Finding all Bessel type solutions for Linear Differential Equations with Rational Function Coefficients

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1 Finding all essel type solutions or Linear Dierential Equations with Rational Function Coeicients Mark van Hoeij & Quan Yuan Florida State University, Tallahassee, FL 06-07, USA & ASTRACT A linear dierential equation with rational unction coeicients has a essel type solution when it is solvable in terms o ν(), ν+1(). For second order equations, with rational unction coeicients, must be a rational unction or the square root o a rational unction. An algorithm was given by Debeerst, van Hoeij, and Koep, that can compute essel type solutions i and only i is a rational unction. In this paper we ext this work to the square root case, resulting in a complete algorithm to ind all essel type solutions. Categories and Subject Descriptors G.4 [Mathematical Sotware]: Algorithm design and analysis; I.1. [Symbolic and Algebraic Manipulation]: Algorithms Algebraic algorithms General Terms Algorithms 1. INTRODUCTION Let a 0, a 1, a C(x) and let L = a + a 1 + a 0 be a dierential operator o order two. The corresponding dierential equation is L(y) = 0, i.e. a y + a 1y + a 0y = 0. Let ν(x) denote one o the essel unctions (one o essel I, J, K, or Y unctions). The question studied in [6, 7] is the ollowing: Given L, decide i there exists a rational unction C(x) such that L has a solution y that can be expressed 1 in terms o ν(). I so, then ind, ν, and the corresponding solutions o L. The same problem was also solved or Kummer/Whittaker unctions, see [6]. This means that or second order L, with rational unction coeicients, there is an almost-complete algorithm in [6] to decide i L(y) = 0 is solvable in terms o 0F 1 or 1F 1 unctions, and i so, to ind the solutions. Supported by NSF grant using sums, products, dierentiation, and exponential integrals (see Deinition ) Permission to make digital or hard copies o all or part o this work or personal or classroom use is granted without ee provided that copies are not made or distributed or proit or commercial advantage and that copies bear this notice and the ull citation on the irst page. To copy otherwise, to republish, to post on servers or to redistribute to lists, requires prior speciic permission and/or a ee. ISSAC 010, 5 8 July 010, Munich, Germany. Copyright 010 ACM /10/ $ The reason this almost-complete algorithm is not complete is the ollowing: I ν() satisies a second order linear dierential equation with rational unction coeicients, then either: C(x), or (square root case): C(x) but C(x). However, only the C(x) case was handled in [6, 7], the square-root case was listed in the conclusion o [7] as a task or uture work. This meant that [6, 7] is not yet a complete solver or 0F 1 and 1F 1 type solutions. In this paper, we treat the square-root case or essel unctions. The combination o this paper with the treatment o Kummer/Whittaker unctions in [6] is then a complete algorithm to ind 0F 1 and 1F 1 type solutions whenever they exist. The reason why the square-root case was not yet treated in [7] will be explained in the next two paragraphs. I is a rational unction = A/, then rom the generalized exponents at the irregular singularities, we can compute, as well as deg(a) linear equations or the coeicients o A, see [7], or see [6] which contains more details and examples. Since a polynomial A o degree deg(a) has deg(a) + 1 coeicients, this meant that only one more equation was needed to reconstruct A, and in each o the various cases in [6, 7] there was a way to compute such an equation. In the square-root case, we can not write as a quotient o polynomials, but we can write = A/. The same method as in [6, 7] will still produce, and linear equations or the coeicients o A. The number o linear equations or the coeicients o A is still the same as it was in the C(x) case. Unortunately, by squaring to make it a rational unction, we doubled the degree o A, but we do not get more linear equations, which means that in the square-root case the number o linear equations is only 1 deg(a) (plus an additional 0 equations coming rom regular singularities). So in the worst case, the number o equations is only hal o the degree o A. This is why the square-root case was not solved in [7] but only mentioned as a uture task. Our approach is as ollows: One can rewrite A = CA 1A d where A 1 can be computed rom the regular singularities, but A can not. The problem is that while the degree o A is only 1 times the degree o A/A1, the linear equations on the coeicients o A translate into polynomial equa- d tions (with degree d) or the coeicients o A. Solving systems o polynomial equations can take too much CPU time. Other 0F 1 and 1F 1 type unctions can be rewritten in terms o essel, or Kummer/Whittaker unctions. For instance, Airy type unctions orm a subclass o essel type unctions (provided that the square-root case is treated!)

