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Research article Special Issues

Complex-balanced equilibria of generalized mass-action systems: necessary conditions for linear stability

  • It is well known that, for mass-action systems, complex-balanced equilibria are asymptotically stable. For generalized mass-action systems, even if there exists a unique complex-balanced equilibrium (in every stoichiometric class and for all rate constants), it need not be stable. We first discuss several notions of matrix stability (on a linear subspace) such as D-stability and diagonal stability, and then we apply abstract results on matrix stability to complex-balanced equilibria of generalized mass-action systems. In particular, we show that linear stability (on the stoichiometric subspace and for all rate constants) implies uniqueness. For cyclic networks, we characterize linear stability (in terms of D-stability of the Jacobian matrix); and for weakly reversible networks, we give necessary conditions for linear stability (in terms of D-semistability of the Jacobian matrices of all cycles in the network). Moreover, we show that, for classical mass-action systems, complex-balanced equilibria are not just asymptotically stable, but even diagonally stable (and hence linearly stable). Finally, we recall and extend characterizations of D-stability and diagonal stability for matrices of dimension up to three, and we illustrate our results by examples of irreversible cycles (of dimension up to three) and of reversible chains and S-systems (of arbitrary dimension).

    Citation: Balázs Boros, Stefan Müller, Georg Regensburger. Complex-balanced equilibria of generalized mass-action systems: necessary conditions for linear stability[J]. Mathematical Biosciences and Engineering, 2020, 17(1): 442-459. doi: 10.3934/mbe.2020024

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  • It is well known that, for mass-action systems, complex-balanced equilibria are asymptotically stable. For generalized mass-action systems, even if there exists a unique complex-balanced equilibrium (in every stoichiometric class and for all rate constants), it need not be stable. We first discuss several notions of matrix stability (on a linear subspace) such as D-stability and diagonal stability, and then we apply abstract results on matrix stability to complex-balanced equilibria of generalized mass-action systems. In particular, we show that linear stability (on the stoichiometric subspace and for all rate constants) implies uniqueness. For cyclic networks, we characterize linear stability (in terms of D-stability of the Jacobian matrix); and for weakly reversible networks, we give necessary conditions for linear stability (in terms of D-semistability of the Jacobian matrices of all cycles in the network). Moreover, we show that, for classical mass-action systems, complex-balanced equilibria are not just asymptotically stable, but even diagonally stable (and hence linearly stable). Finally, we recall and extend characterizations of D-stability and diagonal stability for matrices of dimension up to three, and we illustrate our results by examples of irreversible cycles (of dimension up to three) and of reversible chains and S-systems (of arbitrary dimension).


    The porous media fluid defined in a two-dimensional semi-infinite pipe has received extensive attention. Liu et al. [1] defined a semi-infinite strip pipe whose generatrix is parallel to the coordinate axis (see Figure 1) and obtained the Phragmén-Lindelöf alternative result of shallow water equations.

    Figure 1.  Cylindrical pipe 1.

    Payne and Schaefer [2] considered the case that the generatrix is not parallel to the coordinate axis (see Figure 2) and obtained the Phragmén-Lindelöf alternative for the biharmonic equation.

    Figure 2.  Cylindrical pipe 2.

    Recently, Li and Chen [3] considered the Darcy equations on a semi-infinite channel which was defined as

    R={(x1,x2)|x1>a,0<x2<h(x1)},

    where a>0 and h(x1) is a smooth curve in the plane. They proved that the solutions of Darcy equations grow polynomially or decay exponentially as a spatial variable x1. For more on such studies, one can see [2,4,5,6].

    In this paper, we consider the convergence result on the Soret coefficient of the Darcy model in R. The importance of this type of convergence result was discussed by Hirsch and Smale [7] and there have been a lot of results. At first, people mainly focused on the structural stability of the solutions of various systems of partial differential equations in bounded domains (see [8,9,10,11,12,13,14]). Later, many scholars extended the study of structural stability to the case that there are two kinds of interface links in a bounded region (see [15,16,17,18]). Li et al. [19,20] considered the structural stability of Brinkman-Forechheimer equations and the thermoelastic equations of type III on a three-dimensional semi-infinite cylinder, respectively. The generatrix of the cylinder was parallel to the coordinate axis. However, the stability of partial differential equations on a two-dimensional pipe has not received enough attention. It is especially emphasized that the generatrix of the pipe considered in this paper is not parallel to the coordinate axis.

    We investigate the following double-diffusive Darcy flow of a fluid through a porous medium in R which can be written as (see [21])

    uα=p,α+gαT+hαC, in R×(0,τ), (1.1)
    uα,α=0, in R×(0,τ), (1.2)
    tT+uαT,α=ΔT, in R×(0,τ), (1.3)
    tC+uαC,α=ΔC+σΔT, in R×(0,τ), (1.4)

    where α=1,2. uα,p,T and C represent the velocity, pressure, temperature, and concentration of the flow, respectively. gα and hα are bounded functions. σ>0 is the Soret coefficient. For simplicity, we assume g and h satisfy |g|, |h|1. In this paper, we also use the summation convention summed from 1 to 2, and a comma is used to indicate differentiation. e.g., uα,βuα,β=2α,β=1(uαxβ)2.

    The initial-boundary conditions can be written as

    uα(x1,0,t)=uα(x1,h(x1),t)=0, x1a,0<t<τ, (1.5)
    T(x1,0,t)=T(x1,h(x1),t)=0, x1a,0<t<τ, (1.6)
    C(x1,0,t)=C(x1,h(x1),t)=0, x1a,0<t<τ, (1.7)
    uα(a,x2,t)=Fα(x2,t), 0<x2<h(a),0<t<τ, (1.8)
    T(a,x2,t)=H(x2,t),C(a,x2,t)=˜H(x2,t), 0<x2<h(a),0<t<τ, (1.9)
    T(x1,x2,0)=T0(x1,x2), C(x1,x2,0)=C0(x1,x2), (x1,x2)R, (1.10)

    where T0 and C0 are given functions. Fα, H and ˜H are differentiable functions which are assumed to satisfy the appropriate compatibility conditions

    Fα(h(a),t)=H(h(a),t)=˜H(h(a),t)=0.

    Now, we let v(x1,x2,t) denote a stream function which satisfies

    u1=v,2,u2=v,1.

    Equations (1.1)–(1.8) can be converted to

    Δv=gTgThChC, in R×(0,τ), (1.11)
    tT+vT=ΔT, in R×(0,τ), (1.12)
    tC+vC=ΔC+σΔT, in R×(0,τ), (1.13)
    v(x1,0,t)=v(x1,h(x1),t)=0, x1a,0<t<τ, (1.14)
    vn(x1,0,t)=vn(x1,h(x1),t)=0, x1a,0<t<τ, (1.15)
    T(x1,0,t)=T(x1,h(x1),t)=0, x1a,0<t<τ, (1.16)
    C(x1,0,t)=C(x1,h(x1),t)=0, x1a,0<t<τ, (1.17)
    v(a,x2,t)=˜F1(x2,t)=x20F1(s,t)ds, 0<x2<h(a),0<t<τ, (1.18)
    v,1(a,x2,t)=˜F2(x2,t)=F2(x2,t), 0<x2<h(a),0<t<τ, (1.19)
    T(a,x2,t)=H(x2,t), C(a,x2,t)=˜H(x2,t), 0<x2<h(a),0<t<τ, (1.20)
    T(x1,x2,0)=T0(x1,x2), C(x1,x2,0)=C0(x1,x2), (x1,x2)R, (1.21)

    where vn is the outward normal derivative of v, g=(g1,g2), h=(h1,h2) and =(x2,x1).

