + (g_1 + y_{21})(g_3 - y_{23})\\&\quad + (g_2 + y_{22})(g_2 - y_{22})+ (g_3 + y_{23})(g_1 - y_{21}))^2+ 2(y_{12}(g_0 - y_{20}) - y_{11}(g_1 - y_{21}) \\&\quad + y_{10}(g_2 - y_{22}))(y_{11} - g_4y_{12} - y_{14}(g_2 - y_{22}) + y_{13}(g_3 - y_{23}))\\&\quad + 2(g_0 + y_{20})(g_0 - y_{20})(g_4^2 + 2g_3)\\&\quad + 2y_{10}(g_0 - y_{20})(y_{13} - g_4y_{14}),\\ s_3&= 2(y_{12}(g_0 - y_{20}) - y_{11}(g_1 - y_{21}) + y_{10}(g_2 - y_{22}))(g_4y_{10} + y_{14}(g_0 - y_{20}) - y_{13}(g_1 - y_{21})\\&\quad + y_{12}(g_2 - y_{22}) - y_{11}(g_3 - y_{23})) - 2((g_0 + y_{20})(g_2 - y_{22}) + (g_1 + y_{21})(g_1 - y_{21}) \\&\quad + (g_2 + y_{22})(g_0 - y_{20}))(g_4(g_0 - y_{20}) + g_4(g_0 + y_{20}) + (g_1 + y_{21})(g_3 - y_{23})\\&\quad + (g_2 + y_{22})(g_2 - y_{22}) + (g_3 + y_{23})(g_1 - y_{21})) - (y_{13}(g_0 + y_{20}) + y_{12}(g_1 + y_{21})\\&\quad + y_{11}(g_2 + y_{22}) + y_{10}(g_3 + y_{23}))^2 - 2(y_{11}(g_0 + y_{20}) \\&\quad + y_{10}(g_1 + y_{21}))(y_{10} + g_4y_{11} + y_{14}(g_1 + y_{21})\\&\quad + y_{13}(g_2 + y_{22}) + y_{12}(g_3 + y_{23})) - 2y_{10}(g_0 - y_{20})(y_{11} - g_4y_{12} \\&\quad - y_{14}(g_2 - y_{22}) + y_{13}(g_3 - y_{23})) \\&\quad - 2(g_0 + y_{20})(g_0 - y_{20})(2g_1 + g_4(g_2 - y_{22}) + g_4(g_2 + y_{22}) + (g_3 + y_{23})(g_3 - y_{23})),\\ s_2&= 2(y_{11}(g_0 + y_{20}) + y_{10}(g_1 + y_{21}))(y_{13}(g_0 + y_{20}) + y_{12}(g_1 + y_{21}) + y_{11}(g_2 + y_{22})\\&\quad + y_{10}(g_3 + y_{23})) - (y_{12}(g_0 - y_{20}) - y_{11}(g_1 - y_{21}) \\&\quad + y_{10}(g_2 - y_{22}))^2 + ((g_0 + y_{20})(g_2 - y_{22}) \\&\quad + (g_1 + y_{21})(g_1 - y_{21}) + (g_2 + y_{22})(g_0 - y_{20}))^2 + 2(g_0 + y_{20})(g_0 - y_{20})(g_4(g_0 - y_{20}) \\&\quad + g_4(g_0 + y_{20})+ (g_1 + y_{21})(g_3 - y_{23}) + (g_2 + y_{22})(g_2 - y_{22}) + (g_3 + y_{23})(g_1 - y_{21})) \\&\quad -2y_{10}(g_0 - y_{20})(g_4y_{10} + y_{14}(g_0 - y_{20}) - y_{13}(g_1 - y_{21}) \\&\quad + y_{12}(g_2 - y_{22}) - y_{11}(g_3 - y_{23})),\\ s_1&=2y_{10}(g_0 - y_{20})(y_{12}(g_0 - y_{20}) - y_{11}(g_1 - y_{21}) + y_{10}(g_2 - y_{22})) \\&\quad - 2(g_0 + y_{20})(g_0 - y_{20})((g_0 + y_{20})(g_2 - y_{22})+ (g_1 + y_{21})(g_1 - y_{21}) \\&\quad + (g_2 + y_{22})(g_0 - y_{20})) \\&\quad - (y_{11}(g_0 + y_{20}) + y_{10}(g_1 + y_{21}))^2,\ s_0=(g_0 + y_{20})^2(g_0 - y_{20})^2 - y_{10}^2(g_0 - y_{20})^2. \end{aligned}$$

For simplicity denote $\nu =\kappa ^2$, then Eq. (5.14) turns into

$$\begin{aligned} g(\nu ):=\nu ^{10}+\sum _{j=9}^0s_j\nu ^{j}=0. \end{aligned}$$

(5.15)

If the assumption: $({\textbf {H}}_5):$ Eq. (5.15) has a positive root $\nu _2$ is satisfied. then Eq. (5.14) has a positive root $\kappa _2=\sqrt{\nu _2}$. Letting $\kappa =\kappa _2$ in Eq. (5.13), we can obtain that

$$\begin{aligned} \tau ^*=\frac{1}{\kappa _2}\arccos \left( \frac{e_{13}(\kappa _2)e_{21}(\kappa _2)-e_{11}(\kappa _2)e_{23}(\kappa _2)}{e_{11}(\kappa _2)e_{22}(\kappa _2)-e_{12}(\kappa _2)e_{21}(\kappa _2)}\right) . \end{aligned}$$

