Convert second order PDE $u_{tt} = u_{xx} + u$ to a system of first order PDE's I am attempting to convert 
$$ u_{tt} = u_{xx} + u$$
to a system of first order PDE's.  I believe that the system will require 3 equations, one for each of $u, u_t,$ and $u_x.$  Here is my attempt:
\begin{equation} \frac{\partial}{\partial t} \begin{pmatrix} u \\ u_t \\ u_x \end{pmatrix} -  \frac{\partial}{\partial x} \begin{pmatrix} u \\ u_x \\ u_t \end{pmatrix} = \begin{pmatrix} u_t - u_x \\ u \\ 0 \end{pmatrix} \end{equation}
This doesn't feel correct.  If anyone is familiar with a standard way to do this, any help would be appreciated.  
 A: This is a wave equation and you need to change variables as follows: p = x+t, q = x-t. You will obtain:
$$ -4\frac{d^2u}{dpdq}=u(p,q)$$
And then set $$ v= \frac{du}{dq} $$ and another equation $$\frac{dv}{dp} = -u(p, q)/4 $$
A: You should introduce two more functions $v$ and $w$, so that your system of three equations does indeed contain three unknown functions. Then you specify that $v$ and $w$ are, respectively, the first time-derivative and  the first space-derivative of $u$. So I would rather write something like this
$$
\begin{cases}
v=\partial_{t}u\\
w=\partial_{x}u\\
u=\partial_{t}v-\partial_{x}w.
\end{cases}
$$
A: The proposed first-order recast is absolutely correct. The equation $u_{tt} = u_{xx} + u$ rewrites indeed as a linear system of balance laws ${\bf q}_t + {\bf A}\, {\bf q}_x = {\bf S}\, {\bf q}$, where ${\bf q} = (u,u_t,u_x)^\top$ and
$$
{\bf A} = \begin{pmatrix}
1 & 0 & 0\\
0 & 0 & -1\\
0 & -1 & 0
\end{pmatrix}, \qquad
{\bf S} = \begin{pmatrix}
0 & 1 & 1\\
1 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix} .
$$
Here, $\bf A$ and the matrix exponential $\exp(-{\bf S} t)$ do not commute. Hence, the system cannot be decoupled. One notes that the present system is symmetric hyperbolic (${\bf A}$ is symmetric) but not strictly hyperbolic (two eigenvalues of ${\bf A}$ are equal).
