# A Four-Zone Model and Nonlinear Dynamic Analysis of Solution Multiplicity of Buoyancy Ventilation in Underground Building

Authors: Yuxing Wang; Chunyu Wei
Journal: Complexity
 (2020)

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## Abstract

The solution multiplicity of natural ventilation in buildings is very important to personnel safety and ventilation design. In this paper, a four-zone model of buoyancy ventilation in typical underground building is proposed. The underground structure is divided to four zones, a differential equation is established in each zone, and therefore, there are four differential equations in the underground structure. By solving and analyzing the equilibrium points and characteristic roots of the differential equations, we analyze the stability of three scenarios and obtain the criterions to determine the stability and existence of solutions for two scenarios. According to these criterions, the multiple steady states of buoyancy ventilation in any four-zone underground buildings for different stack height ratios and the strength ratios of the heat sources can be obtained. These criteria can be used to design buoyancy ventilation or natural exhaust ventilation systems in underground buildings. Compared with the two-zone model in (Liu et al. 2020), the results of the proposed four-zone model are more consistent with CFD results in (Liu et al. 2018). In addition, the results of proposed four-zone model are more specific and more detailed in the unstable equilibrium point interval. We find that the unstable equilibrium point interval is divided into two different subintervals corresponding to the saddle point of index 2 and the saddle focal equilibrium point of index 2, respectively. Finally, the phase portraits and vector field diagrams for the two scenarios are given.


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## Body

## 1. Introduction

Nonlinear characteristics exist in systems with various research directions, such as neural network [1–3], chaotic circuit [4–12], and information security [13–15]. Building natural ventilation also has nonlinear characteristics. The solution multiplicity of natural ventilation in buildings is very important to personnel safety and ventilation design [16]. In the last decade, the multisolution problem of natural ventilation in buildings has attracted wide attention. Many papers have been published on the problem of multiple solutions to building natural ventilation [17–23]. These studies can be divided into three categories according to the number of building zones and vents. The first kind research is about the solution multiplicity of single-zone and double-opening buildings under the combined effect of wind pressure and thermal pressure [19, 24–27]. For example, Heiselberg et al. [24] studied the multiple steady-state properties of single-zone and double-opening buildings under the action of wind pressure and buoyancy through a salt water experiment and CFD simulation; Lishman and Woods [25] reported solution multiplicity in both inclined tunnels and two-story aboveground buildings, and the study focused on the confrontation of wind pressure and buoyancy in a single-story building; Yuan and Glicksman [19, 26] researched the effects of different initial conditions on the generation of multiple steady states in single-zone buildings under the combined action of wind pressure and buoyancy; Pulat and Ersan [27] found that different turbulence parameters may produce multiple solutions through CFD simulation. The second kind research is about the solution multiplicity of single-zone and multiple opening buildings [28, 29]. For example, Durrani et al. [28] researched the multiple solutions in a typical aboveground building with one zone and three openings by CFD simulations; through theoretical analysis, Chen and Li [29] researched the buoyancy ventilation of a single-zone building with three horizontal openings. The third kind research is about the solution multiplicity of two-zone buildings [30–33]. For example, Yang et al. [30, 31] analyzed in detail multiple steady states in a two-zone building by theoretical analyses and CFD simulations; Li et al. [32] studied the buoyancy ventilation in a two-story building with two heat sources and three openings by establishing a mathematical model of nonlinear ordinary differential equation; Yang et al. [33] analyzed the smoke exhaust spread of the three entrance tunnel in the fire scenarios and obtained six possible equilibrium states.Recently, there are some reports on the multiplicity of ventilation solutions of underground structures [34, 35]. The study of the multiplicity of solutions for the natural ventilation of underground buildings is different from that of overground buildings. Because the underground buildings are not exposed to the outdoor environment, it is mainly thermal pressure ventilation (or buoyancy ventilation) rather than wind pressure ventilation which exist in the under structure. Liu et al. [34] investigated the formation process of multiple steady states in an underground building with two tunnel connecting to the outdoor environment. The heights of two tunnels are same, and one heat source is located in the corner of the buried space. Using the CFD method reproduces the two steady states of the buoyancy ventilation in the underground building. In 2020, Liu et al. [35] made nonlinear dynamic analysis of solution multiplicity of buoyancy ventilation in a typical underground structure. They gave a description of mathematical model of second-order nonlinear differential equation to underground structure with two zones and two tunnel connecting to the outdoor environment. Two heat sources are located in the two zones, respectively, and the heights of the two tunnels are not necessarily the same. By solving the equilibrium points and characteristic roots of the second order differential equation, the solution multiplicity can be determined when the strength ratio of two heat sources and height ratio of two tunnels were changed. However, in the literatures [34, 35], the underground structure has only two zones. The deeply burred room and tunnel were treated as one zone, which is just a very simple model. In the real underground structure, the deeply burred room and tunnel are different in temperature, mass flow rate, and air density. So, the deeply burred room and tunnel should not be treated as one zone if a more accurate model is needed. In this paper, we propose a new model, where the buried rooms and tunnels are treated as different zones. Thus, the underground structure consists of four zones, two zones for two tunnels, and two zones for deeply burred room. In this paper, the nomenclature of every variables and constants are shown in abbreviation section.


## 2. Proposed New Mathematical Model and Analysis

In order to study the nonlinear dynamics of typical deep-buried underground buildings with two openings, we first established the mathematical mode. Some assumptions were made: (i) each zone is well mixed; (ii) thermal mass is 1; (iii)E1 > 0; (iv) the mass flow impedance coefficient of the geometry is constant. Because the driving force is from thermal pressure, as illustrated in Figure 1, we divide the building into four zones: we can assume the heat source in the left side as positive, and the heat source of the right side could be either negative or positive, such that we could discuss the scenarios of two heat sources (the heat source in the left and right sides both are positive and the heat source in the left side is positive, the heat source in the right side is negative) and the scenario of one heat source.Figure 1
Proposed schematics of four-zone underground structure.


### 2.1. Description of Mathematical Model

The proposed new schematics of underground structure is shown as Figure1, which contains four zones, zone 1 and zone 2 for deeply burred room and zone 3 and zone 4 for two tunnels. The heights of two tunnels are H1 and H2 respectively, the Ti, qi (i = 1∼4) denote the temperature, and mass flow rate of four zones, respectively. Heat sources E1 and E2 are located in zone 1 and zone 2. As shown in Figure 1, two realizations are considered: for realization 1, the air flow enters from zone 1 to zone 2; for realization 2, the air flow enters from zone 2 to zone 1.For realization 1, the heat gain of the internal thermal mass should be equal to the heat released by the heat sources minus the heat loss through airflows. Therefore, we can obtain the heat balance equations of the four zones shown in the following equations:(1)M1CpdT1dt=−q1CPT1−T3+E1,(2)M2CpdT2dt=−q2CPT2−T1+E2,(3)M3CpdT3dt=−q3CPT3−Ta,(4)M4CpdT4dt=−q4CPT4−T2.According to the flow loop method, the total pressure loss at ventilation vents on the flow loop is balanced by sum of the buoyancy pressure [35]. The pressure balance equation can be obtained as follows:(5)−T1−TaTaρagH1+T2−TaTaρagH2=S1q12+S2q22+S3q32+S4q42.According to the conservation of mass, we can obtain(6)q1=q2=q3=q4.Combining equations (5) and (6), mass flow rate can be obtained as follows:(7)q1=q2=q3=q4=q=−T1−Ta/TaρagH1+T2−Ta/TaρagH2S1+S2+S3+S4.For realization 2, similarly, we can obtain the heat balance equations of the four zones as follows:(8)M1CpdT1dt=−q1CPT1−T2+E1,(9)M2CpdT2dt=−q2CPT2−T4+E2,(10)M3CpdT3dt=−q3CPT3−T1,(11)M4CpdT4dt=−q4CPT4−Ta.Based on the flow loop method, the total pressure loss at ventilation vents on the flow loop is balanced by sum of the buoyancy pressure, and we can obtain the pressure balance equation as follows:(12)T1−TaTaρagH1−T2−TaTaρagH2=S1q12+S2q22+S3q32+S4q42.According to the conservation of mass, we can obtain following equations about air flow rate:(13)q1=q2=q3=q4=q.By combining equations (11) and (12), we can obtain(14)q1=q2=q3=q4=T1−Ta/TaρagH1−T2−Ta/TaρagH2S1+S2+S3+S4.


### 2.2. Transformation of Variables and Parameters in Differential Equations

Assumingk = E2/E1, ΔT1 = T1 − Ta, ΔT2 = T2 − Ta, ΔT3 = T3 − Ta, ΔT4 = T4 − Ta, n=ρagH1/S1+S2+S3+S4, α = H2/H1, M1 = M2 = M, M3 = M4 = M/ξ, and τ = t/(MCp), we can obtain simplified nonlinear differential equations.For realization 1, according to equations (1)∼(4), (7) and above assumptions about variables and parameters, the simplified differential equations can be obtained as follows:(15)f1=dΔT1dτ=−nαΔT2−ΔT1ΔT1−ΔT3+E1,(16)f2=dΔT2dτ=−nαΔT2−ΔT1ΔT2−ΔT1+E2,(17)f3=dΔT3dτ=−ξnαΔT2−ΔT1ΔT3,(18)f4=dΔT4dτ=−ξnαΔT2−ΔT1ΔT4−ΔT2.For realization 2, according to equations (8)∼(11) and (14) and above assumptions about variables and parameters, the simplified differential equations can be obtained as follows:(19)f1=dΔT1dτ=−nΔT1−αΔT2ΔT1−ΔT2+E1,(20)f2=dΔT2dτ=−nΔT1−αΔT2ΔT2−ΔT4+E2,(21)f3=dΔT3dτ=−ξnΔT1−αΔT2ΔT3−ΔT1,(22)f4=dΔT4dτ=−ξnΔT1−αΔT2ΔT4.


### 2.3. Nonlinear Dynamic Analysis

We can assume the heat source in the left side as positive and the heat source of the right side could be either negative or positive, such that we could discuss the scenarios of two heat sources (the heat source in the left and right sides both are positive and the heat source in the left side is positive, the heat source in the right side is negative) and the scenario of one heat source.


