Solve Linear Differential Equations in terms of ... · Note: we are interested in a class of equations that can be solved in terms of Hypergeometric Functions. Tingting Fang Solving

Introduction 2-descent C2× C2-descent Tables Methods

Solve Linear Differential Equations in terms ofHypergeometric Functions

Tingting Fang & Mark van HoeijFlorida State University

Boston, USA

January 6th, 2012

Tingting Fang Solving Differential Equations January 6th, 2012 Slide 1/ 23

Introduction

Differential operator and differential equation

LetL = an∂

n + an−1∂n−1 + · · ·+ a1∂ + a0

be a differential operator, with an, an−1, · · · , a1, a0 ∈ C(x) and npositive integer. The corresponding differential equation is

any (n) + an−1y (n−1) + · · ·+ a1y ′ + a0y = 0.

Note: we are interested in a class of equations that can be solvedin terms of Hypergeometric Functions.

Introduction

Differential operator and differential equation

LetL = an∂

n + an−1∂n−1 + · · ·+ a1∂ + a0

be a differential operator, with an, an−1, · · · , a1, a0 ∈ C(x) and npositive integer. The corresponding differential equation is

any (n) + an−1y (n−1) + · · ·+ a1y ′ + a0y = 0.

Note: we are interested in a class of equations that can be solvedin terms of Hypergeometric Functions.

Introduction

Traditional Methods of Solving Differential Operator L

Direct solving by the existing techniques.

Factor L as a product of lower order differential operators,then solve L by solving the lower order ones.

Solve L in terms of lower order differential operator.

My talk here focuses on second order linear differential equations(differential operators) which are irreducible and have noLiouvillian solutions.Question: For the equations that we can’t solve by the abovetechniques, what should we do?

Introduction

My talk here focuses on second order linear differential equations(differential operators) which are irreducible and have noLiouvillian solutions.

Question: For the equations that we can’t solve by the abovetechniques, what should we do?

Introduction

Overview of the methods

We consider to reduce the differential operator L, if possible, toanother differential operator L that is easier to solve (with sameorder, but with fewer true singularities) by using the 2-descentmethod or other descent methods.

1 If the above 2-descent exists, we find L.

2 If the number of true singularities of L drops to 3, we find its

2F1-type solutions, furthermore, find the 2F1 solution of L interms of L’s.

3 If the number of true singularities of L drops to 4, we candecide if L, furthermore L, ∃ 2F1-type solutions by building alarge table that covers the differential operators with 4 truesingularities.

Introduction

Transformations

When we talk about that L can be solved in terms of the solutionsof L, we mean that L can be transformed to L.

There are three types of transformations that preserve order 2:

1 change of variables: y(x)→ y(f (x)), f (x) ∈ C(x) \ C.

2 exp-product: y → e∫r dx · y , r ∈ C(x).

3 gauge transformation: y → r0y + r1y ′, r0, r1 ∈ C(x).

Given L1, L2 ∈ C(x)[∂] with order 2:

If L12&3→ L2, then L1 ∼p L2 (projectively equivalent)

If L13→ L2, then L1 ∼g L2 (gauge equivalent).

Introduction

Transformations

When we talk about that L can be solved in terms of the solutionsof L, we mean that L can be transformed to L.There are three types of transformations that preserve order 2:

Introduction

Transformations

Introduction

Transformations

2-descent Method

Example 1

L = x2(36x2 − 1)(4x2 − 1)(12x2 − 1)∂2+

4x(2x − 1)(1296x5 + 576x4 − 144x3 − 72x2 + x + 1)∂+

2(5184x6 − 864x5 − 1656x4 + 48x3 + 162x2 + 6x − 1)

Question: How to find the 2F1 hypergeometric solution of L?

