# Some Qualitative Properties of Traveling Wave Fronts of Nonlocal Diffusive Competition-Cooperation Systems of Three Species with Delays

Authors: Meiping Yao; Pengzhi Qiao; Yang Wang
Journal: Complexity
 (2020)

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## Abstract

This paper is concerned with nonlocal diffusion systems of three species with delays. By modified version of Ikehara’s theorem, we prove that the traveling wave fronts of such system decay exponentially at negative infinity, and one component of such solutions also decays exponentially at positive infinity. In order to obtain more information of the asymptotic behavior of such solutions at positive infinity, for the special kernels, we discuss the asymptotic behavior of such solutions of such system without delays, via the stable manifold theorem. In addition, by using the sliding method, the strict monotonicity and uniqueness of traveling wave fronts are also obtained.


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## Body

## 1. Introduction

In this paper, we are concerned with the following nonlocal diffusive competition-cooperation systems of three species with delays:(1)∂u1x,t∂t=d1J1∗u1x,t−d1u1x,t+r1u1x,t1−u1x,t−τ11+a12u2x,t−τ12−a13u3x,t−τ13,∂u2x,t∂t=d2J2∗u2x,t−d2u2x,t+r2u2x,t1+a21u1x,t−τ21−u2x,t−τ22−a23u3x,t−τ23,∂u3x,t∂t=d3J3∗u3x,t−d3u3x,t+r3u3x,t1−a31u1x,t−τ31−a32u2x,t−τ32−u3x,t−τ33,where Ji∗uix,t=∫ℝJix−yuiy,tdy,i=1,2,3, Ji denote the diffusive kernel functions of three species, respectively, and uix,t represent the density of three species. We assume that di, ri>0, 0<aij<1, i≠j, i,j=1,2,3 and τij≥0, i,j=1,2,3.By directly computing, (1) has eight equilibria as follows, which are also the equilibria of the corresponding ordinary differential system of (1):(2)e0=0,0,0,e1=1,0,0,e2=0,1,0,e3=0,0,1,e4=1+a121−a12a21,1+a211−a12a21,0,≔k1,k2,0,e5=1−a131−a13a31,0,1−a311−a13a31,e6=0,1−a231−a23a32,1−a321−a23a32,,e7=D1D,D2D,D3D,where(3)D=1−a12a13−a211a23a31a321,D1=1−a12a1311a231a321,D2=11a13−a211a23a3111,D3=1−a121−a2111a31a321=1−a12a211−a31k1−a32k2,where ei, i=0,1,2,3,4,5,6, are non-negative equilibria. Moreover, by directly computing, D3>0 gives D>0. Thus, when D1,D2>0, e7 is the unique positive equilibrium if and only if(4)1−a31k1−a32k2>0.On the other hand, for (1) without diffusion and delays, by direct calculation, e3 is unstable, and e4 is stable, if(5)1−a31k1−a32k2<0.Furthermore, if (5) and D>0, then the third component of e7 is negative, which implies that e7 is not in the first octant.The existence and other properties of traveling wave solutions are important research fields in diffusive equations including reaction diffusion equations and nonlocal diffusion equation. For (1), the vector value function(6)u1x,t,u2x,t,u3x,t=φ1ξ,φ2ξ,φ3ξ≔φξ,ξ=x+ct,is called a traveling wave solution connecting e3 and e4, if it satisfies(7)d1J1∗φ1ξ−d1φ1ξ−cφ1′ξ+r1φ1ξ1−φ1ξ−cτ11+a12φ2ξ−cτ12−a13φ3ξ−cτ13=0,d2J2∗φ2ξ−d2φ2ξ−cφ2′ξ+r2φ2ξ1+a21φ1ξ−cτ21−φ2ξ−cτ22−a23φ3ξ−cτ23=0,d3J3∗φ3ξ−d3φ3ξ−cφ3′ξ+r3φ3ξ1−a31φ1ξ−cτ31−a32φ2ξ−cτ32−φ3ξ−cτ33=0,(8)limξ⟶−∞ϕξ=e3,limξ⟶+∞ϕξ=e4.Moreover, ifϕiξ, i=1,2,3, are monotone in ξ∈ℝ, then we call it the traveling wave front.If the diffusive kernel functionsJix=δx−δ′′x,i=1,2,3, where δ is the Dirac delta function and τij=0, i,j=1,2,3, under some assumptions on the parameters, Hung in [9] firstly proved the existence of traveling wave fronts of (1) connecting e3 with e4 and gave the exact solutions. Then, in [16], when the wave speed c is large, Ma et al. invested the nonlinear stability of such traveling wave fronts of (1) via the weighted energy method and comparison principle. In addition, Liu and Weng obtained the relevant conclusions of the asymptotic speeds of (1) by using the method of super-sub solutions and comparison principle in [12]. Very recently, in [21], Tian and Zhao proved the existence and the global stability of bistable traveling wave fronts of (1) with finite or infinite delays by using the theory of monotone semiflows, the squeezing technique, and the dynamical systems approach. When delays are taken into consideration and D,D1,D2>0 holds, in [10], Huang and Weng proved the existence of traveling wave fronts of (1) connecting e3 with e7 if (4) exists and connecting e3 with e4 if (5) exists.Besides the aforementioned papers, for the research on the existence of traveling wave solutions, we can refer to [2–4, 7, 8, 13, 15, 17, 19–21, 23, 24]. Specially, Li et al. in [13] introduced the nonlocal diffusion and time delays into the classical Lotka–Volterra reaction diffusion system and proved the existence, asymptotic behavior, and uniqueness of the traveling wave front of this system connecting the equilibria on the two axes. From the above statements, as we know, there is no conclusion about the existence, asymptotic behavior, and other properties of the traveling wave solutions of (1), thus, inspired by [13], we mainly consider the asymptotic behavior, strict monotonicity, and uniqueness of the traveling wave fronts of (1), connecting e3 and e4. Firstly, we make a variable substitution u˜1=u1,u˜2=u2,u˜3=1−u3. Thus, (1) is changed into the following cooperative system, by dropping the tildes for simplicity:(9)∂u1x,t∂t=d1J1∗u1x,t−d1u1x,t+r1u1x,t1−a13−u1x,t−τ11+a12u2x,t−τ12+a13u3x,t−τ13,∂u2x,t∂t=d2J2∗u2x,t−d2u2x,t+r2u2x,t1−a23+a21u1x,t−τ21−u2x,t−τ22+a23u3x,t−τ23,∂u3x,t∂t=d3J3∗u3x,t−d3u3x,t+r31−u3x,ta31u1x,t−τ31+a32u2x,t−τ32−u3x,t−τ33.Correspondingly, system (9) also has eight corresponding equilibria as follows:(10)E0=0,0,1,E1=1,0,1,E2=0,1,1,E3=0,0,0,E4=1+a121−a12a21,1+a211−a12a21,1,E5=1−a131−a13a31,0,a311−a131−a13a31,E6=0,1−a231−a23a32,a321−a231−a23a32,E7=D1D,D2D,1−D3D.From the above statements, the traveling wave fronts of (1) connecting e3 with e4 are equivalent to the traveling wave fronts of (9) connecting E3 with E4. And, the boundary conditions (7) and (8) are correspondingly changed into(11)d1J1∗φ1ξ−d1φ1ξ−cφ1′ξ+r1φ1ξf1φ1,τ11,φ2,τ12,φ3,τ13=0,d2J2∗φ2ξ−d2φ2ξ−cφ2′ξ+r2φ2ξf2φ1,τ21,φ2,τ22,φ3,τ23=0,d3J3∗φ3ξ−d3φ3ξ−cφ3′ξ+r31−φ3ξf3φ1,τ31,φ2,τ32,φ3,τ33=0,(12)limξ⟶−∞φξ=E3,limξ⟶+∞φξ=E4,where(13)f1φ1,τ11,φ2,τ12,φ3,τ13=1−a13−φ1ξ−cτ11+a12φ2ξ−cτ12+a13φ3ξ−cτ13,f2φ1,τ21,φ2,τ22,φ3,τ23=1−a23+a21φ1ξ−cτ21−φ2ξ−cτ22+a23φ3ξ−cτ23,f3φ1,τ31,φ2,τ32,φ3,τ33=a31φ1ξ−cτ31+a32φ2ξ−cτ32−φ3ξ−cτ33.In the sequel, the traveling wave front of (9) as we mention is the nondecreasing solution to (11) and (12). We give the basic assumptions of this paper to end this section. Firstly, we always assume that (5) holds. Secondly, the kernels Ji, i=1,2,3, satisfy(P1)Ji∈Cℝ,Jix=Ji−x≥0,∫ℝJixdx=1(P2) For everyλ>0, ∫ℝxjJixe−λxdx<+∞,j=0,1,2InSection 2, we introduce some notations and the main results. In Section 3, we discuss the asymptotic behavior of the nondecreasing solutions to (11) and (12). In Section 4, we give the strict monotonicity and uniqueness of the nondecreasing solutions to (11) and (12).


