Study on the Stability and Entropy Complexity of

entropy Article

Study on the Stability and Entropy Complexity of an Energy-Saving and Emission-Reduction Model with Two Delays Jing Wang 1, * and Yuling Wang 2 1 2

*

Department of Mathematics and Physics, Bengbu University, Bengbu 233030, China School of Economics, South-Central University for Nationalities, Wuhan 430074, China; [email protected] Correspondence: [email protected]; Tel.: +86-552-3170-920

Academic Editor: J. A. Tenreiro Machado Received: 2 August 2016; Accepted: 13 October 2016; Published: 19 October 2016

Abstract: In this paper, we build a model of energy-savings and emission-reductions with two delays. In this model, it is assumed that the interaction between energy-savings and emission-reduction and that between carbon emissions and economic growth are delayed. We examine the local stability and the existence of a Hopf bifurcation at the equilibrium point of the system. By employing System Complexity Theory, we also analyze the impact of delays and the feedback control on stability and entropy of the system are analyzed from two aspects: single delay and double delays. In numerical simulation section, we test the theoretical analysis by using means bifurcation diagram, the largest Lyapunov exponent diagrams, attractor, time-domain plot, Poincare section plot, power spectrum, entropy diagram, 3-D surface chart and 4-D graph, the simulation results demonstrating that the inappropriate changes of delays and the feedback control will result in instability and fluctuation of carbon emissions. Finally, the bifurcation control is achieved by using the method of variable feedback control. Hence, we conclude that the greater the value of the control parameter, the better the effect of the bifurcation control. The results will provide for the development of energy-saving and emission-reduction policies. Keywords: energy-savings; emission-reduction; entropy; stability; two delays; Hopf bifurcation

1. Introduction Energy resources are the backbone of any national economy. Scholars worldwide have been studying energy prices, the low-carbon economy and energy-savings and emission-reduction. Lv and Zhou [1] studied the dynamic behavior of the energy price model with time delay. They explained the reasons why the energy price model was developed and maintained periodically by the bifurcation theory. Tian et al. [2] developed a nonlinear model for the development of oil, gas and other energy resources against the backdrop of the energy structure in Jiangsu, China, one that predominantly relies on the coal consumption, and evidenced empirically the feasibility of the corresponding energy countermeasures. Despite the fact that energy promotes economic development and improves people’s living standards, excessive energy consumption undoubtedly damages the environment, so it is necessary to adopt a low-carbon economy and reduce the possible environmental damage. In fact, the energy-saving and emission-reduction system is a complex nonlinear system, which involves the interactions between a variety of factors such as energy-saving and emission-reduction, carbon emissions, economic growth, energy efficiency, carbon tax and energy intensity, and so forth [3–5]. In view of this, a novel three-dimensional energy-saving and emission-reduction chaotic system is proposed [6].

Entropy 2016, 18, 371; doi:10.3390/e18100371

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The system is established in accordance with the complicated relationship among energy-savings and emission-reduction, carbon emissions and economic growth. This system displays a very complex phenomenon by including a special chaotic attractor named the energy-saving and emission-reduction attractor (the EE attractor for short), which is different from the previous chaotic attractor, such as Lorenz attractor [7], Chen attractor [8], Lü attractor [9], Energy resource attractor [10–12] and so on. Moreover, drawing up rational and effective policies and laws will ensure a better progress in energy-saving and emission-reduction [13,14]. Wang and Xu [15] reported a new four-dimensional energy-saving and emission-reduction chaotic system. The system is obtained in accordance with the complicated relationship among energy-saving and emission-reduction, carbon emission, economic growth and new energy development. The model is described as follows:  . y(t)  x (t) = a1 x (t)( M − 1) − a2 y(t) + a3 z(t),    y(t) z(t)  . y(t) = −b1 x (t) + b2 y(t)(1 − C ) + b3 z(t)(1 − E ) − b4 u(t), . x (t) u(t)   z(t) = c1 x (t)( N − 1) − c2 y(t) − c3 z(t) + c4 u(t)( L − 1),    . z(t) u(t) = d1 y(t) + d2 z(t)( K − 1) − d3 u(t),

(1)

where x (t) is the time-dependent variable of energy-saving and emission-reduction; y(t) the time-dependent variable of carbon emissions; z(t) the time-dependent variable of economic growth (GDP) and u(t) the time-dependent variable of new energy development. ai , di , b j , c j , (i = 1, 2, 3; j = 1, 2, 3, 4) are coefficients and M, N, L, K, C, E positive constants [15]. Model (1) stands for the development and utilization of new energy sources. Therefore, the four-dimensional model of energy-saving and emission-reduction is a step closer to the actual conditions. In view of the current mode of economic development and technological conditions, economic development will increase the consumption of coal and oil, and in turn this will cause an increase in carbon emissions. However, there are many other approaches to promote economic growth, such as those in the modern service sector that are low energy-consuming but high value-added, so economic development may not immediately cause a substantial energy consumption load. In other words, it might not lead to an obvious increase in carbon emissions in the short term. In this essay, we have included the delay between economic development and the significant growth of carbon emission in this kind of economic model. Moreover, energy-savings and emission-reduction reduce carbon emissions by saving or reducing energy consumption. However, due to the complexity of the economic system, the implementation of energy-saving and emission-reduction measures normally will not lead to a significant reduction, showing a certain lag. Meanwhile, energy-saving and emission-reduction will slow down the pace of economic development. This will force enterprises to change their mode of economic development, strengthen energy conservation and efficient use, and actively comply with a recycling economy and low carbon economy, so energy-savings and emission-reduction will promote the sound and fast development of economy, but it still demonstrates an obvious delay. Therefore, the impact of energy-saving and emission-reduction on carbon emission and economic development is estimated as delayed. On the basis of Model (1), we propose a two-delay model as follows:  . y(t)  x (t) = a1 x (t)( M − 1) − a2 y(t) + a3 z(t − τ1 ),    y(t) z(t−τ )  . y(t) = −b1 x (t − τ2 ) + b2 y(t)(1 − C ) + b3 z(t − τ1 )(1 − E 1 ) − b4 u(t), . x (t−τ ) u(t)  z(t) = c1 x (t − τ2 )( N 2 − 1) − c2 y(t) − c3 z(t) + c4 u(t)( L − 1),     . z(t) u(t) = d1 y(t) + d2 z(t)( K − 1) − d3 u(t),

(2)

where τ1 represents the delay time between economic development and carbon emission, and τ2 represents the delay time among energy-saving and emission-reduction, economic development and carbon emission. The rest of this paper is organized as follows: in Section 2, we focus on the local stability and the existence of Hopf bifurcation at the equilibrium point. In Section 3, we study the effects of delays and

