Second Order Differential Equations with Hypergeometric Solutions of Degree Three

Second Order Differential Equations with Hypergeometric Solutions of Degree Three Vijay Jung Kunwar, Mark van Hoeij Department of Mathematics, Florida State University Tallahassee, FL 0-07, USA vkunwar@mathfsuedu, hoeij@mathfsuedu ABSTRACT Let L be a second order linear homogeneous differential equation with rational function coefficients The goal in this paper is to solve L in terms of hypergeometric function F (a, b; c f) where f is a rational function of degree Categories and Subject Descriptors I [Symbolic and Algebraic Manipulation]: Algorithms; G4 [Mathematics of Computing]: Mathematical Software General Terms Algorithms Keywords Symbolic Computation, Differential Equations, Closed Form Solutions INTRODUCTION A linear differential equation with rational function coefficients corresponds to a differential operator L C(x)[ ] where = d For example, if L = dx a +a +a 0 is a differential operator with a, a, a 0 C(x), then the corresponding differential equation L(y) = 0 is a y + a y + a 0y = 0 We assume that L has no Liouvillian solutions, otherwise L can be solved quickly using Kovacic s algorithm [7] Definition If S(x) is a special function that satisfies a differential operator L S (called a base equation) of order n, then a function y is called a linear S-expression if there exist algebraic functions f, r, r 0, r, such that y = ( exp( r dx) r 0S(f)+r S(f) + +r n S(f) (n )) () Supported by NSF grant 07880 Permission to make digital or hard copies of all or part of this work for personal or classroom use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear this notice and the full citation on the first page To copy otherwise, to republish, to post on servers or to redistribute to lists, requires prior specific permission and/or a fee ISSAC, June 9, 0, Boston, Massachusetts, USA Copyright 0 ACM 978--450-059-7//0 $500 More generally, we say that y can be expressed in terms of S if it can be written in terms of expressions of the form (), using field operations and integrals Higher derivatives are not needed in () since they are linear combinations of S(f), S(f),, S(f) (n ) If L S C(x)[ ] is of order n and k = C(x, r, f, r 0, r, ) C(x) then y satisfies an equation L k[ ] of order n If L C(x)[ ] has order or 4, and S is a special function that satisfies a second order equation, then the problem of solving L in terms of S can be reduced, with an algorithm and implementation [], to the problem of solving second order equations This reduction of order motivates a focus on second order equations If y and S satisfy second order operators, then products of () are not needed, and the form reduces to ( y = exp( r dx) r 0S(f) + r S(f) ) () Although form () looks technical, it is the most natural form to consider, because it is closed under the known transformations that send irreducible second order operators in C(x)[ ] to second order linear operators Given an input operator L inp of order, finding a solution of the form () corresponds to finding a sequence of transformations that sends L S to L inp (or a right hand factor of L inp, but we assume L inp to be irreducible): (i) Change of variables: y(x) y(f) (ii) Gauge transformation: y r 0y + r y (iii) Exponential product: y exp( r dx) The function f in (i) above is called the pullback function These transformations are denoted as f r C, 0,r G and r E respectively They send expressions in terms of S to expressions in terms of S So any solver for finding solutions in terms of S, if it is complete, then it must be able to deal with all three transformations In other words, it must be able to find any solution of the form () The goal in this paper is the following: Given L inp C(x)[ ], irreducible, order, find, if it exists, a nonzero solution of form () where S(x) = F (a, b; c x), f, r, r 0, r C(x) and f has degree Given L inp, our task is to find: L S f C M r 0,r G r E L inp

