Q-Difference Equations - Maple Help

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 Q-Difference Equations

The QDifferenceEquations package provides tools for studying equations of the form:

and their solutions $y\left(x\right)$, where  are polynomials in the indeterminates $x$ and $q$. The indeterminate $q$ is considered to be a constant.  is the associated q-difference operator of order $n$, where $Q$ represents the q-shift operator .

For example, the solutions of the first order q-difference equation $\mathrm{Ly}=0$, where , are given by:

$y\left(x\right)=C\cdot \left({x}^{2}-1\right)$,

where $C$ is an arbitrary constant that is allowed to depend on $q,$ but not on $x$.

In Maple 18, two new commands were added to this package:

 • Closure computes the closure in the ring of linear q-difference operators with polynomial coefficients.
 • Desingularize computes a multiple of a given q-difference operator with fewer singularities.

As an example, let's look at the operator $L$ from above.

 > $\mathrm{with}\left(\mathrm{QDifferenceEquations}\right)$
 $\left[{\mathrm{AccurateQSummation}}{,}{\mathrm{AreSameSolution}}{,}{\mathrm{Closure}}{,}{\mathrm{Desingularize}}{,}{\mathrm{ExtendSeries}}{,}{\mathrm{IsQHypergeometricTerm}}{,}{\mathrm{IsSolution}}{,}{\mathrm{PolynomialSolution}}{,}{\mathrm{QBinomial}}{,}{\mathrm{QBrackets}}{,}{\mathrm{QDispersion}}{,}{\mathrm{QECreate}}{,}{\mathrm{QEfficientRepresentation}}{,}{\mathrm{QFactorial}}{,}{\mathrm{QGAMMA}}{,}{\mathrm{QHypergeometricSolution}}{,}{\mathrm{QMultiplicativeDecomposition}}{,}{\mathrm{QPochhammer}}{,}{\mathrm{QPolynomialNormalForm}}{,}{\mathrm{QRationalCanonicalForm}}{,}{\mathrm{QSimpComb}}{,}{\mathrm{QSimplify}}{,}{\mathrm{RationalSolution}}{,}{\mathrm{RegularQPochhammerForm}}{,}{\mathrm{SeriesSolution}}{,}{\mathrm{UniversalDenominator}}{,}{\mathrm{Zeilberger}}\right]$ (1)
 > $L≔\left({x}^{2}-1\right)\cdot Q-\left({q}^{2}\cdot {x}^{2}-1\right):$

This operator has singularities at $x=±1$, where its leading coefficient vanishes. However, the solutions $y\left(x\right)=C\cdot \left({x}^{2}-1\right)$ satisfying $\mathrm{Ly}=0$ are non-singular at both points, so $x=±1$ are two apparent singularities. It is possible to remove such apparent singularities by finding a higher order operator $M$ that has the same solutions as $L$, plus some additional ones. This is what the command Desingularize does.

 > $M≔\mathrm{Desingularize}\left(L,Q,x,q\right)$
 ${M}{:=}{{Q}}^{{2}}{+}\left({-}{{q}}^{{2}}{-}{1}\right){}{Q}{+}{{q}}^{{2}}$ (2)

Let us verify that $y\left(x\right)$ is actually a solution of $M$.

 > $\mathrm{with}\left(\mathrm{OreTools}\right):\mathrm{with}\left(\mathrm{Converters}\right):$
 > $A≔\mathrm{SetOreRing}\left(\left[x,q\right],\mathrm{qshift}\right):$
 > $\mathrm{OM}≔\mathrm{FromPolyToOrePoly}\left(M,Q\right)$
 ${\mathrm{OM}}{:=}{\mathrm{OrePoly}}{}\left({{q}}^{{2}}{,}{-}{{q}}^{{2}}{-}{1}{,}{1}\right)$ (3)
 > $\mathrm{Apply}\left(\mathrm{OM},y\left(x\right),A\right);$
 ${{q}}^{{2}}{}{y}{}\left({x}\right){+}\left({-}{{q}}^{{2}}{-}{1}\right){}{y}{}\left({q}{}{x}\right){+}{y}{}\left({{q}}^{{2}}{}{x}\right)$ (4)
 > $\mathrm{expand}\left(\mathrm{eval}\left(,y=\left(x\to C\cdot \left({x}^{2}-1\right)\right)\right)\right)$
 ${0}$ (5)

The closure of an operator $L$ consists of all left "pseudo"-multiples of $L$, i.e., all operators $R$ for which there exists an operator, $P$ (in $Q,x,q$) and a polynomial $f$ (in $x,q$ only), such that the following torsion relation holds true:

Basically, this means that $\mathrm{PL}$ is a genuine left multiple of $L$ of which one can factor out the content $f$. Both $\mathrm{PL}$ and $R$ have exactly the same solutions, which include all solutions of $L$. In particular, the desingularizing operator $M$ from above is an element of the closure of $L$.

The command Closure computes a basis of the closure.

 > $C≔\mathrm{Closure}\left(L,Q,x,q\right)$
 ${C}{:=}\left[\left({{x}}^{{2}}{-}{1}\right){}{Q}{-}{{q}}^{{2}}{}{{x}}^{{2}}{+}{1}{,}\left({-}{q}{}{x}{-}{1}\right){}{{Q}}^{{2}}{+}\left({{q}}^{{3}}{}{x}{+}{{q}}^{{2}}{+}{q}{}{x}{+}{1}\right){}{Q}{-}\left({q}{}{x}{+}{1}\right){}{{q}}^{{2}}{,}\left({q}{}{x}{-}{1}\right){}{{Q}}^{{2}}{+}\left({-}{{q}}^{{3}}{}{x}{+}{{q}}^{{2}}{-}{q}{}{x}{+}{1}\right){}{Q}{+}\left({q}{}{x}{-}{1}\right){}{{q}}^{{2}}\right]$ (6)

We see that, trivially, $L$ itself belongs to its closure. In addition, the basis contains two second order operators, both of which have fewer and different singularities than $L$ itself, namely, $x=-{q}^{-1}$ and $x={q}^{-1}$, respectively. Since these two singularities are different, the two leading coefficients are coprime as polynomials in $x$, and we can find a linear combination that is monic:

 >
 ${1}{,}{-}\frac{{1}}{{2}}{,}{-}\frac{{1}}{{2}}$ (7)
 > $\mathrm{collect}\left(s\cdot {C}_{2}+t\cdot {C}_{3},Q,\mathrm{normal}\right)$
 ${{Q}}^{{2}}{+}\left({-}{{q}}^{{2}}{-}{1}\right){}{Q}{+}{{q}}^{{2}}$ (8)

This, in fact, is exactly the desingularizing operator from above.