# Dynamics of Lotka-Volterra Competition Systems with Fokker-Planck Diffusion

Authors: Jaywan Chung; Ohsang Kwon
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We consider competition systems of two species which have different dispersal strategies and interspecific competitive strengths. One of the dispersal strategies is random dispersal and the other is a Fokker-Planck diffusion whose motility is piecewise constant and jumps up when the resource is not enough. In this paper, first we show the Fokker-Planck diffusion allows ideal free distribution. Next we show the linear stability of semitrivial steady states is determined exactly by a threshold on the interspecific competitive strengths. Some conditions for coexistence and global asymptotic stability are also provided.


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## Body

## 1. Introduction and Main Result

In this paper, we study Lotka-Volterra competition systems with Fokker-Planck diffusion of the following form:(1)ut=ΔD1s1u+umx-u-b2vinΩ,vt=ΔD2s2v+vmx-b1u-vinΩ,∂∂νD1s1u=∂∂νD2s2v=0on∂Ω,ux,0=u0x≥0,vx,0=v0x≥0inΩ,where Ω is a smooth bounded domain in Rn and a given resource distribution m(x) is a nonconstant C2-function satisfying minΩ¯m>0. The motility functions D1,D2 of Fokker-Planck diffusion are assumed to depend only on(2)s1≔u+b2vm,s2≔vm,respectively. High values of s1,s2 mean a scarcity of resources; the difference between them is that s1 depends on the competitive strength of other species while s2 does not. The nonnegative constants b1,b2 are interspecific competitive strengths. The intraspecific competitive strengths are assumed to be one.We consider two types of motility functions. The first one is the random dispersal (RD) whereDi’s are constant. The other is based on the following step function: (3)γ0s≔l,0≤s≤1,h,1<s<∞,where l and h are positive constants satisfying l<h. For a single species case, if we let s=u/m, the motility function γ(0) models the situation where organisms drastically change their departing probability in random walk from l to h exactly when the resource becomes insufficient. This motility function is calledstarvation-driven diffusion (SDD) [1]. Unfortunately, due to the jump discontinuity, the Laplacian of γ(0)(s) is not well-defined. So we consider a regularized function γ(ϵ) which is anondecreasing smooth function such that l≤γ(ϵ)≤h and(4)γϵs=l,0≤s≤1,h,1+ϵ<s<∞for given ϵ>0. For example, we may choose γ(ϵ)≔γ(0)∗η(ϵ) where η(ϵ) is a standard mollifier supported in (0,ϵ).The second motility function is given by thisγ(ϵ). In such cases we may choose D1(s1)≔γ(ϵ)(s1) or D2(s2)≔γ(ϵ)(s2). For simplicity we will call the first one SDD1 and the second one SDD2.Our first main result shows why the SDD is worthy of notice; it is an efficient dispersal strategy helping the steady state be theideal free distribution (IFD).Theorem 1 (approximate IFD).
Suppose there is a unique positive solutionθγ(ϵ) of(5)Δγϵsu+um-u=0inΩ,∂∂νγϵsu=0on∂Ω,where(6)s≔um.If the resource distribution m satisfies(7)maxΩ¯⁡mminΩ¯⁡m<hl,then(8)m-θγϵL∞Ω¯=Oϵ1/n+12asϵ→0.The uniqueness assumption in Theorem1 can be replaced by some sufficient conditions. For example, a condition on the resource distribution,(9)maxΩ¯⁡m≤2minΩ¯⁡m,guarantees the uniqueness [2, Theorem 3(i)]. There are other sufficient conditions restricting the shape of γ(ϵ) [2, Theorem 2] or the shape of m [2, Theorem 3(ii)].Condition (7) seems not to be a real restriction because it is necessary even when there is no logistic growth term. In [2, Theorem 5], with no logistic growth term, it was shown that(10)m-θγϵL∞Ω¯=Oϵasϵ→0under assumption (7) although their choice of γ(ϵ) is explicit and slightly different from ours; for example, they assume γ(ϵ)(1)=(h+l)/2 but we assume γ(ϵ)(1)=l. Furthermore it was suggested that condition (7) seems almost optimal [2, Remark 3].With the logistic term the same IFD property (10) is proved in [2, Theorem 6]. But it needs a restrictive condition on m [2, equation (62)] and relies on the explicit form of γ(ϵ). Although our decay order in (8) is worse than the order in (10), our theorem does not have such restrictions.In the sense of Theorem1, the motility function γ(ϵ) gives a best possible dispersal strategy in single species case; the steady state population distribution θγ(ϵ) is very close to the resource distribution and furthermore the steady state is globally asymptotically stable [2, Theorem 2].Next we consider competitions between two dispersal strategies among RD,SDD1, and