2.1 Preliminaries: well-posedness and comparison principles

We first present the well-posedness and comparison principles of (1.4).

Proof. $(i)$ We introduce the coordinate transformation $(t,x)\rightarrow (t,y)=(t,y(t,x))$ as follows

(2.1) \begin{equation} y=\frac{2x-s_{1}(t)-s_{2}(t)}{s_{2}(t)-s_{1}(t)}s_{0}\quad \mbox{for}\quad t\gt 0,\quad y=x\quad \mbox{for}\quad -\max \{\tau _1,\tau _2\}\leqslant t\leqslant 0, \end{equation}

and define

\begin{eqnarray*} &&w(t,y)\;:\!=\;u\Big (t,\frac{s_{2}(t)-s_{1}(t)}{2s_{0}}y+\frac{s_{1}(t)+s_{2}(t)}{2}\Big )=u(t,x)\quad \mbox{for } t\gt 0,\\[5pt] &&w(\theta,y)\;:\!=\;u_{0}(\theta,y)\quad \mbox{for} -\tau _2\leqslant \theta \leqslant 0,\\[5pt] &&z(t,y)\;:\!=\;v\Big (t,\frac{s_{2}(t)-s_{1}(t)}{2s_{0}}y+\frac{s_{1}(t)+s_{2}(t)}{2}\Big )=v(t,x)\quad \mbox{for } t\gt 0,\\[5pt] &&z(\theta,y)\;:\!=\;v_{0}(\theta,y)\quad \mbox{for} -\tau _1\leqslant \theta \leqslant 0. \end{eqnarray*}

Denote

\begin{eqnarray*} &&A=A(s_1,s_2)=\frac{4s_{0}^{2}}{(s_{2}(t)-s_{1}(t))^{2}}, \\[5pt] &&B=B(s_1,s_2,y)=\frac{s_{2}^{\prime }(t)-s_{1}^{\prime }(t)}{s_{2}(t)-s_{1}(t)}y +\frac{s_{2}^{\prime }(t)+s_{1}^{\prime }(t)}{s_{2}(t)-s_{1}(t)}s_{0},\\[5pt] &&h(z(t-\tau _{1},y))=h(v(t-\tau _{1},x)), \quad g(w(t-\tau _{2},y))=g(u(t-\tau _{2},x)). \end{eqnarray*}

Then, (1.4) can be transformed into a fixed boundary problem

(2.2) \begin{equation} \left \{\begin{array}{l} \dfrac{\partial w}{\partial t} =d_{1}A\dfrac{\partial ^2 w}{\partial y^2} +d_{1}B\dfrac{\partial w}{\partial y} -b_{1}w+h(z(t-\tau _{1},y)),\quad t\gt 0, -s_0\lt y\lt s_0,\\[10pt] \dfrac{\partial z}{\partial t} =d_{2}A\dfrac{\partial ^2 z}{\partial y^2} +d_{2}B\dfrac{\partial z}{\partial y} -b_{2}z+g(w(t-\tau _{2},y)),\quad t\gt 0, -s_0\lt y\lt s_0,\\[10pt] w(t,y)=z(t,y)=0,\quad t\gt 0, y\geqslant s_0 \mbox{ or } y\leqslant -s_0,\\[5pt] w(\theta,y)=u_0(\theta,y),\quad -\tau _{2}\leqslant \theta \leqslant 0, -s_0\leqslant y\leqslant s_0,\\[5pt] z(\theta,y)=v_0(\theta,y),\quad -\tau _{1}\leqslant \theta \leqslant 0, -s_0\leqslant y\leqslant s_0 \end{array}\right. \end{equation}

and

(2.3) \begin{equation} s_1^{\prime }(t)=-\frac{2s_{0}\mu }{s_{2}(t)-s_{1}(t)}\frac{\partial w}{\partial y}(t, -s_0),\quad s_2^{\prime }(t)=-\frac{2s_{0}\mu }{s_{2}(t)-s_{1}(t)}\frac{\partial w}{\partial y}(t, s_0). \end{equation}

Let $k_{1}=-\mu \frac{\partial u_{0}}{\partial y}(0,-s_{0})$ and $k_{2}=-\mu \frac{\partial u_{0}}{\partial y}(0,s_{0})$ . For $0\lt T_{0}\leqslant \min \{\frac{s_{0}}{2+k_{1}+k_{2}},\tau _{1},\tau _{2}\}$ , we define

\begin{eqnarray*} &&D_{T_{0}}^{s_{1}}=\{s_{1}\in C^{1}([0,T_{0}])\;:\; s_{1}(0)=-s_{0}, s_{1}^{\prime }(0)=k_{1}, \|s_{1}^{\prime }-k_{1}\|_{C([0,T_{0}])}\leqslant 1\},\\[5pt] &&D_{T_{0}}^{s_{2}}=\{s_{2}\in C^{1}([0,T_{0}])\;:\; s_{2}(0)=s_{0}, s_{2}^{\prime }(0)=k_{2}, \|s_{2}^{\prime }-k_{2}\|_{C([0,T_{0}])}\leqslant 1\}. \end{eqnarray*}

Taking any fixed $(s_{1},s_{2})\in D_{T_{0}}^{s_{1}}\times D_{T_{0}}^{s_{2}}$ , we have

\begin{eqnarray*} |s_{2}(t)-s_{1}(t)-2s_{0}| &&=|(s_{2}(t)-s_{0})-(s_{1}(t)+s_{0})|\\[5pt] &&\leqslant \|s_{2}^{\prime }\|_{C([0,T_0])}T_0+\|s_{1}^{\prime }\|_{C([0,T_0])}T_0\\[5pt] &&\leqslant (2+k_1+k_2)T_0\leqslant s_0, \end{eqnarray*}

which implies $s_{2}(t)-s_{1}(t)\geqslant s_0$ . Then, the above coordinate transformation $(t,x)\rightarrow (t,y)$ is a diffeomorphism from $[0,T_{0}]\times [s_{1}(t),s_{2}(t)]$ to $[0,T_{0}]\times [\!-\!s_0,s_0]$ . Note that the parabolic equations in (2.2) are linear, since $z(t-\tau _{1},y)$ and $w(t-\tau _{2},y)$ contained in the right-hand sides of equations are given initial functions when $t\in [0,T_{0}]$ . By the $L^{p}$ theory of parabolic equations and the Sobolev embedding theorem, we can prove that (2.2) admits a unique $(w(t,y),z(t,y))\in [C^{(1+\alpha )/2, 1+\alpha }([0,T_{0}]\times [\!-\!s_{0},s_{0}])]^{2}$ .

Denote

\begin{equation*} \hat {s}_{1}(t)=-s_{0}-\int _{0}^{t}\frac {2s_{0}\mu }{s_{2}(\zeta )-s_{1}(\zeta )} \frac {\partial w}{\partial y}(\zeta,-s_{0})d\zeta \end{equation*}

and

\begin{equation*} \hat {s}_{2}(t)=s_{0}-\int _{0}^{t}\frac {2s_{0}\mu }{s_{2}(\zeta )-s_{1}(\zeta )} \frac {\partial w}{\partial y}(\zeta,s_{0})d\zeta. \end{equation*}

We define an operator $\mathcal{L}$ by

\begin{equation*} \mathcal {L}(s_1,s_2)=(\hat {s}_1,\hat {s}_2). \end{equation*}

Similar to the proof of Theorem 2.1 in [Reference Du and Lin15], we can show that for $T_0\gt 0$ sufficiently small, $\mathcal{L}$ maps $D_{T_{0}}^{s_{1}}\times D_{T_{0}}^{s_{2}}$ into itself and $\mathcal{L}$ is a contraction mapping on $D_{T_{0}}^{s_{1}}\times D_{T_{0}}^{s_{2}}$ . The contraction mapping theorem gives that $\mathcal{L}$ has a unique fixed point in $D_{T_{0}}^{s_{1}}\times D_{T_{0}}^{s_{2}}$ . Then, (2.2) and (2.3) have a unique local classical solution $(w(t,y),z(t,y),s_1(t),s_2(t))$ . We should mention that the local well-posedness can also be established similarly as the proof of Theorem 1.1 in [Reference Wang33], where the existence was proved by the Schauder fixed point theorem.

$(ii)$ We only prove $u(t,x)\leqslant K_{1}$ and $v(t,x)\leqslant K_{2}$ , the remaining part can be obtained by similar arguments as in the proof of Lemma 2.3 in [Reference Ahn, Beak and Lin1].

For any $z\gt 0$ , by Taylor’s formula and the concavity of $h$ we have

\begin{equation*} h(z)=h(z)-h(0)=h^{\prime }(0)z+\frac {1}{2}h^{\prime \prime }(\xi )z^{2}\leqslant h^{\prime }(0)z \end{equation*}

with some $\xi \in (0,z)$ . Since $g$ is bounded, we can choose $K_{i}$ $(i=1,2)$ sufficiently large such that

\begin{equation*} \frac {\|g\|_{L^{\infty }}}{K_{2}}\leqslant b_{2},\quad h^{\prime }(0)\frac {K_{2}}{K_{1}}\leqslant b_{1}, \end{equation*}

which imply

\begin{equation*} \frac {g(K_{1})}{K_{2}}\leqslant b_{2},\quad \frac {h(K_{2})}{K_{1}}=\frac {h(K_{2})}{K_{2}}\cdot \frac {K_{2}}{K_{1}}\leqslant h^{\prime }(0)\frac {K_{2}}{K_{1}}\leqslant b_{1}. \end{equation*}

We may assume that

\begin{eqnarray*} &&u_{0}(\theta,x)\leqslant K_{1} \quad \mbox{for } (\theta,x)\in [\!-\!\tau _{2},0]\times [\!-\!s_{0},s_{0}],\\[5pt] &&v_{0}(\theta,x)\leqslant K_{2} \quad \mbox{for } (\theta,x)\in [\!-\!\tau _{1},0]\times [\!-\!s_{0},s_{0}]. \end{eqnarray*}

Let

\begin{equation*} (U(t,x),V(t,x))\;:\!=\;(K_{1}-u(t,x),K_{2}-v(t,x))e^{-kt} \end{equation*}

with some constant $k$ to be determined later, then $(U,V)$ satisfies

(2.4) \begin{equation} \left \{\begin{array}{l} \dfrac{\partial U}{\partial t} =d_{1}\dfrac{\partial ^{2} U}{\partial x^{2}} -(b_{1}+k)U+e^{-kt}[b_{1}K_{1}-h(v(t-\tau _{1},x))]\\[10pt] \geqslant d_{1}\dfrac{\partial ^{2} U}{\partial x^{2}}-(b_{1}+k)U+e^{-k\tau _{1}}h^{\prime }(\xi )V(t-\tau _{1},x),\quad t\gt 0, s_1(t)\lt x\lt s_2(t),\\[10pt] \dfrac{\partial V}{\partial t} =d_{2}\dfrac{\partial ^{2} V}{\partial x^{2}} -(b_{2}+k)V+e^{-kt}[b_{2}K_{2}-g(u(t-\tau _{2},x))]\\[10pt] \geqslant d_{2}\dfrac{\partial ^{2} V}{\partial x^{2}}-(b_{2}+k)V+e^{-k\tau _{2}}g^{\prime }(\eta )U(t-\tau _{2},x),\quad t\gt 0, s_1(t)\lt x\lt s_2(t),\\[10pt] U(t,x)=K_{1}e^{-kt},\quad t\gt 0, x\geqslant s_2(t) \mbox{ or } x\leqslant s_1(t),\\[5pt] V(t,x)=K_{2}e^{-kt},\quad t\gt 0, x\geqslant s_2(t) \mbox{ or } x\leqslant s_1(t),\\[5pt] U(\theta,x)\geqslant 0,\quad -\tau _{2}\leqslant \theta \leqslant 0, s_1(\theta )\leqslant x\leqslant s_2(\theta ),\\[5pt] V(\theta,x)\geqslant 0,\quad -\tau _{1}\leqslant \theta \leqslant 0, s_1(\theta )\leqslant x\leqslant s_2(\theta ), \end{array}\right. \end{equation}

where $\xi$ lies between $K_{2}$ and $v(t-\tau _{1},x)$ , $\eta$ lies between $K_{1}$ and $u(t-\tau _{2},x)$ .

We claim that $U(t,x),V(t,x)\geqslant 0$ in $(0,+\infty )\times (s_{1}(t),s_{2}(t))$ . Assume by contraction that there exist some $T_0$ and $(t_{0},x_{0})\in (0,T_0]\times (s_1(t),s_2(t))$ such that

\begin{equation*} \min \{U(t_0,x_0),V(t_0,x_0)\} =\min _{(t,x)\in [0,T_0]\times [s_{1}(t),s_{2}(t)]}\min \{U(t,x),V(t,x)\}\lt 0. \end{equation*}

If $U(t_0,x_0)=\min \{U(t_0,x_0),V(t_0,x_0)\}\lt 0$ , then $U(t_0,x_0)$ is the minimum of $U(t,x)$ in $[0,T_0]\times [s_{1}(t),s_{2}(t)]$ . It follows that $\frac{\partial U}{\partial t}(t_0,x_0)\leqslant 0$ and $\frac{\partial ^{2} U}{\partial x^{2}}(t_0,x_0)\geqslant 0$ . On the other hand,

\begin{eqnarray*} -(b_{1}+k)U(t_0,x_0)+e^{-k\tau _{1}}h^{\prime }(\xi )V(t_{0}-\tau _{1},x_{0}) &&\geqslant -(b_{1}+k)U(t_0,x_0)+e^{-k\tau _{1}}h^{\prime }(\xi )U(t_0,x_0)\\[5pt] &&\geqslant (\!-\!b_{1}-k+h^{\prime }(\xi ))U(t_0,x_0). \end{eqnarray*}

Choose

\begin{equation*} k=\max \Big \{\|h^{\prime }\|_{L^{\infty }([0,\max \{K_{2},M_{2}\}])}, \|g^{\prime }\|_{L^{\infty }([0,\max \{K_{1},M_{1}\}])}\Big \} \end{equation*}

with

\begin{eqnarray*} &M_1=\|u\|_{L^{\infty }([\!-\!\tau _{2},T_{0}-\tau _{2}]\times [s_{1}(t),s_{2}(t)])},\\[5pt] &M_2=\|v\|_{L^{\infty }([\!-\!\tau _{1},T_{0}-\tau _{1}]\times [s_{1}(t),s_{2}(t)])}. \end{eqnarray*}

Thus,

\begin{equation*} -(b_{1}+k)U(t_0,x_0)+e^{-k\tau _{1}}h^{\prime }(\xi )V(t_{0}-\tau _{1},x_{0})\gt 0, \end{equation*}

which contradicts with the first equation in (2.4). If $V(t_0,x_0)=\min \{U(t_0,x_0),V(t_0,x_0)\}\lt 0$ , we can similarly prove the claim. This completes the proof of $(ii)$ .

$(iii)$ Since $u, v$ and $s_{1}^{\prime }(t), s_{2}^{\prime }(t)$ are bounded in $(0,T_0]\times (s_{1}(t),s_{2}(t))$ by constants independent of $T_0$ , the global solution is guaranteed.

Next, we provide two comparison principles for the free boundary problem (1.4). The first one is used for comparing the solution $(u(t,x), v(t,x), s_{1}(t), s_{2}(t))$ with a upper solution $(\bar{u}(t,x), \bar{v}(t,x), \bar{s}_{1}(t), \bar{s}_{2}(t))$ in the spatial domain $(s_{1}(t),s_{2}(t))$ , and the second one is used for comparing $(u(t,x), v(t,x), s_{2}(t))$ with $(\bar{u}(t,x), \bar{v}(t,x), \bar{s}_{2}(t))$ in one-side interval $(0,s_{2}(t))$ . The proofs are similar as that of Lemma 2.5 in [Reference Ahn, Beak and Lin1], here we omit the details. Moreover, we can also obtain the corresponding conclusions for lower solution by minor modification.

Lemma 2.1. Suppose that $T\in (0,\infty )$ , $\bar{s}_1, \bar{s}_2\in C([\!-\!\max \{\tau _{1}, \tau _{2}\}, T])\cap C^{1}((0, T])$ , $\bar{u}_{0}(\theta, x)\in C^{1,2}([\!-\!\tau _{2}, 0]\times [\bar{s}_{1}(\theta ), \bar{s}_{2}(\theta )])$ , $\bar{v}_{0}(\theta, x)\in C^{1,2}([\!-\!\tau _{1}, 0]\times [\bar{s}_{1}(\theta ), \bar{s}_{2}(\theta )])$ , $\bar{u}(t,x),\bar{v}(t,x)\in C^{1,2}((0, T]\times (\bar{s}_{1}(t), \bar{s}_{2}(t)))$ , and

\begin{equation*} \left \{\begin {array}{l} \dfrac {\partial \bar {u}}{\partial t} \geqslant d_{1}\dfrac {\partial ^{2}\bar {u}}{\partial x^{2}} -b_{1} \bar {u}+h(\bar {v}(t-\tau _{1},x)), \quad 0\lt t\leqslant T, \bar {s}_1(t)\lt x\lt \bar {s}_2(t),\\[10pt] \dfrac {\partial \bar {v}}{\partial t} \geqslant d_{2}\dfrac {\partial ^{2}\bar {v}}{\partial x^{2}} -b_{2} \bar {v}+g(\bar {u}(t-\tau _{2},x)), \quad 0\lt t\leqslant T, \bar {s}_1(t)\lt x\lt \bar {s}_2(t),\\[10pt] \bar {u}(t,x)=\bar {v}(t,x)=0, \quad 0\lt t\leqslant T, x\geqslant \bar {s}_2(t)\ \ \textit{or}\ \ x\leqslant \bar {s}_1(t),\\[5pt] \bar {s}_2^{\prime }(t)\geqslant -\mu \dfrac {\partial \bar {u}}{\partial x}(t, \bar {s}_2(t)),\quad 0\lt t\leqslant T,\\[10pt] \bar {s}_1^{\prime }(t)\leqslant -\mu \dfrac {\partial \bar {u}}{\partial x}(t, \bar {s}_1(t)),\quad 0\lt t\leqslant T,\\[10pt] \bar {u}(\theta,x)=\bar {u}_0(\theta,x), \quad -\tau _{2}\leqslant \theta \leqslant 0, \bar {s}_1(\theta )\leqslant x\leqslant \bar {s}_2(\theta ),\\[5pt] \bar {v}(\theta,x)=\bar {v}_0(\theta,x), \quad -\tau _{1}\leqslant \theta \leqslant 0, \bar {s}_1(\theta )\leqslant x\leqslant \bar {s}_2(\theta ). \end {array}\right. \end{equation*}

If $\bar{u}_{0}(\theta,x)\geqslant u_0(\theta,x)$ for $(\theta,x)\in [\!-\!\tau _{2},0]\times [s_1(\theta ), s_2(\theta )]$ , $\bar{v}_{0}(\theta,x)\geqslant v_0(\theta,x)$ for $(\theta,x)\in [\!-\!\tau _{1},0]\times [s_1(\theta ), s_2(\theta )]$ and $[\bar{s}_1(\theta ), \bar{s}_2(\theta )]\supseteq [s_1(\theta ), s_2(\theta )]$ for $\theta \in [\!-\!\max \{\tau _{1},\tau _{2}\},0]$ , then we have $\bar{s}_1(t)\geqslant s_1(t)$ , $\bar{s}_2(t)\geqslant s_2(t)$ for $t\in (0, T]$ and $(\bar{u}(t,x), \bar{v}(t,x))\geqslant (u(t,x), v(t,x))$ for $(t,x)\in (0,T]\times (s_1(t), s_2(t))$ .

Lemma 2.2. Suppose that $T\in (0,\infty )$ , $\bar{s}_2\in C([\!-\!\max \{\tau _{1}, \tau _{2}\}, T])\cap C^{1}((0, T])$ , $\bar{u}_{0}(\theta, x)\in C^{1,2}([\!-\!\tau _{2}, 0]\times [0, \bar{s}_{2}(\theta )])$ , $\bar{v}_{0}(\theta, x)\in C^{1,2}([\!-\!\tau _{1}, 0]\times [0, \bar{s}_{2}(\theta )])$ , $\bar{u}(t,x),\bar{v}(t,x)\in C^{1,2}((0, T]\times (0, \bar{s}_{2}(t)))$ , and

\begin{equation*} \left \{\begin {array}{l} \dfrac {\partial \bar {u}}{\partial t} \geqslant d_{1}\dfrac {\partial ^{2}\bar {u}}{\partial x^{2}} -b_{1} \bar {u}+h(\bar {v}(t-\tau _{1},x)), \quad 0\lt t\leqslant T, 0\lt x\lt \bar {s}_2(t),\\[10pt] \dfrac {\partial \bar {v}}{\partial t} \geqslant d_{2}\dfrac {\partial ^{2}\bar {v}}{\partial x^{2}} -b_{2} \bar {v}+g(\bar {u}(t-\tau _{2},x)), \quad 0\lt t\leqslant T, 0\lt x\lt \bar {s}_2(t),\\[10pt] \bar {u}(t,x)=\bar {v}(t,x)=0,\quad 0\lt t\leqslant T, x\geqslant \bar {s}_2(t),\\[5pt] \bar {u}(t,0)\geqslant u(t,0), \bar {v}(t,0)\geqslant v(t,0),\quad 0\lt t\leqslant T,\\[5pt] \bar {s}_{2}^{\prime }(t)\geqslant -\mu \dfrac {\partial \bar {u}}{\partial x}(t, \bar {s}_2(t)),\quad 0\lt t\leqslant T,\\[10pt] \bar {u}(\theta,x)=\bar {u}_0(\theta,x),\quad -\tau _{2}\leqslant \theta \leqslant 0, 0\leqslant x\leqslant \bar {s}_2(\theta ),\\[5pt] \bar {v}(\theta,x)=\bar {v}_0(\theta,x),\quad -\tau _{1}\leqslant \theta \leqslant 0, 0\leqslant x\leqslant \bar {s}_2(\theta ). \end {array}\right. \end{equation*}

If $\bar{u}_{0}(\theta,x)\geqslant u_0(\theta,x)$ for $(\theta,x)\in [\!-\!\tau _{2},0]\times [0, s_2(\theta )]$ , $\bar{v}_{0}(\theta,x)\geqslant v_0(\theta,x)$ for $(\theta,x)\in [\!-\!\tau _{1},0]\times [0, s_2(\theta )]$ and $[0, \bar{s}_2(\theta )]\supseteq [0, s_2(\theta )]$ for $\theta \in [\!-\!\max \{\tau _{1},\tau _{2}\},0]$ , then we have $\bar{s}_2(t)\geqslant s_2(t)$ for $t\in (0, T]$ and $(\bar{u}(t,x), \bar{v}(t,x))\geqslant (u(t,x), v(t,x))$ for $(t,x)\in (0,T]\times (0, s_2(t))$ .

