Electronic Journal of Differential Equations, Vol. 2025 (2025), No. 01, pp. 1–17. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu DOI: 10.58997/ejde.2025.01 EXISTENCE AND BOUNDEDNESS OF SOLUTIONS FOR A PARABOLIC-PARABOLIC PREDATOR-PREY MODEL FENGXIANG ZHAO, HAOTIAN TANG, JIASHAN ZHENG, KAIQIANG LI Abstract. This article concerns the fully parabolic pursuit-prey chemotaxis system ut = ∆u− χ∇ · (u∇w) + u ( λ1 − µ1u r1−1 + av ) , x ∈ Ω, t > 0, vt = ∆v + ξ∇ · (v∇z) + v ( λ2 − µ2v r2−1 − bu ) , x ∈ Ω, t > 0, wt = ∆w − w + v, x ∈ Ω, t > 0, zt = ∆z − z + u, x ∈ Ω, t > 0, in a bounded domain Ω ⊂ RN (N ≥ 1) with homogeneous Neumann boundary conditions, where χ, ξ, λi, µi, a, b are positive constants and ri > 1 (i = 1, 2). We show that if (r1 − 1)(r2 − 1) ≥ 1, the above system exists a unique global and bounded classical solution for all appropriately regular nonnegative initial data, which extends the previous global existence result in Qi and Ke [13]). 1. Introduction In this article, we investigate the indirect pursuit-evasion model (parabolic- parabolic system) ut = ∆u− χ∇ · (u∇w) + u ( λ1 − µ1u r1−1 + av ) , x ∈ Ω, t > 0, vt = ∆v + ξ∇ · (v∇z) + v ( λ2 − µ2v r2−1 − bu ) , x ∈ Ω, t > 0, wt = ∆w − w + v, x ∈ Ω, t > 0, zt = ∆z − z + u, x ∈ Ω, t > 0, ∂u ∂ν = ∂v ∂ν = ∂w ∂ν = ∂z ∂ν = 0, x ∈ ∂Ω, t > 0, u(x, 0) = u0(x), v(x, 0) = v0(x), w(x, 0) = w0(x), z(x, 0) = z0(x), x ∈ Ω, (1.1) where Ω ⊂ RN (N ≥ 1) is a bounded domain with smooth boundary, λ1, λ2, µ1, µ2, r1, r2, χ, ξ, a, b are positive constants. The initial data (u0, v0, w0, z0) satisfies 2020 Mathematics Subject Classification. 35K20, 35K55, 92C17. Key words and phrases. Pursuit-evasion; parabolic-parabolic; boundedness; classical solution. ©2025. This work is licensed under a CC BY 4.0 license. Submitted April 11, 2024. Published January 4, 2025. 1 2 F. ZHAO, H. TANG, J. ZHENG, K. LI EJDE-2025/01 u0 ∈ C0(Ω̄), u0 ≥ 0 in Ω̄, v0 ∈ C0(Ω̄), v0 ≥ 0 in Ω̄, w0 ∈ W 1,ϑ(Ω), w0 ≥ 0 in Ω̄ and some ϑ > N, z0 ∈ W 1,ϑ(Ω), z0 ≥ 0 in Ω̄ and some ϑ > N. (1.2) In this system, u(x, t), v(x, t) are the densities of the predators and the prey, respec- tively, meanwhile w(x, t) and z(x, t) represent concentrations of chemical signals emitted by v(x, t) and u(x, t). χ, ξ measure the strength of attractive and repulsive directed migration, respectively. To understand the model (1.1) better, we need to introduce some results about the classical Keller-Segel model [9], which presents the aggregate and collective behavior of cells due to chemotaxis by means of a coupled system of two equations ut = ∆u−∇ · (u∇v), vt = ∆v − v + u. (1.3) Many works are dedicated to this model and its variants, the most important results are about the existence and boundedness of classical global solutions, the occurrence of the finite-time blow-up for the solutions and large time behavior under some appropriate assumptions. See for example [4, 6, 8, 12, 20, 21, 22, 23, 24, 26] and the references therein. Supposing predators and prey to exert species-characteristic substances, such as pheromones or scent marks, the following pursuit-evasion model was proposed ut = ∆u− χ∇ · (u∇w) + f(u, v), x ∈ Ω, t > 0, vt = ∆v + ξ∇ · (v∇z) + g(u, v), x ∈ Ω, t > 0, τ1wt = ∆w − w + v, x ∈ Ω, t > 0, τ2zt = ∆z − z + u, x ∈ Ω, t > 0, (1.4) where τi ∈ {0, 1} (i = 1, 2). When τ1 = τ2 = 0, it has been proved that the above system possesses a unique non-negative bounded weak solution in two-dimensional space for the case f = g = 0 [5]. The weak solution to (1.4) with f = uv − u and g = v(1− v − u) was also considered in [1]. Turning to the case f = u(λ+ av − u) and g = v(µ− v − bu), Li, Tao and Winkler [10] proved that the parabolic-elliptic system (τ1 = τ2 = 0) admits a global and bounded solution for any given suitably regular initial data when N ≤ 3. Then Liu and Zheng [11] establish the existence of classical global solutions, and the asymptotic behavior for this system in N - dimensional domains. Recently, Zhang and Zheng [27] considered the above system with r1 > 1, r2 > 1, f = u(λ1 − µ1u r1−1 + av) and g = v(λ2 − µ2v r2−1 − bu), and proved that if (r1 − 1)(r2 − 1) > (N−2)+ N , the system exists a unique global and bounded classical solution for all appropriately regular nonnegative initial data. When τ1 = τ2 = 1, since accounting for two taxis mechanisms, system (1.4) can not be regarded as a triangular cross-diffusion model. Due to the complexity of this problem, there are few related works. Recently, Qi and Ke [13] showed that the system with f = u(µ − u + av) and g = v(λ − v − bu) possesses a global bounded classical solution with a positive constant CN/2+1 if a < 2 and N(2− a) 2(CN/2+1) 1 N/2+1 (N − 2)+ > max{χ, ξ}. EJDE-2025/01 GLOBAL SOLUTION FOR A PREDATOR-PREY MODEL 3 Additional relevant results can be found in [2, 14, 15, 16, 19, 28] and the references therein. Inspired by the works mentioned above, we are interested in the model (1.1), and try to establish the existence and boundedness of classical global solution under the Neumann boundary conditions. 1.1. Statements of main results. In this paper, we will prove that if r1 > 1, r2 > 1, and (r2 − 1)(r1 − 1) ≥ 1, then the solution (u, v, w, z) of (1.1) is global in time and bounded for any N ≥ 1. Theorem 1.1. Let Ω ⊂ RN (N ≥ 1) be bounded domain with smooth boundary. Suppose that χ, ξ, λi, µi are positive constants, ri > 1 (i = 1, 2) and the initial data (u0, v0, w0, z0) fulfills (1.2). If (r2 − 1)(r1 − 1) > 1, for all µi (i = 1, 2) > 0, or (r2 − 1)(r1 − 1) = 1, µ1 and µ2 are appropriately large, then system (1.1)-(1.2) possesses a unique global classical solution u ∈ C0(Ω× [0,∞) ∩ C2,1(Ω× (0,∞)), v ∈ C0(Ω× (0,∞) ∩ C2,1(Ω× (0,∞)), w ∈ C0(Ω× [0,∞) ∩ C2,0(Ω× (0,∞)), z ∈ C0(Ω× (0,∞) ∩ C2,0(Ω× (0,∞)). Moreover, there exists a constant C > 0 such that ∥u(·, t)∥L∞(Ω) + ∥v(·, t)∥L∞(Ω) + ∥w(·, t)∥W 1,∞(Ω) + ∥z(·, t)∥W 1,∞(Ω) ≤ C for all t > 0. Remark 1.2. For the case of r1 = r2 = 2, our condition turns into (r1−1)(r2−1) = 1, for which the global and bounded solution has been obtained by Qi and Ke [13]. In this article, we will use C and Ci (i = 1, 2, · · ·) to represent different positive constants which may vary in the context. Besides, we write u(x, t) as u and ∫ Ω udx as ∫ Ω u for convenience. The rest of this article is arranged as follows. In Section 2, we give some prelim- inary works, such as the local existence of solutions and some classical inequalities. In Section 3, the Lp estimates of u and v are given by using elementary energy method. Finally, the proof of our main result is given in Section 4 by using Neu- mann heat semigroup theory. 