1 Introduction

Chemotaxis, the property such that species move toward or away from the higher concentration of the chemical substance, has attracted many mathematicians’ interests in recent years. So far, there is a considerable amount of research on the classical Keller–Segel system [10]

$$\begin{aligned} \left\{ \begin{array}{ll} \partial _tu=\Delta u-\chi \nabla \cdot (u\nabla v)+f(u), &{}\quad x\in \Omega ,\quad t>0,\\ \tau \partial _tv=\Delta v-\lambda v+\alpha u^k,&{}\quad x\in \Omega ,\quad t>0\\ \end{array} \right. \end{aligned}$$
(1.1)

with \(\tau \in \{0, 1\}\). For the case \(f(u)=0, k=1\), there is no possibility that the solution of (1.1) blows up when \(n = 1\); when \(n = 2\), the system of (1.1) exhibits a remarkable feature: critical mass [6, 7, 23, 24]. As for the case \(n\ge 3\), one can refer to [42, 44] for the blowup results. For the case \(f(u)=ru-\mu u^2, k=1\), when \(n=2\), it was proved that the solution of (1.1) with \(\tau =1\) is globally bounded for any \(r\in {\mathbb {R}}\) and \(\mu >0\) [27]; in the smooth and bounded convex domain \(\Omega \subset {\mathbb {R}}^n (n\ge 3)\), Lankeit obtained the existence of global weak solutions to (1.1) with \(\tau =1\) for arbitrary \(\mu >0\) [12], especially, in the three-dimensional setting, they further proved that these solutions become classical solutions after some time; what is more, under the condition that \(\mu \) is sufficiently large, the large time behavior of solutions to (1.1) with \(\tau =1\) was considered in the convex domain [45]. For the case \(f(u)\le a-bu^2, k=1\), in the convex domain \(\Omega \subset {\mathbb {R}}^n (n\ge 3)\), the global solvability for \(\tau =1\) was obtained when b is sufficiently large [43]; when it is supposed that either \(n\le 2\) or \(b>\frac{n-2}{n}\chi _{11}\), Tello and Winkler [32] showed that the system (1.1) with \(\tau =0\) admits globally bounded classical solutions; afterward, their result was extended to the borderline case \(b=\frac{n-2}{n}\chi _{11}\) by Xiang [50]. Besides these, one can refer to [5, 8, 9] for more global boundedness results of (1.1) when \(f(u)\ne 0\). Recently, some new progress on the blowup phenomenon has been made, and Winkler [47] investigated the blowup result of (1.1) for low-dimensional spatial settings; in addition, replacing the second equation by \(0=\Delta v-{\tilde{\mu }}(t)+g(u)\) (\({\tilde{\mu }}(t)=\frac{1}{|\Omega |}\int _{\Omega }g(u(\cdot , t))dx\)), they also obtained a finite-time blowup result without considering the logistic source [48].

After the pioneering works above, the following two-species and one-stimuli chemotaxis model [49]

$$\begin{aligned} \left\{ \begin{array}{ll} \partial _{t} u_{1}=\tau _{1}\Delta u_{1}-\chi _{1}\nabla \cdot (u_{1}\nabla v)+\mu _{1}u_{1}(1-u_{1}-a_{1}u_{2}),&{}\quad x\in \Omega ,\quad t>0,\\ \partial _{t} u_{2}=\tau _{2}\Delta u_{2}-\chi _{2}\nabla \cdot (u_{2}\nabla v)+\mu _{2}u_{2}(1-u_{2}-a_{2}u_{1}),&{}\quad x\in \Omega ,\quad t>0,\\ \tau \partial _{t}v=\Delta v+h(u_1, u_2, v),&{}\quad x\in \Omega ,\quad t>0,\\ \end{array} \right. \nonumber \\ \end{aligned}$$
(1.2)

where \(\tau \in (0, 1)\), which is a variant of the classical Keller–Segel chemotaxis system, has been studied intensively. For the case \(\tau =1\), when \(h(u_1, u_2, v)\) is some function under the explicit conditions, the works related to global dynamics can be found in [18, 22, 25, 26, 39]. When \(h(u_1, u_2, v)=-\gamma v+\alpha _{1}u_{1}+\alpha _{2}u_{2}\), Bai and Winkler [1] proved the global boundedness of solutions for (1.2) when \(n\le 2\), moreover, they gave the asymptotic stabilization of globally bounded solutions when \(n\ge 1\); then the global boundedness result was further extended to the case \(n=3\) in [13]; recently, the conditions for asymptotic stabilization assumed in [1] were improved by Mizukami [19, 21]. Without the kinetic terms, the finite-time blowup of (1.2) in higher dimension was considered in [14]. When considering the strong logistic diffusion, Tu et al. [37] investigated the finite time blowup of solutions. Very recently, considering the competitive kinetics in (1.2), Li [15] established the large densities and simultaneous blowup result for the chemotactic species when \(n\ge 3\). As for the parabolic–parabolic–elliptic case, the global solvability and stabilization of (1.2) was shown in [2, 20, 30, 33].

To better understand the chemotactic interaction when there are two chemicals, the two-species and two-stimuli chemotaxis system is considered,

$$\begin{aligned} \left\{ \begin{array}{ll} \partial _{t} u_{1}=\Delta u_{1}-\chi _{1}\nabla \cdot (u_{1}\nabla v_{1}) +\mu _{1}u_{1}(1-u_{1}-a_{1}u_{2}),&{}\quad x\in \Omega ,\quad t>0,\\ \partial _{t} u_{2}=\Delta u_{2} -\chi _{2}\nabla \cdot (u_{2}\nabla v_{2}) +\mu _{2}u_{2}(1-u_{2}-a_{2}u_{1}),&{}\quad x\in \Omega ,\quad t>0,\\ \tau _1\partial _{t} v_{1}=\Delta v_{1}- v_{1}+u_{2},&{}\quad x\in \Omega ,\quad t>0,\\ \tau _2 \partial _{t} v_{2}=\Delta v_{2}- v_{2}+u_{1},&{}\quad x\in \Omega ,\quad t>0,\\ \end{array}\right. \nonumber \\ \end{aligned}$$
(1.3)

for the case \(\tau _1=\tau _2=0\), the boundedness versus blowup was systematically investigated in [31, 52] when \(\mu _1=\mu _2=0\); when \(\mu _1, \mu _2>0\), the global dynamics of solutions were established in [34, 41, 54, 55]. For the case \(\tau _1=\tau _2=1\), in the absence of the Lotka–Volterra-type competition, Xie and Wang [51] investigated the global boundedness of solutions under the condition that both \(\Vert u_{10}\Vert _{L^1(\Omega )}\) and \(\Vert u_{20}\Vert _{L^1(\Omega )}\) are appropriately small for \(n=2\); when \(\mu _1, \mu _2>0\), the uniform boundedness and asymptotic behavior of solutions for \(n\le 2\) were detected by Black [3]; thereafter, this result was improved to the 3D case [28, 40].

In the present study, we consider the following two-species and two-stimuli system with loop,

$$\begin{aligned} \left\{ \begin{array}{ll} \partial _{t} u_{1}=\Delta u_{1}-\chi _{11}\nabla \cdot (u_{1}\nabla v_{1}) -\chi _{12}\nabla \cdot (u_{1}\nabla v_{2}) +\mu _{1}u_{1}(1-u_{1}-a_{1}u_{2}),&{}\quad x\in \Omega ,\quad t>0,\\ \partial _{t} u_{2}=\Delta u_{2}-\chi _{21}\nabla \cdot (u_{2}\nabla v_{1}) -\chi _{22}\nabla \cdot (u_{2}\nabla v_{2}) +\mu _{2}u_{2}(1-u_{2}-a_{2}u_{1}),&{}\quad x\in \Omega ,\quad t>0,\\ \partial _{t} v_{1}=\Delta v_{1}- v_{1}+u_{1}+u_{2},&{}\quad x\in \Omega ,\quad t>0,\\ \partial _{t} v_{2}=\Delta v_{2}- v_{2}+u_{1}+u_{2},&{}\quad x\in \Omega ,\quad t>0,\\ \frac{\partial u_{1}}{\partial \nu }=\frac{\partial u_{2}}{\partial \nu } =\frac{\partial v_{1}}{\partial \nu }=\frac{\partial v_{2}}{\partial \nu }=0, &{}\quad x\in \Omega ,\quad t>0,\\ u_{1}(x,0)=u_{10}(x), u_{2}(x,0)=u_{20}(x), v_{1}(x,0)=v_{10}(x), v_{2}(x,0) =v_{20}(x),&{}\quad x\in \Omega ,\quad t>0,\\ \end{array} \right. \nonumber \\ \end{aligned}$$
(1.4)

which stems from [11] by Knútsdóttir et al., as a mathematical biological model to describe the metastatic process for the breast cancer. Here, \(\Omega \subset {\mathbb {R}}^n (n\ge 3)\) is a bounded domain with smooth boundary \(\partial \Omega \), \(\frac{\partial }{\partial \nu }\) represents differentiation with respect to the outward normal on \(\partial \Omega \), \(\chi _{ij}>0\), \(\mu _{i}>0\), \(a_i>0\), \(\alpha _{ij}>0\), \(\lambda _{i}>0\) \((i, j=1, 2)\). \(u_1, u_2\) denote the densities of macrophages and tumor cells, and the concentration of the chemical signals \(v_i (i=1, 2)\) is secreted by \(u_1\) and \(u_2\). Let the initial data \(u_{10}, u_{20}, v_{10}\) and \(v_{20}\) satisfy

$$\begin{aligned} 0\le u_{10},\quad u_{20}\in C^{0}({\bar{\Omega }}),\quad 0\le v_{10},\quad v_{20}\in W^{1, \infty }(\Omega ). \end{aligned}$$
(1.5)

For the parabolic–elliptic case, the global dynamics of (1.4) was established for sufficiently large \(\mu _i (i=1, 2)\) [35], and regardless of the Lotka–Volterra-type competition, the simultaneous blowup, global boundedness, gradient estimates of solutions in \({\mathbb {R}}^2\) were investigated in [4, 16, 17, 53]. Nevertheless, the studies on the fully parabolic system of (1.4) are only devoted to the global solvability, and it was shown that the solution exists globally without any restriction on \(\mu _i (i=1, 2)\) in the 2D setting [36]; and the validity of global boundedness was guaranteed for the 3D setting under the condition that \(\mu _i (i=1, 2)\) are sufficiently large [38]; apart from the global boundedness, the large time behavior of solution was also given in both studies.

In this paper, we investigate a phenomenon which is weaker than blowup but still significant, that is, the solution exhibits unbounded peculiarity for the proper choice of initial data. Apart from that, we also show the simultaneous blowup for both chemotactic species once the blowup happens. Our work is enlightened by the one-specie one-stimuli and two-specie one-stimuli case in [14, 15, 46], we shall construct an appropriate functional and use the contradictory argument to address this issue. Compared with the previous works, our difficulty consists in two aspects: (1) the disposal of the functional \(\mathrm {F}(u_1, u_2, v_1, v_2)\) denoted in the following (1.6), the functional in our paper contains more terms than the previous one, this leads to the emergence of some extra cross-terms (such as \(u_i|\nabla v_j|^2, \nabla u_i\cdot \nabla v_j, \frac{|\nabla u_i|^2}{u_i}\), \(i, j=1, 2\)) in the later calculations, while the previous functional given in [15, 46] can be directly handled by completing the square, it is not so trivial to handle these additional terms; (2) the complicated interaction among the loop, multi-stimuli and competitive kinetics, which makes the computation and analysis more delicate.

To this end, for \((u_1, u_2, v_1, v_2)\in \left( C^{2,1}( \overline{\Omega }\times (0,T_{\max }))\right) ^4\), we define the following functional

$$\begin{aligned} \begin{aligned} \mathrm {F}(u_1, u_2, v_1, v_2)&=\frac{1}{2}\int _{\Omega }(|v_1|^2+|v_2|^2+|\nabla v_1|^2+|\nabla v_2|^2)\mathrm{d}x -\int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x\\&\quad +\frac{1}{\chi _1}\int _{\Omega }u_1\ln u_1\mathrm{d}x +\frac{1}{\chi _2}\int _{\Omega }u_2\ln u_2\mathrm{d}x \end{aligned} \end{aligned}$$
(1.6)

with \(\chi _1=\min \{\chi _{11}, \chi _{12}\}\), \(\chi _2=\min \{\chi _{21}, \chi _{22}\}\), which plays an important role in our later proof.

Now we state our main results as follows.

Theorem 1.1

Let \(\Omega =B_R(0)\subset {\mathbb {R}}^n(n\ge 3, R>0)\), \(\chi _{ij}>0\), \(\mu _{i}>0\), \(a_i>0\) \((i, j=1, 2)\). Then for all \({\mathcal {M}}>0, {\mathcal {N}}>0\) and each \(L>0, T>0\), one can find some positive constant \({\mathcal {P}}= {\mathcal {P}}( {\mathcal {M}}, {\mathcal {N}}, L, T, R, n, \chi _{11}, \chi _{12}, \chi _{21}, \chi _{22}, \mu _1, \mu _2, a_1, a_2)\), whenever \((u_{10}, u_{20}, v_{10}, v_{20})\) is from the set

$$\begin{aligned}&{\mathcal {A}}_1({\mathcal {M}}, {\mathcal {N}}, L): = \Big \{(u_{10}, u_{20}, v_{10}, v_{20})\in (C^0({\overline{\Omega }}))^2\nonumber \\&\quad \times (W^{1, \infty }(\Omega ))^2| u_{10}, u_{20}, v_{10}\quad \text {and}\quad v_{20}\quad \text {are positive}\nonumber \\&\quad \text {and radially symmetric in}~{\overline{\Omega }}\quad \text {with}\quad \Vert u_{i0}\Vert _{L^1(\Omega )}\le {\mathcal {M}}, \Vert v_{i0}\Vert _{W^{1, 2}(\Omega )}\nonumber \\&\quad \le {\mathcal {N}}, \mathrm {F}(u_{10}, u_{20}, v_{10}, v_{20}) \le -{\mathcal {P}} \Big \}, \end{aligned}$$
(1.7)

there exists some \((x_{L}, t_{L})\in \Omega \times (0, T)\) which makes the corresponding classical solution \((u_1, u_2, v_1, v_2)\) of (1.4) fulfill

$$\begin{aligned} u_1(x_{L}, t_{L})>L\quad \text {or}\quad u_2(x_{L}, t_{L})>L. \end{aligned}$$

Corollary 1.1

Let \(\Omega =B_R(0)\subset {\mathbb {R}}^n(n\ge 3, R>0)\), \(\chi _{11}=\chi _{12}>0, \chi _{21}=\chi _{22}>0\), \(a_i>0\), \(\mu _{i}\in (0, 1)\) \((i=1, 2)\), \(\mu =\max \{\mu _1, \mu _2\}\). Then for all \({\mathcal {M}}>0, {\mathcal {N}}>0\) and each \(L>0, T>0\), there exists positive constant \(K= K( {\mathcal {M}}, {\mathcal {N}}, L, T, R, n, \chi _{11}, \chi _{12}, \chi _{21}, \chi _{22}, a_1, a_2)\), provided that the initial data is from the set \({\mathcal {A}}_2\) denoted by

$$\begin{aligned}&{\mathcal {A}}_2({\mathcal {M}}, {\mathcal {N}}, L): = \Big \{(u_{10}, u_{20}, v_{10}, v_{20})\in (C^0({\overline{\Omega }}))^2\times (W^{1, \infty }(\Omega ))^2| u_{10}, u_{20}, v_{10}\nonumber \\&\quad \text {and}\quad v_{20}\quad \text {are positive}\;\text {and radially symmetric in}\quad {\overline{\Omega }}\quad \text {with}\quad \Vert u_{i0}\Vert _{L^1(\Omega )}\nonumber \\&\quad \le {\mathcal {M}}, \Vert v_{i0}\Vert _{W^{1, 2}(\Omega )}\le {\mathcal {N}},\quad \mathrm {F}(u_{10}, u_{20}, v_{10}, v_{20}) \le -K \Big \}, \end{aligned}$$
(1.8)

for all \(\mu \in (0, 1) \), one can find \( (x_{\mu }, t_{\mu })\in \Omega \times (0, T)\) such that the corresponding classical solution \((u_{1, \mu }, u_{2, \mu }, v_{1, \mu }, v_{2, \mu })\) of (1.4) satisfies

$$\begin{aligned} u_{1, \mu }(x_{\mu }, t_{\mu })>\frac{L}{\mu }\quad \text {or}\quad u_{2, \mu }(x_{\mu }, t_{\mu })>\frac{L}{\mu }. \end{aligned}$$

Remark 1.1

The above sets \({\mathcal {A}}_i({\mathcal {M}}, {\mathcal {N}}, {\mathcal {L}}) \ne \emptyset (i=1, 2)\), which is a straightforward result from Lemma 6.1 in [44].

Remark 1.2

From Corollary 1.1, we notice that the lower bound of \( u_{1, \mu }(x_{\mu }, t_{\mu }) \) or \( u_{2, \mu }(x_{\mu }, t_{\mu })\) can be sufficiently large as long as we control the value of \(\mu \). In addition, for the case \( \chi _{11}=\chi _{12}, \chi _{21}=\chi _{22}\), problem (1.4) can be reduced to the two-species and one-stimuli case, so our Corollary 1.1 extends the main result in [15].

The following consequence underlines that once the blowup happens, it is simultaneous for the two species.

Theorem 1.2

Let \(\Omega \subset {\mathbb {R}}^n(n\ge 3)\) be a smooth, bounded, and convex domain, \(\chi _{ij}>0\), \(\mu _{i}>0\), \(a_i>0\) \((i, j=1, 2)\), and let \(u_{10}, u_{20}, v_{10}\) and \(v_{20}\) satisfy (1.5). If the corresponding solution \((u_1, u_2, v_1, v_2)\) of (1.4) satisfies

$$\begin{aligned} \displaystyle \lim _{t\rightarrow T}\sup (\Vert u_1\Vert _{L^{\infty }(\Omega )}+\Vert u_2\Vert _{L^{\infty }(\Omega )})=\infty \quad \text {for some}\quad T>0. \end{aligned}$$

Then

$$\begin{aligned} \displaystyle \lim _{t\rightarrow T}\sup \Vert u_1\Vert _{L^{\infty }(\Omega )} =\displaystyle \lim _{t\rightarrow T}\sup \Vert u_2\Vert _{L^{\infty }(\Omega )}=\infty \quad \text {for some}\quad T>0. \end{aligned}$$

This paper is organized as follows. In Sect. 2, we derive the local existence and uniqueness result, as well as some important inequalities. In Sect. 3, we focus on deriving a differential inequality related to the function \(\mathrm {F}(u_1, u_2, v_1, v_2)\) denoted by (1.6), and in light of the contradictory arguments, we accomplish the proof of Theorem 1.1 and Corollary 1.1. Section 4 is devoted to investigating the simultaneous blowup for both species.

2 Local existence and basic estimates

Our goal in this section is to present some useful inequalities and the local well-posedness results for (1.4). First of all, we recall the following lemma, which is a direct result of Lemma 2.1 in [36].

Lemma 2.1

Let \(\Omega \subset {\mathbb {R}}^n(n\ge 1)\) be a smooth and bounded domain, \(\chi _{ij}>0\), \(\mu _{i}>0\), \(a_{i}>0\) (\(i,j=1,2\)), and assume that \((u_{10},u_{20}, v_{10}, v_{20})\) satisfies (1.5). Then there exists a maximal \(T_{\max }\in (0,\infty ]\) such that the system (1.4) has a uniquely determined nonnegative solution \((u_1,u_2,v_1,v_2)\)

$$\begin{aligned} \begin{aligned}&u_1,\quad u_2\in C^0(\overline{\Omega }\times {[0,T_{\max })})\cap C^{2,1}( \overline{\Omega }\times (0,T_{\max })),\\&v_1,\quad v_2\in C^0(\overline{\Omega }\times {[0,T_{\max })})\cap C^{2,1}( \overline{\Omega }\times (0,T_{\max }))\cap L_{loc}^{\infty }( [0, T_{\max }); W^{1, \infty }(\Omega ) ), \end{aligned} \end{aligned}$$

which solves (1.4) classically and satisfies

$$\begin{aligned} \text {either}\quad T_{\max }=\infty , \quad \text {or}\quad \Vert u_1(\cdot ,t)\Vert _{L^{\infty }(\Omega )}+||u_2(\cdot ,t)||_{L^{\infty }(\Omega )}\rightarrow \infty \quad \text {as}\quad t\rightarrow T_{\max }. \end{aligned}$$

Moreover, the classical solution of (1.4) fulfills

$$\begin{aligned} \int _{\Omega }u_1(x, t)\mathrm{d}x\le m_1:= \max \left\{ \int _{\Omega }u_{10}(x)\mathrm{d}x, |\Omega |\right\} \quad \text {for all}\quad t\in (0, T_{\mathrm{max}}) \end{aligned}$$
(2.1)

and

$$\begin{aligned} \int _{\Omega }u_2(x, t)\mathrm{d}x\le m_2:= \max \left\{ \int _{\Omega }u_{20}(x)\mathrm{d}x, |\Omega |\right\} \quad \text {for all}\quad t\in (0, T_{\mathrm{max}}) \end{aligned}$$
(2.2)

as well as

$$\begin{aligned} \int _t^{t+\tau } \int _{\Omega }u_1^2(x, t)\mathrm{d}x\mathrm{d}s\le {\mathbf {k}}_1:= m_1+\frac{m_1}{\mu _1}\quad \text {for all}\quad t\in [0, T_{\mathrm{max}}-\tau ) \end{aligned}$$
(2.3)

and

$$\begin{aligned} \int _t^{t+\tau } \int _{\Omega }u_2^2(x, t)\mathrm{d}x\mathrm{d}s\le {\mathbf {k}}_2:= m_2+\frac{m_2}{\mu _2}\quad \text {for all}\quad t\in [0, T_{\mathrm{max}}-\tau ) \end{aligned}$$
(2.4)

with \(\tau :=\min \{1, \frac{T_{\mathrm{max}}}{2}\}\).