y1 =r1 · 2F1

(1/4, 1/4

3/2144x4+24x2+1

)+r2 · 2F1

(5/4, 5/4

5/2144x4+24x2+1

)(with r1, r2 ∈ C(x))

y2 = · · ·

2-descent Method

Example 1

L = x2(36x2 − 1)(4x2 − 1)(12x2 − 1)∂2+

4x(2x − 1)(1296x5 + 576x4 − 144x3 − 72x2 + x + 1)∂+

2(5184x6 − 864x5 − 1656x4 + 48x3 + 162x2 + 6x − 1)

y1 =r1 · 2F1

(1/4, 1/4

3/2144x4+24x2+1

)+r2 · 2F1

(5/4, 5/4

5/2144x4+24x2+1

)(with r1, r2 ∈ C(x))

y2 = · · ·

2-descent Method

Example 1

L = x2(36x2 − 1)(4x2 − 1)(12x2 − 1)∂2+

4x(2x − 1)(1296x5 + 576x4 − 144x3 − 72x2 + x + 1)∂+

2(5184x6 − 864x5 − 1656x4 + 48x3 + 162x2 + 6x − 1)

y1 =r1 · 2F1

(1/4, 1/4

3/2144x4+24x2+1

)+r2 · 2F1

(5/4, 5/4

5/2144x4+24x2+1

)(with r1, r2 ∈ C(x))

y2 = · · ·

2-descent Method

Definition for 2-descent

Given a second order differential operator L defined over C(x), wesay that L has 2-descent if ∃f ∈ C(x) with degree(f )= 2, and∃L ∈ C(f )[∂f ] such that L ∼p L.

Note: ∂f = ddf = 1

f ′∂

Two steps to find the 2-descent L

1 Finding the subfield C(f ) with [C(x) : C(f )] = 2, i.e. findingf ∈ C(x) of degree 2.

2 Finding the projectively equivalent differential operatorL ∈ C(f )[∂f ].

2-descent Method

Given a second order differential operator L defined over C(x), wesay that L has 2-descent if ∃f ∈ C(x) with degree(f )= 2, and∃L ∈ C(f )[∂f ] such that L ∼p L.Note: ∂f = d

df = 1f ′∂

2-descent Method

df = 1f ′∂

2-descent Method

df = 1f ′∂

2-descent Method

df = 1f ′∂

2-descent Method

Find the subfield L descending to

Since every extension of degree 2 is Galois, so by Luroth’s theorm,we have the following relationship:

Remark

A subfield C(f ) ⊂ C(x) with [C(x) : C(f )] = 2

⇐⇒

σ ∈ Aut(C(x)/C) with degree 2

Necessary Requirements for σ

σ = ax+bcx+d with d = −a;

σ should preserve the set of true singularities of L and theirexponent-difference mod Z.

For each such σ, we compute a candidate subfield C(f ) ⊆ C(x).

2-descent Method

Remark

⇐⇒

2-descent Method

Remark

⇐⇒

2-descent Method

Remark

⇐⇒

Finding the projectively equivalent operator L

Theoretical support

The following σ and C(f ) represent the Mobius transformationfound previously and the corresponding fixed field, respectively.Suppose L descends to L ∈ C(f )[∂f ], we have

L ∼p L = σ(L) ∼p σ(L), and so L ∼p σ(L)

which means we can find the projective equivalence:

y → e∫r dx · (r0y + r1y ′)

from the solution space of L to the solution space of σ(L).Question: How to compute L from it?

Theoretical support

y → e∫r dx · (r0y + r1y ′)

from the solution space of L to the solution space of σ(L).

Question: How to compute L from it?

Theoretical support

y → e∫r dx · (r0y + r1y ′)

from the solution space of L to the solution space of σ(L).Question: How to compute L from it?

Case A

Case A is when L ∼g σ(L), in other words, there exists G = r0+r1 ∂ ∈ C(x)[∂] with G (V (L)) = V (σ(L)). Then ∃ L ∈ C(f )[∂f ]with L ∼g L.

Question: Given G , how to find L?

V (σ(L))

V (σ(L))�

Case A

Case A is when L ∼g σ(L), in other words, there exists G = r0+r1 ∂ ∈ C(x)[∂] with G (V (L)) = V (σ(L)). Then ∃ L ∈ C(f )[∂f ]with L ∼g L.Question: Given G , how to find L?