2.2 Spreading and vanishing

In this subsection, we investigate the long-time behaviour of solution.

Figure 1. Two curves $y=p(x)$ and $y=q(x)$ .

Let

\begin{equation*} \mathcal {R}_0\;:\!=\;\sqrt {\frac {h^{\prime }(0)g^{\prime }(0)}{b_{1}b_{2}}}. \end{equation*}

We claim that (1.3) admits a unique positive equilibrium for $\mathcal{R}_0\gt 1$ and has no positive equilibrium for $0\lt \mathcal{R}_0\leqslant 1$ . Indeed, let

\begin{equation*} p(x)=g\Big (\frac {h(x)}{b_{1}}\Big )\quad \mbox {and} \quad q(x)=b_{2}x, \end{equation*}

we conclude that the curves $y=p(x)$ and $y=q(x)$ have at most one intersection point in the interior of the first quadrant $\mathbb{R}_{+}^{2}$ . Otherwise, the mean value theorem yields that there exists a $\xi \gt 0$ such that

\begin{equation*} p^{\prime \prime }(\xi )=g^{\prime \prime }\Big (\frac {h(\xi )}{b_{1}}\Big )\Big (\frac {h^{\prime }(\xi )}{b_{1}}\Big )^{2} +g^{\prime }\Big (\frac {h(\xi )}{b_{1}}\Big )\frac {h^{\prime \prime }(\xi )}{b_{1}}=0, \end{equation*}

which contradicts with the condition $(\textrm{M})$ . When $\mathcal{R}_0\gt 1$ , we have $p^{\prime }(0)=\frac{h^{\prime }(0)g^{\prime }(0)}{b_{1}}\gt b_{2}=q^{\prime }(0)$ . Then, $p(x)\gt q(x)$ for small $x\gt 0$ due to $p(0)=q(0)=0$ . Moreover, since $g\in L^{\infty }$ , we know that $p(x)\lt q(x)$ for large $x\gt 0$ . Thus, as showed in Figure 1 the curves $y=p(x)$ and $y=q(x)$ have exactly one intersection point $(x^{*}, y^{*})\gt (0,0)$ for $\mathcal{R}_0\gt 1$ , which implies that $(\frac{h(x^{*})}{b_{1}},x^{*})$ is the unique positive equilibrium of (1.3). For $0\lt \mathcal{R}_0\leqslant 1$ , we have $p^{\prime }(0)\leqslant q^{\prime }(0)$ , then the claim can be easily proved by the concavity of $p(x)$ .

Theorem 2.2. If $0\lt \mathcal{R}_0\leqslant 1$ , then the solution $(u, v, s_1, s_2)$ of $(1.4)$ satisfies

\begin{equation*} \lim _{t\rightarrow +\infty }(\|u(t,\cdot )\|_{C([s_1(t), s_2(t)])} +\|v(t,\cdot )\|_{C([s_1(t), s_2(t)])})=0. \end{equation*}

Proof. Let $(w_1(t), w_2(t))$ be the unique solution of

(2.5) \begin{equation} \left \{\begin{array}{l} w_1^{\prime }=-b_{1}w_1+h(w_2(t-\tau _1)),\quad t\gt 0,\\[5pt] w_2^{\prime }=-b_{2}w_2+g(w_1(t-\tau _2)),\quad t\gt 0,\\[5pt] w_1(\theta )=\|u_0\|_{C([\!-\!\tau _{2},0]\times [s_1(\theta ), s_2(\theta )])}, \quad \theta \in [\!-\!\tau _2,0],\\[5pt] w_2(\theta )=\|v_0\|_{C([\!-\!\tau _{1},0]\times [s_1(\theta ), s_2(\theta )])}, \quad \theta \in [\!-\!\tau _1,0]. \end{array}\right. \end{equation}

From the comparison principle, we know that $(u(t,x), v(t,x))\leqslant (w_1(t), w_2(t))$ in $[0, +\infty )\times [s_1(t), s_2(t)]$ .

Denote $\bar{\mathcal{C}}\;:\!=\;C([\!-\!\tau _{2},0], \mathbb{R})\times C([\!-\!\tau _{1},0], \mathbb{R})$ , then $\bar{\mathcal{C}}$ is a Banach space with the usual norm. For any given $w=(w_1, w_2)\in \bar{\mathcal{C}}$ defined on $[\!-\!\tau _{2}, \sigma )\times [\!-\!\tau _{1}, \sigma )$ with $\sigma \gt 0$ , we define $w_{t}\;:\!=\;(w^{1}_{t}, w^{2}_{t})\in \bar{\mathcal{C}}$ for $0\leqslant t\lt \sigma$ , where $w^{1}_{t}(\theta )=w_1(t+\theta )$ for $\theta \in [\!-\!\tau _{2}, 0]$ and $w^{2}_{t}(\theta )=w_2(t+\theta )$ for $\theta \in [\!-\!\tau _{1}, 0]$ . Let

\begin{equation*} \bar {\mathcal {C}}_{+}=\Big \{(\varphi _{1},\varphi _{2})\in \bar {\mathcal {C}}; \varphi _{1}(\theta )\geqslant 0 \mbox { on } [\!-\!\tau _{2},0], \varphi _{2}(\theta )\geqslant 0 \mbox { on } [\!-\!\tau _{1},0]\Big \} \end{equation*}

and define $f\;:\; \bar{\mathcal{C}}_{+}\longrightarrow \mathbb{R}^{2}$ by

\begin{eqnarray*} f(\varphi _{1},\varphi _{2}) &&=(f_{1}(\varphi _{1},\varphi _{2}),f_{2}(\varphi _{1},\varphi _{2}))\\[5pt] &&=\Big (\!-\!b_{1}\varphi _{1}(0)+h(\varphi _{2}(\!-\!\tau _{1})),-b_{2}\varphi _{2}(0)+g(\varphi _{1}(\!-\!\tau _{2}))\Big ), \end{eqnarray*}

then the equations in (2.5) can be rewritten as

(2.6) \begin{equation} w^{\prime }(t)=f(w_{t}). \end{equation}

For any $(y_{1}, y_{2})\in \mathbb{R}^{2}$ , we write $(\hat{y}_{1},\hat{y}_{2})$ for the element of $\bar{\mathcal{C}}$ satisfying $\hat{y}_{1}(\theta )\equiv y_{1}$ for $\theta \in [\!-\!\tau _{2}, 0]$ and $\hat{y}_{2}(\theta )\equiv y_{2}$ for $\theta \in [\!-\!\tau _{1}, 0]$ , and define $\hat{f}\;:\; \mathbb{R}_{+}^{2}\rightarrow \mathbb{R}^{2}$ by $\hat{f}(y_{1}, y_{2})=f(\hat{y}_{1},\hat{y}_{2})$ . Since $f(\hat{0},\hat{0})=(0,0)$ and $df(\hat{0},\hat{0})(\varphi _{1},\varphi _{2})=(\!-\!b_{1}\varphi _{1}(0)+h^{\prime }(0)\varphi _{2}(\!-\!\tau _{1}), -b_{2}\varphi _{2}(0)+g^{\prime }(0)\varphi _{1}(\!-\!\tau _{2}))$ for any $(\varphi _{1},\varphi _{2})\in \bar{\mathcal{C}}$ , we can check that $df(\hat{0},\hat{0})$ satisfies the condition $(\textrm{R})$ in [Reference Zhao and Jing42].

Note that

\begin{equation*} \hat {f}(y_{1}, y_{2})=f(\hat {y}_{1}, \hat {y}_{2}) =\Big (\!-\!b_{1}y_{1}+h(y_{2}), -b_{2}y_{2}+g(y_{1})\Big ) \end{equation*}

and the Fréchet derivative is

(2.7) \begin{equation} D\hat{f}(y_{1},y_{2})=\left ( \begin{array}{c@{\quad}c} -b_{1} & h^{\prime }(y_{2}) \\[5pt] g^{\prime }(y_{1}) & -b_{2} \\[5pt] \end{array} \right ), \quad D\hat{f}(0,0)=\left ( \begin{array}{c@{\quad}c} -b_{1} & h^{\prime }(0) \\[5pt] g^{\prime }(0) & -b_{2} \\[5pt] \end{array} \right ). \end{equation}

From $0\lt \mathcal{R}_{0}\leqslant 1$ , we know that the stability modulus

\begin{equation*} s(D\hat {f}(0,0))=\max \Big \{\textrm {Re}\; \lambda ;\; \det (\lambda I-D\hat {f}(0,0))=0\Big \}\leqslant 0. \end{equation*}

Moreover, since $h^{\prime \prime }(z)\leqslant 0$ , $g^{\prime \prime }(z)\lt 0$ for all $z\gt 0$ , we have

\begin{equation*} \frac {h(\lambda y_2)}{\lambda y_2}=\frac {h(\lambda y_2)-h(0)}{\lambda y_2}\geqslant \frac {h(y_2)-h(0)}{y_2} =\frac {h(y_2)}{y_2} \end{equation*}

and

\begin{equation*} \frac {g(\lambda y_1)}{\lambda y_1}=\frac {g(\lambda y_1)-g(0)}{\lambda y_1}\gt \frac {g(y_1)-g(0)}{y_1} =\frac {g(y_1)}{y_1} \end{equation*}

for $y_1,y_2\gt 0$ and $\lambda \in (0,1)$ . It follows that $\hat{f}\;:\; \mathbb{R}_{+}^{2}\rightarrow \mathbb{R}^{2}$ is strictly subhomogeneous. Similarly, we can check that $f\;:\; \bar{\mathcal{C}}_{+}\rightarrow \mathbb{R}^{2}$ is subhomogeneous. From Theorem 3.2 in [Reference Zhao and Jing42], we know that $(0,0)$ is globally asymptotically stable for (2.6) with respect to $\bar{\mathcal{C}}_{+}$ . That is, the solution $(w_1,w_2)$ of (2.5) satisfies $\lim _{t\rightarrow +\infty }(w_{1}(t), w_{2}(t))=(0,0)$ , which implies $\lim _{t\rightarrow +\infty }(\|u(t,\cdot )\|_{C([s_1(t), s_2(t)])} +\|v(t,\cdot )\|_{C([s_1(t), s_2(t)])})=0$ . This completes the proof.

Due to Theorem 2.1, there exist $s_{1,\infty }, s_{2,\infty }\in (0,+\infty ]$ such that $\lim _{t\rightarrow +\infty }s_1(t)=s_{1,\infty }$ and $\lim _{t\rightarrow +\infty }s_2(t)=s_{2,\infty }$ . We call that the bacteria are spreading if $s_{2,\infty }-s_{1,\infty }=+\infty$ and $\limsup _{t\rightarrow +\infty }(\|u(t,\cdot )\|_{C([s_1(t), s_2(t)])}+\|v(t,\cdot )\|_{C([s_1(t), s_2(t)])})\gt 0$ ; the bacteria are vanishing if $s_{2,\infty }-s_{1,\infty }\lt +\infty$ and $\lim _{t\rightarrow +\infty }(\|u(t,\cdot )\|_{C([s_1(t), s_2(t)])}+\|v(t,\cdot )\|_{C([s_1(t), s_2(t)])})=0$ .

Next, we discuss the spreading and vanishing of bacteria for $\mathcal{R}_{0}\gt 1$ . For epidemic models governed by differential equations, the threshold dynamics of bacteria or virus are usually established in terms of the basic reproduction number $\mathcal{R}_0$ . However, for (1.4), the infected area is changing with time $t$ , and therefore, the basic reproduction number is not a constant and should be a function of $t$ . As in [Reference Lin and Zhu24], we introduce the spatial-temporal risk index, which is expressed by

\begin{equation*} \mathcal {R}_0^F(s_1(t), s_2(t)) =\sqrt {\frac {h^{\prime }(0)g^{\prime }(0)}{[d_1(\frac {\pi }{s_2(t)-s_1(t)})^2+b_1][d_2(\frac {\pi }{s_2(t)-s_1(t)})^2+b_2]}}. \end{equation*}

According to Lemma 4.1 in [Reference Lin and Zhu24], there exist $\lambda _{1}$ and $\kappa \gt 0$ such that $\textrm{sign} (1-\mathcal{R}_0^F(s_1(t), s_2(t))) =\textrm{sign} \lambda _1$ and $(\phi,\psi )\;:\!=\;(\kappa \psi ^*,\psi ^*)$ solves the following problem

(2.8) \begin{equation} \left \{\begin{array}{l} -d_1\phi _{xx}=h^{\prime }(0)\psi -b_1\phi +\lambda _1\phi,\quad x\in (s_1(t), s_2(t)),\\[5pt] -d_2\psi _{xx}=g^{\prime }(0)\phi -b_2\psi +\lambda _1\psi,\quad x\in (s_1(t), s_2(t)),\\[5pt] (\phi (x), \psi (x))=(0,0),\quad x=s_1(t) \mbox{ or } x=s_2(t), \end{array}\right. \end{equation}

where $\psi ^*$ is the principal eigenfunction of $-\Delta$ in $(s_1(t),s_2(t))$ with null Dirichlet boundary condition. It is easy to check that $\mathcal{R}_0^F(s_1(t), s_2(t))$ is increasing in $t$ and satisfies $\mathcal{R}_0^F(s_1(t), s_2(t))\rightarrow \mathcal{R}_0$ as $s_2(t)-s_1(t)\rightarrow +\infty$ .

Proof. Similar as the proof of Lemma 3.1 in [Reference Ahn, Beak and Lin1], we can show that $-2s_{0}\lt s_{1}(t)+s_{2}(t)\lt 2s_{0}$ for $t\geqslant 0$ , which implies $(i)$ and $(ii)$ .

For $(iii)$ , we only prove $\lim _{t\rightarrow +\infty }\|u(t,\cdot )\|_{C([s_1(t), s_2(t)])}=0$ , since the result for $v$ can be obtained similarly. Let $w(t,y)$ and $z(t,y)$ be the functions transformed from $u(t,x)$ and $v(t,x)$ by (2.1). Then, it is sufficient to show $\lim _{t\rightarrow +\infty }\|w(t,\cdot )\|_{C([\!-\!s_0, s_0])}=0$ . We assume by contradiction that

\begin{equation*} \limsup _{t\rightarrow +\infty }\|w(t,\cdot )\|_{C([\!-\!s_0, s_0])}=\delta \gt 0. \end{equation*}

It follows that there exists a sequence $\{(t_{k}, y_{k})\}_{k=1}^{\infty }$ in $(0,+\infty )\times (\!-\!s_0,s_0)$ such that $t_k\rightarrow +\infty$ as $k\rightarrow \infty$ and $w(t_k,y_k)\geqslant \frac{\delta }{2}$ for all $k\in \mathbb{N}$ . Since $\{y_k\}_{k=1}^{\infty }$ is bounded, we may assume that $y_{k}\rightarrow y_{0}$ as $k\rightarrow \infty$ . Similar as the proof of Lemma 3.2 in [Reference Ahn, Beak and Lin1], we can obtain $y_{0}\neq \pm s_{0}$ .

Denote $w_{k}(t,y)=w(t+t_{k},y)$ and $z_{k}(t,y)=z(t+t_{k},y)$ for any $(t,y)\in (t_{0}-t_{k}, +\infty )\times [\!-\!s_0, s_0]$ . Note that $s_{1}^{\prime }(t), s_{2}^{\prime }(t)\rightarrow 0$ as $t\rightarrow +\infty$ . It follows from Theorem 2.1 and the parabolic regularity theory that $\{(w_{k},z_{k})\}$ has a subsequence, denoted by itself, such that $(w_{k}, z_{k})\rightarrow (\tilde{w}, \tilde{z})$ in $C_{loc}^{1,2}(\mathbb{R}\times [\!-\!s_0,s_0])$ as $k\rightarrow \infty$ and $(\tilde{w}, \tilde{z})$ satisfies

(2.9) \begin{equation} \left \{\begin{array}{l} \dfrac{\partial \tilde{w}}{\partial t} =d_{1}\tilde{A}\dfrac{\partial ^{2}\tilde{w}}{\partial y^{2}} -b_{1}\tilde{w}+h(\tilde{z}(t-\tau _{1},y)),\quad -\infty \lt t\lt +\infty, -s_0\lt y\lt s_0,\\[10pt] \dfrac{\partial \tilde{z}}{\partial t} =d_{2}\tilde{A}\dfrac{\partial ^{2}\tilde{z}}{\partial y^{2}} -b_{2}\tilde{z}+g(\tilde{w}(t-\tau _{2},y)),\quad -\infty \lt t\lt +\infty, -s_0\lt y\lt s_0,\\[10pt] \tilde{w}(t,y)=\tilde{z}(t,y)=0,\quad -\infty \lt t\lt +\infty, y\geqslant s_0 \mbox{ or } y\leqslant -s_0,\\[5pt] \tilde{w}(t,y)\geqslant 0, \tilde{z}(t,y)\geqslant 0,\quad -\infty \lt t\lt +\infty, -s_0\lt y\lt s_0, \end{array}\right. \end{equation}

where $\tilde{A}=\frac{4s_{0}^{2}}{(s_{2,\infty }-s_{1,\infty })^{2}}$ . Since $w_{k}(0, y_{k})=w(t_{k},y_{k})\geqslant \frac{\delta }{2}$ , we have $\tilde{w}(0, y_{0})\geqslant \frac{\delta }{2}$ . By the strong maximum principle, we can deduce that $\tilde{w}(t,y)\gt 0$ for $(t,y)\in \mathbb{R}\times (\!-\!s_0,s_0)$ . Applying the Hopf boundary lemma, we have $\frac{\partial \tilde{w}}{\partial y}(t, s_0)\lt 0$ .

On the other hand, by the Stefan condition $s_{2}^{\prime }(t_k)=-\mu \frac{\partial u}{\partial x}(t_{k}, s_2(t_{k}))$ and the fact that $s_{2}^{\prime }(t)\rightarrow 0$ as $t\rightarrow +\infty$ , we have $\frac{\partial u}{\partial x}(t_{k}, s_2(t_{k}))\rightarrow 0$ as $k\rightarrow \infty$ . Note that $\frac{\partial u}{\partial x}(t_{k}, s_2(t_{k})) =\sqrt{A}\frac{\partial w_k}{\partial y}(0,s_0)\rightarrow \sqrt{\tilde{A}}\frac{\partial \tilde{w}}{\partial y}(0,s_0)$ . It follows that $\frac{\partial \tilde{w}}{\partial y}(0,s_0)=0$ , which is a contradiction. Thus, $(iii)$ holds true.

Next, we prove $(iv)$ . Since $\mathcal{R}_{0}\gt 1$ , we have

\begin{equation*} s(D\hat {f}(0,0))=\max \{\textrm {Re } \lambda ; \det (\lambda I-D\hat {f}(0,0))=0\}\gt 0, \end{equation*}

where $D\hat{f}(0,0)$ is defined in (2.7). By Theorem 3.2 in [Reference Zhao and Jing42], we can deduce that the solution of (2.5) satisfies

(2.10) \begin{equation} \lim _{t\rightarrow +\infty }(w_1(t),w_2(t))=(u^{*},v^{*}). \end{equation}

Moreover, by the comparison principle, we know that

(2.11) \begin{equation} \begin{array}{l} u(t,x)\leqslant w_1(t)\quad \mbox{for } (t,x)\in [\!-\!\tau _{2},+\infty )\times [s_{1}(t),s_{2}(t)],\\[5pt] v(t,x)\leqslant w_2(t)\quad \mbox{for } (t,x)\in [\!-\!\tau _{1},+\infty )\times [s_{1}(t),s_{2}(t)]. \end{array} \end{equation}

Hence,

\begin{equation*} \limsup _{t\rightarrow +\infty }(u(t,x), v(t,x))\leqslant (u^{*},v^{*}) \end{equation*}

uniformly for $x\in [s_{1}(t), s_{2}(t)]$ .

Next, we prove

\begin{equation*} \liminf _{t\rightarrow +\infty }(u(t,x), v(t,x))\geqslant (u^{*},v^{*}) \end{equation*}

uniformly in any compact subset of $\mathbb{R}$ .

Since $\mathcal{R}_0\gt 1$ , we can choose a sufficiently large $L_0\gt 0$ such that $\mathcal{R}_0^F(\!-\!L_0, L_0)\gt 1$ . Then, there exist $\lambda _{1}\lt 0$ and $\kappa \gt 0$ such that $(\phi,\psi )\;:\!=\;(\kappa \psi ^*,\psi ^*)$ solves the problem (2.8) with $(s_{1}(t), s_{2}(t))$ replaced by $(\!-\!L_0, L_0)$ , where $\psi ^*$ is the principal eigenfunction of $-\Delta$ in $(\!-\!L_0,L_0)$ with null Dirichlet boundary condition. Due to $s_{2,\infty }-s_{1,\infty }=+\infty$ , we have $s_{2,\infty }=-s_{1,\infty }=+\infty$ by $(ii)$ . Thus, for any $L\geqslant L_0$ , there exists $t_L\gt 0$ such that $s_{2}(t)\geqslant L$ and $s_{1}(t)\leqslant -L$ for all $t\gt t_L$ .