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the feedback control on the stability of the system. In Section 4, the effective Hopf bifurcation control of the system is examined. Finally, conclusions are drawn in Section 5. 2. Equilibrium Points and Local Stability The calculation formulas of equilibrium points of Equation (2) are fairly complicated, so the equilibrium points can be calculated directly by using specific parameters value in numerical simulation section. We assume that Equation (2) has an equilibrium point, called hereafter by E( x ∗ , y∗ , z∗ , u∗ ). It means that x (t), y(t), z(t) and u(t) can be in an equilibrium state through dynamic game. In this paper, we focus on the influence of τ1 , τ2 and the feedback control on the stability of system (2) at the equilibrium point. Next, Equation (2) is linearized at the equilibrium point E( x ∗ , y∗ , z∗ , u∗ ) by Jacobian matrix as follows:  . ∗ y∗ x (t) = ( a1 M − a1 ) x (t) + ( a1 xM − a2 )y(t) + a3 z(t − τ1 ),    ∗ .  2b y∗ y(t) = −b1 x (t − τ2 ) + (b2 − C2 )y(t) + (b3 − 2bE3 z )z(t − τ1 ) − b4 u(t), (3) ∗ ∗ . 1x  − c1 ) x (t − τ2 ) − c2 y(t) − c3 z(t) + ( 2c4Lu − c4 )u(t), z(t) = ( 2cN   ∗  . u(t) = d1 y(t) + ( 2dK2 z − d2 )z(t) − d3 u(t), The characteristic equation of Equation (3) is:

λ − J11 − J21 e−λτ2 − J31 e−λτ2 − J41

where: J11 = ( a1

− J12 λ − J22 − J32 − J42

− J14 − J24 − J34 λ − J44

=0

y∗ x∗ − a1 ), J12 = a1 − a2 , J13 = a3 , J14 = 0, M M

J21 = −b1 , J22 = b2 − J31 = (

− J13 e−λτ1 − J23 e−λτ1 λ − J33 − J43

2b2 y∗ 2b3 z∗ , J23 = (b3 − ), J24 = −b4 , C E

2c u∗ 2c1 x ∗ − c1 ), J32 = −c2 , J33 = −c3 , J34 = 4 − c4 , N L

J41 = 0, J42 = d1 , J43 =

2d2 z∗ − d2 , J44 = −d3 . K

We further get: λ4 + A3 λ3 + A2 λ2 + A1 λ + A0 + ( B2 λ2 + B1 λ + B0 )e−λτ1 + (C2 λ2 + C1 λ + C0 )e−λτ2 + ( D2 λ2 + D1 λ + D0 )e−λ(τ1 +τ2 ) = 0 where: A3 = − J11 − J22 − J33 − J44 , A2 = J11 J22 + J11 J33 + J11 J44 − J14 J41 + J22 J33 + J22 J44 − J24 J42 + J33 J44 − J34 J43 , A1 = − J11 J22 J33 − J11 J22 J44 + J11 J24 J42 − J12 J24 J41 + J14 J22 J41 − J11 J33 J44 , + J11 J34 J43 + J14 J33 J41 − J22 J33 J44 + J22 J34 J43 − J24 J32 J43 + J24 J33 J42 A0 = J11 J22 J33 J44 − J11 J22 J34 J43 + J11 J24 J32 J43 − J11 J24 J33 J42 + J12 J24 J33 J41 − J14 J22 J33 J41 , B2 = J23 J32 , B1 = J11 J23 J32 − J13 J34 J41 + J23 J32 J44 − J23 J34 J42 , B0 = − J11 J23 J32 J44 + J11 J23 J34 J42 − J12 J23 J34 J41 + J13 J22 J34 J41 − J13 J24 J32 J41 + J14 J23 J32 J41 , 2 2 2 C2 = J12 , C1 = J12 J33 + J12 J44 − J12 J14 J42 − J14 J31 J43 , 2 2 C0 = J12 J33 J44 + J12 J34 J43 − J12 J14 J32 J43 + J12 J14 J33 J42 − J12 J24 J31 J43 + J14 J22 J31 J43 ,

(4)

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D2 = J13 J31 , D1 = J12 J13 J32 − J12 J23 J31 + J13 J22 J31 + J13 J31 J44 , D0 = J12 J13 J32 J44 − J12 J13 J34 J42 + J12 J23 J31 J44 − J13 J22 J31 J44 + J13 J24 J31 J42 − J14 J23 J31 J42 . 2.1. Case 1 τ1 > 0, τ2 = 0 For τ2 = 0, Equation (4) can be simplified as follows: λ4 + M3 λ3 + M2 λ2 + M1 λ + M0 + ( N2 λ2 + N1 λ + N0 )e−λτ1 = 0,

(5)

where: M3 = A3 , M2 = A2 + C2 , M1 = A1 + C1 , M0 = A0 + C0 , N2 = B2 + D2 , N1 = B1 + D1 , N0 = B0 + D0 , Let λ = iω1 (ω1 > 0) be the root of Equation (5). Separating the real and imaginary parts, we obtain the following: (

N1 ω1 cosω1 τ1 + ( N2 ω12 − N0 )sinω1 τ1 = M3 ω13 − M1 ω1 . N1 ω1 sinω1 τ1 − ( N2 ω12 − N0 )cosω1 τ1 = M2 ω12 − ω14 − M0

(6)

From Equation (6), we can get: cosω1 τ1 =

N2 ω16 + ( N1 M3 − N2 M2 − N0 )ω14 + ( N2 M0 + N0 M2 − N1 M1 )ω12 − N0 M0 N12 ω12 + ( N2 ω12 − N0 )

2

.

(7)

Squaring both sides, adding both equations and regrouping by powers of ω1 , we obtain that ω1 (ω1 > 0) satisfies the following polynomial: ω18 + ( M36 − 2M2 )ω16 + ( M22 − 2M1 M3 + 2M0 − N22 )ω14 + ( M12 − 2M0 M2 − N12 + 2N0 N2 )ω12 + M02 − N02 = 0. (8)

Let r1 = ω12 , Equation (8) transformed into: r14 + q3 r13 + q2 r12 + q1 r1 + q0 = 0,

(9)

where: q3 = M36 − 2M2 , q2 = M22 − 2M1 M3 + 2M0 − N22 , q1 = M12 − 2M0 M2 − N12 + 2N0 N2 , q0 = M02 − N02 . In the following, we need to seek conditions under which Equation (9) has at least one positive root. Denote: h(r1 ) = r14 + q3 r13 + q2 r12 + q1 r1 + q0 . Since lim h(r1 ) = ∞, we conclude that if q0 < 0, then Equation (9) has at least one positive r1 → ∞

root [16]. Suppose that Equation (9) has positive roots. Without loss of generality, we assume that it has four positive roots, defined by r11 , r12 , r13 and r14 , respectively. Then Equation (8) has four positive roots: ω11 =

√

r11 , ω12 =

√

r12 , ω13 =

√

r13 , ω14 =

√

r14 .

( j)

For each fixed ω1k (k = 1, 2, 3, 4), there exists a sequence {τ1k |k = 1, 2, 3, 4; j = 0, 1, 2, ...} that satisfies Equation (6). According to Equation (7), we can get: ( j)

τ1k =

6 +( N M − N M − N ) ω 4 +( N M + N M − N M ) ω 2 − N M N2 ω1k 2 2 0 2 0 0 2 0 0 1 3 1 1 1 1k 1k w1k arccos 2 +( N ω 2 − N )2 N12 ω1k 2 1k 0

k = 1, 2, 3, 4; j = 0, 1, 2, 3, ...