    In the next section, we give several lemmas that have been derived in the literature. In Section 3, we derive an important lemma that can be used to derive our main result. In Section 4, we obtain the convergence result on the Soret coefficient and give some concrete examples. Section 5 shows the summary and outlook of this paper.

    We also introduce the notations

    Rz={(x1,x2)|x1z>a,0<x2<h(x1)},
    Lz={(x1,x2)|x1=za,0<x2<h(z)},

    where z is a running variable along the x1 axis.

    Here are some lemmas that will be often used in this paper.

    Lemma 2.1 (see [5,22]) If w(x1,0)=w(x1,h)=0 and wn(x1,0)=wn(x1,h)=0, then the following Wirtinger type inequality holds

    Lzw2dx2h2π2Lz(w,2)2dx2,

    where z>0 is a moving point on the x1 axis.

    Lemma 2.2 If φ(x1,0)=φ(x1,h)=0 and φ0, as x1, then

    Rzφ4dx2dξ(Rzφ2dx2dξ)(Rzφ,αφ,αdx2dξ).

    Proof. Since φ(x1,0)=φ(x1,h)=0, we have

    φ2(x1,x2)=2x20φζφ(x1,ζ)dζ=2hx2φζφ(x1,ζ)dζh0|φx2φ(x1,x2)|dx2. (2.1)

    Since φ0, as x1, we have

    φ2(x1,x2)=2x1φξφ(ξ,x2)dξ2x1|φξφ(ξ,x2)|dξ. (2.2)

    Combining Eqs (2.1) and (2.2) and integrating over Rz, we obtain

    Rzφ4(ξ,x2)dx2dξ2[Rz|φx2φ(ξ,x2)|dx2dξ][Rz|φξφ(ξ,x2)|dx2dξ]2Rzφ2dx2dξ(Rz(φξ)2dx2dξ)12(Rz(φx2)2dx2dξ)12.

    Using Young's inequality, we can obtain Lemma 2.2.

    Using a similar method of papers [9,23,24], we can have the following lemma.

    Lemma 2.3 Assume that T0,HL, then

    sup[0,τ]||T||Tm,

    where Tm=max{||T0||, sup[0,τ]H(η)}.

    If C0,˜HL and σ=0 in (1.4), we can also have

    sup[0,τ]||C||Cm, (2.3)

    where Cm=max{||C0||, sup[0,τ]˜H(η)}.

    To obtain our main result, we shall use the following result which can be written as follows.

    Lemma 2.4 (see [3]) Let (v,T,C) be a solution of the Eqs (1.1)–(1.10) in R and  za such that F(z,t)<0. Then for any fixed t

    t0Rzeωη[14β2ωT2+14ωC2+12β1v,αv,α+12β2T,αT,α+12C,αC,α]dx2dξdη+12eωtzLξ[β2T2+C2]dx2dξm1e2m3za1h(ζ)dζ+m2em3za1h(ζ)dζ

    holds, where β1,β2,ω,m1,m2 and m3 are positive constants which depends on the middle parameter σ and boundary conditions of the equation; also F(z,t) has been defined as

    F(z,t)=t0Lzeωη[β1vv,1+β2TT,1+CC,1]dx2dη+β1t0Lzeωη[g2Tv+h2Cv]dx2dη+t0Lzeωη[12β2T2v,212C2v,2+σCT,1]dx2dη.

    Remark 2.1 Lemma 2.4 shows that the solution of Eqs (1.11)–(1.21) decays exponentially with z. Only in this case, the study of structural stability is meaningful. Lemma 2.4 will also provide a priori bounds for the estimate of nonlinear terms (see e.g., (3.55) and (3.56)).

    Remark 2.2 Li and Chen [3] considered several special cases of h(z), e.g., h(z)=h a constant, h(z)k1zτ1 and h(z)k2z(lnz)τ2, where k1,k2>0 and 0<τ1,τ21. In the fourth section, we will also consider several special cases with h(z)=h as a constant, τ1=12,k1=1 and τ1=23,k1=1.

    Lemma 2.5 (see [3]) Assume that (v,T,C) are solutions of Eqs (1.1)–(1.10). If F(z,t)<0 for any za, then

    t0Reωη[14β2ωT2+14ωC2+12β1v,αv,α+12β2T,αT,α+12C,αC,α]dx2dξdη+12eωtR[β2T2+C2]dx2dξr(t),

    where r(t) is a positive known function which depends only on t.

    Now, we derive a bound of Rv,αv,αdx2dξ.

    Lemma 2.6 Assume that ˜F1,˜F2,H,˜HL(R), then

    Rv,αv,αdx2dξϵ1, (2.4)

    where

    ϵ1=2sup[0,τ]{La|˜F1˜F2|dx2dξ+La|g2H˜F1|dx2dξ+La|h2˜H˜F1|dx2dξ}+4max{1β2,1}reωτ.

    Additionally, r is the maximum value of r(t) in [0,τ].

    Proof. Using Eq (1.11), we have

    R[Δv+gT+gT+hC+hC]vdx2dξ=0.

    Using Eqs (1.18)–(1.20) it follows that

    Rv,αv,αdx2dξ=La˜F1˜F2dx2dξLag2H˜F1dx2dξLah2˜H˜F1dx2dξRvgTdx2dξRvhCdx2dξLa|˜F1˜F2|dx2dξ+La|g2H˜F1|dx2dξ+La|h2˜H˜F1|dx2dξ+12Rv,αv,αdx2dξ+RT2dx2dξ+RC2dx2dξ. (2.5)

    Using Lemma 2.5 in Eq (2.5), we can get Lemma 2.6.

    In this section, we derive the convergence result when the Soret coefficient σ0. To do this, we let (v,T,C) be the solution of Eqs (1.11)–(1.21). Furthermore, let (v,T,C) be the solution to the following equations

    Δv=gTgThChC, in R×(0,τ), (3.1)
    tT+vT=ΔT, in R×(0,τ), (3.2)
    tC+vC=ΔC, in R×(0,τ), (3.3)
    v(x1,0,t)=v(x1,h(x1),t)=0, x1a,0<t<τ, (3.4)
    vn(x1,0,t)=vn(x1,h(x1),t)=0, x1a,0<t<τ, (3.5)
    T(x1,0,t)=T(x1,h(x1),t)=0, x1a,0<t<τ, (3.6)
    C(x1,0,t)=C(x1,h(x1),t)=0, x1a,0<t<τ, (3.7)
    v(a,x2,t)=˜F1(x2,t)=x20F1(s,t)ds, 0<x2<h(a),0<t<τ, (3.8)
    v,1(a,x2,t)=˜F2(x2,t)=F2(x2,t), 0<x2<h(a),0<t<τ, (3.9)
    T(a,x2,t)=H(x2,t), C(a,x2,t)=˜H(x2,t), 0<x2<h(a),0<t<τ, (3.10)
    T(x1,x2,0)=T0(x1,x2), C(x1,x2,0)=C0(x1,x2), (x1,x2)R. (3.11)

    Remark 3.1 We note that Lemmas 2.4 and 2.5 also hold for (v,T,C).

