For the existence of interior periodic solutions of model (2), we can investigate the bifurcation near the pest free periodic solution, i.e. (*x*_{p}(*t*), *y*_{p}(*t*)). To do this, for computation convenience we first exchange the variables *x*(*t*) and *y*(*t*), and denote *u*(*t*) = *y*(*t*) and *v*(*t*) = *x*(*t*), then system (2) becomes as follows:

$$\begin{array}{}{\displaystyle \left\{\begin{array}{l}\left.\begin{array}{l}\frac{du(t)}{dt}=cp(v(t))u(t)-Du(t),\\ \frac{dv(t)}{dt}=v(t)g(v(t))-p(v(t))u(t),\end{array}\right\}t\ne nT,\\ \left.\begin{array}{l}u({t}^{+})=u(t)+\frac{\lambda}{1+\theta u(t)},\\ v({t}^{+})=(1-\frac{\delta v(t)}{v(t)+h})v(t),\end{array}\right\}\phantom{\rule{1em}{0ex}}t=nT.\end{array}\right.}\end{array}$$(10)

We let *Φ* be the flow associated to the first two equations of (10), and the fundamental solution matrix of (10) is *X*(*t*) = *Φ*(*t*, *X*_{0}) with *X*_{0} = *X*(0) = (*u*(0^{+}), *v*(0^{+})) and *Φ* = (*Φ*_{1}, *Φ*_{2}). We employ the notations used in this section as those in [33], then we can define the mapping *Θ*_{1}, *Θ*_{2}: *R*^{2} → *R*^{2} as follows

$$\begin{array}{}{\displaystyle \begin{array}{l}{\mathit{\Theta}}_{1}(u,v)=u+\frac{\lambda}{1+\theta u},{\mathit{\Theta}}_{2}(u,v)=(1-\frac{\delta v}{v+h})v\end{array}}\end{array}$$

and the mapping *F*_{1}, *F*_{2}: *R*^{2} → *R*^{2} by

$$\begin{array}{}{\displaystyle {F}_{1}(u,v)=cp(v)u-Du,{F}_{2}(u,v)=vg(v)-p(v)u.}\end{array}$$

Furthermore, we define *Ψ* : [0, +∞) × *R*^{2} → *R*^{2} by

$$\begin{array}{}{\displaystyle \mathit{\Psi}(T,{X}_{0})=\mathit{\Theta}(\mathit{\Phi}(T,{X}_{0}));\mathit{\Psi}(T,{X}_{0})=({\mathit{\Psi}}_{1}(T,{X}_{0}),{\mathit{\Psi}}_{2}(T,{X}_{0})).}\end{array}$$

Based on the above notations we can see that *Ψ* is determined by the values of solutions at impulsive points 0 and *T*, which is called as stroboscopic map of model (10) and *T* is the stroboscopic time snapshot. We know that *X* = (*u*, *v*) is a periodic solution of (10) with period *T* if and only if its initial value *X*_{0} = *X*(0) is a fixed point for map *Ψ*(*T*, ⋅). Therefore, in order to establish the existence of nontrivial periodic solutions of (10), we should prove the existence of the nontrivial fixed points of *Ψ*.

For model (10), it follows from the discussion in the previous section that model (10) has a stable boundary *T* periodic solution, denoted by

$$\begin{array}{}{\displaystyle \zeta (t)=({u}_{s}(t),0)=({y}^{\ast}\mathrm{exp}(-D(t-nT)),0),\phantom{\rule{thinmathspace}{0ex}}t\in (nT,(n+1)T],}\end{array}$$

where
$\begin{array}{}{\displaystyle {y}^{\ast}=\frac{-1+\sqrt{1+4\lambda \theta \mathrm{exp}(-DT)(1-\mathrm{exp}(-DT){)}^{-1}}}{2\theta \mathrm{exp}(-DT)}}\end{array}$ is a positive constant. In order to employ the analytical methods developed in [33, 34], we now consider the bifurcation of nontrivial periodic solutions near (*u*_{s}(*t*), 0) with initial value *X*(0) = (*u*_{s}(0^{+}), 0).

In order to obtain a nontrivial periodic solution of period *τ* with initial value *X*(0), we have only to find the fixed point problem *X* = *Ψ*(*τ*, *u*). Denoting *τ* = *T* + *τ̃*, *X* = *X*_{0} + *X̃*, and the fixed point problem *X* = *Ψ*(*τ*, *u*) equals to

$$\begin{array}{}{\displaystyle {X}_{0}+\stackrel{~}{X}=\mathit{\Psi}(T+\stackrel{~}{\tau},{X}_{0}+\stackrel{~}{X}).}\end{array}$$

Defining

$$\begin{array}{}{\displaystyle N(\stackrel{~}{\tau},\stackrel{~}{X})=({N}_{1}(\stackrel{~}{\tau},\stackrel{~}{X}),{N}_{2}(\stackrel{~}{\tau},\stackrel{~}{X}))={X}_{0}+\stackrel{~}{X}-\mathit{\Psi}(T+\stackrel{~}{\tau},{X}_{0}+\stackrel{~}{X}),}\end{array}$$(11)

so *X*_{0} + *X̃* is a fixed point of *Ψ*(*T*, ⋅) if *N*(*τ̃*, *X̃*) = 0.

