
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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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.
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.
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(∂uα∂xβ)2.
The initial-boundary conditions can be written as
uα(x1,0,t)=uα(x1,h(x1),t)=0, x1≥a,0<t<τ, | (1.5) |
T(x1,0,t)=T(x1,h(x1),t)=0, x1≥a,0<t<τ, | (1.6) |
C(x1,0,t)=C(x1,h(x1),t)=0, x1≥a,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=−∇⊥⋅gT−g⋅∇⊥T−∇⊥⋅hC−h⋅∇⊥C, in R×(0,τ), | (1.11) |
∂tT+∇⊥v⋅∇T=ΔT, in R×(0,τ), | (1.12) |
∂tC+∇⊥v⋅∇C=ΔC+σΔT, in R×(0,τ), | (1.13) |
v(x1,0,t)=v(x1,h(x1),t)=0, x1≥a,0<t<τ, | (1.14) |
vn(x1,0,t)=vn(x1,h(x1),t)=0, x1≥a,0<t<τ, | (1.15) |
T(x1,0,t)=T(x1,h(x1),t)=0, x1≥a,0<t<τ, | (1.16) |
C(x1,0,t)=C(x1,h(x1),t)=0, x1≥a,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)|x1≥z>a,0<x2<h(x1)}, |
Lz={(x1,x2)|x1=z≥a,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
∫Lzw2dx2≤h2π2∫Lz(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 φ→0, as x1→∞, then
∫Rzφ4dx2dξ≤(∫Rzφ2dx2dξ)(∫Rzφ,αφ,αdx2dξ). |
Proof. Since φ(x1,0)=φ(x1,h)=0, we have
φ2(x1,x2)=2∫x20φ∂∂ζφ(x1,ζ)dζ=−2∫hx2φ∂∂ζφ(x1,ζ)dζ≤∫h0|φ∂∂x2φ(x1,x2)|dx2. | (2.1) |
Since φ→0, as x1→∞, we have
φ2(x1,x2)=−2∫∞x1φ∂∂ξφ(ξ,x2)dξ≤2∫∞x1|φ∂∂ξφ(ξ,x2)|dξ. | (2.2) |
Combining Eqs (2.1) and (2.2) and integrating over Rz, we obtain
∫Rzφ4(ξ,x2)dx2dξ≤2[∫Rz|φ∂∂x2φ(ξ,x2)|dx2dξ][∫Rz|φ∂∂ξφ(ξ,x2)|dx2dξ]≤2∫Rzφ2dx2dξ(∫Rz(∂φ∂ξ)2dx2dξ)12(∫Rz(∂φ∂x2)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,H∈L∞, then
sup[0,τ]||T||∞≤Tm, |
where Tm=max{||T0||∞, sup[0,τ]H∞(η)}.
If C0,˜H∈L∞ 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 ∀ z≥a such that F(z,t)<0. Then for any fixed t
∫t0∫Rze−ωη[14β2ωT2+14ωC2+12β1v,αv,α+12β2T,αT,α+12C,αC,α]dx2dξdη+12e−ωt∫∞z∫Lξ[β2T2+C2]dx2dξ≤m1e−2m3∫za1h(ζ)dζ+m2e−m3∫za1h(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)=∫t0∫Lze−ωη[β1vv,1+β2TT,1+CC,1]dx2dη+β1∫t0∫Lze−ωη[g2Tv+h2Cv]dx2dη+∫t0∫Lze−ωη[−12β2T2v,2−12C2v,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,τ2≤1. 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 z≥a, then
∫t0∫Re−ωη[14β2ωT2+14ωC2+12β1v,αv,α+12β2T,αT,α+12C,αC,α]dx2dξdη+12e−ωt∫R[β2T2+C2]dx2dξ≤r(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,˜H∈L∞(R), then
∫Rv,αv,αdx2dξ≤ϵ1, | (2.4) |
where
ϵ1=2sup[0,τ]{∫La|˜F1˜F2|dx2dξ+∫La|g2H˜F1|dx2dξ+∫La|h2˜H˜F1|dx2dξ}+4max{1β2,1}r∗eωτ. |
Additionally, r∗ is the maximum value of r(t) in [0,τ].
