# Bifurcation Behaviors of Steady-State Solution to a Discrete General Brusselator Model

Authors: Ruyun Ma; Zhongzi Zhao
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

We study the local and global bifurcation of nonnegative nonconstant solutions of a discrete general Brusselator model. We generalize the linearu in the standard Brusselator model to the nonlinear fu. Assume that f∈C0,∞,0,∞ is a strictly increasing function, and f′f−1a∈0,∞. Taking b as the bifurcation parameter, we obtain that the solution set of the problem constitutes a constant solution curve and a nonconstant solution curve in a small neighborhood of the bifurcation point b0j,f−1a,ab/f−1a2. Moreover, via the Rabinowitz bifurcation theorem, we obtain the global structure of the set of nonconstant solutions under the condition that fs/s2 is nonincreasing in 0,∞. In this process, we also make a priori estimation for the nonnegative nonconstant solutions of the problem.


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## Body

## 1. Introduction

In 1968, Prigogine and Lefever [1] introduced first the Brusselator model for a chemical reaction-diffusion of self-catalysis as follows:(1)∂u∂t−d1Δu=a−b+1u+u2v,x∈Ω,t>0,∂v∂t−d2Δv=bu−u2v,x∈Ω,t>0,∂u∂n=∂v∂n=0,x∈∂Ω,t>0,where Ω⊂ℝNN≥1 is a smooth and bounded domain, n denotes the outward unit normal vector on ∂Ω, u and v represent the concentration of two intermediary reactants having the diffusion rates d1,d2∈0,∞ with d2>d1, and a,b>0 are the fixed concentrations. This chemical reaction plays an important role due to its similarities with neuronal and biological networks. Therefore, (1) has been extensively investigated in the last decades from both analytical and numerical point of view (see [2–12]). Most of them are interested in finding spatially nonconstant solutions of the equilibrium problem(2)−d1Δu=a−b+1u+u2v,x∈Ω,−d2Δv=bu−u2v,x∈Ω,∂u∂n=∂v∂n=0,x∈∂Ω.From the definition of Strogatz [13], chaos sensitivity depends on initial conditions, which shows that nearby trajectories diverge exponentially. Continuous systems in a 2-dimensional phase space cannot experience such divergence; hence, chaotic behaviors can only be observed in deterministic continuous systems with a phase space of dimension 3, at least. On the contrary, in a discrete map, it is well known that chaos occurs also in one dimension. Therefore, discrete chaotic systems exhibit chaos whatever their dimension is.It is worth to note that discrete models governed by difference equations are more appropriate than the continuous one due to their efficient computational results and rich dynamical behavior (see [14, 15]). Therefore, the discrete Brusselator model has been studied by several authors, and they got some results ([16–18] and the references therein). In particular, Din [16] applied forward Euler’s method to one-dimensional model (1) as follows:(3)xn+1=xn+hα−1+βxn+xn2yn,yn+1=yn+hβxn−xn2yn,where 0<h<1 represents the step size for Euler’s method. The local dynamical behaviors are obtained for (3).Note that [16–18] only studied the dynamical behaviors of the discrete-time Brusselator model. The reason is that the partial difference equation is very difficult for us. Indeed, the discrete-space Brusselator model is also worth studying due to the discontinuity of the space.Therefore, we will consider the discrete space, more general form of (2) with N=1:(4)−d1Δ2ux−1=a−b+1fux+u2xvx,x∈T,−d2Δ2vx−1=bfux−u2xvx,x∈T,Δu0=ΔuT=Δv0=ΔvT=0,where Δux=ux+1−ux, T≔1,Tℤ=1,2,…,T, T≥2 is an integer, d1,d2,a>0 are fixed parameters and d2>d1, and b>0 is a bifurcation parameter. Clearly, fu=fu/u⋅u; then, (4) is seen to be equivalent to(5)−d1Δ2ux−1=a−b+1fuxux⋅ux+u2xvx,x∈T,−d2Δ2vx−1=bfuxux⋅ux−u2xvx,x∈T,Δu0=ΔuT=Δv0=ΔvT=0,where fu/u can be regarded as a variable