Next, our attention is turned to deriving the estimate for \(v_1, v_2\), which plays a significant role in our later proof.

Lemma 2.2

Suppose that the assumptions in Lemma 2.1 are satisfied. Then for any classical solution \((u_1, u_2, v_1, v_2)\) of (1.4), one can find \(M_1, M_2, C_1, {\hat{C}}_1>0\) such that

$$\begin{aligned} \int _{\Omega }v_1\mathrm{d}x\le M_1,\quad \int _{\Omega }v_2\mathrm{d}x\le M_2\quad \text {for all}\quad t\in (0, T_\mathrm{max}) \end{aligned}$$
(2.5)

and

$$\begin{aligned} \int _{\Omega }|\nabla v_1(x, t)|^2\mathrm{d}x\le C_1\quad \text {for all}\quad t\in (0, T_\mathrm{max}) \end{aligned}$$
(2.6)

as well as

$$\begin{aligned} \int _{\Omega }|\nabla v_2(x, t)|^2\mathrm{d}x\le {\hat{C}}_1\quad \text {for all}\quad t\in (0, T_\mathrm{max}). \end{aligned}$$
(2.7)

Proof

Using the fact that \(\int _{\Omega }u_i\mathrm{d}x (i=1, 2)\) are bounded, integrating the third and fourth equations of (1.4) over \(\Omega \) we find

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\int _{\Omega }v_1\mathrm{d}x+\int _{\Omega }v_1\mathrm{d}x \le m_1+m_2,\quad \frac{\mathrm{d}}{\mathrm{d}t}\int _{\Omega }v_2\mathrm{d}x+\int _{\Omega }v_2\mathrm{d}x \le m_1+m_2\quad \text {for all}\quad t\in (0, T_\mathrm{max}), \end{aligned}$$

by a direct computation we can obtain (2.5). Testing the third equation of (1.4) by \(-\Delta v_1\), and in light of the Young inequality, it is clear that

$$\begin{aligned} \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\int _{\Omega }|\nabla v_1|^2\mathrm{d}x+\int _{\Omega }|\nabla v_1|^2\mathrm{d}x+\frac{1}{2}\int _{\Omega }|\Delta v_1|^2\mathrm{d}x \le \int _{\Omega } (u_1^2+u_2^2)\mathrm{d}x \quad \text {for all}\quad t\in (0, T_\mathrm{max}), \end{aligned}$$

then utilizing (2.3) and (2.4), it follows from Lemma 3.4 in [29] that there exists a positive constant \(C_1=C_1(m_1, m_2, \int _{\Omega }|\nabla v_{10}(x, t)|^2\mathrm{d}x)\) such that

$$\begin{aligned} \int _{\Omega }|\nabla v_1(x, t)|^2\mathrm{d}x\le C_1(m_1, m_2, \int _{\Omega }|\nabla v_{10}(x, t)|^2\mathrm{d}x). \end{aligned}$$

In quite a similar manner, we can derive (2.7). \(\square \)

With the aid of Lemmas 2.1 and 2.2, we now establish the pointwise upper bounds for radial solutions \(v_1, v_2\) in the following lemma. The proof of the following lemma is similar to [14, 44], and hence we omit it here for brevity.

Lemma 2.3

Let \(\Omega =B_R(0)\subset {\mathbb {R}}^n(n\ge 3, R>0)\), \(p\in (1, \frac{n}{n-1})\), \(\chi _{ij}>0\), \(\mu _{i}>0\), \(a_{i}>0\) (\(i,j=1,2\)). Then for any radially symmetric initial data \((u_{10},u_{20}, v_{10}, v_{20})\) satisfies (1.5), there exists \(C(p)>0\) such that the corresponding classical solution of (1.4) satisfies

$$\begin{aligned} v_1(r, t)\le C(p)\cdot \left( \Vert u_{10}\Vert _{L^1(\Omega )}+ \Vert u_{20}\Vert _{L^1(\Omega )}+ \Vert v_{10}\Vert _{W^{1, 2}(\Omega )}+\Vert v_{20}\Vert _{W^{1, 2}(\Omega )} \right) \cdot r^{-\frac{n-p}{p}}\nonumber \\ \end{aligned}$$
(2.8)

and

$$\begin{aligned} v_2(r, t)\le C(p)\cdot \left( \Vert u_{10}\Vert _{L^1(\Omega )}+ \Vert u_{20}\Vert _{L^1(\Omega )}+ \Vert v_{10}\Vert _{W^{1, 2}(\Omega )}+\Vert v_{20}\Vert _{W^{1, 2}(\Omega )} \right) \cdot r^{-\frac{n-p}{p}}\nonumber \\ \end{aligned}$$
(2.9)

for all \((r, t)\in (0, R)\times (0, T_{\mathrm{max}})\).

Remark 2.1

From Lemmas 2.12.3, we find that any radially symmetric classical solutions \((u_1, u_2, v_1, v_2)\) from Lemma 2.1 has the property

$$\begin{aligned} \int _{\Omega }u_i\mathrm{d}x\le m\quad \hbox {and}\quad \int _{\Omega }v_i\mathrm{d}x\le M\quad \hbox {for}\quad i=1, 2, \end{aligned}$$
(2.10)

and

$$\begin{aligned} v_1(x, t), \quad v_2(x, t)\le B|x|^{-k}\quad \text {for all}\quad x\in \Omega \end{aligned}$$
(2.11)

with \(m=\max \{m_1, m_2\}\), \(M=\max \{M_1, M_2\}\),

$$\begin{aligned} B:=C(p)\cdot \left( \Vert u_{10}\Vert _{L^1(\Omega )}+ \Vert u_{20}\Vert _{L^1(\Omega )}+ \Vert v_{10}\Vert _{W^{1, 2}(\Omega )}+\Vert v_{20}\Vert _{W^{1, 2}(\Omega )} \right) ,\quad k:=\frac{n-p}{p},\nonumber \\ \end{aligned}$$
(2.12)

where \(p\in (1, \frac{n}{n-1})\), this implies \(k>n-2\).

Moreover, we introduce the space S(MBk), which is defined by

$$\begin{aligned}&S(M, B, k):=\Big \{ (u_1, u_2, v_1, v_2)\in (C^1({\overline{\Omega }}))\times (C^1({\overline{\Omega }})) \times (C^2({\overline{\Omega }}))\nonumber \\&\quad \times (C^2({\overline{\Omega }}))| u_1, u_2, v_1, v_2\quad \hbox {are positive}\;\hbox {and radially symmetric satisfying}\nonumber \\&\quad \frac{\partial v_1}{\partial \nu }=\frac{\partial v_2}{\partial \nu }=0~\text {on}~\partial \Omega ,~ and (2.10), (2.11)~\hbox {hold}. \Big \}. \end{aligned}$$
(2.13)

This thus implies that any radially symmetric classical solutions from Lemma 2.1 have the property \((u_1, u_2, v_1, v_2)\in S(M, B, k)\) for all \(t\in (0, T_{\mathrm{max}})\).

3 Emergence of large population densities

Motivated by the contradictory strategy in [14, 15, 44], for \((u_1, u_2, v_1, v_2)\in \left( C^{2,1}(\overline{\Omega }\times (0,T_{\max }))\right) ^4\), we introduce the energy function \(\mathrm {F}(u_1, u_2, v_1, v_2)\),

$$\begin{aligned} \begin{aligned} \mathrm {F}(u_1, u_2, v_1, v_2)&=\frac{1}{2}\int _{\Omega }(|v_1|^2+|v_2|^2+|\nabla v_1|^2+|\nabla v_2|^2)\mathrm{d}x -\int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x\\&\quad +\frac{1}{\chi _1}\int _{\Omega }u_1\ln u_1\mathrm{d}x +\frac{1}{\chi _2}\int _{\Omega }u_2\ln u_2\mathrm{d}x \end{aligned} \end{aligned}$$
(3.1)

with \(\chi _1=\min \{\chi _{11}, \chi _{12}\}\), \(\chi _2=\min \{\chi _{21}, \chi _{22}\}\). Our aim in this section is to establish the following differential inequality

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}(-\mathrm {F}(u_1, u_2, v_1, v_2))\ge C(-\mathrm {F}(u_1, u_2, v_1, v_2))^{\delta }\quad \text {for some}\quad \delta >1 \end{aligned}$$

under the hypothesis that \(u_1, u_2\) are bounded, this would lead to an absurd conclusion, then we can obtain Theorem 1.1 and Corollary 1.1. To this end, for \((u_1, u_2, v_1, v_2)\in \left( C^{2,1}( \overline{\Omega }\times (0,T_{\max }))\right) ^4\), we introduce the following functions,

$$\begin{aligned}&\mathrm {D}(u_1, u_2, v_1, v_2):=\int _{\Omega }(|v_{1t}|^2+|v_{2t}|^2)\mathrm{d}x +\int _{\Omega }\left( \sqrt{{\hat{\chi }}_1u_1}(\nabla v_1+\nabla v_2) -\frac{\nabla u_1}{\sqrt{{\hat{\chi }}_1u_1}} \right) ^2\mathrm{d}x\nonumber \\&\quad +\int _{\Omega }\left( \sqrt{{\hat{\chi }}_2u_2}(\nabla v_1+\nabla v_2) -\frac{\nabla u_2}{\sqrt{{\hat{\chi }}_2u_2}} \right) ^2\mathrm{d}x,\end{aligned}$$
(3.2)
$$\begin{aligned}&f_1(u_1, u_2, v_1, v_2):=-\Delta v_1+v_1-u_1-u_2,\quad f_2(u_1, u_2, v_1, v_2):=-\Delta v_2+v_2-u_1-u_2,\end{aligned}$$
(3.3)
$$\begin{aligned}&g_1(u_1, v_1, v_2):=\left( \frac{\nabla u_1}{\sqrt{{\hat{\chi }}_1u_1}}-\sqrt{{\hat{\chi }}_1u_1}(\nabla v_1+\nabla v_2) \right) \cdot \frac{x}{|x|},\quad x\ne 0,\end{aligned}$$
(3.4)
$$\begin{aligned}&g_2(u_2, v_1, v_2):=\left( \frac{\nabla u_2}{\sqrt{{\hat{\chi }}_2u_2}}-\sqrt{{\hat{\chi }}_2u_2}(\nabla v_1+\nabla v_2) \right) \cdot \frac{x}{|x|},\quad x\ne 0, \end{aligned}$$
(3.5)

where \({\hat{\chi }}_1= \frac{\chi _{11}+\chi _{12}}{2}\), \({\hat{\chi }}_2=\frac{\chi _{21}+\chi _{22}}{2}\).

The following differential inequality of \(\mathrm {F}(u_1, u_2, v_1, v_2)\) is essential for our proof.

Lemma 3.1

Let \(\Omega =B_R(0)\subset {\mathbb {R}}^n(n\ge 3, R>0)\), \(\chi _{ij}>0\), \(\mu _{i}>0\), \(a_{i}>0\) (\(i,j=1,2\)). Suppose that \((u_1, u_2, v_1, v_2)\in \left( C^{2,1}( \overline{\Omega }\times (0,T_{\max }))\right) ^4\) is a classical solution of (1.4), and \((u_1, u_2, v_1, v_2)\in S(M, B, k)\). Then for all \(t\in (0, T_{\mathrm{max}})\), one obtains

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\mathrm {F}(u_1, u_2, v_1, v_2)\nonumber \\&\le -\mathrm {D}(u_1, u_2, v_1, v_2) +\frac{\chi _{12}-\chi _{11}}{2}\int _{\Omega }u_1|\nabla v_1|^2\mathrm{d}x +\frac{\chi _{11}-\chi _{12}}{2}\int _{\Omega }u_1|\nabla v_2|^2\mathrm{d}x\nonumber \\&\quad +\frac{\chi _{22}-\chi _{21}}{2}\int _{\Omega }u_2|\nabla v_1|^2\mathrm{d}x +\frac{\chi _{21}-\chi _{22}}{2}\int _{\Omega }u_2|\nabla v_2|^2\mathrm{d}x +\left( \frac{\chi _{11}}{\chi _1}-1\right) \int _{\Omega }\nabla u_1\nabla v_1\mathrm{d}x\nonumber \\&\quad +\left( \frac{\chi _{12}}{\chi _1}-1\right) \int _{\Omega }\nabla u_1\nabla v_2\mathrm{d}x + \left( \frac{\chi _{21}}{\chi _2}-1\right) \int _{\Omega }\nabla u_2\nabla v_1\mathrm{d}x + \left( \frac{\chi _{22}}{\chi _2}-1\right) \int _{\Omega }\nabla u_2\nabla v_2\mathrm{d}x\nonumber \\&\quad -\left( \frac{1}{\chi _1}-\frac{2}{\chi _{11}+\chi _{12}}\right) \int _{\Omega }\frac{|\nabla u_1|^2}{u_1}\mathrm{d}x -\left( \frac{1}{\chi _2}-\frac{2}{\chi _{21}+\chi _{22}}\right) \int _{\Omega }\frac{|\nabla u_2|^2}{u_2}\mathrm{d}x\nonumber \\&\quad +(a_1\mu _1+a_2\mu _2)\int _{\Omega }u_1u_2(v_1+v_2)\mathrm{d}x +\int _{\Omega }(\mu _1u_1^2v_1+\mu _2u_2^2v_1+\mu _1u_1^2v_2+\mu _2u_2^2v_2)\mathrm{d}x\nonumber \\&\quad +\int _{\Omega }\left( \frac{\mu _1}{\chi _1}u_1\ln u_1 +\frac{\mu _2}{\chi _2}u_2\ln u_2 +\frac{\mu _1}{\chi _1}u_1\right. \nonumber \\&\quad \left. +\frac{\mu _2}{\chi _2}u_2\right) \mathrm{d}x -\int _{\Omega }(\mu _1u_1v_1+\mu _2u_2v_1+\mu _1u_1v_2+\mu _2u_2v_2)\mathrm{d}x\nonumber \\&\quad -\int _{\Omega }\left( \frac{\mu _1}{\chi _1}u_1^2\ln u_1+\frac{\mu _2}{\chi _2}u_2^2\ln u_2+\frac{a_1\mu _1}{\chi _1}u_1u_2\ln u_1 +\frac{a_2\mu _2}{\chi _2}u_1u_2\ln u_2\right) \mathrm{d}x, \end{aligned}$$
(3.6)

where \(\mathrm {F}\) and \(\mathrm {D}\) are defined in (3.1) and (3.2), \(\chi _1=\min \{\chi _{11}, \chi _{12}\}\), \(\chi _2=\min \{\chi _{21}, \chi _{22}\}\).

Proof

Testing the third and fourth equations of (1.4) by \(v_{1t}, v_{2t}\), respectively, to obtain

$$\begin{aligned} \int _{\Omega }|v_{1t}|^2\mathrm{d}x=-\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\int _{\Omega }(|\nabla v_1|^2+| v_1|^2)\mathrm{d}x +\int _{\Omega }(u_1+u_2)v_{1t}\mathrm{d}x\quad \text {for all}\quad t\in (0, T_\mathrm{max})\nonumber \\ \end{aligned}$$
(3.7)

and

$$\begin{aligned} \int _{\Omega }|v_{2t}|^2\mathrm{d}x=-\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\int _{\Omega }(|\nabla v_2|^2+| v_2|^2)\mathrm{d}x +\int _{\Omega }(u_1+u_2)v_{2t}\mathrm{d}x\quad \text {for all}\quad t\in (0, T_\mathrm{max}).\nonumber \\ \end{aligned}$$
(3.8)

Together with (3.7) and (3.8), we find

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\int _{\Omega }(| v_1|^2+|\nabla v_1|^2+| v_2|^2+|\nabla v_2|^2)\mathrm{d}x\\&\quad =-\int _{\Omega }|v_{1t}|^2\mathrm{d}x-\int _{\Omega }|v_{2t}|^2\mathrm{d}x +\int _{\Omega }(u_1+u_2)v_{1t}\mathrm{d}x\\&\qquad +\int _{\Omega }(u_1+u_2)v_{2t}\mathrm{d}x\quad \text {for all}\quad t\in (0, T_\mathrm{max}). \end{aligned} \end{aligned}$$
(3.9)

For the last two terms of (3.9), utilizing the first and second equations of (1.4), it follows that

$$\begin{aligned} \begin{aligned}&\int _{\Omega }(u_1+u_2)v_{1t}\mathrm{d}x +\int _{\Omega }(u_1+u_2)v_{2t}\mathrm{d}x\\&\quad =\frac{\mathrm{d}}{\mathrm{d}t}\int _{\Omega }(u_1+u_2)v_1\mathrm{d}x-\int _{\Omega }v_{1}(u_1+u_2)_t\mathrm{d}x +\frac{\mathrm{d}}{\mathrm{d}t}\int _{\Omega }(u_1+u_2)v_2\mathrm{d}x-\int _{\Omega }v_{2}(u_1+u_2)_t\mathrm{d}x\\&\quad =\frac{\mathrm{d}}{\mathrm{d}t}\int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x -\int _{\Omega }\left( \sqrt{\frac{\chi _{11}+\chi _{12}}{2}u_1}(\nabla v_1+\nabla v_2) -\frac{\nabla u_1}{\sqrt{\frac{\chi _{11}+\chi _{12}}{2}u_1}}\right) ^2\mathrm{d}x\\&\qquad -\int _{\Omega }\left( \sqrt{\frac{\chi _{21}+\chi _{22}}{2}u_2}(\nabla v_1+\nabla v_2) -\frac{\nabla u_2}{\sqrt{\frac{\chi _{21}+\chi _{22}}{2}u_2}} \right) ^2\mathrm{d}x\\&\qquad -\int _{\Omega }\nabla u_1(\nabla v_1+\nabla v_2)\mathrm{d}x -\int _{\Omega }\nabla u_2(\nabla v_1+\nabla v_2)\mathrm{d}x +\frac{\chi _{12}-\chi _{11}}{2}\int _{\Omega }u_1|\nabla v_1|^2\mathrm{d}x\\&\qquad +\frac{\chi _{11}-\chi _{12}}{2}\int _{\Omega }u_1|\nabla v_2|^2\mathrm{d}x +\frac{\chi _{22}-\chi _{21}}{2}\int _{\Omega }u_2|\nabla v_1|^2\mathrm{d}x +\frac{\chi _{21}-\chi _{22}}{2}\int _{\Omega }u_2|\nabla v_2|^2\mathrm{d}x\\&\qquad +\int _{\Omega }\frac{2|\nabla u_1|^2}{(\chi _{11}+\chi _{12})u_1}\mathrm{d}x +\int _{\Omega }\frac{2|\nabla u_2|^2}{(\chi _{21}+\chi _{22})u_2}\mathrm{d}x\\&\qquad -\mu _1\int _{\Omega }u_1(v_1+v_2)(1-u_1-a_1u_2)\mathrm{d}x -\mu _2\int _{\Omega }u_2(v_1+v_2)(1-u_2-a_2u_1)\mathrm{d}x \end{aligned} \end{aligned}$$
(3.10)

for \(t\in (0, T_{\mathrm{max}})\). For \(\chi _1=\min \{\chi _{11}, \chi _{12}\}\), \(\chi _2=\min \{\chi _{21}, \chi _{22}\}\), a direct calculation reveals that

$$\begin{aligned} \begin{aligned} \frac{1}{\chi _1}\frac{\mathrm{d}}{\mathrm{d}t}\int _{\Omega }u_1\ln u_1\mathrm{d}x&= -\int _{\Omega }\frac{|\nabla u_1|^2}{\chi _1u_1}\mathrm{d}x +\frac{\chi _{11}}{\chi _1}\int _{\Omega }\nabla u_1\nabla v_1\mathrm{d}x +\frac{\chi _{12}}{\chi _1}\int _{\Omega }\nabla u_1\nabla v_2\mathrm{d}x\\&\quad +\frac{\mu _1}{\chi _1}\int _{\Omega }u_1\ln u_1(1-u_1-a_1u_2)\mathrm{d}x +\frac{\mu _1}{\chi _1}\int _{\Omega }u_1(1-u_1-a_1u_2)\mathrm{d}x \end{aligned}\nonumber \\ \end{aligned}$$
(3.11)

for \(t\in (0, T_{\mathrm{max}})\). Similarly, we can deduce that

$$\begin{aligned} \begin{aligned} \frac{1}{\chi _2}\frac{\mathrm{d}}{\mathrm{d}t}\int _{\Omega }u_2\ln u_2\mathrm{d}x&= -\int _{\Omega }\frac{|\nabla u_2|^2}{\chi _2u_2}\mathrm{d}x +\frac{\chi _{21}}{\chi _2}\int _{\Omega }\nabla u_2\nabla v_1\mathrm{d}x +\frac{\chi _{22}}{\chi _2}\int _{\Omega }\nabla u_2\nabla v_2\mathrm{d}x\\&\quad +\frac{\mu _2}{\chi _2}\int _{\Omega }u_2\ln u_2(1-u_2-a_2u_1)\mathrm{d}x +\frac{\mu _2}{\chi _2}\int _{\Omega }u_2(1-u_2-a_2u_1)\mathrm{d}x \end{aligned}\nonumber \\ \end{aligned}$$
(3.12)

for \(t\in (0, T_{\mathrm{max}})\). Thus, a combination of (3.9)–(3.12) yields (3.6). \(\square \)

In the next four lemmas, we focus on dealing with the term \(\int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x\) in \(\mathrm {F}(u_1, u_2, v_1, v_2)\), our aim is to seek the constants \(\vartheta \in (\frac{1}{2}, 1)\) and \(C>0\) such that

$$\begin{aligned} \int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x\le C(D^\vartheta (u_1, u_2, v_1, v_2)+1)\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}), \end{aligned}$$

where D is introduced in (3.2), which acts a pivotal part in the latter proof.

