#### 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.

#### 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) *E*_{1} > 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.

##### 2.1. Description of Mathematical Model

The proposed new schematics of underground structure is shown as Figure 1, 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 *H*_{1} and *H*_{2} respectively, the *T*_{i}, *q*_{i} (*i* = 1∼4) denote the temperature, and mass flow rate of four zones, respectively. Heat sources *E*_{1} and *E*_{2} 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:

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:

According to the conservation of mass, we can obtain

Combining equations (5) and (6), mass flow rate can be obtained as follows:

For realization 2, similarly, we can obtain the heat balance equations of the four zones as follows:

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:

According to the conservation of mass, we can obtain following equations about air flow rate:

By combining equations (11) and (12), we can obtain

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

Assuming *k* = *E*_{2}/*E*_{1}, Δ*T*_{1} = *T*_{1} − *T*_{a}, Δ*T*_{2} = *T*_{2} − *T*_{a}, Δ*T*_{3} = *T*_{3} − *T*_{a}, Δ*T*_{4} = *T*_{4} − *T*_{a}, , *α* = *H*_{2}/*H*_{1}, *M*_{1} = *M*_{2} = *M*, *M*_{3} = *M*_{4} = *M*/*ξ*, and *τ* = *t*/(*MC*_{p}), 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:

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:

##### 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 conditions *k* > 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 . 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:

By solving equations (23)∼(26), we can obtain the equilibrium points as follows:

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

Combining equations (15)∼(18), we can obtain the following expression:where is the Jacobian matrix of equilibrium point *Q*, *λ* is the characteristic root, *E* is the identity matrix, and

Combining equations (31)∼(34), we can obtain characteristic roots as follows:

Assuming *k* = *E*_{2}/*E*_{1}, *α* = *H*_{2}/*H*_{1}, to ensure that a real solution exists for equations (27) and (28), we have

Two other characteristic roots can be obtained according to a quadratic formula:where

For convenience, the characteristic roots are set as follows: the real root are *γ*_{i} (*i* = 1,2,3,4); the complex roots are *σ*_{i} + *jω*_{i} (*i* = 1,2,3,4). It is easy to know that *λ*_{1,2} are always real number, *λ*_{1} = *γ*_{1} = *λ*_{2} = *γ*_{2.}

If the discriminant Δ is greater than 0, both characteristic roots are real number, namely, *λ*_{3} = *γ*_{3} and *λ*_{4} = *γ*_{4}. From equations (37) and (38), we can know that when Δ > 0, the following expression holds:

It is clear that

When the discriminant Δ is less than 0, conjugate complex roots exist, namely, *λ*_{3,4} = *σ*_{3,4} ± *jω*_{3,4}. From equations (37) and (38), we can know that when Δ < 0, the following expression holds:

*(2) Stability Analysis for Realization 1 (k Is Fixed and α Is Control Parameter)*. (1)No equilibrium point exists in realization 1 If no equilibrium point exists in realization 1, from equation (36), it is known that the following expression is satisfied: Therefore from expression (42), when *k* < −1, it is clear that and when *k* > −1, we know that .(2)A stable equilibrium point exists in realization 1 (2.1) Assuming there are all negative real characteristic roots In this case, *λ*_{1}, *λ*_{2}, *λ*_{3}, and *λ*_{4} are all less than 0, and the system has a stable solution. We know that owing to *αk* − 1 + *α* > 0, from equation (35), it is clear that *λ*_{1} and *λ*_{2} are always less than 0. To make *λ*_{3,4} < 0, according to Vieta theorem, we know that *λ*_{3} + *λ*_{4} = −*b* and *λ*_{3}*λ*_{4} = *c*. From equation (38), it is known that always *c* > 0, and owing to *αk* − 1 + *α* > 0, from equation (38), if 5*αk* − 5 + 4*α* > 0, we have −*b* < 0, then *λ*_{3,4} < 0. In this case, the absolute value sign in equation (40) can be removed; therefore, the following expression (43) can be obtained: From expression (43), we can obtain the following expression: In summary, if there are all negative real characteristic roots, the system is stable when the following inequalities are satisfied: (2.2) Assuming two negative real roots and two conjugate complex roots In this case, the four characteristic roots behave as *λ*_{1} = *γ*_{1} = *λ*_{2} = *γ*_{2}, *λ*_{3,4} = *σ*_{3,4} ± *jω*_{3,4}, where *λ*_{1} and *λ*_{2} are two negative real roots and *λ*_{3}, *λ*_{4} are two conjugate complex roots. When *λ*_{1}, *λ*_{2}, *σ*_{3}, and *σ*_{4} are all less than 0, the system has a stable solution. From equation (37), we know *σ*_{3,4} = −*b*/2. Owing to *λ*_{1,2} < 0, to make *σ*_{3,4} < 0, we need −*b* < 0, and from equation (38), we can obtain 5*αk* + 4*α* − 5>0. In this case, the absolute value sign in expression (41) can be removed; therefore, the following expression can be obtained: From (46), we can obtain that In summary, *λ*_{1}, *λ*_{2}, *σ*_{3}, and *σ*_{4} are all less than 0, the system is stable when the following inequalities are satisfied: Combining expressions (45) and (48), a stable equilibrium point exists in realization 1, when the following expression is satisfied: Case 1: when *k* > −4/5, it is easy to know that 5*k* + 4 > 0 and *k* + 1 > 0, then expression (49) can be simplified to From 5*k* + 4 > 0 and *k* + 1 > 0, it is clear that and , Therefore, from expression (49), it can be known that, when , Case 2: when , it is easy to know that 5*k* + 4 < 0 and *k* + 1 > 0. Expression (49) can be simplified to Owing to 5*k* + 4 < 0 and *k* + 1 > 0, it is clear that and . Therefore, when , *α* does not exist. Case 3: when *k* < −1, it is easy to know that 5*k* + 4 < 0 and *k* + 1 < 0. Expression (49) can be simplified to Owing to 5*k* + 4 < 0 and *k* + 1 < 0, it is clear that . Thus we can obtain that . Therefore, when , *α* does not exist. In conclusion, we can obtain following several cases: Case 1: and Case 2: and *α* does not exist Case 3: 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 positive As long as one characteristic root is greater than 0, the equilibrium point of the system is unstable. Owing to *αk* − 1 + *α* > 0, *λ*_{1} and *λ*_{2} are always less than 0 and only *λ*_{3} or *λ*_{4} is greater than 0. According to Vieta theorem, *λ*_{3} + *λ*_{4} = −*b* and *λ*_{3}*λ*_{4} = *c*; it is known that *c* > 0, and owing to *αk* − 1 + *α* > 0, from equation (38), if 5*αk* − 5 + 4*α* < 0, we can obtain −*b* > 0, then *λ*_{3,4} > 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: From expression (53), we can obtain In summary, *λ*_{1} and *λ*_{2} are less than 0 and *λ*_{3} and *λ*_{4} 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: Firstly, we consider the coefficient of *α* on the left side of third equation in expression (55) is greater than 0, namely, , we can obtain . Then, we consider the coefficient of *α* on the left side of the third equation in expression (55) is less than 0, namely, , we can obtain . It is easy to know that . We discuss following several cases: Case 1: when , owing to , it is easy to know that , and . Expression (55) is simplified as From , , , it is clear that and . So we have and . Therefore, when , it can be obtained that . Case 2: when , owing to , it is easy to know that 5*k* + 4 > 0, *k* + 1 > 0 and . Expression (55) can be simplified to From , , , it is clear that and . So we have . Therefore, when , we have Case 3: when , owing to , it is easy to know that , , and . Expression (55) can be simplified to From , , , it is easy to know that , so we have and Therefore, when , we can obtain that . Case 4: when , owing to , it is easy to know that , , and . Equation (55) can be simplified to From , and , it is clear that and , so we have . Therefore, when , *α* does not exist. From what has been discussed above, we can obtain following relation between *k* and *α*: Case 1: and Case 2: and Case 3: and Case 4: and *α* does not exist. (3.2) Assuming two negative real roots and two conjugate complex roots In this case, the four characteristic roots behave as *λ*_{1} = *γ*_{1} = *λ*_{2} = *γ*_{2}, *λ*_{3,4} = *σ*_{3,4} ± *jω*_{3,4}, where *λ*_{1} and *λ*_{2} are two negative real roots and *λ*_{3} and *λ*_{4} are two conjugate complex roots. Owing to *λ*_{1} and *λ*_{2} are always less than 0, when *σ*_{3,4} 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 *σ*_{3,4} = −*b*/2. To make *σ*_{3,4} > 0, we need −*b* > 0, and from equation (38), we can obtain 5*αk* + 4*α* − 5 < 0. In this case, the absolute value sign in Expression (41) can be removed; therefore, the following expression can be obtained: From expression (60), we can obtain that In summary, *λ*_{1} and *λ*_{2} are less than 0, *σ*_{3} and *σ*_{4} 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: According to above discussion, we can obtain the following several cases: Case 1: and Case 2: and Case 3: and *α* does not exist Case 4: and *α* does not exist

