# Nonlinear Dynamic Analysis of Solution Multiplicity of Buoyancy Ventilation in Two Vertically Connected Open Cavities with Unequal Heights

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

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## Abstract

Solution multiplicity of natural ventilation in buildings is of much importance for personnel safety and ventilation design. In this paper, a new mathematical model of buoyancy pressure ventilation for two vertically connected open cavities is presented. Compared with the previous published papers studying two vertically connected open cavities with equal heights and hot sourceE2 < 0 in the upper room, we study two vertically connected open cavities with unequal heights and hot source E2 < 0 or E2 > 0 in the upper room. By solving and analyzing the equilibrium points and characteristic roots of the differential equations, we analyze the stability of two systems with upward flow pattern and downward pattern and obtain the criteria to determine the stability and existence of solutions for two scenarios. According to these criteria, the multiple steady states of buoyancy ventilation in two vertically connected open cavities with unequal heights and variable strength of hot sources can be obtained. These criteria can be used to design buoyancy ventilation or natural exhaust ventilation systems in two vertically connected open cavities. Compared with two stable states of buoyancy ventilation existing in two vertically connected open cavities with equal heights in the previously published papers, we find that more stable states and unstable states of buoyancy ventilation exist in two vertically connected open cavities with unequal heights in our paper. Finally, bifurcation diagrams and the phase portraits for the two scenarios are given.


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## Body

## 1. Introduction

Nonlinear characteristics exist in systems with various research directions, such as chaotic circuit [1–15], neural network [16–21], image encryption [22–25]. Natural ventilation also has nonlinear characteristics. Natural ventilation is increasingly recognized as an important energy saving technology that provides thermal comfort and a healthy indoor environment for buildings [26]. The main objectives of natural ventilation include the provision of adequate outdoor air, effective removal of indoor air, heat, smoke, and/or odour pollution, and comfortable range of air temperature and humidity. In many cases, buoyancy due to internal sources, such as people, heat generating equipment/processes, or fire, may be properly utilized to facilitate ventilation flow and help achieve these objectives [27]. Through good architectural design and opening arrangement, the ideal airflow can be achieved to avoid the diffusion of pollutants or smoke from the original indoor source into the adjacent rooms/zones. Inappropriate airflow patterns in buildings can endanger human safety and comfort. In modern buildings with many rooms and complex configurations, smoke or pollutants in a portion of the building may pose a hazard to occupants and may migrate from the building’s source to nearby or remote locations due to airflow [28]. Especially, in building fires, smoke is a serious hazard. Most deaths in building fires are due to smoke inhalation rather than burns. Fire smoke safety has always been a very important issue in architectural design, and smoke movement and control management have been paid more and more attention [29]. The airflow pattern that drives natural ventilation due to buoyancy is basically the same as that of smoke extraction [30], and similar principles and analytical methods used to predict natural ventilation can also be used to predict smoke or pollutant movement in building fires. For example, CONTAM, a multizone airflow and pollutant transport program, has been used in smoke movement and control design of buildings [29]. Awbi believes that ventilation designers are faced with two major problems: airflow control in space and indoor air movement [31, 32]. When pollutants or smoke sources appear in a certain zone of the building, the ventilation direction or ventilation mode of the building plays an important role in controlling and avoiding bad airflow.It is found that the air flow and temperature of indoor building thermal ventilation behave as nonlinear multiple solution states, which provide a reference for building designers to design building ventilation. Multiplicity of solutions means that under the same boundary conditions, there is more than one solution to the overall flow pattern in the room. Different initial conditions can lead to different solutions [33]. Zheng et al. investigated the nonlinear equations for ventilation flows in an enclosure with a single opening [34]. Yang et al. carried out theoretical and computational nonlinear dynamic analysis on natural ventilation of two-zone buildings [35, 36]. Van Schijndel carried out a numerical study on the chaotic behavior of airflow in a normally ventilated room [37]. Xin et al. made a numerical study to reveal some bifurcated sequences that may lead to chaotic mixed convection in finite space [38]. Dillon et al. studied chaotic natural convection in a high rectangular cavity with nonisothermal walls [39]. Yee et al. paid particular attention to the separation of different nonlinear behaviors and spurious dynamics caused by some numerical uncertainties observed in actual CFD calculations [40]. Liu et al. studied the formation process of multiple steady states of the underground building of two tunnels connected with the outdoor environment [41]. The two tunnels are of the same height, and one of the heat sources is located in a corner of the underground space. The CFD method is used to reproduce two steady states of buoyancy ventilation of underground building. Liu et al. conducted a nonlinear dynamic analysis of the multiplicity of buoyancy ventilation solutions for typical underground structures [42]. The mathematical model of second-order nonlinear differential equation for underground structure with two zones and two tunnels connecting to outdoor environment is described.As a special architectural structure, the cavity can exist in buildings, electronic equipment, etc [43]. Many papers about multiple solutions of air flow and temperature in cavity were published [39, 44, 45]. However, in the published papers, computational fluid mechanics method is adopted to solve the problem, which is relatively complex. Based on the equations of air flow equilibrium and heat equilibrium, Li established dynamic differential equations about temperature with two-layer open cavities and obtained two stable equilibrium points and two corresponding stable manifolds [46]. The hysteresis and Hopf bifurcation between stable manifolds are studied. However, the literature [46] assumes that the height of two cavities is the same. In practice, the height of the two cavities can be different; for example, the height of the first floor is often greater than that of the second floor in a building. Therefore, it is of practical significance and novelty to study the air flow and temperature solutions of buoyancy ventilation in the cavities when the height of the cavity is not equal. In addition, the literature [46] only studies one situation corresponding to hot sources E1 > 0 and E2 < 0. In fact, there is another situation corresponding to hot sources E1 > 0 and E2 > 0. The situation occurs on fine day in winter, when sun shines into the second floor of the cavity which is equivalent of E2 > 0. In this paper, we study the two-layer open cavities with unequal heights and hot sources E1 > 0 and E2<0 or E2>0, the mathematical model is established, and multiple solutions are obtained. Compared with two stable solutions in literature [46], in our paper, more stable states and unstable states are obtained. The criterion for scenario 1(α is fixed and k is the control parameter) corresponds to solutions of three stable states and one unstable state; and criterion for scenario 2 ((k is fixed, α is the control parameter, and k > 0) is corresponding to solutions of ten stable states and three unstable states. The nomenclature of every variables and constants is shown in Figure 1.Figure 1
Nomenclature of variables and constants.


## 2. Proposed Mathematical Model of Two Vertically Connected Open Cavities with Unequal Heights

Two vertically connected cavities with a top opening and a bottom opening are considered as shown in Figure2. All walls and the partition between the two cavities are adiabatic. It is clear that two flow patterns, i. e., an upward flow shown in Figure 2(a) or a downward flow shown in Figure 2(b), are possible depending on the direction of the overall buoyancy force. We assume that the fluid temperature in each cavity is uniform. The magnitude of the heat source in the lower cavity is E1 (>0), and the heat source in the upper cavity is E2 (>0 or <0). Compared with [46], the height of two rooms is equal each other and heat source in the upper cavity E2 is just less than zero, while in this paper, the heights of two rooms is unequal and heat source in the upper cavity E2 is less than zero or larger than zero.Figure 2
Two vertically connected cavities with a top opening and a bottom opening. (a) An upward flow pattern. (b) A downward pattern.
(a)(b)In Figure2(a), 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 two cavities as shown in the following equations:(1)M1C1dT1dt=−q1C1T1−Ta+E1,(2)M2C2dT2dt=−q2C2T2−T1+E2.According to the conservation of mass, we can obtain(3)q1=q2=q.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 [42, 47]. The pressure balance equation can be obtained as follows:(4)1A122q2=g1′H1+g2′H2,where 1/A122=1/A12+1/A22 and gi′=Ti−Ta/Tag.From equation (4), it is clear that(5)q=A12g1′H1+g2′H2=A12T1−TaH1+T2−TaH2Tag.In Figure2(b), 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 two cavities shown in the following equations:(6)M1C1dT1dt=−q1C1T1−T2+E1,(7)M2C2dT2dt=−q2C2T2−Ta+E2.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 [42, 47]. The pressure balance equation can be obtained as follows:(8)1A122q2=−g1′H1−g2′H2,where 1/A122=1/A12+1/A22 and gi=Ti−Ta/Tag.From equation (8), it is clear that(9)q=A12−g1′H1−g2′H2=A123Ta−T1H1+Ta−T2H2Tag.Assuming thatk = E2/E1, ∆T1 = T1 − Ta, ∆T2 = T2 − Ta, M1C1 = M2C2 = MC, τ = t/(MC), b=A12CH1g/Ta, and α = H2/H1, we can obtain simplified nonlinear differential equations as follows.From Figure2(a), according to equations (1), (2), and (5) and above assumptions about variables and parameters, the simplified differential equations can be obtained as follows:(10)f1=dΔT1dτ=−bΔT1+αΔT2ΔT1+E1,(11)f2=dΔT2dτ=−bΔT1+αΔT2ΔT2−ΔT1+E2.From Figure2(b), according to equations (6), (7), and (9) and above assumptions about variables and parameters, the simplified differential equations can be obtained as follows:(12)f1=dΔT1dτ=−b−ΔT1−αΔT2ΔT1−ΔT2+E1,(13)f2=dΔT2dτ=−b−ΔT1−αΔT2ΔT2+E2.


## 3. Nonlinear Dynamic Analysis of Solution Multiplicity

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

The equilibrium point and the characteristic root should be obtained before the stability analysis.


