Proof. Let $\tilde{z}_1= x-Ct$ . When $\tilde{z}_1\le 0$ , we have $\bar{\phi }(\tilde{z}_1)=1$ , and then

(2.8) \begin{equation} \begin{aligned} &\bar{\phi }^{\prime\prime}+C\bar{\phi }^{\prime}+\bar{\phi }\big \{1-r-\bar{\phi }+r(K\star _C\mathcal{L}_{b,c}\bar{\phi })\big \}\\ &\le 1-r-1+r=0. \end{aligned} \end{equation}

When $\tilde{z}_1\gt 0$ , we obtain $\bar{\phi }(\tilde{z}_1)=e^{-\sqrt{r-1} \tilde{z}_1}$ . For any $C\ge \tilde{C}_1$ , where $\tilde{C}_1$ satisfies $-\sqrt{r-1}\tilde{C}_1+r=0$ , it gives

(2.9) \begin{equation} \begin{aligned} &\bar{\phi }^{\prime\prime}+C\bar{\phi }^{\prime}+\bar{\phi }\big \{1-r-\bar{\phi }+r(K\star _C\mathcal{L}_{b,c}\bar{\phi })\big \}\\ &=e^{-\sqrt{r-1} \tilde{z}_1}\big \{-\sqrt{r-1} C-e^{-\sqrt{r-1} \tilde{z}_1}+r(K\star _C\mathcal{L}_{b,c}\bar{\phi })\big \}\\ &\le e^{-\sqrt{r-1} \tilde{z}_1}\big \{-\sqrt{r-1} C+r\big \} \le 0. \end{aligned} \end{equation}

Hence, we conclude that $Q[\bar{\phi }](\theta, x)\le \bar{\phi }(\theta, x-C)$ . In order to prove $Q[\underline{\phi }](\theta, x)\ge \underline{\phi }(\theta, x+C)$ , we let $\tilde{z}_2= x+Ct$ . Choose sufficiently small $\tilde{\epsilon }\gt 0$ and $\tilde{\epsilon }_1\gt 0$ such that

(2.10) \begin{equation} \tilde{\delta }-r(\tilde{\epsilon }+\tilde{\epsilon }_1)\gt 0. \end{equation}

Since $\int _{0}^{\infty }\int _{\mathbb{R}}K(s,y)dyds=1$ , there exist a large enough $\tilde{M}_1\gt 0$ such that

(2.11) \begin{equation} \int _{0}^{\infty }\int _{-\tilde{M}_1}^\infty K(s,y)dyds\gt 1-\tilde{\epsilon }. \end{equation}

Note that we can still derive the existence of $\Psi$ in (2.1) when we choose $\Phi (z)=\underline{\phi }(z)$ (Here, $\Phi ({-}\infty )=1-\tilde{\delta }$ ). There exists $M_0(c)\lt 0$ so that

(2.12) \begin{equation} \mathcal{L}_{b,c}\underline{\phi }(\tilde{z}_2)\gt 1-\tilde{\epsilon }_1 \quad \text{when} \ \tilde{z}_2\lt M_0(c). \end{equation}

For $\tilde{z}_2\lt M_0(c)-\tilde{M}_1\lt 0$ , if $C\ge \tilde{\mu }$ , then by (2.10), (2.11) and (2.12), we have

(2.13) \begin{equation} \begin{aligned} &\underline{\phi }^{\prime\prime}-C\underline{\phi }^{\prime}+\underline{\phi }\big \{1-r-\underline{\phi }+r(K\star _{-C}\mathcal{L}_{b,c}\underline{\phi })\big \}\\ &=\underline{\phi }^{\prime\prime}-C\underline{\phi }^{\prime}+\underline{\phi }\bigg \{1-r-\underline{\phi }+r\int _{0}^{\infty }\int _{-\infty }^{-\tilde{M}_1}K(s,y)\mathcal{L}_{b,c}\underline{\phi }(\tilde{z}_2-Cs-y)dyds\\ &+r\int _{0}^{\infty }\int _{-\tilde{M}_1}^{\infty }K(s,y)\mathcal{L}_{b,c}\underline{\phi }(\tilde{z}_2-Cs-y)dyds \bigg \} \\ &\ge \underline{\phi }^{\prime\prime}-C\underline{\phi }^{\prime}+\underline{\phi }\bigg \{1-r-\underline{\phi }+r(1-\tilde{\epsilon }_1)\int _{0}^{\infty }\int _{-\tilde{M}_1}^{\infty }K(s,y)dyds \bigg \}\\ &\ge \underline{\phi }^{\prime\prime}-C\underline{\phi }^{\prime}+\underline{\phi }\{1-r-\underline{\phi }+r(1-\tilde{\epsilon })(1-\tilde{\epsilon }_1)\}\\ &=(1-\tilde{\delta })\bigg \{ \tilde{\mu }e^{\tilde{\mu } \tilde{z}_2} (C-\tilde{\mu }) +(1-e^{\tilde{\mu } \tilde{z}_2})[\tilde{\delta }-r(\tilde{\epsilon }+\tilde{\epsilon }_1) +r\tilde{\epsilon }\tilde{\epsilon }_1+(1-\tilde{\delta })e^{\tilde{\mu } \tilde{z}_2}] \bigg \}\ge 0.\\ \end{aligned} \end{equation}

Take $\tilde{\eta }\,:\!=\,\frac{\underline{\phi }(M_0(c)-\tilde{M}_1)}{1-\tilde{\delta }}\in (0,1)$ . For $\tilde{z}_2\in [M_0(c)-\tilde{M}_1,0)$ , we have $\underline{\phi }(\tilde{z}_2)\in (0, \tilde{\eta }(1-\tilde{\delta })]$ and $e^{\tilde{\mu } \tilde{z}_2}\in [1-\tilde{\eta },1)$ . If $C\ge \frac{\tilde{\mu }^2+\tilde{\eta }[\tilde{\eta }(1-\tilde{\delta })+r-1]}{\tilde{\mu }(1-\tilde{\eta })}$ , then

(2.14) \begin{equation} \begin{aligned} &\underline{\phi }^{\prime\prime}-C\underline{\phi }^{\prime}+\underline{\phi }\big \{1-r-\underline{\phi }+r(K\star _{-C}\mathcal{L}_{b,c}\underline{\phi })\big \}\\ &\ge \underline{\phi }^{\prime\prime}-C\underline{\phi }^{\prime}+\underline{\phi }\big \{1-r-\underline{\phi }\big \}\\ &\ge -(1-\tilde{\delta })\tilde{\mu }^2e^{\tilde{\mu } \tilde{z}_2}+(1-\tilde{\delta })C\tilde{\mu }e^{\tilde{\mu } \tilde{z}_2}+\tilde{\eta }(1-\tilde{\delta })[1-r-\tilde{\eta }(1-\tilde{\delta })]\\ &\ge (1-\tilde{\delta })\bigg \{ -\tilde{\mu }^2+C\tilde{\mu }(1-\tilde{\eta })+\tilde{\eta }[1-r-\tilde{\eta }(1-\tilde{\delta })] \bigg \}\ge 0.\\ \end{aligned} \end{equation}

For $\tilde{z}_2\ge 0$ , we have $\underline{\phi }(\tilde{z}_2)=0$ , and thus

(2.15) \begin{equation} \underline{\phi }^{\prime\prime}-C\underline{\phi }^{\prime}+\underline{\phi }\big \{1-r-\underline{\phi }+r(K\star _{-C}\mathcal{L}_{b,c}\underline{\phi })\big \}=0. \end{equation}

Therefore, $Q[\underline{\phi }](\theta, x)\ge \underline{\phi }(\theta, x+C)$ for all $C\ge \tilde{C}_2\,:\!=\,\max \{\frac{\tilde{\mu }^2+\tilde{\eta }[\tilde{\eta }(1-\tilde{\delta })+r-1]}{\tilde{\mu }(1-\tilde{\eta })}, \tilde{\mu } \}$ . Let $\tilde{C}=\max \{\tilde{C}_1, \tilde{C}_2\}$ . Thus, the proof is complete.

Proof. Choose $\tilde{M}\gt 0$ such that $e^{-\sqrt{r-1}\tilde{M}}\lt \frac{1}{2}(1-\tilde{\delta })$ . It follows

\begin{align*} e^{-\sqrt {r-1}\tilde {M}} & = \bar {\phi }(\theta, \tilde {M})=\bar {\phi }_n(\theta, \tilde {M}+n+\tilde {C})\ge \phi _n(\theta, \tilde {M}+n+\tilde {C}),\\\frac {1}{2}(1-\tilde {\delta })=\underline {\phi }(\theta, -\frac {1}{\tilde {\mu }}\ln 2) & = \underline {\phi }_n(\theta, -\frac {1}{\tilde {\mu }}\ln 2-(n+\tilde {C}))\le \phi _n(\theta, -\frac {1}{\tilde {\mu }}\ln 2-(n+\tilde {C})). \end{align*}

Define $b_n\,:\!=\,\sup \limits _{}\{x\ :\ \phi _n(\theta, x)\in [\frac{1}{2}(1-\tilde{\delta }),1]\}$ . In other word, $b_n$ is a point so that $\phi _n(\theta, b_n)=\frac 12 (1-\tilde{\delta })$ . Then $-\frac{1}{\tilde{\mu }}\ln 2-(n+\tilde{C})\le b_n\le \tilde{M}+n+\tilde{C}$ . Let $\phi _{+,n}(\theta, x)\,:\!=\,\phi _n(\theta, x+b_n)$ . Then we have

\begin{align*} \phi _{+,n}=\phi _n(\theta, {\cdot}+b_n)=G_n[\phi _n](\theta, {\cdot}+b_n)=Q[\phi _n(\theta, \kappa _n({\cdot}+b_n))]\in Q[C_{\beta }]. \end{align*}

By the solution of the integral form of system (2.3), we have

\begin{align*} \phi _{+,n}(\theta, x)=\int _{\mathbb {R}}\bar {G}(x-y,1)\phi _n(\theta, \kappa _n(y+b_n) )dy+\int _{0}^{1}\int _{\mathbb {R}}\bar {G}(x-y,1-s)\bar {f}(\phi (1,y))dyds, \end{align*}

where $\bar{f}(\phi )=\phi (1-r-\phi +r(K*\mathcal{L}_{b,c}\phi ))$ is the reaction term of (2.3) and $\bar{G}$ is the fundamental solution of $u_t=u_{xx}$ . By calculation, we have $|(\phi _{+,n})_{x}(\theta, x)|\lt \frac{1}{\sqrt{2\pi }}+\sqrt{\frac{2}{\pi }}$ . Therefore, $\phi _{+,n}$ is equicontinuous and uniformly bounded. According to the Ascoli–Arzela theorem, there exist a subsequence (denoted by $n$ ), a non-increasing function $\phi _+$ , and $\xi _+$ such that

\begin{align*} \lim \limits _{n\rightarrow \infty }\frac {b_n}{n}=\xi _+, \quad \lim \limits _{n\rightarrow \infty } \phi _{+,n}=\phi _+. \end{align*}

Define $C_+\,:\!=\,\tilde{C}\xi _+.$ Observe that

\begin{align*} \lim \limits _{n\rightarrow \infty }\kappa _n(x+b_n)-b_n=\lim \limits _{n\rightarrow \infty }(x+\tilde {C}{\cdot}\frac {x+b_n}{n})=x+C_+ \end{align*}

holds uniformly for $x$ in any bounded subset of $\mathbb{R}$ . For any $x\in \mathbb{R}$ , we have

(2.16) \begin{equation} \begin{aligned} \phi _+(\theta, x-C_+)&=\lim \limits _{n\rightarrow \infty } \phi _{+,n}(\theta, x-C_+)=\lim \limits _{n\rightarrow \infty }\phi _n(\theta, x-C_++b_n)=\lim \limits _{n\rightarrow \infty }G_n[\phi _n](\theta, x-C_++b_n)\\ &=\lim \limits _{n\rightarrow \infty }Q[\phi _n(\theta, \kappa _n({\cdot}+b_n))](x-C_+)=\lim \limits _{n\rightarrow \infty }Q[\phi _{+,n}(\theta, \kappa _n({\cdot}+b_n)-b_n)](x-C_+)\\ &=Q[\phi _+](\theta, x). \end{aligned} \end{equation}

Consequently, $\phi _+(\theta, -\infty )$ and $\phi _+(\theta, \infty )$ are fixed points of $Q$ . Observing that $\frac{1}{2}(1-\tilde{\delta })\le \lim \limits _{n\rightarrow \infty }\phi _n(\theta, b_n)=\phi _+(\theta, 0)\lt 1$ . It follows that $\phi _+(\theta, -\infty )=1$ and $\phi _+(\theta, \infty )=0$ . Hence, the proof is complete.

Proof. The existence of $\Phi$ in (2.3) connecting $\beta$ to $0$ can be guaranteed by Theorem2.4. We now prove $\Phi _z(\theta, z)\lt 0$ by a contradiction argument. Assume $\Phi _z(\theta, z_0)=0$ for some $z_0\in \mathbb{R}$ . The strong maximum principle implies that $\Phi _z(\theta, z)\equiv 0$ for $z\in \mathbb{R}$ . It contradicts the boundary condition (2.5). Next, we will show the monotonicity of $C(c)$ with respect to $c$ . Let $0\lt c_1\le c_2$ . By taking the maximum value, we can set a same $\tilde{C}$ in Lemma 2.2 for both $c=c_1$ and $c=c_2$ . According to Lemma 2.1 $c)$ , we conclude that the reaction term in (2.4) is non-increasing with respect to $c$ . From Lemma 2.3, we have $\phi _n(c_1,\theta, \tilde{x})\ge \phi _n(c_2, \theta, \tilde{x})$ (here $\phi _n(c, \theta, \tilde{x})$ implies that $\phi _n$ is dependent on $c$ ). Based on the proof of Theorem2.4, we know that $b_n(c_1)\ge b_n(c_2)$ (Here $b_n(c)$ indicates that $b_n$ is dependent on $c$ ). Consequently, $C(c_1)\ge C(c_2)$ . Therefore, the proof is complete.

Proof. Theorem2.5 gives the existence of the monotone bistable travelling wave solution of (2.4)-(2.5). We also know that $C(c)$ is a non-increasing function of $c$ . To return to (1.5)-(1.6) for the existence of waves, we need to show that $C(c)=c$ has a unique positive root. To this end, it suffices to show $C(0_+)\gt 0$ . Let

\begin{align*} f(\Phi )=\Phi (1-r-\Phi +r(K\star _C\mathcal {L}_{b,c}\Phi )). \end{align*}

Then (2.4) can be written as

(2.17) \begin{equation} \Phi ^{\prime\prime}+C(c,f)\Phi ^{\prime}+f(\Phi )=0. \end{equation}

Here we just use $C(c,f)$ to indicate that the speed is dependent on the reaction term.