According to the variational equations of the first two equations of (10), we have

$$\begin{array}{}{\displaystyle \frac{d}{dt}(\mathit{\Phi}(t,{X}_{0}))=F(\mathit{\Phi}(t,{X}_{0})),}\end{array}$$

which relates to the dynamics of the first two equations in (10). So we obtain that

$$\begin{array}{}{\displaystyle \frac{d}{dt}({D}_{X}(\mathit{\Phi}(t,{X}_{0})))={D}_{X}F(\mathit{\Phi}(t,{X}_{0}))({D}_{X}(\mathit{\Phi}(t,{X}_{0})))}\end{array}$$(12)

with the condition *D*_{X}(*Φ*(0, *X*_{0})) = *I*_{2}, which is the identity matrix in *M*_{2}(*R*). Thus, it follows from equation (10) that we have the particular form

$$\begin{array}{}{\displaystyle \frac{d}{dt}\left(\begin{array}{cc}\frac{\mathrm{\partial}{\mathit{\Phi}}_{1}}{\mathrm{\partial}u}& \frac{\mathrm{\partial}{\mathit{\Phi}}_{1}}{\mathrm{\partial}v}\\ \frac{\mathrm{\partial}{\mathit{\Phi}}_{2}}{\mathrm{\partial}u}& \frac{\mathrm{\partial}{\mathit{\Phi}}_{2}}{\mathrm{\partial}v}\end{array}\right)\phantom{\rule{negativethinmathspace}{0ex}}(t,{X}_{0})\phantom{\rule{negativethinmathspace}{0ex}}=\phantom{\rule{negativethinmathspace}{0ex}}\left(\begin{array}{cc}\frac{\mathrm{\partial}{F}_{1}(\zeta (t))}{\mathrm{\partial}u}& \frac{\mathrm{\partial}{F}_{1}(\zeta (t))}{\mathrm{\partial}v}\\ \frac{\mathrm{\partial}{F}_{2}(\zeta (t))}{\mathrm{\partial}u}& \frac{\mathrm{\partial}{F}_{2}(\zeta (t))}{\mathrm{\partial}v}\end{array}\right)\phantom{\rule{negativethinmathspace}{0ex}}\left(\begin{array}{cc}\frac{\mathrm{\partial}{\mathit{\Phi}}_{1}}{\mathrm{\partial}u}& \frac{\mathrm{\partial}{\mathit{\Phi}}_{1}}{\mathrm{\partial}v}\\ \frac{\mathrm{\partial}{\mathit{\Phi}}_{2}}{\mathrm{\partial}u}& \frac{\mathrm{\partial}{\mathit{\Phi}}_{2}}{\mathrm{\partial}v}\end{array}\right)\phantom{\rule{negativethinmathspace}{0ex}}(t,{X}_{0})}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}=\phantom{\rule{negativethinmathspace}{0ex}}\left(\phantom{\rule{negativethinmathspace}{0ex}}\begin{array}{cc}-D& c{p}^{\prime}(0){u}_{s}(t)\\ 0& g(0)-{p}^{\prime}(0){u}_{s}(t)\end{array}\phantom{\rule{negativethinmathspace}{0ex}}\right)\phantom{\rule{negativethinmathspace}{0ex}}\left(\begin{array}{cc}\frac{\mathrm{\partial}{\mathit{\Phi}}_{1}}{\mathrm{\partial}u}& \frac{\mathrm{\partial}{\mathit{\Phi}}_{1}}{\mathrm{\partial}v}\\ \frac{\mathrm{\partial}{\mathit{\Phi}}_{2}}{\mathrm{\partial}u}& \frac{\mathrm{\partial}{\mathit{\Phi}}_{2}}{\mathrm{\partial}v}\end{array}\right)\phantom{\rule{negativethinmathspace}{0ex}}(t,{X}_{0})}\end{array}$$

with initial value *D*_{X}(*Φ*(0, *X*_{0})) = *I*_{2}.

According to the initial value
$\begin{array}{}{\displaystyle \frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(0,{X}_{0})}{\mathrm{\partial}u}=0,}\end{array}$ we obtain the following

$$\begin{array}{}{\displaystyle \frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(t,{X}_{0})}{\mathrm{\partial}u}=\mathrm{exp}\left(\underset{0}{\overset{t}{\int}}(g(0)-{p}^{\prime}(0){u}_{s}(v))dv\right)\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(0,{X}_{0})}{\mathrm{\partial}u},}\end{array}$$

i.e.
$\begin{array}{}{\displaystyle \frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(t,{u}_{0})}{\mathrm{\partial}u}=0}\end{array}$ for all *t* > 0. Further, we obtain

$$\begin{array}{}{\displaystyle \frac{d}{dt}\frac{\mathrm{\partial}{\mathit{\Phi}}_{1}(t,{u}_{0})}{\mathrm{\partial}u}=-D\frac{\mathrm{\partial}{\mathit{\Phi}}_{1}(t,{u}_{0})}{\mathrm{\partial}u},}\\ {\displaystyle \frac{d}{dt}\frac{\mathrm{\partial}{\mathit{\Phi}}_{1}(t,{u}_{0})}{\mathrm{\partial}v}=-D\frac{\mathrm{\partial}{\mathit{\Phi}}_{1}(t,{u}_{0})}{\mathrm{\partial}v}+c{p}^{\prime}(0){u}_{s}(t)\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(t,{u}_{0})}{\mathrm{\partial}v}}\\ {\displaystyle \frac{d}{dt}\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(t,{u}_{0})}{\mathrm{\partial}v}=(g(0)-{p}^{\prime}(0){u}_{s}(t))\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(t,{u}_{0})}{\mathrm{\partial}v}.}\end{array}$$

According to the initial condition *D*_{X}(*Φ*(0, *X*_{0})) = *I*_{2}, we obtain that

$$\begin{array}{}{\displaystyle \frac{\mathrm{\partial}{\mathit{\Phi}}_{1}(t,{u}_{0})}{\mathrm{\partial}u}=\mathrm{exp}(-Dt),}\\ {\displaystyle \frac{\mathrm{\partial}{\mathit{\Phi}}_{1}(t,{u}_{0})}{\mathrm{\partial}v}=c{p}^{\prime}(0)\underset{0}{\overset{t}{\int}}\mathrm{exp}(-D(t-v)){u}_{s}(v)\rho (v))dv,}\\ {\displaystyle \frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(t,{u}_{0})}{\mathrm{\partial}v}=\rho (t)}\end{array}$$

for all 0 ≤ *t* ≤ *T*, where *ρ*(*t*) = exp
$\begin{array}{}{\int}_{0}^{t}(g(0)-{p}^{\prime}(0){u}_{s}(v))dv.\end{array}$