+

2jπ ω1k

(10)

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( j)

Let τ10 = min{τ1k |k = 1, 2, 3, 4; j = 0, 1, 2, ...} =

τ10 =

(0)

min

k ∈{1,2,3,4}

{τ1k } = τ1k0 , ω10 = ω1k0 . So the τ10 is:

6 + ( N M − N M − N )ω 4 + ( N M + N M − N M )ω 2 − N M N2 ω10 1 2 2 0 2 0 0 2 0 0 1 3 1 1 10 10 arccos 2 2 2 2 ω10 N ω + ( N2 ω − N0 ) 1

10

(11)

10

Based on the above analysis, the main results are presented as below: Lemma 1. If q0 < 0, Equation (5) has a pair of pure imaginary roots ±iω10 when τ1 = τ10 . Next, take the derivative with respect to τ1 in Equation (5) and we can obtain: 

dλ dτ1

 −1

=

(4λ3 + 3M3 λ2 + 2M2 λ + M1 )eλτ1 + (2N2 λ + N1 ) τ1 − λ ( N2 λ3 + N1 λ2 + N0 λ) 

dλ(τ10 ) Re dτ1

 −1

= λ=iω10

P1 P3 + P2 P4 P12 + P22

where: 2 3 P1 = − N1 ω10 , P2 = N0 ω10 − N2 ω10 , 3 2 P3 = 4ω10 sinω10 τ10 − 3M3 ω10 cosω10 τ10 − 2M2 ω10 sinω10 τ10 + M1 cosω10 τ10 , 3 2 P4 = −4ω10 cosω10 τ10 − 3M3 ω10 sinω10 τ10 + 2M2 ω10 cosω10 τ10 + M1 sinω10 τ10 + 2N2 ω10 + N1 ,

Lemma 2. Suppose that P1 P3 + P2 P4 6= 0, then the transversality condition.



dReλ(τ10 ) dτ1 λ=iω

= Re 10

h

i dλ(τ10 ) −1 dτ1 λ=iω10

6= 0. So it satisfies

According to the Lemmas 1–2 and the Hopf bifurcation theorem in [17], we obtain the following results: Theorem 1: The equilibrium point E( x ∗ , y∗ , z∗ , u∗ ) of Equation (2) is asymptotically stable for τ1 ∈ [0, τ10 ) and unstable for τ1 > τ10 ; Equation (2) undergoes a Hopf bifurcation when τ1 = τ10 . 2.2. Case 2 τ1 > 0, τ2 > 0 For τ1 > 0, τ2 > 0, we consider the characteristic Equation (4) with τ1 in its stable intervals (the range of τ1 when the stability of the system is not affected), that is to say, τ1 ∈ [0, τ10 ) [18]. We study the influence of τ2 on the stability of the system when τ1 is fixed. Proposition 1. Suppose that L0 < 0, then the characteristic Equation (4) has a pair of pure imaginary roots ±iω20 for τ2 = τ20 . where: τ20 =

6 + h ω5 + h ω4 + h ω3 + h ω2 + h ω + h h6 ω20 1 5 20 2 20 3 20 0 4 20 1 20 arccos 4 + f ω3 + f ω2 + f ω + f ω20 f 4 ω20 3 20 2 20 0 1 20

(12)

Proof. Let λ = iω2 (ω2 > 0) be a root of Equation (4). Then we get:  (C2 ω22 − C0 + D2 ω22 cosω2 τ1 − D1 ω2 sinω2 τ1 − D0 cosω2 τ1 )sinω2 τ2      +(C1 ω2 + D2 ω22 sinω2 τ1 + D1 ω2 cosω2 τ1 − D0 sinω2 τ1 )cosω2 τ2    = A ω 3 − A ω − B ω 2 sinω τ − B ω cosω τ + B sinω τ , 3 2 2 2 2 1 2 1 0 2 1 1 2 1 2  −(C2 ω22 − C0 + D2 ω22 cosω2 τ1 − D1 ω2 sinω2 τ1 − D0 cosω2 τ1 )cosω2 τ2     +(C1 ω2 + D2 ω22 sinω2 τ1 + D1 ω2 cosω2 τ1 − D0 sinω2 τ1 )sinω2 τ2    = A2 ω22 − ω24 − A0 + B2 ω22 cosω2 τ1 − B1 ω2 sinω2 τ1 − B0 cosω2 τ1

(13)

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From Equation (13), we can obtain: cosω2 τ2 =

h6 ω26 + h5 ω25 + h4 ω24 + h3 ω23 + h2 ω22 + h1 ω2 + h0 f 4 ω24 + f 3 ω23 + f 2 ω22 + f 1 ω2 + f 0

(14)

where: h6 = C2 + D2 cosω2 τ1 , h5 = D2 A3 sinω2 τ1 − D1 sinω2 τ1 h4 = C1 A3 − D2 B2 + D1 A3 cosω2 τ1 − C2 A2 − C2 B2 cosω2 τ1 − C0 − D2 A2 cosω2 τ1 − D0 cosω2 τ1 , h3 = C2 B1 sinω2 τ1 − C1 B2 sinω2 τ1 − D2 A1 sinω2 τ1 − D0 A3 sinω2 τ1 + D1 A2 sinω2 τ1 , h2 = −C1 A1 − C1 B1 cosω2 τ1 + D2 B0 − D1 A1 cosω2 τ1 − D1 B1 + D0 B2 + C2 A0 , +C2 B0 cosω2 τ1 + C0 A2 + C0 B2 cosω2 τ1 + D2 A0 cosω2 τ1 + D0 A2 cosω2 τ1 h1 = C1 B0 sinω2 τ1 + D1 B0 cosω2 τ1 sinω2 τ1 + D0 A1 sinω2 τ1 , −C0 B1 sinω2 τ1 − D1 A0 sinω2 τ1 − D1 B0 sinω2 τ1 cosω2 τ1 h0 = − D0 B0 − C0 A0 − C0 B0 cosω2 τ1 − D0 A0 cosω2 τ1 , f 4 = C22 + D22 + 2D2 C2 cosω2 τ1 , f 3 = −2D1 C2 sinω2 τ1 + 2C1 D2 sinω2 τ1 , f 2 = 2D2 C0 cosω2 τ1 − 2C0 C2 + D12 − 2D0 C2 cosω2 τ1 − 2D0 D2 + C12 + 2D1 C1 cosω2 τ1 , f 1 = 2D1 C0 sinω2 τ1 − 2D0 C1 sinω2 τ1 , f 0 = C02 + D02 + 2C0 D0 cosω2 τ1 . Similar to Case 1, from Equation (13), we have: ω28 + L7 ω27 + L6 ω26 + L5 ω25 + L4 ω24 + L3 ω23 + L2 ω22 + L1 ω2 + L0 = 0

(15)

where L7 = 0, L6 = A23 − 2A2 − 2B2 cosω2 τ1 , L5 = 2B2 A3 sinω2 τ1 − 2B1 sinω2 τ1 , L4 = B22 − 2A1 A3 − 2A3 B1 cosω2 τ1 + A22 + 2A0 + 2A2 B2 cosω2 τ1 + 2B0 cosω2 τ1 − C22 − D22 − 2D2 C2 cosω2 τ1 ,