coefficient. It is well known that the linear terms b+1u and bu in (2) cannot withstand any small perturbation.In fact, (5) has an important application value in biology and chemistry. Xu et al. [19] said that model (1) includes a basic assumption: the cells always live in a continuous patch environment. However, this may not be the case in reality, and the motion of individuals of given cells is random and isotropic, i.e., without any preferred direction, the cells are also absolute individuals. The cells or units are also absolute individuals in microscopic sense, and each isolated cell exchanges materials by diffusion with its neighbors. Thus, it is reasonable to consider a 1D or 2D spatially discrete reaction-diffusion system in order to explain the chemical system.Kang [20] discussed the dynamics of the local map of a discrete version of the Brusselator model. To discretize system (1), he employed the following discretizations.For the derivative in time, he used(6)∂ux,t∂t⟶ux,t+Δt−ux,tΔt.For the space derivative, he used(7)∂ux,t∂x⟶ux+Δx,t−ux,tΔx,(8)∂2ux,t∂x2⟶ux+Δx,t−2ux,t+ux−Δx,tΔx2.It is important to note thatΔ in (6)–(8) is different from Δ in this paper. Our discretization is consistent with Kang’s, and we chose the step size to be 1. When fu=u, (5) is the steady-state form of the problem studied in [19, 20].On the contrary, the Brusselator system has been investigated from the numerical point of view (see [21] and references therein). Most modern texts on numerical analysis give an introduction to numerical solutions of partial differential equations using the finite-difference approach. Twizell et al. [22] had given a second-order finite-difference scheme for the Brusselator reaction-diffusion system. It is well known that (2) is an important mathematical dynamics model in biology and chemistry. In some ways, (5) is even more practical than (2).We will study the local and global bifurcation of nonnegative nonconstant solutions of (4) under the following assumptions:(H1)f∈C0,∞,0,∞ is a strictly increasing function(H2)f′f−1a∈0,∞(H3)fs/s2 is nonincreasing in 0,∞Remark 1.
Iffu=u, then (4) is the discrete version of (2) with N=1. Obviously, discrete Brusselator model (4) is a second-order difference boundary value problem.
The rest of the paper is organized as follows: in Section2, we give a priori estimate and some preliminary results. Section 3 is devoted to studying the local bifurcation of nonnegative nonconstant solutions of (4) under conditions (H1) and (H2). Finally, in Section 4, we add condition (H3) to obtain the global bifurcation of nonnegative nonconstant solutions of (4).


## 2. Preliminary Results

At first, let us look for the constant solution of (4). To get it, it suffices to look for the constant solution of the following problem:(9)a−b+1fux+u2xvx=0,x∈T,bfux−u2xvx=0,x∈T.By (H1), problem (4) has a unique constant solution f−1a,ab/f−1a2.We can easily obtain the following a priori estimate of the nonnegative nonconstant solutions of (4).Lemma 1.
Let (H1), (H2), and (H3) hold. Then, any nonnegative nonconstant solutionu,v of (4) satisfies(10)f−1ab+1≤ux≤f−1a+d2d1⋅abb+1f−1a/b+12,x∈T,bff−1a+d2/d1⋅ab/b+1f−1a/b+12f−1a+d2/d1⋅ab/b+1f−1a/b+122≤vx≤abb+1f−1a/b+12,x∈T.Proof.
Letx0∈T be the minimum point of u. We have(11)ux0+1≥ux0,ux0−1≥ux0,−d1Δ2ux0−1=−d1ux0+1−2ux0+ux0−1≤0.
Then,(12)a−b+1fux0+u2x0vx0≤0,a−b+1fux0≤0,fux0≥ab+1.
Then,ux0≥f−1a/b+1 by (H1), and so(13)ux≥ux0≥f−1ab+1,x∈T.
Letx1∈T be the maximum point of v. Similarly, we can get that(14)vx1+1≤vx1,vx1−1≤vx,−d2Δvx1−1=−d1vx1+1−2vx1+vx1−1≥0,(15)bfux1−u2x1vx1≥0.
Then,(16)vx≤vx1≤bfux1u2x1,x∈T.
Combining this with (13), from (H3), we show(17)vx≤bfux1u2x1≤abb+1f−1a/b+12,x∈T.
Letw=d1u+d2v. Then, it follows from (4) that(18)−Δ2wx−1=a−fux,x∈T,Δw0=ΔwT=0.
