Proof (i) In view of Lemma 2.2, $U(T,0)$ admits the principal eigenvalue $r_{P}(U(T,0))$ corresponding to an eigenvector $\boldsymbol {\phi }^*=(\phi _A^*,\phi _M^*,\phi _F^*)^T \in P\setminus \{ 0\}$ . It suffices to show that $\boldsymbol {\phi }^* \in \mathrm {Int}(P)$ . Write $\lambda ^*=-\frac {\ln r_{P}(U(T,0))}{T} \leq 0$ and $\boldsymbol {w}^*(x,t)=[e^{\lambda ^*t}U(t,0)\boldsymbol {\phi }^*](x)$ , $\forall (x,t) \in \overline {\Omega } \times \mathbb {R}$ . It is easy to see that $\boldsymbol {w}^*=(w_A^*,w_M^*,w_F^*)^T$ is T-periodic in $t \in \mathbb {R}$ .

We first show that $\phi _M^*(x)>0$ , $\forall x \in \overline {\Omega }$ . Suppose that $\phi _M^*(x_M)=0$ for some $x_M \in \overline {\Omega }$ . We claim that $w_M^*(x,t) =0$ , $\forall (x,t) \in \overline {\Omega } \times \mathbb {R}$ . Otherwise, $w_M^*(x_0,t_0)>0$ for some $x_0 \in \overline {\Omega }$ and $t_0 \in [0,T)$ , it then follows from $[e^{\lambda ^*T}U(T,0)\boldsymbol {\phi }^*](x)=\boldsymbol {\phi }^*(x)$ and the strong maximum principle on the second equation of (3.2) that $\phi _M^*(x)>0$ , $\forall x \in \overline {\Omega }$ , which contradicts $\phi _M^*(x_M)=0$ . Thus, $w_M^*(x,t) =0$ , $\forall (x,t) \in \overline {\Omega } \times \mathbb {R}$ . Thanks to the second equation of (3.2) again with $\lambda $ replaced by $\lambda ^*$ , we have $w_A^*(x,t)=0$ , $\forall (x,t) \in \overline {\Omega } \times \mathbb {R}$ . This further implies that

$$ \begin{align*}\begin{cases} \dfrac{\partial w_F^*}{\partial t}=D\Delta w_F^*-\hat{d}(x,t) w_F^*+\lambda^* w_F^*,\quad \qquad x\in \Omega,\ t \in \mathbb{R},\\[3pt] \dfrac{\partial w_F^*}{\partial \nu}=0, \quad \qquad x\in \partial \Omega,\ t \in \mathbb{R}. \end{cases} \end{align*} $$

Integrating both sides of the above equation over $\Omega \times [0,T]$ and using the T-periodicity in t together with the Neumann boundary condition, we obtain $0=\int _0^T\int _\Omega (-\hat d(x,t)+\lambda ^*)w_F^*(x,t)\,\mathrm {d}x\mathrm {d}t$ . Since $w_F^*\geq 0$ , $\lambda ^*\leq 0$ , and $\hat d(x,t)>0$ , it follows that $w_F^*\equiv 0$ on $\overline \Omega \times \mathbb R$ , which implies $\boldsymbol \phi ^*\equiv 0$ . This contradicts $\boldsymbol \phi ^*\in P\setminus \{0\}$ . Therefore, $\phi _M^*(x)>0$ for all $x\in \overline \Omega $ .

Similarly, we can prove that $\phi _F^*(x)>0$ , $\forall x \in \overline {\Omega }$ . It suffices to show that $\phi _A^*(x)>0$ , $\forall x\in \overline {\Omega }$ . Suppose that $\phi _A^*(x_A)=0$ for some $x_A \in \overline {\Omega }$ . If $\phi _M^*(x_A) \phi _F^*(x_A)>0$ , then $w_A^*(x_A,t)>0$ , $\forall t>0$ , which contradicts $\phi _A^*(x_A)=w_A^*(x_A,T)=0$ . If $\phi _M^*(x_A) \phi _F^*(x_A)=0$ , then $\phi _M^*(x_A)=0$ or $\phi _F^*(x_A)=0$ , which is impossible. Therefore, $\phi _A^*(x)>0$ , $\forall x \in \overline {\Omega }$ , and hence, $\boldsymbol {\phi }^*=(\phi _A^*,\phi _M^*,\phi _F^*)^T \in \mathrm {Int}(P)$ .

(ii) Define

$$ \begin{align*}C_T(\overline{\Omega} \times \mathbb{R},\mathbb{R}_+^2)= \{\boldsymbol{w} \in C(\overline{\Omega} \times \mathbb{R},\mathbb{R}_+^2): \boldsymbol{w}(x,t)=\boldsymbol{w}(x,t+T), ~\forall (x,t) \in \overline{\Omega} \times \mathbb{R} \}. \end{align*} $$

For any $\lambda < \eta $ , define $p_{\lambda }(x,t)= p(x,t)+\lambda $ , $\forall (x,t) \in \overline {\Omega } \times \mathbb {R}$ , $\tilde {p}_{\lambda }=\tilde {p} +\lambda $ and let $R_{\lambda }[\boldsymbol {w}]$ be the unique nonnegative T-periodic solution of

$$ \begin{align*}\frac{\partial u_A}{\partial t}=\rho(x,t)H(u_M,u_F)-d(x,t)u_A-\gamma(x,t) u_A+\lambda u_A,\ x\in \overline{\Omega},~ t \in \mathbb{R}, \end{align*} $$

where $\boldsymbol {w}=(w_M,w_F)^T \in C_T(\overline {\Omega } \times \mathbb {R},\mathbb {R}_+^2)$ . It is easy to verify that for any $(x,t) \in \overline {\Omega } \times \mathbb {R}$ and $\lambda < \eta $ , there holds

$$ \begin{align*}\begin{aligned} \left[R_{\lambda}\boldsymbol{w}\right](x,t)=& e^{\int_{0}^{t} p_{\lambda}(x,\tau) \mathrm{d} \tau} (u_A)_0+ \int_{0}^{t} e^{\int_{s}^{t} p_{\lambda}(x,\tau) \mathrm{d} \tau} \rho(x,s)H(w_M(x,s),w_F(x,s)) \mathrm{d} s, \end{aligned} \end{align*} $$

where

$$ \begin{align*}(u_A)_0=\frac{\int_{0}^{T} e^{\int_{s}^{T} p_{\lambda}(x,\tau) \mathrm{d} \tau} \rho(x,s)H(w_M(x,s),w_F(x,s)) \mathrm{d} s }{1 - e^{\int_{0}^{T} p_{\lambda}(x,\tau) \mathrm{d} \tau }}. \end{align*} $$

Let $G=C_{0}^{1}(\overline \Omega _0)$ and $G_+=\{w \in G: w(x) \geq 0,~ \forall x \in \overline {\Omega }_0\}$ . Then $(G, G_+)$ is an ordered Banach space with

$$ \begin{align*}\mathrm{Int}(G_+)=\{w \in G: w(x)> 0,~ \forall x \in \Omega_0,~ \frac{\partial w}{\partial \nu}<0,~ \forall x \in \partial \Omega_0\}. \end{align*} $$

We denote by $C_T(\mathbb {R},G)$ the Banach space of continuous T-periodic functions from $\mathbb {R}$ to G, which is equipped with the maximum norm and the positive cone $C_T(\mathbb {R},G_+)$ . It is easy to see that $\mathrm {Int}(C_T(\mathbb {R},G_+))$ is nonempty.

Define two linear operators $\mathcal {L}_M$ and $\mathcal {L}_F$ by

$$ \begin{align*}\mathcal{L}_M u_M = \frac{\partial u_M}{\partial t} - D\Delta u_M + \hat{d} u_M,\quad \mathcal{L}_F u_F = \frac{\partial u_F}{\partial t} - D\Delta u_F + \hat{d} u_F, \end{align*} $$

with their domains $\mathcal {D}(\mathcal {L}_M)= \mathcal {D}(\mathcal {L}_F)\subset C_T(\overline {\Omega }_0 \times \mathbb {R},\mathbb {R})$ being

$$ \begin{align*}C_T^{2,1}(\overline{\Omega}_0 \times \mathbb{R},\mathbb{R}):=\{u \in C^{2,1}(\overline{\Omega}_0 \times \mathbb{R},\mathbb{R}):u(x,t)=u(x,t+T),~\forall (x,t) \in \overline{\Omega}_0 \times \mathbb{R} \}. \end{align*} $$

For convenience, let $q_M=\theta \gamma $ and $q_F=(1-\theta )\gamma $ . Next, we find a suitable $\overline {\lambda }< \eta $ such that

(3.3) $$ \begin{align} \begin{cases} \mathcal{L}_M \tilde{w}_M -q_M R_{\overline{\lambda}}\boldsymbol{\tilde{w}} \leq \overline{\lambda} \tilde{w}_M,\quad \qquad x \in \Omega_0, ~ t \in \mathbb{R}, \\[3pt] \mathcal{L}_F \tilde{w}_F -q_F R_{\overline{\lambda}}\boldsymbol{\tilde{w}} \leq \overline{\lambda} \tilde{w}_F,\quad \qquad x \in \Omega_0, ~ t \in \mathbb{R},\\[3pt] \tilde{w}_M=0, \tilde{w}_F=0,\quad \qquad x \in \partial \Omega_0,~ t \in \mathbb{R},\\[3pt] \tilde{w}_M \text{ and } \tilde{w}_F \text{ are } T \text{-periodic}, \end{cases} \end{align} $$

where $\boldsymbol {\tilde {w}}=(\tilde {w}_M,\tilde {w}_F)^T \in C_T(\overline {\Omega }_0 \times \mathbb {R},\mathbb {R}^2_+)$ .