## 2. Preliminaries and Main Results

In this section, we will discuss the eigenvalue problems and introduce the main results in this paper. First of all, let(14)Δ1λ,c=d1∫ℝJ1ye−λydy−d1−cλ+r11−a13,Δ2λ,c=d2∫ℝJ2ye−λydy−d2−cλ+r21−a23,Δ3λ,c,τ33=d3∫ℝJ3ye−λydy−d3−cλ−r3e−λcτ33.By some simple computations, we have the following lemma.Lemma 1.
Assume that (P1) and (P2) hold. Then, there existci∗, λi∗, such that Δiλi∗,ci∗=0 and ∂Δiλ,ci∗/∂λλ=λi∗=0, i=1,2. For c>c1∗, the equation Δ1λ,c=0 has two positive roots λ1, λ2 with 0<λ1<λ1∗<λ2, while for 0<c<c1∗, Δ1λ,c>0 for λ∈ℝ. Similarly, for c>c2∗, the equation Δ2λ,c=0 has two positive roots λ3, λ4 with 0<λ3<λ2∗<λ4, while for 0<c<c2∗, Δ2λ,c>0 for λ∈ℝ.Then, for anyc>0, we will discuss the root of Δ3λ,c,τ33=0.Lemma 2.
Assume that (P1) and (P2) hold andτ33∈0,δ1 with small δ1>0, Δ3λ,c,τ33=0 has a unique positive bounded root λ5τ33 for c>0, and Δ3λ,c,τ33<0 for λ∈0,λ5τ33. Moreover, λ=λ5τ33 is a unique root with Reλ=λ5τ33 of Δ3λ,c,τ33=0.Proof.
Let(15)fλ=d3∫ℝJ3ye−λydy−d3−cλ,gλ=r3e−λcτ33.
Note that(16)f′λ=−d3∫ℝyJ3ye−λydy−c,f″λ=d3∫ℝy2J3ye−λydy>0,and f0=0, f′0=−c<0. Thus, f′λ is increasing in 0,+∞. Then, we remark that there are y1∈−∞,0 and δ2>0, such that y1+δ2<0 and J3y>0 on y1−δ2,y1+δ2; otherwise, J3y≡0 by the symmetry of J3, which contradicts (P1). Therefore, by (P2),(17)limλ⟶+∞f′λ≥limλ⟶+∞−d3∫0+∞yJ3ye−λydy−d3∫y1−δ5y1+δ5yJ3ye−λydy−c=+∞,which implies f′λ=0 has a unique zero. Hence, in 0,+∞, there is only λ0 such that(18)fλ0=infλ∈0,+∞fλ<0,and f′λ>0, for λ∈λ0,+∞. Thus, by further noting f+∞=+∞, g′λ<0 and g+∞=0, there is only λ5τ33>λ0>0 such that fλ=gλ.
Ifτ33=0, then let λ50 be the corresponding root, which in fact satisfies fλ=r3. Suppose λ50<λ5τ33, for τ33∈0,δ1. Since λ5τ33, λ50∈λ0,+∞, then fλ50<fλ5τ33. Hence, we have(19)r3=fλ50<fλ5τ33=r3e−cτ33λ5τ33,which is a contradiction. Thus, λ5τ33≤λ50. The rest of the proof is similar to the proof of Lemma 2.8 in [13], so we omit it here.
Letc∗=maxc1∗,c2∗, Λ1∈λ1,λ2 and Λ2∈λ3,λ4. Now, we give the following the exponential decay rates of φξ at the minus infinity.Theorem 1.
Assume that (P1), (P2), and (5) hold. For c≥c∗, assume that φξ is any nondecreasing solution of (11) and (12) for either τii=0 or sufficiently small positive τii, i=1,2,3. Then,(i)
There existθi=θiφ1,φ2,φ3,1≤i≤8,i∈ℤ such that(20)limξ⟶−∞φ1ξ+θ1eΛ1ξ=1,limξ⟶−∞φ2ξ+θ2eΛ2ξ=1,whenc>c∗,limξ⟶−∞φ1ξ+θ3ξμeλ1ξ=1,limξ⟶−∞φ2ξ+θ4ξμeΛ2ξ=1,whenc=c∗=c1∗>c2∗,limξ⟶−∞φ1ξ+θ5ξμeΛ1ξ=1,limξ⟶−∞φ2ξ+θ6ξμeλ3ξ=1,whenc=c∗=c2∗>c1∗,limξ⟶−∞φ1ξ+θ7ξμeλ1ξ=1,limξ⟶−∞φ2ξ+θ8ξμeλ3ξ=1,whenc=c∗=c1∗=c2∗.(i)
There existθi=θiφ1,φ2,φ39≤i≤22,i∈ℤ such thatCase 1.
Whenλ5τ33≤minΛ1,Λ2 and c≥c∗,(21)limξ⟶−∞φ3ξ+θ9eλ5τ33ξ=1.Case 2.
Whenλ5τ33>maxΛ1,Λ2.
Ifc>c∗, then(22)limξ⟶−∞φ3ξ+θ10eλ1ξ=1,whenλ4>λ3≥λ2>λ1,limξ⟶−∞φ3ξ+θ11eλ3ξ=1,whenλ2>λ1≥λ4>λ3,limξ⟶−∞φ3ξ+θ12eλ−ξ=1,whenλ1=λ3,limξ⟶−∞φ3ξ+θ13eλ+ξ=1,whenλ1≠λ3,λ2=λ4,where λ−=λ1=λ3, λ+=minλ1,λ3.
Ifc=c∗=c1∗>c2∗, then(23)limξ⟶−∞φ3ξ+θ14eλ3ξ=1,whenλ3<λ4≤λ1=λ2orλ3<λ1=λ2≤λ4,limξ⟶−∞φ3ξ+θ15ξμeλ1ξ=1,whenλ1=λ2<λ3<λ4,limξ⟶−∞φ3ξ+θ16ξμeλ^ξ=1,whenλ1=λ2=λ3<λ4,where λ^=λ1=λ2=λ3.