#### 3.1.1. Analysis of Equilibrium Points and Characteristic Roots in Figure2(a)

The steady state solutions (equilibrium points) of equations (10)∼(11) in Figure 2(a) can be denoted as ΔT01,ΔT02. In order to solve the equilibrium points, the right sides of equations (10)∼(11) can be equal to zero, and thus we can obtain the following equations:(14)bΔT1+αΔT2ΔT1=E1,(15)bΔT1+αΔT2ΔT2−ΔT1=E2.After equation (14) plus equation (15), the following equation can be easily obtained:(16)ΔT02=1+kΔT01.Substituting equation (16) into equation (14), we can obtain equilibrium points as follows:(17)ΔT01=E12/3b2/31+α+αk1/3,(18)ΔT02=1+kE12/3b2/31+α+αk1/3.For convenience, we can makeM=1+α+αk. To ensure that a real solution exists for equations (17)∼(18) and equilibrium points exist, we have(19)M=1+α+αk>0.When the characteristic determinant is equal to 0, the solution of the differential equation can be discussed by characteristic roots. The expression of characteristic determinant is shown as follows:(20)JQ−λE=∂f1∂ΔT1−λ∂f1∂ΔT2∂f2∂ΔT1∂f1∂ΔT2=0.Combining equations (10), (11), and (20), the following equation can be obtained:(21)−bΔT012A−bA−λ−bαΔT012A−bB2A+bA−bαB2A−bA−λ=0,where J(Q) is the Jacobian matrix of equilibrium point Q, λ is the characteristic root, E is the identity matrix, and for convenience, we can make A=ΔT01+αΔT02 and B=ΔT02−ΔT01.According to equations (17)∼(18), we can obtain the following equations:(22)A=1+α+αk1/3E11/3b1/3,(23)B=kE12/3b2/31+α+αk1/3=kΔT01,(24)ΔT01A=E11/3b1/31+α+αk2/3,(25)BA=kE11/3b1/31+α+αk2/3.Combining equations (21)∼(25), we can obtain the characteristic equation of Figure 2(a) as follows:(26)λ2+b2/3E11/35+4α+5αk21+α+αk2/3λ+3b4/3E12/31+α+αk2/32=0.For convenience, we can make(27)β=b2/3E11/35+4α+5αk21+α+αk2/3.(28)γ=3b4/3E12/31+α+αk2/32.Two characteristic roots of equation (26) can be obtained according to a quadratic formula:(29)λ1,2=−β±β2−4γ2.Sinceb > 0 and E1 > 0, from equations (19) and (28), it is clear that γ is always greater than 0.According to Vieta theorem, we know thatλ1 + λ2 = −β and λ1λ2 = γ. Owing to γ > 0, when β > 0, we know that λ1 and λ2 are all less than 0, or when β < 0, it is clear that λ1 and λ2 are all greater than 0.In summary, according to equation (27), if the following expression is satisfied, λ1 and λ2 are all less than 0 and the equilibrium point is stable:(30)5+4α+5αk>0.If the following expression is satisfied,λ1 and λ2 are all greater than 0 and the equilibrium point is unstable:(31)5+4α+5αk<0.


#### 3.1.2. Stability Analysis in Figure2(a) (α Is Fixed and k Is Control Parameter)

(1)
No equilibrium points exist in Figure2(a)If no equilibrium points exist in Figure2(a), from expression (19), we know that the following expression is satisfied:(32)1+α+αk<0.Solving inequality (32), when α > 0, we can obtain that k≤1+α/α.(2)
Unstable equilibrium points exist in Figure2(a)If unstable equilibrium points exist in Figure2(a), from expressions (19) and (31), we know that the following expression is satisfied:(33)1+α+αk>0,5+4α+5αk<0,α>0.It is clear that expression (33) can be simplified as follows:(34)k>−1+αα,k<−5+4α5α,α>0.Whenα > 0, it is clear that −1+α/α−−5+4α/5α<0 and −1+α/α<−5+4α/5α.Solving expression (34), when α > 0, we can obtain that −1+α/α<k<−5+4α/5α.(3)
Stable equilibrium points exist in Figure2(a)If stable equilibrium points exist in Figure2(a), from expressions (19) and (30), we know that the following expression is satisfied:(35)1+α+αk>0,5+4α+5αk>0,α>0.It is clear that expression (35) can be simplified as follows:(36)k≥1+αα,k≥5+4α5α,α>0.Solving expression (36), when α > 0, we can obtain that k≥5+4α/5α.


#### 3.1.3. Analysis of Equilibrium Points and Characteristic Roots in Figure2(b)

The steady state solutions (equilibrium points) of equations (12)∼(13) in Figure 2(b) can be denoted as ΔT01,ΔT02. In order to solve the equilibrium points, the right sides of equations (12)∼(13) can be equal to zero, and thus we can obtain the following equations:(37)b−ΔT1−αΔT2ΔT1−ΔT2=E1,(38)b−ΔT1−αΔT2ΔT2=E2.After equation (38) plus equation (37), the following equation can be easily obtained:(39)ΔT01=1+kkΔT02.Substituting equation (39) into equation (38), we can obtain equilibrium points as follows:(40)ΔT023=−k3E12b21+k+αk,(41)ΔT013=−1+k3E12b21+k+αk.According to equation (38), since b > 0, when E2 > 0, it is clear that ΔT02 > 0, and when E2 < 0, we have ΔT02 < 0.Besides, to ensure that solutions exist for equations (37)∼(38), we have(42)ΔT01+αΔT02=1+k+kαkΔT02<0.Owing toE1 > 0 and k = E2/E1, when E2 > 0, it is clear from equation (40) that k > 0 and ΔT02 > 0. From expression (42), we known that if 1 + k + αk < 0, equilibrium points exist in Figure 2(b).WhenE2 < 0, it is clear that k < 0 and ΔT02 < 0, we known that if 1 + k + αk < 0, equilibrium points exist in Figure 2(b).In conclusion, when the following expression is met, equilibrium points exist in Figure2(b):(43)M=1+k+αk<0.From expression (43), equations (40)∼(41) can be simplified as follows:(44)ΔT02=kE12/3b2/3−1−k−αk1/3,(45)ΔT01=1+kE12/3b2/3−1−k−αk1/3.When the characteristic determinant is equal to 0, the solution of the differential equation can be discussed. The expression of characteristic determinant is shown in the following equation:(46)JQ−λE=∂f1∂ΔT1−λ∂f1∂ΔT2∂f2∂ΔT1∂f1∂ΔT2=0.Combining equations (12)∼(13), (46), the following equation can be obtained:(47)bB2A−bA−λbαB2A+bAbT022AbαT022A−bA−λ=0.For convenience, we can makeA=−ΔT01−αΔT02 and B=ΔT01−ΔT02.According to equations (44)∼(45), we can obtain the following equations:(48)A=−1−k−αk1/3E11/3b1/3,(49)B=E12/3b2/3−1−k−αk1/3=ΔT02k,(50)ΔT02A=kE11/3b1/3−1−k−αk2/3,(51)BA=E11/3b1/3−1−k−αk2/3.Combining equations (47)∼(51), we can obtain characteristic equation of Figure 2(b) as follows:(52)λ2−b2/3E11/3k1/35+4k+5αk2−1−k−αk2/3λ+3−1−k−αk2/3b4/3E12/32=0.For convenience, we can make(53)β=−b2/3E11/3k1/35+4k+5αk2−1−k−αk2/3,(54)γ=3−1−k−αk2/3b4/3E12/32.Two characteristic roots of equation (52) can be obtained according to a quadratic formula:(55)λ1,2=−β±β2−4γ2.Sinceb > 0 and E1 > 0, from equations (43) and (54), it is clear that γ is always greater than 0.According to Vieta theorem, we know thatλ1 + λ2 = −β and λ1λ2 = γ. Owing to γ > 0, when β > 0, we know that λ1 and λ2 are all less than 0, or when β < 0, we know that λ1 and λ2 are all greater than 0.In summary, according to equation (53), if the following expression is satisfied, λ1 and λ2 are all less than 0 and the equilibrium point is stable:(56)5+4k+5αk<0.If the following expression is satisfied,λ1 and λ2 are all greater than 0 and the equilibrium point is unstable:(57)5+4k+5αk>0.


#### 3.1.4. Stability Analysis in Figure2(b) (α Is Fixed and k Is Control Parameter)

(1)
No equilibrium points exist in Figure2(b)If no equilibrium points exist in Figure2(b), from discussion between expressions (42)∼(43), we know that the following expression is satisfied:(58)1+k+αk>0.Solving expression (58), when α > 0, we can obtain that k≥1/1+α.(2)
Unstable equilibrium points exist in Figure2(b)If unstable equilibrium points exist in Figure2(b), from expression (43) and expression (57), we know that the following expression is satisfied:(59)1+k+αk<0,5+4k+5αk>0,α>0.The coefficient in front ofk is extracted, and expression (59) can be simplified as follows:(60)k<−11+α,k>−54+5α,α>0.Whenα > 0, it is clear that −5/4+5α−−1/1+α<0 and −5/4+5α<−1/1+α.From expression (60), when α > 0, we can obtain that −5/4+5α<k<−1/1+α.(3)
Stable equilibrium points exist in Figure2(b)If stable equilibrium points exist in Figure2(b), from expression (43) and expression (56), we know that the following expression is satisfied:(61)1+k+αk<0,5+4k+5αk<0,α>0.The coefficient in front ofk is extracted, and expression (61) can be simplified as follows:(62)k<−11+α,k<−54+5α,α>0.From expression (62), when α > 0, we can obtain that k<−5/4+5α.In a word, from the above discussion, we can form a criterion shown in Table1 to determine the stability of the system shown in Figure 2 when α is fixed and k is the control parameter. For any α > 0, according to the values of k in Table 1, we can evaluate whether there is an equilibrium point and the equilibrium point is stable or unstable in scenario 1(α is fixed and k is the control parameter).Table 1
Stability criterion for scenario 1(α is fixed and k is the control parameter).
αkFigure2(a)Figure2(b)0,+∞−∞,−1+α/αNoStable−1+α/α,−5+4α/5αUnstableStable−5+4α/5α,−5/4+5αStableStable−5/4+5α,−1/1+αStableUnstable−1/1+α,+∞StableNo


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

The characteristic equation of scenario 2 is the same as that of scenario 1.