L3 = 2B2 A1 sinω2 τ1 + 2B0 A3 sinω2 τ1 − 2B1 A2 sinω2 τ1 + 2D1 C2 sinω2 τ1 − 2C1 D2 sinω2 τ1 , L2 = A21 + B12 + 2A1 B1 cosω2 τ1 − 2B0 B2 − 2A0 A2 − 2B2 A0 cosω2 τ1 − 2B0 A2 cosω2 τ1 , −2D2 C0 cosω2 τ1 + 2C0 C2 + D12 + 2D0 C2 cosω2 τ1 + 2D0 D2 − C12 − 2D1 C1 cosω2 τ1 L1 = −2B0 A1 sinω2 τ1 + 2B1 A0 sinω2 τ1 − 2D1 C0 sinω2 τ1 + 2D0 C1 sinω2 τ1 , L0 = A20 + 2B0 A0 cosω2 τ1 − C02 − D02 − 2D0 C0 cosω2 τ1 , We turn Equation (15) into the following form: f (ω2 ) = ω28 + L7 ω27 + L6 ω26 + L5 ω25 + L4 ω24 + L3 ω23 + L2 ω22 + L1 ω2 + L0

(16)

Since lim f (ω2 ) = ∞, we conclude that if L0 < 0, then Equation (15) has at least one positive ω2 → ∞

root. Without loss of generality, we assume that Equation (15) has a finite number of positive roots defined by {ω21 , ω22 , ..., ω2s }. From Equation (14), we denote: ( j)

τ2k =

6

5

4

3

2k

2k

2k

2

h6 ω2k +h5 ω2k +h4 ω2k +h3 ω2k +h2 ω2k +h1 ω2k +h0 1 ω2k arccos f 4 ω 4 + f 3 ω 3 + f 2 ω 2 + f 1 ω2k + f 0

+

2jπ ω2k , k

= 1, 2, ..., s; j = 0, 1, 2, ...

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( j)

( j)

Then (τ2k , ω2k ) be a root of Equation (13). Therefore, when τ2 = τ2k , the characteristic Equation (4) has a pair of pure imaginary roots ±iω2k . Define: ( j)

τ20 = min{τ2k |k = 1, 2, ..., s; j = 0, 1, 2, ...} =

min

k ∈{1,2,...,s}

(0)

{τ2k } = τ2k0 , ω20 = ω2k0 . ( j)

Let λ(τ2 ) = α(τ2 ) + iω (τ2 ) be the root of the characteristic Equation (4) near τ2 = τ2k satisfying:     ( j) ( j) α τ2k = 0, ω τ2k = ω2k . Then on the basis of above analysis, we can get the following conclusion: τ20 =

6 + h ω5 + h ω4 + h ω3 + h ω2 + h ω + h h6 ω20 1 5 20 3 20 2 20 0 4 20 1 20 arccos 4 + f ω3 + f ω2 + f ω + f ω20 f 4 ω20 2 20 0 3 20 1 20

(17)

The characteristic Equation (4) has a pair of pure imaginary roots ±iω20 when τ2 = τ20 .  Proposition 2. Suppose that ∆ 6= 0, then Re sign, where:

h

i dλ(τ20 ) −1 dτ2 λ=iω20

6= 0, Re

h

i dλ(τ20 ) −1 dτ2 λ=iω20

and ∆ have the same

∆ = Q1 Q3 + Q2 Q4 Proof. Substituting λ(τ2 ) (the root of the characteristic Equation (4), which is defined above) into Equation (4) and taking the derivative with respect to τ2 , we obtain: 

dλ dτ2

 −1

=

Q10 + Q20 + Q30 + Q40 τ − 2 λ λe−λ(τ1 +τ2 ) ( D2 λ2 + D1 λ + D0 ) + λe−λτ2 (C2 λ2 + C1 λ + C0 )

(18)

Q10 = 4λ3 + 3A3 λ2 + 2A2 λ + A1 , Q20 = (2B2 λ + B1 − τ1 B2 λ2 − τ1 B1 λ − τ1 B0 )e−λτ1 , Q30 = (2C2 λ + C1 )e−λτ2 , Q40 = (2D2 λ + D1 − τ1 D2 λ2 − τ1 D1 λ − τ1 D0 )e−λ(τ1 +τ2 ) , From Equation (18), we have: 

dλ(τ20 ) Re dτ2 where:

 −1

= λ=iω20

∆ Q1 Q3 + Q2 Q4 = 2 2 2 Q1 + Q2 Q1 + Q22

2 cos( τ + τ ) − D ω 3 sin( τ + τ ) + D ω sin( τ + τ ) Q1 = − D1 ω20 20 2 20 20 0 20 20 1 1 1 2 cosω τ − C ω 3 sinω τ + C ω sinω τ −C1 ω20 20 20 2 20 20 20 0 20 20 20 3 cos( τ + τ ) + D ω cos( τ + τ ) + D ω 2 sin( τ + τ ) Q2 = − D2 ω20 20 0 20 20 20 1 1 1 20 1 3 cosω τ + C ω cosω τ + C ω 2 sinω τ −C2 ω20 20 20 0 20 20 20 20 20 1 20

2 + A + 2B ω sinω τ + B cosω τ + τ B ω 2 cosω τ − τ B ω sinω τ Q3 = −3A3 ω20 2 20 20 1 20 1 20 1 20 1 1 1 1 2 20 1 1 20 −τ1 B0 cosω20 τ1 + 2C2 ω20 sinω20 τ20 + C1 cosω20 τ20 + 2D2 ω20 sin(τ1 + τ20 ) 2 cos( τ + τ ) − τ D ω sin( τ + τ ) − τ D cos( τ + τ ) + D1 cos(τ1 + τ20 ) + τ1 D2 ω20 20 20 20 1 1 1 20 1 1 0 1 3 + 2A ω + 2B ω cosω τ − B sinω τ − τ B ω 2 sinω τ − τ B ω cosω τ Q4 = −4ω20 2 20 2 20 20 1 20 1 20 1 20 1 1 1 2 20 1 1 20 +τ1 B0 sinω20 τ1 + 2C2 ω20 cosω20 τ20 − C1 sinω20 τ20 + 2D2 ω20 cos(τ1 + τ20 ) 2 sin( τ + τ ) − τ D ω cos( τ + τ ) + τ D sin( τ + τ ) − D1 sin(τ1 + τ20 ) − τ1 D2 ω20 20 20 20 1 1 1 20 1 1 0 1 h i −1 h i −1 dλ(τ20 ) (τ20 ) Since Q21 + Q22 > 0, thus Re 6= 0. We conclude that Re dλdτ and ∆ have dτ 2

the same sign. 