Lemma 3.2

Let \(\Omega =B_R(0)\subset {\mathbb {R}}^n(n\ge 3, R>0)\), \(\chi _{ij}>0\), \(\mu _{i}>0\), \(a_{i}>0\). Suppose that \((u_1, u_2, v_1, v_2)\in \left( C^{2,1}( \overline{\Omega }\times (0,T_{\max }))\right) ^4\) is a classical solution of (1.4), and \((u_1, u_2, v_1, v_2)\in S(M, B, k)\). Let \(f_i\) be given by (3.3) (\(i,j=1,2\)), then for any \(\epsilon \in (0, 1)\), there exist \({\mathcal {C}}_1(M), \hat{{\mathcal {C}}}_1(M)>0\) such that

$$\begin{aligned} \int _{\Omega }(u_1+u_2)v_1\mathrm{d}x \le (1+\epsilon )\int _{\Omega } |\nabla v_1|^2\mathrm{d}x +{\mathcal {C}}_1(M)(\Vert f_1\Vert _{L^2(\Omega )}^{\frac{2n+4}{n+4}}+1) \quad \text {for all}\quad t\in (0, T_{\mathrm{max}})\nonumber \\ \end{aligned}$$
(3.13)

and

$$\begin{aligned} \int _{\Omega }(u_1+u_2)v_2\mathrm{d}x \le (1+\epsilon )\int _{\Omega } |\nabla v_2|^2\mathrm{d}x +\hat{ {\mathcal {C}}}_1(M)(\Vert f_2\Vert _{L^2(\Omega )}^{\frac{2n+4}{n+4}}+1)\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}).\nonumber \\ \end{aligned}$$
(3.14)

Proof

From (3.3), for \((u_1, u_2, v_1, v_2)\in \left( C^{2,1}( \overline{\Omega }\times (0,T_{\max }))\right) ^4\), we have

$$\begin{aligned} f_1:=-\Delta v_1+v_1-u_1-u_2, \end{aligned}$$

testing the above equation by \(v_1\), we find

$$\begin{aligned} \int _{\Omega }(u_1+u_2)v_1\mathrm{d}x = \int _{\Omega }|\nabla v_1|^2\mathrm{d}x+\int _{\Omega }|v_1|^2\mathrm{d}x -\int _{\Omega }f_1v_1\mathrm{d}x \quad \text {for all}\quad t\in (0, T_{\mathrm{max}}),\nonumber \\ \end{aligned}$$
(3.15)

it follows from the Gagliardo–Nirenberg inequality that

$$\begin{aligned} \Vert v_1\Vert _{L^2(\Omega )} \le C_{GN} \Vert \nabla v_1\Vert _{L^2(\Omega )}^a \cdot \Vert v_1\Vert _{L^1(\Omega )}^{1-a}+ \Vert v_1\Vert _{L^1(\Omega )} \end{aligned}$$
(3.16)

with \(\frac{1}{2}=a(\frac{1}{2}-\frac{1}{n})+\frac{1-a}{1}\), i.e., \(a=\frac{n}{n+2}\). In light of (2.10), one can find \(c_1(M)>0\) such that

$$\begin{aligned} \Vert v_1\Vert _{L^2(\Omega )}\le c_1(M)(\Vert \nabla v_1\Vert _{L^2(\Omega )}^{\frac{n}{n+2}}+1)\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}), \end{aligned}$$
(3.17)

thereupon, invoking the Young inequality, we find

$$\begin{aligned} c_1(M)\Vert \nabla v_1\Vert _{L^2(\Omega )}^{\frac{n}{n+2}} \le \frac{\epsilon }{2}\Vert \nabla v_1\Vert _{L^2(\Omega )}^2+c_2(M)\quad \text {for all}\quad \epsilon >0,\quad t\in (0, T_{\mathrm{max}}), \end{aligned}$$

with \(c_2(M)>0\), which combined with (3.17) result in

$$\begin{aligned} \Vert v_1\Vert _{L^2(\Omega )} \le \frac{\epsilon }{2}\Vert \nabla v_1\Vert _{L^2(\Omega )}^2+c_1(M)+c_2(M)\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}). \end{aligned}$$
(3.18)

In view of (3.17) and the Young inequality, for any \(\epsilon >0\), we see that

$$\begin{aligned} \begin{aligned} -\int _{\Omega }f_1v_1\mathrm{d}x&\le \Vert f_1\Vert _{L^2(\Omega )}\cdot \Vert v_1\Vert _{L^2(\Omega )} \le c_1(M)(\Vert \nabla v_1\Vert _{L^2(\Omega )}^{\frac{n}{n+2}}+1)\Vert f_1\Vert _{L^2(\Omega )}\\&\le \frac{\epsilon }{2}\Vert \nabla v_1\Vert _{L^2(\Omega )}^2 +c_3(M)\Vert f_1\Vert _{L^2(\Omega )}^{\frac{2n+4}{n+4}} +c_1(M)\Vert f_1\Vert _{L^2(\Omega )}\\&\le \frac{\epsilon }{2}\Vert \nabla v_1\Vert _{L^2(\Omega )}^2 +c_4(M)\Vert f_1\Vert _{L^2(\Omega )}^{\frac{2n+4}{n+4}}+c_4(M)\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}) \end{aligned} \end{aligned}$$
(3.19)

with \(c_3(M), c_4(M)>0\). Therefore, a combination of (3.15), (3.18) and (3.19) yields (3.13). In a similar way, we can derive (3.14). \(\square \)

The following lemma is an extension of Lemma 4.3 in [44].

Lemma 3.3

Let \(\Omega =B_R(0)\subset {\mathbb {R}}^n(n\ge 3, R>0)\), \(\chi _{ij}>0\), \(\mu _{i}>0\), \(a_{i}>0\). Suppose that \((u_1, u_2, v_1, v_2)\in \left( C^{2,1}( \overline{\Omega }\times (0,T_{\max }))\right) ^4\) is a classical solution of (1.4), and \((u_1, u_2, v_1, v_2)\in S(M, B, k)\). Let \(f_i\) be as in (3.3) (\(i,j=1,2\)), then for any \(\epsilon \in (0, 1)\) and \(r_0\in (0, R)\), there exists \(C=C(\epsilon , m_1, m_2, M, B, k)\) such that

$$\begin{aligned} \int _{\Omega \setminus B_{r_0}}|\nabla v_1|^2\mathrm{d}x \le \epsilon \int _{\Omega }(u_1+u_2)v_1\mathrm{d}x +\epsilon \Vert \nabla v_1\Vert _{L^2(\Omega )}^2 +C\cdot \left( r_0^{-\frac{2n+4}{n}\cdot k}+\Vert f_1\Vert _{L^2(\Omega )}^{\frac{2n+4}{n+4}}\right) \nonumber \\ \end{aligned}$$
(3.20)

and

$$\begin{aligned} \int _{\Omega \setminus B_{r_0}}|\nabla v_2|^2\mathrm{d}x \le \epsilon \int _{\Omega }(u_1+u_2)v_2\mathrm{d}x +\epsilon \Vert \nabla v_2\Vert _{L^2(\Omega )}^2 +C\cdot \left( r_0^{-\frac{2n+4}{n}\cdot k}+\Vert f_2\Vert _{L^2(\Omega )}^{\frac{2n+4}{n+4}}\right) \nonumber \\ \end{aligned}$$
(3.21)

for all \(t\in (0, T_{\mathrm{max}})\).

Proof

Fixing any \(\xi \in (0, 1)\), testing \(f_1\) by \(v_1^{\xi }\), it is easy to obtain that

$$\begin{aligned} \xi \int _{\Omega } v_1^{\xi -1}|\nabla v_1|^2\mathrm{d}x \le \xi \int _{\Omega } v_1^{\xi -1}|\nabla v_1|^2\mathrm{d}x +\int _{\Omega }v_1^{\xi +1}\mathrm{d}x = \int _{\Omega }(u_1+u_2)v_1^{\xi }\mathrm{d}x +\int _{\Omega }f_1v_1^{\xi }\mathrm{d}x\nonumber \\ \end{aligned}$$
(3.22)

for all \(t\in (0, T_{\mathrm{max}})\). In light of (2.11), one obtains

$$\begin{aligned} \xi \cdot B^{\xi -1}\cdot r_0^{(1-\xi )k} \int _{\Omega \setminus B_{r_0}}|\nabla v_1|^2\mathrm{d}x \le \xi \int _{\Omega } v_1^{\xi -1} |\nabla v_1|^2\mathrm{d}x\quad \text {for all}\quad t\in (0, T_\mathrm{max}).\nonumber \\ \end{aligned}$$
(3.23)

Based on the above two inequalities, we arrive at

$$\begin{aligned} \int _{\Omega \setminus B_{r_0}}|\nabla v_1|^2\mathrm{d}x \le \frac{B^{1-\xi }}{\xi }\cdot r_0^{-(1-\xi )k} \int _{\Omega }(u_1+u_2)v_1^{\xi }\mathrm{d}x +\frac{B^{1-\xi }}{\xi }\cdot r_0^{-(1-\xi )k}\int _{\Omega }f_1v_1^{\xi }\mathrm{d}x\nonumber \\ \end{aligned}$$
(3.24)

for all \(t\in (0, T_\mathrm{max})\). To achieve our goal, we need to handle the right hand side of (3.24), for \(\frac{B^{1-\xi }}{\xi }\cdot r_0^{-(1-\xi )k} \int _{\Omega }(u_1+u_2)v_1^{\xi }\mathrm{d}x\), it follows from the Young inequality that for any \(\epsilon >0\), one can find \(C(\epsilon , B)\) satisfies

$$\begin{aligned} \frac{B^{1-\xi }}{\xi }\cdot r_0^{-(1-\xi )k}v_1^{\xi }(r) \le \epsilon \cdot v_1(r) +C(\epsilon , B)r_0^{-k}\quad \text {for all}\quad r\in (0, R), \end{aligned}$$
(3.25)

then the upper bound of \(\int _{\Omega }u_1\mathrm{d}x, \int _{\Omega }u_2\mathrm{d}x\) guarantees that

$$\begin{aligned}&\frac{B^{1-\xi }}{\xi }\cdot r_0^{-(1-\xi )k} \int _{\Omega }(u_1+u_2)v_1^{\xi }\mathrm{d}x \le \int _{\Omega }(u_1+u_2) (\epsilon \cdot v_1+C(\epsilon , B)r_0^{-k}) \mathrm{d}x\nonumber \\&\quad \le \epsilon \int _{\Omega }(u_1+u_2)\cdot v_1\mathrm{d}x +C(\epsilon , B)r_0^{-k}(m_1+m_2)\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}).\nonumber \\ \end{aligned}$$
(3.26)

As for \(\frac{B^{1-\xi }}{\xi }\cdot r_0^{-(1-\xi )k}\int _{\Omega }f_1v_1^{\xi }\mathrm{d}x\), utilizing the estimates (3.17) and (3.25), it is clear that

$$\begin{aligned} \begin{aligned}&\frac{B^{1-\xi }}{\xi }\cdot r_0^{-(1-\xi )k}\int _{\Omega }f_1v_1^{\xi }\mathrm{d}x \le \epsilon \int _{\Omega } f_1v_1\mathrm{d}x +C(\epsilon , B)r_0^{-k}\sqrt{|\Omega |} \cdot \Vert f_1\Vert _{L^2(\Omega )}\\&\quad \le \epsilon \Vert f_1\Vert _{L^2(\Omega )} \Vert v_1\Vert _{L^2(\Omega )}+ \frac{1}{2}\Vert f_1\Vert _{L^2(\Omega )}^{\frac{2n+4}{n+4}} +{\hat{C}}(\epsilon , B)r_0^{-\frac{2n+4}{n}\cdot k}\\&\quad \le c_1(M)\epsilon \cdot \Vert f_1\Vert _{L^2(\Omega )}(\Vert \nabla v_1\Vert _{L^2(\Omega )}^{\frac{n}{n+2}}+1) + \frac{1}{2}\Vert f_1\Vert _{L^2(\Omega )}^{\frac{2n+4}{n+4}} +{\hat{C}}(\epsilon , B)r_0^{-\frac{2n+4}{n}\cdot k}\\&\quad \le \epsilon \Vert \nabla v_1\Vert _{L^2(\Omega )}^2 +c(\epsilon , M)\Vert f_1\Vert _{L^2(\Omega )}^{\frac{2n+4}{n+4}} + c_1(M)\epsilon \Vert f_1\Vert _{L^2(\Omega )} \\&\qquad + \frac{1}{2}\Vert f_1\Vert _{L^2(\Omega )}^{\frac{2n+4}{n+4}} +{\hat{C}}(\epsilon , B)r_0^{-\frac{2n+4}{n}\cdot k} \end{aligned}\nonumber \\ \end{aligned}$$
(3.27)

for all \(t\in (0, T_{\mathrm{max}})\), and applying the Young inequality, we get

$$\begin{aligned}&c_1(M)\epsilon \Vert f_1\Vert _{L^2(\Omega )} \le c_1(M)\epsilon R^kr_0^{-k}\Vert f_1\Vert _{L^2(\Omega )} \nonumber \\&\quad \le \frac{1}{2}\Vert f_1\Vert _{L^2(\Omega )}^{\frac{2n+4}{n+4}} +c(M, \epsilon , k, R)r_0^{-\frac{2n+4}{n}\cdot k}\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}). \end{aligned}$$
(3.28)

Thus, plugging (3.26)–(3.28) into (3.24), it is inferred that

$$\begin{aligned} \begin{aligned} \int _{\Omega \setminus B_{r_0}}|\nabla v_1|^2\mathrm{d}x&\le \epsilon \int _{\Omega }(u_1+u_2)\cdot v_1\mathrm{d}x +\epsilon \Vert \nabla v_1\Vert _{L^2(\Omega )}^2 +C(\epsilon , B)r_0^{-k}(m_1+m_2)\\&\quad +(c(\epsilon , M)+1)\Vert f_1\Vert _{L^2(\Omega )}^{\frac{2n+4}{n+4}} +({\hat{C}}(\epsilon , B)+c(M, \epsilon , k, R))r_0^{-\frac{2n+4}{n}\cdot k} \end{aligned}\nonumber \\ \end{aligned}$$
(3.29)

for all \(t\in (0, T_{\mathrm{max}})\). And we notice that

$$\begin{aligned} r_0^{-k}\le R^{\frac{n+4}{n} k}\cdot r_0^{-\frac{2n+4}{n} k}, \end{aligned}$$

this implies (3.20) holds. In view of a similar proof, one can attain (3.21). \(\square \)

Inspired by the idea in Lemma 4.4 of [44], we generalize the above lemma to the whole spherical region.

Lemma 3.4

Let \(\Omega =B_R(0)\subset {\mathbb {R}}^n(n\ge 3, R>0)\), \(\chi _{ij}>0\), \(\mu _{i}>0\), \(a_{i}>0\), \(i, j=1, 2\), \({\hat{\chi }}_1= \frac{\chi _{11}+\chi _{12}}{2}\), \({\hat{\chi }}_2=\frac{\chi _{21}+\chi _{22}}{2}\). Suppose that \((u_1, u_2, v_1, v_2)\in \left( C^{2,1}( \overline{\Omega }\times (0,T_{\max }))\right) ^4\) is a classical solution of (1.4), and \((u_1, u_2, v_1, v_2)\in S(M, B, k)\). Let \(f_i\) and \(g_i (i=1,2)\) be as in (3.3)–(3.5), then one can find \(C=C(m_1, m_2, R, n, {\hat{\chi }}_1, {\hat{\chi }}_2)>0\) such that

$$\begin{aligned} \int _{ B_{r_0}}(|\nabla v_1|^2+|\nabla v_2|^2)\mathrm{d}x\le & {} C \left( r_0\cdot (\Vert f_1\Vert _{L^2(\Omega )}^2+\Vert f_2\Vert _{L^2(\Omega )}^2) +\Vert g_1\Vert _{L^2(\Omega )} \right. \nonumber \\&\left. +\Vert g_2\Vert _{L^2(\Omega )} +\Vert v_1\Vert _{L^2(\Omega )}^2+\Vert v_2\Vert _{L^2(\Omega )}^2+1 \right) \end{aligned}$$
(3.30)

holds for any \(r_0\in (0, R)\) and \(t\in (0, T_{\mathrm{max}})\).