Let $\underline{U}(t,x)=\delta \kappa \psi ^*(x)$ and $\underline{V}(t,x)=\delta \psi ^*(x)$ . We can choose a sufficiently small $\delta \gt 0$ such that $(\underline{U}, \underline{V})$ satisfies

\begin{eqnarray*} \left \{\begin{array}{l} \dfrac{\partial \underline{U}}{\partial t} \leqslant d_{1}\dfrac{\partial ^{2}\underline{U}}{\partial x^{2}} -b_{1}\underline{U}+h(\underline{V}(t-\tau _{1},x)),\quad t\gt t_{L_{0}}, -L_0\lt x\lt L_0,\\[10pt] \dfrac{\partial \underline{V}}{\partial t} \leqslant d_{2}\dfrac{\partial ^{2}\underline{V}}{\partial x^{2}} -b_{2}\underline{V}+g(\underline{U}(t-\tau _{2},x)),\quad t\gt t_{L_{0}}, -L_0\lt x\lt L_0,\\[10pt] \underline{U}(t,x)=\underline{V}(t,x)=0,\quad t\gt t_{L_{0}}, x=\pm L_0,\\[5pt] \underline{U}(\theta,x)\leqslant u(\theta,x), \quad \theta \in [t_{L_{0}}-\tau _2, t_{L_{0}}], x\in [\!-\!L_0,L_0],\\[5pt] \underline{V}(\theta,x)\leqslant v(\theta,x), \quad \theta \in [t_{L_{0}}-\tau _1, t_{L_{0}}], x\in [\!-\!L_0,L_0]. \end{array}\right. \end{eqnarray*}

For any $L\geqslant L_0$ , we extend $(\underline{U}, \underline{V})$ by defining $(\underline{U}(t,x), \underline{V}(t,x))=(0,0)$ for $(t,x)\in \mathbb{R}\times ([\!-\!L,L]\setminus [\!-\!L_0,L_0])$ . Consider the following problem

(2.12) \begin{equation} \left \{\begin{array}{l} \dfrac{\partial \underline{u}}{\partial t} =d_{1}\dfrac{\partial ^{2}\underline{u}}{\partial x^{2}} -b_{1}\underline{u}+h(\underline{v}(t-\tau _{1},x)),\quad t\gt t_{L}, -L\lt x\lt L,\\[10pt] \dfrac{\partial \underline{v}}{\partial t} =d_{2}\dfrac{\partial ^{2}\underline{v}}{\partial x^{2}} -b_{2}\underline{v}+g(\underline{u}(t-\tau _{2},x)),\quad t\gt t_{L}, -L\lt x\lt L,\\[10pt] \underline{u}(t,x)=\underline{v}(t,x)=0,\quad t\gt t_{L}, x=\pm L,\\[5pt] \underline{u}(\theta,x)=\underline{U}(\theta,x)=\delta \kappa \psi ^*(x), \quad \theta \in [t_{L}-\tau _2, t_{L}], x\in [\!-\!L,L],\\[5pt] \underline{v}(\theta,x)=\underline{V}(\theta,x)=\delta \psi ^*(x), \quad \theta \in [t_{L}-\tau _1, t_{L}], x\in [\!-\!L,L]. \end{array}\right. \end{equation}

By the comparison principle, we have

\begin{eqnarray*} \underline{u}(t,x)\leqslant u(t,x)\quad \mbox{for } (t,x)\in [t_{L}-\tau _2, +\infty )\times [\!-\!L,L],\\[5pt] \underline{v}(t,x)\leqslant v(t,x)\quad \mbox{for } (t,x)\in [t_{L}-\tau _1, +\infty )\times [\!-\!L,L]. \end{eqnarray*}

Moreover, since $(\underline{U}(t,x), \underline{V}(t,x))$ is a lower solution of (2.12), the solution of (2.12) is increasing in $t$ for $x\in [\!-\!L,L]$ . Thus, $(\underline{u}(t,x),\underline{v}(t,x))\rightarrow (\underline{u}^{L},\underline{v}^{L})$ in $C^{2}([\!-\!L,L])$ as $t\rightarrow +\infty$ , where $(\underline{u}^{L},\underline{v}^{L})$ solves

\begin{eqnarray*} \left \{\begin{array}{l} -d_{1}\underline{u}^{L}_{xx}=-b_{1}\underline{u}^{L}+h(\underline{v}^{L}(x)),\quad -L\lt x\lt L,\\[5pt] -d_{2}\underline{v}^{L}_{xx}=-b_{2}\underline{v}^{L}+g(\underline{u}^{L}(x)),\quad -L\lt x\lt L,\\[5pt] \underline{u}^{L}(\pm L)=\underline{v}^{L}(\pm L)=0. \end{array}\right. \end{eqnarray*}

Note that $(\underline{u}^{L},\underline{v}^{L})$ is increasing in $L$ . By classical elliptic regularity theory and a diagonal procedure, we obtain that as $L\rightarrow +\infty$ , $(\underline{u}^{L},\underline{v}^{L})$ converges to $(\underline{u}^{\infty },\underline{v}^{\infty })$ in $[C_{loc}^{2}(\mathbb{R})]^{2}$ , and $(\underline{u}^{\infty },\underline{v}^{\infty })$ solves

(2.13) \begin{equation} \left \{\begin{array}{l} -d_{1}\underline{u}^{\infty }_{xx}=-b_{1}\underline{u}^{\infty }+h(\underline{v}^{\infty }(x)),\quad -\infty \lt x\lt +\infty,\\[5pt] -d_{2}\underline{v}^{\infty }_{xx}=-b_{2}\underline{v}^{\infty }+g(\underline{u}^{\infty }(x)),\quad -\infty \lt x\lt +\infty. \end{array}\right. \end{equation}

Similarly as the proof of Theorem 4.5 in [Reference Ahn, Beak and Lin1] and Lemma 3.5 in [Reference Wang and Du34], we can prove $(\underline{u}^{\infty },\underline{v}^{\infty })=(u^*,v^*)$ and then get the desired result.

Next, we exhibit some sufficient conditions to determine whether the bacteria are vanishing or spreading.

Proof. Since $\mathcal{R}_0^F(\!-\!s_0,s_0)\lt 1$ , we know that there exist $\lambda _{1}\gt 0$ and $\kappa \gt 0$ such that $(\phi,\psi )\;:\!=\;(\kappa \psi ^*,\psi ^*)$ satisfies the problem (2.8) with $(s_{1}(t), s_{2}(t))$ replaced by $(\!-\!s_0, s_0)$ , where $\psi ^*$ is the principal eigenfunction of $-\Delta$ in $(\!-\!s_0,s_0)$ with null Dirichlet boundary condition. Moreover, we have $\psi ^{*}(x)\gt 0$ in $(\!-\!s_0,s_0)$ , $(\psi ^*)^{\prime }(x)\lt 0$ in $(0, s_0]$ and $(\psi ^*)^{\prime }(x)\gt 0$ in $[\!-\!s_0, 0)$ . Since $\lambda _1\gt 0$ , there exists a small $\delta \gt 0$ such that

(2.14) \begin{equation} -\delta \kappa +\frac{1}{(1+\delta )^{2}}(h^{\prime }(0)-b_{1}\kappa +\lambda _{1}\kappa ) +b_1\kappa -h^{\prime }(0)e^{\delta \tau _{1}}\geqslant 0 \end{equation}

and

(2.15) \begin{equation} -\delta +\frac{1}{(1+\delta )^2}(\kappa g^{\prime }(0)-b_{2}+\lambda _{1})+b_{2}-\kappa g^{\prime }(0)e^{\delta \tau _{2}}\geqslant 0. \end{equation}

Define

\begin{eqnarray*} &\sigma (t)= \left \{\begin{array}{l} s_0\!\left(1+\delta -\dfrac{\delta }{2}e^{-\delta t}\right),\quad t\in [0,+\infty ),\\[5pt] s_0\left(1+\dfrac{\delta }{2}\right),\quad t\in [\!-\!\max \{\tau _{1},\tau _{2}\},0], \end{array}\right. \\[3pt] &\bar{u}(t,x)= \left \{\begin{array}{l} \kappa Ce^{-\delta t}\psi ^{*}\!\left(\dfrac{xs_0}{\sigma (t)}\right), \quad (t,x)\in [0,+\infty )\times [\!-\!\sigma (t), \sigma (t)],\\[10pt] \kappa C\psi ^{*}\!\left(\dfrac{2x}{2+\delta }\right),\quad (t,x)\in [\!-\!\tau _2,0]\times [\!-\!\sigma (t), \sigma (t)], \end{array}\right. \\[3pt] &\bar{v}(t,x)= \left \{\begin{array}{l} Ce^{-\delta t}\psi ^{*}\!\left(\dfrac{xs_0}{\sigma (t)}\right), \quad (t,x)\in [0,+\infty )\times [\!-\!\sigma (t), \sigma (t)],\\[10pt] C\psi ^{*}\!\left(\dfrac{2x}{2+\delta }\right), \quad (t,x)\in [\!-\!\tau _1,0]\times [\!-\!\sigma (t), \sigma (t)]. \end{array}\right. \end{eqnarray*}

For $(t,x)\in (0,+\infty )\times (\!-\!\sigma (t), \sigma (t))$ , we can deduce

\begin{eqnarray*} &&\frac{\partial \bar{u}}{\partial t} -d_1\frac{\partial ^{2}\bar{u}}{\partial x^{2}} +b_1\bar{u}-h(\bar{v}(t-\tau _{1},x))\\[3pt] &&=-\delta \bar{u} -\kappa Ce^{-\delta t}\frac{xs_0\sigma '(t)}{\sigma ^2(t)}(\psi ^{*})^{\prime }\Big (\frac{xs_{0}}{\sigma (t)}\Big ) -d_{1}\kappa Ce^{-\delta t}\frac{s_0^2}{\sigma ^2(t)}(\psi ^{*})^{\prime \prime }\Big (\frac{xs_{0}}{\sigma (t)}\Big ) +b_1\bar{u}-h(\bar{v}(t-\tau _{1},x))\\[3pt] &&\geqslant -\delta \bar{u} -\kappa Ce^{-\delta t}\frac{xs_0\sigma '(t)}{\sigma ^2(t)}(\psi ^{*})^{\prime }\Big (\frac{xs_{0}}{\sigma (t)}\Big ) +Ce^{-\delta t}\frac{s_0^2}{\sigma ^2(t)}(h^{\prime }(0)-b_{1}\kappa +\lambda _{1}\kappa )\psi ^{*}\Big (\frac{xs_{0}}{\sigma (t)}\Big ) +b_1\bar{u}\\[3pt] &&\quad -h^{\prime }(0)\bar{v}(t-\tau _{1},x)\\[4pt] &&=-\kappa Ce^{-\delta t}\frac{xs_0\sigma '(t)}{\sigma ^2(t)}(\psi ^{*})^{\prime }\Big (\frac{xs_{0}}{\sigma (t)}\Big )\\[4pt] &&\quad +\Big [\!-\!\delta \kappa +\frac{s_0^2}{\sigma ^2(t)}(h^{\prime }(0)-b_{1}\kappa +\lambda _{1}\kappa ) +b_1\kappa -h^{\prime }(0)\frac{\bar{v}(t-\tau _{1},x)}{\bar{v}(t,x)}\Big ]\bar{v}(t,x). \end{eqnarray*}

Case 1. If $(t,x)\in [\tau _{1},+\infty )\times (\!-\!\sigma (t), \sigma (t))$ , then

\begin{equation*} \frac {\bar {v}(t-\tau _{1},x)}{\bar {v}(t,x)}= e^{\delta \tau _{1}}\frac {\psi ^{*}\Big (\frac {xs_0}{\sigma (t-\tau _{1})}\Big )}{\psi ^{*}\Big (\frac {xs_0}{\sigma (t)}\Big )}. \end{equation*}

When $x\in [0, \sigma (t))$ , we have $\frac{xs_0}{\sigma (t-\tau _{1})}\gt \frac{xs_0}{\sigma (t)}\gt 0$ . Since $(\psi ^*)^{\prime }(x)\lt 0$ in $(0, s_0]$ , we have

\begin{equation*} \frac {\psi ^{*}\Big (\frac {xs_0}{\sigma (t-\tau _{1})}\Big )}{\psi ^{*}\Big (\frac {xs_0}{\sigma (t)}\Big )}\leqslant 1, \quad -\kappa Ce^{-\delta t}\frac {xs_0\sigma '(t)}{\sigma ^2(t)}(\psi ^{*})^{\prime }\Big (\frac {xs_{0}}{\sigma (t)}\Big ) \geqslant 0. \end{equation*}

When $x\in (\!-\!\sigma (t), 0]$ , we can get the same inequalities by the fact $(\psi ^*)^{\prime }(x)\gt 0$ . Consequently,

(2.16) \begin{eqnarray} &&\frac{\partial \bar{u}}{\partial t} -d_1\frac{\partial ^{2}\bar{u}}{\partial x^{2}} +b_1\bar{u}-h(\bar{v}(t-\tau _{1},x))\nonumber \\[4pt] &&\quad\geqslant \Big [\!-\!\delta \kappa +\frac{s_0^2}{\sigma ^2(t)}(h^{\prime }(0)-b_{1}\kappa +\lambda _{1}\kappa ) +b_1\kappa -h^{\prime }(0)e^{\delta \tau _{1}}\Big ]\bar{v}(t,x)\nonumber \\[4pt] &&\quad\geqslant \Big [\!-\!\delta \kappa +\frac{1}{(1+\delta )^{2}}(h^{\prime }(0)-b_{1}\kappa +\lambda _{1}\kappa ) +b_1\kappa -h^{\prime }(0)e^{\delta \tau _{1}}\Big ]\bar{v}(t,x)\nonumber \\[4pt] &&\quad\geqslant 0, \end{eqnarray}

where the last inequality uses (2.14).

Case 2. If $(t,x)\in (0,\tau _{1})\times (\!-\!\sigma (t), \sigma (t))$ , by using similar arguments as above, we can deduce

\begin{equation*} \frac {\bar {v}(t-\tau _{1},x)}{\bar {v}(t,x)} =e^{\delta \tau _{1}}\frac {\psi ^{*}\Big (\frac {2x}{2+\delta }\Big )}{\psi ^{*}\Big (\frac {xs_{0}}{\sigma (t)}\Big )}\leqslant e^{\delta \tau _{1}}, \quad -\kappa Ce^{-\delta t}\frac {xs_0\sigma '(t)}{\sigma ^2(t)}(\psi ^{*})^{\prime }\Big (\frac {xs_{0}}{\sigma (t)}\Big ) \geqslant 0, \end{equation*}

which implies that (2.16) also holds.

Similarly, by (2.15), we have

\begin{equation*} \frac {\partial \bar {v}}{\partial t} -d_2\frac {\partial ^{2}\bar {v}}{\partial x^{2}} +b_2\bar {v}-g(\bar {u}(t-\tau _{2},x))\geqslant 0 \quad \mbox {for } (t,x)\in (0,+\infty )\times (\!-\!\sigma (t), \sigma (t)). \end{equation*}

Besides, choose $C$ large enough such that

\begin{equation*} \bar {u}_{0}(\theta,x)=\kappa C\psi ^{*}\Big (\frac {2x}{2+\delta }\Big ) \geqslant \kappa C\psi ^{*}\Big (\frac {2s_0}{2+\delta }\Big ) \geqslant \|u_0\|_{L^{\infty }([\!-\!\tau _{2},0]\times [s_1(\theta ), s_2(\theta )])} \geqslant u_0(\theta,x) \end{equation*}

for $(\theta,x)\in [\!-\!\tau _{2},0]\times [s_1(\theta ), s_2(\theta )]$ and

\begin{equation*} \bar {v}_{0}(\theta,x)=C\psi ^{*}\Big (\frac {2x}{2+\delta }\Big ) \geqslant C\psi ^{*}\Big (\frac {2s_0}{2+\delta }\Big ) \geqslant \|v_0\|_{L^{\infty }([\!-\!\tau _{1},0]\times [s_1(\theta ), s_2(\theta )])} \geqslant v_0(\theta,x) \end{equation*}

for $(\theta,x)\in [\!-\!\tau _{1},0]\times [s_1(\theta ), s_2(\theta )]$ . Then, take $\mu \gt 0$ sufficiently small such that

\begin{equation*} \sigma '(t)=s_0\frac {\delta ^2}{2}e^{-\delta t}\geqslant -\mu \bar {u}_x(t, \sigma (t)), -\sigma '(t)=-s_0\frac {\delta ^2}{2}e^{-\delta t}\leqslant -\mu \bar {u}_x(t, -\sigma (t))\quad \mbox {for } t\gt 0. \end{equation*}

Since

\begin{equation*}[\!-\!\sigma (\theta ), \sigma (\theta )] =\Big [\!-\!s_0(1+\frac {\delta }{2}),s_0(1+\frac {\delta }{2})\Big ] \supseteq [\!-\!s_0,s_0]\supseteq [s_1(\theta ), s_2(\theta )] \end{equation*}

for $\theta \in [\!-\!\max \{\tau _{1},\tau _{2}\},0]$ , we can apply Lemma 2.1 to conclude that $s_{1}(t)\geqslant -\sigma (t)$ and $s_{2}(t)\leqslant \sigma (t)$ for $t\gt 0$ . Thus, $s_{2,\infty }-s_{1,\infty }\leqslant s_0(2+2\delta )$ , which completes the proof.

Proof. Since $\mathcal{R}_0^F(s_{1}(t_0), s_{2}(t_0))\gt \mathcal{R}_0^F(\!-\!s_0, s_0)\geqslant 1$ holds for any $t_0\gt 0$ , we can choose $t_0$ as initial time when $\mathcal{R}_0^F(\!-\!s_0, s_0)=1$ , so it is sufficient to consider the case $\mathcal{R}_0^F(\!-\!s_0, s_0)\gt 1$ . In such case, we know that there exist $\lambda _{1}\lt 0$ and $\kappa \gt 0$ such that $(\phi,\psi )\;:\!=\;(\kappa \psi ^*,\psi ^*)$ satisfies the problem (2.8) with $(s_{1}(t), s_{2}(t))$ replaced by $(\!-\!s_0, s_0)$ , where $\psi ^*$ is the principal eigenfunction of $-\Delta$ in $(\!-\!s_0,s_0)$ with null Dirichlet boundary condition.

Let $\varepsilon \gt 0$ be a sufficiently small constant and define

\begin{equation*} \underline {u}(t,x)=\varepsilon \kappa \psi ^{*}(x), \underline {v}(t,x)=\varepsilon \psi ^{*}(x) \quad \mbox {for } (t,x)\in [\max \{\tau _{1},\tau _{2}\},+\infty )\times [\!-\!s_0, s_0]. \end{equation*}

By direct calculations, we have

\begin{eqnarray*} \frac{\partial \underline{u}}{\partial t} -d_1\frac{\partial ^{2}\underline{u}}{\partial x^{2}} -h(\underline{v}(t-\tau _{1},x))+b_1\underline{u} &&=\varepsilon \psi ^{*}(x)(h^{\prime }(0)+\lambda _{1}\kappa )-h(\varepsilon \psi ^{*}(x))\\[5pt] &&=\varepsilon \psi ^{*}(x)(h^{\prime }(0)+\lambda _{1}\kappa )-\varepsilon \psi ^{*}(x)h^{\prime }(\xi _{1}(x)) \end{eqnarray*}

and

\begin{eqnarray*} \frac{\partial \underline{v}}{\partial t} -d_2\frac{\partial ^{2}\underline{v}}{\partial x^{2}} -g(\underline{u}(t-\tau _{2},x))+b_2\underline{v} &&=\varepsilon \psi ^{*}(x)(g^{\prime }(0)\kappa +\lambda _{1})-g(\varepsilon \kappa \psi ^{*}(x))\\[5pt] &&=\varepsilon \psi ^{*}(x)(g^{\prime }(0)\kappa +\lambda _{1})-\varepsilon \kappa \psi ^{*}(x)g^{\prime }(\xi _{2}(x)), \end{eqnarray*}

where $\xi _{1}(x)\in (0, \varepsilon \psi ^{*}(x))$ and $\xi _{2}(x)\in (0, \varepsilon \kappa \psi ^{*}(x))$ . We may choose $\varepsilon$ sufficiently small such that

\begin{equation*} \left \{\begin {array}{l} \dfrac {\partial \underline {u}}{\partial t} -d_1\dfrac {\partial ^{2}\underline {u}}{\partial x^{2}} -h(\underline{v}(t-\tau _{1},x))+b_1\underline {u}\leqslant 0, \quad t\gt \max \{\tau _{1},\tau _{2}\}, -s_0\lt x\lt s_0,\\[10pt] \dfrac {\partial \underline {v}}{\partial t} -d_2\dfrac {\partial ^{2}\underline {v}}{\partial x^{2}} -g(\underline{u}(t-\tau _{2},x))+b_2\underline {v}\leqslant 0, \quad t\gt \max \{\tau _{1},\tau _{2}\}, -s_0\lt x\lt s_0,\\[10pt] \underline {u}(\theta,x)\leqslant u(\theta,x), \quad \max \{\tau _{1},\tau _{2}\}-\tau _{2}\leqslant \theta \leqslant \max \{\tau _{1},\tau _{2}\}, -s_0\leqslant x\leqslant s_0,\\[5pt] \underline {v}(\theta,x)\leqslant v(\theta,x), \quad \max \{\tau _{1},\tau _{2}\}-\tau _{1}\leqslant \theta \leqslant \max \{\tau _{1},\tau _{2}\}, -s_0\leqslant x\leqslant s_0. \end {array}\right. \end{equation*}

Moreover, it is easy to check that

\begin{eqnarray*} &\underline{u}(t,x)=\underline{v}(t,x)=0, \quad t\geqslant \max \{\tau _{1},\tau _{2}\}, x\geqslant s_{0} \mbox{ or } x\leqslant -s_{0},\\[10pt] &0=s_{0}^{\prime }\leqslant -\mu \dfrac{\partial \underline{u}}{\partial x}(t, s_0),\quad t\gt \max \{\tau _{1},\tau _{2}\},\\[10pt] &0=-s_{0}^{\prime }\geqslant -\mu \dfrac{\partial \underline{u}}{\partial x}(t, -s_0),\quad t\gt \max \{\tau _{1},\tau _{2}\},\\[5pt] &[\!-\!s_0,s_0]\subseteq [s_1(\theta ), s_2(\theta )],\quad t\gt \max \{\tau _{1},\tau _{2}\}. \end{eqnarray*}

It follows from the comparison principle that $(u(t,x), v(t,x))\geqslant (\underline{u}(t,x), \underline{v}(t,x))$ for $(t,x)\in [\max \{\tau _{1},\tau _{2}\}, +\infty )\times [\!-\!s_0, s_0]$ , thus

\begin{equation*} \liminf _{t\rightarrow \infty }(\|u(t,\cdot )\|_{C([s_1(t), s_2(t)])} +\|v(t,\cdot )\|_{C([s_1(t), s_2(t)])})\geqslant \varepsilon (\kappa +1)\psi ^{*}(0)\gt 0. \end{equation*}

By Theorem 2.3 $(iii)$ , we have $s_{2,\infty }-s_{1,\infty }=+\infty$ .

Proof. Since $\mathcal{R}_0^F(\!-\!L,L)\rightarrow \mathcal{R}_0\gt 1$ as $L\rightarrow +\infty$ , there exits $L_0\gt 0$ such that $\mathcal{R}_0^F(\!-\!L_0, L_0)\gt 1$ . Note that

\begin{equation*} \frac {\partial u}{\partial t} -d_{1}\frac {\partial ^{2} u}{\partial x^{2}} \geqslant -b_{1}u\quad \mbox {for}\ (t,x)\in (0,+\infty )\times (s_1(t), s_2(t)). \end{equation*}

It follows from Lemma 4.3 in [Reference Tarboush, Lin and Zhang30] that there exists $\mu _{L_0}\gt 0$ such that for any $\mu \gt \mu _{L_0}$ , the corresponding solution $(u(t,x), v(t,x), s_1(t), s_2(t))$ of (1.1) satisfies

\begin{equation*} \limsup _{t\rightarrow +\infty }s_1(t)\lt -L_0,\quad \liminf _{t\rightarrow +\infty }s_2(t)\gt L_0. \end{equation*}

Since $-s_1(t)$ and $s_2(t)$ are strictly increasing in $t$ , there exists $T_0\gt 0$ such that $s_1(T_0)\lt -L_0$ and $s_2(T_0)\gt L_0$ . Hence, $\mathcal{R}_0^F(s_1(T_0), s_2(T_0))\gt \mathcal{R}_0^F(\!-\!L_0, L_0)\gt 1$ . By Theorem 2.5, we have $s_{2,\infty }-s_{1,\infty }=+\infty$ for any $\mu \gt \mu _{L_0}$ .