λ=iω20

2

λ=iω20

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Thus, according to propositions 1 and 2 and the Hopf bifurcation theorem in [17], we have the following theorem. Theorem 2: For τ1 ∈ [0, τ10 ), τ10 is defined by Equation (11). The equilibrium point E( x ∗ , y∗ , z∗ , u∗ ) of Equation (2) is asymptotically stable for τ2 ∈ [0, τ20 ) and unstable when τ2 > τ20 . Equation (2) has a Hopf bifurcation at τ2 = τ20 . 3. Numerical Simulation and Analysis In this section, the numerical simulation is carried out in order to support the theoretical analysis. Let x (0) = 0.3; y(0) = 0.5; z(0) = 0.7; u(0) = 0.9; a1 = 4.5; a2 = 1.3; a3 = 20; b1 = 0.25; b2 = 0.85; b3 = 0.35; b4 = 0.04; c1 = 0.5; c2 = 0.3; c3 = 0.2; c4 = 0.1; d1 = 0.01; d2 = 0.02; d3 = 0.06; M = 10; C = 1; E = 8; N = 12; L = 2; K = 2. We consider the following system with given parameter values:  . y(t)  x (t) = 4.5x (t)( 10 − 1) − 1.3y(t) + 20z(t − τ1 ),     y. (t) = −0.25x (t − τ ) + 0.85y(t)(1 − y(t)) + 0.35z(t − τ )(1 − z(t−τ1 ) ) − 0.04u(t), 2 1 8 . x (t−τ2 ) u(t)  z ( t ) = 0.5x ( t − τ )( − 1 ) − 0.3y ( t ) − 0.2z ( t ) + 0.1u ( t )( − 1),  2 12 2    . z(t) u(t) = 0.01y(t) + 0.02z(t)( 2 − 1) − 0.06u(t),

(19)

The following equilibrium points can be obtained by simple calculation: E0 (0, 0, 0, 0), E1 (−0.6625, 1.1410, −0.0578, 0.2100), E2 (−5.2219 ± 54.1330i, −3.8795 ± 0.7677i, −0.9477 ± 17.0453i, −48.6048 ± 10.9389i ), E3 (13.3977 ± 0.7680i, 0.4248 ± 1.7180i, 2.9437 ± 0.2407i, 0.5241 ± 0.1303i ), E4 (−251.5640 ± 371.2492i, 14.7184 ± 1.4202i, 15.8006 ± 47.5443i, −337.9472 ± 234.7986i ), E5 (−48.6048 ± 10.9389i, 6.9068 ± 6.8940i, −28.3090 ± 13.8694i, 112.0947 ± 134.3508i ), E6 (131.6126 ± 3.3325i, 3.3618 ± 6.3095i, 19.4028 ± 18.7720i, −1.8941 ± 114.1013i ), According to actual economic meaning, only E1 is Nash Equilibrium point. So we examine the system stability of the system at the equilibrium point E1 (−0.6625, 1.1410, −0.0578, 0.2100). For Case 1, from Equations (8) and (11), we can obtain ω10 = 4.5329, τ10 = 0.4618, so Equation (5) has a pair of pure imaginary roots ±iω10 when τ1 = τ10 and τ2 = 0. We also get P1 P3 + P2 P4 = 37.4226 6= 0. By theorem 1, the equilibrium point E1 of Equation (2) is asymptotically stable when τ1 ∈ [0, 0.4618) and unstable when τ1 > 0.4618. It has a Hopf bifurcation at τ1 = 0.4618. For Case 2, Let τ1 = 0.4 ∈ [0, τ10 ), from Equations (16) and (17), we can obtain ω20 = 39.6736, τ20 = 0.0622, so Equation (4) has a pair of pure imaginary roots ±iω20 when τ1 = 0.4 and τ2 = τ20 . At this point, Q1 Q3 + Q2 Q4 = 23.8153 6= 0. By theorem 2, the equilibrium point E1 of Equation (2) is asymptotically stable when τ2 ∈ [0, 0.0622) for τ1 = 0.4 and unstable when τ2 > 0.0622 for τ1 = 0.4. Equation (2) undergoes a Hopf bifurcation when τ2 = 0.0622 for τ1 = 0.4. 3.1. The Influence of τ1 on the Stability of Equation (19) Equation (19) moves from stable to unstable with the increase in τ1 and undergoes Hopf bifurcation at τ1 = 0.4618 when τ2 = 0. Figure 1a shows the dynamic evolution process of Equation (19). In this paper, we calculate the largest Lyapunov exponent (LLE) by the method of Wolf reconstruction [19]. We judge whether the system is stable according to the exponent value. If it is less than 0, the system is stable. If it is greater than 0, the system is unstable. If it is equal to 0, the system appears bifurcation. In Figure 1b, we can know that the system has a bifurcation at τ1 = 0.4618. This is consistent with the conclusion of theoretical analysis.

Equation (19). In this paper, we calculate the largest Lyapunov exponent (LLE) by the method of Wolf reconstruction [19]. We judge whether the system is stable according to the exponent value. If it is less than 0, the system is stable. If it is greater than 0, the system is unstable. If it is equal to 0, the system appears bifurcation. In Figure 1b, we can know that the system has a bifurcation at τ 1 = 0.4618 .

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This is consistent with the conclusion of theoretical analysis.

Equation (19). In this paper, we calculate the largest Lyapunov exponent (LLE) by the method of Wolf reconstruction [19]. We judge whether the system is stable according to the exponent value. If it is less than 0, the system is stable. If it is greater than 0, the system is unstable. If it is equal to 0, the system appears bifurcation. In Figure 1b, we can know that the system has a bifurcation at τ 1 = 0.4618 . This is consistent with the conclusion of theoretical analysis. (a)

(b)

Figure onthe thestability stability of of Equation Equation (19) (a) bifurcation bifurcation diagram; = 00.. (a) Figure 1. 1. The The influence influenceof of ττ11 on (19) when when ττ22 = diagram; (b) the largest Lyapunov Lyapunov exponent exponent plot. plot.

According is is stable when and τ 2and this τ 1 τ=10.4 τ 10 < = 0.4618 = 0 .τIn= According to to Theorem Theorem1,1,Equation Equation(19) (19) stable when = τ10 = 0.4618 and τ2 = 0 by Theorem 1. At this point, the frequency spectrum plot is discrete, which means that the system has a periodic solution. The x (t), y(t), z(t) and u(t) will lie on the basin of attraction [20] through the game. These analyses can be (b) (a)5. illustrated by Figures 4 and We discover that economic growth (slowdown) will lead to an increase (reduction) in carbon Figure 1. The influence of τ 1 on the stability of Equation (19) when τ 2 = 0 . (a) bifurcation diagram; emissions when other parameters are fixed. The two variables share a consistent pattern, but with (b) the largest exponent a certain delay in Lyapunov time. Based on theplot. above simulation, we find out that when the delay is greater than the bifurcation value, the carbon emissions pattern will not be consistent with the economic According to Theorem 1, Equation (19) is stable when τ 1 = 0.4 < τ 10 = 0.4618 and τ 2 = 0 . In this growth trend and volatility surfaces. Therefore, we should guarantee the timing of achieving the goal x(t ) , emissions y(t ) , z(t )and and u(t )it less will than converge to the equilibrium through the game. case, of reducing make the bifurcation value. Onlypoint in suchE1a stable environment, the implementation ofthe energy-saving and emission-reduction policies will intend a favorable effect. Figures 2 and 3 show above properties.

Figure 2. Time-domain plot when τ 1 = 0.4 < τ 10 = 0.4618 , τ 2 = 0 .