Now, letx2∈T be the maximum point of w. Observe that(19)wx2+1≥wx2,wx2−1≥wx2,−Δ2wx2−1=a−fux2≥0.
Then, from (H1), it is easy to seeux2≤f−1a. Combining this with (17), we know that, for any x∈T,(20)d1ux≤wx≤wx2≤d1f−1a+d2⋅abb+1f−1a/b+12.
Then,(21)ux≤f−1a+d2d1⋅abb+1f−1a/b+12,x∈T.
Ifx3∈T is the minimum point of v, then(22)bfux3−u2x3vx3≤0,and so(23)vx≥vx3≥bfux3u2x3≥bff−1a+d2/d1⋅ab/b+1f−1a/b+12f−1a+d2/d1⋅ab/b+1f−1a/b+122,x∈T.
Consequently, the proof is completed.Lemma 2.
(see [23]). Assume T≥2 is an integer. Then, the discrete second-order linear Neumann eigenvalue problem(24)−Δ2φx−1=μφx,x∈T,Δφ0=ΔφT=0,has T real and simple eigenvalues, which can be ordered as follows:(25)0=μ0<μ1<μ2<⋯<μT−1.Moreover, forj∈1,…,T−1, the eigenfunction φj corresponding to the eigenvalue μj has exactly j−1 simple generalized zeros.For any fixedT≥2, it is well known that(26)μj=0,j=0,2−cosjπT,j∈1,T−1ℤ,and the corresponding eigenfunctions are(27)φjx=1,j=0,cosjπTx,j∈1,T−1ℤ.Lemma 3.
(see [18], Theorem 2.5). Let a be a constant. Then, for ΔCi=0,(28)∑cosai=sinai−1/22sina/2+Ci,a≠2nπ.Lemma 4.
(see [18], Theorem 2.7). If zn is an indefinite sum of yn, then(29)∑k=mn−1yk=zkmn=zn−zm.
Let(30)X≔u,vu,v:T^⟶ℝ,Δu0=ΔuT=Δv0=ΔvT=0,X1≔uu:T⟶ℝ,Y≔X1×X1,w1,w2Y=∑x=1Tu1xu2x+∑x=1Tv1xv2x,where w1=u1,v1,w2=u2,v2∈Y and T^≔0,T+1ℤ.


## 3. Local Bifurcation

By the second part,w¯≔f−1a,ab/f−1a2 is the unique constant solution of (4).Define the mappingP:0,∞×X⟶Y:(31)Pb,w=d1Δ2ux−1+a−b+1fux+u2xvxd2Δ2vx−1+bfux−u2xvx.For the fixedb>0, w=u,v is a solution of (4) if and only if b,w is a zero-point of P. Note that Pb,w¯=00 since w¯ is the constant solution of (4).Let(32)u=f−1a+∑k=1∞εkuk,v=abf−1a2+∑k=1∞εkvk,(33)b=b0+∑k=1∞εkbk.We also have to Taylor expandf at the point f−1a.The purpose of the rest of this section is to solveb0 and prove that b0,w¯ is the bifurcation point of Pb,w=00.First of all, we substitute (32) and (33) into (4) and let the higher-order term of ε be equal to 0. Then, we can get the problem(34)−d1Δ2u1x−1=−b0f′f−1a−f′f−1a+2ab0f−1au1x+f−1a2v1x,x∈T,−d2Δ2v1x−1=b0f′f−1a−2ab0f−1au1x−f−1a2v1x,x∈T,Δu10=Δu1T=Δv10=Δv1T=0.In (34), by using undetermined coefficient method, it follows that(35)b0=f−1ad1d2μj2+d2f′f−1aμj+d1f−1a2μj+f−1a2f′f−1a2ad2μj−d2f−1af′f−1aμj≔b0j,j=1,2,…T−1.Moreover, it is not difficult to prove (34) has a nontrivial solution u1,v1:(36)u1=c1jcosjπlx=c1jφjx,c1j=−d2μjd1μj+f′f−1a,v1=cosjπlx=φjx.Next, we substitute (32) and (33) into (4) and let the higher-order term of ε2 be equal to 0; then, (4) becomes the following