Citation: Hikaru Nakamiya, Saeka Ijima, Hiromi Nishida. Changes in nucleosome formation at gene promoters in the archiascomycetous yeast Saitoella complicata[J]. AIMS Microbiology, 2017, 3(2): 136-142. doi: 10.3934/microbiol.2017.2.136
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In this paper, we are concerned with the following susceptible-infected-susceptible(SIS) model
{St=(dSSx−a′(x)S)x−β(x)SIS+I+γ(x)I, 0<x<L, t>0,It=(dIIx−a′(x)I)x+β(x)SIS+I−γ(x)I, 0<x<L, t>0,dSSx−a′(x)S=dIIx−a′(x)I=0, x=0,L, t>0,S(x,0)=S0(x), I(x,0)=I0(x),0<x<L. | (1.1) |
Here S(x,t) and I(x,t) denote the density of susceptible and infected individuals in a given spatial interval (0,L), dS and dI are positive constants which stand for the diffusion coefficients for the susceptible and infected populations, a′(x) is a smooth nonnegative function which represents the advection speed rate, while β(x) and γ(x) represent the rates of disease transmission and recovery at location x, which are Hölder continuous functions on (0,L). In addition, S0(x) and I0(x) are continuous and satisfy
(A1)S0(x)≥0 and I0(x)≥0 for x∈(0,L),∫L0I0(x)dx>0. |
We would like to give the survey of some results on SIS model. In [1], Allen et al. investigated a discrete SIS model, in [2], they also proposed the SIS model with no advection in a given spatial region Ω, where they dealt with the existence, uniqueness and asymptotic behaviors of the endemic equilibrium as the diffusion rate of the susceptible individuals approaches to zero. Many authors also considered the SIS reaction–diffusion model, including the global stability of the endemic equilibrium, the effects of large and small diffusion rates of the susceptible and infected population on the persistence and extinction of the disease, discuss how the disease vanish or spreading in high-risk or low-risk domain, and so on. For the dynamics and asymptotic profiles of steady states of an epidemic model in advective environments, we can see[3]. For A SIS reaction-diffusion-advection model in a low-risk and high-risk domain, we can see [4]. For Dynamics of an SIS reaction-diffusion epidemic model for disease transmission, we can see[5], For Concentration profile of endemic equilibrium of a reaction- diffusion-advection SIS epidemic model, we can see [6]. For the varying total population enhances disease persistence, we can see [7]; For the asymptotic profiles of the positive steady state for an SIS epidemic reaction-diffusion model, we can see [8]. For the global stability of the steady states of an SIS epidemic reaction-diffusion model, we can see [9]. For the asymptotic profile of the positive steady state for an SIS epidemic reaction- diffusion model: effects of epidemic risk and population movement, we can see [10]; For reaction-diffusion SIS epidemic model in a time-periodic environment, we can see [11]. For the global dynamics and traveling waves for a periodic and diffusive chemostat model with two nutrients and one microorganism, we can see [12]. For more information about dynamical systems in population biology, we also can refer to see [13] and the references therein. Recently, Cui and Lou studied (1.1) when a′(x)≡q for x∈[0,L] in [14], that is, it is a constant advection. Besides establishing the asymptotic stability of the unique disease-free equilibrium(DFE) when R0<1 and the existence of the endemic equilibrium when R0>1, they found that the DFE changes its stability at most once as dI varies from zero to infinity, which is strong contrast with the case of no advection. Since (1.1) has vary advection, an natural and interesting question is whether we can establish the similar results on (1.1) to those in the case of no advection or not.
Since the functions a′(x), β(x), γ(x), S0(x) and I0(x) are continuous in (0,L), by the standard theory for a system of semilinear parabolic equations, (1.1) is locally wellposedness in (0,Tmax). Noticing (A1), by the maximum principle, S(x,t) and I(x,t) are positive and bounded for x∈[0,L] and t∈(0,Tmax). Hence, by the results in [15], Tmax=∞ and (1.1) posses a unique classical solution (S(x,t),I(x,t)) for all time.
It is easy to verify that
∫L0[S(x,t)+I(x,t)]dx=∫L0[S(x,0)+I(x,0)]dx:=N>0,t>0. | (1.2) |
Inspired by [2] and [14], we say that (0,L) is a low-risk domain if ∫L0β(x)dx<∫L0γ(x)dx and high-risk domain if ∫L0β(x)dx>∫L0γ(x)dx.
The corresponding equilibrium system of (1.1) is
{(dS˜Sx−a′(x)˜S)x−β(x)˜S˜I˜S+˜I+γ(x)˜I=0, 0<x<L,(dI˜Ix−a′(x)˜I)x+β(x)˜S˜I˜S+˜I−γ(x)˜I=0, 0<x<L,dS˜Sx−a′(x)˜S=dI˜Ix−a′(x)˜I=0, x=0,L. | (1.3) |
The half trivial solution (˜S(x),0) of (1.3) is called a disease-free equilibrium(DFE), while the solution (˜S(x),˜I(x)) of (1.3) is called endemic equilibrium(EE) if ˜I(x)>0 for some x∈(0,L).
We also introduce the following basic reproduction number as those in literatures [2] and [14]. We also can refer to [16] and see the definition and the computation of the basic reproduction ratio R0 in models for infectious diseases in heterogeneous populations, refer to [17] and see reproduction numbers and sub-threshold endemic equilibria for compartmental models of disease transmission, see basic reproduction numbers for reaction-diffusion epidemic models [18].
R0=supφ∈H1((0,L)),φ≠0{∫L0β(x)ea(x)dIφ2dxdI∫L0ea(x)dIφ2xdx+∫L0γ(x)ea(x)dIφ2dx}. | (1.4) |
Our first result is concerned with the qualitative properties for R0.
Theorem 1.1. Let ˆR0 be the basic reproduction number when a(x)≡0 which was introduced in [2]. Then the following conclusions hold.
(1) For any given a′(x)>0, R0→β(L)γ(L) as dI→0 and R0→∫L0β(x)dx∫L0γ(x)dx as dI→+∞;
(2) For any given dI>0, R0→ˆR0 as maxx∈[0,L]a′(x)→0 and R0→β(L)γ(L) as minx∈[0,L]a′(x)→+∞;
(3) If β(x)>(<)γ(x) on [0,L], then R0>(<1) for any given dI>0 and a′(x)>0.
Our second result deals with the stability of DFE, which will extend those of [2] and [14].
Theorem 1.2. The DFE is unstable if R0>1 while it is globally asymptotically stable if R0<1.
We will analyze (1.1) under the following assumptions on β(x) and γ(x):
(C1) β(0)−γ(0)<0<β(L)−γ(L), i.e., β(x)−γ(x) changes sign from negative to positive,
or
(C2) β(0)−γ(0)>0>β(L)−γ(L), i.e., β(x)−γ(x) changes sign from positive to negative.
In the point view of biological,
(C1) all lower-risk sites are located at the upstream and all high-risk sites are at the downstream,
or
(C2) all high-risk sites are distributed at the upstream and lower-risk sites are at the downstream.
To state other results, in convenience, let q=maxx∈[0,L]a′(x) and denote a(x)=q˜a(x) sometimes in the sequels.
We can get further properties of R0 when ∫L0β(x)dx>∫L0γ(x)dx.
Theorem 1.3. Assume that ∫L0β(x)dx>∫L0γ(x)dx. Denote R0=R0(dI,q).
(i) If (C1) holds, then the DFE is unstable for any q>minx∈[0,L]a′(x)>0 and dI>0;
(ii) If (C2) holds, then there exists a unique curve in dI–q plane
Γ1={(dI,ρ1(dI)):R0(dI,ρ1(dI))=1,dI∈(0,+∞)} |
with the function ρ1=ρ1(dI):(0,+∞)→(0,+∞) satisfying
limdI→0+ρ1(dI)=0,limdI→+∞ρ1(dI)dI=θ1, |
and such that for every dI>0, the DFE is unstable for 0<minx∈[0,L]a′(x)<q<ρ1(dI) and it is globally and asymptotically stable for q>minx∈[0,L]a′(x)>ρ1(dI).
Here θ1 is the unique positive solution of
∫L0[β(x)−γ(x)]eθ1˜a(x)dx=0. |
Similarly, we can get further properties of R0 when ∫L0β(x)dx<∫L0γ(x)dx.
Theorem 1.4. Assume that ∫L0β(x)dx<∫L0γ(x)dx. Let d∗I is the unique positive root of the equation ˆR0=1, where ˆR0 was introduced in [2].
(1) If (C1) holds, then the DFE is unstable for any q>minx∈[0,L]a′(x)>0 and dI∈(0,d∗I], while for dI∈(d∗I,+∞) there exists a unique curve in dI–q plane
Γ2={(dI,ρ2(dI)):R0(dI,ρ2(dI))=1,dI∈(d∗I,+∞)} |
with the monotone function ρ2=ρ2(dI):(d∗I,+∞)→(0,+∞) satisfying
limdI→d∗I+ρ2(dI)=0,limdI→+∞ρ2(dI)dI=θ2, |
and such that the DFE is unstable for 0<minx∈[0,L]a′(x)<q<ρ2(dI) and it is globally asymptotically stable for q>minx∈[0,L]a′(x)>ρ2(dI).
Here θ2 is the unique positive solution of
∫L0[β(x)−γ(x)]eθ2˜a(x)dx=0. |
(2) If (C2) holds, then for dI∈(0,d∗I), there exists a unique curve in dI−q plane
Γ3={(dI,ρ3(dI)):R0(dI,ρ3(dI))=1,dI∈(0,d∗I)} |
with the function ρ3=ρ3(dI):(0,d∗I)→(0,+∞) satisfying
limdI→0+ρ3(dI)=0,limdI→d∗I−ρ3(dI)=0, |
and such that the DFE is unstable for 0<minx∈[0,L]a′(x)<q<ρ3(dI) and it is globally and asymptotically stable for q>minx∈[0,L]a′(x)>ρ3(dI), while for dI∈(d∗I,+∞), the DFE is globally and asymptotically stable for any q>minx∈[0,L]a′(x)>0.
The following theorem deals with the existence of EE.
Theorem 1.5. Assume that β(x)−γ(x) changes sign once in (0,L). If R0>1, then problem (1.3) possesses at least one EE.
The last theorem will consider the results on (1.1) when β(x)−γ(x) changes sign twice in (0,L).
Theorem 1.6. Assume that β(x)−γ(x) changes sign twice in (0,L).
(1) If ∫L0β(x)dx>∫L0γ(x)dx and β(L)<γ(L), then there exists some positive constant Λ which is independent of dI and q such that for every dI>Λ, we can find a positive constant Q which depends on dI such that R0>1 when 0<minx∈[0,L]a′(x)<q<Q and R0<1 when q>Q.
(2) If ∫L0β(x)dx>∫L0γ(x)dx and β(L)>γ(L), then there exists some positive constant Λ which is independent of dI and q such that for every dI>Λ one of the following conclusions holds:
(i) R0>1 for any q>minx∈[0,L]a′(x)>0;
(ii) There exists a positive constant ˆQ which is independent of dI and satisfies that R0>1 for q≠ˆQ and R0=1 when q=ˆQ;
(iii) There exist two positive constants Q2>Q1 both depending on dI such that R0>1 when q∈(0,Q1)∪(Q2,+∞) while R0<1 when q∈(Q1,Q2).
(3) If ∫L0β(x)dx<∫L0γ(x)dx and β(L)>γ(L), then there exists some positive constant Λ>d∗I which is independent of dI and q such that for every dI>Λ, we can find a positive constant Q which depends on dI such that R0<1 when 0<minx∈[0,L]a′(x)<q<Q and R0>1 when q>Q.
(4) If ∫L0β(x)dx<∫L0γ(x)dx and β(L)<γ(L), then there exists some positive constant Λ>d∗I which is independent of dI and q such that for every dI>Λ one of the following conclusions holds:
(iv) R0<1 for any q>minx∈[0,L]a′(x)>0;
(v) There exists a positive constant ˆQ which is independent of dI and satisfies that R0<1 for q≠ˆQ and R0=1 when q=ˆQ;
(vi) There exist two positive constants Q2>Q1 both depending on dI and satisfy that R0<1 when q∈(0,Q1)∪(Q2,+∞) while R0>1 when q∈(Q1,Q2).
The rest of this paper is organized as follows. In Section 2, we give the proofs of Theorem 1.1 and Theorem 1.2. In Section 3, we will prove Theorem 1.3. In Section 4, we will prove Theorem 1.4. In Section 5, we will prove Theorem 1.5. In Section 6, we will prove Theorem 1.6.
In this section, we first give some qualitative properties of R0, then we deal with the stability of DFE, and we can finish the proofs of Theorem 1.1 and Theorem 1.2.
By the definition of R0, there exits some positive function Φ(x)∈C2([0,L]) such that
{−[dIΦx−a′(x)Φ]x+γ(x)Φ=1R0β(x)Φ,0<x<L,dIΦx(0)−a′(0)Φ(0)=0,dIΦx(L)−a′(L)Φ(L)=0. | (2.1) |
Letting φ(x)=e−a(x)dIΦ(x), we have
{−dIφxx−a′(x)φx+γ(x)φ=1R0β(x)φ,0<x<L,φx(0)=0,φx(L)=0. | (2.2) |
Linearizing (1.1) around (ˆS,0) and letting ˉξ(x,t)=S(x,t)−ˆS(x,t), ˉη(x,t)=I(x,t), we have
{ˉξt=(dSˉξx−a′(x)ˉξ)x−[β(x)−γ(x)]ˉη,0<x<L, t>0,ˉηt=(dIˉηx−a′(x)ˉη)x+[β(x)−γ(x)]ˉη,0<x<L, t>0. |
For the linear system, seeking for the solution which is separation of variables, i.e., ˉξ(x,t)=e−λtξ(x) and ˉη(x,t)=e−λtη(x), we have
{(dSξx−a′(x)ξ)x−[β(x)−γ(x)]η+λξ=0,0<x<L,(dIηx−a′(x)η)x+[β(x)−γ(x)]η+λη=0,0<x<L, | (2.3) |
subject to boundary conditions
{dSξx(0)−a′(0)ξ(0)=0,dSξx(L)−a′(L)ξ(L)=0,dSηx(0)−a′(0)η(0)=0,dSηx(L)−a′(L)η(L)=0. | (2.4) |
By the conservation of total population, we need to impose that
∫L0[ξ(x)+η(x)]dx=0. | (2.5) |
Noticing that the second equation of (2.3) is independent of ξ, letting ζ(x)=e−a(x)dIη(x), we only need to consider the following eigenvalue problem
{dIζxx+a′(x)ζx+[β(x)−γ(x)]ζ(x)+λζ(x)=0,0<x<L,ζx(0)=ζx(L)=0. | (2.6) |
By the results of [19], all the eigenvalues are real, the smallest eigenvalue λ1(dI,q) is simple, and its corresponding eigenfunction ϕ1 can be chosen positive.