We then compute the derivation of *N* according to (11), and observe the following matrix

$$\begin{array}{}{\displaystyle {D}_{X}N(\stackrel{~}{\tau},\stackrel{~}{X})\phantom{\rule{negativethinmathspace}{0ex}}=\phantom{\rule{negativethinmathspace}{0ex}}\left(\phantom{\rule{negativethinmathspace}{0ex}}\begin{array}{cc}{a}^{\prime}& {b}^{\prime}\\ {c}^{\prime}& {d}^{\prime}\end{array}\right)}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}=\phantom{\rule{negativethinmathspace}{0ex}}{\left(\phantom{\rule{negativethinmathspace}{0ex}}\begin{array}{cc}\phantom{\rule{negativethinmathspace}{0ex}}1-\phantom{\rule{negativethinmathspace}{0ex}}(\phantom{\rule{negativethinmathspace}{0ex}}\frac{\mathrm{\partial}{\mathit{\Theta}}_{1}}{\mathrm{\partial}u}\frac{\mathrm{\partial}{\mathit{\Phi}}_{1}}{\mathrm{\partial}u}\phantom{\rule{negativethinmathspace}{0ex}}+\phantom{\rule{negativethinmathspace}{0ex}}\frac{\mathrm{\partial}{\mathit{\Theta}}_{1}}{\mathrm{\partial}v}\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}}{\mathrm{\partial}u}\phantom{\rule{negativethinmathspace}{0ex}})& -(\phantom{\rule{negativethinmathspace}{0ex}}\frac{\mathrm{\partial}{\mathit{\Theta}}_{1}}{\mathrm{\partial}u}\frac{\mathrm{\partial}{\mathit{\Phi}}_{1}}{\mathrm{\partial}v}\phantom{\rule{negativethinmathspace}{0ex}}+\phantom{\rule{negativethinmathspace}{0ex}}\frac{\mathrm{\partial}{\mathit{\Theta}}_{1}}{\mathrm{\partial}v}\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}}{\mathrm{\partial}v}\phantom{\rule{negativethinmathspace}{0ex}})\\ \phantom{\rule{negativethinmathspace}{0ex}}-(\phantom{\rule{negativethinmathspace}{0ex}}\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}u}\frac{\mathrm{\partial}{\mathit{\Phi}}_{1}}{\mathrm{\partial}u}\phantom{\rule{negativethinmathspace}{0ex}}+\phantom{\rule{negativethinmathspace}{0ex}}\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}v}\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}}{\mathrm{\partial}u}\phantom{\rule{negativethinmathspace}{0ex}})& \phantom{\rule{negativethinmathspace}{0ex}}1\phantom{\rule{negativethinmathspace}{0ex}}-\phantom{\rule{negativethinmathspace}{0ex}}(\phantom{\rule{negativethinmathspace}{0ex}}\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}u}\frac{\mathrm{\partial}{\mathit{\Phi}}_{1}}{\mathrm{\partial}v}\phantom{\rule{negativethinmathspace}{0ex}}+\phantom{\rule{negativethinmathspace}{0ex}}\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}v}\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}}{\mathrm{\partial}v}\phantom{\rule{negativethinmathspace}{0ex}})\end{array}\right)}_{(T+\stackrel{~}{\tau},{X}_{0}+\stackrel{~}{X})}.}\end{array}$$

Letting
$\begin{array}{}{\displaystyle {a}^{\prime}={a}_{0}^{\prime},{b}^{\prime}={b}_{0}^{\prime},{c}^{\prime}={c}_{0}^{\prime}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}{d}^{\prime}={d}_{0}^{\prime}}\end{array}$ when (*τ̃*, *X̃*) = (0, (0, 0)), at which
$\begin{array}{}{\displaystyle \frac{\mathrm{\partial}{\mathit{\Theta}}_{1}}{\mathrm{\partial}v}=\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}u}=\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}}{\mathrm{\partial}u}=0,}\end{array}$ then we have

$$\begin{array}{}{\displaystyle {D}_{X}N(0,(0,0))={\left(\begin{array}{cc}1-\frac{\mathrm{\partial}{\mathit{\Theta}}_{1}}{\mathrm{\partial}u}\frac{\mathrm{\partial}{\mathit{\Phi}}_{1}}{\mathrm{\partial}u}& -\frac{\mathrm{\partial}{\mathit{\Theta}}_{1}}{\mathrm{\partial}u}\frac{\mathrm{\partial}{\mathit{\Phi}}_{1}}{\mathrm{\partial}v}\phantom{\rule{negativethinmathspace}{0ex}}\\ 0& 1-\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}v}\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}}{\mathrm{\partial}v}\end{array}\right)}_{(T,{X}_{0})}.}\end{array}$$(13)

Thus, by simple calculations we have

$$\begin{array}{}{\displaystyle {a}_{0}^{\prime}=1-(1-\frac{\theta \lambda}{(1+\theta {y}^{\ast}\mathrm{exp}(-DT){)}^{2}})\mathrm{exp}(-DT)>0,}\\ {\displaystyle {b}_{0}^{\prime}=-(1-\frac{\theta \lambda}{(1+\theta {y}^{\ast}\mathrm{exp}(-DT){)}^{2}})c{p}^{\prime}(0){y}^{\ast}\mathrm{exp}(-DT)\underset{0}{\overset{T}{\int}}\rho (v)dv,}\\ {\displaystyle {c}_{0}^{\prime}=0}\end{array}$$

and

$$\begin{array}{}{\displaystyle {d}_{0}^{\prime}=1-\mathrm{exp}(\underset{0}{\overset{T}{\int}}(g(0)-{p}^{\prime}(0){u}_{s}(t))dt}\\ {\displaystyle \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=1-\rho (T).}\end{array}$$

Based on the above equations, we can see that *D*_{X}N(0, (0, 0)) is an upper triangular matrix with
$\begin{array}{}{\displaystyle {a}_{0}^{\prime}>0}\end{array}$. Thus, the necessary condition for the bifurcation of nontrivial solution is

$$\begin{array}{}{\displaystyle det[{D}_{X}N(0,(0,0))]=0,}\end{array}$$

which reduces to
$\begin{array}{}{\displaystyle {d}_{0}^{\prime}=0}\end{array}$. By simple calculation, we can see that
$\begin{array}{}{\displaystyle {d}_{0}^{\prime}=0}\end{array}$ is equivalent to *R*_{0} = 1. Therefore, in the following we focus on
$\begin{array}{}{\displaystyle {d}_{0}^{\prime}=0}\end{array}$ and address the sufficient condition for the bifurcation of nontrivial solution.