V (σ(L))

V (σ(L))�

Case A

V (σ(L))

V (σ(L))�

Case A

V (σ(L))

V (σ(L))�

Case A

V (σ(L))

V (σ(L))�

Question arising in the above diagram

Question: When does the above diagram commute?

Theorem

Let L and σ be as before, and G : V (L)→ V (σ(L)) be a gaugetransformation. Suppose L1, L2 ∈ C(f )[∂f ] and Ai : V (L)→ V (Li )are gauge transformations. Then:

1 For each i = 1, 2, there is exactly one λi ∈ C∗ such thatthe following diagram commutes

2 If L1 ∼g L2 over C(f ), then λ1 = λ2; Otherwise, λ1 = −λ2.

3 In particular, {λ1,−λ1} depends only on (L, σ,G ).

Question arising in the above diagram

Question: When does the above diagram commute?

Theorem

Let L and σ be as before, and G : V (L)→ V (σ(L)) be a gaugetransformation. Suppose L1, L2 ∈ C(f )[∂f ] and Ai : V (L)→ V (Li )are gauge transformations. Then:

1 For each i = 1, 2, there is exactly one λi ∈ C∗ such thatthe following diagram commutes

2 If L1 ∼g L2 over C(f ), then λ1 = λ2; Otherwise, λ1 = −λ2.

3 In particular, {λ1,−λ1} depends only on (L, σ,G ).

Diagram

V (L)λi G - V (σ(L))

V (Li )�

σ(A i)

Case B

Case B is when L ∼p σ(L), in other words, there exists G = e∫r ·

(r0 + r1 ∂) such that G (V (L)) = V (σ(L)).

Difficulty

We have an exponential part in G comparing with Case A. Thealgorithm mentioned above fails.

Solution

After multiplying solution of L by a suitable e∫s , we can reduce

this case to Case A.

Case B

Difficulty

Solution

Case B

Difficulty

Solution

Application for this result, Example 2

Start with L = ∂2 + 8(8x+1)(4x+1)(4x−1)∂ + 4(8x+1)

x(4x−1)(4x+1) ;

End with 2-descent L = (16x1 − 1)x1∂2 + (32x1 − 1)∂ + 4.

Solution of L :

y1 =2F1

(1/2, 1/2

1−16x + 1

); y2 = 2F1

(1/2, 1/2

)Solution of L :

2· 2F1

(1/2, 1/2

2−16x2 + 1

x(4x + 1)

2· 2F1

(3/2, 3/2

3−16x2 + 1

y2 =4x

4x − 1· 2F1

(1/2, 1/2

8x2(4x + 1)

4x − 1· 2F1

(3/2, 3/2

C2× C2-descent

C(f1)�

?C(f3)

C(f )?�

[C(x) : C(f )] = 4 and the Galois group is C 2× C 2.

C2× C2-descent

C(f1)�

?C(f3)

C(f )?�

[C(x) : C(f )] = 4 and the Galois group is C 2× C 2.

Solve L with 4 true singularities by constructing tables

Main Idea

After implementing 2-decent,C 2× C 2-descent or some otherdescent methods, We may end up with L with 4 true singularities,not 3 true singularities. This time we consider to solve such L byconstructing tables.

Given L with 4 true singularities, we are trying to find the followingform of solutions of L:

e∫r · (r0y + r1y

here r , r0, r1 ∈ C(x), and y = 2F1(a, b; c |f ).Main idea: Given L with 4 true singularities, we first figure outthe ramification type of the singularities, and then match this withthe table we have constructed. If we have the matched ones, wefind the corresponding f in the table. For the left exponential partand the r0 and r1, we implement the equivalence programs to findthem [9].

Main Idea

After implementing 2-decent,C 2× C 2-descent or some otherdescent methods, We may end up with L with 4 true singularities,not 3 true singularities. This time we consider to solve such L byconstructing tables.Given L with 4 true singularities, we are trying to find the followingform of solutions of L:

e∫r · (r0y + r1y

here r , r0, r1 ∈ C(x), and y = 2F1(a, b; c |f ).

