Exceptional solutions to the Painlevé VI equation associated with the generalized Jacobi weight

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1 Random Matrices: Theory and Applications Vol. 6, No. (07) ( pages) c World Scientific Publishing Company DOI: 0.4/S Exceptional solutions to the Painlevé VI equation associated with the generalized Jacobi weight Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. Shulin Lyu and Yang Chen Department of Mathematics, University of Macau Avenida da Universidade, Taipa, Macau, P. R. China lvshulin989@6.com yangbrookchen@yahoo.co.uk Received 7 October 06 Revised December 06 Accepted 4 December 06 Published 6 February 07 We consider the generalized Jacobi weight x α ( x) β x t γ, x 0,, t(t ) > 0, α>, β >, γ R. As is shown in D. Dai and L. Zhang, Painlevé VI and Henkel determinants for the generalized Jocobi weight, J. Phys. A: Math. Theor. 4 (00), Article ID:05507, 4pp., the corresponding Hankel determinant is the τ-function of a particular Painlevé VI. We present all the possible asymptotic expansions of the solution of the Painlevé VI equation near 0, and for generic (α, β, γ). For four special cases of (α, β, γ) which are related to the dimension of the Hankel determinant, we can find the exceptional solutions of the Painlevé VI equation according to the results of A. Eremenko, A. Gabrielov and A. Hinkkanen, Exceptional solutions to the Painlevé VI equation, preprint (06), arxiv: , and thus give another characterization of the Hankel determinant. Keywords: Hankel determinants; asymptotic expansions; Painléve equations. Mathematics Subject Classification 00: 47B5, 4E05, 4M55. Introduction The Hankel determinant plays an important role in random matrix theory 8. It may represent the moment generating function of linear statistics, the partition function, the largest or smallest eigenvalue distribution function of an ensemble of matrices, etc. See, for instance,, 6. We can investigate Hankel determinants by employing ladder operators and orthogonal polynomials (see e.g. 5, 6, 7), or by applying linear statistics technique based on Dyson s coulomb fluid method (refertoe.g., 4, 7, 8)

2 S. Lyu & Y. Chen Define a unitary ensemble of Hermitian matrices M =(M ij ) N N with probability density p(m)dm e tr v(m) vol(dm), Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. vol(dm) = N dm ii i= j<k N d(re M jk )d(im M jk ). Here v(m) is a matrix function 4 defined via Jordan canonical form and vol(dm) is called the volume element 5. The joint probability density function of the eigenvalues {x j } N j= of this ensemble is given in 8 by D N N! i<j N x j x i N w(x k ), (.) where w(x) =e v(x) is a positive weight function supported on a, b with finite moments µ k := b a k= x k w(x)dx, k =0,,,. Here, the normalization constant D N is defined by D N := x j x i N w(x k )dx k. N! a,b N i<j N We consider the generalized Jacobi weight w(x; t) =x α ( x) β x t γ, x 0,, with α>, β>, γ R, andt(t ) > 0. It arises in the application of multiuser multiple-input multiple-output antenna wireless communication systems 0 and is studied in. See also 9. Define monic polynomials P n (x; t) orthogonal to w(x; t) as follows: h n (t)δ mn := 0 k= P m (x; t)p n (x; t)w(x; t)dx, m, n =0,,,..., P n (z; t) :=z n + p (n, t)z n +. Then, it is shown in 9 thatd N = D N (t) can be evaluated as the determinant of the Hankel (or moment) matrix and also as the product of {h n (t)} 0 n N,thatis, D N (t) := x j x i N w(x k ; t)dx k N! 0, N i<j N k= ( ) N =det x i+j w(x; t)dx 0 i,j=0 = N n=0 h n (t)

3 Exceptional solutions to the Painlevé VI equation In addition, Heine s formula gives us P N (z; t) = D N (t) N! 0, N i<j N The next proposition is established in. x j x i N (z x k )w(x k ; t)dx k. k= Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. Proposition.. Define (t )R N (t) W N (t) := N ++α + β + γ +, R N (t) := β PN (y; t)w(y; t) dy h N 0 y. Then W N (t) satisfies the Painlevé VIequation(see (.) below) with (N ++α + β + γ) µ =, µ = α, µ = β, µ 4 = γ. The Painlevé VI(P VI ) equation has the following form: d y dt = ( y + y + )( ) ( dy y t dt t + t + ) dy y t dt ( y(y )(y t) + t (t ) µ + µ t y + µ (t ) (y ) + µ ) 4t(t ) (y t). (.) Eremenko et al. have found all exceptional solutions of P VI,whichisgiven later by Proposition.. Recall that a solution y(t) is called exceptional if y(t) {0,,,t} for all t C\{0, }. Based on their results, our goal is to obtain exceptional solutions for W N (t) and thus propose explicit form or asymptotic expressions near 0, ±, ford N (t). This paper is organized as follows. All possible asymptotic expansions of W N (t) near 0, and for generic (α, β, γ) are developed in Sec.. We devote Sec. to a discussion of the exceptional solutions for W N (t) for four special cases of (α, β, γ).. The Behavior of W N (t) Near0,, and for Generic (α, β, γ) According to the definitions of W N and R N we have (t )R N (t) W N (t) := N ++α + β + γ +, R N (t) := β h N 0 P N (y)w(y; t) dy y, (i) for β =0, R N =0, ; (ii) for γ = 0, it is clear that P N (x) are the Jacobi polynomials