Proof

From the definition of \(f_1, f_2\), we have \(f_1:=-\Delta v_1+v_1-u_1-u_2, f_2:=-\Delta v_2+v_2-u_1-u_2\), which implies that

$$\begin{aligned} (r^{n-1}v_{1r})_r = r^{n-1}v_1- r^{n-1}u_1-r^{n-1}u_2-r^{n-1}f_1 \quad \text {for all}\quad r\in (0, R)\nonumber \\ \end{aligned}$$
(3.31)

and

$$\begin{aligned} (r^{n-1}v_{2r})_r = r^{n-1}v_2- r^{n-1}u_1-r^{n-1}u_2-r^{n-1}f_2 \quad \text {for all}\quad r\in (0, R).\nonumber \\ \end{aligned}$$
(3.32)

Applying (3.4) and (3.5), there appears the relation

$$\begin{aligned} v_{1r}+v_{2r}= \frac{(u_{1})_r}{{\hat{\chi }}_1u_1}-\frac{g_1}{\sqrt{{\hat{\chi }}_1u_1}} = \frac{(u_{2})_r}{{\hat{\chi }}_2u_2}-\frac{g_2}{\sqrt{{\hat{\chi }}_2u_2}} \quad \text {for all}\quad r\in (0, R). \end{aligned}$$
(3.33)

Multiplying (3.31), (3.32) with \(r^{n-1}v_{1r}\) and \(r^{n-1}v_{2r}\) respectively, and putting them together, one can select \(\delta \in (0, \frac{2n-2}{R}]\) such that

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\left( (r^{n-1}v_{1r})^2+ (r^{n-1}v_{2r})^2 \right) _r\\&\quad =-r^{2n-2}u_1(v_{1r}+v_{2r}) -r^{2n-2}u_2(v_{1r}+v_{2r})\\&\qquad -r^{2n-2}f_1\cdot v_{1r}-r^{2n-2}f_2\cdot v_{2r} +\frac{r^{2n-2}}{2}(v_1^2 +v_2^2)_r\\&\quad \le -r^{2n-2}\frac{(u_1)_r}{{\hat{\chi }}_1} +r^{2n-2}\sqrt{u_1}\cdot \frac{g_1}{\sqrt{{\hat{\chi }}_1}}\\&\qquad -r^{2n-2}\frac{(u_2)_r}{{\hat{\chi }}_2} +r^{2n-2}\sqrt{u_2}\cdot \frac{g_2}{\sqrt{{\hat{\chi }}_2 }} +\frac{r^{2n-2}}{2}(v_1^2 +v_2^2)_r\\&\qquad +\frac{\delta }{2}r^{2n-2}(v_{1r}^2 +v_{2r}^2) +\frac{1}{2\delta }r^{2n-2}(f_1^2+f_2^2) \quad \text {for all}\quad r\in (0,R), \end{aligned} \end{aligned}$$
(3.34)

where we have used the Young inequality and (3.33). Let

$$\begin{aligned} y(r) = (r^{n-1}v_{1r})^2+(r^{n-1}v_{2r})^2, \end{aligned}$$

along with (3.34), it is clear that

$$\begin{aligned} \begin{aligned} y^{'}(r)&\le -\frac{2}{{\hat{\chi }}_1}r^{2n-2}(u_1)_r+ \frac{2}{\sqrt{{\hat{\chi }}_1}}r^{2n-2}\sqrt{u_1}g_1 -\frac{2}{{\hat{\chi }}_2}r^{2n-2}(u_2)_r+ \frac{2}{\sqrt{{\hat{\chi }}_2}}r^{2n-2}\sqrt{u_2}g_2\\&\quad +\frac{1}{\delta }r^{2n-2}f_1^2+\frac{1}{\delta }r^{2n-2}f_2^2 +r^{2n-2}(v_1^2)_r+r^{2n-2}(v_2^2)_r +\delta r^{2n-2}((v_{1r})^2+(v_{2r})^2) \end{aligned}\nonumber \\ \end{aligned}$$
(3.35)

for all \(r\in (0, R)\). Since \(y(0)=0\), then integrating (3.35) over (0, r), one achieves

$$\begin{aligned} \begin{aligned} y(r)&\le -\frac{2}{{\hat{\chi }}_1}\int _0^re^{\delta (r-\rho )}\rho ^{2n-2}(u_1)_r(\rho )d\rho +\frac{2}{\sqrt{{\hat{\chi }}_1}}\int _0^re^{\delta (r-\rho )}\rho ^{2n-2}\sqrt{u_1}g_1d\rho \\&\quad -\frac{2}{{\hat{\chi }}_2} \int _0^re^{\delta (r-\rho )}\rho ^{2n-2}(u_2)_r(\rho )d\rho +\frac{2}{\sqrt{{\hat{\chi }}_2}}\int _0^re^{\delta (r-\rho )}\rho ^{2n-2}\sqrt{u_2}g_2d\rho \\&\quad +\frac{1}{\delta }\int _0^re^{\delta (r-\rho )}\rho ^{2n-2}f_1^2(\rho )d\rho +\frac{1}{\delta }\int _0^re^{\delta (r-\rho )}\rho ^{2n-2}f_2^2(\rho )d\rho \\&\quad +\int _0^re^{\delta (r-\rho )}\rho ^{2n-2}(v_1^2)_r(\rho )d\rho +\int _0^re^{\delta (r-\rho )}\rho ^{2n-2}(v_2^2)_r(\rho )d\rho \\&=I_1+I_2+I_3+I_4+I_5+I_6+I_7+I_8 \quad \text {for all}\quad r\in (0, R). \end{aligned} \end{aligned}$$
(3.36)

For \(I_1\), in view of integration by parts, we find

$$\begin{aligned}&I_1= -\frac{2}{{\hat{\chi }}_1}r^{2n-2}u_1(r) +\frac{2}{{\hat{\chi }}_1} \int _0^ru_1e^{\delta (r-\rho )}\left( (2n-2)\rho ^{2n-3} -\delta \rho ^{2n-2}\right) d\rho \nonumber \\&\quad \le \frac{4n-4}{{\hat{\chi }}_1}e^{\delta R}\int _0^r\rho ^{2n-3}u_1(\rho )d\rho \end{aligned}$$
(3.37)

for all \(r\in (0, R)\). In a similar way, we have

$$\begin{aligned} I_3 \le \frac{4n-4}{{\hat{\chi }}_2}e^{\delta R}\int _0^r\rho ^{2n-3}u_2(\rho )d\rho \quad \text {for all}\quad r\in (0, R). \end{aligned}$$
(3.38)

For \(I_2\), using the fact that

$$\begin{aligned} \int _0^r\rho ^{n-1}h(\rho )\mathrm{d}\rho \le \int _0^R\rho ^{n-1}h(\rho )\mathrm{d}\rho =\frac{\Vert h\Vert _{L^1(\Omega )}}{w_n} \quad \text {for all}\quad h\in L^1(\Omega ), \end{aligned}$$
(3.39)

where the abbreviation \(w_n:=n |B_1(0)|\), then we can derive

$$\begin{aligned}&I_2\le \frac{2}{\sqrt{{\hat{\chi }}_1}}\left( \int _0^r\rho ^{n-1}u_1(\rho )d\rho \right) ^{\frac{1}{2}}\cdot \left( \int _0^r e^{2\delta (r-\rho )}\rho ^{3n-3}g_1^2d\rho \right) ^{\frac{1}{2}} \nonumber \\&\quad \le \frac{2e^{\delta R}\sqrt{m_1}}{w_n\sqrt{{\hat{\chi }}_1}}r^{n-1}\Vert g_1\Vert _{L^2(\Omega )} \end{aligned}$$
(3.40)

for all \(r\in (0, R)\), where we have used the fact that \(\Vert u_1\Vert _{L^1(\Omega )}\le m_1\). Analogously, one can deduce

$$\begin{aligned} I_4\le \frac{2e^{\delta R}\sqrt{m_2}}{w_n\sqrt{{\hat{\chi }}_2}}r^{n-1}\Vert g_2\Vert _{L^2(\Omega )}\quad \text {for all}\quad r\in (0, R). \end{aligned}$$
(3.41)

And a simple computation leads to

$$\begin{aligned} I_5\le \frac{e^{\delta R}}{\delta w_n}r^{n-1}\Vert f_1\Vert ^2_{L^2(\Omega )},\quad I_6\le \frac{e^{\delta R}}{\delta w_n}r^{n-1}\Vert f_2\Vert ^2_{L^2(\Omega )}\quad \text {for all}\quad r\in (0, R).\qquad \end{aligned}$$
(3.42)

Once more, applying a integration by parts, utilizing the fact that \(\delta \in (0, \frac{2n-2}{R}]\), we can see that

$$\begin{aligned} I_7= r^{2n-2}v_1^2(r) -\int _0^re^{\delta (r-\rho )}v_1^2\left( (2n-2)\rho ^{2n-3}- \delta \cdot \rho ^{2n-2} \right) d\rho \le r^{2n-2}v_1^2(r)\nonumber \\ \end{aligned}$$
(3.43)

for all \(r\in (0, R)\), and

$$\begin{aligned} I_8\le r^{2n-2}v_2^2(r)\quad \text {for all}\quad r\in (0, R). \end{aligned}$$
(3.44)

From (3.36)–(3.44), it is now clear that there exists \(c_5=c_5(m_1, m_2, R, n, {\hat{\chi }}_1, {\hat{\chi }}_2)>0\) such that

$$\begin{aligned} \begin{aligned} y(r)&\le c_5\Big ( \int _0^r\rho ^{2n-3}(u_1+u_2)d\rho +r^{n-1}(\Vert g_1\Vert _{L^2(\Omega )}+\Vert g_2\Vert _{L^2(\Omega )})\\&\quad +r^{n-1}(\Vert f_1\Vert _{L^2(\Omega )}^2+\Vert f_2\Vert _{L^2(\Omega )}^2) +r^{2n-2}(v_1^2+v_1^2) \Big )\quad \text {for all}\quad r\in (0, R). \end{aligned}\nonumber \\ \end{aligned}$$
(3.45)

Multiplying (3.45) with \(\frac{1}{r^{n-1}}\) and integrating it over \((0, r_0)\), it follows that

$$\begin{aligned} \begin{aligned}&\int _0^{r_0}r^{n-1}(v_{1r}^2+v_{2r}^2)(r)\mathrm{d}r \le c_5\int _0^{r_0}\frac{1}{r^{n-1}}\int _0^r\rho ^{2n-3}(u_1+u_2)(\rho )d\rho \mathrm{d}r\\&\quad +c_5R(\Vert g_1\Vert _{L^2(\Omega )}+\Vert g_2\Vert _{L^2(\Omega )}) +c_5r_0(\Vert f_1\Vert _{L^2(\Omega )}^2+\Vert f_2\Vert _{L^2(\Omega )}^2)\\&\quad +\frac{c_5}{w_n}(\Vert v_1\Vert _{L^2(\Omega )}^2+\Vert v_2\Vert _{L^2(\Omega )}^2) \end{aligned} \end{aligned}$$
(3.46)

for any \(r_0\in (0, R)\), here, we can use Fubini’s theorem to handle the first term on the right side of (3.46),

$$\begin{aligned} \begin{aligned}&c_5\int _0^{r_0}\frac{1}{r^{n-1}}\int _0^r\rho ^{2n-3}(u_1+u_2)(\rho )\mathrm{d}\rho \mathrm{d}r\\&\quad =c_5\int _0^{r_0}\left( \int _{\rho }^{r_0}\frac{1}{r^{n-1}}\mathrm{d}r\right) \rho ^{2n-3}(u_1+u_2)(\rho )\mathrm{d}\rho \\&\quad \le \frac{c_5}{n-2}\int _0^{r_0}\frac{1}{\rho ^{n-2}}\rho ^{2n-3}(u_1+u_2)(\rho )\mathrm{d}\rho \le c_5\frac{m_1+m_2}{(n-2)w_n}, \end{aligned} \end{aligned}$$
(3.47)

which yields (3.30). \(\square \)

Thereupon, utilizing the above three lemmas, we deduce the following relation between \(\int _{\Omega }(u_1+u_2)(v_1+v_2)dx\) and \(\mathrm {D}(u_1, u_2, v_1, v_2)\).

Lemma 3.5

Let \(\Omega =B_R(0)\subset {\mathbb {R}}^n(n\ge 3, R>0)\), \(\chi _{ij}>0\), \(\mu _{i}>0\), \(a_{i}>0\) (\(i,j=1,2\)). Suppose that \((u_1, u_2, v_1, v_2)\in \left( C^{2,1}( \overline{\Omega }\times (0,T_{\max }))\right) ^4\) is a classical solution of (1.4), and \((u_1, u_2, v_1, v_2)\in S(M, B, k)\). Then for \(\mathrm {D}(u_1, u_2, v_1, v_2)\) given by (3.2), one can find \(C=C( m_1, m_2, M, B, k, R, n, {\hat{\chi }}_1, {\hat{\chi }}_2)\) and \(\vartheta = \frac{1}{1+\frac{n}{(2n+4)k}} \in (\frac{1}{2}, 1)\) such that

$$\begin{aligned} \begin{aligned} \int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x \le C\cdot (\mathrm {D}^\vartheta (u_1, u_2, v_1, v_2)+1)\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}). \end{aligned}\nonumber \\ \end{aligned}$$
(3.48)

Proof

In view of Lemmas 3.3 and 3.4, for all \(t\in (0, T_{\mathrm{max}})\), there exist \({\mathfrak {C}}_1={\mathfrak {C}}_1(\epsilon , m_1, m_2, M, B, k)\) and \({\mathfrak {C}}_2={\mathfrak {C}}_2(m_1, m_2, R, n, {\hat{\chi }}_1, {\hat{\chi }}_2)\) satisfying

$$\begin{aligned} \begin{aligned} \int _{\Omega \setminus B_{r_0}}(|\nabla v_1|^2+|\nabla v_2|^2)\mathrm{d}x&\le \epsilon \int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x +\epsilon (\Vert \nabla v_1\Vert _{L^2(\Omega )}^2+\Vert \nabla v_2\Vert _{L^2(\Omega )}^2)\\&\quad +2{\mathfrak {C}}_1r_0^{-\frac{2n+4}{n}\cdot k} +{\mathfrak {C}}_1(\Vert f_1\Vert _{L^2(\Omega )}^{\frac{2n+4}{n+4}}+\Vert f_2\Vert _{L^2(\Omega )}^{\frac{2n+4}{n+4}})\quad \text {for any}\quad \epsilon >0, \end{aligned}\nonumber \\ \end{aligned}$$
(3.49)

and

$$\begin{aligned} \begin{aligned} \int _{ B_{r_0}}(|\nabla v_1|^2+|\nabla v_2|^2)\mathrm{d}x&\le {\mathfrak {C}}_2 \left( r_0\cdot (\Vert f_1\Vert _{L^2(\Omega )}^2+\Vert f_2\Vert _{L^2(\Omega )}^2) +\Vert g_1\Vert _{L^2(\Omega )} \right. \\&\quad \left. +\Vert g_2\Vert _{L^2(\Omega )} +\Vert v_1\Vert _{L^2(\Omega )}^2+\Vert v_2\Vert _{L^2(\Omega )}^2+1 \right) , \end{aligned}\nonumber \\ \end{aligned}$$
(3.50)

putting together the above two estimates we arrive at

$$\begin{aligned} \begin{aligned} (1-\epsilon ) \int _{ \Omega }(|\nabla v_1|^2+|\nabla v_2|^2)\mathrm{d}x&\le \epsilon \int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x +{\mathfrak {C}}_2(\Vert v_1\Vert _{L^2(\Omega )}^2+\Vert v_2\Vert _{L^2(\Omega )}^2)\\&\qquad +{\mathfrak {C}}_2 \left( \Vert g_1\Vert _{L^2(\Omega )} +\Vert g_2\Vert _{L^2(\Omega )} +1 \right) +\phi \quad \text {for all}\quad t\in (0, T_{\mathrm{max}}), \end{aligned}\nonumber \\ \end{aligned}$$
(3.51)

where

$$\begin{aligned} \begin{aligned} \phi =2{\mathfrak {C}}_1r_0^{-\varrho } +{\mathfrak {C}}_1(\Vert f_1\Vert _{L^2(\Omega )}^{\frac{2n+4}{n+4}}+\Vert f_2\Vert _{L^2(\Omega )}^{\frac{2n+4}{n+4}}) +{\mathfrak {C}}_2 r_0\cdot (\Vert f_1\Vert _{L^2(\Omega )}^2+\Vert f_2\Vert _{L^2(\Omega )}^2),\quad \varrho =\frac{2n+4}{n}\cdot k, \end{aligned}\nonumber \\ \end{aligned}$$
(3.52)

and it follows from (3.18) that

$$\begin{aligned} {\mathfrak {C}}_2(\Vert v_1\Vert _{L^2(\Omega )}^2+\Vert v_2\Vert _{L^2(\Omega )}^2) \le \epsilon (\Vert \nabla v_1\Vert _{L^2(\Omega )}^2+\Vert \nabla v_2\Vert _{L^2(\Omega )}^2) +{\mathfrak {C}}_3\quad \text {for all}\quad t\in (0, T_{\mathrm{max}})\nonumber \\ \end{aligned}$$
(3.53)

with \({\mathfrak {C}}_3={\mathfrak {C}}_3(m_1, m_2, R, n, {\hat{\chi }}_1, {\hat{\chi }}_2)>0\). For \(\phi \), it is easy to obtain that

$$\begin{aligned} \phi\le & {} 2{\mathfrak {C}}_1r_0^{-\varrho }+ {\mathfrak {C}}_4 (\Vert f_1\Vert _{L^2(\Omega )}+\Vert f_2\Vert _{L^2(\Omega )})^{\frac{2n+4}{n+4}}\nonumber \\&\quad +{\mathfrak {C}}_5\cdot r_0 (\Vert f_1\Vert _{L^2(\Omega )}+\Vert f_2\Vert _{L^2(\Omega )})^2\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}),\nonumber \\ \end{aligned}$$
(3.54)

where \({\mathfrak {C}}_4={\mathfrak {C}}_4(\epsilon , m_1, m_2, M, B, k), {\mathfrak {C}}_5={\mathfrak {C}}_5(m_1, m_2, R, n, {\hat{\chi }}_1, {\hat{\chi }}_2)>0\). To further estimate \(\phi \), we select \(r_0=\min \{ \frac{R}{2}, (\Vert f_1\Vert _{L^2(\Omega )}+\Vert f_2\Vert _{L^2(\Omega )})^{-\frac{2}{\varrho +1}} \}\), and we divide the following discussion into two cases.

For the case \(\frac{R}{2}\le (\Vert f_1\Vert _{L^2(\Omega )}+\Vert f_2\Vert _{L^2(\Omega )})^{-\frac{2}{\varrho +1}}\), we have

$$\begin{aligned} r_0=\frac{R}{2}\quad \hbox {and}\quad \Vert f_1\Vert _{L^2(\Omega )}+\Vert f_2\Vert _{L^2(\Omega )}\le (\frac{2}{R})^{\frac{\varrho +1}{2}}, \end{aligned}$$

which implies

$$\begin{aligned} \phi \le 2{\mathfrak {C}}_1(\frac{R}{2})^{-\varrho }+ {\mathfrak {C}}_4 (\frac{2}{R})^{\frac{\varrho +1}{2}\cdot \frac{2n+4}{n+4}} +{\mathfrak {C}}_5\cdot \frac{R}{2}\cdot (\frac{2}{R})^{\varrho +1}\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}).\nonumber \\ \end{aligned}$$
(3.55)

For the case \(\frac{R}{2}> (\Vert f_1\Vert _{L^2(\Omega )}+\Vert f_2\Vert _{L^2(\Omega )})^{-\frac{2}{\varrho +1}}\), it means that \(r_0=(\Vert f_1\Vert _{L^2(\Omega )}+\Vert f_2\Vert _{L^2(\Omega )})^{-\frac{2}{\varrho +1}}\), then

$$\begin{aligned} \phi \le (2{\mathfrak {C}}_1+{\mathfrak {C}}_5)(\Vert f_1\Vert _{L^2(\Omega )}+\Vert f_2\Vert _{L^2(\Omega )})^{\frac{2\varrho }{\varrho +1}}+ {\mathfrak {C}}_4 (\Vert f_1\Vert _{L^2(\Omega )}+\Vert f_2\Vert _{L^2(\Omega )})^{\frac{2n+4}{n+4}}\nonumber \\ \end{aligned}$$
(3.56)

for all \(t\in (0, T_{\mathrm{max}})\), therefore, in light of (2.12) and \(p\in (1, \frac{n}{n-1})\), it is inferred that

$$\begin{aligned} k=\frac{n-p}{p}>n-2\ge 1,\quad \varrho =\frac{2n+4}{n}k>1, \end{aligned}$$

and

$$\begin{aligned} \frac{4k}{n}> \frac{4(n-2)}{n} \ge \frac{4}{3}>1, \end{aligned}$$

which guarantees that

$$\begin{aligned} \frac{2\varrho }{\varrho +1}- \frac{2n+4}{n+4} = \frac{2(n+2)(4k-n)}{((2n+4)k+n)(n+4)}>0, \end{aligned}$$
(3.57)

this makes it possible to use Young’s inequality in (3.56), that is,

$$\begin{aligned} \phi \le {\mathfrak {C}}_6(\Vert f_1\Vert _{L^2(\Omega )}^{\frac{2\varrho }{\varrho +1}}+\Vert f_2\Vert _{L^2(\Omega )}^{\frac{2\varrho }{\varrho +1}}) +{\mathfrak {C}}_7\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}) \end{aligned}$$
(3.58)

with \({\mathfrak {C}}_6={\mathfrak {C}}_6(\epsilon , m_1, m_2, M, B, k, R, n, {\hat{\chi }}_1, {\hat{\chi }}_2), {\mathfrak {C}}_7={\mathfrak {C}}_7(\epsilon , m_1, m_2, M, B, k, R, n, {\hat{\chi }}_1, {\hat{\chi }}_2)>0\). Let \(\vartheta =\frac{\varrho }{\varrho +1}=\frac{1}{1+\frac{n}{(2n+4)k}}\in (\frac{1}{2}, 1)\), by adding up (3.51), (3.53), (3.56) and (3.58), for any \(\epsilon >0\), one finds \({\mathfrak {C}}_8={\mathfrak {C}}_8(\epsilon , m_1, m_2, M, B, k, R, n, {\hat{\chi }}_1, {\hat{\chi }}_2)>0\) such that

$$\begin{aligned} \begin{aligned} \int _{ \Omega }(|\nabla v_1|^2+|\nabla v_2|^2)\mathrm{d}x&\le {\mathfrak {C}}_8\Big ( \Vert f_1\Vert _{L^2(\Omega )}^{2\vartheta }{+} \Vert f_2\Vert _{L^2(\Omega )}^{2\vartheta } +\Vert g_1\Vert _{L^2(\Omega )}{+} \Vert g_2\Vert _{L^2(\Omega )}{+}1 \Big )\\&\quad +\frac{\epsilon }{1-2\epsilon }\int _{ \Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x \quad \text {for all}\quad t\in (0, T_{\mathrm{max}}). \end{aligned}\nonumber \\ \end{aligned}$$
(3.59)

It follows from Lemma 3.2 and \(2\vartheta >\frac{2n+4}{n+4}\) that

$$\begin{aligned} \int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x \le (1+\epsilon )\int _{\Omega } (|\nabla v_1|^2+|\nabla v_2|^2)\mathrm{d}x {+}{\mathfrak {C}}_9(M)\left( \Vert f_1\Vert _{L^2(\Omega )}^{2\vartheta } {+}\Vert f_2\Vert _{L^2(\Omega )}^{2\vartheta }{+}1\right) \nonumber \\ \end{aligned}$$
(3.60)

for all \(t\in (0, T_{\mathrm{max}})\), with \({\mathfrak {C}}_9(M)>0\). Selecting \(\epsilon \in (0, \frac{1}{4})\), inserting (3.60) into (3.59) we attain

$$\begin{aligned}&\Big ( 1-\frac{\epsilon (\epsilon +1)}{1-2\epsilon } \Big ) \int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x \le (\epsilon +1){\mathfrak {C}}_{10}\left( \Vert f_1\Vert _{L^2(\Omega )}^{2\vartheta }+\Vert f_2\Vert _{L^2(\Omega )}^{2\vartheta }\right. \\&\quad \left. +\Vert g_1\Vert _{L^2(\Omega )}+\Vert g_2\Vert _{L^2(\Omega )} +1 \right) \end{aligned}$$

for all \(t\in (0, T_{\mathrm{max}})\), with \({\mathfrak {C}}_{10} ={\mathfrak {C}}_{10}( m_1, m_2, M, B, k, R, n, {\hat{\chi }}_1, {\hat{\chi }}_2)>0\), which directly yields (3.48). \(\square \)

With (3.48) at hand, now our eyes are turned to constructing the differential inequality for \(\mathrm {F}(u_1, u_2, v_1, v_2)\).