By Lemma 2.1 $c)$ , there exist a sufficiently small $\delta \gt 0$ such that $\mathcal{L}_{b,c}\Phi (z)\gt 1-\delta$ for sufficiently small $c\lt \delta$ and $\Phi \gt 0$ . Then, for small $\hat{\varepsilon }\gt 0$ , we have

(2.18) \begin{equation} f(\Phi )\ge g(\Phi )\,:\!=\,\left \{ \begin{array}{cc} \begin{aligned} &\Phi (1-r\delta -\Phi ),\quad & \Phi \gt \hat{\varepsilon }, \\[3pt] &\Phi (1-r-\Phi ),\ &\Phi \lt \hat{\varepsilon }.\\ \end{aligned} \end{array} \right . \end{equation}

Consider

(2.19) \begin{equation} \Phi ^{\prime\prime}+C(c,g)\Phi ^{\prime}+g(\Phi )=0. \end{equation}

By choosing $\tilde{\delta }$ in (2.6) such that $\tilde{\delta }-r\delta \gt 0$ , we can derive the existence of $\Phi (z)$ in (2.19) connecting $1-r\delta$ to $0$ according to the similar arguments in Lemmas 2.2 and 2.3, and Theorem2.4. We multiply the left and right sides of (2.19) by $\Phi ^{\prime}$ and integrate both sides from $-\infty$ to $\infty$ with respect to $z$ . It follows that the wave speed $C(c,g)$ satisfies

\begin{align*} C(c,g)=\frac {\int _{0}^{1-r\delta }g(\Phi )d\Phi }{\int _{-\infty }^{\infty }(\Phi ^{\prime})^2dz}. \end{align*}

Since $\hat{\varepsilon }$ and $\delta$ are small enough, we have

\begin{align*}\int _{0}^{1-r\delta }g(\Phi )d\Phi =\frac {1}{6}(1-r \delta )^3-\frac {1}{2}r(1-\delta )\hat {\varepsilon }^2\gt 0.\end{align*}

It means $C(c,g)\gt 0$ . Applying equations (2.17), (2.18), (2.19), and a similar proof as presented in Theorem2.5(concerning the monotonicity of $C(c)$ with respect to $c$ ), we can deduce that $C(c,f)\ge C(c,g)\gt 0$ when $c \rightarrow 0$ . According to Theorem2.5, the equation $C(c)=c$ has a unique positive solution. Consequently, system (2.2) has a monotonic solution $(c,\Phi (z))$ , that is, there exists a monotone bistable travelling wave $(c, \Phi (z), \Psi (z))$ of system (1.5)-(1.6). Based on Lemma 2.1 and Theorem2.5, we know $\Phi ^{\prime}(z)\lt 0$ and $\Psi ^{\prime}(z)\lt 0$ for $z\in \mathbb{R}$ . It completes the proof.

Proof. We only prove $(\phi ^+(t,z), \psi ^+(t,z))$ is an upper solution of (1.2), while the lower solution $(\phi ^-(t,z), \psi ^-(t,z))$ can be verified using a similar argument. Let $\xi =z+ \xi _0+ \sigma _1\delta (1-e^{-\rho t})$ . We choose $\delta$ , $\mu$ and $\rho \lt \min \{\mu ^2,\tilde{\delta }_0\}$ that are small enough, where $\tilde{\delta }_0$ is defined in (K2). We take $R_1^+$ and $R_2^+$ such that

\begin{align*}\frac {R_1^+}{R_2^+\int _{0}^{\infty }\int _{\mathbb {R}}K(s,y)e^{\rho s}dyds}\gt r.\end{align*}

For small enough $\epsilon _1\gt 0$ , $\epsilon _3\gt 0$ and $\epsilon _4\gt 0$ , we take sufficiently large $M_1\gt \frac{1}{\mu ^2}$ such that $\Phi (z), \Psi (z)\lt \epsilon _1$ for $\xi \gt \frac{M_1}{2}$ and

\begin{align*}\int _{0}^{+\infty }\int _{\frac {1}{2}M_1}^\infty K(s,y)dyds\lt \epsilon _3,\ \int _{0}^{+\infty }\int _{\frac {1}{2}M_1}^\infty K(s,y)e^{\rho s}dyds\lt \epsilon _4,\end{align*}

by using (K1) and (K2). Take $M_2\gt 0$ and sufficiently small $\epsilon _2\gt 0$ such that $\Phi (z), \Psi (z)\gt 1-\epsilon _2$ for $\xi \lt -M_2$ .

Let $M\gt \max \{\eta _1, \eta _2, M_1, M_2\}$ . Set

(2.21) \begin{equation} \begin{aligned} &L_1\,:\!=\,\max \{\max _{-M\le \xi \le M}\Phi ^{\prime}(\xi ), \max _{-M\le \xi \le M}\Psi ^{\prime}(\xi )\},\\ &L_2\,:\!=\,\max \{ \max _{\xi \in [{-}M, M]}R_i^{\prime}(\xi ), \max _{\xi \in [{-}M, M]}R_i^{\prime\prime}(\xi ) \}, i=1,2,\\ &\sigma _1\lt \frac{L_2+cL_2+R_1^+\rho +R_2^+\rho + rR_2^+\int _{0}^{\infty }\int _{\mathbb{R}}K(s,y)e^{\rho s}dyds+(1+b)R_1^+}{\rho L_1}. \end{aligned} \end{equation}

Substituting $(\phi ^+,\psi ^+)$ into (1.2), we get

(2.22) \begin{equation} \begin{aligned} &\phi ^+_{zz}+c\phi ^+_{z}-\phi ^+_{t}+\phi ^+(1-r-\phi ^++r(K*\psi ^+))\\ & =\Phi ^{\prime\prime}+R_1^{\prime\prime}\delta e^{-\rho t} +c\Phi ^{\prime}-\sigma _1 \rho \Phi ^{\prime}\delta e^{-\rho t} +cR_1^{\prime}\delta e^{-\rho t} -\sigma _1\rho R_1^{\prime}\delta ^2 e^{-2\rho t} +R_1\rho \delta e^{-\rho t}\\ &+(\Phi +R_1\delta e^{-\rho t})\bigg \{ 1-r-\Phi -R_1\delta e^{-\rho t}+r(K*\psi ^+)\bigg \}\\ &=R_1^{\prime\prime}\delta e^{-\rho t}-\sigma _1 \rho \Phi ^{\prime}\delta e^{-\rho t} +cR_1^{\prime}\delta e^{-\rho t} -\sigma _1\rho R_1^{\prime}\delta ^2 e^{-2\rho t} +R_1\rho \delta e^{-\rho t}\\ &+\Phi \bigg \{-R_1\delta e^{-\rho t}+J(\Psi )+J_1(R_2)\bigg \}\\ &+R_1\delta e^{-\rho t}\bigg \{ 1-r-\Phi -R_1\delta e^{-\rho t}+r(K*\psi ^+)\bigg \}\\ &\,:\!=\,I+II+III, \quad \forall t\gt 0, \end{aligned} \end{equation}

where

\begin{align*} \begin {aligned} K*\psi ^+ & =\int _{0}^{t}\int _{\mathbb {R}}K(s,y)[\Psi (\xi +cs-y+\sigma _1\delta e^{-\rho t}(1-e^{\rho s})) \nonumber\\ & +R_2(\xi +cs-y+\sigma _1\delta e^{-\rho t}(1-e^{\rho s}))\delta e^{-\rho (t-s)}]dyds\\ &+\int _{t}^{\infty }\int _{\mathbb {R}}K(s,y)[\Psi (\xi +cs-y-\sigma _1\delta (1-e^{-\rho t}))+R_2(\xi +cs-y-\sigma _1\delta (1-e^{-\rho t}))\delta ]dyds, \end {aligned} \end{align*}

\begin{align*} \begin {aligned} J(\Psi )&=\int _{0}^{t}\int _{\mathbb {R}}K(s,y)[\Psi (\xi +cs-y+\sigma _1\delta e^{-\rho t}(1-e^{\rho s}))-\Psi (\xi +cs-y)]dyds\\ &+\int _{t}^{\infty }\int _{\mathbb {R}}K(s,y)[\Psi (\xi +cs-y-\sigma _1\delta (1-e^{-\rho t}))-\Psi (\xi +cs-y)]dyds\le 0, \end {aligned} \end{align*}

\begin{align*} \begin {aligned} J_1(R_2)=& r\int _{0}^{t}\int _{\mathbb {R}}K(s,y)R_2(\xi +cs-y+\sigma _1\delta e^{-\rho t}(1-e^{\rho s}))\delta e^{-\rho (t-s)}dyds\\ &+ r\int _{t}^{\infty }\int _{\mathbb {R}}K(s,y)R_2(\xi +cs-y-\sigma _1\delta (1-e^{-\rho t}))\delta dyds\\ &\le r\int _{0}^{t}\int _{\mathbb {R}}K(s,y)R_2(\xi +cs-y+\sigma _1\delta e^{-\rho t}(1-e^{\rho s}))\delta e^{-\rho (t-s)}dyds\\ &+ r\int _{t}^{\infty }\int _{\mathbb {R}}K(s,y)R_2(\xi +cs-y-\sigma _1\delta (1-e^{-\rho t}))\delta e^{-\rho (t-s)}dyds.\\ \end {aligned} \end{align*}

and

(2.23) \begin{equation} \begin{aligned} &\psi ^+_{zz}+c\psi ^+_{z}-\psi ^+_{t}+b\phi ^+(1-\psi ^+)\\ &=\Psi ^{\prime\prime}+R_2^{\prime\prime}\delta e^{-\rho t}+c\Psi ^{\prime}-\sigma _1\rho \Psi ^{\prime}\delta e^{-\rho t}+cR_2^{\prime}\delta e^{-\rho t}-\sigma _1\rho R_2^{\prime}\delta ^2 e^{-2\rho t}+R_2\rho \delta e^{-\rho t} \\ &+b(\Phi +R_1\delta e^{-\rho t})(1-\Psi -R_2\delta e^{-\rho t})\\ &=R_2^{\prime\prime}\delta e^{-\rho t}-\sigma _1\rho \Psi ^{\prime}\delta e^{-\rho t}+cR_2^{\prime}\delta e^{-\rho t}-\sigma _1\rho R_2^{\prime}\delta ^2 e^{-2\rho t}+R_2\rho \delta e^{-\rho t} \\ &-b\Phi R_2\delta e^{-\rho t}+bR_1\delta e^{-\rho t}(1-\Psi -R_2\delta e^{-\rho t}), \quad \forall t\gt 0. \end{aligned} \end{equation}

Here we use the fact $(\Phi, \Psi )$ is the bistable travelling wave of (1.2) and satisfies (1.5). In order to prove $(\phi ^+(z,t),\psi ^+(z,t))$ is an upper solution of (1.2), we consider three cases:

Case (i): $|\xi |\le M$ . According to (2.21) and (2.22), we have

\begin{align*} I\le \delta e^{-\rho t}[L_2-\sigma _1\rho L_1+cL_2+R_1^+\rho ], \end{align*}

\begin{align*} II\le rR_2^+ \delta e^{-\rho t}\int _{0}^{\infty }\int _{\mathbb {R}}K(s,y)e^{\rho s}dyds, \end{align*}

and

\begin{align*} III\le R_1^+\delta e^{-\rho t}[1-r+r(1+R_2^+\delta e^{-\rho t}\int _{0}^{\infty }\int _{\mathbb {R}}K(s,y)e^{\rho s}dyds)]. \end{align*}

Then it gives

(2.24) \begin{equation} \begin{aligned} &\phi ^+_{zz}+c\phi ^+_{z}-\phi ^+_{t}+\phi ^+(1-r-\phi ^++r(K*\psi ^+))\\ &\le \delta e^{-\rho t}[L_2-\sigma _1\rho L_1+cL_2+R_1^+\rho +rR_2^+\int _{0}^{\infty }\int _{\mathbb{R}}K(s,y)e^{\rho s}dyds+R_1^+\\ &+rR_1^+R_2^+\delta \int _{0}^{\infty }\int _{\mathbb{R}}K(s,y)e^{\rho s}dyds]\le 0. \end{aligned} \end{equation}

The last inequality sign in (2.24) holds by (K2), (2.21) and $\delta$ is sufficiently small. By (2.21) and (2.23), we also have

(2.25) \begin{equation} \begin{aligned} &\psi ^+_{zz}+c\psi ^+_{z}-\psi ^+_{t}+b\phi ^+(1-\psi ^+)\\ &\le \delta e^{-\rho t}[L_2-\sigma _1\rho L_1+cL_2+R_2^+\rho +bR_1^+]\le 0. \end{aligned} \end{equation}

Case (ii): $\xi \lt -M$ . Note that $\xi \lt -M_2$ , $R_1(\xi )=R_1^+$ and $R_2(\xi )=R_2^+$ . According to (2.22) and the fact $\frac{R_1^+}{R_2^+\int _{0}^{\infty }\int _{\mathbb{R}}K(s,y)e^{\rho s}dyds}\gt r$ , we have

\begin{align*} I\le R_1^+\rho \delta e^{-\rho t}, \end{align*}

\begin{align*} II\le \Phi \delta e^{-\rho t}({-}R_1^++rR_2^+\int _{0}^{\infty }\int _{\mathbb {R}}K(s,y)e^{\rho s}dyds)\le \delta e^{-\rho t}(1-\epsilon _2)({-}R_1^++rR_2^+\int _{0}^{\infty }\int _{\mathbb {R}}K(s,y)e^{\rho s}dyds), \end{align*}

and

\begin{align*} III\le R_1^+\delta e^{-\rho t}[1-r-(1-\epsilon _2)+r(1+R_2^+\delta e^{-\rho t}\int _{0}^{\infty }\int _{\mathbb {R}}K(s,y)e^{\rho s}dyds)]. \end{align*}

Then

(2.26) \begin{equation} \begin{aligned} &\phi ^+_{zz}+c\phi ^+_{z}-\phi ^+_{t}+\phi ^+(1-r-\phi ^++r(K*\psi ^+))\\ &\le \delta e^{-\rho t}[R_1^+\rho -(1-\epsilon _2)(R_1^+-rR_2^+\int _{0}^{\infty }\int _{\mathbb{R}}K(s,y)e^{\rho s}dyds)\\ &+R_1^+(\epsilon _2+rR_2^+\delta \int _{0}^{\infty }\int _{\mathbb{R}}K(s,y)e^{\rho s}dyds)]\le 0, \end{aligned} \end{equation}

since $\frac{R_1^+}{R_2^+\int _{0}^{\infty }\int _{\mathbb{R}}K(s,y)e^{\rho s}dyds}\gt r$ , and $\rho$ , $\epsilon _2$ , $\delta$ are small enough. By (2.23), we have

(2.27) \begin{equation} \begin{aligned} &\psi ^+_{zz}+c\psi ^+_{z}-\psi ^+_{t}+b\phi ^+(1-\psi ^+)\\ &\le \delta e^{-\rho t}[R_2^+\rho -bR_2^+(1-\epsilon _2)+bR_1^+(1-(1-\epsilon _2)-R_2^+\delta e^{-\rho t})]\\ &\le \delta e^{-\rho t}[R_2^+\rho -bR_2^+(1-\epsilon _2)+bR_1^+\epsilon _2]\le 0. \end{aligned} \end{equation}

The last inequality in (2.27) holds since $\rho$ and $\epsilon _2$ are small enough.