Similar to the proof of Theorem 3.3, differentiating Eq. (5.11) with respect $\tau$ and substituting $\lambda =i\kappa _2$, we can get

$$\begin{aligned} \text {Re}\left( \frac{d\lambda }{d\tau }\right) ^{-1}_{\lambda =i\kappa _2}=\frac{e_2e_3-e_1e_4}{\kappa _2(e_3^2+e_4^2)}, \end{aligned}$$

where $e_1=-3y_{13}\kappa _2^3+y_{11}+(5\kappa _2^4-3g_3\kappa _2^2+g_1-3y_{23}\kappa _2^2+y_{21})\cos \kappa _2\tau ^*\ +(2y_{22}\kappa _2+4g_4\kappa _2^3-2g_2\kappa _2)\sin \kappa _2\tau ^*,\ e_2=-4y_{14}\kappa _2^3+2y_{12}\kappa _2+(5\kappa _2^4-3g_3\kappa _2^2+g_1+3y_{23}\kappa _2^2-y_{21})\sin \kappa _2\tau ^*\ +(2y_{22}\kappa _2-4g_4\kappa _2^3+2g_2\kappa _2)\sin \kappa _2\tau ^*,\ e_3=e_{11}(\kappa _2)\cos \kappa _2\tau ^*-e_{21}(\kappa _2)\sin \kappa _2\tau ^*,\;e_4=e_{12}(\kappa _2)\cos \kappa _2\tau ^*+e_{22}(\kappa _2)\sin \kappa _2\tau ^*.$ Therefore, $\text {Re}\left( \frac{d\lambda }{d\tau }\right) ^{-1}_{\lambda =i\kappa _2}\ne 0$ if the assumption: $({\textbf {H}}_6): e_2e_3-e_1e_4\ne 0$ is satisfied. Consequently, by the Hopf bifurcation theorem [25], Theorem 3.4 can be obtained if $({\textbf {H}}_5)$ and $({\textbf {H}}_6)$ hold. $\square$

Proof of Theorem 3.6

For $\tau _m>0$ and $\tau _h\in (0,\tau _h^*)$, Eq. (3.3) can be written as

$$\begin{aligned} \begin{array}{lll} \displaystyle \lambda ^5 +\sum _{j=4}^0\left( p_{ 0j}+p_{ 2j}e^{-\lambda \tau _h}\right) \lambda ^j +\left( p_{14}\lambda ^4+\sum _{j=3}^0\left( p_{ 1j}+p_{ 3j}e^{-\lambda \tau _h}\right) \lambda ^j\right) e^{-\lambda \tau _m}=0. \end{array} \end{aligned}$$

Considering $\tau _h$ as a parameter and letting $\lambda =i\kappa$, we can obtain

$$\begin{aligned} \begin{array}{l} f_{11}(\kappa )\sin \kappa \tau _m+f_{12}(\kappa )\cos \kappa \tau _m=f_{13}(\kappa ),\ f_{12}(\kappa )\sin \kappa \tau _m-f_{11}(\kappa )\cos \kappa \tau _m=f_{23}(\kappa ), \end{array} \end{aligned}$$

with $f_{11}(\kappa )= -p_{13}\kappa ^3+p_{11}\kappa -(p_{33}\kappa ^3-p_{31}\kappa )\cos \kappa \tau _h-(-p_{32}\kappa ^2+p_{30})\sin \kappa \tau _h,\$ $f_{12}(\kappa )= p_{14}\kappa ^4-p_{12}\kappa ^2+p_{10}-(p_{33}\kappa ^3-p_{31}\kappa )\sin \kappa \tau _h+(-p_{32}\kappa ^2+p_{30})\cos \kappa \tau _h,\ f_{13}(\kappa )= -(p_{04}\kappa ^4-p_{02}\kappa ^2+p_{00})+(p_{23}\kappa ^3-p_{21}\kappa )\sin \kappa \tau _h-(p_{24}\kappa ^4-p_{22}\kappa ^2+p_{20})\cos \kappa \tau _h,\ f_{23}(\kappa )= (\kappa ^5-p_{03}\kappa ^3+p_{01}\kappa )-(p_{23}\kappa ^3-p_{21}\kappa )\cos \kappa \tau _h-(p_{24}\kappa ^4-p_{22}\kappa ^2+p_{20})\sin \kappa \tau _h,$ which implies

$$\begin{aligned} \kappa ^{10}+\sum _{i=4}^0q_{0j}\kappa ^{2j}+\left( \sum _{i=4}^0q_{1j}\kappa ^{2j}\right) \cos \kappa \tau _h+\left( \sum _{i=4}^0q_{2j}\kappa ^{2j}\right) \kappa \sin \kappa \tau _h=0\nonumber \\ \end{aligned}$$

(5.16)

