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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.

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    $\begingroup$ The first row in your equations set seems to be an identity, so you can ignore it, as well as the third one is a restatement of derivative exchange. $\endgroup$ Dec 18, 2016 at 16:22

3 Answers 3

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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 $$\frac{dv}{dp} = -u(p, q)/4 $$

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  • $\begingroup$ Missed 4 out of my solution, indeed, $$u_{tt} = u_{qq}-2u_{qp}+u_{pp}$$ and $$u_{xx} = u_{qq} + 2u_{qp} + u_{pp}$$ $\endgroup$ Dec 18, 2016 at 16:42
  • $\begingroup$ That makes sense. Thanks! $\endgroup$
    – Merkh
    Dec 19, 2016 at 12:45
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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 $$ \begin{cases} v=\partial_{t}u\\ w=\partial_{x}u\\ u=\partial_{t}v-\partial_{x}w. \end{cases} $$

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  • $\begingroup$ Calling $u_t = v$ and $w = u_x$ in my formulation adds nothing to whether or not it is "correct", it is just a relabeling. I was treating $u_t$ and $u_x$ and $u$ as my three unknown functions. $\endgroup$
    – Merkh
    Dec 19, 2016 at 12:43
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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).

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