SDD2. For instance, to observe the competition between SDD1 and RD, we let D1≔γ(ϵ)(s1) and D2≔d for some positive constant d in (1). Then system (1) can have several steady states such as the trivial steady state (u≡0,v≡0) and semitrivial steady states (θγ(ϵ),0) and (0,θd). Here θγ(ϵ) is a positive solution of (5) and θd is a unique positive solution of(11)dΔθd+θdm-θd=0inΩ,∂∂νθd=0on∂Ω.Inspecting the linear stability of each steady state, we can understand which dispersal strategy is more advantageous.In the competition betweenSDD1 and RD, the linear stability of the semitrivial steady state (0,θd), which means RD prevails and SDD1 goes extinct, changes exactly across a threshold b2∗>1 on RD’s interspecific competitive strength. Such a threshold exists for the linear stability of (θγ(ϵ),0) but it is one so less than b2∗. This suggests that SDD1 gains an advantage over RD since the RD needs more interspecific competitive strength to prevail. In the competition between RDs, the RD could prevail with smaller interspecific competitive strength b2<1 [3, Theorem 1.1].Theorem 2 (SDD1 versus RD).
Assume the hypotheses in Theorem1 and let D1≔γ(ϵ)(s1) with s1≔u+b2v/m and D2≔d>0. Then we obtain the following:(i)
Assumeϵ is sufficiently small. There exists a constant b2∗ such that the semitrivial steady state (0,θd) is linearly stable if b2>b2∗ and is linearly unstable if b2<b2∗. Furthermore,(12)1<b2∗≤maxΩ¯⁡mminΩ¯⁡m.(ii)
Assumeϵ is sufficiently small. If b1>1, then (θγ(ϵ),0) is linearly stable. If b1<1, then (θγ(ϵ),0) is linearly unstable.(iii)
Whenb1≥1≥b2,  b1>b2, and l≤d, there is no positive coexistence steady state for any ϵ.In the competition between RDs, the one with smaller diffusivity always prevails [4, 5] if b1=b2=1. Also even if b2 is slightly less than one (v has weak strength), a lower diffusivity of v (v has strong dispersal strategy) can overcome the disadvantage and make v prevail [3].But in the competition betweenSDD1 and RD, Theorem 2 (i) shows that RD cannot prevail no matter how small diffusivity it has if b2<b2∗. When b1=b2=1, the same observation was given in [6, Theorem 1]. Furthermore Theorem 2 (ii) claims that SDD1 loses its advantage as soon as b1 is smaller than one. This means that low interspecific competitive strength of SDD1 cannot be overcome by the dispersal strategy, contrary to the RD case. This is unexpected since the SDD1 seems to be more competitive than RD; the SDD1 can build up the ideal free distribution while the RD cannot.In the competition between RD andSDD2, the dynamics is similar to the case of Theorem 2 but SDD2 is less competitive than SDD1.Theorem 3 (RD versus SDD2).
Assume the hypotheses in Theorem1 and let D1≔d>0 and D2≔γ(ϵ)(s2) with s2≔v/m. Then we obtain the following:(i)
Assumeϵ is sufficiently small. There exists a constant b1∗ such that the semitrivial steady state (θd,0) is linearly stable if b1>b1∗ and is linearly unstable if b1<b1∗. Furthermore,(13)∫Ωm∫Ωθd≤b1∗≤maxΩ¯⁡mminΩ¯⁡m.(ii)
Assumeϵ is sufficiently small. If b2>1, then (0,θγ(ϵ)) is linearly stable. If b2<1, then (0,θγ(ϵ)) is linearly unstable.(iii)
With theb1∗ in (i), suppose either (a)b1<b1∗ and b2<1 or (b)b1>b1∗ and b2>1. Then there is a coexistence steady state for all sufficiently small ϵ.(iv)
Ifb1<1<b2 and l≤d, then (0,θγ(ϵ)) is globally asymptotically stable for all sufficiently small ϵ.Since∫Ωm/∫Ωθd<1 for any d>0 [7, Theorem 1.2], b1∗ is not necessarily bigger than one. So smaller interspecific competitive strength is enough for RD to prevail against SDD2 than the strength required against SDD1 is.By Theorem3 (iv), we can see that b1∗≥1 when l≤d. When b1=b2=1, it is known that if l<d, (0,θγ(ϵ)) is globally asymptotically stable; if l>d, (θd,0) is linearly stable [8, Theorem 2.1].Finally, we consider the competition betweenSDD1 and SDD2.Theorem 4 (SDD withs1 versus SDD with s2).
Assume the hypotheses in Theorem1 and let D1≔γ(ϵ)(s1)>0 and D2≔γ(ϵ)(s2) where s1≔u+b2v/m;  s2≔v/m. Then for all sufficiently small ϵ, we obtain the following:(i)
Ifb1>1, then (θγ(ϵ),0) is linearly stable. If b1<1, then (θγ(ϵ),0) is linearly unstable.(ii)
Ifb2>1, then (0,θγ(ϵ)) is linearly stable. If b2<1, then (0,θγ(ϵ)) is linearly unstable.(iii)
Ifb1≥1>b2, then there is no coexistence steady state.Although we expectSDD1 to be more competitive than SDD2, our linear stability analysis does not indicate it. We solve system (1) numerically to observe it in the next section (see Figure 3).The proofs of the theorems are given in the subsequent sections.