2. Preliminaries In this section, we give the following lemmas which will be used in the later proofs. Firstly, we recall the well-known result about the existence of local solutions to model (1.1). Readers can refer to [18] for a detailed proof. Lemma 2.1. Let Ω ⊂ RN (N ≥ 1) be a bounded domain with smooth boundary. Assume that the initial data satisfies (1.2). Then system (1.1)-(1.2) has a unique nonnegative classical solution u ∈ C0(Ω̄× [0, Tmax)) ∩ C2,1(Ω̄× (0, Tmax)), v ∈ C0(Ω̄× [0, Tmax)) ∩ C2,1(Ω̄× (0, Tmax)), w ∈ C0(Ω̄× [0, Tmax)) ∩ C2,1(Ω̄× (0, Tmax)) ∩ L∞ loc((0, Tmax);W 1,ϑ(Ω)), z ∈ C0(Ω̄× [0, Tmax)) ∩ C2,1(Ω̄× (0, Tmax)) ∩ L∞ loc((0, Tmax);W 1,ϑ(Ω)), 4 F. ZHAO, H. TANG, J. ZHENG, K. LI EJDE-2025/01 where ϑ > N and Tmax ∈ (0,+∞] denotes the maximal existence time. Moreover, if Tmax < ∞, then as t ↗ Tmax, we have ∥u(·, t)∥L∞(Ω) + ∥v(·, t)∥L∞(Ω) + ∥w(·, t)∥W 1,∞(Ω) + ∥z(·, t)∥W 1,∞(Ω) → ∞ . In view of Lemma 2.1, we know that for any s0 ∈ (0, Tmax), s0 ≤ 1 and p > 1, for all τ ∈ [0, s0), there exists K > 0 satisfying ∥u(·, τ)∥L∞(Ω) + ∥v(·, τ)∥L∞(Ω) + ∥w(·, τ)∥W 2,p(Ω) + ∥z(·, τ)∥W 2,p(Ω) ≤ K. (2.1) To prove the main theorem, we apply the following lemmas. Lemma 2.2 ([3, 7]). ( Suppose that γ ∈ (1,∞), g ∈ Lγ((0, T );Lγ(Ω)), and c is a solution to the initial boundary value problem ct = ∆c− c+ g, ∂c ∂ν = 0, c(x, 0) = c0(x). Then there exists a positive constant Cγ such that if s0 ∈ [0, T ) and c(·, s0) ∈ W 2,γ(Ω) with ∂c(·,s0) ∂ν = 0, one has∫ T s0 eγs∥∆c(·, s)∥γLγ(Ω) ds ≤ Cγ (∫ T s0 eγs∥g(·, s)∥γLγ(Ω) ds+ eγs0(∥c(·, s0)∥γLγ(Ω) + ∥∆c(·, s0)∥γLγ(Ω)) ) . Lemma 2.3 ([20]). Let (et∆)t ≥ 0 be the Neumann heat semigroup in Ω, λ1 > 0 de- note the first nonzero eigenvalue of −∆ in Ω under Neumann boundary conditions. Then there exists a constant C, if 1 ≤ q ≤ p ≤ ∞, then ∥∇et∆w∥Lp(Ω) ≤ C2(1 + t− n 2 ( 1 q− 1 p ))e−λ1t∥w∥Lq(Ω) ∀t > 0 holds for each w ∈ Lq(Ω). Lemma 2.4 ([25]). Let y(t) ≥ 0 be a solution of problem y′(t) +Ayp ≤ B, t > 0, y(0) = y0, with A > 0, p > 0 and B ≥ 0. Then for any t > 0, we have y(t) ≤ max{y0, ( B A )1/p}. 3. A priori estimates In this section, according to the Lp estimates of the parabolic equations, we obtain the a priori estimates which plays a vital role in proving our main result. The readers can find a proof of the following lemma in Zheng and Zhang [27]. Lemma 3.1. Under the condition of Lemma 2.1, then there exists a nonnegative constant C such that the solution of (1.1) satisfies∫ Ω u+ ∫ Ω v ≤ C ∀t ∈ (0, Tmax). (3.1) EJDE-2025/01 GLOBAL SOLUTION FOR A PREDATOR-PREY MODEL 5 To address the difference betweenthe parabolic equation and the elliptic equation, we use lemma 2.1 and obtain the following conclusion. To prove the main lemma, is based on the different forms of r1 and r2, so we can prove the theorem in two different forms. Lemma 3.2. Suppose that r1 > 1, r2 > 1 and (r1 − 1)(r2 − 1) > 1. Then for any p > 1 and q > 1, there exists a constant C > 0 such that∫ Ω uq + ∫ Ω vp ≤ C ∀t ∈ (0, Tmax). (3.2) Proof. Assume any p > 1. Multiplying the second equation in (1.1) by vp−1 and integrating by parts, we have 1 p d dt ∫ Ω vp + (p− 1) ∫ Ω vp−2|∇v|2 = p− 1 p ξ ∫ Ω vp∆z + λ2 ∫ Ω vp − µ2 ∫ Ω vp+r2−1 − b ∫ Ω vpu ≤ p− 1 p ξ ∫ Ω vp|∆z| − µ2 ∫ Ω vp+r2−1 + λ2 ∫ Ω vp ∀t ∈ (0, Tmax). (3.3) Now, using the Young inequality, for all t ∈ (0, Tmax), there are constants C1 > 0 and C2 > 0 such that (p− 1)ξ p ∫ Ω vp|∆z| ≤ µ2 4 ∫ Ω vp+r2−1 + C1 ∫ Ω |∆z| p+r2−1 r2−1 , (3.4) λ2 ∫ Ω vp ≤ µ2 2 ∫ Ω vp+r2−1 + C2. (3.5) Then combining (3.3), (3.4) with (3.5), we arrive at d dt ∫ Ω vp ≤ −µ2p 4 ∫ Ω vp+r2−1 + C1p ∫ Ω |∆z| p+r2−1 r2−1 + C2p ∀t ∈ (0, Tmax). (3.6) Furthermore, adding p+r2−1 r2−1 ∫ Ω vp at the both sides of (3.6), since r2 > 1 and t ∈ (0, Tmax), we derive that d dt ∫ Ω vp + p+2 −1 r2 − 1 ∫ Ω vp ≤ −µ2p 4 ∫ Ω vp+r2−1 + C1p ∫ Ω |∆z| p+r2−1 r2−1 + p+ r2 − 1 r2 − 1 ∫ Ω vp + C2p. (3.7) Here, we estimate the third term on the right-hand side of (3.7). Then by using the Young inequality, there exists a constant C3 > 0 such that p+ r2 − 1 r2 − 1 ∫ Ω vp ≤ pµ2 8 ∫ Ω vp+r2−1 + C3 ∀t ∈ (0, Tmax). (3.8) Combining (3.7) with (3.8), for all t ∈ (0, Tmax), we easily obtain d dt ∫ Ω vp + p+ r2 − 1 r2 − 1 ∫ Ω vp ≤ −pµ2 8 ∫ Ω vp+r2−1 + C1p ∫ Ω |∆z| p+r2−1 r2−1 + C4 (3.9) 6 F. ZHAO, H. TANG, J. ZHENG, K. LI EJDE-2025/01 with C4 = pC2 + C3 > 0. Consequently, for all t ∈ (0, Tmax), the inequality (3.9) can be rewritten as follows d dt ( e p+r2−1 r2−1 t∥v∥pLp(Ω) ) ≤ ( − µ2p 8 ∫ Ω vp+r2−1 + C1p ∫ Ω |∆z| p+r2−1 r2−1 + C4 ) e p+r2−1 r2−1 t. (3.10) Let s0 be the same as (2.1). Integrating (3.10) on (s0, t), together with Lemma 2.2, (2.1), for all t ∈ (0, Tmax), one has ∥v(·, t)∥pLp(Ω) ≤ e− p+r2−1 r2−1 (t−s0)∥v(·, s0)∥pLp(Ω) − µ2p 8 e− p+r2−1 r2−1 t ∫ t s0 e p+r2−1 r2−1 s ∫ Ω vp+r2−1 dx ds + C1pe − p+r2−1 r2−1 t ∫ t s0 e p+r2−1 r2−1 s ∫ Ω |∆z| p+r2−1 r2−1 dx ds+ C4 ∫ t s0 e− p+r2−1 r2−1 (t−s)ds ≤ −µ2p 8 e− p+r2−1 r2−1 t ∫ t s0 e p+r2−1 r2−1 s ∫ Ω vp+r2−1 dx ds + C1Cγpe − p+r2−1 r2−1 t ∫ t s0 e p+r2−1 r2−1 s∥u(·, s)∥ p+r2−1 r2−1 L p+r2−1 r2−1 (Ω) ds+ C5 + C1Cγpe − p+r2−1 r2−1 (t−s0) ( ∥z(·, s0)∥ p+r2−1 r2−1 L p+r2−1 r2−1 (Ω) + ∥∆z(·, s0)∥ p+r2−1 r2−1 L p+r2−1 r2−1 (Ω) ) ≤ −µ2p 8 e− p+r2−1 r2−1 t ∫ t s0 e p+r2−1 r2−1 s ∫ Ω vp+r2−1 dx ds + C1Cγpe − p+r2−1 r2−1 t ∫ t s0 e p+r2−1 r2−1 s∥u(·, s)∥ p+r2−1 r2−1 L p+r2−1 r2−1 (Ω) ds+ C6 (3.11) with positive constants C5 > 0 and C6 > 0 as well as Cγ is the same as Lemma 2.2. Multiplying the first equation of (1.1) by uq−1 (q > 1) and integrating by parts over Ω, according to the Young inequality, for all t ∈ (0, Tmax), there exist positive constants Ci (i = 7, 8, 9) such that 1 q d dt ∫ Ω uq + (q − 1) ∫ Ω uq−2|∇u|2 = −q − 1 q χ ∫ Ω uq∆w + ∫ Ω uq(λ1 − µ1u r1−1 + av) ≤ −µ1 4 ∫ Ω uq+r1−1 + C8 ∫ Ω |∆w| q+r1−1 r1−1 + C7 ∫ Ω v q+r1−1 r1−1 + C9. (3.12) Similarly, adding q+r1−1 r1−1 ∫ Ω uq at both sides of (3.12) and using the Young inequal- ity again, for all t ∈ (0, Tmax), there is a constant C10 > 0 such that d dt ∫ Ω uq + q + r1 − 1 r1 − 1 ∫ Ω uq ≤ −µ1q 8 ∫ Ω uq+r1−1 + C8q ∫ Ω |∆w| q+r1−1 r1−1 + C7q ∫ Ω v q+r1−1 r1−1 + C10. (3.13) EJDE-2025/01 GLOBAL SOLUTION FOR A PREDATOR-PREY MODEL 7 Next, multiplying e q+r1−1 r1−1 t on both sides of (3.13). For all t ∈ (s0, Tmax), we see that d dt ( e q+r1−1 r1−1 t∥u∥qLq(Ω) ) ≤ ( − µ1q 8 ∫ Ω uq+r1−1 + C7q ∫ Ω v q+r1−1 r1−1 + C8q ∫ Ω |∆w| q+r1−1 r1−1 + C10 ) e q+r1−1 r1−1 t. (3.14) Integrating (3.14) over (s0, t), for all t ∈ (0, Tmax), there exist positive constants C11 and C12 such that ∥u(·, t)∥qLq(Ω) ≤ e− q+r1−1 r1−1 (t−s0)∥u(·, s0)∥qLq(Ω) − µ1q 8 e− q+r1−1 r1−1 t ∫ t s0 e q+r1−1 r1−1 s ∫ Ω uq+r1−1 dx ds + C7qe − q+r1−1 r1−1 t ∫ t s0 e q+r1−1 r1−1 s ∫ Ω v q+r1−1 r1−1 dx ds + C8qe − q+r1−1 r1−1 t ∫ t s0 e q+r1−1 r1−1 s∥∆w(·, s)∥ q+r1−1 r1−1 L q+r1−1 r1−1 (Ω) ds + C10 ∫ t s0 e− q+r1−1 r1−1 (t−s)ds ≤ −µ1q 8 e− q+r1−1 r1−1 t ∫ t s0 e q+r1−1 r1−1 s ∫ Ω uq+r1−1 dx ds + C11qe − q+r1−1 r1−1 t ∫ t s0 e q+r1−1 r1−1 s∥v(·, s)∥ q+r1−1 r1−1 L q+r1−1 r1−1 (Ω) ds+ C12. (3.15) Since r1 > 1, r2 > 1 and (r1 − 1)(r2 − 1) > 1, we can choose the appropriate numbers q and p such that q + r1 − 1 r1 − 1 < p+ r2 − 1, (3.16) p+ r2 − 1 r2 − 1 < q + r1 − 1. (3.17) Combining (3.11) with (3.15), for all p > 1 and q > 1, there exist positive constants C13 and C14 such that ∥u∥qLq(Ω) + ∥v∥pLp(Ω) ≤ −µ2p 8 e− p+r2−1 r2−1 t ∫ t s0 e p+r2−1 r2−1 s ∫ Ω vp+r2−1 dx ds + C1Cγpe − p+r2−1 r2−1 t ∫ t s0 e p+r2−1 r2−1 s∥u∥ p+r2−1 r2−1 L p+r2−1 r2−1 (Ω) ds − µ1q 8 e− q+r1−1 r1−1 t ∫ t s0 e q+r1−1 r1−1 s ∫ Ω uq+r1−1 dx ds + C11qe − q+r1−1 r1−1 t ∫ t s0 e q+r1−1 r1−1 s ∫ Ω v q+r1−1 r1−1 dx ds+ C6 + C12 ≤ −µ2p 16 e− p+r2−1 r2−1 t ∫ t s0 e p+r2−1 r2−1 s ∫ Ω vp+r2−1 dx ds 8 