The energy efficiency model under the market response and the evolutionary path under its regulation policy in China

Abstract

According to the causal relationship among energy efficiency, energy price, and economic growth in China, a network structure of mutual transmission is constructed and a novel model of nonlinear dynamic system is also established. With the help of numerical simulation, the impacts of the parameters on the motion state of the system and the subsystem are analyzed. Then, the parameters are identified by means of BP neural network, and the model of novel dynamic system is given, which has a practical significance to reflect the actual situation of the system in China. According to the state of the actual evolution, we analyze the regulatory effect of the policy on the state of the system and find that four policies can all make the unstable system become stable. Comparing and analyzing the regulatory effects of the single and combined policy on the system show that the combined policy has a shorter time to reach the stable state and it is more conducive to the reduction of energy price. However, the single policy has a better effect on improving the level of energy efficiency and economic growth. In addition, the evolutionary paths of energy efficiency are analyzed at different regulatory levels under the same policy, and the results show that the higher the regulatory level is and the shorter time it takes to reach steady state for energy efficiency, the lower the level of energy efficiency is.

Appendix 1

Verify

The subsystem (5) has a unique stable equilibrium point when $\left\{\begin{array}{c}\frac{a_2{b}_1}{a_1{b}_2}>{E}^2- EC\\ {}\frac{a_1}{b_2}<\frac{E}{E-C}\end{array}\right.,$ which has only one unstable equilibrium point when $\frac{a_2{b}_1}{a_1{b}_2}<{E}^2- EC$ or $\left\{\begin{array}{c}\frac{a_2{b}_1}{a_1{b}_2}>{E}^2- EC\\ {}\frac{a_1}{b_2}\ge \frac{E}{E-C}\end{array}\right..$

Proof

For the subsystem (5), there is a coefficient matrix $A=\left(\begin{array}{cc}{a}_1E-{a}_1C& {a}_2\\ {}-{b}_1& -{b}_2E\end{array}\right)$, conclusions are as follows when |A| = a₂b₁ − a₁b₂E(E − C) ≠ 0.

The characteristic equation of the coefficient matrix A is

$$ {\lambda}^2+\left[{a}_1\left(E-C\right)-{b}_2E\right]\lambda +{a}_2{b}_1-{a}_1{b}_2E\left(E-C\right)=0\kern0.3em $$

(10)

The eigenvalues of the Eq. (10) are

$$ \frac{T\pm \sqrt{\Delta}}{2}. $$

Where $\frac{T\pm \sqrt{\Delta}}{2},T={a}_1\left(E-C\right)-{b}_2E,D={a}_2{b}_1-{a}_1{b}_2E\left(E-C\right),\Delta ={T}^2-4D.$

According to the equilibrium point, stability theorem and classification of differential equations:

• (1) When D = a₂b₁ − a₁b₂E(E − C) < 0, i.e. $\frac{a_2{b}_1}{a_1{b}_2}<{E}^2- EC,$ then there is bound to be: $\Delta ={T}^2-4D>0\Rightarrow \left\{\begin{array}{c}{\lambda}_1=\frac{T+\sqrt{\Delta}}{2}>0\\ {}{\lambda}_2=\frac{T+\sqrt{\Delta}}{2}<0\end{array}\right.,$ so the equilibrium point is the unstable saddle point.

• (2) When $\left\{\begin{array}{c}D={a}_2{b}_1-{a}_1{b}_2E\left(E-C\right)>0\kern0.1em \\ {}T={a}_1\left(E-C\right)-{b}_2E<0\kern2.1em \end{array}\right.,$ i.e. $\left\{\begin{array}{c}\frac{a_2{b}_1}{a_1{b}_2}>{E}^2- EC\\ {}\frac{a_1}{b_2}<\frac{E}{E-C}\kern0.1em \end{array}\right.,$ there are three situations:

• 1) When $\Delta ={T}^2-4D>0\Rightarrow \left\{\begin{array}{c}{\lambda}_1=\frac{T+\sqrt{\Delta}}{2}<0\\ {}{\lambda}_2=\frac{T+\sqrt{\Delta}}{2}<0\end{array}\right.,$ so the equilibrium point is the node.

• 2) When $\Delta ={T}^2-4D=0\Rightarrow {\lambda}_1={\lambda}_2=\frac{T}{2}<0,$ so the equilibrium point is an abnormal node or a critical node.

• 3) When $\Delta ={T}^2-4D<0\Rightarrow \left\{\begin{array}{c}{\lambda}_1=\frac{T+\sqrt{\Delta}}{2}\\ {}{\lambda}_2=\frac{T+\sqrt{\Delta}}{2}\end{array}\right.,$ the real parts of the eigenvalues are $\frac{T}{2}<0,$ so the equilibrium point is a focus point.

• (3) When $\left\{\begin{array}{c}D={a}_2{b}_1-{a}_1{b}_2E\left(E-C\right)>0\\ {}T={a}_1\left(E-C\right)-{b}_2E=0\kern2.1em \end{array}\right.,$ i.e. $\left\{\begin{array}{c}\frac{a_2{b}_1}{a_1{b}_2}>{E}^2- EC\\ {}\frac{a_1}{b_2}=\frac{E}{E-C}\end{array}\right.,$ so the equilibrium point is the center and is unstable.

• (4) When $\left\{\begin{array}{c}D={a}_2{b}_1-{a}_1{b}_2E\left(E-C\right)>0\\ {}T={a}_1\left(E-C\right)-{b}_2E>0\kern2.1em \end{array}\right.,$ i.e. $\left\{\begin{array}{c}\frac{a_2{b}_1}{a_1{b}_2}>{E}^2- EC\\ {}\frac{a_1}{b_2}>\frac{E}{E-C}\end{array}\right.,$ there are three situations:

• 1) When $\Delta ={T}^2-4D>0\Rightarrow \left\{\begin{array}{c}{\lambda}_1=\frac{T+\sqrt{\Delta}}{2}>0\\ {}{\lambda}_2=\frac{T+\sqrt{\Delta}}{2}>0\end{array}\right.,$ so the equilibrium point is the node.

• 2) When $\Delta ={T}^2-4D=0\Rightarrow {\lambda}_1={\lambda}_2=\frac{T}{2}>0,$ so the equilibrium point is an abnormal node or a critical node.

• 3) When $\Delta ={T}^2-4D<0\Rightarrow \left\{\begin{array}{c}{\lambda}_1=\frac{T+\sqrt{\Delta}}{2}\\ {}{\lambda}_2=\frac{T+\sqrt{\Delta}}{2}\end{array}\right.,$ the real parts of the eigenvalues are $\frac{T}{2}>0,$ so the equilibrium point is a focus point.

In summary, only when the following conditions are satisfied $\left\{\begin{array}{l}D={a}_2{b}_1-{a}_1{b}_2E\left(E-C\right)>0\\ {}T={a}_1\left(E-C\right)-{b}_2E<0\kern2.1em \end{array}\right.,$ i.e. $\left\{\begin{array}{c}\frac{a_2{b}_1}{a_1{b}_2}>{E}^2- EC\\ {}\frac{a_1}{b_2}<\frac{E}{E-C}\end{array}\right.,$ the equilibrium point is stable.