## 2. Numerical Simulation and Conjecture

In this section, by numerically solving the competition system (1), we observe how competitive the dispersal strategies RD, SDD1, and SDD2 are.We fix the following parameters:(14)Ω=0,π⊂R,mx=2sin⁡x+1,l=1,h=4,ϵ=0.1.To construct the motility function γ(ϵ), we choose the following mollifier whose support is [0,1]: (15)ηy≔exp⁡-1/1-1-2y2Iif0<y<1,0otherwise,where I≃0.221997 is the normalization constant which makes ∫η=1. Let η(ϵ)(y)≔η(y/ϵ)/ϵ and a(s)≔min⁡{(s-1)+/ϵ,1}. Then our motility function is given by(16)γϵs≔γ0∗ηϵs=h∫0asη+l∫as1ηand d/dsγ(ϵ)(s)=(h-l)η(ϵ)(s-1).To numerically solve the competition system (1), we use the pdepe function of MATLAB [9] which implements a method of lines [10] with 100 evenly spaced spatial grid points. The mass ratio of u, ∫u(t)dx/∫(u+v)(t)dx, at t=1000 is plotted in Figures 1, 2, and 3 for the situations of Theorems 2, 3, and 4, respectively.Figure 1
S D D 1 versus RD: in the situation of Theorem 2, the mass ratio ∫u(t)dx/∫(u+v)(t)dx of numerical solutions is plotted at t=1000 with the parameters in (14). In the red and blue region SDD1 and RD prevail, respectively. In any cases the SDD1 survives near b1=b2=1 even if RD is stronger than SDD1 (b2>b1).
(a)
d = 2 > l ,   u 0 ≡ m ,   v 0 ≡ 0.1 (b)
d = 2 > l ,   u 0 = v 0 ≡ m / 2 (c)
d = 2 > l ,   u 0 ≡ 0.1 ,   v 0 ≡ m (d)
d = l ,   u 0 ≡ m ,   v 0 ≡ 0.1 (e)
d = l ,   u 0 = v 0 ≡ m / 2 (f)
d = l ,   u 0 ≡ 0.1 ,   v 0 ≡ m (g)
d = 0.5 < l ,   u 0 ≡ m ,   v 0 ≡ 0.1 (h)
d = 0.5 < l ,   u 0 = v 0 ≡ m / 2 (i)
d = 0.5 < l ,   u 0 ≡ 0.1 ,   v 0 ≡ mFigure 2
RD versusSDD2: in the situation of Theorem 3, the mass ratio ∫v(t)dx/∫(u+v)(t)dx of numerical solutions is plotted at t=1000 with the parameters in (14). In the red and blue region SDD2 and RD prevail, respectively. Unlike Figure 1, the SDD2 can go extinct near b1=b2=1 if d<l.
(a)
d = 2 > l ,   v 0 ≡ m ,   u 0 ≡ 0.1 (b)
d = 2 > l ,   v 0 = u 0 ≡ m / 2 (c)
d = 2 > l , v 0 ≡ 0.1 , u 0 ≡ m (d)
d = l ,   v 0 ≡ m ,   u 0 ≡ 0.1 (e)
d = l ,   v 0 = u 0 ≡ m / 2 (f)
d = l , v 0 ≡ 0.1 ,   u 0 ≡ m (g)
d = 0.5 < l , v 0 ≡ m ,   u 0 ≡ 0.1 (h)
d = 0.5 < l ,   v 0 = u 0 ≡ m / 2 (i)
d = 0.5 < l ,   v 0 ≡ 0.1 ,   u 0 ≡ mFigure 3
S D D 1 versus SDD2: in the situation of Theorem 4, the mass ratio ∫u(t)dx/∫(u+v)(t)dx of numerical solutions is plotted at t=1000 with the parameters in (14). In the red and blue region SDD1 and SDD2 prevail, respectively. The SDD1 is advantageous since it survives near b1=b2=1.
(a)
u 0 ≡ m ,   v 0 ≡ 0.1 (b)
u 0 = v 0 ≡ m / 2 (c)
u 0 ≡ 0.1 ,   v 0 ≡ mWhen the interspecific competitive strengths are small (b1,b2≪1), the opponents are relatively weak so two species can coexist (see [3, 7, 11, 12] for other models). Numerical simulation shows in our case the theoretically claimed constants b2∗, b1∗ or one play as the “smallness” thresholds of the interspecific competitive strengths for coexistence. For example, in Figure 1, two species coexist if b1<1 and b2<b2∗.Combining our analysis and numerical observation, we give some conjectures.Conjecture.For simplicity, we ignore unstable coexistence steady states. Assume ϵ is sufficiently small. Then we conjecture the following:(1)
WhenSDD1 competes with RD as in Theorem 2 (see Table 1).(2)
When RD competes withSDD2 as in Theorem 3 (see Table 2).(3)
WhenSDD1 competes with SDD2 as in Theorem 4 (see Table 3).Table 1 b 1 < 1 b 1 > 1 b2>b2∗ RD prevails 1<b2<b2∗ coexist S D D 1 survives b2<1 coexist S D D 1 prevailsTable 2 b 2 < 1 b 2 > 1 b1>b1∗ RD prevails b1<b1∗ coexist S D D 2 prevailsTable 3 b 1 < 1 b 1 > 1 b2>1 S D D 2 survives b2<1 coexist S D D 1 prevails