Case (iii): $\xi \gt M$ . Note that $\xi \gt M_1$ , $R_1(\xi )=e^{-2\mu \xi }$ and $R_2(\xi )=e^{-\mu \xi }$ . We can calculate

(2.28) \begin{equation} \begin{aligned} &\int _{0}^{\infty }\int _{\mathbb{R}}K(s,y)\Psi (\xi +cs-y)dyds\\ &=\int _{0}^{\infty }\int _{-\infty }^{\frac{1}{2}M_1}K(s,y)\Psi (\xi +cs-y)dyds+\int _{0}^{\infty }\int _{\frac{1}{2}M_1}^{\infty }K(s,y)\Psi (\xi +cs-y)dyds\\ &\le \epsilon _1\int _{0}^{\infty }\int _{-\infty }^{\frac{1}{2}M_1}K(s,y)dyds+\int _{0}^{\infty }\int _{\frac{1}{2}M_1}^{\infty }K(s,y)dyds\\ &\le \epsilon _1+\epsilon _3 \end{aligned} \end{equation}

and

(2.29) \begin{equation} \begin{aligned} &\int _{0}^{\infty }\int _{\mathbb{R}}K(s,y)R_2(\xi +cs-y)e^{\rho s}dyds\\ &=\int _{0}^{\infty }\int _{-\infty }^{\frac{1}{2}M_1}K(s,y)R_2(\xi +cs-y)e^{\rho s}dyds+\int _{0}^{\infty }\int _{\frac{1}{2}M_1}^{\infty }K(s,y)R_2(\xi +cs-y)e^{\rho s}dyds\\ &\le e^{-\frac{1}{2}\mu M_1}\int _{0}^{\infty }\int _{-\infty }^{\frac{1}{2}M_1}K(s,y)e^{\rho s}dyds+R_2^+\int _{0}^{\infty }\int _{\frac{1}{2}M_1}^{\infty }K(s,y)e^{\rho s}dyds\\ &\le e^{-\frac{1}{2\mu }}\int _{0}^{\infty }\int _{-\infty }^{\frac{1}{2}M_1}K(s,y)e^{\rho s}dyds+R_2^+\epsilon _4. \end{aligned} \end{equation}

According to (2.22), we have

\begin{align*} I\le e^{-2\mu \xi }\delta e^{-\rho t}(4\mu ^2-2c\mu +\rho ), \end{align*}

(2.30) \begin{equation} \begin{aligned} & II\le \Phi \bigg \{{-}e^{-2\mu \xi }\delta e^{-\rho t}+r\int _{0}^{\infty }\int _{\mathbb{R}}K(s,y)R_2(\xi +cs-y)\delta e^{-\rho (t-s)}dyds\bigg \}\\ &\le \Phi \delta e^{-\rho t}[{-}e^{-2\mu \xi }+ r(e^{-\frac{1}{2\mu }}\int _{0}^{\infty }\int _{-\infty }^{\frac{1}{2}M_1}K(s,y)e^{\rho s}dyds+R_2^+\epsilon _4) ], \end{aligned} \end{equation}

and

(2.31) \begin{equation} \begin{aligned} &III\le e^{-2\mu \xi } \delta e^{-\rho t}\bigg \{ 1-r+r\int _{0}^{\infty }\int _{\mathbb{R}}K(s,y)\Psi (\xi +cs-y)dyds+rR_2^+\delta e^{-\rho t} \int _{0}^{\infty }\int _{-\infty }^{\infty }K(s,y)e^{\rho s}dyds \bigg \}\\ &\le e^{-2\mu \xi } \delta e^{-\rho t}\bigg \{ 1-r+r (\epsilon _1+\epsilon _3)+rR_2^+\delta \int _{0}^{\infty }\int _{-\infty }^{\infty }K(s,y)e^{\rho s}dyds \bigg \}. \end{aligned} \end{equation}

Then equation (2.22) becomes

(2.32) \begin{equation} \begin{aligned} &\phi ^+_{zz}+c\phi ^+_{z}-\phi ^+_{t}+\phi ^+(1-r-\phi ^++r(K*\psi ^+))\\ &\le e^{-2\mu \xi }\delta e^{-\rho t}(4\mu ^2-2c\mu +\rho )+ \Phi \delta e^{-\rho t}[{-}e^{-2\mu \xi }+ r(e^{-\frac{1}{2\mu }}\int _{0}^{\infty }\int _{-\infty }^{\frac{1}{2}M_1}K(s,y)e^{\rho s}dyds+R_2^+\epsilon _4) ]\\ &+e^{-2\mu \xi } \delta e^{-\rho t}\bigg \{ 1-r+r (\epsilon _1+\epsilon _3)+rR_2^+\delta \int _{0}^{\infty }\int _{-\infty }^{\infty }K(s,y)e^{\rho s}dyds \bigg \}\\ &\le e^{-2\mu \xi }\delta e^{-\rho t}\bigg \{ 4\mu ^2-2c\mu +\rho + r(e^{-\frac{1}{2\mu }}\int _{0}^{\infty }\int _{-\infty }^{\frac{1}{2}M_1}K(s,y)e^{\rho s}dyds+R_2^+\epsilon _4)\\ &+1-r+r (\epsilon _1+\epsilon _3)+rR_2^+\delta \int _{0}^{\infty }\int _{-\infty }^{\infty }K(s,y)e^{\rho s}dyds \bigg \}\le 0. \end{aligned} \end{equation}

The last inequality sign in (2.32) holds by (K2), $r\gt 1$ and $\mu$ , $\rho$ , $\epsilon _1$ , $\epsilon _3$ , $\epsilon _4$ , $\delta$ are small enough. It follows from (2.23) that

(2.33) \begin{equation} \begin{aligned} &\psi ^+_{zz}+c\psi ^+_{z}-\psi ^+_{t}+b\phi ^+(1-\psi ^+)\\ &\le \delta e^{-\rho t}e^{-\mu \xi }[\mu ^2-c\mu +\rho ]+be^{-2\mu \xi }\delta e^{-\rho t} \le \delta e^{-\rho t}e^{-\mu \xi }[\mu ^2-c\mu +\mu ^2+be^{-\frac{1}{\mu }}]\\ &\le \mu \delta e^{-\rho t}e^{-\mu \xi }[2\mu -c+b\frac{1}{\mu }e^{-\frac{1}{\mu }}]\le 0, \end{aligned} \end{equation}

since $\rho \lt \mu ^2$ , $\xi \gt M_1\gt \frac{1}{\mu ^2}$ , and $\mu$ , $\frac{1}{\mu }e^{-\frac{1}{\mu }}$ are sufficiently small.

Combining cases (i)-(iii), we know $(\phi ^+(z, t), \psi ^+(z,t))$ is an upper solution of (1.2).

Proof. By using a similar argument in [Reference Trofimchuk, Pinto and Trofimchuk31, Lemma 13], we can conclude the bistable travelling wave of system (1.2) connecting $e_\beta$ to $e_0$ decays exponentially as $z\rightarrow \infty$ . Therefore, there exist $\xi _1\in \mathbb{R}$ and $h\gg 1$ such that, for $\theta \in ({-}\infty, 0]$ , we have

(2.34) \begin{equation} \begin{aligned} &\Phi (\nu\ {\cdot}\ x-c\theta +\xi _1))-\delta R_1(\nu\ {\cdot}\ x-c\theta )\le \Phi ^*(\nu\ {\cdot}\ x-c^*\theta )\le \Phi (\nu\ {\cdot}\ x-c\theta +\xi _1-h)+\delta R_1(\nu\ {\cdot}\ x-c\theta ),\\ &\Psi (\nu\ {\cdot}\ x-c\theta +\xi _1)-\delta R_2(\nu\ {\cdot}\ x-c\theta )\le \Psi ^*(\nu\ {\cdot}\ x-c^*\theta )\le \Psi (\nu\ {\cdot}\ x-c\theta +\xi _1-h)+\delta R_2(\nu\ {\cdot}\ x-c\theta ), \end{aligned} \end{equation}

thanks to the sufficiently small decay rate $\mu$ in $R_1$ and $R_2$ . Then by Lemma 2.7 and the comparison theorem, we have

(2.35) \begin{equation} \begin{aligned} &\Phi (\nu\ {\cdot}\ x-ct+\xi _1-\sigma _1\delta (1-e^{-\rho t}))-\delta R_1(\nu\ {\cdot}\ x-ct-\sigma _1\delta (1-e^{-\rho t}))e^{-\rho t}\le \Phi ^*(\nu\ {\cdot}\ x-c^*t)\\ &\le \Phi (\nu\ {\cdot}\ x-ct+\xi _1-h+\sigma _1\delta (1-e^{-\rho t}))+\delta R_1(\nu\ {\cdot}\ x-ct+\sigma _1\delta (1-e^{-\rho t}))e^{-\rho t},\\ &\Psi (\nu\ {\cdot}\ x-ct+\xi _1-\sigma _1\delta (1-e^{-\rho t}))-\delta R_2(\nu\ {\cdot}\ x-ct-\sigma _1\delta (1-e^{-\rho t}))e^{-\rho t}\le \Psi ^*(\nu\ {\cdot}\ x-c^*t)\\ &\le \Psi (\nu\ {\cdot}\ x-ct+\xi _1-h+\sigma _1\delta (1-e^{-\rho t}))+\delta R_2(\nu\ {\cdot}\ x-ct+\sigma _1\delta (1-e^{-\rho t}))e^{-\rho t}, \ \ t\gt 0. \end{aligned} \end{equation}

Fix $\bar{z}=\nu {\cdot}x-c^*t$ such that $\Phi ^*(\bar{z})\gt 0$ . If $c^*\gt c$ , then

\begin{align*} \Phi ^*(\bar {z})\le \Phi (\bar {z}+(c^*-c))t+\xi _1-h+\sigma _1\delta (1-e^{-\rho t}))+\delta R_1(z+\sigma _1\delta (1-e^{-\rho t}))e^{-\rho t}\rightarrow 0, \text {as} \ t\rightarrow \infty . \end{align*}

It is in contradiction with the fact that $\Phi ^*(\bar{z})\gt 0$ , implying that $c^* \leq c$ . Likewise, according to the left inequality in (2.35), we have $c^* \geq c$ . Consequently, we can conclude that $c^* = c$ .

By taking the limit as $t\rightarrow \infty$ in (2.35), we obtain the following inequalities:

\begin{align*} \begin {aligned} \Phi (z+\xi _1-\sigma _1\delta )\le \Phi ^*(z)\le \Phi (z+\xi _1-h+\sigma _1\delta ), \ \forall z\in \mathbb {R},\\ \Psi (z+\xi _1-\sigma _1\delta )\le \Psi ^*(z)\le \Psi (z+\xi _1-h+\sigma _1\delta ), \ \forall z\in \mathbb {R}. \end {aligned} \end{align*}

Define

\begin{align*} \begin {aligned} \xi _*=\sup \{\xi \ :\ \ \Phi ^*({\cdot})\le \Phi ({\cdot}+\xi ), \ \Psi ^*({\cdot})\le \Psi ({\cdot}+\xi )\},\\ \xi ^*=\inf \{\xi \ :\ \ \Phi ^*({\cdot})\ge \Phi ({\cdot}+\xi ), \ \Psi ^*({\cdot})\ge \Psi ({\cdot}+\xi )\}. \end {aligned} \end{align*}

By employing a similar approach as the proof in [Reference Chen6, Theorem 2.1], we can prove $\xi _*=\xi ^*$ . Thus, the uniqueness of the wave profile, up to translation, is established.

Proof. We only prove that $c$ is non-increasing in $r$ . The monotonicity of $c$ with respect to $b$ can be proved by a similar method. Take $0\lt r_1\le r_2$ . To the contrary, we assume $c_1\lt c_2$ , where $c_i$ is the bistable wave speed when $r=r_i, i=1,2$ . Let $(\Phi _i,\Psi _i)(z), i=1,2,$ be the travelling wave with speed $c_i$ , satisfying system (1.5)-(1.6). since

\begin{align*}\Phi _1(z)\sim 1-a_1e^{\zeta (c_1) z}, \ \Phi _2(z)\sim 1-a_2e^{\zeta (c_2) z},\ \text {as} \ z\rightarrow -\infty, \\ \Phi _1(z)\sim a_3e^{-\xi (c_1,r_1)z}, \ \Phi _2(z)\sim a_4e^{-\xi (c_2,r_2) z},\ \text {as} \ z\rightarrow \infty, \end{align*}

where $0\lt \zeta (c_2)\lt \zeta (c_1), 0\lt \xi (c_1,r_1)\lt \xi (c_2,r_2)$ and $a_i, i=1,2,3,4,$ are positive constants, and then by translation if necessary, we have $\Phi _2(\nu\ {\cdot}\ x)\le \Phi _1(\nu\ {\cdot}\ x)$ for all $x\in \mathbb{R}^n$ . By the $\Psi$ equation in system (1.5), we know $(\Phi _2(\nu\ {\cdot}\ x),\Psi _2(\nu\ {\cdot}\ x))\le (\Phi _1(\nu\ {\cdot}\ x),\Psi _1(\nu\ {\cdot}\ x))$ for all $x\in \mathbb{R}^n$ . Using the comparison principle, we have

\begin{align*} \Phi _2(\nu\ {\cdot}\ x-c_2t)\le \Phi _1(\nu\ {\cdot}\ x-c_1t), \quad \Psi _2(\nu\ {\cdot}\ x-c_2t)\le \Psi _1(\nu\ {\cdot}\ x-c_1t). \end{align*}

Fix $z^*=\nu\ {\cdot}\ x-c_2t$ such that $\Phi _2(z^*)\gt 0$ . It follows that

\begin{align*} \Phi _2(\nu\ {\cdot}\ x-c_2t) \le \Phi _1(\nu\ {\cdot}\ x-c_1t)=\Phi _1(z^*+(c_2-c_1)t)\rightarrow 0,\ \text {as}\ t\rightarrow \infty, \end{align*}

which contradicts $\Phi _2(z^*)\gt 0$ . Therefore, $c_1\ge c_2$ .