#### 2.3.1. Stability Analysis for Scenario 1 (k Is Fixed and α Is Control Parameter)

Here, two conditionsk > 0 (the heat source in the left and right sides both are positive) and k < 0 (the heat source in the left side is positive, and the heat source in the right side is negative) are discussed at the same time. We should solve equilibrium points and characteristic roots before stability analysis:(1) Analysis of Equilibrium Points and Characteristic Roots for Realization 1. The steady state solutions (equilibrium points) of realization 1 equations (15)∼(18) are denoted as ΔT01,ΔT02,ΔT03,ΔT04. In order to solve the equilibrium points, we can make that the right sides of equations (15)∼(18) equal to zero; and the following equations hold:(23)−nαΔT2−ΔT1ΔT1−ΔT3+E1=0,(24)−nαΔT2−ΔT1ΔT2−ΔT1+E2=0,(25)−ξnαΔT2−ΔT1ΔT3=0,(26)−ξnαΔT2−ΔT1ΔT4−ΔT2=0.By solving equations (23)∼(26), we can obtain the equilibrium points as follows:(27)ΔT01=E1n2/3−E1+E1α+E2α1/3=E12/3n2/3αk−1+α1/3,(28)ΔT02=ΔT04=1+kΔT01=1+kE1n2/3−E1+E1α+E2α1/3=1+kE12/3n2/3αk−1+α1/3,(29)ΔT03=0.According to the characteristic determinant equal to 0, the solution of the differential equation can be discussed. The characteristic determinant equation is shown as follows. The expression of characteristic determinant is that(30)JQ−λE=∂f1∂ΔT1−λ∂f1∂ΔT2∂f1∂ΔT3∂f1∂ΔT4∂f2∂ΔT1∂f2∂ΔT2−λ∂f2∂ΔT3∂f2∂ΔT4∂f3∂ΔT1∂f3∂ΔT2∂f3∂ΔT3−λ∂f3∂ΔT4∂f4∂ΔT1∂f4∂ΔT2∂f4∂ΔT3∂f4∂ΔT4−λ=0.Combining equations (15)∼(18), we can obtain the following expression:(31)nΔT01−ΔT032A−nA−λ−nαΔT01−ΔT032AnA0nΔT02−ΔT012A+nA−nαΔT02−ΔT012A−nA−λ00ξnΔT032A−ξnαΔT032A−ξnA−λ0ξnΔT04−ΔT022A−ξnαΔT04−ΔT022A+ξnA0−ξnA−λ=0,where JQ is the Jacobian matrix of equilibrium point Q, λ is the characteristic root, E is the identity matrix, and(32)A=αΔT02−ΔT01=αk−1+α1/3E11/3n1/3,(33)B=ΔT01−ΔT03=E12/3n2/3αk−1+α1/3,(34)C=ΔT02−ΔT01=kE12/3n2/3αk−1+α1/3.Combining equations (31)∼(34), we can obtain characteristic roots as follows:(35)λ1=λ2=−ξnA=−ξnαΔT02−ΔT01=−ξn2/3E11/3αk−1+α1/3.Assumingk = E2/E1, α = H2/H1, to ensure that a real solution exists for equations (27) and (28), we have(36)αk−1+α>0.Two other characteristic roots can be obtained according to a quadratic formula:(37)λ3,4=−b±b2−4ac2a=−nαC−B/2A−2nA±nαC−B/2A+2nA2−4n2αB+C/2+A2−B/22,where(38)a=1;b=nαC−B2A+2nA=n2/3E11/35αk−5+4α2αk−1+α2/3,c=n2αB+C2+A2−B2=3E12/3αk−1+α2/3n4/32>0,Δ=b2−4ac=n4/3E12/35αk−5+4α24αk−1+α4/3−6E12/3αk−1+α2/3n4/3.For convenience, the characteristic roots are set as follows: the real root areγi (i = 1,2,3,4); the complex roots are σi + jωi (i = 1,2,3,4). It is easy to know that λ1,2 are always real number, λ1 = γ1 = λ2 = γ2.If the discriminant Δ is greater than 0, both characteristic roots are real number, namely,λ3 = γ3 and λ4 = γ4. From equations (37) and (38), we can know that when Δ > 0, the following expression holds:(39)n4/3E12/35αk−5+4α2>24n4/3E12/3αk−1+α2.It is clear that(40)5αk−5+4α>26αk−1+α.When the discriminant Δ is less than 0, conjugate complex roots exist, namely,λ3,4 = σ3,4 ± jω3,4. From equations (37) and (38), we can know that when Δ < 0, the following expression holds:(41)5αk−5+4α<26αk−1+α.(2) Stability Analysis for Realization 1 (k Is Fixed and α Is Control Parameter). (1)
No equilibrium point exists in realization 1If no equilibrium point exists in realization 1, from equation (36), it is known that the following expression is satisfied:(42)αk−1+α<0.Therefore from expression (42), when k < −1, it is clear that α>1/k+1 and when k > −1, we know that 0<α<1/k+1.(2)
A stable equilibrium point exists in realization 1(2.1) Assuming there are all negative real characteristic rootsIn this case,λ1, λ2, λ3, and λ4 are all less than 0, and the system has a stable solution.We know that owing toαk − 1 + α > 0, from equation (35), it is clear that λ1 and λ2 are always less than 0. To make λ3,4 < 0, according to Vieta theorem, we know that λ3 + λ4 = −b and λ3λ4 = c. From equation (38), it is known that always c > 0, and owing to αk − 1 + α > 0, from equation (38), if 5αk − 5 + 4α > 0, we have −b < 0, then λ3,4 < 0. In this case, the absolute value sign in equation (40) can be removed; therefore, the following expression (43) can be obtained:(43)5αk−5+4α>26αk−1+α.From expression (43), we can obtain the following expression:(44)αk5−26+4−26>5−26.In summary, if there are all negative real characteristic roots, the system is stable when the following inequalities are satisfied:(45)αk+1−1>0,α5k+4−5>0,αk5−26+4−26>5−26.(2.2) Assuming two negative real roots and two conjugate complex rootsIn this case, the four characteristic roots behave asλ1 = γ1 = λ2 = γ2, λ3,4 = σ3,4 ± jω3,4, where λ1 and λ2 are two negative real roots and λ3, λ4 are two conjugate complex roots. When λ1, λ2, σ3, and σ4 are all less than 0, the system has a stable solution.From equation (37), we know σ3,4 = −b/2. Owing to λ1,2 < 0, to make σ3,4 < 0, we need −b < 0, and from equation (38), we can obtain 5αk + 4α − 5>0.In this case, the absolute value sign in expression (41) can be removed; therefore, the following expression can be obtained:(46)5αk−5+4α<26αk−1+α.From (46), we can obtain that(47)αk5−26+4−26<5−26.In summary,λ1, λ2, σ3, and σ4 are all less than 0, the system is stable when the following inequalities are satisfied:(48)αk+1−1>0,α5k+4−5>0,αk5−26+4−26<5−26.Combining expressions (45) and (48), a stable equilibrium point exists in realization 1, when the following expression is satisfied:(49)αk+1−1>0,α5k+4−5>0.Case 1: whenk > −4/5, it is easy to know that 5k + 4 > 0 and k + 1 > 0, then expression (49) can be simplified to(50)α>1k+1,α>55k+4.From 5k + 4 > 0 and k + 1 > 0, it is clear that 1/k+1−5/5k+4<0 and 0<1/k+1<5/5k+4,Therefore, from expression (49), it can be known that, when k>−4/5, α>5/5k+4Case 2: when−1<k<−4/5, it is easy to know that 5k + 4 < 0 and k + 1 > 0.Expression (49) can be simplified to(51)α>1k+1,α<55k+4.Owing to 5k + 4 < 0 and k + 1 > 0, it is clear that 5/5k+4<0 and 1/k+1>0. Therefore, when −1<k<−4/5, α does not exist.Case 3: whenk < −1, it is easy to know that 5k + 4 < 0 and k + 1 < 0.Expression (49) can be simplified to(52)α<1k+1,α<55k+4.Owing to 5k + 4 < 0 and k + 1 < 0, it is clear that 1/k+1−5/5k+4<0. Thus we can obtain that 5/5k+4<1/k+1<0. Therefore, when k<−1, α does not exist.In conclusion, we can obtain following several cases:Case 1:k>−4/5 and α>5/5k+4Case 2:−1<k<−4/5 and α does not existCase 3:k<−1 and α does not exist(3)
An unstable equilibrium point exists in realization 1(3.1) Assuming characteristic roots are all real and at least one is a positiveAs long as one characteristic root is greater than 0, the equilibrium point of the system is unstable.Owing toαk − 1 + α > 0, λ1 and λ2 are always less than 0 and only λ3 or λ4 is greater than 0. According to Vieta theorem, λ3 + λ4 = −b and λ3λ4 = c; it is known that c > 0, and owing to αk − 1 + α > 0, from equation (38), if 5αk − 5 + 4α < 0, we can obtain −b > 0, then λ3,4 > 0.In this case, the number in the left absolute value sign of expression (40) is less than zero, right absolute value sign of equation (40) is greater than zero, the absolute value signs are removed, and expression (40) can be transformed as follows:(53)5−4α−5αk>26αk−1+α.From expression (53), we can obtain(54)αk26+5+26+4<5+26.In summary,λ1 and λ2 are less than 0 and λ3 and λ4 are greater than 0. The system equilibrium point is unstable, and it is the saddle point of index 2, when the following inequalities are satisfied:(55)αk+α−1>0,5αk+4α−5<0,αk26+5+26+4<5+26.Firstly, we consider the coefficient ofα on the left side of third equation in expression (55) is greater than 0, namely, k26+5+26+4>0, we can obtain k>−26−4/26+5.Then, we consider the coefficient ofα on the left side of the third equation in expression (55) is less than 0, namely, k26+5+26+4<0, we can obtain k<−26−4/26+5.It is easy to know that−1<−26−4/26+5<−4/5. We discuss following several cases:Case 1: when−26−4/26+5<k<−4/5, owing to −1<−26−4/26+5<−4/5, it is easy to know that 5k+4<0, k+1>0 and k26+5+26+4>0. Expression (55) is simplified as(56)α>1k+1,α>55k+4,α<5+26k26+5+26+4.From5k+4<0, k+1>0, k26+5+26+4>0, it is clear that 1/k+1−5+26/k26+5+26+4<0 and 5/5k+4−1/k+1<0. So we have 0<1/k+1<5+26/k26+5+26+4 and 5/5k+4<0.Therefore, when−26−4/26+5<k<−4/5, it can be obtained that 1/k+1<α<26+5/k26+5+26+4.Case 2: when−4/5<k, owing to −1<−26−4/26+5<−4/5, it is easy to know that 5k + 4 > 0, k + 1 > 0 and k26+5+26+4>0. Expression (55) can be simplified to(57)α>1k+1,α<55k+4,α<5+26k26+5+26+4.From5k+4>0, k+1>0, k26+5+26+4>0, it is clear that 5/5k+4−5+26/k26+5+26+4>0 and 5+26/k26+5+26+4−1/k+1>0. So we have 0<1/k+1<5+26/k26+5+26+4<5/5k+4. Therefore, when −4/5<k, we have 1/k+1<α<26+5/k26+5+26+4.Case 3: when−1<k<−26−4/26+5, owing to −1<−26−4/26+5<−4/5, it is easy to know that 5k+4<0, k+1>0, and k26+5+26+4<0.Expression (55) can be simplified to(58)α>1k+1,α>55k+4,α>5+26k26+5+26+4.From5k+4<0, k+1>0, k26+5+26+4<0, it is easy to know that 5+26/k26+5+26+4−5/5k+4<0, so we have 5+26/k26+5+26+4<5/5k+4<0 and 0<1/k+1Therefore, when−1<k<−26−4/26+5, we can obtain that 1/k+1<α.Case 4: whenk<−1, owing to −1<−26−4/26+5<−4/5, it is easy to know that 5k+4<0, k+1<0, and k26+5+26+4<0. Equation (55) can be simplified to(59)α<1k+1,α>55k+4,α>5+26k26+5+26+4.From5k+4<0, k+1<0 and k26+5+26+4<0, it is clear that 5+26/k26+5+26+4−5/5k+4<0 and 1/k+1−5+26/k26+5+26+4<0, so we have 1/k+1<5+26/k26+5+26+4<5/5k+4<0. Therefore, when k<−1, α does not exist.From what has been discussed above, we can obtain following relation betweenk and α:Case 1:−26−4/26+5<k<−4/5 and 1/k+1<α<26+5/k26+5+26+4Case 2:−4/5<k and 1/k+1<α<26+5/k26+5+26+4Case 3:−1<k<−26−4/26+5 and 1/k+1<αCase 4:k<−1 and α does not exist.(3.2) Assuming two negative real roots and two conjugate complex rootsIn this case, the four characteristic roots behave asλ1 = γ1 = λ2 = γ2, λ3,4 = σ3,4 ± jω3,4, where λ1 and λ2 are two negative real roots and λ3 and λ4 are two conjugate complex roots.Owing toλ1 and λ2 are always less than 0, when σ3,4 is greater than 0, the equilibrium point of the system is unstable and it is the saddle-focus point of index 2. From (37), we know σ3,4 = −b/2. To make σ3,4 > 0, we need −b > 0, and from equation (38), we can obtain 5αk + 4α − 5 < 0.In this case, the absolute value sign in Expression (41) can be removed; therefore, the following expression can be obtained:(60)5−5αk−4α<26αk−1+α.From expression (60), we can obtain that(61)αk5+26+4+26>26+5.In summary,λ1 and λ2 are less than 0, σ3 and σ4 are greater than 0, the system equilibrium point is unstable, and it is the saddle-focus point of index 2, when the following inequalities are satisfied:(62)αk+1−1>0,α5k+4−5<0,αk5+26+4+26>26+5.According to above discussion, we can obtain the following several cases:Case 1:−4−26/5+26<k<−4/5 and α>5+26/k5+26+4+26Case 2:−4/5<k and 5+26/k5+26+4+26<α<5/5k+4Case 3:−1<k<−4−26/5+26 and α does not existCase 4:k<−1 and α does not existIfk and α satisfy the above relation, the system equilibrium point is the saddle-focus point of index 2, which is unstable.(3) Analysis of Equilibrium Points and Characteristic Roots for Realization 2. The steady state solution (equilibrium points) of realization 2, equations (19)∼(22), is denoted as ΔT01,ΔT02,ΔT03,ΔT04. In order to solve the equilibrium points, we can make that the right sides of equations (19)∼(22) equal to zero; the following equations hold:(63)−nΔT1−αΔT2ΔT1−ΔT2+E1=0,(64)−nΔT1−αΔT2ΔT2−ΔT4+E2=0,(65)−ξnΔT1−αΔT2ΔT3−ΔT1=0,(66)−ξnΔT1−αΔT2ΔT4=0.By solving equations (63)∼(66), we can obtain the equilibrium points as follows:(67)ΔT02=E2n2/3E2+E1−E1αk1/3=kE12/3n2/3k+1−αk1/3,(68)ΔT01=ΔT03=1+kkΔT02=1+kE1n2/3E2+E1−E1αk1/3=1+kE12/3n2/3k+1−αk1/3,(69)ΔT04=0.According to the characteristic determinant equal to 0, the solution of the differential equation can be discussed. The characteristic determinant equation is that(70)JQ−λE=∂f1∂ΔT1−λ∂f1∂ΔT2∂f1∂ΔT3∂f1∂ΔT4∂f2∂ΔT1∂f2∂ΔT2−λ∂f2∂ΔT3∂f2∂ΔT4∂f3∂ΔT1∂f3∂ΔT2∂f3∂ΔT3−λ∂f3∂ΔT4∂f4∂ΔT1∂f4∂ΔT2∂f4∂ΔT3∂f4∂ΔT4−λ=0.Combining equations (19)∼(22), we can obtain the following expression as follows:(71)−nΔT01−ΔT022A−nA−λnαΔT01−ΔT022A+nA00−nΔT02−ΔT042AnαΔT02−ΔT042A−nA−λ0nA−ξnΔT03−ΔT012A+ξnAξnαΔT03−ΔT012A−ξnA−λ0−ξnΔT042AξnαΔT042A0−ξnA−λ=0,(72)A=ΔT01−αΔT02=k+1−αk1/3E11/3n1/3,(73)B=ΔT01−ΔT02=E12/3n2/3k+1−αk1/3,(74)C=ΔT02−ΔT04=kE12/3n2/3k+1−αk1/3,Combining equations (71)∼(74), we can obtain characteristic roots as follows:(75)λ1=λ2=−ξnA=−ξnΔT01−αΔT02=−ξn2/3E11/3k+1−αk1/3.To ensure that a real solution exists for equations (67) and (68), we have(76)k+1−αk>0.Two other characteristic roots can be obtained according to the quadratic formula:(77)λ3,4=−b±b2−4ac2a=−nB−αC/2A−2nA±nB−αC/2A+2nA2−4n2B+C/2+A2−αC/22,where(78)a=1,b=nB−αC2A+2nA=n2/3E11/34k+5−5αk2k+1−αk2/3,c=n2B+C2+A2−αC2=3n4/3E12/3k+1−αk2/32,Δ=b2−4ac=n4/3E12/34k+5−5αk24k+1−αk4/3−6n4/3E12/3k+1−αk2/3.Therefore, it is easy to know thatλ1,2 are always a real number, λ1 = γ1 = λ2 = γ2. If the discriminant Δ is greater than 0, both characteristic roots are real number:λ3 = γ3, λ4 = γ4. From equations (77) and (78), we can know that when Δ > 0, the following expression holds:(79)n4/3E12/34k+5−5αk2>24n4/3E12/3k+1−αk2.It is clear that(80)4k+5−5αk>26k+1−αk.When the discriminant Δ is less than 0, conjugate complex roots exist, namely,λ3,4 = σ3,4 ± jω3,4. From equations (77) and (78), we can know that when Δ < 0, the following expression holds:(81)4k+5−5αk<26k+1−αk.(4) Stability Analysis for Realization 2 (k Is Fixed and α Is Control Parameter). (1)
No equilibrium point exists in realization 2If no equilibrium point exists in realization 2, from equations (67)∼(68), it is known that the following expression is satisfied:(82)k+1−αk<0.Therefore, from equation (82), when k < −1, it is clear that 0<α<k+1/k and when −1 < k < 0, we know that α does not exist, or when 0 < k, we know that α>k+1/k.(2)
A stable equilibrium point exists in realization 2(2.1) Assuming there are all negative real characteristic rootsIn this case,λ1, λ2, λ3, and λ4 are all less than 0, and the system has a stable solution. We know that, owing to k + 1 − αk > 0, from equation (75), it is clear that λ1 and λ2 are always less than 0. To make λ3,4 < 0, according to Vieta theorem, we know that λ3 + λ4 = −b and λ3λ4 = c. From equation (78), it is known that always c > 0, and owing to k + 1 − αk > 0, from equation (78), if 4k + 5 − 5αk > 0, we have −b < 0, then λ3,4 < 0. In this case, the absolute value sign in equation (80) can be removed; therefore, the following expression can be obtained:(83)4k+5−5αk>26k+1−αk.From expression (83), we can obtain the following expression:(84)αk5−26<k4−26+5−26.In summary, if there are all negative real characteristic roots, the system is stable when the following inequalities are satisfied:(85)αk<k+1,5αk<4k+5,αk5−26<k4−26+5−26.(2.2) Assuming two real roots and two conjugate complex rootsIn this case, the four characteristic roots behave asλ1 = γ1 = λ2 = γ2 and λ3,4 = σ3,4 ± jω3,4, where λ1 and λ2 are two negative real roots and λ3 and λ4 are two conjugate complex roots. When λ1, λ2, σ3, and σ4 are all less than 0, the system has a stable solution.From equation (77), we know σ3,4 = −b/2. Owing to λ1,2 < 0, to make σ3,4 < 0, we have −b < 0, from equation (78), we can obtain 4k + 5 − 5αk > 0.In this case, the absolute value sign in expression (81) can be removed; therefore, the following expression (86) can be obtained:(86)4k+5−5αk<26k+1−αk.From expression (86), we can obtain that(87)αk5−26>k4−.26+5−26.In summary, the system is stable when the following inequalities are satisfied:(88)αk<k+1,5αk<4k+5,αk5−26>k4−.26+5−26.Combining expressions (85) and (88), a stable equilibrium point exists in realization 2, and the following equations are satisfied:(89)αk<k+1,5αk<4k+5.Case 1: whenk > 0, expression (89) can be simplified to(90)α<k+1k,α<4k+55k.Whenk > 0, it is easy to know that 4k+5/5k>0 and k+1/k>0, and it is clear that k+1/k−4k+5/5k>0 and 0<4k+5/5k<k+1/k. Therefore, from expression (90), when k > 0, we know that 0<α<4k+5/5k.Case 2: whenk < 0, expression (89) can be simplified to(91)α>k+1k,α>4k+55k.It is clear thatk+1/k−4k+5/5k>0 and 4k+5/5k<k+1/k. Owing to α > 0, when −1 < k < 0, it clear that k+1/k<0. Therefore, from expression (91), when −1 < k < 0, we know that α > 0.Whenk < −1, it clear that k+1/k>0. Therefore, from expression (91), when k < −1, we know that α>k+1/k.In conclusion, we can obtain following several cases:Case 1:k>0 and 0<α<4k+5/5kCase 2:−1<k<0 and α>0Case 3:k<−1 and α>k+1/k(3)
An unstable equilibrium point exists in realization 1(3.1)Assuming characteristic roots are all real and at least one is a positiveAs long as one characteristic root is greater than 0, the equilibrium point of the system is unstable. Owing tok + 1 − αk > 0, λ1 and λ2 are always less than 0 and only λ3 or λ4 are greater than 0. According to Vieta theorem, it is known that c > 0, and owing to k + 1 − αk > 0, from equation (78), if 4k + 5 − 5αk < 0, we have −b > 0, then λ3,4 > 0.In this case, the number in the left absolute value sign of expression (80) is less than zero, right absolute value sign of expression (80) is greater than zero, and the absolute value signs are removed; expression (92) can be derived as follows:(92)5αk−5−4k>26k+1−αk.From expression (92), we can obtain(93)αk5+26>k4+26+5+26.In summary,λ1 and λ2 are less than 0 and λ3 and λ4 greater than 0. The system equilibrium point is the saddle point of index 2, when the following inequalities are satisfied:(94)αk<k+1,5αk>4k+5,αk5+26>k4+26+5+26.It is easy to know that for anyk, k4+26+5+26/k5+26−k+1/k<0 and 4k+5/5k−k4+26+5+26/k5+26<0, so we have 4k+5/5k<k4+26+5+26/k5+26<k+1/k.We discuss following several cases:Case 1: whenk > 0, expression (94) is simplified as(95)α<k+1k,α>4k+55k,α>k4+26+5+26k5+26.Therefore, whenk > 0, we have k4+2.6+5+26/k5+26<α<1+k/k.Case 2: whenk < 0, expression (94) is simplified as(96)α>k+1k,α<4k+55k,α<k4+26+5+26k5+26.Therefore, whenk < 0, α does not exist.In conclusion, we can obtain following several cases:Case 1:k > 0 and k4+26+5+26/k5+26<α<1+k/kCase 2:k < 0 and α does not existIfk and α satisfy above relation, the system equilibrium point is the saddle point of index 2, which is unstable.(3.2) Assuming two negative real roots and two conjugate complex rootsIn this case, the four characteristic roots behave asλ1 = γ1 = λ2 = γ2 and λ3,4 = σ3,4 ± jω3,4, where λ1 and λ2 are two negative real roots and λ3 and λ4 are two conjugate complex roots.Owing toλ1 and λ2 are always less than 0, when σ3 and σ4 are greater than 0, the equilibrium point of the system is unstable and it is the saddle-focus point of index 2. From (77), we know σ3,4 = −b/2. To make σ3,4 > 0, we have −b > 0. From equation (78), it is known that 4α − 5 + 5αk < 0.In this case, the absolute value sign in expression (81) can be removed; therefore, the following expression (97) can be obtained:(97)5αk−5−4k<26k+1−αk.From expression (97), we can obtain that(98)αk5+26<k4+26+5+26.In summary,λ1 and λ2 are less than 0 and σ3 and σ4 are greater than 0; the system equilibrium point is unstable and it is the saddle-focus point of index 2 when the following inequalities are satisfied:(99)αk<k+1,5αk>4k+5,αk5+26<k4+26+5+26.By referring the classification results of expression (94), we can obtain the following several cases:Case 1:k > 0 and 4k+5/5k<α<k4+2.6+5+26/k5+26Case 2:k < 0 and α does not existIfk and α satisfy above relation, the system equilibrium point is the saddle-focus point of index 2, which is unstable.In brief, for the scenario of the heat source in the left and right sides both are positive (k > 0), from the above discussion, we can form a criterion shown in Table 1 to determine the stability of the system shown in Figure 1 when K is fixed, α is the control parameter, and k > 0. According to the values of α in Table 1, we can know whether there is an equilibrium point and the equilibrium point is stable or not.Table 1
Criterion for scenario 1 ((k is fixed, α is control parameter, and k > 0).
kαRealization 1Realization 20,+∞0,1/k+1NoStable1/k+1,5+26/k5+26+4+26Saddle point of index 2Stable5+26/k5+26+4+26,5/5k+4Saddle-focus point of index 2Stable5/5k+4,4k+5/5kStableStable4k+5/5k,k4+26+5+26/k5+26StableSaddle-focus point of index 2k4+26+5+26/k5+26,k+1/kStableSaddle point of index 2k+1/k,+∞StableNoWe compared Table1 with the part k > 0 of Table 1 in [35]. From these two tables, we can know that the results (whether the equilibrium point is stable or not) are roughly the same as those of [35]. However, our results are more specific and more detailed when the equilibrium points are unstable. In the unstable equilibrium point interval corresponding [35], we obtain two different subintervals corresponding to the saddle point of index 2 and the saddle focal equilibrium point of index 2, respectively. This is because our model has four characteristic roots and only two in [35]. The brief description is as follows. The equilibrium point is unstable in realization 1, when α is in the interval of 1/1+k,5/5k+4, and this interval can be divided into two parts, namely, α∈1/k+1,5+26/k5+26+4+26, where the unstable equilibrium point is the saddle point of index 2 and α∈5+26/k5+26+4+26,5/5k+4, where the unstable equilibrium point is the saddle-focus point of index 2. The equilibrium point is unstable in realization 2, when α is in the interval of 4k+5/5k,k+1/k, and we divide this interval into two parts, namely, α∈4k+5/5k,k4+26+5+26/k5+26, where the unstable equilibrium point is the saddle point of index 2 and α∈k4+26+5+26/k5+26,k+1/k, where the unstable equilibrium point the saddle-focus point of index 2.For the scenario of the heat source in the left is positive and the right is negative (k < 0), from the above discussion, we can form a criterion as shown in Table 2 to determine the stability of the system shown in Figure 1 when K is fixed, α is the control parameter, and k < 0. According to the values of α in Table 2, we can know whether there is an equilibrium point and the equilibrium point is stable or not.Table 2
Criterion for scenario 1 (k is fixed, α is control parameter and k < 0).
kαRealization 1Realization 2−4/5,00,1/k+1NoStable1/k+1,26+5/k26+5+26+4Saddle point of index 2Stable26+5/k26+5+26+4,5/5k+4Saddle-focus point of index 2Stable5/5k+4,+∞StableStable−26−4/26+5,−4/50,1/k+1NoStable1/k+1,5+26/k5+26+4+26Saddle point of index 2Stable5+26/k5+26+4+26,+∞Saddle-focus point of index 2Stable−1,−26−4/26+50,1/k+1NoStable1/k+1,+∞Saddle point of index 2Stable−∞,−10,1+k/kNoNo1+k/k,+∞NoStableWe compared Table2 with the part k < 0 of Table 1 in [35]. From these two tables, we can know the results (whether the equilibrium point is stable or not) are roughly the same as those in [35]. However, similarly, our results are more specific and more detailed when the equilibrium points are unstable. In the unstable equilibrium interval corresponding to literature [35], we obtain two different subintervals corresponding to the saddle point of index 2 and the saddle focal equilibrium point of index 2, respectively. This is because our model has four characteristic roots and only two in [35].