2 However, we discovered that with some modiications, one can actually obtain linear equations or the coeicients o A. This means that we only need to solve linear systems. The result is an eicient algorithm that can handle complicated inputs. An implementation is available online at qyuan.. PRELIMINARIES.1 Dierential Operators We let K[ ] be the ring o dierential operators with coeicients in a dierential ield K. Let C K be constants, C K be an algebraic closure o C K. Usually, we have K = C K(x) and C K is a inite extension o Q. We call p C K { } a singularity o the dierential operator L K[ ], i p = or p is a zero o the leading coeicient o L or p is a pole o a coeicient o L. I p is not a singularity, p is regular. We say y is a solution o L, i L(y) = 0. The vector space o solutions is denoted as V (L). I p is regular, we can express all solutions around p as power series P biti p where t p denotes the local parameter which is t p = 1 i x p = and t p = x p, otherwise.. Formal Solutions and Generalized Exponents Deinition 1. We say e C[t 1 m p ] is a generalized expo- R e t p dt p S, S R m, nent o L at p i L has a solution exp and S / t 1 m p R m, where m Z and R m = C[[t 1 m p ]][log(t p)] I e C we just get a solution x e S, in this case e is called an exponent. I the solution involves a logarithm, we call it a logarithmic solution. I m = 1, then e is unramiied, otherwise it is ramiied. Remark 1. Since we only consider second order dierential operators, m in the deinition can be only 1 or. I the order o L is n, then at every point p, counting with multiplicity, there are n generalized exponents e 1, e,..., e n, and the solutions exp R ei t p dt p S i, i = 1,..., n are a basis o solution space V (L). I p is regular, then the generalized exponents o L at p are 0, 1,..., n 1. One can compute generalized exponents with the Maple command DEtools[gen exp].. essel Functions essel unctions are the solutions o the operators L 1 = x + x + (x ν ) and L = x + x (x + ν ). The two linearly indepent solutions J ν(x) and Y ν(x) o L 1 are called essel unctions o irst and second kind, respectively. Similarly the solutions I ν(x) and K ν(x) o L are called the modiied essel unctions o irst and second kind. Let ν(x) reer to one o the essel unctions. When ν is hal integer, L 1 and L are reducible. One can get the solutions by actoring the operators. We will exclude this case rom this paper. The change o variables x ix ss V (L 1) to V (L ) and vice versa. Since our algorithm will deal with change o variables, as well as two other transormations (see Section.4), we only need one o L 1, L. We choose L and denote L := L. L has only two singularities, 0 and. The generalized exponents are ±ν at 0 and ±t at. Ater a change o variables y(x) y( x), we get a new operator L = x +x 1 4 (x+ν ), which is still in Q(x)[ ]. Let CV(L, ) denote the operator obtained rom L by change o variables x. For any dierential ield extension K o Q(x), i ν C K, and i K, then CV(L, ) K[ ] since this operator can can be written as CV(L, ). The converse is also true: Lemma 1. Let K be a dierential ield extension o Q(x), let, ν be elements o a dierential ield extension o K, and ν be constant. Then CV(L, ) K[ ] K and ν C K. Proo. It remains to prove =. Let ν be a constant, monic(l) be the dierential operator divided by the leading coeicient o L, and We have to prove M := monic(cv(l, )) = + a 1 + a 0 a 0, a 1 K =, ν K and so we assume a 0, a 1 K. Let g =. y computing M = monic(cv(l, g)) we ind a 1 = ld(ld(g)), a 0 = 1 4 (g + ν )ld(g) where ld denotes the logarithmic derivative, ld(a) = a /a. Let a := ld(ld(a 0) + a 1) + ld(a 0) + a 1, a := 4a 0/a, a 4 := a (a 1 + ld(a 0)) which are in K since a 0, a 1 K. Direct substitution shows that a = ld(g), a = g + ν, and a 4 = g. Hence g = a 4/a K and ν = a g K..4 Transormations Deinition. A transormation between two dierential operators L 1 and L is an onto map rom solution space V (L 1) to V (L ). For an order operator L 1 K[ ], there are three types o transormations or which the resulting L is again in K[ ] with order. They are (notation as in [6, 7]): (i) change o variables: y(x) y((x)), (x) K \ C K. (ii) exp-product: y exp( R r dx) y, r K. (iii) gauge transormation: y r 0y + r 1y, r 0, r 1 K. We denote them by C, E, G respectively. We can switch the order o E and G [6]. So we will denote L 1 EG L i some combination o (ii) and (iii) ss L 1 to L. Likewise we denote L 1 CEG L i some combination o (i), (ii),(iii) ss L 1 to L. Remark. The relation EG is an equivalence relation. ut C is not. Deinition. We say L 1 K[ ] is projectively equivalent to L i and only i L 1 EG L.