Proof. Using Eq (1.11), we have
∫R[Δv+∇⊥⋅gT+g⋅∇⊥T+∇⊥⋅hC+h⋅∇⊥C]vdx2dξ=0. |
Using Eqs (1.18)–(1.20) it follows that
∫Rv,αv,αdx2dξ=−∫La˜F1˜F2dx2dξ−∫Lag2H˜F1dx2dξ−∫Lah2˜H˜F1dx2dξ−∫R∇⊥v⋅gTdx2dξ−∫R∇⊥v⋅hCdx2dξ≤∫La|˜F1˜F2|dx2dξ+∫La|g2H˜F1|dx2dξ+∫La|h2˜H˜F1|dx2dξ+12∫Rv,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 σ→0. 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∗=−∇⊥⋅gT∗−g⋅∇⊥T∗−∇⊥⋅hC∗−h⋅∇⊥C∗, in R×(0,τ), | (3.1) |
∂tT∗+∇⊥v∗⋅∇T∗=ΔT∗, in R×(0,τ), | (3.2) |
∂tC∗+∇⊥v∗⋅∇C∗=ΔC∗, in R×(0,τ), | (3.3) |
v∗(x1,0,t)=v∗(x1,h(x1),t)=0, x1≥a,0<t<τ, | (3.4) |
v∗n(x1,0,t)=v∗n(x1,h(x1),t)=0, x1≥a,0<t<τ, | (3.5) |
T∗(x1,0,t)=T∗(x1,h(x1),t)=0, x1≥a,0<t<τ, | (3.6) |
C∗(x1,0,t)=C∗(x1,h(x1),t)=0, x1≥a,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=v−v∗, θ=T−T∗, Σ=C−C∗. |
Then (w,θ,Σ) satisfies
Δw=−∇⊥⋅gθ−g⋅∇⊥θ−∇⊥⋅hΣ−h⋅∇⊥Σ, in R×(0,τ), | (3.12) |
∂tθ+∇⊥w⋅∇T+∇⊥v∗⋅∇θ=Δθ, in R×(0,τ), | (3.13) |
∂tΣ+∇⊥v⋅∇Σ+∇⊥w⋅∇C∗=ΔΣ+σΔT, in R×(0,τ), | (3.14) |
w(x1,0,t)=w(x1,h(x1),t)=0, x1≥a,0<t<τ, | (3.15) |
wn(x1,0,t)=wn(x1,h(x1),t)=0, x1≥a,0<t<τ, | (3.16) |
θ(x1,0,t)=θ(x1,h(x1),t)=0, x1≥a,0<t<τ, | (3.17) |
Σ(x1,0,t)=Σ(x1,h(x1),t)=0, x1≥a,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)=−∫t0∫Rze−ωηww,1dx2dξdη+∫t0∫Rze−ωηg2θwdx2dξdη+∫t0∫Rze−ωηh2Σwdx2dξdη≐A1(z,t)+A2(z,t)+A3(z,t), | (3.22) |
E2(z,t)=∫t0∫Rze−ωηθθ,1dx2dξdη−12∫t0∫Rze−ωηθ2v∗,2dx2dξdη+∫t0∫Rze−ωηθTw,2dx2dξdη+∫t0∫Rze−ωη(ξ−z)∇⊥w⋅∇θTdx2dξdη≐B1(z,t)+B2(z,t)+B3(z,t)+B4(z,t), | (3.23) |
E3(z,t)=∫t0∫Rze−ωηΣΣ,1dx2dξdη−12∫t0∫Rze−ωηΣ2v,2dx2dξdη−∫t0∫Rze−ωηw,2C∗Σdx2dξdη+σ∫t0∫Rze−ωηΣT,1dx2dξdη+∫t0∫Rze−ωη(ξ−z)∇⊥w⋅∇ΣC∗dx2dξdη+σ∫t0∫Rze−ωη(ξ−z)T,αΣ,αdx2dξdη≐C1(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)=−∫t0∫Rze−ωη(ξ−z),αww,αdx2dξdη+∫t0∫Rze−ωηg2θwdx2dξdη+∫t0∫Rze−ωηh2Σwdx2dξdη=∫t0∫Rze−ωη(ξ−z)w,αw,αdx2dξdη+∫t0∫Rze−ωη(ξ−z)g⋅∇⊥wθdx2dξdη+∫t0∫Rze−ωη(ξ−z)h⋅∇⊥wΣdx2dξdη. | (3.25) |
Similarly, we have
E2(z,t)=∫t0∫Rze−ωη(ξ−z)[12ωθ2+θ,αθ,α]dx2dξdη+12e−ωt∫Rz(ξ−z)θ2dx2dξ, | (3.26) |
and
E3(z,t)=∫t0∫Rze−ωη(ξ−z)[12ωΣ2+Σ,αΣ,α]dx2dξdη+12e−ωt∫Rze−ωη(ξ−z)Σ2dx2dξ. | (3.27) |
We also define
E(z,t)=E1(z,t)+δ2E2(z,t)+δ3E3(z,t)=∫t0∫Rze−ωη(ξ−z)[12δ2ωθ2+12δ3ωΣ2+w,αw,α+δ2θ,αθ,α+δ3Σ,αΣ,α]dx2dξdη+12e−ωt∫Rz(ξ−z)[δ2θ2+δ3Σ2]dx2+∫t0∫Rze−ωη(ξ−z)g⋅∇⊥wθdx2dξdη+∫t0∫Rze−ωη(ξ−z)h⋅∇⊥wΣdx2dξdη, | (3.28) |
where δ2 and δ3 are positive constants.