Ifc=c∗=c2∗>c1∗, then(24)limξ⟶−∞φ3ξ+θ17eλ1ξ=1,whenλ1<λ2≤λ3=λ4orλ1<λ3=λ4≤λ2,limξ⟶−∞φ3ξ+θ18ξμeλ3ξ=1,whenλ3=λ4<λ1<λ2,limξ⟶−∞φ3ξ+θ19ξμeλˇξ=1,whenλ1=λ3=λ4<λ2,where λ^=λ1=λ3=λ4
Ifc=c∗=c1∗=c2∗, then(25)limξ⟶−∞φ3ξ+θ20ξμeλ¯ξ=1,whenλ1=λ2=λ3=λ4,limξ⟶−∞φ3ξ+θ21ξμeλ1ξ=1,whenλ1=λ2<λ3=λ4,limξ⟶−∞φ3ξ+θ22ξμeλ3ξ=1,whenλ3=λ4<λ1=λ2,where λ¯=λ1=λ2=λ3=λ4.Case 3.
WhenminΛ1,Λ2≤λ5τ33≤maxΛ1,Λ2, by choosing different θ, the conclusions are the same as Case 2.
Moreover, whenc=c∗,(26)μ=0,whenQiΛi=0orΛiis a simple root ofΔiλ,c=0,1,whenQiΛi¹0andΛiis a double root ofΔiλ,c=0,i=1,2,Q1λ≔∫−∞+∞φ1ξφ1ξ−cτ11−a12φ2ξ−cτ12−a13φ3ξ−cτ13e−λξdξ,Q2λ≔∫−∞+∞φ2ξ−a21φ1ξ−cτ21+φ2ξ−cτ22−a23φ3ξ−cτ23e−λξdξ.
Next, we investigate the exponential decay rates ofφξ at the plus infinity. Similarly, let(27)Δ4λ,c=d3∫ℝJ3ye−λydy−d3−cλ−r3a31k1+a32k2−1.Lemma 3.
Assume that (P1), (P2), and (5) hold. Then, Δ4λ,c=0 has a unique negative root λ6<0 for c>0.Proof.
Since(28)∂Δ4λ,c∂λ=−d3∫ℝyJ3ye−λydy−c=−d3∫0+∞yJ3ye−λy−eλydy−c<0,for λ∈−∞,0,(29)∂2Δ4λ,c∂λ2=d3∫ℝy2J3ye−λydy>0,Δ4−∞,c=+∞,and Δ40,c=−r3a31k1+a32k2−1<0 from (5), then obviously Δ4λ,c=0 has a unique negative root λ6<0.
Then, we introduce the following theorem.Theorem 2.
Assume that (P1), (P2), and (5) hold. For c≥c∗, assume that φξ is any nondecreasing solution of (11) and (12). Then, there exists θ23=θ23φ1,φ2,φ3 such that(30)limξ⟶+∞φ3′ξ1−φ3ξ+θ23=−λ6.We cannot use the method offered in [13] to obtain the asymptotic behavior similar to Theorem 1. In order to obtain more detailed results of φξ at the plus infinity, we consider a special case, τij=0, i,j=1,2,3, and Jix=δx−δ′′x,i=1,2,3. Moreover, from Lemma 8 in Section 3, we conclude that λ8>λ9 are the negative roots of(31)d1λ2−cλ−r1k1d2λ2−cλ−r2k2−a12a21r1r2k1k2=0.Then, with the methods in [5, 18, 22] and the stable manifold theorem, we give the following asymptotic behavior of φξ at the plus infinity.Theorem 3.
Assume that (P1), (P2), and (5) hold. For c≥c∗, we assume that φξ is any nondecreasing solution of (11) and (12), τij=0, i,j=1,2,3, and(32)Jix=δx−δ″x,i=1,2,3,where δ is the Dirac delta function. In addition, let(33)λ7=c−c2+4d3r3a31k1+a32k2−12d3.
Then, whenξ⟶+∞, φξ has the following asymptotic behavior:
(i) Whenλ7>λ8>λ9,(34)φ1ξ=k1−α1s1eλ7ξ+h.o.t.,φ2ξ=k2−α1s2eλ7ξ+h.o.t.,φ3ξ=1−α1eλ7ξ+h.o.t.,where α1>0.(35)s1=a13r1k1−d2λ72+cλ7+r2k2+a12a23r1r2k1k2d1λ72−cλ7−r1k1d2λ72−cλ7−r2k2−a12a21r1r2k1k2,s2=a23r2k2−d1λ72+cλ7+r1k1+a21a13r1r2k1k2d1λ72−cλ7−r1k1d2λ72−cλ7−r2k2−a12a21r1r2k1k2.
(ii) Whenλ8>λ9>λ7,(36)φ1ξ=k1−α1s1eλ7ξ−α2eλ8ξ+h.o.t.,φ2ξ=k2−α1s2eλ7ξ−α2τ1eλ8ξ+h.o.t.,φ3ξ=1−α1eλ7ξ+h.o.t.,where α1, α2>0, τ1=−d1λ82+cλ8+r1k1/a12r1k1>0.
(iii) Whenλ8>λ7>λ9,(37)φ1ξ=k1−α1s1eλ7ξ−α2eλ8ξ+h.o.t.,φ2ξ=k2−α1s2eλ7ξ−α2τ1eλ8ξ+h.o.t.,φ3ξ=1−α1eλ7ξ+h.o.t.,where α1, α2>0.
Moreover, suppose(38)−cd2+cd22+4r2k2d21+a12a23a13≠−cd3+cd32+4r3d3a31k1+a32k2−1.
(iv) Whenλ7=λ8>λ9,(39)φ1ξ=k1−α1eλ8ξ−α2ξeλ8ξ+h.o.t.,φ2ξ=k2−α1τ1eλ8ξ−α2τ1ξeλ8ξ+h.o.t.,φ3ξ=1−α2μ¯1eλ8ξ+h.o.t.,where α1⩾0, α2>0.(40)τ3=a23r2k2/a13r1k1c−2d1λ8−c−2d2λ8τ1−d2λ82+cλ8+r2k2+a12a23/a13r2k2,μ¯1=c−2d1λ8−a12r1k1τ3a13r1k1>0.
(v) Whenλ8>λ7=λ9,(41)φ1ξ=k1−α1eλ8ξ−α3ξeλ9ξ+h.o.t.,φ2ξ=k2−α1τ1eλ8ξ−α2τ2ξeλ9ξ+h.o.t.,φ3ξ=1−α3μ¯2eλ9ξ+h.o.t.,where α1>0, α3>0μ¯2>0, α3<0μ¯2<0, and(42)τ4=a23r2k2/a13r1k1c−2d1λ9−c−2d2λ9τ2−d2λ92+cλ9+r2k2+a12a23/a13r2k2,μ¯2=c−2d1λ9−a12r1k1τ4a13r1k1.We give a remark at once to show (38) is reasonable.Remark 1.
Choosed2=r2=r3=1, d3=2, a12=a21=a13=a31=a23=a32=1/2; then, k1=k2=2. Thus, the left side of the equation (38) is equal to −c+c2+2, while the right side of the equation (38) is −c/2+c/22+2, which are not equal for c>0.
Finally, we introduce conclusions on the strict monotonicity and the uniqueness.Theorem 4.
Assume that (P1), (P2), and (5) hold, and τii=0, i=1,2,3. For c≥c∗, if φξ is any nondecreasing solution of (11) and (12), then it is strictly monotone.Theorem 5.
Under the assumptions of Theorem3, let φ1ξ and φ2ξ be two nondecreasing solutions of (11) and (12), with the same speed c≥c∗ and the same decay rate at ξ=±∞. Then, they are unique (up to a translation).Remark 2.
Whenτij=0, i,j=1,2,3, and Jix=δx−δ′′x,i=1,2,3, the existence of nondecreasing solutions of (11) and (12) had been discussed in [9]. From Theorems 1–5, we further investigate the asymptotic behavior, strict monotonicity, and uniqueness of such solutions in this paper.Remark 3.