system:(37)d1Δ2u2x−1+−b0f′f−1a−f′f−1a+2ab0f−1au2x+f−1a2v2x=−F1,x∈T,d2Δ2v2x−1+b0f′f−1a−2ab0f−1au2x−f−1a2v2x=F1,x∈T,Δu20=Δu2T=Δv20=Δv2T=0,where(38)F1=2ab1f−1a−b1f′f−1au1+2f−1au1v1+ab0f−1a2u12.In order to solveb1 from (37), let us consider the following adjoint system of the homogeneous system related to (37):(39)d1Δ2y2x−1+−b0f′f−1a−f′f−1a+2ab0f−1ay2x+b0f′f−1a−2ab0f−1az2x=0,x∈T,d2Δ2z2x−1+f−1a2y2x−f−1a2z2x=0,x∈T,Δy20=Δy2T=Δz20=Δz2T=0.It is not difficult to verify that (39) has a solution y2,z2:(40)y2=c2jcosjπlx=c2jφjx,c2j=1+d2μjf−1a2,z2=cosjπlx=φjx.By virtue of the solvability condition for (37), it is obvious that(41)∑x=1Tz2x−y2xF1=0.In fact,(42)∑x=1Tz2x−y2xF1=∑x=1T−d2μjf−1a2⋅φjx2ab1f−1a−b1f′f−1a⋅−d2μjd1μj+f′f−1a⋅φjx+2f−1a−d2μjd1μj+f′f−1aφj2x+ab0f−1a2⋅d22μj2d1μj+f′f−1a2⋅φj2x=0.We know that(43)∑x=1Tφj3x=∑x=1Tcos3jπTx=14∑x=1Tcos3jπTx+3cosjπTx.From Lemmas3 and 4, for any j∈1,2,…,T−1 and j≠2nT/3,n∈1,2,…, we obtain(44)∑x=1Tφj3x=14sin3jπ/Tx−1/22sin3jπ/2T1T+1+34sinjπ/Tx−1/22sinjπ/2T1T+1=14⋅sin2T+13jπ/2T−sin3jπ/2T2sin3jπ/2T+34⋅sin2T+1jπ/2T−sinjπ/2T2sinjπ/2T=14⋅cos2T+23jπ/2T⋅sin2T⋅3jπ/2Tsin3jπ/2T+34⋅cos2T+2jπ/2T⋅sin2T⋅jπ/2Tsinjπ/2T=0.Thenb1j≔b1=0, and so F1 will reduce to(45)F1=2f−1ac1jφj2x+ab0f−1a2c12jφj2x=2f−1ac1jcos2jπlx+ab0f−1a2c12jcos2jπlx=f−1ac1jcos2jπlx+1+12⋅ab0f−1a2c12jcos2jπlx+1=122f−1ac1j+ab0f−1a2c12j+122f−1ac1j+ab0f−1a2c12jφ2jx.Therefore, a particular solutionu2,v2 of (37) can be obtained as follows:(46)u2=a1j+a2jcos2jπlx=a1j+a2jφ2jx,v2=a3j+a4jcos2jπlx=a3j+a4jφ2jx,where(47)a2j=f−1a2c1d2μ2j+ab0d2μ2jc12/2f−1ad1d2μ2j2f−1a+f′f−1af−1af−1a2+d2μ2j1+b0−2ab0d2−d1f−1a3μ2j,a1j=0,a3j=−c1f−1a2f−1a+ab02f−1a2c1,a4j=−d1μ2j+f′f−1ad2μ2ja2j.Sinceb1=0, we have to solve b2. We substitute (32) and (33) into (4) and let the higher-order term of ε3 be equal to 0; then, a problem similar to (37) is obtained:(48)d1Δ2u3x−1+−b0f′f−1a−f′f−1a+2ab0f−1au3x+f−1a2v3x=−F2,x∈T,d2Δ2v3x−1+b0f′f−1a−2ab0f−1au3x−f−1a2v3x=F2,x∈T,Δu30=Δu3T=Δv30=Δv3T,where(49)F2=−b2f′f−1a+2ab2f−1au1+2f−1au1v2+2f−1au2v1+u12v1+2ab0f−1a2u1u2.Clearly, (39) is also the adjoint system of the homogeneous system related to (48); then,(50)∑x=1Tz2x−y2xF2=0.According to values ofu1,u2,v1, and v2, we have(51)∑x=1Tz2x−y2xF2=∑x=1T−d2μjf−1a2φjx2a−f′f−1af−1af−1a⋅b2c1φjx+2f−1ac1φjxa3+a4φ2jx+2f−1aa2φ2jxφjx+c12φj3x+2ab0a2c1f−1a2φjxφ2jx.From Lemmas3 and 4, for any j∈1,2,…,T−1 and j≠nT/2,n∈1,2,…, we know that(52)∑x=1Tφj2xφ2jx=∑x=1Tcos2jπTx⋅cos2jπTx=12∑x=1T12cos4jπTx+cos2jπTx+T4=14⋅sin4jπx−1/2/T2sin4jπ/2T1T+1+12⋅sin2jπx−1/2/T2sin2jπ/2T1T+1+T4=14⋅2cos2T+22jπ/Tsin2jπ⋅2T/T2sin2jπ/T+12⋅2cos2T+2jπ/Tsinjπ⋅2T/T2sin2jπ/T+T4=T4≠0.Thus,(53)b2j≔b2≠0,j=1,2,…T−1.From the above analysis, we obtain the main result of this section.Theorem 1.