By choosing $\mu$ as varying parameter, we can obtain the following sharp criteria from Theorems 2.4–2.6. We shall use the notations $u^{\mu }$ , $v^{\mu }$ , $s_{i}^{\mu }$ , $s_{i,\infty }^{\mu }$ $(i=1,2)$ to emphasise the dependence of $u$ , $v$ , $s_{i}$ , $s_{i,\infty }$ on $\mu$ in the following theorem.

Proof. For $\mathcal{R}_0^F(\!-\!s_0, s_0)\geqslant 1$ , Theorem 2.5 implies $\mu ^{*}=0$ .

For $\mathcal{R}_0^F(\!-\!s_0, s_0)\lt 1\lt \mathcal{R}_0$ , we define

\begin{equation*} \Sigma =\{\mu \gt 0\;:\; s_{2,\infty }^{\mu }-s_{1,\infty }^{\mu }\lt +\infty \} \end{equation*}

and

\begin{equation*} \mu ^{*}=\sup \Sigma. \end{equation*}

By Theorems 2.4 and 2.6, we have $\Sigma \neq \emptyset$ and $0\lt \mu ^{*}\lt +\infty$ . It follows that $s_{2,\infty }^{\mu }-s_{1,\infty }^{\mu }=+\infty$ for $\mu \gt \mu ^*$ .

By Lemma 2.1, if $\mu _{1}\gt \mu _{2}$ and $s_{2,\infty }^{\mu _{1}}-s_{1,\infty }^{\mu _{1}}\lt +\infty$ , then $s_{2,\infty }^{\mu _{2}}-s_{1,\infty }^{\mu _{2}}\lt +\infty$ . Thus, to complete the proof it is sufficient to show $s_{2,\infty }^{\mu ^{*}}-s_{1,\infty }^{\mu ^{*}}\lt +\infty$ . Assume, by contradiction, that $s_{2,\infty }^{\mu ^{*}}-s_{1,\infty }^{\mu ^{*}}=+\infty$ , we have $\lim _{t\rightarrow +\infty }\mathcal{R}_0^F(s_{1}^{\mu ^{*}}(t), s_{2}^{\mu ^{*}}(t))=\mathcal{R}_0\gt 1$ . Then, there exists a sufficiently large $T^{*}\gt 0$ such that $\mathcal{R}_0^F(s_{1}^{\mu ^{*}}(T^{*}), s_{2}^{\mu ^{*}}(T^{*}))\gt 1$ . By the continuous dependence of $(u^{\mu }, v^{\mu }, s_1^{\mu }, s_2^{\mu })$ on $\mu$ , we can find a sufficiently small $\delta \gt 0$ such that $\mathcal{R}_0^F(s_{1}^{\mu }(T^{*}), s_{2}^{\mu }(T^{*}))\gt 1$ for all $\mu \in [\mu ^{*}-\delta, \mu ^{*}+\delta ]$ . Choosing $T^{*}$ as initial time, similarly as the proof of Theorem 2.5, we can deduce that $s_{2,\infty }^{\mu }-s_{1,\infty }^{\mu }=+\infty$ for all $\mu \in [\mu ^{*}-\delta, \mu ^{*}+\delta ]$ . This contradicts with the definition of $\mu ^{*}$ .

3.1 Semi-wave problem with time delays

Consider the following nonlinear semi-wave problem with time delays

(3.1) \begin{equation} \left \{\begin{array}{l} c\phi ^{\prime }(s)-d_1\phi ^{\prime \prime }(s)=h(\psi (s-c\tau _1))-b_1\phi (s),\quad s\gt 0,\\[5pt] c\psi ^{\prime }(s)-d_2\psi ^{\prime \prime }(s)=g(\phi (s-c\tau _2))-b_2\psi (s),\quad s\gt 0,\\[5pt] (\phi (s), \psi (s))=(0, 0),\quad s\leqslant 0,\\[5pt] (\phi (+\infty ), \psi (+\infty ))=(u^*, v^*), \end{array}\right. \end{equation}

where $c\gt 0$ and $\tau _1,\tau _2\gt 0$ . The condition $\mathcal{R}_0\gt 1$ is always assumed in this section, which ensures that (3.1) admits a positive equilibrium $(u^*, v^*)$ . In this subsection, we aim to establish the existence, uniqueness and properties of monotone increasing solutions to (3.1).

For the existence of monotone increasing solutions to (3.1), we first characterise the distribution of roots of the following transcendental equation containing a polynomial of degree four:

(3.2) \begin{equation} \Delta ^c(\lambda, \tau ) \;:\!=\;p_1^{c}(\lambda )p_2^{c}(\lambda )-h^{\prime }(0)g^{\prime }(0)e^{-\lambda c \tau } =0 \end{equation}

with

\begin{equation*} p_1^{c}(\lambda )=d_1\lambda ^2-c\lambda -b_1,\quad p_2^{c}(\lambda )=d_2\lambda ^2-c\lambda -b_2,\quad \tau =\tau _1+\tau _2. \end{equation*}

Obviously, for $i=1,2$ , $p_i^{c}(\lambda )=0$ has two real roots

\begin{equation*} \lambda _i^{\pm }=\frac {c\pm \sqrt {c^2+4d_ib_i}}{2d_i}. \end{equation*}

Denote

\begin{equation*} \lambda _{m}^{c}=\min \{\lambda _{1}^{+},\lambda _{2}^{+}\},\quad \lambda _{M}^{c}=\max \{\lambda _{1}^{+},\lambda _{2}^{+}\} \end{equation*}

and define

\begin{equation*} c_{0}^*=\inf \Big \{c_0\gt 0 \Big | \mbox { all roots of } \Delta ^c(\lambda, 0)=0 \mbox { are real for } c\geqslant c_0\Big \}. \end{equation*}

Next we establish the distribution of roots of $\Delta ^c(\lambda, \tau )=0$ according to the value of $c$ in the following lemma.

Proof. $(i)$ For any given $c\gt 0$ , if $\lambda \in [\lambda _{m}^{c}, \lambda _{M}^{c})$ then we have $p_1^{c}(\lambda )p_2^{c}(\lambda )\leqslant 0$ . It follows that

\begin{equation*} \Delta ^c(\lambda, \tau )=p_1^{c}(\lambda )p_2^{c}(\lambda )-h^{\prime }(0)g^{\prime }(0)e^{-\lambda c\tau }\lt 0 \quad \mbox {for } \lambda \in [\lambda _{m}^{c}, \lambda _{M}^{c}). \end{equation*}

If $\lambda \in [\lambda _{M}^{c},+\infty )$ , then $p_1^{c}(\lambda )p_2^{c}(\lambda )$ is strictly increasing in $\lambda$ , which implies that $\Delta ^c(\lambda, \tau ) =p_1^{c}(\lambda )p_2^{c}(\lambda )-h^{\prime }(0)g^{\prime }(0)e^{-\lambda c\tau }$ is also strictly increasing with respect to $\lambda$ in $[\lambda _{M}^{c},+\infty )$ . Note that

\begin{equation*} \Delta ^c(\lambda _{M}^{c}, \tau ) =p_1^{c}(\lambda _{M}^{c})p_2^{c}(\lambda _{M}^{c}) -h^{\prime }(0)g^{\prime }(0)e^{-\lambda _{M}^{c} c\tau } =-h^{\prime }(0)g^{\prime }(0)e^{-\lambda _{M}^{c} c\tau }\lt 0 \end{equation*}

and

\begin{equation*} \Delta ^c(+\infty, \tau )=+\infty. \end{equation*}

Then, $\Delta ^c(\lambda, \tau )=0$ has only one positive root $\tilde{\lambda }_{\tau }^{c}$ in $[\lambda _{m}^{c}, +\infty )$ and $\tilde{\lambda }_{\tau }^{c}\gt \lambda _{M}^{c}$ . Moreover, for any fixed $c\gt 0$ and $\tau \gt 0$ ,

\begin{align*} \frac{d}{d\lambda }\Delta ^c(\lambda, \tau )\Big |_{\lambda =\tilde{\lambda }_{\tau }^{c}} &=(2d_{1}\tilde{\lambda }_{\tau }^{c}-c)\Big [d_{2}(\tilde{\lambda }_{\tau }^{c})^{2}-c\tilde{\lambda }_{\tau }^{c}-b_{2}\Big ] +(2d_{2}\tilde{\lambda }_{\tau }^{c}-c)\Big [d_{1}(\tilde{\lambda }_{\tau }^{c})^{2}-c\tilde{\lambda }_{\tau }^{c}-b_{1}\Big ] \\[5pt] &\quad +2c\tau h^{\prime }(0)g^{\prime }(0)e^{-\tilde{\lambda }_{\tau }^{c} c\tau }\\[5pt] &\gt 0. \end{align*}

$(ii)$ By the proof of Lemma 2.1 in [Reference Wu and Hsu38], $\Delta ^c(\lambda, \tau )=0$ has at most three distinct positive roots in $(0, +\infty )$ for any given $c\gt 0$ and $\tau \gt 0$ . Thus, from $(i)$ we know that $\Delta ^c(\lambda, \tau )=0$ has at most two distinct positive roots in $(0, \lambda _{m}^{c})$ . Moreover,

\begin{equation*} \Delta ^c\left(\frac {1}{\sqrt {c}},\tau \right) =\Big (\frac {d_1}{c}-\sqrt {c}-b_1\Big )\Big (\frac {d_2}{c}-\sqrt {c}-b_2\Big ) -h^{\prime }(0)g^{\prime }(0)e^{-\sqrt {c}\tau }\quad \mbox {and}\quad \lim _{c\rightarrow +\infty }\Delta ^c\left(\frac {1}{\sqrt {c}},\tau \right)=+\infty, \end{equation*}

which imply that $\Delta ^c(\frac{1}{\sqrt{c}},\tau )\gt 0$ and $0\lt \frac{1}{\sqrt{c}}\lt \lambda _{m}^{c}$ for large $c$ . Note that

\begin{equation*} \Delta ^c(0,\tau )=b_{1}b_{2}-h^{\prime }(0)g^{\prime }(0)\lt 0 \quad \mbox {and}\quad \Delta ^c(\lambda _{m}^{c},\tau ) =-h^{\prime }(0)g^{\prime }(0)e^{-\lambda _{m}^{c} c\tau }\lt 0. \end{equation*}

Therefore, for all large $c$ , $\Delta ^c(\lambda, \tau )=0$ has exactly two distinct positive roots in $(0, \lambda _{m}^{c})$ and then has three distinct positive roots in $(0, +\infty )$ . It follows that the set

\begin{equation*} S(\tau )\;:\!=\;\Big \{\mathfrak {C}\gt 0 \big | \Delta ^c(\lambda, \tau ) \mbox { has three distinct positive zeros in } (0,+\infty ) \mbox { for all } c\geqslant \mathfrak {C}\Big \} \end{equation*}

is not empty. Thus,

\begin{equation*} c_{\tau }^{*}\;:\!=\;\inf S(\tau )\geqslant 0 \end{equation*}

is well-defined, and we know that $\Delta ^c(\lambda, \tau )=0$ has three distinct positive roots in $(0, +\infty )$ for $c\gt c_{\tau }^{*}$ . From $(i)$ , we deduce that $\Delta ^c(\lambda, \tau )=0$ has two distinct positive roots in $(0, \lambda _{m}^{c})$ for $c\gt c_{\tau }^{*}$ . We denote by $\check{\lambda }_{\tau }^{c}$ , $\hat{\lambda }_{\tau }^{c}$ the two positive roots and assume $\check{\lambda }_{\tau }^{c}\lt \hat{\lambda }_{\tau }^{c}$ , then $\Delta ^c(\lambda, \tau )\gt 0$ for $\lambda \in (\check{\lambda }_{\tau }^{c}, \hat{\lambda }_{\tau }^{c})$ , and $\Delta ^c(\lambda, \tau )\lt 0$ for $\lambda \in (0, \check{\lambda }_{\tau }^{c})\cup (\hat{\lambda }_{\tau }^{c}, \lambda _{m}^{c})$ . This completes the proof of $(ii)$ - $(a)$ .

We claim that $c_{\tau }^{*}\gt 0$ . Otherwise, for any sequence $\{c_{n}\}$ satisfying $0\lt c_{n}\lt 1$ and $\lim _{n\rightarrow \infty }c_{n}=0$ , $\Delta ^{c_n}(\lambda, \tau )=0$ has two distinct positive roots $\check{\lambda }_{\tau }^{c_n}$ , $\hat{\lambda }_{\tau }^{c_n}$ in $(0, \lambda _{m}^{c_n})$ and one positive root $\tilde{\lambda }_{\tau }^{c_n}$ in $(\lambda _{M}^{c_n},+\infty )$ . Since

\begin{eqnarray*} 0\lt \lambda _{m}^{c_n} &&=\min \Big \{\frac{c_{n}+\sqrt{c_{n}^{2}+4d_1b_1}}{2d_1},\frac{c_{n}+\sqrt{c_{n}^{2}+4d_2b_2}}{2d_2}\Big \}\\[5pt] &&\lt \min \Big \{\frac{1+\sqrt{1+4d_1b_1}}{2d_1},\frac{1+\sqrt{1+4d_2b_2}}{2d_2}\Big \} \end{eqnarray*}

and $\tilde{\lambda }_{\tau }^{c_n}$ satisfies

\begin{equation*} \Big [d_{1}(\tilde {\lambda }_{\tau }^{c_n})^{2}-c_{n}\tilde {\lambda }_{\tau }^{c_n}-b_{1}\Big ] \Big [d_{2}(\tilde {\lambda }_{\tau }^{c_n})^{2}-c_{n}\tilde {\lambda }_{\tau }^{c_n}-b_{2}\Big ] \leqslant h^{\prime }(0)g^{\prime }(0), \end{equation*}

the above sequences $\{\check{\lambda }_{\tau }^{c_n}\}$ , $\{\hat{\lambda }_{\tau }^{c_n}\}, \{\tilde{\lambda }_{\tau }^{c_n}\}$ are uniformly bounded with respect to $n$ . By extracting convergence subsequences and taking $n\rightarrow \infty$ , we can deduce $(\check{\lambda }_{\tau }^{c_n}, \hat{\lambda }_{\tau }^{c_n}, \tilde{\lambda }_{\tau }^{c_n}) \rightarrow (\check{\lambda }_{\tau }^{0}, \hat{\lambda }_{\tau }^{0},\tilde{\lambda }_{\tau }^{0})$ with some $\check{\lambda }_{\tau }^{0}, \hat{\lambda }_{\tau }^{0}, \tilde{\lambda }_{\tau }^{0}\geqslant 0$ . Note that for any fixed $\tau \gt 0$ , $\Delta ^{c_n}(\lambda, \tau )\rightarrow \Delta ^{0}(\lambda, \tau )$ in $C^{2}_{loc}(\mathbb{R})$ as $n\rightarrow \infty$ . It follows that $\check{\lambda }_{\tau }^{0}, \hat{\lambda }_{\tau }^{0}, \tilde{\lambda }_{\tau }^{0}$ are nonnegative roots of $\Delta ^{0}(\lambda, \tau )=0$ . Assume that one of three roots is $0$ , a contradiction occurs since $\Delta ^{0}(0, \tau )\lt 0$ , and the claim is proved. If all the three roots are positive, then there are two possible cases: $(1)$ at least two of these three roots are not equal; $(2)$ all three roots are equal. On the other hand, we can directly check that $\Delta ^{0}(\lambda, \tau )=0$ has only one positive real root, denoted by $\lambda _{\tau }^{0}$ , and

(3.3) \begin{equation} \frac{d}{d\lambda }\Delta ^{0}(\lambda, \tau )|_{\lambda =\lambda _{\tau }^{0}}\gt 0. \end{equation}

Thus, the first case cannot happen. Next, we consider the second case. Since $\check{\lambda }_{\tau }^{c_n}$ , $\hat{\lambda }_{\tau }^{c_n}$ are two roots of $\Delta ^{c_n}(\lambda, \tau )=0$ , by the mean value theorem, there exists a $\xi _{n}$ between $\check{\lambda }_{\tau }^{c_n}$ and $\hat{\lambda }_{\tau }^{c_n}$ such that $\frac{d}{d\lambda }\Delta ^{c_{n}}(\lambda, \tau )|_{\lambda =\xi _{n}}=0$ . Since $\lim _{n\rightarrow \infty }\check{\lambda }_{\tau }^{c_n}=\lim _{n\rightarrow \infty }\hat{\lambda }_{\tau }^{c_n} =\lambda _{\tau }^{0}$ , we also have $\lim _{n\rightarrow \infty }\xi _{n}=\lambda _{\tau }^{0}$ . By taking $n\rightarrow \infty$ in $\frac{d}{d\lambda }\Delta ^{c_{n}}(\lambda, \tau )|_{\lambda =\xi _{n}}=0$ , we get $\frac{d}{d\lambda }\Delta ^{0}(\lambda, \tau )|_{\lambda =\lambda _{\tau }^{0}}=0$ , which contradicts with (3.3). Then, the claim holds true.

Now we prove the first part of $(ii)$ - $(b)$ . Assume by contradiction that $\Delta ^{c_1}(\lambda, \tau )$ has at least one positive root in $(0, \lambda _{m}^{c_{1}})$ for some $0\lt c_{1}\lt c_{\tau }^{*}$ , we will show that

(3.4) \begin{equation} c_{1}\geqslant \inf S(\tau )=c_{\tau }^{*}, \end{equation}

which leads to a contradiction. To prove (3.4), it is sufficient to show that $\Delta ^{c_2}(\lambda, \tau )$ has two distinct positive roots in $(0, \lambda _{m}^{c_2})$ for any $c_2\gt c_1$ . In fact, since $\lambda _{m}^{c_{2}}\gt \lambda _{m}^{c_{1}}$ , we know that

\begin{equation*} p_{1}^{c_1}(\lambda ), p_{2}^{c_1}(\lambda ), p_{1}^{c_2}(\lambda ), p_{2}^{c_2}(\lambda )\lt 0 \quad \mbox {for } \lambda \in (0,\lambda _{m}^{c_{1}}). \end{equation*}

Note that $p_{1}^{c}(\lambda ), p_{2}^{c}(\lambda )$ are decreasing in $c$ . We conclude that $p_{1}^{c}(\lambda ) p_{2}^{c}(\lambda )$ and then $\Delta ^{c}(\lambda,\tau )$ are increasing with respect to $c$ for $\lambda \in (0,\lambda _{m}^{c_{1}})$ . Denote $\tilde{\lambda }_{1}$ by one positive root of $\Delta ^{c_1}(\lambda, \tau )=0$ in $(0, \lambda _{m}^{c_{1}})$ , that is, $\Delta ^{c_1}(\tilde{\lambda }_{1}, \tau )=0$ . Then,

\begin{equation*} \Delta ^{c_2}(\tilde {\lambda }_{1}, \tau )\gt \Delta ^{c_1}(\tilde {\lambda }_{1}, \tau )=0. \end{equation*}

It follows that $\Delta ^{c_2}(\lambda, \tau )=0$ has exactly two distinct positive root in $(0, \lambda _{m}^{c_2})$ for any $c_2\gt c_1$ . Hence, (3.4) holds and we can get the desired result. Clearly, $\Delta ^c(\lambda, \tau )\lt 0$ in $(0, \lambda _{m}^{c})$ for any $0\lt c\lt c_{\tau }^{*}$ .

Based on the above discussions, it is easy to know that $\Delta ^c(\lambda, \tau )=0$ has only one positive root in $(0, \lambda _{m}^{c})$ for $c=c_{\tau }^{*}$ , i.e. $(ii)$ - $(c)$ holds. We claim that $c_{\tau }^{*}\lt c_{0}^{*}$ . Indeed, by Lemma 3.3 $(iii)$ – $(iv)$ in [Reference Wang, Nie and Du35] and the fact $\Delta ^c(\lambda, \tau )\gt \Delta ^c(\lambda, 0)$ , we know that $\Delta ^c(\lambda, \tau )=0$ has three distinct positive roots in $(0, +\infty )$ for all $c\geqslant c_{0}^{*}$ . Thus, $c_{\tau }^{*}\lt c_{0}^{*}$ .

Next, we prove the second part of $(ii)$ - $(b)$ . We will prove the result by taking $\tau$ as a parameter which continuously increases from $0$ to $+\infty$ . Since the range $(0, c_{\tau }^{*})$ of $c$ is dependent on $\tau$ , we cannot simply fix a $c$ as $\tau$ varies. However, we find that $c_{\tau }^{*}$ is decreasing with respect to $\tau$ . In fact, we have showed that $\Delta ^c(\lambda, \tau )\lt 0$ in $(0,\lambda _{m}^{c})$ for $0\lt c\lt c_{\tau }^{*}$ . If $\tilde{\tau }\gt \tau$ , then we can check that

\begin{equation*} \Delta ^c(\lambda, \tilde {\tau })\gt \Delta ^c(\lambda, \tau ). \end{equation*}

It follows that $\Delta ^c(\lambda, \tau )\lt \Delta ^c(\lambda, \tilde{\tau })\leqslant 0$ in $(0,\lambda _{m}^{c})$ for $0\lt c\leqslant c_{\tilde{\tau }}^{*}$ , which implies that

(3.5) \begin{equation} c_{\tilde{\tau }}^{*}\lt c_{\tau }^{*}. \end{equation}

Otherwise, $\Delta ^c(\lambda, \tau )\lt 0$ in $(0,\lambda _{m}^{c})$ for $c=c_{\tau }^{*}$ , which contradicts with $(ii)$ - $(c)$ . Therefore, for any fixed $T\gt 0$ , we have $c_{T}^{*}\lt c_{\tau }^{*}$ for all $\tau \in (0,T)$ . In what follows, by choosing any fixed $c\in (0,c_{T}^{*})$ and varying $\tau$ continuously from $0$ to $T$ , we prove the existence of complex roots with positive real parts.