The proof of existence of traveling wave fronts is standard. For example, supposeD>0, and (P1), (P2), (5) hold. Since system (11) is the quasimonotone system τii=0,i=1,2,3 or weak quasimonotone system τii>0,i=1,2,3, with the methods in the [11–13, 17, 19, 23] and references therein, the existence of traveling wave fronts to (9) can be proved.


## 3. Asymptotic Behavior of Traveling Wave Solutions

In this section and next section, we always assumeτii=0 or τii, i=1,2,3, small enough, and assume (P1), (P2), and (5) hold; then, we investigate the asymptotic behavior, strict monotonicity, and uniqueness of nondecreasing solutions to (11) and (12).To study the asymptotic behavior, we introduce the following result given in [24].Proposition 1.
Letc>0 be a constant and B⋅ be a continuous function having finite limits at infinity B±∞≔limx⟶±∞Bx. Let z⋅ be a measurable function satisfying(43)∫ℝJye∫xx−yzsdsdy−czx+Bx=0,x∈ℝ.
Then,z is uniformly continuous and bounded. In addition, μ±≔limx⟶±∞zx exist and are real roots of the characteristic equation(44)∫ℝJye−μydy−cμ+B±∞=0.Firstly, we give some properties of the solutionsφξ to (11) and (12).Lemma 4.
Suppose (P1), (P2), and (5) hold, and for c≥c∗, φξ is any nondecreasing solution of (11)-(12). Then,(45)φ1ξ,φ2ξ>0,φ3ξ<1,ξ∈ℝ.
Moreover, ifτii=0, i=1,2,3, then(46)φ1ξ<k1,φ2ξ<k2,φ3ξ>0,ξ∈ℝ.Proof.
From the assumptions, we remark thatφ1 is nondecreasing and limξ⟶+∞φ1ξ=k1. Thus, let ξ1∈ℝ be the right-most point of φ1ξ=0; then, φ1ξ attains the minimum at ξ1, implying φ1′ξ1=0. It follows from the first equation of (11) that J1∗φ1ξ1=0. By (P1), there exist y2>0, δ2>0 such that J1y≠0 for y∈y2−δ2,y2+δ2. Since(47)0=J1∗φ1ξ1=∫ℝJ1ξ1−yφ1ydy=∫ℝJ1yφ1ξ1−ydy=∫ℝJ1yφ1ξ1+ydy,following from J1−y=J1y, y∈ℝ. Then, we obtain φ1ξ1±y=0 for y∈y2−δ2,y2+δ2, which contradicts the choice of ξ1. Thus, φ1ξ>0 in ξ∈ℝ. Similarly, we can prove φ2ξ>0,φ3ξ<1, ξ∈ℝ.
Whenτ11=0, similarly, suppose there exists a point ξ2∈ℝ such that φ1ξ2=k1, implying φ1′ξ2=0. By recalling φ3ξ<1 and φ2ξ≤k2, ξ∈ℝ, we deduce(48)0=d1J1∗φ1ξ2−d1φ1ξ2+r1φ1ξ21−a13−φ1ξ2+a12φ2ξ2−cτ12+a13φ3ξ2−cτ13<d1J1∗φ1ξ2−d1k1+r1k1−a12k2+a12φ2ξ2−cτ12≤d1J1∗φ1ξ2−d1k1.
Thus,J1∗φ1ξ2>k1. On the other hand,(49)J1∗φ1ξ2=∫ℝJ1ξ2−yφ1ydy=∫ℝJ1yφ1ξ2−ydy≤k1∫ℝJ1ydy=k1,which is a contradiction. Thus, φ1ξ<k1. Similarly, when τ22=0, φ2ξ<k2.
Whenτ33=0, we also assume there exists a point ξ3∈ℝ such that φ3ξ3=0, and thus φ3′ξ3=0. Also by noting φ1ξ, φ2ξ>0, ξ∈ℝ, we have(50)0=d3J3∗φ3ξ3−d3φ3ξ3+r31−φ3ξ3a31φ1ξ3−cτ31+a32φ2ξ3−cτ32−φ3ξ3=d3J3∗φ3ξ3+r3a31φ1ξ3−cτ31+a32φ2ξ3−cτ32.
Therefore,(51)0≤d3J3∗φ3ξ3=−r3a31φ1ξ3−cτ31+a32φ2ξ3−cτ32<0,which is a contradiction, and thus φ3ξ>0.
By Lemmas1 and 4 and Proposition 1, it is easy to verify the following theorem.Theorem 6.
Suppose (P1), (P2), and (5) hold, and for c≥c∗, φξ is any nondecreasing solution of (11) and (12). Then,(52)limξ⟶−∞φ1′ξφ1ξ=Λ1∈λ1,λ2,limξ⟶−∞φ2′ξφ2ξ=Λ2∈λ3,λ4,where λi, i=1,2,3,4, are described as Lemma 1.Define the bilateral Laplace transform(53)Lλ,ϕ=∫−∞+∞ϕξe−λξdξ,where ϕ:ℝ⟶ℝ is a continuous function. The following lemma is derived from (12) and Theorem 6.Lemma 5.
Suppose (P1) and (P2) hold, and forc≥c∗, φξ is any nondecreasing solution of (11) and (12). Then,(54)Lλ,φ1<+∞,λ∈0,Λ1,Lλ,φ1=+∞,λ∈ℝ\0,Λ1.Lλ,φ2<+∞,λ∈0,Λ2,Lλ,φ2=+∞,λ∈ℝ\0,Λ2.Then, we discuss the convergence ofLλ,φ3 via the following lemma.Lemma 6.
Suppose (P1), (P2), and (5) hold, and for c≥c∗, suppose φξ is any nondecreasing solution of (11) and (12). Then,(55)Lλ,φ3<+∞,λ∈0,γτ33,Lλ,φ3=+∞,λ∈ℝ\0,γτ33,where γτ33=minΛ1,Λ2,λ5τ33.Proof.
We first searchσ such that Lλ,φ3<+∞, λ∈0,σ. From Lemma 2, we can conclude that d3∫ℝJ3ye−λydy−d3−cλ=0 has two roots 0 and Λ∗0<Λ∗ for c>0, and(56)d3∫ℝJ3ye−λydy−d3−cλ<0,for λ∈0,Λ∗. Since P2 implies that(57)−∞<∫−∞0yJ3ye−2λydy<0,for λ>0, then there exists a small Λ¯>0 such that(58)d3∫−∞0yJ3ye−2λydy−c+r32λe−2>0,for λ∈0,Λ¯. For λ∈0,minΛ1,Λ2,Λ∗,1/cτ33,Λ¯, we take z0<0 small enough such that φ3ξ≤1/2 for ξ≤z0+cτ33+1/λ.
Multiplying the third equation of (11) by e−λξ with λ∈0,minΛ1,Λ2,Λ∗,1/cτ33,Λ¯ and integrating from z≤z0 to +∞ gives(59)r3a31∫z+∞1−φ3ξφ1ξ−cτ31e−λξdξ+r3a32∫z+∞1−φ3ξφ2ξ−cτ32e−λξdξ=−d3∫z+∞∫ℝJ3ξ−yφ3ye−λξdydξ+d3∫z+∞φ3ξe−λξdξ+c∫z+∞φ3′ξe−λξdξ+r3∫z+∞1−φ3ξφ3ξ−cτ33e−λξdξ.
For the left side of (59), obviously,(60)r3a31∫z+∞1−φ3ξφ1ξ−cτ31e−λξdξ≤r3a31Lλ,φ1,(61)r3a32∫z+∞1−φ3ξφ2ξ−cτ32e−λξdξ≤r3a32Lλ,φ2.
By recalling the convergence in Lemma5, from (56)–(61), similar to [13], we deduce(62)r3a31Lλ,φ1+r3a32Lλ,φ2≥d3∫−∞0yJ3ye−2λydy−c+r32λe−2φ3ze−λz,which implies, for any λ∈0,minΛ1,Λ2,Λ∗,1/cτ33,Λ¯,(63)0<supz∈ℝφ3ze−λz<+∞.
We conclude that there existsσ∈0,minΛ1,Λ2,Λ∗,1/cτ33,Λ¯ such that Lλ,φ3<+∞ for λ∈0,σ.
We now prove thatmaxσ=γτ33=minΛ1,Λ2,λ5τ33. Multiplying the third equation of (11) by e−λξ with λ>0 and integrating from −∞ to +∞ gives(64)Δ3λ,c,τ33Lλ,φ3=−r3∫−∞+∞1−φ3ξa31φ1ξ−cτ31+a32φ2ξ−cτ32e−λξdξ−r3∫−∞+∞φ3ξφ3ξ−cτ33e−λξdξ.
We claim thatγτ33≤λ5τ33. Otherwise, if γτ33>λ5τ33, then Lλ5τ33,φ3<+∞. Taking λ=λ5τ33 in (64), the left side of (64) equals 0 by Δ3λ5τ33,c,τ33=0, while the right side of (64) is always negative by φ3ξ≤1, which is a contradiction. It also follows easily from (64) that γτ33=minΛ1,Λ2 if Λ1,Λ2<λ5τ33 or γτ33=λ5τ33 if Λ1,Λ2>λ5τ33. Furthermore, if Λ1,Λ2>λ5τ33, it also follows that limλ⟶λ5−τ33Lλ,φ3λ5τ33−λ exists.
In order to study the asymptotic behavior, we introduce the following modified version of Ikehara’s theorem [1], which has been applied early in [6].Lemma 7 (Ikehara’s theorem).
Letϕ be a positive nondecreasing function on ℝ and define Fλ≔∫−∞0ϕξe−λξdξ. Assume that F can be written as Fλ=Hλ/α−λν+1, where ν>−1,α>0, and H is analytic in the strip 0<Reλ≤α; then,(65)limξ⟶−∞ϕξξνeαξ=HαΓα+1.With the aid of the above lemmas, we will finish the proof of Theorem1.Proof of Theorem1.
The proof is motivated by [13–15]. From Lemmas 5 and 6, Lλ,φi are well defined for λ∈ℂ with Reλ∈0,Λi, i=1,2, and Lλ,φ3 is well defined for λ∈ℂ with Reλ∈0,γτ33, respectively. It follows from (11) that(66)Δiλ,c∫−∞+∞φiξe−λξdξ=riQiλ,for λ∈ℂ with 0<Reλ<Λi, i=1,2, and(67)Δ3λ,c,τ33∫−∞+∞φ3ξe−λξdξ=−r3a31e−λcτ31r1Q1λΔ1λ,c−r3a32e−λcτ32r2Q2λΔ2λ,c+r3Q3λ,for λ∈ℂ with 0<Reλ<γτ33, where Q1, Q2 are defined in Theorem 1 and(68)Q3λ≔∫−∞+∞φ3ξa31φ1ξ−cτ31+a32φ2ξ−cτ32−φ3ξ−cτ33e−λξdξ.
By Lemmas2, 5, and 6 and directly calculating, we conclude the following facts.(1)
λ=Λi is a unique root with Reλ=Λi of Δiλ,c=0, i=1,2, and λ=λ5τ33 is a unique root with Reλ=λ5τ33 of Δ3λ,c,τ33=0.(2)
Qiλ are analytic in the strip 0<Reλ<Λi+γτ33, i=1,2, and Q3λ is analytic in the strip 0<Reλ<2γτ33.
From (66), we define(69)Fiλ=∫−∞0φiξe−λξdξ=riQiλΔiλ,c−∫0+∞φiξe−λξdξ,i=1,2,(70)Hiλ=riQiλΔiλ,c/Λi−λν+1−Λi−λν+1∫0+∞φiξe−λξdξ,i=1,2,where ν=0 when c>c∗ and ν=μ when c=c∗.(71)μ=0,whenQiΛi=0orΛiis a simple root ofΔiλ,c=0,1,whenQiΛi¹0andΛiis a double root ofΔiλ,c=0,i=1,2.
Then, from (1) and (2), we deduce thatHiλ are analytic in the strip 0<Reλ≤Λi, i=1,2. Thus, from Lemma 7,(72)limξ⟶−∞φiξξνeΛiξ=HiΛiΓΛi+1,i=1,2,which implies (i) in Theorem 1 holds if HiΛi≠0. In the following, we will prove that HiΛi≠0, i=1,2, in two cases.
Forc>c∗, i=1,2; since ν=0 and Λi is a simple root of Δiλ,c=0, then we have Δiλ,c/Λi−λ≠0 at λ=Λi. For i=1,2, if we suppose QiΛi=0, then from (69), LΛi,φi exist, which contradicts Lemma 5. Thus, QiΛi≠0, i=1,2. Therefore, from (70), we arrive at(73)HiΛi=riQiλΔiλ,c/Λi−λλ=Λi≠0,i=1,2.
Forc=c∗, in this case, there are three subcases, c=c∗=c1∗>c2∗, c=c∗=c2∗>c1∗, and c=c∗=c1∗=c2∗. The first two subcases are equivalent by changing the roles of φ1 and φ2. Thus, without loss of generality, we consider c=c∗=c1∗>c2∗. In this subcase, Λ2 is a simple root of Δ2λ,c=0; repeating the above proofs of HiΛi≠0 with c>c∗, i=1,2, take ν=μ=0, and we directly have(74)H2Λ2≠0.
SinceΛ1=λ1=λ2 is a double root of Δ1λ,c=0, then Δ1λ,c/λ1−λ2≠0 at λ=λ1. When Q1λ1≠0, we can take ν=μ=1 such that(75)H1λ1=r1Q1λΔ1λ,c/λ1−λ2λ=λ1≠0.
WhileQ1λ1=0, λ1 must be a simple root of Q1λ=0; otherwise, Lλ1,φ1 exists by (69), which contradicts Lemma 5. Thus, we can take ν=μ=0 such that(76)H1λ1=r1Q1λΔ1λ,c/λ1−λλ=λ1≠0.
Whenc=c∗=c1∗=c2∗, repeating the above proofs gives HiΛi≠0, i=1,2, where Λ1=λ1=λ2, Λ2=λ3=λ4.
Now we prove (ii) in Theorem1. From (67), we define(77)F3λ=∫−∞0φ3ξe−λξdξ=−∫0+∞φ3ξe−λξdξ−r3a31e−λcτ31r1Q1λΔ1λ,cΔ3λ,c−r3a32e−λcτ32r2Q2λΔ2λ,cΔ3λ,c+r3Q3λΔ3λ,c,(78)H3λ=γτ33−λν+1F3λ=−r3a31e−λcτ31r1Q1λΔ1λ,cΔ3λ,c/γτ33−λν+1−r3a32e−λcτ32r2Q2λΔ2λ,cΔ3λ,c/γτ33−λν+1+r3Q3λΔ3λ,c/γτ33−λν+1−γτ33−λν+1∫0+∞φ3ξe−λξdξ,where ν is the same as above. By using a similar argument as (i), H3λ is analytic in the strip 0<Reλ≤γτ33. Hence, also from Lemma 7,(79)limξ⟶−∞φ3ξξνeγτ33ξ=H3γτ33Γγτ33+1,which implies (ii) in Theorem 1 holds if H3γτ33≠0. In the following, we will prove that H3γτ33≠0 in three cases.Case 1.
λ5τ33≤minΛ1,Λ2. In this case, γτ33=λ5τ33 and(80)H3λ=−r3∫−∞+∞φ3ξφ3ξ−cτ33+a311−φ3ξφ1ξ−cτ31+a321−φ3ξφ2ξ−cτ32e−λξdξλ5τ33−λΔ3λ,c−1−λ5τ33−λ∫0+∞φ3ξe−λξdξ.