3 Lemma. (Lemma in [7]) I L 1 CEG L, then there exist an operator M K[ ] such that L 1 C M EG L We can apply Lemma to L 1 = L and L which is the operator L we want to solve. I M is known (i.e i the change o variables x is known), then the map rom V (M) to V (L) can be computed with existing algorithms [1], [6]. That means L CEG L can be computed i we can ind the change o variables. The goal o this paper is to solve dierential equations in terms o essel unctions. This means: i L CEG L, then solve L. Main problem: Let C K be a ield, C K C, and let K = C K(x). Let L K[ ] be irreducible and o order. The question we will solve in this paper is the ollowing: Does there exist an operator M K[ ] such that 1. L is projectively equivalent to M, and. L g C M or some g K and some constant ν. I so, ind g, ν and solve L. Note that L g C M is the same as L C M, where = g K. The reason we use the second orm is because we can then use the same notation as in [6] [7].. THE CHANGE OF VARIALES.1 The Exponent Dierence To recover in the transormation C, we need some inormation about and this inormation should be invariant under projective equivalence. Since the order o L is, we have two exponents e 1, e at a point p. We consider the exponent dierence (L, p) = ±(e 1 e ). One can veriy that modulo 1 Z is invariant under projective equivalence m [6] (Here m is as in Section.). We use a ± sign because we don t know the order o the generalized exponents. We also deine: Deinition 4. A singularity p o L K[ ] is called: (i) apparent singularity i and only i (L, p) Z and L is not logarithmic at p. This is equivalent to saying that L has a basis o solutions y 1, y or which y 1/y is analytic at p. (ii) regular singularity i and only i (L, p) C \ Z or L is logarithmic at p. (iii) irregular singularity i and only i (L, p) C[t 1 p ]\C we also denote the set o regular singularities and irregular singularities by S reg and S irr. Note: Apparent singularities are neither in S reg nor S irr. The main work in this paper is to construct a inite set o candidates or (, ν) rom the (L, p). Let g = K. I g has a root resp. pole at p o order k N, then we say that has a root resp. pole at p o order m p := k. Theorem 1. Let K = C K(x), and L C M EG L, where K. Note: L is the input to our algorithm, and and M are to be computed. (i) i p is a zero o with multiplicity m p 1 Z+, then p is an apparent singularity or p S reg, and (M, p) = m pν. (ii) p is a pole o with pole order m p 1 Z+ such that = P i= m p it i p, i and only i p S irr and (M, p) = P i<0 iiti p. I p S reg, then (L, p) (M, p) mod Z which means that we can compute m pν mod Z. I p S irr, then (L, p) (M, p) mod 1 m Z. Then P i<0 iti p can be computed rom (L, p) by dividing coeicients by i (the congruence only aects the t 0 p-term o, but that term is not used when p S irr). Proo. We can use the same proo in [6]. Deinition 5. Let = Σ i=n a ix i, N Z, a N 0. We say that we have a k-term truncated power series or when the coeicient o x N,..., x N+k 1 are known. Remark. I a k-term truncated series or is known, then we can compute a k-term truncated series or. According to Theorem 1 (ii), rom (M, p), we can get a m p -term truncated series o at p. In [7], was assumed to be in K, in which case the truncated series is exactly the polar part o at p. ut in this paper, we have to compute g = K. Theorem 1 (ii) gives us the polar part o, i.e. a truncated series or. We square it to obtain a truncated series o g. ut this truncated series or g has m p terms (the same number o terms as the one or, see Remark ). So it is only hal (rounded up) o the polar part o g. For instance, i has a pole o order at x = 0, rom (L, p) we can obtain a truncated series Σ 1 i= aixi o at 0. Squaring this series, we can