Assume that (H1) and (H2) hold. Then, for any positive integerj<T,j≠nT/2 and j≠2nT/3,n∈1,2,…, b0j,w¯ is a bifurcation point of Pb,w=00. Moreover, there is a nontrivial solution ϕε=bε,uε,vε of (4) if ε is small enough, where b,u, and v are continuous with respect to ε:(54)uε=f−1a+εc1jφj+ε2a1j+a2jφ2j+oε2,vε=abf−1a2+εφj+ε2a3j+a4jφ2j+oε2,bε=b0j+ε2b2j+oε2.The set of zero-points ofP constitutes two curves in a neighborhood of bifurcation point b0j,w¯.Letℂ be the closure of the nonconstant solution set of Pb,w=00 and Γj be a connected component of ℂ∪b0j,w¯ and b0j,w¯∈Γj. In a small neighborhood of bifurcation point b0j,w¯, the curve Γj is determined by the eigenfunction φj, where φj has exactly j−1 simple generalized zeros.


## 4. Global Bifurcation

Theorem 2.
Let (H1), (H2), and (H3) hold. Ifμj≠f−1a2/d2,j=1,2,…,T, then projection of continuum Γj is unbounded on the b-axis.Proof.
(4) can be written as follows:(55)−Δ2ux−1=gux,vx,x∈T,−Δ2vx−1=hux,vx,x∈T,Δu0=ΔuT=Δv0=ΔvT=0,where(56)gu,v=1d1a−b+1fu+u2v,hu,v=1d2bfu−u2v.
Letu˜=u−f−1a,v˜=v−ab/f−1a2. Then, (55) is equivalent to the following problem:(57)−Δ2u˜x−1=g0u˜x+g1v˜x+g˜u˜x,v˜x,x∈T,−Δ2v˜x−1=h0u˜x+h1v˜x+h˜u˜x,v˜x,x∈T,Δu˜0=Δu˜T=Δv˜0=Δv˜T=0,where g˜ and h˜ are higher-order terms of u˜,v˜ and(58)g0=guu,vf−1a,ab/f−1a2=1d1−b+1f′f−1a+2abf−1a,g1=gvu,vf−1a,ab/f−1a2=f−1a2d1,h0=huu,vf−1a,ab/f−1a2=1d2bf′f−1a−2abf−1a,h1=hvu,vf−1a,ab/f−1a2=−f−1a2d2.
In this way, we convert the constant solutionw¯=f−1a,ab/f−1a2 of (4) to the trivial solution θ=0,0 of (57).
LetH1:Y⟶X and H2:Y⟶X be the inverse of operators f′f−1a/d1I−A and f−1a2/d2I−A with Neumann boundary conditions, respectively, where A:0,∞⟶−∞,0 and Aux=Δ2ux−1. Set U=u˜,v˜:(59)KbU=1d1−bf′f−1a+2abf−1aH1u˜+f−1a2d1H1v˜,1d2bf′f−1a−2abf−1aH2u˜,WU=H1g˜u˜,v˜,H2h˜u˜,v˜.