## 3. Proof of Theorem1

To prove Theorem1, we need some lemmas. In the rest of the paper, we omit the ϵ-dependence in proofs if there is no confusion.Lemma 5 (approximation from above).
Assume the hypotheses in Theorem1. Then(17)θγϵ≤1+ϵm.Proof.
Define a mapU by U(u)≔γ(u/m)u. Because γ is nondecreasing, U isstrictly increasing in u so it is invertible. Now define U¯≔lmaxΩ¯⁡m. Then(18)Um=γmmm=lm≤U¯.Let u¯≔U-1(U¯)>0. Then by the monotonicity of U, the above inequality gives m≤u¯. Furthermore u¯ is a supersolution of (5) because(19)ΔU¯+u¯m-u¯=u¯m-u¯≤0and ∂/∂ν(U¯)=0. Similarly, we can show that if we choose U_≔δ for a sufficiently small δ>0 then u_≔U-1(U_) is a subsolution of (5) and u_≤u¯. Because the positive solution θγ is unique, by the super- and subsolution method,(20)u_≤θγ≤u¯.By the second inequality,(21)γθγmθγ=Uθγ≤U¯=lmaxΩ¯⁡m.Also from [2, Lemma 1(i)] we know(22)minΩ¯⁡m≤θγ≤maxΩ¯⁡m.Now suppose θγ(x0)>(1+ϵ)m(x0) at some point x0∈Ω¯. Then, at that point, θγ/m>1+ϵ so by (4),(23)γθγmθγ=hθγ≥hminΩ¯⁡m.Combining (21) and (23), we have(24)maxΩ¯⁡mminΩ¯⁡m≥hl,which contradicts to assumption (7). Therefore θγ≤(1+ϵ)m.Lemma 6 (approximation from below).
Assume the hypotheses in Theorem1 and fix a number δ such that 0<δ<1. Then the Lebesgue measure of the set(25)Ωϵ≔x∈Ω:1-ϵδm>θγϵhas a bound(26)Ωϵ≤Cϵ1-δfor some positive constant C independent of ϵ.Proof.
By [2, Lemma 1(ii)], we know(27)∫Ωm≤∫Ωθγ.By definition of Ω(ϵ) and Lemma 5, the right-hand side is estimated by(28)∫Ωθγ=∫Ωϵθγ+∫Ω∖Ωϵθγ<∫Ωϵ1-ϵδm+∫Ω∖Ωϵ1+ϵm.Combining the above inequalities, we have(29)∫Ωϵm<ϵ1-δ∫Ω∖Ωϵm.Now the conclusion follows easily.Now we are ready to prove the IFD property.Proof of Theorem1.
(1) First we claim that∇θγ has a uniform bound independent of ϵ. Because θγ is a solution of Neumann problem (5), by the global Lipschitz regularity for Poisson’s equation (see, e.g., [13, Theorem 1.3]), we have(30)∇γsθγL∞Ω≤Cθγm-θγL∞Ω,s≔θγm,where C is a positive constant independent of ϵ. Because θγ has a uniform bound independent of ϵ, so does ∇(γ(s)θγ). Furthermore, from the computation(31)∇γsθγ=γ∇θγ+γ′θγ∇θγm,we have(32)γ+γ′θγm︸≥l∇θγ=∇γθγ+γ′θγm2∇m.From this relation, we can observe that ∇θγ also has a uniform bound as claimed; if we divide the both sides by γ+γ′θγ/m, then we can easily see that the gradient has a bound.
(2) Next we claim that(33)m-θγLpΩp=Oϵp/p+1asϵ→0.By Lemmas 5 and 6, for sufficiently small ϵ, we have(34)m-θγ≤ϵδminΩ∖Ωϵ.Hence(35)∫Ωm-θγp=∫Ωϵm-θγp+∫Ω∖Ωϵm-θγp≤2maxΩ¯⁡mpΩϵ+ϵδp∫Ωmp≤Cϵ1-δ+ϵδp.By choosing 1-δ=δp or δ=1/(p+1), we obtain the claim.
(3) Finally by Gagliardo-Nirenberg interpolation inequality [14, Theorem 5.9],(36)m-θγL∞Ω≤Cm-θγW1,∞Ωn/n+pm-θγLpΩp/n+pfor any p≥1. Applying the previous claims, we have(37)m-θγL∞Ω≤Cϵp/p+1n+p.The exponent of ϵ is maximized when p=n and the proof is complete.