Proof. Substituting $\underline{\Phi }_0(z)$ into $\Phi$ -equation of (1.5), we obtain

\begin{align*} \begin {aligned} &\underline {\Phi }_0^{\prime\prime}+c\underline {\Phi }_0^{\prime}+\underline {\Phi }_0(1-r-\underline {\Phi }_0+r(K\star _c\underline {\Psi }_0))\\ &=e^{-\mu _1 z}[\mu _1^2-c\mu _1+(1-r)]-Me^{-(\mu _1+\delta _1)z}[(\mu _1+\delta _1)^2-c(\mu _1+\delta _1)+1-r]\\ &-e^{-2\mu _1 z}(1-Me^{-\delta _1 z})^2+r(K\star _c\underline {\Psi }_0)e^{-\mu _1 z}(1-Me^{-\delta _1 z}). \end {aligned} \end{align*}

It is observed that the second term is positive. We can choose a sufficiently large $M$ such that the sum of the second and the third terms is positive. The first term is 0, and the last term is positive. Therefore, the proof is complete.

Proof. we assume that there exists $c_2\gt c_0$ such that the travelling wave solution $(\Phi (z), \Psi (z))$ satisfies

\begin{align*}\Phi (z)\sim A_2e^{-\mu _2(c_2) z},\ \text {as} \ z\rightarrow \infty, \ z=\nu {\cdot}x-c_2t,\end{align*}

with $A_2\gt 0$ . In this case, we claim that (1.5)-(1.6) have no travelling wave solutions for any $c\in [c_0, c_2)$ . To prove this by contradiction, assume that there exists $\hat{c}\in (c_0,c_2)$ for which the system (1.2) has a decreasing travelling wave solution $(\bar{\Phi },\bar{\Psi })(\nu\ {\cdot}\ x-\hat{c}t)$ with initial conditions

\begin{align*} \phi (0,x)=\bar {\Phi }(\nu\ {\cdot}\ x) \quad \text {and} \quad \psi (0,x)=\bar {\Psi }(\nu\ {\cdot}\ x). \end{align*}

Clearly, $(\bar{\Phi },\bar{\Psi })(z)$ satisfies system (1.5) with $c=\hat{c}$ . Since $\mu _2(c)$ is non-decreasing with respect to $c$ , we get $\mu _1(\hat{c})\le \mu _2(\hat{c})\le \mu _2(c_2)$ . Recall that $\Phi (z)\sim A_2 e^{-\mu _2(c_2) z}$ as $z\rightarrow \infty$ . It follows that $\Phi (z)\le \bar{\Phi }(z)$ as $z\rightarrow \infty$ . Note that $\Phi (z)\sim 1-A_3e^{\gamma z}$ as $z\rightarrow -\infty$ for positive $A_3$ and $\gamma$ , and $\gamma$ is non-increasing with respect to $c$ . Therefore, $\Phi (z)\le \bar{\Phi }(z)$ as $z\rightarrow -\infty$ . Making a translation if necessary, we can always ensure that $\Phi (z)\le \bar{\Phi }(z)$ for all $z$ . Considering the second equation of (1.2), we then have $(\Phi, \Psi )(\nu\ {\cdot}\ x)\le (\bar{\Phi },\bar{\Psi })(\nu\ {\cdot}\ x)$ for all $x\in \mathbb{R}^n$ (by shift if necessary). By comparison, it gives that

(3.4) \begin{equation} \Phi (\nu\ {\cdot}\ x-c_2t)\le \bar{\Phi }(\nu\ {\cdot}\ x-\hat{c}t). \end{equation}

Fix $\hat{z}=\nu\ {\cdot}\ x-c_2t$ such that $\Phi (\hat{z})\gt 0$ . Observe that

\begin{align*} \bar {\Phi }(\nu\ {\cdot}\ x-\hat {c}t)=\bar {\Phi }(\hat {z}+(c_2-\hat {c})t))\rightarrow 0, \quad \text {as}\quad t \rightarrow \infty . \end{align*}

Thus, we obtain

\begin{align*} \Phi (\hat {z})\le \bar {\Phi }(\nu\ {\cdot}\ x-\hat {c}t)\rightarrow 0, \quad \text {as}\quad t \rightarrow \infty, \end{align*}

which contradicts the fact $\Phi (\hat{z})\gt 0$ . Therefore, there is no travelling wave solution of system (1.5)-(1.6) for any $\hat{c}\in (c_0,c_2)$ . When $r\in (0,1)$ , if $\hat{c}=c_0$ in our assumption, then the travelling waves exist if and only if $\hat{c}\ge c_0$ [Reference Hasík, Kopfová, Nábělková, Trofymchuk and Trofimchuk12, Theorem 5]. We can still derive a contradiction by the above argument. If $r=1$ , then $c_0=0$ . The second equation of (1.5) and the boundary condition (1.6) mean $\hat{c}\gt c_0$ . Therefore, the proof is complete.

Proof. When $r\lt 1$ , the existence of the minimal wave speed $c_{\min }$ , ensuring that monotone and positive travelling wave solutions of (1.5)-(1.6) exist if and only if $c\ge c_{\min }$ , is guaranteed by [Reference Hasík, Kopfová, Nábělková, Trofymchuk and Trofimchuk12, Theorem 5]. We claim that $c_{\min }$ is non-increasing with respect to $r$ . Indeed, let $r_a\lt r_b\lt 1$ . There exists the minimal-speed travelling wave solution $(c_{\min }^{r_a}, \Phi ^{r_a}(z))$ satisfying system (2.2) with $r=r_a$ (here, $c_{\min }^{r_a}$ represents the minimal wave speed when $r=r_a$ ). Note that $(c_{\min }^{r_a}, \Phi ^{r_a}(z))$ is an upper solution of the system with $r=r_b$ . A lower solution is defined as in (3.3). Then there exists a travelling wave solution of the system for $r=r_b$ when $c=c_{\min }^{r_a}$ . It follows that $c_{\min }^{r_b}\le c_{\min }^{r_a}$ . Therefore, $c_{\min }^{r}$ is non-increasing with respect to $r$ .

For the existence of monotone travelling wave solutions of (1.5)-(1.6) when $r=1$ , we divide the proof into four steps.

Step 1. In this step, we shall prove that when $r=1$ , there exists a minimal speed so that (1.5)-(1.6) has a monotone travelling wave solution. Take a decreasing real sequence $\{r_n\}$ such that $\lim \limits _{n\rightarrow \infty }r_n=\bar{r}=1$ and $r_0=2$ . By Theorem2.6, for each $r_n$ , there exists monotone travelling wave $(c^{r_n}, \Phi ^{r_n}(z))$ of (2.2) with $r=r_n$ (here, $c^{r_n}$ represents the unique positive bistable wave speed when $r=r_n$ ), and $\Psi ^{r_n}(z)=\mathcal{L}_{b,c^{r_n}}\Phi ^{r_n}(z)$ . By shifting, we can choose a large $L_1\gt 0$ to fix $\Phi ^{r_n}(L_1)=\epsilon _1$ for each $n$ , where $\epsilon _1$ is sufficiently small.

Claim 1: The bistable wave speeds $c^{r_n}$ have a finite upper bound.

Let $\hat{r}=0.5$ . Assume that $\Phi ^{\hat{r}}(z)$ is a travelling wave of (2.2) with speed $c=c_1=2c_{\min }^{\hat{r}}$ and $r=\hat{r}$ , where $c_{\min }^{\hat{r}}$ represents the minimal wave speed when $r=\hat{r}$ . We shall prove $c^{r_n}\le c_1$ for all $n\in \mathbb{N}^+$ . To the contrary, assume there exists $j\in \mathbb{N}^+$ such that $c_2\,:\!=\,c^{r_j}\gt c_1$ . By asymptotic analysis, we have

(3.5) \begin{equation} \begin{aligned} &\Phi ^{\hat{r}}(z) \sim A_1 e^{-\frac{c_1-\sqrt{c_1^2-4(1-\hat{r})}}{2} z}, \ \ \Phi ^{r_j}(z) \sim B_j e^{-\frac{c_2+\sqrt{c_2^2-4(1-r_j)}}{2} z}, \ \text{as} \ z\rightarrow \infty, \ A_1,B_j\gt 0, \\ &\Phi ^{\hat{r}}(z) \sim 1-\bar{A}_1 e^{\frac{-c_1+\sqrt{c_1^2+4}}{2} z}, \ \ \Phi ^{r_j}(z) \sim 1-\bar{B}_j e^{\frac{-c_2+\sqrt{c_2^2+4}}{2} z}, \ \text{as} \ z\rightarrow -\infty, \ \ \bar{A}_1, \bar{B}_j\gt 0. \end{aligned} \end{equation}

Thus, by translation if necessary, we obtain $\Phi ^{\hat{r}}(\nu\ {\cdot}\ x)\ge \Phi ^{r_j}(\nu\ {\cdot}\ x)$ for $x\in \mathbb{R}^n$ . A similar argument as in Theorem2.9 leads to a contradiction. It means $c^{r_n}\le c_1$ for all $n\in \mathbb{N}^+$ .

Since $c^{r_n}$ is non-increasing with respect to $r_n$ as stated in Theorem2.9, we know $c^{r_n}$ is non-decreasing with respect to $n$ . Together with Claim 1, we conclude that $c^{r_n}$ has a limit when $n\rightarrow \infty$ , denoted by $c^{\bar{r}}$ . Let $\bar{\alpha }$ be large enough so that

\begin{align*} \bar {F}(\Phi )\,:\!=\,\bar {\alpha }\Phi +\Phi (1-r-\Phi +r(K\star _c\mathcal {L}_{b,c}\Phi )) \end{align*}

is non-decreasing in $\Phi$ . Then we can express $\Phi$ -equation in (2.2) as

(3.6) \begin{equation} \begin{aligned} \Phi ^{\prime\prime}+c\Phi ^{\prime}-\bar{\alpha } \Phi =-\bar{F}(\Phi ).\\ \end{aligned} \end{equation}

Define $\bar{\beta }_1$ and $\bar{\beta }_2$ as

(3.7) \begin{equation} \bar{\beta }_1=\frac{-c-\sqrt{c^2+4\bar{\alpha }}}{2}\lt 0, \quad \bar{\beta }_2=\frac{-c+\sqrt{c^2+4\bar{\alpha }}}{2}\gt 0. \end{equation}

Then, the integral form of (3.6) is

(3.8) \begin{equation} \begin{aligned} \Phi (z)=\frac{1}{\bar{\beta }_2-\bar{\beta }_1}\bigg \{\int _{-\infty }^{z}e^{\bar{\beta }_1(z-t)}\bar{F}(\Phi )(t)dt+\int _{z}^{\infty }e^{\bar{\beta }_2(z-t)}\bar{F}(\Phi )(t)dt\bigg \}. \end{aligned} \end{equation}

Note that $(c^{r_n}, \Phi ^{r_n}(z))$ satisfies the integral system (3.8). Then we have

\begin{align*} \begin{aligned} &\bigg |\frac{d\Phi ^{r_n}(z)}{dz}\bigg |\le \frac{\beta _2}{\beta _2-\beta _1}\bigg |\bigg \{\int _{-\infty }^{z}e^{\beta _1(z-t)}\bar F(\Phi )(t)dt+\int _{z}^{\infty }e^{\beta _2(z-t)}\bar F(\Phi )(t)dt\bigg \}\bigg |\le \beta _2\lt \infty, \\ \end{aligned} \end{align*}

which means $(\Phi ^{r_n}(z))$ is equicontinuous and uniformly bounded. According to the Ascoli–Arzela theorem, there exists a subsequence $\{r_{n_j}\}$ such that $(c^{r_{n_j}}, \Phi ^{r_{n_j}}(z))$ converges to a limit $(c^{\bar{r}}, \Phi ^{\bar{r}}(z))$ uniformly on any compact interval and pointwise on $\mathbb{R}$ . Take $\bar{\lambda }\,:\!=\,\inf _{n\in \mathbb{N}^+} \{\frac{c^{r_n}+\sqrt{(c^{r_n})^2-4(1-r_n)}}{2}\}\gt 0$ . Since $\epsilon _1$ is sufficiently small, and by the asymptotic behaviour as in (3.5), we have $\Phi ^{r_n}(z)\le \epsilon _1 e^{-\bar{\lambda } (z-L_1)}$ when $z\ge L_1$ . It follows that

(3.9) \begin{equation} \Phi ^{\bar{r}}(z)\le \epsilon _1 e^{-\bar{\lambda } (z-L_1)}, \ \text{when}\ \ z\ge L_1, \end{equation}

which means $\int _{0}^{\infty }\Phi ^{\bar{r}}(z)dz$ is finite. Let $j\rightarrow \infty$ . By the dominated converges theorem, we know $(c^{\bar{r}}, \Phi ^{\bar{r}}(z))$ satisfies the integral system (3.8) and $\Psi ^{\bar{r}}(z)=\mathcal{L}_{b,c^{\bar{r}}}\Phi ^{\bar{r}}(z)$ . Using the monotone convergence theorem, we know $(c^{\bar{r}}, \Phi ^{\bar{r}}(\pm \infty ))$ satisfy system (3.8). Therefore, $\Phi ^{\bar{r}}({-}\infty )=1$ and $\Phi ^{\bar{r}}(\infty )=0$ . By Lemma 2.1, we have $\Psi ^{\bar{r}}({-}\infty )=1$ and $\Psi ^{\bar{r}}(\infty )=0$ . Notice that $\mu _1=0$ when $r=\bar{r}=1$ . Thus, inequality (3.9) shows that $\Phi ^{\bar{r}}(z)$ decays exponentially to $0$ with decay rate $\mu _2(c^{\bar{r}})$ , as $z\rightarrow \infty$ . It follows from Theorem3.2 that $c^{\bar{r}}$ is the minimal speed when $r=1$ , that is, $c_{\min }^{\bar{r}}=c^{\bar{r}}$ .