F. ZHAO, H. TANG, J. ZHENG, K. LI EJDE-2025/01 − µ1q 16 e− q+r1−1 r1−1 t ∫ t s0 e q+r1−1 r1−1 s ∫ Ω uq+r1−1 dx ds+ C13 ≤ C14 ∀t ∈ (s0, Tmax). The above estimate together with (2.1) implies Lemma 3.2. □ Lemma 3.3. Suppose that r1 > 1, r2 > 1, if (r1 − 1)(r2 − 1) = 1, and µ1, µ2 are arbitrarily big. Then for each p > 1, there exists C > 0 such that∫ Ω up + ∫ Ω vp ≤ C ∀t ∈ (0, Tmax). (3.18) Proof. Since (r1 − 1)(r2 − 1) = 1, we can choose r > 1 and r̃ > 1 are appropriately big such that r̃ + r1 − 1 = r + r2 − 1 r2 − 1 , r + r2 − 1 = r̃ + r1 − 1 r1 − 1 . (3.19) Multiplying both sides of the second equation in (1.1) by vr−1 and integrating by parts, we have 1 r d dt ∫ Ω vr + (r − 1) ∫ Ω vr−2|∇v|2 = r − 1 r ξ ∫ Ω vr∆z + λ2 ∫ Ω vr − µ2 ∫ Ω vr+r2−1 − b ∫ Ω vru ≤ r − 1 r κ ∫ Ω vr|∆z| − µ2 ∫ Ω vr+r2−1 + λ2 ∫ Ω vr ∀t ∈ (0, Tmax). (3.20) Define κ := max{ξ, χ}. Then, adding r+r2−1 r2−1 ∫ Ω vr at both sides of the above inequality, we have 1 r d dt ∫ Ω vr ≤ r − 1 r κ ∫ Ω vr|∆z| − µ2 ∫ Ω vr+r2−1 + λ2 ∫ Ω vr + r + r2 − 1 r2 − 1 ∫ Ω vr − r + r2 − 1 r2 − 1 ∫ Ω vr = ∫ Ω [ (λ2 + r + r2 − 1 r2 − 1 )vr − µ2v r+r2−1 ] + r − 1 r κ ∫ Ω vr|∆z| − r + r2 − 1 r2 − 1 ∫ Ω vr ∀t ∈ (0, Tmax). (3.21) Along with Young’s inequality, this concludes that∫ Ω (λ2 + r + r2 − 1 r2 − 1 )vr − ∫ Ω µ2v r+r2−1 ≤ (ε2 − µ2) ∫ Ω vr+r2−1 + C1(ε2, r) (3.22) with some constants ε2 > 0 and C1(ε2, r) = r2 − 1 r + r2 − 1 (ε2 r + r2 − 1 r )− r r2−1 (λ2 + r + r2 − 1 r2 − 1 ) r+r2−1 r2−1 |Ω|. EJDE-2025/01 GLOBAL SOLUTION FOR A PREDATOR-PREY MODEL 9 Next, for each λ0 > 0, applying Young’s inequality again, we obtain r − 1 r κ ∫ Ω vr|∆z| ≤ λ0 ∫ Ω vr+r2−1 + r2 − 1 r + r2 − 1 ( λ0 r + r2 − 1 r )− r r2−1 × (r − 1 r κ ) r+r2−1 r2−1 ∫ Ω |∆z| r+r2−1 r2−1 = λ0 ∫ Ω vr+r2−1 +A1λ − r r2−1 0 κ r+r2−1 r2−1 ∫ Ω |∆z| r+r2−1 r2−1 ∀t ∈ (0, Tmax), (3.23) where A1 = r2 − 1 r + r2 − 1 ( r + r2 − 1 r )− r r2−1 ( r − 1 r ) r+r2−1 r2−1 . (3.24) Combining (3.22), (3.23) with (3.21), for all t ∈ (0, Tmax), we easily obtain 1 r d dt ∫ Ω vr ≤ (ε2 + λ0 − µ2) ∫ Ω vr+r2−1 − r + r2 − 1 r2 − 1 ∫ Ω vr +A1λ − r r2−1 0 κ r+r2−1 r2−1 ∫ Ω |∆z| r+r2−1 r2−1 + C1(ε2, r). (3.25) Let s0 be the same as (2.1), then for any t ∈ (s0, Tmax), applying the variation-of- constants formula to the above inequality, we have 1 r ∫ Ω vrdx ≤ 1 r e− r+r2−1 r2−1 (t−s)∥v(·, s0)∥rLr(Ω) + (ε2 + λ0 − µ2) ∫ t s0 e− r+r2−1 r2−1 (t−s) ∫ Ω vr+r2−1 +A1λ − r r2−1 0 κ r+r2−1 r2−1 ∫ t s0 e− r+r2−1 r2−1 (t−s) ∫ Ω |∆z| r+r2−1 r2−1 + C1(ε2, r) ∫ t s0 e− r+r2−1 r2−1 (t−s) ≤ (ε2 + λ0 − µ2) ∫ t s0 e− r+r2−1 r2−1 (t−s) ∫ Ω vr+r2−1 +A1λ − r r2−1 0 κ r+r2−1 r2−1 ∫ t s0 e− r+r2−1 r2−1 (t−s) ∫ Ω |∆z| r+r2−1 r2−1 + C2(ε2, r), (3.26) where C2 := C2(ε2, r) = 1 r ∥v(·, s0)∥rLr(Ω) + C1(ε2, r) r2 − 1 r + r2 − 1 . In view of Lemma 2.2, it follows that A1λ − r r2−1 0 κ r+r2−1 r2−1 ∫ t s0 e− r+r2−1 r2−1 (t−s) ∫ Ω |∆z| r+r2−1 r2−1 = A1λ − r r2−1 0 κ r+r2−1 r2−1 e− r+r2−1 r2−1 t ∫ t s0 e r+r2−1 r2−1 s ∫ Ω |∆z| r+r2−1 r2−1 ≤ A1λ − r r2−1 0 κ r+r2−1 r2−1 e− r+r2−1 r2−1 tCr+1 [ ∫ t s0 ∫ Ω e r+r2−1 r2−1 su r+r2−1 r2−1 + e r+r2−1 r2−1 s0 ( ∥z(·, s0)∥ r+r2−1 r2−1 L r+r2−1 r2−1 (Ω) + ∥∆z(·, s0)∥ r+r2−1 r2−1 L r+r2−1 r2−1 (Ω) )] (3.27) 10 F. ZHAO, H. TANG, J. ZHENG, K. LI EJDE-2025/01 with Cr+1 > 0 for all t ∈ (s0, Tmax). Substituting (3.27) into (3.26), for all t ∈ (s0, Tmax), we obtain 1 r ∫ Ω vr ≤ (ε2 + λ0 − µ2) ∫ t s0 e− r+r2−1 r2−1 (t−s) ∫ Ω vr+r2−1 +A1λ − r r2−1 0 κ r+r2−1 r2−1 Cr+1 ∫ t s0 e− r+r2−1 r2−1 (t−s) ∫ Ω u r+r2−1 r2−1 +A1λ − r r2−1 0 κ r+r2−1 r2−1 e− r+r2−1 r2−1 (t−s0)Cr+1M̃ + C2(ε2, r), (3.28) where M̃ = ∥z(·, s0)∥ r+r2−1 r2−1 L r+r2−1 r2−1 (Ω) + ∥∆z(·, s0)∥ r+r2−1 r2−1 L r+r2−1 r2−1 (Ω) . To deal with u, multiplying both sides of the second equation in (1.1) by ur̃−1 and integrating by parts, for any small ε ∈ (0, 1), we derive from Young’s inequality that 1 r̃ d dt ∫ Ω ur̃ + (r̃ − 1) ∫ Ω ur̃−2|∇u|2 = − r̃ − 1 r̃ χ ∫ Ω ur̃∆w + λ1 ∫ Ω ur̃ − µ1 ∫ Ω ur̃+r1−1 + a ∫ Ω ur̃v ≤ r̃ − 1 r̃ κ ∫ Ω ur̃|∆w|+ λ1 ∫ Ω ur̃ − µ1 ∫ Ω ur̃+r1−1 + ε ∫ Ω ur̃+r1−1 + C3 ∫ Ω v r̃+r1−1 r1−1 ∀t ∈ (0, Tmax), (3.29) where C3 = r̃ + r1 − 1 r1 − 1 (ε r̃ + r1 − 1 r̃ )− r̃ r1−1 a r̃+r1−1 r1−1 . Then we conclude that 1 r̃ d dt ∫ Ω ur̃ ≤ − r̃ + r1 − 1 r1 − 1 ∫ Ω ur̃ + r̃ − 1 r̃ κ ∫ Ω ur̃|∆w| + ∫ Ω ( λ1u r̃ + r̃ + r1 − 1 r1 − 1 ur̃ − µ1u r̃+r1−1 ) + ε ∫ Ω ur̃+r1−1 + C3 ∫ Ω v r̃+r1−1 r1−1 ∀t ∈ (0, Tmax). (3.30) For each ε1 ∈ (0, 1), using Young’s inequality again, we have∫ Ω ( λ1u r̃ + r̃ + r1 − 1 r1 − 1 ur̃ − µ1u r̃+r1−1 ) ≤ (ε1 − µ1) ∫ Ω ur̃+r1−1 + C4(ε1, r̃) ∀t ∈ (0, Tmax) (3.31) with C4 := C4(ε1, r̃) = r1 − 1 r̃ + r1 − 1 (ε1 r̃ + r1 − 1 r̃ )− r̃ r1−1 (λ1 + r̃ + r1 − 1 r1 − 1 ) r̃+r1−1 r1−1 |Ω|. EJDE-2025/01 GLOBAL SOLUTION FOR A PREDATOR-PREY MODEL 11 By applying Young’s inequality again, we obtain r̃ − 1 r̃ κ ∫ Ω ur̃|∆w| ≤ λ0 ∫ Ω ur̃+r1−1 + r1 − 1 r̃ + r1 − 1 ( λ0 r̃ + r1 − 1 r̃ )− r̃ r1−1 × ( r̃ − 1 r̃ κ ) r̃+r1−1 r1−1 ∫ Ω |∆w| r̃+r1−1 r1−1 = λ0 ∫ Ω ur̃+r1−1 + Ã1λ − r̃ r1−1 0 κ r̃+r1−1 r1−1 ∫ Ω |∆w| r̃+r1−1 r1−1 ∀t ∈ (0, Tmax), (3.32) where Ã1 = r1 − 1 r̃ + r1 − 1 ( r̃ + r1 − 1 r̃ )− r̃ r1−1 ( r̃ − 1 r̃ ) r̃+r1−1 r1−1 . (3.33) Combing (3.31), (3.32) with (3.30), for all t ∈ (0, Tmax), it follows that 1 r̃ d dt ∫ Ω ur̃ ≤ (ε1 + λ0 − µ1) ∫ Ω ur̃+r1−1 + Ã1λ − r̃ r1−1 0 κ r̃+r1−1 r1−1 ∫ Ω |∆w| r̃+r1−1 r1−1 + ε ∫ Ω ur̃+r1−1 + C3 ∫ Ω v r̃+r1−1 r1−1 + C4(ε1, r̃). (3.34) Applying the variation-of-constants formula to the above inequality again, for all t ∈ (0, Tmax), there exist C5 > 0 and C̃r̃+1 > 0 such that 1 r̃ ∫ Ω ur̃ ≤ (ε1 + λ0 − µ1) ∫ t s0 e− r̃+r1−1 r1−1 (t−s) ∫ Ω ur̃+r1−1 + Ã1λ − r̃ r1−1 0 κ r̃+r1−1 r1−1 C̃r̃+1 ∫ t s0 e− r̃+r1−1 r1−1 (t−s) ∫ Ω v r̃+r1−1 r1−1 + Ã1λ − r̃ r1−1 0 κ r̃+r1−1 r1−1 e− r̃+r1−1 r1−1 (t−s0)C̃r̃+1Ñ + ε ∫ t s0 e− r̃+r1−1 r1−1 (t−s) ∫ Ω ur̃+r1−1 + C3 ∫ t s0 e− r̃+r1−1 r1−1 (t−s) ∫ Ω v r̃+r1−1 r1−1 + C5 ≤ (ε1 + ε+ λ0 − µ1) ∫ t s0 e− r̃+r1−1 r1−1 (t−s) ∫ Ω ur̃+r1−1 + Ã1λ − r̃ r1−1 0 κ r̃+r1−1 r1−1 C̃r̃+1 ∫ t s0 e− r̃+r1−1 r1−1 (t−s) ∫ Ω v r̃+r1−1 r1−1 + Ã1λ − r̃ r1−1 0 κ r̃+r1−1 r1−1 e− r̃+r1−1 r1−1 (t−s0)C̃r̃+1Ñ + C5 (3.35) with Ñ = ∥w(·, s0)∥ r̃+r1−1 r1−1 L r̃+r1−1 r1−1 (Ω) + ∥∆w(·, s0)∥ r̃+r1−1 r1−1 L r̃+r1−1 r1−1 (Ω) . 12 F. ZHAO, H. TANG, J. ZHENG, K. LI EJDE-2025/01 Adding (3.28) to (3.35) yields 1 r̃ ∫ Ω ur̃ + 1 r ∫ Ω vr ≤ (ε1 + ε+ λ0 − µ1) ∫ t s0 e− r̃+r1−1 r1−1 (t−s) ∫ Ω ur̃+r1−1 + Ã1λ − r̃ r1−1 0 κ r̃+r1−1 r1−1 C̃r̃+1 ∫ t s0 e− r̃+r1−1 r1−1 (t−s) ∫ Ω v r̃+r1−1 r1−1 + (ε2 + λ0 − µ2) ∫ t s0 e− r+r2−1 r2−1 (t−s) ∫ Ω vr+r2−1 +A1λ − r r2−1 0 κ r+r2−1 r2−1 Cr+1 ∫ t s0 e− r+r2−1 r2−1 (t−s) ∫ Ω u r+r2−1 r2−1 + Ã1λ − r̃ r1−1 0 κ r̃+r1−1 r1−1 e− r̃+r1−1 r1−1 (t−s0)C̃r̃+1Ñ +A1λ − r r2−1 0 κ r+r2−1 r2−1 e− r+r2−1 r2−1 (t−s0)Cr+1M̃ + C6 (3.36) with C6 > 0 and for all t ∈ (0, Tmax). Recalling (3.19), we can rewrite (3.36) as 1 r̃ ∫ Ω ur̃ + 1 r ∫ Ω vr ≤ (ε1 + ε+ λ0 +A1λ − r r2−1 0 κ r+r2−1 r2−1 Cr+1 − µ1) ∫ t s0 e− r̃+r1−1 r1−1 (t−s) ∫ Ω ur̃+r1−1 + (ε2 + λ0 + Ã1λ − r̃ r1−1 0 κ r̃+r1−1 r1−1 C̃r̃+1 − µ2) ∫ t s0 e− r+r2−1 r2−1 (t−s) ∫ Ω vr+r2−1 + Ã1λ − r̃ r1−1 0 κ r̃+r1−1 r1−1 e− r̃+r1−1 r1−1 (t−s0)C̃r̃+1Ñ +A1λ − r r2−1 0 κ r+r2−1 r2−1 e− r+r2−1 r2−1 (t−s0)Cr+1M̃ + C6 ∀t ∈ (s0, Tmax). Moreover, we define g1(λ0, r) = λ0 +max{A1κ r+r2−1 r2−1 Cr+1, Ã1κ r̃+r1−1 r1−1 C̃r̃+1}[λ − r r2−1 0 + λ − r̃ r1−1 0 ]. After performing some basic calculations, we can obtain there is η0 > 0 such that g1(η0, r) := min λ0>0 g1(λ0, r). For the above λ0 := η0, we choose µ1 and µ2 is appropriately large such that λ0 +A1λ − N 2 r2−1 0 κ N 2 +r2−1 r2−1 CN 2 +1 < µ1, λ0 + Ã1λ − N 2 r1−1 0 κ N 2 +r1−1 r1−1 C̃N 2 +1 < µ2. Therefore, one can pick r > N 2 and r̃ > N 2 and ε1, ε, ε2 small enough such that ε1 + ε+ λ0 +A1λ − r r2−1 0 κ r+r2−1 r2−1 Cr+1 < µ1, ε2 + λ0 + Ã1λ − r̃ r1−1 0 κ r̃+r1−1 r1−1 C̃r̃+1 < µ2. Then in light of (3.36), there exist positive constants C7 and C8 such that∫ Ω ur̃ ≤ C7 ∀t ∈ (s0, Tmax), (3.37) EJDE-2025/01 GLOBAL SOLUTION FOR A PREDATOR-PREY MODEL 13∫ Ω vr ≤ C8 ∀t ∈ (s0, Tmax). (3.38) Let q0 = min{r, r̃}, since r > N 2 , r̃ > N 2 , then we can get q0 > N 2 . Now, we fix q < Nq0 (N−q0)+ and choose some α > 1 2 such that q < 1 1 q0 − 1 N + 2 N (α− 1 2 ) ≤ Nq0 (N − q0)+ . (3.39) Applying the variation-of-constants formula to w, we obtain w(·, t) = e−(t−s0)(A+1)w(·, s0) + ∫ t s0 e−(t−s)(A+1)v(·, s)ds (3.40) for all t ∈ (s0, Tmax), where A := Ap denotes the sectorial operator defined by Apw := −∆w for all w ∈ D(Ap) := {φ ∈ W 2,p(Ω)|∂φ ∂ν |∂Ω=0}. According to (2.1) and (3.40), there is a positive constant C9 such that ∥(A+ 1)αw(·, t)∥Lq(Ω) ≤ C9 ∫ t s0 (t− s)−α−N 2 ( 1 q0 − 1 q )e−λ1(t−s)∥v(·, s)∥Lq0 (Ω)ds + C9s −α−N 2 (1− 1 q ) 0 ∥w(·, s0)∥L1(Ω)ds ≤ C8C9 ∫ ∞ s0 δ−α−N 2 ( 1 q0 − 1 q )e−λ1δdδ + C9s −α−N 2 (1− 1 q ) 0 K, (3.41) where s0 is same as the parameter in (2.1). From (3.39) and (3.41), for a positive constant C10, we have∫ Ω |∇w|q ≤ C10 ∀t ∈ (0, Tmax) and q ∈ [1, Nq0 (N − q0)+ ). (3.42) Similarly, for C11 > 0, we obtain∫ Ω |∇z|q ≤ C11 ∀t ∈ (0, Tmax) and q ∈ [1, Nq0 (N − q0)+ ). (3.43) Multiplying the second equation of (1.1) by vp−1 and using Young’s inequality, for all t ∈ (0, Tmax), we have 1 p d dt ∫ Ω vp + (p− 1) ∫ Ω vp−2|∇v|2 = −(p− 1)ξ ∫ Ω vp−1∇v · ∇z + λ2 ∫ Ω vp − µ2 ∫ Ω vp+r2−1 − b ∫ Ω vpu ≤ p− 1 2 ∫ Ω vp−2|∇v|2 + ξ2(p− 1) 2 ∫ Ω vp|∇z|2 − µ2 ∫ Ω vp+r2−1 + λ2 ∫ Ω vp ≤ p− 1 2 ∫ Ω vp−2|∇v|2 + ξ2(p− 1) 2 ∫ Ω vp|∇z|2 − µ2 2 ∫ Ω vp+r2−1 + C12 (3.44) 14 F. ZHAO, H. TANG, J. ZHENG, K. LI EJDE-2025/01 with a constant C12 > 0. Since q0 > N 2 implies q0 < Nq0 2(N−q0)+ , there is a constant C13 > 0 such that ξ2(p− 1) 2 ∫ Ω vp|∇z|2 ≤ ξ2(p− 1) 2 (∫ Ω v q0 q0−1p ) q0−1 q0 (∫ Ω |∇z|2q0 )1/q0 ≤ C13∥v p 2 ∥2 L 2q0 q0−1 (Ω) ∀t ∈ (0, Tmax). (3.45) Let p > q0 − 1 and it follows that q0 p < q0 q0 − 1 < N (N − 2)+ . Together with the Gagliardo-Nirenberg inequality and (3.38) implies that C13∥v p 2 ∥2 L 2q0 q0−1 (Ω) ≤ C14(∥∇v p 2 ∥2θL2(Ω)∥v p 2 ∥2(1−θ) L 2q0 p (Ω) + ∥v p 2 ∥2 L 2q0 p (Ω) ) ≤ C15(∥∇v p 2 ∥2L2(Ω)∥+ 1) ∀t ∈ (0, Tmax) (3.46) with some C14 > 0, C15 > 0 and θ = Np 2q0 − N(q0−1) 2q0 1 + Np 2q0 − N 2 ∈ (0, 1). In summary, there is a C16 > 0 such that ξ2(p− 1) 2 ∫ Ω vp|∇z|2 ≤ p− 1 4 ∫ Ω vp−2|∇v|2 + C16 ∀t ∈ (0, Tmax). (3.47) Combining (3.47) with (3.44), for some C17 > 0 and for all t ∈ (0, Tmax), we have 1 p d dt ∫ Ω vp + (p− 1) 4 ∫ Ω vp−2|∇v|2 + µ2 2 ∫ Ω vp+r2−1 ≤ C17. (3.48) According to a standard ODE comparison argument, it implies ∥v(·, t)∥Lp(Ω) ≤ C18 ∀p > 1 and t ∈ (0, Tmax) (3.49) with C18 > 0. As for u, the only difference is the term av in the first equation of (1.1). Recalling the Young inequality and (3.49), there exist positive constants C19 and C20 such that a ∫ Ω upv ≤ µ1 4 ∫ Ω up+r1−1 + C19 ∫ Ω v p+r1−1 r1−1 ≤ µ1 4 ∫ Ω up+r1−1 + C20 ∀t ∈ (0, Tmax). (3.50) We repeat (3.44)-(3.48) and Lemma 2.4, for a constant C21 > 0, we conclude that ∥u(·, t)∥Lp(Ω) ≤ C21 ∀p ≥ 1 and t ∈ (0, Tmax). (3.51) This completes the proof. □ EJDE-2025/01 GLOBAL SOLUTION FOR A PREDATOR-PREY MODEL 15 4. Proof of main result Now a standard procedure enables us to prove the finial step from L1(Ω) to L∞(Ω). Next, we will give the proof of the existence and boundedness of global solutions to system (1.1) by using Neumann heat semigroup theory. Lemma 4.1. Let (u, v, w, z) be a classical nonnegative solution of system (1.1) in Ω× (0, Tmax). Then we can conclude that there exists C > 0 such that sup t∈(0,Tmax) ( ∥u(·, t)∥L∞(Ω) + ∥v(·, t)∥L∞(Ω) + ∥w(·, t)∥W 1,∞(Ω) + ∥z(·, t)∥W 1,∞(Ω) ) ≤ C. Proof. According to the third equation in (1.1) and an associated variation-of- constants formula, we can represent the formula of w ∇w(·, t) = ∇et(∆−1)w(·, s0) +∇ ∫ t 0 e(t−s)(∆−1)v(·, s). (4.1) Then using Lemma 2.4 and − 1 2 − N 2 ( 1 2N − 1 ∞ ) > −1, we have ∥∇w(·, t)∥L∞ (Ω) ≤ ∥∇et(∆−1)w(·, s0)∥L∞ (Ω) + ∫ t 0 ∥∇e(t−s)(∆−1)v(·, s)∥L∞ (Ω) ≤ C1 + ∫ t 0 (1 + (t− s)− 1 2− N 2 ( 1 2N − 1 ∞ ))∥v(·, s)∥L2N (Ω) ≤ C1 + ∫ +∞ 0 (1 + σ− 1 2− N 2 ( 1 2N − 1 ∞ ))∥v(·, s)∥L2N (Ω)) ≤ C2 ∀t ∈ (0, Tmax) (4.2) with some constants C1 > 0 and C2 > 0. This implies that sup t∈(0,Tmax) ∥w(·, t)∥W 1,∞(Ω) ≤ C3 (4.3) with C3 > 0. Similarly, there exists a positive constant C4 such that sup t∈(0,Tmax) ∥z(·, t)∥W 1,∞(Ω) ≤ C4. (4.4) To derive the L∞ estimates of u and v, for convenience, we use the corresponding general result from Tao and Winkler [17] rather than the standard Moser-type iteration technique. According to the [17, Lemma A.1], the first equation in the system (1.1) can be rewritten as ut = ∇ · (D(x, t, u)∇u)− χ∇ · f(x, t) + g(x, t), x ∈ Ω, t > 0, (4.5) where D(x, t, u) = 1, f(x, t) = u∇w and g(x, t) = u(λ1−µ1u r1−1+av). Combining the Lp estimate of u, (4.3), (4.4) and [17, Lemma A.1], we can obtain the L∞ estimate of u directly, i.e., there exists a constant C5 > 0 such that ∥u(·, t)∥L∞(Ω) ≤ C5. (4.6) We can also obtain the L∞ estimate of v in the similar way ∥v(·, t)∥L∞(Ω) ≤ C6 (4.7) 16 F. ZHAO, H. TANG, J. ZHENG, K. LI EJDE-2025/01 with a constant C6 > 0. Thus, the proof is complete. □ 5. Proof of Theorem 1.1 On the basis of the extensibility criterion in Lemma 2.1, we can assert Tmax = ∞ according to the estimates of Lemmas 3.2 and Lemma 4.1. Thus the solution (u, v, w, z) of the model (1.1)-(1.2) is global-in-time and bounded. Acknowledgments. The authors were partially supported by National Natural Science Foundation of China (No. 12101534) and by the Shandong Provincial Natural Science Foundation (No. ZR2022JQ06 and No. ZR2021QA052). References [1] P. Amorim, B. Telch, L. Villada; A reaction-diffusion predator-prey model with pursuit, evasion, and nonlocal sensing, Math. Biosci. 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Fengxiang Zhao School of Mathematical and Informational Sciences, Yantai University, Yantai, 264005, Shandong, China Email address: zfx2037@163.com Haotian Tang Department of Mathematics, Faculty of Science and Technology, University of Macau, Taipa, Macau, China Email address: haotiantang2022@163.com Jiashan Zheng School of Mathematical and Informational Sciences, Yantai University, Yantai, 264005, Shandong, China Email address: zhengjiashan2008@163.com Kaiqiang Li (corresponding author) School of Mathematical and Informational Sciences, Yantai University, Yantai, 264005, Shandong, China Email address: kaiqiangli19@163.com 1. Introduction 1.1. Statements of main results 2. Preliminaries 3. A priori estimates 4. Proof of main result 5. Proof of Theorem ?? Acknowledgments References