## 4. Proof of Theorem2

First we introduce the diffusion pressureU for the first species u:(38)U≔γs1u,s1=u+b2vm.Define(39)Gm,u,v,U≔γu+b2vmu-U.Then for given m,v, and U, the function u can be completely determined by the equation G≡0 since γ is a nondecreasing function. Hence, we may write u=u(m,v,U), and (1) is rewritten as(40)Ut=∂G∂uΔU+um-u-b2v+∂G∂vdΔv+vm-b1u-vinΩ,vt=dΔv+vm-b1u-vinΩ,∂∂νU=∂∂νv=0on∂Ω,Ux,0=U0x≥0,vx,0=v0x≥0inΩ,where U0≔γ(u0+b2v0/m)u0.Let(ue,ve) be a steady state solution of (1). The stability of (ue,ve) is equivalent to that of the steady state (Ue,ve) of (40) for Ue≔γ(ue+b2ve/m)ue. Let U=Ue+Φ and v=ve+ψ with |Φ| and |ψ| small. Then,(41)u=um,ve+ψ,Ue+Φ=ue+∂u∂vψ+∂u∂UΦ+higher  order  termsand we have the following linearized eigenvalue problem:(42)λΦ=∂G∂uΔΦ+m-2ue-b2ve∂u∂UΦ+m-2ue-b2ve∂u∂v-b2ueψ+∂G∂vdΔψ+m-b1ue-2ve-b1ve∂u∂vψ-b1ve∂u∂UΦinΩ,λψ=dΔψ+m-b1ue-2ve-b1ve∂u∂vψ-b1ve∂u∂UΦinΩ,∂∂νΦ=∂∂νψ=0on∂Ω.At the semitrivial steady state (0,θd),(43)∂u∂U=1γb2sd,∂u∂v=0,∂G∂u=γb2sd,∂G∂v=0,where(44)sd≔θdm.So the eigenvalue value problem for (40) at (0,θd) is(45)λΦ=γb2sdΔΦ+m-b2θdγb2sdΦinΩ,λψ=dΔψ+m-2θdψ-b1θdγb2sdΦinΩ,∂∂νΦ=∂∂νψ=0on∂Ω.Also consider the other semitrivial steady state(θγ,0) and define(46)Uγ≔γsγθγ,sγ≔θγm.Then, similarly, the eigenvalue problem linearized at (Uγ,0) is(47)λΦ=γsγ+γ′sγsγΔΦ+m-2θγγsγ+γ′sγsγΦ-b2θγψ+b2γ′sγsγdΔψ+m-b1θγψinΩ,λψ=dΔψ+m-b1θγψinΩ,∂∂νΦ=∂∂νψ=0on∂Ω.We will use the following stability criteria for semitrivial steady states. The proof is similar to that of [6, Lemma 1], but we give it here for completeness.Lemma 7.
Letθγ be unique positive steady states of (5).(i)
Letμ1 be the first eigenvalue of the eigenvalue problem(48)μϕ=Δϕ+m-b2θdγb2sdϕinΩ,∂∂νϕ=0on∂Ω.Ifμ1>0, then the semitrivial steady state (0,θd) of (1) is linearly unstable. If μ1<0, then (0,θd) is linearly stable.(ii)
Letν1 be the first eigenvalue of the eigenvalue problem(49)νψ=dΔψ+m-b1θγψinΩ,∂∂νψ=0on∂Ω.Ifν1>0, then the semitrivial steady state (θγ,0) is linearly unstable. If ν1<0,  (θγ,0) is linearly stable.Proof.
(i) First supposeμ1>0. For a given λ, consider the eigenvalue problem(50)σΦ=ΔΦ+m-b2θdγb2sdΦ-λγb2sdΦinΩ,∂∂νΦ=0on∂Ωand denote its first eigenvalue by σ1λ. Then σ10=μ1>0 by (48). Moreover, σ1λ will be negative when λ is large enough. Thus, there exists λ1>0 such that σ1λ1=0 and the corresponding eigenfunction Φ1 satisfies the first equation in (45) with λ=λ1. Now due to (11), it is easy to see that the operator(51)L≔dΔ+m-2θd-λ1:C2+αΩ¯→CαΩ¯is invertible. Hence there is a solution of Lψ=b1θd/γ(b2sd)Φ1 and it satisfies the second equation in (45). This yields that λ1>0 is an eigenvalue of the linearized problem (45), which implies that (0,θd) is linearly unstable.
Next, assume thatμ1<0. Suppose that the linearized problem (45) has a nonnegative eigenvalue λ1≥0 with the corresponding eigenfunctions (Φ1,ψ1). Since θd is a positive solution of (11), the operator L≔dΔ+(m-2θd) has only strictly negative eigenvalues. But if Φ1≡0, by the second equation in (45), the operator L has a nonnegative eigenvalue λ1, which is a contradiction. Hence Φ1≢0. However, by the Rayleigh quotient of (48) and (45), this yields that(52)μ1=supϕ∈H1Ω∖0⁡∫Ω-∇ϕ2+m-b2θd/γb2sdϕ2∫Ωϕ2≥∫Ω-∇Φ12+m-b2θd/γb2sdΦ12∫ΩΦ12=∫Ωλ1/γb2sdΦ12dx∫ΩΦ12dx≥0.This is a contradiction. Therefore, (0,θd) is linearly stable.