We will show a fact below.
Lemma 2.1.1. For any dI and q>minx∈[0,L]a′(x)>0, λ1(dI,q)<0 if R0>1, λ1(dI,q)=0 if R0=1 and λ1(dI,q)>0 if R0<1.
Proof. Note that (λ1(dI,q),ϕ1) satisfies
{−dI(ϕ1)xx−a′(x)(ϕ1)x+[γ(x)−β(x)]ϕ1(x)=λ1(dI,q)ϕ1(x), 0<x<L,(ϕ1)x(0)=(ϕ1)x(L)=0. | (2.7) |
Multiplying (2.1) by ea(x)dIϕ1 and (2.7) by ea(x)dIΦ, integrating by parts in (0,L), and subtracting the resulting equations, we get
∫L0(1R0−1)β(x)Φ(x)ϕ1(x)dx=∫L0λ1(dI,q)Φ(x)ϕ1(x)dx. |
Using the mean value theorem of integrating, we have
(1R0−1)β(x1)Φ(x1)ϕ1(x1)=λ1(dI,q)Φ(x2)ϕ1(x2) |
for some 0≤x1≤L and 0≤x2≤L. Using β(x1)Φ(x1)ϕ1(x1)>0 and Φ(x2)ϕ1(x2)>0, we know that
(1R0−1)has the same sign ofλ1(dI,q), |
which implies the conclusions are true.
Lemma 2.1.2. If dIq→0 and dIq2→0, ˜a′(x)>δ>0 for some constant δ, then R0→β(L)γ(L).
Proof. Let w(x)=e−qdIA˜a(x)Φ(x), where Φ(x) is the solution of (2.1), A is a constant which will be chosen later. It is easy to verify that w satisfies
{ [q2A(A−1)dI(˜a′(x))2+q(A−1)˜a″(x)+1R0β(x)−γ(x)]w=−dIwxx+(1−2A)a′(x)wx,0<x<L, t>0,dIwx(0)=a′(0)(1−A)w(0),dIwx(L)=a′(L)(1−A)w(L). | (2.8) |
First we chose A=1+C1dIq2, where C1 is a positive constant to be chosen later. Then (2.8) becomes
{ [C1(1+C1dIq2)(˜a′(x))2+q(1+C1dIq2)˜a″(x)+1R0β(x)−γ(x)]w=−dIwxx−(1+2C1dIq2)a′(x)wx,0<x<L, t>0,dIwx(0)=−C1dIq˜a′(0)w(0),dIwx(L)=−C1dIq˜a′(L)w(L). |
Assume that w(x∗)=minx∈[0,L]w(x). We will show that x∗=L below. wx(0)<0 implies that x∗≠0. If x∗∈(0,L), then wxx(x∗)≥0 and wx(x∗)=0, (2.9) means that
[C1(1+C1dIq2)(˜a′(x∗))2+q(1+C1dIq2)˜a″(x∗)+1R0β(x∗)−γ(x∗)]≤0 |
Taking C1=Kq with K large enough, we can get a contradiction. Therefore, x∗=L and w(x)≥w(L) for x∈[0,L], which implies that
Φ(x)Φ(L)≥e−qdI(1+C1dIq2)[˜a(L)−˜a(x)]. | (2.9) |
Next, we chose A=1−C2dIq2, where C2 is a positive constant to be chosen later. Then (2.8) becomes
{ [C2(1−C2dIq2)(˜a′(x))2+q(1−C2dIq2)˜a″(x)+1R0β(x)−γ(x)]w=−dIwxx−(1−2C2dIq2)a′(x)wx,0<x<L, t>0,dIwx(0)=C2dIq˜a′(0)w(0),dIwx(L)=C2dIq˜a′(L)w(L). |
Assume that w(x∗)=maxx∈[0,L]w(x). We will show that x∗=L below. wx(0)>0 implies that x∗≠0. If x∗∈(0,L), then wxx(x∗)≥0 and wx(x∗)=0, (2.10) means that
[C2(1−C2dIq2)(˜a′(x∗))2+q(1−C2dIq2)˜a″(x∗)+1R0β(x∗)−γ(x∗)]≤0 |
Taking C2=K′q with K′ large enough, we can get a contradiction. Therefore, x∗=L and w(x)≤w(L) for x∈[0,L], which implies that
Φ(x)Φ(L)≤e−qdI(1−C2dIq2)[˜a(L)−˜a(x)]. | (2.10) |
Dividing (2.1) by Φ(L) and integrating the result in (0,L), we have
∫L0γ(x)Φ(x)Φ(L)dx=1R0∫L0β(x)Φ(x)Φ(L)dx. | (2.11) |
Letting y=q[˜a(L)−˜a(x)]dI, i.e., x=˜a−1[˜a(L)−dIyq], we have
e−(1+C1dIq)y≤Φ(˜a−1[˜a(L)−dIyq])Φ(L)≤e−(1−C2dIq)y | (2.12) |
and
∫q[˜a(L)−˜a(0)]dI0γ(˜a−1[˜a(L)−dIyq])Φ(˜a−1[˜a(L)−dIyq])˜a′(˜a−1[˜a(L)−dIyq])Φ(L)dy=1R0∫q[˜a(L)−˜a(0)]dI0β(˜a−1[˜a(L)−dIyq])Φ(˜a−1[˜a(L)−dIyq])˜a′(˜a−1[˜a(L)−dIyq])Φ(L)dy. | (2.13) |
Using (2.12), by Lebesgue dominant convergence theorem, then passing to the limit in (2.13), we get
limdI/q→0,dI/q2→0R0=limdI/q→0,dI/q2→0∫q[˜a(L)−˜a(0)]dI0β(˜a−1[˜a(L)−dIyq])Φ(˜a−1[˜a(L)−dIyq])˜a′(˜a−1[˜a(L)−dIyq])Φ(L)dy∫q[˜a(L)−˜a(0)]dI0γ(˜a−1[˜a(L)−dIyq])Φ(˜a−1[˜a(L)−dIyq])˜a′(˜a−1[˜a(L)−dIyq])Φ(L)dy=∫∞0β(L)˜a′(L)e−ydy∫∞0γ(L)˜a′(L)e−ydy=β(L)γ(L). | (2.14) |
We have the following corollary.
Corollary 2.1.1. The following statements hold.
(i) Given dI>0, R0→ˆR0 as q→0;
(ii) Given dI>0, R0→β(L)γ(L) as q→+∞;
(iii) Given q>0, R0→β(L)γ(L) as dI→0;
(iv) Given q>0, R0→∫L0β(x)dx∫L0γ(x)dx as dI→+∞.
Proof. (i) For any fixed φ∈H1((0,L)), φ≠0, we have
limq→0dI∫L0ea(x)dIφ2xdx+∫L0γ(x)ea(x)dIφ2dx∫L0β(x)ea(x)dIφ2dx=dI∫L0φ2xdx+∫L0γ(x)φ2dx∫L0β(x)φ2dx. |
Taking infφ∈H1((0,L)),φ≠0 both sides, we have 1R0→1ˆR0 as q→0.
(ii) and (iii) are the direct conclusions of Lemma 2.2.
(iv) By the definition of 1R0, for φ≡1, we have
1R0≤∫L0γ(x)ea(x)dIdx∫L0β(x)ea(x)dIdx≤maxx∈[0,L]γ(x)minx∈[0,L]β(x), |
which implies that 1R0 is uniformly bounded for dI>0, passing to a subsequence if necessary, it has a finite limit 1ˉR0 as dI→∞.
On the other hand, by the standard elliptic regularity and the Sobolev embedding theorem, Φ is uniformly bounded for all dI≥1. Dividing both sides of (2.1) by dI and letting dI→+∞, we have Φxx→0 for x∈(0,L) and Φx(0)→0, Φx(L)→0. Consequently, there exists a positive constant ˉΦ such that Φ(x)→ˉΦ as dI→+∞. Integrating (2.1) by parts over (0,L), we can get
qdI∫L0e−a(x)dI[dIΦx−a′(x)Φ(x)]dx+∫L0e−a(x)dIγ(x)Φ(x)dx=1R0∫L0e−a(x)dIβ(x)Φ(x)dx. |
Letting dI→+∞, we obtain ˉR0=∫L0β(x)dx∫L0γ(x)dx.
Lemma 2.1.3. The following statements hold.
(i) If β(x)>γ(x) on [0,L], then R0>1 for any dI>0 and q>minx∈[0,L]a′(x)>0;
(i) If β(x)<γ(x) on [0,L], then R0<1 for any dI>0 and q>minx∈[0,L]a′(x)>0.
Proof. (i) If β(x)>γ(x) on [0,L], by the definition of 1R0, for φ≡1, we have
1R0≤∫L0γ(x)ea(x)dIdx∫L0β(x)ea(x)dIdx<1, |
i.e., R0>1.
(ii) Subtracting both sides of (2.2) by β(x)φ, multiplying by ea(x)dIφ, we have
−dIφxxea(x)dIφ−a′(x)φxea(x)dIφ+[γ(x)−β(x)]ea(x)dIφ2=(1R0−1)β(x)ea(x)dIφ2. |
Integrating it by parts over (0,L), using φx(0)=φx(L)=0, we obtain
dI∫L0ea(x)dI(φx)2dx+∫L0[γ(x)−β(x)]ea(x)dIφ2dx=(1R0−1)∫L0β(x)ea(x)dIφ2dx. |
Since β(x)<γ(x) on [0,L], the left side of the above equality is positive, and
(1R0−1)∫L0β(x)ea(x)dIφ2dx>0, |
which implies that R0<1.
Proof. Theorem 1.1 is the direct results of Lemma 2.1.2, Corollary 2.1.1 and Lemma 2.1.3.
Next we will consider the stability of DFE.
Lemma 2.1.4. The DFE is stable if R0<1, while it is unstable if R0>1.
Proof. 1. Assume contradictorily the DFE is unstable if R0<1. Then we can find (λ,ξ,η) which is a solution of (2.3)–(2.4) subject to (2.5), with at least one of ξ and η is not identical zero, and ℜ(λ)≤0. Suppose that η≡0, then ξ≢ on [0, L] . Using (2.3)–(2.4), we have
\begin{align} \left\{ \begin{array}{ll} -(d_S\xi_x-a'(x)\xi)_x = \lambda \xi, \quad 0 < x < L, \\ d_S\xi_x(0)-a'(0)\xi(0) = 0, \quad d_S\xi_x(L)-a'(L)\xi(L) = 0. \end{array} \right. \end{align} | (2.15) |
It is easy to see that \lambda is real and nonnegative, and therefore \lambda = 0 . We find that \xi = \xi_0e^{\frac{q}{d_I}\tilde{a}(x)} , where \xi_0 is some constant to be determined later. By (1.2), we impose that \int_0^L[\xi(x)+\eta(x)]dx = 0 , \xi_0 = 0 , i.e., \xi\equiv 0 on [0, L] . This is a contradiction. Then we conclude that \eta\equiv 0 on [0, L] . From (2.6), \lambda must be real and \lambda\leq 0 . Since \lambda_1(d_I, q) is the principal eigenvalue, then \lambda_1(d_I, q)\leq \lambda\leq 0 . Lemma 2.1 implies that \mathcal{R}_0\geq 1 , which is a contradiction. Then we conclude that if (\lambda, \xi, \eta) is a solution of (2.3)–(2.4), with at least one of \xi and \eta not identical zero on [0, L] , then \Re(\lambda) > 0 . This proves the linear stability of the DFE.
2. Suppose that \mathcal{R}_0 > 1 . Since (\lambda_1(d_I, q), \phi_1) is the principal eigen-pair of (2.6), (\lambda_1(d_I, q), e^{\frac{a(x)}{d_I}}\phi_1) satisfies
\begin{align*} \left\{ \begin{array}{ll} &[d_I (\phi_1)_x-a'(x)\phi_1]_x+[\beta(x)-\gamma(x)]\phi_1+\lambda_1(d_I, q) \phi_1 = 0, \quad 0 < x < L, \\ &d_I (\phi_1)_x-a'(x)\phi_1 = 0, \quad x = 0, \ L. \end{array} \right. \end{align*} |
By the result of Lemma 2.1.1, \lambda_1(d_I, q) < 0 . On the other hand,
\begin{align} \left\{ \begin{array}{ll} &(d_S \xi_x-a'(x)\xi)_x+\lambda \xi = [\beta(x)-\gamma(x)]e^{\frac{a(x)}{d_I}}\phi_1, \quad 0 < x < L, \\ &d_S \xi_x(0)-a'(0)\xi(0) = 0, \quad d_S \xi_x(L)-a'(L)\xi(L) = 0. \end{array} \right. \end{align} | (2.16) |
There exists a unique solution \xi_1 of (2.16). And (2.5) becomes
\int_0^L[\xi_1(x)+e^{\frac{a(x)}{d_I}}\phi_1(x)]dx = 0, |
which implies that (2.3)–(2.4) has a solution (\lambda_1(d_I, q), \xi_1, e^{\frac{a(x)}{d_I}}\phi_1(x)) satisfying \lambda_1(d_I, q) < 0 and e^{\frac{a(x)}{d_I}}\phi_1(x) > 0 in (0, L) . Therefore, the DFE is linearly unstable.
