1 Introduction and the Statement of the Main Results

Chemotaxis describes the oriented movement of biological cells or organism in response to chemical gradients. The oriented movement of cells has a crucial role in a wide range of biological phenomena. At the beginning of 1970s, Keller and Segel (see [23, 24]) introduced systems of partial differential equations of the following form to model the time evolution of both the density u(xt) of a mobile species and the density v(xt) of a chemoattractant,

$$\begin{aligned} {\left\{ \begin{array}{ll} u_{t}=\nabla \cdot (m(u)\nabla u- \chi (u,v)\nabla v) + f(u,v),\quad x\in \Omega , \\ \tau v_t=\Delta v + g(u,v),\quad x\in \Omega , \end{array}\right. } \end{aligned}$$
(1.1)

complemented with certain boundary condition on \(\partial \Omega \) if \(\Omega \) is bounded, where \(\Omega \subset {\mathbb R}^N\) is an open domain; \(\tau \ge 0\) is a non-negative constant linked to the speed of diffusion of the chemical; the function \(\chi (u,v)\) represents the sensitivity with respect to chemotaxis; and the functions f and g model the growth of the mobile species and the chemoattractant, respectively. In literature, (1.1) is called the Keller–Segel (KS) model or a chemotaxis model.

Since the works by Keller and Segel, a rich variety of mathematical models for studying chemotaxis has appeared (see [1, 6, 7, 13, 17, 18, 22, 33, 41,42,43, 46, 49,50,51,52,53,54, 57], and the references therein). The reader is referred to [16, 19] for some detailed introduction into the mathematics of KS models. In the current paper, we consider chemoattraction-repulsion process in which cells undergo random motion and chemotaxis towards attractant and away from repellent [31]. Moreover, we consider the model with proliferation and death of cells and assume that chemicals diffuse very quickly. These lead to the model of partial differential equations as follows:

$$\begin{aligned} {\left\{ \begin{array}{ll} u_{t}=\Delta u -\chi _1\nabla ( u\nabla v_1)+\chi _2 \nabla (u\nabla v_2 )+ u(a -b u), \qquad \ x\in \Omega ,\ t>0, \\ 0=(\Delta - \lambda _1 I)v_1+ \mu _1 u , \qquad \ x\in \Omega ,\ t>0, \\ 0=(\Delta - \lambda _2 I)v_2+ \mu _2 u , \qquad \text {in}\ \ x\in \Omega ,\ t>0 , \end{array}\right. } \end{aligned}$$
(1.2)

complemented with certain boundary condition on \(\partial \Omega \) if \(\Omega \) is bounded.

When \(\Omega \) is a smooth bounded domain, it is seen that (1.2) complemented with Neumann boundary conditions

$$\begin{aligned} \frac{\partial u}{\partial n}=\frac{\partial v_1}{\partial n}=\frac{\partial v_2}{\partial n}=0, \end{aligned}$$
(1.3)

has a unique nonzero constant equilibrium \((\frac{a}{b},\frac{\mu _1}{\lambda _1}\frac{a}{b},\frac{\mu _2}{\lambda _2}\frac{a}{b})\). The global existence of classical solutions and the stability of the above equilibrium solution of (1.2)+(1.3) are among central dynamical issues. They have been studied in many papers (see [8, 20, 21, 27, 29,30,31, 44, 45, 55, 56] and the references therein). For example, in [55], amount others, the authors proved that

  • If\(b>\chi _{1}\mu _1-\chi _2\mu _2\), or\(N\le 2\), or\(\frac{N-2}{N}(\chi _1\mu _1-\chi _2\mu _2)<b\)and\(N\ge 3\), then for every nonnegative initial\(u_0\in C^{0}(\overline{\Omega }),\) (1.2)+(1.3) has a unique global classical solution\((u(\cdot ,\cdot ),v_1(\cdot ,\cdot ),v_2(\cdot ,\cdot ))\)which is uniformly bounded.

  • If\(a=b>2\chi _1\mu _1\), then for every nonnegative initial\(u_0\in C^{0}(\overline{\Omega }),\)\(u_0\ne 0\), the global classical solution\((u(\cdot ,\cdot ),v_1(\cdot ,\cdot ),v_2(\cdot ,\cdot ))\)of (1.2)+(1.3) satisfies

    $$\begin{aligned}\lim _{t\rightarrow \infty }\Big [\Vert u(\cdot ,t)-1\Vert _{C^{0}(\Omega )}+\Vert v_1(\cdot ,t)-\frac{\mu _1}{\lambda _1}\Vert _{C^{0}(\Omega )}+\Vert v_2(\cdot ,t)-\frac{\mu _2}{\lambda _2}\Vert _{C^{0}(\Omega )} \Big ] =0.\end{aligned}$$

While attraction–repulsion chemotaxis systems on bounded domains have been studied in many papers, there is little study of such systems on unbounded domains. The objective of this paper is to study the dynamics of (1.2) with \(\Omega ={\mathbb R}^N\), that is,

$$\begin{aligned} {\left\{ \begin{array}{ll} u_{t}=\Delta u -\chi _1\nabla ( u\nabla v_1)+\chi _2 \nabla (u\nabla v_2 )+ u(a -b u) , \qquad \ x\in {\mathbb R}^N,\ t>0, \\ 0=(\Delta - \lambda _1 I)v_1+ \mu _1 u , \qquad \ x\in {\mathbb R}^N,\ t>0, \\ 0=(\Delta - \lambda _2 I)v_2+ \mu _2 u , \qquad \ x\in {\mathbb R}^N,\ t>0, \\ u(\cdot ,0)=u_{0} , \qquad x\in \mathbb {R}^N . \end{array}\right. } \end{aligned}$$
(1.4)

In the case that the chemorepellent is absent, that is, \(\chi _2=0\), the authors of the current paper studied in [36] the global existence of classical solutions and asymptotic behavior of bounded global classical solutions of (1.4). In the current paper, we investigate the global existence of classical solutions, stability of constant equilibria, and spreading speeds of (1.4) when both chemoattractant and chemorepellent are present. More precisely, we identify the circumstances under which positive classical solutions of (1.4) with nonnegative, bounded, and uniformly continuous initial functions exist globally; investigate the asymptotic stability of the nonzero constant equilibrium \((\frac{a}{b},\frac{\mu _1}{\lambda _1}\frac{a}{b},\frac{\mu _2}{\lambda _2}\frac{a}{b})\); and explore the spreading properties of the global solutions with compactly supported initial functions. We pay special attention to the combined effect of the chemoattractant and chemorepellent on the above dynamical issues.

Note that, due to biological interpretations, only nonnegative initial functions will be of interest. We call \((u(x,t),v_1(x,t),v_2(x,t))\) a classical solution of (1.4) on [0, T) if \(u,v_1,v_2\in C({\mathbb R}^N\times [0,T))\cap C^{2,1}({\mathbb R}^N\times (0,T))\) and satisfies (1.4) for \((x,t)\in {\mathbb R}^N\times (0,T)\) in the classical sense. A classical solution \((u(x,t),v_1(x,t),v_2(x,t))\) of (1.4) on [0, T) is called nonnegative if \(u(x,t)\ge 0\), \(v_1(x,t)\ge 0\) and \(v_2(x,t)\ge 0\) for all \((x,t)\in {\mathbb R}^N\times [0,T)\). A global classical solution of (1.4) is a classical solution on \( [0,\infty )\).

Let

$$\begin{aligned} C_\mathrm{unif}^b({\mathbb R}^N)=\{u\in C({\mathbb R}^N)\,|\, u(x)\,\,\text {is uniformly continuous in}\,\, x\in {\mathbb R}^N\,\, \mathrm{and}\,\, \sup _{x\in {\mathbb R}^N}|u(x)|<\infty \} \end{aligned}$$
(1.5)

equipped with the norm \(\Vert u\Vert _\infty =\sup _{x\in {\mathbb R}^N}|u(x)|\). We have the following result on the global existence of classical solutions of (1.4) for initial functions belonging to \(C^{b}_\mathrm{unif}({\mathbb R}^N)\).

Theorem A

Suppose that

$$\begin{aligned} \chi _1=a=b=0 \end{aligned}$$
(1.6)

or

$$\begin{aligned} b > \chi _1\mu _1-\chi _2\mu _2 + M, \end{aligned}$$
(1.7)

where

$$\begin{aligned} M:= \min \Big \{&\frac{1}{\lambda _2}\big ( (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_+ + \chi _1\mu _1(\lambda _1-\lambda _2)_+ \big ),\nonumber \\&\qquad \frac{1}{\lambda _1}\big ( (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_+ + \chi _2\mu _2(\lambda _1-\lambda _2)_+ \big ) \Big \}. \end{aligned}$$
(1.8)

Then for every nonnegative initial function \(u_{0}\in C^{b}_\mathrm{unif}({\mathbb R}^N)\), (1.4) has a unique nonnegative global classical solution \((u(\cdot ,\cdot ;u_0),v_1(\cdot ,\cdot ;u_0),v_2(\cdot ,\cdot ;u_0))\) with \(u(\cdot ,0;u_0) =u_0\). Furthermore, it holds that

$$\begin{aligned} \Vert u(\cdot ,t;u_0)\Vert _{\infty }\le {\left\{ \begin{array}{ll}\Vert u_0\Vert _{\infty }&{} \text {if (1.6) holds}\\ \max \{\Vert u_0\Vert _{\infty }, \frac{a}{b+\chi _2\mu _2-\chi _1\mu _1-M}\} &{} \text {if (1.7) holds}. \end{array}\right. } \end{aligned}$$
(1.9)

Remark 1.1

\(M\le \chi _2\mu _2\). (1.6) and (1.7) provide explicit conditions for the global existence of classical solutions. The following special and important conditions follow from (1.7).

  1. (i)

    If \(b>\chi _1\mu _1\), (1.4) always has global bounded classical solution for any initial \(u_0\in C^{b}_\mathrm{unif}({\mathbb R}^N)\) with \(u_0\ge 0\).

  2. (ii)

    If \(\lambda _1\le \lambda _2\) and \( \chi _2\mu _2\lambda _2\ge \chi _1\mu _1\lambda _1\), we have that \(M=\chi _2\mu _2-\frac{\lambda _1}{\lambda _2}\chi _1\mu _1\). In this case, it follows from Theorem A that for every nonnegative bounded and uniformly continuous initial data \(u_0\), (1.4) has a unique bounded global classical solution \((u(\cdot ,\cdot ;u_0),v_1(\cdot ,\cdot ;u_0),v_2(\cdot ,\cdot ;u_0))\), whenever \(b>\chi _1\mu _1(1-\frac{\lambda _1}{\lambda _2})\). Thus, in the absence of chemoattractant, i.e \(\chi _1=0\), for every nonnegative bounded and uniformly continuous initial data \(u_0\), (1.4) has a unique bounded global classical solution \((u(\cdot ,\cdot ;u_0),v_1(\cdot ,\cdot ;u_0),v_2(\cdot ,\cdot ;u_0))\), whenever \(b>0\).

  3. (iii)

    If \(\lambda _1\le \lambda _2\) and \( \chi _2\mu _2\lambda _2\le \chi _1\mu _1\lambda _1\), we have that \(M=0\). In this case, it follows from Theorem A that for every nonnegative bounded and uniformly continuous initial data \(u_0\), (1.4) has a unique bounded global classical solution \((u(\cdot ,\cdot ;u_0),v_1(\cdot ,\cdot ;u_0),v_2(\cdot ,\cdot ;u_0))\), whenever \(b>\chi _1\mu _1-\chi _2\mu _2\).

  4. (iv)

    We note that if \(\lambda _1\ge \lambda _2\) and \( \chi _2\mu _2\lambda _2\ge \chi _1\mu _1\lambda _1 \), then \(M= \chi _2\mu _2-\chi _1\mu _1\). Thus, if \(\lambda _1\ge \lambda _2\) and \( \chi _2\mu _2\lambda _2\ge \chi _1\mu _1\lambda _1 \), it follows from Theorem A that for every \(b>0\) and for every nonnegative bounded and uniformly continuous initial data \(u_0\), (1.4) has a unique bounded global classical solution \((u(\cdot ,\cdot ;u_0),v_1(\cdot ,\cdot ;u_0),v_2(\cdot ,\cdot ;u_0))\).

  5. (v)

    If \(\lambda _1\ge \lambda _2\) and \( \chi _2\mu _2\lambda _2\le \chi _1\mu _1\lambda _1\), we have that \(M=\frac{(\lambda _1-\lambda _2)\chi _2\mu _2}{\lambda _1}\). In this case, it follows from Theorem A that for every nonnegative bounded and uniformly continuous initial data \(u_0\), (1.4) has a unique bounded global classical solution \((u(\cdot ,\cdot ;u_0),v_1(\cdot ,\cdot ;u_0),v_2(\cdot ,\cdot ;u_0))\), whenever \(b>\chi _1\mu _1-\frac{\lambda _2}{\lambda _1}\chi _2\mu _2\).

It follows from Remark 1.1 (iii)&(v), that when \(\chi _2=0\), we recover as a special case Theorem 1.5 in [36] for the case \(b>\chi _1\) and \(\mu _1=1\). When (1.7) does not hold, we leave it open whether for any nonnegative initial function \(u_{0}\in C^{b}_\mathrm{unif}({\mathbb R}^N)\) global solution to (1.4) exists.

Theorem A is fundamental. Assume the conditions in Theorem A. Then (1.4) generates a dynamical system on the infinite dimensional space \(X^+=\{u\in C_\mathrm{unif}^b({\mathbb R}^N)\,|\, u\ge 0\}\). Methods and theorems for general infinite dimensional dynamical systems in literature (e.g. [14, 38]) may then be utilized for the further study of many important dynamical aspects, including the long time behavior of bounded solutions, stability of certain special solutions, existence of global attractor, etc. In the following, we explore the stability of the nonzero constant equilibrium \((\frac{a}{b},\frac{\mu _1}{\lambda _1}\frac{a}{b},\frac{\mu _2}{\lambda _2}\frac{a}{b})\).

We first study the stability of \((\frac{a}{b},\frac{\mu _1}{\lambda _1}\frac{a}{b},\frac{\mu _2}{\lambda _2}\frac{a}{b})\) with respect to strictly positive initial functions. From now on, we shall always suppose that \(a>0\), unless otherwise specified. We prove

Theorem B

Suppose that

$$\begin{aligned} b>\chi _1\mu _1-\chi _2\mu _2+K, \end{aligned}$$
(1.10)

where

$$\begin{aligned} K:=\min \Big \{&\frac{1}{\lambda _2}\Big (|\chi _1\mu _1\lambda _1-\chi _2\mu _2\lambda _2|+\chi _1\mu _1|\lambda _1-\lambda _2|\Big ),\nonumber \\&\quad \frac{1}{\lambda _1}\Big (|\chi _1\mu _1\lambda _1-\chi _2\mu _2\lambda _2|+\chi _2\mu _2|\lambda _1-\lambda _2|\Big ) \Big \}. \end{aligned}$$
(1.11)

Then for every initial function \(u_{0}\in C^{b}_\mathrm{unif}({\mathbb R}^N)\) with \(\inf _{x\in {\mathbb R}^N}u_0(x)>0\), (1.4) has a unique bounded global classical solution \((u(\cdot ,\cdot ;u_0),v_1(\cdot ,\cdot ;u_0),v_2(\cdot ,\cdot ;u_0))\) with \(u(\cdot ,0;u_0)=u_{0}\). Furthermore we have that

$$\begin{aligned} \lim _{t\rightarrow \infty }\Vert u(\cdot ,t;u_0)-\frac{a}{b}\Vert _{\infty }=0 \end{aligned}$$
(1.12)

and

$$\begin{aligned} \lim _{t\rightarrow \infty }\Vert \lambda _iv_i(\cdot ,t;u_0)-\frac{a}{b}\mu _i\Vert _{\infty }=0 , \ \forall \ i=1,2. \end{aligned}$$
(1.13)

Remark 1.2

  1. (1)

    (1.10) provides explicit conditions for the global stability of the constant equilibrium \((\frac{a}{b},\frac{\mu _1}{\lambda _1}\frac{a}{b},\frac{\mu _2}{\lambda _2}\frac{a}{b})\) with respect to strictly positive initial functions. We point out the following special and important equivalent conditions of (1.10).

    1. (i)

      If \(\lambda _{1}\le \lambda _2\), and \(\chi _2\mu _2\lambda _2\ge \chi _1\mu _1\lambda _1\), then (1.10) holds if and only if \( b>2 \chi _1\mu _1-2\frac{\lambda _1}{\lambda _2}\chi _1\mu _1\).

    2. (ii)

      If \(\lambda _{1}\le \lambda _2\), and \(\chi _2\mu _2\lambda _2\le \chi _1\mu _1\lambda _1,\) then (1.10) holds if and only if \(b>2\chi _1\mu _1-2\chi _2\mu _2\).

    3. (iii)

      If \(\lambda _{1}\ge \lambda _2\), and \(\chi _2\mu _2\lambda _2\ge \chi _1\mu _1\lambda _1\), then (1.10) holds if and only if \(b>0\).

    4. (iv)

      If \(\lambda _{1}\ge \lambda _2\), and \(\chi _1\mu _1\lambda _1\ge \chi _2\mu _2\lambda _2\), then (1.10) holds if and only if \( b>2 \chi _1\mu _1-2\frac{\lambda _2}{\lambda _1}\chi _2\mu _2\).

  2. (2)

    By (i)-(iv), if \(b>2\chi _1\mu _1\), then (1.10) holds. Hence the hypothesis (1.10) is weaker than the known result on bounded domain.

  3. (3)

    If \(\chi _2=0\), then (ii) and (iv) extend [36, Theorem 1.7].

  4. (4)

    By (i) and (iii), if \(\chi _1=0\), then the constant solution \(\frac{a}{b}\) is stable with respect to strictly positive perturbation whenever \(b>0\).

  5. (5)

    It is interesting to know whether hypothesis (1.7) is enough to have the stability of the constant steady solution \((\frac{a}{b},\frac{a\mu _1}{b\lambda _1},\frac{a\mu _2}{b\lambda _2})\) with respect to strictly positive perturbation. We plan to study this question in our future work.

Next, we study the attraction of \((\frac{a}{b},\frac{\mu _1}{\lambda _1}\frac{a}{b},\frac{\mu _2}{\lambda _2}\frac{a}{b})\) with respect to global classical solutions of (1.4) with compactly supported initial functions, or equivalently, the spreading properties of global classical solutions of (1.4) with compactly supported initial functions. For \(x=(x_1,x_2,\cdots ,x_N)\in \mathbb {R}^N\), let \(|x|=\big (\sum _{i=1}^N x_i^2\big )^{\frac{1}{2}}\). We obtain the following main results.

Theorem C

Suppose that (1.7) holds and define

$$\begin{aligned} D:=\min \Big \{\frac{|\chi _2\mu _2-\chi _1\mu _1|}{2\sqrt{\lambda _2}}+\frac{\chi _1\mu _1|\sqrt{\lambda _1}-\sqrt{\lambda _2}|}{2\sqrt{\lambda _1\lambda _2}}, \frac{|\chi _1\mu _1-\chi _2\mu _2|}{2\sqrt{\lambda _1}}+\frac{\chi _2\mu _2|\sqrt{\lambda _2}-\sqrt{\lambda _1}|}{2\sqrt{\lambda _1\lambda _2}}\Big \}. \end{aligned}$$
(1.14)

Then for every \(u_{0}\in C^{b}_\mathrm{unif}({\mathbb R}^N)\) with \(u_0\ge 0\) and \(supp(u_0)\) being compact and non-empty, we have that

$$\begin{aligned} \lim _{t\rightarrow \infty }\left[ \sup _{|x|\ge ct}|u(x,t;u_0)| + \sup _{|x|\ge ct}| v_1(x,t;u_0)| + \sup _{|x|\ge ct}|v_2(x,t;u_0)|\right] =0 \end{aligned}$$
(1.15)

for every \(c> c_+^{*}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)\), where

$$\begin{aligned} c_{+}^{*}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)=2\sqrt{a}+\frac{\sqrt{a}(D\sqrt{Na}+\chi _2\mu _2)}{b+\chi _2\mu _2-\chi _1\mu _1-M}, \end{aligned}$$
(1.16)

and M is given by (1.8).

Remark 1.3

  1. (i)

    If \(\lambda _1\le \lambda _2\) and \(\chi _2\mu _2\lambda _2\ge \chi _1\mu _1\lambda _1\), then

    $$\begin{aligned} c_{+}^{*}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)=2\sqrt{a}+\frac{\sqrt{a}(D\sqrt{Na}+\chi _2\mu _2)}{b-\left( 1-\frac{\lambda _1}{\lambda _2}\right) \chi _1\mu _1}. \end{aligned}$$
  2. (ii)

    If \(\lambda _1\le \lambda _2\) and \(\chi _2\mu _2\lambda _2\le \chi _1\mu _1\lambda _1\), then

    $$\begin{aligned} c^{*}_+(\chi _1,\mu _1,\lambda _1,\chi _1,\mu _2,\lambda _2)=2\sqrt{a}+\frac{\sqrt{a}(D\sqrt{Na}+\chi _2\mu _2)}{b+\chi _2\mu _2-\chi _1\mu _1}. \end{aligned}$$
  3. (iii)

    If \(\lambda _1\ge \lambda _2 \) and \(\chi _{2}\mu _2\lambda _2\ge \chi _1\mu _1\lambda _1\) then

    $$\begin{aligned} c^{*}_+(\chi _1,\mu _1,\lambda _1,\chi _1,\mu _2,\lambda _2)=2\sqrt{a}+\frac{\sqrt{a}(D\sqrt{Na}+\chi _2\mu _2)}{b}. \end{aligned}$$
  4. (iv)

    If \(\lambda _1\ge \lambda _2 \) and \(\chi _{2}\mu _2\lambda _2\le \chi _1\mu _1\lambda _1\), then

    $$\begin{aligned} c^{*}_+(\chi _1,\mu _1,\lambda _1,\chi _1,\mu _2,\lambda _2)=2\sqrt{a}+\frac{\sqrt{a}(D\sqrt{Na}+\chi _2\mu _2)}{b-\frac{1}{\lambda _1}(\chi _1\mu _1\lambda _1-\chi _2\mu _2\lambda _2) }. \end{aligned}$$
  5. (v)

    Note that \(\chi _2=0\) implies that \(D=\frac{\chi _1\mu _1}{2\sqrt{\lambda _1}}\) and \(M=0\). Hence if \(\chi _2=0\), it follows from Theorem C that \(c^{*}_{+}(\chi _1,\mu _1,\lambda _1,0,\mu _2,\lambda _2)=2\sqrt{a}+\frac{a\chi _1\mu _1\sqrt{N}}{2(b-\chi _1\mu _1)\sqrt{\lambda _1}}\). Thus, in the case \(\chi _2=0\), and \(\mu _1=\lambda _1=1\), we obtain a better estimate for \(c^{*}_{+}(\chi _1,\mu _1,\lambda _1,0,\mu _2,\lambda _2)\) compare to the one giving by [37, Remark 1.2(iii)].

Theorem D

Suppose that (1.10) holds and

$$\begin{aligned} 4a(1-L)-\frac{Na^2D^2}{(b+\chi _2\mu _2-\chi _1\mu _1-M)^2}>0, \end{aligned}$$
(1.17)

where M is given by (1.8) and

$$\begin{aligned} L:=\min \Big \{&\frac{(\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}}{\lambda _2(b+\chi _2\mu _2-\chi _1\mu _1-M)}, \nonumber \\&\quad \frac{(\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _2\mu _2(\lambda _1-\lambda _2)_{-}}{\lambda _1(b+\chi _2\mu _2-\chi _1\mu _1-M)} \Big \}. \end{aligned}$$
(1.18)

Then for every \(u_{0}\in C^{b}_\mathrm{unif}({\mathbb R}^N)\) with \(u_0\ge 0\) and \(supp(u_0)\) being non-empty, we have that

$$\begin{aligned} \lim _{t\rightarrow \infty }\left[ \sup _{|x|\le ct}|u(x,t;u_0)-\frac{a}{b}| + \sup _{|x|\le ct}|\lambda _1 v_1(x,t;u_0)-\frac{a}{b}\mu _1| + \sup _{|x|\le ct}|\lambda _2 v_2(x,t;u_0)-\frac{a}{b}\mu _2|\right] =0 \end{aligned}$$
(1.19)

for every \(0\le c< c_{-}^{*}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)\), where

$$\begin{aligned} c^*_{-}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)=2\sqrt{a(1-L)}-\frac{aD\sqrt{N}}{b+\chi _2\mu _2-\chi _1\mu _1-M}. \end{aligned}$$

Remark 1.4

  1. (i)

    If \(\lambda _1\le \lambda _2\) and \(\chi _2\mu _2\lambda _2\ge \chi _1\mu _1\lambda _1\), then \(L=\frac{\chi _1\mu _1(1-\frac{\lambda _1}{\lambda _2})}{b-\chi _1\mu _1(1-\frac{\lambda _1}{\lambda _2})}\) and

    $$\begin{aligned} c^{*}_{-}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)= 2\sqrt{\frac{a\left( b-2\chi _1\mu _1\left( 1-\frac{\lambda _1}{\lambda _2}\right) \right) }{b-\chi _1\mu _1\left( 1-\frac{\lambda _1}{\lambda _2}\right) }} -\frac{aD\sqrt{N}}{b-\chi _1\mu _1\left( 1-\frac{\lambda _1}{\lambda _2}\right) }. \end{aligned}$$
  2. (ii)

    If \(\lambda _1\le \lambda _2\) and \(\chi _2\mu _2\lambda _2\le \chi _1\mu _1\lambda _1\), then \(L=\frac{\chi _1\mu _1-\chi _2\mu _2}{b+\chi _2\mu _2-\chi _1\mu _1}\) and

    $$\begin{aligned} c^{*}_{-}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)=2\sqrt{\frac{a(b-2(\chi _1\mu _1-\chi _2\mu _2))}{{b+\chi _2\mu _2-\chi _1\mu _1}}} -\frac{aD\sqrt{N}}{{b+\chi _2\mu _2-\chi _1\mu _1}}. \end{aligned}$$
  3. (iii)

    If \(\lambda _1\ge \lambda _2\) and \(\chi _2\mu _2\lambda _2\ge \chi _1\mu _1\lambda _1\), then \(L=0\) and

    $$\begin{aligned} c^{*}_{-}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)= 2\sqrt{a} -\frac{aD\sqrt{N}}{b}. \end{aligned}$$
  4. (iv)

    If \(\lambda _1\ge \lambda _2\) and \(\chi _2\mu _2\lambda _2\le \chi _1\mu _1\lambda _1\), then \(L=\frac{\chi _1\mu _1\lambda _1-\chi _2\mu _2\lambda _2}{\lambda _1(b+\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)}\) and

    $$\begin{aligned} c^{*}_{-}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)= & {} 2\sqrt{\frac{a(b-\frac{2}{\lambda _1}(\chi _1\mu _1\lambda _1-\chi _2\mu _2\lambda _2))}{{b-\frac{1}{\lambda _1}(\chi _1\mu _1\lambda _1-\chi _2\mu _2\lambda _2)}}}\\&-\frac{aD\sqrt{N}}{{b-\frac{1}{\lambda _1}(\chi _1\mu _1\lambda _1-\chi _2\mu _2\lambda _2)}}. \end{aligned}$$
  5. (v)

    If \(\chi _2=0\), by (ii) and (iv), we have that \(c^{*}_{-}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)= 2\sqrt{\frac{a(b-2\chi _1\mu _1)}{b-\chi _1\mu _1}}-\frac{a\chi _1\mu _1\sqrt{N}}{2(b-\chi _1\mu _1)\sqrt{\lambda _1}}.\) Hence in the case \(\chi _2=0,\)\(\mu _1=\lambda _1=1\), we obtain a better estimate on \(c^{*}_{-}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)\) than the ones obtained in [37] and [36].

Observe that, if either \(\chi _1=\chi _2=0\) or \(\chi _1-\chi _2=\mu _1-\mu _2=\lambda _1-\lambda _2=0\), the first equation in (1.4) becomes the following scalar reaction diffusion equation,

$$\begin{aligned} u_{t}=\Delta u + u(a-bu),\quad x\in {\mathbb R}^N,\,\, t>0, \end{aligned}$$
(1.20)

which is referred to as Fisher or KPP equations due to the pioneering works by Fisher ([9]) and Kolmogorov, Petrowsky, Piscunov ([25]) on the spreading properties of (1.20). It follows from the works [9, 25], and [47] that \(c^*_{-}\) and \(c^*_{+}\) in Theorem C and Theorem D, respectively, can be chosen so that \(c^{*}_{-}=c^{*}_{+}=2\sqrt{a}\) (\(c^*:=2\sqrt{a}\) is called the spatial spreading speed of (1.20) in literature), and that (1.20) has traveling wave solutions \(u(t,x)=\phi (x-ct)\) connecting \(\frac{a}{b}\) and 0 (i.e. \((\phi (-\infty )=\frac{a}{b},\phi (\infty )=0)\)) for all speeds \(c\ge c^*\) and has no such traveling wave solutions of slower speed. Since the pioneering works by Fisher [9] and Kolmogorov, Petrowsky, Piscunov [25], a huge amount research has been carried out toward the spreading properties of reaction diffusion equations of the form,

$$\begin{aligned} u_t=\Delta u+u f(t,x,u),\quad x\in {\mathbb R}^N, \end{aligned}$$
(1.21)

where \(f(t,x,u)<0\) for \(u\gg 1\), \(\partial _u f(t,x,u)<0\) for \(u\ge 0\) (see [2,3,4,5, 10, 11, 26, 28, 32, 34, 35, 39, 40, 47, 48, 58], etc.).

Remark 1.5

  1. (i)

    It is clear from Theorem C and Theorem D that

    $$\begin{aligned} \lim _{(\chi _1,\chi _2)\rightarrow (0,0)}c^*_-(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)=\lim _{(\chi _1,\chi _2)\rightarrow (0,0)}c^*_+(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)=2\sqrt{a} \end{aligned}$$

    and

    $$\begin{aligned}&\lim _{(\delta _1,\delta _2,\delta _3)\rightarrow (0,0,0)}c^*_-(\chi +\delta _1,\mu +\delta _2,\lambda +\delta _3,\chi ,\mu ,\lambda )\\&=\lim _{(\delta _1,\delta _2,\delta _3)\rightarrow (0,0,0)}c^*_+(\chi +\delta _1,\mu +\delta _2,\lambda +\delta _3,\chi ,\mu ,\lambda )\\&\lim _{(\delta _1,\delta _2,\delta _3)\rightarrow (0,0,0)}c^*_-(\chi ,\mu ,\lambda ,\chi +\delta _1,\mu +\delta _2,\lambda +\delta _3)\\&=\lim _{(\delta _1,\delta _2,\delta _3)\rightarrow (0,0,0)}c^*_+(\chi ,\mu ,\lambda ,\chi +\delta _1,\mu +\delta _2,\lambda +\delta _3)\\&=2\sqrt{a}, \quad \forall \ \chi>0,\mu>0\ \text {and} \ \lambda >0. \end{aligned}$$

    Hence we recover the know results in the literature when \(\chi _1=\chi _2=0\) or \(\chi _1-\chi _2=\mu _1-\mu _2=\lambda _1-\lambda _2=0\).

  2. (ii)

    For every \(\chi _i\ge 0,\ \mu _i>0,\ \lambda _i>0\), let

    $$\begin{aligned} c^*_{up}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)=\inf \{ c^*>0\ | \ (1.15)\ \text {holds}\} \end{aligned}$$

    and

    $$\begin{aligned} c^*_{low}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)=\sup \{ c^*\ge 0 \ | \ (1.19)\ \text {holds}\}. \end{aligned}$$

    \([c^*_{low}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2),c^*_{up}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)]\) is called the spreading speed interval of (1.4). Theorem C implies that if (1.7) holds, then

    $$\begin{aligned} c^*_{up}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)\le c^*_{+}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)<\infty . \end{aligned}$$

    Under the hypotheses of Theorem D, we have that

    $$\begin{aligned} c^*_{low}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)\ge c^*_{-}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)>0. \end{aligned}$$

    It is interesting to know the relationship between \(c^*_{up}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)\) and \(2\sqrt{a}\) as well as the relationship between \(c^*_{low}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)\) and \(2\sqrt{a}\). It is also interesting to know whether \(c^*_{low}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)=c^*_{up}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)\). We plan to study these questions in our future works.

  3. (iii)

    When \(\chi _2=0\), \(\lambda _1=\mu _1=1\), and \(0< \chi _1<\frac{b}{2}\), in a very recent work [37] it was shown that there is a positive constant \(c^{*}(\chi _1)\ge 2\sqrt{a}\) such that for every \(c\ge c^{*}(\chi _1)\) and \(\xi \in S^{N-1}\), (1.4) has a traveling wave solution \((u(x,t),v(x,t))=(u(x\cdot \xi -ct),v(x\cdot \xi -ct))\) connecting the trivial solutions \((\frac{a}{b},\frac{a}{b})\) and (0, 0) and propagating in the direction of \(\xi \) with speed c, and no such traveling wave solution exists for speed less than \(2\sqrt{a}\). We plan to study these questions for (1.4) when both \(\chi _1>0\) and \(\chi _2>0\).

We end up the introduction with the following remarks. First, our study is based on many techniques developed in [36]. But, to apply these techniques to (1.4) with non-zeros \(\chi _1\) and \(\chi _2\), nontrivial modifications are needed and made in the current paper. The modified techniques would be useful for the further study of attraction–repulsion chemotaxis systems. Second, most results obtained in [36] for the special case \(\chi _2=0\) are recovered and extended further in the current paper. Third, conditions explicitly depending on the sensitivity parameters \(\chi _1\) and \(\chi _2\) of the chemoattractant and chemorepellent are provided in the current paper for the global existence of classical solutions of (1.4) and stability of the nonzero constant equilibrium \((\frac{a}{b},\frac{\mu _1}{\lambda _1}\frac{a}{b},\frac{\mu _2}{\lambda _2}\frac{a}{b})\), and lower and upper bounds explicitly depending on \(\chi _1\) and \(\chi _2\) are established for the spreading speeds of positive solutions with compactly supported initial distributions. These conditions and lower and upper bounds would be of great practical importance.

The rest of the paper is organized as follows. Section 2 is devoted to the study of global existence of classical solutions. It is here that we prove Theorem A. In Sect. 3, we study the asymptotic stability of the constant equilibrium \((\frac{a}{b},\frac{\mu _1}{\lambda _1}\frac{a}{b},\frac{\mu _2}{\lambda _2}\frac{a}{b})\) and prove Theorem B. We study the spreading properties of global classical solutions of (1.4) with compactly supported initial functions and prove Theorems C and D in Sect. 4.

2 Global Existence

In this section, we discuss the existence of global/bounded classical solutions and prove Theorem A. We start with the following result which guarantees the existence of a unique local in time classical solution of (1.4) for any nonnegative bounded and uniformly continuous initial data.

Lemma 2.1

For any \(u_0 \in C_\mathrm{unif}^{b}({\mathbb R}^N)\) with \(u_0 \ge 0\), there exists \(T_{max} \in (0,\infty ]\) such that (1.4) has a unique non-negative classical solution \((u(x,t;u_0),v_1(x,t;u_0), v_2(x,t,u_0))\) on \([0,T_{\max })\) with \(\lim _{t\rightarrow 0}u(\cdot ,t;u_0)=u_0\) in \(C_\mathrm{uinf}^b({\mathbb R}^N)\)-norm. Moreover, if \(T_{max}< \infty ,\) then

$$\begin{aligned} \limsup _{t \rightarrow T_{max}}\Vert u(\cdot ,t;u_0)\Vert _{\infty }=\infty . \end{aligned}$$
(2.1)

Proof

It follows from the similar arguments used in the proof of [36, Theorem 1.1]. \(\square \)

Proof of Theorem A

Let \(u_{0}\in C^{b}_\mathrm{unif}({\mathbb R}^N)\) with \(u_0\ge 0\) be given and let \((u(\cdot ,\cdot ;u_0),v_1(\cdot ,\cdot ;u_0)\), \(v_2(\cdot ,\cdot ;u_0))\) be the classical solution of (1.4) with initial function \(u_0\) defined on the maximal interval \([0, T_{\max })\) of existence. Then,

$$\begin{aligned} u_{t}= & {} \Delta u-\chi _1\nabla (u\nabla v_1)+\chi _2\nabla (u\nabla v_2) +u(a-bu)\nonumber \\= & {} \Delta u +\nabla (\chi _2 v_2-\chi _1 v_1)\nabla u+u(a-\chi _1\Delta v_1 +\chi _2\Delta v_2 -bu),\quad x\in {\mathbb R}^N. \end{aligned}$$
(2.2)

The second and third equations of (1.4) yield that \(\Delta v_i=\lambda _i v_i-\mu _i u\), \(i=1,2\). Hence equation (2.2) becomes

$$\begin{aligned} u_{t}= \Delta u +\nabla (\chi _2 v_2-\chi _1 v_1)\nabla u+u\Big (a+ (\chi _{2}\lambda _2v_2-\chi _1\lambda _1v_1)-(b+\chi _2\mu _2-\chi _1\mu _1 )u\Big ),\,\,\, x\in {\mathbb R}^N. \end{aligned}$$
(2.3)

Let

$$\begin{aligned} C_0:={\left\{ \begin{array}{ll} \Vert u_0\Vert _{\infty }&{}\text {if }\ \chi _1=a=b=0,\\ \max \{\Vert u_0\Vert _{\infty }, \frac{a}{b+\chi _2\mu _2-\chi _1\mu _1-M}\}&{} \text {if} \ b+\chi _2\mu _2-\chi _1\mu _1-M>0 \end{array}\right. } \end{aligned}$$
(2.4)

where M is given by (1.8). Let \(T>0\) be a given positive real number and consider \(\mathcal {E}^{T}:=C^{b}_\mathrm{unif}({\mathbb R}^N\times [0, T])\) endowed with the norm

$$\begin{aligned} \Vert u\Vert _{\mathcal {E}^T}:=\sum _{k=1}^{\infty }\frac{1}{2^k} \Vert u\Vert _{L^{\infty }([-k,k]\times [0,T])}. \end{aligned}$$
(2.5)

We note that the convergence in \((\mathcal {E}^T,\Vert .\Vert _{\mathcal {E}^T})\) is equivalent to the uniform convergence on compact subsets on \({\mathbb R}^N\times [0,T]\). Next, we consider the subset \(\mathcal {E}\) of \(\mathcal {E}^T\) defined by

$$\begin{aligned} \mathcal {E}:=\{u\in C_\mathrm{unif}^b({\mathbb R}^N\times [0, T])\,|\, u(\cdot ,0)=u_0, 0\le u(x,t)\le C_0, x\in {\mathbb R}^N, 0\le t\le T\}. \end{aligned}$$

It is clear that

$$\begin{aligned} \Vert u\Vert _{\mathcal {E}^T}\le C_{0}, \quad \forall \ u\in \mathcal {E}. \end{aligned}$$
(2.6)

It readily follows from the definition of \(\mathcal {E}\) and (2.6) that \(\mathcal {E}\) is a closed bounded and convex subset of \(\mathcal {E}^T\). We shall show that \(u(\cdot ,\cdot ;u_0)\in \mathcal {E}\).

For every \(u\in \mathcal {E}\) let us define \(v_{i}(\cdot ,\cdot ;u)\), \(i=1,2\) by

$$\begin{aligned} v_{i}(x,t;u)=\mu _{i}\int _{0}^{\infty }\int _{{\mathbb R}^N}\frac{e^{-\lambda _i s}}{(4\pi s)^{\frac{N}{2}}}e^{-\frac{|z-x|^2}{4s}}u(z,t)dzds, \quad x\in {\mathbb R}^N,\ t\in [0, T]. \end{aligned}$$
(2.7)

and let U(xtu) be the solution of the initial value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} U_{t}=\Delta U+\nabla \big (\chi _2 v_2(x,t;u)-\chi _1v_1(x,t;u)\big )\nabla U\\ \qquad \,\,\, +U\Big (a+(\chi _2\lambda _2v_2(x,t;u) -\chi _1\lambda _1v_1(x,t;u)) -(b+\chi _2\mu _2-\chi _1\mu _1)U\Big ),\quad x\in {\mathbb R}^N\\ U(\cdot ,0,u)=u_0(\cdot ). \end{array}\right. } \end{aligned}$$
(2.8)

For every \(u\in \mathcal {E}\), using (2.7), we have that

$$\begin{aligned} ( \chi _2\lambda _2v_2 -\chi _1\lambda _1v_1)(x,t;u)= & {} \int _{0}^{\infty }\int _{{\mathbb R}^N}\left[ \chi _2\lambda _2\mu _2e^{-\lambda _{2}s} - \chi _1\lambda _1\mu _1e^{-\lambda _{1}s}\right] \frac{e^{-\frac{|x-z|^2}{4s}}}{(4\pi s)^{\frac{N}{2}}}u(z,t)dzds \nonumber \\= & {} (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)\int _0^{\infty }\int _{{\mathbb R}^N}e^{-\lambda _2 s}\frac{e^{-\frac{|x-z|^2}{4s}}}{(4\pi s)^{\frac{N}{2}}}u(z,t)dzds \nonumber \\&+\chi _1\mu _1\lambda _1 \int _{0}^{\infty }\int _{{\mathbb R}^N}(e^{-\lambda _{2}s}-e^{-\lambda _{1}s})\frac{e^{-\frac{|x-z|^2}{4s}}}{(4\pi s)^{\frac{N}{2}}}u(z,t)dzds \nonumber \\\le & {} (\chi _2\lambda _2\mu _2-\chi _1\lambda _1\mu _1)_{+}C_0\int _{0}^{\infty }\int _{{\mathbb R}^N}e^{-\lambda _{2}s}\frac{e^{-\frac{|x-z|^2}{4s}}}{(4\pi s)^{\frac{N}{2}}}dzds\nonumber \\&+ \chi _1\mu _1\lambda _1 C_{0} \int _{0}^{\infty }\int _{{\mathbb R}^N}(e^{-\lambda _{2}s}-e^{-\lambda _{1}s})_{+}\frac{e^{-\frac{|x-z|^2}{4s}}}{(4\pi s)^{\frac{N}{2}}}dzds \nonumber \\= & {} \frac{C_0}{\lambda _2}\Big ((\chi _2\lambda _2\mu _2-\chi _1\lambda _1\mu _1)_{+}+\chi _1\mu _1(\lambda _1-\lambda _2)_{+}\Big ). \end{aligned}$$
(2.9)

Similarly, we have that

$$\begin{aligned} ( \chi _2\lambda _2v_2 -\chi _1\lambda _1v_1)(x,t;u)= & {} \int _{0}^{\infty }\int _{{\mathbb R}^N}\left[ \chi _2\lambda _2\mu _2e^{-\lambda _{2}s} - \chi _1\lambda _1\mu _1e^{-\lambda _{1}s}\right] \frac{e^{-\frac{|x-z|^2}{4s}}}{(4\pi s)^{\frac{N}{2}}}u(z,t)dzds \nonumber \\= & {} \chi _2\mu _2\lambda _2\int _0^{\infty }\int _{{\mathbb R}^N}(e^{-\lambda _2 s}-e^{-\lambda _1 s})\frac{e^{-\frac{|x-z|^2}{4s}}}{(4\pi s)^{\frac{N}{2}}}u(z,t)dzds \nonumber \\&+(\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)\int _{0}^{\infty }\int _{{\mathbb R}^N}e^{-\lambda _{1}s}\frac{e^{-\frac{|x-z|^2}{4s}}}{(4\pi s)^{\frac{N}{2}}}u(z,t)dzds \nonumber \\\le & {} \chi _2\mu _2\lambda _2 C_0\int _0^{\infty }\int _{{\mathbb R}^N}(e^{-\lambda _2 s}-e^{-\lambda _1 s})_{+}\frac{e^{-\frac{|x-z|^2}{4s}}}{(4\pi s)^{\frac{N}{2}}}dzds \nonumber \\&+(\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}C_{0}\int _{0}^{\infty }\int _{{\mathbb R}^N}e^{-\lambda _{1}s}\frac{e^{-\frac{|x-z|^2}{4s}}}{(4\pi s)^{\frac{N}{2}}}dzds \nonumber \\= & {} \frac{C_0}{\lambda _{1}}\Big ( \chi _2\mu _2(\lambda _1-\lambda _2)_{+} + (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+} \Big ). \end{aligned}$$
(2.10)

Thus, it follows from (2.9) and (2.10) that for every \(u\in \mathcal {E}\), we have that

$$\begin{aligned} (\chi _2 \lambda _2 v_2-\chi _1 \lambda _1 v_1)(x,t;u)\le M C_{0} \end{aligned}$$
(2.11)

where M is given by (1.8). Thus for every \(u\in \mathcal {E}\), we have that

$$\begin{aligned} U_t(x,t;u)\le&\Delta U(x,t;u)+\nabla (\chi _2v_2-\chi _1v_1)\nabla U(x,t;u)\nonumber \\&+\underbrace{\Big (a+M C_0-(b+\chi _2\mu _2-\chi _1\mu _1)U(x,t;u)\Big )U(x,t;u)}_{\mathcal {L}(U)}. \end{aligned}$$
(2.12)

Note that

$$\begin{aligned} \mathcal {L}(C_0)=\Big (a-(b+\chi _2\mu _2-\chi _1\mu _1-M)C_0\Big )C_0\le 0. \end{aligned}$$

Thus, using comparison principle for parabolic equations, we obtain that

$$\begin{aligned} U(x,t;u)\le C_0,\quad \forall \ x\in {\mathbb R}^N,\ \forall \ t\in [0,T],\ \forall \ u\in \mathcal {E}. \end{aligned}$$
(2.13)

Thus \( U(\cdot ,\cdot ;u)\in \mathcal {E}\) for every \(u\in \mathcal {E}\). By the arguments in [37, Lemma 4.3], the mapping \(\mathcal {E}\ni u\mapsto U(\cdot ,\cdot ;u)\in \mathcal {E}\) is continuous and compact, and then by Schauder’s fixed theorem, it has a fixed point \(u^*\). Clearly \((u^*,v_1(\cdot ,\cdot ;u^*),v_2(\cdot ,\cdot ;u^*))\) is a classical solution of (1.4). Thus, by Lemma 2.1, we have that \(T_{\max }\ge T\) and \(u(\cdot ,\cdot ;u_0)=u^*\). Since \(T>0\) is arbitrary chosen, Theorem A follows. \(\square \)

3 Asymptotic Stability of the Constant Equilibrium \((\frac{a}{b},\frac{\mu _1}{\lambda _1}\frac{a}{b},\frac{\mu _2}{\lambda _2}\frac{a}{b})\)

In this section, we discuss the asymptotic stability of the constant equilibrium \((\frac{a}{b},\frac{\mu _1}{\lambda _1}\frac{a}{b},\frac{\mu _2}{\lambda _2}\frac{a}{b})\) of (1.4) and prove Theorem B. Throughout this section we suppose that (1.7) holds, so that for every nonnegative, bounded, and uniformly continuous initial function \(u_0\), (1.4) has a nonnegative bounded global classical solution \((u(x,t;u_0),v_1(x,t;u_0),v_2(x,t;u_0))\).

For given \(u_0\in C^{b}_\mathrm{unif}({\mathbb R}^N)\) with \(u_0\ge 0 \), define

$$\begin{aligned} \underline{u}:=\liminf _{t\rightarrow \infty }\inf _{x\in {\mathbb R}^N}u(x,t;u_0)\quad \text {and}\quad \overline{u}:=\limsup _{t\rightarrow \infty }\sup _{x\in {\mathbb R}^N}u(x,t;u_0). \end{aligned}$$

Using the definition of limsup and liminf, we have that for every \(\varepsilon >0\), there is \(T_{\varepsilon }>0\) such that

$$\begin{aligned} \underline{u}-\varepsilon \le u(x,t;u_0)\le \overline{u}+\varepsilon \quad \forall \ x\in {\mathbb R}^N,\ \forall \ t\ge T_{\varepsilon }. \end{aligned}$$

Hence, it follows from comparison principle for elliptic equations, that

$$\begin{aligned} \mu _i(\underline{u}-\varepsilon )\le \lambda _iv_i(x,t;u_0)\le \mu _i(\overline{u}+\varepsilon ), \forall \ x\in {\mathbb R}^N,\ \ \forall \ t\ge T_{\varepsilon },\ i=1,2. \end{aligned}$$
(3.1)

We first show the following important result.

Lemma 3.1

Suppose that (1.7) holds. If \(\inf _{x\in {\mathbb R}^N}u_0(x)>0\), then

$$\begin{aligned} \inf _{x\in {\mathbb R}^N}u(x,t;u_0)>0,\ \quad \forall \ t>0. \end{aligned}$$
(3.2)

Proof

Let \(K:=\chi _{1}\lambda _{1}\sup _{x\in {\mathbb R}^N}v_{1}(x,t;u_0)\). Thus, it follows from (2.3) that

$$\begin{aligned} u_{t}(\cdot ,\cdot ;u_0)\ge&\Delta u +\nabla (\chi _{2}v_2(\cdot ,\cdot ;u_0)-\chi _{1}v_1(\cdot ,\cdot ;u_0))\nabla u(\cdot ,\cdot ;u_0)\\&\,\,\, +(a-K-(b+\chi _2\mu _2-\chi _1\mu _1)u(\cdot ,\cdot ;u_0))u(\cdot ,\cdot ;u_0). \end{aligned}$$

Hence, comparison principle for parabolic equations implies that

$$\begin{aligned} u(x,t;u_0)\ge W(t),\quad \forall \ t\ge 0, \ x\in {\mathbb R}^N, \end{aligned}$$

where W is the solution of the ODE

$$\begin{aligned} {\left\{ \begin{array}{ll} W_t=W(a-K-(b+\chi _2\mu _2-\chi _{1}\mu _1)W),\ t>0,\\ W(0)=\inf _{x\in {\mathbb R}^N}u_0(x). \end{array}\right. } \end{aligned}$$

Since \(b+\chi _{2}\mu _2-\chi _1\mu _1>0\), and \(\inf _{x\in {\mathbb R}^{N}}u_{0}(x)>0\), we have that W(t) is defined for all time and satisfies \(W(t)>0\) for every \(t\ge 0\). Hence, we obtain that \(0<W(t)\le \inf _{x\in {\mathbb R}^N}u(x,t;u_0)\) for all \( t\ge 0\).

Proof of Theorem B

We divide the proof into two cases.

Case I. Assume that \(b+\chi _2\mu _2-\chi _1\mu _1-\frac{1}{\lambda _2}\Big [|\chi _1\mu _1\lambda _1- \chi _2\mu _2\lambda _2|+\chi _1\mu _1|\lambda _1-\lambda _2| \Big ]>0\).

