[物理学与PDEs]第3章习题7 快、慢及Alfv\'en 特征速度的比较

证明: 当 $H_1\neq 0$ 及 $H_2^2+H_3^2\neq 0$ 时, 快、慢及 Alfv\'en 特征速度 $C_f$, $C_s$ 及 $C_a$ 满足 $$\bex 0<C_s^2<C_a^2<C_f^2.  \eex$$

 

证明: 显然有 $0<C_s^2$. 往证 $C_a^2>C_s^2$: $$\beex \bea \cfrac{\mu_0}{\rho}H_1^2&\quad\wedge\quad \cfrac{1}{2}\sez{ \tilde c^2+\cfrac{\mu_0}{\rho}H^2 -\sqrt{\sex{\tilde c +\cfrac{\mu_0}{\rho}H^2}^2-\cfrac{4\mu_0\tilde c^2}{\rho} H_1^2}}\\ \cfrac{2\mu_0}{\rho}H_1^2-\sex{\tilde c^2+\cfrac{\mu_0}{\rho}H^2}&\quad\wedge\quad -\sqrt{\sex{\tilde c +\cfrac{\mu_0}{\rho}H^2}^2-\cfrac{4\mu_0\tilde c^2}{\rho}H_1^2}\\ \sez{\cfrac{2\mu_0}{\rho}H_1^2-\sex{\tilde c^2+\cfrac{\mu_0}{\rho}H^2}}^2&\quad\vee\quad \sex{\tilde c +\cfrac{\mu_0}{\rho}H^2}^2-\cfrac{4\mu_0\tilde c^2}{\rho}H_1^2\\ \cfrac{4\mu_0^2}{\rho^2}H_1^4 -\cfrac{4\mu_0}{\rho}H_1^2\sex{\tilde c^2+\cfrac{\mu_0}{\rho}H^2} &\quad\vee\quad -\cfrac{4\mu_0\tilde c^2}{\rho}H_1^2\\ \cfrac{4\mu_0^2}{\rho^2}H_1^4&\quad\vee\quad \cfrac{4\mu_0^2}{\rho^2}H_1^2H^2\\ H_1^2&\quad\vee\quad H^2\quad(H_1\neq 0)\\ 0&\quad\vee\quad H_2^2+H_3^2.  \eea \eeex$$ 同理有 $C_a^2<C_f^2$.