(ii) Consider the linearized eigenvalue problem of (5) at Uγ, which is written as(53)λΦ=Δ+m-2θγ∂u∂UΦ.Since θγ is a unique positive steady state, it is linearly neutrally stable (in fact it is globally asymptotically stable [2, Theorem 2]). Hence every eigenvalue of the linear operator Δ+(m-2θγ)∂u/∂U is nonpositive. Thus, if ν1>0, because of ∂G/∂u=γ(sγ)+γ′(sγ)sγ>0, the operator(54)L≔Δ+m-2θγ∂u∂U-ν1∂G∂u-1has strictly negative eigenvalues and thus it is invertible. Let {ν1,ψ1} be an eigenpair of (49) and Φ1 be the solution of(55)LΦ1=b2θγψ1-∂G∂u-1∂G∂vdΔψ1+m-b1θγψ1.Then, {ν1,(ϕ1,Ψ1)} is an eigenpair of the linearized problem (47), which implies that (Uγ,0) is linearly unstable.
Since an eigenvalue of (47) is also an eigenvalue of (49), eigenvalues of (47) are all strictly negative if ν1<0 and hence (Uγ,0) is linearly stable.Now we are ready to prove Theorem2. Due to Lemma 7, it is sufficient to compute the sign of the first eigenvalues μ1 and ν1.Proof of Theorem2.
(i) Recall(56)μ1=supϕ∈H1Ω∖0⁡∫Ω-∇ϕ2+m-b2θd/γb2sdϕ2∫Ωϕ2.When b2=1, by assumption (7), we can show that μ1>0 (see the proof of [6, Theorem 1]). On the other hand, because γ(s) is a nondecreasing function, m-b2θd/γ(b2sd) and μ1 decrease strictly as b2 increases. By the continuity of μ1, there exists a constant b2∗>1 such that (0,θd) is linearly unstable for all b2<b2∗ and (0,θd) is linearly stable for all b2>b2∗.
Furthermore, sinceminΩ¯⁡m≤minΩ¯⁡θd<maxΩ¯⁡θd≤maxΩ¯⁡m (one may check this using the maximum principle), when b2≥maxΩ¯⁡m/minΩ¯⁡m, m-b2θd≤0 in Ω. Thus, μ1≤0 for all b2≥maxΩ¯⁡m/minΩ¯⁡m. This implies that b2∗≤maxΩ¯⁡m/minΩ¯⁡m.
(ii) Recall(57)ν1=supψ∈H1Ω∖0⁡∫Ω-d∇ψ2+m-b1θγψ2∫Ωψ2.By Theorem 1, ‖m-θγ‖L∞(Ω¯)→0 as ϵ→0. Therefore, for a given b1>1, there is a δ>0 such that(58)m-b1θγ=m-θγ-b1-1θγ≤-δfor  any  sufficiently  smallϵ.Hence for any ψ∈H1(Ω)∖{0},(59)∫Ω-d∇ψ2+m-b1θγψ2∫Ωψ2≤-δ,which implies ν1<0.
Similarly, for a givenb1<1, there is a δ>0 such that(60)m-b1θγ=m-θγ+1-b1θγ≥δfor  any  sufficiently  smallϵ.Then with ψ1≔1,(61)ν1=supψ∈H1Ω∖0⁡∫Ω-d∇ψ2+m-b1θγψ2∫Ωψ2≥∫Ω-d∇ψ12+m-b1θγψ12∫Ωψ12≥δ>0.
(iii) Suppose there is a positive coexistence steady state(u,v). Then by (40),(62)ΔU+m-u-b2vγu+b2v/mU=0inΩ,∂∂νU=0on∂Ω.Considering the Rayleigh quotient of this equation, we have(63)0=supϕ∈H1Ω∖0⁡∫Ω-∇ϕ2+m-u-b2v/γu+b2v/mϕ2∫Ωϕ2.Similarly we also have(64)∫Ωm-b1u-vv2=∫Ωd∇v2.Choosing ϕ=v in (63),(65)0≥∫Ω-∇v2+m-u-b2vγu+b2v/mv2=∫Ω-1dm-b1u-vv2+1γu+b2v/mm-u-b2vv2by64=∫Ω1l-1dm-b1u-vv2+1γu+b2v/m-1lm-u-b2vv2+1lb1-1u+1-b2vv2.>∫Ω1γu+b2v/m-1lm-u-b2vv2by  the  assumption  and64.Now observe the sign of the integrand. If s1=u+b2v/m≤1,  γ(s1)=l so the integrand vanishes. Hence the integral is actually taken over the set where u+b2v/m>1, and then the last integral is positive. This is a contradiction so there is no coexistence steady state.