Main idea: Given L with 4 true singularities, we first figure outthe ramification type of the singularities, and then match this withthe table we have constructed. If we have the matched ones, wefind the corresponding f in the table. For the left exponential partand the r0 and r1, we implement the equivalence programs to findthem [9].

Main Idea

After implementing 2-decent,C 2× C 2-descent or some otherdescent methods, We may end up with L with 4 true singularities,not 3 true singularities. This time we consider to solve such L byconstructing tables.Given L with 4 true singularities, we are trying to find the followingform of solutions of L:

e∫r · (r0y + r1y

here r , r0, r1 ∈ C(x), and y = 2F1(a, b; c |f ).Main idea: Given L with 4 true singularities, we first figure outthe ramification type of the singularities, and then match this withthe table we have constructed. If we have the matched ones, wefind the corresponding f in the table. For the left exponential partand the r0 and r1, we implement the equivalence programs to findthem [9].

Riemann-Hurwitz Theorem

Question: How do we construct these f in the table?

By Riemann-Hurwitz Theorem, the ramification type of thesingularities should satisfy the equation

2n − 2 =∑p∈P1

(ep − 1)

Riemann-Hurwitz Theorem

Question: How do we construct these f in the table?By Riemann-Hurwitz Theorem, the ramification type of thesingularities should satisfy the equation

2n − 2 =∑p∈P1

(ep − 1)

Belyi Maps

Table for Belyi Maps

When f is a Belyi Map, that means all the ramification points p off occur over the set {0, 1,∞}. This part of classification of f s isdone by Dr. van Hoeij and Dr. Vidunas [7].

Degree bound for f Number of Cases

Non Parametric Case 60 366

Parametric Case 12 48

Belyi Maps

Table for Belyi Maps

When f is a Belyi Map, that means all the ramification points p off occur over the set {0, 1,∞}. This part of classification of f s isdone by Dr. van Hoeij and Dr. Vidunas [7].

Non Parametric Case 60 366

Parametric Case 12 48

Near Belyi Maps

Table for Near Belyi Maps

When f is a near Belyi Map, we mean that f ramifies only above 4points, and such that there is a simple ramified point in one ofthese fibers. (The other ramified points occur over {0, 1,∞})

Non Parametric Case 18 ≥ 10

Parametric Case 6 ≥ 7

Near Belyi Maps

Table for Near Belyi Maps

When f is a near Belyi Map, we mean that f ramifies only above 4points, and such that there is a simple ramified point in one ofthese fibers. (The other ramified points occur over {0, 1,∞})

Non Parametric Case 18 ≥ 10

Parametric Case 6 ≥ 7

Near Belyi Maps

Parametric Case for near Belyi Maps

Exponent-Difference Ramification type Deg ftype (Pattern)

(1/2, a, b) (2),(1,1),(1,1) 2 Has 2-descent

(1/2, 1/3, a) (2),(1,1),(1,1) 2 Has 2-descent

(1/2, 1/3, a) (2,1),(3),(1,1,1) 3 f ∈ C(x)

(1/2, 1/3, a) (22),(3,1),(1,1,2) 4 f ∈ C(x)

(1/2, 1/3, a) (23),(32),(1,1,22) 6 f ∈ C(x)(23),(32),(13, 3) f ∈ C(x)

(1/2, 1/4, a) (2),(2),(1,1) 2 Has 2-descent

(1/2, 1/4, a) (2,2),(4),(14) 4 Has 2-descent

Near Belyi Maps

Non parametric Case for near Belyi Maps

Exponent-Difference Ramification type Deg ftype (Pattern)

(1/2, 1/3, 1/7) (25),(33,1),(7,13) 10 f ∈ C(x)(1/2, 1/3, 2/7) (25),(33,1),(7,13) 10 f ∈ C(x)(1/2, 1/3, 3/7) (25),(33,1),(7,13) 10 f ∈ C(x)

(1/2, 1/3, 1/7) (26),(34),(7,13,2) 12 f ∈ C(x)(1/2, 1/3, 2/7) (26),(34),(7,13,2) 12 f ∈ C(x)(1/2, 1/3, 3/7) (26),(34),(7,13,2) 12 f ∈ C(x)