If *k* 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 . 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:

By solving equations (63)∼(66), we can obtain the equilibrium points as follows:

According to the characteristic determinant equal to 0, the solution of the differential equation can be discussed. The characteristic determinant equation is that

Combining equations (19)∼(22), we can obtain the following expression as follows:

Combining equations (71)∼(74), we can obtain characteristic roots as follows:

To ensure that a real solution exists for equations (67) and (68), we have

Two other characteristic roots can be obtained according to the quadratic formula:where

Therefore, it is easy to know that *λ*_{1,2} are always a real number, *λ*_{1} = *γ*_{1} = *λ*_{2} = *γ*_{2.} If the discriminant Δ is greater than 0, both characteristic roots are real number:*λ*_{3} = *γ*_{3}, *λ*_{4} = *γ*_{4}. From equations (77) and (78), we can know that when Δ > 0, the following expression holds:

It is clear that

When the discriminant Δ is less than 0, conjugate complex roots exist, namely, *λ*_{3,4} = *σ*_{3,4} ± j*ω*_{3,4}. From equations (77) and (78), we can know that when Δ < 0, the following expression holds:

*(4) Stability Analysis for Realization 2 ( k Is Fixed and α Is Control Parameter)*. (1)No equilibrium point exists in realization 2 If no equilibrium point exists in realization 2, from equations (67)∼(68), it is known that the following expression is satisfied: Therefore, from equation (82), when

*k*< −1, it is clear that and when −1 <

*k*< 0, we know that

*α*does not exist, or when 0 <

*k*, we know that .(2)A stable equilibrium point exists in realization 2 (2.1) Assuming there are all negative real characteristic roots In this case,

*λ*

_{1},

*λ*

_{2},

*λ*

_{3}, and

*λ*

_{4}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

*λ*

_{1}and

*λ*

_{2}are always less than 0. To make

*λ*

_{3,4}< 0, according to Vieta theorem, we know that

*λ*

_{3}+

*λ*

_{4}= −

*b*and

*λ*

_{3}

*λ*

_{4}=

*c*. From equation (78), it is known that always

*c*> 0, and owing to

*k*+ 1 −

*αk*> 0, from equation (78), if 4

*k*+ 5 − 5

*αk*> 0, we have −

*b*< 0, then

*λ*

_{3,4}< 0. In this case, the absolute value sign in equation (80) can be removed; therefore, the following expression can be obtained: From expression (83), we can obtain the following expression: In summary, if there are all negative real characteristic roots, the system is stable when the following inequalities are satisfied: (2.2) Assuming two real roots and two conjugate complex roots In this case, the four characteristic roots behave as

*λ*

_{1}=

*γ*

_{1}=

*λ*

_{2}=

*γ*

_{2}and

*λ*

_{3,4}=

*σ*

_{3,4}±

*jω*

_{3,4}, where

*λ*

_{1}and

*λ*

_{2}are two negative real roots and

*λ*

_{3}and

*λ*

_{4}are two conjugate complex roots. When

*λ*

_{1},

*λ*

_{2},

*σ*

_{3}, and

*σ*

_{4}are all less than 0, the system has a stable solution. From equation (77), we know