r There are algorithms [] to find the transformations 0,r G and r E but to apply them we first need M (or equivalently, f and L S) Thus the crucial part is to compute f We compute f from the singularities of M Since we do not yet know M, the only singularities of M that we know are those singularities of L inp that can not disappear (turn into r regular points) under transformations 0,r G and r E Definition A singularity is called non-removable if it stays singular under any combination of r 0,r G and r E A singularity x = p of L inp that can become a regular point r under 0,r r G and/or E need not be a singularity of M Such singularities (removable singularities) provide no information about f They include apparent singularities (singularities p where all solutions are analytic at x = p, such singularities can disappear under r 0,r G ) More generally, if there exist functions u, y, y with y, y analytic at x = p such that uy, uy is a basis of local solutions of L at x = p, then x = p is removable (such p can be sent to an apparant singularity with r E ) Motivation Equations with a F -type solution are common We examined integer sequences u(0), u(), u(), from the Online Encyclopedia of Integer Sequences (oeisorg) for which y = n u(n)xn Z[x] is (a) convergent, and (b) holonomic, meaning that y satisfies a linear differential operator L Among the L s obtained this way, all second order L s (including dozens that had no Liouvillian solutions) turned out to have F -type solutions For third order operators we used order reduction [] to find solutions of the form exp( ) rdx) (r 0(S(f)) + r S(f)S(f) + r (S(f) ), where S(x) = F (a, b; c x) The key step to find F -type solutions is to find the pullback function f, and the F -parameters a, b, c Classifying all rational functions f C(x) that can occur as a pullback function for some L with d non-removable singular points is ongoing work, see [8] for d = 4 and [9] for d = 5 (with at least one logarithmic singularity) For a fixed d, a large table is needed to ensure that we can solve every second order L with d singularities that has a F -type solution But the table can be greatly reduced by developing algorithms such as -descent [] If f is a rational function with a degree decomposition, then we can apply the -descent algorithm to L to reduce the degree of f in half After trying -descent, the next case is to solve every L that has a F -type solution where f is a rational function of degree This is useful in its own right because it solves many equations, but it also significantly reduces the tabulation work that is needed (many f s from [8],[9] have a decomposition factor of degree or ) Hypergeometric solutions, an example Consider the operator L = ( x ) ( 8x ) + 4x ( 4x 7 ) + 4x () L can be solved in terms of F (a, b; c f) where f is a rational function of degree We give one such solution (a second independent solution looks similar): sol L = ( x ( ) ( + x ) F, ; 5 ) f (4) where f = (x )(4x+ ) (x+ )(4x ) L is defined over Q, ie L Q(x)[ ] but f Q(x); instead f Q(, x) Such a field extension can only occur when f is not unique (replacing by gives another solution) The non-uniqueness of f in this example is explained by the fact that L has a symmetry x x (The change of variables x x produces an operator L x that equals L) The change of variables x x produces an operator L x that is still in Q(x)[ ] (this is a trivial case of -descent) L x = x(x )(8x ) +(x x+ ) +x 8 (5) Our program produces the following solution of L x: sol L x = F (, 4 ; x (8x ) ) () Applying x x to () produces another solution of L: Sol L = F (, 4 ; x ( 8x ) ) (7) The pullback function f in (7) has degree, but is nicer than the degree function in (4) So we chose in our implementation to search only for f s which are defined over the field of constants specified in the input, and to apply -descent for the cases where f requires a field extension The following diagrams show the impact of the change of variables x f Notation: (see section for more details and definitions) p: non-removable singularity with exponent-difference Hc,x: a,b hypergeometric differential operator L: p ( x )(4 x+ ) ( x+ )(4 x ) H, 5 II p 0 :,x I Diagram L: L x: H p p 0, 4 :,x III II x x(8x ) 8 p 0 I Diagram The diagrams give the singularity structures (the p s and s, see section for definitions) of L, L x and Hc,x a,b The hypergeometric function F (a, b; c x) is a solution of Hc,x a,b (see Section ) Choosing (a, b, c) = (,, 5 ) makes the exponent-differences of Hc,x a,b equal to (,, ) Then the pullback function f = (x )(4x+ ) (x+ )(4x in Diagram sends ) the singularity structure of Hc,x a,b to that of L

