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I have a system of two quasilinear equations as follows:

$\frac{\partial{w}}{\partial{t}}+D\frac{\partial{w}}{\partial{x}}=0$

with $w=[u , \rho]^T$ and $D=\begin{bmatrix} u & c^2/\rho \\ \rho & u \end{bmatrix}$

The two characteristics of this hyperbolic system are given by $\frac{dx}{dt}=\lambda$ where $\lambda$ are the eigenvalues of $D$; $det(D-\lambda I)=0 \to (u-\lambda)^2=c^2$, and $\lambda_{+}=u+c$ and $\lambda_{-}=u-c$.

The eigenvectors are $[c,-\rho]^T$ for $\lambda{-}$ and $[c,\rho]^T$ for $\lambda{+}$ , such that the usual matrices are

$T=\begin{bmatrix} c & c \\ -\rho & \rho \end{bmatrix}$ and $T^{-1}=\frac{1}{2\rho c}\begin{bmatrix} \rho & -c \\ \rho & c \end{bmatrix}$, such that $\Lambda=T^{-1}DT=\begin{bmatrix} u-c & 0 \\ 0 & u+c \end{bmatrix}$.

Put $\alpha$ and $\beta$ the curvilinear coordinates along the characteristics $\frac{dx}{dt} = u-c$ and $\frac{dx}{dt} = u+c$ respectively; then the system transforms to the canonical form.

$\frac{dt}{d\alpha} =\frac{dt}{d\beta} =1$,

$\frac{dx}{d\alpha} =u-c$,

$\frac{dx}{d\beta} =u+c$

$\rho \frac{du}{d\alpha}-c\frac{d\rho}{d\alpha} =0$

$\rho \frac{du}{d\beta}+c\frac{d\rho}{d\beta} =0$

. My question is how to drive the solution from the above equations? If an initial conditon is necessary, consider an arbitrary initial condition. thank you for your help.

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    $\begingroup$ I don't understand why you want to do it the way you have here? Your original problem gives you two PDEs, one of them is the inhomogeneous Burgers equation in $u$ only and the other is the continuity of mass equation. Both are solvable in their own right. $\endgroup$ – Mattos Sep 6 '16 at 16:19
  • $\begingroup$ The two equations are $\rho_t + (\rho u)_x = 0$ (continuity / mass conservation equation) and $u_t + uu_x = - \frac{\rho_x}{\rho}$ (momentum conservation). Togetaher these form the one dimensional (compressible) Euler equations with pressure term $P = \rho$. There are many books on this subject or you could try searching for something like "how to solve Euler equations". $\endgroup$ – Winther Sep 14 '16 at 13:43

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