Note that dim (ker(*D*_{X}N(0, (0, 0)))) = 1 and a basis of ker(*D*_{X}N(0, (0, 0))) is
$\begin{array}{}{\displaystyle (-\frac{{b}_{0}^{\prime}}{{a}_{0}^{\prime}},1).}\end{array}$ Thus, the equation *N*(*τ̃*, *X̃*) = 0 is equivalent to

$$\begin{array}{}{\displaystyle {N}_{1}(\stackrel{~}{\tau},a{Y}_{0}+z{E}_{0})=0,\phantom{\rule{thinmathspace}{0ex}}{N}_{2}(\stackrel{~}{\tau},a{Y}_{0}+z{E}_{0})=0,}\end{array}$$

where *E*_{0} = (1, 0), *Y*_{0} =
$\begin{array}{}(-\frac{{b}_{0}^{\prime}}{{a}_{0}^{\prime}},1)\end{array}$
and *X̃* = *aY*_{0} + *zE*_{0} represents the direct summation decomposition of *X̃* using the projections onto ker(*D*_{X}N(0, (0, 0))) (i.e. the central manifold) and *Im*(*D*_{X}N(0, (0, 0))) (i.e. the stable manifold).

Now, we define

$$\begin{array}{}{\displaystyle {f}_{1}(\stackrel{~}{\tau},a,z)={N}_{1}(\stackrel{~}{\tau},a{Y}_{0}+z{E}_{0}),{f}_{2}(\stackrel{~}{\tau},a,z)={N}_{2}(\stackrel{~}{\tau},a{Y}_{0}+z{E}_{0}),}\end{array}$$

and consequently we have

$$\begin{array}{}{\displaystyle \frac{\mathrm{\partial}{f}_{1}}{\mathrm{\partial}z}(0,0,0)=\frac{\mathrm{\partial}{N}_{1}}{\mathrm{\partial}u}(0,(0,0))\frac{\mathrm{\partial}u}{\mathrm{\partial}z}={a}_{0}^{\prime}>0.}\end{array}$$

Therefore, based on the implicit function theorem, we confirm that there exists a unique continuous *z* as a function of *τ̃* and *a* such that *z* = *z*(*τ̃*, *a*) and *z*(0, 0) = 0, which can be solved from the equation *f*_{1}(*τ̃*, *a*, *z*) = 0 near (0, (0, 0)). Furthermore, we have

$$\begin{array}{}{\displaystyle {f}_{1}(\stackrel{~}{\tau},a,z(\stackrel{~}{\tau},a))={N}_{1}(\stackrel{~}{\tau},a{Y}_{0}+z(\stackrel{~}{\tau},a){E}_{0})=0}\end{array}$$

and

$$\begin{array}{}{\displaystyle \frac{\mathrm{\partial}{N}_{1}(0,0)}{\mathrm{\partial}u}(-\frac{{b}_{0}^{\prime}}{{a}_{0}^{\prime}})+\frac{\mathrm{\partial}{N}_{1}(0,0)}{\mathrm{\partial}u}\frac{\mathrm{\partial}z}{\mathrm{\partial}a}(0,0)+\frac{\mathrm{\partial}{N}_{2}(0,0)}{\mathrm{\partial}v}=0.}\end{array}$$

Thus, we have

$$\begin{array}{}{\displaystyle \frac{\mathrm{\partial}z}{\mathrm{\partial}a}(0,0)=-(\frac{\mathrm{\partial}{N}_{1}(0,0)}{\mathrm{\partial}u}{)}^{-1}\frac{\mathrm{\partial}{N}_{1}(0,0)}{\mathrm{\partial}v}+\frac{{b}_{0}^{\prime}}{{a}_{0}^{\prime}}=0.}\end{array}$$

It follows from (11) that

$$\begin{array}{}{\displaystyle \frac{\mathrm{\partial}z}{\mathrm{\partial}\stackrel{~}{\tau}}(0,0)=\frac{1}{{a}_{0}^{\prime}}\frac{\mathrm{\partial}{\mathit{\Theta}}_{1}}{\mathrm{\partial}u}\frac{\mathrm{\partial}{\mathit{\Phi}}_{1}(T,{X}_{0})}{\mathrm{\partial}\stackrel{~}{\tau}}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}=\frac{1}{{a}_{0}^{\prime}}(1-\frac{\theta \lambda}{(1+\theta {u}_{s}(T){)}^{2}}){\dot{u}}_{s}(T).}\end{array}$$

Therefore, we conclude that *N*(*τ̃*, *X̃*) = 0 if and only if

$$\begin{array}{}{\displaystyle {f}_{2}(\stackrel{~}{\tau},a)={N}_{2}(\stackrel{~}{\tau},a{Y}_{0}+z(\stackrel{~}{\tau},a){E}_{0})=0,}\end{array}$$(14)

and the number of its roots equals the number of periodic solutions of model (10).

For convenience, we denote

$$\begin{array}{}{\displaystyle f(\stackrel{~}{\tau},a)={f}_{2}(\stackrel{~}{\tau},a)}\end{array}$$

with *f*(0, 0) = *N*_{2}(0, (0, 0)) = 0. In order to study the properties of the function *f*, we first compute the derivatives of *f* around (0, 0). To do this, we compute the first order partial derivatives
$\begin{array}{}\frac{\mathrm{\partial}f}{\mathrm{\partial}\stackrel{~}{\tau}}(0,0)\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\frac{\mathrm{\partial}f}{\mathrm{\partial}a}(0,0)\end{array}$.