Note that the zeros of $\Delta ^c(\lambda, \tau )$ are continuous in $\tau \in (0, T)$ for fixed $c\in (0, c_{T}^{*})$ . Define $\lambda =\alpha (\tau )+i\beta (\tau )$ , where $\alpha (\tau )$ and $\beta (\tau )$ are continuous in $\tau \in (0, T)$ . Separating the real and imaginary parts of $\Delta ^c(\lambda,\tau )=0$ , we derive

(3.6) \begin{equation} \left \{\begin{array}{rl} F_1(\alpha,\beta,\tau ) &=\Big [d_1(\alpha ^2-\beta ^2)-c\alpha -b_1\Big ]\Big [d_2(\alpha ^2-\beta ^2)-c\alpha -b_2\Big ]\\[5pt] &\quad -(2d_1\alpha \beta -c\beta )(2d_2\alpha \beta -c\beta ) -h^{\prime }(0)g^{\prime }(0)e^{-c\tau \alpha }\cos c\tau \beta =0,\\[5pt] F_2(\alpha,\beta,\tau ) &=(2d_1\alpha \beta -c\beta )\Big [d_2(\alpha ^2-\beta ^2)-c\alpha -b_2\Big ]\\[5pt] &\quad +(2d_2\alpha \beta -c\beta )\Big [d_1(\alpha ^2-\beta ^2)-c\alpha -b_1\Big ] +h^{\prime }(0)g^{\prime }(0)e^{-c\tau \alpha }\sin c\tau \beta =0. \end{array}\right. \end{equation}

We divide the rest of proof into four steps.

Step 1. There is a complex root in $D$ provided that $\tau$ is small enough.

For $\tau =0, \Delta ^c(\lambda,0)=p_1^{c}(\lambda )p_2^{c}(\lambda )-h^{\prime }(0)g^{\prime }(0)$ , then it follows from Lemma 3.3 in [Reference Wang, Nie and Du35] that $\Delta ^c(\lambda,0)=0$ has a pair of conjugate complex roots for $c\in (0, c_{T}^{*})\subset (0, c_{0}^*)$ . We denote one of the complex roots in $D$ by $\lambda =\alpha +i\beta$ . By direct calculations,

\begin{equation*} \begin {array}{rl} &\det \left ( \begin {array}{c@{\quad}c} \partial _{\alpha }F_1 & \partial _{\beta } F_1 \\[10pt] \partial _{\alpha }F_2 & \partial _{\beta } F_2 \\[5pt] \end {array} \right )\Big |_{\tau =0}\\[5pt] &\quad=\Big \{(2d_1\alpha -c)\Big [d_2(\alpha ^2-\beta ^2)-c\alpha -b_2\Big ] +(2d_2\alpha -c)\Big [d_1(\alpha ^2-\beta ^2)-c\alpha -b_1\Big ] \\[5pt] &\qquad -\,2d_1\beta ^2(2d_2\alpha -c) -2d_2\beta ^2(2d_1\alpha -c)\Big \}^2 \\[5pt] &\quad\quad +\,\Big \{2d_1\beta \Big [d_2(\alpha ^2-\beta ^2)-c\alpha -b_2\Big ] +2d_2\beta \Big [d_1(\alpha ^2-\beta ^2)-c\alpha -b_1\Big ]\\[5pt] &\quad\quad +\,(2d_1\alpha \beta -c\beta )(2d_2\alpha -c) +(2d_2\alpha \beta -c\beta )(2d_1\alpha -c)\Big \}^2\geqslant 0. \end {array} \end{equation*}

Since $\tau =0, \lambda =\alpha +i\beta$ satisfies (3.6), it follows from the second equation of (3.6) that

\begin{equation*} (2d_1\alpha -c)\Big [d_2(\alpha ^2-\beta ^2)-c\alpha -b_2\Big ] +(2d_2\alpha -c)\Big [d_1(\alpha ^2-\beta ^2)-c\alpha -b_1\Big ]=0. \end{equation*}

Thus,

\begin{equation*} \begin {array}{rl} &\det \left ( \begin {array}{c@{\quad}c} \partial _{\alpha }F_1 & \partial _{\beta } F_1 \\[5pt] \partial _{\alpha }F_2 & \partial _{\beta } F_2 \\[5pt] \end {array} \right )\Big |_{\tau =0}\\[5pt] &\quad=4\beta ^4\Big [d_1(2d_2\alpha -c)+d_2(2d_1\alpha -c)\Big ]^2 +4\beta ^2\Big \{d_1\Big [d_2(\alpha ^2-\beta ^2)-c\alpha -b_2\Big ]\\[5pt] &\quad\quad +d_2\Big [d_1(\alpha ^2-\beta ^2)-c\alpha -b_1\Big ]+ (2d_1\alpha -c)(2d_2\alpha -c)\Big \}^2\geqslant 0, \end {array} \end{equation*}

where the equality holds if and only if

(3.7) \begin{equation} \left \{\begin{array}{l} d_1(2d_2\alpha -c)+d_2(2d_1\alpha -c)=0,\\[5pt] d_1\Big [d_2(\alpha ^2-\beta ^2)-c\alpha -b_2\Big ]+d_2\Big [d_1(\alpha ^2-\beta ^2)-c\alpha -b_1\Big ]+ (2d_1\alpha -c)(2d_2\alpha -c)=0. \end{array}\right. \end{equation}

In view of the first equation of (3.7), we have $\alpha =\frac{cd_1+cd_2}{4d_1d_2}\gt 0$ . Substituting this $\alpha$ into the second equation of (3.7) gives

\begin{equation*} \begin {array}{l} 6d_1d_2\alpha ^2+d_1(\!-\!d_2\beta ^2-b_2)+d_2(\!-\!d_1\beta ^2-b_1)-3cd_1\alpha -3cd_2\alpha \\[5pt] \quad=\dfrac {3}{2}\alpha (cd_1+cd_2)+d_1(\!-\!d_2\beta ^2-b_2)+d_2(\!-\!d_1\beta ^2-b_1) -3cd_1\alpha -3cd_2\alpha \\[5pt] \quad =-\dfrac {3}{2}cd_1\alpha -\dfrac {3}{2}cd_2\alpha +d_1(\!-\!d_2\beta ^2-b_2)+d_2(\!-\!d_1\beta ^2-b_1)=0, \end {array} \end{equation*}

which is not solvable for $\beta$ . Therefore, we get

\begin{equation*} \begin {array}{l} \det \left ( \begin {array}{c@{\quad}c} \partial _{\alpha }F_1 & \partial _{\beta } F_1 \\[5pt] \partial _{\alpha }F_2 & \partial _{\beta } F_2 \\[5pt] \end {array} \right )\Big |_{\tau =0}\gt 0. \end {array} \end{equation*}

Then, the implicit function theorem indicates that, for small $\tau \gt 0$ , $\Delta ^c(\lambda, \tau )=0$ with $c\in (0, c_{T}^{*})$ admits a pair of complex solutions near $\alpha +i\beta$ , and thus in the open domain $D$ .

Step 2. If $\lambda =\alpha (\tau )+i\beta (\tau )$ touches the pure imaginary axis at some $\tau =\tau _0\in (0,T)$ , i.e. $\alpha (\tau _0)=0$ , then $\alpha '(\tau _0)\gt 0$ . Moreover, $\beta _0\;:\!=\;\beta (\tau _0)\gt 0$ .

Since $\alpha (\tau _0)$ and $\beta (\tau _0)$ satisfy (3.6), we get

(3.8) \begin{equation} \left \{\begin{array}{l} (d_1\beta _0^2+b_1)(d_2\beta _0^2+b_2)-c^2\beta _0^2 -h^{\prime }(0)g^{\prime }(0)\cos c\tau _{0}\beta _0=0,\\[5pt] c\beta _0(d_2\beta _0^2+b_2)+c\beta _0(d_1\beta _0^2+b_1) +h^{\prime }(0)g^{\prime }(0)\sin c\tau _{0}\beta _0=0. \end{array}\right. \end{equation}

Besides,

\begin{equation*} \begin {array}{rl} &\det \left ( \begin {array}{c@{\quad}c} \partial _{\alpha }F_1 & \partial _{\beta } F_1 \\[5pt] \partial _{\alpha }F_2 & \partial _{\beta } F_2 \\[5pt] \end {array} \right )\Big |_{\tau =\tau _0}\\[5pt] &\quad=\Big \{c(d_2\beta _0^2+b_2)+c(d_1\beta _0^2+b_1)+2d_1c\beta _0^2+2d_2c\beta _0^2+2c\tau h^{\prime }(0)g^{\prime }(0)\cos c\tau _{0}\beta _0\Big \}^2\\[5pt] &\quad\quad +\Big \{2d_1\beta _0(d_2\beta _0^2+b_2)+2d_2\beta _0(d_1\beta _0^2+b_1)-2c^2\beta _0+2c\tau h^{\prime }(0)g^{\prime }(0)\sin c\tau _{0}\beta _0\Big \}^2\geqslant 0, \end {array} \end{equation*}

where the equality holds if and only if

(3.9) \begin{equation} \left \{\begin{array}{l} 3d_1\beta _0^2+3d_2\beta _0^2+b_1+b_2 +2c\tau h^{\prime }(0)g^{\prime }(0)\cos c\tau _{0}\beta _0=0,\\[5pt] 2d_1\beta _0^2(d_2\beta _0^2+b_2)+2d_2\beta _0^2(d_1\beta _0^2+b_1)-2c^2\beta _0^2 +2c\beta \tau h^{\prime }(0)g^{\prime }(0)\sin c\tau _{0}\beta _0=0. \end{array}\right. \end{equation}

Substituting (3.8) into (3.9), we give

(3.10) \begin{equation} \left \{\begin{array}{l} 3d_1\beta _0^2+3d_2\beta _0^2+b_1+b_2-2\tau c^2\beta _0^2+2\tau (d_1\beta _0^2+b_1)(d_2\beta _0^2+b_2)=0,\\[5pt] 2d_1\beta _0^2(d_2\beta _0^2+b_2)+2d_2\beta _0^2(d_1\beta _0^2+b_1)-2c^2\beta _0^2-2\tau c^2\beta _0^2(d_1\beta _0^2+d_2\beta _0^2+b_1+b_2)=0. \end{array}\right. \end{equation}

Multiplying the first equation in (3.10) by $d_1\beta _0^2+d_2\beta _0^2+b_1+b_2$ , and then subtracting it from the second equation in (3.10), we get

\begin{equation*} \begin {array}{l} 2d_1d_2\beta _0^4+3d_2^2\beta _0^4+3d_1^2\beta _0^4+d_2b_1\beta _0^2+d_1b_2\beta _0^2+3d_1^2\beta _0^2 +3d_2^2\beta _0^2+2c^2\beta _0^2+b_1d_1\beta _0^2\\[5pt] \quad+b_2d_2\beta _0^2+(b_1+b_2)^2 +2\tau (d_1\beta _0^2+b_1)(d_2\beta _0^2+b_2)(d_1\beta _0^2+d_2\beta _0^2+b_1+b_2)=0, \end {array} \end{equation*}

which is not solvable for $\beta _0\gt 0$ . Thus,

\begin{equation*} \begin {array}{l} \det \left ( \begin {array}{c@{\quad}c} \partial _{\alpha }F_1 & \partial _{\beta } F_1 \\[5pt] \partial _{\alpha }F_2 & \partial _{\beta } F_2 \\[5pt] \end {array} \right )\Big |_{\tau =\tau _0}\gt 0. \end {array} \end{equation*}

Moreover,

\begin{equation*} \begin {array}{c} \left ( \begin {array}{c} \partial _{\tau }F_1 \\[5pt] \partial _{\tau }F_2 \\[5pt] \end {array} \right )\Big |_{\tau =\tau _0}=2h^{\prime }(0)g^{\prime }(0)c\beta _0\left ( \begin {array}{c} \sin c\tau _0\beta _0 \\[5pt] \cos c\tau _0\beta _0 \\[5pt] \end {array} \right ). \end {array} \end{equation*}

Consequently, the implicit function theorem indicates that

\begin{equation*} \left ( \begin {array}{c} \alpha '(\tau ) \\[5pt] \beta '(\tau ) \\[5pt] \end {array} \right )\Big |_{\tau =\tau _0}=-\left ( \begin {array}{c@{\quad}c} \partial _{\alpha }F_1 & \partial _{\beta } F_1 \\[5pt] \partial _{\alpha }F_2 & \partial _{\beta } F_2 \\[5pt] \end {array} \right )^{-1}\Big |_{\tau =\tau _0}\left ( \begin {array}{c} \partial _{\tau }F_1 \\[5pt] \partial _{\tau }F_2 \\[5pt] \end {array} \right )\Big |_{\tau =\tau _0}. \end{equation*}

Direct computation induces that

\begin{equation*} \begin {array}{l} \alpha '(\tau _0)=-\dfrac {(\partial _\beta F_2\partial _\tau F_1-\partial _\beta F_1\partial _\tau F_2)|_{\tau =\tau _0}} {\det \left ( \begin {array}{c@{\quad}c} \partial _{\alpha }F_1 & \partial _{\beta } F_1 \\[5pt] \partial _{\alpha }F_2 & \partial _{\beta } F_2 \\[5pt] \end {array} \right )\Big |_{\tau =\tau _0}}. \end {array} \end{equation*}

Besides, by (3.8), we have

\begin{equation*} \begin {array}{rl} &(\partial _\beta F_2\partial _\tau F_1-\partial _\beta F_1\partial _\tau F_2)|_{\tau =\tau _0}\\[5pt] &\quad=\Big [c(d_2\beta _0^2+b_2)+c(d_1\beta _0^2+b_1)+2d_1c\beta _0^2+2d_2c\beta _0^2+2c\tau h^{\prime }(0)g^{\prime }(0)\cos c\tau _{0}\beta _0\Big ]\\[5pt] &\quad \quad \times \Big (2h^{\prime }(0)g^{\prime }(0)c\beta _0\sin c\tau _{0}\beta _0\Big )\\[5pt] &\quad \quad+\Big [2d_1\beta _0(\!-\!d_2\beta _0^2-b_2)+2d_2\beta _0(\!-\!d_1\beta _0^2-b_1)+2c^2\beta _0-2c\tau h^{\prime }(0)g^{\prime }(0)c\beta _0\sin c\tau _{0}\beta _0\Big ]\\[5pt] &\quad \qquad \times \Big (2h^{\prime }(0)g^{\prime }(0)c\beta _0\cos c\tau _{0}\beta _0\Big )\\[5pt] &\quad=\Big [c(d_2\beta _0^2+b_2)+c(d_1\beta _0^2+b_1)+2d_1c\beta _0^2+2d_2c\beta _0^2+ 2c\tau (d_1\beta _0^2+b_1)(d_2\beta _0^2+b_2)\\[5pt] &\quad \quad -\,2\tau c^3\beta _0^2\Big ] \times \Big [\!-\!2c^2\beta _0^2(d_2\beta _0^2+b_2)-2c^2\beta _0^2(d_1\beta _0^2+b_1)\Big ]\\[5pt] &\qquad +\, \Big [2d_1\beta _0(\!-\!d_2\beta _0^2-b_2)+2d_2\beta _0(\!-\!d_1\beta _0^2-b_1) +2c^2\beta _0+2\tau c^2\beta _0(d_2\beta _0^2+b_2)\\[5pt] &\qquad \quad +\,2\tau c^2\beta _0(d_1\beta _0^2+b_1)\Big ] \times \Big [2c\beta _0(d_1\beta _0^2+b_1)(d_2\beta _0^2+b_2)-2c^3\beta _0^3\Big ]\\[5pt] &\quad=-2c^3\beta _0^2\Big [(d_2\beta _0^2+b_2)+(d_1\beta _0^2+b_1)+2d_1\beta _0^2+2d_2\beta _0^2+ 2\tau (d_1\beta _0^2+b_1)(d_2\beta _0^2+b_2)\Big ]\\[5pt] &\quad \quad \times \Big [(d_2\beta _0^2+b_2)+(d_1\beta _0^2+b_1)\Big ] +4\tau c^5\beta _0^4\Big [(d_2\beta _0^2+b_2)+(d_1\beta _0^2+b_1)\Big ]\\[5pt] &\qquad -\,4c\beta _0^{2}\Big [d_1(d_2\beta _0^2+b_2)+d_2(d_1\beta _0^2+b_1)\Big ] \times \Big [(d_1\beta _0^2+b_1)(d_2\beta _0^2+b_2)\Big ]\\[5pt] &\qquad -\,4c^5\beta _0^4\Big [1+\tau (d_2\beta _0^2+b_2) +\tau (d_1\beta _0^2+b_1)\Big ] +4c^3\beta _0^4\Big [d_1(d_2\beta _0^2+b_2)+d_2(d_1\beta _0^2+b_1)\Big ]\\[5pt] &\qquad +\,4c^{3}\beta _0^{2}\Big [1+\tau (d_2\beta _0^2+b_2) +\tau (d_1\beta _0^2+b_1)\Big ]\times \Big [(d_1\beta _0^2+b_1)(d_2\beta _0^2+b_2)\Big ]\\[5pt] &\quad=-\Big [2c^3\beta _0^2(d_2\beta _0^2+b_2)^2+2c^3\beta _0^2(d_1\beta _0^2+b_1)^2 +4d_2c^3\beta _0^4(d_2\beta _0^2+b_2)\\[5pt] &\qquad \quad +\, 4d_1c^3\beta _0^4(d_1\beta _0^2+b_1) +4d_1c\beta _0^2(d_1\beta _0^2+b_1)(d_2\beta _0^2+b_2)^2\\[5pt] &\qquad \quad +\, 4d_2c\beta _0^2(d_1\beta _0^2+b_1)^2(d_2\beta _0^2+b_2)+4c^5\beta _0^4\Big ]\lt 0. \end {array} \end{equation*}

Due to $ \det \left ( \begin{array}{c@{\quad}c} \partial _{\alpha }F_1 & \partial _{\beta } F_1 \\[5pt] \partial _{\alpha }F_2 & \partial _{\beta } F_2 \\[5pt] \end{array} \right )\Big |_{\tau =\tau _0}\gt 0,$ hence $\alpha '(\tau _0)\gt 0$ .

Step 3. As $\tau$ varies from $0$ to $T$ , the complex root cannot touch the real axis at some $\tau _1\in (0,T)$ for $c\in (0, c_{T}^{*})$ .

If not, then the complex root changes into $\tilde{\lambda }\gt 0$ at $\tau =\tau _1$ . Since complex roots appear in conjugate pairs, the multiplicity of $\tilde{\lambda }$ is larger than two. However, due to $\Delta ^c(\lambda, \tau _1)\lt 0$ in $(0, \lambda _{m}^{c})$ for $c\in (0, c_{T}^{*})\subset (0,c_{\tau _1}^{*})$ , we know that $\tilde{\lambda }$ must satisfies $\tilde{\lambda }\in [\lambda _{m}^{c}, +\infty )$ . From $(i)$ , $\tilde{\lambda }$ is a simple root. This is a contradiction.

Step 4. Since $T\gt 0$ is arbitrary, by Steps 1–3 we can obtain that as $\tau$ continuously increases from $0$ to $+\infty$ , the complex root cannot escape the first quadrant of the complex plane. Thus, it always stays in $D$ . This completes the proof of this lemma.

Proof. Assume on the contrary that (3.1) has a monotone solution $(\phi, \psi )$ for some $c\geqslant c_{\tau }^{*}$ . Clearly, $(\phi (s), \psi (s))\leqslant (u^*, v^*)$ for $s\in \mathbb{R}$ . From the first equation in (3.1), we have

(3.11) \begin{eqnarray} \phi (s)= &&\frac{1}{d_1(\lambda _1^{+}-\lambda _1^{-})} \Big [\int _0^s(e^{\lambda _1^{-}(s-\xi )}- e^{\lambda _1^{-}s-\lambda _1^{+}\xi })h(\psi (\xi -c\tau _{1}))d\xi \nonumber \\[5pt] &&\quad \qquad \qquad \qquad +\int _s^{+\infty }(e^{\lambda _1^{+}(s-\xi )} -e^{\lambda _1^{-}s-\lambda _1^{+}\xi })h(\psi (\xi -c\tau _{1}))d\xi \Big ] \end{eqnarray}

for all $s\gt 0$ . By direct calculations,

\begin{eqnarray*} |\phi ^{\prime }(s)| &&\leqslant \frac{1}{d_1(\lambda _1^{+}-\lambda _1^{-})} \Big |\int _{0}^{s}\lambda _1^{-}(e^{\lambda _1^{-}(s-\xi )}- e^{\lambda _1^{-}s-\lambda _1^{+}\xi })h(\psi (\xi -c\tau _{1}))d\xi \\[5pt] &&\qquad \qquad \qquad \qquad +\int _{s}^{+\infty }(\lambda _1^{+}e^{\lambda _1^{+}(s-\xi )} -\lambda _1^{-}e^{\lambda _1^{-}s-\lambda _1^{+}\xi })h(\psi (\xi -c\tau _{1}))d\xi \Big |\\[5pt] &&\leqslant \frac{h(v^{*})}{d_{1}(\lambda _1^{+}-\lambda _1^{-})}(1-\frac{\lambda _1^{-}}{\lambda _1^{+}}) e^{\lambda _1^{-}s} \leqslant \frac{h(v^{*})}{d_{1}\lambda _1^{+}}=:C_1 \end{eqnarray*}

and

\begin{equation*} |\phi ^{\prime \prime }(s)|=\frac {|c\phi ^{\prime }-h(\psi (s-c\tau _{1}))+b_1\phi |}{d_1} \leqslant \frac {cC_1+h(v^*)+b_{1}u^*}{d_1}=:C_2. \end{equation*}

Similarly, there exist $C_3\gt 0$ and $C_4\gt 0$ such that $|\psi ^{\prime }(s)|\leqslant C_3$ and $|\psi ^{\prime \prime }(s)|\leqslant C_4$ . Denote the derivatives of $(\phi, \psi )$ by $(\phi ^{(i)}, \psi ^{(i)})$ for $i=1,2$ , then

\begin{eqnarray*} &\int _0^{\infty }\phi ^{(i)}(s)e^{-\lambda _0s}ds\lt \infty, \int _0^{\infty }\psi ^{(i)}(s)e^{-\lambda _0s}ds\lt \infty, i=0,1,2,\\[5pt] &\int _0^{\infty }\psi (s-c\tau _{1})e^{-\lambda _0s}ds\lt \infty, \int _0^{\infty }\phi (s-c\tau _{2})e^{-\lambda _0s}ds\lt \infty, \end{eqnarray*}

for each $\lambda \gt 0$ . From Lemma 3.1 $(ii)$ , $\Delta ^c(\lambda, \tau )=0$ has a positive root $\lambda _0\in (0, \lambda _{m}^{c})$ for $c\geqslant c_{\tau }^{*}$ (when $c\gt c_{\tau }^{*}$ we only choose one of positive roots). Next, multiplying the equations in (3.1) by $e^{-\lambda _0s}$ and integrating from $0$ to $+\infty$ , we obtain

\begin{eqnarray*} &0\lt d_1\phi ^{\prime }(0) =p_1^{c}(\lambda _0)\int _0^{+\infty }\phi (s)e^{-\lambda _0s}ds +\int _0^{+\infty }h(\psi (s-c\tau _{1}))e^{-\lambda _0s}ds,\\[5pt] &0\lt d_2\psi ^{\prime }(0) =p_2^{c}(\lambda _0)\int _0^{+\infty }\psi (s)e^{-\lambda _0s}ds +\int _0^{+\infty }g(\phi (s-c\tau _{2}))e^{-\lambda _0s}ds. \end{eqnarray*}