Step 2. In this step, we aim to prove the existence of monostable monotone travelling waves of system (1.5)-(1.6) with $r=1$ when $c$ is sufficiently large. By letting

\begin{align*} \tilde {z}=c z, \ \ \ \ \tilde {\Phi }(\tilde {z})=\Phi (z), \ \ \ \ \tilde {\Psi }(\tilde {z})=\Psi (z), \end{align*}

system (1.5)-(1.6) can be transformed into

(3.10) \begin{equation} \left \{ \begin{array}{cc} \begin{aligned} &\frac{1}{c^2}\tilde{\Phi }^{\prime\prime}(\tilde{z})+\tilde{\Phi }^{\prime}(\tilde{z})+\tilde{\Phi }(\tilde{z})\bigg [{-}\tilde{\Phi }(\tilde{z})+\int _{0}^{\infty }\int _{\mathbb{R}} K(s,y)\tilde{\Psi }(\tilde{z}+s-\frac{1}{c}y)dyds\bigg ] =0, \\[3pt] &\frac{1}{c^2}\tilde{\Psi }^{\prime\prime}(\tilde{z})+\tilde{\Psi }^{\prime}(\tilde{z})+b\tilde{\Phi }(\tilde{z})(1-\tilde{\Psi }(\tilde{z}))=0, \\[3pt] &(\tilde{\Phi }, \tilde{\Psi })({-}\infty )=e_\beta, \ \ (\tilde{\Phi }, \tilde{\Psi })(+\infty )=e_0, \end{aligned} \end{array}\right . \end{equation}

when $r=1$ . Let $c=\frac{1}{\varepsilon }$ , where $\varepsilon$ is small enough. We first consider

(3.11) \begin{equation} \left \{ \begin{array}{cc} \begin{aligned} &\tilde{\Phi }^{\prime}(\tilde{z})+\tilde{\Phi }(\tilde{z})\bigg [{-}\tilde{\Phi }(\tilde{z})+\int _{0}^{\infty }\int _{\mathbb{R}} K(s,y)\tilde{\Psi }(\tilde{z}+s)dyds\bigg ] =0, \\[3pt] &\tilde{\Psi }^{\prime}(\tilde{z})+b\tilde{\Phi }(\tilde{z})(1-\tilde{\Psi }(\tilde{z}))=0, \\[3pt] &(\tilde{\Phi }, \tilde{\Psi })({-}\infty )=e_\beta, \ \ (\tilde{\Phi }, \tilde{\Psi })(+\infty )=e_0. \end{aligned} \end{array}\right . \end{equation}

From the second equation in (3.11) and $(\tilde{\Phi }, \tilde{\Psi })({+}\infty )=e_0$ , we have

\begin{align*} \tilde {\Psi }(\tilde {z})=1-e^{-\int _{\tilde {z}}^{\infty }b\tilde {\Phi }(\tau )d\tau }. \end{align*}

Thus, system (3.11) becomes

(3.12) \begin{equation} \tilde{\Phi }^{\prime}(\tilde{z})+\tilde{\Phi }(\tilde{z})\bigg [{-}\tilde{\Phi }(\tilde{z})+\int _{0}^{\infty }\int _{\mathbb{R}} K(s,y)(1-e^{-\int _{\tilde{z}+s}^{\infty }b\tilde{\Phi }(\tau )d\tau })dyds\bigg ] =0 \end{equation}

with the boundary condition

(3.13) \begin{equation} \tilde{\Phi }({-}\infty )=1, \ \ \ \tilde{\Phi }(+\infty )=0. \end{equation}

The integral form of (3.12) is

\begin{align*} \tilde {T}_1(\tilde {\Phi })=\int _{\tilde {z}}^{\infty }e^{4(\tilde {z}-\tau _1)}\tilde {\Phi }(\tau _1)\bigg [4-\tilde {\Phi }(\tau _1)+\int _{0}^{\infty }\int _{\mathbb {R}} K(s,y)(1-e^{-\int _{\tau _1+s}^{\infty }b\tilde {\Phi }(\tau )d\tau })dyds\bigg ] d\tau _1. \end{align*}

By calculation, we can establish $\int _{0}^{\infty } \tilde{T}_1(\tilde{\Phi })(\tilde{z})d\tilde{z}$ is finite when $\tilde{\Phi }(\tilde{z}) \in L^1[0,\infty )$ . Note that $\tilde{T}_1$ has only two fixed points $0$ and $1$ . Then $\tilde{T}_1\ :\ \mathcal{M}_\beta \cap L^1[0,\infty )\rightarrow \mathcal{M}_\beta \cap L^1[0,\infty )$ , where $\mathcal{M}_\beta$ is defined in Section 2 with $\beta =1$ , is a strictly monotone continuous operator. By the Dancer–Hess Lemma (see [Reference Dancer and Hess7, Proposition 1], [Reference Zhao45, p. 45]), there exists an entire orbit of $\tilde{T}_1$ connecting $1$ to $0$ . Therefore, system (3.12)-(3.13) has a monotone solution, denoted by $\tilde{\Phi }_1(\tilde{z})$ . From (3.12), $\tilde{\Phi }_1(\tilde{z})$ does not decay exponentially to $0$ as $z\rightarrow \infty$ . Consider $z\rightarrow \infty$ . We let

(3.14) \begin{equation} \tilde{\Phi }_1(\tilde{z})=a_1 \tilde{z}^{-\gamma }+o(\tilde{z}^{-\gamma }), \ \ \ \ a_1, \gamma \gt 0. \end{equation}

Substituting (3.14) into (3.12) yields

(3.15) \begin{equation} \begin{aligned} &-a_1\gamma \tilde{z}^{-(\gamma +1)}+a_1 \tilde{z}^{-\gamma }\bigg [{-}a_1 \tilde{z}^{-\gamma }+ \int _{0}^{\infty }\int _{\mathbb{R}} k(s, y) (1-e^{-\int _{\tilde{z}+s}^{\infty }ba_1\tau ^{-\gamma } d\tau })\bigg ]+o(\tilde{z}^{-(\gamma +1)}) \\ &=-a_1\gamma \tilde{z}^{-(\gamma +1)}-a_1^2 \tilde{z}^{-2\gamma }+a_1 \tilde{z}^{-\gamma }\int _{0}^{\infty }\int _{\mathbb{R}} k(s, y)\frac{ba_1}{\gamma -1}(\tilde{z}+s)^{-\gamma +1} dyds+o(\tilde{z}^{-2\gamma +1}) +o(\tilde{z}^{-(\gamma +1)})\\ &=-a_1\gamma \tilde{z}^{-(\gamma +1)}-a_1^2 \tilde{z}^{-2\gamma }+\frac{ba_1^2}{\gamma -1} \tilde{z}^{-2\gamma +1}\int _{0}^{\infty }\int _{\mathbb{R}} k(s, y)(\frac{\tilde{z}+s}{\tilde{z}})^{-\gamma +1} dyds+o(\tilde{z}^{-2\gamma +1}) +o(\tilde{z}^{-(\gamma +1)}) \\ &=-a_1\gamma \tilde{z}^{-(\gamma +1)}-a_1^2 \tilde{z}^{-2\gamma }+\frac{ba_1^2}{\gamma -1} \tilde{z}^{-2\gamma +1}+o(\tilde{z}^{-2\gamma +1})+o(\tilde{z}^{-(\gamma +1)}) =0. \end{aligned} \end{equation}

According to the leading terms in (3.15), we derive that $\gamma =2$ and $a_1=\frac{2}{b}$ . Therefore, system (3.11) has a monotone solution $(\tilde{\Phi }_1, \tilde{\Psi }_1)(\tilde{z})$ , where $\tilde{\Psi }_1(\tilde{z})=1-e^{-\int _{\tilde{z}}^{\infty }b\tilde{\Phi }_1(\tau )d\tau }$ , with decay behaviour

(3.16) \begin{equation} \tilde{\Phi }_1(\tilde{z})=\frac{2}{b} \tilde{z}^{-2}, \ \ \tilde{\Psi }_1(\tilde{z})=2\tilde{z}^{-1}, \ \ \text{as} \ \ \tilde{z}\rightarrow \infty . \end{equation}

From (3.11), we know $\tilde{\Phi }^{\prime\prime}_1$ and $\tilde{\Psi }^{\prime\prime}_1$ are smooth. Next, we want to apply a perturbation argument to prove that solutions of (3.10) exist for large speed $c$ by using $(\tilde{\Phi }_1, \tilde{\Psi }_1)$ . Let $\tilde{\alpha }_1$ be large enough so that

\begin{align*} F_3(\tilde {\Phi }, \tilde {\Psi })\,:\!=\,\tilde {\alpha }_1 \tilde {\Phi }+\tilde {\Phi }(\tilde {z})\bigg [{-}\tilde {\Phi }(\tilde {z})+\int _{0}^{\infty }\int _{\mathbb {R}} K(s,y)\tilde {\Psi }(\tilde {z}+s-\frac {1}{c}y)dyds\bigg ] \end{align*}

and

\begin{align*} F_4(\tilde {\Phi }, \tilde {\Psi })\,:\!=\,\tilde {\alpha }_1 \tilde {\Psi }+b\tilde {\Phi }(\tilde {z})(1-\tilde {\Psi }(\tilde {z})) \end{align*}

are non-decreasing functions of $\tilde{\Phi }$ and $\tilde{\Psi }$ . Then we can write system (3.10) as

(3.17) \begin{equation} \begin{aligned} \frac{1}{c^2}\tilde{\Phi }^{\prime\prime}(\tilde{z})+\tilde{\Phi }^{\prime}(\tilde{z})-\tilde{\alpha }_1 \tilde{\Phi }=-F_3(\tilde{\Phi }, \tilde{\Psi }),\\ \frac{1}{c^2}\tilde{\Psi }^{\prime\prime}(\tilde{z})+\tilde{\Psi }^{\prime}(\tilde{z})-\tilde{\alpha }_1 \tilde{\Psi }=-F_4(\tilde{\Phi }, \tilde{\Psi }).\\ \end{aligned} \end{equation}

Define $\tilde{\beta }_1$ and $\tilde{\beta }_2$ as

(3.18) \begin{equation} \tilde{\beta }_1(c)=\frac{-c^2-c\sqrt{c^2+4\tilde{\alpha }_1}}{2}\lt 0, \quad \tilde{\beta }_2(c)=\frac{-c^2+c\sqrt{c^2+4\tilde{\alpha }_1}}{2}\gt 0. \end{equation}

The integral form of (3.17) is

(3.19) \begin{equation} \begin{aligned} \tilde{\Phi }(\tilde{z})=\frac{c^2}{\tilde{\beta }_2-\tilde{\beta }_1}\bigg \{\int _{-\infty }^{\tilde{z}}e^{\tilde{\beta }_1(\tilde{z}-t)}F_3(\tilde{\Phi },\tilde{\Psi })(t)dt+\int _{\tilde{z}}^{\infty }e^{\tilde{\beta }_2(\tilde{z}-t)}F_3(\tilde{\Phi }, \tilde{\Psi })(t)dt\bigg \} \,=\!:\,P_{1,c}(F_3)(\tilde{z}),\\ \tilde{\Psi }(\tilde{z})=\frac{c^2}{\tilde{\beta }_2-\tilde{\beta }_1}\bigg \{\int _{-\infty }^{\tilde{z}}e^{\tilde{\beta }_1(\tilde{z}-t)}F_4(\tilde{\Phi },\tilde{\Psi })(t)dt+\int _{\tilde{z}}^{\infty }e^{\tilde{\beta }_2(\tilde{z}-t)}F_4(\tilde{\Phi }, \tilde{\Psi })(t)dt\bigg \} \,=\!:\,P_{2,c}(F_4)(\tilde{z}).\\ \end{aligned} \end{equation}

When $c\rightarrow \infty$ , we have

\begin{align*} \begin {aligned} P_{1,\infty }(F_3)(\tilde {z})=\int _{\tilde {z}}^{\infty }e^{\tilde {\alpha }_1(\tilde {z}-t)}F_3(\tilde {\Phi }, \tilde {\Psi })(t)dt,\\ P_{2,\infty }(F_4)(\tilde {z})=\int _{\tilde {z}}^{\infty }e^{\tilde {\alpha }_1(\tilde {z}-t)}F_4(\tilde {\Phi }, \tilde {\Psi })(t)dt.\\ \end {aligned} \end{align*}

Assume that $(\tilde{\Phi }^*(\tilde{z}), \tilde{\Psi }^*(\tilde{z}) )\,:\!=\,(\tilde{\Phi }_1(\tilde{z})+\tilde{W_1}(\tilde{z}), \tilde{\Psi }_1(\tilde{z})+\tilde{W_2}(\tilde{z}))$ , where $\tilde{W_1}(\tilde{z})$ and $\tilde{W_2}(\tilde{z})$ are functions in $B_0$ that need to be determined. Here, $B_0$ is defined as

\begin{align*}B_0=\{u\in C({-}\infty, \infty )\ :\ u(\pm \infty )=0\}.\end{align*}

Therefore, we need to prove the existence of $\tilde{W_1}(\tilde{z})$ and $\tilde{W_2}(\tilde{z})$ in $B_0$ such that $(\tilde{\Phi }^*(\tilde{z}), \tilde{\Psi }^*(\tilde{z}) )$ satisfies (3.10) with $c=\frac{1}{\varepsilon }$ . Plugging $(\tilde{\Phi }^*(\tilde{z}), \tilde{\Psi }^*(\tilde{z}) )$ into (3.10) and utilising (3.11) lead to

