Algorithms for Solving Linear Differential Equations with Rational Function Coefficients

Transcription

1 Algorithms for Solving Linear Differential Equations with Rational Function Coefficients Erdal Imamoglu Department of Mathematics Florida State University May 25, / 37

2 1 Introduction 2 Formal Solutions, Exponents 3 Quotient Method 4 Integral Bases 5 Applications 2 / 37

3 Introduction Closed form solutions of differential equations are solutions that can be written in terms of well-studied functions (exponential, logarithmic, Airy, Bessel, Whittaker, hypergeometric,...). There are powerful algorithms for some type of closed form solutions. We are interested to find hypergeometric solutions. Hypergeometric solutions are common, but current solvers in computer algebra systems often fail to find such solutions. 3 / 37

4 Introduction Consider the second order linear differential equation L inp (y) = 0 where L inp = A A 1 + A 0 Q(x)[ ] ( = d ) dx A hypergeometric solution of L inp is a solution of the form ( ) (r0 S(x) = exp r dx 2F 1 (a 1, a 2 ; b 1 ; f) + r 1 2F 1(a 1, a 2 ; b 1 ; f) ) where f, r, r 0, r 1 Q(x), a 1, a 2, b 1 Q. We are interested in finding such solutions of a given regular singular L inp Q(x)[ ]. 4 / 37

5 Introduction There is also a conjecture about hypergeometric solutions. Conjecture (van Hoeij, Kunwar) Every second order globally bounded equation has a hypergeometric solution or an algebraic solution. A globally bounded equation is a differential equation which (after a simple scaling) admits a Convergent Integer power Series solution (CIS solution). Example (Franel Numbers, OEIS A000172) sequence = 1, 2, 10, 56, 346, 2252, 15184, ,... y(x) = x + 10 x x x x L inp = x (1 + x) (8 x 1) 2 + ( 24 x x 1 ) + (2 + 8 x) ( 1 1 y(x) = 1 2 x 2F 1 3, 2 3 ; 1; 27 x 2 ) (1 2 x) 3 5 / 37

6 Introduction Question How can we find hypergeometric solutions of second order regular singular differential operators? There are powerful algorithms for specific tasks: Fang, van Hoeij (2012) Kunwar, van Hoeij (2013) Kunwar (2014) van Hoeij, Vidunas (2015) We want to develop a general algorithm. 6 / 37

7 Introduction Contributions We have developed two (heuristic) effective algorithms to find hypergeometric solutions (if they exist) of second order regular singular differential operators in Q(x)[ ]. One of our algorithms is the most general algorithm in the literature. We have developed fast algorithms to simplify n-th order regular singular differential operators in Q(x)[ ]. Simplifying Solving. 7 / 37

8 Introduction Value to the Scientific Society Hypergeometric solutions are common in physics and combinatorics. Our implementations have been already used by physicists. Example Feynman Diagrams. One of our algorithms simplify a differential operator to another operator which is easier to solve (Simplifying Solving). Example A 3rd order operator. 8 / 37

9 1 Introduction 2 Formal Solutions, Exponents 3 Quotient Method 4 Integral Bases 5 Applications 9 / 37

10 Formal Solutions, Exponents Let L inp = A A 1 + A 0 Q(x)[ ]. s Q is a singularity if it is a root of A 2, or a pole A 1 or A 0. s is a regular singularity if A 1 A 2 (x s) 1, A 0 A 2 (x s) 0 are analytic at s. L inp is regular singular if it has only regular singularities. Formal solutions of L inp at x = s: y 1 = (x s) e 1 ( ) y 2 = (x s) e 2 ( ) + ky 1 log(x s) (k might be 0) Exponents of L inp at x = s: e 1, e 2. Exponent-difference: (L inp, s) = e 1 e / 37

11 1 Introduction 2 Formal Solutions, Exponents 3 Quotient Method 4 Integral Bases 5 Applications 11 / 37