Since $h(z)\leqslant h^{\prime }(0)z$ and $g(z)\lt g^{\prime }(0)z$ for $z\gt 0$ by Taylor’s formula and the concavity of $h,g$ , we have

\begin{equation*} h(\psi (s-c\tau _{1}))\leqslant h^{\prime }(0)\psi (s-c\tau _{1}),\quad g(\phi (s-c\tau _{2}))\leqslant g^{\prime }(0)\phi (s-c\tau _{2}), \end{equation*}

which imply that

(3.12) \begin{eqnarray} &p_1^{c}(\lambda _0)\int _0^{+\infty }\phi (s)e^{-\lambda _0s}ds +h^{\prime }(0)\int _0^{+\infty }\psi (s-c\tau _{1})e^{-\lambda _0s}ds\gt 0,\nonumber \\[5pt] &p_2^{c}(\lambda _0)\int _0^{+\infty }\psi (s)e^{-\lambda _0s}ds +g^{\prime }(0)\int _0^{+\infty }\phi (s-c\tau _{2})e^{-\lambda _0s}ds\gt 0. \end{eqnarray}

Hence,

(3.13) \begin{eqnarray} 0 &&\lt p_1^{c}(\lambda _0)\int _0^{+\infty }\phi (s)e^{-\lambda _0s}ds +h^{\prime }(0)\int _0^{+\infty }\psi (s-c\tau _{1})e^{-\lambda _0s}ds \nonumber \\[5pt] &&=\frac{p_1^{c}(\lambda _0)e^{\lambda _{0}c\tau }}{g^{\prime }(0)} \Big [g^{\prime }(0)e^{-\lambda _{0}c\tau }\int _0^{+\infty }\phi (s)e^{-\lambda _0s}ds +\frac{h^{\prime }(0)g^{\prime }(0)e^{-\lambda _0c\tau }}{p_1^{c}(\lambda _0)} \int _0^{+\infty }\psi (s-c\tau _{1})e^{-\lambda _0s}ds\Big ] \nonumber \\[5pt] &&=\frac{p_1^{c}(\lambda _0)e^{\lambda _{0}c\tau }}{g^{\prime }(0)} \Big [g^{\prime }(0)e^{-\lambda _0c\tau _{1}}\int _{c\tau _{2}}^{+\infty }\phi (s^{\prime }-c\tau _{2})e^{-\lambda _0s^{\prime }}ds^{\prime } +p_2^{c}(\lambda _0)e^{-\lambda _0c\tau _{1}}\int _{-c\tau _{1}}^{+\infty }\psi (s^{\prime })e^{-\lambda _0s^{\prime }}ds^{\prime }\Big ] \nonumber \\[5pt] &&=\frac{p_1^{c}(\lambda _0)e^{\lambda _{0}c\tau _{2}}}{g^{\prime }(0)} \Big [g^{\prime }(0)\int _{0}^{+\infty }\phi (s^{\prime }-c\tau _{2})e^{-\lambda _0s^{\prime }}ds^{\prime } +p_2^{c}(\lambda _0)\int _{0}^{+\infty }\psi (s^{\prime })e^{-\lambda _0s^{\prime }}ds^{\prime }\Big ]. \end{eqnarray}

By (3.12) and (3.13), we have $p_1^{c}(\lambda _0)\gt 0$ . On the other hand, since $\lambda _0\in (0, \lambda _{m}^{c})$ , we can deduce that $p_1^{c}(\lambda _0)\lt 0$ , which is a contradiction.

Next, we establish the existence of monotone solutions to (3.1) for $0\lt c\lt c_{\tau }^{*}$ .

In [Reference Chen, Li, Teng and Wang10, Reference Sun and Fang28], to get the monotone increasing solutions of semi-wave problems with one time delay $\tau \gt 0$ , lower solutions were constructed by complex roots with imaginary part $\textrm{Im} \lambda \in (0, \frac{\pi }{c\tau })$ . However, we cannot derive a suitable upper bound of $\textrm{Im} \lambda$ for the transcendental equation (3.2) in the above Lemma 3.1 $(ii)$ - $b$ , which together with the influence of two time delays $(\tau _1,\tau _2)$ bring some difficulties in constructing lower solution of (3.1).

To get the monotone increasing solutions of (3.1), we first consider the corresponding perturbed problem

(3.14) \begin{equation} \left \{\begin{array}{l} c\phi ^{\prime }(s)-d_1\phi ^{\prime \prime }(s)=h(\psi (s-c\tau _1))-b_1\phi (s),\quad s\gt 0,\\[5pt] c\psi ^{\prime }(s)-d_2\psi ^{\prime \prime }(s)=g(\phi (s-c\tau _2))-b_2\psi (s),\quad s\gt 0,\\[5pt] (\phi (s), \psi (s))=(\delta u^*, \delta v^*),\quad s\leqslant 0,\\[5pt] (\phi (+\infty ), \psi (+\infty ))=(u^*, v^*), \end{array}\right. \end{equation}

where $\delta \in (0,\frac{1}{2})$ is a small parameter, and then get the desired solution of (3.1) by taking $\delta \rightarrow 0$ . The idea of approximation is motivated by the works on semi-wave problems with nonlocal diffusion (see [Reference Du, Li and Zhou14]).

In the following, we establish sufficient conditions for the existence of monotone solutions to (3.14) by employing the lower-upper solutions technique and the Schauder fixed point theorem, which was proposed in [Reference Ma25].

For a fixed

\begin{equation*} \sigma \in (0,\min _{i=1,2}\{-\lambda _{i}^{-},\lambda _{i}^{+}\}), \end{equation*}

we introduce the Banach space $(\mathbb{X}_{\sigma }(\mathbb{R}, \mathbb{R}^{2}), |\cdot |_{\sigma })$ with

\begin{equation*} \mathbb {X}_{\sigma }(\mathbb {R}, \mathbb {R}^{2}) \;:\!=\;\Big \{\Phi \in C(\mathbb {R}, \mathbb {R}^{2});\; |\Phi |_{\sigma }\;:\!=\;\sup _{s\in \mathbb {R}}|\Phi (s)|e^{-\sigma |s|}\lt \infty \Big \}. \end{equation*}

For $i=1,2$ , we define the mapping

\begin{eqnarray*} F_i\;:\; &\mathbb{X}_{\sigma }(\mathbb{R}, \mathbb{R}^{2})\rightarrow C(\mathbb{R})\\[5pt] &\Phi =(\phi,\psi )\mapsto F_i(\Phi )(s) \end{eqnarray*}

by

(3.15) \begin{equation} F_1(\Phi )(s) =\left \{\begin{array}{l} \delta u^{*}e^{\lambda _1^{-}s} +\dfrac{1}{d_1(\lambda _1^{+}-\lambda _1^{-})} \Big [\int _0^s(e^{\lambda _1^{-}(s-\xi )}- e^{\lambda _1^{-}s-\lambda _1^{+}\xi })h(\psi (\xi -c\tau _{1}))d\xi \\[5pt] \qquad \qquad \qquad \qquad +\int _s^{+\infty }(e^{\lambda _1^{+}(s-\xi )} -e^{\lambda _1^{-}s-\lambda _1^{+}\xi })h(\psi (\xi -c\tau _{1}))d\xi \Big ],\quad s\gt 0,\\[5pt] \delta u^*,\quad s\leqslant 0 \end{array}\right. \end{equation}

and

(3.16) \begin{equation} F_2(\Phi )(s) =\left \{\begin{array}{l} \delta v^{*}e^{\lambda _1^{-}s} +\dfrac{1}{d_2(\lambda _2^{+}-\lambda _2^{-})} \Big [\int _0^s(e^{\lambda _2^{-}(s-\xi )}- e^{\lambda _2^{-}s-\lambda _2^{+}\xi })g(\phi (\xi -c\tau _{2}))d\xi \\[5pt] \qquad \qquad \qquad \qquad +\int _s^{+\infty }(e^{\lambda _2^{+}(s-\xi )} -e^{\lambda _2^{-}s-\lambda _2^{+}\xi })g(\phi (\xi -c\tau _{2}))d\xi \Big ],\quad s\gt 0,\\[5pt] \delta v^*,\quad s\leqslant 0. \end{array}\right. \end{equation}

By direct calculations, we see that $(F_1, F_2)$ is well-defined and satisfies

\begin{eqnarray*} \left \{\begin{array}{l} c(F_1(\Phi ))^{\prime }(s)-d_1(F_1(\Phi ))^{\prime \prime }(s) =h(\psi (s-c\tau _1))-b_{1} F_1(\Phi )(s),\quad s\gt 0,\\[5pt] c(F_2(\Phi ))^{\prime }(s)-d_2(F_2(\Phi ))^{\prime \prime }(s) =g(\phi (s-c\tau _2))-b_{2} F_2(\Phi )(s),\quad s\gt 0,\\[5pt] (F_1(\Phi )(s), F_2(\Phi )(s))=(\delta u^*, \delta v^*),\quad s\leqslant 0. \end{array}\right. \end{eqnarray*}

Clearly, any fixed point of $(F_{1},F_{2})$ in $\mathbb{X}_{\sigma }(\mathbb{R}, \mathbb{R}^{2})$ is a solution of (3.14). Moreover, the mapping $(F_1, F_2)$ satisfies the following monotonicity properties.

Proof. Since $h,g$ are strictly increasing functions, we can directly get $(i)$ from (3.15) and (3.16). Now we prove $(ii)$ . It is sufficient to consider the case $s\gt 0$ . Differentiating (3.15) with respect to $s$ , we have

\begin{eqnarray*} (F_1(\Phi ))^{\prime }(s) =\delta \lambda _1^{-}u^{*}e^{\lambda _1^{-}s} +\frac{1}{d_1(\lambda _1^{+}-\lambda _1^{-})} &&\Big [\int _{0}^{s}\lambda _1^{-}(e^{\lambda _1^{-}(s-\xi )}- e^{\lambda _1^{-}s-\lambda _1^{+}\xi })h(\psi (\xi -c\tau _{1}))d\xi \\[5pt] &&\quad+\int _s^{+\infty }(\lambda _1^{+}e^{\lambda _1^{+}(s-\xi )} -\lambda _1^{-}e^{\lambda _1^{-}s-\lambda _1^{+}\xi })h(\psi (\xi -c\tau _{1}))d\xi \Big ]. \end{eqnarray*}

Since $h(\psi (s-c\tau _{1}))$ is monotone increasing in $s\in \mathbb{R}$ and $h(\psi (s-c\tau _{1}))\geqslant h(\delta v^{*})\gt 0$ , we have

\begin{eqnarray*} (F_1(\Phi ))^{\prime }(s) &&\geqslant \delta \lambda _1^{-}u^{*}e^{\lambda _1^{-}s} +\frac{1}{d_1(\lambda _1^{+}-\lambda _1^{-})} \Big [\int _{0}^{s}\lambda _1^{-}e^{\lambda _1^{-}s}(e^{-\lambda _1^{-}\xi }- e^{-\lambda _1^{+}\xi })h(\psi (s-c\tau _{1}))d\xi \\[5pt] &&\quad\qquad\qquad \qquad \qquad \qquad \qquad +\int _s^{+\infty }(\lambda _1^{+}e^{\lambda _1^{+}(s-\xi )} -\lambda _1^{-}e^{\lambda _1^{-}s-\lambda _1^{+}\xi })h(\psi (s-c\tau _{1}))d\xi \Big ]\\[5pt] &&=\delta \lambda _1^{-}u^{*}e^{\lambda _1^{-}s} +\frac{h(\psi (s-c\tau _{1}))}{d_1(\lambda _1^{+}-\lambda _1^{-})} \Big [\lambda _1^{-}e^{\lambda _1^{-}s} \int _{0}^{s}(e^{-\lambda _1^{-}\xi }- e^{-\lambda _1^{+}\xi })d\xi \\[5pt] &&\quad\qquad\qquad \qquad \qquad \qquad \qquad +\int _s^{+\infty }(\lambda _1^{+}e^{\lambda _1^{+}(s-\xi )} -\lambda _1^{-}e^{\lambda _1^{-}s-\lambda _1^{+}\xi })d\xi \Big ]\\[5pt] &&=\delta \lambda _1^{-}u^{*}e^{\lambda _1^{-}s} +\frac{h(\psi (s-c\tau _{1}))}{d_1\lambda _1^{+}} e^{\lambda _1^{-}s}\\[5pt] &&\geqslant \delta \lambda _1^{-}u^{*}e^{\lambda _1^{-}s} +\frac{h(\delta v^{*})}{d_1\lambda _1^{+}} e^{\lambda _1^{-}s}. \end{eqnarray*}

Since $\delta \in (0,1)$ , we have $h(\delta v^*)\geqslant \delta h(v^*)$ , which together with the fact $b_{1}u^{*}=h(v^{*})$ imply that

\begin{eqnarray*} (F_1(\Phi ))^{\prime }(s) &&\geqslant \delta \lambda _1^{-}u^{*}e^{\lambda _1^{-}s} +\delta \frac{h(v^{*})}{d_1\lambda _1^{+}} e^{\lambda _1^{-}s} =\delta \lambda _1^{-}u^{*}e^{\lambda _1^{-}s} +\delta \frac{b_{1}u^{*}}{d_1\lambda _1^{+}} e^{\lambda _1^{-}s}\\[5pt] &&=\delta u^{*}e^{\lambda _1^{-}s} (\lambda _1^{-}+\frac{b_{1}}{d_1\lambda _1^{+}}) =0. \end{eqnarray*}

This completes the proof.

Now we introduce the definition of lower and upper solutions to (3.14). If $(\underline{\phi },\underline{\psi })$ and $(\bar{\phi }, \bar{\psi })$ are continuous from $\mathbb{R}$ into $[\delta u^{*},u^{*}]\times [\delta v^{*}, v^{*}]$ , twice continuously differentiable on $\mathbb{R}_{+}\backslash \{\xi _{i}\}_{i=1}^{n}$ , and satisfy

\begin{eqnarray*} \left \{\begin{array}{l} c\underline{\phi }^{\prime }(s)-d_1\underline{\phi }^{\prime \prime }(s) \leqslant h(\underline{\psi }(s-c\tau _1))-b_1\underline{\phi }(s),\quad s\gt 0, s\notin \{\xi _{i}\}_{i=1}^{n},\\[5pt] c\underline{\psi }^{\prime }(s)-d_2\underline{\psi }^{\prime \prime }(s) \leqslant g(\underline{\phi }(s-c\tau _2))-b_2\underline{\psi }(s),\quad s\gt 0, s\notin \{\xi _{i}\}_{i=1}^{n},\\[5pt] \underline{\phi }^{\prime }_{+}(\xi _{i})\geqslant \underline{\phi }^{\prime }_{-}(\xi _{i}),\quad \underline{\psi }^{\prime }_{+}(\xi _{i})\geqslant \underline{\psi }^{\prime }_{-}(\xi _{i}),\quad 1\leqslant i\leqslant n,\\[5pt] (\underline{\phi }(s), \underline{\psi }(s))=(\delta u^*, \delta v^*),\quad s\leqslant 0,\\[5pt] (\underline{\phi }(+\infty ), \underline{\psi }(+\infty ))\leqslant (u^*, v^*) \end{array}\right. \end{eqnarray*}

and

\begin{eqnarray*} \left \{\begin{array}{l} c\bar{\phi }^{\prime }(s)-d_1\bar{\phi }^{\prime \prime }(s) \geqslant h(\bar{\psi }(s-c\tau _1))-b_1\bar{\phi }(s),\quad s\gt 0, s\notin \{\xi _{i}\}_{i=1}^{n},\\[5pt] c\bar{\psi }^{\prime }(s)-d_2\bar{\psi }^{\prime \prime }(s) \geqslant g(\bar{\phi }(s-c\tau _2))-b_2\bar{\psi }(s),\quad s\gt 0, s\notin \{\xi _{i}\}_{i=1}^{n},\\[5pt] \bar{\phi }^{\prime }_{+}(\xi _{i})\leqslant \bar{\phi }^{\prime }_{-}(\xi _{i}),\quad \bar{\psi }^{\prime }_{+}(\xi _{i})\leqslant \bar{\psi }^{\prime }_{-}(\xi _{i}),\quad 1\leqslant i\leqslant n,\\[5pt] (\bar{\phi }(s), \bar{\psi }(s))=(\delta u^*, \delta v^*),\quad s\leqslant 0,\\[5pt] (\bar{\phi }(+\infty ), \bar{\psi }(+\infty ))=(u^*, v^*), \end{array}\right. \end{eqnarray*}

then $(\underline{\phi },\underline{\psi })$ and $(\bar{\phi }, \bar{\psi })$ are called a lower solution and an upper solution of (3.14), respectively.

Given a pair of lower and upper solutions $(\underline{\phi },\underline{\psi })$ and $(\bar{\phi }, \bar{\psi })$ satisfying

\begin{equation*} \sup _{t\leqslant s}(\underline {\phi }(t), \underline {\psi }(t)) \leqslant (\bar {\phi }(s), \bar {\psi }(s)),\quad \forall s\in \mathbb {R}, \end{equation*}

then the set

\begin{eqnarray*} \Gamma &&\;:\!=\;\Big \{(\phi, \psi )\in \mathbb{X}_{\sigma }(\mathbb{R}, \mathbb{R}^{2})\;:\; (\underline{\phi }, \underline{\psi })\leqslant (\phi, \psi )\leqslant (\bar{\phi }, \bar{\psi }) \mbox{ and } (\phi, \psi ) \mbox{ is monotone increasing on } \mathbb{R}_{+}\Big \}\\[5pt] &&\neq \emptyset \end{eqnarray*}

since $(\phi (s), \psi (s))=\sup _{t\leqslant s}(\underline{\phi }(t), \underline{\psi }(t))\in \Gamma$ . Next, we prove that $(F_{1}, F_{2})$ maps $\Gamma$ into itself and is completely continuous.

Proof. We first prove $(F_{1}, F_{2})(\Gamma )\subset \Gamma$ . Let $\underline{\Phi }(s)=(\underline{\phi }(s),\underline{\psi }(s))$ and $\bar{\Phi }(s)=(\bar{\phi }(s), \bar{\psi }(s))$ . By Lemma 3.3, it is sufficient to prove that $\underline{\Phi }(s) \leqslant (F_{1}(\underline{\Phi })(s), F_{2}(\underline{\Phi })(s)) \leqslant (F_{1}(\bar{\Phi })(s), F_{2}(\bar{\Phi })(s)) \leqslant \bar{\Phi }(s)$ for all $s\in \mathbb{R}_{+}$ . Let $\xi _{0}=0$ and $\xi _{m+1}=\infty$ . Assume that $s\in (\xi _{i},\xi _{i+1})$ for some $i$ , then we have

\begin{eqnarray*} F_{1}(\underline{\Phi })(s)&&=\delta u^{*}e^{\lambda _1^{-}s} +\frac{1}{d_1(\lambda _1^{+}-\lambda _1^{-})} \Big [\int _0^s(e^{\lambda _1^{-}(s-\xi )}- e^{\lambda _1^{-}s-\lambda _1^{+}\xi })h(\underline{\psi }(\xi -c\tau _{1}))d\xi \\[5pt] &&\qquad\qquad \qquad \qquad \qquad \qquad +\int _s^{+\infty }(e^{\lambda _1^{+}(s-\xi )} -e^{\lambda _1^{-}s-\lambda _1^{+}\xi })h(\underline{\psi }(\xi -c\tau _{1}))d\xi \Big ]\\[5pt] &&\geqslant \delta u^{*}e^{\lambda _1^{-}s} +\frac{1}{d_1(\lambda _1^{+}-\lambda _1^{-})} \Big [\int _{0}^{s}(e^{\lambda _1^{-}(s-\xi )}- e^{\lambda _1^{-}s-\lambda _1^{+}\xi })(b_{1}\underline{\phi }(\xi ) -d_{1}\underline{\phi }^{\prime \prime }(\xi )+c\underline{\phi }^{\prime }(\xi ))d\xi \\[5pt] &&\qquad\qquad \qquad \qquad \qquad \qquad +\int _s^{+\infty }(e^{\lambda _1^{+}(s-\xi )} -e^{\lambda _1^{-}s-\lambda _1^{+}\xi })(b_{1}\underline{\phi }(\xi ) -d_{1}\underline{\phi }^{\prime \prime }(\xi )+c\underline{\phi }^{\prime }(\xi ))d\xi \Big ]\\[5pt] &&=\delta u^{*}e^{\lambda _1^{-}s}-e^{\lambda _1^{-}s}\underline{\phi }(0) +\underline{\phi }(s) +\frac{1}{(\lambda _1^{+}-\lambda _1^{-})} \Big [\sum _{k=1}^{i}(e^{\lambda _1^{-}(s-\xi _{k})}- e^{\lambda _1^{-}s-\lambda _1^{+}\xi _{k}})(\underline{\phi }_{+}^{\prime }(\xi _{k}) -\underline{\phi }_{-}^{\prime }(\xi _{k}))\\[5pt] &&\qquad\qquad \qquad \qquad \qquad \qquad +\sum _{k=i+1}^{m}(e^{\lambda _1^{+}(s-\xi _{k})} -e^{\lambda _1^{-}s-\lambda _1^{+}\xi _{k}})(\underline{\phi }_{+}^{\prime }(\xi _{k}) -\underline{\phi }_{-}^{\prime }(\xi _{k}))\Big ]\\[5pt] &&\geqslant \underline{\phi }(s). \end{eqnarray*}

Similarly, we can prove that $F_{2}(\underline{\Phi })(s)\geqslant \underline{\psi }(s)$ and $(F_{1}(\bar{\Phi })(s), F_{2}(\bar{\Phi })(s)) \leqslant (\bar{\phi }(s), \bar{\psi }(s))$ for all $s\in (\xi _{i},\xi _{i+1})$ . By the continuity, we can get the same results for the endpoints $\xi _{i}$ $(i=1,\cdots, m)$ . Thus, $(F_{1}, F_{2})(\Gamma )\subset \Gamma$ .

The continuity and compactness of $(F_{1}, F_{2})$ can be obtained by similar arguments in Lemmas 2.4–2.5 of [Reference Wang, Nie and Du35], here we omit the details.