    Now, we let

    w=vv, θ=TT, Σ=CC.

    Then (w,θ,Σ) satisfies

    Δw=gθgθhΣhΣ, in R×(0,τ), (3.12)
    tθ+wT+vθ=Δθ, in R×(0,τ), (3.13)
    tΣ+vΣ+wC=ΔΣ+σΔT, in R×(0,τ), (3.14)
    w(x1,0,t)=w(x1,h(x1),t)=0, x1a,0<t<τ, (3.15)
    wn(x1,0,t)=wn(x1,h(x1),t)=0, x1a,0<t<τ, (3.16)
    θ(x1,0,t)=θ(x1,h(x1),t)=0, x1a,0<t<τ, (3.17)
    Σ(x1,0,t)=Σ(x1,h(x1),t)=0, x1a,0<t<τ, (3.18)
    w(a,x2,t)=w,1(a,x2,t)=0, 0<x2<h(a),0<t<τ, (3.19)
    θ(a,x2,t)=Σ(a,x2,t)=0, 0<x2<h(a),0<t<τ, (3.20)
    θ(x1,x2,0)=Σ(x1,x2,0)=0, (x1,x2)R. (3.21)

    We establish the following auxiliary functions

    E1(z,t)=t0Rzeωηww,1dx2dξdη+t0Rzeωηg2θwdx2dξdη+t0Rzeωηh2Σwdx2dξdηA1(z,t)+A2(z,t)+A3(z,t), (3.22)
    E2(z,t)=t0Rzeωηθθ,1dx2dξdη12t0Rzeωηθ2v,2dx2dξdη+t0RzeωηθTw,2dx2dξdη+t0Rzeωη(ξz)wθTdx2dξdηB1(z,t)+B2(z,t)+B3(z,t)+B4(z,t), (3.23)
    E3(z,t)=t0RzeωηΣΣ,1dx2dξdη12t0RzeωηΣ2v,2dx2dξdηt0Rzeωηw,2CΣdx2dξdη+σt0RzeωηΣT,1dx2dξdη+t0Rzeωη(ξz)wΣCdx2dξdη+σt0Rzeωη(ξz)T,αΣ,αdx2dξdηC1(z,t)+C2(z,t)+C3(z,t)+C4(z,t)+C5(z,t)+C6(z,t), (3.24)

    where ω is an arbitrary positive constant.

    Using the divergence theorem and Eqs (3.12)-(3.21), we have

    E1(z,t)=t0Rzeωη(ξz),αww,αdx2dξdη+t0Rzeωηg2θwdx2dξdη+t0Rzeωηh2Σwdx2dξdη=t0Rzeωη(ξz)w,αw,αdx2dξdη+t0Rzeωη(ξz)gwθdx2dξdη+t0Rzeωη(ξz)hwΣdx2dξdη. (3.25)

    Similarly, we have

    E2(z,t)=t0Rzeωη(ξz)[12ωθ2+θ,αθ,α]dx2dξdη+12eωtRz(ξz)θ2dx2dξ, (3.26)

    and

    E3(z,t)=t0Rzeωη(ξz)[12ωΣ2+Σ,αΣ,α]dx2dξdη+12eωtRzeωη(ξz)Σ2dx2dξ. (3.27)

    We also define

    E(z,t)=E1(z,t)+δ2E2(z,t)+δ3E3(z,t)=t0Rzeωη(ξz)[12δ2ωθ2+12δ3ωΣ2+w,αw,α+δ2θ,αθ,α+δ3Σ,αΣ,α]dx2dξdη+12eωtRz(ξz)[δ2θ2+δ3Σ2]dx2+t0Rzeωη(ξz)gwθdx2dξdη+t0Rzeωη(ξz)hwΣdx2dξdη, (3.28)

    where δ2 and δ3 are positive constants.

    Using the Cauchy-Schwarz inequality, we have

    |t0Rzeωη(ξz)gwθdx2dξdη|t0Rzeωη(ξz)θ2dx2dξdη+14t0Rzeωη(ξz)w,αw,αdx2dξdη, (3.29)
    |t0Rzeωη(ξz)hwΣdx2dξdη|t0Rzeωη(ξz)Σ2dx2dξdη+14t0Rzeωη(ξz)w,αw,αdx2dξdη. (3.30)

    Inserting Eqs (3.29) and (3.30) into Eq (3.28) and choosing ω=max{4δ3,4δ2}, we have

    E(z,t)t0Rzeωη(ξz)[14δ2ωθ2+14ωδ3Σ2+12w,αw,α+δ2θ,αθ,α+δ3Σ,αΣ,α]dx2dξdη+12eωtRz(ξz)[δ2θ2+δ3Σ2]dx2dξ. (3.31)

    From Eq (3.28), we have

    zE(z,t)=t0Rzeωη[12δ2ωθ2+12δ3ωΣ2+w,αw,α+δ2θ,αθ,α+δ3Σ,αΣ,α]dx2dξdη+12eωtRz[δ2θ2+δ3Σ2]dx2dξ+t0Rzeωηgwθdx2dξdη+t0RzeωηhwΣdx2dξdη. (3.32)

    Using the Cauchy-Schwarz inequality, we have

    |t0Rzeωηgwθdx2dξdη|t0Rzeωηθ2dx2dξdη+14t0Rzeωηw,αw,αdx2dξdη, (3.33)
    |t0RzeωηhwΣdx2dξdη|t0RzeωηΣ2dx2dξdη+14t0Rzeωηw,αw,αdx2dξdη. (3.34)

    Inserting Eqs (3.33) and (3.34) into Eq (3.32), we have

    zE(z,t)t0Rzeωη[34δ2ωθ2+34ωδ3Σ2+32w,αw,α+δ2θ,αθ,α+δ3Σ,αΣ,α]dx2dξdη+12eωtRz[δ2θ2+δ3Σ2]dx2dξ, (3.35)

    and

    zE(z,t)t0Rzeωη[14δ2ωθ2+14ωδ3Σ2+12w,αw,α+δ2θ,αθ,α+δ3Σ,αΣ,α]dx2dξdη+12eωtRz[δ2θ2+δ3Σ2]dx2dξ. (3.36)

    Based on Eqs (3.22)–(3.24) and using Eq (3.36), we have the following lemma.

    Lemma 3.1. Assume that T0,C0,H,˜HC, 1h(z)C(R) and the function E(z,t) is defined in Eq (3.28). Then E(z,t) satisfies

    E(z,t)n1h(z)[zE(z,t)]+4σ2δ3β3β2ωm1e2m3za1h(ζ)dζ+4σ2δ3β3β2ωm2em3za1h(ζ)dζ+4β2σ2m1ze2m3ξa1h(ζ)dζdξ+4β2σ2m2zem3ξa1h(ζ)dζdξ, (3.37)

    where n1 is a positive constant.