For every \(t\ge T_\varepsilon \) (\(T_\epsilon \) is such that (3.1) holds), and \(x\in {\mathbb R}^N\), we have that

$$\begin{aligned} (\chi _2\lambda _2v_2-\chi _1\lambda _1v_1)(x,t;u_0)= & {} (\chi _2\mu _2\lambda _2{-}\chi _1\mu _1\lambda _1)\int _0^\infty \int _{{\mathbb R}^N}e^{-\lambda _2 s}\frac{e^{-\frac{|x-z|^2}{4s}}}{(4\pi s)^{\frac{N}{2}}}u(z,t;u_0)dzds\nonumber \\&+ \chi _1\mu _1\lambda _1\int _0^\infty \int _{{\mathbb R}^N}(e^{-\lambda _2 s}-e^{-\lambda _1 s})\frac{e^{-\frac{|x-z|^2}{4s}}}{(4\pi s)^{\frac{N}{2}}}u(z,t;u_0)dzdt\nonumber \\\le & {} \frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _{1}\mu _{1}(\lambda _1-\lambda _2)_{+}\right] (\overline{u}+\varepsilon )\nonumber \\&-\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] (\underline{u}-\varepsilon )\nonumber \\ \end{aligned}$$
(3.3)

and

$$\begin{aligned} (\chi _2\lambda _2v_2-\chi _1\lambda _1v_1)(x,t;u_0)= & {} (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)\int _0^\infty \int _{{\mathbb R}^N}e^{-\lambda _2 s}\frac{e^{-\frac{|x-z|^2}{4s}}}{(4\pi s)^{\frac{N}{2}}}u(z,t;u_0)dzds\nonumber \\+ & {} \chi _1\mu _1\lambda _1\int _0^\infty \int _{{\mathbb R}^N}(e^{-\lambda _2 s}-e^{-\lambda _1 s})\frac{e^{-\frac{|x-z|^2}{4s}}}{(4\pi s)^{\frac{N}{2}}}u(z,t;u_0)dzdt\nonumber \\\ge & {} \frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _{1}\mu _{1}(\lambda _1-\lambda _2)_{+}\right] (\underline{u}-\varepsilon )\nonumber \\&-\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] (\overline{u}+\varepsilon ).\nonumber \\ \end{aligned}$$
(3.4)

Hence, for every \(t\ge T_\varepsilon \), \(x\in {\mathbb R}^N\), it follows from (2.3), (3.1) and (3.3) that

$$\begin{aligned} u_t\le & {} \Delta u+\nabla (\chi _2v_2-\chi _1v_1)\nabla u + \left( a +\frac{1}{\lambda _2} \left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+} +\chi _{1}\mu _{1}(\lambda _1-\lambda _2)_{+}\right] (\overline{u}+\varepsilon ) \right) u\nonumber \\&-\left( \frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] (\underline{u}-\varepsilon ) +(b+\chi _2\mu _2-\chi _1\mu _1)u\right) u. \end{aligned}$$
(3.5)

Thus, by comparison principle for parabolic equations, we have that

$$\begin{aligned} u(x,t;u_0)\le U_{\varepsilon }(t), \quad \forall x\in {\mathbb R}^N,\ t\ge T_\varepsilon , \end{aligned}$$
(3.6)

where \(U_\varepsilon (t)\) is the solution of the ODE

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t U=\Big (a +\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _{1}\mu _{1}(\lambda _1-\lambda _2)_{+}\right] (\overline{u}+\varepsilon ) \Big )U\\ \ \ \ \ \ -\Big (\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] (\underline{u}-\varepsilon ) +(b+\chi _2\mu _2-\chi _1\mu _1)U\Big )U \quad t>T_\varepsilon ,\\ U(T_\varepsilon )=\Vert u(\cdot ,T_\varepsilon ;u_0)\Vert _{\infty }. \end{array}\right. } \end{aligned}$$

Since \(b+\chi _2\mu _2-\chi _1\mu _1>0\) and \(\Vert u(\cdot ,T_{\infty };u_0)\Vert _{\infty }>0\), we have that \(U_{\varepsilon }(t)\) is defined for all time \(t\ge T_\varepsilon \) and satisfies

$$\begin{aligned} \lim _{t\rightarrow \infty }U_{\varepsilon }=&\frac{1}{b+\chi _2\mu _2-\chi _1\mu _1}\Big \{a+\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _{1} \mu _{1}(\lambda _1-\lambda _2)_{+}\right] (\overline{u}+\varepsilon )\\&\qquad \qquad \qquad \qquad -\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] (\underline{u}-\varepsilon )\Big \}_+ \end{aligned}$$

This combined with (3.6) yield that

$$\begin{aligned} \overline{u}&\le \frac{1}{b+\chi _2\mu _2-\chi _1\mu _1} \Big \{a+\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _{1}\mu _{1}(\lambda _1-\lambda _2)_{+}\right] (\overline{u}+\varepsilon ) \\&\qquad \qquad \qquad \qquad \quad -\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] (\underline{u}-\varepsilon )\Big \}_+. \end{aligned}$$

Letting \(\varepsilon \) goes to 0 in the last inequality, we obtain that

$$\begin{aligned} \overline{u}&\le \frac{1}{b+\chi _2\mu _2-\chi _1\mu _1} \Big \{a+\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _{1}\mu _{1}(\lambda _1-\lambda _2)_{+}\right] \overline{u} \\&\qquad \qquad \qquad \quad \qquad -\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] \underline{u}\Big \}_+. \end{aligned}$$

If

$$\begin{aligned} \Big \{&a+\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _{1}\mu _{1}(\lambda _1-\lambda _2)_{+}\right] \overline{u}\\&- \frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] \underline{u}\Big \}_+=0, \end{aligned}$$

then \(\overline{u}=\underline{u}=0\). This in turn yields that

$$\begin{aligned} 0=&\Big \{a+\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _{1}\mu _{1}(\lambda _1-\lambda _2)_{+}\right] \overline{u}\\&-\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] \underline{u}\Big \}_+=a, \end{aligned}$$

which is impossible, since \(a>0\). Hence

$$\begin{aligned} \overline{u}\le&\frac{1}{b+\chi _2\mu _2-\chi _1\mu _1}\Big [ a+\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _{1}\mu _{1}(\lambda _1-\lambda _2)_{+}\right] \overline{u}\nonumber \\&\qquad \qquad \qquad \qquad -\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] \underline{u}\Big ]. \end{aligned}$$
(3.7)

On the other hand, for every \(t\ge T_\varepsilon \), \(x\in {\mathbb R}^N\), it follows from (2.3), (3.1) and (3.3) that

$$\begin{aligned} u_t\ge & {} \Delta u+\nabla (\chi _2v_2-\chi _1v_1)\nabla u + \Big (a +\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _{1}\mu _{1}(\lambda _1-\lambda _2)_{+}\right] (\underline{u}-\varepsilon ) \Big )u\nonumber \\&-\Big (\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] (\overline{u}+\varepsilon ) +(b+\chi _2\mu _2-\chi _1\mu _1)u\Big )u. \end{aligned}$$
(3.8)

Thus, by comparison principle for parabolic equations, we have that

$$\begin{aligned} u(x,t;u_0)\ge U^{\varepsilon }(t), \quad \forall x\in {\mathbb R}^N,\ t\ge T_\varepsilon , \end{aligned}$$
(3.9)

where \(U^\varepsilon (t)\) is the solution of the ODE

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t U=\Big (a +\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _{1}\mu _{1}(\lambda _1-\lambda _2)_{+}\right] (\underline{u}-\varepsilon ) \Big )U\\ \ \ \ \ \ -\Big (\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] (\overline{u}+\varepsilon ) +(b+\chi _2\mu _2-\chi _1\mu _1)U\Big )U, \quad t>T_\varepsilon \\ U(T_\varepsilon )=\inf _{x\in {\mathbb R}^N}u(x,T_\varepsilon ). \end{array}\right. } \end{aligned}$$

But, by Lemma 3.1 we have that \(\inf _{x\in {\mathbb R}^N}u(x,T_\varepsilon ;u_0) >0\). Since \(b+\chi _2\mu _2-\chi _1\mu _1>0\), we have that \(U^{\varepsilon }(t)\) is defined for all time \(t\ge T_\varepsilon \) and satisfies

$$\begin{aligned} \lim _{t\rightarrow \infty }U^{\varepsilon }&= \frac{1}{b+\chi _2\mu _2-\chi _1\mu _1}\Big \{a +\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _{1}\mu _{1}(\lambda _1-\lambda _2)_{+}\right] (\underline{u}-\varepsilon ) \\&\quad \qquad \qquad \qquad \qquad -(\frac{1}{\lambda _2}\left[ \chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] (\overline{u}+\varepsilon )\Big \}_{+}. \end{aligned}$$

This combined with (3.9) yield that

$$\begin{aligned} \underline{u}&\ge \frac{1}{b+\chi _2\mu _2-\chi _1\mu _1} \Big \{a +\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _{1}\mu _{1}(\lambda _1-\lambda _2)_{+}\right] (\underline{u}-\varepsilon )\\&\qquad \qquad \qquad \qquad \qquad -(\frac{1}{\lambda _2}\left[ \chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] (\overline{u}+\varepsilon )\Big \}_{+}. \end{aligned}$$

Letting \(\varepsilon \) goes to 0 in the last inequality, we obtain that

$$\begin{aligned} \underline{u}&\ge \frac{1}{b+\chi _2\mu _2-\chi _1\mu _1} \Big \{a +\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _{1}\mu _{1}(\lambda _1-\lambda _2)_{+}\right] \underline{u}\nonumber \\&\quad -(\frac{1}{\lambda _2}\left[ \chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] \overline{u}\Big \}_{+}. \end{aligned}$$
(3.10)

It follows from inequalities (3.7) and (3.10) that

$$\begin{aligned} \Big (b+\chi _2\mu _2-\chi _1\mu _1-\frac{1}{\lambda _2}\big [|\chi _1\mu _1\lambda _1- \chi _2\mu _2\lambda _2|+\chi _1\mu _1|\lambda _1-\lambda _2| \big ]\Big )(\overline{u}-\underline{u})\le 0. \end{aligned}$$
(3.11)

In this case, it follows from inequality (3.11) that \(\overline{u}=\underline{u}\). Combining this with (3.7) and (3.10), we obtain that \(\overline{u}=\underline{u}=\frac{a}{b}\). This ends the first case.

Case II. Assume that \(b+\chi _2 \mu _2 -\chi _1\mu _1 -\frac{1}{\lambda _{1}}\big [|\chi _1\mu _1\lambda _1-\chi _2\mu _2\lambda _2|+\chi _2\mu _2|\lambda _1-\lambda _2|\big ]>0\).

Rewrite \(\chi _2\lambda _2v_2-\chi _1\lambda _1v_1\) in the form

$$\begin{aligned} (\chi _2 \lambda _2 v_2-\chi _1 \lambda _1 v_1)(x,t;u_0)= & {} \chi _2\mu _2\lambda _2\int _0^\infty \int _{{\mathbb R}^N}(e^{-\lambda _2 s}-e^{-\lambda _{1}s})\frac{-\frac{|x-z|^2}{4s}}{(4\pi s)^{\frac{N}{2}}}u(z,t;u_0)dzds\nonumber \\&+ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)\int _0^\infty \int _{{\mathbb R}^N}e^{-\lambda _1 s}\frac{e^{-\frac{|x-z|^2}{4s}}}{(4\pi s)^{\frac{N}{2}}}u(z,t;u_0)dzds.\nonumber \\ \end{aligned}$$
(3.12)

It follows from the arguments used to establish inequalities (3.7) and (3.10) that

$$\begin{aligned} \overline{u}\,\, \le \,\, \frac{1}{ b+\chi _2\mu _2-\chi _1\mu _1} \Big \{&a+\frac{1}{\lambda _1}\left[ (\chi _2\mu _2(\lambda _1-\lambda _2)_{+}+\chi _2\mu _2(\lambda _1-\lambda _2)_{+}\right] \overline{u} \nonumber \\&-\frac{1}{\lambda _1}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _2\mu _2(\lambda _1-\lambda _2)_+\right] \underline{u}\Big \} \end{aligned}$$
(3.13)

and

$$\begin{aligned} \underline{u}\ge \frac{1}{ b+\chi _2\mu _2-\chi _1\mu _1} \Big \{&a+\frac{1}{\lambda _1}\left[ (\chi _2\mu _2(\lambda _1-\lambda _2)_{+}+\chi _2\mu _2(\lambda _1-\lambda _2)_{+}\right] \underline{u}\nonumber \\&-\frac{1}{\lambda _1}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _2\mu _2(\lambda _1-\lambda _2)_+\right] \overline{u}\Big \} \end{aligned}$$
(3.14)

hold respectively. It follows from (3.13) and (3.14) that

$$\begin{aligned} \Big ( b+\chi _2 \mu _2 -\chi _1\mu _1 -\frac{1}{\lambda _{1}}\big [|\chi _1\mu _1\lambda _1-\chi _2\mu _2\lambda _2|+\chi _2\mu _2|\lambda _1-\lambda _2|\big ]\Big )(\overline{u}-\underline{u})\le 0. \end{aligned}$$
(3.15)

Since \(b+\chi _2 \mu _2 -\chi _1\mu _1 -\frac{1}{\lambda _{1}}\left[ |\chi _1\mu _1\lambda _1-\chi _2\mu _2\lambda _2|+\chi _2\mu _2|\lambda _1-\lambda _2|\right] >0\), it follows from inequality (3.15) that \(\overline{u}=\underline{u}\). Combining this with (3.13) and (3.14), we obtain that \(\overline{u}=\underline{u}=\frac{a}{b}\). This end the second case.