get the coeicients o x 6, x 5, x 4 o g, but not more. So we have: Corollary 1. I L C M EG L and g = then we have: (i) i p S reg then p is a zero o g. (ii) p S irr i and only i p is a pole o g. We can also get a m p -term truncated series o g rom (L, p), where m p is the pole order o.. The Parameter ν The exponent dierence is also associated with the essel parameter ν. As in [6], we have: Theorem. I L CEG L, then (i) i S reg = then ν Q \ Z. The ollowing hold or any p S reg: (ii) L logarithmic at p i and only i ν Z. (iii) i L is not logarithmic at p and (L, p) Q then ν Q \ Z. (iv) (L, p) C K \ Q i and only i ν C K \ Q. (v) (L, p) / C K i and only i ν / C K. We will divide our algorithm into dierent cases by dierent situations in Theorem. We call (ii) logarithmic case. (i) and (iii) rational case, and (iv) and (v) irrational case. We also have easy case which will be deined later. For the logarithmic case, we have

4 Remark 4. I any p S reg is logarithmic then by Theorem (ii), ν Z, then by again Theorem (ii), we have every p S reg must be logarithmic. I not, then L has no essel type solutions. Also by the act C(x) ν(x) + C(x) ν(x) is invariant under ν ν + 1 and ν 1 ν, or the logarithmic case, we can let ν = 0. For the rational case: Remark 5. Since C(x) ν(x) + C(x) ν(x) is invariant under ν ν + 1 and ν 1 ν, i ν Q then we can just ocus on ν [0, 1 ]. I ν = 1 the operator will be reducible, it is easy to solve the operator by actoring. So we just consider ν [0, 1 ). We will give a method to ind a inite list o candidates or ν and later. I we ix, then we have: Lemma. Let Z be the set o all zeroes o, or p Z let m p be the multiplicity at p. (i) I (L, p) C K, then let j (L, p) + i N p := 0 i m p 1, i Z m p We can make the rational part o each element in N p belong to [0, 1 ]. Let the new set be Np. Then ν N := p Sreg N p. (ii) I (L, p) / C K, we can write (L, p) as a 1 k + a where k C K and a 1, a C K. Then ν = a 1 k m p (i or dierent p, we get dierent ν then there are no essel type solutions.) Proo. The lemma ollows rom the act that we know the number (M, p) = m pν mod Z, the act that ν C K, and the act that C(x) ν(x) + C(x) ν(x) is invariant under ν ν + 1 and ν 1 ν.. Easy, Logarithmic and Irrational Cases To retrieve, we need enough linear equations. We assume L C M EG L. We want to get inormation about rom L. Since might not in K, but g = is in K, we can assume g = A, A, CK[x], is monic and gcd(a, ) = 1. We want to get inormation about A, rom L. Since Maple can compute generalized exponents o L, we can compute the exponent dierence at singularity p. Then by Corollary 1 we can get the set S reg, which give us some zeroes o g, and S irr, which give the truncated series at each p S irr. The ollowing two lemmas are true or all cases: Lemma 4. We can retrieve rom S irr. Proo. According to Theorem 1 (ii), i p S irr then p is a pole o. Let m p 1 Z+ be the pole order o (M, p). g has a pole order m p. The Theorem implies = Q p S irr \{ } (x p)mp. Lemma 5. Let j deg() + m d A = deg() (i) I S reg then deg(a) < d A; (ii) i S irr then deg(a) = d A; (iii) otherwise deg(a) d A. i S irr otherwise In all cases, we can write A = Σ d A a ix i, so we have d A + 1 unknowns. Proo. According to Corollary 1 (i), i S reg then we have deg(a) < deg(). I S irr with pole order m, then deg(a) = deg() + m (see Corollary 1 (ii)). I / S irr then does not have a pole at, so that deg(a) deg(). Lemma 6. Assume p C K, i p S reg, we will get one linear equation or the coeicients o A. I p S irr with m p as pole order o (L, p), we will get m p linear equations. Proo. According to Corollary 1 (i), i p S reg, p is a zero o A. Then we will get a linear equation o {a i},...,da by setting rem(a, x p) = 0. In addition, or each p S irr with with