For $N=M,F$ , let $\lambda _N$ be the principal eigenvalue of

$$ \begin{align*}\begin{cases} \mathcal{L}_{N} w =\lambda_N w ,\quad \qquad x \in \Omega_0, ~ t \in \mathbb{R},\\[3pt] w=0,\quad \qquad x \in \partial \Omega_0,~ t \in \mathbb{R},\\[3pt] w \text{ is } T \text{-periodic}, \end{cases} \end{align*} $$

with the positive eigenvector $\tilde {w}_N \in \mathrm {Int}(C_T(\mathbb {R},G_+)),$ where $\max _{(x,t) \in \overline {\Omega }_0 \times \mathbb {R}}\vert \tilde {w}_N(x,t) \vert =~1 $ . It is easy to verify that there exist $a>0$ and $b>0$ such that $a \tilde {w}_M \leq \tilde {w}_F \leq b \tilde {w}_M$ . Hence, there exist $c_M>0$ , $c_F>0$ , $\hat {c}_M>0,$ and $\hat {c}_F>0$ such that

$$ \begin{align*} &H(\tilde{w}_M(x,t),\tilde{w}_F(x,t)) \geq c_M \tilde{w}_M(x,t), \quad \\ & H(\tilde{w}_M(x,t),\tilde{w}_F(x,t)) \geq c_F \tilde{w}_F(x,t),~ \forall (x,t) \in \overline{\Omega}_0 \times \mathbb{R}, \end{align*} $$

and

$$ \begin{align*}\frac{1}{T} \int_{0}^{T}\tilde{w}_M(x,s) \mathrm{d} s \geq \hat{c}_M \tilde{w}_M(x,t),\quad \!\frac{1}{T} \!\int_{0}^{T}\tilde{w}_F(x,s) \mathrm{d} s \geq \hat{c}_F \tilde{w}_F(x,t),~ \forall (x,t) \in \overline{\Omega}_0 \times \mathbb{R}. \end{align*} $$

Choose $\hat {\eta }<\eta $ such that $\lambda _M> \hat {\eta }$ and $\lambda _F> \hat {\eta }$ . For any $\lambda \in [\hat {\eta },\eta )$ , there exists $\hat {c}_{1}>0$ independent of $\lambda $ such that $e^{\int _{s}^{t} p_{\lambda }(x,\tau )\mathrm {d} \tau }> \hat {c}_{1}$ , $\forall s,t \in \mathbb {R}$ with $0 \leq s \leq t \leq T$ , $x \in \overline {\Omega }_0$ . Moreover, there exist $\hat {c}_2>0$ , $\tilde {c}_M>0,$ and $\tilde {c}_F>0$ such that $\rho (x,t)> \hat {c}_2$ , $\theta \gamma (x,t) \geq \tilde {c}_M$ , $(1-\theta ) \gamma (x,t) \geq \tilde {c}_F$ , $\forall (x,t) \in \overline {\Omega }_0 \times \mathbb {R}$ . An easy computation yields that

$$ \begin{align*}\begin{aligned} \left[R_{\lambda} \boldsymbol{\tilde{w}}\right](x,t) &\geq \frac{\int_{0}^{T} \hat{c}_1^2 \hat{c}_2 c_M \tilde{w}_M(x,s) \mathrm{d}s} {1-e^{-(\eta-\lambda)T}} \geq \hat{c}_1^2 \hat{c}_2 c_M \frac{\int_{0}^{T}\tilde{w}_M(x,s)\mathrm{d}s} {(\eta-\lambda)T} \\ &\geq \frac{\hat{c}_1^2 \hat{c}_2 c_M\hat{c}_M} {\eta-\lambda}\tilde{w}_M(x,t), \quad (x,t)\in\overline{\Omega}_0\times\mathbb R,\ \lambda\in[\hat\eta,\eta), \end{aligned} \end{align*} $$

and

$$ \begin{align*}\left[R_{\lambda} \boldsymbol{\tilde{w}}\right](x,t) \geq \frac{\hat{c}_1^2 \hat{c}_2 c_F \hat{c}_F}{\eta-\lambda} \tilde{w}_F(x,t),~\forall (x,t) \in \overline{\Omega}_0 \times \mathbb{R},~ \lambda \in [\hat{\eta},\eta). \end{align*} $$

Therefore,

$$ \begin{align*}q_N R_{\lambda} \boldsymbol{\tilde{w}} \geq \frac{b_N}{\eta- \lambda} \tilde{w}_N, ~\forall \lambda \in [\hat{\eta},\eta),\end{align*} $$

where $b_N=\hat {c}_1^2 \hat {c}_2 \tilde {c}_N c_N \hat {c}_N$ for $N=M,F$ . We set $\tilde {\lambda }_N=\frac {b_N}{\hat {\eta }-\lambda _N} +\eta $ for $N=M,F$ , $\overline {\lambda }= \max (\tilde {\lambda }_M,\tilde {\lambda }_F,\hat {\eta })$ . It is easy to see that $\overline {\lambda }<\eta $ . A straightforward computation gives rise to

$$ \begin{align*}\begin{aligned} \mathcal{L}_M \tilde{w}_M -q_M R_{\overline{\lambda}} \boldsymbol{\tilde{w}} &\leq \lambda_M \tilde{w}_M - \frac{b_M}{\eta -\overline{\lambda}} \tilde{w}_M\leq \lambda_M \tilde{w}_M- \frac{b_M}{\eta - \tilde{\lambda}_M} \tilde{w}_M\\ &=\hat{\eta}\tilde{w}_M\leq \overline{\lambda} \tilde{w}_M,~\forall(x,t) \in \Omega_0 \times \mathbb{R}. \end{aligned} \end{align*} $$

By similar arguments, we also get

$$ \begin{align*}\mathcal{L}_F \tilde{w}_F -q_F R_{\overline{\lambda}} \boldsymbol{\tilde{w}} \leq \overline{\lambda} \tilde{w}_F,~\forall(x,t) \in \Omega_0 \times \mathbb{R}. \end{align*} $$

Therefore, (3.3) holds true.

Letting $\tilde {w}_A= R_{\overline {\lambda }}\boldsymbol {\tilde {w}}$ , we then have

$$ \begin{align*}\begin{cases} \dfrac{\partial \tilde{w}_A}{\partial t}=\rho(x,t)H(\tilde{w}_M,\tilde{w}_F)-d(x,t)\tilde{w}_A-\gamma(x,t) \tilde{w}_A+\overline{\lambda} \tilde{w}_A ,\quad \qquad x\in \Omega_0,\ t \in \mathbb{R},\\[4pt] \dfrac{\partial \tilde{w}_M}{\partial t} \leq D\Delta \tilde{w}_M+\theta\gamma(x,t) \tilde{w}_A-\hat{d}(x,t) \tilde{w}_M+ \overline{\lambda} \tilde{w}_M,\quad \qquad x\in \Omega_0,\ t \in \mathbb{R},\\[4pt] \dfrac{\partial \tilde w_F}{\partial t}\leq D\Delta \tilde{w}_F+(1-\theta)\gamma(x,t) \tilde{w}_A-\hat{d}(x,t) \tilde{w}_F+ \overline{\lambda} \tilde{w}_F,\quad \qquad x\in \Omega_0,\ t \in \mathbb{R},\\[4pt] \tilde{w}_M=0,~ \tilde{w}_F=0,\quad \qquad x\in \partial\Omega_0,\ t \in \mathbb{R},\\[4pt] \tilde{w}_N\ \text{ is }T\text{-periodic in }t, \ N=A,M,F. \end{cases} \end{align*} $$

For $N=A,M,F$ , define

$$ \begin{align*}\tilde{\phi}_N(x)= \begin{cases} \tilde{w}_N(x,0),\quad\qquad x\in \overline{\Omega}_0,\\[4pt] 0,\quad\qquad x \in \overline{\Omega} \setminus \overline{\Omega}_0. \end{cases} \text{ and } \boldsymbol{\tilde{\phi}}=(\tilde{\phi}_A,\tilde{\phi}_M,\tilde{\phi}_F)^T. \end{align*} $$

We then obtain that $e^{\overline {\lambda } T} U(T,0) \boldsymbol {\tilde {\phi }} \geq \boldsymbol {\tilde {\phi }}$ . Choose $\sigma \in (0,\eta -\overline {\lambda })$ small enough. It follows from [Reference Wang, Zhang and Zhao20, Lemma 2.1] that $r_{P}(e^{\overline {\lambda }T} U(T,0)) \geq 1$ , and hence, $r_{P}(e^{(\eta -\sigma ) T}U(T,0))>r_{P}(e^{\overline {\lambda }T} U(T,0)) \geq 1$ . With Lemma 3.1, we have $\alpha (e^{(\eta -\sigma )T}U(T,0)B) \leq e^{-\sigma T}\alpha (B)$ for any bounded set $B \subseteq P$ . It then follows from Lemma 2.2 that $e^{(\eta -\sigma ) T}U(T,0)$ , and hence, $U(T,0)$ admits the principal eigenvalue with an eigenvector $\phi ^*\in P \setminus \{0\}$ .

