Parametric Stokes phenomena of Gauss s hypergeometric differential equation with a large parameter

Transcription

1 Parametric Stokes phenomena of Gauss s hypergeometric differential equation with a large parameter Mika TANDA (Kinki Univ) Collaborator :Takashi AOKI (Kinki Univ) Global Study of Differential Equations in the Complex Domain Sept / 36

2 1 Hypergeometric differential equations 2 Stokes graphs 3 Voros coefficients 4 Borel sums of Voros coefficients 5 Parametric Stokes phenomena 2 / 36

3 Introduce We consider the following differential equation with a large parameter η: ( d 2 dx 2 + η2 Q ) ψ = 0 ( ) 3 / 36

4 Introduce We consider the following differential equation with a large parameter η: ( d 2 dx 2 + η2 Q ) ψ = 0 ( ) with Q = Q 0 + η 2 Q 1, Q 0 = (α β)2 x 2 + 2(2αβ αγ βγ)x + γ 2 4x 2 (x 1) 2, Q 1 = x2 x + 1 4x 2 (x 1) 2 3 / 36

5 Introduce We consider the following differential equation with a large parameter η: ( d 2 dx 2 + η2 Q ) ψ = 0 ( ) with Q = Q 0 + η 2 Q 1, Q 0 = (α β)2 x 2 + 2(2αβ αγ βγ)x + γ 2 4x 2 (x 1) 2, Q 1 = x2 x + 1 4x 2 (x 1) 2 The equation ( ) comes from the Classical HGDE (Gauss s hypergeometric differential equation): x(1 x) d2 w + (c (a + b + 1)x)dw abw = 0 dx2 dx 3 / 36

6 Introduce We consider the following differential equation with a large parameter η: ( d 2 dx 2 + η2 Q ) ψ = 0 ( ) with Q = Q 0 + η 2 Q 1, Q 0 = (α β)2 x 2 + 2(2αβ αγ βγ)x + γ 2 4x 2 (x 1) 2, Q 1 = x2 x + 1 4x 2 (x 1) 2 The equation ( ) comes from the Classical HGDE (Gauss s hypergeometric differential equation): x(1 x) d2 w + (c (a + b + 1)x)dw abw = 0 dx2 dx Introduce a large parameter η by setting a = ηα, b = 1 + ηβ, c = 1 + ηγ 2 3 / 36

7 Introduce We consider the following differential equation with a large parameter η: ( d 2 dx 2 + η2 Q ) ψ = 0 ( ) with Q = Q 0 + η 2 Q 1, Q 0 = (α β)2 x 2 + 2(2αβ αγ βγ)x + γ 2 4x 2 (x 1) 2, Q 1 = x2 x + 1 4x 2 (x 1) 2 The equation ( ) comes from the Classical HGDE (Gauss s hypergeometric differential equation): x(1 x) d2 w + (c (a + b + 1)x)dw abw = 0 dx2 dx Introduce a large parameter η by setting a = ηα, b = 1 + ηβ, c = 1 + ηγ 2 and eliminate the first-order term by ψ = x ηγ 2 (1 x) η(α+β γ) 2 w 3 / 36

8 Our equation ( ) has the following formal solutions (WKB solutions) ψ ± = 1 Sodd exp(± x a S odd dx) a is a zero of Q 0 dx (a is a turning point) a formal solution S = S odd + S even = j= 1 η j S j to Riccati equation ds dx + S2 = η 2 Q S 1 = Q 0, 2S 1 S 0 + ds 1 dx = 0, 4 / 36

9 Stokes graphs A Stokes curve is an integral curve of Im Q 0 dx = 0 emanating from a turning point A Stokes graph of our equation is a collection of all Stokes curves, turning points a k (k = 0, 1) and singular points b 0 = 0, b 1 = 1, b 2 = 5 / 36

10 Stokes graphs A Stokes curve is an integral curve of Im Q 0 dx = 0 emanating from a turning point A Stokes graph of our equation is a collection of all Stokes curves, turning points a k (k = 0, 1) and singular points b 0 = 0, b 1 = 1, b 2 = We assume (i) αβγ(α β)(α γ)(α + β γ) 0 (ii) Re α Re β Re (γ α) Re (γ β) 0 (iii) Re (α β) Re (α + β γ) Re γ 0 5 / 36

11 Stokes graphs A Stokes curve is an integral curve of Im Q 0 dx = 0 emanating from a turning point A Stokes graph of our equation is a collection of all Stokes curves, turning points a k (k = 0, 1) and singular points b 0 = 0, b 1 = 1, b 2 = We assume (i) αβγ(α β)(α γ)(α + β γ) 0 (ii) Re α Re β Re (γ α) Re (γ β) 0 (iii) Re (α β) Re (α + β γ) Re γ 0 Assumption (i) = There are two distinct turning points a 0, a 1 and a 0, a 1 0, 1, 5 / 36

12 Stokes graphs A Stokes curve is an integral curve of Im Q 0 dx = 0 emanating from a turning point A Stokes graph of our equation is a collection of all Stokes curves, turning points a k (k = 0, 1) and singular points b 0 = 0, b 1 = 1, b 2 = We assume (i) αβγ(α β)(α γ)(α + β γ) 0 (ii) Re α Re β Re (γ α) Re (γ β) 0 (iii) Re (α β) Re (α + β γ) Re γ 0 Assumption (i) = There are two distinct turning points a 0, a 1 and a 0, a 1 0, 1, Assumptions (ii) and (iii) = There is no Stokes curves which connect turning point(s) 5 / 36

13 We assume that (α, β, γ) are contained in (i) Let n 0, n 1 and n 2 be numbers of Stokes curves that flow into 0, 1 and, respectively ˆn will denote (n 0, n 1, n 2 ) 6 / 36

14 We assume that (α, β, γ) are contained in (i) Let n 0, n 1 and n 2 be numbers of Stokes curves that flow into 0, 1 and, respectively ˆn will denote (n 0, n 1, n 2 ) ˆn characterizes topological configration of Stokes graphs ˆn is constant on a connected component of the set of all (α, β, γ) satisfying (ii) and (iii) We defined ω 1 = {(α, β, γ) C 3 0 < Reα < Reγ < Reβ}, ω 2 = {(α, β, γ) C 3 0 < Reα < Reβ < Reγ < Reα + Reβ}, ω 3 = {(α, β, γ) C 3 0 < Reγ < Reα < Reβ}, ω 4 = {(α, β, γ) C 3 0 < Reγ < Reα + Reβ < Reβ} If (α, β, γ) are contained in ω h (h = 1, 2, 3, 4) respectively, we give a characterization of the Stokes geometry of our equation 6 / 36

15 Re γ > 0 fixed Reα-Reβ plane Re β ω 4 ω 1 ω 3 ω 2 0 Re α 7 / 36

16 Examples of Stokes graphs for (α, β, γ) of values (01, 2, 1) ω 1, (2, 39, 4) ω 2, (11, 2, 1) ω 3, ( 01, 2, 1) ω 4 ˆn = (2, 2, 2), ˆn = (4, 1, 1), ˆn = (1, 4, 1), ˆn = (1, 1, 4) 8 / 36

17 Examples of Stokes graphs for (α, β, γ) of values (01, 2, 1) ω 1, (2, 39, 4) ω 2, (11, 2, 1) ω 3, ( 01, 2, 1) ω 4 ˆn = (2, 2, 2), ˆn = (4, 1, 1), ˆn = (1, 4, 1), ˆn = (1, 1, 4) We denote by ι j (j = 0, 1, 2) the following mappings We have to keep in mind that Q is invariant under these involutions ι 0 : (α, β, γ) ( α, β, γ) ι 1 : (γ α, γ β, γ) ι 2 : (β, α, γ) Each domain is covered by one of ω h (h = 1, 2, 3, 4) via involutions in the configuration 8 / 36

18 and set G = the group generated by ι k (k = 0, 1, 2) Π j = r(ω j ) r G 9 / 36

19 G = the group generated by ι k (k = 0, 1, 2) and set Π j = r(ω j ) Theorem 1 (Aoki T and Tanda M [2] 2013) (1) If (α, β, γ) Π 1, then ˆn = (2, 2, 2) (2) If (α, β, γ) Π 2, then ˆn = (4, 1, 1) (3) If (α, β, γ) Π 3, then ˆn = (1, 4, 1) (4) If (α, β, γ) Π 4, then ˆn = (1, 1, 4) r G 9 / 36

20 and set G = the group generated by ι k (k = 0, 1, 2) Π j = r(ω j ) Theorem 1 (Aoki T and Tanda M [2] 2013) (1) If (α, β, γ) Π 1, then ˆn = (2, 2, 2) (2) If (α, β, γ) Π 2, then ˆn = (4, 1, 1) (3) If (α, β, γ) Π 3, then ˆn = (1, 4, 1) (4) If (α, β, γ) Π 4, then ˆn = (1, 1, 4) Re β Reα-Reβ planes Re β (1,1,4) (2,2,2) (1,4,1) Π 4 Π 1 Π 3 0 Π 2 (4,1,1) Re α r G 0 Re α Re γ > 0 Re γ < 0 9 / 36

21 Voros coefficients Q0 γ at x = 0, 2x Q0 α + β γ at x = 1, 2(x 1) Q0 β α at x =, 2x V j for (b j, a) (b 0 = 0, b 1 = 1, b 2 = ) has following form: the Voros coefficient V j = V j (α, β, γ; η) := a b j (S odd ηs 1 )dx, Since residues of S odd and ηs 1 as the singular points coincide, V j are well defined and we have a formal power series V j in η 1 The Voros coefficient V j (α, β, γ) describes the discrepancy between WKB solutions normalized at a and those normalized at b j 10 / 36

22 Voros coefficients Q0 γ at x = 0, 2x Q0 α + β γ at x = 1, 2(x 1) Q0 β α at x =, 2x V j for (b j, a) (b 0 = 0, b 1 = 1, b 2 = ) has following form: the Voros coefficient V j = V j (α, β, γ; η) := a b j (S odd ηs 1 )dx, Since residues of S odd and ηs 1 as the singular points coincide, V j are well defined and we have a formal power series V j in η 1 The Voros coefficient V j (α, β, γ) describes the discrepancy between WKB solutions normalized at a and those normalized at b j 10 / 36

23 Voros coefficients Q0 γ at x = 0, 2x Q0 α + β γ at x = 1, 2(x 1) Q0 β α at x =, 2x V j for (b j, a) (b 0 = 0, b 1 = 1, b 2 = ) has following form: the Voros coefficient V j = V j (α, β, γ; η) := a b j (S odd ηs 1 )dx, Since residues of S odd and ηs 1 as the singular points coincide, V j are well defined and we have a formal power series V j in η 1 The Voros coefficient V j (α, β, γ) describes the discrepancy between WKB solutions normalized at a and those normalized at b j 10 / 36

24 Theorem 2 (Aoki T and Tanda M [2] 2013) V j (α, β, γ; η) for (j, a) (j = 0, 1, 2) has following forms: V 0 = 1 B n η 1 n { ( (1 2 1 n ) 2 n(n 1) n=2 V 1 = 1 2 V 2 = 1 2 n=2 n=2 B n η 1 n n(n 1) 1 α + 1 n 1 β + 1 n 1 (γ α) + 1 n 1 (γ β) n 1 { ( (1 2 1 n 1 ) α + 1 n 1 β n 1 ) + 2 }, γ n 1 ) 1 (γ α) 1 n 1 (γ β) n 1 } 2 +, (α + β γ) n 1 B n η 1 n { ( (1 2 1 n 1 ) n(n 1) α 1 ) n 1 β n 1 (γ α) n 1 (γ β) n 1 Here, B n are Bernoulli numbers defined by 2 (β α) n 1 } te t e t 1 = n=0 B n n! tn 11 / 36

25 Borel sums of Voros coefficients Re γ > 0 fixed Reα-Reβ plane Re β ω 4 ω 1 ω 3 ω h (h = 1, 2, 3, 4): Stokes regions of Voros coefficients ω 2 0 Re α V j are Borel summable in each ω h (h = 1, 2, 3, 4) V h : The Borel sums of V j in ω h (h = 1, 2, 3, 4) j 12 / 36

26 Theorem 3 (Aoki T and Tanda M [4] to appear) (i) The Borel sums V 1 of V j in ω 1 have following forms: j V 1 = log Γ( 1 + αη)γ( 1 + βη)γ( 1 + (γ α)η)(β γ)(β γ)η γ 2γη 1 Γ( 1 + (β, 2 γ)η)γ2 (γη)α αη β βη (γ α) (γ α)η η V 1 1 = 1 2 log Γ( (γ α)η)γ2 ((α + β γ)η)α αη β βη (β γ) (β γ)η η Γ( αη)γ( βη)γ( (β γ)η)(γ α)(γ α)η (α + β γ) 2(α+β γ)η 1, V 1 = log Γ( 1 + βη)γ( 1 + (γ α)η)γ( 1 + (β γ)η)ααη (β α) 2(β α)η 1 Γ( αη)γ2 ((β α)η)β βη (γ α) (γ α)η (β γ) (β γ)η η 13 / 36

27 (ii) The Borel sums V 2 j of V j in ω 2 have following forms: V 2 = log Γ( αη)γ( βη)γ( (γ α)η)γ( 1 + (γ β)η)γ2ηγ 1 2, Γ 2 (γη)α αη β βη (γ α) (γ α)η (γ β) (γ β)η 2πη V 2 1 = 1 2 log Γ( (γ α)η)γ( (γ β)η)γ2 ((α + β γ)η)α αη β βη η Γ( αη)γ( βη)(γ α)(γ α)η (γ β) (γ β)η (α + β γ) 2(α+β γ)η 1 2π, V 2 2 = 1 2 log Γ( βη)γ( (γ α)η)ααη (γ β) (γ β)η (β α) 2(β α)η 1 2π Γ( αη)γ( (γ β)η)γ2 ((β α)η)β βη (γ α) (γ α)η η 14 / 36

28 (iii)the Borel sums V 3 j of V j in ω 3 have following forms: V 3 = log Γ( αη)γ( βη)(α γ)(α γ)η (β γ) (β γ)η γ 2γη 1 2π Γ( (α γ)η)γ( (β γ)η)γ2 (γη)α αη β βη η V 3 = log 2πΓ 2 ((α + β γ)η)α αη β βη (α γ) (α γ)η (β γ) (β γ)η η Γ( αη)γ( βη)γ( (α γ)η)γ(, 1 + (β γ)η)(α + β γ)2(α+β γ)η 1 2 V 3 2 = 1 2 log 2πΓ( βη)γ( (β γ)η)ααη (α γ) (α γ)η (β α) 2(β α)η 1 Γ( αη)γ( (α γ)η)γ2 ((β α)η)β βη (β γ) (β γ)η η 15 / 36

29 (iv) The Borel sums V 4 j of V j in ω 4 have following forms: V 4 = log Γ( 1 + βη)γ( 1 + (γ 2 2 α)η)( α) αη (β γ) (β γ)η γ 2γη 1 2π Γ( 1 αη)γ( 1 + (β, 2 2 γ)η)γ2 (γη)β βη (γ α) (γ α)η η V 4 1 = 1 2 log Γ( 1 2 αη)γ( (γ α)η)γ2 ((α + β γ)η)β βη (β γ) (β γ)η η Γ( βη)γ( (β γ)η)( α) αη (γ α) (γ α)η (α + β γ) 2(α+β γ)η 1 2π, V 4 = log Γ( 1 αη)γ( 1 + βη)γ( 1 + (γ α)η)γ( 1 + (β γ)η)(β α)2(β α)η Γ 2 ((β α)η)( α) αη β βη (γ α) (γ α)η (β γ) (β γ)η 2πη 16 / 36

30 Proof We consider the poof of the Borel sum V 1 and 0 V2 of Voros 0 coefficient V 0 in ω 1 and ω 2, respectively The Borel sums V 1 and 0 V2 of V 0 0 in ω 1 and ω 2 have following forms: V 1 = log Γ( 1 + αη)γ( 1 + βη)γ( 1 + (γ α)η)(β γ)(β γ)η γ 2γη 1 Γ( 1 + (β, 2 γ)η)γ2 (γη)α αη β βη (γ α) (γ α)η η V 2 = log Γ( 1 + αη)γ( 1 + βη)γ( 1 + (γ α)η)γ( 1 + (γ β)η)γ2ηγ Γ 2 (γη)α αη β βη (γ α) (γ α)η (γ β) (γ β)η 2πη The idea of proof of the Theorem 3 is to use the method developed by Takei [10] (2008) To find the Borel sums V 1 and 0 V2, we first take 0 the Borel transform V 0,B (α, β, γ; y) of V 0 17 / 36

31 By the definition, we have the Borel transform V 0,B (α, β, γ; y) as follows V 0,B = 1 B n y n 2 { ( (1 2 1 n ) 1 2 n! α 1 ) n 1 β 1 n 1 (γ α) 1 2 } n 1 (γ β) n 1 γ n 1 n=2 Bernoulli number g(t) = n=2 B n n! tn = tet e t 1 1 t 2 1 e 2x 1 = 1 2 ( 1 e x 1 1 e x + 1 ) g 0 (t) = g(t) t g 1 (t) = y 2 = 1 y ( 1 exp y t t y ), 1 exp y 2t exp y 2t + 1 2t y 18 / 36

32 Hence the Borel transform V 0,B (α, β, γ; y) is written in the form V 0,B (α, β, γ; y) = 1 4y {g 1(γ α) + g 1 (γ β) + g 1 (α) + g 1 (β)} g 0 (γ) We introduce the following auxiliary infinite series: where Ṽ 0 = V 0 + µ(γ α) + µ(γ β) + µ(α) + µ(β), µ(t) = 1 4 tη 2 log( tη ) = 1 4 The Borel transform µ B (t; y) of µ(t) is µ B (t; y) = 1 1 ( y ) n t = 8t (n + 2)n! 2t 4y 2 n=0 n=0 The Borel transform Ṽ 0,B of Ṽ 0 is related to V 0,B by g(t) = 1 ( 2y e 1 2t y e 1 t y 1 2 t ) Then we have y 1 n + 2 ( 2tη) (n+1) {( y ) ( t 2 exp y ) } + 2 2t Ṽ 0,B = g(γ α) + g(γ β) + g(α) + g(β) g 0 (γ) 19 / 36

33 Next we consider the Borel sums V 1 0 and V2 0 of V 0 We use the following integral formula representation of the logarithm of the Γ-function ([5] BMP, 1997) L(θ) = 0 ( 1 e s ) e θs s s where Re θ is positive We can compute the inverse Laplace transform 0 ds = log Γ(θ) (θ 1 ) log θ + θ, 2π 2 g(α)e yη dy of g(α) by using Ṽ 0,B if Re α is positive If Re α is negative, we make use of the relation g(t) = g( t) 20 / 36

34 If (α, β, γ) are contained in ω 1, Re α, Re β, Re γ α and Re γ are positive and Re γ β is negative If (α, β, γ) are contained in ω 2, Re α, Re β, Re γ α, Re γ β and Re γ are positive We use the Borel transforms V 1 and 0,B V2 of Voros coefficients V 0,B 0 V 1 =g(α) + g(β) + g(γ α) g(β γ) g 0,B 0(γ), V 2 0,B =g(α) + g(β) + g(γ α) + g(γ β) g 0(γ) Similarly we can compute the inverse Laplace transforms and we have: Ṽ 1 = L( (γ α)η) L(1 2 + (β γ)η) + L(1 2 + αη) + L(1 + βη) L(γη), 2 Ṽ 2 = L( (γ α)η) + L(1 2 + (γ β)η) + L(1 2 + αη) + L(1 + βη) L(γη) 2 21 / 36

35 Since µ is a convergent power series of η 1, we have V 1 0 =Ṽ1 + µ(γ α) µ(β γ) + µ(α) + µ(β), 0 V 2 0 =Ṽ2 + µ(γ α) + µ(γ β) + µ(α) + µ(β) 0 Then we obtain The Borel sums V 1 and 0 V2 of V 0 0 in ω 1 and ω 2 have following forms: V 1 = log Γ( 1 + αη)γ( 1 + βη)γ( 1 + (γ α)η)(β γ)(β γ)η γ 2γη 1 Γ( 1 + (β 2 γ)η)γ2 (γη)α αη β βη (γ α) (γ α)η η V 2 = log Γ( 1 + αη)γ( 1 + βη)γ( 1 + (γ α)η)γ( 1 + (γ β)η)γ2ηγ , Γ 2 (γη)α αη β βη (γ α) (γ α)η (γ β) (γ β)η 2πη 22 / 36

36 Relation between Borel sums of V j in two regions Re γ > 0 fixed Reα-Reβ plane Re β ω 4 ω 1 ω 3 Re (β γ) ω 2 0 Re α We compare V 2 j (j = 0, 1, 2) with the analytic continuation V 1 j to ω 2 23 / 36

37 Theorem 4 The Borel sums V 1 of Voros coefficients V j can be analytically j continued over ω 2 (j = 0, 1, 2) The analytic continuations of the Borel sums V 1 to ω 2 are related to V 2 as follows: j j V 1 2 =V2 1 log(exp 2(γ β)ηπi + 1), 2 2 V 1 j =V 2 j + 1 log(exp 2(γ β)ηπi + 1) (j = 0, 1) 2 Here β γ = (γ β)e πi 24 / 36

38 Proof We compare V 2 with the analytic continuation 0 V1 to ω 0 2 Let us recall the Borel sums V 1 and 0 V2 of V 0 0 in ω 1 and ω 2 have the following forms, respectively: V 2 = log Γ( αη)γ( βη)γ( (γ α)η)γ( 1 + (γ β)η)γ2ηγ 1 2 Γ 2 (γη)α αη β βη (γ α) (γ α)η (γ β) (γ β)η 2πη V 1 = log Γ( αη)γ( βη)γ( (γ α)η)(β γ)(β γ)η γ 2γη 1 Γ( (β, γ)η)γ2 (γη)α αη β βη (γ α) (γ α)η η = 1 2 log Γ( αη)γ( βη)γ( (γ α)η)(γ β)(β γ)η γ 2γη 1 (β γ)ηπi Γ( (β, γ)η)γ2 (γη)α αη β βη (γ α) (γ α)η η 2 Subtracting V 2 from the analytic continuation 0 V1 to ω 0 2, we have V 1 0 V2 =1 0 2 log 2π Γ( (γ β)η)γ( 1 2 (γ β)ηπi + (β γ)η) 2 25 / 36

39 Proof We compare V 2 with the analytic continuation 0 V1 to ω 0 2 Let us recall the Borel sums V 1 and 0 V2 of V 0 0 in ω 1 and ω 2 have the following forms, respectively: V 2 = log Γ( αη)γ( βη)γ( (γ α)η)γ( 1 + (γ β)η)γ2ηγ 1 2 Γ 2 (γη)α αη β βη (γ α) (γ α)η (γ β) (γ β)η 2πη V 1 = log Γ( αη)γ( βη)γ( (γ α)η)(β γ)(β γ)η γ 2γη 1 Γ( (β, γ)η)γ2 (γη)α αη β βη (γ α) (γ α)η η = 1 2 log Γ( αη)γ( βη)γ( (γ α)η)(γ β)(β γ)η γ 2γη 1 (β γ)ηπi Γ( (β, γ)η)γ2 (γη)α αη β βη (γ α) (γ α)η η 2 Subtracting V 2 from the analytic continuation 0 V1 to ω 0 2, we have V 1 0 V2 =1 0 2 log 2π Γ( (γ β)η)γ( 1 2 (γ β)ηπi + (β γ)η) 2 25 / 36

40 Proof We compare V 2 with the analytic continuation 0 V1 to ω 0 2 Let us recall the Borel sums V 1 and 0 V2 of V 0 0 in ω 1 and ω 2 have the following forms, respectively: V 2 = log Γ( αη)γ( βη)γ( (γ α)η)γ( 1 + (γ β)η)γ2ηγ 1 2 Γ 2 (γη)α αη β βη (γ α) (γ α)η (γ β) (γ β)η 2πη V 1 = log Γ( αη)γ( βη)γ( (γ α)η)(β γ)(β γ)η γ 2γη 1 Γ( (β, γ)η)γ2 (γη)α αη β βη (γ α) (γ α)η η = 1 2 log Γ( αη)γ( βη)γ( (γ α)η)(γ β)(β γ)η γ 2γη 1 (β γ)ηπi Γ( (β, γ)η)γ2 (γη)α αη β βη (γ α) (γ α)η η 2 Subtracting V 2 from the analytic continuation 0 V1 to ω 0 2, we have V 1 0 V2 =1 0 2 log 2π Γ( (γ β)η)γ( 1 2 (γ β)ηπi + (β γ)η) 2 25 / 36

41 Proof We compare V 2 with the analytic continuation 0 V1 to ω 0 2 Let us recall the Borel sums V 1 and 0 V2 of V 0 0 in ω 1 and ω 2 have the following forms, respectively: V 2 = log Γ( αη)γ( βη)γ( (γ α)η)γ( 1 + (γ β)η)γ2ηγ 1 2 Γ 2 (γη)α αη β βη (γ α) (γ α)η (γ β) (γ β)η 2πη V 1 = log Γ( αη)γ( βη)γ( (γ α)η)(β γ)(β γ)η γ 2γη 1 Γ( (β, γ)η)γ2 (γη)α αη β βη (γ α) (γ α)η η = 1 2 log Γ( αη)γ( βη)γ( (γ α)η)(γ β)(β γ)η γ 2γη 1 (β γ)ηπi Γ( (β, γ)η)γ2 (γη)α αη β βη (γ α) (γ α)η η 2 Subtracting V 2 from the analytic continuation 0 V1 to ω 0 2, we have V 1 0 V2 =1 0 2 log 2π Γ( (γ β)η)γ( 1 2 (γ β)ηπi + (β γ)η) 2 25 / 36

42 Well-known Hence we obtain Γ( t)γ(1 2 t) = π cos πt, cos t = eit + e it 2 V 1 0 V2 =1 log(exp 2(γ β)ηπi + 1) 0 2 In the same way, we can compute the relations between V 1 and 1 V2 1 and V 1 and 2 V2 Hence we have 2 V 1 1 V2 =1 log(exp 2(γ β)ηπi + 1), 1 2 V 1 2 V2 = 1 log(exp 2(γ β)ηπi + 1) / 36

43 Parametric Stokes phenomena for the WKB solution Re γ > 0 fixed Reα-Reβ plane Re β ω 4 ω 1 ω 3 Re (β γ) ω 2 0 Re α 27 / 36

44 Parametric Stokes phenomena for the WKB solution Re γ > 0 fixed Reα-Reβ plane Re β ω 4 ω 1 ω 3 R 1 and R 2 : The Stokes regions Re (β γ) ω 2 0 Re α 0 1 R 1 a 0 R 2 a 0 a a a 0 a 1 (α, β, γ) = (05, 105, 1) in ω 1 (05, i, 1) (05, 095, 1) in ω 2 The WKB solutions are Borel summable in R 1 and R 2 (Koike and R Schäfke, [7] to appear) 27 / 36

45 Let ψ 1 and ψ 2 denote the Borel sum of the WKB solution: ψ = 1 Sodd exp( x a 0 S odd dx) in the Stokes region R 1 and R 2, respectively Theorem 4 We obtain the following relation between ψ 1 and ψ 2 ψ 1 = (1 + exp(2πi(γ β)η)) 1 2 ψ 2 28 / 36

46 Let ψ 1 and ψ 2 denote the Borel sum of the WKB solution: ψ = 1 Sodd exp( x a 0 S odd dx) in the Stokes region R 1 and R 2, respectively Theorem 4 We obtain the following relation between ψ 1 and ψ 2 ψ 1 = (1 + exp(2πi(γ β)η)) 1 2 ψ 2 Out of Proof We use the following relation: ψ = exp( V 0 )ψ (0), where ψ (0) = ( 1 x exp (S odd ηs 1 )dx ± η Sodd 0 x a 0 ) S 1 dx 28 / 36

47 Moreover we use the following result by Koike and Schäfke ([7] to appear): ψ (0),1 = ψ (0),2 and the following relation by Theorem 4: V 1 0 =V2 + 1 log(exp 2(γ β)ηπi + 1) 0 2 Here β γ = (γ β)e πi Hence we have ψ 1 = (exp( V 1 0 ))ψ(0),1 = (1 + exp(2πi(γ β)η)) 1 2 (exp( V 2 0 ))ψ(0),2 = (1 + exp(2πi(γ β)η)) 1 2 ψ 2 In the similar manner, we can compute other relations 29 / 36

48 Q : :invariant under involutions ι m (m = 0, 1,, 6) ι 0 : (α, β, γ) ( α, β, γ) ι 1 : (γ β, γ α, γ) ι 2 : (β, α, γ) ι 3 = ι 1 ι 2 : (γ α, γ β, γ) ι 4 = ι 0 ι 2 : ( β, α, γ) ι 5 = ι 0 ι 1 : (β γ, α γ, γ) ι 6 = ι 0 ι 1 ι 2 : (α γ, β γ, γ) ω hm = ι m (ω h ) : Images in ω h by ι m Here, h = 1, 2, 3, 4, m = 0, 1,, 6 30 / 36

49 Reα-Reβ planes Re β Re β ω 4 ω 1 ω 3 ω 46 ω 16 ω 36 ω 41 ω 32 ω 44 ω 35 ω 2 ω 26 ω 11 ω 21 ω 22 ω 12 ω 14 ω 24 ω 25 ω 15 ω 23 ω 20 Reα ω 31 ω 42 ω 34 ω 45 Reα ω 33 ω 13 ω 43 ω 30 ω 10 ω 40 Re γ > 0 Re γ < 0 31 / 36

50 Reα-Reβ planes Re β Re β ω 4 ω 1 ω 3 ω 46 ω 16 ω 36 ω 41 ω 32 ω 44 ω 35 ω 2 ω 26 ω 11 ω 21 ω 22 ω 12 ω 14 ω 24 ω 25 ω 15 ω 23 ω 20 Reα ω 31 ω 42 ω 34 ω 45 Reα ω 33 ω 13 ω 43 ω 30 ω 10 ω 40 Re γ > 0 Re γ < 0 31 / 36

51 0 1 R 1 a 0 R 2 a 0 a a a 0 a 1 (α, β, γ) = (05, 105, 1) in ω 1 (05, i, 1) (05, 095, 1) in ω 2 ι 1 : (α, β, γ) (γ β, γ α, γ) 32 / 36

52 0 1 R 1 a 0 R 2 a 0 a a a 0 a 1 (α, β, γ) = (05, 105, 1) in ω 1 (05, i, 1) (05, 095, 1) in ω 2 ι 1 : (α, β, γ) (γ β, γ α, γ) 0 1 R 11 a 0 R 21 a 0 a a a 0 a 1 (α, β, γ) = ( 005, 05, 1) in ω 11 ( 001i, 05, 1) (005, 05, 1) in ω 21 The WKB solutions are Borel summable in R 11 and R 21 The top and bottom Stokes graphs are same graphs, respectively 32 / 36

53 Let ψ 11 and ψ 21 denote the Borel sum of the WKB solution: ψ = 1 Sodd exp( x a 0 S odd dx) in the Stokes region R 11 and R 21, respectively Theorem 4 We obtain the following relation between ψ 1 and ψ 2 ψ 1 = (1 + exp(2πi(γ β)η)) 1 2 ψ 2 ι 1 : (α, β, γ) (γ β, γ α, γ) Theorem 5 We obtain the following relation between ψ 11 and ψ 21 ψ 11 = (1 + exp(2πiαη)) 1 2 ψ / 36

54 References [1] Aoki, T, Takahashi, T and Tanda, M, Exact WKB analysis of confluent hypergeometric differential equations with a large parameter, in preparation [2] Aoki, T and Tanda, M, Characterization of Stokes graphs and Voros coefficients of hypergeometric differential equations with a large parameter, RIMS Kôkyûroku Bessatsu B40 (2013), [3] Aoki, T and Tanda, M, Some concrete shapes of degenerate Stokes curves of hypergeometric differential equations with a large parameter, J School Sci Eng Kinki Univ, 47 (2011), 5 8 [4] Aoki, T and Tanda, M, Borel sums of Voros coefficients of hypergeometric differential equations with a large parameter, to appear in RIMS Kôkyûroku [5] B Candelpergher, M A Coppo and E Delabaere, La sommation de Ramanujan, L Enseignement Mathématique, 43 (1997), / 36

55 [6]Kawai, T and Takei, Y, Algebraic Analysis of Singular Perturbation Theory, Translation of Mathematical Monographs, vol 227, AMS, 2005 [7]Koike,T and R Schäfke, On the Borel summability of WKB solutions of Schrödinger equations with polynomial potential and its application, to appear in RIMS Kôkyûroku Bessatsu [8] Koike T and Takei Y, On the Voros coefficient for the Whittaker equation with a large parameter, Some progress around Sato s conjecture in exact WKB analysis, Publ RIMS, Kyoto Univ 47 (2011), [9] Tanda, M Exact WKB analysis of hypergeometric differential equations, to appear in RIMS Kôkyûroku Bessatsu [10] Takei Y, Sato s conjecture for the Weber equation and transformation theory for Schrödinger equations with a merging pair of turning points, RIMS Kôkyûroku Bessatsu B10 (2008), / 36

56 END Thank you for your attention 36 / 36