$$u_t+A(x,t,u)u_x=b(x,t,u) \tag 1$$ $$u=(u_1, \dots, u_n), b=(b_1, \dots, b_n)$$ $$A=[a_{ij}], i,j = 1, \dots, n$$ $$$$ We set the question if there are characteristic directions at the path of which the PDE system $(1)$ is reduced to an ODE system.
So we take linear combinations of the above equations.
We use the vector $$\gamma=(\gamma_1, \dots, \gamma_n). \tag 2$$
We take $$\gamma^T(u_t+Au_x)=\gamma^Tb. \tag 3 $$
We want to conclude to a form of total derivative of the linear combination of $u$.
That means:(we consider that $u_j=u_j(x(t),t)$)
$$\frac{d}{dt}(\gamma_1u_1+\gamma_2u_2 +\dots+\gamma_n u_n)=\frac{\partial}{\partial{t}}(\gamma_1u_1+\gamma_2u_2 +\dots+\gamma_n u_n)+\frac{dx}{dt} \frac{\partial}{\partial{x}}(\gamma_1u_1+\gamma_2u_2 +\dots+\gamma_n u_n)$$
and we define $\frac{dx}{dt}=\lambda$
The relation above means: $$m^T\left(\frac{\partial{u}}{\partial{t}}+\lambda\frac{\partial{u}}{\partial{x}}\right)=\gamma^Tb \tag 4$$
From the relation $(3)$ and $(4)$, we see that $\gamma=m=(\gamma_1, \dots, \gamma_n)$ and that $\gamma^TA=\lambda \gamma^T$
$\lambda:$ eigenvalue of the matrix $A$
$\gamma^T:$ left eigenvector of $A$
$$$$ We define the characteristic directions:
$$\frac{dx}{dt}=\lambda, \text{ so } \gamma^T(\frac{\partial{u}}{\partial{t}}+\frac{dx}{dt} \frac{\partial{u}}{\partial{x}})=\gamma^Tb$$
$$\gamma^T \frac{du}{dt}=\gamma^Tb$$
We conclude that to reduce the PDE system to an ODE system, the matrix $A$ should have $n$ discrete real eigenvalues.
In this case we say that the system is hyperbolic. $$$$
- I haven't understood the relation $(4)$... Could you explain it to me?? What is $m$??
Equation 4 is unusual - I believe that $m$ is just a stand in for an arbitrary left handed eigenvector and so the right hand side should be $m^Tb$
Usually I have seen this done where we explicitly know we are going after the left handed eigenvectors (for example - for Euler's equation from gas dynamics there are two eigenvalues and two eigenvectors). Exploiting the left handed eigenvectors $\underline{is}$ the big trick.
So, if I know that $E_i$ is the $ith$ left handed eigenvector of $A$ corresponding to the eigenvalue $\lambda_i$ then hitting $\frac{\partial u}{\partial t} + A\frac{\partial u}{\partial x} = b$ on the left with $E_i$ yields $E_i^T(\frac{\partial u}{\partial t} + \lambda_i\frac{\partial u}{\partial x}) = E_i^Tb$.
Then we can define a new variable (the $ith$ Riemann invariant) $R_i$ such that $\frac{\partial R_i}{\partial t} + \lambda_i\frac{\partial R_i}{\partial x} = E_i^Tb$. Where $\frac{\partial R_i}{\partial t} = E_i^T\frac{\partial u}{\partial t}$.
(We aren't always so lucky as to have $E_i$ be a bunch of constants - usually, they depend on the primitive variables of the system)
Now we define the characteristic curve by $\frac{dx}{dt} = \lambda_i$ and we can track the evolution of $R_i$ along the characteristic curves via $\frac{d R_i}{dt} = E_i^Tb$