Lemma 2.1.5. If \mathcal{R}_0 < 1 , then (S, I)\rightarrow (\hat{S}, 0) in C([0, L]) as t\rightarrow +\infty .
Proof. If \mathcal{R}_0 < 1 , letting u(x, t) = Me^{-\lambda_1(d_I, q)t}e^{\frac{a(x)}{d_I}}\phi_1(x) , then we have
\begin{align*} \left\{ \begin{array}{ll} &u_t = [d_I u_x-a'(x)u]_x+[\beta(x)-\gamma(x)]u, \quad 0 < x < L, \quad t > 0, \\ &d_I u_x(0, t)-a'(0)u(0, t) = 0, \quad d_I u_x(L, t)-a'(L)u(L, t) = 0, \ t > 0. \end{array} \right. \end{align*} |
Here (\lambda_1(d_I, q), \phi_1) is the principal eigen-pair, \lambda_1(d_I, q) > 0 and \phi_1(x) > 0 on [0, L] . M is large enough such that I(x, 0)\leq u(x, 0) for every x\in (0, L) . Noticing that
\begin{align*} \left\{ \begin{array}{ll} &I_t = [d_I I_x-a'(x)I]_x+[\beta(x)-\gamma(x)]I, \quad 0 < x < L, \quad t > 0, \\ &d_I u_x(0, t)-a'(0)u(0, t) = 0, \quad d_I u_x(L, t)-a'(L)u(L, t) = 0, \ t > 0. \end{array} \right. \end{align*} |
By the comparison principle, we have I(x, t)\leq u(x, t) for every x\in (0, L) and t\geq 0 . Obviously, u(x, t)\rightarrow 0 for every x\in (0, L) as t\rightarrow \infty , which implies that I(x, t)\rightarrow 0 for every x\in (0, L) as t\rightarrow \infty .
Now we will show that S\rightarrow \hat{S} as t\rightarrow +\infty . Since
S_t = (d_S S_x-a'(x)S)_x-\beta(x)\frac{SI}{S+I}+\gamma(x)I, \ 0 < x < L, \ t > 0, |
we have
|S_t-(d_S S_x-a'(x)S)_x|\leq (\|\beta\|_{\infty}+\|\gamma\|_{\infty})I\leq Ce^{-\lambda_1(d_I, q)t}, |
for 0 < x < L , t > 0 . Noticing that
\lim\limits_{t\rightarrow +\infty} e^{-\lambda_1(d_I, q)t}\rightarrow 0 |
as t\rightarrow +\infty , we know that there exists a positive function \tilde{S}(x) such that
\lim\limits_{t\rightarrow +\infty} S(x, t) = \tilde{S}(x), \quad \int_0^L \tilde{S}(x)dx = N. |
Therefore, \lim_{t\rightarrow +\infty} S(x, t) = \tilde{S}(x) = \hat{S}(x) .
Proof. Theorem 1.2 is the direct results of Lemma 2.1.4 and Lemma 2.1.5.
In this section, we will study further properties of \mathcal{R}_0 in the case of \beta(x)-\gamma(x) changing sign once.
Lemma 2.2.1. Assume that \phi_1 is a positive eigenfunction corresponding to \mathcal{R}_0 = 1 , \beta(x)-\gamma(x) changes sign once in (0, L) . If assumption (C1)(or (C2)) holds, then (\phi_1)_x > 0 (or (\phi_1)_x < 0 ) in (0, L) .
Proof. If \beta(x)-\gamma(x) changes sign once in (0, L) and assumption (C1) holds, then there exists some x_0\in (0, L) such that \beta(x)-\gamma(x) < 0 in (0, x_0) , \beta(x_0) = \gamma(x_0) and \beta(x)-\gamma(x) > 0 in (x_0, L) .
By the definition of \phi_1 , we have
\begin{equation} \left\{ \begin{array}{ll} -d_I(\phi_1)_{xx}-a'(x) (\phi_1)_x = [\beta(x)-\gamma(x)]\phi_1, \quad 0 < x < L, \\ (\phi_1)_x(0) = (\phi_1)_x(L) = 0. \end{array} \right. \end{equation} | (2.17) |
Multiplying (2.17) by e^{\frac{a(x)}{d_I}} , we obtain
-d_I(e^{\frac{a(x)}{d_I}}(\phi_1)_{x})_x = [\beta(x)-\gamma(x)]e^{\frac{a(x)}{d_I}}\phi_1. |
Under the assumptions on \beta(x) and \gamma(x) , we can obtain (e^{\frac{a(x)}{d_I}}(\phi_1)_{x})_x > 0 in (0, x_0) , (e^{\frac{a(x)}{d_I}}(\phi_1)_{x})_x = 0 at x_0 and (e^{\frac{a(x)}{d_I}}(\phi_1)_{x})_x < 0 in (x_0, L) . That is, e^{\frac{a(x)}{d_I}}(\phi_1)_{x} is strictly increasing in (0, x_0) and strictly decreasing in (x_0, L) . Noticing that (\phi_1)_x(0) = (\phi_1)_x(L) = 0 , we can get e^{\frac{a(x)}{d_I}}(\phi_1)_{x} > 0 in (0, L) . So (\phi_1)_x > 0 in (0, L) .
Similarly, if \beta(x)-\gamma(x) changes sign once in (0, L) and assumption (C2) holds, (\phi_1)_x < 0 in (0, L) . We omit the details here.
Now we prove two general lemmas below.
For any continuous function m(x) on [0, L] , define
F(\eta) = \int_0^L\tilde{a}'(x)e^{\eta \tilde{a}(x)}m(x)dx, \quad 0\leq \eta < \infty. |
Lemma 2.2.2. Assume that m(x)\in C^1([0, L]) and m(L) > 0 (or m(L) < 0 ). Then there exists some positive constant M such that F(\eta) > 0 (or F(\eta) < 0 ) for any \eta > M .
Proof. Since m'(x) and \tilde{a}'(x) is uniformly bounded independent of \eta , we have
\begin{align*} \lim\limits_{\eta\rightarrow +\infty}\eta e^{-\eta \tilde{a}(L)}F(\eta)& = \lim\limits_{\eta\rightarrow +\infty}\int_0^L\eta \tilde{a}'(x)e^{-\eta[\tilde{a}(x)-\tilde{a}(L)]}m(x)dx\\ & = m(L)-\lim\limits_{\eta\rightarrow +\infty}\left(m(0)e^{\eta[\tilde{a}(0)-\tilde{a}(L)]}+\int_0^L m'(x)e^{\eta[\tilde{a}(x)-\tilde{a}(L)]}dx\right)\\ & = m(L)-\lim\limits_{\eta\rightarrow +\infty}\left(m(0)e^{\eta[\tilde{a}(0)-\tilde{a}(L)]}+\int_0^Lm'(x)e^{\tilde{a}'(\xi)[x-L]}dx\right)\\ & = m(L) > 0( < 0). \end{align*} |
Therefore, there exists some positive constant M such that F(\eta) > 0(<0) for \eta > M .
Lemma 2.2.3. Assume that m(x) changes sign once in (0, L) . Then
(i) If m(L) > 0 and \int_0^L \tilde{a}'(x)m(x)dx > 0 , then F(\eta) > 0 for any \eta > 0 ;
(ii) If m(L) < 0 and \int_0^L \tilde{a}'(x)m(x)dx < 0 , then F(\eta) < 0 for any \eta > 0 ;
(iii) If m(L) > 0 and \int_0^L \tilde{a}'(x)m(x)dx < 0 , then there exists a unique \eta_1\in (0, +\infty) such that F(\eta_1) = 0 and F'(\eta_1) > 0 ;
(iv) If m(L) < 0 and \int_0^L \tilde{a}'(x)m(x)dx > 0 , then there exists a unique \eta_1\in (0, +\infty) such that F(\eta_1) = 0 and F'(\eta_1) < 0 .
Proof. We only prove part (i) and part (iii). The proofs of part (ii) and part (iv) are similar.
(i) If m(L) > 0 and m(x) changes sign once in (0, L) , then there exists x_1\in (0, L) such that m(x) < 0 for x\in (0, x_1) and m(x) > 0 for x\in (x_1, L) . Since \tilde{a}(x) is increasing, we have m(x)[\tilde{a}(x)-\tilde{a}(x_1)] > 0 for x\in (0, L) and x\neq x_1 . And
\begin{align} [e^{-\tilde{a}(x_1)\eta}F(\eta)]'& = e^{-\tilde{a}(x_1)\eta}[F'(\eta)-\tilde{a}(x_1)F(\eta)]\\ & = e^{-\tilde{a}(x_1)\eta}\int_0^L[\tilde{a}(x)-\tilde{a}(x_1)]m(x)\tilde{a}'(x)e^{\eta \tilde{a}(x)}dx > 0, \end{align} | (2.18) |
which implies that e^{-\tilde{a}(x_1)\eta}F(\eta) is strictly increasing in \eta\in (0, \infty) , e^{-\tilde{a}(x_1)\eta}F(\eta) > F(0) = \int_0^L\tilde{a}'(x)m(x)dx > 0 . Consequently, F(\eta) > 0 for any \eta > 0 . Here the prime notation denotes differentiation by \eta . Part (i) is proved.
(iii) \int_0^L\tilde{a}'(x)m(x)dx < 0 means that F(0) < 0 , while, by the result of Lemma 2.2.2, m(L) > 0 means that F(\eta) > 0 for \eta > M with M large enough. By continuity, there at least exists a positive root for F(\eta) = 0 . But e^{-\tilde{a}(x_1)\eta}F(\eta) is increasing in \eta\in (0, \infty) , so F(\eta) = 0 only has a unique positive root \eta_1 . By (2.18), we have F'(\eta_1) > a(x_1)F(\eta_1) = 0 . Part (iii) is proved.
In this section, we consider the stability of DFE. First we have
Lemma 2.3.1. Assume that \beta(x)-\gamma(x) changes sign once in (0, L) and \int_0^L\beta(x)dx > \int_0^L\gamma(x)dx .
(i) If \beta(x) and \gamma(x) satisfy (C1), then \mathcal{R}_0 > 1 for d_I > 0 and q > \min_{x\in [0, L]}a'(x) > 0 ;
(ii) If \beta(x) and \gamma(x) satisfy (C2), then for every d_I > 0 , there exists a unique \bar{q} = \bar{q}(d_I) such that \mathcal{R}_0 > 1 for 0 < \min_{x\in [0, L]}a'(x) < q < \bar{q} , \mathcal{R}_0 = 1 for q = \bar{q} and \mathcal{R}_0 < 1 for q > \bar{q} .
Proof. (i) Subtracting both sides of (2.2) by \beta(x)\varphi , multiplying by \frac{e^{\frac{a(x)}{d_I}}}{\varphi} , we have
\begin{align*} [-d_I\varphi_{xx}-a'(x)\varphi_x]\frac{e^{\frac{a(x)}{d_I}}}{\varphi}+[\gamma(x)-\beta(x)]e^{\frac{a(x)}{d_I}} = (\frac{1}{\mathcal{R}_0}-1)\beta(x)e^{\frac{a(x)}{d_I}}. \end{align*} |
Integrating it by parts over (0, L) , using \varphi_x(0) = \varphi_x(L) = 0 , we obtain
d_I\int_0^L\frac{e^{\frac{a(x)}{d_I}}(\varphi_x)^2}{\varphi^2}dx+\int_0^L[\beta(x)-\gamma(x)]e^{\frac{a(x)}{d_I}}dx = (1-\frac{1}{\mathcal{R}_0})\int_0^L\beta(x)e^{\frac{a(x)}{d_I}}dx. |
Using Lemma 2.2.3(i) with m(x) = \frac{[\beta(x)-\gamma(x)]}{\tilde{a}'(x)} , \int_0^L[\beta(x)-\gamma(x)]e^{\frac{a(x)}{d_I}}dx > 0 , and
(1-\frac{1}{\mathcal{R}_0})\int_0^L\beta(x)e^{\frac{a(x)}{d_I}}\varphi^2dx > 0, |
which implies that \mathcal{R}_0 > 1 .
