1
$\begingroup$

Consider the system of first order PDEs

$ \left\{ \begin{eqnarray} \frac{\partial}{\partial t} v_1 + \frac{\partial}{\partial x_1} p_1 + \eta(x_1) v_1 = 0 \\ \frac{\partial}{\partial t} v_2 + \frac{\partial}{\partial x_2} p_2 + \eta(x_2) v_2 = 0 \\ \frac{\partial}{\partial t} p_1 + \frac{\partial}{\partial x_1} v_1 + \eta(x_1) p_1 = f(t) \\ \frac{\partial}{\partial t} p_2 + \frac{\partial}{\partial x_2} v_2 + \eta(x_2) p_2 = 0 \end{eqnarray} \right. $

with initial condition $v_1 = v_2 = p_1 = p_2 = 0$ at $t=0$, where $v_1,v_2,p_1,p_2 : \mathbb{R}^3 \to \mathbb{R} $ are the unknown functions, $\eta, f: \mathbb{R} \to \mathbb{R}$ are given smooth functions.

I would like to consider this system of PDEs with the method of characteristics. I know the method for a single PDE, but I don't know how to find the characteristics of such a system.

Could anyone please give some advice?

Regards.

I could understand that this system is a hyperbolic one.

Take $U=(v_1,v_2,p_1,p_2)'$, then the system can be rewritten as

$\frac{\partial}{\partial t} U + \left( \begin{array}{cccc} ~0 ~ & 0 ~ & 1 ~ & 0 \\ 0 & 0 & 0 & 0 \\ 1 ~ & 0 ~ & 0 ~ & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \frac{\partial}{\partial x_1} U + \left( \begin{array}{cccc} ~0 ~ & 0 ~ & 0 ~ & 0 \\ 0 & 0 & 0 & 1 \\ 0 ~ & 0 ~ & 0 ~ & 0 \\ 0 & 1 & 0 & 0 \end{array} \right) \frac{\partial}{\partial x_2} U + \left( \begin{array}{cccc} ~\eta(x_1) ~ & 0 ~ & 0 ~ & 0 \\ 0 & \eta(x_2) & 0 & 0 \\ 0 ~ & 0 ~ & \eta(x_1) ~ & 0 \\ 0 & 0 & 0 & \eta(x_2) \end{array} \right) U =\left( \begin{array}{cccc} 0 \\ 0 \\ f(t) \\ 0 \end{array} \right) . $

Take $B_1 = \left( \begin{array}{cccc} ~0 ~ & 0 ~ & 1 ~ & 0 \\ 0 & 0 & 0 & 0 \\ 1 ~ & 0 ~ & 0 ~ & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) $, $B_2 = \left( \begin{array}{cccc} ~0 ~ & 0 ~ & 0 ~ & 0 \\ 0 & 0 & 0 & 1 \\ 0 ~ & 0 ~ & 0 ~ & 0 \\ 0 & 1 & 0 & 0 \end{array} \right) $, then define a matrix

$B (y_1,y_2)= y_1 B_1 + y_2 B_2 = \left( \begin{array}{cccc} ~0 ~ & 0 ~ & y_1 ~ & 0 \\ 0 & 0 & 0 & y_2 \\ y_1 ~ & 0 ~ & 0 ~ & 0 \\ 0 & y_2 & 0 & 0 \end{array} \right)$.

The eigenvalues of $B$ are $-y_1, y_1, -y_2, y_2$, and the corresponding eigenvectors are $(-1, 0, 1, 0)', (1, 0, 1, 0)', (0, -1, 0, 1)', (0, 1, 0, 1)'$.

Since the eigenvalues of $B$ are distinct and all real and the four eigenvectors are linearly independent, the system of first order PDEs is hyperbolic.

$\endgroup$
  • $\begingroup$ This question has been answered on http://mathoverflow.net/. I would like to express my gratitude to Robert Bryant for giving the answer and explaining the method of characteristics, and to Igor Khavkine for discussing the method of characteristics. $\endgroup$ – William Wu Jan 8 '15 at 1:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.