#### 3.2.1. Stability Analysis in Figure2(a) (k Is Fixed and α Is Control Parameter)

(1)
No equilibrium points exist in Figure2(a)In Section3.1.2 ordinal (1), we know that no equilibrium points exist in Figure 2(a), when the following inequalities (taking out the coefficient of α in expression (32)) are satisfied:(63)1+kα+1<0,α>0.Whenk > −1, solving expression (63), we can obtain that α<−1/1+k. Owing to α > 0, it is clear that −1/1+k<0 which contradicts k > −1. Therefore, from expression (63), when k > −1, we know that α does not exist.Whenk < −1, solving expression (63), it is clear that α>−1/1+k. Since −1/1+k>0, from expression (63), when k < −1, we know that α>−1/1+k.In conclusion, we can obtain the following two cases:Case 1:k > −1 and α does not existCase 2:k < −1 and α>−1/1+k(2)
Unstable equilibrium points exist in Figure2(a)In Section3.1.2 ordinal (2), we know that unstable equilibrium points exist in Figure 2(a), when the following inequalities (taking out the coefficient of α in expression (33)) are satisfied:(64)1+kα+1>0,4+5kα+5<0,α>0,whenk < −1, it is clear that 1 + k < 0 and 4 + 5k < 0. Expression (64) is simplified as(65)α<−11+k,α>−54+5k,α>0.We know that fork < −1, since 1 + k < 0 and 4 + 5k < 0, we have −1/1+k−−5/4+5k>0, and it is clear that −1/1+k>−5/4+5k, −1/1+k>0, and −5/4+5k>0.Therefore, from expression (65), when k < −1, we know that −5/4+5k<α<−1/1+k.When −1<k < −4/5, it is clear that 1 + k > 0 and 4 + 5k < 0. Expression (64) is simplified as(66)α>−11+k,α>−54+5k,α>0.We know that, for −1 <k < −4/5, since 1 + k > 0 and 4 + 5k < 0, we have −1/1+k−−5/4+5k<0, and it is clear that −1/1+k<−5/4+5k and −1/1+k<0, −5/4+5k>0.Therefore, from expression (66), when −1 < k < −4/5, we know that −5/4+5k<α.Whenk > −4/5, it is clear that 1 + k > 0 and 4 + 5k > 0. Expression (64) is simplified as(67)α>−11+k,α<−54+5k,α>0.We know that, fork > −4/5, since 1 + k > 0 and 4 + 5k > 0, we have −1/1+k−−5/4+5k>0, and it is clear that −1/1+k>−5/4+5k and −1/1+k>0, −5/4+5k>0.Therefore, from expression (67), when k > −4/5, we know that α does not exist.In conclusion, we can obtain following several cases:Case 1:k < −1 and −5/4+5k<α<−1/1+kCase 2: 1 <k < −4/5 and −5/4+5k<αCase 3: −4/5 <k and α does not exist(3)
Stable equilibrium points exist in Figure2(a)In Section3.1.2 ordinal (3), we know that stable equilibrium points exist in Figure 2(a), when the following inequalities (taking out the coefficient of α in expression (35)) are satisfied:(68)1+kα+1>0,4+5kα+5>0,α>0.Referring to the classification results of unstable equilibrium points about expression (64), we can obtain the following several cases about expression (68).Case 1:k < −1 and 0<α<−5/4+5kCase 2: 1 <k < −4/5 and 0<α<−5/4+5kCase 3: −4/5 <k and α>0


#### 3.2.2. Stability Analysis in Figure2(b) (k Is Fixed and α Is Control Parameter)

(1)
No equilibrium points exist in Figure2(b)In Section3.1.4 ordinal (1), we know that no equilibrium points exist in Figure 2(b), when the following inequalities (derived from expression (58)) are satisfied:(69)αk+1+k>0,α>0.Whenk > 0, solving expression (69), we can obtain that α>−1+k/k. Owing to α > 0, it is clear that −1+k/k<0. Therefore, from expression (69), when k > 0, we know that α > 0.Whenk < 0, solving expression (69), it is clear that α<−1+k/k. To ensure −1+k/k>0, we have −1 < k < 0; therefore, from expression (69), when −1 < k < 0, we know that 0<α<−1+k/k, and when k < −1, α does not exist.In conclusion, we can obtain following several cases:Case 1:k > 0 and α > 0Case 2: −1<k < 0 and 0<α<−1+k/kCase 3:k < −1 and α does not exist(2)
Unstable equilibrium points exist in Figure2(b)In Section3.1.4 ordinal (2), we know that unstable equilibrium points exist in Figure 2(b), when the following inequalities (derived from expression (59) are satisfied:(70)αk+1+k<0,5αk+5+4k>0,α>0.Whenk > 0, expression is simplified as(71)α<−1+kk,α>−5+4k5k,α>0.We know that, for anyk > 0, we have −1+k/k−−5+4k/5k<0, and it is clear that −1+k/k<−5+4k/5k which is in contradiction with expression (71).Therefore, from expression (71), when k > 0, we know that α does not exist.Whenk < 0, expression (70) is simplified as(72)α>−1+kk,α<−5+4k5k,α>0.It is clear that, for anyk < 0, we have −1+k/k<−5+4k/5k.Whenk < −5/4, we know that −1+k/k<0 and −5+4k/5k<0. Therefore, from expression (72), when k < −5/4, we know that α does not exist.When −5/4 <k < −1, we know that −1+k/k<0 and −5+4k/5k>0. Therefore, from expression (72), when −5/4 < k < −1, we know that 0<α<−5+4k/5k.When −1 <k < 0, we know that −1+k/k>0 and −5+4k/5k>0. Therefore, from expression (72), when −1 < k < 0, we know that −1+k/k<α<−5+4k/5k.In conclusion, we can obtain following several cases:Case 1:k > 0 and α does not existCase 2:k < −5/4 and α does not existCase 3: −5/4<k < −1 and 0<α<−5+4k/5kCase 4: −1<k < 0 and −1+k/k<α<−5+4k/5k(3)
Stable equilibrium points exist in Figure2(b)In Section3.1.4 ordinal (3), we know that stable equilibrium points exist in Figure 2(b), when the following inequalities (derived from expression (61) are satisfied:(73)αk+1+k<0,5αk+5+4k<0,α>0.Whenk > 0, expression (73) is simplified as(74)α<−1+kk,α<−5+4k5k,α>0.We know that, for anyk > 0, we have −1+k/k<−5+4k/5k. Owing to α > 0, when k > 0, it is clear that −1+k/k<0 and −5+4k/5k<0.Therefore, from expression (74), when k > 0, we know that α does not exist.Whenk < 0, expression (73) is simplified as(75)α>−1+kk,α>−5+4k5k,α>0.For anyk < 0, we know that −1+k/k<−5+4k/5k.Whenk < −5/4, we know that −1+k/k<0 and −5+4k/5k<0. Therefore, from expression (75), when k < −5/4, we know that α > 0.When −5/4 <k < −1, we know that −1+k/k<0 and −5+4k/5k>0. Therefore, from expression (75), when −5/4 < k < −1, we know that α>−5+4k/5k.When −1 <k < 0, we know that −1+k/k>0 and −5+4k/5k>0. Therefore, from expression (75), when −1 < k < 0, we know that α>−5+4k/5k.In conclusion, we can obtain following several cases:Case 1:k > 0 and α does not existCase 2:k < −5/4 and α > 0Case 3: −5/4 <k < −1 and α>−5+4k/5kCase 4: −1 <k < 0 and α>−5+4k/5kIn brief, for the scenario 2,k can be discussed in the following intervals: k<−5/4, −5/4<k<−1, −1<k<−4/5, −4/5<k<0, and 0<k.From the above discussion, whenK is fixed and α is the control parameter, we can form a criterion shown in Table 2 to determine the stability of the system shown in Figure 2. According to the values of α in Table 2, we can know whether there is an equilibrium point and whether the equilibrium point is stable or not.Table 2
Stability criterion for scenario 2(k is fixed and α is the control parameter).
kαFigure2(a)Figure2(b)−∞,−5/40,−5/4+5kStableStable−5/4+5k,−1/1+kUnstableStable−1/1+k,+∞NoStable−5/4,−10,−5+4k/5kStableUnstable−5+4k/5k,−5/4+5kStableStable−5/4+5k,−1/1+kUnstableStable−1/1+k,+∞NoStable−1,−4/50,−1+k/kStableNo−1+k/k,−5+4k/5kStableUnstable−5+4k/5k,−5/4+5kStableStable−5/4+5k,+∞UnstableStable−4/5,00,−1+k/kStableNo−1+k/k,−5+4k/5kStableUnstable−5+4k/5k,+∞StableStable0,+∞0,+∞StableNoCompared with Tables1–2 in this paper and theoretical analysis results in [44], it can be seen that more stable states and unstable states exist in two vertically connected open cavities with unequal heights in our paper.


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

The equilibrium point and the characteristic root should be obtained before the stability analysis.


### 3.1.1. Analysis of Equilibrium Points and Characteristic Roots in Figure2(a)

The steady state solutions (equilibrium points) of equations (10)∼(11) in Figure 2(a) can be denoted as ΔT01,ΔT02. In order to solve the equilibrium points, the right sides of equations (10)∼(11) can be equal to zero, and thus we can obtain the following equations:(14)bΔT1+αΔT2ΔT1=E1,(15)bΔT1+αΔT2ΔT2−ΔT1=E2.After equation (14) plus equation (15), the following equation can be easily obtained:(16)ΔT02=1+kΔT01.Substituting equation (16) into equation (14), we can obtain equilibrium points as follows:(17)ΔT01=E12/3b2/31+α+αk1/3,(18)ΔT02=1+kE12/3b2/31+α+αk1/3.For convenience, we can makeM=1+α+αk. To ensure that a real solution exists for equations (17)∼(18) and equilibrium points exist, we have(19)M=1+α+αk>0.When the characteristic determinant is equal to 0, the solution of the differential equation can be discussed by characteristic roots. The expression of characteristic determinant is shown as follows:(20)JQ−λE=∂f1∂ΔT1−λ∂f1∂ΔT2∂f2∂ΔT1∂f1∂ΔT2=0.Combining equations (10), (11), and (20), the following equation can be obtained:(21)−bΔT012A−bA−λ−bαΔT012A−bB2A+bA−bαB2A−bA−λ=0,where J(Q) is the Jacobian matrix of equilibrium point Q, λ is the characteristic root, E is the identity matrix, and for convenience, we can make A=ΔT01+αΔT02 and B=ΔT02−ΔT01.According to equations (17)∼(18), we can obtain the following equations:(22)A=1+α+αk1/3E11/3b1/3,(23)B=kE12/3b2/31+α+αk1/3=kΔT01,(24)ΔT01A=E11/3b1/31+α+αk2/3,(25)BA=kE11/3b1/31+α+αk2/3.Combining equations (21)∼(25), we can obtain the characteristic equation of Figure 2(a) as follows:(26)λ2+b2/3E11/35+4α+5αk21+α+αk2/3λ+3b4/3E12/31+α+αk2/32=0.For convenience, we can make(27)β=b2/3E11/35+4α+5αk21+α+αk2/3.(28)γ=3b4/3E12/31+α+αk2/32.Two characteristic roots of equation (26) can be obtained according to a quadratic formula:(29)λ1,2=−β±β2−4γ2.Sinceb > 0 and E1 > 0, from equations (19) and (28), it is clear that γ is always greater than 0.According to Vieta theorem, we know thatλ1 + λ2 = −β and λ1λ2 = γ. Owing to γ > 0, when β > 0, we know that λ1 and λ2 are all less than 0, or when β < 0, it is clear that λ1 and λ2 are all greater than 0.In summary, according to equation (27), if the following expression is satisfied, λ1 and λ2 are all less than 0 and the equilibrium point is stable:(30)5+4α+5αk>0.If the following expression is satisfied,λ1 and λ2 are all greater than 0 and the equilibrium point is unstable:(31)5+4α+5αk<0.