Denote *η*(*τ̃*) = *T* + *τ̃*, *η*_{1}(*τ̃*, *a*) = *x*_{0} −$\begin{array}{}\frac{{b}_{0}^{\prime}}{{a}_{0}^{\prime}}a\end{array}$
+ *z*(*τ̃*, *a*) and *η*_{2}(*τ̃*, *a*) = *a*, then we have

$$\begin{array}{}{\displaystyle \frac{\mathrm{\partial}f(\stackrel{~}{\tau},a)}{\mathrm{\partial}a}=\frac{\mathrm{\partial}}{\mathrm{\partial}a}({\eta}_{2}-{\mathit{\Theta}}_{2}(\mathit{\Phi}(\eta ,{\eta}_{1},{\eta}_{2})))}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}=1-\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}v}(\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}u}(\phantom{\rule{negativethinmathspace}{0ex}}-\frac{{b}_{0}^{\prime}}{{a}_{0}^{\prime}}\phantom{\rule{negativethinmathspace}{0ex}}+\phantom{\rule{negativethinmathspace}{0ex}}\frac{\mathrm{\partial}z(\stackrel{~}{\tau},a)}{\mathrm{\partial}a}\phantom{\rule{negativethinmathspace}{0ex}})\phantom{\rule{negativethinmathspace}{0ex}}+\phantom{\rule{negativethinmathspace}{0ex}}\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}v})}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}=1-\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}v}.}\end{array}$$

So, we have
$\begin{array}{}\frac{\mathrm{\partial}f(0,0)}{\mathrm{\partial}a}=1-\rho (T)\end{array}$
. It follows from
$\begin{array}{}{d}_{0}^{\prime}\end{array}$
= 1 − *ρ*(*T*) that
$\begin{array}{}{d}_{0}^{\prime}\end{array}$
= 0 indicates that
$\begin{array}{}\frac{\mathrm{\partial}f}{\mathrm{\partial}a}\end{array}$
(0, 0) = 0.

Similarly, we obtain that

$$\begin{array}{}{\displaystyle \frac{\mathrm{\partial}f}{\mathrm{\partial}\stackrel{~}{\tau}}(\stackrel{~}{\tau},a)=\frac{\mathrm{\partial}}{\mathrm{\partial}\stackrel{~}{\tau}}({\eta}_{2}-{\mathit{\Theta}}_{2}(\mathit{\Phi}(\eta ,{\eta}_{1},{\eta}_{2})))(\stackrel{~}{\tau},a)}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=-\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}v}(\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}\stackrel{~}{\tau}}+\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}u}\frac{\mathrm{\partial}z}{\mathrm{\partial}\stackrel{~}{\tau}}).}\end{array}$$

It follows from
$\begin{array}{}\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}\stackrel{~}{\tau}}\end{array}$
= 0 at (*τ̃*, *a*) = (0, 0) that

$$\begin{array}{}{\displaystyle \frac{\mathrm{\partial}f}{\mathrm{\partial}\stackrel{~}{\tau}}(0,0)=\frac{\mathrm{\partial}f}{\mathrm{\partial}a}(0,0)=0.}\end{array}$$

Furthermore, denote
$\begin{array}{}A=\frac{{\mathrm{\partial}}^{2}f(0,0)}{\mathrm{\partial}{\stackrel{~}{\tau}}^{2}},B=\frac{{\mathrm{\partial}}^{2}f(0,0)}{\mathrm{\partial}\stackrel{~}{\tau}\mathrm{\partial}a}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}C=\frac{{\mathrm{\partial}}^{2}f(0,0)}{\mathrm{\partial}{a}^{2}}\end{array}$
. In the following, we should calculate the second-order partial derivatives in term of the parameters of the equation.

$$\begin{array}{}{\displaystyle \frac{{\mathrm{\partial}}^{2}f(\stackrel{~}{\tau},a)}{\mathrm{\partial}{\stackrel{~}{\tau}}^{2}}=\frac{\mathrm{\partial}}{\mathrm{\partial}\stackrel{~}{\tau}}(-\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}v}(\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}\stackrel{~}{\tau}}+\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}u}\frac{\mathrm{\partial}z}{\mathrm{\partial}\stackrel{~}{\tau}}))}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\text{\hspace{0.17em}}=-\frac{{\mathrm{\partial}}^{2}{\mathit{\Theta}}_{2}}{\mathrm{\partial}{v}^{2}}(\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}\stackrel{~}{\tau}}+\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}u}\frac{\mathrm{\partial}z}{\mathrm{\partial}\stackrel{~}{\tau}}{)}^{2}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\text{\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}-\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}v}(\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}{\stackrel{~}{\tau}}^{2}}+2\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}u\mathrm{\partial}\stackrel{~}{\tau}}\frac{\mathrm{\partial}z}{\mathrm{\partial}\stackrel{~}{\tau}}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\text{\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}+\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}{u}^{2}}(\frac{\mathrm{\partial}z}{\mathrm{\partial}\stackrel{~}{\tau}}{)}^{2}+\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}u}\frac{{\mathrm{\partial}}^{2}z}{\mathrm{\partial}{\stackrel{~}{\tau}}^{2}}).}\end{array}$$

Since
$\begin{array}{}\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}}{\mathrm{\partial}{u}^{2}}=\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}}{\mathrm{\partial}\stackrel{~}{\tau}}=\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}}{\mathrm{\partial}u}=\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}}{\mathrm{\partial}u\mathrm{\partial}\stackrel{~}{\tau}}\end{array}$
= 0 for (*τ̃*, *a*) = (0, 0), we have

$$\begin{array}{}{\displaystyle A=\frac{{\mathrm{\partial}}^{2}f(0,0)}{\mathrm{\partial}{\stackrel{~}{\tau}}^{2}}=-\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(T,{X}_{0})}{\mathrm{\partial}{\stackrel{~}{\tau}}^{2}}.}\end{array}$$

According to
$\begin{array}{}\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(t,{X}_{0})}{\mathrm{\partial}{t}^{2}}\end{array}$
= 0 for all 0 ≤ *t* ≤ *T*, we have

$$\begin{array}{}{\displaystyle \frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(T,{X}_{0})}{\mathrm{\partial}{\stackrel{~}{\tau}}^{2}}=0,}\end{array}$$

which indicates that *A* = 0. By the same methods as shown above, we have

$$\begin{array}{}{\displaystyle \frac{{\mathrm{\partial}}^{2}f(\stackrel{~}{\tau},a)}{\mathrm{\partial}{a}^{2}}=\frac{\mathrm{\partial}}{\mathrm{\partial}a}\phantom{\rule{negativethinmathspace}{0ex}}(\phantom{\rule{negativethinmathspace}{0ex}}1-\phantom{\rule{negativethinmathspace}{0ex}}\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}v}(\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}u}(\phantom{\rule{negativethinmathspace}{0ex}}-\frac{{b}_{0}^{\prime}}{{a}_{0}^{\prime}}\phantom{\rule{negativethinmathspace}{0ex}}+\phantom{\rule{negativethinmathspace}{0ex}}\frac{\mathrm{\partial}z(\stackrel{~}{\tau},a)}{\mathrm{\partial}a}\phantom{\rule{negativethinmathspace}{0ex}})\phantom{\rule{negativethinmathspace}{0ex}}+\phantom{\rule{negativethinmathspace}{0ex}}\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}v}))}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{{\mathrm{\partial}}^{2}{\mathit{\Theta}}_{2}}{\mathrm{\partial}{v}^{2}}(\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}u}(-\frac{{b}_{0}^{\prime}}{{a}_{0}^{\prime}}+\frac{\mathrm{\partial}z}{\mathrm{\partial}a})+\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}v}{)}^{2}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}-\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}v}(\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}{u}^{2}}(-\frac{{b}_{0}^{\prime}}{{a}_{0}^{\prime}}+\frac{\mathrm{\partial}z}{\mathrm{\partial}a}{)}^{2})}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-2\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}v}\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}v\mathrm{\partial}u}(-\frac{{b}_{0}^{\prime}}{{a}_{0}^{\prime}}+\frac{\mathrm{\partial}z}{\mathrm{\partial}a})}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}-\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}v}(\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}u}\frac{{\mathrm{\partial}}^{2}z}{\mathrm{\partial}{a}^{2}}+\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}{v}^{2}}).}\end{array}$$

According to
$\begin{array}{}\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}v}=1,\frac{{\mathrm{\partial}}^{2}{\mathit{\Theta}}_{2}}{\mathrm{\partial}{v}^{2}}=-\frac{2\delta}{h}\end{array}$
for (*τ̃*, *a*) = (0, 0), in order to calculate *C*, we only need to calculate two terms, i.e.
$\begin{array}{}\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(t,{X}_{0})}{\mathrm{\partial}u\mathrm{\partial}v}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(t,{X}_{0})}{{\mathrm{\partial}}^{2}v}.\end{array}$.

Then, it follows that

$$\begin{array}{}{\displaystyle \frac{d}{dt}(\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(t,{X}_{0})}{\mathrm{\partial}u\mathrm{\partial}v})=(\frac{\mathrm{\partial}{F}_{2}(\zeta (t))}{\mathrm{\partial}v}+\frac{\mathrm{\partial}{F}_{2}(\zeta (t))}{\mathrm{\partial}u})(\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(t,{X}_{0})}{\mathrm{\partial}u\mathrm{\partial}v})}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}+\frac{{\mathrm{\partial}}^{2}{F}_{2}(\phantom{\rule{negativethinmathspace}{0ex}}\zeta (t)\phantom{\rule{negativethinmathspace}{0ex}})}{\mathrm{\partial}u\mathrm{\partial}v}\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(t,\phantom{\rule{negativethinmathspace}{0ex}}{X}_{0}\phantom{\rule{negativethinmathspace}{0ex}})}{\mathrm{\partial}v}\phantom{\rule{negativethinmathspace}{0ex}}+\phantom{\rule{negativethinmathspace}{0ex}}\frac{{\mathrm{\partial}}^{2}{F}_{2}(\zeta (t))}{\mathrm{\partial}{u}^{2}}\frac{\mathrm{\partial}{\mathit{\Phi}}_{1}(t,{X}_{0})}{\mathrm{\partial}v}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}=(g(0)-{p}^{\prime}(0){u}_{s}(t))\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(t,{X}_{0})}{\mathrm{\partial}u\mathrm{\partial}v}-{p}^{\prime}(0)\rho (t)}\end{array}$$

with initial condition
$\begin{array}{}\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(0,{X}_{0})}{\mathrm{\partial}u\mathrm{\partial}v}\end{array}$
= 0. So, we have

$$\begin{array}{}{\displaystyle \frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(T,{X}_{0})}{\mathrm{\partial}u\mathrm{\partial}v}=-{p}^{\prime}(0)\underset{0}{\overset{T}{\int}}\rho (u)\mathrm{exp}(\underset{u}{\overset{T}{\int}}(g(0)-{p}^{\prime}(0){u}_{s}(v))dv)du}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}=-{p}^{\prime}(0)\rho (T)T.}\end{array}$$