12 Quotient Method Special Case Let L inp Q(x)[ ] with ord(l inp ) = 2. Assume that we want to find hypergeometric solutions this form: ( ) exp r dx 2F 1 (a 1, a 2 ; b 1 ; f) (f, r Q(x), a 1, a 2, b 1 Q) If they exist, then there exists a GHDO such that because L B = 2 + b 1 (a 1 + a 2 + 1) a 1a 2 x(1 x) x(1 x) Q(x)[ ] L B f C M r E L inp 2F 1 (a 1, a 2 ; b 1 ; x) f C 2 F 1 (a 1, a 2 ; b 1 ; f) r E ( exp ) r dx 2F 1 (a 1, a 2 ; b 1 ; f) 12 / 37

13 Quotient Method Problem We know L inp, we do not know L B (i.e., a 1, a 2, b 1 ), f, r. Assume that we have a (right) candidate L B. Then L B f C M r E L inp. We can compute formal solutions y i of L B and Y i of L inp at a non-removable singularity (order of formal solutions = a). ( ) y i (x) f C y i (f(x)) r E Y i (x) = exp r dx y i (f(x)) (i = 1, 2) Let q(x) = y 1(x) y 2 (x), Q(x) = Y 1(x) Y 2 (x). This gives us q(f(x)) = c Q(x) where c is an unknown constant, so f(x) q 1 (c Q(x)) mod x a 13 / 37

14 Quotient Method Problem We do not know c. Idea Choose a suitable prime l and compute f(x) q 1 (c Q(x)) mod (x a, l) and try c = 1,..., l 1. For each c try rational function reconstruction to find f F l (x). If this succeeds for at least one c, then try l-adic Hensel lifting and rational number reconstruction to find f Q(x). f Compute M s.t. L B C M r E L inp and find r (there is a formula). Problem We do not know L B (i.e., a 1, a 2, b 1 Q). 14 / 37

15 Quotient Method Question How to find candidates for L B (i.e., a 1, a 2, b 1 Q)? α 0 = 1 b 1, α 1 = b 1 a 1 a 2, α = a 2 a 1. Assume that L B f C M r E L inp. L B L inp singularities 0, 1, s 1,..., s r exponent-differences α 0, α 1, α (L inp, s 1 ),..., (L inp, s r ) { a } α 0, α 1, α kb : a {1, (L inp, s i )}, 1 b d f, 1 k a f where d f = deg(f) and a f = [Q(x, f) : Q(x)]. We have a bound on d f and we consider a f 2. Problem This set might have too many elements. 15 / 37

16 Quotient Method Theorem (Special Case of Riemann-Hurwitz Formula for DEs) If f : X P 1 and L B f C M r E L inp, then 2 + (1 (L inp, s i )) = d f 2 + a s f i where d f = deg(f) and a f = [C(x, f) : C(x)]. i {0,1, } (1 α i ) This formula eliminates vast majority of the candidates: { a } α 0, α 1, α kb : a {1, (L inp, s i )}, 1 b d f, 1 k a f 16 / 37

17 Quotient Method Algorithm Outline (find 2f1) - Compute candidates L B s (if any). - Compute quotients of formal solutions at a non-removable singularity. - Use modular reduction, Hensel lifting, rational reconstruction to find f (if any). - Find r. - Return a basis of hypergeometric solutions (if they exist) or an empty list. Algorithm find 2f1 is very effective to find solutions of type ( ) (r0 exp r dx 2F 1 (a 1, a 2 ; b 1 ; f) + r 1 2F 1(a 1, a 2 ; b 1 ; f) ) where r 1 = 0. Question What if r 1 0? Idea Simplifying Solving 17 / 37

18 1 Introduction 2 Formal Solutions, Exponents 3 Quotient Method 4 Integral Bases 5 Applications 18 / 37