### 3.1.2. Stability Analysis in Figure2(a) (α Is Fixed and k Is Control Parameter)

(1)
No equilibrium points exist in Figure2(a)If no equilibrium points exist in Figure2(a), from expression (19), we know that the following expression is satisfied:(32)1+α+αk<0.Solving inequality (32), when α > 0, we can obtain that k≤1+α/α.(2)
Unstable equilibrium points exist in Figure2(a)If unstable equilibrium points exist in Figure2(a), from expressions (19) and (31), we know that the following expression is satisfied:(33)1+α+αk>0,5+4α+5αk<0,α>0.It is clear that expression (33) can be simplified as follows:(34)k>−1+αα,k<−5+4α5α,α>0.Whenα > 0, it is clear that −1+α/α−−5+4α/5α<0 and −1+α/α<−5+4α/5α.Solving expression (34), when α > 0, we can obtain that −1+α/α<k<−5+4α/5α.(3)
Stable equilibrium points exist in Figure2(a)If stable equilibrium points exist in Figure2(a), from expressions (19) and (30), we know that the following expression is satisfied:(35)1+α+αk>0,5+4α+5αk>0,α>0.It is clear that expression (35) can be simplified as follows:(36)k≥1+αα,k≥5+4α5α,α>0.Solving expression (36), when α > 0, we can obtain that k≥5+4α/5α.


### 3.1.3. Analysis of Equilibrium Points and Characteristic Roots in Figure2(b)

The steady state solutions (equilibrium points) of equations (12)∼(13) in Figure 2(b) can be denoted as ΔT01,ΔT02. In order to solve the equilibrium points, the right sides of equations (12)∼(13) can be equal to zero, and thus we can obtain the following equations:(37)b−ΔT1−αΔT2ΔT1−ΔT2=E1,(38)b−ΔT1−αΔT2ΔT2=E2.After equation (38) plus equation (37), the following equation can be easily obtained:(39)ΔT01=1+kkΔT02.Substituting equation (39) into equation (38), we can obtain equilibrium points as follows:(40)ΔT023=−k3E12b21+k+αk,(41)ΔT013=−1+k3E12b21+k+αk.According to equation (38), since b > 0, when E2 > 0, it is clear that ΔT02 > 0, and when E2 < 0, we have ΔT02 < 0.Besides, to ensure that solutions exist for equations (37)∼(38), we have(42)ΔT01+αΔT02=1+k+kαkΔT02<0.Owing toE1 > 0 and k = E2/E1, when E2 > 0, it is clear from equation (40) that k > 0 and ΔT02 > 0. From expression (42), we known that if 1 + k + αk < 0, equilibrium points exist in Figure 2(b).WhenE2 < 0, it is clear that k < 0 and ΔT02 < 0, we known that if 1 + k + αk < 0, equilibrium points exist in Figure 2(b).In conclusion, when the following expression is met, equilibrium points exist in Figure2(b):(43)M=1+k+αk<0.From expression (43), equations (40)∼(41) can be simplified as follows:(44)ΔT02=kE12/3b2/3−1−k−αk1/3,(45)ΔT01=1+kE12/3b2/3−1−k−αk1/3.When the characteristic determinant is equal to 0, the solution of the differential equation can be discussed. The expression of characteristic determinant is shown in the following equation:(46)JQ−λE=∂f1∂ΔT1−λ∂f1∂ΔT2∂f2∂ΔT1∂f1∂ΔT2=0.Combining equations (12)∼(13), (46), the following equation can be obtained:(47)bB2A−bA−λbαB2A+bAbT022AbαT022A−bA−λ=0.For convenience, we can makeA=−ΔT01−αΔT02 and B=ΔT01−ΔT02.According to equations (44)∼(45), we can obtain the following equations:(48)A=−1−k−αk1/3E11/3b1/3,(49)B=E12/3b2/3−1−k−αk1/3=ΔT02k,(50)ΔT02A=kE11/3b1/3−1−k−αk2/3,(51)BA=E11/3b1/3−1−k−αk2/3.Combining equations (47)∼(51), we can obtain characteristic equation of Figure 2(b) as follows:(52)λ2−b2/3E11/3k1/35+4k+5αk2−1−k−αk2/3λ+3−1−k−αk2/3b4/3E12/32=0.For convenience, we can make(53)β=−b2/3E11/3k1/35+4k+5αk2−1−k−αk2/3,(54)γ=3−1−k−αk2/3b4/3E12/32.Two characteristic roots of equation (52) can be obtained according to a quadratic formula:(55)λ1,2=−β±β2−4γ2.Sinceb > 0 and E1 > 0, from equations (43) and (54), it is clear that γ is always greater than 0.According to Vieta theorem, we know thatλ1 + λ2 = −β and λ1λ2 = γ. Owing to γ > 0, when β > 0, we know that λ1 and λ2 are all less than 0, or when β < 0, we know that λ1 and λ2 are all greater than 0.In summary, according to equation (53), if the following expression is satisfied, λ1 and λ2 are all less than 0 and the equilibrium point is stable:(56)5+4k+5αk<0.If the following expression is satisfied,λ1 and λ2 are all greater than 0 and the equilibrium point is unstable:(57)5+4k+5αk>0.


### 3.1.4. Stability Analysis in Figure2(b) (α Is Fixed and k Is Control Parameter)

(1)
No equilibrium points exist in Figure2(b)If no equilibrium points exist in Figure2(b), from discussion between expressions (42)∼(43), we know that the following expression is satisfied:(58)1+k+αk>0.Solving expression (58), when α > 0, we can obtain that k≥1/1+α.(2)
Unstable equilibrium points exist in Figure2(b)If unstable equilibrium points exist in Figure2(b), from expression (43) and expression (57), we know that the following expression is satisfied:(59)1+k+αk<0,5+4k+5αk>0,α>0.The coefficient in front ofk is extracted, and expression (59) can be simplified as follows:(60)k<−11+α,k>−54+5α,α>0.Whenα > 0, it is clear that −5/4+5α−−1/1+α<0 and −5/4+5α<−1/1+α.From expression (60), when α > 0, we can obtain that −5/4+5α<k<−1/1+α.(3)
Stable equilibrium points exist in Figure2(b)If stable equilibrium points exist in Figure2(b), from expression (43) and expression (56), we know that the following expression is satisfied:(61)1+k+αk<0,5+4k+5αk<0,α>0.The coefficient in front ofk is extracted, and expression (61) can be simplified as follows:(62)k<−11+α,k<−54+5α,α>0.From expression (62), when α > 0, we can obtain that k<−5/4+5α.In a word, from the above discussion, we can form a criterion shown in Table1 to determine the stability of the system shown in Figure 2 when α is fixed and k is the control parameter. For any α > 0, according to the values of k in Table 1, we can evaluate whether there is an equilibrium point and the equilibrium point is stable or unstable in scenario 1(α is fixed and k is the control parameter).Table 1
Stability criterion for scenario 1(α is fixed and k is the control parameter).
αkFigure2(a)Figure2(b)0,+∞−∞,−1+α/αNoStable−1+α/α,−5+4α/5αUnstableStable−5+4α/5α,−5/4+5αStableStable−5/4+5α,−1/1+αStableUnstable−1/1+α,+∞StableNo


## 3.1.1. Analysis of Equilibrium Points and Characteristic Roots in Figure2(a)

The steady state solutions (equilibrium points) of equations (10)∼(11) in Figure 2(a) can be denoted as ΔT01,ΔT02. In order to solve the equilibrium points, the right sides of equations (10)∼(11) can be equal to zero, and thus we can obtain the following equations:(14)bΔT1+αΔT2ΔT1=E1,(15)bΔT1+αΔT2ΔT2−ΔT1=E2.After equation (14) plus equation (15), the following equation can be easily obtained:(16)ΔT02=1+kΔT01.Substituting equation (16) into equation (14), we can obtain equilibrium points as follows:(17)ΔT01=E12/3b2/31+α+αk1/3,(18)ΔT02=1+kE12/3b2/31+α+αk1/3.For convenience, we can makeM=1+α+αk. To ensure that a real solution exists for equations (17)∼(18) and equilibrium points exist, we have(19)M=1+α+αk>0.When the characteristic determinant is equal to 0, the solution of the differential equation can be discussed by characteristic roots. The expression of characteristic determinant is shown as follows:(20)JQ−λE=∂f1∂ΔT1−λ∂f1∂ΔT2∂f2∂ΔT1∂f1∂ΔT2=0.Combining equations (10), (11), and (20), the following equation can be obtained:(21)−bΔT012A−bA−λ−bαΔT012A−bB2A+bA−bαB2A−bA−λ=0,where J(Q) is the Jacobian matrix of equilibrium point Q, λ is the characteristic root, E is the identity matrix, and for convenience, we can make A=ΔT01+αΔT02 and B=ΔT02−ΔT01.According to equations (17)∼(18), we can obtain the following equations:(22)A=1+α+αk1/3E11/3b1/3,(23)B=kE12/3b2/31+α+αk1/3=kΔT01,(24)ΔT01A=E11/3b1/31+α+αk2/3,(25)BA=kE11/3b1/31+α+αk2/3.Combining equations (21)∼(25), we can obtain the characteristic equation of Figure 2(a) as follows:(26)λ2+b2/3E11/35+4α+5αk21+α+αk2/3λ+3b4/3E12/31+α+αk2/32=0.For convenience, we can make(27)β=b2/3E11/35+4α+5αk21+α+αk2/3.(28)γ=3b4/3E12/31+α+αk2/32.Two characteristic roots of equation (26) can be obtained according to a quadratic formula:(29)λ1,2=−β±β2−4γ2.Sinceb > 0 and E1 > 0, from equations (19) and (28), it is clear that γ is always greater than 0.According to Vieta theorem, we know thatλ1 + λ2 = −β and λ1λ2 = γ. Owing to γ > 0, when β > 0, we know that λ1 and λ2 are all less than 0, or when β < 0, it is clear that λ1 and λ2 are all greater than 0.In summary, according to equation (27), if the following expression is satisfied, λ1 and λ2 are all less than 0 and the equilibrium point is stable:(30)5+4α+5αk>0.If the following expression is satisfied,λ1 and λ2 are all greater than 0 and the equilibrium point is unstable:(31)5+4α+5αk<0.


## 3.1.2. Stability Analysis in Figure2(a) (α Is Fixed and k Is Control Parameter)

(1)
No equilibrium points exist in Figure2(a)If no equilibrium points exist in Figure2(a), from expression (19), we know that the following expression is satisfied:(32)1+α+αk<0.Solving inequality (32), when α > 0, we can obtain that k≤1+α/α.(2)
Unstable equilibrium points exist in Figure2(a)If unstable equilibrium points exist in Figure2(a), from expressions (19) and (31), we know that the following expression is satisfied:(33)1+α+αk>0,5+4α+5αk<0,α>0.It is clear that expression (33) can be simplified as follows:(34)k>−1+αα,k<−5+4α5α,α>0.Whenα > 0, it is clear that −1+α/α−−5+4α/5α<0 and −1+α/α<−5+4α/5α.Solving expression (34), when α > 0, we can obtain that −1+α/α<k<−5+4α/5α.(3)
Stable equilibrium points exist in Figure2(a)If stable equilibrium points exist in Figure2(a), from expressions (19) and (30), we know that the following expression is satisfied:(35)1+α+αk>0,5+4α+5αk>0,α>0.It is clear that expression (35) can be simplified as follows:(36)k≥1+αα,k≥5+4α5α,α>0.Solving expression (36), when α > 0, we can obtain that k≥5+4α/5α.