pole order m p, by Corollary 1 (ii) we will have a m p -term truncated series o g at p. Then we can get the truncated series o A = g. On the other hand, we can rewrite A = Σ d A a ix i as a truncated series at p (by Taylor or Laurent series). Since the terms in a Taylor series or Laurent series dep linearly on the coeicients o A, by comparing the coeicients, each term will give a linear equation o a i. Example 1. L = + 10x 1x + 1x 4 x(x 1)(x )(5x ) 1 8x 4 89x + 105x 59x + 16)(5x ) 6 x (x 1) (x ) 6 and K = Q(x). Then we can compute S reg = {0}, S irr = {} and the truncated series o g at x = is 6t 4 + 1t + O(t ), so = (x )4 and d A = 4. We assume A = Σ 4 a ix i. Then rem(a, x) = a 0 = 0 give us one linear equation. And since we can rewrite A = (a0 + a1 + 4a + 8a + 16a 4)t 4 + (a 1 + 4a + 1a + a 4)t + O(t ). y comparing the coeicient o two truncated series, we can get linear equations a 0 + a 1 + 4a + 8a + 16a 4 = 6 and a 1 + 4a + 1a + a 4 = 1. For this example, we have 5 unknowns and we only have linear equations. ut we can still solve it (see Example ). For p C K, p / C K, we have: Lemma 7. I p / C K, let l(x) be the minimal polynomial o p over C K. I p S reg, we will have deg(l) linear equations. I p S irr, we will have deg(l) m p linear equations. Proo. I p S reg, then p is a zero o g. Then all the conjugates in C K o p are zeroes o g. There are deg(l) conjugated zeroes and by setting rem(a, l(x)) = 0, we will get deg(l) linear equations with coeicient in C K. I p S irr with m p as pole order o (L, p), we can irst consider it in the ield C K(p). Then according to lemma 6 we will get m p linear equations with coeicients in C K(p). Let c + P n ciai = 0 be such an equation, where {ai} are unknowns and {c i} are coeicients in C K(p). We can rewrite the equation as P deg(l) 1 e ip i = 0 where {e i} are linear unctions with coeicients in C K. Now p is algebraic over C K o degree deg(l), so 1, p,..., p deg(l) 1 are linearly indepent over C K. Hence the e i are 0; we get deg(l) linear equations over C K. We can do this or each o the m p linear equations. Then we get deg(l) m p equations. the data is rom examples at qyuan

5 Example. Suppose S reg. I C K, we get one equation by rem(a, x ) = 0. I / C K, we get two equations by rem(a, x ) = 0. Suppose S irr, and that one o the m linear equations is + (1 )a 1 + (1 + )a = 0. I / C K, we can rewrite that equation as ( + a 1 + a ) + (a a 1) = 0. Then we can get two equations {+a 1 +a = 0, a a 1 = 0}. So ar, we get at least #S reg + 1 da linear equations or the coeicients o A. I this number is greater than deg(a), then we can solve them and ind A. We call this case the easy case. This case is very similar to the case in [7]. For the logarithmic case and irrational case, we have: Lemma 8. In the logarithmic and the irrational case, we know all zeroes o A. In the irrational case, we know their multiplicities as well. Proo. y Theorem 1 (i), a change o variables can transer a regular singularity to an apparent singularity only i ν Q \ Z. So in the logarithmic and irrational cases, S reg contains all zeroes. In the irrational case, or each p S reg, let a p be the coeicient o the irrational part o the exponent dierence. Then there exists k, such that kσ p Sreg a p = d A. Then ap will k give the multiplicity o p. In the irrational case, there is only one unknown coeicient, the leading coeicient o A. ut we have 1 da linear equations, enough to get A. In the logarithmic case, we have to do a combinatorial search: try all possible combinations o multiplicities o zeroes o A. Ater that the only unknown is also leading coeicient o A. We have enough equations to ind it. Once we get, we can get a list o ν by Lemma and Remark 4..4 Rational Case The hardest case is the rational