4 S. Lyu & Y. Chen Hence, here and in the subsequent sections, we assume N ++α + β + γ 0, β 0,γ 0,andN. As is given in Proposition., W N satisfies a particular P VI. By multiplying both sides of this equation by t (t ) W N (W N )(W N t) togetridofthe numerator, we come to t (t ) W N (W N )(W N t)w N + t (t ) (WN (t +)W N + t) (W N ) t(t )W N (W N )((t )W N t )W N Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. +µ WN (W N ) (W N t) + µ t(w N ) (W N t) + µ (t )WN (W N t) + µ 4 t(t )WN (W N ) =0, (.) ( ) with (µ,µ,µ,µ 4 )= (N++α+β+γ), α, β, γ. We first develop the asymptotic expansion of W N as t 0. Putting t = 0 in the non-derivative term of (.) gives rise to finite solutions for W N (0). From this fact and the series expansions for W N near 0 given by Corollaries. and.5, we realize that W N (t) has this formal expansion near 0 which starts from a 0 t 0 and for consistency the higher terms are of the form a j t j, i.e. a j t j, t 0. j=0 On substituting it into the left-hand side of (.) and setting the coefficients of its series expansion at 0 to be 0 term by term, we can find a 0, a, a, a,etc. Specifically, the first term gives us which has three solutions a 4 0(N ++α + β + γ)a 0 (N ++α +β + γ) (N ++α + β + γ)a 0 (N ++α + γ) = 0, a 0 = N ++α +β + γ N ++α + β + γ, N ++α + γ N ++α + β + γ, 0. Now we discuss these three cases of a 0 one by one. Note that the first, second, third, or fourth term which will appear in the following discussion refers to the corresponding term of the series expansion of the left-hand side of (.) at 0. () If a 0 = N++α+β+γ N++α+β+γ 0, then the second term gives us 4 +N + α +β + γ +N + α +β + γ a =

5 Exceptional solutions to the Painlevé VI equation (a) In case γ = 4 N α β, wehave β + + a t + a (β +)t + (β +) a β + α + β + ) t + O(t 4 ), a R. ( ( ) (β + N) (N + α + β)+ 4 Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. (b) In case γ = N α β, wehave β + a t + a ( β)t + ( β) a + +α ) t + O(t 4 ), a R. β ((β + N)((N + α + β)+) 4( β) (c) In case γ 4 N α β or N α β, wehavea =0and the third term gives us 5 +N + α +β + γ +N + α +β + γ a =0. (i) For γ = 5 N α β, wehave β + + a t β β + ( ) 6(β + N) (N + α + β)+ +α + t 5 + a (β +)t 4 + O(t 5 ), a R. (ii) For γ = N α β, wehave β + a t + β 6(β + N)((N + α + β)+)+α t 5( β) + a ( β)t 4 + O(t 5 ), a R

6 S. Lyu & Y. Chen (iii) For γ 5 N α β or N α β, wehavea =0and the fourth term gives us where δ a + δ =0, (.) Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. δ := (N + α +β + γ)(n ++α +β + γ), β(n + β)(n ++α + β + γ)+(α + γ)( + γ) δ :=. N ++α + β + γ (A) In case δ =0,from(.) weseethatδ =0.Solvingforγ from δ = 0 and inserting it into δ =0,weget(N +β)(n +α+β) =0 or (N ++β)(n ++α + β) = 0, which is impossible for α>, β>, and N. (B) In case δ 0,wehave N ++α +β + γ N ++α + β + γ δ δ t + O(t 4 ). () If a 0 = N++α+γ N++α+β+γ 0, then the second term gives rise to 4 +N + α + γ +N + α + γ a =0. (a) In case γ = 4 N α, wehave β + a t + a ( β)t + ( β) a + ( + a ( β) a +(β +) +β + β (9N(N + α)+6(n + α)+) t 4( β) ) t 4 + O(t 5 ), a R. N(N + α)+n + α (b) In case γ = N α, wehave β + + a t + a (β +)t + (β +) a ( + a (β +) a +( β) β 6 β (9N(N + α)+(n + α) ) t 4(β +) ) t 4 + O(t 5 ), a R. N(N + α)+n + α