## 3.1.3. Analysis of Equilibrium Points and Characteristic Roots in Figure2(b)

The steady state solutions (equilibrium points) of equations (12)∼(13) in Figure 2(b) can be denoted as ΔT01,ΔT02. In order to solve the equilibrium points, the right sides of equations (12)∼(13) can be equal to zero, and thus we can obtain the following equations:(37)b−ΔT1−αΔT2ΔT1−ΔT2=E1,(38)b−ΔT1−αΔT2ΔT2=E2.After equation (38) plus equation (37), the following equation can be easily obtained:(39)ΔT01=1+kkΔT02.Substituting equation (39) into equation (38), we can obtain equilibrium points as follows:(40)ΔT023=−k3E12b21+k+αk,(41)ΔT013=−1+k3E12b21+k+αk.According to equation (38), since b > 0, when E2 > 0, it is clear that ΔT02 > 0, and when E2 < 0, we have ΔT02 < 0.Besides, to ensure that solutions exist for equations (37)∼(38), we have(42)ΔT01+αΔT02=1+k+kαkΔT02<0.Owing toE1 > 0 and k = E2/E1, when E2 > 0, it is clear from equation (40) that k > 0 and ΔT02 > 0. From expression (42), we known that if 1 + k + αk < 0, equilibrium points exist in Figure 2(b).WhenE2 < 0, it is clear that k < 0 and ΔT02 < 0, we known that if 1 + k + αk < 0, equilibrium points exist in Figure 2(b).In conclusion, when the following expression is met, equilibrium points exist in Figure2(b):(43)M=1+k+αk<0.From expression (43), equations (40)∼(41) can be simplified as follows:(44)ΔT02=kE12/3b2/3−1−k−αk1/3,(45)ΔT01=1+kE12/3b2/3−1−k−αk1/3.When the characteristic determinant is equal to 0, the solution of the differential equation can be discussed. The expression of characteristic determinant is shown in the following equation:(46)JQ−λE=∂f1∂ΔT1−λ∂f1∂ΔT2∂f2∂ΔT1∂f1∂ΔT2=0.Combining equations (12)∼(13), (46), the following equation can be obtained:(47)bB2A−bA−λbαB2A+bAbT022AbαT022A−bA−λ=0.For convenience, we can makeA=−ΔT01−αΔT02 and B=ΔT01−ΔT02.According to equations (44)∼(45), we can obtain the following equations:(48)A=−1−k−αk1/3E11/3b1/3,(49)B=E12/3b2/3−1−k−αk1/3=ΔT02k,(50)ΔT02A=kE11/3b1/3−1−k−αk2/3,(51)BA=E11/3b1/3−1−k−αk2/3.Combining equations (47)∼(51), we can obtain characteristic equation of Figure 2(b) as follows:(52)λ2−b2/3E11/3k1/35+4k+5αk2−1−k−αk2/3λ+3−1−k−αk2/3b4/3E12/32=0.For convenience, we can make(53)β=−b2/3E11/3k1/35+4k+5αk2−1−k−αk2/3,(54)γ=3−1−k−αk2/3b4/3E12/32.Two characteristic roots of equation (52) can be obtained according to a quadratic formula:(55)λ1,2=−β±β2−4γ2.Sinceb > 0 and E1 > 0, from equations (43) and (54), it is clear that γ is always greater than 0.According to Vieta theorem, we know thatλ1 + λ2 = −β and λ1λ2 = γ. Owing to γ > 0, when β > 0, we know that λ1 and λ2 are all less than 0, or when β < 0, we know that λ1 and λ2 are all greater than 0.In summary, according to equation (53), if the following expression is satisfied, λ1 and λ2 are all less than 0 and the equilibrium point is stable:(56)5+4k+5αk<0.If the following expression is satisfied,λ1 and λ2 are all greater than 0 and the equilibrium point is unstable:(57)5+4k+5αk>0.


## 3.1.4. Stability Analysis in Figure2(b) (α Is Fixed and k Is Control Parameter)

(1)
No equilibrium points exist in Figure2(b)If no equilibrium points exist in Figure2(b), from discussion between expressions (42)∼(43), we know that the following expression is satisfied:(58)1+k+αk>0.Solving expression (58), when α > 0, we can obtain that k≥1/1+α.(2)
Unstable equilibrium points exist in Figure2(b)If unstable equilibrium points exist in Figure2(b), from expression (43) and expression (57), we know that the following expression is satisfied:(59)1+k+αk<0,5+4k+5αk>0,α>0.The coefficient in front ofk is extracted, and expression (59) can be simplified as follows:(60)k<−11+α,k>−54+5α,α>0.Whenα > 0, it is clear that −5/4+5α−−1/1+α<0 and −5/4+5α<−1/1+α.From expression (60), when α > 0, we can obtain that −5/4+5α<k<−1/1+α.(3)
Stable equilibrium points exist in Figure2(b)If stable equilibrium points exist in Figure2(b), from expression (43) and expression (56), we know that the following expression is satisfied:(61)1+k+αk<0,5+4k+5αk<0,α>0.The coefficient in front ofk is extracted, and expression (61) can be simplified as follows:(62)k<−11+α,k<−54+5α,α>0.From expression (62), when α > 0, we can obtain that k<−5/4+5α.In a word, from the above discussion, we can form a criterion shown in Table1 to determine the stability of the system shown in Figure 2 when α is fixed and k is the control parameter. For any α > 0, according to the values of k in Table 1, we can evaluate whether there is an equilibrium point and the equilibrium point is stable or unstable in scenario 1(α is fixed and k is the control parameter).Table 1
Stability criterion for scenario 1(α is fixed and k is the control parameter).
αkFigure2(a)Figure2(b)0,+∞−∞,−1+α/αNoStable−1+α/α,−5+4α/5αUnstableStable−5+4α/5α,−5/4+5αStableStable−5/4+5α,−1/1+αStableUnstable−1/1+α,+∞StableNo


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

The characteristic equation of scenario 2 is the same as that of scenario 1.


### 3.2.1. Stability Analysis in Figure2(a) (k Is Fixed and α Is Control Parameter)

(1)
No equilibrium points exist in Figure2(a)In Section3.1.2 ordinal (1), we know that no equilibrium points exist in Figure 2(a), when the following inequalities (taking out the coefficient of α in expression (32)) are satisfied:(63)1+kα+1<0,α>0.Whenk > −1, solving expression (63), we can obtain that α<−1/1+k. Owing to α > 0, it is clear that −1/1+k<0 which contradicts k > −1. Therefore, from expression (63), when k > −1, we know that α does not exist.Whenk < −1, solving expression (63), it is clear that α>−1/1+k. Since −1/1+k>0, from expression (63), when k < −1, we know that α>−1/1+k.In conclusion, we can obtain the following two cases:Case 1:k > −1 and α does not existCase 2:k < −1 and α>−1/1+k(2)
Unstable equilibrium points exist in Figure2(a)In Section3.1.2 ordinal (2), we know that unstable equilibrium points exist in Figure 2(a), when the following inequalities (taking out the coefficient of α in expression (33)) are satisfied:(64)1+kα+1>0,4+5kα+5<0,α>0,whenk < −1, it is clear that 1 + k < 0 and 4 + 5k < 0. Expression (64) is simplified as(65)α<−11+k,α>−54+5k,α>0.We know that fork < −1, since 1 + k < 0 and 4 + 5k < 0, we have −1/1+k−−5/4+5k>0, and it is clear that −1/1+k>−5/4+5k, −1/1+k>0, and −5/4+5k>0.Therefore, from expression (65), when k < −1, we know that −5/4+5k<α<−1/1+k.When −1<k < −4/5, it is clear that 1 + k > 0 and 4 + 5k < 0. Expression (64) is simplified as(66)α>−11+k,α>−54+5k,α>0.We know that, for −1 <k < −4/5, since 1 + k > 0 and 4 + 5k < 0, we have −1/1+k−−5/4+5k<0, and it is clear that −1/1+k<−5/4+5k and −1/1+k<0, −5/4+5k>0.Therefore, from expression (66), when −1 < k < −4/5, we know that −5/4+5k<α.Whenk > −4/5, it is clear that 1 + k > 0 and 4 + 5k > 0. Expression (64) is simplified as(67)α>−11+k,α<−54+5k,α>0.We know that, fork > −4/5, since 1 + k > 0 and 4 + 5k > 0, we have −1/1+k−−5/4+5k>0, and it is clear that −1/1+k>−5/4+5k and −1/1+k>0, −5/4+5k>0.Therefore, from expression (67), when k > −4/5, we know that α does not exist.In conclusion, we can obtain following several cases:Case 1:k < −1 and −5/4+5k<α<−1/1+kCase 2: 1 <k < −4/5 and −5/4+5k<αCase 3: −4/5 <k and α does not exist(3)
Stable equilibrium points exist in Figure2(a)In Section3.1.2 ordinal (3), we know that stable equilibrium points exist in Figure 2(a), when the following inequalities (taking out the coefficient of α in expression (35)) are satisfied:(68)1+kα+1>0,4+5kα+5>0,α>0.Referring to the classification results of unstable equilibrium points about expression (64), we can obtain the following several cases about expression (68).Case 1:k < −1 and 0<α<−5/4+5kCase 2: 1 <k < −4/5 and 0<α<−5/4+5kCase 3: −4/5 <k and α>0


