试将一维理想磁流体力学方程组 (5. 10)-(5. 16) 化为一阶拟线性对称双曲组的形式.
解答: 由 (5. 12),(5. 16) 知 $$\beex \bea 0&=\cfrac{\p p}{\p \rho}\sex{\cfrac{\p \rho}{\p t}+u_1\cfrac{\p \rho}{\p x}+\rho \cfrac{\p u_1}{\p x}} +\cfrac{\p\rho}{\p S}\sex{\cfrac{\p S}{\p t}+u_1\cfrac{\p S}{\p x}}\\ &=\cfrac{\p p}{\p t}+u_1\cfrac{\p p}{\p x}+\tilde c^2\rho \cfrac{\p u_1}{\p x}, \eea \eeex$$ 而 $$\bex \cfrac{1}{\tilde c^2\rho }\cfrac{\p p}{\p t}+\cfrac{u_1}{\tilde c^2\rho}\cfrac{\p p}{\p x}+\cfrac{\p u_1}{\p x}=0. \eex$$ 于是 (5. 10)-(5. 16) 为 $$\beex \bea \mu_0\cfrac{\p H_2}{\p t}+\mu_0u_1\cfrac{\p H_2}{\p x} +\mu_0H_2\cfrac{\p u_1}{\p x} -\mu_0H_1\cfrac{\p u_2}{\p x}&=0,\\ \mu_0\cfrac{\p H_3}{\p t} +\mu_0u_1\cfrac{\p H_3}{\p x} +\mu_0H_3\cfrac{\p u_1}{\p x} -\mu_0H_1\cfrac{\p u_3}{\p x}&=0,\\ \cfrac{1}{\tilde c^2\rho }\cfrac{\p p}{\p t}+\cfrac{u_1}{\tilde c^2\rho}\cfrac{\p p}{\p x}+\cfrac{\p u_1}{\p x}&=0,\\ \cfrac{1}{\rho}\cfrac{\p u_1}{\p t} +\cfrac{u_1}{\rho}\cfrac{\p u_1}{\p t} +\cfrac{\p p}{\p x} +\mu_0\sex{H_2\cfrac{\p H_2}{\p x}+H_3\cfrac{\p H_3}{\p x}}&=F_1,\\ \rho \cfrac{\p u_2}{\p t} +\rho u_1\cfrac{\p u_2}{\p x}-\mu_0H_1\cfrac{\p H_2}{\p x}&=F_2,\\ \rho \cfrac{\p u_3}{\p t}+\rho u_1\cfrac{\p u_3}{\p x} -\mu_0H_1\cfrac{\p H_3}{\p x}&=F_3,\\ \cfrac{\p S}{\p t}+u_1\cfrac{\p S}{\p x}&=0; \eea \eeex$$ 可写成 $$\bex A(U)\cfrac{\p U}{\p t}+B(U)\cfrac{\p U}{\p x}=C, \eex$$ 其中 $$\beex \bea U&=(H_2,H_3,p,u_1,u_2,u_3,S)^T,\\ A(U)&=\diag(\mu_0,\mu_0,\cfrac{1}{\tilde c^2\rho},\cfrac{1}{\rho},\rho,\rho,1),\\ B(U)&=\sex{\ba{ccccccc} \mu_0u_1&0&0&\mu_0H_2&-\mu_0H_1&0&0\\ 0&\mu_0u_1&0&\mu_0H_3&0&-\mu_0H_1&0\\ 0&0&\cfrac{u_1}{\tilde c^2 \rho}&1&0&0&0\\ \mu_0H_2&\mu_0H_3&1&\cfrac{u_1}{\rho}&0&0&0\\ -\mu_0H_1&0&0&0&\rho u_1&0&0\\ 0&-\mu_0H_1&0&0&0&\rho u_1&0\\ 0&0&0&0&0&0&u_1 \ea},\\ C&=(0,0,0,F_1,F_2,F_3,0)^T. \eea \eeex$$