Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2013, Article ID 851028, 8 pages http://dx.doi.org/10.1155/2013/851028 Research Article Positive Steady States of a Strongly Coupled Predator-Prey System with Holling-(π + 1) Functional Response Xiao-zhou Feng1,2 and Zhi-guo Wang2 1 2 College of Science, Xi’an Technological University, Xi’an 710032, China Institute of Mathematics, Shaanxi Normal University, Xi’an 710062, China Correspondence should be addressed to Xiao-zhou Feng; flxzfxz8@163.com Received 25 December 2012; Accepted 2 March 2013 Academic Editor: Junjie Wei Copyright © 2013 X.-z. Feng and Z.-g. Wang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. This paper discusses a predator-prey system with Holling-(π + 1) functional response and the fractional type nonlinear diffusion term in a bounded domain under homogeneous Neumann boundary condition. The existence and nonexistence results concerning nonconstant positive steady states of the system were obtained. In particular, we prove that the positive constant solution (Μ π’, ΜV) is asymptotically stable when the parameter k satisfies some conditions. 1. Introduction In this paper, we are interested in the positive steady states of the strongly coupled predator-prey system with Holling(π + 1) functional response. The specific system is as follows: −π1 Δπ’ = π’ (1 − −π2 Δ (V + π’ π’π V )− π π + π’π π3 V ππ’π V ) = −πV + 1+π’ π + π’π ππ’ πV = =0 π] π] in Ω, in Ω, (1) on πΩ, where 1 ≤ π < +∞; Ω is a bounded domain in π π with smooth boundary πΩ; π/π] is the outward directional derivative normal to πΩ; π’ and V stand for the densities of the prey and predator; the given coefficients ππ (π = 1, 2), π, π, π, and π are positive constants. The term π’π /(π + π’π ) is named Holling-(π + 1) functional response [1, 2]. In the second equation, the fractional type nonlinear diffusion term Δπ2 (π3 V/(1 + π’)) models a situation in which the population pressure of the predator species weakens in high-density areas of the prey species. For more precise details, we can refer to [3, 4]. Paper [3] discusses a strongly coupled predatorprey system with nonmonotonic functional response, the existence and nonexistence results concerning nonconstant positive steady states of the system were proved by degree theory. Paper [4] considers the positive steady states for a prey-predator model with some nonlinear diffusion terms, and the sufficient conditions for the existence of positive steady state solutions were obtained by bifurcation theory. In recent years, there has been considerable interest in the dynamics of strongly coupled reaction-diffusion systems with cross-diffusion. We point out that most efforts have concentrated on the Lotka-Volterra competition system which was proposed first by Shigesada et al. [5]. Since their pioneering work, many authors have studied population models with cross-diffusion terms from various mathematical viewpoints, for example, the global existence of time-depending solutions [6–11], the stability analysis for steady states [12–14], and the steady state problems [15–21]. In this paper, we mainly consider the existence of solutions of (1). The