with

$$\begin{aligned} q_{04}&= p_{04}^2 - p_{14}^2 + p_{24}^2 - 2p_{03},\ q_{03}=p_{03}^2 - p_{13}^2 + p_{23}^2 - p_{33}^2 + 2p_{01} \\&\quad - 2p_{02}p_{04} + 2p_{12}p_{14} - 2p_{22}p_{24},\\ q_{02}&=p_{02}^2 - p_{12}^2 + p_{22}^2 - p_{32}^2 + 2(p_{00}p_{04} - p_{01}p_{03} - p_{10}p_{14} + p_{11}p_{13} \\&\quad + p_{20}p_{24} - p_{21}p_{23} + p_{31}p_{33}),\\ q_{01}&= p_{01}^2 - p_{11}^2 + p_{21}^2 -p_{31}^2 - 2p_{00}p_{02} + 2p_{10}p_{12} - 2p_{20}p_{22} + 2p_{30}p_{32},\\ q_{00}&= p_{00}^2 - p_{10}^2 + p_{20}^2 - p_{30}^2,\ q_{14}=2p_{04}p_{24} - 2p_{23},\\ q_{13}&=2p_{21} - 2p_{02}p_{24} + 2p_{03}p_{23} - 2p_{04}p_{22} - 2p_{13}p_{33} + 2p_{14}p_{32},\\ q_{12}&=2p_{00}p_{24} - 2p_{01}p_{23} + 2p_{02}p_{22} - 2p_{03}p_{21} + 2p_{04}p_{20} + 2p_{11}p_{33} \\&\quad - 2p_{12}p_{32} + 2p_{13}p_{31} - 2p_{14}p_{30},\\ q_{11}&= 2(p_{01}p_{21} - p_{00}p_{22} - p_{02}p_{20} + p_{10}p_{32} - p_{11}p_{31} + p_{12}p_{30}),\\ q_{10}&= 2p_{00}p_{20} - 2p_{10}p_{30},\ q_{24}= - 2p_{24},\ q_{23}= 2p_{22} + 2p_{03}p_{24} - 2p_{04}p_{23} + 2p_{14}p_{33},\\ q_{22}&= 2p_{02}p_{23} - 2p_{01}p_{24} - 2p_{20} - 2p_{03}p_{22} + 2p_{04}p_{21} - 2p_{12}p_{33} + 2p_{13}p_{32} - 2p_{14}p_{31},\\ q_{21}&=2p_{01}p_{22} - 2p_{00}p_{23} - 2p_{02}p_{21} + 2p_{03}p_{20} + 2p_{10}p_{33} - 2p_{11}p_{32} + 2p_{12}p_{31} - 2p_{13}p_{30},\\ q_{20}&= 2p_{00}p_{21} - 2p_{01}p_{20} - 2p_{10}p_{31} + 2p_{11}p_{30}. \end{aligned}$$

If the assumption: $({\textbf {H}}_7):$ Eq. (5.16) has a positive root $\hat{\kappa }$ is satisfied, then, from Eq. (5.8) we can obtain that $\hat{\tau }_m =\frac{1}{\hat{\kappa }}\arccos \left( \frac{f_{13}(\hat{\kappa })f_{12}(\hat{\kappa })-f_{23}(\hat{\kappa })f_{11}(\hat{\kappa })}{f_{12}^2(\hat{\kappa })+f_{11}^2(\hat{\kappa })}\right) .$

Similar to the proof of Theorem 3.3, differentiating Eq. (5.7) with respect $\tau _m$ and substituting $\lambda =i\hat{\kappa }$, we can get $\text {Re}\left( \frac{d\lambda }{d\tau _m}\right) ^{-1}_{\lambda =i\hat{\kappa }}=\frac{q_1q_4-q_2q_3}{\hat{\kappa }(q_1^2+q_2^2)},$ where $q_1=(p_{14}\hat{\kappa }^4-p_{12}\hat{\kappa }^2+p_{10}+p_{30}-p_{32}\hat{\kappa }^2)\cos \hat{\kappa }\hat{\tau }_m+(p_{11}\hat{\kappa }-p_{13}\hat{\kappa }^3 + p_{33}\hat{\kappa }^3-p_{31}\hat{\kappa })\sin \hat{\kappa }\hat{\tau }_m,\ q_2=-(p_{14}\hat{\kappa }^4-p_{12}\hat{\kappa }^2+p_{10}+p_{30}-p_{32}\hat{\kappa }^2)\sin \hat{\kappa }\hat{\tau }_m+(p_{11}\hat{\kappa }-p_{13}\hat{\kappa }^3 +p_{33}\hat{\kappa }^3-p_{31}\hat{\kappa })\cos \hat{\kappa }\hat{\tau }_m,\ q_3=5\hat{\kappa }^4-3p_{03}\hat{\kappa }^2+p_{01}+(p_{21}-3p_{23}\hat{\kappa }^2-\tau _h((p_{24}\hat{\kappa }^4-p_{22}\hat{\kappa }^2+p_{20})))\cos \hat{\kappa }\tau _h +(2p_{22}\hat{\kappa }-4p_{24}\hat{\kappa }^3-\tau _h(p_{23}\hat{\kappa }^3-p_{21}\hat{\kappa }))\sin \hat{\kappa }\tau _h +(p_{11}-3p_{13}\hat{\kappa }^2)\cos \hat{\kappa }\hat{\tau }_m +(2p_{12}\hat{\kappa }-4p_{14}\hat{\kappa }^3)\sin \hat{\kappa }\hat{\tau }_m +(p_{31}-3p_{33}\hat{\kappa }^2-\tau _h(-p_{32}\hat{\kappa }^2+p_{30}))\cos \hat{\kappa }(\tau _h+\hat{\tau }_m) +(2p_{32}\hat{\kappa }-\tau _h(p_{31}\hat{\kappa }-p_{33}\hat{\kappa }^3))\sin \hat{\kappa }(\tau _h+\hat{\tau }_m),\ q_4=-4p_{04}\hat{\kappa }^3+2p_{02}\hat{\kappa }+(3p_{23}\hat{\kappa }^2-p_{21}+\tau _h(p_{24}\hat{\kappa }^4-p_{22}\hat{\kappa }^2+p_{20}))\sin \hat{\kappa }\tau _h +(2p_{22}\hat{\kappa }-4p_{24}\hat{\kappa }^3-\tau _h(p_{23}\hat{\kappa }^3-p_{21}\hat{\kappa }))\cos \hat{\kappa }\tau _h +(3p_{13}\hat{\kappa }^2-p_{11})\sin \hat{\kappa }\hat{\tau }_m$