Using the Cauchy-Schwarz inequality, we have
|∫t0∫Rze−ωη(ξ−z)g⋅∇⊥wθdx2dξdη|≤∫t0∫Rze−ωη(ξ−z)θ2dx2dξdη+14∫t0∫Rze−ωη(ξ−z)w,αw,αdx2dξdη, | (3.29) |
|∫t0∫Rze−ωη(ξ−z)h⋅∇⊥wΣdx2dξdη|≤∫t0∫Rze−ωη(ξ−z)Σ2dx2dξdη+14∫t0∫Rze−ωη(ξ−z)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)≥∫t0∫Rze−ωη(ξ−z)[14δ2ωθ2+14ωδ3Σ2+12w,αw,α+δ2θ,αθ,α+δ3Σ,αΣ,α]dx2dξdη+12e−ωt∫Rz(ξ−z)[δ2θ2+δ3Σ2]dx2dξ. | (3.31) |
From Eq (3.28), we have
−∂∂zE(z,t)=∫t0∫Rze−ωη[12δ2ωθ2+12δ3ωΣ2+w,αw,α+δ2θ,αθ,α+δ3Σ,αΣ,α]dx2dξdη+12e−ωt∫Rz[δ2θ2+δ3Σ2]dx2dξ+∫t0∫Rze−ωηg⋅∇⊥wθdx2dξdη+∫t0∫Rze−ωηh⋅∇⊥wΣdx2dξdη. | (3.32) |
Using the Cauchy-Schwarz inequality, we have
|∫t0∫Rze−ωηg⋅∇⊥wθdx2dξdη|≤∫t0∫Rze−ωηθ2dx2dξdη+14∫t0∫Rze−ωηw,αw,αdx2dξdη, | (3.33) |
|∫t0∫Rze−ωηh⋅∇⊥wΣdx2dξdη|≤∫t0∫Rze−ωηΣ2dx2dξdη+14∫t0∫Rze−ωηw,αw,αdx2dξdη. | (3.34) |
Inserting Eqs (3.33) and (3.34) into Eq (3.32), we have
−∂∂zE(z,t)≤∫t0∫Rze−ωη[34δ2ωθ2+34ωδ3Σ2+32w,αw,α+δ2θ,αθ,α+δ3Σ,αΣ,α]dx2dξdη+12e−ωt∫Rz[δ2θ2+δ3Σ2]dx2dξ, | (3.35) |
and
−∂∂zE(z,t)≥∫t0∫Rze−ωη[14δ2ωθ2+14ωδ3Σ2+12w,αw,α+δ2θ,αθ,α+δ3Σ,αΣ,α]dx2dξdη+12e−ωt∫Rz[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,˜H∈C∞, 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)]+4σ2δ3β3β2ωm1e−2m3∫za1h(ζ)dζ+4σ2δ3β3β2ωm2e−m3∫za1h(ζ)dζ+4β2σ2m1∫∞ze−2m3∫ξa1h(ζ)dζdξ+4β2σ2m2∫∞ze−m3∫ξa1h(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)≤[∫t0∫Rze−ωηw2dx2dξdη∫t0∫Rze−ωη(w,1)2dx2dξdη]12≤hπ[∫t0∫Rze−ωη(w,2)2dx2dξdη∫t0∫Rze−ωη(w,1)2dx2dξdη]12≤h2π∫t0∫Rze−ωηw,αw,αdx2dξdη, | (3.38) |
A2(z,t)≤[∫t0∫Rze−ωηw2dx2dξdη∫t0∫Rze−ωηθ2dx2dξdη]12≤2√2hπ√δ2ω[12∫t0∫Rze−ωη(w,2)2dx2dξdη⋅14δ2ω∫t0∫Rze−ωηθ2dx2dξdη]12≤√2h√δ2ωπ[12∫t0∫Rze−ωηw,αw,αdx2dξdη+14δ2ω∫t0∫Rze−ωηθ2dx2dξdη], | (3.39) |