Diagram has two changes of variables, x x(8x ) produces the singularity structure of L x from a hypergeometric equation with exponent-differences (,, ) After that, x x produces the singularity structure of L Presenting (4) in the general form (), we get r = u Q(x, ) where u = ( + x ) u ( x ), r 0 = and r = 0 For (7), we also have r = 0 (with r = 0, r 0 = ) The case r 0 is more complicated as this involves integer shifts of the exponent-differences, see section 4 for an example Goal of the paper The examples in the above diagrams illustrate the effect of change of variables x f on the singularities and their exponent-differences; this paper will use that to reconstruct f whenever it has degree, a problem that turns out to consist of 8 cases One of those cases (case sing5 in section 4) was treated previously in section in [4] The goal in this paper is to cover all cases with f of degree Combining this work with the -descent algorithm from [] and the tables from [8, 9] leads to a solver that can find F -type solutions for many second order equations The combined solver is very effective; it appears that closed form solutions exist for every second order L that has a non-zero convergent solution of the form n=0 u(n)xn with u(n) Z PRELIMINARIES AND NOTATIONS This section gives a brief summary of prior results needed for this paper Notations used throughout the paper: C: a subfield of C L inp C(x)[ ]: input differential operator Hc,x: a,b Gauss hypergeometric differential operator S(x) = F (a, b; c x): hypergeometric function (a solution of Hc,x) a,b e 0, e, e : exponent differences of Hc,x a,b at 0,, f C(x): a rational function of degree : obtained from Ha,b c,x by a change of variables x f S(f) = F (a, b; c f): a solution of Differential Operators and Singularities A derivation = d on C(x) produces a non-commutative dx ring C(x)[ ] A differential operator of order n is an element L C(x)[ ] of the form L = n i=0 ai i, with a i C(x) and a n 0 The solution space (set of all solutions in a universal extension) of L is denoted by V (L) = {y L(y) = 0} After clearing denominators, we may assume a i C[x] Then p C is called a regular (or non-singular) point when a n(p) 0 Otherwise it is called a singular point (or a singularity) The point p = is called regular if the change of variable x /x produces an operator L /x with a regular point at x = 0 Given { p P = C { }, we define the local x p if x parameter as t p = if x = x Definition L C(x)[ ] is called Fuchsian (or regular singular) if all its singularities are regular singularities A singularity p is a regular singularity when: () (if p ): t i p an i a n is analytic at p for i n () (if p = ): L /x has a regular singularity at x = 0 Otherwise p is called irregular singular This paper considers only Fuchsian operators, of order For closed form solutions of non-fuchsian equations (L having at least one irregular singularity) see [] Theorem ([0], Sections 4) Suppose L has order and p P If x = p is a regular singularity or a regular point of L, then there exists the following basis of V (L) at x = p y = t e p y = t e p i=0 aiti p, a 0 0 and i=0 biti p + cy log(t p), b 0 0 where e, e, a i, b i, c C such that: (i) If e = e then c must be non zero (ii) Conversely, if c 0 then e e must be in Z Notation In Theorem : If c 0 then x = p is called a logarithmic singularity e, e are called the exponents of L at x = p (they are computed as the roots of the indicial equation) (L) := ±(e e ) is the exponent difference at p 4 If (L ) (L ) mod Z then we say that (L ) matches (L ) Remark 4 Logarithmic singularities are always nonremovable (they stay logarithmic under the transformations in Definition ) If e e Z and x = p is non logarithmic then x = p is either a regular point or a removable singularity Proofs and more details can be found in [5] The relation with Theorem is as follows: () x = p is non-singular {e, e } = {0, } and c = 0 () x = p is non-removable c 0 or e e Z Definition 5 The singularity structure of L is: Sing(L) = {(p, (L) mod Z) : p is non-removable} It is often more convenient to express singularities in terms of monic irreducible polynomials Definition Let C be a field of characteristic 0 places(c):= {f C[x] f is monic and irreducible} { } The degree of a place p is if p = and deg(p) otherwise Example 7 Consider L in section {( Sing(L) =, ) (,, ) (,, ) (,, )} In terms of places(q) it is written as: {( Sing(L) = x, ), ( x 8, )} Gauss Hypergeometric Equation The Gauss hypergeometric differential equation (GHE) is: x( x)y + (c (a + b + )x)y aby = 0 (8) It has three regular singularities, with exponents {0, c} at x = 0, {0, c a b} at x = and {a, b} at x = The corresponding differential operator is denoted by: c,x = x( x) + (c (a + b + )x) ab (9) One of the solutions at x = 0 is F (a, b; c x) Computing a F -type solution of a second order L inp corresponds to computing transformations from Hc,x a,b to L inp

Remark 8 The exponent differences of c,x can be obtained from the parameters a, b, c and vice versa: (e 0, e, e ) = ( c, c a b, b a) Remark 9 We assume that Hc,x a,b has no Liouvillian solutions For such Hc,x, a,b the points 0,, are never nonsingular or removable singularities So if Hc,x a,b has e p Z (with p {0,, }) then p is a logarithmic singularity Properties of Transformations For second order operators, we use the notation L L if L can be transformed to L with any combination of the three transformations from section If L L then f r L 0,r r C G E L More details can be found in [] Remark 0 r 0,r G and r E are equivalence relations remains same under r E but may change by an integer under r 0,r G f So if L C M r 0,r r G E L inp for some input L inp with L, M unknown, then (M) can be ( mod Z and up to ±) read from (L inp), Sing(L inp) = Sing(M) Hence L, f, M should be reconstructed from Sing(L inp) If one of e 0, e, e is in +Z then Ha,b c,x is determined, up to the equivalence relation r 0,r r G E, by the triple (e 0, e, e ) up to ± and mod Z If {e 0, e, e } ( + Z) = then the triple leaves two separate cases for Hc,x a,b up to r 0,r r G E ; we need to consider (e 0, e, e ) up to ± and mod Z, and (e 0 +, e, e ) up to ± See Theorem 8, section 5 in [4] for details Because of the transformation M r 0,r r G E L inp in remark 0 only non-removable singularities of L inp provide usable data for M and f Definition Two operators L, L are called projectively equivalent (notation: L p L ) if L r 0,r G r E L The following lemma gives the effect of C on the singularities and their exponent differences: (see [4] for more details) Lemma Let e 0, e, e be the exponent differences of Hc,x a,b at 0,, Let be the operator obtained from Ha,b c,x by applying x f Let d = be the exponent difference of at x = p Then: If p is a root of f with multiplicity m, then d = me 0 If p is a root of f with multiplicity m, then d = me If p is a pole of f of order m, then d = me f 4 An Example Involving All Three Transformations Let u(0) =, u() = 88, u(n + ) = 4(59(n ) 977)u(n + ) 8 (n 9)u(n)) (n + ) (0) This defines a sequence, 88, -, that this is an integer sequence? Consider the following differential operator: How to prove L = (x 7) ( x + ) + ( x + ) 9 (x + 9) () Our implementation solves this equation One solution is: ( sol L = s (g F, 5 ) ( 5 ; f + h F, )) ; f where s = 98 4 (x ) 5 4 and f = 7(x 7)(x +) (x ) (), g = (x+)(x ), h = x+40 One can convert between differential equations and recurrences (see gfun package in Maple) and find: sol L = n=0 ( x 7 ) n u(n) () 7 7 where u(n) are given by the recurrence relation in (0) The explicit expression () can be used to prove u(n) Z for n = 0,, (it is not clear if there is a different way to prove that for this example) The following diagram shows the effects of f C and r 0,r G on the set of non-removable singularities and their exponent differences ( r E does not affect them): H L:, 5 :,f p 7 y r 0y + r y 0 p 7 0 0 0 f = 7(x 7)(x +) (x ) H, 5 :,x p 0 0 I Diagram III Suppose y(x) = n=0 u(n)xn is convergent with u(n) Z, (n = 0,,, ) and satisfies a second order differential operator L Q(x)[ ] In all known examples such y(x) is either II

algebraic or expressible in terms of F hypergeometric functions Hence algorithms for finding F -type solutions are useful for integer sequences PROBLEM DISCUSSION Starting with a Fuchsian linear differential operator L inp, of order, and no Liouvillian solutions, we want to find a solution of such L inp in the form : y = e r (r 0S(f) + r S(f) ) (4) where S(x) = F (a, b; c x), f, r, r 0, r C(x) and f has degree There are two key steps: Compute f and (e 0, e, e ) such that Sing( ) = Sing(Linp) (5) (see remark 8 for the relation between (e 0, e, e ) and (a, b, c)) Compute projective equivalence p between and which sends solutions S(f) = F (a, b; c f) of L inp to solutions of Linp of the form (4) If we find f then [] takes care of the second step Hence the crucial part is to compute f (as well as a, b, c, or equivalently, e 0, e, e ) Let f = A/B where