Figure 2. 2. Time-domain Time-domain plot plot when when ττ1 1== 0.4 , ττ22 = 0.4 < < ττ1010==0.4618 Figure 0.4618, = 00..

(a)

(b)

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Figure 2. Time-domain plot when τ 1 = 0.4 < τ 10 = 0.4618 , τ 2 = 0 .

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Figure

3.

The

(a)

(b)

(c)

(d)

EE

attractor

when

τ 1 = 0.4 < τ 10 = 0.4618

11 of 21

,

τ2 = 0

and

( x(0), y(0), z(0), u(0)) = (0.3,0.5,0.7,0.9) . (a) x(t ), y(t ), z(t ) ; (b) x(t ), y(t ), u(t ) ; (c) x(t ), z(t ), u(t ) ; (d) y(t ), z(t ), u(t ) .

Equation (19) is unstable when τ 1 = 0.5 > τ 10 = 0.4618 and τ 2 = 0 by Theorem 1. At this point, (c) (d) the frequency spectrum plot is discrete, which means that the system has a periodic solution. The The attractor , game. and τ 1 =τ0.4 τ 10 = 0.4618 and τ 2 = 0 by Theorem 1. At this point, the frequency spectrum plot is discrete, which means that the system has a periodic solution. The x(t ) , y(t ) , z(t ) and u(t ) will lie on the basin of attraction [20] through the game. These analyses can be illustrated by Figures 4 and 5.

(a)

(b)

Figure 4. 4. Equation (a)0.time-domain plot;plot; (b) 0.50.5 > τ> = 0.4618,τ 2 =τ02 . = 10 τ10 = 0.4618, Figure Equation(19) (19)isisunstable unstablewhen whenτ 1τ1= = (a) time-domain (b) frequency spectrumplot. plot. frequency spectrum

(a)

(b)

Figure 4. Equation (19) is unstable when τ 1 = 0.5 > τ 10 = 0.4618,τ 2 = 0 . (a) time-domain plot; (b)

frequency spectrum plot.

(a)

(b)

(a)

(b)

Figure 4. Equation (19) is unstable when τ 1 = 0.5 > τ 10 = 0.4618,τ 2 = 0 . (a) time-domain plot; (b)

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frequency spectrum plot. Entropy 2016, 18, 371

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Figure

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5.

The

(a)

(b)

(c)

(d) EE

attractor

when

τ 1 = 0.5 > τ 10 = 0.4618,τ 2 = 0

and

( x(0), y(0), z(0), u(0)) = (0.3,0.5,0.7,0.9) . (a) x(t ), y(t ), z(t ) ; (b) x(t ), y(t ), u(t ) ; (c) x(t ), z(t ), u(t ) ; (d) y(t ), z(t ), u(t ) .

We discover that economic growth (slowdown) will lead to an increase (reduction) in carbon emissions when other parameters are fixed. The two variables share a consistent pattern, but with a certain delay in time. Based on the above simulation, we find out that when the delay is greater than the bifurcation value, the carbon emissions pattern will not be consistent with the economic growth trend and volatility surfaces. (c) Therefore, we should guarantee the timing (d) of achieving the goal of reducing emissions and make it less than the bifurcation value. Only in such a stable environment, Figure 5. EE of The EE τ = attractor τ 1 0policies =and 0.5 > τ 10will =y(0), 0.4618, = 0= (0.3,and the implementation energy-saving intend a2 favorable effect. Figure 5. The attractor when 0.5 emission-reduction > τ10 =when 0.4618, τ2 = (x(0), z(0),τu(0)) 0.5, 1 and 0.7, 0.9). (a) zx(0), (t), uy(0)) (t), z=(t(0.3,0.5,0.7,0.9) ); (b) x (t), y(t), u(.t)(a) ; (c)xx((tt),),yz((tt),),zu((t t));; (b) (d) yx((tt)), , zy(t()t,),uu((tt)). ; (c) x(t ), z(t ), u(t ) ; ( x(0), y(0), 3.2. The (d) Influence y(t ), z(t ), of u(tτ) .1 on the Entropy of Equation (19) 3.2. The Influence of τ1 on the Entropy of Equation (19) Kolmogorov entropy can be used to measure the degree of complexity of the system. Let k be We discover entropy that economic growth (slowdown) will lead to an increase (reduction) in kcarbon Kolmogorov can be used to measure the degree of complexity of the system. Let be the the value of Kolmogorov entropy. If the system is stable and in regular motion, then k = 0 , emissions when otherentropy. parameters fixed.isThe two variables share a consistent but with a value of Kolmogorov If theare system stable and in regular motion, then k =pattern, 0, otherwise k > 0. k >in0 time. k is, the more otherwise . The Based greateronthe value ofsimulation, complex the system is. Weishave already certain delay the above we find out that when the delay greater than The greater the value of k is, the more complex the system is. We have already found that Equation (19) , as is shownwith in Figure 1. Basedgrowth on the found that Equation (19) displaysemissions bifurcation at τ 1 will = 0.4618be the bifurcation value,at the pattern consistent economic displays bifurcation τ1 carbon = 0.4618, as is shown in Figurenot 1. Based on the abovethe analysis, we can get k = 0complex k > 0of above we can get that if τ 1 < 0.4618 thenmore , otherwise . The more complex the trend volatility surfaces. Therefore, guarantee the the timing achieving the and goalthe of that if and τanalysis, kwe > should 0., The system is, the longer 1 < 0.4618, then k = 0, otherwise reducing emissions and make it less than the bifurcation value. Only in such a stable environment, system is, theitlonger the system more difficult be forThe theentropy system properties to return are to stability. more difficult will beand for the to returnittowill stability. displayedThe in the implementation of energy-saving and emission-reduction policies will intend a favorable effect. entropy displayed in Figure 6. Figure 6.properties are 3.2. The Influence of τ

1

on the Entropy of Equation (19)

Kolmogorov entropy can be used to measure the degree of complexity of the system. Let k be the value of Kolmogorov entropy. If the system is stable and in regular motion, then k = 0 , otherwise k > 0 . The greater the value of k is, the more complex the system is. We have already found that Equation (19) displays bifurcation at τ 1 = 0.4618 , as is shown in Figure 1. Based on the above analysis, we can get that if τ 1 < 0.4618 , then k = 0 , otherwise k > 0 . The more complex the system is, the longer and the more difficult it will be for the system to return to stability. The entropy properties are displayed in Figure 6.