(ii) Differentiating both sides of (2.2) with respect to q , denoting the differentiation with respect to q by the dot notation, we obtain
\begin{equation} \left\{ \begin{array}{ll} -d_I\dot{\varphi}_{xx}-\tilde{a}'(x) \varphi_x-\tilde{a}'(x)\dot{\varphi}_x+\gamma(x)\dot{\varphi} = -\frac{\dot{\mathcal{R}}_0}{\mathcal{R}_0^2}\beta(x)\varphi+\frac{1}{\mathcal{R}_0}\beta(x)\dot{\varphi}, \quad 0 < x < L, \\ \dot{\varphi}_x(0) = \dot{\varphi}_x(L) = 0. \end{array} \right. \end{equation} | (2.19) |
Multiplying (2.19) by e^{\frac{a(x)}{d_I}}\varphi and integrating the resulting equation in (0, L) , we have
\begin{align} &\quad d_I\int_0^Le^{\frac{a(x)}{d_I}}\dot{\varphi}_x\varphi_xdx-\int_0^Le^{\frac{a(x)}{d_I}}\varphi_x\varphi \tilde{a}'(x)dx+\int_0^L\gamma(x)e^{\frac{a(x)}{d_I}}\dot{\varphi}\varphi dx\\ & = -\frac{\dot{\mathcal{R}}_0}{\mathcal{R}_0^2}\int_0^L\beta(x)e^{\frac{a(x)}{d_I}}\varphi^2dx +\frac{1}{\mathcal{R}_0}\int_0^L\beta(x)e^{\frac{a(x)}{d_I}}\dot{\varphi}\varphi dx. \end{align} | (2.20) |
Multiplying (2.2) by e^{\frac{a(x)}{d_I}}\dot{\varphi} and integrating the resulting equation in (0, L) , we get
\begin{align} d_I\int_0^Le^{\frac{a(x)}{d_I}}\dot{\varphi}_x\varphi_xdx+\int_0^L\gamma(x)e^{\frac{a(x)}{d_I}}\dot{\varphi}\varphi dx = \frac{1}{\mathcal{R}_0}\int_0^L\beta(x)e^{\frac{a(x)}{d_I}}\dot{\varphi}\varphi dx. \end{align} | (2.21) |
Subtracting (2.20) and (2.21), we obtain
\begin{align} \frac{\partial \mathcal{R}_0}{\partial q} = \frac{\mathcal{R}_0^2\int_0^Le^{\frac{a(x)}{d_I}}\varphi_x\varphi \tilde{a}'(x)dx}{\int_0^L\beta(x)e^{\frac{a(x)}{d_I}}\varphi^2dx}. \end{align} | (2.22) |
By the result of Corollary 2.1.1, we know that
\lim\limits_{q\rightarrow \infty} \mathcal{R}_0 = \frac{\beta(L)}{\gamma(L)} < 1. |
Meanwhile, we have
\lim\limits_{q\rightarrow 0} \mathcal{R}_0 = \hat{\mathcal{R}}_0 > 1 |
for any d_I . Then there must exist at least some \bar{q} such that \mathcal{R}_0(\bar{q}) = 1 . By Lemma 2.1.1, for any \bar{q} > 0 satisfying \mathcal{R}_0(\bar{q}) = 1 , (\phi_1)_x < 0 in (0, L) . Recalling (2.22), we have
\frac{\partial \mathcal{R}_0}{\partial \bar{q}} = \frac{\int_0^Le^{\frac{\bar{q}}{d_I}\tilde{a}(x)}(\phi_1)_x\phi_1 dx} {\int_0^L\beta(x)e^{\frac{\bar{q}}{d_I}\tilde{a}(x)}(\phi_1)^2dx} < 0, |
which implies that \bar{q} is the unique point satisfying \mathcal{R}_0(\bar{q}) = 1 .
The following lemma will tell us that there exists a function q = \rho_1(d_I) such that \mathcal{R}_0(d_I, \rho_1(d_I)) = 1 and give the asymptotic profile of \rho_1(d_I) if \int_0^L\beta(x)dx > \int_0^L\gamma(x)dx .
Lemma 2.3.2. Assume that \beta(x)-\gamma(x) changes sign once in (0, L) , \int_0^L\beta(x)dx > \int_0^L\gamma(x)dx , and \theta_1 is the unique solution of
\int_0^L[\beta(x)-\gamma(x)]e^{\theta_1\tilde{a}(x)}dx = 0. |
Suppose that \beta(x) and \gamma(x) satisfy (C2). Then there exists a function \rho_1:(0, \infty)\rightarrow (0, \infty) such that \mathcal{R}_0(d_I, \rho_1(d_I)) = 1 . And \rho_1 satisfies
\lim\limits_{d_I\rightarrow 0} \rho_1(d_I) = 0, \quad \lim\limits_{d_I\rightarrow \infty} \frac{\rho_1(d_I)}{d_I} = \theta_1. |
Proof. 1. Let's first consider the limit of \frac{\rho_1(d_I)}{d_I} as d_I\rightarrow \infty . Assume that \frac{\rho_1(d_I)}{d_I}\rightarrow \infty as d_I\rightarrow \infty . Under the assumption (C2), by Lemma 2.1.4, we have
\lim\limits_{\rho_1(d_I)\rightarrow \infty, \frac{\rho_1(d_I)}{d_I}\rightarrow \infty} \mathcal{R}_0(d_I, \rho_1(d_I)) = \frac{\beta(L)}{\gamma(L)} < 1, |
which is a contradiction to \mathcal{R}_0(d_I, \rho_1(d_I)) = 1 .
Next, we will prove that \frac{\rho_1(d_I)}{d_I}\rightarrow \theta_1 as d_I\rightarrow \infty . Here \theta_1 is the unique positive root of \int_0^L[\beta(x)-\gamma(x)]e^{\theta_1\tilde{a}(x)}dx = 0 . By the discussions above, we know that \frac{\rho_1(d_I)}{d_I} is bounded for large d_I . Passing to a subsequence if necessary, we suppose that \frac{\rho_1(d_I)}{d_I}\rightarrow \theta_* for some nonnegative number \theta_* as d_I\rightarrow \infty . Let \tilde{\varphi} be the unique normalized eigenfunction of the eigenvalue \mathcal{R}_0(d_I, \rho_1(d_I)) = 1 . Then
\begin{equation} \left\{ \begin{array}{ll} -d_I(e^{\frac{\rho_1(d_I)}{d_I}\tilde{a}(x)}\tilde{\varphi}_x)_x+[\gamma(x)-\beta(x)]e^{\frac{\rho_1(d_I)}{d_I}\tilde{a}(x)}\tilde{\varphi} = 0, \quad 0 < x < L, \\ \tilde{\varphi}_x(0) = \tilde{\varphi}_x(L) = 0. \end{array} \right. \end{equation} | (2.23) |
Integrating (2.23) in (0, L) , we get
\begin{align} \int_0^L[\beta(x)-\gamma(x)]e^{\frac{\rho_1(d_I)}{d_I}\tilde{a}(x)}\tilde{\varphi}dx = 0. \end{align} | (2.24) |
Recalling that, up to a subsequence if necessary, \tilde{\varphi}\rightarrow 1 in C([0, 1]) as d_I\rightarrow \infty . Letting d_I\rightarrow \infty in (2.24), we have
\int_0^L[\beta(x)-\gamma(x)]e^{\theta_*\tilde{a}(x)}dx = 0. |
By Lemma 2.2.3 with m(x) = \frac{[\beta(x)-\gamma(x)]}{\tilde{a}'(x)} , F(\eta) has a unique positive root, i.e., \theta_* = \theta_1 .
2. Contradictorily, assume that q = \rho_1(d_I)\rightarrow q^* > 0 or q = \rho_1(d_I)\rightarrow \infty as d_I\rightarrow 0 . By Lemma 2.1.4, we know that
\lim\limits_{\rho_1(d_I)\rightarrow q^*, \frac{\rho_1(d_I)}{d_I}\rightarrow \infty} \mathcal{R}_0(d_I, \rho_1(d_I)) = \frac{\beta(L)}{\gamma(L)} < 1 |
or
\lim\limits_{\rho_1(d_I)\rightarrow \infty, \frac{\rho_1(d_I)}{d_I}\rightarrow \infty} \mathcal{R}_0(d_I, \rho_1(d_I)) = \frac{\beta(L)}{\gamma(L)} < 1, |
which is a contradiction to \mathcal{R}_0(d_I, \rho_1(d_I)) = 1 . Therefore, we have \lim_{d_I\rightarrow 0}\rho_1(d_I) = 0 .
To study the properties of \mathcal{R}_0 when \int_0^L\beta(x)dx < \int_0^L\gamma(x)dx , we need the following results which were stated in [2]:
Proposition 2.3.1. Assume that \beta(x)-\gamma(x) changes sign in (0, L) .
(i) \hat{\mathcal{R}}_0 is a monotone decreasing function of d_I with \hat{\mathcal{R}}_0\rightarrow \max\{\beta(x)/\gamma(x):x\in[0, L]\} as d_I\rightarrow 0 and \hat{\mathcal{R}}_0\rightarrow \int_0^L\beta(x)dx/\int_0^L\gamma(x)dx as d_I\rightarrow +\infty ;
(ii) \hat{\mathcal{R}}_0 > 1 for all d_I > 0 if \int_0^L\beta(x)dx\geq \int_0^L\gamma(x)dx ;
(iii) There exists a threshold value d^*_I\in (0, +\infty) such that \hat{\mathcal{R}}_0 > 1 for d_I < d^*_I and \hat{\mathcal{R}}_0 < 1 for d_I > d^*_I if \int_0^L\beta(x)dx < \int_0^L\gamma(x)dx .
Lemma 2.3.3. Assume that \beta(x)-\gamma(x) changes sign once in (0, L) and \int_0^L\beta(x)dx < \int_0^L\gamma(x)dx . Then there exists some constant d^*_I > 0 such that d^*_I is the unique positive root of the equation \hat{\mathcal{R}_0}(d_I) = 1 and the following statements hold.
1. If \beta(x) and \gamma(x) satisfy (C1), then
(i) for d_I\in (0, d^*_I] , \mathcal{R}_0 > 1 for any q > \min_{x\in [0, L]}a'(x) > 0 ;
(ii) for d_I\in (d^*_I, \infty) , there exists a unique \bar{q} = \bar{q}(d_I) such that \mathcal{R}_0 < 1 for any 0 < \min_{x\in [0, L]}a'(x) < q < \bar{q} and \mathcal{R}_0 > 1 for any q > \bar{q} .
2. If \beta(x) and \gamma(x) satisfy (C2), then
(iii) for d_I\in (0, d^*_I] , there exists a unique \bar{q} = \bar{q}(d_I) such that \mathcal{R}_0 > 1 for any 0 < \min_{x\in [0, L]}a'(x) < q < \bar{q} and \mathcal{R}_0 < 1 for any q > \bar{q} ;
(iv) for d_I\in (d^*_I, \infty) , \mathcal{R}_0 < 1 for any q > \min_{x\in [0, L]}a'(x) > 0 .
Proof. (i) Noticing that \beta(x) and \gamma(x) satisfy (C1), similar to the proof of (ii) in Lemma 2.1.4, we can prove that there exists a unique \bar{q} > 0 satisfying \mathcal{R}_0(\bar{q}) = 1 and \mathcal{R}'_0(\bar{q}) > 0 . Hence, the conclusion is true for d_I\in (d^*_I, +\infty) .
For d_I\in (0, d^*_I] , by the results of Proposition 2.3.1, we have \lim_{q\rightarrow 0}\mathcal{R}_0 = \hat{\mathcal{R}}_0\geq 1 . By the results of Corollary 2.1.1, \lim_{q\rightarrow +\infty}\mathcal{R}_0 = \beta(L)/\gamma(L) > 1 under the condition (C1). Hence \mathcal{R}_0 > 1 for any q > 0 .
(ii) The proof of Lemma 2.3.3 under the condition (C2) is similar to that of Lemma 2.1.4, we omit the details here.
Lemma 2.3.4. Assume that \beta(x)-\gamma(x) changes sign once in (0, L) and \int_0^L\beta(x)dx < \int_0^L\gamma(x)dx . Then there exists a constant d^*_I > 0 such that d^*_I is the unique positive root of the equation \hat{\mathcal{R}_0}(d_I) = 1 and the following statements hold.
1. If \beta(x) and \gamma(x) satisfy (C1), then there exists a function \rho_2:(d^*_I, \infty)\rightarrow (0, \infty) such that \rho_2 is a monotone increasing function of d_I and \mathcal{R}_0(d_I, \rho_2(d_I)) = 1 . Let \theta_2 be the unique solution of
\int_0^L[\beta(x)-\gamma(x)]e^{\theta_2\tilde{a}(x)}dx = 0. |
Then
\lim\limits_{d_I\rightarrow d^*_I+}\rho_2(d_I) = 0, \quad \lim\limits_{d_I\rightarrow \infty}\frac{\rho_2(d_I)}{d_I} = \theta_2. |
2. If \beta(x) and \gamma(x) satisfy (C2), then there exists a function \rho_3:(0, d^*_I)\rightarrow (0, \infty) such that \mathcal{R}_0(d_I, \rho_3(d_I)) = 1 and
\lim\limits_{d_I\rightarrow 0+}\rho_3(d_I) = 0, \quad \lim\limits_{d_I\rightarrow d^*_I-}\frac{\rho_3(d_I)}{d_I} = 0. |
Proof. 1. If we can prove that \rho'_2(d_I) > 0 for d_I\in (d^*_I, \infty) , then \rho_2(d_I) is a monotone increasing function of d_I . Here the prime notation denotes differentiation by d_I . Since \mathcal{R}_0(d_I, \rho_2(d_I)) = 1 , we can get
\begin{align} \frac{\partial \mathcal{R}_0}{\partial q}\rho'_2(d_I)+\frac{\partial \mathcal{R}_0}{\partial d_I} = 0. \end{align} | (2.25) |
By Lemma 2.3.1, \frac{\partial \mathcal{R}_0}{\partial q} > 0 for \mathcal{R}_0(d_I, \rho_2(d_I)) = 1 . So we need to prove that \frac{\partial \mathcal{R}_0}{\partial d_I} < 0 .