$+(2p_{12}\hat{\kappa }-4p_{14}\hat{\kappa }^3)\cos \hat{\kappa }\hat{\tau }_m +(3p_{33}\hat{\kappa }^2-p_{31}+\tau _h(p_{30}-p_{32}\hat{\kappa }^2))\sin \hat{\kappa }(\tau _h+\hat{\tau }_m) +(2p_{32}\hat{\kappa }-\tau _h(p_{31}\hat{\kappa }-p_{33}\hat{\kappa }^3))\cos \hat{\kappa }(\tau _h+\hat{\tau }_m).$ Thus, $\text {Re}\left( \frac{d\lambda }{d\tau _m}\right) ^{-1}_{\lambda =i\hat{\kappa }}\ne 0$ if the assumption: $({\textbf {H}}_8): q_1q_4-q_2q_3\ne 0$ is satisfied. Therefore, by the Hopf bifurcation theorem [25], Theorem 3.4 can be obtained if $({\textbf {H}}_7)$ and $({\textbf {H}}_8)$ hold. $\square$

Proof of Theorem 3.7

Define $C= C([-1,0],\mathbb {R}^5)$ the space of continuous real valued functions. Let $\tau _m=\hat{\tau }_m+\varrho$ and make time-scaling $t\rightarrow t/{\hat{\tau }_m}$. Let $x_1(t)=S_h(t)-S_h^*$, $x_2(t)=V(t)-V^*$, $x_3(t)=I_h(t)-I_h^*$, $x_4(t)=R_h(t)-R_h^*$, $x_5(t)=I_m(t)-I_m^*$, then model (1.2) is transformed into

$$\begin{aligned} \frac{dx(t)}{dt}=L_{ \varrho }(x_t)+F(\varrho ,x_t), \end{aligned}$$

(5.17)

where $x(t)=(x_1(t),x_2(t),x_3(t),x_4(t),x_5(t))^T\in \mathbb {R}^5$ and $x_t(\theta )=x(t+\theta )\in C([-1,0],\mathbb {R}^5)$. In (5.17), $L_{\varrho }: C \rightarrow \mathbb {R}^5$ and $F:\mathbb { R}\times C \rightarrow \mathbb {R}^5$ are given by

$$\begin{aligned} L_{\varrho }(\varphi )= (\hat{\tau }_m+\varrho )\left( A^{'}\varphi (0)+C^{'}\varphi (-\frac{\tau _h^*}{\hat{\tau }_m})+B^{'}\varphi (-1)\right) , \end{aligned}$$

where

$$\begin{aligned} \small A^{'}= \begin{pmatrix} a_{11} &{} 0 &{} 0 &{} 0 &{} 0\\ a_{21} &{} a_{22} &{} 0 &{} 0 &{}0\\ 0 &{} 0 &{} a_{33} &{} 0 &{} 0\\ 0 &{} 0 &{} a_{43} &{} a_{22} &{} 0\\ 0 &{} 0 &{} 0 &{} 0 &{} a_{55} \end{pmatrix},B^{'}= \begin{pmatrix} b_{11} &{} 0 &{} 0 &{} 0 &{} b_{15}\\ 0 &{} 0 &{} 0 &{} 0 &{}0\\ b_{31} &{} 0 &{} 0 &{} 0 &{} b_{35}\\ 0 &{} 0 &{} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 &{} 0 &{} 0 \end{pmatrix}, C^{'}= \begin{pmatrix} 0 &{} 0 &{} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 &{} 0 &{}0\\ 0 &{} 0 &{} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} c_{53} &{} 0 &{} c_{55} \end{pmatrix}. \end{aligned}$$

with $a_{11}=-\mu _h-\eta ,\ a_{21}=\eta ,\ a_{22}=-\mu _h,\ a_{33}=-(\mu _h+\alpha +\gamma ),\ a_{43}=\gamma ,\ \ a_{55}=-\mu _m,\ b_{11}=-\beta _h I_m^*,\ b_{15}=-\beta _h S_h^*,\ b_{31}=\beta _h I_m^*,\ b_{35}=\beta _hS_h^*,\ c_{53}=\beta _m (\frac{b_m}{\mu _m}-I_m^*),\ c_{55}=-\beta _m I_h^*,$ and

$$\begin{aligned} \small F= \begin{pmatrix} \displaystyle -\beta _hx_1(-\frac{\tau _h^*}{\hat{\tau }_m})x_5(-\frac{\tau _h^*}{\hat{\tau }_m}) \\ \displaystyle 0 \\ \displaystyle \beta _hx_1(-\frac{\tau _h^*}{\hat{\tau }_m})x_5(-\frac{\tau _h^*}{\hat{\tau }_m})\\ \displaystyle 0\\ \displaystyle -\beta _mx_3(-1)x_5(-1) \end{pmatrix}. \end{aligned}$$

According to the Riesz representation theorem, there exists a bounded variation function $\zeta (\theta ,\mu )$ in $\theta \in [-1,0]$ such that $\small L_{\varrho }(\varphi )=\int _{-1}^0d\zeta (\theta ,\varrho )\varphi (\theta ),\quad \varphi \in C.$ We select

$$\begin{aligned} \zeta (\theta ,\varrho )= {\left\{ \begin{array}{ll} \displaystyle (\hat{\tau }_m+\varrho )(A^{'}+B^{'}+C^{'}) ,&{}\theta =0, \\ \displaystyle (\hat{\tau }_m+\varrho )(B^{'}+C^{'}) ,&{}\theta \in [-\frac{\hat{\tau }_m}{\tau _h^*},0),\\ \displaystyle (\hat{\tau }_m+\varrho )B^{'} ,&{}\theta \in (-1,-\frac{\hat{\tau }_m}{\tau _h^*}),\\ \displaystyle 0 ,&{}\theta =-1,\\ \end{array}\right. } \end{aligned}$$

For $\varphi \in C$, define

$$\begin{aligned} \small A(\varrho )\varphi = {\left\{ \begin{array}{ll} \displaystyle \frac{d{\varphi }(\theta )}{d\theta } ,&{}\theta \in [-1,0), \\ \displaystyle \int _{-1}^0d\zeta (\varrho ,\theta )\varphi (\theta ) ,&{}\theta =0, \end{array}\right. } \end{aligned}$$

(5.18)

and $\small R(\mu )\varphi = {\left\{ \begin{array}{ll} 0,&{} \theta \in [-1,0) \\ F(\varrho ,\varphi ),&{} \theta =0 \end{array}\right. }.$ Then model (5.17) is equivalent to

$$\begin{aligned} \small \frac{dx_t}{dt}=A(\varrho )x_t+R(\varrho )x_t, \end{aligned}$$

(5.19)

where $x_t=u(t+\theta )$ for $\theta \in [-1,0]$.