#### 2.3.2. Stability Analysis for Scenario 2 (α Is Fixed and k Is Control Parameter)

(1) Stability Analysis for Realization 1 (α Is Fixed and k Is Control Parameter). In this scenario, the characteristic equation is the same as that of scenario 1.(1)
No equilibrium point exists in realization 1In point (1) of Section 2.3.1, we already know that no equilibrium point exists in realization 1, when the following inequalities (equation (42) in Section 2.3.1) are satisfied:(100)αk−1+α<0.Therefore, from expression (98), when α > 0, it is clear that k<1−α/α.(2)
A stable equilibrium point exists in realization 1In point (2) of Section 2.3.1, we already know that the system was stable, when the following inequalities (expression (49) in Section 2.3.1) are satisfied:(101)αk+1−1>0,α5k+4−5>0.Taking out the coefficientk of expression (101), we can obtain expression (102):(102)kα>1−α,5kα>5−4α.Owing toα > 0, expression (102) are simplified as(103)k>1−αα,k>5−4α5α.Owing toα > 0, it is clear that 5−4α/5α−1−α/α>0 and 5−4α/5α>1−α/α; therefore, from expression (101), when α > 0, we know that 5−4α/5α<k.(3)
An unstable equilibrium point exists in realization 1(3.1) Assuming characteristic roots are all real and at least one is a positiveIn point (3.1) of Section 2.3.1, we already know that the equilibrium point of the system is unstable and it is the saddle point of index 2, when the following inequalities (expression (55) in Section 2.3.1) are satisfied:(104)αk+α−1>0,5αk+4α−5<0,αk26+5+26+4<5+26.Taking out the coefficientk of expression (104), we can obtain the following expression:(105)kα>1−α,5kα<5−4α,5+26kα<5+26−α4+26.Owing toα > 0, expression (105) is simplified:(106)k>1−αα,k<5−4α5α,k<5+26−α4+265+26α.It is easy to know that, for anyα > 0, 1−α/α−5+26−α4+26/α5+26<0 and 5+26−α4+26/α5+26−5−4α/5α<0, so we have 1−α/α<5+26−α4+26/α5+26<5−4α/5α. Therefore, from expression (106), it can be known that, when α > 0, we know that 1−α/α<k<5+26−α4+26/α5+26.Ifα and k satisfy above relation, the system equilibrium point is the saddle point of index 2, which is unstable.(3.2) Assuming two negative real roots and two conjugate complex rootsIn point (3.2) of Section 2.3.1, we already know that the equilibrium point of the system is unstable and it is the saddle-focus point of index 2, when the following inequalities (expression (62) in Section 2.3.1) are satisfied:(107)αk+1−1>0,α5k+4−5<0,αk5+26+4+26>26+5.Taking out the coefficientk of expression (107), we can obtain the following expression:(108)kα>1−α,5kα<5−4α,5+26kα>5+26−α4+26.Owing toα > 0, expression (108) is simplified:(109)k>1−αα,k<5−4α5α,k>5+26−α4+265+26α.By referring the solution procedure of expression (104), we can obtain the following result. When α > 0, we know that 5+26−α4+26/α5+26<k<5−4α/5α.Ifα and k satisfy the above relation, the system equilibrium point is the saddle-focus point of index 2, which is unstable.(2) Stability Analysis for Realization 2 (α Is Fixed, k Is Control Parameter). (1)
No equilibrium point exists in realization 2In point (1) of (4) of Section 2.3.1, we already know that no equilibrium point exists in realization 2, when the following inequalities (expression (82) in Section 2.3.1) are satisfied:(110)k+1−αk<0.Therefore, from expression (110), when 0 < α < 1, it is clear that k<1/α−1 and when α > 1, we know that k>1/α−1.(2)
A stable equilibrium point exists in realization 2In point (2) of (4) of Section 2.3.1, we already know that the system was stable, when the following inequalities (expression (89) in (4) of Section 2.3.1) are satisfied:(111)αk<k+1,5αk<4k+5.Taking out the coefficientk of expression (111), we can obtain the following expression:(112)kα−1<1,k5α−4<5.Case 1: whenα > 1, it is easy to know that α − 1 > 0 and 5α − 4 > 0, and then expression (112) can be simplified to(113)k<1α−1,k<55α−4.Fromα − 1 > 0 and 5α − 4 > 0, it is clear that 5/5α−4−1/α−1<0 and 0<5/5α−4<1/α−1. Therefore, from expression (113), when α > 1, we know that k<−5/4−5α.Case 2: when4/5<α<1, it is easy to know that α − 1 < 0 and 5α − 4 > 0; then expression (112) can be simplified to(114)k>1α−1,k<55α−4.Fromα − 1 < 0 and 5α − 4 > 0, it is clear that 5/5α−4−1/α−1>0, 1/α−1<0 and 0<5/5α−4. Therefore, from expression (112), when 4/5<α<1, we know that 1/α−1<k<5/5α−4.Case 3: when0<α<4/5, it is easy to know that α − 1 < 0 and 5α − 4 < 0, then expression (112) can be simplified to(115)k>1α−1,k>55α−4.Fromα − 1 < 0 and 5α − 4 < 0, it is clear that 5/5α−4−1/α−1<0 and 5/5α−4<1/α−1<0, Therefore, from expression (115), when 0<α<4/5, we know that 1/α−1<k. In conclusion, we can obtain following several cases:Case 1:α > 1 and k<−5/4−5αCase 2:4/5<α<1 and 1/α−1<k<5/5α−4Case 3:0<α<4/5 and 1/α−1<k(3)
An unstable equilibrium point exists in realization 2(3.1) Assuming characteristic roots are all real and at least one is a positiveIn point (3.1) of Section 2.3.1, we already know that the equilibrium point of the system is unstable and it is the saddle point of index 2, when the following inequalities (expression (94) in Section 2.3.1) are satisfied:(116)αk<k+1,5αk>4k+5,αk5+26>k4+26+5+26.Taking out the coefficientk of expression (116), we can obtain the following expression:(117)k1−α+1>0,k4−5α+5<0,kα5+26+−26−4>5+26.Firstly, we consider the coefficient ofk on the left side of third inequality in expression (117) is greater than 0, namely, α5+26+−26−4>0, and we can obtain α>4+26/5+26. Then, we consider the coefficient of α on the left side of third inequality in (117) is less than 0, namely, α5+26+−26−4<0, and we can obtain α<4+26/5+26. It is easy to know that 0<4/5<4+26/5+26<1. We discuss following several cases:Case 1: when4+26/5+26<α<1, owing to 0<4/5<4+26/5+26<1, it is easy to know that 4 − 5α < 0, 1 − α > 0, and α5+26+−26−4>0; equation (117) is simplified as(118)k>−11−α,k>−54−5α,k>5+26α5+26−4−26.From 4 − 5α < 0, 1 − α > 0, and α5+26+−26−4>0, it is clear that −1/1−α−−5/4−5α<0, −5/4−5α−5+26/α5+26−4−26<0, and −1/1−α<−5/4−5α<5+26/α5+26−4−26. Therefore, when 4+26/5+26<α<1, we know that k>5+26/α5+26−4−26.Case 2: whenα > 1, owing to 0<4/5<4+26/5+26<1, it is easy to know that 4 − 5α < 0, 1 − α < 0, and α5+26+−26−4>0; expression (117) is simplified as(119)k<−11−α,k>−54−5α,k>5+26α5+26−4−26.From 4 − 5α < 0, 1 − α < 0, and α5+26+−26−4>0, it is clear that −5/4−5α−5+26/α5+26−4−26<0, 5+26/α5+26−4−26−−1/1−α<0, and −5/4−5α<5+26/α5+26−4−26<−1/1−α. Therefore, when α > 1, we know that 5+26/α5+26−4−26<k<−1/1−α.Case 3: when4/5<α<4+26/5+26, owing to 0<4/5<4+26/5+26<1, it is easy to know that 4 − 5α < 0, 1 − α > 0, and α5+26+−26−4<0; expression (117) are simplified as(120)k>−11−α,k>−54−5α,k<5+26α5+26−4−26.From 4 − 5α < 0, 1 − α > 0, and α5+26+−26−4<0, it is clear that 5+26/α5+26−4−26−−1/1−α<0, −1/1−α−−5/4−5α<00<−5/4−5α, and 5+26/α5+26−4−26<−1/1−α<0. Therefore, when 4/5<α<4+26/5+26, we know that k does not exist.Case 4: when0<α<4/5, owing to 0<4/5<4+26/5+26<1, it is easy to know that 4 − 5α > 0, 1 − α > 0, and α5+26+−26−4<0; expression (117) is simplified as(121)k>−11−α,k<−54−5α,k<5+26α5+26−4−26.From 4 − 5α > 0, 1 − α > 0, and α5+26+−26−4<0, it is clear that −5/4−5α−5+26/α5+26−4−26<0, 5+26/α5+26−4−26−−1/1−α<0, and −5/4−5α<5+26/α5+26−4−26<−1/1−α<0. Therefore, when 0<α<4/5, we know that k does not exist.From what has been discussed above, we can obtain following relations betweenα and k.Case 1:4+26/5+26<α<1 and k>5+26/α5+26−4−26Case 2:α > 1 and 5+26/α5+26−4−26<k<−1/1−αCase 3:0<α<4/5 and k does not exist.Case 4:4/5<α<4+26/5+26 and k does not exist.Ifα and k satisfy above relation, the system equilibrium point is the saddle point of index 2, which is unstable.(3.2) Assuming two negative real roots and two conjugate complex rootsIn point (3.2) of Section 2.3.1, we already know that the equilibrium point of the system is unstable and it is the saddle-focus point of index 2, when the following inequalities (expression (96) in Section 2.3.1) are satisfied:(122)αk<k+1,5αk>4k+5,αk5+26<k4+26+5+26.Taking out the coefficientk of expression (122), we can obtain the following expression:(123)k1−α+1>0,k4−5α+5<0,kα5+26+−26−4<5+26.By referring the classification results of expression (117), we can obtain the following several cases:Case 1:4+26/5+26<α<1 and −5/4−5α<k<5+26/α5+26−4−26Case 2:α > 1 and −5/4−5α<k<5+26/α5+26−4−26Case 3:0<α<4/5 and k does not existCase 4:4/5<α<4+26/5+26 and k>−5/4−5αIfk and α satisfy above relation, the system equilibrium point is the saddle-focus point of index 2, which is unstable.In brief, for the scenario 2,α can be discussed in the following sections: 0<α<4/5, 4/5<α<4+26/5+26, and 4+26/5+26<α<1 and 1<α.From the above discussion, we can form a criterion shown in Table3 to determine the stability of the system shown in Figure 1; when α is fixed, k is the control parameter. According to the values of k in Table 3, we can know whether there is an equilibrium point and the equilibrium point is stable or not.Table 3
Criterion for scenario 2 (α is fixed and k is control parameter).
αkRealization 1Realization 20,4/5−∞,−1/1−αNoNo−1/1−α,5−26/α5−26+26−4NoStable5−26/α5−26+26−4,1−α/αNoStable1−α/α,5+26−α4+26/α5+26Saddle point of index 2Stable5+26−α4+26/α5+26,5−4α/5αSaddle-focus point of index 2Stable5−4α/5α,+∞StableStable4/5,4+26/5+26−∞,−1/1−αNoNo−1/1−α,1−α/αNoStable1−α/α,5+26−α4+26/α5+26Saddle point of index 2Stable5+26−α4+26/α5+26,5−4α/5αSaddle-focus point of index 2Stable5−4α/5α,4−5α/−5StableStable−5/4−5α,+∞StableSaddle-focus point of index 24+26/5+26,1−∞,−1/1−αNoNo−1/1−α,1−α/αNoStable1−α/α,5+26−α4+26/α5+26Saddle point of index 2Stable5+26−α4+26/α5+26,5−4α/5αSaddle-focus point of index 2Stable5−4α/5α,−5/4−5αStableStable−5/4−5α,5+26/α5+26−4−26StableSaddle-focus point of index 25+26/α5+26−4−26,+∞StableSaddle point of index 21,+∞−∞,1−α/αNoStable1−α/α,5+26−α4+26/α5+26Saddle point of index 2Stable5+26−α4+26/α5+26,5−4α/5αSaddle-focus point of index 2Stable5−4α/5α,−5/4−5αStableStable−5/4−5α,5+26/α5+26−4−26StableSaddle-focus point of index 25+26/α5+26−4−26,−1/1−αStableSaddle point of index 2−1/1−α,+∞StableNoWe compared Table3 with Table 2 in [35]. From these two tables, we can know the results (whether the equilibrium point is stable or not) are roughly the same as those in [35]. However, our results are more specific and more detailed when the equilibrium points are unstable. In the unstable equilibrium point interval corresponding to [35], we can obtain two different subintervals corresponding to the saddle point of index 2 and the saddle focal equilibrium point of index 2, respectively. This is because our model has four characteristic roots and only two in [35].


#### 2.3.3. Stability Analysis for Scenario 3 (One Heat Sink and One Heat Source)

For realization 1, the heat at the bottom releases to the right side. Therefore, we can obtain following equations:(124)f1=dΔT1dτ=−nαΔT2−ΔT1ΔT1−ΔT3,(125)f2=dΔT2dτ=−nαΔT2−ΔT1ΔT2−ΔT1+E1,(126)f3=dΔT3dτ=−ξnαΔT2−ΔT1ΔT3,(127)f4=dΔT4dτ=−ξnαΔT2−ΔT1ΔT4−ΔT2.For realization 2, the heat at the bottom releases to zone 1. Therefore,(128)f1=dΔT1dτ=−nΔT1−αΔT2ΔT1−ΔT2+E1,(129)f2=dΔT2dτ=−nΔT1−αΔT2ΔT2−ΔT4,(130)f3=dΔT3dτ=−ξnΔT1−αΔT2ΔT3−ΔT1,(131)f4=dΔT4dτ=−ξnΔT1−αΔT2ΔT4.(1) Stability Analysis for Realization 1. The steady state solution (equilibrium points) of realization 1 with equations (124)∼(127) is denoted as ΔT01,ΔT02,ΔT03,ΔT04; in order to solve the equilibrium points, we can make that the right sides of equations (124)∼(127) equal to zero; the following equations hold:(132)−nαΔT2−ΔT1ΔT1−ΔT3=0,(133)−nαΔT2−ΔT1ΔT2−ΔT1+E1=0,(134)−ξnαΔT2−ΔT1ΔT3=0,(135)−ξnαΔT2−ΔT1ΔT4−ΔT2=0.By solving equations (132)∼(135), we can obtain the value of equilibrium points as follows:(136)ΔT01=ΔT03=0,(137)ΔT02=ΔT04=E12/3n2/3α1/3.According to the characteristic determinant equal to 0, the solution of the differential equation can be discussed. The characteristic determinant equation is that(138)JQ−λE=∂f1∂ΔT1−λ∂f1∂ΔT2∂f1∂ΔT3∂f1∂ΔT4∂f2∂ΔT1∂f2∂ΔT2−λ∂f2∂ΔT3∂f2∂ΔT4∂f3∂ΔT1∂f3∂ΔT2∂f3∂ΔT3−λ∂f3∂ΔT4∂f4∂ΔT1∂f4∂ΔT2∂f4∂ΔT3∂f4∂ΔT4−λ=0.Combining equations (124)∼(127), we can obtain the following expression:(139)nΔT01−ΔT032A−nA−λ−nαΔT01−ΔT032AnA0nΔT02−ΔT012A+nA−nαΔT02−ΔT012A−nA−λ00ξnΔT032A−ξnαΔT032A−ξnA−λ0ξnΔT04−ΔT022A−ξnαΔT04−ΔT022A+ξnA0−ξnA−λ=0,(140)A=αΔT02−ΔT01=α1/3E11/3n1/3,(141)B=ΔT01−ΔT03=0,(142)C=ΔT02−ΔT01=E12/3n2/3α1/3.Combining equations (139)∼(142), we can obtain characteristic root as follows:(143)λ1=λ2=−ξnA=−ξnαΔT02−ΔT01=−ξn2/3E11/3α1/3.Two other eigenvalues can be obtained according to the quadratic formula:(144)λ3,4=−b±b2−4ac2a=−nαC/2A−2nA±nαC/2A+2nA2−4n2αC/2+A22,where(145)a=1,b=nαC2A+2nA=3n2/3E11/3α1/32,c=n2αC2+A2=3E12/3α2/3n4/32,Δ=b2−4ac=9n4/3E12/3α2/34−6E12/3α2/3n4/3=−15n4/3E12/3α2/34.It is easy to know thatλ1,2 are always real number λ1 = γ1 = λ2 = γ2. Owing to α > 0, E1 > 0, and n > 0, from equation (143), it is clear that λ1,2 are always less than 0. From equation (145), we know that Δ < 0. Since the discriminant Δ < 0, we know that λ3 and λ4 are always conjugate complex numbers λ3,4 = σ3,4 ± jω3,4. Hence, the stability related with λ3,4 is determined by the real part σ3,4. Owing to α > 0, E1 > 0, n > 0, from equations (144) and (145), we know that σ3,4 = −b < 0. As a result, λ1, λ2, and σ3,4 are negative. It is clear that the equilibrium point is stable in realization 1.(2) Stability Analysis for Realization 2. The steady state solution (equilibrium points) of realization 2 with equations (127)∼(130) is denoted as ΔT01,ΔT02,ΔT03,ΔT04. In order to solve the equilibrium points, we can make that the right sides of equations (127)∼(130) equal to zero; the following equations hold:(146)−nΔT1−αΔT2ΔT1−ΔT2+E1=0,(147)−nΔT1−αΔT2ΔT2−ΔT4=0,(148)−ξnΔT1−αΔT2ΔT3−ΔT1=0,(149)−ξnΔT1−αΔT2ΔT4=0.By solving equations (146)∼(149), we can obtain the value of equilibrium point as follows:(150)ΔT01=ΔT03=E12/3n2/3,(151)ΔT02=ΔT04=0.According to the characteristic determinant equal to 0, the solution of the differential equation can be discussed. The characteristic determinant equation is that(152)JQ−λE=∂f1∂ΔT1−λ∂f1∂ΔT2∂f1∂ΔT3∂f1∂ΔT4∂f2∂ΔT1∂f2∂ΔT2−λ∂f2∂ΔT3∂f2∂ΔT4∂f3∂ΔT1∂f3∂ΔT2∂f3∂ΔT3−λ∂f3∂ΔT4∂f4∂ΔT1∂f4∂ΔT2∂f4∂ΔT3∂f4∂ΔT4−λ=0.Combining equations (128)∼(131), we can obtain the following expression:(153)−nΔT01−ΔT022A−nA−λnαΔT01−ΔT022A+nA00−nΔT02−ΔT042AnαΔT02−ΔT042A−nA−λ0nA−ξnΔT03−ΔT012A+ξnAξnαΔT03−ΔT012A−ξnA−λ0−ξnΔT042AξnαΔT042A0−ξnA−λ=0,(154)A=ΔT01−αΔT02=E11/3n1/3,(155)B=ΔT01−ΔT02=E12/3n2/3,(156)C=ΔT02−ΔT04=0.Combining equations (153)∼(156), we can obtain characteristic roots as follows:(157)λ1=λ2=−ξnA=−ξnΔT01−αΔT02=−ξn2/3E11/3.Two other eigenvalues can be obtained according to the quadratic formula:(158)λ3,4=−b±b2−4ac2a=−nB/2A−2nA±nB/2A+2nA2−4n2B/2+A22,where(159)a=1,b=nB2A+2nA=5n2/3E11/32,c=n2B2+A2=3n4/3E12/32,Δ=b2−4ac=25n4/3E12/34−6n4/3E12/3=n4/3E12/34.It is easy to know thatλ1,2 are always real number λ1 = γ1 = λ2 = γ2. Owing to E1 > 0 and n > 0, from equation (157), we know that λ1,2 are always less than 0; from equation (159), we know that Δ > 0, and thus λ3 and λ4 are always real numbers λ3 = γ3 and λ4 = γ4. Combining equations (158) and (159), we can obtain that λ3=−n2/3E11/3 and λ4=−3n2/3E11/3/2. Owing to E1 > 0 and n > 0, we can know that λ3 and λ4 are less than 0. In conclusion, four eigenvalues λ1, λ2, λ3, and λ4 are less than 0, and it is clear that the equilibrium point is stable in realization 2.In brief, for the scenario 3 of one heat source at the bottom of the building, because the heat can enter either the left or the right side, two steady states always exist for this scenario. The parameterα will not affect the stability and existence of solution of buoyancy ventilation of this building configuration.


## 2.1. Description of Mathematical Model

The proposed new schematics of underground structure is shown as Figure1, which contains four zones, zone 1 and zone 2 for deeply burred room and zone 3 and zone 4 for two tunnels. The heights of two tunnels are H1 and H2 respectively, the Ti, qi (i = 1∼4) denote the temperature, and mass flow rate of four zones, respectively. Heat sources E1 and E2 are located in zone 1 and zone 2. As shown in Figure 1, two realizations are considered: for realization 1, the air flow enters from zone 1 to zone 2; for realization 2, the air flow enters from zone 2 to zone 1.For realization 1, the heat gain of the internal thermal mass should be equal to the heat released by the heat sources minus the heat loss through airflows. Therefore, we can obtain the heat balance equations of the four zones shown in the following equations:(1)M1CpdT1dt=−q1CPT1−T3+E1,(2)M2CpdT2dt=−q2CPT2−T1+E2,(3)M3CpdT3dt=−q3CPT3−Ta,(4)M4CpdT4dt=−q4CPT4−T2.According to the flow loop method, the total pressure loss at ventilation vents on the flow loop is balanced by sum of the buoyancy pressure [35]. The pressure balance equation can be obtained as follows:(5)−T1−TaTaρagH1+T2−TaTaρagH2=S1q12+S2q22+S3q32+S4q42.According to the conservation of mass, we can obtain(6)q1=q2=q3=q4.Combining equations (5) and (6), mass flow rate can be obtained as follows:(7)q1=q2=q3=q4=q=−T1−Ta/TaρagH1+T2−Ta/TaρagH2S1+S2+S3+S4.For realization 2, similarly, we can obtain the heat balance equations of the four zones as follows:(8)M1CpdT1dt=−q1CPT1−T2+E1,(9)M2CpdT2dt=−q2CPT2−T4+E2,(10)M3CpdT3dt=−q3CPT3−T1,(11)M4CpdT4dt=−q4CPT4−Ta.Based on the flow loop method, the total pressure loss at ventilation vents on the flow loop is balanced by sum of the buoyancy pressure, and we can obtain the pressure balance equation as follows:(12)T1−TaTaρagH1−T2−TaTaρagH2=S1q12+S2q22+S3q32+S4q42.According to the conservation of mass, we can obtain following equations about air flow rate:(13)q1=q2=q3=q4=q.By combining equations (11) and (12), we can obtain(14)q1=q2=q3=q4=T1−Ta/TaρagH1−T2−Ta/TaρagH2S1+S2+S3+S4.


## 2.2. Transformation of Variables and Parameters in Differential Equations

Assumingk = E2/E1, ΔT1 = T1 − Ta, ΔT2 = T2 − Ta, ΔT3 = T3 − Ta, ΔT4 = T4 − Ta, n=ρagH1/S1+S2+S3+S4, α = H2/H1, M1 = M2 = M, M3 = M4 = M/ξ, and τ = t/(MCp), we can obtain simplified nonlinear differential equations.For realization 1, according to equations (1)∼(4), (7) and above assumptions about variables and parameters, the simplified differential equations can be obtained as follows:(15)f1=dΔT1dτ=−nαΔT2−ΔT1ΔT1−ΔT3+E1,(16)f2=dΔT2dτ=−nαΔT2−ΔT1ΔT2−ΔT1+E2,(17)f3=dΔT3dτ=−ξnαΔT2−ΔT1ΔT3,(18)f4=dΔT4dτ=−ξnαΔT2−ΔT1ΔT4−ΔT2.For realization 2, according to equations (8)∼(11) and (14) and above assumptions about variables and parameters, the simplified differential equations can be obtained as follows:(19)f1=dΔT1dτ=−nΔT1−αΔT2ΔT1−ΔT2+E1,(20)f2=dΔT2dτ=−nΔT1−αΔT2ΔT2−ΔT4+E2,(21)f3=dΔT3dτ=−ξnΔT1−αΔT2ΔT3−ΔT1,(22)f4=dΔT4dτ=−ξnΔT1−αΔT2ΔT4.


## 2.3. Nonlinear Dynamic Analysis

We can assume the heat source in the left side as positive and the heat source of the right side could be either negative or positive, such that we could discuss the scenarios of two heat sources (the heat source in the left and right sides both are positive and the heat source in the left side is positive, the heat source in the right side is negative) and the scenario of one heat source.


### 2.3.1. Stability Analysis for Scenario 1 (k Is Fixed and α Is Control Parameter)

Here, two conditionsk > 0 (the heat source in the left and right sides both are positive) and k < 0 (the heat source in the left side is positive, and the heat source in the right side is negative) are discussed at the same time. We should solve equilibrium points and characteristic roots before stability analysis:(1) Analysis of Equilibrium Points and Characteristic Roots for Realization 1. The steady state solutions (equilibrium points) of realization 1 equations (15)∼(18) are denoted as ΔT01,ΔT02,ΔT03,ΔT04. In order to solve the equilibrium points, we can make that the right sides of equations (15)∼(18) equal to zero; and the following equations hold:(23)−nαΔT2−ΔT1ΔT1−ΔT3+E1=0,(24)−nαΔT2−ΔT1ΔT2−ΔT1+E2=0,(25)−ξnαΔT2−ΔT1ΔT3=0,(26)−ξnαΔT2−ΔT1ΔT4−ΔT2=0.By solving equations (23)∼(26), we can obtain the equilibrium points as follows:(27)ΔT01=E1n2/3−E1+E1α+E2α1/3=E12/3n2/3αk−1+α1/3,(28)ΔT02=ΔT04=1+kΔT01=1+kE1n2/3−E1+E1α+E2α1/3=1+kE12/3n2/3αk−1+α1/3,(29)ΔT03=0.According to the characteristic determinant equal to 0, the solution of the differential equation can be discussed. The characteristic determinant equation is shown as follows. The expression of characteristic determinant is that(30)JQ−λE=∂f1∂ΔT1−λ∂f1∂ΔT2∂f1∂ΔT3∂f1∂ΔT4∂f2∂ΔT1∂f2∂ΔT2−λ∂f2∂ΔT3∂f2∂ΔT4∂f3∂ΔT1∂f3∂ΔT2∂f3∂ΔT3−λ∂f3∂ΔT4∂f4∂ΔT1∂f4∂ΔT2∂f4∂ΔT3∂f4∂ΔT4−λ=0.Combining equations (15)∼(18), we can obtain the following expression:(31)nΔT01−ΔT032A−nA−λ−nαΔT01−ΔT032AnA0nΔT02−ΔT012A+nA−nαΔT02−ΔT012A−nA−λ00ξnΔT032A−ξnαΔT032A−ξnA−λ0ξnΔT04−ΔT022A−ξnαΔT04−ΔT022A+ξnA0−ξnA−λ=0,where JQ is the Jacobian matrix of equilibrium point Q, λ is the characteristic root, E is the identity matrix, and(32)A=αΔT02−ΔT01=αk−1+α1/3E11/3n1/3,(33)B=ΔT01−ΔT03=E12/3n2/3αk−1+α1/3,(34)C=ΔT02−ΔT01=kE12/3n2/3αk−1+α1/3.Combining equations (31)∼(34), we can obtain characteristic roots as follows:(35)λ1=λ2=−ξnA=−ξnαΔT02−ΔT01=−ξn2/3E11/3αk−1+α1/3.Assumingk = E2/E1, α = H2/H1, to ensure that a real solution exists for equations (27) and (28), we have(36)αk−1+α>0.Two other characteristic roots can be obtained according to a quadratic formula:(37)λ3,4=−b±b2−4ac2a=−nαC−B/2A−2nA±nαC−B/2A+2nA2−4n2αB+C/2+A2−B/22,where(38)a=1;b=nαC−B2A+2nA=n2/3E11/35αk−5+4α2αk−1+α2/3,c=n2αB+C2+A2−B2=3E12/3αk−1+α2/3n4/32>0,Δ=b2−4ac=n4/3E12/35αk−5+4α24αk−1+α4/3−6E12/3αk−1+α2/3n4/3.For convenience, the characteristic roots are set as follows: the real root areγi (i = 1,2,3,4); the complex roots are σi + jωi (i = 1,2,3,4). It is easy to know that λ1,2 are always real number, λ1 = γ1 = λ2 = γ2.If the discriminant Δ is greater than 0, both characteristic roots are real number, namely,λ3 = γ3 and λ4 = γ4. From equations (37) and (38), we can know that when Δ > 0, the following expression holds:(39)n4/3E12/35αk−5+4α2>24n4/3E12/3αk−1+α2.It is clear that(40)5αk−5+4α>26αk−1+α.When the discriminant Δ is less than 0, conjugate complex roots exist, namely,λ3,4 = σ3,4 ± jω3,4. From equations (37) and (38), we can know that when Δ < 0, the following expression holds:(41)5αk−5+4α<26αk−1+α.(2) Stability Analysis for Realization 1 (k Is Fixed and α Is Control Parameter). (1)
No equilibrium point exists in realization 1If no equilibrium point exists in realization 1, from equation (36), it is known that the following expression is satisfied:(42)αk−1+α<0.Therefore from expression (42), when k < −1, it is clear that α>1/k+1 and when k > −1, we know that 0<α<1/k+1.(2)
A stable equilibrium point exists in realization 1(2.1) Assuming there are all negative real characteristic rootsIn this case,λ1, λ2, λ3, and λ4 are all less than 0, and the system has a stable solution.We know that owing toαk − 1 + α > 0, from equation (35), it is clear that λ1 and λ2 are always less than 0. To make λ3,4 < 0, according to Vieta theorem, we know that λ3 + λ4 = −b and λ3λ4 = c. From equation (38), it is known that always c > 0, and owing to αk − 1 + α > 0, from equation (38), if 5αk − 5 + 4α > 0, we have −b < 0, then λ3,4 < 0. In this case, the absolute value sign in equation (40) can be removed; therefore, the following expression (43) can be obtained:(43)5αk−5+4α>26αk−1+α.From expression (43), we can obtain the following expression:(44)αk5−26+4−26>5−26.In summary, if there are all negative real characteristic roots, the system is stable when the following inequalities are satisfied:(45)αk+1−1>0,α5k+4−5>0,αk5−26+4−26>5−26.(2.2) Assuming two negative real roots and two conjugate complex rootsIn this case, the four characteristic roots behave asλ1 = γ1 = λ2 = γ2, λ3,4 = σ3,4 ± jω3,4, where λ1 and λ2 are two negative real roots and λ3, λ4 are two conjugate complex roots. When λ1, λ2, σ3, and σ4 are all less than 0, the system has a stable solution.From equation (37), we know σ3,4 = −b/2. Owing to λ1,2 < 0, to make σ3,4 < 0, we need −b < 0, and from equation (38), we can obtain 5αk + 4α − 5>0.In this case, the absolute value sign in expression (41) can be removed; therefore, the following expression can be obtained:(46)5αk−5+4α<26αk−1+α.From (46), we can obtain that(47)αk5−26+4−26<5−26.In summary,λ1, λ2, σ3, and σ4 are all less than 0, the system is stable when the following inequalities are satisfied:(48)αk+1−1>0,α5k+4−5>0,αk5−26+4−26<5−26.Combining expressions (45) and (48), a stable equilibrium point exists in realization 1, when the following expression is satisfied:(49)αk+1−1>0,α5k+4−5>0.Case 1: whenk > −4/5, it is easy to know that 5k + 4 > 0 and k + 1 > 0, then expression (49) can be simplified to(50)α>1k+1,α>55k+4.From 5k + 4 > 0 and k + 1 > 0, it is clear that 1/k+1−5/5k+4<0 and 0<1/k+1<5/5k+4,Therefore, from expression (49), it can be known that, when k>−4/5, α>5/5k+4Case 2: when−1<k<−4/5, it is easy to know that 5k + 4 < 0 and k + 1 > 0.Expression (49) can be simplified to(51)α>1k+1,α<55k+4.Owing to 5k + 4 < 0 and k + 1 > 0, it is clear that 5/5k+4<0 and 1/k+1>0. Therefore, when −1<k<−4/5, α does not exist.Case 3: whenk < −1, it is easy to know that 5k + 4 < 0 and k + 1 < 0.Expression (49) can be simplified to(52)α<1k+1,α<55k+4.Owing to 5k + 4 < 0 and k + 1 < 0, it is clear that 1/k+1−5/5k+4<0. Thus we can obtain that 5/5k+4<1/k+1<0. Therefore, when k<−1, α does not exist.In conclusion, we can obtain following several cases:Case 1:k>−4/5 and α>5/5k+4Case 2:−1<k<−4/5 and α does not existCase 3:k<−1 and α does not exist(3)
An unstable equilibrium point exists in realization 1(3.1) Assuming characteristic roots are all real and at least one is a positiveAs long as one characteristic root is greater than 0, the equilibrium point of the system is unstable.Owing toαk − 1 + α > 0, λ1 and λ2 are always less than 0 and only λ3 or λ4 is greater than 0. According to Vieta theorem, λ3 + λ4 = −b and λ3λ4 = c; it is known that c > 0, and owing to αk − 1 + α > 0, from equation (38), if 5αk − 5 + 4α < 0, we can obtain −b > 0, then λ3,4 > 0.In this case, the number in the left absolute value sign of expression (40) is less than zero, right absolute value sign of equation (40) is greater than zero, the absolute value signs are removed, and expression (40) can be transformed as follows:(53)5−4α−5αk>26αk−1+α.From expression (53), we can obtain(54)αk26+5+26+4<5+26.In summary,λ1 and λ2 are less than 0 and λ3 and λ4 are greater than 0. The system equilibrium point is unstable, and it is the saddle point of index 2, when the following inequalities are satisfied:(55)αk+α−1>0,5αk+4α−5<0,αk26+5+26+4<5+26.Firstly, we consider the coefficient ofα on the left side of third equation in expression (55) is greater than 0, namely, k26+5+26+4>0, we can obtain k>−26−4/26+5.Then, we consider the coefficient ofα on the left side of the third equation in expression (55) is less than 0, namely, k26+5+26+4<0, we can obtain k<−26−4/26+5.It is easy to know that−1<−26−4/26+5<−4/5. We discuss following several cases:Case 1: when−26−4/26+5<k<−4/5, owing to −1<−26−4/26+5<−4/5, it is easy to know that 5k+4<0, k+1>0 and k26+5+26+4>0. Expression (55) is simplified as(56)α>1k+1,α>55k+4,α<5+26k26+5+26+4.From5k+4<0, k+1>0, k26+5+26+4>0, it is clear that 1/k+1−5+26/k26+5+26+4<0 and 5/5k+4−1/k+1<0. So we have 0<1/k+1<5+26/k26+5+26+4 and 5/5k+4<0.Therefore, when−26−4/26+5<k<−4/5, it can be obtained that 1/k+1<α<26+5/k26+5+26+4.Case 2: when−4/5<k, owing to −1<−26−4/26+5<−4/5, it is easy to know that 5k + 4 > 0, k + 1 > 0 and k26+5+26+4>0. Expression (55) can be simplified to(57)α>1k+1,α<55k+4,α<5+26k26+5+26+4.From5k+4>0, k+1>0, k26+5+26+4>0, it is clear that 5/5k+4−5+26/k26+5+26+4>0 and 5+26/k26+5+26+4−1/k+1>0. So we have 0<1/k+1<5+26/k26+5+26+4<5/5k+4. Therefore, when −4/5<k, we have 1/k+1<α<26+5/k26+5+26+4.Case 3: when−1<k<−26−4/26+5, owing to −1<−26−4/26+5<−4/5, it is easy to know that 5k+4<0, k+1>0, and k26+5+26+4<0.Expression (55) can be simplified to(58)α>1k+1,α>55k+4,α>5+26k26+5+26+4.From5k+4<0, k+1>0, k26+5+26+4<0, it is easy to know that 5+26/k26+5+26+4−5/5k+4<0, so we have 5+26/k26+5+26+4<5/5k+4<0 and 0<1/k+1Therefore, when−1<k<−26−4/26+5, we can obtain that 1/k+1<α.Case 4: whenk<−1, owing to −1<−26−4/26+5<−4/5, it is easy to know that 5k+4<0, k+1<0, and k26+5+26+4<0. Equation (55) can be simplified to(59)α<1k+1,α>55k+4,α>5+26k26+5+26+4.From5k+4<0, k+1<0 and k26+5+26+4<0, it is clear that 5+26/k26+5+26+4−5/5k+4<0 and 1/k+1−5+26/k26+5+26+4<0, so we have 1/k+1<5+26/k26+5+26+4<5/5k+4<0. Therefore, when k<−1, α does not exist.From what has been discussed above, we can obtain following relation betweenk and α:Case 1:−26−4/26+5<k<−4/5 and 1/k+1<α<26+5/k26+5+26+4Case 2:−4/5<k and 1/k+1<α<26+5/k26+5+26+4Case 3:−1<k<−26−4/26+5 and 1/k+1<αCase 4:k<−1 and α does not exist.(3.2) Assuming two negative real roots and two conjugate complex rootsIn this case, the four characteristic roots behave asλ1 = γ1 = λ2 = γ2, λ3,4 = σ3,4 ± jω3,4, where λ1 and λ2 are two negative real roots and λ3 and λ4 are two conjugate complex roots.Owing toλ1 and λ2 are always less than 0, when σ3,4 is greater than 0, the equilibrium point of the system is unstable and it is the saddle-focus point of index 2. From (37), we know σ3,4 = −b/2. To make σ3,4 > 0, we need −b > 0, and from equation (38), we can obtain 5αk + 4α − 5 < 0.In this case, the absolute value sign in Expression (41) can be removed; therefore, the following expression can be obtained:(60)5−5αk−4α<26αk−1+α.From expression (60), we can obtain that(61)αk5+26+4+26>26+5.In summary,λ1 and λ2 are less than 0, σ3 and σ4 are greater than 0, the system equilibrium point is unstable, and it is the saddle-focus point of index 2, when the following inequalities are satisfied:(62)αk+1−1>0,α5k+4−5<0,αk5+26+4+26>26+5.According to above discussion, we can obtain the following several cases:Case 1:−4−26/5+26<k<−4/5 and α>5+26/k5+26+4+26Case 2:−4/5<k and 5+26/k5+26+4+26<α<5/5k+4Case 3:−1<k<−4−26/5+26 and α does not existCase 4:k<−1 and α does not existIfk and α satisfy the above relation, the system equilibrium point is the saddle-focus point of index 2, which is unstable.(3) Analysis of Equilibrium Points and Characteristic Roots for Realization 2. The steady state solution (equilibrium points) of realization 2, equations (19)∼(22), is denoted as ΔT01,ΔT02,ΔT03,ΔT04. In order to solve the equilibrium points, we can make that the right sides of equations (19)∼(22) equal to zero; the following equations hold:(63)−nΔT1−αΔT2ΔT1−ΔT2+E1=0,(64)−nΔT1−αΔT2ΔT2−ΔT4+E2=0,(65)−ξnΔT1−αΔT2ΔT3−ΔT1=0,(66)−ξnΔT1−αΔT2ΔT4=0.By solving equations (63)∼(66), we can obtain the equilibrium points as follows:(67)ΔT02=E2n2/3E2+E1−E1αk1/3=kE12/3n2/3k+1−αk1/3,(68)ΔT01=ΔT03=1+kkΔT02=1+kE1n2/3E2+E1−E1αk1/3=1+kE12/3n2/3k+1−αk1/3,(69)ΔT04=0.According to the characteristic determinant equal to 0, the solution of the differential equation can be discussed. The characteristic determinant equation is that(70)JQ−λE=∂f1∂ΔT1−λ∂f1∂ΔT2∂f1∂ΔT3∂f1∂ΔT4∂f2∂ΔT1∂f2∂ΔT2−λ∂f2∂ΔT3∂f2∂ΔT4∂f3∂ΔT1∂f3∂ΔT2∂f3∂ΔT3−λ∂f3∂ΔT4∂f4∂ΔT1∂f4∂ΔT2∂f4∂ΔT3∂f4∂ΔT4−λ=0.Combining equations (19)∼(22), we can obtain the following expression as follows:(71)−nΔT01−ΔT022A−nA−λnαΔT01−ΔT022A+nA00−nΔT02−ΔT042AnαΔT02−ΔT042A−nA−λ0nA−ξnΔT03−ΔT012A+ξnAξnαΔT03−ΔT012A−ξnA−λ0−ξnΔT042AξnαΔT042A0−ξnA−λ=0,(72)A=ΔT01−αΔT02=k+1−αk1/3E11/3n1/3,(73)B=ΔT01−ΔT02=E12/3n2/3k+1−αk1/3,(74)C=ΔT02−ΔT04=kE12/3n2/3k+1−αk1/3,Combining equations (71)∼(74), we can obtain characteristic roots as follows:(75)λ1=λ2=−ξnA=−ξnΔT01−αΔT02=−ξn2/3E11/3k+1−αk1/3.To ensure that a real solution exists for equations (67) and (68), we have(76)k+1−αk>0.Two other characteristic roots can be obtained according to the quadratic formula:(77)λ3,4=−b±b2−4ac2a=−nB−αC/2A−2nA±nB−αC/2A+2nA2−4n2B+C/2+A2−αC/22,where(78)a=1,b=nB−αC2A+2nA=n2/3E11/34k+5−5αk2k+1−αk2/3,c=n2B+C2+A2−αC2=3n4/3E12/3k+1−αk2/32,Δ=b2−4ac=n4/3E12/34k+5−5αk24k+1−αk4/3−6n4/3E12/3k+1−αk2/3.Therefore, it is easy to know thatλ1,2 are always a real number, λ1 = γ1 = λ2 = γ2. If the discriminant Δ is greater than 0, both characteristic roots are real number:λ3 = γ3, λ4 = γ4. From equations (77) and (78), we can know that when Δ > 0, the following expression holds:(79)n4/3E12/34k+5−5αk2>24n4/3E12/3k+1−αk2.It is clear that(80)4k+5−5αk>26k+1−αk.When the discriminant Δ is less than 0, conjugate complex roots exist, namely,λ3,4 = σ3,4 ± jω3,4. From equations (77) and (78), we can know that when Δ < 0, the following expression holds:(81)4k+5−5αk<26k+1−αk.(4) Stability Analysis for Realization 2 (k Is Fixed and α Is Control Parameter). (1)
No equilibrium point exists in realization 2If no equilibrium point exists in realization 2, from equations (67)∼(68), it is known that the following expression is satisfied:(82)k+1−αk<0.Therefore, from equation (82), when k < −1, it is clear that 0<α<k+1/k and when −1 < k < 0, we know that α does not exist, or when 0 < k, we know that α>k+1/k.(2)
A stable equilibrium point exists in realization 2(2.1) Assuming there are all negative real characteristic rootsIn this case,λ1, λ2, λ3, and λ4 are all less than 0, and the system has a stable solution. We know that, owing to k + 1 − αk > 0, from equation (75), it is clear that λ1 and λ2 are always less than 0. To make λ3,4 < 0, according to Vieta theorem, we know that λ3 + λ4 = −b and λ3λ4 = c. From equation (78), it is known that always c > 0, and owing to k + 1 − αk > 0, from equation (78), if 4k + 5 − 5αk > 0, we have −b < 0, then λ3,4 < 0. In this case, the absolute value sign in equation (80) can be removed; therefore, the following expression can be obtained:(83)4k+5−5αk>26k+1−αk.From expression (83), we can obtain the following expression:(84)αk5−26<k4−26+5−26.In summary, if there are all negative real characteristic roots, the system is stable when the following inequalities are satisfied:(85)αk<k+1,5αk<4k+5,αk5−26<k4−26+5−26.(2.2) Assuming two real roots and two conjugate complex rootsIn this case, the four characteristic roots behave asλ1 = γ1 = λ2 = γ2 and λ3,4 = σ3,4 ± jω3,4, where λ1 and λ2 are two negative real roots and λ3 and λ4 are two conjugate complex roots. When λ1, λ2, σ3, and σ4 are all less than 0, the system has a stable solution.From equation (77), we know σ3,4 = −b/2. Owing to λ1,2 < 0, to make σ3,4 < 0, we have −b < 0, from equation (78), we can obtain 4k + 5 − 5αk > 0.In this case, the absolute value sign in expression (81) can be removed; therefore, the following expression (86) can be obtained:(86)4k+5−5αk<26k+1−αk.From expression (86), we can obtain that(87)αk5−26>k4−.26+5−26.In summary, the system is stable when the following inequalities are satisfied:(88)αk<k+1,5αk<4k+5,αk5−26>k4−.26+5−26.Combining expressions (85) and (88), a stable equilibrium point exists in realization 2, and the following equations are satisfied:(89)αk<k+1,5αk<4k+5.Case 1: whenk > 0, expression (89) can be simplified to(90)α<k+1k,α<4k+55k.Whenk > 0, it is easy to know that 4k+5/5k>0 and k+1/k>0, and it is clear that k+1/k−4k+5/5k>0 and 0<4k+5/5k<k+1/k. Therefore, from expression (90), when k > 0, we know that 0<α<4k+5/5k.Case 2: whenk < 0, expression (89) can be simplified to(91)α>k+1k,α>4k+55k.It is clear thatk+1/k−4k+5/5k>0 and 4k+5/5k<k+1/k. Owing to α > 0, when −1 < k < 0, it clear that k+1/k<0. Therefore, from expression (91), when −1 < k < 0, we know that α > 0.Whenk < −1, it clear that k+1/k>0. Therefore, from expression (91), when k < −1, we know that α>k+1/k.In conclusion, we can obtain following several cases:Case 1:k>0 and 0<α<4k+5/5kCase 2:−1<k<0 and α>0Case 3:k<−1 and α>k+1/k(3)
An unstable equilibrium point exists in realization 1(3.1)Assuming characteristic roots are all real and at least one is a positiveAs long as one characteristic root is greater than 0, the equilibrium point of the system is unstable. Owing tok + 1 − αk > 0, λ1 and λ2 are always less than 0 and only λ3 or λ4 are greater than 0. According to Vieta theorem, it is known that c > 0, and owing to k + 1 − αk > 0, from equation (78), if 4k + 5 − 5αk < 0, we have −b > 0, then λ3,4 > 0.In this case, the number in the left absolute value sign of expression (80) is less than zero, right absolute value sign of expression (80) is greater than zero, and the absolute value signs are removed; expression (92) can be derived as follows:(92)5αk−5−4k>26k+1−αk.From expression (92), we can obtain(93)αk5+26>k4+26+5+26.In summary,λ1 and λ2 are less than 0 and λ3 and λ4 greater than 0. The system equilibrium point is the saddle point of index 2, when the following inequalities are satisfied:(94)αk<k+1,5αk>4k+5,αk5+26>k4+26+5+26.It is easy to know that for anyk, k4+26+5+26/k5+26−k+1/k<0 and 4k+5/5k−k4+26+5+26/k5+26<0, so we have 4k+5/5k<k4+26+5+26/k5+26<k+1/k.We discuss following several cases:Case 1: whenk > 0, expression (94) is simplified as(95)α<k+1k,α>4k+55k,α>k4+26+5+26k5+26.Therefore, whenk > 0, we have k4+2.6+5+26/k5+26<α<1+k/k.Case 2: whenk < 0, expression (94) is simplified as(96)α>k+1k,α<4k+55k,α<k4+26+5+26k5+26.Therefore, whenk < 0, α does not exist.In conclusion, we can obtain following several cases:Case 1:k > 0 and k4+26+5+26/k5+26<α<1+k/kCase 2:k < 0 and α does not existIfk and α satisfy above relation, the system equilibrium point is the saddle point of index 2, which is unstable.(3.2) Assuming two negative real roots and two conjugate complex rootsIn this case, the four characteristic roots behave asλ1 = γ1 = λ2 = γ2 and λ3,4 = σ3,4 ± jω3,4, where λ1 and λ2 are two negative real roots and λ3 and λ4 are two conjugate complex roots.Owing toλ1 and λ2 are always less than 0, when σ3 and σ4 are greater than 0, the equilibrium point of the system is unstable and it is the saddle-focus point of index 2. From (77), we know σ3,4 = −b/2. To make σ3,4 > 0, we have −b > 0. From equation (78), it is known that 4α − 5 + 5αk < 0.In this case, the absolute value sign in expression (81) can be removed; therefore, the following expression (97) can be obtained:(97)5αk−5−4k<26k+1−αk.From expression (97), we can obtain that(98)αk5+26<k4+26+5+26.In summary,λ1 and λ2 are less than 0 and σ3 and σ4 are greater than 0; the system equilibrium point is unstable and it is the saddle-focus point of index 2 when the following inequalities are satisfied:(99)αk<k+1,5αk>4k+5,αk5+26<k4+26+5+26.By referring the classification results of expression (94), we can obtain the following several cases:Case 1:k > 0 and 4k+5/5k<α<k4+2.6+5+26/k5+26Case 2:k < 0 and α does not existIfk and α satisfy above relation, the system equilibrium point is the saddle-focus point of index 2, which is unstable.In brief, for the scenario of the heat source in the left and right sides both are positive (k > 0), from the above discussion, we can form a criterion shown in Table 1 to determine the stability of the system shown in Figure 1 when K is fixed, α is the control parameter, and k > 0. According to the values of α in Table 1, we can know whether there is an equilibrium point and the equilibrium point is stable or not.Table 1
Criterion for scenario 1 ((k is fixed, α is control parameter, and k > 0).
kαRealization 1Realization 20,+∞0,1/k+1NoStable1/k+1,5+26/k5+26+4+26Saddle point of index 2Stable5+26/k5+26+4+26,5/5k+4Saddle-focus point of index 2Stable5/5k+4,4k+5/5kStableStable4k+5/5k,k4+26+5+26/k5+26StableSaddle-focus point of index 2k4+26+5+26/k5+26,k+1/kStableSaddle point of index 2k+1/k,+∞StableNoWe compared Table1 with the part k > 0 of Table 1 in [35]. From these two tables, we can know that the results (whether the equilibrium point is stable or not) are roughly the same as those of [35]. However, our results are more specific and more detailed when the equilibrium points are unstable. In the unstable equilibrium point interval corresponding [35], we obtain two different subintervals corresponding to the saddle point of index 2 and the saddle focal equilibrium point of index 2, respectively. This is because our model has four characteristic roots and only two in [35]. The brief description is as follows. The equilibrium point is unstable in realization 1, when α is in the interval of 1/1+k,5/5k+4, and this interval can be divided into two parts, namely, α∈1/k+1,5+26/k5+26+4+26, where the unstable equilibrium point is the saddle point of index 2 and α∈5+26/k5+26+4+26,5/5k+4, where the unstable equilibrium point is the saddle-focus point of index 2. The equilibrium point is unstable in realization 2, when α is in the interval of 4k+5/5k,k+1/k, and we divide this interval into two parts, namely, α∈4k+5/5k,k4+26+5+26/k5+26, where the unstable equilibrium point is the saddle point of index 2 and α∈k4+26+5+26/k5+26,k+1/k, where the unstable equilibrium point the saddle-focus point of index 2.For the scenario of the heat source in the left is positive and the right is negative (k < 0), from the above discussion, we can form a criterion as shown in Table 2 to determine the stability of the system shown in Figure 1 when K is fixed, α is the control parameter, and k < 0. According to the values of α in Table 2, we can know whether there is an equilibrium point and the equilibrium point is stable or not.Table 2
Criterion for scenario 1 (k is fixed, α is control parameter and k < 0).
kαRealization 1Realization 2−4/5,00,1/k+1NoStable1/k+1,26+5/k26+5+26+4Saddle point of index 2Stable26+5/k26+5+26+4,5/5k+4Saddle-focus point of index 2Stable5/5k+4,+∞StableStable−26−4/26+5,−4/50,1/k+1NoStable1/k+1,5+26/k5+26+4+26Saddle point of index 2Stable5+26/k5+26+4+26,+∞Saddle-focus point of index 2Stable−1,−26−4/26+50,1/k+1NoStable1/k+1,+∞Saddle point of index 2Stable−∞,−10,1+k/kNoNo1+k/k,+∞NoStableWe compared Table2 with the part k < 0 of Table 1 in [35]. From these two tables, we can know the results (whether the equilibrium point is stable or not) are roughly the same as those in [35]. However, similarly, our results are more specific and more detailed when the equilibrium points are unstable. In the unstable equilibrium interval corresponding to literature [35], we obtain two different subintervals corresponding to the saddle point of index 2 and the saddle focal equilibrium point of index 2, respectively. This is because our model has four characteristic roots and only two in [35].


### 2.3.2. Stability Analysis for Scenario 2 (α Is Fixed and k Is Control Parameter)

(1) Stability Analysis for Realization 1 (α Is Fixed and k Is Control Parameter). In this scenario, the characteristic equation is the same as that of scenario 1.(1)
No equilibrium point exists in realization 1In point (1) of Section 2.3.1, we already know that no equilibrium point exists in realization 1, when the following inequalities (equation (42) in Section 2.3.1) are satisfied:(100)αk−1+α<0.Therefore, from expression (98), when α > 0, it is clear that k<1−α/α.(2)
A stable equilibrium point exists in realization 1In point (2) of Section 2.3.1, we already know that the system was stable, when the following inequalities (expression (49) in Section 2.3.1) are satisfied:(101)αk+1−1>0,α5k+4−5>0.Taking out the coefficientk of expression (101), we can obtain expression (102):(102)kα>1−α,5kα>5−4α.Owing toα > 0, expression (102) are simplified as(103)k>1−αα,k>5−4α5α.Owing toα > 0, it is clear that 5−4α/5α−1−α/α>0 and 5−4α/5α>1−α/α; therefore, from expression (101), when α > 0, we know that 5−4α/5α<k.(3)
An unstable equilibrium point exists in realization 1(3.1) Assuming characteristic roots are all real and at least one is a positiveIn point (3.1) of Section 2.3.1, we already know that the equilibrium point of the system is unstable and it is the saddle point of index 2, when the following inequalities (expression (55) in Section 2.3.1) are satisfied:(104)αk+α−1>0,5αk+4α−5<0,αk26+5+26+4<5+26.Taking out the coefficientk of expression (104), we can obtain the following expression:(105)kα>1−α,5kα<5−4α,5+26kα<5+26−α4+26.Owing toα > 0, expression (105) is simplified:(106)k>1−αα,k<5−4α5α,k<5+26−α4+265+26α.It is easy to know that, for anyα > 0, 1−α/α−5+26−α4+26/α5+26<0 and 5+26−α4+26/α5+26−5−4α/5α<0, so we have 1−α/α<5+26−α4+26/α5+26<5−4α/5α. Therefore, from expression (106), it can be known that, when α > 0, we know that 1−α/α<k<5+26−α4+26/α5+26.Ifα and k satisfy above relation, the system equilibrium point is the saddle point of index 2, which is unstable.(3.2) Assuming two negative real roots and two conjugate complex rootsIn point (3.2) of Section 2.3.1, we already know that the equilibrium point of the system is unstable and it is the saddle-focus point of index 2, when the following inequalities (expression (62) in Section 2.3.1) are satisfied:(107)αk+1−1>0,α5k+4−5<0,αk5+26+4+26>26+5.Taking out the coefficientk of expression (107), we can obtain the following expression:(108)kα>1−α,5kα<5−4α,5+26kα>5+26−α4+26.Owing toα > 0, expression (108) is simplified:(109)k>1−αα,k<5−4α5α,k>5+26−α4+265+26α.By referring the solution procedure of expression (104), we can obtain the following result. When α > 0, we know that 5+26−α4+26/α5+26<k<5−4α/5α.Ifα and k satisfy the above relation, the system equilibrium point is the saddle-focus point of index 2, which is unstable.(2) Stability Analysis for Realization 2 (α Is Fixed, k Is Control Parameter). (1)
No equilibrium point exists in realization 2In point (1) of (4) of Section 2.3.1, we already know that no equilibrium point exists in realization 2, when the following inequalities (expression (82) in Section 2.3.1) are satisfied:(110)k+1−αk<0.Therefore, from expression (110), when 0 < α < 1, it is clear that k<1/α−1 and when α > 1, we know that k>1/α−1.(2)
A stable equilibrium point exists in realization 2In point (2) of (4) of Section 2.3.1, we already know that the system was stable, when the following inequalities (expression (89) in (4) of Section 2.3.1) are satisfied:(111)αk<k+1,5αk<4k+5.Taking out the coefficientk of expression (111), we can obtain the following expression:(112)kα−1<1,k5α−4<5.Case 1: whenα > 1, it is easy to know that α − 1 > 0 and 5α − 4 > 0, and then expression (112) can be simplified to(113)k<1α−1,k<55α−4.Fromα − 1 > 0 and 5α − 4 > 0, it is clear that 5/5α−4−1/α−1<0 and 0<5/5α−4<1/α−1. Therefore, from expression (113), when α > 1, we know that k<−5/4−5α.Case 2: when4/5<α<1, it is easy to know that α − 1 < 0 and 5α − 4 > 0; then expression (112) can be simplified to(114)k>1α−1,k<55α−4.Fromα − 1 < 0 and 5α − 4 > 0, it is clear that 5/5α−4−1/α−1>0, 1/α−1<0 and 0<5/5α−4. Therefore, from expression (112), when 4/5<α<1, we know that 1/α−1<k<5/5α−4.Case 3: when0<α<4/5, it is easy to know that α − 1 < 0 and 5α − 4 < 0, then expression (112) can be simplified to(115)k>1α−1,k>55α−4.Fromα − 1 < 0 and 5α − 4 < 0, it is clear that 5/5α−4−1/α−1<0 and 5/5α−4<1/α−1<0, Therefore, from expression (115), when 0<α<4/5, we know that 1/α−1<k. In conclusion, we can obtain following several cases:Case 1:α > 1 and k<−5/4−5αCase 2:4/5<α<1 and 1/α−1<k<5/5α−4Case 3:0<α<4/5 and 1/α−1<k(3)
An unstable equilibrium point exists in realization 2(3.1) Assuming characteristic roots are all real and at least one is a positiveIn point (3.1) of Section 2.3.1, we already know that the equilibrium point of the system is unstable and it is the saddle point of index 2, when the following inequalities (expression (94) in Section 2.3.1) are satisfied:(116)αk<k+1,5αk>4k+5,αk5+26>k4+26+5+26.Taking out the coefficientk of expression (116), we can obtain the following expression:(117)k1−α+1>0,k4−5α+5<0,kα5+26+−26−4>5+26.Firstly, we consider the coefficient ofk on the left side of third inequality in expression (117) is greater than 0, namely, α5+26+−26−4>0, and we can obtain α>4+26/5+26. Then, we consider the coefficient of α on the left side of third inequality in (117) is less than 0, namely, α5+26+−26−4<0, and we can obtain α<4+26/5+26. It is easy to know that 0<4/5<4+26/5+26<1. We discuss following several cases:Case 1: when4+26/5+26<α<1, owing to 0<4/5<4+26/5+26<1, it is easy to know that 4 − 5α < 0, 1 − α > 0, and α5+26+−26−4>0; equation (117) is simplified as(118)k>−11−α,k>−54−5α,k>5+26α5+26−4−26.From 4 − 5α < 0, 1 − α > 0, and α5+26+−26−4>0, it is clear that −1/1−α−−5/4−5α<0, −5/4−5α−5+26/α5+26−4−26<0, and −1/1−α<−5/4−5α<5+26/α5+26−4−26. Therefore, when 4+26/5+26<α<1, we know that k>5+26/α5+26−4−26.Case 2: whenα > 1, owing to 0<4/5<4+26/5+26<1, it is easy to know that 4 − 5α < 0, 1 − α < 0, and α5+26+−26−4>0; expression (117) is simplified as(119)k<−11−α,k>−54−5α,k>5+26α5+26−4−26.From 4 − 5α < 0, 1 − α < 0, and α5+26+−26−4>0, it is clear that −5/4−5α−5+26/α5+26−4−26<0, 5+26/α5+26−4−26−−1/1−α<0, and −5/4−5α<5+26/α5+26−4−26<−1/1−α. Therefore, when α > 1, we know that 5+26/α5+26−4−26<k<−1/1−α.Case 3: when4/5<α<4+26/5+26, owing to 0<4/5<4+26/5+26<1, it is easy to know that 4 − 5α < 0, 1 − α > 0, and α5+26+−26−4<0; expression (117) are simplified as(120)k>−11−α,k>−54−5α,k<5+26α5+26−4−26.From 4 − 5α < 0, 1 − α > 0, and α5+26+−26−4<0, it is clear that 5+26/α5+26−4−26−−1/1−α<0, −1/1−α−−5/4−5α<00<−5/4−5α, and 5+26/α5+26−4−26<−1/1−α<0. Therefore, when 4/5<α<4+26/5+26, we know that k does not exist.Case 4: when0<α<4/5, owing to 0<4/5<4+26/5+26<1, it is easy to know that 4 − 5α > 0, 1 − α > 0, and α5+26+−26−4<0; expression (117) is simplified as(121)k>−11−α,k<−54−5α,k<5+26α5+26−4−26.From 4 − 5α > 0, 1 − α > 0, and α5+26+−26−4<0, it is clear that −5/4−5α−5+26/α5+26−4−26<0, 5+26/α5+26−4−26−−1/1−α<0, and −5/4−5α<5+26/α5+26−4−26<−1/1−α<0. Therefore, when 0<α<4/5, we know that k does not exist.From what has been discussed above, we can obtain following relations betweenα and k.Case 1:4+26/5+26<α<1 and k>5+26/α5+26−4−26Case 2:α > 1 and 5+26/α5+26−4−26<k<−1/1−αCase 3:0<α<4/5 and k does not exist.Case 4:4/5<α<4+26/5+26 and k does not exist.Ifα and k satisfy above relation, the system equilibrium point is the saddle point of index 2, which is unstable.(3.2) Assuming two negative real roots and two conjugate complex rootsIn point (3.2) of Section 2.3.1, we already know that the equilibrium point of the system is unstable and it is the saddle-focus point of index 2, when the following inequalities (expression (96) in Section 2.3.1) are satisfied:(122)αk<k+1,5αk>4k+5,αk5+26<k4+26+5+26.Taking out the coefficientk of expression (122), we can obtain the following expression:(123)k1−α+1>0,k4−5α+5<0,kα5+26+−26−4<5+26.By referring the classification results of expression (117), we can obtain the following several cases:Case 1:4+26/5+26<α<1 and −5/4−5α<k<5+26/α5+26−4−26Case 2:α > 1 and −5/4−5α<k<5+26/α5+26−4−26Case 3:0<α<4/5 and k does not existCase 4:4/5<α<4+26/5+26 and k>−5/4−5αIfk and α satisfy above relation, the system equilibrium point is the saddle-focus point of index 2, which is unstable.In brief, for the scenario 2,α can be discussed in the following sections: 0<α<4/5, 4/5<α<4+26/5+26, and 4+26/5+26<α<1 and 1<α.From the above discussion, we can form a criterion shown in Table3 to determine the stability of the system shown in Figure 1; when α is fixed, k is the control parameter. According to the values of k in Table 3, we can know whether there is an equilibrium point and the equilibrium point is stable or not.Table 3
Criterion for scenario 2 (α is fixed and k is control parameter).
αkRealization 1Realization 20,4/5−∞,−1/1−αNoNo−1/1−α,5−26/α5−26+26−4NoStable5−26/α5−26+26−4,1−α/αNoStable1−α/α,5+26−α4+26/α5+26Saddle point of index 2Stable5+26−α4+26/α5+26,5−4α/5αSaddle-focus point of index 2Stable5−4α/5α,+∞StableStable4/5,4+26/5+26−∞,−1/1−αNoNo−1/1−α,1−α/αNoStable1−α/α,5+26−α4+26/α5+26Saddle point of index 2Stable5+26−α4+26/α5+26,5−4α/5αSaddle-focus point of index 2Stable5−4α/5α,4−5α/−5StableStable−5/4−5α,+∞StableSaddle-focus point of index 24+26/5+26,1−∞,−1/1−αNoNo−1/1−α,1−α/αNoStable1−α/α,5+26−α4+26/α5+26Saddle point of index 2Stable5+26−α4+26/α5+26,5−4α/5αSaddle-focus point of index 2Stable5−4α/5α,−5/4−5αStableStable−5/4−5α,5+26/α5+26−4−26StableSaddle-focus point of index 25+26/α5+26−4−26,+∞StableSaddle point of index 21,+∞−∞,1−α/αNoStable1−α/α,5+26−α4+26/α5+26Saddle point of index 2Stable5+26−α4+26/α5+26,5−4α/5αSaddle-focus point of index 2Stable5−4α/5α,−5/4−5αStableStable−5/4−5α,5+26/α5+26−4−26StableSaddle-focus point of index 25+26/α5+26−4−26,−1/1−αStableSaddle point of index 2−1/1−α,+∞StableNoWe compared Table3 with Table 2 in [35]. From these two tables, we can know the results (whether the equilibrium point is stable or not) are roughly the same as those in [35]. However, our results are more specific and more detailed when the equilibrium points are unstable. In the unstable equilibrium point interval corresponding to [35], we can obtain two different subintervals corresponding to the saddle point of index 2 and the saddle focal equilibrium point of index 2, respectively. This is because our model has four characteristic roots and only two in [35].


### 2.3.3. Stability Analysis for Scenario 3 (One Heat Sink and One Heat Source)

For realization 1, the heat at the bottom releases to the right side. Therefore, we can obtain following equations:(124)f1=dΔT1dτ=−nαΔT2−ΔT1ΔT1−ΔT3,(125)f2=dΔT2dτ=−nαΔT2−ΔT1ΔT2−ΔT1+E1,(126)f3=dΔT3dτ=−ξnαΔT2−ΔT1ΔT3,(127)f4=dΔT4dτ=−ξnαΔT2−ΔT1ΔT4−ΔT2.For realization 2, the heat at the bottom releases to zone 1. Therefore,(128)f1=dΔT1dτ=−nΔT1−αΔT2ΔT1−ΔT2+E1,(129)f2=dΔT2dτ=−nΔT1−αΔT2ΔT2−ΔT4,(130)f3=dΔT3dτ=−ξnΔT1−αΔT2ΔT3−ΔT1,(131)f4=dΔT4dτ=−ξnΔT1−αΔT2ΔT4.(1) Stability Analysis for Realization 1. The steady state solution (equilibrium points) of realization 1 with equations (124)∼(127) is denoted as ΔT01,ΔT02,ΔT03,ΔT04; in order to solve the equilibrium points, we can make that the right sides of equations (124)∼(127) equal to zero; the following equations hold:(132)−nαΔT2−ΔT1ΔT1−ΔT3=0,(133)−nαΔT2−ΔT1ΔT2−ΔT1+E1=0,(134)−ξnαΔT2−ΔT1ΔT3=0,(135)−ξnαΔT2−ΔT1ΔT4−ΔT2=0.By solving equations (132)∼(135), we can obtain the value of equilibrium points as follows:(136)ΔT01=ΔT03=0,(137)ΔT02=ΔT04=E12/3n2/3α1/3.According to the characteristic determinant equal to 0, the solution of the differential equation can be discussed. The characteristic determinant equation is that(138)JQ−λE=∂f1∂ΔT1−λ∂f1∂ΔT2∂f1∂ΔT3∂f1∂ΔT4∂f2∂ΔT1∂f2∂ΔT2−λ∂f2∂ΔT3∂f2∂ΔT4∂f3∂ΔT1∂f3∂ΔT2∂f3∂ΔT3−λ∂f3∂ΔT4∂f4∂ΔT1∂f4∂ΔT2∂f4∂ΔT3∂f4∂ΔT4−λ=0.Combining equations (124)∼(127), we can obtain the following expression:(139)nΔT01−ΔT032A−nA−λ−nαΔT01−ΔT032AnA0nΔT02−ΔT012A+nA−nαΔT02−ΔT012A−nA−λ00ξnΔT032A−ξnαΔT032A−ξnA−λ0ξnΔT04−ΔT022A−ξnαΔT04−ΔT022A+ξnA0−ξnA−λ=0,(140)A=αΔT02−ΔT01=α1/3E11/3n1/3,(141)B=ΔT01−ΔT03=0,(142)C=ΔT02−ΔT01=E12/3n2/3α1/3.Combining equations (139)∼(142), we can obtain characteristic root as follows:(143)λ1=λ2=−ξnA=−ξnαΔT02−ΔT01=−ξn2/3E11/3α1/3.Two other eigenvalues can be obtained according to the quadratic formula:(144)λ3,4=−b±b2−4ac2a=−nαC/2A−2nA±nαC/2A+2nA2−4n2αC/2+A22,where(145)a=1,b=nαC2A+2nA=3n2/3E11/3α1/32,c=n2αC2+A2=3E12/3α2/3n4/32,Δ=b2−4ac=9n4/3E12/3α2/34−6E12/3α2/3n4/3=−15n4/3E12/3α2/34.It is easy to know thatλ1,2 are always real number λ1 = γ1 = λ2 = γ2. Owing to α > 0, E1 > 0, and n > 0, from equation (143), it is clear that λ1,2 are always less than 0. From equation (145), we know that Δ < 0. Since the discriminant Δ < 0, we know that λ3 and λ4 are always conjugate complex numbers λ3,4 = σ3,4 ± jω3,4. Hence, the stability related with λ3,4 is determined by the real part σ3,4. Owing to α > 0, E1 > 0, n > 0, from equations (144) and (145), we know that σ3,4 = −b < 0. As a result, λ1, λ2, and σ3,4 are negative. It is clear that the equilibrium point is stable in realization 1.(2) Stability Analysis for Realization 2. The steady state solution (equilibrium points) of realization 2 with equations (127)∼(130) is denoted as ΔT01,ΔT02,ΔT03,ΔT04. In order to solve the equilibrium points, we can make that the right sides of equations (127)∼(130) equal to zero; the following equations hold:(146)−nΔT1−αΔT2ΔT1−ΔT2+E1=0,(147)−nΔT1−αΔT2ΔT2−ΔT4=0,(148)−ξnΔT1−αΔT2ΔT3−ΔT1=0,(149)−ξnΔT1−αΔT2ΔT4=0.By solving equations (146)∼(149), we can obtain the value of equilibrium point as follows:(150)ΔT01=ΔT03=E12/3n2/3,(151)ΔT02=ΔT04=0.According to the characteristic determinant equal to 0, the solution of the differential equation can be discussed. The characteristic determinant equation is that(152)JQ−λE=∂f1∂ΔT1−λ∂f1∂ΔT2∂f1∂ΔT3∂f1∂ΔT4∂f2∂ΔT1∂f2∂ΔT2−λ∂f2∂ΔT3∂f2∂ΔT4∂f3∂ΔT1∂f3∂ΔT2∂f3∂ΔT3−λ∂f3∂ΔT4∂f4∂ΔT1∂f4∂ΔT2∂f4∂ΔT3∂f4∂ΔT4−λ=0.Combining equations (128)∼(131), we can obtain the following expression:(153)−nΔT01−ΔT022A−nA−λnαΔT01−ΔT022A+nA00−nΔT02−ΔT042AnαΔT02−ΔT042A−nA−λ0nA−ξnΔT03−ΔT012A+ξnAξnαΔT03−ΔT012A−ξnA−λ0−ξnΔT042AξnαΔT042A0−ξnA−λ=0,(154)A=ΔT01−αΔT02=E11/3n1/3,(155)B=ΔT01−ΔT02=E12/3n2/3,(156)C=ΔT02−ΔT04=0.Combining equations (153)∼(156), we can obtain characteristic roots as follows:(157)λ1=λ2=−ξnA=−ξnΔT01−αΔT02=−ξn2/3E11/3.Two other eigenvalues can be obtained according to the quadratic formula:(158)λ3,4=−b±b2−4ac2a=−nB/2A−2nA±nB/2A+2nA2−4n2B/2+A22,where(159)a=1,b=nB2A+2nA=5n2/3E11/32,c=n2B2+A2=3n4/3E12/32,Δ=b2−4ac=25n4/3E12/34−6n4/3E12/3=n4/3E12/34.It is easy to know thatλ1,2 are always real number λ1 = γ1 = λ2 = γ2. Owing to E1 > 0 and n > 0, from equation (157), we know that λ1,2 are always less than 0; from equation (159), we know that Δ > 0, and thus λ3 and λ4 are always real numbers λ3 = γ3 and λ4 = γ4. Combining equations (158) and (159), we can obtain that λ3=−n2/3E11/3 and λ4=−3n2/3E11/3/2. Owing to E1 > 0 and n > 0, we can know that λ3 and λ4 are less than 0. In conclusion, four eigenvalues λ1, λ2, λ3, and λ4 are less than 0, and it is clear that the equilibrium point is stable in realization 2.In brief, for the scenario 3 of one heat source at the bottom of the building, because the heat can enter either the left or the right side, two steady states always exist for this scenario. The parameterα will not affect the stability and existence of solution of buoyancy ventilation of this building configuration.


## 2.3.1. Stability Analysis for Scenario 1 (k Is Fixed and α Is Control Parameter)

Here, two conditionsk > 0 (the heat source in the left and right sides both are positive) and k < 0 (the heat source in the left side is positive, and the heat source in the right side is negative) are discussed at the same time. We should solve equilibrium points and characteristic roots before stability analysis:(1) Analysis of Equilibrium Points and Characteristic Roots for Realization 1. The steady state solutions (equilibrium points) of realization 1 equations (15)∼(18) are denoted as ΔT01,ΔT02,ΔT03,ΔT04. In order to solve the equilibrium points, we can make that the right sides of equations (15)∼(18) equal to zero; and the following equations hold:(23)−nαΔT2−ΔT1ΔT1−ΔT3+E1=0,(24)−nαΔT2−ΔT1ΔT2−ΔT1+E2=0,(25)−ξnαΔT2−ΔT1ΔT3=0,(26)−ξnαΔT2−ΔT1ΔT4−ΔT2=0.By solving equations (23)∼(26), we can obtain the equilibrium points as follows:(27)ΔT01=E1n2/3−E1+E1α+E2α1/3=E12/3n2/3αk−1+α1/3,(28)ΔT02=ΔT04=1+kΔT01=1+kE1n2/3−E1+E1α+E2α1/3=1+kE12/3n2/3αk−1+α1/3,(29)ΔT03=0.According to the characteristic determinant equal to 0, the solution of the differential equation can be discussed. The characteristic determinant equation is shown as follows. The expression of characteristic determinant is that(30)JQ−λE=∂f1∂ΔT1−λ∂f1∂ΔT2∂f1∂ΔT3∂f1∂ΔT4∂f2∂ΔT1∂f2∂ΔT2−λ∂f2∂ΔT3∂f2∂ΔT4∂f3∂ΔT1∂f3∂ΔT2∂f3∂ΔT3−λ∂f3∂ΔT4∂f4∂ΔT1∂f4∂ΔT2∂f4∂ΔT3∂f4∂ΔT4−λ=0.Combining equations (15)∼(18), we can obtain the following expression:(31)nΔT01−ΔT032A−nA−λ−nαΔT01−ΔT032AnA0nΔT02−ΔT012A+nA−nαΔT02−ΔT012A−nA−λ00ξnΔT032A−ξnαΔT032A−ξnA−λ0ξnΔT04−ΔT022A−ξnαΔT04−ΔT022A+ξnA0−ξnA−λ=0,where JQ is the Jacobian matrix of equilibrium point Q, λ is the characteristic root, E is the identity matrix, and(32)A=αΔT02−ΔT01=αk−1+α1/3E11/3n1/3,(33)B=ΔT01−ΔT03=E12/3n2/3αk−1+α1/3,(34)C=ΔT02−ΔT01=kE12/3n2/3αk−1+α1/3.Combining equations (31)∼(34), we can obtain characteristic roots as follows:(35)λ1=λ2=−ξnA=−ξnαΔT02−ΔT01=−ξn2/3E11/3αk−1+α1/3.Assumingk = E2/E1, α = H2/H1, to ensure that a real solution exists for equations (27) and (28), we have(36)αk−1+α>0.Two other characteristic roots can be obtained according to a quadratic formula:(37)λ3,4=−b±b2−4ac2a=−nαC−B/2A−2nA±nαC−B/2A+2nA2−4n2αB+C/2+A2−B/22,where(38)a=1;b=nαC−B2A+2nA=n2/3E11/35αk−5+4α2αk−1+α2/3,c=n2αB+C2+A2−B2=3E12/3αk−1+α2/3n4/32>0,Δ=b2−4ac=n4/3E12/35αk−5+4α24αk−1+α4/3−6E12/3αk−1+α2/3n4/3.For convenience, the characteristic roots are set as follows: the real root areγi (i = 1,2,3,4); the complex roots are σi + jωi (i = 1,2,3,4). It is easy to know that λ1,2 are always real number, λ1 = γ1 = λ2 = γ2.If the discriminant Δ is greater than 0, both characteristic roots are real number, namely,λ3 = γ3 and λ4 = γ4. From equations (37) and (38), we can know that when Δ > 0, the following expression holds:(39)n4/3E12/35αk−5+4α2>24n4/3E12/3αk−1+α2.It is clear that(40)5αk−5+4α>26αk−1+α.When the discriminant Δ is less than 0, conjugate complex roots exist, namely,λ3,4 = σ3,4 ± jω3,4. From equations (37) and (38), we can know that when Δ < 0, the following expression holds:(41)5αk−5+4α<26αk−1+α.(2) Stability Analysis for Realization 1 (k Is Fixed and α Is Control Parameter). (1)
No equilibrium point exists in realization 1If no equilibrium point exists in realization 1, from equation (36), it is known that the following expression is satisfied:(42)αk−1+α<0.Therefore from expression (42), when k < −1, it is clear that α>1/k+1 and when k > −1, we know that 0<α<1/k+1.(2)
A stable equilibrium point exists in realization 1(2.1) Assuming there are all negative real characteristic rootsIn this case,λ1, λ2, λ3, and λ4 are all less than 0, and the system has a stable solution.We know that owing toαk − 1 + α > 0, from equation (35), it is clear that λ1 and λ2 are always less than 0. To make λ3,4 < 0, according to Vieta theorem, we know that λ3 + λ4 = −b and λ3λ4 = c. From equation (38), it is known that always c > 0, and owing to αk − 1 + α > 0, from equation (38), if 5αk − 5 + 4α > 0, we have −b < 0, then λ3,4 < 0. In this case, the absolute value sign in equation (40) can be removed; therefore, the following expression (43) can be obtained:(43)5αk−5+4α>26αk−1+α.From expression (43), we can obtain the following expression:(44)αk5−26+4−26>5−26.In summary, if there are all negative real characteristic roots, the system is stable when the following inequalities are satisfied:(45)αk+1−1>0,α5k+4−5>0,αk5−26+4−26>5−26.(2.2) Assuming two negative real roots and two conjugate complex rootsIn this case, the four characteristic roots behave asλ1 = γ1 = λ2 = γ2, λ3,4 = σ3,4 ± jω3,4, where λ1 and λ2 are two negative real roots and λ3, λ4 are two conjugate complex roots. When λ1, λ2, σ3, and σ4 are all less than 0, the system has a stable solution.From equation (37), we know σ3,4 = −b/2. Owing to λ1,2 < 0, to make σ3,4 < 0, we need −b < 0, and from equation (38), we can obtain 5αk + 4α − 5>0.In this case, the absolute value sign in expression (41) can be removed; therefore, the following expression can be obtained:(46)5αk−5+4α<26αk−1+α.From (46), we can obtain that(47)αk5−26+4−26<5−26.In summary,λ1, λ2, σ3, and σ4 are all less than 0, the system is stable when the following inequalities are satisfied:(48)αk+1−1>0,α5k+4−5>0,αk5−26+4−26<5−26.Combining expressions (45) and (48), a stable equilibrium point exists in realization 1, when the following expression is satisfied:(49)αk+1−1>0,α5k+4−5>0.Case 1: whenk > −4/5, it is easy to know that 5k + 4 > 0 and k + 1 > 0, then expression (49) can be simplified to(50)α>1k+1,α>55k+4.From 5k + 4 > 0 and k + 1 > 0, it is clear that 1/k+1−5/5k+4<0 and 0<1/k+1<5/5k+4,Therefore, from expression (49), it can be known that, when k>−4/5, α>5/5k+4Case 2: when−1<k<−4/5, it is easy to know that 5k + 4 < 0 and k + 1 > 0.Expression (49) can be simplified to(51)α>1k+1,α<55k+4.Owing to 5k + 4 < 0 and k + 1 > 0, it is clear that 5/5k+4<0 and 1/k+1>0. Therefore, when −1<k<−4/5, α does not exist.Case 3: whenk < −1, it is easy to know that 5k + 4 < 0 and k + 1 < 0.Expression (49) can be simplified to(52)α<1k+1,α<55k+4.Owing to 5k + 4 < 0 and k + 1 < 0, it is clear that 1/k+1−5/5k+4<0. Thus we can obtain that 5/5k+4<1/k+1<0. Therefore, when k<−1, α does not exist.In conclusion, we can obtain following several cases:Case 1:k>−4/5 and α>5/5k+4Case 2:−1<k<−4/5 and α does not existCase 3:k<−1 and α does not exist(3)
An unstable equilibrium point exists in realization 1(3.1) Assuming characteristic roots are all real and at least one is a positiveAs long as one characteristic root is greater than 0, the equilibrium point of the system is unstable.Owing toαk − 1 + α > 0, λ1 and λ2 are always less than 0 and only λ3 or λ4 is greater than 0. According to Vieta theorem, λ3 + λ4 = −b and λ3λ4 = c; it is known that c > 0, and owing to αk − 1 + α > 0, from equation (38), if 5αk − 5 + 4α < 0, we can obtain −b > 0, then λ3,4 > 0.In this case, the number in the left absolute value sign of expression (40) is less than zero, right absolute value sign of equation (40) is greater than zero, the absolute value signs are removed, and expression (40) can be transformed as follows:(53)5−4α−5αk>26αk−1+α.From expression (53), we can obtain(54)αk26+5+26+4<5+26.In summary,λ1 and λ2 are less than 0 and λ3 and λ4 are greater than 0. The system equilibrium point is unstable, and it is the saddle point of index 2, when the following inequalities are satisfied:(55)αk+α−1>0,5αk+4α−5<0,αk26+5+26+4<5+26.Firstly, we consider the coefficient ofα on the left side of third equation in expression (55) is greater than 0, namely, k26+5+26+4>0, we can obtain k>−26−4/26+5.Then, we consider the coefficient ofα on the left side of the third equation in expression (55) is less than 0, namely, k26+5+26+4<0, we can obtain k<−26−4/26+5.It is easy to know that−1<−26−4/26+5<−4/5. We discuss following several cases:Case 1: when−26−4/26+5<k<−4/5, owing to −1<−26−4/26+5<−4/5, it is easy to know that 5k+4<0, k+1>0 and k26+5+26+4>0. Expression (55) is simplified as(56)α>1k+1,α>55k+4,α<5+26k26+5+26+4.From5k+4<0, k+1>0, k26+5+26+4>0, it is clear that 1/k+1−5+26/k26+5+26+4<0 and 5/5k+4−1/k+1<0. So we have 0<1/k+1<5+26/k26+5+26+4 and 5/5k+4<0.Therefore, when−26−4/26+5<k<−4/5, it can be obtained that 1/k+1<α<26+5/k26+5+26+4.Case 2: when−4/5<k, owing to −1<−26−4/26+5<−4/5, it is easy to know that 5k + 4 > 0, k + 1 > 0 and k26+5+26+4>0. Expression (55) can be simplified to(57)α>1k+1,α<55k+4,α<5+26k26+5+26+4.From5k+4>0, k+1>0, k26+5+26+4>0, it is clear that 5/5k+4−5+26/k26+5+26+4>0 and 5+26/k26+5+26+4−1/k+1>0. So we have 0<1/k+1<5+26/k26+5+26+4<5/5k+4. Therefore, when −4/5<k, we have 1/k+1<α<26+5/k26+5+26+4.Case 3: when−1<k<−26−4/26+5, owing to −1<−26−4/26+5<−4/5, it is easy to know that 5k+4<0, k+1>0, and k26+5+26+4<0.Expression (55) can be simplified to(58)α>1k+1,α>55k+4,α>5+26k26+5+26+4.From5k+4<0, k+1>0, k26+5+26+4<0, it is easy to know that 5+26/k26+5+26+4−5/5k+4<0, so we have 5+26/k26+5+26+4<5/5k+4<0 and 0<1/k+1Therefore, when−1<k<−26−4/26+5, we can obtain that 1/k+1<α.Case 4: whenk<−1, owing to −1<−26−4/26+5<−4/5, it is easy to know that 5k+4<0, k+1<0, and k26+5+26+4<0. Equation (55) can be simplified to(59)α<1k+1,α>55k+4,α>5+26k26+5+26+4.From5k+4<0, k+1<0 and k26+5+26+4<0, it is clear that 5+26/k26+5+26+4−5/5k+4<0 and 1/k+1−5+26/k26+5+26+4<0, so we have 1/k+1<5+26/k26+5+26+4<5/5k+4<0. Therefore, when k<−1, α does not exist.From what has been discussed above, we can obtain following relation betweenk and α:Case 1:−26−4/26+5<k<−4/5 and 1/k+1<α<26+5/k26+5+26+4Case 2:−4/5<k and 1/k+1<α<26+5/k26+5+26+4Case 3:−1<k<−26−4/26+5 and 1/k+1<αCase 4:k<−1 and α does not exist.(3.2) Assuming two negative real roots and two conjugate complex rootsIn this case, the four characteristic roots behave asλ1 = γ1 = λ2 = γ2, λ3,4 = σ3,4 ± jω3,4, where λ1 and λ2 are two negative real roots and λ3 and λ4 are two conjugate complex roots.Owing toλ1 and λ2 are always less than 0, when σ3,4 is greater than 0, the equilibrium point of the system is unstable and it is the saddle-focus point of index 2. From (37), we know σ3,4 = −b/2. To make σ3,4 > 0, we need −b > 0, and from equation (38), we can obtain 5αk + 4α − 5 < 0.In this case, the absolute value sign in Expression (41) can be removed; therefore, the following expression can be obtained:(60)5−5αk−4α<26αk−1+α.From expression (60), we can obtain that(61)αk5+26+4+26>26+5.In summary,λ1 and λ2 are less than 0, σ3 and σ4 are greater than 0, the system equilibrium point is unstable, and it is the saddle-focus point of index 2, when the following inequalities are satisfied:(62)αk+1−1>0,α5k+4−5<0,αk5+26+4+26>26+5.According to above discussion, we can obtain the following several cases:Case 1:−4−26/5+26<k<−4/5 and α>5+26/k5+26+4+26Case 2:−4/5<k and 5+26/k5+26+4+26<α<5/5k+4Case 3:−1<k<−4−26/5+26 and α does not existCase 4:k<−1 and α does not existIfk and α satisfy the above relation, the system equilibrium point is the saddle-focus point of index 2, which is unstable.(3) Analysis of Equilibrium Points and Characteristic Roots for Realization 2. The steady state solution (equilibrium points) of realization 2, equations (19)∼(22), is denoted as ΔT01,ΔT02,ΔT03,ΔT04. In order to solve the equilibrium points, we can make that the right sides of equations (19)∼(22) equal to zero; the following equations hold:(63)−nΔT1−αΔT2ΔT1−ΔT2+E1=0,(64)−nΔT1−αΔT2ΔT2−ΔT4+E2=0,(65)−ξnΔT1−αΔT2ΔT3−ΔT1=0,(66)−ξnΔT1−αΔT2ΔT4=0.By solving equations (63)∼(66), we can obtain the equilibrium points as follows:(67)ΔT02=E2n2/3E2+E1−E1αk1/3=kE12/3n2/3k+1−αk1/3,(68)ΔT01=ΔT03=1+kkΔT02=1+kE1n2/3E2+E1−E1αk1/3=1+kE12/3n2/3k+1−αk1/3,(69)ΔT04=0.According to the characteristic determinant equal to 0, the solution of the differential equation can be discussed. The characteristic determinant equation is that(70)JQ−λE=∂f1∂ΔT1−λ∂f1∂ΔT2∂f1∂ΔT3∂f1∂ΔT4∂f2∂ΔT1∂f2∂ΔT2−λ∂f2∂ΔT3∂f2∂ΔT4∂f3∂ΔT1∂f3∂ΔT2∂f3∂ΔT3−λ∂f3∂ΔT4∂f4∂ΔT1∂f4∂ΔT2∂f4∂ΔT3∂f4∂ΔT4−λ=0.Combining equations (19)∼(22), we can obtain the following expression as follows:(71)−nΔT01−ΔT022A−nA−λnαΔT01−ΔT022A+nA00−nΔT02−ΔT042AnαΔT02−ΔT042A−nA−λ0nA−ξnΔT03−ΔT012A+ξnAξnαΔT03−ΔT012A−ξnA−λ0−ξnΔT042AξnαΔT042A0−ξnA−λ=0,(72)A=ΔT01−αΔT02=k+1−αk1/3E11/3n1/3,(73)B=ΔT01−ΔT02=E12/3n2/3k+1−αk1/3,(74)C=ΔT02−ΔT04=kE12/3n2/3k+1−αk1/3,Combining equations (71)∼(74), we can obtain characteristic roots as follows:(75)λ1=λ2=−ξnA=−ξnΔT01−αΔT02=−ξn2/3E11/3k+1−αk1/3.To ensure that a real solution exists for equations (67) and (68), we have(76)k+1−αk>0.Two other characteristic roots can be obtained according to the quadratic formula:(77)λ3,4=−b±b2−4ac2a=−nB−αC/2A−2nA±nB−αC/2A+2nA2−4n2B+C/2+A2−αC/22,where(78)a=1,b=nB−αC2A+2nA=n2/3E11/34k+5−5αk2k+1−αk2/3,c=n2B+C2+A2−αC2=3n4/3E12/3k+1−αk2/32,Δ=b2−4ac=n4/3E12/34k+5−5αk24k+1−αk4/3−6n4/3E12/3k+1−αk2/3.Therefore, it is easy to know thatλ1,2 are always a real number, λ1 = γ1 = λ2 = γ2. If the discriminant Δ is greater than 0, both characteristic roots are real number:λ3 = γ3, λ4 = γ4. From equations (77) and (78), we can know that when Δ > 0, the following expression holds:(79)n4/3E12/34k+5−5αk2>24n4/3E12/3k+1−αk2.It is clear that(80)4k+5−5αk>26k+1−αk.When the discriminant Δ is less than 0, conjugate complex roots exist, namely,λ3,4 = σ3,4 ± jω3,4. From equations (77) and (78), we can know that when Δ < 0, the following expression holds:(81)4k+5−5αk<26k+1−αk.(4) Stability Analysis for Realization 2 (k Is Fixed and α Is Control Parameter). (1)
No equilibrium point exists in realization 2If no equilibrium point exists in realization 2, from equations (67)∼(68), it is known that the following expression is satisfied:(82)k+1−αk<0.Therefore, from equation (82), when k < −1, it is clear that 0<α<k+1/k and when −1 < k < 0, we know that α does not exist, or when 0 < k, we know that α>k+1/k.(2)
A stable equilibrium point exists in realization 2(2.1) Assuming there are all negative real characteristic rootsIn this case,λ1, λ2, λ3, and λ4 are all less than 0, and the system has a stable solution. We know that, owing to k + 1 − αk > 0, from equation (75), it is clear that λ1 and λ2 are always less than 0. To make λ3,4 < 0, according to Vieta theorem, we know that λ3 + λ4 = −b and λ3λ4 = c. From equation (78), it is known that always c > 0, and owing to k + 1 − αk > 0, from equation (78), if 4k + 5 − 5αk > 0, we have −b < 0, then λ3,4 < 0. In this case, the absolute value sign in equation (80) can be removed; therefore, the following expression can be obtained:(83)4k+5−5αk>26k+1−αk.From expression (83), we can obtain the following expression:(84)αk5−26<k4−26+5−26.In summary, if there are all negative real characteristic roots, the system is stable when the following inequalities are satisfied:(85)αk<k+1,5αk<4k+5,αk5−26<k4−26+5−26.(2.2) Assuming two real roots and two conjugate complex rootsIn this case, the four characteristic roots behave asλ1 = γ1 = λ2 = γ2 and λ3,4 = σ3,4 ± jω3,4, where λ1 and λ2 are two negative real roots and λ3 and λ4 are two conjugate complex roots. When λ1, λ2, σ3, and σ4 are all less than 0, the system has a stable solution.From equation (77), we know σ3,4 = −b/2. Owing to λ1,2 < 0, to make σ3,4 < 0, we have −b < 0, from equation (78), we can obtain 4k + 5 − 5αk > 0.In this case, the absolute value sign in expression (81) can be removed; therefore, the following expression (86) can be obtained:(86)4k+5−5αk<26k+1−αk.From expression (86), we can obtain that(87)αk5−26>k4−.26+5−26.In summary, the system is stable when the following inequalities are satisfied:(88)αk<k+1,5αk<4k+5,αk5−26>k4−.26+5−26.Combining expressions (85) and (88), a stable equilibrium point exists in realization 2, and the following equations are satisfied:(89)αk<k+1,5αk<4k+5.Case 1: whenk > 0, expression (89) can be simplified to(90)α<k+1k,α<4k+55k.Whenk > 0, it is easy to know that 4k+5/5k>0 and k+1/k>0, and it is clear that k+1/k−4k+5/5k>0 and 0<4k+5/5k<k+1/k. Therefore, from expression (90), when k > 0, we know that 0<α<4k+5/5k.Case 2: whenk < 0, expression (89) can be simplified to(91)α>k+1k,α>4k+55k.It is clear thatk+1/k−4k+5/5k>0 and 4k+5/5k<k+1/k. Owing to α > 0, when −1 < k < 0, it clear that k+1/k<0. Therefore, from expression (91), when −1 < k < 0, we know that α > 0.Whenk < −1, it clear that k+1/k>0. Therefore, from expression (91), when k < −1, we know that α>k+1/k.In conclusion, we can obtain following several cases:Case 1:k>0 and 0<α<4k+5/5kCase 2:−1<k<0 and α>0Case 3:k<−1 and α>k+1/k(3)
An unstable equilibrium point exists in realization 1(3.1)Assuming characteristic roots are all real and at least one is a positiveAs long as one characteristic root is greater than 0, the equilibrium point of the system is unstable. Owing tok + 1 − αk > 0, λ1 and λ2 are always less than 0 and only λ3 or λ4 are greater than 0. According to Vieta theorem, it is known that c > 0, and owing to k + 1 − αk > 0, from equation (78), if 4k + 5 − 5αk < 0, we have −b > 0, then λ3,4 > 0.In this case, the number in the left absolute value sign of expression (80) is less than zero, right absolute value sign of expression (80) is greater than zero, and the absolute value signs are removed; expression (92) can be derived as follows:(92)5αk−5−4k>26k+1−αk.From expression (92), we can obtain(93)αk5+26>k4+26+5+26.In summary,λ1 and λ2 are less than 0 and λ3 and λ4 greater than 0. The system equilibrium point is the saddle point of index 2, when the following inequalities are satisfied:(94)αk<k+1,5αk>4k+5,αk5+26>k4+26+5+26.It is easy to know that for anyk, k4+26+5+26/k5+26−k+1/k<0 and 4k+5/5k−k4+26+5+26/k5+26<0, so we have 4k+5/5k<k4+26+5+26/k5+26<k+1/k.We discuss following several cases:Case 1: whenk > 0, expression (94) is simplified as(95)α<k+1k,α>4k+55k,α>k4+26+5+26k5+26.Therefore, whenk > 0, we have k4+2.6+5+26/k5+26<α<1+k/k.Case 2: whenk < 0, expression (94) is simplified as(96)α>k+1k,α<4k+55k,α<k4+26+5+26k5+26.Therefore, whenk < 0, α does not exist.In conclusion, we can obtain following several cases:Case 1:k > 0 and k4+26+5+26/k5+26<α<1+k/kCase 2:k < 0 and α does not existIfk and α satisfy above relation, the system equilibrium point is the saddle point of index 2, which is unstable.(3.2) Assuming two negative real roots and two conjugate complex rootsIn this case, the four characteristic roots behave asλ1 = γ1 = λ2 = γ2 and λ3,4 = σ3,4 ± jω3,4, where λ1 and λ2 are two negative real roots and λ3 and λ4 are two conjugate complex roots.Owing toλ1 and λ2 are always less than 0, when σ3 and σ4 are greater than 0, the equilibrium point of the system is unstable and it is the saddle-focus point of index 2. From (77), we know σ3,4 = −b/2. To make σ3,4 > 0, we have −b > 0. From equation (78), it is known that 4α − 5 + 5αk < 0.In this case, the absolute value sign in expression (81) can be removed; therefore, the following expression (97) can be obtained:(97)5αk−5−4k<26k+1−αk.From expression (97), we can obtain that(98)αk5+26<k4+26+5+26.In summary,λ1 and λ2 are less than 0 and σ3 and σ4 are greater than 0; the system equilibrium point is unstable and it is the saddle-focus point of index 2 when the following inequalities are satisfied:(99)αk<k+1,5αk>4k+5,αk5+26<k4+26+5+26.By referring the classification results of expression (94), we can obtain the following several cases:Case 1:k > 0 and 4k+5/5k<α<k4+2.6+5+26/k5+26Case 2:k < 0 and α does not existIfk and α satisfy above relation, the system equilibrium point is the saddle-focus point of index 2, which is unstable.In brief, for the scenario of the heat source in the left and right sides both are positive (k > 0), from the above discussion, we can form a criterion shown in Table 1 to determine the stability of the system shown in Figure 1 when K is fixed, α is the control parameter, and k > 0. According to the values of α in Table 1, we can know whether there is an equilibrium point and the equilibrium point is stable or not.Table 1
Criterion for scenario 1 ((k is fixed, α is control parameter, and k > 0).
kαRealization 1Realization 20,+∞0,1/k+1NoStable1/k+1,5+26/k5+26+4+26Saddle point of index 2Stable5+26/k5+26+4+26,5/5k+4Saddle-focus point of index 2Stable5/5k+4,4k+5/5kStableStable4k+5/5k,k4+26+5+26/k5+26StableSaddle-focus point of index 2k4+26+5+26/k5+26,k+1/kStableSaddle point of index 2k+1/k,+∞StableNoWe compared Table1 with the part k > 0 of Table 1 in [35]. From these two tables, we can know that the results (whether the equilibrium point is stable or not) are roughly the same as those of [35]. However, our results are more specific and more detailed when the equilibrium points are unstable. In the unstable equilibrium point interval corresponding [35], we obtain two different subintervals corresponding to the saddle point of index 2 and the saddle focal equilibrium point of index 2, respectively. This is because our model has four characteristic roots and only two in [35]. The brief description is as follows. The equilibrium point is unstable in realization 1, when α is in the interval of 1/1+k,5/5k+4, and this interval can be divided into two parts, namely, α∈1/k+1,5+26/k5+26+4+26, where the unstable equilibrium point is the saddle point of index 2 and α∈5+26/k5+26+4+26,5/5k+4, where the unstable equilibrium point is the saddle-focus point of index 2. The equilibrium point is unstable in realization 2, when α is in the interval of 4k+5/5k,k+1/k, and we divide this interval into two parts, namely, α∈4k+5/5k,k4+26+5+26/k5+26, where the unstable equilibrium point is the saddle point of index 2 and α∈k4+26+5+26/k5+26,k+1/k, where the unstable equilibrium point the saddle-focus point of index 2.For the scenario of the heat source in the left is positive and the right is negative (k < 0), from the above discussion, we can form a criterion as shown in Table 2 to determine the stability of the system shown in Figure 1 when K is fixed, α is the control parameter, and k < 0. According to the values of α in Table 2, we can know whether there is an equilibrium point and the equilibrium point is stable or not.Table 2
Criterion for scenario 1 (k is fixed, α is control parameter and k < 0).
kαRealization 1Realization 2−4/5,00,1/k+1NoStable1/k+1,26+5/k26+5+26+4Saddle point of index 2Stable26+5/k26+5+26+4,5/5k+4Saddle-focus point of index 2Stable5/5k+4,+∞StableStable−26−4/26+5,−4/50,1/k+1NoStable1/k+1,5+26/k5+26+4+26Saddle point of index 2Stable5+26/k5+26+4+26,+∞Saddle-focus point of index 2Stable−1,−26−4/26+50,1/k+1NoStable1/k+1,+∞Saddle point of index 2Stable−∞,−10,1+k/kNoNo1+k/k,+∞NoStableWe compared Table2 with the part k < 0 of Table 1 in [35]. From these two tables, we can know the results (whether the equilibrium point is stable or not) are roughly the same as those in [35]. However, similarly, our results are more specific and more detailed when the equilibrium points are unstable. In the unstable equilibrium interval corresponding to literature [35], we obtain two different subintervals corresponding to the saddle point of index 2 and the saddle focal equilibrium point of index 2, respectively. This is because our model has four characteristic roots and only two in [35].


## 2.3.2. Stability Analysis for Scenario 2 (α Is Fixed and k Is Control Parameter)

(1) Stability Analysis for Realization 1 (α Is Fixed and k Is Control Parameter). In this scenario, the characteristic equation is the same as that of scenario 1.(1)
No equilibrium point exists in realization 1In point (1) of Section 2.3.1, we already know that no equilibrium point exists in realization 1, when the following inequalities (equation (42) in Section 2.3.1) are satisfied:(100)αk−1+α<0.Therefore, from expression (98), when α > 0, it is clear that k<1−α/α.(2)
A stable equilibrium point exists in realization 1In point (2) of Section 2.3.1, we already know that the system was stable, when the following inequalities (expression (49) in Section 2.3.1) are satisfied:(101)αk+1−1>0,α5k+4−5>0.Taking out the coefficientk of expression (101), we can obtain expression (102):(102)kα>1−α,5kα>5−4α.Owing toα > 0, expression (102) are simplified as(103)k>1−αα,k>5−4α5α.Owing toα > 0, it is clear that 5−4α/5α−1−α/α>0 and 5−4α/5α>1−α/α; therefore, from expression (101), when α > 0, we know that 5−4α/5α<k.(3)
An unstable equilibrium point exists in realization 1(3.1) Assuming characteristic roots are all real and at least one is a positiveIn point (3.1) of Section 2.3.1, we already know that the equilibrium point of the system is unstable and it is the saddle point of index 2, when the following inequalities (expression (55) in Section 2.3.1) are satisfied:(104)αk+α−1>0,5αk+4α−5<0,αk26+5+26+4<5+26.Taking out the coefficientk of expression (104), we can obtain the following expression:(105)kα>1−α,5kα<5−4α,5+26kα<5+26−α4+26.Owing toα > 0, expression (105) is simplified:(106)k>1−αα,k<5−4α5α,k<5+26−α4+265+26α.It is easy to know that, for anyα > 0, 1−α/α−5+26−α4+26/α5+26<0 and 5+26−α4+26/α5+26−5−4α/5α<0, so we have 1−α/α<5+26−α4+26/α5+26<5−4α/5α. Therefore, from expression (106), it can be known that, when α > 0, we know that 1−α/α<k<5+26−α4+26/α5+26.Ifα and k satisfy above relation, the system equilibrium point is the saddle point of index 2, which is unstable.(3.2) Assuming two negative real roots and two conjugate complex rootsIn point (3.2) of Section 2.3.1, we already know that the equilibrium point of the system is unstable and it is the saddle-focus point of index 2, when the following inequalities (expression (62) in Section 2.3.1) are satisfied:(107)αk+1−1>0,α5k+4−5<0,αk5+26+4+26>26+5.Taking out the coefficientk of expression (107), we can obtain the following expression:(108)kα>1−α,5kα<5−4α,5+26kα>5+26−α4+26.Owing toα > 0, expression (108) is simplified:(109)k>1−αα,k<5−4α5α,k>5+26−α4+265+26α.By referring the solution procedure of expression (104), we can obtain the following result. When α > 0, we know that 5+26−α4+26/α5+26<k<5−4α/5α.Ifα and k satisfy the above relation, the system equilibrium point is the saddle-focus point of index 2, which is unstable.(2) Stability Analysis for Realization 2 (α Is Fixed, k Is Control Parameter). (1)
No equilibrium point exists in realization 2In point (1) of (4) of Section 2.3.1, we already know that no equilibrium point exists in realization 2, when the following inequalities (expression (82) in Section 2.3.1) are satisfied:(110)k+1−αk<0.Therefore, from expression (110), when 0 < α < 1, it is clear that k<1/α−1 and when α > 1, we know that k>1/α−1.(2)
A stable equilibrium point exists in realization 2In point (2) of (4) of Section 2.3.1, we already know that the system was stable, when the following inequalities (expression (89) in (4) of Section 2.3.1) are satisfied:(111)αk<k+1,5αk<4k+5.Taking out the coefficientk of expression (111), we can obtain the following expression:(112)kα−1<1,k5α−4<5.Case 1: whenα > 1, it is easy to know that α − 1 > 0 and 5α − 4 > 0, and then expression (112) can be simplified to(113)k<1α−1,k<55α−4.Fromα − 1 > 0 and 5α − 4 > 0, it is clear that 5/5α−4−1/α−1<0 and 0<5/5α−4<1/α−1. Therefore, from expression (113), when α > 1, we know that k<−5/4−5α.Case 2: when4/5<α<1, it is easy to know that α − 1 < 0 and 5α − 4 > 0; then expression (112) can be simplified to(114)k>1α−1,k<55α−4.Fromα − 1 < 0 and 5α − 4 > 0, it is clear that 5/5α−4−1/α−1>0, 1/α−1<0 and 0<5/5α−4. Therefore, from expression (112), when 4/5<α<1, we know that 1/α−1<k<5/5α−4.Case 3: when0<α<4/5, it is easy to know that α − 1 < 0 and 5α − 4 < 0, then expression (112) can be simplified to(115)k>1α−1,k>55α−4.Fromα − 1 < 0 and 5α − 4 < 0, it is clear that 5/5α−4−1/α−1<0 and 5/5α−4<1/α−1<0, Therefore, from expression (115), when 0<α<4/5, we know that 1/α−1<k. In conclusion, we can obtain following several cases:Case 1:α > 1 and k<−5/4−5αCase 2:4/5<α<1 and 1/α−1<k<5/5α−4Case 3:0<α<4/5 and 1/α−1<k(3)
An unstable equilibrium point exists in realization 2(3.1) Assuming characteristic roots are all real and at least one is a positiveIn point (3.1) of Section 2.3.1, we already know that the equilibrium point of the system is unstable and it is the saddle point of index 2, when the following inequalities (expression (94) in Section 2.3.1) are satisfied:(116)αk<k+1,5αk>4k+5,αk5+26>k4+26+5+26.Taking out the coefficientk of expression (116), we can obtain the following expression:(117)k1−α+1>0,k4−5α+5<0,kα5+26+−26−4>5+26.Firstly, we consider the coefficient ofk on the left side of third inequality in expression (117) is greater than 0, namely, α5+26+−26−4>0, and we can obtain α>4+26/5+26. Then, we consider the coefficient of α on the left side of third inequality in (117) is less than 0, namely, α5+26+−26−4<0, and we can obtain α<4+26/5+26. It is easy to know that 0<4/5<4+26/5+26<1. We discuss following several cases:Case 1: when4+26/5+26<α<1, owing to 0<4/5<4+26/5+26<1, it is easy to know that 4 − 5α < 0, 1 − α > 0, and α5+26+−26−4>0; equation (117) is simplified as(118)k>−11−α,k>−54−5α,k>5+26α5+26−4−26.From 4 − 5α < 0, 1 − α > 0, and α5+26+−26−4>0, it is clear that −1/1−α−−5/4−5α<0, −5/4−5α−5+26/α5+26−4−26<0, and −1/1−α<−5/4−5α<5+26/α5+26−4−26. Therefore, when 4+26/5+26<α<1, we know that k>5+26/α5+26−4−26.Case 2: whenα > 1, owing to 0<4/5<4+26/5+26<1, it is easy to know that 4 − 5α < 0, 1 − α < 0, and α5+26+−26−4>0; expression (117) is simplified as(119)k<−11−α,k>−54−5α,k>5+26α5+26−4−26.From 4 − 5α < 0, 1 − α < 0, and α5+26+−26−4>0, it is clear that −5/4−5α−5+26/α5+26−4−26<0, 5+26/α5+26−4−26−−1/1−α<0, and −5/4−5α<5+26/α5+26−4−26<−1/1−α. Therefore, when α > 1, we know that 5+26/α5+26−4−26<k<−1/1−α.Case 3: when4/5<α<4+26/5+26, owing to 0<4/5<4+26/5+26<1, it is easy to know that 4 − 5α < 0, 1 − α > 0, and α5+26+−26−4<0; expression (117) are simplified as(120)k>−11−α,k>−54−5α,k<5+26α5+26−4−26.From 4 − 5α < 0, 1 − α > 0, and α5+26+−26−4<0, it is clear that 5+26/α5+26−4−26−−1/1−α<0, −1/1−α−−5/4−5α<00<−5/4−5α, and 5+26/α5+26−4−26<−1/1−α<0. Therefore, when 4/5<α<4+26/5+26, we know that k does not exist.Case 4: when0<α<4/5, owing to 0<4/5<4+26/5+26<1, it is easy to know that 4 − 5α > 0, 1 − α > 0, and α5+26+−26−4<0; expression (117) is simplified as(121)k>−11−α,k<−54−5α,k<5+26α5+26−4−26.From 4 − 5α > 0, 1 − α > 0, and α5+26+−26−4<0, it is clear that −5/4−5α−5+26/α5+26−4−26<0, 5+26/α5+26−4−26−−1/1−α<0, and −5/4−5α<5+26/α5+26−4−26<−1/1−α<0. Therefore, when 0<α<4/5, we know that k does not exist.From what has been discussed above, we can obtain following relations betweenα and k.Case 1:4+26/5+26<α<1 and k>5+26/α5+26−4−26Case 2:α > 1 and 5+26/α5+26−4−26<k<−1/1−αCase 3:0<α<4/5 and k does not exist.Case 4:4/5<α<4+26/5+26 and k does not exist.Ifα and k satisfy above relation, the system equilibrium point is the saddle point of index 2, which is unstable.(3.2) Assuming two negative real roots and two conjugate complex rootsIn point (3.2) of Section 2.3.1, we already know that the equilibrium point of the system is unstable and it is the saddle-focus point of index 2, when the following inequalities (expression (96) in Section 2.3.1) are satisfied:(122)αk<k+1,5αk>4k+5,αk5+26<k4+26+5+26.Taking out the coefficientk of expression (122), we can obtain the following expression:(123)k1−α+1>0,k4−5α+5<0,kα5+26+−26−4<5+26.By referring the classification results of expression (117), we can obtain the following several cases:Case 1:4+26/5+26<α<1 and −5/4−5α<k<5+26/α5+26−4−26Case 2:α > 1 and −5/4−5α<k<5+26/α5+26−4−26Case 3:0<α<4/5 and k does not existCase 4:4/5<α<4+26/5+26 and k>−5/4−5αIfk and α satisfy above relation, the system equilibrium point is the saddle-focus point of index 2, which is unstable.In brief, for the scenario 2,α can be discussed in the following sections: 0<α<4/5, 4/5<α<4+26/5+26, and 4+26/5+26<α<1 and 1<α.From the above discussion, we can form a criterion shown in Table3 to determine the stability of the system shown in Figure 1; when α is fixed, k is the control parameter. According to the values of k in Table 3, we can know whether there is an equilibrium point and the equilibrium point is stable or not.Table 3
Criterion for scenario 2 (α is fixed and k is control parameter).
αkRealization 1Realization 20,4/5−∞,−1/1−αNoNo−1/1−α,5−26/α5−26+26−4NoStable5−26/α5−26+26−4,1−α/αNoStable1−α/α,5+26−α4+26/α5+26Saddle point of index 2Stable5+26−α4+26/α5+26,5−4α/5αSaddle-focus point of index 2Stable5−4α/5α,+∞StableStable4/5,4+26/5+26−∞,−1/1−αNoNo−1/1−α,1−α/αNoStable1−α/α,5+26−α4+26/α5+26Saddle point of index 2Stable5+26−α4+26/α5+26,5−4α/5αSaddle-focus point of index 2Stable5−4α/5α,4−5α/−5StableStable−5/4−5α,+∞StableSaddle-focus point of index 24+26/5+26,1−∞,−1/1−αNoNo−1/1−α,1−α/αNoStable1−α/α,5+26−α4+26/α5+26Saddle point of index 2Stable5+26−α4+26/α5+26,5−4α/5αSaddle-focus point of index 2Stable5−4α/5α,−5/4−5αStableStable−5/4−5α,5+26/α5+26−4−26StableSaddle-focus point of index 25+26/α5+26−4−26,+∞StableSaddle point of index 21,+∞−∞,1−α/αNoStable1−α/α,5+26−α4+26/α5+26Saddle point of index 2Stable5+26−α4+26/α5+26,5−4α/5αSaddle-focus point of index 2Stable5−4α/5α,−5/4−5αStableStable−5/4−5α,5+26/α5+26−4−26StableSaddle-focus point of index 25+26/α5+26−4−26,−1/1−αStableSaddle point of index 2−1/1−α,+∞StableNoWe compared Table3 with Table 2 in [35]. From these two tables, we can know the results (whether the equilibrium point is stable or not) are roughly the same as those in [35]. However, our results are more specific and more detailed when the equilibrium points are unstable. In the unstable equilibrium point interval corresponding to [35], we can obtain two different subintervals corresponding to the saddle point of index 2 and the saddle focal equilibrium point of index 2, respectively. This is because our model has four characteristic roots and only two in [35].


## 2.3.3. Stability Analysis for Scenario 3 (One Heat Sink and One Heat Source)

For realization 1, the heat at the bottom releases to the right side. Therefore, we can obtain following equations:(124)f1=dΔT1dτ=−nαΔT2−ΔT1ΔT1−ΔT3,(125)f2=dΔT2dτ=−nαΔT2−ΔT1ΔT2−ΔT1+E1,(126)f3=dΔT3dτ=−ξnαΔT2−ΔT1ΔT3,(127)f4=dΔT4dτ=−ξnαΔT2−ΔT1ΔT4−ΔT2.For realization 2, the heat at the bottom releases to zone 1. Therefore,(128)f1=dΔT1dτ=−nΔT1−αΔT2ΔT1−ΔT2+E1,(129)f2=dΔT2dτ=−nΔT1−αΔT2ΔT2−ΔT4,(130)f3=dΔT3dτ=−ξnΔT1−αΔT2ΔT3−ΔT1,(131)f4=dΔT4dτ=−ξnΔT1−αΔT2ΔT4.(1) Stability Analysis for Realization 1. The steady state solution (equilibrium points) of realization 1 with equations (124)∼(127) is denoted as ΔT01,ΔT02,ΔT03,ΔT04; in order to solve the equilibrium points, we can make that the right sides of equations (124)∼(127) equal to zero; the following equations hold:(132)−nαΔT2−ΔT1ΔT1−ΔT3=0,(133)−nαΔT2−ΔT1ΔT2−ΔT1+E1=0,(134)−ξnαΔT2−ΔT1ΔT3=0,(135)−ξnαΔT2−ΔT1ΔT4−ΔT2=0.By solving equations (132)∼(135), we can obtain the value of equilibrium points as follows:(136)ΔT01=ΔT03=0,(137)ΔT02=ΔT04=E12/3n2/3α1/3.According to the characteristic determinant equal to 0, the solution of the differential equation can be discussed. The characteristic determinant equation is that(138)JQ−λE=∂f1∂ΔT1−λ∂f1∂ΔT2∂f1∂ΔT3∂f1∂ΔT4∂f2∂ΔT1∂f2∂ΔT2−λ∂f2∂ΔT3∂f2∂ΔT4∂f3∂ΔT1∂f3∂ΔT2∂f3∂ΔT3−λ∂f3∂ΔT4∂f4∂ΔT1∂f4∂ΔT2∂f4∂ΔT3∂f4∂ΔT4−λ=0.Combining equations (124)∼(127), we can obtain the following expression:(139)nΔT01−ΔT032A−nA−λ−nαΔT01−ΔT032AnA0nΔT02−ΔT012A+nA−nαΔT02−ΔT012A−nA−λ00ξnΔT032A−ξnαΔT032A−ξnA−λ0ξnΔT04−ΔT022A−ξnαΔT04−ΔT022A+ξnA0−ξnA−λ=0,(140)A=αΔT02−ΔT01=α1/3E11/3n1/3,(141)B=ΔT01−ΔT03=0,(142)C=ΔT02−ΔT01=E12/3n2/3α1/3.Combining equations (139)∼(142), we can obtain characteristic root as follows:(143)λ1=λ2=−ξnA=−ξnαΔT02−ΔT01=−ξn2/3E11/3α1/3.Two other eigenvalues can be obtained according to the quadratic formula:(144)λ3,4=−b±b2−4ac2a=−nαC/2A−2nA±nαC/2A+2nA2−4n2αC/2+A22,where(145)a=1,b=nαC2A+2nA=3n2/3E11/3α1/32,c=n2αC2+A2=3E12/3α2/3n4/32,Δ=b2−4ac=9n4/3E12/3α2/34−6E12/3α2/3n4/3=−15n4/3E12/3α2/34.It is easy to know thatλ1,2 are always real number λ1 = γ1 = λ2 = γ2. Owing to α > 0, E1 > 0, and n > 0, from equation (143), it is clear that λ1,2 are always less than 0. From equation (145), we know that Δ < 0. Since the discriminant Δ < 0, we know that λ3 and λ4 are always conjugate complex numbers λ3,4 = σ3,4 ± jω3,4. Hence, the stability related with λ3,4 is determined by the real part σ3,4. Owing to α > 0, E1 > 0, n > 0, from equations (144) and (145), we know that σ3,4 = −b < 0. As a result, λ1, λ2, and σ3,4 are negative. It is clear that the equilibrium point is stable in realization 1.(2) Stability Analysis for Realization 2. The steady state solution (equilibrium points) of realization 2 with equations (127)∼(130) is denoted as ΔT01,ΔT02,ΔT03,ΔT04. In order to solve the equilibrium points, we can make that the right sides of equations (127)∼(130) equal to zero; the following equations hold:(146)−nΔT1−αΔT2ΔT1−ΔT2+E1=0,(147)−nΔT1−αΔT2ΔT2−ΔT4=0,(148)−ξnΔT1−αΔT2ΔT3−ΔT1=0,(149)−ξnΔT1−αΔT2ΔT4=0.By solving equations (146)∼(149), we can obtain the value of equilibrium point as follows:(150)ΔT01=ΔT03=E12/3n2/3,(151)ΔT02=ΔT04=0.According to the characteristic determinant equal to 0, the solution of the differential equation can be discussed. The characteristic determinant equation is that(152)JQ−λE=∂f1∂ΔT1−λ∂f1∂ΔT2∂f1∂ΔT3∂f1∂ΔT4∂f2∂ΔT1∂f2∂ΔT2−λ∂f2∂ΔT3∂f2∂ΔT4∂f3∂ΔT1∂f3∂ΔT2∂f3∂ΔT3−λ∂f3∂ΔT4∂f4∂ΔT1∂f4∂ΔT2∂f4∂ΔT3∂f4∂ΔT4−λ=0.Combining equations (128)∼(131), we can obtain the following expression:(153)−nΔT01−ΔT022A−nA−λnαΔT01−ΔT022A+nA00−nΔT02−ΔT042AnαΔT02−ΔT042A−nA−λ0nA−ξnΔT03−ΔT012A+ξnAξnαΔT03−ΔT012A−ξnA−λ0−ξnΔT042AξnαΔT042A0−ξnA−λ=0,(154)A=ΔT01−αΔT02=E11/3n1/3,(155)B=ΔT01−ΔT02=E12/3n2/3,(156)C=ΔT02−ΔT04=0.Combining equations (153)∼(156), we can obtain characteristic roots as follows:(157)λ1=λ2=−ξnA=−ξnΔT01−αΔT02=−ξn2/3E11/3.Two other eigenvalues can be obtained according to the quadratic formula:(158)λ3,4=−b±b2−4ac2a=−nB/2A−2nA±nB/2A+2nA2−4n2B/2+A22,where(159)a=1,b=nB2A+2nA=5n2/3E11/32,c=n2B2+A2=3n4/3E12/32,Δ=b2−4ac=25n4/3E12/34−6n4/3E12/3=n4/3E12/34.It is easy to know thatλ1,2 are always real number λ1 = γ1 = λ2 = γ2. Owing to E1 > 0 and n > 0, from equation (157), we know that λ1,2 are always less than 0; from equation (159), we know that Δ > 0, and thus λ3 and λ4 are always real numbers λ3 = γ3 and λ4 = γ4. Combining equations (158) and (159), we can obtain that λ3=−n2/3E11/3 and λ4=−3n2/3E11/3/2. Owing to E1 > 0 and n > 0, we can know that λ3 and λ4 are less than 0. In conclusion, four eigenvalues λ1, λ2, λ3, and λ4 are less than 0, and it is clear that the equilibrium point is stable in realization 2.In brief, for the scenario 3 of one heat source at the bottom of the building, because the heat can enter either the left or the right side, two steady states always exist for this scenario. The parameterα will not affect the stability and existence of solution of buoyancy ventilation of this building configuration.


## 3. Model Validation

Reference [35] showed the comparison results with [34], which is the literature studying the structure of one heat source at the bottom with two adiabatic tunnels. In order to verify the validity and accuracy of our model, similarly, we compared the scenario 3 results with those of [34]. The outdoor temperature is 288 K, the air density 1.225 kg/m3, Cp 1.0 kJ/(kg·K), heat source 1 kW, H1 5.5 m, H2 5.5 m, S1+2+3+4 37.2933 kg−1/m−1, and gravity acceleration g 9.81 m/s2.The comparison in temperature difference between the proposed model and the validated results from [34] is illustrated in Figure 2(a). The comparison in flow rate between the proposed model and the validated results from [34] is illustrated in Figure 2(b). In [35], we know the maximum relative error for the temperature difference is 13.2% and the maximum relative error for the flow rate is 15.9%, when the strength of the local heat source is 100 W. While in our results, the maximum relative error for the temperature difference is 2.16% and the maximum relative error for the flow rate is 3.23%. These show that our results are closer to the CFD simulation results in [34] than in [35], which means that the proposed 4-zone model in our paper is more reasonable than 2-zone model in [35].Figure 2
Modeled results validation. (a) Temperature comparison between the four-zone model and previous CFD results. (b) Mass flow rates comparison between the four-zone model and previous CFD results (S1 indicates status1/realization 1; 1, 2, etc. indicates zone serial number).
(a)(b)Referring to method of [35], we rectify our model and obtain the rectified results of temperature and flow rate. Here, we give rectified results in zone 2 for the realization 1. Tunnel H1 on the left remains unchanged to 5.5 m, and tunnel H2 on the right is subtracted from room height. In order to compare with [34, 35], the height of the room is 0.5 m, so after adjustment, H2 = 5 m. After improvement, the maximum relative error of flow rate is 1.0% and the maximum relative error of temperature is 1.58% when our results are compared with the ones in [34]. While in [35], the maximum relative error of flow rate is 12.31% and the maximum relative error of temperature is 10.39%. Therefore, the results of the model we proposed are closer to the CFD results of [34].


## 4. Representation of Phase Portraits

We compare the formation process of the system from different initial conditions to the steady state under different control parametera by means of the vector field and phase portrait. Based on the analysis of stability of equilibrium points for scenario 1, we fix control parameter k to 10 and four typical values of α are selected, which are 0.05, 0.5, 1, and 2. We make phase portrait and vector field for scenario 1 (k > 0) through MATLAB, and the results are shown in Figure 3. Figure 3(a) denotes the phase portrait and vector field for k = 10 and α = 0.05 of realization 2; Figure 3(b) denotes phase portrait and vector field for k = 10 and α = 0.5, where the red point and blue point denote the equilibrium point of realization 1 and realization 2, respectively. Figures 3(c) and 3(d) display the vector field and phase portrait of realization 1 and realization 2, respectively, when the control parameter α is equal to 1. The red dot in Figure 3(c) represents the stable equilibrium point of realization 1, and the initial condition around the equilibrium point converges to stable equilibrium point. The blue dot in Figure 3(d) represents the unstable equilibrium point of realization 2, and the initial condition around the unstable equilibrium point forms the periodic motion. Figure 3(e) denotes the phase portrait and vector field for k = 10 and α = 2 of realization 1. In Figure 3(e), there is only one stable equilibrium point of realization 1, and all different initial states converge to the same steady state.Figure 3
Phase portrait and vector field for scenario 1 (k > 0): (a) k = 10 and α = 0.05 of realization 2; (b) k = 10 and α = 0. 5, where the red point and blue point denote the equilibrium point of realization 1 and realization 2, respectively; (c) k = 10 and α = 1 of realization 1; (d) k = 10 and α = 1 of realization 2; (e) k = 10 and α = 2 of realization 1.
(a)(b)(c)(d)(e)Based on the analysis of stability of equilibrium points for scenario 1, we fix control parameterk to −0.5, and three typical values of α are selected, which are 1, 3, and 10. We make phase portrait and vector field for scenario 1 (k < 0), and the results are shown in Figure 4. Figure 4(a) denotes the phase portrait and vector field for k = −0.5 and α = 1 of realization 2. In Figure 4(a), only one stable equilibrium point appeared, and all different initial states converge to the same steady state. Figure 4(b) denotes the phase portrait and vector field for k = −0.5 and α = 3. In Figure 4(b), the red dot represents the unstable equilibrium point of realization 1, and the initial condition around the unstable equilibrium point converges to periodic motion. And, the blue dot represents stable equilibrium point of realization 2, and the initial conditions around the stable equilibrium point converge to the equilibrium point. Figure 4(c) denotes phase portrait and vector field for k = −0.5, α = 10. The red dot and blue dot in Figure 4(c) represent the stable equilibrium point of realization 1 and realization 2, respectively. The initial condition around the red dot converges to the stable equilibrium point of realization 1, and the initial condition around the blue dot converges to the equilibrium point of realization 2.Figure 4
Phase portrait and vector field for scenario 1 (k < 0). (a) k = −0.5 and α = 1 of realization 2; (b) k = −0.5 and α = 3, where the red dot indicates the equilibrium point of realization 1 and the blue dot represents the equilibrium point of realization 2; (c) k = −0.5 and α = 10, where the red dot indicates the equilibrium point of realization 1 and the blue dot represents the equilibrium point of realization 2.
(a)(b)(c)Based on the analysis of stability of equilibrium points for scenario 2, we fixed control parameterα to 2, and five typical values of k were selected, which were −1, −0.35, 0.5, 0.85, and 2. We make phase portrait and vector field for scenario 2, and the results are shown in Figure 5. In Figure 5(a), only one stable equilibrium point of realization 2 appeared, when k is equal to −1, and all different initial states converge to the same steady state. Figures 5(b) and 5(c) display the vector field and phase portrait of realization 1 and realization 2, respectively, when the control parameter k is equal to −0.35. The red dot in Figure 5(b) represents the unstable equilibrium point of realization 1, and the initial condition around the unstable equilibrium point forms the periodic motion. The blue dot in Figure 5(c) represents the stable equilibrium point of realization 2, and the initial condition around the equilibrium point converges to stable equilibrium point. Figure 5(d) denotes the phase portrait and vector field for α = 2 and k = 0.5, where the red dot indicates the equilibrium point of realization 1 and the blue dot represents the equilibrium point of realization 2. Figures 5(e) and 5(f) display the vector field and phase portrait of realization 1 and realization 2, respectively, when the control parameter k is equal to 0.85. The red dot in Figure 5(e) represents the stable equilibrium point of realization 1, and the initial condition around the equilibrium point converges to stable equilibrium point. The blue dot in Figure 5(f) represents the unstable equilibrium point of realization 2, and the initial condition around the unstable equilibrium point forms the periodic motion. In Figure 5(g), there is only one stable equilibrium point of realization 1, and all different initial states converge to the same steady state.Figure 5
Phase portrait and vector field for scenario 2. (a)α = 2 and k = −1 of realization 2; (b) α = 2 and k = −0.35 of realization 1; (c) α = 2 and k = −0.35 of realization 2; (d) α = 2 and k = 0.5, where the red dot indicates the equilibrium point of realization 1 and the blue dot represents the equilibrium point of realization 2; (e) α = 2 and k = 0.85 of realization 1; (f) α = 2 and k = 0.85 of realization 2; (g) α = 2 and k = 2 of realization 1.
(a)(b)(c)(d)(e)(f)(g)


## 5. Conclusions

Nonlinear dynamical analysis was performed to study the buoyancy ventilation of a typical four-zone underground building. A new model with four zones that described the buoyancy ventilation of the underground buildings was proposed and validated by results from the previous studies. The new model contained four nonlinear ordinary differential equations. The criteria for the stability and existence of equilibrium points were derived mathematically in detail about three different scenarios. The criterion for scenario 1 (k was fixed and control α) is summarized in Tables 1 and 2, which correspond k > 0 and k < 0, respectively; the criterion for scenario 2 (α was fixed and control κ) is summarized in Table 3. Two stable equilibrium points existed in scenario 3 (one heat source at the bottom of the building and control α). Finally, the phase portraits and vector field diagrams obtained by applying the fourth-order Runge–Kutta method are given.



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