## 5. Proof of Theorem3

In this section, we consider the following system:(66)ut=dΔu+umx-u-b2vinΩ,vt=Δγs2v+vmx-b1u-vinΩ,∂∂νu=∂∂νγs2v=0on∂Ω,ux,0=u0x≥0,vx,0=v0x≥0inΩ,where s2≔v/m. We introduce the diffusion pressure V for the second species v:(67)V≔γs2v,Vγ≔γsγθγ,sγ≔θγmwhere θγ is a unique positive solution of (5). Define(68)Gm,v,V≔γs2v-Vthen(69)∂G∂v=γs2+γ′s2s2>0and (66) is rewritten as(70)ut=dΔu+um-u-b2vinΩ,Vt=∂G∂vΔV+vm-b1u-vinΩ,∂∂νu=∂∂νV=0on∂Ω,ux,0=u0x≥0,Vx,0=V0x≥0inΩ,where V0≔γ(v0/m)v0. The corresponding eigenvalue value problems linearized at the semitrivial steady states (θd,0) and (0,Vγ) are(71)λϕ=dΔϕ+m-2θdϕ-b2lθdΨinΩ,λΨ=lΔΨ+m-b1θdΨinΩ,∂∂νϕ=∂∂νΨ=0on∂Ω,and(72)λϕ=dΔϕ+m-b2θγϕinΩ,λΨ=γsγ+γ′sγsγΔΨ+m-2θγγsγ+γ′sγsγΨ-b1θγϕinΩ,∂∂νϕ=∂∂νΨ=0on∂Ω,respectively.Omitting the proof, we use the following stability criteria for semitrivial steady states; the proof is similar to the one of Lemma7.Lemma 8.
Letθγ be unique positive steady states of (5).(i)
Letμ1 be the first eigenvalue of(73)μΨ=lΔΨ+m-b1θdΨinΩ,∂∂νΨ=0on∂Ω,Ifμ1>0, then the semitrivial steady state (θd,0) of (66) is linearly unstable. If μ1<0, then (θd,0) is linearly stable.(ii)
Letν1 be the first eigenvalue of(74)νϕ=dΔϕ+m-b2θγϕinΩ,∂∂νϕ=0on∂Ω,Ifν1>0, then (0,θγ) is linearly unstable. If ν1<0, (0,θγ) is linearly stable.Because system (66) has no cross-diffusion (a cross-diffusion is a Fokker-Planck diffusion depending on both u and v), we can prove that it is a strongly monotone dynamical system.Lemma 9 (monotonicity).
Let(u,v) and (u~,v~) be solutions of (66) with initial data (u0,v0) and (u~0,v~0), respectively. If u0(x)≤u~0(x) and v0(x)≥v~0(x), then u(x,t)<u~(x,t) and v(x,t)>v~(x,t).We omit the proof; it is a slight modification of the proof in [8, pp.266–267].Now we give a proof Theorem3. Due to Lemma 8, it suffices to check the sign of μ1 and ν1 for the linear stability.Proof of Theorem3.
(i) Since(75)μ1=supΨ∈H1Ω∖0⁡∫Ω-l∇Ψ2+m-b1θdΨ2∫ΩΨ2,μ1 decreases strictly as b1 increases. If b1≥maxΩ¯⁡m/minΩ¯⁡m, then m-b1θd≤0 and μ1≤0.
Ifb1<∫Ωm/∫Ωθd, choosing Ψ≡1 in the above Rayleigh quotient, we obtain(76)μ1≥1Ω∫Ωm-b1θd>0.Therefore, there exists a constant b1∗ such that ∫Ωm/∫Ωθd≤b1∗≤maxΩ¯⁡m/minΩ¯⁡m, μ1<0 when b1>b1∗, and μ1>0 when b1<b1∗.
(ii) The proof is the same as the one for Theorem2 (ii).
(iii) By (i), (ii), and the assumption, the semitrivial steady states(θd,0) and (0,θγ) are either both stable or both unstable. Hence there is a coexistence steady state by virtue of the monotonicity of system [15].
(iv) By an argument almost similar to the proof of Theorem2 (iii), we can show there is no coexistence steady state. Also (0,θγ) is linearly stable by (ii). Therefore by the theory of monotone dynamical system [16, Proposition 9.1 and Theorem 9.2], the linearly stable semitrivial steady state (0,θγ) is globally asymptotically stable as in [15, 17].


