## Riemann Solution and Stability of Coupled Aw-Rascle-Zhang Model

Pan Lijun,*, Lv Shun,*, Weng Shasha,*

School of Mathematics, Nanjing University of Aeronautics and Astronautics, Nanjing 211106; Key Laboratory of Mathematical Modelling and High Performance Computing of Air Vehicles (NUAA), MIIT, Nanjing 211106

 基金资助: 中国留学基金(201506835005)

 Fund supported: China Scholarship Fund(201506835005)

Abstract

This paper studies the Riemann problem of the coupled Aw-Rascle-Zhang traffic model with different pressure laws on the connected roads. Using the method of characteristic analysis and theories of phase transition, we construct the Riemann solution to the coupled Aw-Rascle-Zhang model for the ommitted case $v_- + \eta (\rho_-)^{\gamma} = v_+$ in reference [5], and correct Riemann solution for the case $v_+ + \eta(\rho_-)^{\gamma} < v_+$, which complete the work of Herty, et al. Furthermore, when the parameter of the pressure term $\mu \to \eta$, the uniqueness and stability of the Riemann solution of the coupled Aw-Rascle model are proved.

Keywords： Coupled Aw-Rascle-Zhang model; Riemann problem; Uniqueness; Stability

Pan Lijun, Lv Shun, Weng Shasha. Riemann Solution and Stability of Coupled Aw-Rascle-Zhang Model[J]. Acta Mathematica Scientia, 2024, 44(4): 885-895

## 1 引言

2000 年, Aw、Rascle[1]和 Zhang[10]提出了一种二阶交通流模型

\left\{\begin{array}{l}\begin{aligned}& \partial_t \rho + \partial_x(\rho v) = 0, \\ & \partial_t (\rho^{} w^{}) + \partial_x (\rho^{} v^{} w^{} ) = 0, \\& w^{} = v^{} + \eta p(\rho^{} ), \\\end{aligned}\end{array}\right.

$s(\, 0;\{ w^{\mu} = w_-\} \, ) = \rho_{\alpha}^{\mu} v_{\alpha}^{\mu} > d( \, \rho_-, \{ w^{\eta} = w_- \} \, ).$

### 图3

$\rho^{\eta} v^{\eta} = \rho^{\mu} v^{\mu} = d( \, \rho_-, \{ w^{\eta} = w_- \} \, ),$

$d( \rho_-, \{ w^{\eta} = w_- \} )$ 的值取决于 $\rho_- $$\rho_{\alpha}^{\eta} 的值的大小关系, 下分 \rho_- \geqslant \rho_{\alpha}^{\eta}$$ \rho_- < \rho_{\alpha}^{\eta}$ 两种情况讨论.

$R_1^{\eta}: x = \sigma^{\eta} t, \quad \, \lambda_1^{\eta}(\rho^{}_-, v^{}_-) \leqslant \sigma^{\eta} \leqslant \lambda_1^{\eta}(\rho^{\eta}_{\alpha}, v^{\eta}_{\alpha}) = 0,$
$R_1^{\mu}: x = \sigma^{\mu} t, \quad \, \lambda_1^{\mu}(\rho^{\mu}, v^{\mu}) \leqslant \sigma^{\mu} \leqslant \lambda_1^{\mu}(0, v_- + \eta (\rho_-)^{\gamma}) = v_+, \qquad J_2^{\mu}: x = v_+ t.$

“+”意味着“随后”.

$\rho^{\eta} v^{\eta} = \rho_{}^{\mu} v_{}^{\mu} = d ( \rho_-, \{ w^{\eta} = w_- \}) = \rho_{\alpha}^{\eta} v_{\alpha}^{\eta}.$