case. We will compute (see Lemma 9 and 10) a inite set o possible values or d = denom(ν). We notice that d > because the case ν Z has already been treated (logarithmic case) and i ν 1 Z then L is reducible. Let L C M EG L, g = = A. Let p be a root o A and (L, p) mpν mod Z. I d m p, change o variables x will s p to an apparent singularity. This is hard because i p is apparent, then p / S reg, which means that not all roots o A are known (not all roots o A are in S reg). ut i a zero p o A becomes an apparent singularity, the multiplicity m p 4 must be a multiple o d. So we can rewrite A = CA 1A d, where A 1, A C K[x] and C C K, A 1 is monic and the roots o A 1 are the known roots o A (the elements o S reg). For S reg =, we can let A 1 = 1 and ix d by the ollowing lemma [7]: Lemma 9. I S reg =, then d d A. For S reg, we have: Lemma 10. I S reg, we can ind a list o candidate pairs (d, A 1) by solving an equation. 4 I m p is multiplicity o at p, then m p is multiplicity o A. Proo. We assume N = #S reg, S reg = {p 1,..., p N } and (L, p) is the exponent dierence at p. Let A 1 = Π N i=1(x p i) mp i, 1 mpi < d and d p = denom( (L, p)). For each point p S reg, d p d. So we have l d where l := lcm p Sreg d p. So d can only be a multiple o l, and it must be d A. So there are d A/l possibilities or d. Once we ix d, then or each p S irr we have d d p m p. So solve ( P N i=1 mp ) + deg(a)d = da, 1 i mp < d and d i d pi m pi. It will give initely many candidates or A 1. Lemma 11. We can choose C, such that the dth root o the coeicient o the initial term o the truncated series o A/(CA 1) at p is in C K. Proo. I (C K ) S irr =, then ext C K so that it contains at least one element o S irr. Choose p (C K ) S irr. From (L, p), we can compute a truncated series or = CA 1A d. From it, we can compute a truncated series or /A 1. Let C be the coeicient o the irst term o this series, which will inish the proo (note that /A 1 = CA d ). Now the only unknown part o A is A. We can assume A = P deg(a ) b ix i. Since deg(a ) 1 d da 1 da, we have Lemma 1. For the rational case, we only need 1 da + 1 equations to recover A. We can not get the equations by the same methods as in Lemma 6 and [6, 7]. I we do so, the equations we get or {b i} will not be linear. The solution to this problem is as ollows: Theorem. In the rational case, or A = CA 1A d, and A = P deg(a ) b ix i, or each p S irr with m p as pole order o exponent dierence, i p C K, we will get m p linear equations o {b i}. Proo. Since the exponent dierence at p will give a m p -term truncated series o g = A at x = p, we can also write and CA 1 as a series at p. Then we can get the m p term truncated series o A d = g CA 1. We assume the series is Pm p<i m p c it i p where t p is the local parameter at p. We can rewrite the series as c mp t mp p S, where S is a power series with the initial term 1. Let S 1/d be a power series with irst term 1 such that S d 1/d = S. Write S 1/d = 1 + Σ i>0a it i p where a 1,..., a mp 1 are computed by Hensel liting. Let µ d = {r r C K, r d = 1}. y Lemma 11 there should be a dth root o c mp in C K. Let c be such a root. Then or each r µ d, let S r = ct mp/d p rs 1/d. Then S r is a truncated series at p whose dth power is the truncated series o g CA 1 at p. Then we can also rewrite A = P deg(a ) b ix i as a truncated series at p. y comparing the coeicients o S r and A, we will get m p linear equations. Doing this or every p S irr provides enough linear equations to ind A. Note that we have to try all combinations o r µ d at every p S irr. Remark 6. I p / C K, we can use the results rom Lemma 7 to get equations. So we can always obtain 1 da linear equations, while 1 da + 1 equations are suicient. So we always get enough linear equations.