A, B C[x] with gcd(a, B) = The hypergeometric operator Hc,x a,b has singularities at x = 0,, So one might expect L inp to have singularities whenever f = 0, or ; ie at the roots of A, A B and B If all roots of A, A B, B would appear among the singularities of L inp, then it would be fairly easy to reconstruct f = A/B However, that is not true in general (it is true for 8 out of the 8 cases in Tab in Section 4) For example; if f has a root p with multiplicity and e 0 is a half-integer (an odd integer divided by ), then p will be a removable singularity or a non-singular point of Such p does not appear in Sing(L inp) 4 COMPUTING PULLBACK FROM THE SINGULARITY STRUCTURE 4 Relating singularities to f Let [[a,, a i], [b,, b j], [c,, c k ]] denote the branching pattern of f It contains the branching orders of f above 0, and respectively (So f has i distinct roots with multiplicities a,, a i Likewise f and have j resp k f distinct roots) Using lemma and remark 4, the singularities of are as follows: P 0 = {x : f = 0 and (e 0 Z or a l e 0 / Z) for l i} P = {x : f = 0 and (e Z or b l e / Z) for l j} P = {x : f = 0 and (e Z or c l e / Z) for l k} where (e 0, e, e ) are the exponent differences of Hc,x a,b at (0,, ) respectively The union of P 0, P and P are the non-removable singularities of (or Linp, by eq (5)) Since f has degree, L inp could have at most 9 singularities The least we could have is when we choose the branching pattern of f as [[],[,],[,]] and (e 0, e, e ) (±,, ) mod Z But a hypergeometric equation with two exponentdifferences in + Z has Liouvillian solutions, so we do not treat this case here If L inp has non-removable singularities, then we can transform these to 0,, via a Möbius transformation and express the solution accordingly (with a rational function of degree ) This case is already treated in [4] So we exclude these cases (Liouvillian and nonremovable singularities) from our consideration 4 Tabulating cases Let d be the total number of non-removable singularities in L inp From Section 4 we have 4 d 9 The first step is to enumerate all possibilities for exponent differences e 0, e, e and branching patterns above {0,, } for each d We express all such possibilities for degree rational function f in the following table: Notation 4 d: number of non-removable singularities in L inp, E, E, E : elements of C : an element of + Z : an element of ( + Z) ( + Z) d Case Exponent difference Branching pattern at 0,, resp above 0,, resp case4, E [,], [], [,,] case4,, E [], [], [,,] 4 case4,, [,], [,], [] case44,, [], [,], [,] Liouv,, [,], [,], [,] case5,, E [], [], [,,] Liouv, E [,], [,], [,,] 5 case5,, E [,], [], [,,] case5 [,], [,,], [] case54,, [,], [,], [,] case55,, [], [,], [,] case, E, E [], [,,], [,,] case,, E [,], [,], [,,] case,, E [], [,], [,,] case4,, [,], [,], [,] case7, E, E [], [,,], [,,] 7 case7, E, E [,], [,,], [,,] case7,, E [,], [,], [,,] 8 case8, E, E [,], [,,], [,,] 9 case9 E, E, E [,,], [,,], [,,] Table : Cases for degree pullback up to permutation of 0,, Two cases (denoted Liouv) in Tab correspond to the hypergeometric equations with two singularities having a halfinteger exponent difference Such equations have Liouvillian solutions (this follows from Kovacic algorithm and also from Theorem 8, section 5 in [4]) Now the main task is to compute f for the remaining 8 cases Recall that non removable singularities of come from (form a subset of) the roots of f, f and poles of f We will use the singularity structure of L inp to recover f 4 Treating one case The main algorithm in Section 5 takes as input C, L inp, x where C is a field of characteristic 0, and L inp C(x)[ ] has

order and no Liouvillian solutions It computes Sing(L inp) and d Then it loops over the corresponding cases in Tab For example; if d = 4 then it loops over cases 4 44 in Tab Each case in Tab is a subprogram Each of these subprograms takes C, Sing(L inp) as input, checks if Sing(L inp) is compatible with that particular case, and if so, returns a set of candidates for f, (e 0, e, e ) that are compatible with that particular case We give details for only one case, namely Algorithm[5] (notation: casei, j is handled by Algorithm[ij]) The other cases are treated by similar programs (details can be found in our implementation []) Let L inp C(x)[ ] be input differential operator with 5 nonremovable singularities In terms of places(c), there are 7 