### 3.2.2. Stability Analysis in Figure2(b) (k Is Fixed and α Is Control Parameter)

(1)
No equilibrium points exist in Figure2(b)In Section3.1.4 ordinal (1), we know that no equilibrium points exist in Figure 2(b), when the following inequalities (derived from expression (58)) are satisfied:(69)αk+1+k>0,α>0.Whenk > 0, solving expression (69), we can obtain that α>−1+k/k. Owing to α > 0, it is clear that −1+k/k<0. Therefore, from expression (69), when k > 0, we know that α > 0.Whenk < 0, solving expression (69), it is clear that α<−1+k/k. To ensure −1+k/k>0, we have −1 < k < 0; therefore, from expression (69), when −1 < k < 0, we know that 0<α<−1+k/k, and when k < −1, α does not exist.In conclusion, we can obtain following several cases:Case 1:k > 0 and α > 0Case 2: −1<k < 0 and 0<α<−1+k/kCase 3:k < −1 and α does not exist(2)
Unstable equilibrium points exist in Figure2(b)In Section3.1.4 ordinal (2), we know that unstable equilibrium points exist in Figure 2(b), when the following inequalities (derived from expression (59) are satisfied:(70)αk+1+k<0,5αk+5+4k>0,α>0.Whenk > 0, expression is simplified as(71)α<−1+kk,α>−5+4k5k,α>0.We know that, for anyk > 0, we have −1+k/k−−5+4k/5k<0, and it is clear that −1+k/k<−5+4k/5k which is in contradiction with expression (71).Therefore, from expression (71), when k > 0, we know that α does not exist.Whenk < 0, expression (70) is simplified as(72)α>−1+kk,α<−5+4k5k,α>0.It is clear that, for anyk < 0, we have −1+k/k<−5+4k/5k.Whenk < −5/4, we know that −1+k/k<0 and −5+4k/5k<0. Therefore, from expression (72), when k < −5/4, we know that α does not exist.When −5/4 <k < −1, we know that −1+k/k<0 and −5+4k/5k>0. Therefore, from expression (72), when −5/4 < k < −1, we know that 0<α<−5+4k/5k.When −1 <k < 0, we know that −1+k/k>0 and −5+4k/5k>0. Therefore, from expression (72), when −1 < k < 0, we know that −1+k/k<α<−5+4k/5k.In conclusion, we can obtain following several cases:Case 1:k > 0 and α does not existCase 2:k < −5/4 and α does not existCase 3: −5/4<k < −1 and 0<α<−5+4k/5kCase 4: −1<k < 0 and −1+k/k<α<−5+4k/5k(3)
Stable equilibrium points exist in Figure2(b)In Section3.1.4 ordinal (3), we know that stable equilibrium points exist in Figure 2(b), when the following inequalities (derived from expression (61) are satisfied:(73)αk+1+k<0,5αk+5+4k<0,α>0.Whenk > 0, expression (73) is simplified as(74)α<−1+kk,α<−5+4k5k,α>0.We know that, for anyk > 0, we have −1+k/k<−5+4k/5k. Owing to α > 0, when k > 0, it is clear that −1+k/k<0 and −5+4k/5k<0.Therefore, from expression (74), when k > 0, we know that α does not exist.Whenk < 0, expression (73) is simplified as(75)α>−1+kk,α>−5+4k5k,α>0.For anyk < 0, we know that −1+k/k<−5+4k/5k.Whenk < −5/4, we know that −1+k/k<0 and −5+4k/5k<0. Therefore, from expression (75), when k < −5/4, we know that α > 0.When −5/4 <k < −1, we know that −1+k/k<0 and −5+4k/5k>0. Therefore, from expression (75), when −5/4 < k < −1, we know that α>−5+4k/5k.When −1 <k < 0, we know that −1+k/k>0 and −5+4k/5k>0. Therefore, from expression (75), when −1 < k < 0, we know that α>−5+4k/5k.In conclusion, we can obtain following several cases:Case 1:k > 0 and α does not existCase 2:k < −5/4 and α > 0Case 3: −5/4 <k < −1 and α>−5+4k/5kCase 4: −1 <k < 0 and α>−5+4k/5kIn brief, for the scenario 2,k can be discussed in the following intervals: k<−5/4, −5/4<k<−1, −1<k<−4/5, −4/5<k<0, and 0<k.From the above discussion, whenK is fixed and α is the control parameter, we can form a criterion shown in Table 2 to determine the stability of the system shown in Figure 2. According to the values of α in Table 2, we can know whether there is an equilibrium point and whether the equilibrium point is stable or not.Table 2
Stability criterion for scenario 2(k is fixed and α is the control parameter).
kαFigure2(a)Figure2(b)−∞,−5/40,−5/4+5kStableStable−5/4+5k,−1/1+kUnstableStable−1/1+k,+∞NoStable−5/4,−10,−5+4k/5kStableUnstable−5+4k/5k,−5/4+5kStableStable−5/4+5k,−1/1+kUnstableStable−1/1+k,+∞NoStable−1,−4/50,−1+k/kStableNo−1+k/k,−5+4k/5kStableUnstable−5+4k/5k,−5/4+5kStableStable−5/4+5k,+∞UnstableStable−4/5,00,−1+k/kStableNo−1+k/k,−5+4k/5kStableUnstable−5+4k/5k,+∞StableStable0,+∞0,+∞StableNoCompared with Tables1–2 in this paper and theoretical analysis results in [44], it can be seen that more stable states and unstable states exist in two vertically connected open cavities with unequal heights in our paper.


## 3.2.1. Stability Analysis in Figure2(a) (k Is Fixed and α Is Control Parameter)

(1)
No equilibrium points exist in Figure2(a)In Section3.1.2 ordinal (1), we know that no equilibrium points exist in Figure 2(a), when the following inequalities (taking out the coefficient of α in expression (32)) are satisfied:(63)1+kα+1<0,α>0.Whenk > −1, solving expression (63), we can obtain that α<−1/1+k. Owing to α > 0, it is clear that −1/1+k<0 which contradicts k > −1. Therefore, from expression (63), when k > −1, we know that α does not exist.Whenk < −1, solving expression (63), it is clear that α>−1/1+k. Since −1/1+k>0, from expression (63), when k < −1, we know that α>−1/1+k.In conclusion, we can obtain the following two cases:Case 1:k > −1 and α does not existCase 2:k < −1 and α>−1/1+k(2)
Unstable equilibrium points exist in Figure2(a)In Section3.1.2 ordinal (2), we know that unstable equilibrium points exist in Figure 2(a), when the following inequalities (taking out the coefficient of α in expression (33)) are satisfied:(64)1+kα+1>0,4+5kα+5<0,α>0,whenk < −1, it is clear that 1 + k < 0 and 4 + 5k < 0. Expression (64) is simplified as(65)α<−11+k,α>−54+5k,α>0.We know that fork < −1, since 1 + k < 0 and 4 + 5k < 0, we have −1/1+k−−5/4+5k>0, and it is clear that −1/1+k>−5/4+5k, −1/1+k>0, and −5/4+5k>0.Therefore, from expression (65), when k < −1, we know that −5/4+5k<α<−1/1+k.When −1<k < −4/5, it is clear that 1 + k > 0 and 4 + 5k < 0. Expression (64) is simplified as(66)α>−11+k,α>−54+5k,α>0.We know that, for −1 <k < −4/5, since 1 + k > 0 and 4 + 5k < 0, we have −1/1+k−−5/4+5k<0, and it is clear that −1/1+k<−5/4+5k and −1/1+k<0, −5/4+5k>0.Therefore, from expression (66), when −1 < k < −4/5, we know that −5/4+5k<α.Whenk > −4/5, it is clear that 1 + k > 0 and 4 + 5k > 0. Expression (64) is simplified as(67)α>−11+k,α<−54+5k,α>0.We know that, fork > −4/5, since 1 + k > 0 and 4 + 5k > 0, we have −1/1+k−−5/4+5k>0, and it is clear that −1/1+k>−5/4+5k and −1/1+k>0, −5/4+5k>0.Therefore, from expression (67), when k > −4/5, we know that α does not exist.In conclusion, we can obtain following several cases:Case 1:k < −1 and −5/4+5k<α<−1/1+kCase 2: 1 <k < −4/5 and −5/4+5k<αCase 3: −4/5 <k and α does not exist(3)
Stable equilibrium points exist in Figure2(a)In Section3.1.2 ordinal (3), we know that stable equilibrium points exist in Figure 2(a), when the following inequalities (taking out the coefficient of α in expression (35)) are satisfied:(68)1+kα+1>0,4+5kα+5>0,α>0.Referring to the classification results of unstable equilibrium points about expression (64), we can obtain the following several cases about expression (68).Case 1:k < −1 and 0<α<−5/4+5kCase 2: 1 <k < −4/5 and 0<α<−5/4+5kCase 3: −4/5 <k and α>0