research method refers to [3, 22, 23]. For convenience of the research, we write (1) as the following form: −ΔΦ (π) = πΊ (π) , π’ π = [ ], V (2) 2 Journal of Applied Mathematics where boundary condition. Then, there exists a positive constant πΆ = πΆ(Ω, βπβ∞ ) such that maxΩ π ≤ πΆminΩ π. Maximum Principle (see [25]). Suppose that π ∈ πΆ(Ω × π 1 ). If π€ ∈ πΆ2 (Ω) ∩ πΆ1 (Ω) satisfies π1 π’ π (π) [ ] Φ (π) = [ 1 ]=[ π3 ] , π2 (π) π V (1 + ) [ 2 1+π’ ] πΊ (π) = [ π (π’) = π’ (1 − π’ ), π Δπ€ (π₯) + π (π₯, π€ (π₯)) ≥ 0 in Ω, (3) ππ€ ≤0 π] π (π’) − Vπ (π’) ], −πV + πVπ (π’) π (π’) = π’π , π + π’π π ∈ π+ . (4) The main work of this paper is to study the effects of the fractional type nonlinear diffusion pressures on the existence of nonconstant positive steady states of (1). Here, a positive solution means a smooth solution (π’, V) with both π’ and V being positive. We will demonstrate that the cross-diffusion pressure π3 may help forming more patterns. Obviously, for system (1), one notes that when π ≤ π, there holds supπ ≥0 {−π+ ππ π /(π + π π )} ≤ 0, so that the only nonnegative solutions to (1) are π = (0, 0) and π = (π, 0). Consequently, (1) does not have any positive solution. On the other hand, when π > π, the unique positive constant solution to (1) is (Μ π’, ΜV); that is, and π€(π₯0 ) = maxΩ π€, then π(π₯0 , π€(π₯0 )) ≥ 0. Theorem 1. Let π = (π, π, π, π) ∈ (0, ∞)4 and π, πΜ > 0 be fixed constants. Assume that π1 , π2 ≥ π and 0 ≤ π3 ≤ πΜ Μ Ω), such that any there exists a positive constant πΆ = πΆ(π, π, π, positive solution (π’, V) of (1) satisfies max π’ (π₯) + max V (π₯) + Ω Ω maxΩ π’ (π₯) maxΩ V (π₯) + ≤ πΆ. minΩ π’ (π₯) minΩ V (π₯) (7) Proof. Let π1 (π) and π2 (π) be defined as in (3), and denote that π1 (π) = π’ (1 − π’Μ = √ π ππ , π−π (1 − π’Μ/π) (π + π’Μπ ) (π − π’Μ) πΜ π’ ΜV = = , ππ π’Μπ−1 π2 (π) = V ( (5) if π > π’Μ. The organization of our paper is as follows. In Section 2, we establish a priori upper and lower bounds for positive solutions of (1). In Section 3, we use a degree theory to develop a general result to enable one to conclude the existence or nonexistence of nonconstant steady-state solutions or patterns as the index of positive constant steady states changes. In Section 4, we establish the existence of nonconstant positive solutions to (1) for a large range of diffusion and cross-diffusion coefficients. Meanwhile, we prove that the positive constant solution (Μ π’, ΜV) is asymptotically stable for different ranges of parameters. 2. Upper and Lower Bounds for Positive Solutions The main purpose of this section is to give a priori upper and lower positive bounds for positive solutions of (1). Firstly, we cite two known results. Harnack Inequality (see [24]). Let π ∈ πΆ(Ω) and π ∈ πΆ2 (Ω) β πΆ1 (Ω) be a positive classical solution to Δπ(π₯) + π(π₯)π(π₯) = 0 in Ω subject to the homogeneous Neumann (6) on πΩ, π’ π’π−1 V ), − π π + π’π ππ’π − π) , π + π’π (8) π πΊ (π) = (π1 (π) , π2 (π)) . Then, (1) becomes −ΔΦ (π) = πΊ (π) in Ω, ππ =0 π] on πΩ. (9) For the first equation of (9), by the Maximum Principle, we have max π’ (π₯) ≤ Ω 1 max π (π) < π. π1 Ω 1 (10) The function π2 (π) satisfies Δπ2 (π) + π (π₯) π2 (π) = 0 in Ω, ππ2 (π) =0 π] on πΩ, (11) where π(π₯) = π2 (π)/π2 (π) ∈ πΆ(Ω). It is easy to verify that the norm σ΅©σ΅© V (ππ’π / (π + π’π ) − π) σ΅©σ΅© πππ / (π + ππ ) + π σ΅© σ΅©σ΅© σ΅©σ΅© ≤ βπβ∞ = σ΅©σ΅©σ΅©σ΅© π2 σ΅©σ΅© π2 (V + π3 V/ (1 + π’)) σ΅©σ΅©σ΅©∞ πππ + π (π + ππ ) ; ≤ π (π + ππ ) (12) Journal of Applied Mathematics 3 then, βπβ∞ is bounded by a constant depending only on π and π (π2 ≥ π). By the Harnack inequality, we have Proof. Since ∫Ω π2 (π)ππ₯ = 0, there exists π₯1 ∈ Ω, such that π2 (π(π₯1 )) = 0; that is, max π2 (π (π₯)) ≤ πΆ1 min π2 (π (π₯)) , ππ’π (π₯1 ) = ππ + ππ’π (π₯1 ) . Ω Ω (13) It follows that maxΩ π’ ≥ √π aπ/π. Consequently, by maxΩ π’/minΩ π’ ≤ πΆ in Theorem 1, we have where πΆ1 = πΆ1 (Ω, βπβ∞ ) = πΆ1 (π, π, Ω). It follows that −1 maxΩ V maxΩ π2 (π (π₯)) maxΩ {1 + π3 (1 + π’) } × ≤ minΩ V minΩ π2 (π (π₯)) minΩ {1 + π3 (1 + π’)−1 } Μ . ≤ πΆ1 (1 + π3 ) ≤ πΆ1 (1 + π) (14) Integrating the equations of (1) over Ω by parts and making use of the boundary conditions, we have π ∫ π’ (1 − Ω π’ ) ππ₯ = π ∫ V ππ₯. π Ω π π’ π ∫ π’ (1 − ) ππ₯ < π. π π |Ω| π Ω min π’ ≥ Ω (16) π’π σ³¨→ π’, −Δπ’π = Μ π π. max V ≤ πΆ1 (1 + π) π Ω βπβ∞ σ΅© σ΅© π’ π’π−1 V σ΅©σ΅©σ΅© 1 σ΅©σ΅©σ΅© σ΅©σ΅©1 − − σ΅©σ΅© π1 σ΅©σ΅©σ΅© π π + π’π σ΅©σ΅©σ΅©∞ 1 1 ≤ [1 + βπ’ (π₯)β∞ + πβV (π₯)β∞ ] , π1 π (18) maxΩ π’ maxΩ π1 (π (π₯)) ≤ πΆ2 . = minΩ π’ minΩ π1 (π (π₯)) (19) Thus, along with (10), (14), and (17), we can complete the proof. Theorem 2. Assume that πππ =ΜΈ π(π + ππ ). Let π, πΜ > 0 be fixed Μ π, Ω) > 0, such constants. There exists a constant πΆ = πΆ(π, π, that any positive solution (π’, V) of (1) satisfies min π’ (π₯) > πΆ, Ω min V (π₯) > πΆ, Ω Μ provided that π1 , π2 ≥ π and π3 ≤ π. (20) as π σ³¨→ ∞. (23) π’π−1 V π’ 1 (1 − π − π ππ ) π’π π1,π π π + π’π in Ω, (24) on πΩ. Μ π, Ω) such that According to (7), there exists πΆ = πΆ(π, π, βπ’π β∞ ≤ πΆ, β(1 − π’π /π) − (π’ππ−1 Vπ /(π + π’ππ ))/π1,π (1 + π3,π Vπ )β∞ ≤ πΆ, where πΆ is a positive constant which does not depend on π. For each π given in problem (24), it follows from πΏπ estimate that βπ’π βπ2,π (Ω) ≤ πΆ2 |Ω|1/π , where π > 1. Let π > π; then, by Sobolev Imbedding Theorems, we get σ΅©σ΅© σ΅©σ΅© ∗σ΅© σ΅© ∗ 2 1/π σ΅©σ΅©π’π σ΅©σ΅©πΆ1,πΌ (Ω) ≤ πΆ σ΅©σ΅©σ΅©π’π σ΅©σ΅©σ΅©π2,π (Ω) ≤ πΆ πΆ |Ω| , (π−1)/π where π = [(π − 1)π] /ππ. Then, βπβ∞ is bounded by a constant depending only on π and π (π1 ≥ π). By the Harnack Inequality, there exists a positive constant πΆ2 = Μ Ω) such that max π (π(π₯)) ≤ πΆ min π (π(π₯)). πΆ2 (π, π, π, 2 Ω 1 Ω 1 Then, Ω ππ’π =0 π] (17) Similarly, consider the equation of π1 as follows: minVπ (π₯) σ³¨→ 0, Μ and By (14), we have maxΩ Vπ (π₯)/minΩ Vπ (π₯) ≤ πΆ1 (1 + π), then, maxΩ Vπ (π₯) → 0; furthermore, Vπ (π₯) → 0 uniformly holds as π → ∞. Then, the first equation of (1) becomes the following: Along with (14), we have = √π ππ/π 1 Μ π, Ω) , (22) max π’ ≥ := πΆ3 = πΆ3 (π, π, πΆ Ω πΆ where π3 ≤ πΜ and π1 , π2 ≥ π. In the following, we mainly prove that minΩ V(π₯) > πΆ as πππ =ΜΈ π(π + ππ ). Now, we suppose that claim is not true; then, there exists a sequence (π1,π , π2,π , π3,π ) → (π1 , π2 , π3 ) Μ And the positive solution with π1,π , π2,π ≥ π and π3,π ≤ π. (π’π , Vπ ) of (1) corresponding to (π1,π , π2,π , π3,π ) is such that (15) For any π ∈ πΏ1 (Ω), we denote that π = (1/|Ω|) ∫Ω π(π₯)ππ₯. Then, from (15) and (10), we have V= (21) (25) where πΌ ∈ (0, 1). We choose πΌ < 1−π/π such that Imbedding is compact, and along with elliptic equation regularity theory, there exists the subsequence of {π’π }, which is still denoted by {π’π }, and exists π’ such that π’π → π’ uniformly holds in πΆ2,πΌ (Ω). On the other hand, Vπ → 0, and when π1,π → π1 ∈ [π, ∞), the limit of (24) becomes the following problem: −Δπ’ = π’ 1 π’ (1 − ) π1 π in Ω, ππ’ =0 π] on πΩ. (26) Applying Maximum Principle to problem (26) and noting that minΩ π’π ≥ πΆ3 , we have π’ = π. However, when π1,π → ∞, the limit of (24) becomes the following problem: −Δπ’ = 0 in Ω, ππ’ =0 π] on πΩ, (27) Μ for some nonnegative constant πΆ. Μ which implies that π’ = πΆ π−1 π Since ∫Ω π’π (1−π’π /π−π’π Vπ /(π+π’π ))ππ₯ = 0, by letting π → ∞ 4 Journal of Applied Mathematics Μ we also and noting that minΩ π’π ≥ πΆ3 , Vπ → 0, and π’π → πΆ, have π’ = π. By a similar argument as that in (24), for the second equation of (1), we can prove that there exists a subsequence in πΆ2,πΌ (Ω), such that Vπ (π₯)/maxΩ Vπ (π₯) → V0 . Since Μ then Vπ /maxΩ Vπ (π₯) ≥ minΩ Vπ (π₯)/maxΩ Vπ (π₯) ≥ 1/πΆ(1 + π), Μ V0 ≥ 1/πΆ(1 + π). Dividing the second equation of (1) by maxΩ Vπ (π₯), and integrating over Ω, we have ππ’ππ Vπ ( − π) ππ₯ = 0. maxΩ Vπ (π₯) π + π’ππ ∫ Ω (28) Let π → ∞, and note that Vπ /maxΩ Vπ (π₯) → V0 and π’π → π; then, πππ /(π + ππ ) = π. This contradiction to the assumption completes the proof. 3. A Result on Degree Theory In this section, we obtain nonexistence of nonconstant positive solutions to (1) as π3 = 0. Meanwhile, by degree theory, a general result to establish the existence of nonconstant positive solutions to (1) in the next section is proved. Denote d = (π1 , π2 , π3 ) and π = (π, π, π, π). We will fix π ∈ (0, ∞)4 and take d ∈ (0, ∞)2 × [0, ∞) as bifurcation parameters, the dependence of π will often be suppressed. Define ππ π = {π ∈ [πΆ (Ω)] | = 0 on πΩ} , π] 2 2 π+ = {(π’, V) ∈ π | π’ > 0, V > 0 on Ω} , π΅ (πΆ) = {(π’, V) ∈ π | does not depend on d if πΆ > πΆ0 (d, π), and it also does not depend on π, if π changes continuously in (0, ∞)4 without touching the surface π = ππ/ππ + π. For the case π > ππ/ππ + π, by degree invariance, we need only consider a special d; say d = (π, π, 0) with large π. For this we can use the following nonexistence result. To compare the existence regions of (1) with and without cross-diffusion, we give a nonexistence result stronger than what is needed here. Theorem 3. Let (π2 , π) ∈ (0, ∞)5 be given. Then, there exists a positive constant πΜ1 (π2 , π) such that when d = (π1 , π2 , 0) and π1 ≥ πΜ1 (π2 , π), (1) does not have any nonconstant positive solution. Proof. First, by (7), there exists a positive constant πΆ = πΆ(π, πΜ2 ) such that a classical solution to (1) satisfies π’(π₯), V(π₯) ≤ πΆ provided that π2 ≥ πΜ2 . Now, we write π as average of π over Ω, where π (π’, V) = π’ (1 − π’ π’π−1 V ), − π π + π’π π (π’, V) = ππ’π V . π + π’π (32) Multiplying the equation for π’ of (1) by (π’−π’) and integrating