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: $$\frac{\partial}{\partial t} \begin{pmatrix} u \\ u_t \\ u_x \end{pmatrix} - \frac{\partial}{\partial x} \begin{pmatrix} u \\ u_x \\ u_t \end{pmatrix} = \begin{pmatrix} u_t - u_x \\ u \\ 0 \end{pmatrix}$$

This doesn't feel correct. If anyone is familiar with a standard way to do this, any help would be appreciated.

• 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. Dec 18, 2016 at 16:22

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

• Missed 4 out of my solution, indeed, $$u_{tt} = u_{qq}-2u_{qp}+u_{pp}$$ and $$u_{xx} = u_{qq} + 2u_{qp} + u_{pp}$$ Dec 18, 2016 at 16:42
• That makes sense. Thanks! Dec 19, 2016 at 12:45

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 $$\begin{cases} v=\partial_{t}u\\ w=\partial_{x}u\\ u=\partial_{t}v-\partial_{x}w. \end{cases}$$

• 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. Dec 19, 2016 at 12:43

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).