Analysis of dynamic properties of carbon emission–carbon absorption model with time delay based on China

Abstract

Since the proposal of China’s double carbon goal, it has attracted wide attention from all walks of life. In order to study the impact of energy, economy and the time it takes for carbon dioxide released into the atmosphere to be completely absorbed on China’s future carbon emission and carbon absorption trend, and to provide estimates and suggestions for China’s carbon neutrality in future, we establish a carbon emission–carbon absorption model with time delay. First, we calculate the equilibrium of the model and analyze the stability of the equilibrium and then further analyze the existence of the Hopf bifurcation of the equilibrium, the Hopf bifurcation normal form of the model is derived by using the multiple time scales method, and the Hopf bifurcation direction and the stability of the bifurcation periodic solution are determined. Finally, we fit and analyze the official data, give the actual model parameters and use MATLAB to carry out numerical simulations to verify the correctness of the theoretical analysis. We find that numerically periodic solutions exist in a wide range around time delay. According to our model, the final results show that China will reach the carbon peak in 2027, but according to the current development model and carbon absorption level of China, carbon neutrality cannot be achieved before 2060. In this regard, we put forward relevant policies and suggestions to accelerate the realization of carbon neutrality.

Appendix

We regard the delay $\tau$ as a bifurcation parameter, let $\tau =\tau _c+\varepsilon \mu$, where $\tau _c=\tau ^{(j)}(j = 0, 1, 2, \cdots )$ is the critical value of Hopf bifurcation given in (7), $\mu$ is the disturbance parameter and $\varepsilon$ is the dimensionless scale parameter. Suppose system (10) undergoes a Hopf bifurcation from the trivial equilibrium E at the critical point $\tau = \tau _c$, and then, the solution of (10) is assumed as follows:

$$\begin{aligned} Z(t)= & {} Z(T_0,T_1,T_2,\cdots .)\nonumber \\= & {} \sum \limits _{k=1}^{+\infty }\varepsilon ^k Z_k (T_0,T_1,T_2,\cdots .), \end{aligned}$$

(15)

where $T_0=t, T_1=\varepsilon t, T_2=\varepsilon ^2 t,\cdots$, and we can get

$$\begin{aligned}&Z(T_0,T_1,T_2,\cdots )\\&\quad =(x(T_0,T_1,T_2,\cdots ),y(T_0,T_1,T_2,\cdots ),\\&\qquad G(T_0,T_1,T_2,\cdots ),S(T_0,T_1,T_2,\cdots ))^T,\\&Z_k (T_0,T_1,T_2,\cdots )\\&\quad =(x_k (T_0,T_1,T_2,\cdots ),y_k(T_0,T_1,T_2,\cdots ),\\&\qquad G_k(T_0,T_1,T_2,\cdots ),S_k(T_0,T_1,T_2,\cdots ))^T, \end{aligned}$$

and the derivative with regard to t is transformed into

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}t}= & {} \frac{\partial }{\partial T_0}+\varepsilon \frac{\partial }{\partial T_1}+\varepsilon ^2 \frac{\partial }{\partial T_2}+ \cdots \\= & {} D_0+\varepsilon D_1+\varepsilon ^2 D_2+ \cdots , \end{aligned}$$

where $D_i$ is differential operator, and

$$\begin{aligned} D_i=\frac{\partial }{\partial T_i},i=0,1,2,\cdots . \end{aligned}$$

Then, we obtain

$$\begin{aligned} {\dot{Z}}(t)= & {} \varepsilon D_0 Z_1+\varepsilon ^2 D_1 Z_1 +\varepsilon ^3 D_2 Z_1+\varepsilon ^2 D_0 Z_2\nonumber \\&+\varepsilon ^3 D_1 Z_2+\varepsilon ^3 D_0 Z_3+\cdots . \end{aligned}$$

(16)

We expand $x(T_0 -1,\varepsilon (T_0 -1),\varepsilon ^2 (T_0 -1),\cdots )$ at $x(T_0 -1,T_1,T_2,\cdots )$ by the Taylor-series expansion, and we get

$$\begin{aligned} x(t-1)= & {} \varepsilon x_{1,\tau _c}+\varepsilon ^2 x_{2,\tau _c}+\varepsilon ^3 x_{3,\tau _c}-\varepsilon ^2 D_1 x_{1,\tau _c}\nonumber \\&-\varepsilon ^3 D_2 x_{1,\tau _c}-\varepsilon ^3 D_1 x_{2,\tau _c}+\cdots , \end{aligned}$$

(17)

where $x_{j,\tau _c} = x_j (T_0 -1,T_1,T_2,\cdots ), j=1,2,3,\cdots .$

Substituting Eqs. (15)–(17) into Eq. (10), after obtaining the equation, let the coefficients of $\varepsilon$, $\varepsilon ^2$ and $\varepsilon ^3$ on both sides of the equation be equal. So, we get the following expression:

$$\begin{aligned} \left\{ \begin{aligned}&D_0 x_1-\tau _c[r_1 x_1-a_1(x_1 y^*+x^* y_1)-\frac{b n_1 G_1}{M}+n_2 S_1]=0,\\&\quad D_0 y_1-\tau _c[-r_2 y_1+a_2 x_{1,\tau _c} y^*+a_2 x^* y_1]=0,\\&\quad D_0 G_1+\tau _c r_3 G_1=0,\\&\quad D_0 S_1-\tau _c[r_4 S_1-\frac{2 r_4 S^* S_1}{N_2}+c G_1 S^*+c S_1 G^*]=0. \end{aligned} \right. \nonumber \\ \end{aligned}$$

(18)

$$\begin{aligned} \left\{ \begin{aligned}&D_0 x_2-\tau _c [r_1 x_2 - a_1(x_2 y^*+x^* y_2)- \frac{b n_1 G_2}{M} + n_2 S_2]\\&\quad =-D_1 x_1-\tau _c a_1 x_1 y_1+\mu [r_1 x_1\\&\quad -a_1(x_1 y^*+x^* y_1)-\frac{b n_1 G_1}{M}+n_2 S_1],\\&\quad D_0 y_2-\tau _c[-r_2 y_2+a_2 x_{2,\tau _c} y^*+a_2 x^* y_2]\\&\quad =-D_1 y_1+\tau _c a_2(x_{1,\tau _c} y_1-D_1 x_{1,\tau _c} y^*)\\&\quad +\mu [-r_2 y_1+a_2(x_{1,\tau _c} y^*+x^* y_1)],\\&\quad D_0 G_2+\tau _c r_3 G_2=-D_1 G_1-\frac{\tau _c r_3}{N_1} G_1^2-\mu r_3 G_1,\\&\quad D_0 S_2-\tau _c(r_4 S_2-\frac{2 r_4 S^* S_2}{N_2}+c G_2 S^*+c S_2 G^*)\\&\quad =-D_1 S_1+\tau _c(-\frac{r_4}{N_2} S_1^2+c G_1 S_1)\\&\quad +\mu [r_4 S_1(1-\frac{2 S^*}{N_2})+c(G_1 S^*+G^* S_1)]. \end{aligned} \right. \nonumber \\ \end{aligned}$$

(19)

$$\begin{aligned} \left\{ \begin{aligned}&D_0 x_3-\tau _c[r_1 x_3-a_1(x_3 y^*+x^* y_3)-\frac{b n_1 G_3}{M}+n_2 S_3]\\&\quad =-D_2 x_1-D_1 x_2-a_1 \tau _c (x_1 y_2+x_2 y_1)\\&\quad +\mu [r_1 x_2-a_1(x_1 y_1+x_2 y^*+x^* y_2)\\&\quad -\frac{b n_1 G_2}{M}+n_2 S_2],\\&\quad D_0 y_3-\tau _c(-r_2 y_3+a_2 x_{3,\tau _c} y^*+a_2 x^* y_3)\\&\quad =-D_2 y_1-D_1 y_2+a_2 \tau _c[(x_{2,\tau _c}-D_1 x_{1,\tau _c}) y_1\\&\quad -(D_2 x_{1,\tau _c}+D_1 x_{2,\tau _c})y^*+x_{1,\tau _c} y_2]\\&\quad +\mu [-r_2 y_2 + a_2 (x_{1,\tau _c} y_1 + x^* y_2)\\&\quad +a_2(x_{2,\tau _c} - D_1 x_{1,\tau _c}) y^*],\\&\quad D_0 G_3+\tau _c r_3 G_3=-D_2 G_1-D_1 G_2-\frac{2 \tau _c G_1 G_2 r_3}{N_1}\\&\quad -\mu r_3(G_2+\frac{G_1^2}{N_1}),\\&\quad D_0 S_3-\tau _c [r_4 S_3-\frac{2 r_4 S^* S_3}{N_2}+c(G_3 S^*+G^* S_3)]\\&\quad =-D_1 S_2-D_2 S_1+\tau _c[-\frac{2r_4 S_1 S_2}{N_2}+c G_1 S_2\\&\quad +c G_2 S_1]+\mu [r_4 S_2(1-\frac{2S^*}{N_2})-\frac{r_4 S_1^2}{N_2}\\&\quad +c(G_1 S_1+G_2 S^*+G^* S_2)]. \end{aligned} \right. \nonumber \\ \end{aligned}$$

(20)

The solution of (18) is given as follows:

$$\begin{aligned} Z_1=Q h \mathrm {e}^{\mathrm {i} \omega \tau _c T_0} +{\bar{Q}} {\bar{h}} \mathrm {e}^{-\mathrm {i} \omega \tau _c T_0}, \end{aligned}$$

(21)

where $Q = Q(T_1, T_2,\cdots )$, h is obtained from (15). Thus, substituting solution (21) into the right part of Eq. (19) and assuming the coefficient vector of $\mathrm {e}^{\mathrm {i} \omega \tau _c T_0}$ is noted as $m_1$, by $\langle h^*, m_1 \rangle =0$, we can solve $\frac{\partial Q}{\partial T_1}$, namely

$$\begin{aligned} \frac{\partial Q}{\partial T_1}=L \mu Q, \end{aligned}$$

(22)

where

$$\begin{aligned} L=-\frac{\omega ^2+J}{P -\tau _c J}, \end{aligned}$$

with

$$\begin{aligned} P=2 \mathrm {i} \omega - r_1 + a_1 y^*,J=a_1 a_2 x^* y^* \mathrm {e}^{-\mathrm {i} \omega \tau _c}. \end{aligned}$$

Solving Eq. (19), we get its solution in the following form:

$$\begin{aligned} \begin{aligned} x_2&=f_1 \mathrm {e}^{\mathrm {i} \omega \tau _c T_0}+g_1 \mathrm {e}^{2\mathrm {i} \omega \tau _c T_0}+\bar{f_1} \mathrm {e}^{-\mathrm {i} \omega \tau _c T_0}\\&+\bar{g_1} \mathrm {e}^{-2\mathrm {i} \omega \tau _c T_0}+l_1,\\ y_2&=f_2 \mathrm {e}^{\mathrm {i} \omega \tau _c T_0}+g_2 \mathrm {e}^{2\mathrm {i} \omega \tau _c T_0}+\bar{f_2} \mathrm {e}^{-\mathrm {i} \omega \tau _c T_0}\\&+\bar{g_2} \mathrm {e}^{-2\mathrm {i} \omega \tau _c T_0}+l_2,\\ G_2&=0,\\ S_2&=0. \end{aligned} \end{aligned}$$

(23)

