Forecasting fuel combustion-related CO₂ emissions by a novel continuous fractional nonlinear grey Bernoulli model with grey wolf optimizer

Abstract

Foresight of CO₂ emissions from fuel combustion is essential for policy-makers to identify ready targets for effective reduction plans and to further improve energy policies and plans. A new method for forecasting the future development of China’s CO₂ emissions from fuel combustion is proposed in this paper by using grey forecasting theory. Although the existing fractional nonlinear grey Bernoulli model (denoted as FNGBM(1,1)) has been theoretically proven to enhance the adaptability to diverse sequences, its fixed integer-order differential derivative still impairs the performance to some extent. To this end, a varying-order differential derivative is introduced into the existing differential equation to enable a more flexible structure, thus improving the prediction ability of FNGBM(1,1). Specifically, because of the advantages of conformable fractional accumulation, the traditional differential derivative is first replaced by the conformable fractional differential derivative. As a consequence, the continuous conformable fractional nonlinear grey Bernoulli model (hereinafter referred to as CCFNGBM(1,1)) is proposed. To further increase the validity of the model, a metaheuristic algorithm, namely Grey Wolf Optimizer (GWO), is then applied to search for the optimal emerging coefficients for the proposed model. Two real examples and China’s CO₂ emissions from fuel combustion are considered to verify the effectiveness of the newly proposed model, the experimental results show that the newly proposed model outperforms other benchmark models in terms of forecasting accuracy. The proposed model is finally employed to forecast the future China’s CO₂ emissions from fuel combustion by 2023, accounting for 10,039.80 million tons. Based on the forecasts, several policy suggestions are provided to curb CO₂ emissions.

Appendix: A. Proof of Theorem 1

Proof

Multiplying both sides of Eq. (10) by $\left (x^{(r)}(t)\right )^{\gamma }$, we have

$$ \begin{array}{@{}rcl@{}} \frac{d^{\alpha}x^{(r)}(t)}{dt^{\alpha}}\left( x^{(r)}(t)\right)^{\gamma}+a\left( x^{(r)}(t)\right)^{1-\gamma}=b \end{array} $$

(26)

Setting $y^{(r)}(t)=\left (x^{(r)}(t)\right )^{1-\gamma }$, Eq. (26) can be rewritten as

$$ \begin{array}{@{}rcl@{}} \frac{d^{\alpha}y^{(r)}(t)}{dt^{\alpha}}+a\left( 1-\gamma\right)y^{(r)}(t)=b\left( 1-\gamma\right) \end{array} $$

(27)

According to the features of the conformable fractional derivative (), $T_{\alpha }f(t)=t^{1-\alpha }\frac {df(t)}{dt}$, where T_αf(t) and $\frac {df(t)}{dt}$ represent the α −order conformable fractional derivative and the integer-order derivative in Riemannian geometry, respectively. Equation (27) becomes

$$ \begin{array}{@{}rcl@{}} \left( y^{(r)}(t)\right)^{1-\alpha}\frac{dy^{(r)}(t)}{dt}+a\left( 1-\gamma\right)y^{(r)}(t)=b\left( 1-\gamma\right) \end{array} $$

(28)

Solving Eq. (28) yields that

$$ \begin{array}{@{}rcl@{}} y^{(r)}(t)&=&exp\left( -\int\left( 1-\gamma\right)\frac{a}{t^{1-\alpha}}dt\right)\left\{ \int\left( 1-\gamma\right)\frac{b}{t^{1-\alpha}}\right.\\&&\left.\times exp\left( -\int\left( 1-\gamma\right)\frac{a}{t^{1-\alpha}}dt\right)dt+C \right\} \end{array} $$

(29)

Further,

$$ \begin{array}{@{}rcl@{}} x^{(r)}(t)&=&\left\{ exp\left( -\int\left( 1-\gamma\right)\frac{a}{t^{1-\alpha}}dt\right)\left\{ \int\left( 1-\gamma\right)\frac{b}{t^{1-\alpha}}\right.\right.\\ &&\left.\left.\times exp\left( -\int\left( 1-\gamma\right)\frac{a}{t^{1-\alpha}}dt\right)dt+C \right\} \right\}^{\frac{1}{1-\gamma}} \end{array} $$

(30)

Assuming that x^(r)(1) = x⁽⁰⁾(1), we obtain

$$ \begin{array}{@{}rcl@{}} C=\left( x^{(0)}(1)^{1-\gamma}-\frac{b}{a}\right)\times exp\left\{\frac{a}{\alpha} \left( 1-\gamma\right)\right\} \end{array} $$

(31)

Substituting Eq. (31) into Eq. (30), we have

$$ \begin{array}{@{}rcl@{}} x^{(r)}(t)=\left\{\left( x^{(0)}(1)^{1-\gamma}-\frac{b}{a}\right)\times exp \left\{-\frac{a}{\alpha}\left( 1-\gamma\right)\left( t^{\alpha}-1\right)\right\} \right\}^{\frac{1}{1-\gamma}} \end{array} $$

(32)

Setting t = k, this completes the proof. □