over Ω by parts, we have π1 ∫ ∇|π’ − π’|2 ππ₯ = ∫ π (π’, V) (π’ − π’) ππ₯ Ω Ω = ∫ {π (π’, V) − π (π’, V)} (π’ − π’) ππ₯ Ω (29) = ∫ {ππ’ (π, π) (π’ − π’)2 1 < π’, V < πΆ on Ω} . πΆ (33) Ω +πV (π, π) (π’ − π’) (V − V) } ππ₯ Since [ Φπ (π) = [ [ π1 ππV − 2 3 2 [ (1 + π’) 0 ] ] π3 ] ] π2 [1 + (1 + π’) ] 2 ≤ ∫ {(πΆ1 π’ − π’) + π(V − V)2 } ππ₯ (30) and det Φπ(π) is positive for all nonnegative π, Φ−1 π exists. Hence, π is a positive solution to (1) if and only if =0 π2 ∫ ∇|V − V|2 ππ₯ = ∫ [−πV + π (π’, V)] (V − V) ππ₯ Ω πΉ (d; π) := π − (πΌ − Δ)−1 × {Φπ(π)−1 [πΊ (π) + ∇πΦππ∇π] + π} Ω for some positive constant πΆ1 = πΆ1 (π, πΜ2 ) and π = π(π, πΜ2 ) βͺ 1, where π and π lie between π’ and π’, V and V, respectively. Similarly, we have (31) Ω = ∫ [−π (V − V) + π (π’, V) − π (π’, V)] Ω × (V − V) ππ₯ in π+ , where (πΌ − Δ)−1 is the inverse of πΌ − Δ in π. As πΉ(d; ⋅) is a compact perturbation of the identity operator πΌ, the LeraySchauder deg(πΉ(d; ⋅), 0, π΅) is well defined if πΉ(d; π) =ΜΈ 0 for all π ∈ ππ΅. By Theorems 1 and 2, there exists a positive constant πΆ0 (d, π) such that if πΉ(d; π) = 0 in π+ , then π ∈ π΅(πΆ0 ). Note that πΆ0 (d, π) can be taken as a continuous function for π =ΜΈ ππ/ππ + π. By the invariance property of the LeraySchauder degree, we then conclude that deg(πΉ(d; ⋅), 0, π΅(πΆ)) = ∫ {−π(V − V)2 + πV (π, π) (V − V)2 Ω + ππ’ (π, π) (π’ − π’) (V − V) } ππ₯ ≤ ∫ {(π − π) (V − V)2 + π1 (V − V)2 Ω +πΆ2 (π’ − π’)2 } ππ₯ (34) Journal of Applied Mathematics 5 for some positive constant πΆ2 = πΆ2 (π, πΜ2 ) and π1 = π1 (π, dΜ2 ) βͺ 1, where π and π are the same as in (33). Adding (33) and (34), we obtain ∫ {π1 ∇|π’ − π’|2 + π2 ∇|V − V|2 } ππ₯ ππ΅. Since for each integer π ≥ 1, ππ is invariant under π·π(πΉ(d; π∗ )), and π is an eigenvalue of π·ππΉ on ππ if and only if π is an eigenvalue of following matrix: πΌ− Ω 2 2 ≤ ∫ {(πΆ1 + πΆ2 ) (π’ − π’) + (π − π + π + π1 ) (V − V) } ππ₯. Ω (35) It follows from the PoincareΜ Inequality that π1 ∫ {π1 (π’ − π’)2 + π2 (V − V)2 } ππ₯ Ω 2 1 −1 [Φπ(π∗ ) πΊπ (π∗ ) + πΌ] 1 + ππ 1 −1 [π πΌ − Φπ(π∗ ) πΊπ (π∗ )] . = 1 + ππ π Since π»(π, π∗ ; π) =ΜΈ 0, π·ππΉ(d; π∗ ) is invertible. Therefore, the number of eigenvalues with negative real parts of π·ππΉ(d; π∗ ) on ππ is odd if and only if π»(d, π∗ ; ππ ) < 0, and therefore index (πΉ (d; ⋅) , π∗ ) = (−1)π , 2 π= ∑ 1. π≥1, π»(d,ππ )<0 ≤ ∫ {(πΆ1 + πΆ2 ) (π’ − π’) + (π − π + π + π1 ) (V − V) } ππ₯. Ω (40) (41) (36) Since we can choose π2 ≥ πΜ2 such that π1 π2 > π − π, we may also choose π and π1 sufficiently small such that π1 π2 > π − π + π + π1 . Consequently, by (36), π1 π1 ∫ (π’ − π’)2 ππ₯ ≤ (πΆ1 + πΆ2 ) ∫ (π’ − π’)2 ππ₯, Ω Ω (37) which implies that π’ = π’ = constant, and, hence, V = V = constant if π1 ≥ πΜ1 (π2 , π) := (πΆ1 (π, πΜ2 ) + πΆ2 (π, πΜ2 ))/π1 . Thus, we complete the proof of the theorem. In the following, we only calculate deg(πΉ, 0, π΅) when all solutions to πΉ = 0 are positive constant solutions in π΅(πΆ). Let 0 = π1 < π2 < π3 < ⋅ ⋅ ⋅ be the eigenvalues of the operator −Δ in Ω with zero flux boundary condition, πΈ(ππ ) be the eigenspace corresponding to ππ in πΆ1 (Ω), πππ , π = 1, 2, 3, . . ., dim πΈ(ππ ) an orthonormal basis