*σ*

_{3,4}= −

*b*/2. Owing to

*λ*

_{1,2}< 0, to make

*σ*

_{3,4}< 0, we have −

*b*< 0, from equation (78), we can obtain 4

*k*+ 5 − 5

*αk*> 0. In this case, the absolute value sign in expression (81) can be removed; therefore, the following expression (86) can be obtained: From expression (86), we can obtain that In summary, the system is stable when the following inequalities are satisfied: Combining expressions (85) and (88), a stable equilibrium point exists in realization 2, and the following equations are satisfied: Case 1: when

*k*> 0, expression (89) can be simplified to When

*k*> 0, it is easy to know that and , and it is clear that and . Therefore, from expression (90), when

*k*> 0, we know that . Case 2: when

*k*< 0, expression (89) can be simplified to It is clear that and . Owing to

*α*> 0, when −1 <

*k*< 0, it clear that . Therefore, from expression (91), when −1 <

*k*< 0, we know that

*α*> 0. When

*k*< −1, it clear that . Therefore, from expression (91), when

*k*< −1, we know that . In conclusion, we can obtain following several cases: Case 1: and Case 2: and Case 3: and (3)An unstable equilibrium point exists in realization 1 (3.1)Assuming characteristic roots are all real and at least one is a positive As long as one characteristic root is greater than 0, the equilibrium point of the system is unstable. Owing to

*k*+ 1 −

*αk*> 0,

*λ*

_{1}and

*λ*

_{2}are always less than 0 and only

*λ*

_{3}or

*λ*

_{4}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 4

*k*+ 5 − 5

*αk*< 0, we have −

*b*> 0, then

*λ*

_{3,4}> 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: From expression (92), we can obtain In summary,

*λ*

_{1}and

*λ*

_{2}are less than 0 and

*λ*

_{3}and

*λ*

_{4}greater than 0. The system equilibrium point is the saddle point of index 2, when the following inequalities are satisfied: It is easy to know that for any

*k*, and , so we have .We discuss following several cases: Case 1: when

*k*> 0, expression (94) is simplified as Therefore, when

*k*> 0, we have . Case 2: when

*k*< 0, expression (94) is simplified as Therefore, when

*k*< 0,

*α*does not exist. In conclusion, we can obtain following several cases: Case 1:

*k*> 0 and Case 2:

*k*< 0 and

*α*does not exist If

*k*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 roots In this case, the four characteristic roots behave as

*λ*

_{1}=

*γ*

_{1}=

*λ*

_{2}=

*γ*

_{2}and

*λ*

_{3,4}=

*σ*

_{3,4}±

*jω*

_{3,4}, where

*λ*

_{1}and

*λ*

_{2}are two negative real roots and

*λ*

_{3}and

*λ*

_{4}are two conjugate complex roots. Owing to

*λ*

_{1}and

*λ*

_{2}are always less than 0, when

*σ*

_{3}and

*σ*

_{4}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

*σ*

_{3,4}= −

*b*/2. To make

*σ*

_{3,4}> 0, we have −

*b*> 0. From equation (78), it is known that 4

*α*− 5 + 5

*α*k < 0. In this case, the absolute value sign in expression (81) can be removed; therefore, the following expression (97) can be obtained: From expression (97), we can obtain that In summary,

*λ*

_{1}and

*λ*

_{2}are less than 0 and

*σ*

_{3}and

*σ*

_{4}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: By referring the classification results of expression (94), we can obtain the following several cases: Case 1:

*k*> 0 and Case 2:

*k*< 0 and

*α*does not exist If

*k*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.

We compared Table 1 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 , and this interval can be divided into two parts, namely, , where the unstable equilibrium point is the saddle point of index 2 and , 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 , and we divide this interval into two parts, namely, , where the unstable equilibrium point is the saddle point of index 2 and , 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.

We compared Table 2 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 1 In 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: Therefore, from expression (98), when *α* > 0, it is clear that .(2)A stable equilibrium point exists in realization 1 In 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: Taking out the coefficient *k* of expression (101), we can obtain expression (102): Owing to *α* > 0, expression (102) are simplified as Owing to *α* > 0, it is clear that and ; therefore, from expression (101), when *α* > 0, we know that .(3)An unstable equilibrium point exists in realization 1 (3.1) Assuming characteristic roots are all real and at least one is a positive In 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: Taking out the coefficient *k* of expression (104), we can obtain the following expression: Owing to *α* > 0, expression (105) is simplified: