Introduction 2-descent C 2 × C 2-descent Tables Methods Solve Linear Differential Equations in terms of Hypergeometric Functions Tingting Fang & Mark van Hoeij Florida State University Boston, USA January 6th, 2012 Tingting Fang Solving Differential Equations January 6th, 2012 Slide 1/ 23
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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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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?
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?
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?
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?
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?
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.
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.
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.
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.
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.
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.
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.
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.
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 ′∂
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.
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 ′∂
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.
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 ′∂
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.
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 ′∂
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.
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?
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).
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?
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?
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?
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?
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?
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 ).
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 ).
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].
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].
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].
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].
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].
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,∞})
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,∞})
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:
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:
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:
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van der Put, M., and Singer, M. F.Tingting Fang Solving Differential Equations January 6th, 2012 Slide 23/ 23