For $N=M,F$ , let $\hat {\lambda }_N$ be the principal eigenvalue of

$$ \begin{align*}\begin{cases} \mathcal{L}_{N} w =\hat\lambda_N w ,\quad \qquad x \in \Omega, ~ t \in \mathbb{R},\\[4pt] \frac{\partial w}{\partial \nu}=0,\quad \qquad x \in \partial \Omega,~ t \in \mathbb{R},\\[4pt] w \text{ is } T \text{-periodic}, \end{cases} \end{align*} $$

with the positive eigenvector $\hat {w}_N \in \mathrm {Int}(C_T(\overline {\Omega }\times \mathbb {R},\mathbb {R}_+))$ satisfying $\max _{(x,t) \in \overline {\Omega } \times \mathbb {R}}\vert \hat {w}_N(x,t) \vert =1 $ . Note that $\lambda ^*=- \frac {\ln r_{P}( U(T,0))}{T}\leq \min (\eta ,\hat {\lambda }_M,\hat {\lambda }_F)$ . Since $r_{P}(e^{(\eta -\sigma ) T}U(T,0))>1$ , we have $\lambda ^*<\eta -\sigma $ . In view of $\lambda _0:=\hat {\lambda }_M=\hat {\lambda }_F$ , choosing $\hat {w}_A=0$ , we then have

$$ \begin{align*}\begin{cases} \dfrac{\partial \hat{w}_A}{\partial t}\leq\rho(x,t)H(\hat{w}_M,\hat{w}_F)-d(x,t)\hat{w}_A-\gamma(x,t) \hat{w}_A+\lambda_0 \hat{w}_A ,\quad \qquad x\in \Omega,\ t\in \mathbb{R},\\[4pt] \dfrac{\partial \hat{w}_M}{\partial t}=D\Delta \hat{w}_M+\theta\gamma(x,t) \hat{w}_A-\hat{d}(x,t) \hat{w}_M+\lambda_0 \hat{w}_M,\quad \qquad x\in \Omega,\ t\in \mathbb{R},\\[4pt] \dfrac{\partial \hat{w}_F}{\partial t}=D\Delta \hat{w}_F+(1-\theta)\gamma(x,t) \hat{w}_A-\hat{d}(x,t) \hat{w}_F+\lambda_0 \hat{w}_F,\quad \qquad x\in \Omega,\ t\in \mathbb{R},\\[4pt] \dfrac{\partial \hat{w}_M(x,t)}{\partial \nu}=0,\frac{\partial \hat{w}_F(x,t)}{\partial \nu}=0,\quad \qquad \ x\in \partial\Omega,\ t\in \mathbb{R},\\[4pt] \hat{w}_N\ \text{ is }T\text{-periodic in }t, \ N=A,M,F. \end{cases} \end{align*} $$

Define

$$ \begin{align*}\hat{\psi}_A(x)=0, \ \hat{\psi}_M(x)=\hat{w}_M(x,0), \ \hat{\psi}_F(x)=\hat{w}_F(x,0), \text{ and } \boldsymbol{\hat{\psi}}=(\hat\psi_A,\hat\psi_M,\hat\psi_F)^T. \end{align*} $$

By the comparison arguments, it is not hard to verify that $e^{\lambda _0 T} U(T,0) \boldsymbol {\hat {\psi }} \gg \boldsymbol {\hat {\psi }}$ . Using [Reference Wang, Zhang and Zhao20, Lemma 2.1] again, we have $r_{P}(e^{\lambda _0 T} U(T,0))>1$ , which implies that $\lambda ^*<\lambda _0$ . Therefore, $\lambda ^*<\min (\eta ,\hat {\lambda }_M,\hat {\lambda }_F)$ . Now one can easily obtain $\boldsymbol {\phi }^* \in \mathrm {Int}(P)$ by using the arguments similar to those in case (i).

Proof It is easy to see that (H1)–(H4) in [Reference Wang, Zhang and Zhao20, Section 3] hold true. Thanks to Lemma 3.2, if there exists a domain $\Omega _0 \subseteq \Omega $ such that $\tilde {p}(x)=-\eta $ for all $x \in \Omega _0$ , then (H5) in [Reference Wang, Zhang and Zhao20, Section 3] is true. Thus, the desired conclusion follows from Lemma 2.1 and [Reference Wang, Zhang and Zhao20, Theorem 3.4].

To prove the conclusion in the general case, we first take a series of functions $d_{\epsilon }$ and domains $\Omega _{\epsilon } \subseteq \Omega $ such that $d_{\epsilon } \leq d \leq d_{\epsilon }+\epsilon $ and $\tilde {p}^{\epsilon }(x)=\max _{x \in \Omega } \tilde {p}^{\epsilon } (x)$ , $\forall x \in \overline {\Omega }_{\epsilon }$ , where $\tilde {p}^{\epsilon }(x)=T^{-1}\int _{0}^{T} [-d_{\epsilon }(x,t)-\gamma (x,t)] \mathrm {d} t$ , $\forall x \in \overline {\Omega }$ .

We can define $\mathcal {R}_0^{\epsilon }$ and $\{U^{\epsilon }(t,s): t \geq s \}$ (resp. $\mathcal {R}_{0,\epsilon }$ and $\{U_{\epsilon }(t,s): t \geq s \}$ ) in the same way as $\mathcal {R}_0$ and $\{U(t,s): t \geq s \}$ with d replaced by $d_{\epsilon }$ (resp. $d_{\epsilon }+\epsilon $ ). It is easy to see that $\mathcal {R}_{0,\epsilon } \leq \mathcal {R}_{0} \leq \mathcal {R}_0^{\epsilon }$ and $r_{P}(U_{\epsilon }(T,0)) \leq r_{P}(U(T,0)) \leq r_{P}(U^{\epsilon }(T,0))\leq r_{P}(U_{\epsilon }(T,0)) e^{\epsilon T}$ . Moreover, $\mathcal {R}_{0,\epsilon }-1$ and $r_{P}(U_{\epsilon }(T,0)) -1$ have the same sign, and $\mathcal {R}_{0}^{\epsilon }-1$ and $r_{P}(U^{\epsilon }(T,0)) -1$ share the same sign.

If $r_{P}(U(T,0))=1$ , then $\mathcal {R}_0=1$ by [Reference Wang, Zhang and Zhao20, Lemma 3.2(i)]. If $r_{P}(U(T,0))>1$ , one can choose $\epsilon>0$ small enough such that $r_{P}(U(T,0)) e^{-\epsilon T}>1$ . We then have $r_{P}(U_{\epsilon }(T,0))\geq r_{P}(U(T,0)) e^{-\epsilon T}>1$ , and hence, $\mathcal {R}_0\geq \mathcal {R}_{0,\epsilon }>1$ . By the similar arguments, we can also prove that $\mathcal {R}_0<1$ whenever $r_{P}(U(T,0))<1$ .

Repeating the above arguments, we see that $\mu \mathcal {R}_0-1$ and $r_{P}(U(T,0;\mu ))-1$ have the same sign for all $\mu>0$ . By the comparison arguments, it is not hard to prove that $\mathcal {R}_0\geq \theta (1-\theta )\underline {\rho }\underline {\gamma } [{\overline {d}}(\overline {d}+\overline {\gamma })]^{-1}>0$ , where $\underline {\rho }=\min _{(x,t)\in \overline {\Omega } \times \mathbb {R}} \rho (x,t)$ , $\underline {\gamma }=\min _{(x,t)\in \overline {\Omega } \times \mathbb {R}} \gamma (x,t)$ , $\overline {\gamma }=\max _{(x,t)\in \overline {\Omega } \times \mathbb {R}} \gamma (x,t)$ , and $\overline {d}=\max _{(x,t)\in \overline {\Omega } \times \mathbb {R}} \max (d(x,t),\hat {d}(x,t))$ . Therefore, $\mathcal {R}_0^{-1}$ is the unique solution of $r_{P}(U(T,0;\mu ))=1$ .

Proof (i) It suffices to prove that $\mathcal {R}_0^{-1}$ is an eigenvalue of the weighted eigenvalue problem

(3.4) $$ \begin{align} \begin{cases} \dfrac{\partial w_A}{\partial t}=\lambda\rho(x,t) \theta(1-\theta)w_S-d(x,t)w_A-\gamma(x,t) w_A ,\quad \qquad x\in \Omega,\ t\in \mathbb{R},\\[4pt] \dfrac{\partial w_S}{\partial t}=D\Delta w_S+\gamma(x,t) w_A-\hat{d}(x,t) w_S,\quad \qquad x\in \Omega,\ t\in \mathbb{R},\\[4pt] \dfrac{\partial w_S(x,t)}{\partial \nu}=0,\quad \qquad \ x\in \partial\Omega,\ t\in \mathbb{R},\\[4pt] w_A \text{ and } w_S\ \text{are }T\text{-periodic in }t, \end{cases} \end{align} $$

with a strongly positive eigenvector. With Proposition 3.1, we have $r_{P}(U(T,0;\mathcal {R}_0^{-1}))=1$ . Thanks to Lemma 3.2, there exists a strongly positive eigenvector $(w_A^*,w_M^*,w_F^*)^T$ such that

(3.5) $$ \begin{align} \begin{cases} \dfrac{\partial w_A^*}{\partial t}=\mathcal{R}_0^{-1}\rho(x,t)H(w_M^*,w_F^*)-d(x,t)w_A^*-\gamma(x,t) w_A^* ,\quad \qquad x\in \Omega,\ t>0,\\[4pt] \dfrac{\partial w_M^*}{\partial t}=D\Delta w_M^*+\theta\gamma(x,t) w_A^*-\hat{d}(x,t) w_M^*,\quad \qquad x\in \Omega,\ t>0,\\[4pt] \dfrac{\partial w_F^*}{\partial t}=D\Delta w_F^*+(1-\theta)\gamma(x,t) w_A^*-\hat{d}(x,t) w_F^*,\quad \qquad x\in \Omega,\ t>0,\\ \frac{\partial w_M^*(x,t)}{\partial \nu}=0,\frac{\partial w_F^*(x,t)}{\partial \nu}=0,\quad \qquad \ x\in \partial\Omega,\ t>0,\\[4pt] w_N^*\ \text{ is }T\text{-periodic in }t, \ N=A,M,F. \end{cases} \end{align} $$

Multiplying the second and third equations by $1-\theta $ and $\theta $ , respectively, and then making difference, we obtain

$$ \begin{align*}\begin{cases} \dfrac{\partial z}{\partial t}=D\Delta z-\hat{d}(x,t) z,\quad \qquad x\in \Omega,\ t>0,\\[4pt] \dfrac{\partial z(x,t)}{\partial \nu}=0\quad \qquad \ x\in \partial\Omega,\ t>0,\\[4pt] z\ \text{ is }T\text{-periodic in }t, \end{cases} \end{align*} $$

where $z=(1-\theta ) w_M^* - \theta w_F^*$ . It is easy to verify that $z \equiv 0$ due to the Krein–Rutman theorem. Let $w_S^*=\theta ^{-1}w_M^*=(1-\theta )^{-1}w_F^*$ . Then (3.4) holds with $\mu =\mathcal {R}_0^{-1}$ , $w_A=w_A^*$ , and $w_S=w_S^*$ .