    Proof. Using the Hölder inequality, Young's inequality and Lemma 2.1, we have

    A1(z,t)[t0Rzeωηw2dx2dξdηt0Rzeωη(w,1)2dx2dξdη]12hπ[t0Rzeωη(w,2)2dx2dξdηt0Rzeωη(w,1)2dx2dξdη]12h2πt0Rzeωηw,αw,αdx2dξdη, (3.38)
    A2(z,t)[t0Rzeωηw2dx2dξdηt0Rzeωηθ2dx2dξdη]1222hπδ2ω[12t0Rzeωη(w,2)2dx2dξdη14δ2ωt0Rzeωηθ2dx2dξdη]122hδ2ωπ[12t0Rzeωηw,αw,αdx2dξdη+14δ2ωt0Rzeωηθ2dx2dξdη], (3.39)
    A3(z,t)2hδ3ωπ[12t0Rzeωηw,αw,αdx2dξdη+14ωδ3t0RzeωηΣ2dx2dξdη]. (3.40)

    Combining Eqs (3.22) and (3.38)–(3.40), we obtain

    E1(z,t)hπ[1+2δ2ω+2δ3ω]t0Rzeωη12w,αw,αdx2dξdη+2hδ2ωπt0Rzeωη14δ2ωθ2dx2dξdη+2hδ3ωπt0Rzeωη14ωδ3Σ2dx2dξdη. (3.41)

    Using the Hölder inequality, Young's inequality, Lemmas 2.1, 2.3, 2.5 and Eq (2.5), we have

    B1(z,t)[t0Rzeωηθ2dx2dξdηt0Rzeωη(θ,1)2dx2dξdη]12hπ[t0Rzeωη(θ,2)2dx2dξdηt0Rzeωη(θ,1)2dx2dξdη]12hπδ212δ2t0Rzeωηθ,αθ,αdx2dξdη, (3.42)
    B2(z,t)12t0[Rzeωη(v,2)2dx2dξRzeωηθ4dx2dξ]12dη12t0[Reωη(v,2)2dx2dξRzeωηθ2dx2dξ]12dηϵ12δ2πt0[Rzeωηθ2dx2dξRzeωηθ,αθ,αdx2dξ]12dηhϵ12δ2πt0Rzeωηδ2θ,αθ,αdx2dξdη, (3.43)
    B3(z,t)Tm[t0Rzeωη(w,2)2dx2dξdη]12[t0Rzeωηθ2dx2dξdη]12hπδ2Tm[12t0Rzeωηw,αw,αdx2dξdη]12[δ2t0Rzeωηθ,αθ,αdx2dξdη]12hπ2δ2Tm[12t0Rzeωηw,αw,αdx2dξdη+δ2t0Rzeωηθ,αθ,αdx2dξdη]. (3.44)

    Using Lemma 2.3, we have

    B4(z,t)12T2mt0Rzeωη(ξz)w,αw,αdx2dξdη+12t0Rzeωη(ξz)θ,αθ,αdx2dξdη. (3.45)

    Inserting Eqs (3.42) and (3.45) into Eq (3.23), we obtain

    E2(z,t)[hπδ2+hπ2δ2Tm+hϵ12δ2π]t0Rzeωηδ2θ,αθ,αdx2dξdη+hπ2δ2Tm[12t0Rzeωηw,αw,αdx2dξdη]+12T2mt0Rzeωη(ξz)w,αw,αdx2dξdη+12t0Rzeωη(ξz)θ,αθ,αdx2dξdη. (3.46)

    Similarly, we have

    C1(z,t)hπ[t0Rzeωη(Σ,2)2dx2dξdηt0Rzeωη(Σ,1)2dx2dξdη]12h2δ3πδ3t0RzeωηΣ,αΣ,αdx2dξdη, (3.47)
    C2(z,t)12t0[Rzeωη(v,2)2dx2dξRzeωηΣ4dx2dξ]12dηhϵ12δ3πδ3t0RzeωηΣ,αΣ,αdx2dξdη, (3.48)
    C3(z,t)Cm[t0Rzeωη(w,2)2dx2dξdη]12[t0RzeωηΣ2dx2dξdη]12hπ2δ3Cm[12t0Rzeωηw,αw,αdx2dξdη+δ3t0RzeωηΣ,αΣ,αdx2dξdη], (3.49)
    C4(z,t)14ωδ3t0RzeωηΣ2dx2dξdη+σ2β3ωt0Rzeωη(T,1)2dx2dξdη, (3.50)
    C5(z,t)14δ3t0Rzeωη(ξz)w,αw,αdx2dξdη+C2mδ3t0Rzeωη(ξz)Σ,αΣ,αdx2dξdη (3.51)

    and

    C6(z,t)1δ3σ2t0Rzeωη(ξz)T,αT,αdx2dξdη+14δ3t0Rzeωη(ξz)Σ,αΣ,αdx2dξdη. (3.52)

    Inserting Eqs (3.47)–(3.52) into Eq (3.24), we obtain

    E3(z,t)[h2δ3π+hπ2δ3Cm+hϵ12δ3π]δ3t0RzeωηΣ,αΣ,αdx2dξdη+hπ2δ3Cm[12t0Rzeωηw,αw,αdx2dξdη]+14ωδ3t0RzeωηΣ2dx2dξdη+[14δ3+C2mδ3]t0Rzeωη(ξz)Σ,αΣ,αdx2dξdη+14δ3t0Rzeωη(ξz)w,αw,αdx2dξdη+σ2β3ωt0Rzeωη(T,1)2dx2dξdη+1δ3σ2t0Rzeωη(ξz)T,αT,αdx2dξdη. (3.53)

    Now, inserting Eqs (3.41), (3.46) and (3.53) into Eq (3.28), choosing δ2<12T2m,δ3<14C2m and noting Eqs (3.36) and (3.31), we obtain

    E(z,t)12E(z,t)+12n1h(z)[zE(z,t)]+σ2δ3β3ωt0Rzeωη(T,1)2dx2dξdη+σ2t0Rzeωη(ξz)T,αT,αdx2dξdη, (3.54)

    where

    12n1max{1π[1+2δ2ω+2δ3ω]+Tmπ2δ2+Cmπ2δ3,1+2δ2ωπ,12δ3π+1π2δ3Cm+ϵ12δ3π,hπδ2+1π2δ2Tm+ϵ12δ2π}.

    From Lemma 2.4, we have

    t0RzeωηT,αT,αdx2dξdη2β2m1e2m3za1h(ζ)dζ+2β2m2em3za1h(ζ)dζ. (3.55)

    Integrating Eq (3.55) from z to , we get

    t0Rzeωη(ξz)T,αT,αdx2dξdη2β2m1ze2m3ξa1h(ζ)dζdξ+2β2m2zem3ξa1h(ζ)dζdξ. (3.56)

    Combining Eqs (3.54)–(3.56), we can obtain Lemma 3.1.