Then, substituting solution (23) into Eq. (19), we get

$$\begin{aligned} f_1= & {} \frac{I_1}{I} \mu Q, \quad f_2=\frac{I_2}{I} \mu Q, \quad g_1=-\frac{K_1}{K} Q^2 h_2, \\ g_2= & {} \frac{K_2}{K} Q^2 h_2, \quad l_1 =-\frac{V_1}{y^*} Q {\bar{Q}}, \quad l_2=\frac{V_2}{a_1 x^* y^*} Q {\bar{Q}}, \end{aligned}$$

with

$$\begin{aligned} T= & {} -L h_2 + (1 - \tau _c M) a_2 y^* \mathrm {e}^{-\mathrm {i} \omega \tau _c}, \\ I_1= & {} \mathrm {i} \omega (-L + \mathrm {i}\omega ) - a_1 x^* T, \\ I_2= & {} (\mathrm {i}\omega -r_1+a_1 y^*)T + a_2 y^* \mathrm {e}^{-\mathrm {i} \omega \tau _c} (-M + \mathrm {i} \omega ), \\ I= & {} \tau _c [\mathrm {i}\omega (\mathrm {i}\omega -r_1+a_1 y^*)+J],\\ K= & {} 2\mathrm {i} \omega P - a_1 a_2 x^* y^* \mathrm {e}^{-2\mathrm {i} \omega \tau _c}, \\ K_1= & {} a_1 (2\mathrm {i} \omega +r_2 \mathrm {e}^{-\mathrm {i}\omega \tau _c}),\\ K_2= & {} a_2 (P \mathrm {e}^{-\mathrm {i}\omega \tau _c}-a_1 y^* \mathrm {e}^{-2\mathrm {i}\omega \tau _c}), \\ V_1= & {} h_2 \mathrm {e}^{\mathrm {i} \omega \tau _c}+\bar{h_2} \mathrm {e}^{ -\mathrm {i} \omega \tau _c}, \\ V_2= & {} (-r_1+a_1 y^*) V_1 - a_1 y^*(h_2+\bar{h_2}). \end{aligned}$$

Next, substituting solutions (21) and (23) into (20) and assuming the coefficient vector of $\mathrm {e}^{\mathrm {i} \omega \tau _c T_0}$ is noted as $m_2$, by solvability condition, we get $\langle h^*, m_2\rangle =0$. Note that $\mu$ is disturbance parameter, and $\mu ^2$ has little influence for small unfolding parameter, thus, we ignore the $\mu ^2 G$ term, and then, $\frac{\partial G}{\partial T_2}$ can be solved in the following form:

$$\begin{aligned} \frac{\partial Q}{\partial T_2}=H Q^2 {\bar{Q}}, \end{aligned}$$

(24)

where

$$\begin{aligned} H=-\frac{\mathrm {i} \omega a_1 \tau _c R_1 + a_1 a_2 x^* \tau _c R_2}{P - \tau _c J}, \end{aligned}$$

with

$$\begin{aligned} R_1= & {} \frac{V_2}{a_1 x^* y^*} + h_2 (\frac{K_2}{K} - \frac{V_1}{y^*} - \frac{K_1}{K} \bar{h_2}), \\ R_2= & {} h_2 (-\frac{V_1}{y^*} - \frac{K_1}{K} \bar{h_2} \mathrm {e}^{-2\mathrm {i} \omega \tau _c} + \frac{K_2}{K} \mathrm {e}^{\mathrm {i}\omega \tau _c})\\&+ \frac{V_2}{a_1 x^* y^*} \mathrm {e}^{-\mathrm {i}\omega \tau _c}. \end{aligned}$$

Let $Q \rightarrow Q/\varepsilon$, we obtain the normal form of Hopf bifurcation of system (1) truncated at the cubic order terms

$$\begin{aligned} {\dot{Q}} = L \mu Q + H Q^2 {\bar{Q}}, \end{aligned}$$

(25)

where L is given in (22) and H is given in (24).