(ii) If all parameters are independent of t, then so are $w_A^*$ and $w_S^*$ . By the first equation of (3.4), it then follows that

$$ \begin{align*}w_A^*=\mathcal{R}_0^{-1}\theta(1-\theta)\frac{\rho(x)}{d(x)+\gamma(x)} w_S^*, \end{align*} $$

which yields that $\mathcal {R}_0^{-1}$ is an eigenvalue of the weighted eigenvalue problem

$$ \begin{align*}\begin{cases} D\Delta \hat{w} + \lambda \theta(1-\theta) \frac{\rho(x)\gamma(x)}{d(x)+\gamma(x)}\hat{w} -\hat{d}(x) \hat{w}=0,\quad \qquad \ x\in \Omega,\\[4pt] \dfrac{\partial \hat{w}(x)}{\partial \nu}=0,\quad \qquad \ x\in \partial\Omega. \end{cases} \end{align*} $$

Now the desired conclusion follows from the same arguments as in [Reference Allen, Bolker, Lou and Nevai1, Lemma 2.3].

Proof By [Reference Martin and Smith16, Corollary 4] and assumption (A), it follows that for every initial value function $\boldsymbol {\phi }:=(\phi _A,\phi _M,\phi _F)^T\in \Sigma (0)$ , system (1.1) admits a unique non-continuable solution $\boldsymbol {u}(\cdot ,t;\boldsymbol {\phi }):=(u_A,u_M,u_F)^T(\cdot ,t;\boldsymbol {\phi })$ on its maximal interval of existence $[0,t_{\boldsymbol {\phi }})$ with $\boldsymbol {u}(\cdot ,0;\boldsymbol {\phi })=\boldsymbol {\phi }$ , where $t_{\boldsymbol {\phi }}\leq \infty $ , and $\boldsymbol {u}(\cdot ,t;\boldsymbol {\phi })\in \Sigma (t)$ for all $t\in [0,t_{\boldsymbol {\phi }})$ . Thus, $0\leq u_A(x,t;\boldsymbol {\phi })\leq K(x,t),\ \forall \ x \in \bar {\Omega },\ t\geq 0$ . Then we can use standard comparison arguments to further show that solutions of system (1.1) are ultimately bounded and uniformly bounded. Hence, we can define the solution maps $S(t):\Sigma (0)\rightarrow \Sigma (t)$ associated with system (1.1) by

$$ \begin{align*}S(t)\boldsymbol{\phi}=\boldsymbol{u}(\cdot,t;\boldsymbol{\phi}),\ \forall \ \boldsymbol{\phi}\in \Sigma(0), \end{align*} $$

where $\boldsymbol {u}(x,t;\boldsymbol {\phi })$ is the unique solution of system (1.1) with $\boldsymbol {u}(x,0;\boldsymbol {\phi })=\boldsymbol {\phi }(x),\ \forall \ x \in \bar {\Omega }$ . Let $S:=S(T):\Sigma (0)\rightarrow \Sigma (0)$ be the Poincaré map associated with system (1.1) on $\Sigma (0)$ . By the theory of periodic semiflows (see [Reference Zhao22]), it suffices to prove a threshold-type result on the global dynamics for S on $\Sigma (0)$ in terms of $\mathcal {R}_0$ . By using the arguments similar to those in [Reference Hsu, Wang and Zhao9], we see that S is $\alpha $ -contracting in the sense that $\lim _{n\to \infty }\alpha (S^n(B))=0\ \text { for any bounded set}\ B \subset \Sigma (0)$ . It follows from [Reference Magal and Zhao14, Theorem 2.6] that S admits a strong global attractor that attracts each bounded set in $\Sigma (0)$ . By the essentially same arguments as those for [Reference Bai and Zhao3, Theorem 3.1], we can easily show that the desired statements are valid in the case where $\mathcal {R}_0<1$ or $\mathcal {R}_0>1$ .

It remains to consider the critical case where $\mathcal {R}_0=1$ . In view of Proposition 3.1, we have $r_{P}(U(T,0))=1$ . It follows from Lemma 3.2 that 0 is the principal eigenvalue of system (3.2), which corresponds to an eigenvector $\boldsymbol {\psi }=(\psi _A,\psi _M,\psi _F)^T \in \mathrm {Int}(P)$ . Motivated by [Reference Cui, Lam and Lou5, Reference Li and Zhao10], we define

$$ \begin{align*}c(t;\boldsymbol{\phi}):=\max\left\{\max_{x\in \bar{\Omega}}\frac{u_A(x,t;\boldsymbol{\phi})}{\psi_A(x,t)},\ \max_{x\in \bar{\Omega}}\frac{u_M(x,t;\boldsymbol{\phi})}{\psi_M(x,t)},\ \max_{x\in \bar{\Omega}}\frac{u_F(x,t;\boldsymbol{\phi})}{\psi_F(x,t)}\right\},\ t\geq0.\end{align*} $$

Then we have the following observation.

Claim. $c(nT;\boldsymbol {\phi })$ is non-increasing in n, and $c(nT;\boldsymbol {\phi })$ is not a constant sequence whenever $\boldsymbol {\phi } \in \tilde {\Sigma }$ .

Indeed, since $\phi _A(\cdot )\leq K(\cdot ,0)$ , we have $u_A(\cdot ,t,\boldsymbol {\phi })\leq K(\cdot ,t)$ for all $t\geq 0$ . Fix $n_0>0$ , we let

$$ \begin{align*}\begin{cases} \bar{u}_A(x,t;\boldsymbol{\phi}):=c(n_0 T;\boldsymbol{\phi})\psi_A(x,t),\ x \in \bar{\Omega},\ t\geq n_0 T,\\[4pt] \bar{u}_M(x,t;\boldsymbol{\phi}):=c(n_0 T;\boldsymbol{\phi})\psi_M(x,t),\ x \in \bar{\Omega},\ t\geq n_0 T,\\[4pt] \bar{u}_F(x,t;\boldsymbol{\phi}):=c(n_0 T;\boldsymbol{\phi})\psi_F(x,t),\ x \in \bar{\Omega},\ t\geq n_0 T. \end{cases} \end{align*} $$

Then $(\bar {u}_A(x,t;\boldsymbol {\phi }),\bar {u}_M(x,t;\boldsymbol {\phi }),\bar {u}_F(x,t;\boldsymbol {\phi }))$ satisfies (3.2) with $\lambda =0$ . Since $(u_A,u_M,u_F)(x,t;\boldsymbol {\phi })$ satisfies system (1.1), it follows that

(3.6) $$ \begin{align} \begin{cases} \dfrac{\partial u_A}{\partial t}\leq\rho(x,t)H(u_M,u_F)-d(x,t)u_A-\gamma(x,t) u_A,\quad \qquad x\in \Omega,\ t>0,\\[4pt] \dfrac{\partial u_M}{\partial t}=D\Delta u_M+\theta\gamma(x,t) u_A-\hat{d}(x,t) u_M,\quad \quad \qquad x\in \Omega,\ t>0,\\[4pt] \dfrac{\partial u_F}{\partial t}=D\Delta u_F+(1-\theta)\gamma(x,t) u_A-\hat{d}(x,t) u_F,\quad \qquad x\in \Omega,\ t>0,\\[4pt] \dfrac{\partial u_M(x,t,\boldsymbol{\phi})}{\partial \nu}=0,\frac{\partial u_F(x,t,\boldsymbol{\phi})}{\partial \nu}=0,\quad \qquad \ x\in \partial\Omega,\ t>0, \end{cases} \end{align} $$

and

$$ \begin{align*}\begin{cases} u_A(x,n_0 T;\boldsymbol{\phi})\leq \bar{u}_A(x,n_0 T;\boldsymbol{\phi}),\quad \qquad x\in \overline{\Omega},\\[4pt] u_M(x,n_0 T;\boldsymbol{\phi})\leq \bar{u}_M(x,n_0 T;\boldsymbol{\phi}),\quad \qquad x\in \overline{\Omega},\\[4pt] u_F(x,n_0 T;\boldsymbol{\phi})\leq \bar{u}_F(x,n_0 T;\boldsymbol{\phi}),\quad \qquad x\in \overline{\Omega}. \end{cases} \end{align*} $$

By the comparison principle, we have

$$ \begin{align*}(u_A(\cdot,t;\boldsymbol{\phi}),u_M(\cdot,t;\boldsymbol{\phi}),u_F(\cdot,t;\boldsymbol{\phi}))\leq (\bar{u}_A(\cdot,t;\boldsymbol{\phi}),\bar{u}_M(\cdot,t;\boldsymbol{\phi}),\bar{u}_F(\cdot,t;\boldsymbol{\phi})) ,\ \forall\ t\geq n_0 T.\end{align*} $$