19 Integral Bases Example (Simplifying Solving) Consider the following number field Q[x]/(f 1 ) where f 1 = 98818x x x x x x We can reduce this to Q[x]/(f 1 ) = Q[x]/(f 2 ) where f 2 = x 6 5x 4 21x 3 12x 2. Key step to find an isomorphic number field is computation of an integral basis. 19 / 37

20 Integral Bases Let A = {f Q[x] f is irreducible}. f 1 f 2 Q[x]/(f 1 ) = Q[x]/(f 2 ) (f 1, f 2 Q[x]) Goal Given f 1, find f 2 A with small bit-size such that f 1 f 2. f 2 = a standard form of f 1 Solution POLRED algorithm (Cohen and Diaz Y Diaz, 1991). - Compute a basis for the algebraic integers of Q[x]/(f 1 ). - Apply LLL (Lenstra, Lenstra, and Lovasz, 1982) to this basis. Application Reduce computations in Q[x]/(f 1 ) to computations in Q[x]/(f 2 ). 20 / 37

21 Integral Bases Let D = Q(x)[ ] and A = {L D L is irreducible}. L 1 L 2 D/DL 1 = D/DL2 as D-modules (L 1, L 2 D) Goal Given L 1, find L 2 A with small bit-size such that L 1 L 2. L 2 = a standard form of L 1 Solution Idea Imitate POLRED. Application Reduce solving L 1 with many apparent singularities to solving L 2 with few apparent singularities. 21 / 37

22 Integral Bases Computation of integral bases for (algebraic) number fields (algebraic) function fields are well studied: Trager (1984) Cohen and Diaz Y Diaz (1991) van Hoeij (1994) de Jong (1998) Montes (1999) van Hoeij and Stillman (2015) Differential analogue of integral bases is new (Kauers and Koutschan, 2015). We give an integral basis algorithm that is much faster. In addition, we normalize the basis at infinity, which is the analogue of phase 2 in POLRED. 22 / 37

23 Integral Bases L C(x)[ ] be regular singular. L has a basis of formal solutions at x = s in the form { y = t νs s P i t i x s, if s s t s = if s = i=0 where ν s C and P i C[log (t s )] with deg(p i ) < ord(l) and P x, The valuation of y at x = s is v s (y) = Re(ν s ) 23 / 37

24 Integral Bases Fix L C(x)[ ], ord(l) = n, and let G C(x)[ ]. G is called integral for L at s if v s (G) = inf{v s (G(y)) y is a solution of L at x = s} 0 G is called integral for L if v s (G) 0 for all s C. Consider the C[x]-module O L O L = {G C(x)[ ] G is integral for L and ord(g) < n} A basis of O L is called an (global) integral basis for L. 24 / 37

25 Integral Bases Let L C(x)[ ], ord(l) = n, and P C[x]. The set {b 1,..., b n } is a local integral basis for L at P when { A1 b A } n b n A i, B i C[x] and gcd(p, B i ) = 1 B 1 B n = {G G is integral for L at every root of P } A local integral basis at a finite singularity s C is the local integral basis at P = x s. 25 / 37

26 Integral Bases Question How to compute a local integral basis for an operator at one point? Theorem {b 1,..., b n } is a local integral basis for L C(x)[ ] at x = 0 1 i, j {1,..., n} we have v 0 (b i (y j )) 0 (y j = a solution of L at x = 0) and 2 (c 1,..., c n ) C n \ (0,..., 0) there exists j {1,..., n} such that v 0 ((c 1 b c n b n )(y j )) < 1 26 / 37

27 Integral Bases Let L inp Q(x)[ ] be regular singular. Algorithm Outline (local basis at 0) - Compute v j = v 0 (y j ) of the all formal solutions y j of L inp at x = 0. - Let m = min(v j ). - Let b 1 = x m 0. - For i from 2 to n do: 1 Let b i = x b i 1 2 Make the ansatz A = 1 x (u 1 b u i 1 b i 1 + b i ). 3 Evaluate A(y j ) and equate coefficients of the non-integral terms to 0. 4 If there is a solution, find it, update b i = A and return to Step 2. - Return b 1,..., b n. 27 / 37