    Based on Lemma 3.1, we derive our main results in this section. To do this, integrating Eq (3.37) from a to z, we obtain

    E(z,t)E(a,t)e1n1za1h(ζ)dζ+4m1δ3σ2n1β3β2ωe1n1za1h(ζ)dζza1h(ξ)eξa{1n1h(ζ)2m3h(ζ)}dζdξ+4m2σ2δ3n1β3β2ωe1n1za1h(ζ)dζza1h(ξ)eξa{1n1h(ζ)m3h(ζ)}dζdξ+4m1σ2n1β2e1n1za1h(ζ)dζza1h(τ)e1n1τa1h(ζ)dζτe2m3ξa1h(ζ)dζdξdτ+4m2σ2n1β2e1n1za1h(ζ)dζza1h(τ)e1n1τa1h(ζ)dζτem3ξa1h(ζ)dζdξdτ. (4.1)

    To get the convergence results on the Soret coefficient, we have to derive the upper bound for E(a,t). To do this, we choose z=a in Eq (3.37) to get

    E(a,t)n1h(a)[zE(a,t)]+4δ3σ2β3β2ωm1+4δ3σ2β3β2ωm2+4β2σ2m1ae2m3ξa1h(ζ)dζdξ+4β2σ2m2aem3ξa1h(ζ)dζdξ. (4.2)

    By differentiating Eqs (3.22)–(3.24), then choosing z = a and using Eqs (3.19) and (3.20), it can be obtained that

    \begin{align} -\frac{\partial}{\partial z}E(a, t)& = -\frac{\partial}{\partial z}E_1(a, t)-\delta_2\frac{\partial}{\partial z}E_2(a, t) -\delta_3\frac{\partial}{\partial z}E_3(a, t)\\ & = \delta_2\int_{0}^{t}\int_{R}e^{-\omega\eta}\nabla^\perp w\cdot\nabla\theta Tdx_2d\xi d\eta\\ &+\delta_3\int_{0}^{t}\int_{R}e^{-\omega\eta}\nabla^\perp w\cdot\nabla\Sigma C^*dx_2d\xi d\eta+\delta_3\sigma\int_0^t\int_{R}e^{-\omega\eta}T_{, \alpha}\Sigma_{, \alpha}dx_2d\xi d\eta. \end{align} (4.3)

    Using the Hölder inequality, Young's inequality and Lemmas 2.3 and 2.5, we have

    \begin{align} \delta_2\int_{0}^{t}\int_{R}e^{-\omega\eta}&\nabla^\perp w\cdot\nabla\theta Tdx_2d\xi d\eta\\ &\leq\frac{1}{2}\delta_2T_m^2\int_{0}^{t}\int_{R}e^{-\omega\eta}w_{, \alpha}w_{, \alpha}dx_2d\xi d\eta +\frac{1}{2}\delta_2\int_{0}^{t}\int_{R}e^{-\omega\eta}\theta_{, \alpha}\theta_{, \alpha}dx_2d\xi d\eta, \end{align} (4.4)
    \begin{align} \delta_3\int_{0}^{t}\int_{R}e^{-\omega\eta}&\nabla^\perp w\cdot\nabla\Sigma C^*dx_2d\xi d\eta\\ &\leq\delta_3C_m^2\int_{0}^{t}\int_{R}e^{-\omega\eta}w_{, \alpha}w_{, \alpha}dx_2d\xi d\eta +\frac{1}{4}\delta_3\int_{0}^{t}\int_{R}e^{-\omega\eta}\Sigma_{, \alpha}\Sigma_{, \alpha}dx_2d\xi d\eta, \end{align} (4.5)
    \begin{align} \delta_3\sigma\int_0^t\int_{R}e^{-\omega\eta}&T_{, \alpha}\Sigma_{, \alpha}dx_2d\xi d\eta\\ &\leq\delta_3\sigma^2\int_{0}^{t}\int_{R}e^{-\omega\eta}T_{, \alpha}T_{, \alpha}dx_2d\xi d\eta +\frac{1}{4}\delta_3\int_{0}^{t}\int_{R}e^{-\omega\eta}\Sigma_{, \alpha}\Sigma_{, \alpha}dx_2d\xi d\eta\\ &\leq\frac{2}{\beta_2}\delta_3\sigma^2r(t) +\frac{1}{4}\delta_3\int_{0}^{t}\int_{R}e^{-\omega\eta}\Sigma_{, \alpha}\Sigma_{, \alpha}dx_2d\xi d\eta. \end{align} (4.6)

    On the other hand, we choose z = a in Eq (3.36) to get

    \begin{align} -\frac{\partial}{\partial z}E(a, t)&\geq\int_0^t\int_{R}e^{-\omega\eta}\Big[\frac{1}{4}\delta_2\omega \theta^2+\frac{1}{4}\omega \delta_3\Sigma^2+\frac{1}{2}w_{, \alpha}w_{, \alpha}+\delta_2\theta_{, \alpha}\theta_{, \alpha} +\delta_3\Sigma_{, \alpha}\Sigma_{, \alpha}\Big]dx_2d\xi d\eta\\ &+\frac{1}{2} e^{-\omega t}\int_{R}\Big[\delta_2 \theta^2+\delta_3\Sigma^2\Big]dx_2d\xi. \end{align} (4.7)

    Inserting Eqs (4.4)–(4.6) into Eq (4.3) and noting Eq (4.7), we get

    \begin{align} -\frac{\partial}{\partial z}E(a, t)\leq\frac{1}{2}\Big[-\frac{\partial}{\partial z}E(a, t)\Big]+\frac{2}{\beta_2}\delta_3\sigma^2r(t), \end{align}

    or

    \begin{align} -\frac{\partial}{\partial z}E(a, t)\leq\frac{4}{\beta_2}\delta_3\sigma^2r(t). \end{align} (4.8)

    Inserting Eq (4.8) into Eq (4.2), we have

    \begin{align} E(a, t)&\leq n_2\sigma^2 +\frac{4}{\beta_2}\sigma^2m_1\int_a^\infty e^{-2m_3\int_a^\xi\frac{1}{h(\zeta)}d\zeta}d\xi +\frac{4}{\beta_2}\sigma^2m_2\int_a^\infty e^{-m_3\int_a^\xi\frac{1}{h(\zeta)}d\zeta}d\xi, \end{align} (4.9)

    where n_2 = n_1(h(a)+h^\frac{3}{2}(a))\frac{4}{\beta_2}\delta_3r(t)+\frac{4\delta_3}{\beta_3\beta_2\omega}m_1 +\frac{4\delta_3}{\beta_3\beta_2\omega}m_2 .

    Now, inserting Eq (4.9) into Eq (4.1) and in light of Eq (3.31), we can obtain the following theorem.

    Theorem 4.1 Letting (w, \theta, \Sigma) is solution of Eqs (3.12)–(3.21) with T_0, C_0, H, \widetilde{H}\in C^\infty , then

    (w, \theta, \Sigma)\rightarrow \boldsymbol{0}, \ as \ \sigma\rightarrow0.