A3(z,t)≤√2h√δ3ωπ[12∫t0∫Rze−ωηw,αw,αdx2dξdη+14ωδ3∫t0∫Rze−ωηΣ2dx2dξdη]. | (3.40) |
Combining Eqs (3.22) and (3.38)–(3.40), we obtain
E1(z,t)≤hπ[1+√2√δ2ω+√2√δ3ω]∫t0∫Rze−ωη12w,αw,αdx2dξdη+√2h√δ2ωπ∫t0∫Rze−ωη14δ2ωθ2dx2dξdη+√2h√δ3ωπ∫t0∫Rze−ωη14ωδ3Σ2dx2dξdη. | (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)≤[∫t0∫Rze−ωηθ2dx2dξdη∫t0∫Rze−ωη(θ,1)2dx2dξdη]12≤hπ[∫t0∫Rze−ωη(θ,2)2dx2dξdη∫t0∫Rze−ωη(θ,1)2dx2dξdη]12≤hπδ2⋅12δ2∫t0∫Rze−ωηθ,αθ,αdx2dξdη, | (3.42) |
B2(z,t)≤12∫t0[∫Rze−ωη(v∗,2)2dx2dξ∫Rze−ωηθ4dx2dξ]12dη≤12∫t0[∫Re−ωη(v∗,2)2dx2dξ∫Rze−ωηθ2dx2dξ]12dη≤√ϵ12δ2π∫t0[∫Rze−ωηθ2dx2dξ∫Rze−ωηθ,αθ,αdx2dξ]12dη≤h√ϵ12δ2π∫t0∫Rze−ωηδ2θ,αθ,αdx2dξdη, | (3.43) |
B3(z,t)≤Tm[∫t0∫Rze−ωη(w,2)2dx2dξdη]12[∫t0∫Rze−ωηθ2dx2dξdη]12≤hπ√δ2Tm[12∫t0∫Rze−ωηw,αw,αdx2dξdη]12[δ2∫t0∫Rze−ωηθ,αθ,αdx2dξdη]12≤hπ√2δ2Tm[12∫t0∫Rze−ωηw,αw,αdx2dξdη+δ2∫t0∫Rze−ωηθ,αθ,αdx2dξdη]. | (3.44) |
Using Lemma 2.3, we have
B4(z,t)≤12T2m∫t0∫Rze−ωη(ξ−z)w,αw,αdx2dξdη+12∫t0∫Rze−ωη(ξ−z)θ,αθ,αdx2dξdη. | (3.45) |
Inserting Eqs (3.42) and (3.45) into Eq (3.23), we obtain
E2(z,t)≤[hπδ2+hπ√2δ2Tm+h√ϵ12δ2π]∫t0∫Rze−ωηδ2θ,αθ,αdx2dξdη+hπ√2δ2Tm[12∫t0∫Rze−ωηw,αw,αdx2dξdη]+12T2m∫t0∫Rze−ωη(ξ−z)w,αw,αdx2dξdη+12∫t0∫Rze−ωη(ξ−z)θ,αθ,αdx2dξdη. | (3.46) |
Similarly, we have
C1(z,t)≤hπ[∫t0∫Rze−ωη(Σ,2)2dx2dξdη∫t0∫Rze−ωη(Σ,1)2dx2dξdη]12≤h2δ3π⋅δ3∫t0∫Rze−ωηΣ,αΣ,αdx2dξdη, | (3.47) |
C2(z,t)≤12∫t0[∫Rze−ωη(v∗,2)2dx2dξ∫Rze−ωηΣ4dx2dξ]12dη≤h√ϵ12δ3π⋅δ3∫t0∫Rze−ωηΣ,αΣ,αdx2dξdη, | (3.48) |
C3(z,t)≤Cm[∫t0∫Rze−ωη(w,2)2dx2dξdη]12[∫t0∫Rze−ωηΣ2dx2dξdη]12≤hπ√2δ3Cm[12∫t0∫Rze−ωηw,αw,αdx2dξdη+δ3∫t0∫Rze−ωηΣ,αΣ,αdx2dξdη], | (3.49) |