Figure 6. 6. The entropy plot respect totoτ τ1 when whenτ τ = = 0. Figure The entropy plot respect 2 2 0. 1

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3.3. The Influence of τ 1 , b 3 , b 4 on the Stability of Equation (19) 12 of 20 of τ 1 , b 3 , b 4 on the Stability of Equation (19) The influence of τ 1 and b 3 on y is shown in Figure 7. The system shows a sign of The influence of τ 1 and b 3 on y is shown in Figure 7. The system shows a sign of instability at τ 1 of no obvious = 0.4618 b 3 ofhas 3.3. The Influence τ1 , b3 , b.4 However, on the Stability Equation (19) impact on y when τ 1 < 0.55 . With the instability at τ 1 = 0.4618 . However, b 3 has no obvious impact on y when τ 1 < 0.55 . With the increase of b 3 , y becomes larger and then decreases when τ 1 > 0 .5 5 . The minimum value of y is Theof influence of τ1 andlarger b3 onand y isthen shown in Figure 7. The shows a sign value of instability increase decreases when minimum of y is b 3 , y becomes τ 1 > system .5 5 . The y is0 5.124 −1.448 the maximum value ofwhen for (τ 1 , b 3 )However, = (0.5 , 0.85)b while (τ 1With , b 3 ) = the (0.6 increase , 0.55) . The at τ1 =for 0.4618. has no obvious impact on y τ < 0.55. of 1 −1.448 for (τ 1 , b 3 ) = (0.5 , 0.85) 3 while the maximum value of y is 5.124 for (τ 1 , b 3 ) = (0.6 , 0.55) . The y y value of is −0.66 approximately when . Thus, the impact of on can be ignored b3 , y becomes larger and then decreases when τ1 > 0.55. The minimum bvalue of y is −1.448 for τ 1 < 0.4618 3 of= y(0.5, is −0.66 approximately when value .5.124 Thus,for the(τimpact of(0.6, on). yThe canvalue be ignored τ 1 < 0.4618 b 3 0.55 (value τ , b ) 0.85 ) while the maximum of y is , b ) = of y is 3 3 1 1 when τ 1 < 0.4618 . when τ 1 < 0.4618 . when τ1 < 0.4618. Thus, the impact of b3 on y can be ignored when τ1 < 0.4618. − 0.66 approximately Entropy 18, 371 3.3. The2016, Influence

Figure 7. The influence of τ 1 and b 3 on y when τ 2 = 0 . Figure 7. The influence of τof1 τand on y when τ = 0 . b Figure 7. The influence 1 and b33 on y when τ2 = 20.

Figure 8 shows that b 4 has a greater impact on y. With an increase in b 4 , the system shifts y. With Figure 88 shows shows that that b b 4has has a greater impact an increase , the system shifts b 4 system Figure abgreater onon y.isWith an increase in b4in, the from 4 from unstable to stable. When the system beginning to become stable. Theshifts minimum = 0 .2 , impact 4 from unstable to stable. system is beginning become The minimum = 0system .2 , the is unstable to stable. When bWhen beginning to becometostable. Thestable. minimum value of y 4 = 0.2,b 4the value of y is −1.838 for (τ 1 , b 4 ) = (0.5 , 0.15) and the maximum value of y is 4.207 for is − 1.838 for ( τ , b ) = ( 0.5, 0.15 ) and the maximum value of y is 4.207 for ( τ , b ) = ( 0.6, 0 ) . The value y y 1 4−1.838 for (τ 1 , b 4 ) = (0.5 , 0.15) and the maximum value 1 4 of value of is is 4.207 for y y . The value of is stable at 2.889, but has little effect on . Therefore, we ( τ , b ) = ( 0 .6 , 0 ) τ 1 4 1 of y is stable at 2.889, but τ has little effect on y. Therefore, we should consider the effect of b on the 1 4 (τ 1 , b 4 ) = ( 0 .6 , 0 ) . The value of y is stable at 2.889, but τ 1 has little effect on y. Therefore, we should consider the effect of b 4 on the stability of the system. stability of the system. should consider the effect of b 4 on the stability of the system.

y when Figure 8. The influence of τof1 τand ony when Figure 8. The influence and bb 4 on τ2 =τ 20.= 0 . Figure 8. The influence of τ 1 1and b 44 on y when τ2 = 0 . τ 1 , b 3 and b 4 on y in Figure 9. The system is stable when We focus focusononthethe influence We influence of τof b4 on by in Figure 9. The system is stable when τ1 < 0.4618, 1 , bτ3 and , b 3 and We focus on the influence of on y in Figure 9. The is stable when 4 τ 1 < 0.4618 , and unstable when τ 1 >1 0.4618 b 3 and b 4 system . The change in only leads only tobut a larger and unstable when τ > 0.4618. The change in b and b leads to a larger fluctuation when 3 1 4 τ 1 < 0.4618 , and unstable τ > 0.4618 b b when . The change in and leads only to a larger 1 3 4 τ > 0.4618 , the fluctuation trend is basically symmetrical. However, fluctuation but when 1 τ1 > 0.4618, the fluctuation trend is basically symmetrical. However, b3 and b4 have no effect onb 3y , the trend is basically symmetrical. However, b 3 fluctuation but when τ 1 > 0.4618 τ 1 stability τ20 = 0.0622. Meanwhile, the Poincare plot has has five discrete discrete points, which when meansτthat that the system has periodic On the the basis of theorem 2, Equation (19)points, is unstable = 0.4 , τ system = 0 .08 has > τ 20aa=periodic 0 .06 22 . Meanwhile, Poincare plot five which means the 1 2 , , and will lie on the basin of attraction through the game. Figures solution. x ( t ) y ( t ) z ( t ) u ( t ) Meanwhile, plot five points, which means thatthe thegame. system has a13periodic solution. x (t)the , y(tPoincare ), z(t) and u(thas ) will lie discrete on the basin of attraction through Figures and 14

13 andthese 14 show solution. xcharacteristics. ( t ) , these y ( t ) ,characteristics. z ( t ) and u ( t ) will lie on the basin of attraction through the game. Figures show 13 and 14 show these characteristics.

(a)

(b)

(a) unstable when τ 1 = 0.4 , τ 2 = 0 .08 > τ 20 = 0 .06 22(b.)(a) Time-domain plot; Figure 13. 13. Equation Equation (19) Figure (19) is is unstable when τ1 = 0.4, τ2 = 0.08 > τ20 = 0.0622. (a) Time-domain plot; (b) Poincare Poincare plot. (19) is unstable when τ = 0.4 , τ = 0 .08 > τ = 0 .06 22 . (a) Time-domain plot; Figure 13. Equation (b) plot. 1 2 20 (b) Poincare plot.

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Figure

14.

16 of 21 15 of 20 16 of 21

(a)

(b)

(a)

(b)

(c)

(d)

The (c) EE

attractor

when

τ 1 = 0.4 , τ 2 = 0 .08 (d)> τ 20 = 0 .06 22

and

( x(0), y(0), z(0), u(0)) = (0.3, 0.5, 0.7, 0.9) . (a) x(t), y(t ), z(t ) ; (b) x(t), y(t ), u(t) ; (c) x(t), z(t), u(t ) ; Figure 14. The EE attractor when and τ 1 = 0.4 , τ 2 = 0 .08 > τ 20 = 0 .06 22 Figure ), z(The t), uEE (t) .attractor when τ1 = 0.4, τ2 = 0.08 > τ20 = 0.0622 and (x(0), y(0), z(0), u(0)) = (0.3, (d) y(t14. t), xy((tt)),, zz((tt)),;u(b) x(t), ( x(0), (0),xu((0)) 0.5, 0.7,y(0), 0.9).z(a) t), y(=t)(0.3, , z(t)0.5, ; (b)0.7, x (t)0.9) , y(t.),(a) u(t)x; ((c) (t); (d) y(yt()t, ), z(ut)(t, )u;(t(c) ). x(t), z(t), u(t ) ; y ( t ), z ( t ), u ( t ) . (d) 3.5. The Influence of τ 2 on the Entropy of Equation (19)