In order to obtain the formula for
$\begin{array}{}\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(t,{X}_{0})}{\mathrm{\partial}{v}^{2}}\end{array}$
, we have the following different equation

$$\begin{array}{}{\displaystyle \frac{d}{dt}(\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(t,{X}_{0})}{\mathrm{\partial}{v}^{2}})=\frac{\mathrm{\partial}{F}_{2}(\zeta (t))}{\mathrm{\partial}v}(\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(t,{X}_{0})}{\mathrm{\partial}{v}^{2}})+\frac{{\mathrm{\partial}}^{2}{F}_{2}(\zeta (t))}{\mathrm{\partial}{v}^{2}}\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(t,{X}_{0})}{\mathrm{\partial}v}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}+\frac{{\mathrm{\partial}}^{2}{F}_{2}(\zeta (t))}{\mathrm{\partial}v\mathrm{\partial}u}\frac{\mathrm{\partial}{\mathit{\Phi}}_{1}(t,{X}_{0})}{\mathrm{\partial}v}+\frac{\mathrm{\partial}{F}_{2}(\zeta (t))}{\mathrm{\partial}u}\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{1}(t,{X}_{0})}{\mathrm{\partial}{v}^{2}}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}=(g(0)-{p}^{\prime}(0){u}_{s}(t))\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(t,{X}_{0})}{\mathrm{\partial}{v}^{2}}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}+(2{g}^{\prime}(0)-{p}^{\u2033}(0){u}_{s}(t))\rho (t)}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}-{p}^{\prime}(0)\underset{0}{\overset{t}{\int}}\mathrm{exp}(-D(t-v))c{p}^{\prime}(0){u}_{s}(v)\rho (v))dv}\end{array}$$

with initial condition
$\begin{array}{}\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(0,{X}_{0})}{\mathrm{\partial}{v}^{2}}\end{array}$
= 0. Integrating the above equation one yields

$$\begin{array}{}{\displaystyle \frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(t,{x}_{0})}{\mathrm{\partial}{v}^{2}}=\underset{0}{\overset{t}{\int}}\phantom{\rule{negativethinmathspace}{0ex}}\mathrm{exp}\phantom{\rule{negativethinmathspace}{0ex}}(\phantom{\rule{negativethinmathspace}{0ex}}\underset{v}{\overset{t}{\int}}\phantom{\rule{negativethinmathspace}{0ex}}(g(0)-{p}^{\prime}(0){u}_{s}(\xi ))d\xi \phantom{\rule{negativethinmathspace}{0ex}})(2{g}^{\prime}(0)-{p}^{\u2033}(0){u}_{s}(v))\rho (v)dv}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}-c{p}^{\prime 2}(0)\underset{0}{\overset{t}{\int}}\{\mathrm{exp}(\underset{v}{\overset{t}{\int}}(g(0)-{p}^{\prime}(0){x}_{s}(\xi ))d\xi )\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\cdot \{\underset{0}{\overset{v}{\int}}\mathrm{exp}(-D(v-\theta )){u}_{s}(\theta )\rho (\theta )d\theta \}dv.}\end{array}$$

Therefore, we already deduce that

$$\begin{array}{}{\displaystyle C=\frac{2\delta}{h}(\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(T,{X}_{0})}{\mathrm{\partial}v}{)}^{2}+2\frac{{b}_{0}^{\prime}}{{a}_{0}^{\prime}}\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(T,{X}_{0})}{\mathrm{\partial}u\mathrm{\partial}v}-\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(T,{X}_{0})}{\mathrm{\partial}{v}^{2}}}\\ {\displaystyle \phantom{\rule{1em}{0ex}}=\rho (T)(\frac{2\delta \rho (T)}{h}\phantom{\rule{negativethinmathspace}{0ex}}-\phantom{\rule{negativethinmathspace}{0ex}}\frac{2{b}_{0}^{\prime}{p}^{\prime}(0)T}{{a}_{0}^{\prime}}\phantom{\rule{negativethinmathspace}{0ex}}-\phantom{\rule{negativethinmathspace}{0ex}}2{g}^{\prime}(0)T\phantom{\rule{negativethinmathspace}{0ex}}+\phantom{\rule{negativethinmathspace}{0ex}}\frac{{y}^{\ast}{p}^{\u2033}(0)(1-\mathrm{exp}(-DT))}{D})}\\ {\displaystyle \phantom{\rule{2em}{0ex}}+c{p}^{\prime 2}(0){y}^{\ast}\underset{0}{\overset{T}{\int}}\{\mathrm{exp}(\underset{v}{\overset{T}{\int}}(g(0)-{u}_{s}(t))dt){e}^{-Dv}\underset{0}{\overset{v}{\int}}\rho (\theta )d\theta \}dv.}\end{array}$$

For calculation of *B*, once again as above we have

$$\begin{array}{}{\displaystyle \frac{{\mathrm{\partial}}^{2}f(\stackrel{~}{\tau},a)}{\mathrm{\partial}\stackrel{~}{\tau}\mathrm{\partial}a}=\frac{\mathrm{\partial}}{\mathrm{\partial}a}(-\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}v}(\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}\stackrel{~}{\tau}}+\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}u}\frac{\mathrm{\partial}z}{\mathrm{\partial}\stackrel{~}{\tau}}))}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\text{\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}=-\frac{{\mathrm{\partial}}^{2}{\mathit{\Theta}}_{2}}{\mathrm{\partial}{v}^{2}}(\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}\stackrel{~}{\tau}}+\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}u}\frac{\mathrm{\partial}z}{\mathrm{\partial}\stackrel{~}{\tau}})}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\times (\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}u}(-\frac{{b}_{0}^{\prime}}{{a}_{0}^{\prime}}+\frac{\mathrm{\partial}z}{\mathrm{\partial}a})+\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}v})}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}-\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}v}(\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}\stackrel{~}{\tau}\mathrm{\partial}u}+\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}{u}^{2}}\frac{\mathrm{\partial}z}{\mathrm{\partial}\stackrel{~}{\tau}})(-\frac{{b}_{0}^{\prime}}{{a}_{0}^{\prime}}+\frac{\mathrm{\partial}z}{\mathrm{\partial}a})}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}-\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}v}(\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}\stackrel{~}{\tau}\mathrm{\partial}v}+\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}u\mathrm{\partial}v}\frac{\mathrm{\partial}z}{\mathrm{\partial}\stackrel{~}{\tau}})}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}-\frac{\mathrm{\partial}{\mathit{\Theta}}_{2}}{\mathrm{\partial}v}\frac{\mathrm{\partial}{\mathit{\Phi}}_{2}(\eta ,{\eta}_{1},{\eta}_{2})}{\mathrm{\partial}u}\frac{{\mathrm{\partial}}^{2}z}{\mathrm{\partial}\stackrel{~}{\tau}\mathrm{\partial}a}.}\end{array}$$