Lemma 3.6

Let \(\Omega =B_R(0)\subset {\mathbb {R}}^n(n\ge 3, R>0)\), \(\chi _{ij}>0\), \(\mu _{i}>0\), \(a_{i}>0\) (\(i,j=1,2\)), \(L>0\), T>0. And let \((u_1, u_2, v_1, v_2)\) be a classical solution of (1.4), \((u_1, u_2, v_1, v_2)\in S(M, B, k)\) with the property

$$\begin{aligned} u_1(x, t)\le L,\quad u_2(x, t)\le L\quad \text {for all}\quad (x, t)\in \Omega \times (0, T). \end{aligned}$$
(3.61)

Then there exist \(T^{*}\in (0, T]\) and positive constants \({\overline{C}}={\overline{C}}( m_1, m_2, M, L, T, B, k, R, n, \chi _{11}, \chi _{12}, \chi _{21}, \chi _{22},\) \( \mu _1, \mu _2, a_1, a_2)\) as well as \(C:=C( m_1, m_2, M, L, B, k, R, n, \chi _{11}, \chi _{12}, \chi _{21}, \chi _{22}, \mu _1, \mu _2, a_1, a_2)\) whenever

$$\begin{aligned} \mathrm {F}(u_{10}, u_{20}, v_{10}, v_{20}) < -{\overline{C}}, \end{aligned}$$
(3.62)

the following inequality holds true for all \(t\in (0, T^{*})\),

$$\begin{aligned} -\frac{\mathrm{d}}{\mathrm{d}t}\mathrm {F}(u_1, u_2, v_1, v_2) \ge C\left( - \mathrm {F}(u_1, u_2, v_1, v_2) \right) ^{\frac{1}{\vartheta }}, \end{aligned}$$
(3.63)

where \(\mathrm {F}(u_1, u_2, v_1, v_2)\), \(\mathrm {D}(u_1, u_2, v_1, v_2)\), \(\vartheta \) are defined in (3.1), (3.2) and Lemma 3.5 respectively. Furthermore, \(-\mathrm {F}(u_1, u_2, v_1, v_2)\) is nondecreasing for all \(t\in (0, T^{*})\).

Proof

Step 1 With the boundedness restriction of \(u_1, u_2\), we claim that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\mathrm {F}(u_1, u_2, v_1, v_2) \le -\mathrm {D}(u_1, u_2, v_1, v_2) +{\mathcal {C}}_1\mathrm {F}(u_1, u_2, v_1, v_2)+{\mathcal {C}}_0\int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x+{\mathcal {C}}_0\nonumber \\ \end{aligned}$$
(3.64)

for all \(t\in (0, T)\), where \({\mathcal {C}}_1=\mu _1+\mu _2\), \({\mathcal {C}}_0={\mathcal {C}}_0( \chi _{11}, \chi _{12}, \chi _{21}, \chi _{22}, \mu _1, \mu _2, m_1, m_2, M, L, a_1, a_2 )\). Indeed, recalling Lemma 3.1, we have

$$\begin{aligned} \begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\mathrm {F}(u_1, u_2, v_1, v_2)\\&\quad \le -\mathrm {D}(u_1, u_2, v_1, v_2) +\frac{\chi _{12}-\chi _{11}}{2}\int _{\Omega }u_1|\nabla v_1|^2\mathrm{d}x +\frac{\chi _{11}-\chi _{12}}{2}\int _{\Omega }u_1|\nabla v_2|^2\mathrm{d}x\\&\qquad +\frac{\chi _{22}-\chi _{21}}{2}\int _{\Omega }u_2|\nabla v_1|^2\mathrm{d}x +\frac{\chi _{21}-\chi _{22}}{2}\int _{\Omega }u_2|\nabla v_2|^2\mathrm{d}x +\left( \frac{\chi _{11}}{\chi _1}-1\right) \int _{\Omega }\nabla u_1\nabla v_1\mathrm{d}x\\&\qquad + \left( \frac{\chi _{12}}{\chi _1}-1\right) \int _{\Omega }\nabla u_1\nabla v_2\mathrm{d}x + \left( \frac{\chi _{21}}{\chi _2}-1\right) \int _{\Omega }\nabla u_2\nabla v_1\mathrm{d}x + \left( \frac{\chi _{22}}{\chi _2}-1\right) \int _{\Omega }\nabla u_2\nabla v_2\mathrm{d}x\\&\qquad -\left( \frac{1}{\chi _1}-\frac{2}{\chi _{11}+\chi _{12}}\right) \int _{\Omega }\frac{|\nabla u_1|^2}{u_1}\mathrm{d}x -\left( \frac{1}{\chi _2}-\frac{2}{\chi _{21}+\chi _{22}}\right) \int _{\Omega }\frac{|\nabla u_2|^2}{u_2}\mathrm{d}x\\&\qquad +(a_1\mu _1+a_2\mu _2)\int _{\Omega }u_1u_2(v_1+v_2)\mathrm{d}x +\int _{\Omega }(\mu _1u_1^2v_1+\mu _2u_2^2v_1+\mu _1u_1^2v_2+\mu _2u_2^2v_2)\mathrm{d}x\\&\qquad +\int _{\Omega }\left( \frac{\mu _1}{\chi _1}u_1\ln u_1 +\frac{\mu _2}{\chi _2}u_2\ln u_2 +\frac{\mu _1}{\chi _1}u_1 +\frac{\mu _2}{\chi _2}u_2\right) \mathrm{d}x\\&\qquad -\int _{\Omega }(\mu _1u_1v_1+\mu _2u_2v_1+\mu _1u_1v_2+\mu _2u_2v_2)\mathrm{d}x\\&\qquad -\int _{\Omega }\left( \frac{\mu _1}{\chi _1}u_1^2\ln u_1+\frac{\mu _2}{\chi _2}u_2^2\ln u_2+\frac{a_1\mu _1}{\chi _1}u_1u_2\ln u_1 +\frac{a_2\mu _2}{\chi _2}u_1u_2\ln u_2\right) \mathrm{d}x\quad \text {for all}\quad t\in (0, T). \end{aligned} \end{aligned}$$
(3.65)

In light of (2.6), (2.7), (2.10) and (3.61), it follows that

$$\begin{aligned} \begin{aligned}&\frac{\chi _{12}-\chi _{11}}{2}\int _{\Omega }u_1|\nabla v_1|^2\mathrm{d}x +\frac{\chi _{11}-\chi _{12}}{2}\int _{\Omega }u_1|\nabla v_2|^2\mathrm{d}x +\frac{\chi _{22}-\chi _{21}}{2}\int _{\Omega }u_2|\nabla v_1|^2\mathrm{d}x\\&\quad +\frac{\chi _{21}-\chi _{22}}{2}\int _{\Omega }u_2|\nabla v_2|^2\mathrm{d}x \le \left( |\chi _{12}-\chi _{11} | +|\chi _{21}-\chi _{22} | \right) L\cdot (C+{\hat{C}})\quad \text {for all}\quad t\in (0, T). \end{aligned}\nonumber \\ \end{aligned}$$
(3.66)

When \(\chi _{11}=\chi _{12}\), we have \(\left( \frac{\chi _{11}}{\chi _1}-1\right) \int _{\Omega }\nabla u_1\nabla v_1\mathrm{d}x + \left( \frac{\chi _{12}}{\chi _1}-1\right) \int _{\Omega }\nabla u_1\nabla v_2\mathrm{d}x -\left( \frac{1}{\chi _1}-\frac{2}{\chi _{11}+\chi _{12}}\right) \int _{\Omega }\frac{|\nabla u_1|^2}{u_1}\mathrm{d}x=0\); when \(\chi _{11}\ne \chi _{12}\), the Young inequality ensures that there exists \(0<\eta \le \min \left\{ \frac{1}{\chi _1}-\frac{2}{\chi _{11}+\chi _{12}}, \frac{1}{\chi _2}-\frac{2}{\chi _{21}+\chi _{22}}\right\} \) such that

$$\begin{aligned} \begin{aligned}&\left( \frac{\chi _{11}}{\chi _1}-1\right) \int _{\Omega }\nabla u_1\nabla v_1\mathrm{d}x + \left( \frac{\chi _{12}}{\chi _1}-1\right) \int _{\Omega }\nabla u_1\nabla v_2\mathrm{d}x -\left( \frac{1}{\chi _1}-\frac{2}{\chi _{11}+\chi _{12}}\right) \int _{\Omega }\frac{|\nabla u_1|^2}{u_1}\mathrm{d}x\\&\quad \le \left( \eta -\frac{1}{\chi _1}+\frac{2}{\chi _{11}+\chi _{12}}\right) \int _{\Omega }\frac{|\nabla u_1|^2}{u_1}dx +C(\eta , \chi _{11}, \chi _{12}, \chi _1)\int _{\Omega } u_1(|\nabla v_1|^2+|\nabla v_2|^2)\mathrm{d}x\\&\quad \le C(\eta , \chi _{11}, \chi _{12}, \chi _1)\cdot L\cdot (C+{\hat{C}}) \quad \text {for all}\quad t\in (0, T). \end{aligned}\nonumber \\ \end{aligned}$$
(3.67)

A similar calculation reveals that

$$\begin{aligned} \begin{aligned}&\left( \frac{\chi _{21}}{\chi _2}-1\right) \int _{\Omega }\nabla u_2\nabla v_1\mathrm{d}x + \left( \frac{\chi _{22}}{\chi _2}-1\right) \int _{\Omega }\nabla u_2\nabla v_2\mathrm{d}x -\left( \frac{1}{\chi _2}-\frac{2}{\chi _{21}+\chi _{22}}\right) \int _{\Omega }\frac{|\nabla u_2|^2}{u_2}\mathrm{d}x\\&\quad \le C(\eta , \chi _{21}, \chi _{22}, \chi _2)\cdot L\cdot (C+{\hat{C}}) \quad \text {for all}\quad t\in (0, T). \end{aligned}\nonumber \\ \end{aligned}$$
(3.68)

And utilizing the \(L^1\) bound for \(u_i, v_i (i=1, 2)\), there appears the relation

$$\begin{aligned} \begin{aligned}&(a_1\mu _1+a_2\mu _2)\int _{\Omega }u_1u_2(v_1+v_2)\mathrm{d}x +\int _{\Omega }(\mu _1u_1^2v_1+\mu _2u_2^2v_1+\mu _1u_1^2v_2+\mu _2u_2^2v_2)\mathrm{d}x\\&\quad + \int _{\Omega } \left( \frac{\mu _1}{\chi _1}u_1 +\frac{\mu _2}{\chi _2}u_2\right) \mathrm{d}x \le 2(a_1\mu _1+a_2\mu _2)L^2M+2(\mu _1+\mu _2)L^2M+m\left( \frac{\mu _1}{\chi _1} +\frac{\mu _2}{\chi _2}\right) \end{aligned}\nonumber \\ \end{aligned}$$
(3.69)

for all \(t\in (0, T)\). Furthermore, the definition of \(\mathrm {F}(u_1, u_2, v_1, v_2)\) in (3.1) implies

$$\begin{aligned} \begin{aligned}&\int _{\Omega }\left( \frac{\mu _1}{\chi _1}u_1\ln u_1 +\frac{\mu _2}{\chi _2}u_2\ln u_2 \right) \mathrm{d}x -\int _{\Omega }(\mu _1u_1v_1+\mu _2u_2v_1+\mu _1u_1v_2+\mu _2u_2v_2)\mathrm{d}x\\&\quad = \mu _1\left( \mathrm {F}(u_1, u_2, v_1, v_2)-\frac{1}{2}\int _{\Omega }(|v_1|^2+|\nabla v_1|^2+|v_2|^2+|\nabla v_2|^2)\mathrm{d}x \right) \\&\qquad -\frac{\mu _1}{\chi _2}\int _{\Omega }u_2\ln u_2\mathrm{d}x+\mu _1\int _{\Omega }u_2(v_1+v_2)\mathrm{d}x\\&\qquad +\mu _2\left( \mathrm {F}(u_1, u_2, v_1, v_2)-\frac{1}{2}\int _{\Omega }(|v_1|^2+|\nabla v_1|^2+|v_2|^2+|\nabla v_2|^2)\mathrm{d}x \right) \\&\qquad -\frac{\mu _2}{\chi _1}\int _{\Omega }u_1\ln u_1\mathrm{d}x+\mu _2\int _{\Omega }u_1(v_1+v_2)\mathrm{d}x\\&\quad \le (\mu _1+\mu _2)\mathrm {F}(u_1, u_2, v_1, v_2) +\left( \frac{\mu _1}{\chi _2}+\frac{\mu _2}{\chi _1}\right) \frac{|\Omega |}{ e} +(\mu _1+\mu _2)\\&\quad \int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x \quad \text {for all}\quad t\in (0, T), \end{aligned} \end{aligned}$$
(3.70)

where we have used the inequality \(\rho \ln \rho \ge -\frac{1}{e}\) for all \(\rho >0\). As for the rest terms in (3.65), the inequality \(\rho ^2\ln \rho \ge -\frac{1}{2e}\) directly yields

$$\begin{aligned} \begin{aligned}&-\int _{\Omega }\left( \frac{\mu _1}{\chi _1}u_1^2\ln u_1+\frac{\mu _2}{\chi _2}u_2^2\ln u_2+\frac{a_1\mu _1}{\chi _1}u_1u_2\ln u_1 +\frac{a_2\mu _2}{\chi _2}u_1u_2\ln u_2\right) \mathrm{d}x\\&\quad \le \frac{|\Omega |}{2e} \left( \frac{\mu _1}{\chi _1}+\frac{\mu _2}{\chi _2}\right) +\frac{L|\Omega |}{e} \left( \frac{a_1\mu _1}{\chi _1}+\frac{a_2\mu _2}{\chi _2} \right) \quad \text {for all}\quad t\in (0, T). \end{aligned} \end{aligned}$$
(3.71)

It then follows from (3.65)–(3.71) that (3.64) holds for all \(t\in (0, T)\).

Step 2 Now our attention is turned to showing that there exists \(T^*\in (0, T]\) such that

$$\begin{aligned} -\frac{\mathrm{d}}{\mathrm{d}t}\mathrm {F}(u_1, u_2, v_1, v_2) \ge C\left( - \mathrm {F}(u_1, u_2, v_1, v_2) \right) ^{\frac{1}{\vartheta }}\quad \text {for all}\quad t\in (0, T^*). \end{aligned}$$

It follows from the definition of \(\mathrm {F}(u_1, u_2, v_1, v_2)\) at (3.1) that

$$\begin{aligned} \mathrm {F}(u_1, u_2, v_1, v_2)\ge -\int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x-\frac{|\Omega |}{e}\left( \frac{1}{\chi _1}+\frac{1}{\chi _2} \right) \quad \text {for all}\quad t\in (0, T),\nonumber \\ \end{aligned}$$
(3.72)

hence, by Lemma 3.5 and the Young inequality, one obtains

$$\begin{aligned} \begin{aligned}&{\mathcal {C}}_0\left( -\mathrm {F}(u_1, u_2, v_1, v_2){-}\frac{|\Omega |}{e}\left( \frac{1}{\chi _1}{+}\frac{1}{\chi _2} \right) \right) {\le } {\mathcal {C}}_0\int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x \le \mathcal {\tilde{ C}}_1(\mathrm {D}^\vartheta (u_1, u_2, v_1, v_2)+1)\\&\quad \le {\overline{\epsilon }}\cdot \mathrm {D}(u_1, u_2, v_1, v_2)+{\mathcal {C}}_2\quad \text {for all}\quad t\in (0, T),\quad {\overline{\epsilon }}>0, \end{aligned}\nonumber \\ \end{aligned}$$
(3.73)

where \(\mathcal {\tilde{ C}}_1=\mathcal {\tilde{ C}}_1( m_1, m_2, M, L, B, k, R, n, \chi _{11}, \chi _{12}, \chi _{21}, \chi _{22}, \mu _1, \mu _2, a_1, a_2)\), \({\mathcal {C}}_2={\mathcal {C}}_2( m_1, m_2, M, L, B, k, R, \) \(n, \chi _{11}, \chi _{12}, \chi _{21}, \chi _{22}, \mu _1, \mu _2, a_1, a_2))\) are positive constants.