$\eta$-路段上, 由上式和 $\rho_- \geqslant \rho_{\alpha}^{\eta}$, 状态 $(\rho^{\eta}, v^{\eta})$ 是唯一确定的且与状态 $(\rho^{\eta}_{\alpha}, v^{\eta}_{\alpha})$ 重合, 即 $(\rho^{\eta}, v^{\eta}) = (\rho_{\alpha}^{\eta}, v_{\alpha}^{\eta})$. 根据 (2.4) 式, 左状态 $(\rho_-, v_-)$ 需通过 1-疏散波 $R_1^{\eta}$ 与右状态 $(\rho^{\eta}_{\alpha}, v^{\eta}_{\alpha})$ 连接. 疏散波 $R_1^{\eta}$ 左边界波速 $\lambda_1^{\eta}(\rho_-, v_-)$ 和右边界波速 $\lambda_1^{\eta}(\rho^{\eta}_{\alpha}, v^{\eta}_{\alpha})$ 满足

$\lambda_1^{\eta} (\rho_-, v_-) < \lambda_1^{\eta} (\rho^{\eta}_{\alpha}, v^{\eta}_{\alpha}) = 0.$

$\mu$-路段上, 由耦合条件 (3.1), 状态 $(\rho^{\mu}, v^{\mu})$ 满足

\left\{\begin{array}{l}\begin{aligned} & \rho^{\mu} v^{\mu} \, = \, \rho^{\eta}_{\alpha} \, v^{\eta}_{\alpha}, \\ & v^{\mu} + \mu (\rho^{\mu})^{\gamma} \, = \, v^{\eta}_{\alpha} + \eta \, ( \rho^{\eta}_{\alpha})^{\gamma}.\end{aligned}\end{array}\right.

$\tau^{} = v_+.$

$0 < \lambda_1^{\mu}(\rho^{\mu}, v^{\mu}) < \lambda_1^{\mu}(0, v_- + \eta (\rho_-)^{\gamma}) = v_- + \eta (\rho_-)^{\gamma},$

$\tau^{} = v_+.$

## 4 CARZ 黎曼解的修正及稳定性

$\mu \to \eta$ 时, 可以证明除 $v_- + \eta (\rho_-)^{\gamma} < v_+$ 情形, 文献[5,9]中的黎曼解(即问题 (1.3)-(1.4) 的解) 均收敛到问题 (1.1)、(1.4) 的解. 因此, 本节只考虑情形 $v_- + \eta (\rho_-)^{\gamma} \leqslant v_+$ 下黎曼解的渐近性态. 在该情形下且 $\eta$ 充分小时, 文献[5]中黎曼问题 (1.3)-(1.4) 的解的结构如下

(1) 当 $\rho_- v_- > \rho_{\alpha}^{\mu} v_{\alpha}^{\mu}$

$(\rho_-, v_-) + S_1^{\eta} + (\rho^{\eta}, v^{\eta}) + PT + (\rho_{\alpha}^{\mu}, v_{\alpha}^{\mu})+ R_1^{\mu} + Vacuum + J_2^{\mu} + (\rho_+, v_+).$

$\rho^{\eta} v^{\eta} = \rho^{\mu}_{\alpha} v^{\mu}_{\alpha}, \quad v^{\eta} + \eta (\rho^{\eta})^{\gamma} = v^{}_{-} + \eta ( \rho^{}_{-})^{\gamma},$

$S_1^{\eta}: x = \frac{\rho^{\mu}_{\alpha} v^{\mu}_{\alpha} - \rho_- v_-}{\rho^{\eta} - \rho_-} t, \quad R_1^{\mu}: x = \sigma_1^{\mu} t, \, 0 \leqslant \sigma_1^{\mu} \leqslant v_- + \eta (\rho_-)^{\gamma}, \quad J_2^{\mu}: x = v_+ t.$

(2) 当 $\rho_- v_- < \rho_{\alpha}^{\mu} v_{\alpha}^{\mu}$

$(\rho_-, v_-) + PT + (\rho^{\mu}, v^{\mu}) + S_1^{\mu} + Vacuum + J_2^{\mu} + (\rho_+, v_+).$