It follows from

$$\begin{array}{}{\displaystyle \frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(t,{X}_{0})}{\mathrm{\partial}v\mathrm{\partial}\stackrel{~}{\tau}}=\frac{\mathrm{\partial}F(\zeta (t))}{\mathrm{\partial}v}\mathrm{exp}(\underset{0}{\overset{t}{\int}}\frac{\mathrm{\partial}F(\zeta (t))}{\mathrm{\partial}v}dt),}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=(g(0)-{p}^{\prime}(0){u}_{s}(t))\rho (t)}\end{array}$$

and

$$\begin{array}{}{\displaystyle \frac{\mathrm{\partial}z(0,0)}{\mathrm{\partial}\stackrel{~}{\tau}}=-\frac{D}{{a}_{0}^{\prime}}(1-\frac{\theta \lambda}{(1+\theta {u}_{s}(T){)}^{2}}){u}_{s}(T)}\end{array}$$

that

$$\begin{array}{}{\displaystyle B=-\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(T,{X}_{0})}{\mathrm{\partial}\stackrel{~}{\tau}\mathrm{\partial}v}-\frac{{\mathrm{\partial}}^{2}{\mathit{\Phi}}_{2}(T,{X}_{0})}{\mathrm{\partial}u\mathrm{\partial}v}\frac{\mathrm{\partial}z(0,0)}{\mathrm{\partial}\stackrel{~}{\tau}}}\\ {\displaystyle \phantom{\rule{1em}{0ex}}=-\rho (T)(g(0)-{p}^{\prime}(0){u}_{s}(T)+\frac{{p}^{\prime}(0)TD{u}_{s}(T)}{{a}_{0}^{\prime}}(1-\frac{\lambda \theta}{(1+\theta {u}_{s}(T){)}^{2}})).}\end{array}$$

Furthermore, we study the Taylor expansion of *f*(*τ̃*, *a*) near (*τ̃*, *a*) = (0, 0). Since
$\begin{array}{}A=\frac{{\mathrm{\partial}}^{2}f(0,0)}{\mathrm{\partial}{\stackrel{~}{\tau}}^{2}},B=\frac{{\mathrm{\partial}}^{2}f(0,0)}{\mathrm{\partial}\stackrel{~}{\tau}\mathrm{\partial}a}\end{array}$
, and
$\begin{array}{}C=\frac{{\mathrm{\partial}}^{2}f(0,0)}{\mathrm{\partial}{a}^{2}}\end{array}$
, we have

$$\begin{array}{}{\displaystyle f(\stackrel{~}{\tau},a)=Ba\stackrel{~}{\tau}+C\frac{{a}^{2}}{2}+o(\stackrel{~}{\tau},a)({\stackrel{~}{\tau}}^{2}+{a}^{2})=\frac{a}{2}\stackrel{~}{f}(\stackrel{~}{\tau},a),}\end{array}$$

where
$\begin{array}{}\stackrel{~}{f}(\stackrel{~}{\tau},a)=2B\stackrel{~}{\tau}+Ca+\frac{1}{a}o(\stackrel{~}{\tau},a)({\stackrel{~}{\tau}}^{2}+{a}^{2}),\frac{\mathrm{\partial}\stackrel{~}{f}(0,0)}{\mathrm{\partial}\stackrel{~}{\tau}}=2B\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\frac{\mathrm{\partial}\stackrel{~}{f}(0,0)}{\mathrm{\partial}a}=C\end{array}$.

For *B* ≠ 0 (*C* ≠ 0), in order to employ the implicit function theorem about *f*(*τ̃*, *a*) = 0, we deduce that there exists a unique function *τ̃* = *σ*(*a*) (*a* = *γ*(*τ̃*)) near 0, which ensures that for all *a* (*τ̃*) near 0 there exists a *σ*(*a*) (*γ*(*τ̃*)) such that *f̃*(*σ*(*a*), *a*) = 0 (*f̃*(*τ̃*, *γ*(*τ̃*)) = 0) and *σ*(0) = 0 ( *γ*(0) = 0).

Therefore, equation (14) is equivalent to

$$\begin{array}{}{\displaystyle 2B\stackrel{~}{\tau}+Ca+\frac{1}{a}o(\stackrel{~}{\tau},a)({\stackrel{~}{\tau}}^{2}+{a}^{2})=0.}\end{array}$$

If *BC* ≠ 0, solving the above equation, we have
$\begin{array}{}\frac{a}{\stackrel{~}{\tau}}\simeq -\frac{2B}{C}\end{array}$
. If *BC* = 0, then the above equation can not be solved with relation to the interesting parameters. It is necessary to expand *f* to the third or a higher order if *BC* = 0, which is challenge for calculations. Finally, we have the following theorem

#### Theorem 3.1

*Assume that*
$\begin{array}{}{d}_{0}^{\prime}\end{array}$
= 0. *If* *BC* ≠ 0, *then in model (2) there occurs bifurcation at the threshold parameter values which satisfy*
$\begin{array}{}{d}_{0}^{\prime}\end{array}$
= 0, *and the bifurcation is supercritical provided* *BC* < 0 *and it is subcritical if* *BC* > 0.

Although Theorem 3.1 reveals the existence and stability of nontrivial periodic solution of model (2), the conditions including the sign *BC* are quite complex. This indicates that it is hard to clarify the effects of impulsive period and nonlinear pulse on the pest control. Therefore, in order to verify our main results we choose the Holling Type II functional response curve as an example in the coming section.

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