This implies that

$$ \begin{align*}c(nT;\boldsymbol{\phi})\leq c(n_0T;\boldsymbol{\phi}),\ \forall\ n\geq n_0.\end{align*} $$

Moreover, $c(nT;\boldsymbol {\phi })$ is non-increasing in n. In addition, if $\boldsymbol {\phi }:=(\phi _A,\phi _M,\phi _F)\in \tilde \Sigma $ , then one can show that

$$ \begin{align*}(u_A(x,t;\boldsymbol{\phi}),u_M(x,t;\boldsymbol{\phi}),u_F(x,t;\boldsymbol{\phi}))\gg (0,0,0),\ \forall x \in \bar{\Omega},\ t> 0,\end{align*} $$

and hence,

$$ \begin{align*}1-\frac{u_A(x,t;\boldsymbol{\phi})}{K(x,t)}<1,\ \forall\ x \in \bar{\Omega},\ t> 0,\end{align*} $$

which implies that the first inequality in (3.6) must be strict. Thus, the strong comparison principle ensures

$$ \begin{align*}(u_A(\cdot,t;\boldsymbol{\phi}),u_M(\cdot,t;\boldsymbol{\phi}),u_F(\cdot,t;\boldsymbol{\phi}))\ll (\bar{u}_A(\cdot,t;\boldsymbol{\phi}),\bar{u}_M(\cdot,t;\boldsymbol{\phi}),\bar{u}_F(\cdot,t;\boldsymbol{\phi})) ,\ \forall\ t> n_0T.\end{align*} $$

It then follows that $c(n T;\boldsymbol {\phi })< c(n_0 T;\boldsymbol {\phi })$ whenever $n> n_0$ , and hence, $c(nT;\boldsymbol {\phi })$ is not a constant sequence whenever $\boldsymbol {\phi } \in \tilde \Sigma $ .

Now we show that $\lim _{n\rightarrow \infty }S^n\boldsymbol {\phi }=0$ for all $\boldsymbol {\phi }\in \tilde \Sigma $ . Clearly, $\lim _{n\rightarrow \infty }c(nT;\boldsymbol {\phi })=c_0\geq 0$ . Let $\omega (\boldsymbol {\phi })$ be the omega limit set of the forward orbit through $\boldsymbol {\phi }$ for S. For any given $\boldsymbol {\varphi } \in \omega (\boldsymbol {\phi })$ , there is a sequence $n_k\to \infty $ such that $\lim _{k\rightarrow \infty }S^{n_k}\boldsymbol {\phi }=\boldsymbol {\varphi }$ . It then follows that

$$ \begin{align*}\lim_{k\rightarrow\infty }S^{m+n_k}\boldsymbol{\phi}=\lim_{k\rightarrow\infty}S^mS^{n_k}\boldsymbol{\phi}=S^m\boldsymbol{\varphi},\ \forall m\geq 0.\end{align*} $$

By the definition of $c(t;\boldsymbol {\varphi })$ , we have

$$ \begin{align*} c(mT;\boldsymbol{\varphi})=\lim_{k\rightarrow\infty }c((m+n_k)T;\boldsymbol{\phi})=c_0, \ \forall m \geq 0. \end{align*} $$

It then follows from the above claim that $\boldsymbol {\varphi }=(0,0,0)$ , and hence, $\omega (\boldsymbol {\phi })=\{(0,0,0)\}$ . The desired conclusion can be derived easily for all $\boldsymbol {\phi }\in \Sigma (0) \setminus \tilde \Sigma $ .

Proof Following the arguments similar to those for[Reference Zhao23, Lemma 4.1], we first rewrite system (1.2) as an abstract integral equation in the Banach space $(\mathcal {C}_{\boldsymbol {u}^*(\cdot ,0)},\|\cdot \|_{\infty })$ , and then apply [Reference Martin and Smith16, Corollary 5] with $\tau =0$ to obtain the desired conclusions.

Proof We first introduce the following evolution system:

(4.5) $$ \begin{align} \begin{cases} \dfrac{\partial u_A}{\partial t}=\rho(x,t)\theta(1-\theta)u_S(1-\frac{u_A}{K(x,t)})-d(x,t)u_A-\gamma(x,t) u_A,\quad \qquad x\in \mathbb{R},\ t>0,\\[4pt] \dfrac{\partial u_S}{\partial t}=D\frac{\partial^2 u_S}{\partial x^2} +\gamma(x,t) u_A-\hat{d}(x,t) u_S,\quad \quad \qquad x\in \mathbb{R},\ t>0,\\[4pt] \end{cases} \end{align} $$

and the associated linear eigenvalue problem:

(4.6) $$ \begin{align} \begin{cases} \dfrac{\partial w_A}{\partial t}=\rho(x,t)\theta(1-\theta)w_S-d(x,t)w_A-\gamma(x,t) w_A+\lambda w_A,\quad \qquad x\in \mathbb{R},\ t \in \mathbb{R},\\[4pt] \dfrac{\partial w_S}{\partial t}=D\frac{\partial^2 w_S}{\partial x^2} - 2 D\mu \frac{\partial w_S}{\partial x} +\gamma(x,t) w_A + D \mu^2 w_S-\hat{d}(x,t) w_S+\lambda w_S,\quad \quad \qquad x\in \mathbb{R},\ t \in \mathbb{R},\\[4pt] w_A \text{ and } w_S \text{ are } T-\text{periodic in } t \text{ and } L-\text{periodic in } x. \end{cases} \end{align} $$

We next prove $c_+^*=\inf _{\mu>0} -\frac {\lambda ^*(\mu )}{\mu }$ by dividing the proof into four steps, and $c_-^*=\inf _{\mu>0} -\frac {\lambda ^*(-\mu )}{\mu }$ can be obtained similarly.

Step 1: The principal eigenvalue of the nonlinear eigenvalue problem (4.4) coincides with that of the linear eigenvalue problem (4.6). It then suffices to show the following two claims.

Claim 1. If $\lambda ^*$ is the principal eigenvalue of eigenvalue problem (4.4), then it is also the principal eigenvalue of (4.6).

Claim 2. If $\tilde {\lambda }$ is the principal eigenvalue of (4.6), then it is also the principal eigenvalue of (4.4).

We only prove Claim 1, and Claim 2 can be derived similarly. For any given $\mu \geq 0$ , let $(w_A^*,w_M^*,w_F^*)^T$ be the principal eigenvector of (4.4) corresponding to $\lambda ^*$ , and let $\hat {\lambda }$ be the principal eigenvalue of

(4.7) $$ \begin{align} \begin{cases} \dfrac{\partial \hat{w}}{\partial t}=D\frac{\partial^2 \hat{w}}{\partial x^2} - 2 D\mu \frac{\partial \hat{w}}{\partial x} + D \mu^2 \hat{w}-\hat{d}(x,t) \hat{w}+\lambda \hat{w},\quad \quad \qquad x\in \mathbb{R},\ t \in \mathbb{R},\\[4pt] \hat{w} \text{ is } T-\text{periodic in } t \text{ and } L-\text{periodic in } x. \end{cases} \end{align} $$

Thanks to the Krein–Rutman theorem, it is easy to verify that $\lambda ^*<\hat {\lambda }$ . Multiplying $1-\theta $ and $\theta $ on the second and third equations of (4.4), respectively, and then making difference, we obtain

$$ \begin{align*}\frac{\partial z}{\partial t}=D\frac{\partial^2 z}{\partial x^2}- 2 D\mu \frac{\partial z}{\partial x}+ D \mu^2 z-\hat{d}(x,t) z+\lambda^* z,\quad \quad \qquad x\in \mathbb{R},\ t \in \mathbb{R}. \end{align*} $$

Here, $z=(1-\theta )w_M^*-\theta w_F^*$ is T-periodic in t and L-periodic in x. In view of $\lambda ^*<\hat {\lambda }$ , we have $z \equiv 0$ due to the Krein–Rutman theorem. Then we have

$$ \begin{align*}\frac{1}{\theta}w_M^*=\frac{1}{1-\theta}w_F^*=w_M^*+w_F^*.\end{align*} $$

An easy computation yields that $\lambda ^*$ and $(w_A^*,w_M^*+w_F^*)$ satisfy (4.6), and the desired conclusion follows from [Reference Wang, Zhang and Zhao20, Corollary 2.6] or the Krein–Rutman theorem again.

Step 2. Systems (1.2) and (4.5) admit the same minimum wave speed of rightward time–space periodic traveling waves.

