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Suppose we are given a system of first order PDEs with constant coefficients. In particular, suppose we are given $k$ PDEs for $u_1,u_2, \dots u_n$ with respect to independent variables $x_1,x_2, \dots x_n$. Label these as $L_j[u,x]=0$ for $j=1,2,\dots k$.

Question: is it possible to combine the system of $k$ first order PDEs into an uncoupled system of $d>1$-degree PDEs which contains no non-polynomial extraneous solutions? Ideally, $u_1,u_2, \dots u_n$ would be solutions of the same PDE.

The problem I'm interested has $k=n^2-n$. I have a few simple examples of success. I'll give them here to illustrate the goal of the question:

  1. $u_x=v_y$ and $u_y=-v_x$ hence $u_{xx}+u_{yy}=0$ and $v_{xx}+v_{yy}=0$
  2. $u_x=v_y$ and $u_y=v_x$ hence $u_{xx}-u_{yy}=0$ and $v_{xx}-v_{yy}=0$

These examples are simple enough that it is obvious how to find the desired uncoupled PDEs.(thanks to joriki pointing out there are extra linear solutions in the unreduced equations, for example $u=x-y, v=x$ solves $u_{xx}+u_{yy}=0$ but not $u_x=v_y$ and $u_y=-v_x$)

It turns out that for my problem there is a natural method to write an $n$-th order PDE which possesses solutions to the system $L_j[u,x]=0$ for $j=1,2,\dots n^2-n$. Unfortunately, the $n$-th order PDE includes many extraneous solutions which are not found in the original $L_j[u,x]=0$ for $j=1,2,\dots n^2-n$ system. I want to find a method to remove these extra solutions and uncover the analog of $u_{xx}+u_{yy}=0$ and $v_{xx}+v_{yy}=0$ for other examples.

Any advice is appreciated, I'm not looking for an authoritative general solution, I'm interested in advice on what techniques I should think about to properly confront this problem.

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But don't the higher-order equations in your "examples of success" also have extraneous solutions? Adding a linear function breaks the first-order equations but not the second-order ones. –  joriki Sep 19 '12 at 7:48
    
@joriki good point. I should allow for those extraneous solutions. I'll edit the question accordingly. –  James S. Cook Sep 19 '12 at 12:28
    
upon breakfast, it occurred to me that polynomials of order $d-1$ will be solutions for the $d$-th order uncoupled PDE. I had $d=2$ in my token examples. Will edit again. –  James S. Cook Sep 19 '12 at 13:03
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