For $\psi \in C^1([0,1],(\mathbb {R}^5)^*)$, being the conjugated space of $C^1([0,1],\mathbb {R}^5)$, define

$$\begin{aligned} \small A^*\psi (s)= {\left\{ \begin{array}{ll} \displaystyle -\frac{d{\psi }(s)}{ds},&{} s\in (0,1], \\ \displaystyle \int _{-1}^0d\zeta ^T(t,0)\psi (-t),&{} s=0, \end{array}\right. } \end{aligned}$$

and the bilinear inner product

$$\begin{aligned} \langle \psi ,\varphi \rangle =\bar{\psi }(0)\varphi (0) -\int _{-1}^0\int _{\xi =0}^{\theta }\bar{\psi }(\xi -\theta )d\zeta (\theta )\varphi (\xi )d\xi , \end{aligned}$$

(5.20)

where $\zeta (\theta )=\zeta (\theta ,0)$. From the discussion in Sect. 4, we know that $\pm i\hat{\kappa }\hat{\tau }_m$ are eigenvalues of A(0). Thus, $\pm i\hat{\kappa }\hat{\tau }_m$ are also eigenvalues of $A^*(0)$. We will calculate the eigenvectors of A(0) and $A^*(0)$ with respond to $\pm i\hat{\kappa }\hat{\tau }_m$.

Assume $q(\theta )=(1,q_2,q_3,q_4,q_5)^Te^{i\hat{\kappa }\hat{\tau }_m\theta }$ is the eigenvector of A(0) corresponding to $i\hat{\kappa }$, namely, $A(0)q(\theta )=i\hat{\kappa }\hat{\tau }_m q(\theta )$ and let $q^*(s)=D(1,q_1^*,q_2^*,q_3^*,q_4^*)^T e^{i\hat{\kappa }\hat{\tau }_ms }$ is the eigenvector corresponding to $-i\hat{\kappa }$, then we have

$$\begin{aligned} \begin{aligned} q_1&=\frac{a_{21}+b_{25}e^{-i\hat{\kappa }\tau _h^*}q_4}{i\hat{\kappa }-a_{22}},\ q_2=\frac{(i\hat{\kappa }-c_{55}e^{-i\hat{\kappa }\hat{\tau }_m})q_4}{c_{53}e^{-i\hat{\kappa }\hat{\tau }_m}},\\ q_3&=\frac{a_{43}(i\hat{\kappa }-c_{55}e^{-i\hat{\kappa }\hat{\tau }_m})}{c_{53}e^{-i\hat{\kappa }\hat{\tau }_m}}q_4,\ q_4=\frac{b_{15}e^{-i\hat{\kappa }\tau _h^*}}{i\hat{\kappa }-a_{11}-b_{11}e^{-i\hat{\kappa }\tau _h^*}},\\ q_1^*&=0,\ q_2^*=\frac{-i\hat{\kappa }-a_{11}-b_{11}e^{-i\hat{\kappa }\tau _h^*} }{b_{31}e^{-i\hat{\kappa }\tau _h^*}},\ q_3^*=0,\\ q_4^*&=\frac{(-i\hat{\kappa }-a_{33})(-i\hat{\kappa }-a_{11}-b_{11}e^{-i\hat{\kappa }\tau _h^*}) }{c_{53}b_{31}e^{-i\hat{\kappa }(\tau _h^*+\hat{\tau }_m)}}, \end{aligned} \end{aligned}$$

where D is a constant satisfying $\langle q^*(s),q(\theta )\rangle =1$. By (5.20), we get

$$\begin{aligned} \begin{aligned} \displaystyle&\langle q^*(s),q(\theta )\rangle \\ \displaystyle&={\bar{D}}(1+q_2{\bar{q}}_2^*+q_4\bar{q}_4^*+\tau _h^*e^{-i\hat{\kappa }\tau _h^*}(b_{11}+b_{31}q_2^*+q_1b_{32}q_2^* +q_4(b_{15}+b_{35}q_2^*))\\&\quad +\hat{\tau }_me^{-i\hat{\kappa }\hat{\tau }_h}q_4^*(c_{53}q_2+c_{55}q_4)) \end{aligned} \end{aligned}$$

Therefore, we can choose

$$\begin{aligned} {\bar{D}}{} & {} =(1+q_2{\bar{q}}_2^*+q_4\bar{q}_4^*+\tau _h^*e^{-i\hat{\kappa }\tau _h^*}(b_{11}+b_{31}q_2^*+q_1b_{32}q_2^* +q_4(b_{15}+b_{35}q_2^*)) \\ {}{} & {} \quad +\hat{\tau }_me^{-i\hat{\kappa }\hat{\tau }_h}q_4^*(c_{53}q_2+c_{55}q_4))^{-1} \end{aligned}$$

such that $\langle q^*,q \rangle =1$ and $\langle q^*,{\bar{q}} \rangle =0$.