Let

$$ \begin{align*}(u_A,u_M,u_F)(x,t):=(U_A,U_M,U_F)(x,t,\xi)\ \text{with}\ \xi=x-ct.\end{align*} $$

An easy computation yields that

$$ \begin{align*}\frac{\partial u_M}{\partial t}=\frac{\partial U_M}{\partial t}-c\frac{\partial U_M}{\partial \xi},\quad &\frac{\partial u_M}{\partial x}=(U_M)_x+\frac{\partial U_M}{\partial \xi},\quad \\ &\frac{\partial^2 u_M}{\partial x^2}=\frac{\partial^2 U_M}{\partial x^2}+2\frac{\partial^2 U_M}{\partial \xi \partial x}+\frac{\partial^2 U_M}{\partial \xi^2}. \end{align*} $$

Then system (1.2) becomes

(4.8) $$ \begin{align} \!\begin{cases} \dfrac{\partial U_A}{\partial t}=\rho(x,t)H(U_M,U_F)(1-\frac{U_A}{K(x,t)})-d(x,t)U_A-\gamma(x,t) U_A+c\frac{\partial U_A}{\partial \xi},\qquad (x,t,\xi) \in \mathbb{R}^3,\\[4pt] \dfrac{\partial U_M}{\partial t}=D(\frac{\partial^2 U_M}{\partial x^2}+2\frac{\partial^2 U_M}{\partial \xi \partial x}+\frac{\partial^2 U_M}{\partial \xi^2})+\theta\gamma(x,t) U_A-\hat{d}(x,t) U_M+c\frac{\partial U_M}{\partial \xi},\qquad (x,t,\xi) \in \mathbb{R}^3,\\[4pt] \dfrac{\partial U_F}{\partial t}=D(\frac{\partial^2 U_F}{\partial x^2}+2\frac{\partial^2 U_F}{\partial \xi \partial x}+\frac{\partial^2 U_F}{\partial \xi^2})+(1-\theta)\gamma(x,t) U_A-\hat{d}(x,t) U_F+c\frac{\partial U_F}{\partial \xi},\qquad (x,t,\xi) \in \mathbb{R}^3,\\[4pt] U_N(x+L,t+T,\xi)=U_N(x+L,t,\xi)=U_N(x,t,\xi), \qquad \ (x,t,\xi)\in\mathbb{R}^3,\ N=A,M,F. \end{cases} \end{align} $$

Substituting $(u_A,u_S)(x,t):=(U_A,U_S)(x,t,\xi )\ \text {with}\ \xi =x-ct$ into (4.5), we obtain

(4.9) $$ \begin{align} \begin{cases} \dfrac{\partial U_A}{\partial t}=\rho(x,t)\theta(1-\theta)U_S(1-\frac{U_A}{K(x,t)})-d(x,t)U_A-\gamma(x,t) U_A+c\frac{\partial U_A}{\partial \xi},\qquad (x,t,\xi)\in\mathbb{R}^3,\\[4pt] \dfrac{\partial U_S}{\partial t}=D(\frac{\partial^2 U_S}{\partial x^2}+2\frac{\partial^2 U_S}{\partial \xi \partial x}+\frac{\partial^2 U_S}{\partial \xi^2})+\gamma(x,t) U_A-\hat{d}(x,t) U_S+c\frac{\partial U_S}{\partial \xi},\qquad (x,t,\xi)\in\mathbb{R}^3,\\[4pt] U_N(x+L,t+T,\xi)=U_N(x+L,t,\xi)=U_N(x,t,\xi), \qquad \ (x,t,\xi)\in\mathbb{R}^3,\ N=A,S. \end{cases} \end{align} $$

We then have the following two claims:

Claim 3. If $(U_A,U_S,c)$ satisfies (4.9), then $(U_A,\theta U_S,(1-\theta )U_S,c)$ satisfies (4.8).

Claim 4. If $(U_A,U_M,U_F,c)$ satisfies (4.8), then $(U_A,U_M+U_F,c)$ satisfies (4.9).

Claim 3 can be obtained via a straightforward computation, and so we only prove Claim 4. Multiplying $1-\theta $ and $\theta $ on the second and third equations of (4.8), respectively, and then making difference, we obtain

(4.10) $$ \begin{align} \frac{\partial Z}{\partial t}=D\left(\frac{\partial^2 Z}{\partial x^2}+2\frac{\partial^2 Z}{\partial \xi \partial x}+\frac{\partial^2 Z}{\partial \xi^2} \right)-\hat{d}(x,t) Z+c\frac{\partial Z}{\partial \xi},\qquad (x,t,\xi)\in\mathbb{R}^3, \end{align} $$

where $Z=(1-\theta )U_M-\theta U_F$ . Since Z is a bounded entire solution of (4.10), it is not hard to verify that Z converges to $0$ uniformly for $(x,\xi ) \in \mathbb {R}^2$ as $t\rightarrow +\infty $ . Notice that Z is T-periodic in t. It then follows that $Z \equiv 0$ for $(x,t,\xi )\in \mathbb {R}^3$ . This implies that $(1-\theta )U_M\equiv \theta U_F$ , $(x,t,\xi )\in \mathbb {R}^3$ . Let $U_S=\frac {1}{\theta }U_M=\frac {1}{1-\theta }U_F=U_M+U_F$ . A straightforward computation shows that Claim 4 holds.

Step 3. Prove the conclusion when $\lambda ^*(\mu )$ is the principal eigenvalue of (4.4).

For any $\mu \in \mathbb {R}$ , let $\{Q_{\mu }(t,s)\}_{t\geq s}$ be the evolution family on $\mathcal {X}$ generated by system (4.6) with $\lambda =0$ and define

$$ \begin{align*}\tilde{\lambda}({\mu})=-\frac{1}{T} \log r(Q_{\mu}(T,0)).\end{align*} $$

By the arguments similar to those for [Reference Fang, Lai and Wang6, Theorem 5.1] (see also [Reference Fang, Yu and Zhao7, Reference Liang and Zhao13]), the rightward spreading speed of (4.5) can be characterized by $\inf _{\mu>0}-\frac {\tilde {\lambda }({\mu })}{\mu }$ , which coincides with the minimum wave speed of rightward almost pulsating waves of (4.5). By Step 2, systems (1.2) and (4.5) have the same minimum wave speed of rightward almost pulsating waves. It then follows from Step 1 that

$$ \begin{align*}c^*_+=\inf_{\mu>0}-\frac{\tilde{\lambda}({\mu})}{\mu}=\inf_{\mu>0}-\frac{\lambda^*({\mu})}{\mu}.\end{align*} $$

Step 4. Complete the proof.

We take a series of functions $d_{\epsilon }$ and domains $\Omega _{\epsilon } \subseteq \Omega :=[0,L]$ such that $d_{\epsilon } \leq d \leq d_{\epsilon }+\epsilon $ and $\tilde {p}^{\epsilon }(x)=\max _{x \in \Omega } \tilde {p}^{\epsilon } (x)$ , $\forall x \in \overline {\Omega }_{\epsilon }$ , where $\tilde {p}^{\epsilon }(x)=T^{-1}\int _{0}^{T} [-d_{\epsilon }(x,t)-\gamma (x,t)] \mathrm {d} t$ , $\forall x \in \overline {\Omega }$ . Then (4.4) admits the principal eigenvalue $\lambda ^\epsilon (\mu )$ and $\lambda _{\epsilon }(\mu )$ with d replaced by $d_{\epsilon }$ and $d_{\epsilon }+\epsilon $ , respectively. It is not hard to verify that

$$ \begin{align*}\vert \lambda^{\epsilon}(\mu) -\lambda_{\epsilon}(\mu) \vert \leq \epsilon, \quad -\lambda_{\epsilon}(\mu) \leq -\lambda^*(\mu) \leq -\lambda^{\epsilon}(\mu), \, \forall \mu>0. \end{align*} $$

Therefore, the two corresponding perturbed systems admit the rightward spreading speeds $c_{+}^{\epsilon }$ and $c_{+,\epsilon }$ , respectively. Repeating the above three steps, we conclude that

$$ \begin{align*}c_+^{\epsilon}=\inf_{\mu>0} -\frac{\lambda^{\epsilon}(\mu)}{\mu},\text{ and } c_{+,\epsilon}=\inf_{\mu>0} -\frac{\lambda_{\epsilon}(\mu)}{\mu}. \end{align*} $$

Notice that $-\frac {\lambda ^{\epsilon }(\mu )}{\mu } \rightarrow +\infty $ and $-\frac {\lambda _{\epsilon }(\mu )}{\mu } \rightarrow +\infty $ as $\mu \rightarrow 0$ and $\mu \rightarrow +\infty $ . Thus, the two infima are attained at some finite values of $\mu $ . It follows that $ c_+^{\epsilon } \rightarrow c^*_+ $ and $ c_{+,\epsilon } \rightarrow c^*_+ $ as $\epsilon \rightarrow 0^+$ , which implies the desired result.

Proof Since $\mathcal {R}_0>1$ and $L_0(T,0)=\mathcal {U}(T,0)$ restricted on ${\mathcal {X}_+}$ , we have $r_{\mathcal {X}_+}(L_{0}(T,0))> 1$ due to Lemma 4.1. In view of Lemma 2.1, there exists $\tilde {\mu }$ such that $r_{\mathcal {X}_+}(L_{\mu }(T,0))> 1$ for $\mu \in (0,\tilde {\mu })$ . Thus, Lemma 4.6 implies that eigenvalue problem (4.4) admits a principal eigenvalue $\lambda ^*(\mu )= -\frac {1}{T} \log r_{\mathcal {X}_+}(L_{\mu }(T,0))$ for any $\mu \in (0,\tilde {\mu })$ . We still let $\{Q_{\mu }(t,s)\}_{t\geq s}$ be the evolution family on $\mathcal {X}$ generated by system (4.6) with $\lambda =0$ and write $\tilde {\lambda }({\mu })=-\frac {1}{T} \log r(Q_{\mu }(T,0))$ . According to Claim 1 in the proof of Theorem 4.3, $\tilde {\lambda }(\mu )=\lambda ^*(\mu )$ is the principal eigenvalue of eigenvalue problem (4.6) for any $\mu \in (0,\tilde {\mu })$ . We only need to prove that $\tilde {\mu }$ can be extended to $+\infty $ . It then suffices to prove the following three claims:

Claim 1. $-\tilde {\lambda }(\mu )$ is convex on $\mathbb {R}$ .

Claim 2. $\lim \limits _{\mu \rightarrow +\infty } -\tilde {\lambda }(\mu )=+\infty $ .