    Specifically

    \begin{align} \int_0^t\int_{R_z}e^{-\omega\eta}(\xi-z)&\Big[\frac{1}{4}\delta_2\omega \theta^2+\frac{1}{4}\omega \delta_3\Sigma^2+\frac{1}{2}w_{, \alpha}w_{, \alpha}+\delta_2\theta_{, \alpha}\theta_{, \alpha} +\delta_3\Sigma_{, \alpha}\Sigma_{, \alpha}\Big]dx_2d\xi d\eta\\ &+\frac{1}{2} e^{-\omega t}\int_{R_z}(\xi-z)\Big[\delta_2 \theta^2+\delta_3\Sigma^2\Big]dx_2d\xi\\ &\leq n_2\sigma^2e^{-\frac{1}{n_1}\int_a^z\frac{1}{h(\zeta)}d\zeta} \\ &+\frac{4}{\beta_2}\sigma^2m_1\int_a^\infty e^{-2m_3\int_a^\xi\frac{1}{h(\zeta)}d\zeta}d\xi e^{-\frac{1}{n_1}\int_a^z\frac{1}{h(\zeta)}d\zeta} \\ &+\frac{4}{\beta_2}\sigma^2m_2\int_a^\infty e^{-m_3\int_a^\xi\frac{1}{h(\zeta)}d\zeta}d\xi e^{-\frac{1}{n_1}\int_a^z\frac{1}{h(\zeta)}d\zeta}\\ &+\frac{4 m_1\delta_3\sigma^2}{n_1\beta_3\beta_2\omega}e^{-\frac{1}{n_1}\int_a^z\frac{1}{h(\zeta)}d\zeta} \int_a^z \frac{1}{h(\xi)}e^{\int_a^\xi\{\frac{1}{n_1h(\zeta)}-\frac{2m_3}{h(\zeta)}\}d\zeta}d\xi\\ &+\frac{4 m_2\sigma^2\delta_3}{n_1\beta_3\beta_2\omega}e^{-\frac{1}{n_1}\int_a^z\frac{1}{h(\zeta)}d\zeta} \int_a^z \frac{1}{h(\xi)}e^{\int_a^\xi\{\frac{1}{n_1h(\zeta)}-\frac{m_3}{h(\zeta)}\}d\zeta}d\xi\\ & +\frac{4m_1\sigma^2}{n_1\beta_2}e^{-\frac{1}{n_1} \int_a^z\frac{1}{h(\zeta)}d\zeta} \int_a^z \frac{1}{h(\tau)}e^{\frac{1}{n_1}\int_a^\tau\frac{1}{h(\zeta)}d\zeta}\int_\tau^\infty e^{-2m_3\int_a^\xi\frac{1}{h(\zeta)}d\zeta}d\xi d\tau\\ &+\frac{4m_2\sigma^2}{n_1\beta_2}e^{-\frac{1}{n_1}\int_a^z\frac{1}{h(\zeta)}d\zeta} \int_a^z \frac{1}{h(\tau)}e^{\frac{1}{n_1}\int_a^\tau\frac{1}{h(\zeta)}d\zeta}\int_\tau^\infty e^{-m_3\int_a^\xi\frac{1}{h(\zeta)}d\zeta}d\xi d\tau. \end{align} (4.10)

    Next, we give some examples.

    Remark 4.1 If h(z) = h is a positive constant, then

    \begin{align*} \int_{a}^z\frac{1}{h(\zeta)}d\zeta = \frac{1}{h}(z-a). \end{align*}

    Therefore, we have

    \begin{align} \int_\tau^\infty e^{-m_3\int_a^\xi\frac{1}{h(\zeta)}d\zeta}d\xi = \frac{h}{m_3}e^{-\frac{m_3}{h}(\tau-a)}, \ \int_\tau^\infty e^{-2m_3\int_a^\xi\frac{1}{h(\zeta)}d\zeta}d\xi = \frac{h}{2m_3}e^{-\frac{2m_3}{h}(\tau-a)}. \end{align} (4.11)

    Choosing n_1 such that \frac{1}{n_1}\neq2m_3 and \frac{1}{n_1}\neq m_3 , then

    \begin{align} \int_a^z \frac{1}{h(\xi)}e^{\int_a^\xi\{\frac{1}{n_1h(\zeta)}-\frac{m_3}{h(\zeta)}\}d\zeta}d\xi & = \frac{1}{h}\int_a^ze^{[\frac{1}{n_1h}-\frac{m_3}{h}](\xi-a)}d\xi \\ & = \frac{1}{[\frac{1}{n_1}-m_3]}\Big[e^{[\frac{1}{n_1h}-\frac{m_3}{h}](z-a)}-1\Big], \end{align} (4.12)
    \begin{align} \int_a^z \frac{1}{h(\xi)}e^{\int_a^\xi\{\frac{1}{n_1h(\zeta)}-\frac{2m_3}{h(\zeta)}\}d\zeta}d\xi & = \frac{1}{[\frac{1}{n_1}-2m_3]}\Big[e^{[\frac{1}{n_1h}-\frac{2m_3}{h}](z-a)}-1\Big], \end{align} (4.13)

    and

    \begin{align} \int_a^z \frac{1}{h(\tau)}&e^{\frac{1}{n_1}\int_a^\tau\frac{1}{h(\zeta)}d\zeta} \int_\tau^\infty e^{-2m_3\int_a^\xi\frac{1}{h(\zeta)}d\zeta}d\xi d\tau = \frac{h}{2m_3[\frac{1}{n_1}-2m_3]}\Big[e^{[\frac{1}{n_1h}-\frac{2m_3}{h}](z-a)}-1\Big], \end{align} (4.14)
    \begin{align} \int_a^z \frac{1}{h(\tau)}&e^{\frac{1}{n_1}\int_a^\tau\frac{1}{h(\zeta)}d\zeta}\int_\tau^\infty e^{-m_3\int_a^\xi\frac{1}{h(\zeta)}d\zeta}d\xi d\tau = \frac{h}{m_3[\frac{1}{n_1}-m_3]}\Big[e^{[\frac{1}{n_1h}-\frac{m_3}{h}](z-a)}-1\Big]. \end{align} (4.15)

    Inserting Eqs (4.11)–(4.15) into Eq (4.10), we get

    \begin{align} \int_0^t\int_{R_z}e^{-\omega\eta}(\xi-z)&\Big[\frac{1}{4}\delta_2\omega \theta^2+\frac{1}{4}\omega \delta_3\Sigma^2+\frac{1}{2}w_{, \alpha}w_{, \alpha}+\delta_2\theta_{, \alpha}\theta_{, \alpha} +\delta_3\Sigma_{, \alpha}\Sigma_{, \alpha}\Big]dx_2d\xi d\eta\\ &+\frac{1}{2} e^{-\omega t}\int_{R_z}(\xi-z)\Big[\delta_2 \theta^2+\delta_3\Sigma^2\Big]dx_2d\xi\\ &\leq n_4\sigma^2e^{-n_3(z-a)} +n_5\sigma^2\Big[e^{-\frac{2m_3}{h}(z-a)}-e^{-n_3(z-a)}\Big]\\ &+n_6\sigma^2\Big[e^{-\frac{m_3}{h}(z-a)}-e^{-n_3(z-a)}\Big], \end{align} (4.16)

    where

    \begin{align*} n_3& = \frac{1}{n_1h}, n_4 = n_2+\frac{2m_1h}{m_3\beta_2}+\frac{4m_2h}{m_3\beta_2}, \\ n_5& = \frac{2 m_1\delta_3}{n_1\beta_3\beta_2\omega} \frac{1}{[\frac{1}{n_1}-m_3]}+\frac{2m_1}{n_1\beta_2}\frac{h}{2m_3 [\frac{1}{n_1}-2m_3]} \\ n_6& = \frac{4 m_2\delta_3}{n_1\beta_3\beta_2\omega} \frac{1}{[\frac{1}{n_1}-m_3]} +\frac{4m_2}{n_1\beta_2} \frac{h}{m_3[\frac{1}{n_1}-m_3]}. \end{align*}

    From Eq (4.16), we can conclude that Theorem 4.1 not only shows the convergence of the solutions of Eqs (3.12)–(3.21) on the coefficient \sigma , but it also shows a exponentially decay result as z\rightarrow \infty .