To compute the center manifold $C_0$ at $\varrho =0$. Define

$$\begin{aligned} z(t)=\langle q^*, x_t \rangle ,\quad W(t,\theta )=x_t(\theta )-z(t)q(\theta )-{\bar{z}}(t){\bar{q}}(\theta ). \end{aligned}$$

(5.21)

On $C_0$, we have

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

(5.22)

Note that W is real if $x_t$ is real, so we deal the real solutions only. For solution $x_t\in C_0$ with $\zeta =0$, we have

$$\begin{aligned} \begin{aligned} \displaystyle \frac{dz(t)}{dt}&= i\omega z+{\bar{q}}^*(0)f(0,W(z(t),\bar{z}(t),0))+z(t)q(\theta )+{\bar{z}}(t){\bar{q}}(\theta )) \triangleq i\omega z\\&\quad +{\bar{q}}^*(0) f_0(z,{\bar{z}}). \end{aligned}\qquad \end{aligned}$$

(5.23)

Denote $f_0(z,{\bar{z}})\triangleq f_{z^2}\frac{z^2}{2}+f_{z\bar{z}}z{\bar{z}}+f_{z^2{\bar{z}}} z^2{\bar{z}}+\cdots ,$ and write equation (5.23) as $\frac{dz(t)}{dt}=i\omega z+g(z,{\bar{z}}).$ Besides, denote

$$\begin{aligned} g(z,{\bar{z}})={\bar{q}}^*(0)f_0(z,{\bar{z}})=g_{20}\frac{z^2}{2}+g_{11} z{\bar{z}}+g_{02}\frac{{\bar{z}}^2}{2}+g_{21}\frac{ z^2{\bar{z}}}{2}+\cdots . \end{aligned}$$

(5.24)

Then we have

$$\begin{aligned} g_{20}={\bar{q}}^*(0)f_{z^2},\ g_{11}={\bar{q}}^*(0)f_{z{\bar{z}}},\ g_{02}={\bar{q}}^*(0)f_{{\bar{z}}^2},\ g_{21}={\bar{q}}^*(0)f_{z^2{\bar{z}}}. \end{aligned}$$

(5.25)

From (5.21) and (5.22), it follows that

$$\begin{aligned} \begin{aligned} x_t(\theta )&=(1,q_1,q_2,q_3,q_4)^Te^{i\hat{\kappa }\hat{\tau }_m \theta }z+(1,{\bar{q}}_1,{\bar{q}}_2,{\bar{q}}_3,\bar{q}_4)^Te^{-i\hat{\kappa }\hat{\tau }_m \theta }{\bar{z}}\\&\quad +W_{20}(\theta )\frac{z^2}{2}+W_{11}(\theta )z\bar{z}+W_{02}(\theta )\frac{{\bar{z}}^2}{2}+\cdots . \end{aligned} \end{aligned}$$

We have

$$\begin{aligned} \begin{aligned} x_{1t}(-\frac{\tau _h^*}{\hat{\tau }_m})&=\displaystyle ze^{-i\hat{\kappa }\tau _h^*}+\bar{z}e^{i\hat{\kappa }\tau _h^*}+W_{20}^{(1)}(-\frac{\tau _h^*}{\hat{\tau }_m})\frac{z^2}{2}+W_{11}^{(1)}(-\frac{\tau _h^*}{\hat{\tau }_m})z\bar{z}+W_{02}^{(1)}(-\frac{\tau _h^*}{\hat{\tau }_m})\frac{\bar{z}^2}{2}+\cdots ,\\ x_{5t}(-\frac{\tau _h^*}{\hat{\tau }_m})&=\displaystyle q_4ze^{-i\hat{\kappa }\tau _h^*}+{\bar{q}}_4\bar{z}e^{i\hat{\kappa }\tau _h^*}+W_{20}^{(5)}(-\frac{\tau _h^*}{\hat{\tau }_m})\frac{z^2}{2}+W_{11}^{(5)}(-\frac{\tau _h^*}{\hat{\tau }_m})z\bar{z}+W_{02}^{(5)}(-\frac{\tau _h^*}{\hat{\tau }_m})\frac{\bar{z}^2}{2}+\cdots ,\\ x_{3t}(-1)&=\displaystyle q_2ze^{-i\hat{\kappa }\hat{\tau }_m}+\bar{q}_2\bar{z}e^{i\hat{\kappa }\hat{\tau }_m}+W_{20}^{(3)}(-1)\frac{z^2}{2}+W_{11}^{(3)}(-1)z\bar{z}+W_{02}^{(3)}(-1)\frac{{\bar{z}}^2}{2}+\cdots ,\\ x_{5t}(-1)&=\displaystyle q_4ze^{-i\hat{\kappa }\hat{\tau }_m}+\bar{q}_4\bar{z}e^{i\hat{\kappa }\hat{\tau }_m}+W_{20}^{(5)}(-1)\frac{z^2}{2}+W_{11}^{(5)}(-1)z\bar{z}+W_{02}^{(5)}(-1)\frac{{\bar{z}}^2}{2}+\cdots , \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{array}{lll} f_{z^2}= \begin{pmatrix} \displaystyle -\beta _hq_4e^{-2i\hat{\kappa }\tau _h^*} \\ \displaystyle 0 \\ \displaystyle \beta _hq_4e^{-2i\hat{\kappa }\tau _h^*}\\ \displaystyle 0\\ \displaystyle -\beta _mq_2q_4e^{-2i\hat{\kappa }\hat{\tau }_m} \end{pmatrix},\ f_{z{\bar{z}}}= \begin{pmatrix} \displaystyle -\beta _h(q_4+{\bar{q}}_4) \\ \displaystyle 0 \\ \displaystyle \beta _h(q_4+{\bar{q}}_4)\\ \displaystyle 0\\ \displaystyle -\beta _m(q_2q_4+{\bar{q}}_2{\bar{q}}_4) \end{pmatrix},\ f_{{\bar{z}}^2}= \begin{pmatrix} \displaystyle -\beta _h{\bar{q}}_4 e^{2i\hat{\kappa }\tau _h^*} \\ \displaystyle 0 \\ \displaystyle \beta _h{\bar{q}}_4e^{2i\hat{\kappa }\tau _h^*} \\ \displaystyle 0\\ \displaystyle -\beta _m{\bar{q}}_2{\bar{q}}_4 e^{2i\hat{\kappa }\hat{\tau }_m} \end{pmatrix},\\ f_{z^2{\bar{z}}}= \begin{pmatrix} \displaystyle -\beta _h(\frac{1}{2} {W_{20}^{(1)}(-\frac{\tau _h^*}{\hat{\tau }_m} )} {\bar{q}}_4e^{i\hat{\kappa }\tau _h^*} + W_{11}^{(5)}(-\frac{\tau _h^*}{\hat{\tau }_m} )e^{-i\hat{\kappa }\tau _h^*} +\frac{1}{2} {W_{20}^{(5)}(-\frac{\tau _h^*}{\hat{\tau }_m} )} e^{-i\hat{\kappa }\tau _h^*}) \\ \displaystyle 0 \\ \displaystyle \beta _h(\frac{1}{2} {W_{20}^{(1)}(-\frac{\tau _h^*}{\hat{\tau }_m} )} e^{i\hat{\kappa }\tau _h^*}{\bar{q}}_4+ W_{11}^{(5)}( -\frac{\tau _h^*}{\hat{\tau }_m})e^{-i\hat{\kappa }\tau _h^*} +\frac{1}{2} {W_{20}^{(5)}(-\frac{\tau _h^*}{\hat{\tau }_m} } ){\bar{q}}_4 e^{-i\hat{\kappa }\tau _h^*}) \\ \displaystyle 0\\ \displaystyle -\beta _m(\frac{1}{2} {W_{20}^{(3)}(-1 )} \bar{q}_4e^{i\hat{\kappa }\hat{\tau }_m} + W_{11}^{(3)}( -1)q_4 e^{-i\hat{\kappa }\hat{\tau }_m}+\frac{1}{2}{W_{20}^{(5)}( -1)} {\bar{q}}_2 e^{-i\hat{\kappa }\hat{\tau }_m}) \end{pmatrix}. \end{array} \end{aligned}$$