(3.20) \begin{equation} \begin{aligned} &\varepsilon ^2 \tilde{W_1}^{\prime\prime}+\tilde{W_1}^{\prime}-\tilde{\alpha }_1 \tilde{W_1}=-(\tilde{F_0}+\tilde{F}_{1,\varepsilon }+\tilde{F}_{2,\varepsilon }+\tilde{F_h}),\\ &\varepsilon ^2\tilde{W_2}^{\prime\prime}+\tilde{W_2}^{\prime}-\tilde{\alpha }_1 \tilde{W_2}=-(\tilde{G_0}+\tilde{G_{\varepsilon }}+\tilde{G_h}), \end{aligned} \end{equation}

where

\begin{align*} \begin {aligned} &\tilde {F_0}=\bigg [\tilde {\alpha }_1-2\tilde {\Phi }_1+ \int _{0}^{\infty }\int _{\mathbb {R}} K(s,y)\tilde {\Psi }_1(\tilde {z}+s)dyds\bigg ] \tilde {W_1}+\tilde {\Phi }_1 \int _{0}^{\infty }\int _{\mathbb {R}} K(s,y)\tilde {W_2}(\tilde {z}+s)dyds,\\ &\tilde {F}_{1, \varepsilon }=\varepsilon ^2\tilde {\Phi }_1^{\prime\prime}+\tilde {\Phi }_1 \int _{0}^{\infty }\int _{\mathbb {R}} K(s,y)\big [\tilde {\Psi }_1(\tilde {z}+s-\varepsilon y)-\tilde {\Psi }_1(\tilde {z}+s)\big ]dyds \\ &\tilde {F}_{2,\varepsilon }=\tilde {\Phi }_1 \int _{0}^{\infty }\int _{\mathbb {R}} K(s,y)\big [\tilde {W_2}(\tilde {z}+s-\varepsilon y)-\tilde {W_2}(\tilde {z}+s)\big ]dyds\\ & \quad \quad +\tilde {W_1} \int _{0}^{\infty }\int _{\mathbb {R}} K(s,y)\big [\tilde {\Psi }_1(\tilde {z}+s-\varepsilon y)-\tilde {\Psi }_1(\tilde {z}+s)\big ]dyds\\ &\tilde {F_h}=\tilde {W_1} \int _{0}^{\infty }\int _{\mathbb {R}} K(s,y)\tilde {W_2}(\tilde {z}+s-\varepsilon y)dyds-\tilde {W_1}^2,\\ &\tilde {G_0}=(\tilde {\alpha }_1-b\tilde {\Phi }_1)\tilde {W_2}+b(1-\tilde {\Psi }_1)\tilde {W_1},\\ &\tilde {G_\varepsilon }=\varepsilon ^2\tilde {\Psi }_1^{\prime\prime},\\ &\tilde {G_h}=-b\tilde {W_1}\tilde {W_2}. \end {aligned} \end{align*}

Therefore, we have

(3.21) \begin{equation} \begin{aligned} &\tilde{W_1}=P_{1,\infty }(\tilde{F_0})+ P_{1,\frac{1}{\varepsilon }}(\tilde{F_0})-P_{1,\infty }(\tilde{F_0})+ P_{1,\frac{1}{\varepsilon }}(\tilde{F}_{1,\varepsilon })+P_{1,\frac{1}{\varepsilon }}(\tilde{F}_{2,\varepsilon })+P_{1,\frac{1}{\varepsilon }}(\tilde{F_h}),\\ &\tilde{W_2}=P_{2,\infty }(G_0)+P_{2,\frac{1}{\varepsilon }}(G_0)-P_{2,\infty }(G_0)+P_{2,\frac{1}{\varepsilon }}(\tilde{G_\varepsilon })+P_{2,\frac{1}{\varepsilon }}(\tilde{G_h}). \end{aligned} \end{equation}

Define

\begin{align*} \tilde {P}(\tilde {W})= \begin {pmatrix} P_{1,\infty }(\tilde {F_0}(\tilde {W}))\\ P_{2,\infty }(\tilde {G_0}(\tilde {W})) \end {pmatrix}, \ \tilde {W}=(\tilde {W_1}, \tilde {W_2}). \end{align*}

By a simple estimate, we have

\begin{align*}P_{1,\frac {1}{\varepsilon }}(\tilde {F_0})-P_{1,\infty }(\tilde {F_0})= O(\varepsilon ^2\tilde {W}), P_{2,\frac {1}{\varepsilon }}(G_0)-P_{2,\infty }(G_0)=O(\varepsilon ^2\tilde {W}),\\ P_{1,\frac {1}{\varepsilon }}(\tilde {F}_{1, \varepsilon })=O(\varepsilon ), P_{1,\frac {1}{\varepsilon }}(\tilde {F}_{2, \varepsilon })=O(\varepsilon \tilde {W}), P_{2,\frac {1}{\varepsilon }}(\tilde {G_\varepsilon })=O(\varepsilon ^2),\end{align*}

and $ P_{1,\frac{1}{\varepsilon }}(\tilde{F}_{1, \varepsilon })=o(\tilde{z}^{-3})$ as $\tilde{z}\rightarrow \infty$ . The linear operator $\tilde{P}$ is both compact and strongly positive. It possesses a simple principal eigenvalue $\lambda =1$ with the corresponding positive eigenfunction $({-}\tilde{\Phi }_1^{\prime},-\tilde{\Psi }_1^{\prime})$ . To eliminate this eigenfunction from $B_0$ , we introduce the weighted functional space as follows

\begin{align*} \mathcal {H}=\{h(\tilde {z})\in B_0\ :\ h\tilde {z}^3=o(1) \quad \rm {as} \quad \tilde {z} \rightarrow \infty \}. \end{align*}

Since the eigenfunction $({-}\tilde{\Phi }_1^{\prime},-\tilde{\Psi }_1^{\prime})$ does not belong to the space $\mathcal{H}\times B_0$ , it implies that $\tilde{P}$ does not possess the eigenvalue $\lambda =1$ in the functional space $\mathcal{H}\times B_0$ . Consequently, $I-\tilde{P}$ has a bounded inverse in $\mathcal{H}\times B_0$ , where $I$ is the identity operator. By applying the abstract implicit function theorem in $\mathcal{H}\times B_0$ , we conclude that there exists $\varepsilon _0$ such that a solution $(\tilde{W_1},\tilde{W_2})$ of the system (3.20) exists for any $\varepsilon \in [0, \varepsilon _0)$ . Consequently, there exists a solution $(\tilde{\Phi }^*(\tilde{z}), \tilde{\Psi }^*(\tilde{z}) )$ of (3.10) for all $c\ge \tilde{M}\gt \frac{1}{\varepsilon _0}$ , with decay behaviour (3.16).

Step 3. By continuation and the abstract implicit function theorem, we further prove the existence of monostable monotone travelling waves of system (3.10) for $c= \tilde{M}-\delta$ , with $\delta$ is a sufficiently small and to-be-determined value. Let $(\tilde{\Phi }_2(\tilde{z}), \tilde{\Psi }_2(\tilde{z}))$ be the solution of (3.10) for $c=c_2= \tilde{M}$ . Thus, we have

(3.22) \begin{equation} \left \{ \begin{array}{cc} \begin{aligned} &\frac{1}{c_2^2}\tilde{\Phi }_2^{\prime\prime}(\tilde{z})+\tilde{\Phi }_2^{\prime}(\tilde{z})+\tilde{\Phi }_2(\tilde{z})\bigg [{-}\tilde{\Phi }_2(\tilde{z})+\int _{0}^{\infty }\int _{\mathbb{R}} K(s,y)\tilde{\Psi }_2(\tilde{z}+s-\frac{1}{c_2}y)dyds\bigg ] =0, \\[3pt] &\frac{1}{c_2^2}\tilde{\Psi }_2^{\prime\prime}(\tilde{z})+\tilde{\Psi }_2^{\prime}(\tilde{z})+b\tilde{\Phi }_2(\tilde{z})(1-\tilde{\Psi }_2(\tilde{z}))=0, \\[3pt] &(\tilde{\Phi }_2, \tilde{\Psi }_2)({-}\infty )=e_\beta, \ \ (\tilde{\Phi }_2, \tilde{\Psi }_2)(+\infty )=e_0, \end{aligned} \end{array}\right . \end{equation}

Assuming $c=c_\delta =c_2-\delta$ . To find the solution of (3.17), say $(\bar{\Phi }_\delta (\tilde{z}), \bar{\Psi }_\delta (\tilde{z}))$ at $c=c_{\delta }$ , we let $\bar{\Phi }_\delta (\tilde{z})=\tilde{\Phi }_2(\tilde{z})+\bar{W}_1(\tilde{z})$ and $\bar{\Psi }_\delta (\tilde{z})=\tilde{\Psi }_2(\tilde{z})+\bar{W}_2(\tilde{z})$ , where $\bar{W}_1(\tilde{z})$ and $\bar{W}_2(\tilde{z})$ are functions in $B_0$ that need to be determined. Thus, we aim to prove the existence of $\bar{W}_1(\tilde{z})$ and $\bar{W}_2(\tilde{z})$ in $B_0$ such that $(\bar{\Phi }_\delta (\tilde{z}), \bar{\Psi }_\delta (\tilde{z}))$ satisfies (3.10) or (3.17) with speed $c_\delta$ . Substituting $(\bar{\Phi }_\delta (\tilde{z}), \bar{\Psi }_\delta (\tilde{z}))$ into (3.10) and using (3.22) give

(3.23) \begin{equation} \begin{aligned} &\bar{W}_1=P_{1,c_2}(\bar{F}_0)+P_{1,c_\delta }(\bar{F}_0)-P_{1,c_2}(\bar{F}_0) + P_{1,c_\delta }(\bar{F}_{1,\delta })+P_{1,c_\delta }(\bar{F}_{2,\delta })+P_{1,c_\delta }(\bar{F}_h),\\ &\bar{W}_2=P_{2,c_2}(\bar{G}_0)+ P_{2,c_\delta }(\bar{G}_0)-P_{2,c_2}(\bar{G}_0)+P_{2,c_\delta }(\bar{G}_\delta )+P_{2,c_\delta }(\bar{G}_h), \end{aligned} \end{equation}

where

\begin{align*} \begin {aligned} &\bar {F}_0=\bigg [\tilde {\alpha }_1-2\tilde {\Phi }_2+\int _{0}^{\infty }\int _{\mathbb {R}} K(s,y)\tilde {\Psi }_2(\tilde {z}+s-\frac {1}{c_2}y)dyds\bigg ] \bar {W}_1+\tilde {\Phi }_2\int _{0}^{\infty }\int _{\mathbb {R}} K(s,y)\bar {W}_2(\tilde {z}+s-\frac {1}{c_2}y)dyds,\\ &\bar {F}_{1, \delta }=\big [\frac {1}{(c_2-\delta )^2}-\frac {1}{c_2^2}\big ]\tilde {\Phi }_2^{\prime\prime}+\tilde {\Phi }_2 \int _{0}^{\infty }\int _{\mathbb {R}} K(s,y)\big [\tilde {\Psi }_2(\tilde {z}+s-\frac {1}{c_2-\delta } y)-\tilde {\Psi }_2(\tilde {z}+s-\frac {1}{c_2}y)\big ]dyds \end {aligned} \end{align*}

\begin{align*} \begin {aligned} &\bar {F}_{2,\delta }=\tilde {\Phi }_2 \int _{0}^{\infty }\int _{\mathbb {R}} K(s,y)\big [\bar {W}_2(\tilde {z}+s-\frac {1}{c_2-\delta }y)-\tilde {W_2}(\tilde {z}+s-\frac {1}{c_2}y)\big ]dyds\\ & \quad \quad +\bar {W}_1 \int _{0}^{\infty }\int _{\mathbb {R}} K(s,y)\big [\tilde {\Psi }_2(\tilde {z}+s-\frac {1}{c_2-\delta }y)-\tilde {\Psi }_2(\tilde {z}+s-\frac {1}{c_2}y)\big ]dyds\\ &\bar {F}_h=\bar {W}_1 \int _{0}^{\infty }\int _{\mathbb {R}} K(s,y)\bar {W}_2(\tilde {z}+s-\frac {1}{c_2-\delta }y)dyds-\bar {W}_1^2,\\ &\bar {G}_0=(\tilde {\alpha }_1-b\tilde {\Phi }_2)\bar {W}_2+b(1-\tilde {\Psi }_2)\bar {W}_1,\\ &\bar {G}_\delta =\big [\frac {1}{(c_2-\delta )^2}-\frac {1}{c_2^2}\big ]\tilde {\Psi }_2^{\prime\prime},\\ &\bar {G}_h=-b\bar {W}_1\bar {W}_2. \end {aligned} \end{align*}

Define

\begin{align*} P(\bar {W})= \begin {pmatrix} P_{1,c_2}(\bar {F}_0(\bar {W}))\\ P_{2,c_2}(\bar {G}_0(\bar {W})) \end {pmatrix}, \ \bar {W}=(\bar {W}_1, \bar {W}_2). \end{align*}

Clearly, we have

\begin{align*}P_{1,c_\delta }(\bar {F_0})-P_{1,c_2}(\bar {F_0})= O(\delta ^2\bar {W}), P_{2,c_\delta }(\bar {G}_0)-P_{2,c_2}(\bar {G}_0)=O(\delta ^2\bar {W}),\\P_{1,c_\delta }(\bar {F}_{1,\delta })=O(\delta ), P_{1,c_\delta }(\bar {F}_{2,\delta })=O(\delta \bar {W}), P_{2,c_\delta }(\bar {G_\delta })=O(\delta ^2),\end{align*}

and $ P_{1,c_\delta }(\bar{F}_{1,\delta })=o(\tilde{z}^{-3})$ as $\tilde{z}\rightarrow \infty$ . The linear operator $P$ is both compact and strongly positive. It possesses a simple principal eigenvalue $\lambda =1$ with the corresponding positive eigenfunction $({-}\tilde{\Phi }_2^{\prime},-\tilde{\Psi }_2^{\prime})$ . By a similar argument as in Step 2, the eigenfunction $({-}\tilde{\Phi }_2^{\prime},-\tilde{\Psi }_2^{\prime})$ does not belong to the space $\mathcal{H}\times B_0$ . It follows from the abstract implicit function theorem in $\mathcal{H}\times B_0$ that there exists $\delta _0$ such that a solution $(\bar{W}_1,\bar{W}_2)$ of the system (3.23) exists for any $\delta \in [0, \delta _0)$ . Consequently, there exists a solution $(\bar{\Phi }_\delta (\tilde{z}), \bar{\Psi }_\delta (\tilde{z}))$ of (3.10) with $c=c_\delta$ .