Proof. From Lemma 3.4, $(F_1, F_2)$ has a fixed point $\Phi =(\phi,\psi )\in \Gamma$ by the Schauder’s fixed point theorem. Note that $(\bar{\phi }(s), \bar{\psi }(s))=(\underline{\phi }(s), \underline{\psi }(s))=(\delta u^*, \delta v^*)$ for $s\leqslant 0$ . We have $(\phi (s),\psi (s))=(\delta u^*, \delta v^*)$ for $s\leqslant 0$ . Since $(\phi,\psi )$ is monotone increasing and bounded, $\lim _{s\rightarrow +\infty }(\phi (s), \psi (s))$ is well-defined. By Lemma 2.3 in [Reference Wu and Zou37], we can deduce that $\lim _{s\rightarrow +\infty }(\phi (s),\psi (s))=(u^*,v^*)$ , which implies $\Phi =(\phi,\psi )\in \Gamma$ is a monotone increasing solution of (3.14).

Proof. By Lemma 3.5, it is sufficient to construct a pair of lower and upper solutions of (3.14).

Take

\begin{equation*} (\underline {\phi }(s), \underline {\psi }(s))=(\delta u^*, \delta v^*),\quad s\in \mathbb {R}. \end{equation*}

Since

\begin{equation*} h(\underline {\psi }(s-c\tau _1))=h(\delta v^*)\geqslant \delta h(v^*),\quad g(\underline {\phi }(s-c\tau _1))=g(\delta u^*)\geqslant \delta g(u^*), \end{equation*}

we can check that $(\underline{\phi }(s), \underline{\psi }(s))$ satisfies

\begin{equation*} \left \{\begin {array}{l} c\underline {\phi }^{\prime }(s)-d_1\underline {\phi }^{\prime \prime }(s)=0 \leqslant h(\underline {\psi }(s-c\tau _1))-b_1\underline {\phi }(s),\quad s\gt 0,\\[5pt] c\underline {\psi }^{\prime }(s)-d_2\underline {\psi }^{\prime \prime }(s)=0 \leqslant g(\underline {\phi }(s-c\tau _2))-b_2\underline {\psi }(s),\quad s\gt 0,\\[5pt] (\underline {\phi }(s), \underline {\psi }(s))=(\delta u^*, \delta v^*),\quad s\leqslant 0,\\[5pt] (\underline {\phi }(+\infty ), \underline {\psi }(+\infty ))\leqslant (u^*, v^*). \end {array}\right. \end{equation*}

Thus, $(\underline{\phi }(s), \underline{\psi }(s))$ is a lower solution of (3.14).

Now, we construct an upper solution of (3.14). Choose $k\gt 0$ large enough such that

\begin{equation*} 0\lt \frac {1}{k}\lt \min \Big \{c\tau _{1}, c\tau _{2}, \frac {c}{b_{1}}, \frac {c}{b_{2}}\Big \}. \end{equation*}

Define

\begin{equation*} \bar {\phi }(s)=\left \{\begin {array}{l} \delta u^{*},\quad s\leqslant 0,\\[1pt] u^{*}\Big [\!-\!k^2(1-\delta )\left(s-\dfrac {1}{k}\right)^2+1\Big ],\quad 0\lt s\leqslant \dfrac {1}{k},\\[3pt] u^{*}, \quad s\gt \dfrac {1}{k} \end {array}\right. \end{equation*}

and

\begin{eqnarray*} \bar{\psi }(s)=\left \{\begin{array}{l} \delta v^{*},\quad s\leqslant 0,\\[1pt] v^*\Big [\!-\!k^2(1-\delta )\left(s-\dfrac{1}{k}\right)^2+1\Big ],\quad 0\lt s\leqslant \dfrac{1}{k},\\[3pt] v^*,\quad s\gt \dfrac{1}{k}. \end{array}\right. \end{eqnarray*}

Obviously, $(\bar{\phi }(s), \bar{\psi }(s))$ satisfies $(\bar{\phi }(s), \bar{\psi }(s))=(\delta u^*, \delta v^*)$ for $s\leqslant 0$ and $(\bar{\phi }(+\infty ), \bar{\psi }(+\infty ))=(u^*, v^*)$ . We claim that $(\bar{\phi }, \bar{\psi })$ is an upper solution of (3.14). In fact, for $0\lt s\lt \frac{1}{k}$ , we have $\bar{\psi }(s-c\tau _{1})=\delta v^{*}$ and

\begin{eqnarray*} &&c\bar{\phi }^{\prime }(s)-d_1\bar{\phi }^{\prime \prime }(s)-h(\bar{\psi }(s-c\tau _{1}))+b_1\bar{\phi }(s)\\[1pt] &&\quad=u^{*}k^{2}(1-\delta )\Big [2d_{1}-2c\!\left(s-\frac{1}{k}\right)-b_{1}\!\left(s-\frac{1}{k}\right)^{2}\Big ] +b_{1}u^{*}-h(\delta v^{*})\\[3pt] &&\quad=u^{*}k^{2}(1-\delta )\Big [2d_{1}-2c\!\left(s-\frac{1}{k}\right)-b_{1}\!\left(s-\frac{1}{k}\right)^{2}\Big ] +h(v^{*})-h(\delta v^{*}). \end{eqnarray*}

Since $-\frac{c}{b_{1}}\lt -\frac{1}{k}\leqslant s-\frac{1}{k}\lt 0$ , we can deduce $2d_{1}-2c(s-\frac{1}{k})-b_{1}(s-\frac{1}{k})^{2}\gt 0$ and then

\begin{equation*} c\bar {\phi }^{\prime }(s)-d_1\bar {\phi }^{\prime \prime }(s)-h(\bar {\psi }(s-c\tau _{1}))+b_1\bar {\phi }(s) \gt h(v^{*})-h(\delta v^{*}) \geqslant 0. \end{equation*}

For $s\gt \frac{1}{k}$ , we have

\begin{eqnarray*} c\bar{\phi }^{\prime }(s)-d_1\bar{\phi }^{\prime \prime }(s)-h(\bar{\psi }(s-c\tau _{1}))+b_1\bar{\phi }(s) &&=-h(\bar{\psi }(s-c\tau _{1}))+b_{1}u^{*}\\[3pt] &&=-h(\bar{\psi }(s-c\tau _{1}))+h(v^{*}) \geqslant 0. \end{eqnarray*}

Thus,

\begin{equation*} c\bar {\phi }^{\prime }(s)-d_1\bar {\phi }^{\prime \prime }(s)-h(\bar {\psi }(s-c\tau _{1}))+b_1\bar {\phi }(s) \geqslant 0 \quad \mbox {for } s\in \mathbb {R}_{+}\setminus \left\{\frac {1}{k}\right\}. \end{equation*}

We can similarly obtain

\begin{equation*} c\bar {\psi }^{\prime }(s)-d_2\bar {\psi }^{\prime \prime }(s)-g(\bar {\phi }(s-c\tau _{2}))+b_2\bar {\psi }(s)\geqslant 0 \quad \mbox {for } s\in \mathbb {R}_{+}\setminus \left\{\frac {1}{k}\right\}. \end{equation*}

Moreover, $\bar{\phi }_{+}^{\prime }(\frac{1}{k})=\bar{\phi }_{-}^{\prime }(\frac{1}{k})=0$ and $\bar{\psi }_{+}^{\prime }(\frac{1}{k})=\bar{\psi }_{-}^{\prime }(\frac{1}{k})=0$ . Therefore, $(\bar{\phi }, \bar{\psi })$ is an upper solution of (3.14). It follows that the perturbed problem (3.14) admits a monotone increasing solution $(\phi ^{\delta }(s), \psi ^{\delta }(s))$ .

Next, we prove that if $\delta _{1},\delta _{2}\in (0,\frac{1}{2})$ and $\delta _{1}\gt \delta _{2}$ , then the solutions obtained in the above way satisfy $(\phi ^{\delta _{1}}(s), \psi ^{\delta _{1}}(s))\gt (\phi ^{\delta _{2}}(s), \psi ^{\delta _{2}}(s))$ for all $s\in \mathbb{R}$ . Let $\tilde{\phi }(s)=\phi ^{\delta _{1}}(s)-\phi ^{\delta _{2}}(s)$ and $\tilde{\psi }(s)=\psi ^{\delta _{1}}(s)-\psi ^{\delta _{2}}(s)$ . Without loss of generality, we assume $\tau _{1}\lt \tau _{2}$ .

First, we consider $s\in [0,c\tau _{1}]$ . For this case, $h(\psi ^{\delta _{i}}(s-c\tau _{1}))=h(\delta _{i}v^{*})$ and $g(\phi ^{\delta _{i}}(s-c\tau _{2}))=g(\delta _{i}u^{*})$ . Then, $(\tilde{\phi }, \tilde{\psi })$ satisfies

\begin{eqnarray*} \left \{\begin{array}{l} c\tilde{\phi }^{\prime }(s)-d_1\tilde{\phi }^{\prime \prime }(s)=h(\delta _{1}v^{*})-h(\delta _{2}v^{*})-b_1\tilde{\phi }(s) \gt -b_1\tilde{\phi }(s),\quad 0\lt s\leqslant c\tau _{1},\\[5pt] c\tilde{\psi }^{\prime }(s)-d_2\tilde{\psi }^{\prime \prime }(s)=g(\delta _{1}u^{*})-g(\delta _{2}u^{*})-b_2\tilde{\psi }(s) \gt -b_2\tilde{\psi }(s),\quad 0\lt s\leqslant c\tau _{1},\\[5pt] (\tilde{\phi }(0), \tilde{\psi }(0))=(\delta _{1} u^*-\delta _{2} u^*, \delta _{1} v^*-\delta _{2} v^*)\gt (0,0). \end{array}\right. \end{eqnarray*}

By the maximum principle, we have $\tilde{\phi }(s)\gt 0$ and $\tilde{\psi }(s)\gt 0$ for $s\in [0, c\tau _{1}]$ . Thus, $\phi ^{\delta _{1}}(s)\gt \phi ^{\delta _{2}}(s)$ and $\psi ^{\delta _{1}}(s)\gt \psi ^{\delta _{2}}(s)$ for $s\in (\!-\!\infty, c\tau _{1}]$ .

For $s\in [c\tau _{1}, 2c\tau _{1}]$ , we have $s-c\tau _{1}, s-c\tau _{2}\in (\!-\!\infty, c\tau _{1}]$ . By the above result, $h(\psi ^{\delta _{1}}(s-c\tau _{1}))\gt h(\psi ^{\delta _{2}}(s-c\tau _{1}))$ and $g(\phi ^{\delta _{1}}(s-c\tau _{2}))\gt g(\phi ^{\delta _{2}}(s-c\tau _{2}))$ . It follows that $(\tilde{\phi }, \tilde{\psi })$ satisfies

\begin{eqnarray*} \left \{\begin{array}{l} c\tilde{\phi }^{\prime }(s)-d_1\tilde{\phi }^{\prime \prime }(s) \gt -b_1\tilde{\phi }(s),\quad c\tau _{1}\lt s\leqslant 2c\tau _{1},\\[5pt] c\tilde{\psi }^{\prime }(s)-d_2\tilde{\psi }^{\prime \prime }(s) \gt -b_2\tilde{\psi }(s),\quad c\tau _{1}\lt s\leqslant 2c\tau _{1},\\[5pt] (\tilde{\phi }(c\tau _{1}), \tilde{\psi }(c\tau _{1}))\gt (0,0). \end{array}\right. \end{eqnarray*}

By the maximum principle, we have $\tilde{\phi }(s)\gt 0$ and $\tilde{\psi }(s)\gt 0$ for $s\in [c\tau _{1},2c\tau _{1}]$ , which imply that $\phi ^{\delta _{1}}(s)\gt \phi ^{\delta _{2}}(s)$ and $\psi ^{\delta _{1}}(s)\gt \psi ^{\delta _{2}}(s)$ for $s\in (\!-\!\infty,2c\tau _{1}]$ .

By repeating the above steps, we conclude that $(\phi ^{\delta _{1}}(s), \psi ^{\delta _{1}}(s))\gt (\phi ^{\delta _{2}}(s), \psi ^{\delta _{2}}(s))$ for $s\in \mathbb{R}$ .

In what follows, we exhibit the existence and uniqueness of monotone increasing solutions to (3.1).

Proof. $(i)$ (Uniqueness of Monotone Increasing Solutions) Suppose that (3.1) has two monotone increasing solutions $(\phi _1, \psi _1)$ and $(\phi _2, \psi _2)$ . Then $(0,0)\lt (\phi _i, \psi _i)\lt (u^*, v^*)$ in $(0, +\infty )$ and $(\phi _i(+\infty ), \psi _i(+\infty ))=(u^*, v^*)$ for $i=1,2$ . The Hopf boundary lemma implies that $((\phi _i)_{+}^{\prime }(0),(\psi _i)_{+}^{\prime }(0))\gt (0,0)$ for $i=1,2$ .

Define

\begin{equation*} \rho _{1}\;:\!=\;\inf \Big \{\frac {\phi _1(s)}{\phi _2(s)} \Big | s\gt 0\Big \},\quad \rho _{2}\;:\!=\;\inf \Big \{\frac {\psi _1(s)}{\psi _2(s)} \Big | s\gt 0\Big \} \end{equation*}

and

\begin{equation*} \rho ^*\;:\!=\;\min \{\rho _{1}, \rho _{2}\}. \end{equation*}

By the L’H $\hat{o}$ pital’s rule, it is easy to see that

\begin{equation*} \lim _{s\rightarrow 0^+}\Big (\frac {\phi _1(s)}{\phi _2(s)}, \frac {\psi _1(s)}{\psi _2(s)}\Big ) =\Big (\frac {(\phi _1)_{+}^{\prime }(0)}{(\phi _2)_{+}^{\prime }(0)}, \frac {(\psi _1)_{+}^{\prime }(0)}{(\psi _2)_{+}^{\prime }(0)}\Big )\gt (0,0). \end{equation*}

Moreover, $\lim _{s\rightarrow +\infty }(\frac{\phi _1(s)}{\phi _2(s)}, \frac{\psi _1(s)}{\psi _2(s)})=(1,1)$ . Thus, $\rho _{1}, \rho _{2}$ are well-defined and $\rho ^*\in (0,1]$ .

We claim that $\rho ^*\equiv 1$ . Assume on the contrary that $\rho ^*\in (0,1)$ . Let

\begin{equation*} \tilde {\phi }(s)=\phi _1(s)-\rho ^*\phi _2(s),\quad \tilde {\psi }(s)=\psi _1(s)-\rho ^*\psi _2(s). \end{equation*}

Obviously, $(\tilde{\phi }(s), \tilde{\psi }(s))\geqslant (0,0)$ for $s\gt 0$ , $(\tilde{\phi }(0), \tilde{\psi }(0))=(0,0)$ , $(\tilde{\phi }(+\infty ), \tilde{\psi }(+\infty ))=((1-\rho ^*)u^*, (1-\rho ^*)v^*)\gt (0,0)$ , and

\begin{equation*} \begin {array}{l} c\tilde {\phi }^{\prime }(s)-d_1\tilde {\phi }^{\prime \prime }(s)+b_{1}\tilde {\phi }(s)\\[5pt] \quad =h(\psi _1(s-c\tau _{1}))-\rho ^{*}h(\psi _2(s-c\tau _{1}))\\[5pt] \quad \geqslant h(\psi _1(s-c\tau _{1}))-h(\rho ^{*}\psi _2(s-c\tau _{1}))\geqslant 0\quad \mbox {for } s\gt 0. \end {array} \end{equation*}

Similarly, we have

\begin{equation*} c\tilde {\psi }^{\prime }(s)-d_2\tilde {\psi }^{\prime \prime }(s)+b_2\tilde {\psi }(s) \geqslant g(\phi _1(s-c\tau _{2}))-g(\rho ^{*}\phi _2(s-c\tau _{2})) \geqslant 0 \quad \mbox {for } s\gt 0. \end{equation*}

From the Hopf boundary lemma, we can deduce that $(0,0)\lt (\tilde{\phi }_{+}^{\prime }(0), \tilde{\psi }_{+}^{\prime }(0))=((\phi _1)_{+}^{\prime }(0)-\rho ^*(\phi _2)_{+}^{\prime }(0), (\psi _1)_{+}^{\prime }(0)-\rho ^*(\psi _2)_{+}^{\prime }(0))$ , which implies

\begin{equation*} \lim _{s\rightarrow 0^+}\Big (\frac {\phi _1(s)}{\phi _2(s)}, \frac {\psi _1(s)}{\psi _2(s)}\Big ) \gt (\rho ^*, \rho ^*). \end{equation*}

Therefore, by the definition of $\rho ^*$ , there exists $s_0\in (0,+\infty )$ such that $\tilde{\phi }(s_0)=0$ or $\tilde{\psi }(s_0)=0$ . By the maximum principle, we know $\tilde{\phi }(s)\equiv 0$ or $\tilde{\psi }(s)\equiv 0$ for $s\geqslant 0$ , which is a contradiction since $(\tilde{\phi }(+\infty ), \tilde{\psi }(+\infty ))=((1-\rho ^*)u^*, (1-\rho ^*)v^*)\gt (0,0)$ . Thus, $\rho ^*\equiv 1$ . It follows that $(\phi _1(s), \psi _1(s))\geqslant (\phi _2(s), \psi _2(s))$ for $s\gt 0$ .

Clearly, the same method can be employed to prove $(\phi _1(s), \psi _1(s))\leqslant (\phi _2(s), \psi _2(s))$ for $s\gt 0$ . Hence, the uniqueness follows directly.

$(ii)$ (Strict Monotonicity of Solution on $\mathbb{R}_{+}$ ) Assume that $(\phi,\psi )$ is a monotone increasing solution of (3.1). For any $\vartheta \gt 0$ , by (3.11), we get

\begin{eqnarray*} \phi (s+\vartheta ) &&=\frac{1}{d_1(\lambda _1^{+}-\lambda _1^{-})} \Big [\int _{0}^{s+\vartheta }(e^{\lambda _1^{-}(s+\vartheta -\xi )}- e^{\lambda _1^{-}(s+\vartheta )-\lambda _1^{+}\xi })h(\psi (\xi -c\tau _{1}))d\xi \\[5pt] &&\qquad\qquad \qquad \qquad +\int _{s+\vartheta }^{+\infty }(e^{\lambda _1^{+}(s+\vartheta -\xi )} -e^{\lambda _1^{-}(s+\vartheta )-\lambda _1^{+}\xi })h(\psi (\xi -c\tau _{1}))d\xi \Big ]\\[5pt] &&=\frac{1}{d_1(\lambda _1^{+}-\lambda _1^{-})} \Big [\int _{-\vartheta }^{s}(e^{\lambda _1^{-}(s-\xi ^{\prime })}- e^{\lambda _1^{-}(s+\vartheta )-\lambda _1^{+}(\xi ^{\prime }+\vartheta )})h(\psi (\xi ^{\prime }+\vartheta -c\tau _{1}))d\xi ^{\prime }\\[5pt] &&\qquad\qquad \qquad \qquad +\int _{s}^{+\infty }(e^{\lambda _1^{+}(s-\xi ^{\prime })} -e^{\lambda _1^{-}(s+\vartheta )-\lambda _1^{+}(\xi ^{\prime }+\vartheta )})h(\psi (\xi ^{\prime }+\vartheta -c\tau _{1}))d\xi ^{\prime }\Big ]\\[5pt] &&\gt \frac{1}{d_1(\lambda _1^{+}-\lambda _1^{-})} \Big [\int _{0}^{s}(e^{\lambda _1^{-}(s-\xi ^{\prime })}- e^{\lambda _1^{-}s-\lambda _1^{+}\xi ^{\prime }})h(\psi (\xi ^{\prime }-c\tau _{1}))d\xi ^{\prime }\\[5pt] &&\qquad\qquad \qquad \qquad +\int _{s}^{+\infty }(e^{\lambda _1^{+}(s-\xi ^{\prime })} -e^{\lambda _1^{-}s-\lambda _1^{+}\xi ^{\prime }})h(\psi (\xi ^{\prime }-c\tau _{1}))d\xi ^{\prime }\Big ]\\[5pt] &&=\phi (s), \quad \forall s\in \mathbb{R}_{+}. \end{eqnarray*}

Similarly, we can prove $\psi (s+\vartheta )\gt \psi (s)$ on $\mathbb{R}_{+}$ for any $\vartheta \gt 0$ .

$(iii)$ (Existence of Monotone Increasing Solution) For the existence, we divide the proof into the following steps.

Step 1. For any fixed $c\gt 0$ , we prove that the equations

\begin{equation*} \left \{\begin {array}{l} c\phi ^{\prime }(s)-d_1\phi ^{\prime \prime }(s)=h(\psi (s-c\tau _1))-b_1\phi (s), \quad s\in \mathbb {R},\\[5pt] c\psi ^{\prime }(s)-d_2\psi ^{\prime \prime }(s)=g(\phi (s-c\tau _2))-b_2\psi (s), \quad s\in \mathbb {R} \end {array}\right. \end{equation*}

have a monotone increasing solution $(\phi,\psi )$ satisfying either

\begin{equation*} (a)\qquad \quad (\phi (\!-\!\infty ),\psi (\!-\!\infty ))=(0,0),\quad (\phi (+\infty ),\psi (+\infty ))=(u^*,v^*), \end{equation*}

or

\begin{equation*} (b)\quad (\phi (s),\psi (s))=(0,0)\quad \mbox {for } s\leqslant 0, \quad (\phi (+\infty ),\psi (+\infty ))=(u^*,v^*). \end{equation*}

For the case $(a)$ , $(u(t,x),v(t,x))\;:\!=\;(\phi (x+ct),\psi (x+ct))$ is a travelling wave solution of

(3.17) \begin{equation} \left \{\begin{array}{l} \dfrac{\partial u}{\partial t}=d_{1}\dfrac{\partial ^{2}u}{\partial x^{2}}-b_{1}u+h(v(t-\tau _{1},x)),\quad t\gt 0, x\in \mathbb{R},\\[10pt] \dfrac{\partial v}{\partial t}=d_{2}\dfrac{\partial ^{2}v}{\partial x^{2}}-b_{2}v+g(u(t-\tau _{2},x)),\quad t\gt 0, x\in \mathbb{R}. \end{array}\right. \end{equation}

For the case $(b)$ , $(\phi (s),\psi (s))$ is a monotone increasing solution of (3.1), and $(u(t,x),v(t,x))\;:\!=\;(\phi (x+ct),\psi (x+ct))$ is called a semi-wave solution of (3.17). Thus, the conclusion of this step implies that (3.17) has either a monotone travelling wave or a monotone semi-wave with speed $c$ for any fixed $c\gt 0$ .