C4(z,t)≤14ωδ3∫t0∫Rze−ωηΣ2dx2dξdη+σ2β3ω∫t0∫Rze−ωη(T,1)2dx2dξdη, | (3.50) |
C5(z,t)≤14δ3∫t0∫Rze−ωη(ξ−z)w,αw,αdx2dξdη+C2mδ3∫t0∫Rze−ωη(ξ−z)Σ,αΣ,αdx2dξdη | (3.51) |
and
C6(z,t)≤1δ3σ2∫t0∫Rze−ωη(ξ−z)T,αT,αdx2dξdη+14δ3∫t0∫Rze−ωη(ξ−z)Σ,αΣ,αdx2dξdη. | (3.52) |
Inserting Eqs (3.47)–(3.52) into Eq (3.24), we obtain
E3(z,t)≤[h2δ3π+hπ√2δ3Cm+h√ϵ12δ3π]⋅δ3∫t0∫Rze−ωηΣ,αΣ,αdx2dξdη+hπ√2δ3Cm[12∫t0∫Rze−ωηw,αw,αdx2dξdη]+14ωδ3∫t0∫Rze−ωηΣ2dx2dξdη+[14δ3+C2mδ3]∫t0∫Rze−ωη(ξ−z)Σ,αΣ,αdx2dξdη+14δ3∫t0∫Rze−ωη(ξ−z)w,αw,αdx2dξdη+σ2β3ω∫t0∫Rze−ωη(T,1)2dx2dξdη+1δ3σ2∫t0∫Rze−ωη(ξ−z)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)]+σ2δ3β3ω∫t0∫Rze−ωη(T,1)2dx2dξdη+σ2∫t0∫Rze−ωη(ξ−z)T,αT,αdx2dξdη, | (3.54) |
where
12n1≥max{1π[1+√2√δ2ω+√2√δ3ω]+Tmπ√2δ2+Cmπ√2δ3,1+√2√δ2ωπ,12δ3π+1π√2δ3Cm+√ϵ12δ3π,hπδ2+1π√2δ2Tm+√ϵ12δ2π}. |
From Lemma 2.4, we have
∫t0∫Rze−ωηT,αT,αdx2dξdη≤2β2m1e−2m3∫za1h(ζ)dζ+2β2m2e−m3∫za1h(ζ)dζ. | (3.55) |
Integrating Eq (3.55) from z to ∞, we get
∫t0∫Rze−ωη(ξ−z)T,αT,αdx2dξdη≤2β2m1∫∞ze−2m3∫ξa1h(ζ)dζdξ+2β2m2∫∞ze−m3∫ξa1h(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)e−1n1∫za1h(ζ)dζ+4m1δ3σ2n1β3β2ωe−1n1∫za1h(ζ)dζ∫za1h(ξ)e∫ξa{1n1h(ζ)−2m3h(ζ)}dζdξ+4m2σ2δ3n1β3β2ωe−1n1∫za1h(ζ)dζ∫za1h(ξ)e∫ξa{1n1h(ζ)−m3h(ζ)}dζdξ+4m1σ2n1β2e−1n1∫za1h(ζ)dζ∫za1h(τ)e1n1∫τa1h(ζ)dζ∫∞τe−2m3∫ξa1h(ζ)dζdξdτ+4m2σ2n1β2e−1n1∫za1h(ζ)dζ∫za1h(τ)e1n1∫τa1h(ζ)dζ∫∞τe−m3∫ξa1h(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+4β2σ2m1∫∞ae−2m3∫ξa1h(ζ)dζdξ+4β2σ2m2∫∞ae−m3∫ξa1h(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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