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The answers to this question https://mathoverflow.net/questions/125639/replacing-large-dimensional-ode-systems-with-one-pde suggest that, in general, one can not hope for "replacing" an ODE system with a single PDE.

On the other hand, this article http://www.atlantis-press.com/php/download_paper.php?id=596 suggests we can do that in some cases. In particular, we read on page 15:

"Corollary 1. A system of n kth-order ODEs which admits a kn-dimensional solvable symmetry algebra $L_{kn}$ for which the (k − 1)th prologation of its symbols as well as" [some diff operator omitted] "are unconnected, is solvable by quadratures. Proof. Write the system of n kth-order ODEs as a first-order linear homogeneous PDE and use Theorem 1."

I'm looking for references on how to "Write the system of n kth-order ODEs as a first-order linear homogeneous PDE".

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  • $\begingroup$ @j.c.would you consider writing your comment as an answer? $\endgroup$ – Sergio Parreiras Jan 15 '14 at 0:26
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I have a feeling you think that statement in the proof of that corollary is saying something which it doesn't. Part of that may be due to some of the notation.

Note that the homogeneous PDE that one uses in the proof of this corollary is for a single multivariable function $f$ (not directly mentioned in the statement of the corollary, confusingly) and not the single variable functions $x_i(t)$. The functions $x_i,x^{(1)}_i=dx_1/dt,x^{(2)}_i=d^2x_1/dt^2,\dots$ are being treated like independent variables that $f$ depends on. You can see this in the calculations done in section 3 of that paper.

To write a system of $n$ $k$-th order ODEs $x_i^{(k)}=f_i(t,x,\dots,x^{(k-1)})$, $i=1,\dots,n$ (using the notation of the paper you cited) as a first-order linear homogeneous PDE, we may use the method of characteristics (in reverse).

We will see that this PDE ends up being $Af=0$ where $A$ is precisely that "some diff operator omitted" as you call it.

A lot of these equations may look unmotivated at first (and probably there is a quicker derivation) but this is one way to see what they mean.

One has $kn+1$ first-order equations (first order in $s$) of which $(k-1)n$ will later result in some pretty silly identities (just imagine replacing $s$ with $t$):

$\frac{d}{ds}x_i=x_i^{(1)}$

$\frac{d}{ds}x_i^{(1)}=x_i^{(2)}$

...

$\frac{d}{ds}x_i^{(k-2)}=x_i^{(k)}$.

And now the original system of $n$ ODEs ($k$th order if we were to try to solve for the $x_i$ directly, but 1st order in the current setup where $x_i$ and its derivatives are taken to be independent variables)

$\frac{d}{ds}x_i^{(k-1)}=x_i^{(k)}=f_i(t,x,\dots,x^{(k-1)})$.

There are two more equations we need to use to complete the magic. Here $f$ (an auxiliary function unrelated to the $f_i$) is going to be the function that will satisfy our eventual 1st order linear homogeneous PDE.

$\frac{d}{ds}t=1$

$\frac{d}{ds}f=0$

The first equation, in particular, assures that the previous $kn$ equations follow from the setup for this corollary.

Let's use a new set of variables $kn$ $y_{(l+1)+k(i-1)}=x_i^{(l)}$ for $i=1,\dots,n$ and $l=0,\dots,k$ so that $y_1=x_1,y_2=x_1^{(1)},\dots,y_k=x_1^{(k)},y_{k+1}=x_2,\dots$. This notation perhaps makes it more clear that these $kn$ variables will all be treated like independent variables from here on out.

Then the above system of $kn+2$ equations takes the form

$dy_j/ds=g_j(t,y_1,\dots,y_{kn})$ for $j=1,\dots,kn$

$dt/ds=1$

$df/ds=0$

Note that by the chain rule we can also write $$\frac{df}{ds}=\frac{\partial f}{\partial t}\frac{dt}{ds}+\frac{\partial f}{\partial y_1}\frac{dy_1}{ds}+\cdots+\frac{\partial f}{\partial y_{kn}}\frac{dy_{kn}}{ds}$$.

Then by replacing each of the $dy_j/ds$ by the expressions above, we see that $Af=0$ where $A$ is precisely the differential operator $$A=\frac{\partial}{\partial t}+\sum_{i=1}^n\left(x_i^{(1)}\frac{\partial}{\partial x_i}+\cdots+x_i^{(k-1)}\frac{\partial}{\partial x_i^{(k-2)}}+f_i(t,x,\dots,x^{(k-1)})\frac{\partial}{\partial x_i^{(k-1)}}\right)$$ as in the full statement of the corollary.

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