Therefore, it follows from the results of cases I and II that if

$$\begin{aligned} b+\chi _2\mu _2-\chi _1\mu _1 > \min \Big \{&\frac{1}{\lambda _{2}}\left[ |\chi _1\mu _1\lambda _1-\chi _2\mu _2\lambda _2|+\chi _1\mu _1|\lambda _1-\lambda _2|\right] ,\\&\frac{1}{\lambda _{1}}\left[ |\chi _1\mu _1\lambda _1-\chi _2\mu _2\lambda _2|+\chi _2\mu _2|\lambda _1-\lambda _2|\right] \Big \}, \end{aligned}$$

then \(\overline{u}=\underline{u}=\frac{a}{b}\). Thus Theorem B follows. \(\square \)

4 Spreading Properties of Classical Solutions

In this section we study how fast the mobiles species spread over time and prove Theorems C and D. Throughout this section, we always suppose that \(u_0\in C^{b}_\mathrm{unif}({\mathbb R}^N)\), \(u_0(x)\ge 0\) has compact and nonempty support. The next three lemmas will be useful in the subsequent.

Lemma 4.1

Let \(u_0\in C_\mathrm{unif}^{b}({\mathbb R}^N)\), \(u_0\ge 0\), and \((u(\cdot ,\cdot ;u_0),v_{1}(\cdot ,\cdot ;u_0),v_2(\cdot ,\cdot ;u_0))\) be the classical solution of (1.4) with \(u(\cdot ,0;u_0)=u_0\). Then for every \(i\in \{1,\cdots ,N\}\), we have that

$$\begin{aligned}&\Vert \partial _{x_i}(\chi _2v_2-\chi _1v_1)(\cdot ,t;u_0)\Vert _{\infty }\nonumber \\&\quad \le \min \Big \{\frac{|\chi _2\mu _2-\chi _1\mu _1|}{2\sqrt{\lambda _2}}+\frac{\chi _1\mu _1|\sqrt{\lambda _1}-\sqrt{\lambda _2}|}{2\sqrt{\lambda _1\lambda _2}}, \frac{|\chi _1\mu _1-\chi _2\mu _2|}{2\sqrt{\lambda _1}}\nonumber \\&\qquad +\frac{\chi _2\mu _2|\sqrt{\lambda _2}-\sqrt{\lambda _1}|}{2\sqrt{\lambda _1\lambda _2}} \Big \}\Vert u(\cdot ,t;u_0)\Vert _{\infty } \end{aligned}$$
(4.1)

for every \(t\ge 0\).

Proof

For every \(i\in \{1,\cdots ,N\}\) and \(k\in \{1,2\}\), we have that

$$\begin{aligned}&\partial _{x_i}(\chi _kv_k)(x,t;u_0)\\&=\frac{\chi _k\mu _k}{\pi ^{\frac{N}{2}}}\int _0^\infty \int _{{\mathbb R}^N}\frac{ e^{-\lambda _k s}}{\sqrt{ s}}z_i e^{-|z|^2}u(x+2\sqrt{s}z,t;u_0)dzds \\&=\frac{\chi _k\mu _k}{\pi ^{\frac{N}{2}}}\int _0^\infty \int _{{\mathbb R}^{N-1}}\frac{ e^{-\lambda _k s}e^{-|y|^2}}{\sqrt{ s}}\left[ \int _{{\mathbb R}}\tau e^{-\tau ^2}u(x+2\sqrt{s}\tau e_i +2\sqrt{s} \pi _i^{-1}(y),t;u_0)d\tau \right] dyds \\&=\frac{\chi _k\mu _k}{\pi ^{\frac{N}{2}}}\int _0^\infty \int _{{\mathbb R}^{N-1}}\frac{ e^{-\lambda _k s}e^{-|y|^2}}{\sqrt{ s}}\left[ \int _0^\infty \tau e^{-\tau ^2}u(x+2\sqrt{s}\tau e_i +2\sqrt{s} \pi _i^{-1}(y),t;u_0)d\tau \right] dyds\\&- \frac{\chi _k\mu _k}{\pi ^{\frac{N}{2}}}\int _0^\infty \int _{{\mathbb R}^{N-1}}\frac{ e^{-\lambda _k s}e^{-|y|^2}}{\sqrt{ s}}\left[ \int _0^\infty \tau e^{-\tau ^2}u(x-2\sqrt{s}\tau e_i +2\sqrt{s} \pi _i^{-1}(y),t;u_0)d\tau \right] dyds, \end{aligned}$$

where \(e_i=(\delta _{1i},\delta _{2i},\cdots ,\delta _{Ni})\) with \(\delta _{ij}=0\) if \(i\not =j\) and \(\delta _{ii}=1\) for \(i,j=1,2,\cdots ,N\), and \(\pi _i^{-1}(y)=(y_1,y_2,\cdots ,y_{i-1},0,y_i,\cdots ,y_{N-1})\). Hence,

$$\begin{aligned}&\partial _{x_i}(\chi _2v_2-\chi _1v_1)(x,t;u_0)\nonumber \\&= \frac{(\chi _2\mu _2-\chi _1\mu _1)}{\pi ^{\frac{N}{2}}}\int _0^\infty \int _{{\mathbb R}^{N-1}}\frac{e^{-\lambda _2 s}e^{-|y|^2}}{\sqrt{s}}\left[ \int _0^\infty \tau e^{-\tau ^2}u(x+2\sqrt{s}\tau e_i +2\sqrt{s}\pi _i^{-1}(y),t;u_0)d\tau \right] dyds\nonumber \\&+ \frac{\chi _1\mu _1}{\pi ^{\frac{N}{2}}}\int _0^\infty \int _{{\mathbb R}^{N-1}}\frac{(e^{-\lambda _2 s}-e^{-\lambda _1 s})e^{-|y|^2}}{\sqrt{s}}\left[ \int _0^\infty \tau e^{-\tau ^2}u(x+2\sqrt{s}\tau e_i +2\sqrt{s} \pi _i^{-1}(y),t;u_0)d\tau \right] dyds\nonumber \\&+ \frac{(\chi _1\mu _1-\chi _2\mu _2)}{\pi ^{\frac{N}{2}}}\int _0^\infty \int _{{\mathbb R}^{N-1}}\frac{e^{-\lambda _2 s}e^{-|y|^2}}{\sqrt{s}}\left[ \int _0^\infty \tau e^{-\tau ^2}u(x-2\sqrt{s}\tau e_i +2\sqrt{s} \pi _i^{-1}(y),t;u_0)d\tau \right] dyds\nonumber \\&+ \frac{\chi _1\mu _1}{\pi ^{\frac{N}{2}}}\int _0^\infty \int _{{\mathbb R}^{N-1}}\frac{(e^{-\lambda _1 s}-e^{-\lambda _2 s})e^{-|y|^2}}{\sqrt{s}}\left[ \int _0^\infty \tau e^{-\tau ^2}u(x-2\sqrt{s}\tau e_i +2\sqrt{s} \pi _i^{-1}(y),t;u_0)d\tau \right] dyds. \end{aligned}$$
(4.2)

Using the fact that \(\int _0^\infty \frac{e^{-\lambda _k s}}{\sqrt{s}}ds=\frac{\sqrt{\pi }}{\sqrt{\lambda _k}}\), \(\int _0^{\infty }\tau e^{-\tau ^2}d\tau =\frac{1}{2}\), \(\int _{{\mathbb R}^{N-1}}e^{-|y|^2}dy=\pi ^{\frac{N-1}{2}}\), it follows from (4.2) that for every \(x\in {\mathbb R}^N,\ t\ge 0\), we have

$$\begin{aligned} |\partial _{x_i}(\chi _2v_2-\chi _1v_1)(x,t;u_0)| \le \left[ \frac{|\chi _2\mu _2-\chi _1\mu _1|}{2\sqrt{\lambda _2}}+\frac{\chi _1\mu _1|\sqrt{\lambda _1}-\sqrt{\lambda _2}|}{2\sqrt{\lambda _1\lambda _2}}\right] \Vert u(\cdot ,t;u_0)\Vert _{\infty }. \end{aligned}$$

Similarly, we have that

$$\begin{aligned} |\partial _{x_i}(\chi _1v_1-\chi _2v_2)(x,t;u_0)| \le \left[ \frac{|\chi _1\mu _1-\chi _2\mu _2|}{2\sqrt{\lambda _1}}+\frac{\chi _2\mu _2|\sqrt{\lambda _2}-\sqrt{\lambda _1}|}{2\sqrt{\lambda _1\lambda _2}}\right] \Vert u(\cdot ,t;u_0)\Vert _{\infty }. \end{aligned}$$

The lemma thus follows. \(\square \)

Lemma 4.2

Suppose that (1.7) holds. Let \(u_0\in C_\mathrm{unif}^{b}({\mathbb R}^N)\), \(u_0\ge 0\), and \((u(\cdot ,\cdot ;u_0),v_{1}(\cdot ,\cdot ;u_0)\), \(v_2(\cdot ,\cdot ;u_0))\) be the classical solution of (1.4) with \(u(\cdot ,0;u_0)=u_0\). Then we have that

$$\begin{aligned} \limsup _{t\rightarrow \infty }\Vert u(\cdot ,t;u_0)\Vert _{\infty }\le \frac{a}{b+\chi _2\mu _2-\chi _1\mu _1-M}, \end{aligned}$$
(4.3)

where M is given by (1.8).

Proof

It follows from inequalities (3.7) and (3.13) that

$$\begin{aligned} \overline{u}\le \frac{a+\frac{1}{\lambda _2}\left[ ((\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _1\mu _1(\lambda _1-\lambda _2)_{+} ) \right] \overline{u}}{b+\chi _2\mu _2-\chi _1\mu _1} \end{aligned}$$

and

$$\begin{aligned} \overline{u}\le \frac{a+\frac{1}{\lambda _1}\left[ ((\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _2\mu _2(\lambda _1-\lambda _2)_{+} ) \right] \overline{u}}{b+\chi _2\mu _2-\chi _1\mu _1}. \end{aligned}$$

Which is equivalent to

$$\begin{aligned} (b+\chi _2\mu _2-\chi _1\mu _1)\overline{u}\le a+\frac{1}{\lambda _2}\left[ ((\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _1\mu _1(\lambda _1-\lambda _2)_{+} ) \right] \overline{u} \end{aligned}$$

and

$$\begin{aligned} (b+\chi _2\mu _2-\chi _1\mu _1)\overline{u}\le a+\frac{1}{\lambda _1}\left[ ((\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _2\mu _2(\lambda _1-\lambda _2)_{+} ) \right] \overline{u}. \end{aligned}$$

Hence

$$\begin{aligned} (b+\chi _2\mu _2-\chi _1\mu _1)\overline{u}\le a+M\overline{u}. \end{aligned}$$

The lemma thus follows. \(\square \)

Lemma 4.3

  1. 1)

    If there is a positive constant \(c^*_{-}(\chi _1,\mu _1,\lambda _2,\chi _2,\mu _2,\lambda _2)\) such that

    $$\begin{aligned} \lim _{t\rightarrow \infty }\sup _{|x|\le ct}|u(x,t;u_0)-\frac{a}{b}|=0\quad \forall \ 0\le c< c^*_{-}(\chi _1,\mu _1,\lambda _2,\chi _2,\mu _2,\lambda _2), \end{aligned}$$
    (4.4)

    then for every \(i=1,2\) we have

    $$\begin{aligned} \lim _{t\rightarrow \infty }\sup _{|x|\le ct}|\lambda _i v_i(x,t;u_0)-\frac{a}{b}\mu _i|=0\quad \forall \ 0\le c< c^*_{-}(\chi _1,\mu _1,\lambda _2,\chi _2,\mu _2,\lambda _2). \end{aligned}$$
    (4.5)
  2. 2)

    If there is a positive constant \( c^*_{+}(\chi _1,\mu _1,\lambda _2,\chi _2,\mu _2,\lambda _2)\) such that

    $$\begin{aligned} \lim _{t\rightarrow \infty }\sup _{|x|\ge ct}u(x,t;u_0)=0 \quad \forall \ c> c^*_{+}(\chi _1,\mu _1,\lambda _2,\chi _2,\mu _2,\lambda _2), \end{aligned}$$
    (4.6)

    then for each \(i=1,2\) we have that

    $$\begin{aligned} \lim _{t\rightarrow \infty }\sup _{|x|\ge ct}v_i(x,t;u_0)=0 \quad \forall \ c> c^*_{+}(\chi _1,\mu _1,\lambda _2,\chi _2,\mu _2,\lambda _2). \end{aligned}$$
    (4.7)

The proof of Lemma 4.3 follows from the proof of Lemma 5.5 [36].

Now, we are ready to prove Theorem C.