(1/2, 1/3, 1/7) (29),(36),(7,32,22) 18 Has 2-descent(1/2, 1/3, 2/7) (29),(36),(7,32,22) 18 Has 2-descent(1/2, 1/3, 3/7) (29),(36),(7,32,22) 18 Has 2-descent

(1/2, 1/3, 1/8) (26),(34),(8,14) 12 f ∈ C(x)

Example

Example 3

L = (x − 37)∂2 + ∂ + 9(9+x)16(x2+3)

L has the ramification type (2,1),(3),(1,1,1). By searching for thetable, we know this corresponds to one of the Parametric case.In this case f has degree 3, and the exponent-difference type is( 1

2 ,13 , 0).

We calculate f and get:

f =8(9x + 10)2

(3x − 13)3

Furthermore, we find the 2F1-type solution of L as follows:

Example

Example 3

L = (x − 37)∂2 + ∂ + 9(9+x)16(x2+3)

2 ,13 , 0).

f =8(9x + 10)2

(3x − 13)3

Example

Example 3

L = (x − 37)∂2 + ∂ + 9(9+x)16(x2+3)

2 ,13 , 0).

f =8(9x + 10)2

(3x − 13)3

Example

Example 3, Continued...

y1 =(x2 + 3)

2(3x − 13)54

· 2F1

(1/12, 5/12

1/28(9x+10)2

(3x−13)3

)− 5(9x + 10)(x − 37)(x2 + 3)

(3x − 13)174

· 2F1

(13/12, 17/12

3/28(9x+10)2

(3x−13)3

)y2 = · · ·

Example

Barkatou, M. A., and Pflugel, E.On the Equivalence Problem of Linear Differential Systemsand its Application for Factoring Completely ReducibleSystems. In ISSAC 1998, 268–275.

Bronstein, M.An improved algorithm for factoring linear ordinary differentialoperators. In ISSAC 1994, 336–340.

Compoint, E.,van der Put, M.,and Weil, J.A.Effective descent for differential operators. J. Algebra.324(2010), 146–158.

Debeerst, R, van Hoeij, M, and Koepf. W.Solving Differential Equations in Terms of Bessel Functions. InISSAC 2008, 39–46.

Fang, T.

Example

Implementation and examples for 2-descent.www.math.fsu.edu/∼tfang/2descentprogram/

Fang, T. van Hoeij, M.2-descent for second order linear differential equations. InISSAC 2011, 107–114.

van Hoeij, M., and Vidunas, R.All non-Liouvillian 2F1-solvable Heun equations with pullbacksin C(x). www.math.fsu.edu/∼hoeij/files/Heun/TextFormat/

van Hoeij, M.Factorization of Linear Differential Operators. PhD thesis,Universiteit Nijmegen, 1996.

van Hoeij, M.Implementation for finding equivalence map.www.math.fsu.edu/∼hoeij/files/equiv.

Example

van Hoeij, M.Solving Third Order Linear Differential Equations in Terms ofSecond Order Equations. In ISSAC 2007, 355–360.Implementation at:www.math.fsu.edu/∼hoeij/files/ReduceOrder

van Hoeij, M, and Cremona, J.Solving conics over function fields. J. de Theories des Nombresde Bordeaux.18(2006), 595–606.

van Hoeij, M, and van der Put, M.Descent for differential modules and skew fields. J. Algebra.296(2006), 18–55.

van der Hoeven, J.Around the Numeric-Symbolic Computation of DifferentialGalois Groups. J. Symb. Comp. 42 (2007), 236–264.

van der Put, M., and Singer, M. F.Tingting Fang Solving Differential Equations January 6th, 2012 Slide 23/ 23

Example

Galois Theory of Linear Differential Equations, vol. 328 of ASeries of Comprehensive Studies in Mathematics. Springer,Berlin, 2003.

Solve Linear Differential Equations in terms of ... · Note: we are interested in a class of equations that can be solved in terms of Hypergeometric Functions. Tingting Fang Solving

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