ways to end up with 5 points: One place of degree 5 (note: a place of degree > is always a monic irreducible polynomial of that degree A place of degree can be either or a monic polynomial of degree ) Places of degrees 4, Places of degrees, 4 Places of degrees,, 5 Places of degrees,, Places of degrees,,, 7 Places of degrees,,,, Algorithm[5]: Compute f C(x) of degree and exponent differences (e 0, e, e ) for Hc,x a,b corresponding to case5 in Tab Input: Field C and Sing(L inp) in terms of places(c) Output: A set of lists [f, (e 0, e, e )] where f C(x) has degree and branching ( ) pattern [,], [,,], [] above 0,, such that Sing = Sing(L inp) where (a, b, c) corresponds to (e 0, e, e ) by remark 8 and (see Tab ) e 0 + Z, e is arbitrary, and e ± + Z Check if Sing(L inp) is consistent with case 5 (if not, return the empty set and stop) as follows: The branching pattern [,] at f = 0 indicates that f has two roots a, a C { } with multiplicities resp Then x = a will have an exponent-difference e 0 + Z but x = a will be a regular point or a removable singularity, and so it does not appear in Sing(L inp) The branching pattern [] at f = indicates that f has precisely one pole b C { }, of order Then x = b will have an exponent-difference ±e mod Z In case 5 we have e ± + Z and hence the point x = b must be a non-removable singularity Combined with x = a we see that case 5 is only possible when Sing(L inp) has at least two places of degree So in the above listed 7 cases (5, 4+, ), we can exit Algorithm[5] immediately if we are not in case 4,, or 7 The branching pattern [,,] at f = indicates that f has three distinct roots, each of multiplicity Thus there must be at least three distinct singularities that match the exponent-difference ±e mod Z If we can not find three singularities (one place of degree, or places of degrees and, or three places of degree ) whose exponent-differences match (up to ± and mod Z) then Algorithm[5] stops This condition determines e (up to ± and mod Z) We know from Kovacic algorithm that if there are two e i + Z then Ha,b c,x will have Liouvillian solutions Since we exclude Liouvillian cases, it follows that only e 0 is in + Z We conclude that Sing(Linp) must have either or singularities in C { } with an exponent-difference in + Z and that such singularities can only occur when e ± + Z So if there are more than, then Algorithm[5] stops (x a Set Cand = and write f = k )(x a ) where (x b) a, a, b C { } and k C We replace any factor x in f by in the implementation Compute the set of places with an exponent-difference in + Z This set may only have or elements that must have degree Now a loops over this set, and e 0 is the exponent-difference at x = a Loop b over the places in Sing(L inp) of degree, skipping a, and only considering a, b for which the remaining three singularities have matching exponentdifferences Let e b be the exponent-difference at x = b Now loop e over e b, (e b ), (e b+) For e one can take the exponent-difference at any of the remaining singularities The reason that there are three cases for e is because we have to determine e mod Z Now e = e b but if a gauge transformation occurred, ie if the r in the form () in Section is non-zero, then e b is only known mod Z, and this leaves in general three candidate values for e mod Z (it suffices to compute the e i mod Z, see section 5 in [4], summarized in Remark 0) 4 Among the remaining singularities, let P C[x] be the product of their places (replacing x by if that is among them) So P has degree if is not among the remaining singularities, and otherwise it has degree In each loop, the a, b appearing in f are known, while k and a are unknown Take the numerator of f and compute its remainder mod P Equate the coefficients of this remainder to 0 This gives deg(p ) equations for k, a If deg(p ) = we obtain one more equation by setting f( ) = (the resulting equation is k = ) Then we have equations for unknowns k, a Compute the solutions k C and a C { } If any solution exists, then add the resulting [f, (e 0, e, e )] to the set Cand 5 Return the set Cand (which could be empty, but could also have one or more members) Example 4 Take C = Q Let Sing(L inp) in terms of places(q) be given by: Sing(L inp) = { [, ], [x, 7 ], [x, ], [x + x + 44, 5 7 ]}