## 3.2.2. Stability Analysis in Figure2(b) (k Is Fixed and α Is Control Parameter)

(1)
No equilibrium points exist in Figure2(b)In Section3.1.4 ordinal (1), we know that no equilibrium points exist in Figure 2(b), when the following inequalities (derived from expression (58)) are satisfied:(69)αk+1+k>0,α>0.Whenk > 0, solving expression (69), we can obtain that α>−1+k/k. Owing to α > 0, it is clear that −1+k/k<0. Therefore, from expression (69), when k > 0, we know that α > 0.Whenk < 0, solving expression (69), it is clear that α<−1+k/k. To ensure −1+k/k>0, we have −1 < k < 0; therefore, from expression (69), when −1 < k < 0, we know that 0<α<−1+k/k, and when k < −1, α does not exist.In conclusion, we can obtain following several cases:Case 1:k > 0 and α > 0Case 2: −1<k < 0 and 0<α<−1+k/kCase 3:k < −1 and α does not exist(2)
Unstable equilibrium points exist in Figure2(b)In Section3.1.4 ordinal (2), we know that unstable equilibrium points exist in Figure 2(b), when the following inequalities (derived from expression (59) are satisfied:(70)αk+1+k<0,5αk+5+4k>0,α>0.Whenk > 0, expression is simplified as(71)α<−1+kk,α>−5+4k5k,α>0.We know that, for anyk > 0, we have −1+k/k−−5+4k/5k<0, and it is clear that −1+k/k<−5+4k/5k which is in contradiction with expression (71).Therefore, from expression (71), when k > 0, we know that α does not exist.Whenk < 0, expression (70) is simplified as(72)α>−1+kk,α<−5+4k5k,α>0.It is clear that, for anyk < 0, we have −1+k/k<−5+4k/5k.Whenk < −5/4, we know that −1+k/k<0 and −5+4k/5k<0. Therefore, from expression (72), when k < −5/4, we know that α does not exist.When −5/4 <k < −1, we know that −1+k/k<0 and −5+4k/5k>0. Therefore, from expression (72), when −5/4 < k < −1, we know that 0<α<−5+4k/5k.When −1 <k < 0, we know that −1+k/k>0 and −5+4k/5k>0. Therefore, from expression (72), when −1 < k < 0, we know that −1+k/k<α<−5+4k/5k.In conclusion, we can obtain following several cases:Case 1:k > 0 and α does not existCase 2:k < −5/4 and α does not existCase 3: −5/4<k < −1 and 0<α<−5+4k/5kCase 4: −1<k < 0 and −1+k/k<α<−5+4k/5k(3)
Stable equilibrium points exist in Figure2(b)In Section3.1.4 ordinal (3), we know that stable equilibrium points exist in Figure 2(b), when the following inequalities (derived from expression (61) are satisfied:(73)αk+1+k<0,5αk+5+4k<0,α>0.Whenk > 0, expression (73) is simplified as(74)α<−1+kk,α<−5+4k5k,α>0.We know that, for anyk > 0, we have −1+k/k<−5+4k/5k. Owing to α > 0, when k > 0, it is clear that −1+k/k<0 and −5+4k/5k<0.Therefore, from expression (74), when k > 0, we know that α does not exist.Whenk < 0, expression (73) is simplified as(75)α>−1+kk,α>−5+4k5k,α>0.For anyk < 0, we know that −1+k/k<−5+4k/5k.Whenk < −5/4, we know that −1+k/k<0 and −5+4k/5k<0. Therefore, from expression (75), when k < −5/4, we know that α > 0.When −5/4 <k < −1, we know that −1+k/k<0 and −5+4k/5k>0. Therefore, from expression (75), when −5/4 < k < −1, we know that α>−5+4k/5k.When −1 <k < 0, we know that −1+k/k>0 and −5+4k/5k>0. Therefore, from expression (75), when −1 < k < 0, we know that α>−5+4k/5k.In conclusion, we can obtain following several cases:Case 1:k > 0 and α does not existCase 2:k < −5/4 and α > 0Case 3: −5/4 <k < −1 and α>−5+4k/5kCase 4: −1 <k < 0 and α>−5+4k/5kIn brief, for the scenario 2,k can be discussed in the following intervals: k<−5/4, −5/4<k<−1, −1<k<−4/5, −4/5<k<0, and 0<k.From the above discussion, whenK is fixed and α is the control parameter, we can form a criterion shown in Table 2 to determine the stability of the system shown in Figure 2. According to the values of α in Table 2, we can know whether there is an equilibrium point and whether the equilibrium point is stable or not.Table 2
Stability criterion for scenario 2(k is fixed and α is the control parameter).
kαFigure2(a)Figure2(b)−∞,−5/40,−5/4+5kStableStable−5/4+5k,−1/1+kUnstableStable−1/1+k,+∞NoStable−5/4,−10,−5+4k/5kStableUnstable−5+4k/5k,−5/4+5kStableStable−5/4+5k,−1/1+kUnstableStable−1/1+k,+∞NoStable−1,−4/50,−1+k/kStableNo−1+k/k,−5+4k/5kStableUnstable−5+4k/5k,−5/4+5kStableStable−5/4+5k,+∞UnstableStable−4/5,00,−1+k/kStableNo−1+k/k,−5+4k/5kStableUnstable−5+4k/5k,+∞StableStable0,+∞0,+∞StableNoCompared with Tables1–2 in this paper and theoretical analysis results in [44], it can be seen that more stable states and unstable states exist in two vertically connected open cavities with unequal heights in our paper.


## 4. Numerical Simulations

To further study the nonlinear dynamical behavior of the proposed model, we solve the differential equation numerically by the forth-order Runge–Kutta method. The area of each room is 3 m (wide) × 3 m (high). The height of lower room is fixed as 4 m. There are three openings of 0.3 m2. One of the middle partitions is used to connect the two rooms, and the top and bottom partitions are used to two connect the two rooms to the outdoors, respectively. Section 4.1 describes the bifurcation diagrams and phase portraits of scenario 1, where the height ratio is fixed and the heat ratio is the control parameter; Section 4.2 describes the bifurcation diagrams and the phase portraits of scenario 2, where the heat strength ratio is fixed and the height ratio is the control parameter.


### 4.1. Fixed Height Ratioα and Different Heat Strength Ratio k

For scenario 1, we useα = 1 as an example to study the effect of control parameter k on the stability of the system. The bifurcation diagram for ventilation flow rates is plotted in Figure 3(a), and the bifurcation diagram for air temperature is plotted in Figure 3(b). In Figure 3(a), there is only one stable equilibrium point for the flow rate in the zone between Ad and Bd, where the heat ratio is k < −2. The air flow in this zone is downward and the temperature in the upper room is less than that in the lower room, as given in Figure 3(b). In the zone between Eu and Fu, there is also one stable equilibrium point for the flow rate. In the zone between Eu and Fu, when k is less than zero, there is a heat source in the lower room and a heat sink in the upper room and the air temperature in the upper room is lower than that in the lower room. The air temperature in the upper room equals to that in the lower room when k = 0. The heat sink in the upper room turns to a heat source when k further increases.Figure 3
Bifurcation diagram. (a) Air flow rate. (b) Temperature. In the symbols,X#∗, X(A–F) indicates different points on the curves, the subscript #(1, 2) indicates the zone number, and ∗d,u refer to downward and upward flows, respectively.
(a)(b)The flow in the zone between Cu and Du has two stable equilibrium points. When the flow rate is positive, the flow is upward and the air temperature in the lower room is higher than that in the upper room. When the flow rate is negative, the flow is downward and the same air temperature trend can be observed in Figure 3(b). The flow in the zones between Bu and Cu and between Dd and Ed has one stable equilibrium point and one unstable equilibrium point, respectively. At the point Bu, the flow is upward and the temperatures in both rooms are infinite, which indicates that the upward flow is unstable. It can also be given that the upward flow between Bu and Cu is unstable. With the same principle, an unstable equilibrium point is generated from infinity at Ed. The equilibrium point between Dd and Ed is also unstable.By solving the system equations numerically using the fourth-order Runge–Kutta method, we plot the phase portraits of the model. Based on the bifurcation diagram, six typical values ofk are selected, which are −2.5, −1.85, −1.75, −0.56, −0.52, and −0.5. In Figure 4, we give the phase portraits for the six heat source ratios. When k = −2.5 and k = −0.5, the orbits from different initial conditions converge to one equilibrium point. When k = −1.85 and k = −0.52, a stable equilibrium point and an unstable equilibrium point can be seen. When k = −1.75 and k = −0.56, we see two stable equilibrium points and unstable periodic orbits may also be seen in both phase portraits. The numerical simulated phase portraits agree well with the theoretical analysis results as shown in Table 1.Figure 4
Phase portraits for different heat ratios. (a)k = −2.5. (b) k = −1.85. (c) k = −1.75. (d) k = −0.65. (e) k = −0.52. (f) k = −0.5.
(a)(b)(c)(d)(e)(f)


### 4.2. Fixed Heat Strength and Different Height Ratio

For scenario 2, we use four typical values ofk, which are −2, −1.2, −0.9, and −0.5 to study the effect of control parameter α on the stability of the system. When k is fixed to −2 and α varies from 0 to 1.5, the bifurcation diagram for the airflow rates and air temperatures are plotted in Figure 5(a) and Figure 5(b), respectively. In Figure 5(a), there are two stable equilibrium points between Au (or Ad) and Bu (or Bd). The air temperature in the lower room is higher than that in the upper room when the flow is upward. When the flow is downward, the same trend can be observed. In Figure 5(a), it is shown that an unstable equilibrium point is generated from infinity at the point Cu. This is shown in Figure 5(b) where the air temperature in both the lower room and the upper room is infinite. It can also be seen that the upward flow between Bu and Cu in Figure 5(a) is also unstable. There is only one stable equilibrium point in the zone between Cd and Dd for the air flow rate. The air flow is downward. The air temperature in the lower room is higher than that in the upper room.Figure 5
Bifurcation diagram. (a) Air flow rate. (b) Temperature.
(a)(b)In Figure6, we give the phase portraits for α selected as 0.5, 0.85, and 1.5, when k is fixed to −2. When α = 0.5, we see two stable equilibrium points in the phase portrait. When α = 0.85, there is a stable equilibrium point and an unstable equilibrium point. When α = 1.5, there is only one stable equilibrium point. The numerical simulated phase portraits agree well with the theoretical analysis results as shown in Table 2.Figure 6
Phase portraits for different height ratios ask = −2. (a) α = 0.5. (b) α = 0.85. (c) α = 1.5.
(a)(b)(c)The bifurcation diagram for the airflow rates and air temperatures, whenk is fixed to −1.2 and α varies from 0 to 6, is plotted in Figures 7(a) and 7(b), respectively. In Figure 7(a), there are two stable equilibrium points between Au (or Ad) and Bu (or Bd). In Figure 7(a), it is shown that an unstable equilibrium point is generated from infinity at the point Cu. This is shown in Figure 7(b) where the air temperature in both the lower room and the upper room is infinite. It can also be seen that the upward flow between Bu and Cu in Figure 7(a) is also unstable. There is only one stable equilibrium point in the zone between Cd and Dd for the air flow rate. The air flow is downward. The air temperature in the lower room is higher than that in the upper room.Figure 7
Bifurcation diagram. (a) Air flow rate. (b) Temperature.
(a)(b)In Figure8, we give the phase portraits for α selected as 0.01, 1, 3, and 6, when k is fixed as −1.2. When α = 0.01, we see a stable equilibrium point and an unstable equilibrium point in the phase portrait. When α = 1, there are two stable equilibrium points. When α = 3, there is one stable equilibrium point and one unstable equilibrium point. When α = 6, there is only one stable equilibrium point in the phase portrait. The numerical simulated phase portraits agree well with the theoretical analysis results as shown in Table 2.Figure 8
Phase portraits for different height ratios ask = −1.2. (a) α = 0.01. (b) α = 1. (c) α = 3. (d) α = 6.
(a)(b)(c)(d)The bifurcation diagram for the airflow rates and air temperatures, whenk is fixed to −0.9 and α varies from 0 to 12, is plotted in Figures 9(a) and 9(b), respectively. In Figure 9(a), there is only one stable equilibrium point between Au and Bu. In Figure 7(a), it is shown that an unstable equilibrium point is generated from infinity at the point Bu. This is shown in Figure 7(b) where the air temperature in both the lower room and the upper room is infinite. It can also be seen that the upward flow between Bu and Cu in Figure 7(a) is also unstable. In the zone between Cu (or Cd) and Du (or Dd), there are two stable equilibrium points. There is one stable equilibrium point and one unstable equilibrium point in the zone between Dd and Ed for the air flow rate.Figure 9
Bifurcation diagram. (a) Air flow rate. (b) Temperature.
(a)(b)In Figure10, we give the phase portraits for α selected as 0.05, 0.2, 1, and 12 and when k is fixed as −0.9. When α = 0.05, we see only one stable equilibrium point in the phase portrait. When α = 0.2, there is one stable equilibrium point and one unstable equilibrium point. When α = 1, there are two stable equilibrium points. When α = 12, there is one stable equilibrium point and one unstable equilibrium point. The numerical simulated phase portraits agree well with the theoretical analysis as shown in Table 2.Figure 10
Phase portraits for different height ratios ask = −0.9. (a) α = 0.05. (b) α = 0.2. (c) α = 1. (d) α = 12.
(a)(b)(c)(d)Figure11 shows the bifurcation diagram for the airflow rates and air temperatures, when k is fixed to −0.5 and α varies from 0 to 2, respectively. In Figure 11(a), there is only one stable equilibrium point between Au and Bu. In Figure 11(a), it is shown that an unstable equilibrium point is generated from infinity at the point Bd. This is shown in Figure 11(b) where the air temperature in both the lower room and the upper room is infinite. It can also be seen that the upward flow between Bu and Cu in Figure 11(a) is also unstable. In the zone between Cu (or Cd) and Du (or Dd), there are two stable equilibrium points.Figure 11
Bifurcation diagram. (a) Air flow rate. (b) Temperature.
(a)(b)In Figure12, we give the phase portraits for α selected as 0.5, 1.1, and 2 and when k is fixed to −0.5. When α = 0.5, we see only one stable equilibrium point in the phase portrait. When α = 1.1 there is one stable equilibrium point and one unstable equilibrium point. When α = 2, there are two stable equilibrium points. The numerical simulated phase portraits agree well with the theoretical analysis results as shown in Table 2.Figure 12
Phase portraits for different height ratios ask = −0.5. (a) α = 0.5. (b) α = 1.1. (c) α = 2.
(a)(b)(c)