For any \({{\overline{C}}}>0\), it follows from Lemma 6.1 in [44] that one can always find radially symmetric initial data such that \(\mathrm {F}(u_{10}, u_{20}, v_{10}, v_{20})< -{{\overline{C}}}\). We fix

$$\begin{aligned} {{\overline{C}}}=2\max \left\{ \frac{{\mathcal {C}}_0+{\mathcal {C}}_2}{{\mathcal {C}}_1}, \frac{2|\Omega |}{e}\left( \frac{1}{\chi _1}+\frac{1}{\chi _2}\right) +\frac{2\mathcal {\tilde{C}}_1}{{\mathcal {C}}_0}, \left[ \frac{8}{3T}\left( \frac{2\mathcal {\tilde{C}}_1}{{\mathcal {C}}_1} \right) ^{\frac{1}{\vartheta }}\frac{\vartheta }{1-\vartheta }\right] ^{\frac{\vartheta }{1-\vartheta }} \right\} ,\nonumber \\ \end{aligned}$$
(3.74)

when \(\mathrm {F}(u_{10}, u_{20}, v_{10}, v_{20})< -{{\overline{C}}}\), the continuous property of \(\mathrm {F}\) guarantees that the set

$$\begin{aligned} \Lambda :=\Big \{ \breve{T}\in (0, T)| \mathrm {F}(u_1(\cdot , t), u_2(\cdot , t), v_1(\cdot , t), v_2(\cdot , t))< -{{\overline{C}}} \quad \text {for all}\quad t\in (0, \breve{T}) \Big \} \ne \emptyset ,\nonumber \\ \end{aligned}$$
(3.75)

then we can define

$$\begin{aligned} T^*=\sup \Lambda . \end{aligned}$$

From the definition of \(T^*\) and (3.74), we have \(-\mathrm {F}(u_1, u_2, v_1, v_2)\ge {{\overline{C}}}> \frac{{\mathcal {C}}_0+{\mathcal {C}}_2}{{\mathcal {C}}_1}\) and \( -\mathrm {F}(u_1, u_2, v_1, v_2)\ge {{\overline{C}}}> \frac{2|\Omega |}{e}\left( \frac{1}{\chi _1}+\frac{1}{\chi _2}\right) +\frac{2\mathcal {\tilde{C}}_1}{{\mathcal {C}}_0} \) for all \(t\in (0, T^*)\), which warrants that

$$\begin{aligned} {\mathcal {C}}_1\mathrm {F}(u_1(\cdot , t), u_2(\cdot , t), v_1(\cdot , t), v_2(\cdot , t))+{\mathcal {C}}_0+{\mathcal {C}}_2< 0 \quad \text {for all}\quad t\in (0, T^*),\nonumber \\ \end{aligned}$$
(3.76)

and

$$\begin{aligned} -\frac{{\mathcal {C}}_0\mathrm {F}(u_1, u_2, v_1, v_2) }{\mathcal {\tilde{ C}}_1} -\frac{{\mathcal {C}}_0|\Omega | }{\mathcal {\tilde{ C}}_1e}\cdot \left( \frac{1}{\chi _1}+\frac{1}{\chi _2}\right) -1> -\frac{{\mathcal {C}}_0\mathrm {F}(u_1, u_2, v_1, v_2)}{2\mathcal {\tilde{ C}}_1}>0\quad \text {for all}\quad t\in (0, T^*),\nonumber \\ \end{aligned}$$
(3.77)

along with (3.73), one obtains

$$\begin{aligned} \mathrm {D}(u_1, u_2, v_1, v_2)\ge \left( -\frac{{\mathcal {C}}_0\mathrm {F}(u_1, u_2, v_1, v_2) }{\mathcal {\tilde{ C}}_1} -\frac{{\mathcal {C}}_0|\Omega | }{\mathcal {\tilde{ C}}_1e}\cdot \left( \frac{1}{\chi _1}+\frac{1}{\chi _2}\right) -1 \right) ^{\frac{1}{\vartheta }}\quad \text {for all}\quad t\in (0, T^*),\nonumber \\ \end{aligned}$$
(3.78)

thereupon, by virtue of (3.64) and (3.73), fixing \({\overline{\epsilon }} \in (0, \frac{1}{4})\), it is clear that

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\mathrm {F}(u_1, u_2, v_1, v_2)&\le -(1-{\overline{\epsilon }})\mathrm {D}(u_1, u_2, v_1, v_2) +{\mathcal {C}}_1\mathrm {F}(u_1, u_2, v_1, v_2) +{\mathcal {C}}_0+{\mathcal {C}}_2\\&\le -\frac{3}{4}\mathrm {D}(u_1, u_2, v_1, v_2) +{\mathcal {C}}_1\mathrm {F}(u_1, u_2, v_1, v_2) +{\mathcal {C}}_0+{\mathcal {C}}_2 \quad \text {for all}\quad t\in (0, T^*), \end{aligned}\nonumber \\ \end{aligned}$$
(3.79)

inserting (3.78) into (3.79), we arrive at

$$\begin{aligned} \begin{aligned}&-\frac{\mathrm{d}}{\mathrm{d}t}\mathrm {F}(u_1, u_2, v_1, v_2)\\&\quad \ge \frac{3}{4}\left( -\frac{{\mathcal {C}}_0\mathrm {F} (u_1, u_2, v_1, v_2) }{\mathcal {\tilde{ C}}_1} -\frac{{\mathcal {C}}_0|\Omega | }{\mathcal {\tilde{ C}}_1e}\cdot \left( \frac{1}{\chi _1}+\frac{1}{\chi _2}\right) -1 \right) ^{\frac{1}{\vartheta }} -{\mathcal {C}}_1\mathrm {F}(u_1, u_2, v_1, v_2) -{\mathcal {C}}_0-{\mathcal {C}}_2 \end{aligned}\nonumber \\ \end{aligned}$$
(3.80)

for all \( t\in (0, T^*).\) In light of (3.76), (3.77) and (3.80), we can estimate

$$\begin{aligned} -\frac{d}{dt}\mathrm {F}(u_1, u_2, v_1, v_2) \ge \frac{3}{4}\left( -\frac{{\mathcal {C}}_1\mathrm {F}(u_1, u_2, v_1, v_2)}{2\mathcal {\tilde{ C}}_1} \right) ^{\frac{1}{\vartheta }}>0 \quad \text {for all}\quad t\in (0, T^*).\nonumber \\ \end{aligned}$$
(3.81)

Thus, we obtain (3.63), and the nondecreasing property of \(\mathrm {F}(u_1, u_2, v_1, v_2)\) is also proved. \(\square \)

In what follows, we shall deduce a contradictory result, which verifies Theorem 1.1.

Proof of Theorem 1.1

To obtain that for any \(0<L<\infty \), \(T>0\), there exists some \(({\hat{x}}, {\hat{t}})\in \Omega \times (0, T)\) such that the corresponding solution \(( u_1, u_2, v_1, v_2)\) of (1.4) satisfies \( u_1({\hat{x}}, {\hat{t}})>L\) or \( u_2({\hat{x}}, {\hat{t}})> L\). We assume, on the contrary,

$$\begin{aligned} u_1(x, t)\le L,\quad u_2(x, t)\le L \quad \text {for all}\quad (x, t)\in \Omega \times (0, T). \end{aligned}$$
(3.82)

Let \(\Psi (t):=-\mathrm {F}( u_1(\cdot , t), u_2(\cdot , t), v_1(\cdot , t), v_2(\cdot , t))\), from Lemma 3.6, we notice that \(\Psi (t)\) is nondecreasing, that is \(\Psi (t)\ge \Psi (0)>{{\overline{C}}}\) for all \(t\in (0, T^*)\), where

$$\begin{aligned} {{\overline{C}}}=2\max \left\{ \frac{{\mathcal {C}}_0+{\mathcal {C}}_2}{{\mathcal {C}}_1}, \frac{2|\Omega |}{e}\left( \frac{1}{\chi _1}+\frac{1}{\chi _2}\right) +\frac{2\mathcal {\tilde{C}}_1}{{\mathcal {C}}_0}, \left[ \frac{8}{3T}\left( \frac{2\mathcal {\tilde{C}}_1}{{\mathcal {C}}_1} \right) ^{\frac{1}{\vartheta }}\frac{\vartheta }{1-\vartheta }\right] ^{\frac{\vartheta }{1-\vartheta }} \right\} ,\nonumber \\ \end{aligned}$$
(3.83)

the continuity and the nondecreasing property of \(\mathrm {F}( u_1, u_2, v_1, v_2)\) actually warrant that \(T^*=T\). Accordingly, from Lemma 3.6, we know that \(\Psi \) satisfies

$$\begin{aligned} \Psi ^{'}(t)\ge C\Psi ^\delta (t)\quad \text {for all}\quad (x, t)\in \Omega \times (0, T) \end{aligned}$$
(3.84)

with \(\delta :=\frac{1}{\vartheta }>1\) and \(C=\frac{3}{4}\left( \frac{{\mathcal {C}}_1}{2\mathcal {\tilde{ C}}_1} \right) ^{\frac{1}{\vartheta }}>0\). For the above ordinary differential equation, it is easy to check that

$$\begin{aligned} \Psi ^{\delta -1}(t) \ge \left( \frac{1}{\Psi ^{1-\delta }(0)-C(\delta -1)t} \right) \quad \text {for all}\quad t\in (0, T), \end{aligned}$$
(3.85)

therefore,

$$\begin{aligned} \Psi ^{1-\delta }(t) \le \Psi ^{1-\delta }(0)-C(\delta -1)t\quad \text {for all}\quad t\in (0, T), \end{aligned}$$
(3.86)

utilizing (3.83), we find

$$\begin{aligned} \Psi ^{1-\delta }(\frac{T}{2}) \le \frac{1}{\Psi ^{\delta -1}(0)}-C(\delta -1)\frac{T}{2} < \frac{3T}{8}\left( \frac{{\mathcal {C}}_1}{2\mathcal {\tilde{ C}}_1} \right) ^{\frac{1}{\vartheta }}\frac{1-\vartheta }{\vartheta } -\frac{3}{4}\left( \frac{{\mathcal {C}}_1}{2\mathcal {\tilde{ C}}_1} \right) ^{\frac{1}{\vartheta }}(\frac{1}{\vartheta }-1)\frac{T}{2}=0,\nonumber \\ \end{aligned}$$
(3.87)

then we achieve a contradiction. Hence, we obtain Theorem 1.1. \(\square \)

Now we extend the method in [46] to prove Corollary 1.1.

Proof of Corollary 1.1

Given any \(L>0\) and \(T>0\), we assume that the claimed property in Corollary  1.1 is false, then there exists \(\mu =\max \{\mu _1, \mu _2\}\in (0, 1)\) such that the corresponding solution \( (u_{1}, u_{2}, v_{1}, v_{2}):= (u_{1, \mu }, u_{2, \mu }, v_{1, \mu }, v_{2, \mu })\) satisfies

$$\begin{aligned} \parallel u_{1, \mu }(x, t)\parallel _{L^{\infty }(\Omega )}\le \frac{L}{\mu }\quad \text {and}\quad \parallel u_{2, \mu }(x, t)\parallel _{L^{\infty } (\Omega )}\le \frac{L}{\mu } \quad \text {for all}\quad (x, t)\in \Omega \times (0, T),\nonumber \\ \end{aligned}$$
(3.88)

where \(\mu _1\in (0, 1), \mu _2\in (0, 1)\).

Recalling Lemma 3.1, in light of the condition \(\chi _{11}=\chi _{12}, \chi _{21}=\chi _{22}\), we have

$$\begin{aligned} \begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\mathrm {F}(u_1, u_2, v_1, v_2)\\&\quad \le -\mathrm {D}(u_1, u_2, v_1, v_2) +(a_1\mu _1{+}a_2\mu _2)\int _{\Omega }u_1u_2(v_1{+}v_2)\mathrm{d}x {+}\int _{\Omega }(\mu _1u_1^2v_1{+}\mu _2u_2^2v_1{+}\mu _1u_1^2v_2+\mu _2u_2^2v_2)\mathrm{d}x\\&\qquad +\int _{\Omega }\left( \frac{\mu _1}{\chi _1}u_1\ln u_1 +\frac{\mu _2}{\chi _2}u_2\ln u_2 +\frac{\mu _1}{\chi _1}u_1 +\frac{\mu _2}{\chi _2}u_2\right) \mathrm{d}x -\int _{\Omega }(\mu _1u_1v_1+\mu _2u_2v_1+\mu _1u_1v_2+\mu _2u_2v_2)\mathrm{d}x\\&\qquad -\int _{\Omega }\left( \frac{\mu _1}{\chi _1}u_1^2\ln u_1+\frac{\mu _2}{\chi _2}u_2^2\ln u_2+\frac{a_1\mu _1}{\chi _1}u_1u_2\ln u_1 +\frac{a_2\mu _2}{\chi _2}u_1u_2\ln u_2\right) \mathrm{d}x\quad \text {for all}\quad t\in (0, T), \end{aligned}\nonumber \\ \end{aligned}$$
(3.89)

where \(\chi _1=\chi _{11}=\chi _{12}, \chi _2=\chi _{21}=\chi _{22}\). Utilizing (2.10), (3.88) and the condition \(\mu =\max \{\mu _1, \mu _2\}\in (0, 1)\), a straightforward computation shows that

$$\begin{aligned} \begin{aligned}&(a_1\mu _1+a_2\mu _2)\int _{\Omega }u_1u_2(v_1+v_2)\mathrm{d}x +\int _{\Omega }(\mu _1u_1^2v_1+\mu _2u_2^2v_1+\mu _1u_1^2v_2+\mu _2u_2^2v_2)\mathrm{d}x\\&\qquad + \int _{\Omega } \left( \frac{\mu _1}{\chi _1}u_1 +\frac{\mu _2}{\chi _2}u_2\right) \mathrm{d}x \le (a_1\mu _1+a_2\mu _2)\frac{L}{\mu }\int _{\Omega }u_2(v_1+v_2)\mathrm{d}x\\&\qquad +\frac{L}{\mu }\int _{\Omega }(\mu _1u_1v_1+\mu _2u_2v_1+\mu _1u_1v_2+\mu _2u_2v_2)\mathrm{d}x +m\left( \frac{\mu _1}{\chi _1}+\frac{\mu _2}{\chi _2}\right) \\&\quad \le (a_1+a_2)L\int _{\Omega }u_2(v_1+v_2)\mathrm{d}x +L\int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x+\left( \frac{m}{\chi _1}+\frac{m}{\chi _2}\right) \end{aligned}\nonumber \\ \end{aligned}$$
(3.90)

for all \(t\in (0, T)\). Making use of the definition of \(\mathrm {F}(u_1, u_2, v_1, v_2)\) in (3.1) and the inequality \(\rho \ln \rho \ge -\frac{1}{e}\) for all \(\rho >0\), we can estimate

$$\begin{aligned}&\int _{\Omega }\left( \frac{\mu _1}{\chi _1}u_1\ln u_1 +\frac{\mu _2}{\chi _2}u_2\ln u_2 \right) \mathrm{d}x -\int _{\Omega }(\mu _1u_1v_1+\mu _2u_2v_1+\mu _1u_1v_2+\mu _2u_2v_2)\mathrm{d}x\nonumber \\&\quad = \mu _1\left( \mathrm {F}-\frac{1}{2}\int _{\Omega }(|v_1|^2+|\nabla v_1|^2+|v_2|^2+|\nabla v_2|^2)\mathrm{d}x \right) -\frac{\mu _1}{\chi _2}\int _{\Omega }u_2\ln u_2\mathrm{d}x+\mu _1\int _{\Omega }u_2(v_1+v_2)\mathrm{d}x\nonumber \\&\quad \quad +\mu _2\left( \mathrm {F}-\frac{1}{2}\int _{\Omega }(|v_1|^2+|\nabla v_1|^2+|v_2|^2+|\nabla v_2|^2)\mathrm{d}x \right) -\frac{\mu _2}{\chi _1}\int _{\Omega }u_1\ln u_1dx+\mu _2\int _{\Omega }u_1(v_1+v_2)\mathrm{d}x\nonumber \\&\quad \le (\mu _1+\mu _2)\mathrm {F} +\left( \frac{\mu _1}{\chi _2}+\frac{\mu _2}{\chi _1}\right) \frac{|\Omega |}{ e} +\mu \int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x\nonumber \\&\quad \le 2\mathrm {F} +\left( \frac{1}{\chi _2}+\frac{1}{\chi _1}\right) \frac{|\Omega |}{ e} +\int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x \quad \text {for all}\quad t\in (0, T). \end{aligned}$$
(3.91)

And it follows from the inequalities \(\rho ^2\ln \rho \ge -\frac{1}{2e}\) and \(\rho \ln \rho \ge -\frac{1}{e}\) that

$$\begin{aligned} \begin{aligned}&-\int _{\Omega }\left( \frac{\mu _1}{\chi _1}u_1^2\ln u_1+\frac{\mu _2}{\chi _2}u_2^2\ln u_2+\frac{a_1\mu _1}{\chi _1}u_1u_2\ln u_1 +\frac{a_2\mu _2}{\chi _2}u_1u_2\ln u_2\right) \mathrm{d}x\\&\quad \le \frac{|\Omega |}{2e} \left( \frac{\mu _1}{\chi _1}+\frac{\mu _2}{\chi _2}\right) +\frac{L|\Omega |}{\mu e} \left( \frac{a_1\mu _1}{\chi _1}+\frac{a_2\mu _2}{\chi _2} \right) \\&\quad \le \frac{|\Omega |}{2e} \left( \frac{1}{\chi _1}+\frac{1}{\chi _2}\right) +\frac{L|\Omega |}{ e} \left( \frac{a_1}{\chi _1}+\frac{a_2}{\chi _2} \right) \quad \text {for all}\quad t\in (0, T). \end{aligned} \end{aligned}$$
(3.92)

Therefore, plugging the above (3.90), (3.91) and (3.92) into (3.89), one obtains

$$\begin{aligned} \begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\mathrm {F}(u_1, u_2, v_1, v_2)\\&\quad \le -\mathrm {D}(u_1, u_2, v_1, v_2) +2\mathrm {F}(u_1, u_2, v_1, v_2) +((a_1+a_2+1)L+1)\int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x\\&\qquad +\left( \frac{m}{\chi _1}+\frac{m}{\chi _2}\right) +\left( \frac{1}{\chi _2} +\frac{1}{\chi _1}\right) \frac{|\Omega |}{ e}+ \frac{|\Omega |}{2e} \left( \frac{1}{\chi _1}+\frac{1}{\chi _2}\right) +\frac{L|\Omega |}{ e} \left( \frac{a_1}{\chi _1}+\frac{a_2}{\chi _2} \right) \\&\le -\mathrm {D}(u_1, u_2, v_1, v_2) +2\mathrm {F}(u_1, u_2, v_1, v_2) + {\check{C}}_1\int _{\Omega }(u_1+u_2)(v_1+v_2)dx+{\check{C}}_1 \quad \text {for all}\quad t\in (0, T), \end{aligned}\nonumber \\ \end{aligned}$$
(3.93)

with

$$\begin{aligned} {\check{C}}_1=(a_1+a_2+1)L+1+\left( \frac{m}{\chi _1}+\frac{m}{\chi _2}\right) +\left( \frac{1}{\chi _2}+\frac{1}{\chi _1}\right) \frac{|\Omega |}{ e}+ \frac{|\Omega |}{2e} \left( \frac{1}{\chi _1}+\frac{1}{\chi _2}\right) +\frac{L|\Omega |}{ e} \left( \frac{a_1}{\chi _1}+\frac{a_2}{\chi _2} \right) . \end{aligned}$$

The definition of \(\mathrm {F}(u_1, u_2, v_1, v_2)\) in (3.1) guarantees that

$$\begin{aligned} \mathrm {F}(u_1, u_2, v_1, v_2)\ge -\int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x-\frac{|\Omega |}{e}\left( \frac{1}{\chi _1}+\frac{1}{\chi _2} \right) \quad \text {for all}\quad t\in (0, T),\nonumber \\ \end{aligned}$$
(3.94)

it is inferred from (3.94) that

$$\begin{aligned} {\check{C}}_1\left( -\mathrm {F}(u_1, u_2, v_1, v_2)-\frac{|\Omega |}{e}\left( \frac{1}{\chi _1}+\frac{1}{\chi _2} \right) \right) \le {\check{C}}_1\int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x \quad \text {for all}\quad t\in (0, T),\nonumber \\ \end{aligned}$$
(3.95)

and from Lemma 3.5, there exist \(C=C( m_1, m_2, M, B, k, R, n, \chi _1, \chi _2)>0\) and \(\vartheta \in (\frac{1}{2}, 1)\) such that

$$\begin{aligned} \int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x \le C\cdot (\mathrm {D}^\vartheta (u_1, u_2, v_1, v_2)+1) \quad \text {for all}\quad t\in (0, T),\nonumber \\ \end{aligned}$$
(3.96)

then together with (3.95) and (3.96), applying the Young inequality, we can derive

$$\begin{aligned} \begin{aligned}&{\check{C}}_1\left( -\mathrm {F}(u_1, u_2, v_1, v_2)-\frac{|\Omega |}{e}\left( \frac{1}{\chi _1}+\frac{1}{\chi _2} \right) \right) \le {\check{C}}_1\int _{\Omega }(u_1+u_2)(v_1+v_2)\mathrm{d}x\\&\quad \le C{\check{C}}_1 (\mathrm {D}^\vartheta (u_1, u_2, v_1, v_2)+1)={\check{C}}_2 (\mathrm {D}^\vartheta (u_1, u_2, v_1, v_2)+1)\\&\quad \le \varepsilon \mathrm {D}(u_1, u_2, v_1, v_2)+{\check{C}}_3 \quad \text {for all}\quad t\in (0, T),\quad \varepsilon >0, \end{aligned} \end{aligned}$$
(3.97)

where \({\check{C}}_2:={\check{C}}_2( m_1, m_2, M, B, k, R, n, \chi _1, \chi _2, a_1, a_2, L ), {\check{C}}_3:={\check{C}}_3( m_1, m_2, M, B, k, R, n, \chi _1, \chi _2, a_1, a_2, L )\) are positive constants.