## 6. Proof of Theorem4

We consider the following system:(77)ut=Δγs1u+umx-u-b2vinΩ,vt=Δγs2v+vmx-b1u-vinΩ,∂∂νγs1u=∂∂νγs2v=0on∂Ω,ux,0=u0x≥0,vx,0=v0x≥0inΩ,where s1≔u+b2v/m and s2≔v/m. Set(78)U≔γs1u,V≔γs2v,Uγ≔γsγθγ,sγ≔θγm.Then the corresponding eigenvalue problems linearized at the semitrivial steady states (Uγ,0) and (0,Uγ) are(79)λΦ=γsγ+γ′sγsγΔΦ+m-2θγγsγ+γ′sγsγΦ-b2lθγΨ+b2γ′sγsγΔΨ+m-b1θγlΨinΩ,λΨ=lΔΨ+m-b1θγΨinΩ,∂∂νΦ=∂∂νΨ=0on∂Ωand(80)λΦ=γb2sγΔΦ+m-b2θγγb2sγΦinΩ,λΨ=γsγ+γ′sγsγΔΨ+m-2θγγsγ+γ′sγsγΨ-b1θγγb2sγΦinΩ,∂∂νΦ=∂∂νΨ=0on∂Ω,respectively.In a similar way to the proofs of Lemmas7 and 8, we obtain the following stability criterion. We omit the proof.Lemma 10.
Letθγ be unique positive steady states of (5).(i)
Letμ1 be the first eigenvalue of(81)μΨ=lΔΨ+m-b1θγΨinΩ,∂∂νΨ=0.on∂Ω.Ifμ1>0, then the semitrivial steady state (θγ,0) of (77) is linearly unstable. If μ1<0, (θγ,0) is linearly stable.(ii)
Letν1 be the first eigenvalue of(82)νϕ=Δϕ+m-b2θγγb2sγϕinΩ,∂∂νϕ=0.on∂Ω.Ifν1>0, then (0,θγ) is linearly unstable. If ν1<0, (0,θγ) is linearly stable.Proof of Theorem4.
(i), (ii) The proof is similar to the proofs of Theorems2 and 3.
(iii) If there is a coexistence steady state(u,v),(83)0=supψ∈H1Ω∖0⁡∫Ω-∇ψ2+m-u-b2v/γu+b2v/mψ2∫Ωψ2.Choosing ψ=γ(v/m)v,(84)0≥∫Ω-∇γvmv2+m-u-b2vγu+b2v/mγ2vmv2=∫Ωγvmv2γv/mγu+b2v/mm-u-b2v-m-b1u-v.If we show(85)A≔γv/mγu+b2v/mm-u-b2v-m-b1u-vis strictly positive, it causes a contradiction so the proof is complete.
Supposem≥u+b2v. Then γ(v/m)/γ(u+b2v/m)=γ(v/m)/l≥1 so(86)A≥m-u-b2v-m-b1u-v=b1-1u+1-b2v>0.Similarly, if m<u+b2v and m≥v, then(87)γv/mγu+b2v/mm-u-b2v=lγu+b2v/mm-u-b2v≥m-u-b2v.Thus A≥(b1-1)u+(1-b2)v>0.
Finally supposem<u+b2v and m<v. Note that(88)-m-b1u-v=-m+u+b2v+b1-1u+1-b2v>b1-1u+1-b2v≥1-b2v>1-b2m.So if (1+ϵ)m≥u+b2v and ϵ<l/h(1-b2), we have(89)A≥-ϵhlm+1-b2m>0.When (1+ϵ)m<u+b2v, by the structure of γ in (4), γ(v/m)/γ(u+b2v/m)=γ(v/m)/h≤1. Hence(90)A≥b1-1u+1-c1v>0.This completes the proof of Theorem 4.



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*Source: 7879598-2018-09-02.xml*