    Remark 4.2. If h(z) = \sqrt{z} , then

    \begin{align*} \int_a^z\frac{1}{h(\zeta)}d\zeta& = \int_a^z\frac{1}{\sqrt{\zeta}}d\zeta = 2(\sqrt{z}-\sqrt{a}). \end{align*}

    Therefore

    \begin{align} \int_z^\infty e^{-m_3\int_{a}^\xi\frac{1}{h(\zeta)}d\zeta}d\xi& = \int_z^\infty e^{-2m_3[\sqrt{\xi}-\sqrt{a}]}d\xi \\ & = -\int_z^\infty\frac{1}{m_3}\sqrt{\xi}d\Big\{e^{-2m_3[\sqrt{\xi}-\sqrt{a}]}\Big\} \\ &\leq\frac{1}{m_3}\sqrt{z}e^{-2m_3[\sqrt{z}-\sqrt{a}]}, \end{align} (4.17)
    \begin{align} \int_z^\infty e^{-2m_3\int_{a}^\xi\frac{1}{h(\zeta)}d\zeta}d\xi&\leq\frac{1}{2m_3}\sqrt{z}e^{-4m_3[\sqrt{z}-\sqrt{a}]}. \end{align} (4.18)

    Moreover, we also have

    \begin{align} \int_a^z \frac{1}{h(\xi)}e^{\int_a^\xi\{\frac{1}{n_1h(\zeta)}-\frac{2m_3}{h(\zeta)}\}d\zeta}d\xi & = \int_a^z \frac{1}{\sqrt{\xi}}e^{2[\frac{1}{n_1} -2m_3][\sqrt{\xi}-\sqrt{a}]}d\xi \\ & = \frac{1}{\frac{1}{n_1} -2m_3} \Big[e^{2[\frac{1}{n_1} -2m_3][\sqrt{z}-\sqrt{a}]}-1\Big], \end{align} (4.19)
    \begin{align} \int_a^z \frac{1}{h(\xi)}e^{\int_a^\xi\{\frac{1}{n_1h(\zeta)}-\frac{m_3}{h(\zeta)}\}d\zeta}d\xi & = \frac{1}{\frac{1}{n_1} -m_3} \Big[e^{2[\frac{1}{n_1} -m_3][\sqrt{z}-\sqrt{a}]}-1\Big], \end{align} (4.20)
    \begin{align} \int_a^z \frac{1}{h(\tau)}e^{\frac{1}{n_1}\int_a^\tau\frac{1}{h(\zeta)}d\zeta}\int_\tau^\infty e^{-2m_3\int_a^\xi\frac{1}{h(\zeta)}d\zeta}d\xi d\tau & = \frac{1}{2m_3}\int_a^z e^{2[\frac{1}{n_1} -2m_3][\sqrt{\tau}-\sqrt{a}]}d\tau \\ &\leq\frac{1}{2m_3[\frac{1}{n_1} -2m_3]}\sqrt{z}\Big[e^{2[\frac{1}{n_1} -2m_3][\sqrt{z}-\sqrt{a}]}-1\Big], \end{align} (4.21)

    and

    \begin{align} \int_a^z \frac{1}{h(\tau)}&e^{\frac{1}{n_1}\int_a^\tau\frac{1}{h(\zeta)}d\zeta}\int_\tau^\infty e^{-m_3\int_a^\xi\frac{1}{h(\zeta)}d\zeta}d\xi d\tau \leq\frac{1}{m_3[\frac{1}{n_1} -m_3]}\sqrt{z}\Big[e^{2[\frac{1}{n_1} -m_3][\sqrt{z}-\sqrt{a}]}-1\Big]. \end{align} (4.22)

    Inserting Eqs (4.19)–(4.22) into Eq (4.10), we obtain

    \begin{align} \int_0^t\int_{R_z}e^{-\omega\eta}(\xi-z)&\Big[\frac{1}{4}\delta_2\omega \theta^2+\frac{1}{4}\omega \delta_3\Sigma^2+\frac{1}{2}w_{, \alpha}w_{, \alpha}+\delta_2\theta_{, \alpha}\theta_{, \alpha} +\delta_3\Sigma_{, \alpha}\Sigma_{, \alpha}\Big]dx_2d\xi d\eta\\ &+\frac{1}{2} e^{-\omega t}\int_{R_z}(\xi-z)\Big[\delta_2 \theta^2+\delta_3\Sigma^2\Big]dx_2d\xi\\ &\leq n_7\sigma^2e^{-\frac{2}{n_1}(\sqrt{z}-\sqrt{a})} \\ &+\frac{4 m_1\delta_3\sigma^2}{n_1\beta_3\beta_2\omega} \frac{1}{\frac{1}{n_1} -2m_3} \Big[e^{ -4m_3[\sqrt{z}-\sqrt{a}]}-e^{-\frac{2}{n_1}(\sqrt{z}-\sqrt{a})}\Big] \\ &+\frac{4 m_2\sigma^2\delta_3}{n_1\beta_3\beta_2\omega} \frac{1}{\frac{1}{n_1} -m_3} \Big[e^{ -2m_3[\sqrt{z}-\sqrt{a}]}-e^{-\frac{2}{n_1}(\sqrt{z}-\sqrt{a})}\Big] \\ & +\frac{4m_1\sigma^2}{n_1\beta_2} \frac{1}{2m_3[\frac{1}{n_1} -2m_3]}\sqrt{z}\Big[e^{ -4m_3[\sqrt{z}-\sqrt{a}]}-e^{-\frac{2}{n_1}(\sqrt{z}-\sqrt{a})}\Big] \\ &+\frac{4m_2\sigma^2}{n_1\beta_2} \frac{1}{m_3[\frac{1}{n_1} -m_3]}\sqrt{z}\Big[e^{ -2m_3[\sqrt{z}-\sqrt{a}]}-e^{-\frac{2}{n_1}(\sqrt{z}-\sqrt{a})}\Big], \end{align} (4.23)

    where

    \begin{align*} n_7& = n_2+\frac{2m_1\sqrt{a}}{m_3\beta_2}+\frac{4 m_2\sqrt{a}}{m_3\beta_2}. \end{align*}

    From Eq (4.23), we can conclude that Theorem 4.1 not only shows the convergence of the solutions of Eqs (3.12)–(3.21) on the coefficient \sigma , but it also shows a exponentially decay result as z\rightarrow \infty . Obviously, the decay rate is slightly slower than that in Remark 3.1.