It can be verified that (57) is equivalent to(60)U=KbU+WU,in X. For any fixed b>0, Kb and WU are linear compact operators in X and WU=oU. By the Rabinowitz global bifurcation theorem [24], we need to verify(i)
1 is an eigenvalue ofKb0j, and its algebraic multiplicity is 1(ii)
The index ofI−Kb−W changes when b crosses b0j
Now, we will prove (i). SupposeΨ=ξψ,ξ=∑ajφj,ψ=∑cjφj. Let(61)Kb−IΨ=00,i.e.,(62)1d1−b+1f′f−1a+2abf−1a+Af−1a2d11d2bf′f−1a−2abf−1a−f−1a2d2+AΨ=00,
Thus,(63)∑j=0∞Ljajcjφj=00,where(64)Lj=1d1−b+1f′f−1a+2abf−1a−μjf−1a2d11d2bf′f−1a−2abf−1a−f−1a2d2−μj.
By computation,detLj=0 if and only if b=b0j; taking b=b0j leads to(65)Ljajcj=00f′f−1a+d1μjd2μjajcj.
Then,kerKb0j−I=spanΨ,Ψ=−d2μjf′f−1a+d1μjφj. This implies that 1 is the eigenvalue of K=Kb0j and dimkerK−I=1. The algebraic multiplicity of eigenvalue 1 is the dimension of the generalized null space ∪i=1∞kerK−Ii; therefore, kerK−I∩ImK−I=θT.
LetKT be the transposed matrix of K:(66)KT=1d1−bf′f−1a+2abf−1aH11d2bf′f−1a−2abf−1aH2f−1a2d1H10,and Ψ∗=ξ∗ψ∗,ξ∗=∑aj∗φj,ψ∗=∑cj∗φj. Suppose Ψ∗∈kerKT−I. Then,(67)1d1−bf′f−1a+2abf−1aH1ξ∗+1d2bf′f−1a−2abf−1aH2ψ∗=ξ∗,f−1a2d1H1ξ∗=ψ∗.
From the definition ofH1andH2, (67) can also be written as(68)−d1d2Δ2ξ∗x−1=d2​−bf′f−1a+2abf−1a−f−1a2d1ξ∗−d2f′f−1af−1a2−bf′f−1a+2abf−1aψ∗,−d1Δ2ψ∗x−1=f−1a2ξ∗x−f′f−1aψ∗x.
That is to say,(69)∑j=0∞Bjaj∗cj∗φj=00,where(70)Bj=d2−bf′f−1a+2abf−1a−f−1a2d1−d1d2μj−d2f′f−1af−1a2−bf′f−1a+2abf−1af−1a2−d1μj−f′f−1a.
Similarly,detBj=0 if and only if b=b0j; taking b=b0j leads to(71)Bjaj∗cj∗=00f−1a2−d1μj−f′f−1aaj∗cj∗.
Then,kerK∗−I=spand1μj+f′f−1af−1a2φj. According to μj≠f−1a2/d2, we obtain(72)Ψ,Ψ∗Y=∑x=1T−d2μjd1μj+f′f−1aφjx+∑x=1Tf′f−1a+d1μjf−1a2φj2x=d1μj+f′f−1af−1a2−d2μj∑x=1Tφj2x=d1μj+f′f−1af−1a2−d2μj∑x=1Tcos2jπTx=d1μj+f′f−1af−1a2−d2μj⋅12∑x=1Tcos2jπTx+1=d1μj+f′f−1af−1a2−d2μj12⋅cos2T+2jπ/Tsin2πsinjπ/T+T2≠0.
This suggests thatΨ∉kerK∗−I⊥=ImK−I, and so (i) is proved.
Now, we will prove (ii). From (i), for anyb>0,b≠b0j and b belongs to a small neighborhood of b0j, and Kb−I:X⟶X is a bijection. Fix b>0; then, θ is a solution of (60), and θ is isolated. From Leray–Schauder fixed point theory, we can get(73)indexI−Kb−W,b,θ=degI−Kb,B,θ=−1γ,where B is a sufficiently small ball centered at θ, γ is the sum of the algebraic multiplicity of the eigenvalues of Kb, and γ>1.
We are going to verify that, forε>0 is small enough,(74)indexI−Kb0j−ε−W,b0j−ε,θ≠indexI−Kb0j+ε−W,b0j+ε,θ.
