Consider the linear differential system

For a rational function matrix $S$ with $\det \left(S\right)\ne 0$ the transformation

leads to a new system $Z\'=BZ$. One has

The matrices A and $B$ are said to be equivalent.

The super_reduce function computes a matrix $B$ equivalent to A in a special form, called super-irreducible at p. The output is a list of two elements. The first is the super-irreducible matrix; the second gives information about the integer slopes and Newton polynomials of the Newton polygon of the computed system. It is a list of couples, the first element of which is an integer (the slope) and the second a polynomial in u (the Newton polynomial associated with the slope).

The moser_reduce function computes a matrix $B$ equivalent to A which is Moser-irreducible at p. This is an equivalent  matrix of minimal pole order. The output is similar to the super_reduce function, but only the Newton polynomials associated with the two biggest slopes are returned.

If T and invT are passed as arguments, they are assigned to the transformation matrix T and their inverse which transforms A to the super-irreducible (Moser-irreducible) matrix.

These functions are part of the DEtools package, and so they can be used in the form super_reduce(..) and moser_reduce(..) only after executing the command with(DEtools).  However, they can always be accessed through the long form of the command by using DEtools[super_reduce](..) or DEtools[moser_reduce](..).

$\mathrm{with}\left(\mathrm{DEtools}\right)\:$

$A≔\mathrm{array}\left(1..4\,1..4\,\left[\left(2\,1\right)=0\,\left(3\,2\right)=-\frac{6}{x^4}\,\left(2\,4\right)=-\frac{1}{x^2}\,\left(4\,2\right)=\frac{3-2x}{x^3}\,\left(1\,3\right)=\frac{5+2x}{x^3}\,\left(4\,1\right)=-\frac{4}{x^2}\,\left(4\,3\right)=1\,\left(4\,4\right)=\frac{2}{x^2}\,\left(1\,1\right)=-1\,\left(2\,2\right)=-1\,\left(2\,3\right)=0\,\left(1\,4\right)=\frac{4x^2-2x}{x^4}\,\left(3\,3\right)=-1\,\left(3\,1\right)=\frac{8}{x^3}\,\left(1\,2\right)=\frac{1}{x}\,\left(3\,4\right)=\frac{2}{x^2}\right]\right)$

$A≔\left[\begin{array}{cccc}\mathrm{-1}& \frac{1}{x}& \frac{5+2x}{x^3}& \frac{4x^2-2x}{x^4}\\ 0& \mathrm{-1}& 0& -\frac{1}{x^2}\\ \frac{8}{x^3}& -\frac{6}{x^4}& \mathrm{-1}& \frac{2}{x^2}\\ -\frac{4}{x^2}& \frac{3-2x}{x^3}& 1& \frac{2}{x^2}\end{array}\right]$

$B≔\mathrm{moser\_reduce}\left(A\,x\,x\,u\,'T'\,'\mathrm{invT}'\right)\:$

$B\left[1\right]$

$\left[\begin{array}{cccc}-\frac{7x+6}{3x}& -\frac{-5+16x}{3x^3}& -\frac{24}{x^2}& -\frac{4\left(4x+3\right)}{9x^2}\\ -\frac{2x^2-3x-24}{x^3}& -\frac{6+x}{x^2}& \frac{5x^2-24}{x^2}& -\frac{8}{3x^2}\\ -\frac{6}{x^3}& \frac{2}{x^2}& -\frac{x^2+x-8}{x^2}& 0\\ x& \frac{2\left(2x-1\right)}{x^2}& \frac{3\left(6x-1\right)}{x^2}& \frac{1}{3}\end{array}\right]$

$B≔\mathrm{super\_reduce}\left(A\,x\,x\,u\,'T'\,'\mathrm{invT}'\right)\:$

$\mathrm{print}\left(B\left[1\right]\right)$

$\left[\begin{array}{cccc}-\frac{7x+6}{3x}& -\frac{-5+16x}{3x^3}& -\frac{24}{x^2}& -\frac{4\left(4x+3\right)}{9x^2}\\ -\frac{2x^2-3x-24}{x^3}& -\frac{6+x}{x^2}& \frac{5x^2-24}{x^2}& -\frac{8}{3x^2}\\ -\frac{6}{x^3}& \frac{2}{x^2}& -\frac{x^2+x-8}{x^2}& 0\\ x& \frac{2\left(2x-1\right)}{x^2}& \frac{3\left(6x-1\right)}{x^2}& \frac{1}{3}\end{array}\right]$

$B\left[2\right]$

$\left[\left[0\,1\right]\,\left[1\,-u^2+2u+2\right]\,\left[2\,u^2-40\right]\right]$

$\mathrm{map}\left(\mathrm{normal}\,\mathrm{evalm}\left(B\left[1\right]-\mathrm{invT}\&*\left(A\&*T-\mathrm{map}\left(\mathrm{diff}\,\mathrm{eval}\left(T\right)\,x\right)\right)\right)\right)$

$\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]$

$A≔\mathrm{array}\left(1..4\,1..4\,\left[\left(2\,1\right)=0\,\left(3\,2\right)=0\,\left(2\,4\right)=\frac{7}{{\left(x^2+1\right)}^2}\,\left(4\,2\right)=\frac{3}{{\left(x^2+1\right)}^2}\,\left(1\,3\right)=-\frac{6}{x^2+1}\,\left(4\,1\right)=0\,\left(4\,3\right)=-\frac{3x^2}{{\left(x^2+1\right)}^3}\,\left(4\,4\right)=1\,\left(1\,1\right)=1\,\left(2\,2\right)=\frac{4}{x^2+1}\,\left(2\,3\right)=0\,\left(1\,4\right)=\frac{9}{x^2+1}\,\left(3\,3\right)=1\,\left(3\,1\right)=\frac{-5x-3}{{\left(x^2+1\right)}^2}\,\left(1\,2\right)=0\,\left(3\,4\right)=0\right]\right)$

$A≔\left[\begin{array}{cccc}1& 0& -\frac{6}{x^2+1}& \frac{9}{x^2+1}\\ 0& \frac{4}{x^2+1}& 0& \frac{7}{{\left(x^2+1\right)}^2}\\ \frac{-5x-3}{{\left(x^2+1\right)}^2}& 0& 1& 0\\ 0& \frac{3}{{\left(x^2+1\right)}^2}& -\frac{3x^2}{{\left(x^2+1\right)}^3}& 1\end{array}\right]$

$B≔\mathrm{super\_reduce}\left(A\,x\,x^2+1\,u\,'T'\,'\mathrm{invT}'\right)\:$

$\mathrm{print}\left(B\left[1\right]\right)$

$\left[\begin{array}{cccc}\frac{x^2-2x+1}{x^2+1}& -\frac{5x+3}{{\left(x^2+1\right)}^2}& \frac{7}{{\left(x^2+1\right)}^2}& -\frac{x^2-2x-3}{x^2+1}\\ -\frac{6}{x^2+1}& \frac{x^2-2x+1}{x^2+1}& \frac{9}{{\left(x^2+1\right)}^2}& \frac{6}{x^2+1}\\ -\frac{3x^2}{{\left(x^2+1\right)}^2}& 0& 1& \frac{3}{x^2+1}\\ 0& 0& \frac{7}{{\left(x^2+1\right)}^2}& \frac{4}{x^2+1}\end{array}\right]$

Pfluegel, E. "An Algorithm For Computing Exponential Solutions of First Order Linear Differential Systems." In Proceedings of ISSAC '97, pp. 164-171. Edited by Wolfgang Kuchlin. New York: ACM Press, 1997.