Our input is the following: Sing(L inp) = { [, ], [x, 7 ], [x, ], [x + x + 44, 5 7 ]} Notations in the steps below come from Algorithm[5] Write f(x) = k (x a )(x a ) (x b) Step : Sing(L inp) satisfies the conditions for sing5 ; () [, ] and [x, ] have degree and both have a half-integer exponent difference () The exponent differences in [x, ] and 7 [x + x+44, 5 ] 7 match, after all, we are working up to ± and mod Z Step : The candidates for x a are and x For (x a the first case, we get f = k ) and e (x b) 0 = (note: we may equally well take ) For the second case, we get (x )(x a f = k ) and e (x b) 0 = Step : For the first case, x b can only be x and e b = (because if we take x b = x then there would not remain three singularities with matching exponent-differences) Likewise, for the second case, x b can only be and e b = (x a First case: f = k ) and e (x ) = consider e { e b, (e b+), (e b ) (note: we should } since e b is determined mod Z, and we have to determine e mod Z However, (e b +) = is discarded since there should not be two e is in + Z And (e b ) = but an exponent-difference is equivalent to an exponent-difference ) Second case: f = k (x )(x a ) and e = Step 4: In both cases P = x (x + x + 44) and e = 7 (we could equally well take 5 ) Dividing the numerator of 7 f by P produces equations in k and a In first case the equations have a solution; {k =, a = }, and in second case they do not Step 5: The output Cand has one element, namely {[ (x+/) (x ), (, 7, )] } 5 MAIN ALGORITHM We have developed the algorithms to compute f s and possible exponent differences (e 0, e, e ) for Hc,x a,b corresponding to all 8 cases as given in Tab Now we give our main algorithm: Let C C be a field and L inp C(x)[ ] be the input differential operator The main algorithm first computes the singularity structure of L inp in terms of places(c) Suppose d is the total number of non-removable singularities of L inp Now we call all algorithms corresponding to d to produce a set of candidates for f C(x) and the exponent differences (e 0, e, e ) = ( c, c a b, b a) For each member from that list we compute Hc,x, a,b and apply projective equivalence [] between and Linp to find ( (if it exists) a nonzero map from V ) to V (L inp) which sends solutions S(f) = F (a, b; c f) of to solutions e r (r 0S(f) + r S(f) ) of L inp Algorithm: Solve an irreducible second order linear differential operator L inp C(x)[ ] in terms of F (a, b; c f), with f C(x) of degree Input: A field C of characteristic 0, L inp C(x)[ ] of order which has no Liouvillian solutions, and a variable x Output: A non zero solution y = e r (r 0S(f) + r S(f) ), if it exists, such that L inp(y) = 0, where S(f) = F (a, b; c f), f, r,r 0,r C(x) and f has degree Step : Find the singularity structure of L inp in terms of places(c) Let d be the total number of non-removable singularities Step : Let k be the total number of cases in Tab for d For example; if d = then k = 4 Let Candidates = Algorithm[na], where a = { k} That produces a set of lists [f, (e 0, e, e )] of all possible rational function f C(x) of degree and corresponding exponent differences (e 0, e, e ) for Hc,x a,b Step : Hc,x a,b = x( x) + (c (a + b + )x) ab where a, b, c come from the relation (e 0, e, e ) = ( c, c a b, b a) For each element [f, (e 0, e, e )] in Candidates (Step ), (i) If {e 0, e, e } +Z then Cand := {[f, (e0, e, e )]} otherwise Cand := {[f, (e 0, e, e )], [f, (e 0+, e, e )]} (That determines Hc,x a,b up to projective equivalence, see remark 0) (ii) From each element in Cand above (a) compute a, b, c, (b) substitute the values of a, b, c in Hc,x a,b and (c) apply the change of variable x f on Hc,x a,b That produces a list of operators Step : Compute the projective equivalence [] between each operator in Step and Linp If the output is zero, then go back to Step and take the next element from Candidates Otherwise, we get a map of the form: G = e r (r 0 + r ), where r, r 0, r C(x) and = d dx Step : S(f) = F (a, b; c f) is a solution of Now compute G(S(f)) That gives a solution of L inp Step 4: Repeat the same procedure for each element in Candidates That gives us a list of solutions of L inp Step 5: Choose the best solution (with shortest length) from the list (to obtain a second independent solution of L inp, just use a second solution of Hc,x) a,b Example 5 Consider the operator in Section 4; L inp = (x 7) ( x + ) + ( x + ) 9 (x + 9) Procedure to solve this equation is the following: Step : Read the file hypergeomdegtxt from the folder hypergeomdeg in wwwmathfsuedu/~vkunwar Step : L inp Q(x)[ ] We want the solution of L inp in the base field Q Type hypergeomdeg({}, L inp, x) (in Maple {} is the code for Q)