Take a sequence $\{\delta _{n}\}_{n=1}^{\infty }$ satisfying $\delta _{n}\in (0,\frac{1}{2})$ and $\delta _{n}\searrow 0$ as $n\rightarrow \infty$ , and let $(\phi _{n}, \psi _{n})$ be the monotone increasing solution of (3.14) with $\delta$ replaced by $\delta _{n}$ , which is obtained in Theorem 3.1. Then, $s_{n}\;:\!=\;\max \{s\;:\; \phi _{n}(s)=\frac{1}{2}u^{*}\}$ is well-defined. By Theorem 3.1, we know that $s_{n}$ is monotone increasing in $n$ .

Define $(\tilde{\phi }_{n}(s), \tilde{\psi }_{n}(s))\;:\!=\;(\phi _{n}(s+s_{n}), \psi _{n}(s+s_{n}))$ for $s\in \mathbb{R}$ . Then, $\tilde{\phi }_{n}(0)=\frac{1}{2}u^{*}$ and $(\tilde{\phi }_{n}(s), \tilde{\psi }_{n}(s))$ satisfies

\begin{equation*} \left \{\begin {array}{l} c\tilde {\phi }_{n}^{\prime }(s)-d_1\tilde {\phi }_{n}^{\prime \prime }(s)=h(\tilde {\psi }_{n}(s-c\tau _1))-b_1\tilde {\phi }_{n}(s),\quad s\gt -s_{n},\\[5pt] c\tilde {\psi }_{n}^{\prime }(s)-d_2\tilde {\psi }_{n}^{\prime \prime }(s)=g(\tilde {\phi }_{n}(s-c\tau _2))-b_2\tilde {\psi }_{n}(s),\quad s\gt -s_{n},\\[5pt] (\tilde {\phi }_{n}(s), \tilde {\psi }_{n}(s))=(\delta _{n} u^*, \delta _{n} v^*),\quad s\leqslant -s_{n},\\[5pt] (\tilde {\phi }_{n}(+\infty ), \tilde {\psi }_{n}(+\infty ))=(u^*, v^*). \end {array}\right. \end{equation*}

Since $s_{n}$ is monotone increasing in $n$ , we have $s_{0}\;:\!=\;\lim _{n\rightarrow \infty }s_{n}\in (0,+\infty ]$ .

From the proof of Lemma 3.2, we know that there exists a positive constant $C$ independent of $n$ such that

\begin{equation*} |\phi _{n}(s)|, |\phi _{n}^{\prime }(s)|, |\phi _{n}^{\prime \prime }(s)|, |\psi _{n}(s)|, |\psi _{n}^{\prime }(s)|, |\psi _{n}^{\prime \prime }(s)|\leqslant C \end{equation*}

for all $s\in \mathbb{R}_{+}$ . That is, $\phi _{n}$ and $\psi _{n}$ are uniformly bounded in $C^{2}(\mathbb{R}_{+})$ , from which we can deduce that $\phi _{n}(s)$ , $\phi _{n}^{\prime }(s)$ , $\psi _{n}(s)$ and $\psi _{n}^{\prime }(s)$ are equicontinuous on $\mathbb{R}_{+}$ . Applying the equations in (3.14), we know that $\phi _{n}^{\prime \prime }(s)$ and $\psi _{n}^{\prime \prime }(s)$ are also equicontinuous on $\mathbb{R}_{+}$ . By the Arzela–Ascoli theorem, there is a subsequence, still denoted by $(\tilde{\phi }_{n}, \tilde{\psi }_{n})$ , converges to $(\tilde{\phi },\tilde{\psi })$ in $C_{loc}^{2}(\mathbb{R})$ . It is easy to know that $(\tilde{\phi }(s),\tilde{\psi }(s))$ is monotone increasing in $s$ and satisfies $\tilde{\phi }(0)=\frac{1}{2}u^{*}$ .

(a) If $s_{0}=+\infty$ , then $(\tilde{\phi }(s),\tilde{\psi }(s))$ satisfies

\begin{equation*} \left \{\begin {array}{l} c\tilde {\phi }^{\prime }(s)-d_1\tilde {\phi }^{\prime \prime }(s)=h(\tilde {\psi }(s-c\tau _1))-b_1\tilde {\phi }(s),\quad s\in \mathbb {R},\\[5pt] c\tilde {\psi }^{\prime }(s)-d_2\tilde {\psi }^{\prime \prime }(s)=g(\tilde {\phi }(s-c\tau _2))-b_2\tilde {\psi }(s),\quad s\in \mathbb {R}. \end {array}\right. \end{equation*}

Since $(\tilde{\phi }(s),\tilde{\psi }(s))$ is monotone increasing and uniformly continuous on $\mathbb{R}_{+}$ , by Lemma 2.3 in [Reference Wu and Zou37] we can deduce that $\lim _{s\rightarrow \infty }\tilde{\phi }^{\prime }(s) =\lim _{s\rightarrow \infty }\tilde{\phi }^{\prime \prime }(s)=0$ and $\lim _{s\rightarrow \infty }\tilde{\psi }^{\prime }(s) =\lim _{s\rightarrow \infty }\tilde{\psi }^{\prime \prime }(s)=0$ , which imply $(\tilde{\phi }(\pm \infty ), \tilde{\psi }(\pm \infty ))=(0, 0)$ or $(u^*, v^*)$ . In view of $\tilde{\phi }(0)=\frac{1}{2}u^{*}$ , we know that $(\tilde{\phi }(\!-\!\infty ), \tilde{\psi }(\!-\!\infty ))=(0, 0)$ and $(\tilde{\phi }(+\infty ), \tilde{\psi }(+\infty ))=(u^*, v^*)$ .

(b) If $s_{0}\in (0,+\infty )$ , then $(\tilde{\phi }(s),\tilde{\psi }(s))$ satisfies

\begin{equation*} \left \{\begin {array}{l} c\tilde {\phi }^{\prime }(s)-d_1\tilde {\phi }^{\prime \prime }(s)=h(\tilde {\psi }(s-c\tau _1))-b_1\tilde {\phi }(s),\quad s\gt -s_{0},\\[5pt] c\tilde {\psi }^{\prime }(s)-d_2\tilde {\psi }^{\prime \prime }(s)=g(\tilde {\phi }(s-c\tau _2))-b_2\tilde {\psi }(s),\quad s\gt -s_{0},\\[5pt] (\tilde {\phi }(s), \tilde {\psi }(s))=(0, 0),\quad s\leqslant -s_{0}. \end {array}\right. \end{equation*}

Let $(\phi (s),\psi (s))=(\tilde{\phi }(s-s_{0}),\tilde{\psi }(s-s_{0}))$ , we can also prove $(\phi (+\infty ), \psi (+\infty ))=(u^*, v^*)$ . Obviously, $(\phi (s), \psi (s))=(0, 0)$ for $s\leqslant 0$ . This completes the proof of Step 1.

Step 2. For any $c\in (0, c_{\tau }^{*})$ , we prove that (3.17) has no travelling wave solution.

We claim that $c_{\tau }^{*}$ is the minimal wave speed for travelling wave solutions of (3.17), that is, (3.17) admits a monotone travelling wave solution for $c\geqslant c_{\tau }^{*}$ and has no monotone travelling wave solution for $c\in (0, c_{\tau }^{*})$ .

Indeed, for the quasi-monotone system (3.17), applying the theory of monotone semiflows developed by Liang and Zhao [Reference Liang and Zhao22], we can prove that $c_{\tau }^{*}$ is the asymptotic spreading speed and coincides with the minimal wave speed. The proof is similar as that of Theorem 5.1 in [Reference Liang and Zhao22], Theorems 2.1–2.2 in [Reference Fang, Wei and Zhao17] and Theorem 3.1 in [Reference Li and Zhang21]. The key step in the proof is verifying the conditions in [Reference Liang and Zhao22].

We first introduce some notations. Let $\mathcal{C}=C([\!-\!\tau _{2},0]\times \mathbb{R}, \mathbb{R})\times C([\!-\!\tau _{1},0]\times \mathbb{R}, \mathbb{R})$ , $\bar{\mathcal{C}}=C([\!-\!\tau _{2},0], \mathbb{R})\times C([\!-\!\tau _{1},0], \mathbb{R})$ , $X=(C\cap L^{\infty })(\mathbb{R}, \mathbb{R}^{2})$ ,

\begin{equation*} \mathcal {C}_{(u^*,v^*)}=\Big \{(\varphi _{1},\varphi _{2})\in \mathcal {C}; 0\leqslant \varphi _{1}(\theta,x)\leqslant u^* \mbox {on } [\!-\!\tau _{2},0]\times \mathbb {R}, 0\leqslant \varphi _{2}(\theta,x)\leqslant v^* \mbox {on } [\!-\!\tau _{1},0]\times \mathbb {R}\Big \} \end{equation*}

and

\begin{equation*} \bar {\mathcal {C}}_{(u^*,v^*)}=\Big \{(\varphi _{1},\varphi _{2})\in \bar {\mathcal {C}}; 0\leqslant \varphi _{1}(\theta )\leqslant u^* \mbox {on } [\!-\!\tau _{2},0], 0\leqslant \varphi _{2}(\theta )\leqslant v^* \mbox {on } [\!-\!\tau _{1},0]\Big \}. \end{equation*}

Clearly, any element in $\bar{\mathcal{C}}$ or $X$ can be regarded as a function in $\mathcal{C}$ . We equip $\mathcal{C}$ with the compact open topology and define the metric function $d(\cdot,\cdot )$ in $\mathcal{C}$ with respect to this topology by

\begin{eqnarray*} &&d((\varphi _{1},\varphi _{2}),(\tilde{\varphi }_{1},\tilde{\varphi }_{2}))\\[5pt] &&\quad=\sum _{k=1}^{\infty }\frac{\max _{\theta \in [\!-\!\tau _2,0],|x|\leqslant k}|\varphi _{1}(\theta,x)-\tilde{\varphi }_{1}(\theta,x)| +\max _{\theta \in [\!-\!\tau _1,0],|x|\leqslant k}|\varphi _{2}(\theta,x)-\tilde{\varphi }_{2}(\theta,x)|}{2^{k}} \end{eqnarray*}

for any $(\varphi _{1},\varphi _{2}),(\tilde{\varphi }_{1},\tilde{\varphi }_{2})\in \mathcal{C}$ , so that $(\mathcal{C}, d)$ is a metric space.

Define $f\;:\; \mathcal{C}\longrightarrow X$ by

\begin{eqnarray*} f(\varphi _{1},\varphi _{2})(x) &&=\Big (f_{1}(\varphi _{1},\varphi _{2})(x),f_{2}(\varphi _{1},\varphi _{2})(x)\Big )\\[5pt] &&=\Big (\!-\!b_{1}\varphi _{1}(0,x)+h(\varphi _{2}(\!-\!\tau _{1},x)),-b_{2}\varphi _{2}(0,x)+g(\varphi _{1}(\!-\!\tau _{2},x))\Big ). \end{eqnarray*}

We can then rewrite (3.17) as follows

(3.18) \begin{equation} \left \{\begin{array}{l} \dfrac{\partial u}{\partial t} =d_{1}\dfrac{\partial ^{2} u}{\partial x^{2}} +f_{1}(u_{t},v_{t})(x),\\[10pt] \dfrac{\partial v}{\partial t} =d_{2}\dfrac{\partial ^{2} v}{\partial x^{2}} +f_{2}(u_{t},v_{t})(x), \end{array}\right. \end{equation}

where $(u_t,v_t)\in \mathcal{C}$ with $u_{t}(\theta,x)=u(t+\theta,x)$ for $\theta \in [\!-\!\tau _{2},0]$ and $v_{t}(\theta,x)=v(t+\theta,x)$ for $\theta \in [\!-\!\tau _{1},0]$ . The condition $(A1)$ in [Reference Liang and Zhao22] is satisfied by the property that both $(u(t,-x),v(t,-x))$ and $(u(t,x+y),v(t,x+y))$ , $\forall y\in \mathbb{R}$ , are also solutions when $(u(t,x),v(t,x))$ is a solution.

By Lemma 2.9 in [Reference Wu and Hsu38], there exists a solution map $Q_{t}=(Q_{t}^{u},Q_{t}^{v})\;:\; \mathcal{C}_{(u^*,v^*)}\rightarrow \mathcal{C}_{(u^*,v^*)}$ for $t\gt 0$ :

\begin{eqnarray*} Q_{t}^{u}[(\varphi _{1},\varphi _{2})](\theta,x) =u_{t}(\theta,x;\varphi _{1},\varphi _{2}),\quad \forall (\theta,x)\in [\!-\!\tau _{2},0]\times \mathbb{R},\\[5pt] Q_{t}^{v}[(\varphi _{1},\varphi _{2})](\theta,x) =v_{t}(\theta,x;\varphi _{1},\varphi _{2}),\quad \forall (\theta,x)\in [\!-\!\tau _{1},0]\times \mathbb{R}, \end{eqnarray*}

and $Q_{t}:\mathcal{C}_{(u^*,v^*)}\rightarrow \mathcal{C}_{(u^*,v^*)}$ is monotone. Thus, the condition $(A4)$ in [Reference Liang and Zhao22] holds. Similar as the proof of Lemma 2.2 in [Reference Fang, Wei and Zhao17] and Lemma 3.2 in [Reference Li and Zhang21], we can show that $Q_{t}$ is continuous in $(\varphi _{1},\varphi _{2})$ with respect to the compact open topology uniformly for $t\in [0,t_0]$ with any $t_0\gt 0$ . Since (3.18) is an autonomous system, $\{Q_{t}\}_{t\geqslant 0}$ is a semiflow on $\mathcal{C}_{(u^*,v^*)}$ . Thus, the condition $(A2)$ in [Reference Liang and Zhao22] holds.

Let $\hat{Q}_{t}$ be the restriction of $Q_{t}$ to $\bar{\mathcal{C}}_{(u^*,v^*)}$ . It is easy to see that $\hat{Q}_{t}\;:\; \bar{\mathcal{C}}_{(u^*,v^*)}\rightarrow \bar{\mathcal{C}}_{(u^*,v^*)}$ is the solution semiflow generated by

(3.19) \begin{equation} \left \{\begin{array}{l} u^{\prime }(t)=f_{1}(u_{t},v_{t}),\\[5pt] v^{\prime }(t)=f_{2}(u_{t},v_{t}) \end{array}\right. \end{equation}

with the initial data $(\varphi _{1},\varphi _{2})\in \bar{\mathcal{C}}_{(u^*,v^*)}$ . By Corollary 5.3.5 in [Reference Smith27], $\hat{Q}_{t}$ is eventually strongly monotone on $\bar{\mathcal{C}}_{(u^*,v^*)}$ . In the proof of Theorem 2.3 $(iv)$ , we have known that $s(D\hat{f}(0,0))=\max \{\textrm{Re} \lambda ; \det (\lambda I-D\hat{f}(0,0))=0\}\gt 0$ . By Corollary 5.5.2 in [Reference Smith27], $(\hat{0},\hat{0})$ is an unstable equilibrium of (3.19). By the Dancer-Hess connecting orbit lemma (see, e.g. Section 2.1 in [Reference Zhao41]), the semiflow $\hat{Q}_{t}$ admits a strongly monotone full orbit connecting $(0,0)$ to $(u^*,v^*)$ . Thus, the condition $(A5)$ holds for each $Q_{t}$ , $t\gt 0$ .

Similar as the proof of Lemma 2.3 in [Reference Fang, Wei and Zhao17] and Lemma 3.3 in [Reference Li and Zhang21], we can prove that the following two statements hold:

$(a)$ for $(t_1,t_2)\in (\tau _{2},+\infty )\times (\tau _{1},+\infty )$ , $(Q_{t_1}^{u}[\mathcal{C}_{(u^*,v^*)}], Q_{t_2}^{v}[\mathcal{C}_{(u^*,v^*)}])$ is precompact in $\mathcal{C}_{(u^*,v^*)}$ ;

$(b)$ for $(t_1,t_2)\in [0,\tau _{2}]\times [0,\tau _{1}]$ , $(Q_{t_1}^{u}[\mathcal{C}_{(u^*,v^*)}](0,\cdot )$ , $Q_{t_2}^{v}[\mathcal{C}_{(u^*,v^*)}](0,\cdot ))$ is precompact in $X$ , and there are positive numbers $(\zeta,\vartheta )\leqslant (\tau _{2},\tau _{1})$ such that $Q_{t_1}^{u}[(u,v)](\theta,\cdot )=u(\theta +\zeta,x)$ for $-\tau _{2}\leqslant \theta \leqslant -\zeta$ , $Q_{t_2}^{v}[(u,v)](\theta,\cdot )=v(\theta +\vartheta,x)$ for $-\tau _{1}\leqslant \theta \leqslant -\vartheta$ , and the operators

\begin{eqnarray*} S_{1}[(u,v)](\theta,x) =\left \{\begin{array}{l} u(0,x),\quad \theta \in [\!-\!\tau _{2},-\zeta ),\\[5pt] Q_{t_1}^{u}[(u,v)](\theta,x),\quad \theta \in [\!-\!\zeta,0], \end{array}\right .\\[5pt] S_{2}[(u,v)](\theta,x) =\left \{\begin{array}{l} v(0,x),\quad \theta \in [\!-\!\tau _{1},-\vartheta ),\\[5pt] Q_{t_2}^{v}[(u,v)](\theta,x),\quad \theta \in [\!-\!\vartheta,0] \end{array}\right. \end{eqnarray*}

have the property that $S_{1}[D], S_{2}[D]$ are precompact in $\mathcal{C}_{(u^*,v^*)}$ for any $T$ -invariant set $D\subset \mathcal{C}_{(u^*,v^*)}$ with $D(0,\cdot )$ precompact in $X$ . Thus, the condition $(A6)(a),(b^{\prime })$ in [Reference Liang and Zhao22] holds. By the theory of monotone semiflows developed in [Reference Liang and Zhao22], there exists $c^{*}\gt 0$ such that $c^{*}$ is the asymptotic spreading speed, which coincides with the minimal wave speed.

Next, we prove $c^{*}=c_{\tau }^{*}$ . Let $M_t=(M_{t}^{u}, M_{t}^{v})\;:\; \mathcal{C}\rightarrow \mathcal{C}$ be the solution map at time $t$ of the following linear equations

(3.20) \begin{equation} \left \{\begin{array}{l} \dfrac{\partial u}{\partial t} =d_{1}\dfrac{\partial ^{2} u}{\partial x^{2}} -b_{1}u+h^{\prime }(0)v_{t},\\[10pt] \dfrac{\partial v}{\partial t} =d_{2}\dfrac{\partial ^{2} v}{\partial x^{2}} -b_{2}v+g^{\prime }(0)u_{t}. \end{array}\right. \end{equation}

For $\lambda \geqslant 0$ , we define the linear map $B_{t}=(B^{u}_{t},B^{v}_{t})\;:\; \bar{\mathcal{C}}\rightarrow \bar{\mathcal{C}}$ by

\begin{eqnarray*} B^{u}_{t}[(\varphi _{1},\varphi _{2})](\theta ) =M_{t}^{u}[(\varphi _{1},\varphi _{2})e^{-\lambda x}](\theta,0), \quad \forall \theta \in [\!-\!\tau _{2},0],\\[5pt] B^{v}_{t}[(\varphi _{1},\varphi _{2})](\theta ) =M_{t}^{v}[(\varphi _{1},\varphi _{2})e^{-\lambda x}](\theta,0), \quad \forall \theta \in [\!-\!\tau _{1},0]. \end{eqnarray*}

Then, $B_{t}=(B^{u}_{t},B^{v}_{t})\;:\; \bar{\mathcal{C}}\rightarrow \bar{\mathcal{C}}$ is the solution map of the following equations

(3.21) \begin{equation} \left \{\begin{array}{l} u^{\prime }(t) =d_{1}\lambda ^{2}u(t) -b_{1}u(t)+h^{\prime }(0)v_{t},\\[5pt] v^{\prime }(t) =d_{2}\lambda ^{2}v(t) -b_{2}v(t)+g^{\prime }(0)u_{t}. \end{array}\right. \end{equation}

Let

\begin{equation*} A(\chi ) =\left ( \begin {array}{c@{\quad}c} d_{1}\lambda ^{2}-b_{1} & h^{\prime }(0)e^{-\chi \tau _{1}} \\[5pt] g^{\prime }(0)e^{-\chi \tau _{2}} & d_{2}\lambda ^{2}-b_{2} \\[5pt] \end {array} \right ). \end{equation*}

Since (3.21) is a cooperative and irreducible delay equations, it follows that

\begin{equation*} \det (\chi I-A(\chi ))=0, \end{equation*}

i.e.

\begin{equation*} \chi ^{2}-[(d_{1}\lambda ^{2}-b_{1})+(d_{2}\lambda ^{2}-b_{2})]\chi +(d_{1}\lambda ^{2}-b_{1})(d_{2}\lambda ^{2}-b_{2}) -h^{\prime }(0)g^{\prime }(0)e^{-\chi (\tau _{1}+\tau _{2})} =0, \end{equation*}

admits a real root $\chi (\lambda )$ which is greater than the real parts of all other ones (see Theorem 5.5.1 in [Reference Smith27]).

Define $\Phi (\lambda )= \frac{\chi (\lambda )}{\lambda }$ . By Theorem 3.10 in [Reference Liang and Zhao22], we know $c^{*}=\inf _{\lambda \gt 0}\Phi (\lambda )=\inf _{\lambda \gt 0}\frac{\chi (\lambda )}{\lambda }$ . Let $c=\frac{\chi (\lambda )}{\lambda }$ , then $c^{*}$ satisfies $c^{*}=\Phi (\lambda ^{*})$ and $\frac{dc}{d \lambda }|_{\lambda =\lambda ^{*}}$ =0. Then, $(c^{*},\lambda ^{*})$ can be determined as the positive solution to the system

(3.22) \begin{equation} \Delta ^{c}(\lambda,\tau )=0,\quad \frac{\partial \Delta ^{c}(\lambda,\tau )}{\partial \lambda }=0, \end{equation}

where $\Delta ^{c}(\lambda,\tau )$ , as a function of $c$ and $\lambda$ , is defined in (3.2). From the conclusions of Lemma 3.1, $c^{*}=c_{\tau }^{*}$ . This completes the proof of the claim.

Figure 2. Dichotomy between semi-waves and travelling waves.

We should mention that the existence of monotone travelling wave solution to (3.17) for $c\geqslant c_{\tau }^{*}$ has also been proved in [Reference Wu and Hsu38] by the method of lower and upper solutions.

Step 3. For any $c\in (0, c_{\tau }^{*})$ , we prove that (3.17) has a monotone semi-wave with speed $c$ , or equivalently, (3.1) has a monotone increasing solution.

By the conclusions in Steps 1–2, we can immediately get the desired result.

Proof. $(i)$ Since $0\lt c_1\lt c_2$ and $((\phi ^{c_2})^{\prime }(s), (\psi ^{c_2})^{\prime }(s))\gt (0,0)$ for $s\in \mathbb{R}_{+}$ , we have