of πΈ(ππ ), and πππ = dim πΈ(π ) π πππ . {πΆπππ | πΆ ∈ π 2 }. Then, π = ⊕∞ π=1 ππ , where ππ = ⊕π=1 This decomposed method is similar to that of [22]. Let π∗ be a positive root to πΊ(π) = 0. We can calculate π·ππΉ (d; π∗ ) = πΌ − (πΌ − Δ)−1 ∗ ∗ × {Φ−1 π (π ) πΊπ (π ) + πΌ} in πΏ (π, π) , (38) where πΏ(π, π) is a linear mapping from π to itself. Denote that π»(d, π∗ ; π) := det[ππΌ − Φπ(π∗ )−1 πΊπ(π∗ )]. ∗ Lemma 4. Let π be a positive root to πΊ(⋅) = 0, and assume that π»(d, π∗ ; ππ ) =ΜΈ 0 for all π. Then, index (πΉ (d; ⋅) , π∗ ) = (−1)π , π= ∑ π≥1, π»(d,ππ )<0 1. (39) Proof. If π·ππΉ is invertible, then the index of πΉ at π∗ is defined as index(πΉ(d; ⋅), π∗ ) = (−1)π , where π is the number of eigenvalues of π·ππΉ with negative real parts. The deg(πΉ(d; ⋅), 0, π΅) is then equal to summation of the indexes over all solutions to πΉ = 0 in π΅, provided that πΉ =ΜΈ 0 on Theorem 5. Assume that π = (π, π, π, π) ∈ (0, ∞)4 with πππ =ΜΈ π(π + ππ ). Then, for every d ∈ (0, ∞)2 × [0, ∞), there exists a constant πΆ0 (d, π) such that for every πΆ > πΆ0 (d, π), { 0, { { deg (πΉ (d; ⋅) , 0, π΅ (πΆ)) = { { { 1, { ππ + π, ππ ππ if π < π + π. π if π < (42) Proof. Since (1) does not have any positive solution in π΅(πΆ), when π ≤ π, we then conclude that deg (πΉ (d; ⋅) , 0, π΅ (πΆ)) = 0, ∀π ∈ {(π, π, π, π) ∈ (0, ∞)4 | π < ππ + π} , ππ (43) πΆ > πΆ0 (d, π) . Next, to complete the proof of Theorem 5. We need only to calculate the degree for the case π > ππ/ππ + π. In this case, by Theorem 3, all positive solutions to πΉ = 0 are the unique positive constant solution to πΊ(⋅) = 0, denoted by Μ = ππΌ so that π»(d, π; Μ π) = Μ when d = (π, π, 0), Φπ(π) π, 2 Μ Μ (1/π) det(πππΌ − πΊπ(π)), det(πΊπ(π)) = ππ (π − π)ΜV/π2 π’Μ > 0. Μ ππ ) > 0 This implies that when π is sufficiently large, π»(d, π; for all π = 1, 2, 3, . . ., and thus π = 0. It follows from Lemma 4 that deg(πΉ(d; ⋅), 0, π΅(πΆ)) = 1. This completes the proof. Remark 6. The change of degree when π passes the borderline π = ππ/ππ + π is due to the appearance (disappearance) of a positive constant steady-state bifurcating from π ≡ (π, 0). 4. Existence of Nonconstant Positive Solutions In this section, we establish the existence of positive nonconstant solutions for (1). In particular, we show that for certain ranges of parameters where (1) does not have any positive nonconstant steady state, our model can still produce patterns. The idea is as follows. First we calculate the index of πΉ(d; ⋅) at positive constant steady states. Suppose that the 6 Journal of Applied Mathematics sum of all these indices is not equal to the degree stated in Theorem 5. Then, πΉ(d; ⋅) = 0 in π΅(πΆ) for πΆ = πΆ0 (d, π) must have a nonconstant positive solution, which also solves (1). In the following, we always assume that π > π, and (1) has Μ = (Μ a unique positive constant solution (Μ π’, ΜV), when π π’, ΜV) is a positive solution to πΊ(π) = 0. We can get the following Μ by simply calculating results at π 1 π − π) − 2π) π’Μ − π [π (π − π) − π]} − [ ππ {(π (π π] ] Μ =[ πΊπ (π) ] [ ππ (π − π) ΜV 0 ] [ πΜ π’ 4.1. The Case π(π−π)−2π>0 (π>2π/(π−π)). In this subsection, we consider local stability of the constant