3.5. The Influence of τ2 on the Entropy of Equation (19) τ 2 we From Figureof 10, know that ofEquation 3.5. The Influence on the Entropy Equation (19) has a bifurcation when τ 1 = 0.4 , τ 2 = 0 .0 62 2 . From Figure 10, we know that Equation (19) has a bifurcation when τ1 = 0.4, τ2 = 0.0622. Equation is , then the entropy value is is more than 0. As in Section 3.2, the. ττ2 >>0.0622 Equation (19) isunstable unstable for 0.0622, then the entropy value more than 0. As in, τSection From(19) Figure 10, wefor know that Equation (19) has a bifurcation when τ 1 = 0.4 = 0 .0 623.2, 2 2 2 growth in the value of entropy shares the same pattern with that in . The change of entropy τ the growth in the value of entropy shares the same pattern with that in τ . The change of is 2 2 Equation (19) is unstable for τ 2 > 0.0622 , then the entropy value is more than 0. As in Section 3.2, the shown in Figure 15. With the increase of entropy value, the system will be more unstable. In this case, shown 15. of With the increase of entropy value,with the system be more unstable. In this growthin inFigure the value entropy shares the same pattern that in will τ 2 . The change of entropy is the effecteffect of energy-saving and emission-reduction is poorly maintained. case,implementation the implementation of energy-saving and emission-reduction is poorly maintained. shown in Figure 15. With the increase of entropy value, the system will be more unstable. In this case, the implementation effect of energy-saving and emission-reduction is poorly maintained.

Figure 15. The entropy plot respect to τ 2 when τ 1 = 0 .4 . Figure 15. The entropy plot respect to τ2 when τ1 = 0.4. Figure 15. The entropy plot respect to τ

2

when τ 1 = 0 .4 .

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3.6. The Influence of τ 22 ,, bb33 ,, bb44 on the Stability of Equation (19) 3.6. The Influence of τ2 ,that b3 , b4the on influence the Stability Figure 16 shows of ofτ Equation on y (19) is greater. With an increase in τ 22 , the system 22 .0 66 . The minimum value moves from stable to unstable. The system is becoming unstable at τ 22 == 00.0 Figure 16 shows that the influence of τ2 on y is greater. With an increase in τ , the system moves y 1.048 for ((τ 22 ,, bb33 )) == ((00.1, y is 1.212 for2 ((τ 22 ,, bb33 )) == (0.09 .1, 00.1) .1) (0.09,, 0.1) 0.1) of the maximum value from is stable to unstable. The system and is becoming unstable at τ2of= 0.06. The minimum value of y is. y y b τ > 0 0 .0 .0 66 , The is stable in the vicinity of 1.1. But b 3 has little effect on . Only when 22 > value 1.048value for (τof 2 , b3 ) = (0.1, 0.1) and the maximum value3of y is 1.212 for ( τ2 , b3 ) = (0.09, 0.1). The y the the of in willvicinity turn from large to bsmall with the increase of bb . Therefore, we have to make of yvalue is stable of 1.1. But 3 has little effect on y. Only 33when τ2 > 0.06, the value of y τ 22 > 0.1689 0.1689 0.1689 .. unstable and is when 0.1689 . We can conclude that the control effect will be larger 19 if ofthe 20 control parameter k stands higher.

control2016, is successful Entropy 18, 371

(a)

(b)

k =0.20.2> >0.1689 0.1689 Figure 22. Equation Equation (20) (20) isisstable stablewhen whenk = . (a) time-domain τ 2 ()0.5, = ( 00).5. ,(a) 0 ) time-domain Figure 22. for for plot; (τ1 , τ(τ2 )1 ,= (b) the attractor. plot; (b)EE the EE attractor. Based onthe theabove above analysis, assume the unstable system can to return to stability if Based on analysis, wewe assume thatthat the unstable system can return stability if effective effectiveis exerted. control This is exerted. Thiswhen means that when problems arise from and control means that problems arise from energy-saving and energy-saving emission-reduction emission-reduction policies and lead to market volatility, we can take control measures to ensure policies and lead to market volatility, we can take control measures to ensure that the system returns to the state. system returns to a stable state. athat stable 5. Conclusions Conclusions

In this this paper, we examine the ofstability of a four dimensional and paper, we examine the stability a four dimensional energy-saving andenergy-saving emission-reduction emission-reduction model delays. Weoffocus onand thethe impacts of delays thestability feedback model with two delays. We with focustwo on the impacts delays feedback control and on the of control on the stability of the system. The system shifts from stable to unstable when the delays are the system. The system shifts from stable to unstable when the delays are larger than the bifurcation larger than thecase, bifurcation In this case, fluctuation shows the up implementation in the system and values. In this a large values. fluctuation shows upainlarge the system and affects of affects the implementation of energy-saving emission reduction policies. However, can energy-saving and emission reduction policies. and However, we can control the bifurcation of thewe system control the bifurcation the system by adopting feedback control and the by adopting the variableoffeedback control approach the andvariable the system will return to aapproach stable state when system willparameters return to a are stable the control set.state when the control parameters are set. The results resultsshow showthat thattime timedelays delays play important in stability the stability the system. Only play an an important rolerole in the of theofsystem. Only when when the energy-savings and emission-reduction is stable, the purpose of reducing the energy-savings and emission-reduction systemsystem is stable, can thecan purpose of reducing carbon carbon emissions be achieved. Therefore, must ensure that parameters the delay parameters are region in a stable emissions be achieved. Therefore, we must we ensure that the delay are in a stable and region and then add other parameters to maintain a long-term stable system. then add other parameters to maintain a long-term stable system. According to the above analysis, if we want to keep the energy-saving and emission-reduction system running in a stable state, we can take the following following strategies strategies when when other other things things are are equal. equal. , we must make sure that or . Second, when , we First, when when ττ2 2== 0, 0 we must make sure that τ1 < τ 1 0.4618 < 0.4618 > 0Second, .5 τ 1 =we 0 .4 must or b4 >b0.5. when τ1 = 0.4, 4 keep must τkeep 2 < 0.0622. τ 2 < 0.0622 . Acknowledgments: The authors thank the reviewers for their careful reading and providing some pertinent suggestions. The research research was was supported supported by by The The Key Key Project Project of of Natural Science Research Research of of Anhui suggestions. The Natural Science Anhui province province (No. KJ2015A144); The Fundamental Research Funds for the Central Universities, South-Central University for (No. KJ2015A144); The Fundamental Research Funds for the Central Universities, South-Central University for Nationalities (CSY13011). Nationalities (CSY13011). Author Contributions: Jing Wang performed mathematical derivation and numerical simulation; Yuling Wang Author Wang performed mathematical ofderivation and numerical built the Contributions: economic model Jing and provided economic interpretation the conclusions; they wrotesimulation; this researchYuling article together. Both havemodel read and final manuscript. Wang built theauthors economic andapproved providedthe economic interpretation of the conclusions; they wrote this research article together. authors have andof approved Conflicts of Interest: TheBoth authors declare noread conflict interest. the final manuscript.

Conflicts of Interest: The authors declare no conflict of interest.

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