## 4.1. Fixed Height Ratioα and Different Heat Strength Ratio k

For scenario 1, we useα = 1 as an example to study the effect of control parameter k on the stability of the system. The bifurcation diagram for ventilation flow rates is plotted in Figure 3(a), and the bifurcation diagram for air temperature is plotted in Figure 3(b). In Figure 3(a), there is only one stable equilibrium point for the flow rate in the zone between Ad and Bd, where the heat ratio is k < −2. The air flow in this zone is downward and the temperature in the upper room is less than that in the lower room, as given in Figure 3(b). In the zone between Eu and Fu, there is also one stable equilibrium point for the flow rate. In the zone between Eu and Fu, when k is less than zero, there is a heat source in the lower room and a heat sink in the upper room and the air temperature in the upper room is lower than that in the lower room. The air temperature in the upper room equals to that in the lower room when k = 0. The heat sink in the upper room turns to a heat source when k further increases.Figure 3
Bifurcation diagram. (a) Air flow rate. (b) Temperature. In the symbols,X#∗, X(A–F) indicates different points on the curves, the subscript #(1, 2) indicates the zone number, and ∗d,u refer to downward and upward flows, respectively.
(a)(b)The flow in the zone between Cu and Du has two stable equilibrium points. When the flow rate is positive, the flow is upward and the air temperature in the lower room is higher than that in the upper room. When the flow rate is negative, the flow is downward and the same air temperature trend can be observed in Figure 3(b). The flow in the zones between Bu and Cu and between Dd and Ed has one stable equilibrium point and one unstable equilibrium point, respectively. At the point Bu, the flow is upward and the temperatures in both rooms are infinite, which indicates that the upward flow is unstable. It can also be given that the upward flow between Bu and Cu is unstable. With the same principle, an unstable equilibrium point is generated from infinity at Ed. The equilibrium point between Dd and Ed is also unstable.By solving the system equations numerically using the fourth-order Runge–Kutta method, we plot the phase portraits of the model. Based on the bifurcation diagram, six typical values ofk are selected, which are −2.5, −1.85, −1.75, −0.56, −0.52, and −0.5. In Figure 4, we give the phase portraits for the six heat source ratios. When k = −2.5 and k = −0.5, the orbits from different initial conditions converge to one equilibrium point. When k = −1.85 and k = −0.52, a stable equilibrium point and an unstable equilibrium point can be seen. When k = −1.75 and k = −0.56, we see two stable equilibrium points and unstable periodic orbits may also be seen in both phase portraits. The numerical simulated phase portraits agree well with the theoretical analysis results as shown in Table 1.Figure 4
Phase portraits for different heat ratios. (a)k = −2.5. (b) k = −1.85. (c) k = −1.75. (d) k = −0.65. (e) k = −0.52. (f) k = −0.5.
(a)(b)(c)(d)(e)(f)


## 4.2. Fixed Heat Strength and Different Height Ratio

For scenario 2, we use four typical values ofk, which are −2, −1.2, −0.9, and −0.5 to study the effect of control parameter α on the stability of the system. When k is fixed to −2 and α varies from 0 to 1.5, the bifurcation diagram for the airflow rates and air temperatures are plotted in Figure 5(a) and Figure 5(b), respectively. In Figure 5(a), there are two stable equilibrium points between Au (or Ad) and Bu (or Bd). The air temperature in the lower room is higher than that in the upper room when the flow is upward. When the flow is downward, the same trend can be observed. In Figure 5(a), it is shown that an unstable equilibrium point is generated from infinity at the point Cu. This is shown in Figure 5(b) where the air temperature in both the lower room and the upper room is infinite. It can also be seen that the upward flow between Bu and Cu in Figure 5(a) is also unstable. There is only one stable equilibrium point in the zone between Cd and Dd for the air flow rate. The air flow is downward. The air temperature in the lower room is higher than that in the upper room.Figure 5
Bifurcation diagram. (a) Air flow rate. (b) Temperature.
(a)(b)In Figure6, we give the phase portraits for α selected as 0.5, 0.85, and 1.5, when k is fixed to −2. When α = 0.5, we see two stable equilibrium points in the phase portrait. When α = 0.85, there is a stable equilibrium point and an unstable equilibrium point. When α = 1.5, there is only one stable equilibrium point. The numerical simulated phase portraits agree well with the theoretical analysis results as shown in Table 2.Figure 6
Phase portraits for different height ratios ask = −2. (a) α = 0.5. (b) α = 0.85. (c) α = 1.5.
(a)(b)(c)The bifurcation diagram for the airflow rates and air temperatures, whenk is fixed to −1.2 and α varies from 0 to 6, is plotted in Figures 7(a) and 7(b), respectively. In Figure 7(a), there are two stable equilibrium points between Au (or Ad) and Bu (or Bd). In Figure 7(a), it is shown that an unstable equilibrium point is generated from infinity at the point Cu. This is shown in Figure 7(b) where the air temperature in both the lower room and the upper room is infinite. It can also be seen that the upward flow between Bu and Cu in Figure 7(a) is also unstable. There is only one stable equilibrium point in the zone between Cd and Dd for the air flow rate. The air flow is downward. The air temperature in the lower room is higher than that in the upper room.Figure 7
Bifurcation diagram. (a) Air flow rate. (b) Temperature.
(a)(b)In Figure8, we give the phase portraits for α selected as 0.01, 1, 3, and 6, when k is fixed as −1.2. When α = 0.01, we see a stable equilibrium point and an unstable equilibrium point in the phase portrait. When α = 1, there are two stable equilibrium points. When α = 3, there is one stable equilibrium point and one unstable equilibrium point. When α = 6, there is only one stable equilibrium point in the phase portrait. The numerical simulated phase portraits agree well with the theoretical analysis results as shown in Table 2.Figure 8
Phase portraits for different height ratios ask = −1.2. (a) α = 0.01. (b) α = 1. (c) α = 3. (d) α = 6.
(a)(b)(c)(d)The bifurcation diagram for the airflow rates and air temperatures, whenk is fixed to −0.9 and α varies from 0 to 12, is plotted in Figures 9(a) and 9(b), respectively. In Figure 9(a), there is only one stable equilibrium point between Au and Bu. In Figure 7(a), it is shown that an unstable equilibrium point is generated from infinity at the point Bu. This is shown in Figure 7(b) where the air temperature in both the lower room and the upper room is infinite. It can also be seen that the upward flow between Bu and Cu in Figure 7(a) is also unstable. In the zone between Cu (or Cd) and Du (or Dd), there are two stable equilibrium points. There is one stable equilibrium point and one unstable equilibrium point in the zone between Dd and Ed for the air flow rate.Figure 9
Bifurcation diagram. (a) Air flow rate. (b) Temperature.
(a)(b)In Figure10, we give the phase portraits for α selected as 0.05, 0.2, 1, and 12 and when k is fixed as −0.9. When α = 0.05, we see only one stable equilibrium point in the phase portrait. When α = 0.2, there is one stable equilibrium point and one unstable equilibrium point. When α = 1, there are two stable equilibrium points. When α = 12, there is one stable equilibrium point and one unstable equilibrium point. The numerical simulated phase portraits agree well with the theoretical analysis as shown in Table 2.Figure 10
Phase portraits for different height ratios ask = −0.9. (a) α = 0.05. (b) α = 0.2. (c) α = 1. (d) α = 12.
(a)(b)(c)(d)Figure11 shows the bifurcation diagram for the airflow rates and air temperatures, when k is fixed to −0.5 and α varies from 0 to 2, respectively. In Figure 11(a), there is only one stable equilibrium point between Au and Bu. In Figure 11(a), it is shown that an unstable equilibrium point is generated from infinity at the point Bd. This is shown in Figure 11(b) where the air temperature in both the lower room and the upper room is infinite. It can also be seen that the upward flow between Bu and Cu in Figure 11(a) is also unstable. In the zone between Cu (or Cd) and Du (or Dd), there are two stable equilibrium points.Figure 11
Bifurcation diagram. (a) Air flow rate. (b) Temperature.
(a)(b)In Figure12, we give the phase portraits for α selected as 0.5, 1.1, and 2 and when k is fixed to −0.5. When α = 0.5, we see only one stable equilibrium point in the phase portrait. When α = 1.1 there is one stable equilibrium point and one unstable equilibrium point. When α = 2, there are two stable equilibrium points. The numerical simulated phase portraits agree well with the theoretical analysis results as shown in Table 2.Figure 12
Phase portraits for different height ratios ask = −0.5. (a) α = 0.5. (b) α = 1.1. (c) α = 2.
(a)(b)(c)


## 5. Conclusions

Nonlinear dynamical analysis was performed to study the solution multiplicity of buoyancy ventilation in two vertically connected open cavities with unequal heights and variable heat source strength. A new mathematical model of multiple solutions of buoyancy pressure ventilation for two vertically connected open cavities with unequal heights and hot sourceE2 < 0 or E2 > 0 in upper room was presented. By solving and analyzing the equilibrium points and characteristic roots of the differential equations, we analyze the stability of two systems with upward flow pattern and downward pattern and obtain the criteria to determine the stability and existence of solutions for two scenarios. The criteria for the stability and existence of equilibrium points were derived mathematically in detail about two scenarios. The criterion for scenario 1 (α = H2/H1 is fixed and k = E2/E1 is control parameter) is summarized in Table 1. The criterion for scenario 2 (k is fixed and α is control parameter) is summarized in Table 2. Finally, bifurcation diagrams and the phase portraits obtained by applying the fourth-order Runge–Kutta method are given.



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