Claim 3. $\tilde {\lambda }(\mu )=\tilde {\lambda }(-\mu )$ , $\forall \mu \in \mathbb {R}$ .

If these claims are valid, then $\tilde {\lambda }(\mu )$ is non-increasing with respect to $\mu \in (0,+\infty )$ , and hence, $r(Q_{\mu }(T,0)) \geq 1$ for all $\mu \geq 0$ . Thus, [Reference Fang, Lai and Wang6, Lemma 3.1] implies that $\tilde {\lambda }(\mu )$ is the principal eigenvalue of (4.6). By Claim 2 in the proof of Theorem 4.3, we see that $\lambda ^*(\mu )=\tilde {\lambda }(\mu )$ is the principal eigenvalue of eigenvalue problem (4.4) for any $\mu \in (0,+\infty )$ . Combining with Claim 3, we can obtain the desired conclusions.

Indeed, Claim 1 follows from the proof of Theorem 4.3. For readers’ convenience, we provide more details. We take a series of functions $d_{\epsilon }$ and define $\lambda ^{\epsilon }(\mu )$ and $\lambda _{\epsilon }(\mu )$ as those in the proof of Theorem 4.3. By Step 1 of Theorem 4.3, it is known that $\lambda ^{\epsilon }(\mu )$ and $\lambda _{\epsilon }(\mu )$ are also the principal eigenvalues of (4.6) with d replaced by $d_{\epsilon }$ and $d_{\epsilon }+\epsilon $ , respectively. According to [Reference Liang and Zhao12], we obtain that $-\lambda ^{\epsilon }(\mu )$ and $-\lambda _{\epsilon }(\mu )$ are convex on $\mathbb {R}$ , so is $-\tilde {\lambda }(\mu )$ by letting $\epsilon \rightarrow 0^+$ .

In order to prove Claim 2, we choose four constants $a_{11}$ , $a_{12}$ , $a_{21}$ , and $a_{22}$ such that

$$ \begin{align*} & 0\geq a_{11} \geq -d(x) -\gamma(x), a_{12} \geq \rho(x)\theta(1-\theta)>0, a_{21} \geq \gamma(x)>0, 0 \geq a_{22} \geq -\hat{d}(x), \forall x \in \mathbb{R}, \end{align*} $$

and four constants $b_{11}$ , $b_{12}$ , $b_{21}$ , and $b_{22}$ such that

$$ \begin{align*}b_{11} \leq -d(x) -\gamma(x), 0<b_{12} \leq \rho(x)\theta(1-\theta),0< b_{21} \leq \gamma(x), b_{22} \leq -\hat{d}(x), \forall x \in \mathbb{R}. \end{align*} $$

By the arguments in Lemma 3.2, it then follows that the following two eigenvalue problems

(4.11) $$ \begin{align} \begin{cases} a_{11}\phi_1+ a_{12} \phi_2=-\lambda \phi_1,\quad \qquad x\in \mathbb{R},\\[4pt] D\dfrac{\partial^2 \phi_2}{\partial x^2} - 2 D\mu \frac{\partial \phi_2}{\partial x} + a_{21} \phi_1 + D \mu^2 \phi_2+ a_{22} \phi_2=-\lambda \phi_2,\quad \quad \qquad x\in \mathbb{R},\\[4pt] \phi_1 \text{ and } \phi_2 \text{ are } L-\text{periodic in } x, \end{cases} \end{align} $$

and

(4.12) $$ \begin{align} \begin{cases} b_{11}\phi_1+ b_{12} \phi_2=-\lambda \phi_1,\quad \qquad x\in \mathbb{R},\\[4pt] D\dfrac{\partial^2 \phi_2}{\partial x^2} - 2 D\mu \frac{\partial \phi_2}{\partial x} + b_{21} \phi_1 + D \mu^2 \phi_2+ b_{22} \phi_2=-\lambda \phi_2,\quad \quad \qquad x\in \mathbb{R},\\[4pt] \phi_1 \text{ and } \phi_2 \text{ are } L-\text{periodic in } x \end{cases} \end{align} $$

have the principal eigenvalues, which are denoted by $\overline {\lambda }(\mu )$ and $\underline {\lambda }(\mu )$ , respectively. By the comparison arguments, we have $\overline {\lambda }(\mu ) \leq \tilde {\lambda }(\mu )\leq \underline {\lambda }(\mu ) $ , $\forall \mu \in \mathbb {R}_+$ . A direct computation shows that $\lim \limits _{\mu \rightarrow +\infty } -\frac {\overline {\lambda }(\mu )}{\mu ^2}=\lim \limits _{\mu \rightarrow +\infty } -\frac {\underline {\lambda }(\mu )}{\mu ^2}=D>0$ . This yields that $\lim \limits _{\mu \rightarrow +\infty } -\tilde {\lambda }(\mu )=+\infty $ .

Finally, we prove Claim 3. Write $m_{11}(x)=-d(x) -\gamma (x)$ , $m_{12}(x)=\rho (x)\theta (1-\theta )$ , $m_{21}(x)=\gamma (x)$ , $m_{22}(x)=-\hat {d}(x) $ , $\tilde {m}_{12}(x)=\tilde {m}_{21}(x)=[m_{12}(x)m_{21}(x)]^{\frac {1}{2}}$ , $\forall x \in \mathbb {R}$ . Clearly, if $\lambda $ , $\phi _1$ , and $\phi _2$ satisfy the following eigenvalue problem:

$$ \begin{align*} \begin{cases} m_{11}\phi_1+ m_{12} \phi_2=-\lambda \phi_1,\quad \qquad x\in \mathbb{R},\\[4pt] D\dfrac{\partial^2 \phi_2}{\partial x^2} - 2 D\mu \frac{\partial \phi_2}{\partial x} + m_{21} \phi_1 + D \mu^2 \phi_2+ m_{22} \phi_2=-\lambda \phi_2,\quad \quad \qquad x\in \mathbb{R},\\[4pt] \phi_1 \text{ and } \phi_2 \text{ are } L-\text{periodic in } x, \end{cases} \end{align*} $$

then $\lambda $ , $\psi _1=[m_{12}(x)]^{-\frac {1}{2}}[m_{21}(x)]^{\frac {1}{2}} \phi _1$ , and $\psi _2=\phi _2$ satisfy the following eigenvalue problem:

(4.13) $$ \begin{align} \begin{cases} m_{11}\psi_1+ \tilde{m}_{12} \psi_2=-\lambda \psi_1,\quad \qquad x\in \mathbb{R},\\[4pt] D\dfrac{\partial^2 \psi_2}{\partial x^2} - 2 D\mu \frac{\partial \psi_2}{\partial x} + \tilde{m}_{21} \psi_1 + D \mu^2 \psi_2+m_{22} \psi_2=-\lambda \psi_2,\quad \quad \qquad x\in \mathbb{R},\\[4pt] \psi_1 \text{ and } \psi_2 \text{ are } L-\text{periodic in } x. \end{cases} \end{align} $$

Now Claim 3 follows from this observation and the adjoint eigenvalue problem.

Proof We denote $\lambda ^*(D,\mu )$ the principal eigenvalue of (4.4) which depends on D and $\mu $ . We remark that $\lambda ^*(D,\mu )$ is also the principal eigenvalue of (4.6) due to Step 1 of Theorem 4.3.

(i) In view of Lemma 4.8(ii), there exists $\underline {D}>0$ such that for all $D\in (0,\underline {D})$ , $\mathcal {R}_0(D)>1$ , and hence, $c^*(D)>0$ due to Proposition 4.2. Let $\overline {\lambda }^*(D,\mu )$ be the principal eigenvalue of eigenvalue problem (4.11). By the comparison arguments, it is easy to see that $\overline {\lambda }^*(D,\mu ) \leq \lambda ^*(D,\mu ) $ , $\forall (D,\mu ) \in \mathbb {R}^2_+$ . Combining with $c^*(D)= \inf _{\mu>0} -\frac {\lambda ^*(D,\mu )}{\mu }$ , we only need to show that

$$ \begin{align*}\lim\limits_{D \rightarrow 0^+}\inf_{\mu>0} -\frac{\overline{\lambda}^*(D,\mu)}{\mu}=0.\end{align*} $$

It is easy to verify that $\lim \limits _{D \rightarrow 0^+}\overline {\lambda }^*(D,\mu )$ exists for any $\mu \geq 0$ , which is the smaller solution of the equation $(\lambda +a_{11})(\lambda +a_{22})-a_{12}a_{21}=0$ . For convenience, we denote this limit by $\overline {\lambda }$ . In view of $\mathcal {R}_0(0)>1$ , it then follows that $\frac {a_{12}a_{21}}{a_{11}a_{22}}>1$ , which implies $\overline {\lambda }<0$ .

For any $\epsilon>0$ , there exists $M>0$ such that $-\epsilon \leq \frac {-\overline {\lambda }+\epsilon }{M} \leq \epsilon $ . Moreover, there exists $D_0>0$ such that $\vert \overline {\lambda }^*(D,M)- \overline {\lambda } \vert \leq \epsilon $ for all $D \in (0,D_0)$ . Therefore,

$$ \begin{align*}-\frac{\overline{\lambda}^*(D,M)}{M}\leq \frac{-\overline{\lambda} +\epsilon}{M}\leq \epsilon, \end{align*} $$

which implies that $\lim \limits _{D \rightarrow 0^+}\inf _{\mu>0} -\frac {\overline {\lambda }^*(D,\mu )}{\mu }=0$ .

(ii) In view of Lemma 4.8, there exists $D^*>0$ such that $\mathcal {R}_0(D^*)=1$ , and $\mathcal {R}_0(D)>1$ for all $D\in (0,D^*)$ and $\mathcal {R}_0(D)<1$ for all $D>D^*$ . Proposition 4.2 implies that $c^*(D)>0$ for all $D\in (0,D^*)$ .

