Constructing a linear ODE for a product of two holonomic functions without introducing additional singularities

A function $f$ is called holonomic if it satisfies some linear differential equation with polynomial coefficients $$p_n(x) f^{(n)}(x)+\dots+p_1(x)f'(x)+p_0(x)f(x)=0.$$ Now if $f,g$ are holonomic then so are their sum and product. To obtain a differential equation for $h=fg$, first observe that $$h^{(k)} = \sum_{i=0}^k {k\choose i} f^{(i)} g^{(k-i)}.$$ Since each $f^{(i)}$ is a linear combintation of $f,f',\dots,f^{(n-1)}$ with rational coefficients (where $n$ is the order of the ODE satisfied by $f$), and analogously each $g^{(i)}$ is a linear combination of $g,g',\dots,g^{(m-1)}$, then $h,h',h'',\dots$ span a finite-dimensional vector space of dimension at most $d=mn$, and hence there exists a nontrivial relation of the form $$r_d(x)h^{(d)}+\dots+r_1(x)h'+r_0(x)h=0$$with $r_i(x)$ rational functions.

Now suppose that the ODE which $f$ satisfies had singular points $u_1,\dots,u_n$, while the equation of $g$ had singular points $w_1,\dots,w_m$. The ODE for $h$ might have additional singular points besides $u_1,\dots,u_n,w_1,\dots,w_m$. For example, if $$(x-a)f''(x)+cf(x)=0\\\\ (x-b)g''(x)+dg(x)=0$$ then $h=fg$ satisfies an ODE of order 4 with leading coefficient $$(x-a)^3 (x-b)^3 \biggl((c-d)x-(bc-ad)\biggr)h^{(4)}+\dots$$ (I calculated this using C.Mallinger's GeneratingFunctions Mathematica package).

Is it possible to construct a holonomic ODE for $h$ (in the above example and in general) without introducing additional singular points?

• You have bound on the other of the differential equation of $h$ by the order of $f$ and $g$. Can we have bound on the degree of $h$ by the degree of the differential equation for $f$ and $g$?
– GGT
Nov 21 '17 at 23:52

Hi Dima,

The operator you obtain by the algorithm you describe has minimal possible order. You pay for the minimality of the order by having (in general) a nonminimal degree of the polynomial coefficients. You can turn the minimal-order operator into a minimal-degree operator if you are willing to pay the price of a higher order. There is in general no operator which has both order and degree as small as possible.

As long as you are only interested in the singularities, i.e., you are only concerned about avoiding extra factors in the leading coefficient, you can use a desingularization algorithm to turn the minimal order operator into one which has no unnecessary singular points. For differential operators, this is a classical technique, explained for example in the ODE book of Ince (Section 16.4). For recurrence operators, there is an algorithm by Abramov and van Hoeij (http://www.math.fsu.edu/~hoeij/papers/issac99/new.pdf).

If you are interested in minimizing the degree not only of the leading coefficient, but of all the polynomial coefficients in the operator simultaneously, see my joint paper with Chen, Jaroschek, and Singer, to appear on this year's ISSAC (http://www.risc.uni-linz.ac.at/people/mkauers/publications/jaroschek13.pdf).

Best regards, Manuel

• Thanks Manuel! That kind of things is exactly what I needed.
– dima
Apr 4 '13 at 11:19

The paper which may be relevant to this question is: MR0320402 Frank, Günter; Wittich, Hans Zur Theorie linearer Differentialgleichungen im Komplexen. Math. Z. 130 (1973), 363–370. (They consider the differential equations with only one singular point. But their results show that the answer may depend on the nature of the singularity).

• Since my German is nonexistent, could you please pinpoint the location in the paper where they talk about the case of one singular point? Also, I couldn't figure out if they always assume that the solutions of the ODEs are entire functions...
– dima
Mar 31 '13 at 12:48
• The paper is about entire solutions. The singular point is at infinity. They prove that the class of solutions of linear differential equations with polynomial coefficients and NO singular points in C (that is the top coefficient is 1) is an algebra. That is this class of functions is closed under addition and multiplication. This is a special case of your conjecture. They also say that if you consider entire coefficients (not necessarily polynomials) the class of solutions you obtain is not even closed with respect to addition. Mar 31 '13 at 13:47