Proof of Theorem C

Combining inequalities (4.1) and (4.3), we obtain that

$$\begin{aligned} \Vert \nabla (\chi _2v_2-\chi _1v_1)(\cdot ,t;u_0)\Vert _{\infty }\le \frac{aD\sqrt{N}}{b+\chi _2\mu _2-\chi _1\mu _1-M} +D\varepsilon \sqrt{N}, \quad \forall \ t\ge T_\varepsilon \end{aligned}$$
(4.8)

where D is given by (1.14) and M given by (1.8). Let

$$\begin{aligned} K_{\varepsilon }:= \sup _{0\le t\le T_\varepsilon }\Vert \nabla (\chi _2 v_2-\chi _1 v_1)(\cdot ,t;u_0)\Vert _{\infty } \quad \text {and}\quad K_{\varepsilon ,2}=:\sup _{0\le t\le T_\varepsilon }\Vert \frac{\chi _2\lambda _2}{\sqrt{a }}v_{2}(\cdot ,t;u_0)\Vert _{\infty }. \end{aligned}$$

Choose \(C>0\) such that

$$\begin{aligned} u_{0}(x)\le C e^{-\sqrt{a}|x|},\quad \forall \ x\in {\mathbb R}^N. \end{aligned}$$

Let \(\xi \in S^{N-1}\) be given and consider

$$\begin{aligned} \overline{U}(x,t;\xi )= Ce^{-\sqrt{a}(x\cdot \xi -( 2\sqrt{a} +K_\varepsilon + K_{\varepsilon ,2})t)}. \end{aligned}$$

We have that

$$\begin{aligned}&\overline{U}_t-\Delta \overline{U}-\nabla ((\chi _2v_2-\chi _1v_1)(\cdot ,\cdot ;u_0))\nabla \overline{U}-(a+(\chi _2\lambda _2v_2-\chi _1\lambda _1v_1)(\cdot ,\cdot ;u_0)\nonumber \\&\quad -(b+\chi _2\mu _2-\chi _1\mu _1)\overline{U})\overline{U}\nonumber \\&=\Big (\sqrt{a}(2\sqrt{a}+K_{\varepsilon }+K_{\varepsilon ,2})-a+\sqrt{a}\nabla ((\chi _2v_2-\chi _1v_1)(\cdot ,\cdot ;u_0))\cdot \xi \Big )\overline{U}\nonumber \\&\qquad - \Big (a +(\chi _2\lambda _2v_2-\chi _1\lambda _1v_1)(\cdot ,\cdot ;u_0)-(b+\chi _2\mu _2-\chi _1\mu _1)\overline{U} \Big )\overline{U}\nonumber \\&= \Big ( \sqrt{a}\big (K_\varepsilon +\nabla ((\chi _2v_2-\chi _1v_1)(\cdot ,\cdot ;u_0))\cdot \xi \big ) +(\sqrt{a}K_{\varepsilon ,2}-\chi _2\lambda _2v_2(\cdot ,\cdot ;u_0))\Big )\overline{U}\nonumber \\&\ \ +\Big (\chi _1\lambda _1v_1(\cdot ,\cdot ;u_0) +(b+\chi _2\mu _2-\chi _1\mu _1)\overline{U}\Big )\overline{U}\nonumber \\&\ge 0 \quad \forall x\in {\mathbb R}^N, \ 0<t\le T_{\varepsilon },\ \forall \ \xi \in S^{N-1}. \end{aligned}$$
(4.9)

Since \(\overline{U}(x,0;\xi )=Ce^{-\sqrt{a}x\cdot \xi }\ge Ce^{-\sqrt{a}|x|}\ge u_0(x)\), by comparison principle for parabolic equations, we obtain that

$$\begin{aligned} u(x,t;u_0)\le \overline{U}(x,t;\xi ),\quad \forall \ x\in {\mathbb R}^N,\ 0\le t\le T_\varepsilon \ \forall \ \xi \in S^{N-1}. \end{aligned}$$
(4.10)

Next, consider

$$\begin{aligned} \overline{W}(x,t;\xi )=Ce^{-\sqrt{a}(x\cdot \xi -(2\sqrt{a}+ L_\varepsilon +L_{\varepsilon ,2})(t-T_\varepsilon ))}e^{\sqrt{a}(2\sqrt{a}+K_\varepsilon +K_{\varepsilon ,2})T_{\varepsilon }}, \quad x\in {\mathbb R}^N, \ t\ge T_{\varepsilon }, \end{aligned}$$
(4.11)

where

$$\begin{aligned} L_{\varepsilon }:=\frac{aD\sqrt{N}}{b+\chi _2\mu _2-\chi _1\mu _1-M}+D\varepsilon \sqrt{N} \end{aligned}$$

and

$$\begin{aligned} L_{\varepsilon ,2}:=\frac{\sqrt{a}\chi _2\mu _2}{b+\chi _2\mu _2-\chi _1\mu _1-M} +\frac{ \chi _2\mu _2\varepsilon }{\sqrt{a}}. \end{aligned}$$

It follows from (3.1), (4.3) and (4.8) that for any \(x\in {\mathbb R}^N\) and \(t\ge T_\epsilon \),

$$\begin{aligned}&\overline{W}_t-\Delta \overline{W}-\nabla ((\chi _2v_2-\chi _1v_1)(\cdot ,\cdot ;u_0)) \nabla \overline{W}-(a+(\chi _2\lambda _2v_2-\chi _1\lambda _1v_1) (\cdot ,\cdot ;u_0)\\&\quad -(b+\chi _2\mu _2-\chi _1\mu _1)\overline{W})\overline{W}\ge 0. \end{aligned}$$

Observe that \(\overline{W}(\cdot ,T_\varepsilon ;\xi )=\overline{U}(\cdot ,T_\varepsilon ;\xi )\ge u(\cdot ,T_\varepsilon )\). Hence, comparison principle for parabolic equations implies that

$$\begin{aligned} 0\le u(x,t;u_0)\le \overline{W}(x,t;\xi ),\quad x\in {\mathbb R}^N,\ t\ge T_{\varepsilon },\ \xi \in S^{N-1}. \end{aligned}$$
(4.12)

Hence, for every \(c>2\sqrt{a}+L_{\varepsilon }+L_{\varepsilon ,2}\), and \(t>T_\varepsilon \), we have

$$\begin{aligned}&\sup _{|x|\ge ct}u(x,t;u_0)\le \sup _{|x|\ge ct }\overline{W}\left( x,t,\frac{1}{|x|}x\right) \\&\quad \le \sup _{|x|\ge ct}Me^{-\sqrt{a}(c-(2\sqrt{a}+ L_\varepsilon +L_{\varepsilon ,2})(t-T_\varepsilon ))}e^{\sqrt{a}(2\sqrt{a}+K_\varepsilon +K_{\varepsilon ,2})T_{\varepsilon }} \rightarrow 0 \end{aligned}$$

as \( t\rightarrow \infty \). Thus by taking

$$\begin{aligned} c_{+}^{*}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2):=2\sqrt{a}+\lim _{\varepsilon \rightarrow 0^+}(L_{\varepsilon }+L_{\varepsilon ,2})=2\sqrt{a}+\frac{\sqrt{a}(D\sqrt{N a}+\chi _2\mu _2)}{b+\chi _2\mu _2-\chi _1\mu _1-M}, \end{aligned}$$
(4.13)

and using Lemma 4.3, the result of Theorem C follows.

In order to prove Theorem D, we first establish the following important Lemma.

Lemma 4.4

Let L be given by (1.18). Then,

$$\begin{aligned}&\lim _{ R\rightarrow \infty } \inf _{ |x| \ge R,\ T\ge R}\left( 4(a+(\chi _2\lambda _2v_2-\chi _1\lambda _1v_1)(x,t;u_0)-|\nabla (\chi _2v_2-\chi _1v_1)(x,t;u_0)|^2\right) \nonumber \\&\ge 4a(1-L)-\frac{Na^2D^2}{(b+\chi _2\mu _2-\chi _1\mu _1)^2} . \end{aligned}$$
(4.14)

Proof

Using inequalities (3.4) and (4.1), we have that for every \(t\ge T_{\varepsilon }, \ x\in {\mathbb R}^N\),

$$\begin{aligned}&4\big (a+(\chi _2\lambda _2v_2-\chi _1\lambda _1v_1)(x,t;u_0)\big ) -|\nabla (\chi _2v_2(x,t;u_0)-\chi _1v_1(x,t;u_0))|^2\nonumber \\&\ge 4\big (a+ \frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _1\mu _1(\lambda _1-\lambda _2)_{+}\right] (\underline{u}-\varepsilon )\big )\nonumber \\&-\frac{4}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] (\overline{u}+\varepsilon )-ND^{2}(\overline{u}+\varepsilon )^2. \end{aligned}$$
(4.15)

Letting first \(R\rightarrow \infty \) and then \(\varepsilon \rightarrow 0\), it follows from (4.15) that

$$\begin{aligned}&\lim _{ R\rightarrow \infty } \inf _{ |x| \ge R,\ T\ge R}\left( 4(a+(\chi _2\lambda _2v_2-\chi _1v_1\lambda _1)(x,t;u_0)-|\nabla (\chi _2v_2-\chi _1v_1)(x,t;u_0)|^2\right) \nonumber \\&\ge 4\big (a+ \frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _1\mu _1(\lambda _1-\lambda _2)_{+}\right] \underline{u}\big )\nonumber \\&-\frac{4}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] \overline{u}-ND^{2}\overline{u}^2. \end{aligned}$$
(4.16)

But Theorem C implies that \(\underline{u}=0\). Hence, inequality (4.16) implies that

$$\begin{aligned}&\lim _{ R\rightarrow \infty } \inf _{ |x| \ge R,\ T\ge R}\left( 4(a+(\chi _2\lambda _2v_2-\chi _1v_1\lambda _1)(x,t;u_0)-|\nabla (\chi _2v_2-\chi _1v_1)(x,t;u_0)|^2\right) \nonumber \\&\ge 4\big (a-\frac{1}{\lambda _2}\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] \overline{u}\big )-ND^{2}\overline{u}^2. \end{aligned}$$
(4.17)

Thus, it follows from (4.17) and (4.3) that

$$\begin{aligned}&\lim _{ R\rightarrow \infty } \inf _{ |x| \ge R,\ T\ge R}\left( 4(a+(\chi _2\lambda _2v_2-\chi _1v_1\lambda _1)(x,t;u_0)-|\nabla (\chi _2v_2-\chi _1v_1)(x,t;u_0)|^2\right) \nonumber \\&\ge 4\big (a-\frac{a\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] }{\lambda _2(b+\chi _2\mu _2-\chi _1\mu _1-M)}\big )-\frac{ND^{2}a^2}{(b+\chi _2\mu _2-\chi _1\mu _1-M)^2}. \end{aligned}$$
(4.18)

Similarly, by rewriting \((\chi _2\mu _2v_2-\chi _1\mu _1v_1)(x,t;u_0)\) in the form given by (3.12), same arguments as above yield that

$$\begin{aligned}&\lim _{ R\rightarrow \infty } \inf _{ |x| \ge R,\ T\ge R}\left( 4(a+(\chi _2\lambda _2v_2-\chi _1v_1\lambda _1)(x,t;u_0)-|\nabla (\chi _2v_2-\chi _1v_1)(x,t;u_0)|^2\right) \nonumber \\&\ge 4\big (a-\frac{a\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _2\mu _2(\lambda _1-\lambda _2)_{-}\right] }{\lambda _1(b+\chi _2\mu _2-\chi _1\mu _1-M)}\big )-\frac{ND^{2}a^2}{(b+\chi _2\mu _2-\chi _1\mu _1-M)^2}. \end{aligned}$$
(4.19)

The Lemma thus follows. \(\square \)

Proof of Theorem D

The arguments used in this proof generalize some of the arguments used in the proof of Theorem 9(i) [36]. Hence some details might be omitted. We refer the reader to [36] for the proofs of the estimates stated below.

Since (1.17) holds, we have

$$\begin{aligned} c^{*}_{-}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2) :=2\sqrt{a(1-L)}-\frac{aD\sqrt{N}}{b+\chi _2\mu _2-\chi _1\mu _1-M}>0, \end{aligned}$$

where M, D and L are given by (1.8) and (1.14) and (1.18) respectively. We first note that, it follows from Lemma 4.4 and the proof of Lemma 5.4 [36] that for every \(0\le c< c^*_{-}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)\) we have

$$\begin{aligned} \liminf _{t\rightarrow \infty }\inf _{|x|\le ct}u(x,t;u_0)>0. \end{aligned}$$
(4.20)

It suffices to prove the following claim.

Claim. For every \(0\le c<c^*_{-}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)\), we have that

$$\begin{aligned} \lim _{t\rightarrow \infty }\sup _{|x|\le ct}|u(x,t;u_{0})-\frac{a}{b} |=0. \end{aligned}$$
(4.21)

Suppose that the claim is not true. Then there is \(0\le c<c^*_{-}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2)\), \(\delta >0\), a sequence \(\{x_n\}_{n\ge 1},\) a sequence of positive numbers \(\{t_n\}_{n\ge 1}\) with \(t_n\rightarrow \infty \) as \(n\rightarrow \infty \) such that

$$\begin{aligned} |x_n|\le ct_{n},\quad \forall \ n\ge 1, \end{aligned}$$
(4.22)

and

$$\begin{aligned} |u(x_n,t_n;u_0)-\frac{a}{b}|\ge \delta , \qquad \forall \ n\ge 1. \end{aligned}$$
(4.23)

For every \(n\ge 1\), let us define

$$\begin{aligned} u_n(x,t)=u(x+x_n,t+t_n;u_0),\, \,\,\, v_{kn}(x,t)=v_k(x+x_n,t_n;u_0)\,\,\, (k=1,2) \end{aligned}$$
(4.24)

for all \(x\in {\mathbb R}^N\) and \(t\ge -t_n\).