(5.26)

In order to get $g_{11}$, we still need to compute $W_{20}(\theta )$ and $W_{11}(\theta )$. From (5.19) and (5.21), we have

$$\begin{aligned} \begin{aligned} \dot{W}&=\displaystyle \dot{x}_t-\dot{z} q(\theta ) -\dot{{\bar{z}}} {\bar{q}}(\theta )\\&=\displaystyle {\left\{ \begin{array}{ll} A(0)W- {\bar{g}}{\bar{q}}(\theta )- gq(\theta ), &{} \theta \in [-1,0), \\ A(0)W- gq(0)- {\bar{g}} q(0) +f_0(z,{\bar{z}}),&{} \theta =0. \end{array}\right. } \end{aligned} \end{aligned}$$

(5.27)

On the other hand, in $C_0$, we can write (5.27) as

$$\begin{aligned} \begin{aligned} \dot{W}&=W_z\dot{z}+W_z\dot{{\bar{z}}}\\&=[W_{20}(\theta )z+W_{11}(\theta ){\bar{z}}](i\hat{\kappa }\hat{\tau }_m+g(z,{\bar{z}}))+[W_{11}(\theta )z+W_{02}(\theta )\bar{z}]\\&\quad (-i\hat{\kappa }\hat{\tau }_m+{\bar{g}}(z,{\bar{z}})). \end{aligned} \end{aligned}$$

(5.28)

Then substituting (5.22) and (5.24) into (5.27) and (5.28), comparing the coefficients of $\frac{z^2}{2}$ and $z{\bar{z}}$, one can get

$$\begin{aligned} (2i\hat{\kappa }\hat{\tau }_mI-A)W_{20}(\theta )= {\left\{ \begin{array}{ll} \displaystyle -g_{20}q(\theta )-{\bar{g}}_{02}{\bar{q}}(\theta ),&{} s\in [-1,0), \\ \displaystyle -g_{20}q(0)-{\bar{g}}_{02}{\bar{q}}(0)+f_{z^2},&{} s=0, \end{array}\right. } \end{aligned}$$

(5.29)

$$\begin{aligned} -AW_{11}(\theta )= {\left\{ \begin{array}{ll} \displaystyle -g_{11}q(\theta )-{\bar{g}}_{11}{\bar{q}}(\theta ),&{} s\in [-1,0), \\ \displaystyle -g_{11}q(0)-{\bar{g}}_{11}{\bar{q}}(0)+f_{z{\bar{z}}},&{} s=0, \end{array}\right. } \end{aligned}$$

(5.30)

From (5.18) and (5.29) we can see that when $\theta \in [-1,0),$ $W_{20}^{'}(\theta )=2i\hat{\kappa }\hat{\tau }_mW_{02}(\tau )+g_{20}q(\theta )+\bar{g}_{02}{\bar{q}}(\theta ),$ which has the solution

$$\begin{aligned} W_{20}(\theta )=\frac{ig_{02}}{\hat{\kappa }\hat{\tau }_m}q(0)e^{i\hat{\kappa }\hat{\tau }_m\theta } +\frac{i{\bar{g}}_{02}}{3\hat{\kappa }\hat{\tau }_m}\bar{q}(0)e^{-i\hat{\kappa }\hat{\tau }_m\theta } +\mathcal {E}_1e^{2i\hat{\kappa }\hat{\tau }_m\theta }. \end{aligned}$$

