[Math] Splitting a 2nd order PDE into a system of first order PDEs/ODEs

ordinary differential equationspartial differential equationssystems of equations

Consider a standard wave equation:
$ \frac{\partial^2 p}{\partial t^2} = c^2 \frac{\partial^2 p}{\partial x^2} $

The question is how to formulate this as a first order system:
$ \frac{\partial \textbf{u}}{\partial t} + A \frac{\partial \textbf{u}}{\partial x} = 0 $

Thankfully, I have the answer to be:
$ \textbf{u} = [ p_t, c p_x]^T, \hspace{10mm} A = \left( \begin{array}{cc}
0 & -c \\
-c & 0 \end{array} \right) $

However!

1) I would like to know how is this obtained – in a formal way.

For example, in the case with wave equation above we are given hints on how to choose $ u_1 $ and $ u_2 $ (in the problem I was solving).

But I would like to know if such procedure have a name and if there is more formal way to express it – how to apply it to more general 2nd order PDE?

Whenever I find something like this, it always starts with "try this variables…" but that sounds like advice from someone who knows the solution. Is it possible that the whole method is based on trial and error of different substitutions which seem reasonable?

2) How do we prove that the system of first order PDE above is indeed the wave equation decomposed?

Answer would be appreciated, but I would also go with good reference book or material which discuss this – the procedure, not particular solutions based on hints which I would never thought of myself.

Best Answer

Generally the idea is to write the vector $u$ as the space-time gradient of the solution to the PDE (be it the wave equation you have posted or a more general one), then you can calculate $u_t$ and $u_x$ using the PDE.

We see that $$u_t=[p_{tt},cp_{xt}]^T=[c^2p_{xx},cp_{xt}]^T,$$ and $$u_x=[p_{xt},cp_{xx}]^T,$$ now we wish to find the matrix $A$ such that $u_t+Au_x=0$, this gives us $$[c^2p_{xx}+A_{11}cp_{xt}+A_{12}cp_{xx},cp_{xt}+A_{21}p_{xt}+A_{22}cp_{xx}]^T=[0,0]^T,$$ choosing $A_{11}=A_{22}=0$ and $A_{12}=A_{21}=-c$, we get the desired equality.