0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you for the answer, that helps a bit. Could you clarify the first sentence - "write the vector u as the space-time gradient of the solution to the PDE", does that mean writing u as $ [p_t, p_x] $, i.e. two variables of which one is gradient in space one is in time? I omitted $ c $ on purpose, for example I go without it (assuming I do not know the solution) and then I "tune" it by playing around and adding $ c $. $\endgroup$ – nevermind Aug 11 '15 at 11:34
  • $\begingroup$ That is what I meant, though it doesn't seem to matter too much if you put the "$c$" in or not, as you have seen, you can omit the $c$ and then $A$ changes. The benefit of keeping the $c$ is that the matrix $A$ is then symmetric, so the system is symmetric hyperbolic, this may not be the case (I have not checked) if you go with the standard gradient. $\endgroup$ – Ellya Aug 11 '15 at 13:05
  • $\begingroup$ If you were looking the $n$ dimensional wave equation $p_{tt}-c^2\Delta p$, then it would make sense to let $u=[p_t,cp_{x_1},\ldots,cp_{x_n}]^T$. $\endgroup$ – Ellya Aug 11 '15 at 13:07
  • $\begingroup$ Thank you very much, it is clear now and hopefully it will help me in the future as well! $\endgroup$ – nevermind Aug 12 '15 at 11:24

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.