We next show that $\frac {\partial }{\partial \mu } \lambda ^*(D^*,0)=0$ . For convenience, we use $\dot {\lambda }$ , $\dot {\psi }_1$ , and $\dot {\psi }_2$ to represent $\frac {\partial }{\partial \mu } \lambda $ , $\frac {\partial }{\partial \mu } \psi _1$ , and $\frac {\partial }{\partial \mu } \psi _2$ , respectively. Differentiating (4.13) with respect to $\mu $ , we obtain

(4.14) $$ \begin{align} \begin{cases} m_{11}\dot{\psi}_1+ \tilde{m}_{12} \dot{\psi}_2=-\lambda \dot{\psi}_1-\dot{\lambda} \psi_1,\quad \qquad x\in \mathbb{R},\\[4pt] D\dfrac{\partial^2 \dot{\psi}_2}{\partial x^2} - 2 D\mu \frac{\partial \dot{\psi}_2}{\partial x} + \tilde{m}_{21} \dot{\psi}_1 + D \mu^2 \dot{\psi}_2+m_{22} \dot{\psi}_2-2D \frac{\partial \psi_2}{\partial x}+2D \mu \psi_2\qquad\\\quad =-\lambda \dot{\psi}_2-\dot{\lambda} \psi_2,\quad \quad \qquad x\in \mathbb{R},\\[4pt] \dot{\psi}_1 \text{ and } \dot{\psi}_2 \text{ are } L-\text{periodic in } x. \end{cases} \end{align} $$

Multiplying the first equation of (4.14) by $\psi _1$ and the second equation of (4.14) by $\psi _2$ and summing them together, and multiplying the first equation of (4.13) by $\dot {\psi }_1$ and the second equation of (4.13) by $\dot {\psi }_2$ and summing them together, and integrating them from $0$ to L, and then making a difference, we further have

$$ \begin{align*}\begin{aligned} -\dot{\lambda} \int_{0}^{L}[\psi_1^2+\psi_2^2] \mathrm{d} x &=- 4 D\mu \int_{0}^{L} \frac{\partial \dot{\psi}_2}{\partial x} \psi_2 \mathrm{d} x -2D \int_{0}^{L} \frac{\partial \psi_2}{\partial x} \psi_2 \mathrm{d} x + 2D \mu \int_{0}^{L} \psi_2^2 \mathrm{d} x \\ &=- 4 D\mu \int_{0}^{L} \frac{\partial \dot{\psi}_2}{\partial x} \psi_2 \mathrm{d} x+ 2D \mu \int_{0}^{L} \psi_2^2 \mathrm{d} x. \end{aligned} \end{align*} $$

Letting $\mu =0$ and $D=D^{*}$ , we see that $\frac {\partial }{\partial \mu } \lambda ^*(D^*,0)=0$ .

Notice that $\lambda ^*(D^*,0)=0$ due to $\mathcal {R}_0(D^*)=1$ . For any sequence $D_n \rightarrow D^{*}$ with $D_n<D^{*}$ as $n \rightarrow \infty $ , there exists $\mu _n \rightarrow 0^+$ such that $\frac {\lambda ^*(D_n,\mu _n)}{\mu _n} \rightarrow 0$ as $n \rightarrow \infty $ . It follows that $\inf _{\mu>0} -\frac {\lambda ^*(D_n,\mu )}{\mu } \leq 0$ as $n \rightarrow \infty $ . Therefore, $c^*(D_n) \rightarrow 0$ as $n \rightarrow \infty $ .

(iii) In view of Lemma 4.8(i) and (iii), we have $\mathcal {R}_0(D)>1$ , which implies that $c^*(D)>0$ for all $D\in (0,+\infty )$ by Proposition 4.2. We next prove the following two claims:

Claim 1. There exist $\rho _0>0$ and $\overline {D}_1>0$ such that $-\lambda ^*(D,\mu ) \geq \rho _0$ holds uniformly for $\mu \in [0,+\infty )$ when $ D>\overline {D}_1$ .

Claim 2. For any given $\epsilon _0>0$ , $\lim \limits _{D \rightarrow +\infty } -\frac {\lambda ^*(D,\mu )}{D}=\mu ^2$ holds uniformly on $[\epsilon _0,+\infty )$ , that is, for every $\epsilon>0$ , there exists $\overline {D}_2>0$ large enough such that $\vert -\frac {\lambda ^*(D,\mu )}{D}-\mu ^2 \vert \leq \epsilon $ for all $\mu \in [\epsilon _0,+\infty )$ when $D>\overline {D}_2$ .

We postpone the proof of these two claims and complete the proof in a few lines. For any given $M>0$ , there exists $\delta _0>0$ such that $\frac {\rho _0}{\delta _0} \geq M$ due to $\rho _0>0$ . Thus, $\frac {\rho _0}{ \mu } \geq M$ , $\mu \in [0,\delta _0]$ . According to Claim 1, there exists $D_1$ such that $\frac {-\lambda ^*(D,\mu )}{\mu } \geq \frac {\rho _0}{ \mu } \geq M$ holds for all $\mu \in (0,\delta _0]$ when $D \geq D_1$ . By Claim 2, we can choose $D_2$ large enough such that $\frac {M}{D_2} \leq \frac {\delta _0}{2}$ , and $-\frac {\lambda ^*(D,\mu )}{D} \geq \frac {\delta _0}{2} \mu \geq \frac {M}{D_2} \mu $ for all $\mu \in [\delta _0,+\infty )$ when $D \geq D_2$ . Therefore, $-\frac {\lambda ^*(D,\mu )}{\mu } \geq \frac {MD}{D_2} \geq M$ for all $\mu \in [\delta _0,+\infty )$ when $D \geq D_2$ . Thus, $\frac {-\lambda ^*(D,\mu )}{\mu } \geq M$ for all $\mu \in (0,\infty )$ when $D \geq \max (D_1,D_2)$ .

Note that $-\lambda ^*(D,0)$ is the principal eigenvalue of (4.13) with $\mu =0$ . It is easy to verify that $-\lambda ^*(D,0)$ is non-increasing with respect to $D \in (0,+\infty )$ . In view of $\tilde {\mathcal {R}}_0>1$ , we have $\theta (1-\theta )\int _{0}^{L} \frac {\rho (x)\gamma (x)}{d(x)+\gamma (x)} \mathrm {d} x>\int _{0}^{L} \hat {d} (x) \mathrm {d} x$ . This implies that $\theta (1-\theta )\int _{0}^{L} \frac {\rho (x)\gamma (x)}{d(x)+\gamma (x)+\varrho } \mathrm {d} x=\int _{0}^{L} \hat {d} (x) \mathrm {d} x +\varrho L$ has a unique positive root $\varrho _1$ . It then follows that $-\lambda ^*(D,0)$ converges to $\varrho _1$ as $D \rightarrow +\infty $ . Thus, there exists $\overline {D}_1$ such that $-\lambda ^*(D,0)>\rho _0$ whenever $D>\overline D_1$ for some $\rho _0>0$ . By the proof of Proposition 4.2, we see that $-\lambda ^*(D,\mu )$ is non-decreasing with respect to $\mu \in (0,+\infty )$ . This proves Claim 1.

Let $\underline {\lambda }^*(D,\mu )$ be the principal eigenvalue of the eigenvalue problem (4.12). By the comparison arguments, it is easy to see that

$$ \begin{align*}\overline{\lambda}^*(D,\mu) \leq \lambda^*(D,\mu) \leq \underline{\lambda}^* (D,\mu), \forall (D,\mu) \in \mathbb{R}^2_+, \end{align*} $$

where $\overline {\lambda }^*(D,\mu )$ is the principal eigenvalue of (4.11). It suffices to show that $\lim \limits _{D \rightarrow +\infty } -\frac {\overline {\lambda }^*(D,\mu )}{D}= \lim \limits _{D \rightarrow +\infty } -\frac {\underline {\lambda }^*(D,\mu )}{D}=\mu ^2 $ holds uniformly on $[\epsilon _0,+\infty )$ . Here, we prove that $\lim \limits _{D \rightarrow +\infty } -\frac {\underline {\lambda }^*(D,\mu )}{D}=\mu ^2 $ holds uniformly on $[\epsilon _0,+\infty )$ , and the other limit can be proved similarly. Since $\underline {\lambda }^*(D,\mu )$ is the smaller solution of the equation

$$ \begin{align*}(\lambda+b_{11})(\lambda+b_{22}+D\mu^2)-b_{12}b_{21}=0, \end{align*} $$

it follows that

$$ \begin{align*}&\underline{\lambda}^*(D,\mu)\\ &\quad =\frac{1}{2}\left(-b_{11}-b_{22}-D\mu^2- \sqrt{(b_{11}+b_{22}+D\mu^2)^2-4b_{11}b_{22}-4b_{11}D\mu^2+4b_{12}b_{21}}\right).\end{align*} $$

A straightforward computation gives rise to

$$ \begin{align*}-\frac{\underline{\lambda}^*(D,\mu)}{D}-\mu^2= \frac{2\frac{b_{12}b_{21}}{D\mu^2}-2\frac{b_{11}b_{22}}{D\mu^2}+2b_{22}}{D\left(\sqrt{(\frac{b_{11}}{D\mu^2}-\frac{b_{22}}{D\mu^2}-1)^2+4\frac{b_{12}b_{21}}{D^2\mu^4}} -\frac{b_{11}}{D\mu^2} -\frac{b_{22}}{D\mu^2}+1\right)}, \end{align*} $$

which implies Claim 2.