steady state Μ for evolution dynamics π≡π ππ‘ = ΔΦ (π) + πΊ (π) ππ =0 π] 0 π1 ] [ ] [ Μ Φπ (π) = [ π π ΜV π3 ] 2 3 ] π [1 + − 2 2 (1 + π’Μ) ] [ (1 + π’Μ) Μ = (Μ where π π’, ΜV) is a positive constant solution to πΊ(⋅) = 0 by (5). Μ takes the form Proof. The linearization of (47) at π π 0 =: [π11 π ] , 21 22 Μ Δπ + πΊπ (π) Μ π ππ‘ = Φπ (π) Μ πΊπ (π)) Μ Μ π) ≡ det (ππΌ − Φ−1 (π) π» (d, π; π ππ =0 π] (44) Μ ⋅) = 0, we will restrict To calculate the roots of π»(d, π; our attention to large |d|. Note that Μ π) = π2 − Λ π π, lim π» (d, π; ππ → ∞ ∀π = 2, 3, (45) where Λ 2 = Λ 2 (π1 , π3 ; π) = (−π3 ππΜV + (1 + π’Μ + π3 ) (1 + π’Μ) × {(π (π − π) − 2π) π’Μ − π (π (π − π) − π)} ) −1 × (πππ1 (1 + π’Μ + π3 ) (1 + π’Μ)) , Λ 3 = Λ 3 (π1 ; π) = (47) on πΩ, Theorem 7. Let π > π’Μ > ((π(π − π) − 2π)/(π(π − π) − π))Μ π’ (π > 2π/(π − π)). Then the positive constant solution π(π₯, π‘) ≡ (Μ π’, ΜV) is asymptotically stable with respect to the Μ dynamics (47). Consequently, in a small neighborhood of π, (1) does not have any nonconstant positive solution. π =: [π11 − π ] , π21 0 σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨π σ΅¨ π/π − π22 π11 π21 π/π = π2 + π σ΅¨ 21 σ΅¨ + . π11 π22 π11 π22 in Ω × (0, ∞) , −ππΜV + {(π (π − π) − 2π) π’Μ − π (π (π − π) − π)} (1 + π’Μ) . πππ1 (1 + π’Μ) (46) The sign of the trace tr(πΊπ) = π11 is determined by πΎ = (π(π−π)−2π)Μ π’−π(π(π−π)−π) and the sign of the determinant det(πΊπ) = ππ2 (π − π)V/π2 π’ > 0. Hence, we will discuss separately the following cases: (i) π(π − π) − 2π > 0(π > 2π/(π − π)), obviously π > π’Μ > ((π(π − π) − 2π)/(π(π − π) − π))Μ π’, then πΎ < 0; (ii) π(π − π) − π < 0(π < π/(π − π)): (iia) if π’Μ < π < ((2π − π(π − π))/(π − π(π − π)))Μ π’, then πΎ < 0; (iib) if π > ((2π − π(π − π))/(π − π(π − π)))Μ π’, then πΎ > 0. in Ω × (0, ∞) , (48) on πΩ. Μ Denote that the corresponding linear operator πΏ := Φπ(π)Δ+ Μ πΊπ(π). For each eigen-pair (ππ , ππ ) (π = 1, 2, 3, . . .) of −Δ on Ω with no flux boundary condition, π = πΆπππ‘ π is a solution to (48) if and only if (π, πΆ) is an eigen-pair of the matrix π΄ π := Μ Μ −ππ Φπ(π)+πΊ π (π). ππ is invariant under the operator πΏ. From (44), we have π11 < 0 (π > π (π − π) − 2π π’Μ) , π (π − π) − π π21 > 0 (π > π) . (49) Then π π det π΄ π := π11 π22 ππ2 − ( π21 + π11 π22 ) ππ + π21 > 0, π π tr π΄ π := − (π11 + π22 ) ππ + π11 < 0. (50) Hence, we conclude that the two eigenvalues π+π and π−π of π΄ π have negative real parts. More specifically, we have the following: 2 (i) For π = 1, π1 = 0, if π11 − 4π21 π/π ≤ 0, then Re π±π = 2 (1/2)π11 < 0; if π11 − 4π21 π/π > 0, then Re π+π = 1[ 2 − 4π21 π ] ≤ 0, π11 + √ π11 2 π [ ] 1 2 − 4π21 π ] < 0. Re π−π = [π11 − √ π11 2 π [ ] (51) Journal of Applied Mathematics 7 (ii) For π ≥ 2, as ππ is increasing with respect to π and ππ → ∞ as π → ∞, it follows that if (tr π΄ π )2 − 4 det π΄ π ≤ 0, then Re π±π = 1 1 tr π΄ π = [− (π11 + π22 ) ππ + π11 ] 2 2 1 ≤ [− (π11 + π22 ) π2 + π11 ] < 0; 2 Μ π− (d, π) Μ = [det (Φπ (π))] Μ −1 det (πΊπ (π)) Μ π+ (d, π) (52) = if (tr π΄ π )2 − 4 det π΄ π > 0, since det π΄ π > 0 and tr π΄ π < 0, then Re π−π = ≤ = π2 → ∞ 2 tr π΄ π − √(tr π΄ π ) − 4 det π΄ π ≤ π3 → ∞ (53) det π΄ π < −Μπ. tr π΄ π