6 Remark 7. I we get a candidate (, d), then {} { a d gcd(a, d) = 1, 1 a < 1 d} is a list o candidates or (, ν). To sum up, or all dierent cases, we have: Theorem 4. From (L, p), we can always get a list o candidates or (, ν). Proo. We always have at least #S reg + 1 da linear equations or the coeicients o A. ut we may have enough equa- tions (easy case), or only need either 1 (logarithmic case and irrational case) or 1 da + 1 equations (rational case) to get g. y Remark 4, Remark 5 Remark 7 and Lemma, we can also get a inite list o ν. The theorem means that we can always ind the change o variables. Ater that, we can compute the projective equivalence to complete the algorithm. Example. Continue with Example 1. We know S reg = {0}, S irr = {} with the truncated series o g is = 6t 4 + 1t + O(t ), = (x )4 and d A = 4. Lemma 6 did not provide suiciently many equations. ut or this case the only possible situation is A = CxA, and A = a 0+a 1x. The truncated series at x = o CA is the series o (x ) 4 /x at, which is + 9t + O(t ). So we can let C =. Then g CA 1 series o is S = 1 + t. Since K = Q(x), the only rd root o 1 is 1. So the only possible truncated series which is rd root o S is 1 + t + O(t ). And comparing it with a 0 + a 1x = a 0 + a 1 + a 1t, we get two linear equations a 0+a 1 = 1 and a 1 = 1. Solve them we get a 0 = 1, a 1 = 1. So g = x(x 1). (x ) 4 4. THE ALGORITHM The input o the algorithm is a dierential operator L o order. We want to ind solutions that can be represented in terms o essel unctions, i such solutions exist. Otherwise the algorithm outputs. Algorithm 1 gives the sketch. Now we will explain the detail how to retrieve, ν in dierent cases. 4.1 Easy Case In this case, we have enough linear equations rom Lemma 6 to recover g. Ater that, we can use Lemma to get ν. See Algorithm or detail. 4. Logarithmic Case y Remark 4, we can let ν = 0. y Lemma 8, we know all the zeroes o g. We do not yet know the leading coeicient and the multiplicity o each zero. So we can try all combinations o possible multiplicities. Algorithm will give the sketch. 4. Irrational Case In this case, by Lemma 8 we have all the zeroes with multiplicities o g. The only unknown part should be the leading coeicient. ut we have at least one linear equations. Algorithm 4 gives the sketch. 4.4 Rational Case This is the most complicated case. Let d = denom(ν) and = g = CA 1A d. Algorithm 5 gives the sketch. Data: an irreducible dierential operator L Result: solutions represented in terms o essel unctions i they exist Find all singularities by actoring the leading coeicient o L over C K; oreach Singularity p do compute the generalized exponents at p, then compute the exponent dierences and then the truncated series o g Get S reg and S irr according to the generalized exponent dierences; Compute, d A (Lemma 4 and 5) and the number o linear equations N (N #S reg + 1 da) ; i N > d A then go to easy case else i L logarithmic at some p S reg then go to logarithmic case else i there is p S reg with (L, p) / Q (i.e ν / Q) then go to irrational case else go to rational case /* It will give us a list o candidates or (, ν), where is the unction o the change o variables, and ν is the parameter o essel unctions */ oreach (, ν) in list o candidates do Compute an operator M (,ν) such that L C M (,ν) ; Use algorithm described in [1] to compute whether M (,ν) EG L and compute the transormation; i such transormation exists then Add the solution to Solutions List Output the solutions list; Algorithm 1: Main Algorithm Data: S reg, S irr with truncated series,, d A Result: potential list o (, ν) Find all linear equations described in Lemma 6; Solve linear equations to ind ; i there is no solution then output else Use Lemma to get a list N o candidate ν s oreach ν N do Add (, ν) to output list Algorithm : Easy Case