$S_1^{\mu}: x = v^{\mu} t, \quad J_2^{\mu}: x = v_+ t.$

### 图4

$R_1^{\eta}: x = \sigma^{\eta} t, \quad \, \lambda_1^{\eta}(\rho^{}_-, v^{}_-) \leqslant \sigma^{\eta} \leqslant \lambda_1^{\eta}(\rho^{\eta}_{}, v^{\eta}_{}) < 0,$
$R_1^{\mu}: x = \sigma^{\mu} t, \quad \, 0 = \lambda_1^{\mu}(\rho^{\mu}_{\alpha}, v^{\mu}_{\alpha}) \leqslant \sigma^{\mu} \leqslant \lambda_1^{\mu}(0, v_- + \eta (\rho_-)^{\gamma}), \qquad J_2^{\mu}: x = v_+ t.$

(2) 如果 $\rho_- v_- < \rho_{\alpha}^{\mu} v_{\alpha}^{\mu} $$\rho_- < \rho_{\alpha}^{\eta} , 则黎曼问题 (1.3)-(1.4) 的解的结构可表示成 (如图5) ### 图5 图5 CARZ 模型 v_- + \eta (\rho_-)^{\gamma} < v_+$$ \rho_- v_- < \rho_{\alpha}^{\mu} v_{\alpha}^{\mu}$ 情形下的黎曼解 ($\rho_- < \rho_{\alpha}^{\eta}$)

$(\rho_-, v_-) + PT + (\rho^{\mu}, v^{\mu}) + R_1^{\mu} + Vacuum + J_2^{\mu} + (\rho_+, v_+).$

$R_1^{\mu}: x = \sigma^{\mu} t, \quad \, \lambda_1^{\mu}(\rho^{\mu}, v^{\mu}) \leqslant \sigma^{\mu} \leqslant \lambda_1^{\mu}(0, v_- + \eta (\rho_-)^{\gamma}), \qquad J_2^{\mu}: x = v_+ t.$

(3) 如果 $\rho_- v_- > \rho_{\alpha}^{\mu} v_{\alpha}^{\mu}$, 则黎曼问题 (1.3)-(1.4) 的解的结构为 (4.1) 式 (如图6).

### 图6

$d( \rho_-, \{ w^{\eta} = w_- \} )\, = \rho_{\alpha}^{\eta} v_{\alpha}^{\eta} > \rho_{\alpha}^{\mu} v_{\alpha}^{\mu} = s(0;\{w^{\mu} = w_- \}).$

$\eta$- 路段上, 状态 $(\rho^{\eta}, v^{\eta})$ 有两种情况: $\rho^{\eta} < \rho^{\eta}_{\alpha} < \rho_- $$\rho^{\eta}_{\alpha} < \rho^{\eta} < \rho_- . 根据 (2.4) 式, 左状态 (\rho_-, v_-) 需通过 1-疏散波 R_1^{\eta} 与右状态 (\rho^{\eta}_{}, v^{\eta}_{}) 连接. 若 \rho^{\eta} < \rho^{\eta}_{\alpha} < \rho_- , 由性质 3.1, 疏散波 R_1^{\eta} 右边界波速 \lambda_1^{\eta}(\rho^{\eta}, v^{\eta}) > 0 , 即波速会出现正值, 矛盾! 若 \rho^{\eta}_{\alpha} < \rho^{\eta} < \rho_- , 疏散波 R_1^{\eta} 左边界波速 \lambda_1^{\eta}(\rho_{-}, v_{-}) 和右边界波速 \lambda_1^{\eta}(\rho^{\eta}, v^{\eta} ) 满足 \lambda_1^{\eta}(\rho_{-}, v_{-}) < \lambda_1^{\eta}(\rho^{\eta}, v^{\eta} ) < \lambda_1^{\eta}(\rho^{\eta}_{\alpha}, v^{\eta}_{\alpha} ) = 0. 疏散波的波速为负, 故 \rho^{\eta}_{\alpha} < \rho^{\eta} . 类似定理 3.1 中的证明可得: 由方程组 (4.2) 可以唯一确定状态 (\rho^{\eta}, v^{\eta}) . \mu -路段上, 左状态 (\rho^{\mu}, v^{\mu})= (\rho^{\mu}_{\alpha}, v^{\mu}_{\alpha}) 需通过 1- 疏散波 R_1^{\mu} 与真空状态连接, 疏散波 R_1^{\mu} 左边界波速 \lambda_1^{\mu}(\rho^{\mu}_{\alpha}, v^{\mu}_{\alpha}) 和右边界波速 \lambda_1^{\mu}(0, v_- + \eta (\rho_-)^{\gamma}) 满足 0 = \lambda_1^{\mu}(\rho^{\mu}_{\alpha}, v^{\mu}_{\alpha}) < \lambda_1^{\mu}(0, v_- + \eta (\rho_-)^{\gamma}) = v_- + \eta (\rho_-)^{\gamma}, 疏散波的波速非负. 最后, 真空状态通过接触间断波 J_2^{\mu}: \tau^{} = v_+ 与右状态 (\rho_+, v_+) 连接. (2) 如图5 所示, 由于 \rho_- v_- < \rho_{\alpha}^{\mu} v_{\alpha}^{\mu}$$ \rho_- < \rho_{\alpha}^{\eta}$, 则有