Choosing

$$\begin{aligned} C^*=2\max \left\{ \frac{{\check{C}}_1+{\check{C}}_3}{2}, \frac{2|\Omega | }{e}\cdot \left( \frac{1}{\chi _1}+\frac{1}{\chi _2}\right) +\frac{2{\check{C}}_2}{{\check{C}}_1}, \left[ \left( \frac{2{\check{C}}_2}{{\check{C}}_1}\right) ^{\frac{1}{\vartheta }} \frac{4\vartheta }{(1-\vartheta )T} \right] ^{\frac{\vartheta }{1-\vartheta }} \right\} , \end{aligned}$$

from Lemma 6.1 in [44], one can always find radially symmetric initial data such that \(\mathrm {F}(u_{10}, u_{20}, v_{10}, v_{20})\) \(< -C^*\). When \(\mathrm {F}(u_{10}, u_{20}, v_{10}, v_{20})< -C^*\), the continuous property of \(\mathrm {F}(u_1, u_2, v_1, v_2)\) guarantees that the set

$$\begin{aligned} {\tilde{\Lambda }}:=\Big \{ \tilde{T}\in (0, T)| \mathrm {F}(u_1(\cdot , t), u_2(\cdot , t), v_1(\cdot , t), v_2(\cdot , t))< -C^* \quad \text {for all}\quad t\in (0, \tilde{T}) \Big \}\nonumber \\ \end{aligned}$$
(3.98)

is not empty, then we can denote

$$\begin{aligned} {\mathcal {T}}^*=\sup {\tilde{\Lambda }}. \end{aligned}$$

Hence,

$$\begin{aligned} \mathrm {F}(u_1(\cdot , t), u_2(\cdot , t), v_1(\cdot , t), v_2(\cdot , t))&\le -C^*\le -2\max \left\{ \frac{{\check{C}}_1+{\check{C}}_3}{2}, \frac{2|\Omega | }{e}\cdot \left( \frac{1}{\chi _1}+\frac{1}{\chi _2}\right) +\frac{2{\check{C}}_2}{{\check{C}}_1}\right\} \nonumber \\&<-\max \left\{ \frac{{\check{C}}_1+{\check{C}}_3}{2}, \frac{2|\Omega |}{e}\cdot \left( \frac{1}{\chi _1}+\frac{1}{\chi _2}\right) +\frac{2{\check{C}}_2}{{\check{C}}_1}\right\} \end{aligned}$$
(3.99)

for all \(t\in (0, {\mathcal {T}}^*)\), which results in

$$\begin{aligned} 2\mathrm {F}(u_1, u_2, v_1, v_2) + {\check{C}}_1+{\check{C}}_3 <0\quad \text {for all}\quad t\in (0, {\mathcal {T}}^*) \end{aligned}$$
(3.100)

and

$$\begin{aligned} \frac{{\check{C}}_1\mathrm {F}(u_1, u_2, v_1, v_2)}{{\check{C}}_2} +\frac{{\check{C}}_1|\Omega | }{{\check{C}}_2e}\cdot \left( \frac{1}{\chi _1}+\frac{1}{\chi _2}\right) +1< \frac{{\check{C}}_1\mathrm {F}(u_1, u_2, v_1, v_2) }{2{\check{C}}_2}<0\quad \text {for all}\quad t\in (0, {\mathcal {T}}^*).\nonumber \\ \end{aligned}$$
(3.101)

Thanks to (3.97), we can derive

$$\begin{aligned} \mathrm {D}(u_1, u_2, v_1, v_2)\ge \left( -\frac{{\check{C}}_1\mathrm {F}(u_1, u_2, v_1, v_2) }{{\check{C}}_2} -\frac{{\check{C}}_1|\Omega | }{{\check{C}}_2e}\cdot \left( \frac{1}{\chi _1}+\frac{1}{\chi _2}\right) -1 \right) ^{\frac{1}{\vartheta }}>0 \quad \text {for all}\quad t\in (0, {\mathcal {T}}^*)\nonumber \\ \end{aligned}$$
(3.102)

and

$$\begin{aligned} {\check{C}}_1\int _{\Omega }(u_1+u_2)(v_1+v_2)dx \le \varepsilon \mathrm {D}(u_1, u_2, v_1, v_2)+{\check{C}}_3 \quad \text {for all}\quad t\in (0, {\mathcal {T}}^*),\quad \varepsilon >0.\nonumber \\ \end{aligned}$$
(3.103)

Fixing \(\varepsilon \in (0, \frac{1}{2})\), using the definition of \({\check{C}}_1\), and by virtue of (3.93) as well as (3.103), one obtains

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\mathrm {F}(u_1, u_2, v_1, v_2) \le -(1-\varepsilon )\mathrm {D}(u_1, u_2, v_1, v_2) +2\mathrm {F}(u_1, u_2, v_1, v_2) + {\check{C}}_1+{\check{C}}_3\quad \text {for all}\quad t\in (0, {\mathcal {T}}^*). \end{aligned}\nonumber \\ \end{aligned}$$
(3.104)

In light of (3.102) and (3.104), we can see that

$$\begin{aligned} \begin{aligned}&-\frac{d}{dt}\mathrm {F}(u_1, u_2, v_1, v_2)\\&\quad \ge \frac{1}{2} \left( -\frac{{\check{C}}_1\mathrm {F}(u_1, u_2, v_1, v_2) }{{\check{C}}_2} -\frac{{\check{C}}_1|\Omega | }{{\check{C}}_2e}\cdot \left( \frac{1}{\chi _1}+\frac{1}{\chi _2}\right) -1 \right) ^{\frac{1}{\vartheta }}-2\mathrm {F}(u_1, u_2, v_1, v_2) -{\check{C}}_1-{\check{C}}_3 \end{aligned}\nonumber \\ \end{aligned}$$
(3.105)

for all \(t\in (0, {\mathcal {T}}^*)\). Therefore, a combination of (3.100), (3.101) and (3.105) yields

$$\begin{aligned} -\frac{\mathrm{d}}{\mathrm{d}t}\mathrm {F}(u_1, u_2, v_1, v_2) \ge \frac{1}{2} \left( -\frac{{\check{C}}_1\mathrm {F}(u_1, u_2, v_1, v_2) }{2{\check{C}}_2} \right) ^{\frac{1}{\vartheta }}>0\quad \text {for all}\quad t\in (0, {\mathcal {T}}^*).\nonumber \\ \end{aligned}$$
(3.106)

The monotonicity and the continuous property of \(\mathrm {F}\) guarantee that \({\mathcal {T}}^*=T\). Let \(Y(t)=-\mathrm {F}\), then

$$\begin{aligned} Y^{'}(t)\ge \frac{1}{2}\left( \frac{Y(t)}{2{\check{C}}} \right) ^{\frac{1}{\vartheta }} \quad \text {for all}\quad t\in (0, T), \end{aligned}$$
(3.107)

where \({\check{C}}=\frac{{\check{C}}_2}{{\check{C}}_1}\), \(\vartheta \in (0, 1)\), a direct calculation warrants that

$$\begin{aligned} Y^{1-\frac{1}{\vartheta }}(t) \le Y^{1-\frac{1}{\vartheta }}(0)- \frac{1-\vartheta }{2(2{\check{C}})^{\frac{1}{\vartheta }}\vartheta }t < \frac{(1-\vartheta )T}{4\left( 2{\check{C}}\right) ^{\frac{1}{\vartheta }}\vartheta } -\frac{1-\vartheta }{2(2{\check{C}})^{\frac{1}{\vartheta }}\vartheta }t \quad \text {for all}\quad t\in (0, T),\nonumber \\ \end{aligned}$$
(3.108)

where we have used \(Y(0)=\mathrm {F}(u_{10}, u_{20}, v_{10},v_{20})>\left[ \left( \frac{2{\check{C}}_2}{{\check{C}}_1}\right) ^{\frac{1}{\vartheta }}\frac{4\vartheta }{(1-\vartheta )T}\right] ^{\frac{\vartheta }{1-\vartheta }}=\left[ \left( 2{\check{C}}\right) ^{\frac{1}{\vartheta }}\frac{4\vartheta }{(1-\vartheta )T} \right] ^{\frac{\vartheta }{1-\vartheta }}\) and \(0<\vartheta <1\), thus,

$$\begin{aligned} Y^{1-\frac{1}{\vartheta }}\left( \frac{T}{2}\right) < \frac{(1-\vartheta )T}{4\left( 2{\check{C}}\right) ^{\frac{1}{\vartheta }}\vartheta } -\frac{1-\vartheta }{2(2{\check{C}})^{\frac{1}{\vartheta }}\vartheta }\frac{T}{2}=0, \end{aligned}$$

providing a contradiction. This completes the proof of Corollary 1.1. \(\square \)

4 Simultaneous blowup

In this section, our attention is turned to investigating whether the blowup of two species occurs at the same time when the blowup happens. We define the blowup time by \(T_{\mathrm{max}}\), to obtain Theorem 1.2, it suffices to prove that the boundedness of \(u_1\) holds when \(u_2\) is bounded, which contradicts with the hypothesis that \(\Vert u_1\Vert _{L^{\infty }(\Omega )}+\Vert u_2\Vert _{L^{\infty }(\Omega )} \rightarrow \infty \) as \(t\rightarrow T_{\mathrm{max}}\). To begin with, we give the following lemma which is essential for our latter proof.

Lemma 4.1

Let \(\Omega \subset {\mathbb {R}}^n(n\ge 3)\) be a smooth and bounded domain, \(\chi _{ij}>0\), \(\mu _{i}>0\), \(a_{i}>0\) (\(i,j=1,2\)). And let \((u_1, u_2, v_1, v_2)\) be a classical solution of (1.4), \(q>\max \{ 1, \frac{n}{2}-1 \}\), \(p>\max \{ 1, \frac{2q(n-2)}{2q+n-2} \}\) with the property

$$\begin{aligned} \frac{q}{q-1}\cdot \frac{\frac{p}{2}-\frac{nq-n+2}{2nq}}{\frac{p}{2}-\frac{1}{2}+\frac{1}{n}}<1, \quad \frac{2q}{p}\cdot \frac{\frac{p}{2}-\frac{p(2q+n-2)}{4nq}}{\frac{p}{2}+\frac{1}{n}-\frac{1}{2}}<1. \end{aligned}$$
(4.1)

Then for any \(\zeta >0\), \(t\in (0, T_{\mathrm{max}})\), one can find \(\tilde{C}:=\tilde{C}(\zeta , p, q)>0\) such that

$$\begin{aligned} \int _{\Omega }u_1^p|\nabla v_i|^2dx \le \zeta \int _{\Omega }|\nabla u_1^{\frac{p}{2}}|^2\mathrm{d}x +\zeta \int _{\Omega }|\nabla |\nabla v_i|^q|^2\mathrm{d}x+\tilde{C}\quad \text {for}\quad i=1, 2,\qquad \end{aligned}$$
(4.2)

and

$$\begin{aligned} \int _{\Omega }u_1^2|\nabla v_i|^{2q-2}\mathrm{d}x \le \zeta \int _{\Omega }|\nabla u_1^{\frac{p}{2}}|^2\mathrm{d}x +\zeta \int _{\Omega }|\nabla |\nabla v_i|^q|^2\mathrm{d}x+\tilde{C} \quad \text {for}\quad i =1, 2.\nonumber \\ \end{aligned}$$
(4.3)

Proof

This lemma is a straightforward result from Lemmas 4.4 and 4.5 in [15], for completeness, we provide the detailed proof. Due to the fact that the estimates for (4.2) and (4.3) are similar, so we just consider the case (4.2). It follows from \(n\ge 3, p>1, q>1\) that

$$\begin{aligned} npq>nq-n+2, \end{aligned}$$
(4.4)

and from \(q>\frac{n}{2}-1\) we see that

$$\begin{aligned} \frac{nq-n+2}{2nq}>\frac{1}{2}-\frac{1}{n}. \end{aligned}$$
(4.5)

Thanks to (4.4), one can apply the H\(\ddot{o}\)lder inequality to estimate

$$\begin{aligned} \int _{\Omega }u_1^p|\nabla v_i|^2\mathrm{d}x \le \left( \int _{\Omega }u^{\frac{npq}{nq-n+2}}\mathrm{d}x \right) ^{\frac{nq-n+2}{nq}} \cdot \left( \int _{\Omega }|\nabla v_i|^{\frac{2nq}{n-2}}\mathrm{d}x \right) ^{\frac{n-2}{nq}}\quad \text {for}\quad i=1, 2,\nonumber \\ \end{aligned}$$
(4.6)

then utilizing (4.5), the Gagliardo–Nirenberg inequality and the boundedness of \(\int _{\Omega }udx\), there exist \(\breve{C}_1, \breve{C}_2>0\) such that

$$\begin{aligned} \begin{aligned} \left( \int _{\Omega }u^{\frac{npq}{nq-n+2}}\mathrm{d}x \right) ^{\frac{nq-n+2}{nq}}&=\Vert u^{\frac{p}{2}} \Vert ^2_{L^{\frac{2nq}{nq-n+2} }(\Omega )}\\&\le \breve{C}_1\left( \Vert \nabla u^{\frac{p}{2}}\Vert ^{2a}_{L^2(\Omega )} \Vert u^{\frac{p}{2}} \Vert ^{2(1-a)}_{L^\frac{2}{p}(\Omega )} + \Vert u^{\frac{p}{2}} \Vert _{L^\frac{2}{p}(\Omega )}^2 \right) \\&\le \breve{C}_2\Vert \nabla u^{\frac{p}{2}}\Vert ^{2a}_{L^2(\Omega )}+\breve{C}_2, \end{aligned} \end{aligned}$$
(4.7)

with

$$\begin{aligned} a=\frac{\frac{p}{2}-\frac{nq-n+2}{2nq}}{\frac{p}{2}-\frac{1}{2}+\frac{1}{n}}\in (0, 1), \end{aligned}$$
(4.8)

applying the Sobolev embedding, the Gagliardo–Nirenberg inequality, (2.6) and (2.7), one obtains

$$\begin{aligned} \begin{aligned} \left( \int _{\Omega }|\nabla v_i|^{\frac{2nq}{n-2}}\mathrm{d}x \right) ^{\frac{n-2}{nq}}&= \Vert |\nabla v_i|^q\Vert ^{\frac{2}{q}}_{L^{\frac{2n}{n-2}}(\Omega )}\\&\le \breve{C}_3 \Vert |\nabla v_i|^q\Vert ^{\frac{2}{q}}_{W^{1,2}(\Omega )}\\&\le \breve{C}_3^{'} \left( \Vert \nabla |\nabla v_i|^q\Vert ^{\frac{2}{q}}_{L^2(\Omega )} + \Vert |\nabla v_i|^q\Vert ^{\frac{2}{q}}_{L^{2}(\Omega )} \right) \\&\le \breve{C}_3^{'} \left( \Vert \nabla |\nabla v_i|^q\Vert ^{\frac{2}{q}}_{L^2(\Omega )} + \breve{C}_4 ( \Vert \nabla |\nabla v_i|^q\Vert ^{\lambda \cdot \frac{2}{q}}_{L^{2}(\Omega )} \Vert |\nabla v_i|^q\Vert _{L^{\frac{2}{q}}(\Omega )}^{(1-\lambda )\frac{2}{q}}+ \Vert |\nabla v_i|^q\Vert _{L^{\frac{2}{q}}(\Omega )}^{\frac{2}{q}} ) \right) \\&\le \breve{C}_5\Vert \nabla |\nabla v_i|^q\Vert ^{\frac{2}{q}}_{L^2(\Omega )} +\breve{C}_5\quad \text {for}\quad i=1, 2, \end{aligned}\nonumber \\ \end{aligned}$$
(4.9)

where \(\breve{C}_3, \breve{C}_3^{'}, \breve{C}_4, \breve{C}_5>0\), \(\lambda =\frac{\frac{q}{2}-\frac{1}{2}}{\frac{q}{2}-\frac{1}{2}+\frac{1}{n}}=\frac{n(q-1)}{n(q-1)+2}\in (0, 1)\). Therefore, combing (4.6), (4.7) with (4.9), and using the Young inequality, for any \(\zeta >0\), we can find \(\breve{C}_6, \breve{C}_7>0\) such that

$$\begin{aligned} \begin{aligned} \int _{\Omega }u_1^p|\nabla v_i|^2\mathrm{d}x&\le \breve{C}_2\breve{C}_5 \left( \Vert \nabla u^{\frac{p}{2}}\Vert ^{2a}_{L^2(\Omega )}\Vert \nabla |\nabla v_i|^q\Vert ^{\frac{2}{q}}_{L^2(\Omega )} +\Vert \nabla u^{\frac{p}{2}}\Vert ^{2a}_{L^2(\Omega )}+\Vert \nabla |\nabla v_i|^q\Vert ^{\frac{2}{q}}_{L^2(\Omega )}+1 \right) \\&\le \breve{C}_2\breve{C}_5 \Vert \nabla u^{\frac{p}{2}}\Vert ^{2a}_{L^2(\Omega )}\Vert \nabla |\nabla v_i|^q\Vert ^{\frac{2}{q}}_{L^2(\Omega )} +\frac{\zeta }{2}\Vert \nabla u^{\frac{p}{2}}\Vert ^{2}_{L^2(\Omega )} +\frac{\zeta }{2}\Vert \nabla |\nabla v_i|^q\Vert ^{2}_{L^2(\Omega )} +\breve{C}_6\\&\le \zeta \Vert \nabla |\nabla v_i|^q\Vert ^{2}_{L^2(\Omega )} + \breve{C}_7\left( \Vert \nabla u^{\frac{p}{2}}\Vert ^{2}_{L^2(\Omega )}\right) ^{a\cdot \frac{q}{q-1}} {+}\frac{\zeta }{2}\Vert \nabla u^{\frac{p}{2}}\Vert ^{2}_{L^2(\Omega )} {+}\breve{C}_6\quad \text {for}\quad i=1, 2. \end{aligned}\nonumber \\ \end{aligned}$$
(4.10)

In light of the first inequality in (4.1), it is clear that

$$\begin{aligned} a\cdot \frac{q}{q-1}<1, \end{aligned}$$

hence, using the Young inequality once again, we arrive at

$$\begin{aligned} \begin{aligned} \int _{\Omega }u_1^p|\nabla v_i|^2dx \le \zeta \Vert \nabla u^{\frac{p}{2}}\Vert ^{2}_{L^2(\Omega )} +\zeta \Vert \nabla |\nabla v_i|^q\Vert ^{2}_{L^2(\Omega )} +\breve{C}_8\quad \text {for}\quad i=1, 2, \end{aligned}\qquad \end{aligned}$$
(4.11)

where \(\breve{C}_8>0\). As for (4.3), it can be derived by a similar method. \(\square \)

In the following lemma, we utilize Lemma 4.1 to derive the \(L^p\) estimate for \(u_1\) under the condition that \(\Vert u_2\Vert _{L^{\infty }(\Omega )}\) is bounded.

Lemma 4.2

Let \(\Omega \subset {\mathbb {R}}^n(n\ge 3)\) be a smooth, bounded and convex domain, \(\chi _{ij}>0\), \(\mu _{i}>0\), \(a_{i}>0\) (\(i,j=1,2\)), \(q>\max \{ 1, \frac{n}{2}-1 \}\) and \(p>\max \{ 1, \frac{2q(n-2)}{2q+n-2} \}\) satisfy (4.1). \((u_1, u_2, v_1, v_2)\) is a classical solution of (1.4). If there exists \(L_1>0\) such that

$$\begin{aligned} \Vert u_2\Vert _{L^{\infty }(\Omega )}\le L_1\quad \text {for all}\quad t\in [0, T_{\mathrm{max}}]. \end{aligned}$$
(4.12)

Then

$$\begin{aligned} \Vert u_1\Vert _{L^{p}(\Omega )}\le L_2\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}), \end{aligned}$$
(4.13)

where \(L_2>0\) depends on \(L_1\).

Proof

Testing the first equation of (1.4) by \(u_1^{p-1}\), it follows that

$$\begin{aligned} \begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\int _{\Omega }u_1^p\mathrm{d}x +\frac{4(p-1)}{p}\int _{\Omega }|\nabla u_1^{\frac{p}{2}}|^2\mathrm{d}x\\&\quad = \chi _{11}p(p-1)\int _{\Omega }u_1^{p-1}\nabla u_1\nabla v_1\mathrm{d}x + \chi _{12}p(p-1)\int _{\Omega }u_1^{p-1}\nabla u_1\nabla v_2\mathrm{d}x\\&\qquad + \mu _{1}p\int _{\Omega }u_1^{p}\mathrm{d}x - \mu _{1}p\int _{\Omega }u_1^{p+1}\mathrm{d}x - a_{1}\mu _{1}p\int _{\Omega }u_1^{p}u_2\mathrm{d}x\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}). \end{aligned} \end{aligned}$$
(4.14)

For \(\chi _{11}p(p-1)\int _{\Omega }u_1^{p-1}\nabla u_1\nabla v_1\mathrm{d}x\), in light of the Young inequality, a simple calculation reveals that

$$\begin{aligned} \chi _{11}p(p-1)\int _{\Omega }u_1^{p-1}\nabla u_1\nabla v_1\mathrm{d}x \le \frac{p-1}{p}\int _{\Omega }|\nabla u_1^{\frac{p}{2}}|^2\mathrm{d}x + \chi _{11}^2p(p-1)\int _{\Omega }u_1^{p}|\nabla v_1|^2\mathrm{d}x\nonumber \\ \end{aligned}$$
(4.15)

for all \(t\in (0, T_{max})\). Similarly, we can deduce that

$$\begin{aligned} \chi _{12}p(p-1)\int _{\Omega }u_1^{p-1}\nabla u_1\nabla v_2\mathrm{d}x \le \frac{p-1}{p}\int _{\Omega }|\nabla u_1^{\frac{p}{2}}|^2\mathrm{d}x + \chi _{12}^2p(p-1)\int _{\Omega }u_1^{p}|\nabla v_2|^2\mathrm{d}x\nonumber \\ \end{aligned}$$
(4.16)

for all \(t\in (0, T_{\mathrm{max}})\). And by virtue of the Young inequality, one can find \(\tilde{C}_1>0\) such that

$$\begin{aligned} \mu _{1}p\int _{\Omega }u_1^{p}dx \le \frac{\mu _1p}{2}\int _{\Omega }u_1^{p+1}\mathrm{d}x+\tilde{C}_1\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}), \end{aligned}$$
(4.17)

then it follows from (4.14)–(4.17) that

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\int _{\Omega }u_1^p\mathrm{d}x +\frac{2(p-1)}{p}&\int _{\Omega }|\nabla u_1^{\frac{p}{2}}|^2\mathrm{d}x+\frac{\mu _1p}{2}\int _{\Omega }u_1^{p+1}\mathrm{d}x \le \chi _{11}^2p(p-1)\int _{\Omega }u_1^{p}|\nabla v_1|^2\mathrm{d}x\\&\quad + \chi _{12}^2p(p-1)\int _{\Omega }u_1^{p}|\nabla v_2|^2\mathrm{d}x+\tilde{C}_1\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}), \end{aligned}\nonumber \\ \end{aligned}$$
(4.18)

and applying (4.2), we can find \(\tilde{C}_2>0\) such that

$$\begin{aligned} \begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\int _{\Omega }u_1^p\mathrm{d}x +\frac{2(p-1)}{p}\int _{\Omega }|\nabla u_1^{\frac{p}{2}}|^2\mathrm{d}x+\frac{\mu _1p}{2}\int _{\Omega }u_1^{p+1}\mathrm{d}x\\&\quad \le \chi _{11}^2p(p-1)\left( \zeta \int _{\Omega }|\nabla u_1^{\frac{p}{2}}|^2\mathrm{d}x +\zeta \int _{\Omega }|\nabla |\nabla v_1|^q|^2\mathrm{d}x\right) \\&\qquad + \chi _{12}^2p(p-1)\left( \zeta \int _{\Omega }|\nabla u_1^{\frac{p}{2}}|^2\mathrm{d}x +\zeta \int _{\Omega }|\nabla |\nabla v_2|^q|^2\mathrm{d}x\right) +\tilde{C}_2\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}). \end{aligned}\nonumber \\ \end{aligned}$$
(4.19)

To estimate the terms \(\int _{\Omega }|\nabla |\nabla v_i|^q|^2\mathrm{d}x (i=1,2)\) in (4.19), one needs to cope with the following energy-type inequality concerning \(\int _{\Omega }|\nabla v_i|^{2q}\mathrm{d}x(i=1, 2)\). We notice that these two terms can be estimated by the similar method, so we just consider the case \(\int _{\Omega }|\nabla |\nabla v_1|^q|^2\mathrm{d}x\).