IfH3λ5τ33=0, then(81)∫−∞+∞φ3ξφ3ξ−cτ33+a311−φ3ξφ1ξ−cτ31+a321−φ3ξφ2ξ−cτ32e−λ5τ33ξdξ=0,implying φ1ξ≡φ2ξ≡φ3ξ≡0 in ℝ, which contradicts the results in Lemma 4.Case 2.
λ5τ33>maxΛ1,Λ2. In this case, γτ33=minΛ1,Λ2, and from Lemma 2, we note that Δ3γτ33,c,τ33≠0. Thus, from (78), in order to prove H3γτ33≠0, it is sufficient for us to prove(82)−r3a31e−λcτ31r1Q1λΔ2λ,c−r3a32e−λcτ32r2Q2λΔ1λ,cΔ1λ,cΔ2λ,cΔ3λ,c/γτ33−λν+1λ=γτ33≠0.
First of all, whenc>c∗, in this case ν=0, and there are four subcases:(83)iλ4>λ3≥λ2>λ1,iiλ2>λ1≥λ4>λ3,iiiλ1=λ3,ivλ1≠λ3,λ2=λ4.
For subcase (i),γτ33=λ1, and λ1 is a simple root of Δ1λ,c=0 and Δ2λ1,c>0; then, by repeating the process to prove H1Λ1≠0 with c>c∗, we easily obtain(84)H3λ1=−r3a31e−λcτ31r1Q1λΔ2λ,cΔ1λ,cΔ2λ,cΔ3λ,c/λ1−λλ=λ1≠0.
For subcase (iii),γτ33=λ1=λ3≔λ−, and λ− is a simple root of Δiλ,c=0, i=1,2. By similar process, we deduce(85)H3λ−=−r3a31e−λcτ31r1Q1λΔ2λ,c−r3a32e−λcτ32r2Q2λΔ1λ,cΔ1λ,cΔ2λ,cΔ3λ,c/λ−−λλ=λ−≠0.
The proofs of subcases (ii) and (iv) are similar.
Secondly, we considerc=c∗. Similarly, we only consider c=c∗=c1∗>c2∗ and c=c∗=c1∗=c2∗. For c=c∗=c2∗>c1∗, we can get the corresponding conclusions by similar steps, and we omit the process here.
Whenc=c∗=c1∗>c2∗, there are also four subcases.(86)iλ3<λ4≤λ1=λ2,iiλ3<λ1=λ2≤λ4,iiiλ1=λ2<λ3<λ4,ivλ1=λ2=λ3<λ4.
For subcases (i) and (ii),γτ33=λ3. Since λ3 is a simple root of Δ2λ,c=0, similar to the proof of subcase (i), when c>c∗, we get(87)H3λ3≠0.For subcase (iii), γτ33=Λ1=λ1, and λ1 is a double root of Δ1λ,c=0, and thus(88)Δ1λ,cΔ2λ,cΔ3λ,cλ1−λ2≠0,at λ=λ1. We can take ν=μ=1 such that(89)H3λ1=−r3a31e−λcτ31r1Q1λΔ2λ,cΔ1λ,cΔ2λ,cΔ3λ,c/λ1−λ2λ=λ1≠0,when Q1λ1≠0. On the other hand, Q1λ1=0; then, λ1 must be a simple root of Q1λ=0. Thus, we can take ν=μ=0 such that(90)H3λ1=−r3a31e−λcτ31r1Q1λΔ2λ,cΔ1λ,cΔ2λ,cΔ3λ,c/λ1−λλ=λ1≠0.
For subcase (iv),γτ33=λ1=λ2=λ3≔λ^ and λ^ is a double root of Δ1λ,c=0, while it is a simple root of Δ2λ,c=0. Since Q2λ^≠0, we can take ν=μ=1 such that(91)H3λ^=−r3a31e−λcτ31r1Q1λΔ2λ,cΔ1λ,cΔ2λ,cΔ3λ,c/λ^−λ2λ=λ^≠0,when Q1λ^≠0. When Q1λ^=0, we can take ν=μ=0 such that(92)H3λ^=−r3a31e−λcτ31r1Q1λΔ2λ,c−r3a32e−λcτ32r2Q2λΔ1λ,cΔ1λ,cΔ2λ,cΔ3λ,c/λ^−λλ=λ^≠0.
Whenc=c∗=c1∗=c2∗, there are three subcases:(93)iλ1=λ2=λ3=λ4,iiλ1=λ2<λ3=λ4,iiiλ3=λ4<λ1=λ2.
Repeating the above proofs, one can easily obtainH3γτ33≠0 in each subcase.Case 3.
minΛ1,Λ2≤λ5τ33≤maxΛ1,Λ2. In this case, the proof is similar.
The following corollary is obvious by Theorem1.Corollary 1.
Ifφξ is described as Theorem 1, then limξ⟶−∞φ3′ξ/φ3ξ=γ.Now, we investigate the exponential decay rates ofφξ at the plus infinity. For convenience, let φ˜1=k1−φ1, φ˜2=k2−φ2, and φ˜3=1−φ3; substituting φ˜1,φ˜2,φ˜3 into (11), we have(94)d1J1∗φ˜1ξ−d1φ˜1ξ−cφ˜1′ξ+r1k1−φ˜1ξf˜1φ˜1,τ11,φ˜2,τ12,φ˜3,τ13=0,d2J2∗φ˜2ξ−d2φ˜2ξ−cφ˜2′ξ+r2k2−φ˜2ξf˜2φ˜1,τ21,φ˜2,τ22,φ˜3,τ23=0,d3J3∗φ˜3ξ−d3φ˜3ξ−cφ˜3′ξ+r3φ˜3ξf˜3φ˜1,τ31,φ˜2,τ32,φ˜3,τ33=0,satisfying(95)limξ⟶−∞φ˜ξ=E4,limξ⟶+∞φ˜ξ=E3,where(96)f˜1φ˜1,τ11,φ˜2,τ12,φ˜3,τ13=−φ˜1ξ−cτ11+a12φ˜2ξ−cτ12+a13φ˜3ξ−cτ13,f˜2φ˜1,τ21,φ˜2,τ22,φ˜3,τ23=a21φ˜1ξ−cτ21−φ˜2ξ−cτ22+a23φ˜3ξ−cτ23,f˜3φ˜1,τ31,φ˜2,τ32,φ˜3,τ33=−a31k1−a32k2+1+a31φ˜1ξ−cτ31+a32φ˜2ξ−cτ32−φ˜3ξ−cτ33.Letz˜ξ=φ˜3′ξ/φ˜3ξ. Since φ3ξ<1 by Lemma 4, Theorem 2 is based on limξ⟶−∞φ˜ξ=0 and Proposition 1.Finally, from Lemma6, the convergence of Lλ,φ3 is determined both by Lλ,φ1 and Lλ,φ2, while for ξ=+∞, we cannot obtain the convergence of Lλ,φ1 and Lλ,φ2 only by Lλ,φ3. Thus, in order to investigate the asymptotic behavior of the nondecreasing solution φξ, we enforce the assumptions and adopt the methods in [5, 18, 22]. Now, we suppose τij=0, i,j=1,2,3, and Jix=δx−δ′′x,i=1,2,3, where δ is the Dirac delta function. With these assumptions, system (11) is transformed into the following system:(97)d1φ1″−cφ1′+r1φ11−a13−φ1+a12φ2+a13φ3=0,d2φ2″−cφ2′+r2φ21−a23+a21φ1−φ2+a23φ3=0,d3φ3″−cφ3′+r31−φ3a31φ1+a32φ2−φ3=0,which is equivalent to (by taking ϕ1=φ1′,ϕ2=φ2′,ϕ3=φ3′)(98)φ1′=ϕ1,ϕ1′=1d1cϕ1−r1φ11−a13−φ1+a12φ2+a13φ3,φ2′=ϕ2,ϕ2′=1d2cϕ2−r2φ21−a23+a21φ1−φ2+a23φ3,φ3′=ϕ3,ϕ3′=1d3cϕ3−r31−φ3a31φ1+a32φ2−φ3.Linearizing the above system at the equilibrium pointk1,0,k2,0,1,0 gives(99)Φ1+′=AΦ1+,where(100)Φ1+=φ1+ϕ1+φ2+ϕ2+φ3+ϕ3+,A=010000r1k1d1cd1−a12r1k1d10−a13r1k1d10000100−a21r2k2d20r2k2d2cd2−a23r2k2d200000010000r3a31k1+a32k2−1d3cd3,and φ1+=φ1−k1, φ2+=φ2−k2, φ3+=φ3−1. Let(101)Bλ=A−λI,and then the characteristic equation of matrix A is(102)Bλ=−λ10000r1k1d1cd1−λ−a12r1k1d10−a13r1k1d1000−λ100−a21r2k2d20r2k2d2cd2−λ−a23r2k2d200000−λ10000r3a31k1+a32k2−1d3cd3−λ=0,namely,(103)d3λ2−cλ−r3a31k1+a32k2−1=0,or(104)d1λ2−cλ−r1k1d2λ2−cλ−r2k2−a12a21r1r2k1k2=0.Since (5) holds, by directly computing, (103) has only a negative root(105)λ7=c−c2+4d3r3a31k1+a32k2−12d3.Although it is hard to get exact expression of the roots of (104), we can judge the real and positive or negative properties of them in the following lemma, which is sufficient for us to discuss the asymptotic behavior.Lemma 8.
Forc≥c∗, (104) has four different real roots, and among the four real roots, two are positive roots while the others are negative roots.Proof.
Let(106)hςλ=d1λ2−cλ−r1k1d2λ2−cλ−r2k2−ςa12a21r1r2k1k2,where ς∈0,1. And(107)h1λ=d1λ2−cλ−r1k1d2λ2−cλ−r2k2−a12a21r1r2k1k2.
Firstly, whenς=0, by simple calculation, the function(108)h0λ=d1λ2−cλ−r1k1d2λ2−cλ−r2k2has the following four zero points:(109)μ1=c+c2+4d1r1k12d1>0,μ2=c−c2+4d1r1k12d1<0,μ3=c+c2+4d2r2k22d2>0,μ4=c−c2+4d2r2k22d2<0.
Whenμ1≠μ3 and μ2≠μ4, then h0λ=0 has four different roots, two positive and two negative. Moreover, we notice that the image of hςλ is strictly decreasing as ς is increasing, and ∀ς∈0,1, hς0=1−ςa12a21r1r2k1k2>0, hς±∞=±∞. Thus, we deduce that h1λ=0 has two different positive roots and two different negative roots.
Forμ1=μ3 and μ2≠μ4, h0λ=0 has one positive root (the algebraic multiplicity is 2) and two different negative roots. Repeating the above process, we can prove that h1λ=0 has two different positive roots and two different negative roots. When μ1≠μ3 and μ2=μ4 or μ1=μ3 and μ2=μ4, the same result still holds.From Lemma8, (104) has two different negative roots, denoted by λ8 and λ9 separately. Without loss of generality, we assume λ9<λ8<0.Now, we give the proof of Theorem3 to finish this section.Proof of Theorem3.
We firstly suppose(110)d1λ72−cλ7−r1k1d2λ72−cλ7−r2k2−a12a21r1r2k1k2≠0,which is equivalent to λ7, λ8, and λ9 which are three different negative real roots of Bλ=0. Then, there are three corresponding eigenfunctions:(111)r11=s1,s1λ7,s2,s2λ7,1,λ7T,r21=1,λ8,τ1,τ1λ8,0,0T,r31=1,λ9,τ2,τ2λ9,0,0T,where(112)τ1=−d1λ82+cλ8+r1k1a12r1k1,τ2=−d1λ92+cλ9+r1k1a12r1k1,s1=a13r1k1−d2λ72+cλ7+r2k2+a12a23r1r2k1k2d1λ72−cλ7−r1k1d2λ72−cλ7−r2k2−a12a21r1r2k1k2,s2=a23r2k2−d1λ72+cλ7+r1k1+a21a13r1r2k1k2d1λ72−cλ7−r1k1d2λ72−cλ7−r2k2−a12a21r1r2k1k2.
From the proofs of Lemma8, it is easy to infer(113)λ8>maxμ2,μ4≥minμ2,μ4>λ9,and thus τ1>0, τ2<0. Therefore, when ξ⟶+∞, (99) has the following solutions:(114)Φ1+′=C1r11eλ7ξ+C2r21eλ8ξ+C3r31eλ9ξ,where Ci, i=1,2,3, are arbitrary constants. Thus, by the stable manifold theorem, we have(115)φ1ξ=k1−α1s1eλ7ξ−α2eλ8ξ−α3eλ9ξ+h.o.t.,φ2ξ=k2−α1s2eλ7ξ−α2τ1eλ8ξ−α3τ2eλ9ξ+h.o.t.,φ3ξ=1−α1eλ7ξ+h.o.t..
Similar to [18], we finally determine the positive or negative of αi, i=1,2,3. Since φ3≤1, then α1>0.
Whenλ9<λ8<λ7<0, we deduce that λ7>maxμ2,μ4. Thus,(116)−d2λ72+cλ7+riki>0,i=1,2,which implies s1,s2>0. Moreover, by noting limξ⟶+∞eλ7ξ/eλiξ=+∞, i=8,9, (115) can be simplified into(117)φ1ξ=k1−α1s1eλ7ξ+h.o.t.,φ2ξ=k2−α1s2eλ7ξ+h.o.t.,φ3ξ=1−α1eλ7ξ+h.o.t..
Whenλ7<λ9<λ8<0, the denominators of si are positive, i=1,2, and limξ⟶+∞eλ7ξ/eλiξ=0, i=8,9. Since the signs of molecules of si may not be positive, i=1,2, α2,α3≥0 and α2,α3≠0,0. If α2=0, we suppose α3>0; then, −α3τ2eλ9ξ>0 for large ξ,φ2>k2. We suppose α3<0; similarly, φ1>k1 for large ξ. Thus, α3=0, which is a contradiction. Hence, α2>0. Thus, (115) can be simplified into(118)φ1ξ=k1−α1s1eλ7ξ−α2eλ8ξ+h.o.t.,φ2ξ=k2−α1s2eλ7ξ−α2τ1eλ8ξ+h.o.t.,φ3ξ=1−α1eλ7ξ+h.o.t..
Whenλ9<λ7<λ8<0, the proof is similar.
In the following, we assume that (110) does not hold, namely, λ7=λ8 or λ7=λ9, and we want to find two independent vectors of B2λr=0 at first. By directly computing, r12=1,λ,τ,τλ,0,0T is a vector of B2λr=0; when λ=λ7=λ8, r12=r21, with τ=τ1>0, and when λ=λ7=λ9, r12=r31, with τ=τ2<0.
Then, we suppose another linearly independent vector isr22=x1,x2,x3,x4,x5,x6T, which satisfies B2λr22=0 and Bλr22=r12, since Bλr12=0. We can solve Bλr22=r12 to find r22. Thus, in fact, r22 satisfies(119)−λ10000r1k1d1cd1−λ−a12r1k1d10−a13r1k1d1000−λ100−a21r2k2d20r2k2d2cd2−λ−a23r2k2d200000−λ10000r3a31k1+a32k2−1d3cd3−λx1x2x3x4x5x6=1λττλ00,which is equivalent to(120)−λx1+x2=1,r1k1d1x1+cd1−λx2−a12r1k1d1x3−a13r1k1d1x5=λ,−λx3+x4=τ,−a21r2k2d2x1+r2k2d2x3+cd2−λx4−a23r2k2d2x5=τλ,−λx5+x6=0,r3a31k1+a32k2−1d3x5+cd3−λx6=0.
Takex6=λx5 into the last formula above, and we get(121)−d3λ2+cλ+r3a31k1+a32k2−1x5=0.
From (103), for any x5, the coefficient of x5 equals zero. If x5=μ¯, then x6=λμ¯, for some μ¯≠0. Thus, we can deduce that(122)−λx1+x2=1,r1k1x1+c−d1λx2−a12r1k1x3−a13r1k1μ¯=d1λ,−λx3+x4=τ,−a21r2k2x1+r2k2x3+c−d2λx4−a23r2k2μ¯=d2τλ.
Takingx1=0, x2=1 gives(123)c−2d1λ−a12r1k1x3=a13r1k1μ¯,−λx3+x4=τ,r2k2x3+c−d2λx4−a23r2k2μ¯=d2τλ.
From the first equation, we get(124)μ¯=c−2d1λ−a12r1k1x3a13r1k1.
Substituting it into the last one gives(125)−λx3+x4=τ,r2k2x3+c−d2λx4−a23r2k2c−2d1λ−a12r1k1x3a13r1k1=d2λ−λx3+x4.
Thus,(126)−λx3+x4=τ,d2λ2+r2k2+a12a23a13r2k2x3+c−2d2λx4=a23r2k2a13r1k1c−2d1λ.
From the first equation, we have(127)−d2λ2+cλ+r2k2+a12a23a13r2k2x3=a23r2k2a13r1k1c−2d1λ−c−2d2λτ.
Whenλ=λ8, let λ˜ be the negative root of(128)d2λ2−cλ−r2k2−a12a23a13r2k2=0.
Sincea12a23/a13r2k2>0, then λ˜<μ4. Moreover, by recalling (113), λ˜<λ8, which implies(129)−d2λ82+cλ8+r2k2+a12a23a13r2k2>0.
Thus,(130)x3=a23r2k2/a13r1k1c−2d1λ8−c−2d2λ8τ1−d2λ82+cλ8+r2k2+a12a23/a13r2k2.
Then, we will proveμ¯>0, which is equivalent to(131)c−2d1λ8a13r1k1>a12a23r2k2c−2d1λ8−a13r1k1c−2d2λ8τ1a132r1k1−d2λ82+cλ8+r2k2+a12a23/a13r2k2.
Sincemaxμ2,μ4<λ8<0, −d2λ82+cλ8+r2k2>0, τ1>0, then(132)a13−d2λ82+cλ8+r2k2c−2d1λ8>0>−a13r1k1c−2d2λ8τ1.
Thus,(133)a13c−2d1λ8−d2λ82+cλ8+r2k2+a12a23a13r2k2>a12a23r2k2c−2d1λ8−a13r1k1c−2d2λ8τ1,which is equivalent to (131).
Whenλ=λ9, from (38),(134)−d2λ92+cλ9+r2k2+a12a23a13r2k2≠0.
Sinceτ2<0, λ9<0, then(135)x3=a23r2k2/a13r1k1c−2d1λ9−c−2d2λ9τ2−d2λ92+cλ9+r2k2+a12a23/a13r2k2≠0.
Ifx3<0, then μ¯>0. If x3>0, then from τ2<0, λ9<0, we have(136)a23r2k2a13r1k1c−2d1λ9−c−2d2λ9τ2>0,which implies(137)−d2λ92+cλ9+r2k2+a12a23a13r2k2>0.
Supposeμ¯≥0; by recalling λ9<minμ2,μ4, from the above proof, we deduce that(138)0>−d2λ92+cλ9+r2k2≥−c−2d2λ9c−2d1λ9a12r1k1τ2>0,which is a contradiction, and thus μ¯<0.
Therefore, whenλ8=λ7>λ9, ξ⟶+∞, (99) has the following solutions:(139)Φ1+′=C1r21eλ8ξ+C2r21ξ+r22eλ8ξ+C3r31eλ9ξ,where Ci, i=1,2,3, are arbitrary constants and r22=0,1,τ3,τ1+λ8τ3,μ¯1,λ8μ¯1T.(140)τ3=a23r2k2/a13r1k1c−2d1λ8−c−2d2λ8τ1−d2λ82+cλ8+r2k2+a12a23/a13r2k2,μ¯1=c−2d1λ8−a12r1k1τ3a13r1k1>0.
By the stable manifold theorem, we have(141)φ1ξ=k1−α1eλ8ξ−α2ξeλ8ξ+h.o.t.,φ2ξ=k2−α1τ1eλ8ξ−α2τ1ξeλ8ξ+h.o.t.,φ3ξ=1−α2μ¯1eλ8ξ+h.o.t.,for some α1≥0, α2>0.
Moreover, whenλ7=λ9<λ8, ξ⟶+∞, (99) has the following solutions:(142)Φ1+′=C1r21eλ8ξ+C2r31eλ9ξ+C3r31ξ+r22eλ9ξ,where Ci, i=1,2,3, are arbitrary constants and r22=0,1,τ4,τ2+λ9τ4,μ¯2,λ9μ¯2T.(143)τ4=a23r2k2/a13r1k1c−2d1λ9−c−2d2λ9τ2−d2λ92+cλ9+r2k2+a12a23/a13r2k2,μ¯2=c−2d1λ9−a12r1k1τ4a13r1k1.
By the stable manifold theorem, we have(144)φ1ξ=k1−α1eλ8ξ−α3ξeλ9ξ+h.o.t.,φ2ξ=k2−α1τ1eλ8ξ−α2τ2ξeλ9ξ+h.o.t.,φ3ξ=1−α3μ¯2eλ9ξ+h.o.t.,for some α1>0, α3>0μ¯2>0, α3<0μ¯2<0.


## 4. Other Properties

In this section, we adopt the sliding method to prove the strict monotonicity and the uniqueness. Similar to [13], we first give the strong comparison principle.Lemma 9.
Assume that (P1), (P2), and (5) hold. For c≥c∗, let φ1ξ and φ2ξ be two nondecreasing solutions of (11) and (12). Then, either φ11<φ12, φ21<φ22, and φ31<φ32 in ℝ or φ11≡φ12, φ21≡φ22, and φ31≡φ32 in ℝ.Finally, we finish the proof of Theorem4.Proof of Theorem4.
Letφξ be a nondecreasing solution of (11) and (12); then, φi′ξ≥0, i=1,2,3. By Theorem 6 and Corollary 1, for a large enough N>0, we have φi′ξ>0, ξ∈−∞,−N, i=1,2,3. We need to prove φi′ξ>0 for ξ∈−N,+∞, i=1,2,3. Let ξ4 be the left-most point such that φ1′ξ4=0; then, φ1′ξ4 is the minimum of φ1′ξ, implying φ1′′ξ4=0. Differentiating the first equation of (11) gives(145)0=d1J1∗φ1′ξ4+r1φ1ξ4a12φ2′ξ4−cτ12+a13φ3′ξ4−cτ13≥0.
Thus, we concludeφ1′ξ4−y=0, y∈ℝ, which contradicts the definition of ξ4. Therefore, φ1′ξ>0. With similar process, φ2′ξ,φ3′ξ>0 in ℝ.
In the end, the proof of Theorem5 is completely similar to [14, 15], so we omit it here.



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