Step : The program first finds the singularity structure; Sing(L inp) = { [, ], [x 7, 0], [x +, ] } (our implementation uses to encode a singularity at, and polynomials to encode finite singularities) Step 4: We get d = 4 The program loops over the four subprograms corresponding to case4, case44 to compute f: Algorithm[4] returns the following: F = {[f, [, 0, ]], [f, [,, ]], [f, [, 0, ]], [f, [,, ]] } (9 x+0) ( x ) where f = 8 Note: this set contains p-duplicates, the four triples (e 0, e, e ) all give projectively equivalent Hc,x a,b so we could delete three and still find a solution (if it exists) The reason they were left in the current version of the implementation is because they may help to find a solution of smaller size In the next version, we plan to make the code more efficient by removing p-duplicates, keeping only those for which the integer-differences between the exponent-differences of and L inp are minimized (in this example, only the second element of F would be kept in this approach) Algorithm[4] returns NULL Algorithm[4] and Algorithm[44] require at least linear polynomials in Q[x] for Sing(L inp) which is not the case here So Sing(L inp) does not qualify the conditions for these algorithms Hence F gives the Candidates Note that we are in the case {e 0, e, e } + Z Step 5: Taking first element i = [8 (9x+0), [, 0, ]] in (x ) Candidates and applying Step and Step of the above main algorithm, we get G = e r (r 0 + r ) with e r = ( 9 0 x+)( x +)( 7 x+) ( x+)( x+) 4, r = + 90 9 x 7 9 x and r 0 = 79x 4 9845x 599x +445x+977 8 (x 7)(x )(9x+0) Step : We have S(f) = F (, 7 ; 5 8 (9x+0) (x ) ) Computing G(S(f)) we get y = e r (r 0S(f) + r S(f) ) as a solution of L inp where e r, r 0, r are given in Step 5 Step 7: Taking second element i = [8 (9x+0), [,, ]] in (x ) Candidates we get another solution y with e r ( 0 = 9 x+), r 0 = (87x +84x 78x+484), 98( x)(9x+0) ( x+)( x+) 7 4 r = x + and S(f) = F ( 7, ; 5 8 (9x+0) (x ) ) Steps 8 and 9: Process the third and fourth element Each produces a solution that looks quite similar to that given in Steps and 7 Step 0: The solution in Step 7 has the shortest length So the implementation returns that as a solution of L inp After minor simplification this leads to the solution given in section 4 REFERENCES [] R Debeerst, M van Hoeij, W Koepf: Solving Differential Equations in Terms of Bessel Functions, ISSAC 08 Proceedings, 9-4 (008) [] M van Hoeij: An implementation for finding equivalence map, wwwmathfsuedu/ hoeij/files/equiv [] Q Yuan, M van Hoeij: Finding all Bessel type solutions for Linear Differential Equations with Rational Function Coefficients, ISSAC 00 Proceedings, 7-44 (00) [4] A Bostan, F Chyzak, M van Hoeij, L Pech: Explicit formula for the generating series of diagonal D rook paths, Seminaire Lotharingien de Combinatoire, Ba (0) [5] R Vidunas: Algebraic transformations of Gauss hypergeometric functions Funkcialaj Ekvacioj, Vol 5 (Aug 009), 9-80 [] T Fang, M van Hoeij: -Descent for Second Order Linear Differential Equations, ISSAC 0 Proceedings, 07-4 (0) [7] J Kovacic: An algorithm for solving second order linear homogeneous equations, J Symbolic Computations,, -4 (98) [8] M van Hoeij, R Vidunas: Belyi functions for hyperbolic hypergeometric-to-heun transformations, arxiv:80 [9] M van Hoeij, V J Kunwar: Hypergeometric solutions of second order differential equations with 5 non-removable singularities, (In Progress) [0] Z X Wang, D R Guo: Special Functions World Scientific Publishing Co Pte Ltd, Singapore, 989 [] V J Kunwar: An implementation for hypergeometric solution of second order differential equation with degree rational function, wwwmathfsuedu/ vkunwar/hypergeomdeg [] M van Hoeij: Solving Third Order Linear Differential Equation in Terms of Second Order Equations, ISSAC 07 Proceedings, 55-0, 007 Implementation: wwwmathfsuedu/ hoeij/files/reduceorder [] Q Yuan: Finding all Bessel type solutions for Linear Differential Equations with Rational Function Coefficients, PhD thesis and implementation, available at wwwmathfsuedu/ qyuan (0) [4] T Fang: Solving Linear Differential Equations in Terms of Hypergeometric Functions by -Descent, PhD thesis and implementation, available at wwwmathfsuedu/ tfang (0)