    Remark 4.3. If h(z) satisfies

    \begin{align} h(z) = z^\frac{2}{3}, \end{align} (4.24)

    then

    \begin{align*} \int_{a}^z\frac{1}{h}d\zeta& = \int_{a}^z\frac{1}{\zeta^\frac{2}{3}}d\zeta = 3\Big[\sqrt[3]{z}-\sqrt[3]{a}\Big]. \end{align*}

    Therefore, we have

    \begin{align} \int_\tau^\infty e^{-2m_3\int_a^\xi\frac{1}{h(\zeta)}d\zeta}d\xi& = \int_\tau^\infty e^{-6m_3\Big(\sqrt[3]{\xi}-\sqrt[3]{a}\Big)}d\xi \\ & = -\frac{1}{2m_3}\int_\tau^\infty \xi^\frac{2}{3}d\Big[e^{-6m_3\Big(\sqrt[3]{\xi}-\sqrt[3]{a}\Big)}\Big] \\ &\leq-\frac{1}{2m_3}\tau^\frac{2}{3}\int_\tau^\infty d\Big[e^{-6m_3\Big(\sqrt[3]{\xi}-\sqrt[3]{a}\Big)}\Big] \\ & = \frac{1}{2m_3}\tau^\frac{2}{3}e^{-6m_3\Big(\sqrt[3]{\tau}-\sqrt[3]{a}\Big)}, \end{align} (4.25)
    \begin{align} \int_\tau^\infty e^{-m_3\int_a^\xi\frac{1}{h(\zeta)}d\zeta}d\xi &\leq\frac{1}{m_3}\tau^\frac{2}{3}e^{-3m_3\Big(\sqrt[3]{\tau}-\sqrt[3]{a}\Big)}. \end{align} (4.26)

    Choosing n_1 > 3 , we get

    \begin{align} \int_a^z \frac{1}{h(\xi)}e^{\int_a^\xi\{\frac{1}{n_1h(\zeta)}-\frac{2m_3}{h(\zeta)}\}d\zeta}d\xi & = \int_a^z \frac{1}{\xi^\frac{2}{3}}e^{[\frac{3}{n_1} -6m_3]\Big(\sqrt[3]{\xi}-\sqrt[3]{a}\Big)}d\xi \\ & = \frac{3}{\frac{3}{n_1} -6m_3} \Big[e^{[\frac{3}{n_1} -6m_3]\Big(\sqrt[3]{\xi}-\sqrt[3]{a}\Big)}-1\Big], \end{align} (4.27)
    \begin{align} \int_a^z \frac{1}{h(\xi)}e^{\int_a^\xi\{\frac{1}{n_1h(\zeta)}-\frac{m_3}{h(\zeta)}\}d\zeta}d\xi &\leq\frac{3}{\frac{3}{n_1} -3m_3} \Big[e^{[\frac{3}{n_1} -3m_3]\Big(\sqrt[3]{\xi}-\sqrt[3]{a}\Big)}-1\Big], \end{align} (4.28)

    and

    \begin{align} \int_a^z \frac{1}{h(\tau)}&e^{\frac{1}{n_1}\int_a^\tau\frac{1}{h(\zeta)}d\zeta}\int_\tau^\infty e^{-2m_3\int_a^\xi\frac{1}{h(\zeta)}d\zeta}d\xi d\tau \\ & = \frac{1}{2m_3}\int_a^z e^{[\frac{3}{n_1}-6m_3]\Big(\sqrt[3]{\tau}-\sqrt[3]{a}\Big)}d\tau \\ &\leq\frac{1}{2m_3[\frac{3}{n_1}-6m_3]}z^\frac{2}{3}\Big[e^{[\frac{3}{n_1}-6m_3]\Big(\sqrt[3]{\tau}-\sqrt[3]{a}\Big)}-1\Big], \end{align} (4.29)
    \begin{align} \int_a^z\frac{1}{h(\tau)}&e^{\frac{1}{n_1}\int_a^\tau\frac{1}{h(\zeta)}d\zeta}\int_\tau^\infty e^{-m_3\int_a^\xi\frac{1}{h(\zeta)}d\zeta}d\xi d\tau \\ &\leq\frac{1}{2m_3[\frac{3}{n_1}-3m_3]}z^\frac{2}{3}\Big[e^{[\frac{3}{n_1}-3m_3]\Big(\sqrt[3]{\tau}-\sqrt[3]{a}\Big)}-1\Big]. \end{align} (4.30)

    Inserting Eqs (4.27)–(4.30) into Eq (4.10), we obtain

    \begin{align} \int_0^t\int_{R_z}e^{-\omega\eta}(\xi-z)&\Big[\frac{1}{4}\delta_2\omega \theta^2+\frac{1}{4}\omega \delta_3\Sigma^2+\frac{1}{2}w_{, \alpha}w_{, \alpha}+\delta_2\theta_{, \alpha}\theta_{, \alpha} +\delta_3\Sigma_{, \alpha}\Sigma_{, \alpha}\Big]dx_2d\xi d\eta\\ &+\frac{1}{2} e^{-\omega t}\int_{R_z}(\xi-z)\Big[\delta_2 \theta^2+\delta_3\Sigma^2\Big]dx_2d\xi\\ &\leq n_8\sigma^2e^{-\frac{3}{n_1}\Big[\sqrt[3]{z}-\sqrt[3]{a}\Big]} \\ &+\frac{4 m_1\delta_3\sigma^2}{n_1\beta_3\beta_2\omega} \frac{3}{\frac{3}{n_1} -6m_3} \Big[e^{ -6m_3\Big(\sqrt[3]{\xi}-\sqrt[3]{a}\Big)}-e^{-\frac{3}{n_1}\Big[\sqrt[3]{z}-\sqrt[3]{a}\Big]}\Big] \\ &+\frac{4 m_2\sigma^2\delta_3}{n_1\beta_3\beta_2\omega} \frac{3}{\frac{3}{n_1} -3m_3} \Big[e^{ -3m_3\Big(\sqrt[3]{\xi}-\sqrt[3]{a}\Big)}-e^{-\frac{3}{n_1}\Big[\sqrt[3]{z}-\sqrt[3]{a}\Big]}\Big] \\ & +\frac{4m_1\sigma^2}{n_1\beta_2} \frac{1}{2m_3[\frac{3}{n_1}-6m_3]}z^\frac{2}{3}\Big[e^{-6m_3\Big(\sqrt[3]{z}-\sqrt[3]{a}\Big)}-e^{-\frac{3}{n_1}\Big[\sqrt[3]{z}-\sqrt[3]{a}\Big]}\Big] \\ &+\frac{4m_2\sigma^2}{n_1\beta_2} \frac{1}{m_3[\frac{3}{n_1}-3m_3]}z^\frac{2}{3}\Big[e^{-3m_3\Big(\sqrt[3]{z}-\sqrt[3]{a}\Big)} -e^{-\frac{3}{n_1}\Big[\sqrt[3]{z}-\sqrt[3]{a}\Big]}\Big], \end{align} (4.31)

    where

    \begin{align*} n_8& = n_2+\frac{2m_1a^\frac{2}{3}}{m_3\beta_2}+\frac{4m_2a^\frac{2}{3}}{\beta_2m_3}. \end{align*}

    In this case, the inequality (Eq 4.31) not only shows the convergence of the solutions of Eqs (3.12)–(3.21) on the coefficient \sigma , but it also shows a exponentially decay result as z\rightarrow \infty .

    In this paper, the convergence of the solutions of Eqs (3.12)–(3.21) on the coefficient \sigma has been obtained and three examples have been given. Obviously, the convergence of various systems of partial differential equations defined on R is rare. But Eq (1.1) is linear. It will be meaningful to study nonlinear equations (e.g., Brinkman equations, Forchheimer equations) by using the method in this paper.

    The authors express their heartfelt thanks to the editors and referees who have provided some important suggestions. This work is supported by the Tutor System Rroject of Guangzhou Huashang College (2021HSDS16).

    All authors declare no conflicts of interest in this paper.



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