7 Data: S reg, S irr with truncated series,, d A Result: list o (, ν) i not every singularity p S reg is logarithmic then output else Let ν = 0, A = aπ p Sreg (x p) ap ; oreach {a p} such that Σ p Sreg a p = d A do Use linear equations described in Lemma 6 to solve a ; i the solution exists then Add ( A, 0) to output list Algorithm : Logarithmic case Data: S reg, S irr with truncated series,, d A Result: list o (, ν) Use Lemma 8 ind all zeroes and multiplicities; Use linear equations given by Lemma 6 to get the leading coeicient; Use Lemma to get a list o candidates or ν s; Add solutions to output list; Algorithm 4: Irrational case Data: S reg, S irr with truncated series,, d A Result: list o (, ν) i S reg = then Let the list o candidates or d be the set o actors o d A; Let A 1 = 1; else Use Lemma 10 to get a list o candidates or d and A 1 oreach candidate (d, A 1) do Fix C by Lemma 11; Use linear equations given by Theorem to compute A ; I a solution exists, add {} { a gcd(a, d) = 1, 1 a < 1 d} to output list d Algorithm 5: Rational case 5. EXAMPLES This section will illustrate the algorithm with a ew examples 5. Example 4. Let L = + 10x+4x 4x 4. K = Q(x) Step 1: We get S reg =. S irr = { } with the truncated series o g at x = is 4 9 t 6 4 t 4 + O(t ). So d A = 6 and =1. Step : It is the rational case with S reg =. So d {, 6} and we can write A = CA d. I d = then A = CA, A = a 0 + a 1x + a x. Since = 1, then the truncated series o g is the same as g. So we can let C = 4. Then 9 the truncated series o A is t 6 +t 4 = t 6 (1 t ). Since the only rd root o 1 in C K is 1, then the only rd root o 1 t is 1 t. So by comparing coeicients o t (1 t ) and A = a 0 + a 1t 1 + a t, we can get A = x 1 and then g = 4 9 (x 1). We can do this process or d = 6, in p this case, we have no solution. So we have ( (x 1), 1 ) as the only possible candidate. Step : We compute L C M, and then the projective equivalence rom M to L. Combining these transormations produces the ollowing solutions o L: C 1( (x4 + x x + x + ) I 1 ( p (x 1) x 1 ) + (x + 1)(x 1)I 4 ( p (x 1) )) + C ( (x4 + x x + x + ) K 1 ( p (x 1) x 1 ) (x + 1)(x 1)K 4 ( p (x 1) )) Example 5. Consider the operator: L := 15x4 0x + x + 8x 4 x(x 1)(15x 10x + 9x 4) 1 6x (15x 10x + 9x 4)(x 1) (075x0 165x x x x x x x x x x x x x x 6 9x x x 161x + 4x 64). Step 1: The non-apparent singularities are, 1, 0. Step : S reg = {1, 0}, with the exponent dierence 5 and 4 respectively. We also have Sirr = { } and the truncated series o g at x = is t 15 5t t 1 5t 1 + 8t 11 46t t 9 8t 8 + O(t 7 ). So = 1 and d A = 15. Step : we can easily veriy that this is a rational case. Since the exponent dierence o at 0 and 1 both have denominator, so d is a multiple o. I d = then A = Cx (x 1)A d or A = Cx(x 1) A d. I d = 6, then the multiplicity o both 1 and 0 should be a multiple o 6 = then it will contradict with d A = 15. Similarly A = Cx (x 1) A 9, A = Cx 5 (x 1) 10 and A = Cx 10 (x 1) 5 are candidates as well. Then we compute each candidate by the method in Theorem. 5 More examples are given at qyuan

8 Finally, we get = p x 4 (x 1) 5 (x + 1) and ν = 1 is the only remaining candidate. Step 4: Let L C M. Now M is already equal to L. So the general solution is: C 1I 1 ( p x 4 (x 1) 5 (x + 1) )+ C K 1 ( p x 4 (x 1) 5 (x + 1) ) 6. CONCLUSION We developed an algorithm to solve second order dierential equations in terms o essel unctions. We exted the algorithm described in [7] which already solved the problem in the C(x) case, but not in the square root case. We implemented the algorithm in Maple (available rom qyuan). A uture task is to try to develop a similar algorithm to ind F 1 type solutions. 7. REFERENCES [1] arkatou, M. A., and Plügel, E. On the Equivalence Problem o Linear Dierential Systems and its Application or Factoring Completely Reducible Systems. In ISSAC 1998, [] ronstein, M. An improved algorithm or actoring linear ordinary dierential operators. In ISSAC 1994, [] ronstein, M., and Laaille, S. Solutions o Linear Ordinary Dierential Equations in Terms o Special Functions. In ISSAC 00, 8. [4] Chan, L., and Cheb-Terrab, E. S. Non-Liouvillian Solutions or Second Order Linear ODEs. In ISSAC 004, [5] Cluzeau, T., and van Hoeij, M. A Modular Algorithm to Compute the Exponential Solutions o a Linear Dierential Operator. J. Symb. Comput. 8 (004), [6] Debeerst, R. Solving Dierential Equations in Terms o essel Functions. Master s thesis, Universität Kassel, 007. [7] Debeerst, R, van Hoeij, M, and Koep. W. Solving Dierential Equations in Terms o essel Functions. In ISSAC 008, 9 46 [8] Everitt, W. N., Smith, D. J., and van Hoeij, M. The Fourth-Order Type Linear Ordinary Dierential Equations. arxiv:math/ (006). [9] van Hoeij, M. Factorization o Linear Dierential Operators. PhD thesis, Universiteit Nijmegen, [10] van der Hoeven, J. Around the Numeric-Symbolic Computation o Dierential Galois Groups. J. Symb. Comp. 4 (007), [11] van der Put, M., and Singer, M. F. Galois Theory o Linear Dierential Equations, Springer, erlin, 00. [1] Willis,. L. An Extensible Dierential Equation Solver. SIGSAM ulletin 5 (001), 7.