Ifτ is an eigenvalue of Kb and Ψ=ξψ is the corresponding eigenfunction, then(75)Kb−IΨ=00,i.e.,(76)−τd1Δ2ξx−1=−bf′f−1a+2abf−1a−τf′f−1aξx+f−1a2ψx,x∈T,−τd2Δ2ψx−1=bf′f−1a−2abf−1aξx−f−1a2τψx,x∈T.
By virtue ofξ=∑ajφj and ψ=∑cjφj, we can get(77)∑j=0∞τμjd1+bf′f−1a−2abf−1a+τf′f−1a−f−1a2−bf′f−1a+2abf−1aτμjd2+τf−1a2ajcjφj=00.
Then, the characteristic equation is(78)d1d2μj2+d2μif′f−1a+d1μjf−1a2+f′f−1af−1a2τ2+bd2f′f−1aμj−2ad2μjf−1a+f′f−1af−1a2−2af−1aτ+f−1a2−bf′f−1a+2abf−1a=0,j=0,1,2,…,T.
Ifτ=1, b can be solved from (78):(79)b=f−1ad1d2μj2+d2f′f−1aμj+d1f−1a2μj+f−1a2f′f−1a2ad2μj−d2f−1af′f−1aμj=b0j.
Therefore, by calculating the corresponding eigenvalues of (78), we can obtain that when b passes through b0j, the number of eigenvalues of Kb which is greater than 1 is the same, and their algebraic multiplicity are equal. By plugging (79) into (78), we have(80)2ad2μj−d2f−1af′f−1aμjf−1aτ2+f′f−1ad2μj+f−1a2−2ad2μjf−1a−2af−1aτ+f−1a2−f′f−1a+2af−1a=0.
Then,(81)d2f′f−1aμj−2ad2μjf−1a+f′f−1af−1a2−2af−1a2−42ad2μj−d2f−1af′f−1aμj⋅−f′f−1af−1a+2a=d2f′f−1aμj−2ad2μjf−1a−f′f−1af−1a2−2af−1a2>0,and so (80) has two different roots τ1=1,τ2=f−1a2/d2μj. Thus, two things will happen:(a)
Ifμj>f−1a2/d2, then τ1b0j=1,τ2b0j<1.(b)
Ifμj<f−1a2/d2, then τ1b0j=1,τ2b0j>1.
When scenario (a) occurs,b passes through b0j and τ2b<1. From (78), τ1b0j+ε>1,τ1b0j−ε<1. Therefore, the matrix Kb0j+ε has exactly one more eigenvalue that is >1 than Kb0j−ε does, and its algebraic multiplicity is 1. Then, (74) holds. That is to say, the index jumps as b goes through b0j.
When scenario (b) occurs,b passes through b0j and τ2b>1. From (78), τ1b0j+ε>1,τ1b0j−ε<1. Similarly, the index jumps as b goes through b0j. Therefore, (ii) is true regardless of (a) or (b).
Thus, by the index jump principle and [24], Theorem 1.3, it follows that there exists a connected component Γ^j of nontrivial solutions of (60), and Γ^j comes from the bifurcation point b0j,θ. We know that Γ^j is also the connected component Γj of the nonconstant solution of (4) from b0j,w¯. Γ^j and Γj are both in ℝ×X. By the Rabinowitz global bifurcation theorem, the connected component Γj joins b0j,w¯ to either ∞ or b0k,w¯ in ℝ×X, where k≠j.
We first prove that the latter situation will not happen. According to Theorem1, the solution on the connected component sent from b0j,w¯ is related to φj, and φj has exactly j−1 simple generalized zeros. In the same way, the solution on the connected component sent from b0k,w¯ is related to φk, and φk has k−1 simple generalized zeros. If the connected component sent Γj joining b0j,w¯ to b0k,w¯, the solution b,w∈Γj is related to both φj and φk, which is impossible. On the contrary, Lemma 1 shows that if b=bc∈0,∞, then the solutions u and v of (4) are both bounded. So, the connected component Γj will not join b0j,w¯ to bc,∞. Therefore, the connected component Γj can only join b0j,w¯ to either ∞,∞ or ∞,m, where m∈0,∞. But, in any case, the projection of continuum Γj is unbounded on the b-axis.



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*Source: 5417218-2020-05-11.xml*