$d( \rho_-, \{ w^{\eta} = w_- \} )\, = \rho_- v_- < \rho_{\alpha}^{\mu} v_{\alpha}^{\mu} = s(0;\{w^{\mu} = w_- \}).$

(3) 如图6 所示, 由于 $\rho_- v_- > \rho_{\alpha}^{\mu} v_{\alpha}^{\mu}$, 则有

$d( \rho_-, \{ w^{\eta} = w_- \} )\, > \rho_{\alpha}^{\mu} v_{\alpha}^{\mu} = s(0;\{w^{\mu} = w_- \}).$

$\frac{{\rm d} m_1^{\eta} }{{\rm d} \rho} = v_- + \eta(\rho_-)^{\gamma} - \eta (1+\gamma)\rho^{\gamma}, \qquad \frac{{\rm d}^2 m_1^{\eta} }{{\rm d} \rho^2} = -\eta \gamma (1+\gamma)\rho^{\gamma-1} < 0,$

$\rho^{\eta} v^{\eta} = \lim\limits_{\mu \to \eta } \rho_{\alpha}^{\mu} v_{\alpha}^{\mu} = \rho_{\alpha}^{\eta} v_{\alpha}^{\eta}.$

\begin{align*}\lim\limits_{\mu \to \eta } \lambda_1^{\eta}(\rho^{\eta}, v^{\eta}) &= \lim\limits_{\mu \to \eta }\lambda_1^{\eta}(\rho^{\eta}_{\alpha}, v^{\eta}_{\alpha}) = v^{\eta}_{\alpha} - \eta \gamma (\rho^{\eta}_{\alpha} )^{\gamma} \\&= \lim\limits_{\mu \to \eta } ( v^{\mu}_{\alpha} - \mu \gamma (\rho^{\mu}_{\alpha} )^{\gamma} )= \lim\limits_{\mu \to \eta } \lambda_1^{\mu}(\rho^{\mu}_{\alpha}, v^{\mu}_{\alpha} )= 0,\end{align*}

$\frac{{\rm d} \rho^{\mu}}{{\rm d} \mu } = \frac{ (\rho^{\mu})^{\gamma+2} } { \rho_- v_- \mu \gamma (\rho^{\mu})^{\gamma+1} } = \frac{ (\rho^{\mu})^{\gamma+2} } { \rho^{\mu} v^{\mu} - \mu \gamma (\rho^{\mu})^{\gamma+1} } = \frac{ (\rho^{\mu})^{\gamma+1} } { v^{\mu} - \mu \gamma (\rho^{\mu})^{\gamma} }.$

$\lim\limits_{\mu \to \eta } \rho_{}^{\mu} v_{}^{\mu} = \rho_- v_-, \qquad \lim\limits_{\mu \to \eta } ( v^{\mu} + \mu \gamma (\rho^{\mu})^{\gamma} ) = v_- + \eta (\rho_-)^{\gamma}.$

$\mu \to \eta$ 时, 耦合 Aw-Rascle-Zhang 交通流模型的黎曼问题 (1.3)-(1.4) 在情形 $v_- + \eta (\rho_-)^{\gamma} = v_+$ 下的黎曼解的极限可类似定理 4.2 讨论, 我们得出如下定理.

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