Step 4. Note that $\mu _1=0$ and $\mu _2=c$ from (3.2) for $r=1$ . When $c\gt c_{\min }^{\bar{r}}$ , Theorem3.2 implies that if travelling wave solutions of (1.5)-(1.6) with $r=1$ exist, they do not decay exponentially to $0$ as $z\rightarrow \infty$ . This indicates that travelling wave solutions of (3.10), if they exist, exhibit algebraic decay behaviour (3.16) for all $c\gt c_{\min }^{\bar{r}}$ . By repeating the continuation process from Step 3, we establish the existence of monostable monotone travelling waves of system (3.10) for all $c\in (c_{\min }^{\bar{r}}, \tilde{M})$ . Therefore, we prove that when $r=1$ , there is a finite $c_{\min }^{\bar{r}}\gt 0$ such that monotone and positive travelling wave solutions of (1.5)-(1.6) exist if and only if $c\ge c_{\min }^{\bar{r}}$ . When $c\gt c_{\min }^{\bar{r}}$ , the travelling wave solution $(\Phi, \Psi )(z)$ of (1.5)-(1.6) has the following algebraic decay behaviour

\begin{align*} \Phi (z) \sim \frac {2c^2}{b} z^{-2}, \ \ \Psi (z) \sim 2c z^{-1}, \ \ \text {as} \ \ z\rightarrow \infty . \end{align*}

Proof. To establish the necessity, we assume that the minimal wave speed is nonlinearly selected, denoted as $c_{\min } \gt c_0$ . Our goal is to demonstrate that the travelling wave $(\Phi _2(z), \Psi _2(z))$ of system (1.5)-(1.6), with a speed $c = c_2 = c_{\min }$ , exhibits the following behaviour

\begin{align*} \Phi _2(z)\sim A_2e^{-\mu _2(c_2) z},\ \text {as} \ z\rightarrow \infty, \ z=\nu {\cdot}x-c_2t,\ ||\nu ||=1, \end{align*}

with $A_2\gt 0$ . To the contrary, we assume that

(3.24) \begin{equation} \Phi _2(z)\sim A_2e^{-\mu _1(c_2) z},\ \text{as} \ z\rightarrow \infty, \ \ \text{for} \ r\in (0,1) \end{equation}

and

(3.25) \begin{equation} \Phi _2(z)\sim \frac{2c_2^2}{b}z^{-2},\ \text{as} \ z\rightarrow \infty, \ \ \text{for} \ r=1. \end{equation}

By a similar argument in Step 3 of Theorem3.3, we can prove that there is a monotone travelling wave solution of (1.5)-(1.6) when $c=c_\delta =c_2-\delta$ , where $\delta$ is a sufficiently small and to-be-determined value. It implies that $c_2$ is not the minimum wave speed. This contradiction establishes the necessity. Clearly, for $r=1$ , Step 3 of Theorem3.3 shows the existence of monotone travelling wave solution of (1.5)-(1.6) when $c=c_2-\delta$ with a small $\delta$ . Thus, we only focus on $r\in (0,1)$ . Let $\alpha$ be large enough so that

\begin{align*} F(\Phi, \Psi )\,:\!=\,\alpha \Phi +\Phi (1-r-\Phi +r(K\star _c\Psi )) \end{align*}

and

\begin{align*} G(\Phi, \Psi )\,:\!=\,\alpha \Psi +b\Phi (1-\Psi ) \end{align*}

are non-decreasing functions of $\Phi$ and $\Psi$ . Then we can express system (1.5) as

(3.26) \begin{equation} \begin{aligned} \Phi ^{\prime\prime}+c\Phi ^{\prime}-\alpha \Phi =-F(\Phi, \Psi ),\\ \Psi ^{\prime\prime}+c\Psi ^{\prime}-\alpha \Psi =-G(\Phi, \Psi ).\\ \end{aligned} \end{equation}

Define $\beta _1$ and $\beta _2$ as

(3.27) \begin{equation} \beta _1=\frac{-c-\sqrt{c^2+4\alpha }}{2}\lt 0, \quad \beta _2=\frac{-c+\sqrt{c^2+4\alpha }}{2}\gt 0. \end{equation}

In view of the variation of parameters, the integral form of (3.26) is given by

(3.28) \begin{equation} \begin{aligned} \Phi (z)=\frac{1}{\beta _2-\beta _1}\bigg \{\int _{-\infty }^{z}e^{\beta _1(z-t)}F(\Phi, \Psi )(t)dt+\int _{z}^{\infty }e^{\beta _2(z-t)}F(\Phi, \Psi )(t)dt\bigg \} \,=\!:\,T_{1,c}(F)(z),\\ \Psi (z)=\frac{1}{\beta _2-\beta _1}\bigg \{\int _{-\infty }^{z}e^{\beta _1(z-t)}G(\Phi, \Psi )(t)dt+\int _{z}^{\infty }e^{\beta _2(z-t)}G(\Phi, \Psi )(t)dt\bigg \} \,=\!:\,T_{2,c}(G)(z).\\ \end{aligned} \end{equation}

Note that $(\Phi _2(z), \Psi _2(z))$ satisfies

(3.29) \begin{equation} \left \{ \begin{array}{cc} \begin{aligned} &\Phi _2^{\prime\prime}+c_2\Phi _2^{\prime}+\Phi _2(1-r-\Phi _2+r(K\star _{c_2}\Psi _2))=0,\\[3pt] &\Psi _2^{\prime\prime}+c_2\Psi _2^{\prime}+b\Phi _2(1-\Psi _2)=0,\\[3pt] &(\Phi _2, \Psi _2)({-}\infty )=e_\beta, \quad (\Phi _2, \Psi _2)(\infty )=e_0. \end{aligned} \end{array} \right . \end{equation}

Define $\bar{\Phi }(z)=\Phi _2(z) \omega (z)$ , where

\begin{align*} \omega = \frac {1}{1+\delta \exp \{(\mu _1(c_\delta )-\mu _1(c_2))z\}}, \end{align*}

Using asymptotic analysis, we can deduce that

\begin{align*} \bar {\Phi }(z)\sim \frac {A_2}{\delta }e^{-\mu _1(c_\delta )z} \ \text {as} \ z\rightarrow \infty . \end{align*}

To find the solution of (3.26), say $(\Phi _\delta (z), \Psi _\delta (z))$ , at $c=c_{\delta }$ , we introduce $\Phi _\delta (z)$ as $\Phi _\delta (z)=\bar{\Phi }(z)+W_1(z)$ and $\Psi _\delta (z)$ as $\Psi _\delta (z)=\Psi _2(z)+W_2(z)$ , where $W_1(z)$ and $W_2(z)$ are functions in $B_0$ that need to be determined. Here, $B_0$ is defined inTheorem3.3. By substituting $(\Phi _\delta (z), \Psi _\delta (z))$ into (1.5) and utilising (3.29), we observe that $W_1$ and $W_2$ fulfill the following equations

(3.30) \begin{equation} \begin{aligned} &W_1=T_{1,c_2}(F_0)+T_{1,c_\delta }(F_0)-T_{1,c_2}(F_0)+ T_{1,c_\delta }(F_{1,\delta })+ T_{1,c_\delta }(F_{2,\delta })+T_{1,c_\delta }(F_h),\\ &W_2=T_{2,c_2}(G_0)+T_{2,c_\delta }(G_0)-T_{2,c_2}(G_0)+T_{2,c_\delta }(G_{1,\delta })+T_{2,c_\delta }(G_{2,\delta })+T_{2,c_\delta }(G_h) \end{aligned} \end{equation}

where

\begin{align*} \begin {aligned} &F_0=\big \{\alpha +1-r-2\Phi _2+r(K\star _{c_2}\Psi _2)\big \} W_1+r\Phi _2(K\star _{c_2} W_2),\\ &F_{1,\delta }=(\omega \Phi _2)^{\prime\prime}+(c_2-\delta )(\omega \Phi _2)^{\prime}-\omega \Phi _2^{\prime\prime}-c_2\omega \Phi _2^{\prime} +(1-\omega )\omega \Phi _2^2+r\Phi _2\omega \big [K\star _{c_\delta }\Psi _2-K\star _{c_2}\Psi _2 \big ],\\ &F_{2,\delta }=(1-\omega )\Phi _2\big [ 2W_1-r(K\star _{c_2} W_2)\big ]+r\Phi _2\omega \big [K\star _{c_\delta }W_2-K\star _{c_2}W_2\big ]+rW_1\big [K\star _{c_\delta }\Psi _2-K\star _{c_2}\Psi _2 \big ], \\ &F_h=rW_1 (K\star _{c_\delta } W_2)-W_1^2,\\ &G_0=(\alpha -b\Phi _2)W_2+b(1-\Psi _2)W_1,\\ &G_{1,\delta }=-\delta \Psi _2^{\prime}-(1-\omega )b\Phi _2(1-\Psi _2),\\ &G_{2,\delta }=(1-\omega )b\Phi _2W_2,\\ &G_h=-bW_1W_2. \end {aligned} \end{align*}

Define

\begin{align*} T(W)= \begin {pmatrix} T_{1,c_2}(F_0(W))\\ T_{2,c_2}(G_0(W)) \end {pmatrix}, \ W=(W_1, W_2). \end{align*}

It is worth noting that $\omega ^{\prime}=-(\mu _1(c_\delta )-\mu _1(c_2))\omega (1-\omega )$ and $\omega ^{\prime\prime}=(\mu _1(c_\delta )-\mu _1(c_2))^2\omega (1-\omega )(1-2\omega )$ . Clearly, we have

\begin{align*}T_{1,c_\delta }(F_0)-T_{1,c_2}(F_0) & = O(\delta W), T_{2,c_\delta }(G_0)-T_{2,c_2}(G_0)=O(\delta W),\\T_{1,c_\delta }(F_{1,\delta })=O(\delta ), T_{1,c_\delta }(F_{2,\delta }) & = O(\delta W), T_{2,c_\delta }(G_{1,\delta })=O(\delta ), T_{2,c_\delta }(G_{2,\delta })=O(\delta W), \end{align*}

and $T_{1,c_\delta }(F_{1,\delta })=o(e^{-\mu _1(c_\delta ) z})$ as $z\rightarrow \infty$ . The linear operator $T$ is both compact and strongly positive. It possesses a simple principal eigenvalue $\lambda =1$ with the corresponding positive eigenfunction $({-}\Phi _2^{\prime},-\Psi _2^{\prime})$ . Here, we introduce another weighted functional space

\begin{align*} \mathcal {H}_1=\{h(\xi )\in B_0\ :\ he^{\mu _1(c_\delta )\xi }=o(1) \quad \rm {as} \quad \xi \rightarrow \infty \} \end{align*}

to remove eigenfunction $({-}\Phi _2^{\prime},-\Psi _2^{\prime})$ from $B_0$ . Similar to Step 3 of Theorem3.3, by applying the abstract implicit function theorem in $\mathcal{H}_1\times B_0$ , we conclude that there exists $\delta _1$ such that a solution $(W_1,W_2)$ of the system (3.30) exists for any $\delta \in [0, \delta _1)$ when $r\in (0,1)$ . Consequently, there exists a solution $(\Phi _\delta (z), \Psi _\delta (z))$ of (1.5)-(1.6) with $c=c_\delta$ when $r\in (0,1)$ .

The sufficiency result follows from Theorem3.2.

Proof. (i) On the contrary, we assume that for $c\in (c_0, c_1)$ , there exists a monotone travelling wave solution $(\Phi, \Psi )(\nu\ {\cdot}\ x-ct)$ of the system (1.2) with the initial conditions

\begin{align*} \phi (0,x)=\Phi (\nu\ {\cdot}\ x),\quad \psi (0,x)=\Psi (\nu\ {\cdot}\ x). \end{align*}

Note that $\underline{\Phi }(\nu\ {\cdot}\ x)\le \Phi (\nu\ {\cdot}\ x)$ for $x\in \mathbb{R}^n$ by shifting if necessary. According to $\Psi$ -equation of (1.5), we have $(\underline{\Phi },\underline{\Psi })(\nu\ {\cdot}\ x)\le (\Phi, \Psi )(\nu\ {\cdot}\ x)$ for $x\in \mathbb{R}^n$ . By the comparison theorem, we obtain

\begin{align*} \underline {\Phi }(\nu\ {\cdot}\ x-c_1t)\le \Phi (\nu\ {\cdot}\ x-ct),\\\underline {\Psi }(\nu\ {\cdot}\ x-c_1t)\le \Psi (\nu\ {\cdot}\ x-ct). \end{align*}

Fix $z^*=\nu {\cdot}x-c_1t$ such that $\underline{\Phi }(z^*)\gt 0$ . Since

\begin{align*} \Phi (\nu\ {\cdot}\ x-ct)=\Phi (z^*+(c_1-c)t)\rightarrow 0, \ \text {as} \ t\rightarrow \infty, \end{align*}

it follows that $\underline{\Phi }(z^*)\le 0$ , which leads to a contradiction.

For part (ii), Lemma 3.1 provides a lower solution. Our result follows by the standard upper-lower solution technique. Hence, the proof is complete.

Proof. Let $r\in (0,1)$ be fixed and $b_2\gt b_1\gt 0$ . According to Theorem3.3, there exists a monotone travelling wave solution $(\Phi _{b_2}, \Psi _{b_2})$ of system (1.5)-(1.6) with speed $c=c_{\min }(b_2)$ for $b=b_2$ . By applying the comparison theorem, it can be shown that $(\Phi _{b_2}, \Psi _{b_2})$ serves as an upper solution of the system for $b=b_1$ . A lower solution is defined as Lemma 3.1. Utilising the upper-lower solution method, it follows that monotone travelling wave solutions of the system exist for $b=b_1$ when $c=c_{\min }(b_2)$ . Consequently, we obtain $c_{\min }(b_1)\le c_{\min }(b_2)$ . Therefore, the minimal speed $c_{\min }$ demonstrates a non-decreasing behaviour with respect to $b$ .

Now we prove that there is a finite value of $b$ so that transition of the speed selection is realised for the fixed $r\in (0,1)$ . Lemma 2.1 $b)$ implies $\Psi (z)\rightarrow 0$ as $b\rightarrow 0$ . Then $\Phi$ -equation of (1.5) becomes the Fisher–KPP equation as $b\rightarrow 0$ . Therefore, $c_{\min }=2\sqrt{1-r}=c_0$ (see [Reference Kolmogoroff, Petrovsky and Piscounoff17]). Similarly, Lemma 2.1 $b)$ leads us to derive $c_{\min }=2\gt c_0$ as $b\rightarrow \infty$ . Consequently, due to the non-decreasing property of $c_{\min }$ with respect to $b$ , there exists a finite $b^*$ such that $c_{\min } = c_0$ for $b \in (0, b^*]$ , and $c_{\min } \in (c_0, 2]$ for $b \gt b^*$ . If $r=1$ , then $c_0=0$ . We know the wave speed $c_{\min }$ of system (1.5)-(1.6) is positive. It follows that $c_{\min }$ is always bigger than $c_0$ . Therefore, the proof is complete.

Proof. We can use part (ii) of Theorem3.5. For illustration, We will alternatively use the upper-lower solution method to prove it. Define a pair of continuous function $(\bar{\Phi }_1, \bar{\Psi }_1)(z)$ as

(3.32) \begin{equation} \bar{\Phi }_1(z)=\frac{1}{1+e^{\mu _1 z}}, \quad \bar{\Psi }_1(z)=\left \{ \begin{array}{cc} \begin{aligned} &1,\quad \quad z\le z_4,\\ &\frac{2(1-r)}{r} \bar{\Phi }_1, \ z\gt z_4,\\ \end{aligned} \end{array}\right . \end{equation}

where $z_4=\frac{1}{\mu _1}\ln \frac{2(1-r)}{r}$ . Let $c=c_0+\varepsilon _3$ , where $ \varepsilon _3$ is sufficiently small. It follows that $\mu _1\sim \bar{\mu }=\sqrt{1-r}$ . We shall prove $(\bar{\Phi }_1, \bar{\Psi }_1)(z)$ is an upper solution of (1.5)-(1.6) under condition (3.32). Substituting $(\bar{\Phi }_1, \bar{\Psi }_1)(z)$ into (1.5) gives

\begin{align*} \bar {\Psi }_1^{\prime\prime}(z)+c\bar {\Psi }_1^{\prime}(z)+b\bar {\Phi }_1(z)(1-\bar {\Psi }_1(z))=0, \ z\le z_4, \end{align*}

\begin{align*} \begin {aligned} &\bar {\Psi }_1^{\prime\prime}(z)+c\bar {\Psi }_1^{\prime}(z)+b\bar {\Phi }_1(z)(1-\bar {\Psi }_1(z))\\ &=\frac {2(1-r)}{r}\bar {\Phi }_1(1-\bar {\Phi }_1)\bigg \{ \mu _1^2-c\mu _1-2\mu _1^2\bar {\Phi }_1+\frac {b(1-\frac {2(1-r)}{r}\bar {\Phi }_1)}{\frac {2(1-r)}{r}(1-\bar {\Phi }_1)} \bigg \}\\ &\le \frac {2(1-r)}{r}\bar {\Phi }_1(1-\bar {\Phi }_1)[{-}2\mu _1^2\bar {\Phi }_1-(1-r)+\frac {br}{2(1-r)}]\\ &\le \frac {2(1-r)}{r}\bar {\Phi }_1(1-\bar {\Phi }_1)[\frac {br}{2(1-r)}-(1-r)]\le 0, \ z\gt z_4, \end {aligned} \end{align*}

by (3.31), and

\begin{align*} \begin {aligned} &\bar {\Phi }_1^{\prime\prime}(z)+c\bar {\Phi }_1^{\prime}(z)+\bar {\Phi }_1(z)(1-r-\bar {\Phi }_1(z)+r(K\star _c\bar {\Psi }_1)(z))\\ &=\bar {\Phi }_1(1-\bar {\Phi }_1)\bigg \{ \mu _1^2-c\mu _1-2\mu _1^2\bar {\Phi }_1+\frac {1-r-\bar {\Phi }_1+r(K\star _c\bar {\Psi }_1)}{1-\bar {\Phi }_1} \bigg \}\\ &=\bar {\Phi }_1^2(1-\bar {\Phi }_1)\bigg \{-2\mu _1^2+\frac {r(K\star _c\bar {\Psi }_1)-r\bar {\Phi }_1}{(1-\bar {\Phi }_1)\bar {\Phi }_1} \bigg \}\\ &\,:\!=\,\bar {\Phi }_1^2(1-\bar {\Phi }_1)\bigg \{-2\mu _1^2+J \bigg \}. \end {aligned} \end{align*}

Note that

(3.33) \begin{equation} J=\left \{ \begin{array}{cc} \begin{aligned} &\frac{r}{\bar{\Phi }_1}\le 2(1-r),\ z\le z_4,\\ &\frac{2(1-r)\Theta _c-r}{1-\bar{\Phi }_1}\le \frac{2(1-r)[2(1-r)\Theta _c-r]}{2-3r}, z\gt z_4, \end{aligned} \end{array}\right . \end{equation}

where

\begin{align*} \Theta _c=\int _{0}^{\infty }\int _{-\infty }^{cs}K(s,y)dyds+\int _{0}^{\infty }\int _{cs}^{\infty }K(s,y)e^{-\mu _1(cs-y)}dyds. \end{align*}

Then $-2\mu _1^2+J \le 0$ by (3.31). Therefore, $(\bar{\Phi }_1, \bar{\Psi }_1)(z)$ is an upper solution of (1.5)-(1.6). By Theorem3.5, we know $c_{\min }=c_0$ when (3.31) holds. The proof is complete.

Proof. We shall use Theorem3.5 to prove this theorem. The key point is to find a suitable lower solution satisfying all conditions in Theorem3.5. Define a pair of continuous function $(\underline{\Phi }_1, \underline{\Psi }_1)(z)$ as

\begin{align*} \underline {\Phi }_1(z)=\frac {k_1}{1+e^{\mu _2 z}},\quad \underline {\Psi }_1(z)=\frac {\underline {\Phi }_1}{k_1}, \end{align*}

where $0\lt k_1\lt 1$ . We need to prove $(\underline{\Phi }_1, \underline{\Psi }_1)(z)$ is the lower solution of system (1.5)-(1.6). According to (3.34), $k_1$ can be chosen such that

(3.35) \begin{equation} \frac{3(1-r)}{b}\lt k_1\lt \min \{r\chi -(1-r), 1\}. \end{equation}

We have

\begin{align*} \begin {aligned} &\underline {\Phi }_1^{\prime\prime}(z)+c\underline {\Phi }_1^{\prime}(z)+\underline {\Phi }_1(z)(1-r-\underline {\Phi }_1(z)+r(K\star _c\underline {\Psi }_1)(z))\\ &=\underline {\Phi }_1(1-\frac {\underline {\Phi }_1}{k_1}) \bigg \{\mu _2^2-c\mu _2+1-r+\frac {\underline {\Phi }_1}{k_1}[{-}2\mu _2^2\\ &+\frac {(1-r)\frac {\underline {\Phi }_1}{k_1}-\underline {\Phi }_1+r\frac {\Phi _1}{k_1}\int _{0}^{\infty }\int _{\mathbb {R}}K(s,y)\frac {1+e^{\mu _2 z}}{1+e^{\mu _2 z}e^{\mu _2(cs-y)}}dyds}{(1-\frac {\underline {\Phi }_1}{k_1})\frac {\underline {\Phi }_1}{k_1}}]\bigg \}\\ &=\frac {\underline {\Phi }_1^2}{k_1}(1-\frac {\underline {\Phi }_1}{k_1})\bigg \{ -2\mu _2^2+\frac {(1-r)-k_1+r\int _{0}^{\infty }\int _{\mathbb {R}}K(s,y)\frac {1+e^{\mu _2 z}}{1+e^{\mu _2 z}e^{\mu _2(cs-y)}}dyds}{1-\frac {\underline {\Phi }_1}{k_1}} \bigg \}\\ &\ge \frac {\underline {\Phi }_1^2}{k_1}(1-\frac {\underline {\Phi }_1}{k_1})\bigg \{ -2\mu _2^2+(1-r)-k_1+r[\int _{0}^{\infty }\int _{-\infty }^{cs}K(s,y)e^{-\mu _2(cs-y)}dyds\\ &+\int _{0}^{\infty }\int _{cs}^{\infty }K(s,y)dyds] \bigg \}\\ &\ge 0, \end {aligned} \end{align*}

and

\begin{align*} \begin {aligned} &\underline {\Psi }_1^{\prime\prime}(z)+c\underline {\Psi }_1^{\prime}(z)+b\underline {\Phi }_1(z)(1-\underline {\Psi }_1(z))\\ &=\frac {\underline {\Phi }_1}{k_1}(1-\frac {\underline {\Phi }_1}{k_1})[\mu _2^2-c\mu _2-2\mu _2^2\frac {\underline {\Phi }_1}{k_1}+bk_1]\\ &\ge \frac {\underline {\Phi }_1}{k_1}(1-\frac {\underline {\Phi }_1}{k_1})[{-}2\mu _2^2+bk_1-(1-r)]\ge 0, \end {aligned} \end{align*}

by using (3.35) and $\mu _2\sim \sqrt{1-r}$ for $c=c_0+\varepsilon _3$ , where $\varepsilon _3\gt 0$ is small. Thus, $(\underline{\Phi }_1, \underline{\Psi }_1)(z)$ is a lower solution of (1.5)-(1.6). It follows that $c_{\min }\gt c_0$ by Theorem3.5. Thus, the proof is complete.

Proof. Note that $c_{\min }\gt c_0=0$ for all $b\gt 0$ when $r=1$ . It follows from Theorem3.4 that when $r=1$ , the result (3) holds for $b\gt 0$ . Thus, we only consider $r\in (0,1)$ . For $b \in (0, b^*]$ , Theorem3.6 implies $c_{\min }=c_0$ . Then we have

(3.37) \begin{equation} \Phi (z)\sim Aze^{-\mu _1(c_0) z}+Be^{-\mu _1(c_0) z}, \ \ \text{as} \ z\rightarrow \infty, \end{equation}

where $A\gt 0$ , or $B\gt 0$ if $A=0$ , see [Reference Aronson and Weinberger4]. To establish result (2), we utilise a method similar to that employed in Theorem3.4. Denote the minimal-speed travelling wave solution $(\Phi (z),\Psi (z))$ by $(\Phi ^*(z), \Psi ^*(z))$ for $b=b^*$ . Then $(\Phi ^*(z), \Psi ^*(z))$ satisfies

(3.38) \begin{equation} \left \{ \begin{array}{cc} \begin{aligned} &\Phi ^{*^{\prime\prime}}+c_0\Phi ^{*^{\prime}}+\Phi ^*(1-r-\Phi ^*+r(K\star _{c_0}\Psi ^*))=0,\\ &\Psi ^{*^{\prime\prime}}+c_0\Psi ^{*^{\prime}}+b^*\Phi ^*(1-\Psi ^*)=0.\\ \end{aligned} \end{array}\right . \end{equation}

To the contrary, we assume that $\Phi ^*(z)\sim Aze^{-\mu _1(c_0) z}$ as $z\rightarrow \infty$ with $A\gt 0$ . Consider $b_\varepsilon =b^*+\varepsilon$ , where $\varepsilon$ is sufficiently small to be determined. Let

\begin{align*}\Phi _\varepsilon (z)=\Phi ^*(z)+W_3(z)\ \text {and} \ \Psi _\varepsilon (z)=\Psi ^*(z)+W_4(z), \end{align*}

where $W_3, W_4 \in B_0=\{u\in C({-}\infty, \infty ), u(\pm \infty )=0\}$ . Our goal is to show the existence of $W_3$ and $W_4$ such that $(\Phi _\varepsilon (z), \Psi _\varepsilon (z))$ satisfies (1.5) with $b = b_\varepsilon$ and $c = c_0$ . If we can demonstrate this, it would contradict the definition of $b^*$ and therefore establish result (2). Suppose that $(\Phi _\varepsilon (z), \Psi _\varepsilon (z))$ is a solution of (1.5) with $b=b_\varepsilon$ and $c=c_0$ . It implies

(3.39) \begin{equation} \left \{ \begin{array}{cc} \begin{aligned} &\Phi _\varepsilon ^{\prime\prime}+c_0\Phi _\varepsilon ^{\prime}+\Phi _\varepsilon (1-r-\Phi _\varepsilon +r(K\star _{c_0}\Psi _\varepsilon ))=0,\\ &\Psi _\varepsilon ^{\prime\prime}+c_0\Psi _\varepsilon ^{\prime}+b_{\varepsilon }\Phi _\varepsilon (1-\Psi _\varepsilon )=0.\\ \end{aligned} \end{array}\right . \end{equation}

By (3.38) and (3.39), $W_3$ and $W_4$ satisfy

(3.40) \begin{equation} \begin{aligned} &W_3^{\prime\prime}+c_0W_3^{\prime}-\alpha W_3=-(P_0+P_h),\\ &W_4^{\prime\prime}+c_0W_4^{\prime}-\alpha W_4=-(R_0+R_{1,\varepsilon }+R_{2,\varepsilon }+R_h), \end{aligned} \end{equation}

where

\begin{align*} \begin {aligned} &P_0=\{ \alpha +1-r-2\Phi ^*+r(K\star _{c_0}\Psi ^*)\}W_3+r\Phi ^*(K\star _{c_0} W_4),\\ &P_h=-W_3^2+rW_3(K\star _{c_0} W_4),\\ &R_0=( \alpha -b^*\Phi ^*)W_4+b^*(1-\Psi ^*)W_3,\\ &R_{1,\varepsilon }=\varepsilon \Phi ^*(1-\Psi ^*),\\ &R_{2,\varepsilon }=\varepsilon (1-\Psi ^*)W_3-\varepsilon \Phi ^*W_4,\\ &R_h=-(b^*+\varepsilon )W_3W_4.\\ \end {aligned} \end{align*}

Therefore, for $c=c_0$ , we have

(3.41) \begin{equation} \begin{aligned} &W_3=T_{1,c_0}(P_0)+T_{1,c_0}(P_h),\\ &W_4=T_{2,c_0}(R_0)+T_{2,c_0}(R_{1,\varepsilon })+T_{2,c_0}(R_{2,\varepsilon })+T_{2,c_0}(R_h), \end{aligned} \end{equation}

where $T_{1,c_0}$ and $T_{2,c_0}$ are defined in (3.28). Define

\begin{align*} \bar {T}(\bar {W})= \begin {pmatrix} T_{1,c_0}(P_0(\bar {W}))\\ T_{2,c_0}(R_0(\bar {W})) \end {pmatrix}, \ \bar {W}=(W_3, W_4). \end{align*}

Clearly, $T_{2,c_0}(R_{1,\varepsilon })=O(\varepsilon )$ and $T_{2,c_0}(R_{2,\varepsilon })=O(\varepsilon \bar{W})$ . The linear operator $\bar{T}$ is both compact and strongly positive, and it possesses a simple principal eigenvalue $\lambda =1$ associated with the positive eigenfunction $({-}\Phi ^{*^{\prime}},-\Psi ^{*^{\prime}})$ . To eliminate this eigenfunction from $B_0$ , we introduce a weighted functional space defined as

\begin{align*}\mathcal {H}_2=\{h(\xi )\in B_0\ :\ he^{\mu _1(c_0)\xi }=o(1) \quad \rm {as} \quad \xi \rightarrow \infty \}.\end{align*}