28 Integral Bases Improvements (Apparent singularities and algebraic singularities) 1 Apparent Singularities A local integral basis at an apparent singularity is given by {b 1 = e 1,..., b n = en } If e i ord(l inp ), then b i = Rem( e i, L inp ). Our implementation only checks for apparent singularities of the most common type where e 1, e 2,..., e n 1, e n = 0, 1,..., n 2, n 28 / 37

29 Integral Bases 2 Algebraic Singularities Let s be an algebraic singularity with minimal polynomial P. We compute a local integral basis {b 1,..., b n } for L at x = s. We want to scale b i in such a way that - {c 1 b 1,..., c n b n } is still an integral basis at x = s, - {c 1 b 1,..., c n b n } have valuations 1 at all roots of P x s. Let for i = 1,..., n and then c i = will be a local integral basis at P. ( P ) vs(b1 )+i x s {Tr(c 1 b 1 ),..., Tr(c n b n )} 29 / 37

30 Integral Bases Let L inp Q(x)[ ] be regular singular. Algorithm Outline (global integral basis) - Let P sing be polynomial whose roots are singularities of L inp. - For each irreducible factor P of P sing, compute a local integral bases for L inp at P. - Combine all local integral bases to form a global integral basis for L inp. 30 / 37

31 Integral Bases Timings Comparison of timings of Kauers and Koutschan s integral basis algorithm and our integral basis algorithm (in seconds) on a computer with a 2.5 GHZ Intel Core i5-3210m CPU and 8 GB RAM. Example Kauers-Koutschan Our algorithm > > > > / 37

32 Integral Bases Now, we know how to compute a global integral basis for L inp. We have control on finite singularities of L inp. Question What about the point x =? Normalization Let L inp C(x)[ ] with ord(l inp ) = n. The set {b 1,..., b n } is called normalized at s, if r i C(x) such that {r 1 b 1,..., r n b n } is a local integral basis for L inp at s. We want to normalize a global integral basis at x =. Normalization of an integral basis at x = for an algebraic function was introduced (Trager, 1984). 32 / 37

33 Integral Bases Let L inp Q(x)[ ] be regular singular and let be a global integral basis for L inp. {B 1,..., B n } Algorithm Outline (normalization at infinity) - Compute a local integral basis {b 1,..., b n } at. - Compute change of basis matrix and follow Trager s method. Basis elements of a normalized integral basis for L inp gives us gauge transformations to simplify L inp (to simplify it to its standard form). Example Linp[9]. 33 / 37

34 1 Introduction 2 Formal Solutions, Exponents 3 Quotient Method 4 Integral Bases 5 Applications 34 / 37

35 Applications Lets return to our problems: 1 Simplify a given differential operator (Simplifying Solving). 2 Find hypergeometric solutions (generalize find 2f1). 1 Simplify a given differential operator: Goal Given L 1, find L 2 with small bit-size such that L 1 is gauge equivalent to L 2. Solution Idea Imitate POLRED. Solution standard forms algorithm: - Compute a normalized integral basis for L 1. - If necessary, search for a better basis element to simplify L / 37

36 Applications 2 Generalize find 2f1: Let L inp Q(x)[ ] be a second order regular singular operator. We want to find hypergeometric solutions of L inp of the form ( ) (r0 exp r dx 2F 1 (a 1, a 2 ; b 1 ; f) + r 1 2F 1(a 1, a 2 ; b 1 ; f) ) Algorithm Outline (hypergeometricsols) - Try find 2f1. - If find 2f1 returns an empty list, use standard forms to simplify L inp to another operator L inp. - Feed find 2f1 with L inp. - If find 2f1 solves L inp, then obtain solutions of L inp from solutions of L inp. 36 / 37

37 Thank You Thanks to my advisor Mark van Hoeij for his continuous support, encouragement, and friendship. Thanks to the members of my committee for their time and efforts. 37 / 37