for some positive constant πΜ that does not depend on π. The previous arguments show that there exists a positive constant π, which does not depend on π, such that Re π π < −π, ∀π. Consequently, the spectrum of πΏ, which consists of eigenvalues, lies in {Re(π) < −π}. It then follows from Theorem 5.1.1 of [26, page 98] that the constant steady-state Μ is asymptotically stable to (47). π(π₯, π‘) ≡ π 4.2. The Case π(π−π)−π<0 (π<π/(π−π)) 4.2.1. (πΎ<0)Μ π’<π<((2π−π(π−π))/(π−π(π−π)))Μ π’ Theorem 8. Let π’Μ < π < ((2π−π(π−π))/(π−π(π−π)))Μ π’. Then the positive constant solution π(π₯, π‘) ≡ (Μ π’, ΜV) is asymptotically stable with respect to the dynamics (47). Theorem 9 (existence with suitable π2 and π3 ). Assume that π’, define π > ππ/ππ + π and π > ((2π − π(π − π))/(π − π(π − π)))Μ Λ 2 (π1 , π3 ; π) and Λ 3 (π1 ; π) as (46). (i) Suppose that π1 and π3 are given such that Μ ∈ (ππ , ππ+1 ) for some positive even integer Λ 2 (π1 , π3 ; π) π. There exists a positive constant π·2 such that if π2 ≥ π·2 , then (1) has at least one nonconstant positive solution. (ii) Suppose that π1 is given such that Λ 3 (π1 ; π) ∈ (ππ , ππ+1 ) for some positive even integer π. Then, for any given π2 > 0, there exists a positive constant π·3 such that if π3 ≥ π·3 , (1) has at least one nonconstant positive solution. Μ = Λ 2 (π1 , π3 ; π) Μ , lim π+ (d, π) Μ = Λ 3 (π1 ; π) Μ . lim π+ (d, π) π3 → ∞ (55) Μ ∈ (ππ , ππ+1 ) for some positive even Suppose that Λ 2 (π1 , π3 ; π) integer π; then, there exists a positive constant π·2 β« 1 such that π2 β« π·2 , and we have Μ < π2 , 0 = π1 < π− (d, π) Μ ∈ (ππ , ππ+1 ) . (56) π+ (d, π) Μ ππ ) < 0 is equivalent to π ∈ Hence, if π2 β« 1, π»(d, π; {2, 3, . . . , π}, since π is even. It follows from Lemma 4 that Μ = (−1)π−1 = −1. index (πΉ ( d ; ⋅) , π) (57) Consequently, πΉ(d; π) = 0 has at least one nonconstant positive solution that is different from the constant function Μ Otherwise, the degree of πΉ = 0 in π΅(πΆ) would be −1 π = π. for all large enough πΆ, which would contradict Theorem 5. This proves, the first assertion of the theorem, and the second assertion is similarly proved. Remark 10. For Λ 3 (π1 ; π) to be positive, it is necessary and sufficient to have π> Proof. Similar to the proof of Theorem 7. 4.2.2. (πΎ>0) π>((2π−π(π−π))/(π−π(π−π)))Μ π’. In this case, Μ Μ π = (Μ π’, V) is the only positive constant solution to πΊ(⋅) = 0. By fixing the diffusion coefficients π1 (for prey) and using the diffusion coefficients π2 and π3 (for predator) as bifurcation parameters, we will show that (1) can create nonconstant positive solutions. We want to emphasize that it is caused by the presence of cross-diffusion which has a more complex role than that of the diffusion coefficients π1 and π2 . (54) π2 → ∞ Μ = 0, lim π− (d, π) 1 1 tr π΄ π ≤ [− (π11 + π22 ) π2 + π11 ] < 0; 2 2 2 det π΄ π π21 π/π > 0. π11 π22 From (45), we see that Μ = 0, lim π− (d, π) 1 2 (tr π΄ π − √(tr π΄ π ) − 4 det π΄ π ) 2 1 2 Re π+π = (tr π΄ π + √(tr π΄ π ) − 4 det π΄ π ) 2 Μ with Re(π− ) ≤ Re(π+ ), the two Proof. Denote by π± (d, π), Μ roots to π»(d, π; π) = 0, and then π’ (π + 1) πΜ + π’Μ. 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