By a direct calculation, we find

$$\begin{aligned} \begin{aligned}&\frac{1}{q}\frac{\mathrm{d}}{\mathrm{d}t}\int _{\Omega }|\nabla v_1|^{2q}\mathrm{d}x +2\int _{\Omega }|\nabla v_1|^{2q}\mathrm{d}x = 2\int _{\Omega }(\nabla v_1)^{2q-1}\cdot \nabla \Delta v_1\mathrm{d}x \\&\quad +2\int _{\Omega }(\nabla v_1)^{2q-1}\cdot \nabla u_1\mathrm{d}x +2\int _{\Omega }(\nabla v_1)^{2q-1}\cdot \nabla u_2\mathrm{d}x=J_1+J_2+J_3\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}). \end{aligned}\nonumber \\ \end{aligned}$$
(4.20)

To deal with \(J_1\), we notice that the domain \(\Omega \subset {\mathbb {R}}^n(n\ge 3)\) is convex, which implies \(\frac{\partial }{\partial \nu }|\nabla v_i|^2\le 0 (i=1, 2)\), then applying the equality \(2\nabla v_i\cdot \nabla \Delta v_i=\Delta |\nabla v_i|^2-2|D^2v_i|^2 (i=1, 2)\), one can infer that

$$\begin{aligned} \begin{aligned} J_1&= -(q-1)\int _{\Omega }(|\nabla v_1|^2)^{q-2}\cdot |\nabla |\nabla v_1|^2|^2\mathrm{d}x +\int _{\partial \Omega }(\nabla v_1)^{2q-2}\cdot \frac{\partial }{\partial \nu }|\nabla v_1|^2 -2\int _{\Omega }(\nabla v_1)^{2q-2}\cdot |D^2 v_1|^2\\&\le -(q-1)\int _{\Omega }|\nabla v_1|^{2q-4}\cdot |\nabla |\nabla v_1|^2|^2\mathrm{d}x -2\int _{\Omega }(\nabla v_1)^{2q-2}\cdot |D^2 v_1|^2\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}). \end{aligned}\nonumber \\ \end{aligned}$$
(4.21)

For \(J_2\), employing Young’s inequality and the inequality \(|\Delta v_i|^2\le n|D^2v_i|^2 (i=1, 2)\), we arrive at

$$\begin{aligned} \begin{aligned} J_2&= -2\int _{\Omega }u_1\Delta v_1|\nabla v_1|^{2q-2}\mathrm{d}x -2(q-1)\int _{\Omega }u_1(\nabla v_1)^{q-2}\cdot (\nabla v_1)^{q-1}\cdot \nabla |\nabla v_1|^2\mathrm{d}x\\&\le \int _{\Omega }|\nabla v_1|^{2q-2}\cdot |D^2v_1|^2\mathrm{d}x+n\int _{\Omega }u_1^2 |\nabla v_1|^{2q-2}\mathrm{d}x\\&\quad + \frac{q-1}{4}\int _{\Omega }|\nabla v_1|^{2q-4}|\nabla |\nabla v_1|^2|^2\mathrm{d}x+4(q-1)\int _{\Omega }u_1^2|\nabla v_1|^{2q-2}\mathrm{d}x \quad \text {for all}\quad t\in (0, T_{\mathrm{max}}). \end{aligned}\nonumber \\ \end{aligned}$$
(4.22)

Similarly, we have

$$\begin{aligned} \begin{aligned} J_3&\le \int _{\Omega }|\nabla v_1|^{2q-2}\cdot |D^2v_1|^2dx+(n+4(q-1))\int _{\Omega }u_2^2 |\nabla v_1|^{2q-2}\mathrm{d}x\\&\quad + \frac{q-1}{4}\int _{\Omega }|\nabla v_1|^{2q-4}|\nabla |\nabla v_1|^2|^2\mathrm{d}x \quad \text {for all}\quad t\in (0, T_{\mathrm{max}}). \end{aligned} \end{aligned}$$
(4.23)

Putting the above four inequalities together, and using the identity \(|\nabla v_1|^{2q-4}|\nabla |\nabla v_1|^2|^2=\frac{4}{q^2}|\nabla |\nabla v_1|^q|^2\), we find

$$\begin{aligned} \begin{aligned}&\frac{1}{q}\frac{\mathrm{d}}{\mathrm{d}t}\int _{\Omega }|\nabla v_1|^{2q}\mathrm{d}x +2\int _{\Omega }|\nabla v_1|^{2q}\mathrm{d}x + \frac{2(q-1)}{q^2}\int _{\Omega }|\nabla |\nabla v_1|^q|^2\mathrm{d}x\\&\quad \le (n+4(q-1))\int _{\Omega }u_1^2|\nabla v_1|^{2q-2}\mathrm{d}x+(n+4(q-1))\int _{\Omega }u_2^2 |\nabla v_1|^{2q-2}\mathrm{d}x\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}). \end{aligned}\nonumber \\ \end{aligned}$$
(4.24)

In the same vein, for \(|\nabla v_2|^{2q}\), one obtains

$$\begin{aligned} \begin{aligned}&\frac{1}{q}\frac{\mathrm{d}}{\mathrm{d}t}\int _{\Omega }|\nabla v_2|^{2q}\mathrm{d}x +2\int _{\Omega }|\nabla v_2|^{2q}\mathrm{d}x + \frac{2(q-1)}{q^2}\int _{\Omega }|\nabla |\nabla v_2|^q|^2\mathrm{d}x\\&\quad \le (n+4(q-1))\int _{\Omega }u_1^2|\nabla v_2|^{2q-2}\mathrm{d}x+(n+4(q-1))\int _{\Omega }u_2^2 |\nabla v_2|^{2q-2}\mathrm{d}x\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}). \end{aligned}\nonumber \\ \end{aligned}$$
(4.25)

Let \(C^*=\max \left\{ \chi _{11}^2p(p-1), \chi _{12}^2p(p-1), n+4(q-1) \right\} \), combining (4.19), (4.24) with (4.25), and utilizing (4.3), we can see that

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t} \left( \int _{\Omega }u_1^p\mathrm{d}x+\frac{1}{q}\int _{\Omega }|\nabla v_1|^{2q}\mathrm{d}x+ \frac{1}{q}\int _{\Omega }|\nabla v_2|^{2q}\mathrm{d}x\right) +\frac{2(p-1)}{p}\int _{\Omega }|\nabla u_1^{\frac{p}{2}}|^2dx+\frac{\mu _1p}{2}\int _{\Omega }u_1^{p+1}\mathrm{d}x\nonumber \\&\qquad + \frac{2(q-1)}{q^2}\int _{\Omega }|\nabla |\nabla v_1|^q|^2\mathrm{d}x + \frac{2(q-1)}{q^2}\int _{\Omega }|\nabla |\nabla v_2|^q|^2\mathrm{d}x +2\int _{\Omega }|\nabla v_1|^{2q}\mathrm{d}x+2\int _{\Omega }|\nabla v_2|^{2q}\mathrm{d}x\nonumber \\&\quad \le 4C^* \zeta \int _{\Omega }|\nabla u_1^{\frac{p}{2}}|^2\mathrm{d}x + 2 C^*\zeta \left( \int _{\Omega }|\nabla |\nabla v_1|^q|^2\mathrm{d}x+\int _{\Omega }|\nabla |\nabla v_2|^q|^2\mathrm{d}x\right) \nonumber \\&\qquad +C^*\left( \int _{\Omega }u_2^2 |\nabla v_1|^{2q-2}\mathrm{d}x +\int _{\Omega }u_2^2 |\nabla v_2|^{2q-2}\mathrm{d}x \right) +\tilde{C}_3 \quad \text {for all}\quad t\in (0, T_{\mathrm{max}}), \end{aligned}$$
(4.26)

applying \(\Vert u_2\Vert _{L^{\infty }(\Omega )}<L_1\) and the Young inequality, we can find \(\tilde{C}_4>0, \tilde{C}_5>0\) such that

$$\begin{aligned}&C^*\int _{\Omega }u_2^2 |\nabla v_i|^{2q-2}\mathrm{d}x \le (2-\frac{1}{q})\int _{\Omega } |\nabla v_i|^{2q}\mathrm{d}x+\tilde{C}_4\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}),\\&\quad i=1, 2, \end{aligned}$$

and

$$\begin{aligned} \frac{\mu _1p}{2}\int _{\Omega }u_1^{p+1}\mathrm{d}x\ge \int _{\Omega }u_1^{p}\mathrm{d}x -\tilde{C}_5\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}). \end{aligned}$$

Hence, choosing \(0<\zeta \le \min \left\{ \frac{p-1}{2pC^*}, \frac{q-1}{q^2C^*} \right\} \), the above three inequalities warrant that

$$\begin{aligned} \begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t} \left( \int _{\Omega }u_1^p\mathrm{d}x+\frac{1}{q}\int _{\Omega }|\nabla v_1|^{2q}\mathrm{d}x+ \frac{1}{q}\int _{\Omega }|\nabla v_2|^{2q}\mathrm{d}x\right) +\int _{\Omega }u_1^{p}\mathrm{d}x\\&\quad +\frac{1}{q}\int _{\Omega }(|\nabla v_1|^{2q}+|\nabla v_2|^{2q})\mathrm{d}x \le \tilde{C}_3 +\tilde{C}_4 +\tilde{C}_5\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}), \end{aligned} \end{aligned}$$
(4.27)

thereupon, with the aid of an application of ODE comparison argument, we can obtain (4.13) immediately. \(\square \)

With the above preparations, now we extend the method in [1, 15] to prove the boundedness of \(\Vert u_1\Vert _{L^{\infty }(\Omega )}\).

Lemma 4.3

Let \(\Omega \subset {\mathbb {R}}^n(n\ge 3)\) be a smooth and bounded domain, \(\chi _{ij}>0\), \(\mu _{i}>0\), \(a_{i}>0\) (\(i,j=1,2\)), \(p>\frac{n}{2}\). Assume that \((u_1, u_2, v_1, v_2)\) is a classical solution of (1.4) with the property

$$\begin{aligned} \displaystyle \sup _{t\in (0, T_{\mathrm{max}})}( \Vert u_1(\cdot , t)\Vert _{L^{p}(\Omega )}+\Vert u_2(\cdot , t)\Vert _{L^{p}(\Omega )} )<\infty . \end{aligned}$$
(4.28)

Then \(T_{\mathrm{max}}=\infty \) and

$$\begin{aligned} \displaystyle \sup _{t>0} \Vert u_1(\cdot , t)\Vert _{L^{\infty }(\Omega )}<\infty . \end{aligned}$$
(4.29)

Proof

We define

$$\begin{aligned} \Upsilon (T):=\displaystyle \sup _{t\in (0, T)} \Vert u_1(\cdot , t)\Vert _{L^{\infty }(\Omega )}\quad \text {for all}\quad T\in (0, T_{\mathrm{max}}), \end{aligned}$$
(4.30)

and let \(m:=\frac{1}{|\Omega |}\int _{\Omega }u_{10}\mathrm{d}x\), \(t_0=(t-1)_+\). For \(u_1\), we find

$$\begin{aligned} \begin{aligned}&\Vert u_1(\cdot , t)-m\Vert _{L^{\infty }(\Omega )}\\&\quad \le \Vert e^{(t-t_0)\Delta }(u_1(t_0)-m)\Vert _{L^{\infty }(\Omega )}+\chi _{11}\int _{t_0}^t\Vert e^{(t-s)\Delta }\nabla \cdot (u_1(\cdot , s)\nabla v_1(\cdot , s))\Vert _{L^{\infty }(\Omega )}\mathrm{d}s\\&\qquad +\chi _{12}\int _{t_0}^t\Vert e^{(t-s)\Delta }\nabla \cdot (u_1(\cdot , s)\nabla v_2(\cdot , s))\Vert _{L^{\infty }(\Omega )}\mathrm{d}s +\mu _1\int _{t_0}^t\Vert e^{(t-s)\Delta }u_1(1-u_1-a_1u_2)\Vert _{L^{\infty }(\Omega )}\mathrm{d}s\\&\quad ={\hat{I}}_1+{\hat{I}}_2+{\hat{I}}_3+{\hat{I}}_4\quad \text {for all}\quad t\in (0, T_{\mathrm{max}}). \end{aligned}\nonumber \\ \end{aligned}$$
(4.31)

In light of the \(L^p-L^q\) estimates for the heat semigroup (see Lemma 1.3 in [42]), for all \(t\in (0, T_{\mathrm{max}})\), it is clear that

$$\begin{aligned} {\hat{I}}_1\le \left\{ \begin{array}{ll} {\mathcal {K}}_1(\Vert u_{10}+m\Vert _{L^{p}(\Omega )}) \le {\mathcal {K}}_1^{'}&{}\quad \hbox {if}\quad t> 1,\\ {\mathcal {K}}_2(\Vert u_{10}\Vert _{L^{\infty }(\Omega )}+m) \le {\mathcal {K}}_2^{'}&{}\quad \hbox {if}\quad t\le 1 \end{array}\right. \end{aligned}$$
(4.32)

with \({\mathcal {K}}_1, {\mathcal {K}}_2, {\mathcal {K}}_1^{'}, {\mathcal {K}}_2^{'}>0\). To bound \(I_2\), we need introduce some new parameters. Since \(p>\frac{n}{2}\), we have \(n<\frac{np}{n-p}\), which guarantees that there exists \(\sigma >1\) and \(r>n\) such that

$$\begin{aligned} n<r\sigma <\frac{np}{n-p}, \end{aligned}$$
(4.33)

where \(r>n\) ensures \(\frac{1}{2}+\frac{n}{2r}<1\). And let \(\iota =1-\frac{\sigma -1}{r\sigma }\), it is obvious that \(0<\iota <1\). Then applying the \(L^p-L^q\) estimates of the heat semigroup and the H\(\ddot{o}\)lder inequality, we conclude

$$\begin{aligned} \begin{aligned} {\hat{I}}_2&\le {\mathcal {K}}_3\int _{0}^t\left( 1+ (t-s)^{-\frac{1}{2}-\frac{n}{2}\cdot \frac{1}{r}} \right) e^{-{\tilde{\lambda }}(t-s)} \Vert u_1(\cdot , s)\nabla v_1(\cdot , s) \Vert _{L^{r}(\Omega )}\mathrm{d}s\\&\le {\mathcal {K}}_3\int _{0}^t\left( 1+ (t-s)^{-\frac{1}{2}-\frac{n}{2}\cdot \frac{1}{r}} \right) e^{-{\tilde{\lambda }}(t-s)} \Vert u_1\Vert _{L^{r\sigma ^{'}}(\Omega )} \Vert \nabla v_1\Vert _{L^{r\sigma }(\Omega )}\mathrm{d}s\\&\le {\mathcal {K}}_3\int _{0}^t\left( 1+ (t-s)^{-\frac{1}{2}-\frac{n}{2}\cdot \frac{1}{r}} \right) e^{-{\tilde{\lambda }}(t-s)} \Vert u_1\Vert _{L^{\infty }(\Omega )}^{\iota } \Vert u_1\Vert _{L^{1}(\Omega )}^{1-\iota } \Vert \nabla v_1\Vert _{L^{r\sigma }(\Omega )}\mathrm{d}s\\&\le {\mathcal {K}}_3 m_1^{1-\iota } \left( \Upsilon (T)\right) ^{\iota } \int _{0}^t\left( 1+ (t-s)^{-\frac{1}{2}-\frac{n}{2}\cdot \frac{1}{r}} \right) e^{-{\tilde{\lambda }}(t-s)} \Vert \nabla v_1\Vert _{L^{r\sigma }(\Omega )}\mathrm{d}s \end{aligned} \end{aligned}$$
(4.34)

for all \(t\in (0, T_{\mathrm{max}})\), where \({\tilde{\lambda }}\) denotes the first nonzero eigenvalue of \(-\Delta \) in \(\Omega \), and \({\mathcal {K}}_3>0\). For \(\Vert \nabla v_1\Vert _{L^{r\sigma }(\Omega )}\), again, making use of the \(L^p-L^q\) estimates, we can see that

$$\begin{aligned} \begin{aligned}&\Vert \nabla v_1(\cdot , t)\Vert _{L^{r\sigma }(\Omega )} \le \Vert \nabla e^{t(\Delta -\lambda _1)}v_{10} \Vert _{L^{r\sigma }(\Omega )} + \int _0^t\Vert \nabla e^{(t-s)(\Delta -\lambda _1)} ( \alpha _{11}u_1+\alpha _{12}u_2 ) \Vert _{L^{r\sigma }(\Omega )}\mathrm{d}s\\&\quad \le {\mathcal {K}}_4 \Vert \nabla v_{10} \Vert _{L^{r\sigma }(\Omega )} +{\mathcal {K}}_4\int _0^t \left( 1+(t-s)^{-\frac{1}{2}-\frac{n}{2}(\frac{1}{p}-\frac{1}{r\sigma })} \right) e^{-\lambda _1(t-s)} (\Vert u_1\Vert _{L^p(\Omega )}+\Vert u_2\Vert _{L^p(\Omega )})\mathrm{d}s\\&\quad \le {\mathcal {K}}_5\quad \text {for all}\quad t\in (0, T_{\mathrm{max}})\quad \text {and some}\quad {\mathcal {K}}_4,\quad {\mathcal {K}}_5>0, \end{aligned}\nonumber \\ \end{aligned}$$
(4.35)

where we have used \(\frac{1}{2}+\frac{n}{2}(\frac{1}{p}-\frac{1}{r\sigma })<1\), which is guaranteed by (4.33). Therefore, (4.34) and (4.35) indicate that

$$\begin{aligned} {\hat{I}}_2\le {\mathcal {K}}_6\left( \Upsilon (T)\right) ^{\iota } \end{aligned}$$
(4.36)

with \({\mathcal {K}}_6>0\), similarly, for \({\hat{I}}_3\), we have

$$\begin{aligned} {\hat{I}}_3\le {\mathcal {K}}_6^{'}\left( \Upsilon (T)\right) ^{\iota } \end{aligned}$$
(4.37)

with \({\mathcal {K}}_6^{'}>0\). As for \({\hat{I}}_4\), we note that \(u_1(1-u_1-a_1u_2)\le \frac{1}{4}\), which warrants the boundedness of \({\hat{I}}_4\). Thus, for \(0<\iota <1\), it follows from (4.31)–(4.37) that

$$\begin{aligned} \Upsilon (T)\le {\mathcal {K}}_7+{\mathcal {K}}_7\left( \Upsilon (T)\right) ^{\iota }\quad \text {for all}\quad T\in (0, T_{\mathrm{max}}), \end{aligned}$$
(4.38)

with \({\mathcal {K}}_7>0\), this implies that \(T_{max}=\infty \) and (4.29) holds. \(\square \)

Proof of Theorem 1.2

Theorem 1.2 is a direct result of Lemma 4.2 and Lemma 4.3.

\(\square \)