Differentiating both sides of (2.2) with respect to d_I , denoting the differentiation with respect to d_I by the dot notation, we obtain
\begin{equation} \left\{ \begin{array}{ll} -\varphi_{xx}-d_I\dot{\varphi}_{xx}-a'(x)\dot{\varphi}_x+\gamma(x)\dot{\varphi} = -\frac{\dot{\mathcal{R}}_0}{\mathcal{R}_0^2}\beta(x)\varphi+\frac{1}{\mathcal{R}_0}\beta(x)\dot{\varphi}, \quad 0 < x < L, \\ \dot{\varphi}_x(0) = \dot{\varphi}_x(L) = 0. \end{array} \right. \end{equation} | (2.26) |
Multiplying (2.26) by e^{\frac{a(x)}{d_I}}\varphi and integrating the resulting equation in (0, L) , we obtain
\begin{align} &\quad -\int_0^Le^{\frac{a(x)}{d_I}}\varphi_{xx}\varphi dx+d_I\int_0^Le^{\frac{a(x)}{d_I}}\dot{\varphi}_x\varphi_xdx+\int_0^L\gamma(x)e^{\frac{a(x)}{d_I}}\dot{\varphi}\varphi dx\\ & = -\frac{\dot{\mathcal{R}}_0}{\mathcal{R}_0^2}\int_0^L\beta(x)e^{\frac{a(x)}{d_I}}\varphi^2dx +\frac{1}{\mathcal{R}_0}\int_0^L\beta(x)e^{\frac{a(x)}{d_I}}\dot{\varphi}\varphi dx. \end{align} | (2.27) |
Multiplying (2.2) by e^{\frac{a(x)}{d_I}}\dot{\varphi} and integrating the resulting equation in (0, L) , we get
\begin{align} d_I\int_0^Le^{\frac{a(x)}{d_I}}\dot{\varphi}_x\varphi_xdx+\int_0^L\gamma(x)e^{\frac{a(x)}{d_I}}\dot{\varphi}\varphi dx = \frac{1}{\mathcal{R}_0}\int_0^L\beta(x)e^{\frac{a(x)}{d_I}}\dot{\varphi}\varphi dx. \end{align} | (2.28) |
Subtracting (2.27) and (2.28), we have
\begin{align} \frac{\partial \mathcal{R}_0}{\partial d_I} = \frac{\mathcal{R}_0^2\int_0^Le^{\frac{a(x)}{d_I}}\varphi_{xx}\varphi dx}{\int_0^L\beta(x)e^{\frac{a(x)}{d_I}}\varphi^2dx} = -\frac{\mathcal{R}_0^2\int_0^Le^{\frac{a(x)}{d_I}}(\varphi_x)^2dx}{\int_0^L\beta(x)e^{\frac{a(x)}{d_I}}\varphi^2dx} -\frac{\mathcal{R}_0^2\int_0^Le^{\frac{a(x)}{d_I}}\varphi_x\varphi a'(x) dx}{d_I\int_0^L\beta(x)e^{\frac{a(x)}{d_I}}\varphi^2dx}. \end{align} | (2.29) |
By Lemma 2.2.1, for any d_I satisfying \mathcal{R}_0(d_I, q) = 1 , (\phi_1)_x > 0 , we can get
\begin{align} \frac{\partial \mathcal{R}_0}{\partial d_I} = -\frac{\mathcal{R}_0^2\int_0^Le^{\frac{a(x)}{d_I}}[(\phi_1)_x]^2dx}{\int_0^L\beta(x)e^{\frac{a(x)}{d_I}}\phi_1^2dx} -\frac{\mathcal{R}_0^2\int_0^Le^{\frac{a(x)}{d_I}}(\phi_1)_x\phi_1 a'(x)dx}{d_I\int_0^L\beta(x)e^{\frac{a(x)}{d_I}}\phi_1^2dx} < 0. \end{align} | (2.30) |
(2.25) and (2.30) imply that \rho'_2(d_I) > 0 for d_I\in (d^*_I, \infty) .
The proof of \lim_{d_I\rightarrow \infty} \frac{\rho_2(d_I)}{d_I} = \theta_2 ( \theta_2 is the unique solution of \int_0^L[\beta(x)-\gamma(x)]e^{\theta_2 a(x)}dx = 0 ) is similar to the proof of Lemma 2.3.2, we omit the details here.
Now we will prove that \lim_{d_I\rightarrow d^*_I+}\rho_2(d_I) = 0 . Assume that there exists q^* such that q = \rho_2(d_I)\rightarrow q^* as d_I\rightarrow d_I^*+ . Then there exists a positive function \phi^*(x)\in C^2([0, L]) such that
\begin{equation} \left\{ \begin{array}{ll} -d^*_I\phi^*_{xx}-q^*\tilde{a}'(x)\phi^*_x+\gamma(x)\phi^* = \beta(x)\phi^*, \quad 0 < x < L, \\ \phi^*_x(0) = \phi^*_x(L) = 0. \end{array} \right. \end{equation} | (2.31) |
Noticing that d^*_I is the unique positive root of \hat{\mathcal{R}}_0 = 1 and the definition of \hat{\mathcal{R}}_0 implies q = 0 , there exists a positive function \hat{\phi}(x)\in C^2([0, L]) such that
\begin{equation} \left\{ \begin{array}{ll} -d^*_I\hat{\phi}_{xx}+\gamma(x)\hat{\phi} = \beta(x)\hat{\phi}, \quad 0 < x < L, \\ \hat{\phi}_x(0) = \hat{\phi}_x(L) = 0. \end{array} \right. \end{equation} | (2.32) |
Multiplying (2.31) by \hat{\phi} , (2.32) by \phi^* , subtracting the two resulting equations, then integrating by parts over (0, L) , we get
q^*\int_0^L\tilde{a}'(x)\phi^*_x\hat{\phi}dx = 0. |
Since \phi^*_x is positive(by Lemma 2.2.1), we have q^* = 0 . Therefore, \lim_{d_I\rightarrow d^*_I+}\rho_2(d_I) = 0 .
2. Using the arguments above, similar to the proof of Lemma 2.3.2, we can obtain the conclusions.
In this section, we will show that: If the disease-free equilibrium is unstable, then we can use the bifurcation analysis and degree theory to study the existence of endemic equilibrium.
Letting \tilde{S} = e^{\frac{a(x)}{d_S}}\bar{S} , \tilde{I} = e^{\frac{a(x)}{d_I}}\bar{I} , we have
\begin{equation} \left\{ \begin{array}{llll} d_S \bar{S}_{xx}+a'(x)\bar{S}_x-\beta(x)\frac{e^{\frac{a(x)}{d_I}}\bar{S}\bar{I}}{e^{\frac{a(x)}{d_S}}\bar{S}+e^{\frac{a(x)}{d_I}}\bar{I}}+ \gamma(x)e^{(\frac{1}{d_I}-\frac{1}{d_S})a(x)}\bar{I} = 0, \ \ & 0 < x < L, \\ d_I \bar{I}_{xx}+a'(x)\bar{I}_x+\beta(x)\frac{e^{\frac{a(x)}{d_S}}\bar{S}\bar{I}}{e^{\frac{a(x)}{d_S}}\bar{S}+e^{\frac{a(x)}{d_I}}\bar{I}}- \gamma(x)\bar{I} = 0, \ \ & 0 < x < L, \\ \bar{S}_x(0) = \bar{S}_x(L) = 0, \quad \bar{I}_x(0) = \bar{I}_x(L) = 0, \ \ & \\ \int_0^L[e^{\frac{a(x)}{d_S}}\bar{S}+e^{\frac{a(x)}{d_I}}\bar{I}]dx = N. \end{array} \right. \end{equation} | (2.33) |
Since the structure of the solution set of (2.33) is the same as that of (1.3), we study (2.33) instead of (1.3). Denote the unique disease-free equilibrium of (2.33) by (\hat{\bar{S}}, 0) = (\frac{N}{\int_0^L e^{\frac{a(x)}{d_S}}}, 0) . We will consider a branch of positive solutions of (2.33) bifurcating from the branch of semi-trivial solutions given by
\Gamma_S: = \{(q, (\hat{\bar{S}}, 0)):0 < \min\limits_{x\in [0, L]}a'(x) < q < \infty\} |
through using the local and global bifurcation theorems. For fixed d_S , d_I > 0 , we take q as the bifurcation parameter. Let
X = \{u\in W^{2, p}((0, L)):u_x(0) = u_x(L) = 0\}, \quad Y = L^p((0, L)) |
for p > 1 and the set of positive solution of (2.33) to be
O = \{(q, (S, I))\in \mathbb{R}^+\times X\times X:q > \min\limits_{x\in [0, L]}a'(x) > 0, S > 0, I > 0, (q, (S, I)) \ {\rm satisfies}\ (2.33)\}. |
Lemma 2.4.1 Assume that d_S , d_I > 0 and \beta(x)-\gamma(x) changes sign once in (0, L) . Then
1. q_* > 0 is a bifurcation point for the positive solutions of (2.33) from the semi-trivial branch \Gamma_S if and only if q_* satisfies R_0(d_I, q_*) = 1 . That is,
(I) If \int^L_0 \beta(x)dx > \int^L_0\gamma(x)dx , then such q_* exists uniquely for any d_I > 0 if and only if assumption (C2) holds;
(II) If \int^L_0 \beta(x)dx < \int^L_0\gamma(x)dx , let d^*_I be the unique positive root of \hat{R}_0 = 1 , then such q_* exists uniquely for any d_I > 0 if and only if either \beta(x) and \gamma(x) satisfy condition (C1) and d > d^*_I or they satisfy condition (C2) and 0 < d < d^*_I .
2. There exits some \delta > 0 such that all positive solutions of (2.33) near (q_*, (\hat{\bar{S}}, 0)))\in\mathbb{R}\times X\times X can be parameterized as
\begin{align} \Gamma = \{(q(\tau), (\hat{\bar{S}}+\bar{S}_1(\tau), \bar{I}_1(\tau))): \tau\in [0, \delta)\}, \end{align} | (2.34) |
where (q(\tau), (\hat{\bar{S}}+\bar{S}_1(\tau), I_1(\tau))) is a smooth curve with respect to \tau and satisfies q(0) = q_* , \hat{S}_1(0) = I_1(0) = 0 .
3. There exists a connected component \Sigma of \bar{O} satisfying \Gamma\subseteq \Sigma , and \Sigma possesses some properties as follows.
Case (I) Assume that \int^L_0\beta(x)dx > \int^L_0\gamma(x)dx and (C2) holds. Then there exists some endemic equilibrium (\hat{S}_*, \hat{I}_*) of (2.33) when q = 0 such that for \Sigma , the projection of \Sigma to the q -axis satisfies Proj_q \Sigma = [0, q_*] and the connected component \Sigma connects to (0, (\hat{S}_*, \hat{I}_*)) .
Case (II) Assume that \int^L_0\beta(x)dx < \int^L_0\gamma(x)dx . Then
(i) If (C1) holds and d_I > d^*_I , then (2.33) has no positive solution for 0 < \min_{x\in [0, L]}a'(x) < q < q_* and for \Sigma , the projection of \Sigma to the q -axis satisfies Proj_q \Sigma = [q_*, \infty) .
(ii) If (C2) holds and 0 < d_I < d^*_I , then there exists some endemic equilibrium (\hat{S}_*, \hat{I}_*) of (2.33) when q = 0 such that for \Sigma , the projection of \Sigma to the q -axis satisfies Proj_q \Sigma = [0, q_*] and the connected component \Sigma connects to (0, (\hat{S}_*, \hat{I}_*)) .
Proof. 1. Let F:\mathbb{R}^+\times X\times X\rightarrow Y\times Y\times \mathbb{R} be the mapping as follows.
\begin{equation*} F(q, (\bar{S}, \bar{I})) = \left( \begin{array}{llll} d_S \bar{S}_{xx}+a'(x)\bar{S}_x-\beta(x)\frac{e^{\frac{a(x)}{d_I}}\bar{S}\bar{I}}{e^{\frac{a(x)}{d_S}}\bar{S}+e^{\frac{a(x)}{d_I}}\bar{I}}+ \gamma(x)e^{(\frac{1}{d_I}-\frac{1}{d_S})a(x)}\bar{I}\\ d_I \bar{I}_{xx}+a'(x)\bar{I}_x+\beta(x)\frac{e^{\frac{a(x)}{d_S}}\bar{S}\bar{I}}{e^{\frac{a(x)}{d_S}}\bar{S}+e^{\frac{a(x)}{d_I}}\bar{I}}- \gamma(x)\bar{I}\\ \int_0^L[e^{\frac{a(x)}{d_S}}\bar{S}+e^{\frac{a(x)}{d_I}}\bar{I}]dx-N \end{array} \right). \end{equation*} |
It is to verify that the pair (\bar{S}, \bar{I}) is a solution of (2.33) if only if F(q, (\bar{S}, \bar{I})) = 0 . Obviously, F(q, (\hat{\bar{S}}, 0)) = 0 for any q > \min_{x\in [0, L]}a'(x) > 0 . The Fr \acute{e} chet derivatives of F at (\hat{\bar{S}}, 0) are given by
\begin{equation*} D_{(\bar{S}, \bar{I})}F(q, (\hat{\bar{S}}, 0))\left[\begin{array}{llll}\Phi\\ \Psi \end{array}\right] = \left( \begin{array}{llll} d_S \Phi_{xx}+\tilde{a}'(x)\Phi_x+[\gamma(x)-\beta(x)]e^{(\frac{1}{d_I}-\frac{1}{d_S})a(x)}\Psi\\ d_I \Psi_{xx}+\tilde{a}'(x)\Psi_x+[\beta(x)-\gamma(x)]\Psi\\ \int_0^L[e^{\frac{a(x)}{d_S}}\Phi+e^{\frac{a(x)}{d_I}}\Psi]dx \end{array} \right), \end{equation*} |
\begin{equation*} D_{q, (\bar{S}, \bar{I})}F(q, (\hat{\bar{S}}, 0))\left[\begin{array}{llll}\Phi\\ \Psi \end{array}\right] = \left( \begin{array}{llll} \tilde{a}'(x)\Phi_x+(\frac{a(x)}{d_I}-\frac{a(x)}{d_S})[\gamma(x)-\beta(x)]e^{(\frac{1}{d_I}-\frac{1}{d_S})a(x)}\Psi\\ \qquad \quad \tilde{a}'(x)\Psi_x\\ \int_0^L[\frac{a(x)}{d_S}e^{\frac{a(x)}{d_S}}\Phi+\frac{a(x)}{d_I}e^{\frac{a(x)}{d_I}}\Psi]dx \end{array} \right), \end{equation*} |
\begin{equation*} D_{(\bar{S}, \bar{I}), (\bar{S}, \bar{I})}F(q, (\hat{\bar{S}}, 0))\left[\begin{array}{llll}\Phi\\ \Psi \end{array}\right]^2 = \left( \begin{array}{llll} \frac{2}{\hat{\bar{S}}} \beta(x)e^{2(\frac{q}{d_I}-\frac{q}{d_S})\tilde{a}(x)}\Psi^2\\ -\frac{2}{\hat{\bar{S}}} \beta(x)e^{(\frac{1}{d_I}-\frac{1}{d_S})a(x)}\Psi^2\\ \qquad \quad 0 \end{array} \right). \end{equation*} |
If (\Phi_1, \Psi_1) is a nontrivial solution of the following problem
\begin{equation} \left\{\begin{array}{llll} d_S \Phi_{xx}+\tilde{a}'(x)\Phi_x+[\gamma(x)-\beta(x)]e^{(\frac{1}{d_I}-\frac{1}{d_S})a(x)}\Psi = 0, & 0 < x < L, \\ d_I \Psi_{xx}+\tilde{a}'(x)\Psi_x+[\beta(x)-\gamma(x)]\Psi = 0, & 0 < x < L, \\ \Phi_x(0) = \Phi_x(L) = \Psi_x(0) = \Psi_x(L) = 0, &\\ \int_0^L[e^{\frac{a(x)}{d_S}}\Phi+e^{\frac{a(x)}{d_I}}\Psi]dx = 0, & \end{array} \right. \end{equation} | (2.35) |
then (q_*, (\hat{\bar{S}}, 0))) is degenerate solution of (2.33). The second equation of (2.33) has a positive solution \Psi_1 only if q = q_* satisfies \mathcal{R}_0(d_I, q_*) = 1 . And \Phi_1 satisfies
\begin{equation} \left\{\begin{array}{llll} d_S (\Phi_1)_{xx}+\tilde{a}'(x)(\Phi_1)_x+[\gamma(x)-\beta(x)]e^{(\frac{1}{d_I}-\frac{1}{d_S})a(x)}\Psi_1 = 0, & 0 < x < L, \\ (\Phi_1)_x(0) = (\Phi_1)_x(L) = 0, &\\ \int_0^L[e^{\frac{a(x)}{d_S}}\Phi_1+e^{\frac{a(x)}{d_I}}\Psi_1]dx = 0, & \end{array} \right. \end{equation} | (2.36) |
Obviously, \Phi_1 is uniquely determined by \Psi_1 in (2.36). Therefore, q = q_* is the only possible bifurcation point along \Gamma_S where positive solutions of (2.33) bifurcates and such q_* exists if and only if \mathcal{R}_0 = 1 . We can obtain the necessary and sufficient conditions for the occurrence of bifurcation by Lemma 2.3.1 and Lemma 2.3.3.
2. At (q, (\bar{S}, \bar{I})) = (q_*, (\hat{\bar{S}}, 0)) , the kernel
{\rm Ker}(D_{(\bar{S}, \bar{I})}F(q_*, (\hat{\bar{S}}, 0))) = {\rm span}\{(\Phi_1, \Psi_1)\}, |
where (\Phi_1, \Psi_1) is the solution of (2.35) with q = q_* . Up to a multiple of constant, (\Phi_1, \Psi_1) is unique. And the range of D_{(\bar{S}, \bar{I})}F(q_*, (\hat{\bar{S}}, 0)) is given by
{\rm Range}(D_{(\bar{S}, \bar{I})}F(q_*, (\hat{\bar{S}}, 0))) = \{(f, g, k)\in Y\times Y\times \mathbb{R}^N:\int_0^Lg\Psi_1e^{\frac{a(x)}{d_I}}dx = 0\}, |
and it is co-dimension one. By the result of Lemma 2.1.1, (\Psi_1)_x keeps one sign in (0, L) and \int_0^L(\Psi_1)_x\Psi_1e^{\frac{a(x)}{d_I}}dx\neq 0 , which implies that
D_{q, (\bar{S}, \bar{I})}F(q_*, (\hat{\bar{S}}, 0))[(\Phi_1, \Psi_1)]\notin {\rm Range}(D_{q, (\bar{S}, \bar{I})}F(q_*, (\hat{\bar{S}}, 0))). |
Therefore, using the local bifurcation theorem in [20] to F(q, (\bar{S}, \bar{I})) at (q_*, (\hat{\bar{S}}, 0)) , we know that the set of positive solutions of (2.33) is a smooth curve
\Gamma = \{(q(\tau), (\hat{\bar{S}}+\bar{S}_1(\tau), \bar{I}_1(\tau))):\tau\in [0, \delta)\} |
satisfying q(0) = q_* , \bar{S}_1(\tau) = \tau \hat{\bar{S}}+o(|\tau|) and I_1(\tau) = o(|\tau|) . Similar to the procedure in [21] and [22], (also see [23]), we can compute
q' = -\frac{ < l, D_{(\bar{S}, \bar{I}), (\bar{S}, \bar{I})}F(q_*, (\hat{\bar{S}}, 0))[\Phi_1, \Psi_1]^2 > }{2 < l, D_{q, (\bar{S}, \bar{I})}F(q_*, (\hat{\bar{S}}, 0))[(\Phi_1, \Psi_1)]} = \frac{\int_0^L\beta(x)e^{(\frac{1}{d_I}-\frac{1}{d_S})a(x)}\phi_1^3dx} {\hat{\bar{S}}\int_0^Le^{\frac{a(x)}{d_I}}\phi_1(\phi_1)_xdx}. |
Here l is the linear functional on Y\times Y\times \mathbb{R} defined by < l, [f, g, k] > = \int_0^Lg\Psi_1e^{\frac{a(x)}{d_I}}dx .
3. By the global bifurcation theorem in [23] and [24], we can get the existence of the connected component \Sigma . Moreover, \Sigma is either unbounded, or connects to another (q, (\hat{\bar{S}}, 0)) , or \Sigma connects to another point on the boundary of O .
Case (I) Assume that \int^L_0\beta(x)dx > \int^L_0\gamma(x)dx and (C2) holds. By Lemma 2.2.1 and the proof of part 2, we see that there exits a unique q_* such that the local bifurcation occurs at (q_*, (\hat{\bar{S}}, 0)) and q'(0) < 0 , which means that the bifurcation direction is subcritical. Therefore, there exists some small \delta > 0 such that (2.33) has a positive solution if q_*-\delta < q < q_* . By Lemma 2.1.4, \mathcal{R}_0 > 1 if q_*-\delta < q < q_* for \delta > 0 small enough. By Lemma 2.1.5, (2.33) has no positive solution if \mathcal{R}_0 < 1 , which implies that (2.33) has no positive solution if q > q_* . Consequently, the projection of \Sigma to the q -axis Proj_q\Sigma \subset [0, q_*] . And \Sigma must be bounded in \bar{O} because the positive solutions are uniformly bounded in L^{\infty} for 0\leq q\leq q_* . So the third option must happen here. Hence \Sigma must connect to (0, (\bar{S}_*, \bar{I}_*)) , so 0\in Proj_q\Sigma . Here (\bar{S}_*, \bar{I}_*) is the unique endemic equilibrium of (2.33) when q = 0 .
Case (II) Assume that \int^L_0\beta(x)dx < \int^L_0\gamma(x)dx .
(i) If (C1) holds and d_I > d^*_I , by Lemma 2.2.1 and the bifurcation analysis above, there exists unique bifurcation point q_* satisfying q'(0) > 0 , which means the bifurcation direction is supercritical. Then there exists some small \delta > 0 such that (2.33) has a positive solution if q_* < q < q_*+\delta . By Lemma 2.3.3, \mathcal{R}_0 > 1 if q_* < q < q_*+\delta for some \delta > 0 small enough. By Lemma 2.1.5, (2.33) has no positive solution if \mathcal{R}_0 < 1 , which implies that (2.33) has no positive solution if 0 < q < q_* . So the first option must happen here. If there exists some finite q^* > q_* such that Proj_q\Sigma = [q_*, q^*) , then it contradicts to the fact that all positive solutions are uniformly bounded in L^{\infty} for q = q^* . Consequently, the projection of \Sigma to the q -axis Proj_q\Sigma = [q_*, \infty) .
(ii) If (C2) holds and 0 < d_I < d^*_I , the proof is similar to that of Case (I), we omit the details here.
We will give the Leray-Schauder degree argument.
Lemma 2.4.2. For any \epsilon > 0 , there exist two constants \underline{C} and \bar{C} which depend on d_I , \epsilon , \|\beta\|_{\infty} , \|\gamma\|_{\infty} and N such that if \mathcal{R}_0\neq 1 , then for any positive solution of (2.33),
\begin{align} \underline{C}\leq \bar{S}(x), \bar{I}(x)\leq \bar{C}\quad {\rm for \ any }\ x\in [0, L] \end{align} | (2.37) |
for any \epsilon\leq d_S\leq \frac{1}{\epsilon} and 0\leq q\leq \frac{1}{\epsilon} .
Proof. \int_0^L[e^{\frac{a(x)}{d_S}}\bar{S}+e^{\frac{a(x)}{d_I}}\bar{I}]dx = N means that \bar{S}(x) and \bar{I}(x) are bounded in L^1 space. Using the standard theory of elliptic equation, it is easy to see that \bar{S} and \bar{I} have the upper bound \bar{C} depending on d_I , \epsilon , \|\beta\|_{\infty} , \|\gamma\|_{\infty} and N .
Therefore, we just need to prove that \bar{S} and \bar{I} have lower bounds.
Suppose contradictorily that there exist a sequence of \{(d_{S, i}, q_i)\}_{i = 1}^{\infty} satisfies \epsilon\leq d_{S, i}\leq \frac{1}{\epsilon} and 0\leq q_i\leq \frac{1}{\epsilon} and \mathcal{R}_0\neq 1 , and \{(\bar{S}_i(x), \bar{I}_i(x))\}_{i = 1}^{\infty} are the corresponding positive solutions of (2.33) satisfying
\max\limits_{x\in [0, L]} I_i(x)\rightarrow 0, \quad {\rm as}\ i\rightarrow \infty, |
and (\bar{S}_i(x), \bar{I}_i(x)) satisfies
\begin{equation} \left\{ \begin{array}{llll} d_{S, i} (\bar{S}_i)_{xx}+q_i \tilde{a}'(x)(\bar{S}_i)_x-\beta(x)\frac{e^{\frac{q_i}{d_I}\tilde{a}(x)}\bar{S}_i\bar{I}_i}{e^{\frac{q_i}{d_{S, i}}\tilde{a}(x)}\bar{S}_i+e^{\frac{q_i}{d_I}\tilde{a}(x)}\bar{I}_i}+ \gamma(x)e^{(\frac{q_i}{d_I}-\frac{q_i}{d_{S, i}})\tilde{a}(x)}\bar{I}_i = 0, \ \ & 0 < x < L, \\ d_I (\bar{I}_i)_{xx}+q_i \tilde{a}'(x)(\bar{I}_i)_x+\beta(x)\frac{e^{\frac{q_i}{d_{S, i}}\tilde{a}(x)}\bar{S}_i\bar{I}_i}{e^{\frac{q_i}{d_{S, i}}\tilde{a}(x)}\bar{S}_i +e^{\frac{q_i}{d_I}\tilde{a}(x)}\bar{I}_i}- \gamma(x)\bar{I}_i = 0, \ \ & 0 < x < L, \\ (\bar{S}_i)_x(0) = (\bar{S}_i)_x(L) = 0, \quad (\bar{I}_i)_x(0) = (\bar{I}_i)_x(L) = 0, \ \ & \\ \int_0^L[e^{\frac{q_i}{d_{S, i}}\tilde{a}(x)}\bar{S}_i+e^{\frac{q_i}{d_I}\tilde{a}(x)}\bar{I}_i]dx = N. \end{array} \right. \end{equation} | (2.38) |
Up to a subsequence, we assume that d_{S, i}\rightarrow d_S > 0 and q_i\rightarrow q\geq 0 . Note that \|\bar{I}_i\|_{\infty} are uniformly bounded. Letting \tilde{\bar{I}}_i = \frac{\bar{I}_i}{\|\bar{I}_i\|_{\infty}} , we have
\begin{equation*} \left\{ \begin{array}{llll} d_I (\tilde{\bar{I}}_i)_{xx}+q_i \tilde{a}'(x)(\tilde{\bar{I}}_i)_x+\beta(x)\tilde{\bar{I}}_i\frac{e^{\frac{q_i}{d_{S, i}}\tilde{a}(x)}\bar{S}_i}{e^{\frac{q_i}{d_{S, i}}\tilde{a}(x)}\bar{S}_i +e^{\frac{q_i}{d_I}\tilde{a}(x)}\bar{I}_i} - \gamma(x)\tilde{\bar{I}}_i = 0, \ \ & 0 < x < L, \\ (\tilde{\bar{I}}_i)_x(0) = (\tilde{\bar{I}}_i)_x(L) = 0.\ \ & \\ \end{array} \right. \end{equation*} |
By standard regularity and Sobolev embedding theorem in [25], up to a subsequence, \bar{I}_i\rightarrow 0 in C^1([0, L]) and there exists I^* > 0 such that \tilde{\bar{I}}_i\rightarrow I^* in C^1([0, L]) and \|I^*\|_{\infty} = 1 . Since \bar{I}_i\rightarrow 0 in C^1([0, L]) and \int_0^L[e^{\frac{q_i}{d_{S, i}}\tilde{a}(x)}\bar{S}_i+e^{\frac{q_i}{d_I}\tilde{a}(x)}\bar{I}_i]dx = N implies that \bar{S}_i is bounded in L^1([0, L]) , using the equation of \bar{S}_i , we get \bar{S}_i\rightarrow \hat{\bar{S}} > 0 in C^1([0, L]) . Letting i\rightarrow \infty in the equation of \bar{I}_i , we have
\begin{equation} \left\{ \begin{array}{llll} d_I I^*_{xx}+a'(x)I^*_x+[\beta(x)- \gamma(x)]I^* = 0, \ \ & 0 < x < L, \\ I^*_x(0) = I^*_x(L) = 0.\ \ & \\ \end{array} \right. \end{equation} | (2.39) |
Since I^* > 0 , (2.39) means that 0 is the principle eigenvalue, which is a contradiction of the assumption of \mathcal{R}_0\neq 1 for any d_I > 0 and 0\leq q \leq \frac{1}{\epsilon} . Therefore, there must exist some positive constant \underline{C} such that \max_{x\in [0, L]}I(x)\geq \underline{C} . Similar to the argument in [26], by Harnack inequality, we have
\max\limits_{x\in[0, L]}\bar{I}(x)\leq C^*\min\limits_{x\in [0, L]}\bar{I}(x) |
for some constant C^* depending on d_I , \epsilon , \|\beta\|_{\infty} , \|\gamma\|_{\infty} and N , which implies that \bar{I}(x) has uniformly positive lower bound.
Now we prove that S(x) has a uniform positive lower bound. Let S(x_0) = \min_{x\in [0, L]}S(x) . Using the minimum principle in [27], we have
\beta(x_0)\frac{e^{\frac{q}{d_I}\tilde{a}(x_0)}\bar{S}(x_0)}{e^{\frac{q}{d_S}\tilde{a}(x_0)}\bar{S}(x_0)+e^{\frac{q}{d_I}\tilde{a}(x_0)}\bar{I}(x_0)}- \gamma(x_0)e^{(\frac{q}{d_I}-\frac{q}{d_S})\tilde{a}(x_0)}\geq 0. |
Consequently,
\beta(x_0)\frac{\bar{S}(x_0)}{\bar{I}(x_0)}\geq \beta(x_0)\frac{e^{\frac{q}{d_I}\tilde{a}(x_0)}\bar{S}(x_0)}{e^{\frac{q}{d_S}\tilde{a}(x_0)}\bar{S}(x_0)+e^{\frac{q}{d_I}\tilde{a}(x_0)}\bar{I}(x_0)}\geq \gamma(x_0)e^{(\frac{q}{d_I}-\frac{q}{d_S})\tilde{a}(x_0)} |
and
\bar{S}(x_0)\geq \frac{\gamma(x_0)e^{(\frac{q}{d_I}-\frac{q}{d_S})\tilde{a}(x_0)}\bar{I}(x_0)}{\beta(x_0)}\bar{I}(x_0)\geq C\min\limits_{x\in [0, L]}\bar{I}(x), |
which completes the proof.
Lemma 2.4.3. Assume that \beta(x)-\gamma(x) changes sign once in (0, L) and one of the following conditions holds:
(i) d_I > 0 , q > \min_{x\in [0, L]}a'(x) > 0 , \int_0^L\beta(x)dx > \int_0^L\gamma(x)dx and (C2) holds;
(ii) 0 < d_I < d^*_I , q > \min_{x\in [0, L]}a'(x) > 0 , \int_0^L\beta(x)dx < \int_0^L\gamma(x)dx and (C1) holds.
Then (2.33) has at least an endemic equilibrium.
Proof. Note that we can extend the ranges of f and g properly for any nonnegative pair (f, g)\in C([0, L])\times C([0, L]) such that the function \frac{fg}{e^{\frac{\tau a(x)}{d_S}}f+e^{\frac{\tau a(x)}{d_I}}g} is Lipschitz continuous for f, g\in \mathbb{R} and \tau\in [0, 1] . Therefore we define the following compact operator family from C([0, L])\times C([0, L]) to C([0, L])\times C([0, L]) :
\begin{equation} \left\{ \begin{array}{llll} \quad (\tau d_S+(1-\tau)d_I) u_{xx}+\tau a'(x)u_x+ \gamma(x)e^{(\frac{\tau }{d_I}-\frac{\tau }{d_S})a(x)}v&\\ = \beta(x)\frac{e^{fg\frac{\tau a(x)}{d_I}}}{fe^{\frac{\tau a(x)}{d_S}}+ge^{\frac{\tau a(x)}{d_I}}}, & 0 < x < L, \\ d_I v_{xx}+\tau a'(x)v_x- \gamma(x)v = -\beta(x)\frac{fge^{\frac{\tau a(x)}{d_S}}}{fe^{\frac{\tau a(x)}{d_S}}+ge^{\frac{\tau a(x)}{d_I}}}, & 0 < x < L, \\ u_x(0) = u_x(L) = 0, \quad v_x(0) = v_x(L) = 0, & \\ \int_0^L[e^{\frac{\tau a(x)}{\tau d_S+(1-\tau)d_I}}u+e^{\frac{\tau a(x)}{d_I}}v]dx = N. \end{array} \right. \end{equation} | (2.40) |
Since the operator d_I\frac{d^2}{dx^2}+\tau a'(x)\frac{d}{dx}-\gamma(x) is invertible, then for any \tau\in [0, 1] and (f, g)\in C([0, L])\times C([0, L]) , by the second equation of (2.40), v is uniquely determined. Substituting this v into the first and last equations of (2.40), u is also uniquely determined. Therefore, we can define \mathcal{G}_{\tau}(f, g): = (u, v) .
Under conditions (i) and (ii), \mathcal{R}_{0, \tau} > 1 for any \tau\in [0, 1] . Here
\mathcal{R}_{0, \tau} = \sup\limits_{\varphi\in H^1((0, L)), \varphi\neq 0}\left\{\frac{\int_0^L\beta(x)e^{\frac{\tau a(x)}{d_I}}\varphi^2dx} {d_I\int_0^L\beta(x)e^{\frac{\tau a(x)}{d_I}}\varphi_x^2dx+\int_0^L\gamma(x)e^{\frac{\tau a(x)}{d_I}}\varphi^2dx}\right\}. |
By the result of Lemma 2.4.2, for any \tau\in [0, 1] , there exist two positive constant \bar{C} and \underline{C} depending on d_S , d_I , q , \|\beta\|_{\infty} , \|\gamma\|_{\infty} and N such that \underline{C}\leq u, v\leq \bar{C} for any solution of (2.40).
Let
D = \{(u, v)\in C([0, L])\times C([0, L]):\frac{\underline{C}}{2}\leq u, v\leq 2\bar{C}\}. |
Then (\bar{S}, \bar{I})\neq \mathcal{G}(\tau, (\bar{S}, \bar{I})) for any \tau\in [0, 1] and (\bar{S}, \bar{I})\in \partial D , which implies that Leray-Schauder degree deg({\bf{I}}-\mathcal{G}(\tau, (\cdot, \cdot)), D, 0) is well defined, and it is independent of \tau . Here {\bf{I}} is the identity map. Moreover, (\bar{S}, \bar{I}) is a solution of (2.33) if and only if (\bar{S}, \bar{I}) satisfies (\bar{S}, \bar{I}) = \mathcal{G}(1, (\bar{S}, \bar{I})) . If (\bar{S}, \bar{I})\in D and ({\bf{I}}-\mathcal{G}(0, (\cdot, \cdot)))(\bar{S}, \bar{I}) = 0 , then (\bar{S}, \bar{I}) is a positive solution of
\begin{equation} \left\{ \begin{array}{llll} d_I\bar{S}_{xx}-\beta(x)\frac{\bar{S}\bar{I}}{\bar{S}+\bar{I}}+\gamma(x)\bar{I} = 0, & 0 < x < L, \\ d_I \bar{I}_{xx}+\beta(x)\frac{\bar{S}\bar{I}}{\bar{S}+\bar{I}}-\gamma(x)\bar{I} = 0, & 0 < x < L, \\ \bar{S}_x(0) = \bar{S}_x(L) = 0, \quad \bar{I}_x(0) = \bar{I}_x(L) = 0, & \\ \int_0^L[\bar{S}+\bar{I}]dx = N. \end{array} \right. \end{equation} | (2.41) |
By the result of [2], (2.41) has a unique positive solution (S_*, I_*) satisfying S_*+I_* = \frac{N}{L} if the basic reproduction number \hat{\mathcal{R}}_0 > 1 . Linearizing (2.41) around (S_*, I_*) , we get
\begin{equation} \left\{ \begin{array}{llll} -d_I\Phi_{xx}+\beta(x)\frac{I_*^2}{(S_*+I_*)^2}\Phi+\beta(x)\frac{S_*^2}{(S_*+I_*)^2}\Psi-\gamma(x)\Psi = \mu\Phi, & 0 < x < L, \\ -d_I \Psi_{xx}-\beta(x)\frac{S_*^2}{(S_*+I_*)^2}\Psi+\gamma(x)\Psi-\beta(x)\frac{I_*^2}{(S_*+I_*)^2}\Phi = \mu\Psi, & 0 < x < L, \\ \Phi_x(0) = \Phi_x(L) = 0, \quad \Psi_x(0) = \Psi_x(L) = 0, & \\ \int_0^L[\Phi+\Psi]dx = N. \end{array} \right. \end{equation} | (2.42) |
Adding the first two equations of (2.42) and using the boundary condition \Phi_x = \Psi_x = 0 , x = 0, L , we get
\begin{align*} &-d_I(\Phi_{xx}+\Psi_{xx}) = \mu (\Phi+\Psi), \quad x\in (0, L), \\ &(\Phi+\Psi)_x = 0, \quad x = 0, L. \end{align*} |
Solving it, we have \Phi = -\Psi . Substituting this relation into the first equation of (2.42), we obtain
-d_I\Phi_{xx}+\left(\frac{2L\beta(x)}{N}I_*+\gamma(x)-\beta(x)\right)\Phi = \mu\Phi. |
Since I_* is a positive solution of (2.40), we know that -d_I\frac{d^2}{dx^2}+\frac{2L}{N}\beta(x)I_*+\gamma(x)-\beta(x) is a positive operator, so \mu > 0 . Hence the unique positive solution (S_*, I_*) is linearly stable. Using Leray-Schauder degree index (see Theorem 1.2.8.1 in [28]), we obtain
deg({\bf{I}}-\mathcal{G}(0, (\cdot, \cdot)), D, 0) = 1. |
Consequently, using the homotopy invariance of Leray-Schauder degree, we have
deg({\bf{I}}-\mathcal{G}(1, (\cdot, \cdot)), D, 0) = deg({\bf{I}}-\mathcal{G}(0, (\cdot, \cdot)), D, 0) = 1 |
for (d_I, q)\in \Omega_{hh}^U\cup \Omega_{lh}^{U_1} . By the properties of degree, \mathcal{G}(1, (\cdot, \cdot) has a fixed point in D if (d_I, q)\in \Omega_{hh}^U\cup \Omega_{lh}^{U_1} , which implies that (2.33) has at least one positive solution.
In this section, we consider the properties of \mathcal{R}_0 when \beta(x)-\gamma(x) changes sign twice. We also need the results on the positive roots of F(\eta) which is defined as
F(\eta) = \int_0^L\tilde{a}'(x)m(x)e^{\eta \tilde{a}(x)}dx, \quad 0\leq \eta < \infty, |
for any given continuous function m(x) on [0, L] .
Lemma 2.5.1. Assume that there exists 0 < x_1 < x_2 < L such that m(x_1) = m(x_2) = 0 , i.e., m(x) change sign twice for x\in [0, L] . Then
(i) If m(L) < 0 and \int_0^L\tilde{a}'(x)m(x)dx > 0 , then F(\eta) has a unique positive root \eta_1 for \eta\in (0, +\infty) satisfying F'(\eta_1) < 0 ;
(ii) If m(L) > 0 and \int_0^L\tilde{a}'(x)m(x)dx < 0 , then F(\eta) has a unique positive root \eta_1 for \eta\in (0, +\infty) satisfying F'(\eta_1) > 0 ;
(iii) If m(L) > 0 and \int_0^L\tilde{a}'(x)m(x)dx > 0 , then F(\eta) has at most two positive roots for \eta\in (0, +\infty) ;
(iv) If m(L) < 0 and \int_0^L\tilde{a}'(x)m(x)dx < 0 , then F(\eta) has at most two positive roots for \eta\in (0, +\infty) .
Proof. We only prove part (i) and part (iii). The proofs of part (ii) and part (iv) are similar.
(i). Let G_1(\eta): = e^{-\tilde{a}(x_2)\eta}[\tilde{a}(x_1)F(\eta)-F'(\eta)] and the prime notation denote differentiation with respect to \eta . Since m(L) < 0 and m(x) changes sign twice, it is easy to see that m(x) < 0 for x\in (0, x_1)\cup (x_2, L) and m(x) > 0 for x\in (x_1, x_2) . Note that \tilde{a}(x) is increasing. We know that
m(x)[\tilde{a}(x)-\tilde{a}(x_1)][\tilde{a}(x)-\tilde{a}(x_2)] < 0 |
for x\in (0, L) and x\neq x_i(i = 1, 2) . As a result, for any \eta > 0 , we have
\begin{align*} G'_1(\eta)& = -e^{-\tilde{a}(x_2)\eta}\left( F''(\eta)-[\tilde{a}(x_1)+\tilde{a}(x_2)]F'(\eta)+\tilde{a}(x_1)\tilde{a}(x_2)F(\eta)\right)\\ & = -\int_0^L e^{\eta[\tilde{a}(x)-\tilde{a}(x_2)]}\tilde{a}'(x)m(x)[\tilde{a}(x)-\tilde{a}(x_1)][\tilde{a}(x)-\tilde{a}(x_2)]dx > 0, \end{align*} |
which implies that G'_1(\eta) is a strictly increasing function for \eta\in (0, \infty) . By Lemma 2.2.2 and m(L) < 0 , F(\eta) < 0 for \eta > M if M is large enough. But F(0) = \int_0^L\tilde{a}'(x)m(x)dx > 0 , so there exits at least a positive root of F(\eta) . Let \eta_1 be the smallest positive one, then F'(\eta_1)\leq 0 .
If F'(\eta_1) = 0 , since
\begin{align*} &\quad F''(\eta)-[\tilde{a}(x_1)+\tilde{a}(x_2)]F'(\eta)+\tilde{a}(x_1)\tilde{a}(x_2)F(\eta)\\ & = \int_0^L e^{\eta[\tilde{a}(x)-\tilde{a}(x_2)]}\tilde{a}'(x)m(x)[\tilde{a}(x)-\tilde{a}(x_1)][\tilde{a}(x)-\tilde{a}(x_2)]dx < 0, \end{align*} |
then
F''(\eta_1)-[\tilde{a}(x_1)+\tilde{a}(x_2)]F'(\eta_1)+\tilde{a}(x_1)\tilde{a}(x_2)F(\eta_1) = F''(\eta_1) < 0. |
That is, \eta_1 is a strict local maximum value point of F(\eta) , which is a contradiction. So F'(\eta_1) < 0 . Now we will prove that \eta_1 is the unique positive root of F(\eta) . Assume contradictorily that \eta_2 > \eta_1 is the first number such that F(\eta_2) = 0 . Since F(\eta_1) = 0 and F'(\eta_1) < 0 , then F(\eta) < 0 in (\eta_1, \eta_2) , which implies that F'(\eta_2)\geq 0 . By the definition of G_1(\eta) , and noticing that F(\eta_1) = F(\eta_2) = 0 , we have G_1(\eta_1) = -\tilde{a}(x_1)e^{\tilde{a}(x_2)\eta_1}F'(\eta_1) > 0 and G_1(\eta_2) = -\tilde{a}(x_1)e^{\tilde{a}(x_2)\eta_2}F'(\eta_2)\leq 0 , which is a contradiction to the fact that G_1(\eta) is strictly increasing.
(iii) By Lemma 2.2.2 and m(L) > 0 , we see that F(\eta) > 0 for \eta > M if M is large enough. Then either F(\eta) > 0 for any \eta > 0 or F(\eta) has positive roots in (0, \infty) . Let G_2(\eta) = e^{-\tilde{a}(x_2)\eta}[F'(\eta)-\tilde{a}(x_1)F(\eta)] and \eta_1 be the first positive root of F(\eta) = 0 . Similar to the proof of part (i), it is easy to prove that G_2(\eta) is strictly monotone increasing in (0, +\infty) and F'(\eta_1)\leq 0 . We discuss in two cases.
Case 1: F'(\eta_1) = 0 . We will show that \eta_1 is the unique positive root of F(\eta) . Since
\begin{align*} &\quad F''(\eta)-[\tilde{a}(x_1)+\tilde{a}(x_2)]F'(\eta)+\tilde{a}(x_1)\tilde{a}(x_2)F(\eta)\\ & = \int_0^Le^{\eta[\tilde{a}(x)-\tilde{a}(x_2)]}\tilde{a}'(x)m(x)[\tilde{a}(x)-\tilde{a}(x_1)][\tilde{a}(x)-\tilde{a}(x_2)]dx > 0 \end{align*} |
then F''(\eta_1)-[\tilde{a}(x_1)+\tilde{a}(x_2)]F'(\eta_1)+\tilde{a}(x_1)\tilde{a}(x_2)F(\eta_1) = F''(\eta_1) > 0 . That is, F(\eta) attains a strict local minimum at \eta_1 . Now we will prove that \eta_1 is the unique positive root of F(\eta) . Assume contradictorily that \eta_2 > \eta_1 is the first number such that F(\eta_2) = 0 . Since \eta_1 is a strict local minimum value point, we have F(\eta) > 0 in (\eta_1, \eta_2) , which implies that F'(\eta_2)\leq 0 . By the definition of G_2(\eta) , and noticing that F(\eta_1) = F(\eta_2) = 0 , we have G_2(\eta_1) = 0 and G_2(\eta_2) = e^{a(x_2)\eta_2}F'(\eta_2)\leq 0 , which is a contradiction to the fact that G_2(\eta) is strictly increasing. So F(\eta) only has a unique positive root \eta_1 in this case.
Case 2. F'(\eta_1) < 0 . Since F(\eta_1) = 0 , so F(\eta) < 0 if \eta > \eta_1 and \eta close to \eta_1 enough. By Lemma 3.2 and m(L) > 0 , F(\eta) > 0 for \eta > M if M is large enough. Therefore, there exists at least a root of F(\eta) = 0 in (\eta_1, \infty) . Assume that \eta_2 is the first root of F(\eta) = 0 in (\eta_1, \infty) . Then F(\eta) < 0 in (\eta_1, \eta_2) and F'(\eta_2)\geq 0 . If F'(\eta_2) = 0 , then
\begin{align*} F''(\eta_2)& = F''(\eta_2)-[\tilde{a}(x_1)+\tilde{a}(x_2)]F'(\eta_2)+\tilde{a}(x_1)\tilde{a}(x_2)F(\eta_2)\\ & = \int_0^Le^{\eta_2[\tilde{a}(x)-\tilde{a}(x_2)]}\tilde{a}'(x)m(x)[\tilde{a}(x)-\tilde{a}(x_1)][\tilde{a}(x)-\tilde{a}(x_2)]dx > 0. \end{align*} |
And F(\eta) attains a strict local minimum at \eta_2 , which is a contradiction. Hence F'(\eta_2) > 0 .
We need to show that there is no positive root of F(\eta) = ) for \eta > \eta_2 . Assume contradictorily that there exists \eta_3 > \eta_2 such that F(\eta_3) = 0 and F(\eta) > 0 in (\eta_2, \eta_3) . Then F'(\eta_3) < 0 . And G_2(\eta_2) = e^{\tilde{a}(x_2)\eta_2}F'(\eta_2) > 0 and G_2(\eta_3) = e^{a(x_2)\eta_3}F'(\eta_3) < 0 , which contradicts the fact that G_2(\eta) is strictly increasing. Therefore we have proved that there exists a unique \eta_2 > \eta_1 such that F(\eta_2) = 0 and F'(\eta_2) > 0 .
Now we give the proof of Theorem 1.6 below.
Proof. We only prove part(i) and (iii). The proofs of (ii) and (iv) are similar.
Part (i): Similar to the proofs of Lemma 2.3.2 and 2.3.3, it is easy to prove that there exists some positive constant \Lambda which is independent of d_I and q and for each d_I > \Lambda , there exists some \tilde{q} = \tilde{q}(d_I) which satisfies \mathcal{R}_0(d_I, \tilde{q}) = 1 and \frac{\tilde{q}}{d_I}\rightarrow \eta_0 as d_I\rightarrow \infty . Here \eta_0 is the unique positive root of F(\eta) = 0 .
Next, we will prove that
\frac{\partial \mathcal{R}_0}{\partial q}(d_I, \tilde{q}) < 0 |
for any \tilde{q} satisfying \mathcal{R}_0(d_I, \tilde{q}) = 1 if d_I is large enough.
Let \tilde{\varphi} be the unique normalized eigenfunction of the eigenvalue \mathcal{R}_0(d_I, \tilde{q}) = 1 , i.e., \max_{[0, L]}\tilde{\varphi} = 1 and
\begin{equation} \left\{ \begin{array}{ll} -d_I(e^{\frac{\tilde{q}}{d_I}\tilde{a}(x)}\tilde{\varphi}_x)_x+[\gamma(x)-\beta(x)]e^{\frac{\tilde{q}}{d_I}\tilde{a}(x)}\tilde{\varphi} = 0, \quad 0 < x < L, \\ \tilde{\varphi}_x(0) = \tilde{\varphi}_x(L) = 0. \end{array} \right. \end{equation} | (2.43) |
By (2.22), we have
\begin{align} \frac{\partial \mathcal{R}_0}{\partial q}(d_I, \tilde{q}) = \frac{\mathcal{R}_0^2\int_0^Le^{\frac{\tilde{q}}{d_I}\tilde{a}(x)}\tilde{\varphi}_x\tilde{\varphi} \tilde{a}'(x)dx}{\int_0^L\beta(x)e^{\frac{\tilde{q}}{d_I}\tilde{a}(x)}\tilde{\varphi}^2dx}. \end{align} | (2.44) |
Multiplying (2.43) by \int_0^x\tilde{\varphi}(s)ds and integrating it over (0, L) , we get
d_I\int_0^Le^{\frac{\tilde{q}}{d_I}\tilde{a}(x)}\tilde{\varphi}_x\tilde{\varphi}\tilde{a}'(x)dx +\int_0^L[\gamma(x)-\beta(x)]e^{\frac{\tilde{q}}{d_I}\tilde{a}(x)}\tilde{\varphi}(\int_0^x\tilde{\varphi}(s)ds)dx = 0. |
Substitute it into (2.44), we obtain
d_I\frac{\partial \mathcal{R}_0}{\partial q}(d_I, \tilde{q}) = \frac{\int_0^L[\beta(x)-\gamma(x)]e^{\frac{\tilde{q}}{d_I}\tilde{a}(x)}\tilde{\varphi}(\int_0^x\tilde{\varphi}(s)ds)dx} {\int_0^L\beta(x) e^{\frac{\tilde{q}}{d_I}\tilde{a}(x)}\tilde{\varphi}^2dx}. |
As d_I\rightarrow \infty , \frac{\tilde{q}}{d_I}\rightarrow \eta_0 and \tilde{\varphi}\rightarrow 1 , we have
\lim\limits_{d_I\rightarrow \infty}d_I\frac{\partial \mathcal{R}_0}{\partial q}(d_I, \tilde{q}) = \frac{\int_0^Lx[\beta(x)-\gamma(x)]e^{\eta_0\tilde{a}(x)}dx}{\int_0^L\beta(x)e^{\eta_0\tilde{a}(x)}dx}. |
By Lemma 2.5.1(i),
\int_0^Lx[\beta(x)-\gamma(x)]e^{\eta_0\tilde{a}(x)}dx = F'(\eta_0) < 0. |
Hence, there exists some constant Q > 0 (dependent on d_I ) such that \mathcal{R}_0 > 1 for 0 < q < Q and \mathcal{R}_0 < 1 for q > Q .
Part (iii). According to the results of Lemma 2.5.1(iii), we divide into three cases to prove it.
Case 1. F(\eta) > 0 for any \eta > 0 . It is easy to show that there exists some positive constant \Lambda independent of d_I and q such that \mathcal{R}_0 > 1 for every d_I > \Lambda and any q > 0 .
Case 2. F(\eta) has a unique positive root \eta_1 for \eta\in(0, +\infty) and F'(\eta_1) = 0 . Similar to the proof of part (i), we can prove that there exists some positive constant \Lambda independent of d_I and q such that for every d_I > \Lambda , there exists some \tilde{q} = \tilde{q}(d_I) such that \mathcal{R}_0(d_I, \tilde{q}) = 1 and \frac{\tilde{q}}{d_I}\rightarrow \eta_0 as d_I\rightarrow \infty , where \eta_0 is the unique positive root of F(\eta) = 0 . Moreover, \frac{\partial \mathcal{R}_0}{\partial q}(d_I, \tilde{q}) = 0 . Therefore there exists some positive constant \Lambda which is independent of d_I and q such that for every d_I > \Lambda , there exists a constant Q > 0 dependent on d_I satisfying \mathcal{R}_0 = 1 for q = Q and \mathcal{R}_0 > 1 for q\in (0, Q)\cup (Q, \infty) .
Case 3. F(\eta) has two positive roots \eta_1 and \eta_2 ( \eta_1 < \eta_2 ) for \eta\in (0, +\infty) and F'(\eta_1) < 0 , F'(\eta_2) > 0 . Similar to the discussion of part (i), for each d_I > 0 , there exist \tilde{q}_1 = \tilde{q}_1(d_I) and \tilde{q}_2 = \tilde{q}_2(d_I) such that \mathcal{R}_0(d_I, \tilde{q}_i) = 1(i = 1, 2) and \frac{\tilde{q}_1}{d_I}\rightarrow \eta_1 , \frac{\tilde{q}_2}{d_I}\rightarrow \eta_2 as d_I\rightarrow \infty . And
\frac{\partial \mathcal{R}_0}{\partial q}(d_I, \tilde{q}_1) < 0, \quad \frac{\partial \mathcal{R}_0}{\partial q}(d_I, \tilde{q}_2) > 0. |
Consequently, there exist two constants Q_2 > Q_1 > 0 which depend on d_I and satisfy that \mathcal{R}_0 > 1 for q\in(0, Q_1)\cup (Q_2, \infty) , \mathcal{R}_0 < 1 for q\in(Q_1, Q_2) .
In this section, we will summarize the main results of this paper.
Theorem 1.1 gives some properties for the basic reproduction number \mathcal{R}_0 and Theorem 1.2 says that \mathcal{R}_0 = 1 is the watershed for judging whether the DFE is stable or not. Theorem 1.3 and Theorem 1.4 deal with the stable and unstable regions of the DFE. Theorem 1.5 establishes the existence of EE. Theorem 1.6 considers the results on (1.1) when \beta(x)-\gamma(x) changes sign twice in (0, L) .
We only establish the results on (1.1) under the assumption of a'(x) > 0 in this paper. However, it is much more difficult to obtain the results on (1.1) if there exists some x_0\in (0, L) satisfying a'(x_0) = 0 .
Biologically, the influence of advection is from the upstream to the downstream, small diffusion or large advection tends to force the individuals to concentrate at the downstream end. Therefore, the disease persists for arbitrary advection rate if the habitat is a high-risk domain and the downstream end is a high-risk site. While the advection transports the individuals to a favorable location and thus it can help eliminate the disease if the downstream end is a low-risk site. In conclusion, when advection is strong or the diffusion is small, the disease will be eliminated if the downstream end is a low–risk site, while the disease will persist if the downstream end is a high–risk site.
The authors thank the anonymous referees for their helpful suggestions.
Xiaowei An was supported by Natural Science Foundation of China People's Police University(No.ZKJJPY201723).
All authors declare no conflicts of interest in this paper.
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