We first show that there is a subsequence of \(\{(u_n,v_{1n},v_{2n})\}\) which converges locally uniformly. To this end, let \(\{T(t)\}_{t\ge 0}\) denote the analytic semigroup generated by the closed linear operator \((\Delta -I)u\) on \(C^{b}_\mathrm{unif}({\mathbb R}^N)\). Then the variation of constant formula yield that

$$\begin{aligned} u(x,t;u_0)=&T(t)u_0+\int _0^t T(t-s)\nabla \cdot ((\chi _2u\nabla v_2-\chi _1 u\nabla v_1)(\cdot ,s;u_0))ds\nonumber \\&+\int _0^t T(t-s)(((a+1)u-bu^2)(\cdot ,s;u_0))ds. \end{aligned}$$
(4.25)

Let \(0<\alpha <\frac{1}{2}\) be fixed and let \(X^{\alpha }\) denotes the fractional powers associated to the semigroup \(\{T(t)\}_{t\ge 0}\). Thus, there is a constant \(C_{\alpha }\)(see [15]) depending only on \(\alpha \) and the dimension N such that

$$\begin{aligned}&\Vert u_{n}(\cdot ,0)\Vert _{X^{\alpha }} \le C_{\alpha }t_n^{-\alpha }\Vert u_0\Vert _{\infty } +C_{\alpha }\int _0^{t_n}e^{-(t_n-s)}(t_n-s)^{-\frac{1}{2}-\alpha }\Vert (\chi _2u\nabla v_{2}-\chi _1u\nabla v_1)(\cdot ,s;u_0)\Vert _{\infty }ds\nonumber \\&\quad + C_{\alpha }\int _{0}^{t_n}e^{-(t_n-s)}(t_n-s)^{-\alpha }\Vert ( (a+1)u-bu^2)(\cdot ,s;u_0)\Vert _{\infty }ds. \end{aligned}$$
(4.26)

Using the facts that \(\sup _{t\ge 0}\Vert u(\cdot ,t)\Vert _{\infty }<\infty \), \(t_n\rightarrow \infty \) as \(n\rightarrow \infty \), \(\int _0^{\infty }e^{-\tau }\tau ^{-\frac{1}{2}-\alpha }d\tau =\Gamma (\frac{1}{2}-\alpha )<\infty \) and \(\int _{0}^\infty e^{-\tau }\tau ^{-\alpha }d\tau =\Gamma (1-\alpha )<\infty \), it follows from (4.26) that

$$\begin{aligned} \sup _{n\ge 1}\Vert u_{n}(\cdot ,0)\Vert _{X^\alpha }<\infty . \end{aligned}$$
(4.27)

Similar arguments as those used in the proof of Theorem 1.1 [36] yield that the functions \(u_n : [-T\ , \ T]\rightarrow X^{\alpha } \) are equicontinuous for every \(T>0\). Hence Arzela-Ascili’s Theorem and Theorem 15 (page 80 of [12]) imply that there is a function \((\tilde{u},\tilde{v}_1,\tilde{v}_2)\in \left[ C^{2,1}({\mathbb R}^N\times {\mathbb R})\right] ^3\) and a subsequence \(\{(u_{n'},v_{1n'},v_{2n'})\}_{n\ge 1}\) of \(\{(u_{n},v_{1n},v_{2n})\}_{n\ge 1}\) such that \((u_{n'},v_{1n'},v_{2n'})\rightarrow (\tilde{u},\tilde{v}_{1},\tilde{v}_{2})\) in \(C_{loc}^{1+\delta ',\delta '}({\mathbb R}^N\times {\mathbb R}) \) for some \(\delta '>0\). Moreover \(\mu _i\tilde{u}=(\lambda _i I-\Delta )\tilde{v}_{i}\) for every \(i=1,2\). Note that

$$\begin{aligned} \tilde{u}(x,t)=\lim _{n\rightarrow \infty }u(x+x_{n'},t+t_{n'};u_0),\quad \forall \ x\in {\mathbb R}^N,\ \ t\in {\mathbb R}. \end{aligned}$$

Hence

$$\begin{aligned} |\tilde{u}(0,0)-\frac{a}{b}|\ge \delta . \end{aligned}$$
(4.28)

Choose \(\tilde{c}\in (c\ ,\ c^*_{-}(\chi _1,\mu _1,\lambda _1,\chi _2,\mu _2,\lambda _2))\). For every \(x\in {\mathbb R}^N, t\in {\mathbb R}\) and \(t_{n'}\ge \frac{|x|+\tilde{c}|t|}{\tilde{c}-c}\), we have

$$\begin{aligned} |x+x_{n'}|\le |x|+ct_{n'}\le \tilde{c}(t_{n'}+t). \end{aligned}$$

It follows from last inequality and (4.20) that

$$\begin{aligned} \tilde{u}(x,t)=\lim _{n\rightarrow \infty }u(x+x_{n'},t+t_{n'};u_0)\ge \liminf _{s\rightarrow \infty }\inf _{|y|\le \tilde{c}s}u(y,s;u_0)>0,\quad \forall \ x\in {\mathbb R}^N, \ \ t\in {\mathbb R}. \end{aligned}$$

Hence \(\inf _{(x,t)\in {\mathbb R}^{N+1}}\tilde{u}(x,t)>0\).

Next, we claim that \(\tilde{u}(x,t)=\frac{a}{b}\) for every \(x\in {\mathbb R}^N,\ t\in {\mathbb R}\). Indeed, let \(\underline{u}_0=\inf _{(x,t)\in {\mathbb R}^{N+1}}\tilde{u}(x,t)\) and \(\overline{u}_0(x,t)=\sup _{(x,t)\in {\mathbb R}^{N+1}}\tilde{u}(x,t)\). For every \(t_0\in {\mathbb R}\), let \(\overline{U}(t,t_0)\) and \(\underline{U}(t,t_0)\) be the solution of the ODEs

$$\begin{aligned} {\left\{ \begin{array}{ll} \overline{U}_t= \big (a-(b+\chi _2\mu _2-\chi _1\mu _1)\overline{U}\big )\overline{U}\\ \qquad \,\,\, +\frac{1}{\lambda _2}\Big (\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _1\mu _1(\lambda _1-\lambda _2)_{+}\right] \overline{u}_0 \\ \qquad \,\,\, -\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] \underline{u}_0\Big )\overline{U},\quad t>t_0\\ \overline{U}(t_0,t_0)=\overline{u}_0 \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} {\left\{ \begin{array}{ll} \underline{U}_t=\big (a-(b+\chi _2\mu _2-\chi _1\mu _1)\underline{U}\big )\underline{U}\\ \qquad \,\,\, +\frac{1}{\lambda _2}\Big (\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _1\mu _1(\lambda _1-\lambda _2)_{+}\right] \underline{u}_0 \\ \qquad \,\,\, -\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\lambda _2)_{-}\right] \overline{u}_0\Big )\underline{U},\quad t>t_0\\ \underline{U}(t_0,t_0)=\underline{u}_0 \end{array}\right. } \end{aligned}$$

respectively. It follows from the arguments used to establish (3.6) and (3.9) that

$$\begin{aligned} \underline{U}(t-t_0,0)=\underline{U}(t,t_0)\le \tilde{u}(x,t),\quad \forall \ x\in {\mathbb R}^N,\ t\ge t_0 \end{aligned}$$
(4.29)

and

$$\begin{aligned} \overline{U}(t-t_0,0)=\overline{U}(t,t_0)\ge \tilde{u}(x,t),\quad \forall \ x\in {\mathbb R}^N,\ \ t\ge t_0 \end{aligned}$$
(4.30)

respectively. Note that for every \(t\in {\mathbb R}\) fixed, we have that

$$\begin{aligned} \lim _{t_0\rightarrow -\infty }\overline{U}(t,t_0)=\frac{1}{b+\chi _2\mu _2-\chi _1\mu _1} \Big \{&a+\frac{1}{\lambda _2}\big (\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _1\mu _1(\lambda _1-\lambda _2)_{+}\right] \overline{u}_{0} \nonumber \\&-\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\chi _2)_{-}\right] \underline{u}_0\big )\Big \} \end{aligned}$$
(4.31)

and

$$\begin{aligned} \lim _{t_0\rightarrow -\infty }\underline{U}(t,t_0)=\frac{1}{b+\chi _2\mu _2-\chi _1\mu _1} \Big \{&a+\frac{1}{\lambda _2}\big (\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _1\mu _1(\lambda _1-\lambda _2)_{+}\right] \underline{u}_{0} \nonumber \\&-\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\chi _2)_{-}\right] \overline{u}_0\big )\Big \}. \end{aligned}$$
(4.32)

Combining (4.29) and (4.32), we have that

$$\begin{aligned} \frac{1}{b+\chi _2\mu _2-\chi _1\mu _1} \Big \{&a+\frac{1}{\lambda _2}\big (\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _1\mu _1(\lambda _1-\lambda _2)_{+}\right] \underline{u}_{0} \nonumber \\&-\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\chi _2)_{-}\right] \overline{u}_0\big )\Big \} \le \underline{u}_0. \end{aligned}$$
(4.33)

Combining (4.30) and (4.31), we have that

$$\begin{aligned} \frac{1}{b+\chi _2\mu _2-\chi _1\mu _1} \Big \{&a+\frac{1}{\lambda _2}\big (\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _1\mu _1(\lambda _1-\lambda _2)_{+}\right] \overline{u}_{0} \nonumber \\&-\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _1\mu _1(\lambda _1-\chi _2)_{-}\right] \underline{u}_0\big )\Big \} \ge \overline{u}_0. \end{aligned}$$
(4.34)

Thus, it follows from inequalities (4.33) and (4.34) that

$$\begin{aligned} \left( b+\chi _2\mu _2-\chi _1\mu _1-\frac{1}{\lambda _2}\Big ( |\chi _1\mu _1\lambda _1-\chi _2\mu _2\lambda _2|+\chi _1\mu _1|\lambda _1-\lambda _2| \Big ) \right) (\overline{u}_0-\underline{u}_0)\le 0. \end{aligned}$$
(4.35)

Similarly, for every \(t_0\in {\mathbb R}\), by considering \(\overline{V}(t,t_0)\) and \(\underline{V}(t,t_0)\) to be the solutions of the ODEs

$$\begin{aligned} {\left\{ \begin{array}{ll} \overline{V}_t= \big (a-(b+\chi _2\mu _2-\chi _1\mu _1)\overline{V}\big )\overline{V}\\ \qquad \,\, +\frac{1}{\lambda _1}\Big (\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _2\mu _2(\lambda _1-\lambda _2)_{+}\right] \overline{u}_0 \\ \qquad \,\, -\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _2\mu _2(\lambda _1-\lambda _2)_{-}\right] \underline{u}_0\Big )\overline{V},\quad t>t_0\\ \overline{V}(t_0,t_0)=\overline{u}_0 \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} {\left\{ \begin{array}{ll} \underline{V}_t=\big (a-(b+\chi _2\mu _2-\chi _1\mu _1)\underline{V}\big )\underline{V}\\ \qquad \,\, +\frac{1}{\lambda _1}\Big (\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{+}+\chi _2\mu _2(\lambda _1-\lambda _2)_{+}\right] \underline{u}_0 \\ \qquad \,\, -\left[ (\chi _2\mu _2\lambda _2-\chi _1\mu _1\lambda _1)_{-}+\chi _2\mu _2(\lambda _1-\lambda _2)_{-}\right] \overline{u}_0\Big )\underline{V},\quad t>t_0\\ \underline{V}(t_0,t_0)=\underline{u}_0 \end{array}\right. } \end{aligned}$$

respectively. Using systems (4.36) and (4.36), similar arguments used to establish (4.35) yield that

$$\begin{aligned} \left( b+\chi _2\mu _2-\chi _1\mu _1-\frac{1}{\lambda _1}\Big ( |\chi _1\mu _1\lambda _1-\chi _2\mu _2\lambda _2|+\chi _2\mu _2|\lambda _1-\lambda _2| \Big ) \right) (\overline{u}_0-\underline{u}_0)\le 0. \end{aligned}$$
(4.36)

It follows from inequalities (4.35) and (4.36) that

$$\begin{aligned} (b+\chi _2\mu _2-\chi _1\mu _1-K)(\overline{u}_0-\underline{u}_0)\le 0. \end{aligned}$$
(4.37)

Since (1.10) holds, it follows from the last inequality that \(\overline{u}_{0}=\underline{u}_0\). Combining this with inequalities (4.33) and (4.34) we obtain that \(\overline{u}_0=\underline{u}_0=\frac{a}{b}\). Hence, we have that \(\tilde{u}(x,t)=\frac{a}{b}\) for every \(x\in {\mathbb R}^N\) and \(t\in {\mathbb R}\). In particular, we have that \(\tilde{u}(0,0)=\frac{a}{b}\), which contradicts (4.28).

Hence the claim is true and Theorem D is thus proved. \(\square \)