(5.31)

When $\theta =0$

$$\begin{aligned} \int _{-1}^0d\zeta (\theta )W_{20}(\theta )=2i\hat{\kappa }\hat{\tau }_mW_{20}+g_{02}q(0)+\bar{g}_{02}{\bar{q}}(0)-f_{z^2}. \end{aligned}$$

(5.32)

Substituting equation (5.31) into (5.32), one can obtain

$$\begin{aligned} \mathcal {E}_1=\left( 2i \hat{\kappa }\hat{\tau }_m I- \int _{-1}^0e^{2i\hat{\kappa }\hat{\tau }_m\theta }d\zeta (\theta )\right) f_{z^2}. \end{aligned}$$

(5.33)

From (5.18) and (5.30) we can see that when $\theta \in [-1,0),$ $W_{11}^{'}(\theta )= g_{11}q(\theta )+{\bar{g}}_{11}\bar{q}(\theta ),$ which has the solution

$$\begin{aligned} W_{11}(\theta )=-\frac{ig_{11}}{\hat{\kappa }\hat{\tau }_m}q(0)e^{i\hat{\kappa }\hat{\tau }_m\theta } +\frac{i{\bar{g}}_{11}}{\hat{\kappa }\hat{\tau }_m}\bar{q}(0)e^{-i\hat{\kappa }\hat{\tau }_m\theta } +\mathcal {E}_2. \end{aligned}$$

(5.34)

When $\theta =0$

$$\begin{aligned} \int _{-1}^0d\zeta (\theta )W_{11}(\theta )= g_{11}q(0)+{\bar{g}}_{11}\bar{q}(0)-f_{z{\bar{z}}}. \end{aligned}$$

(5.35)

Substituting equation (5.34) into (5.35) one can obtain

$$\begin{aligned} \mathcal {E}_2=\left( \int _{-1}^0d\zeta (\theta )\right) f_{z{\bar{z}}}. \end{aligned}$$

(5.36)

Therefore, from (5.25), (5.26), (5.33), (5.36) we can obtain

$$\begin{aligned} g_{20}&=2\hat{\tau }_m{\bar{D}}((-\beta _h q_4+{\bar{q}}_2^*\beta _h q_4)e^{-2\hat{\kappa }\tau _h^*}-\beta _m {\bar{q}}_4^*q_2q_4e^{-2\hat{\kappa }\hat{\tau }_m}),\\ g_{11}&=\hat{\tau }_m{\bar{D}}(\beta _h({\bar{q}}_4+q_4)(-1+{\bar{q}}_2^*) e^{-2\hat{\kappa }\tau _h^*}-\beta _m {\bar{q}}_4^*(q_2{\bar{q}}_4+{\bar{q}}_2q_4)e^{-2\hat{\kappa }\hat{\tau }_m}),\\ g_{02}&=2\hat{\tau }_m{\bar{D}}((-\beta _h {\bar{q}}_4+{\bar{q}}_2^*\beta _h q_4)e^{-2\hat{\kappa }\tau _h^*}-\beta _m {\bar{q}}_4^*{\bar{q}}_2{\bar{q}}_4e^{-2\hat{\kappa }\hat{\tau }_m})\\ g_{21}&=2\hat{\tau }_m{\bar{D}}[-\beta _h(\frac{1}{2}{W_{20}^{(1)}(-\frac{\tau _h^*}{\hat{\tau }_m} )} {\bar{q}}_4e^{i\hat{\kappa }\tau _h^*} + W_{11}^{(5)}(-\frac{\tau _h^*}{\hat{\tau }_m} )e^{-i\hat{\kappa }\tau _h^*} +\frac{1}{2} {W_{20}^{(5)}(-\frac{\tau _h^*}{\hat{\tau }_m} )} e^{-i\hat{\kappa }\tau _h^*}) \\&\quad +{\bar{q}}_2^*\beta _h(\frac{1}{2} {W_{20}^{(1)}(-\frac{\tau _h^*}{\hat{\tau }_m} )} e^{i\hat{\kappa }\tau _h^*}{\bar{q}}_4+ W_{11}^{(5)}( -\frac{\tau _h^*}{\hat{\tau }_m})e^{-i\hat{\kappa }\tau _h^*} +\frac{1}{2} {W_{20}^{(5)}(-\frac{\tau _h^*}{\hat{\tau }_m} } ){\bar{q}}_4 e^{-i\hat{\kappa }\tau _h^*})\\&\quad -{\bar{q}}_4^*\beta _m( \frac{1}{2}{W_{20}^{(3)}(-1 )} \bar{q}_4e^{i\hat{\kappa }\hat{\tau }_m} + W_{11}^{(3)}( -1)q_4 e^{-i\hat{\kappa }\hat{\tau }_m}+\frac{1}{2} {W_{20}^{(5)}( -1)} {\bar{q}}_2 e^{-i\hat{\kappa }\hat{\tau }_m})]. \end{aligned}$$

After analysis and computation, we have the following quantities:

$$\begin{aligned} \begin{aligned} C_1(0)&=\displaystyle \frac{i}{2\omega }\big (g_{20}g_{11}-2|g_{11}|^2-\frac{1}{3}|g_{02}|^2\big )+\frac{g_{21}}{2},\ \displaystyle \mu _2=\displaystyle -\frac{Re\{C_1(0)\}}{Re\{\lambda '(\hat{\tau }_m)\}},\\ \beta _2&=\displaystyle 2Re\{C_1(0)\},\ \displaystyle T_2=\displaystyle -\frac{Im(C_1(0))+\mu _2Im\{\lambda '(\hat{\tau }_m)\}}{\omega }. \end{aligned} \end{aligned}$$

$\square$