7 Exceptional solutions to the Painlevé VI equation Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. (c) In case γ 4 N α or N α, wehavea =0andthethird term gives rise to 5 +N + α + γ +N + α + γ a =0. (i) For γ = 5 N α, wehave β + a t β 8 + N(N + α)+(n + α)+ t β 5 + a ( β)t 4 + a (6β +) t 5 + O(t ), a R. +5β + (ii) For γ = N α, wehave ( ) 6 N(N + α)+n + α 5 β + + a t β 8N(N + α)+(n + α) t 5(β +) + a (β +)t 4 a (6β )(6N(N + α)+n + α) 5 +β +t 5 + O(t ), a R. (iii) For γ 5 N α or N α, wehavea = 0 and the fourth term gives rise to where δ a δ 4 =0, (.) δ := (N + α + γ)(n ++α + γ), βn(n +)+(N ++γ)(α + γ) δ 4 :=. N ++α + β + γ (A) If δ =0,thenfrom(.) we readily get δ 4 = 0. A combination of δ =0andδ 4 =0givesα = γ = N or α = γ = N, which is impossible for α>, β>, and N. (B) If δ 0,thenwehave N ++α + γ N ++α + β + γ + δ 4 t + O(t 4 ). δ () If a 0 = 0, then the second term yields where a (δ 5 a + a )=0, δ 5 := 9 (N + α + γ)(n ++α +β + γ)+9β

8 S. Lyu & Y. Chen (a) In case a 0,wehave a t δ5 a t + O(t). (b) In case a = 0, the third term yields a (9α 9γ +) =0. a Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. (i) For a 0,wehave a t + (9α 9γ +)t + ( 9 8 (N + α + γ)(n ++α +β + γ)+ 9 4 β a (9(α + γ) )(9(α γ) ) (ii) For a = 0, the fourth term yields (A) If γ = ±α, α 0,wehave (B) If γ ±α, wehave (α γ)a α(α + γ)a α =0. t + O(t 4 ). αt α γ + O(t 4 ), t O(t ). ) a or αt α + γ + O(t 4 ). Now we investigate the behavior of W N as t by an argument analogous to the one used for W N as t 0. Replacing W N (t) byty (t) andt by T in the non-derivative term of (.), and setting T = 0, we obtain finite solutions for Y. This indicates that series expansion for W N (t) starts from B t.weplug B t + B t + B t b j +, t, j=0 t j into the left-hand side of (.) and take its series expansion at. The first term suggests N ++α + β N ++α + β +γ B B B 4 N ++α + β + γ N ++α + β + γ =

9 Exceptional solutions to the Painlevé VI equation (4) If B = N++α+β N++α+β+γ 0,then N ++α + β N ++α + β + γ t + γ(n(n +)+(N ++β)(α + β)) (N ++α + β + γ)(n + α + β)(n ++α + β) + O(t ). Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. (4) If B = N++α+β+γ N++α+β+γ 0, then the second term suggests + N + α + β + γ + N + α + β + γ B =0. (5a) In case γ = N α+β,wehave t N ++ (α + β) + B t B N ++ (α + β) t + O(), B R. (5b) In case γ = N α+β,wehave t N ++ (α + β) + B t + B N ++ (α + β) t + O(), B R. (5c) In case γ N α+β or N α+β,wehaveb =0andthe third term suggests N + α + β + γ 6 + N + α + β + γ B =0. (i) For γ = 5 6 N α+β,wehave t N + + (α + β) + B t (9(α β )+5) (6N +5+(α + β)) + 6N ++(α + β) B 0(6N ++(α + β)) 4t 9 0 (α β )(N ++α + β)+ B + O(t ), B R. 6 t

10 S. Lyu & Y. Chen (ii) For γ = 6 N α+β,wehave t N (α + β) + B t Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. (9(α β )+5) (6N ++(α + β)) + +6N +5+(α + β) B 0(6N +5+(α + β)) 4t (α β )(N + α + β) B + O(t ), B 6 t R. (iii) For γ 5 6 N α+β or 6 N α+β,wehaveb =0andthe fourth term suggests δ 6 b 0 + δ 7 =0, (.4) where δ 6 := (N + α + β +γ)(n ++α + β +γ), γ(n ++β +γ)(α + β)+(n + γ)(n ++γ) δ 7 :=. N ++α + β + γ (A) If δ 6 = 0, i.e. γ = N α+β or N α+β,then(.4) gives us δ 7 = 0. On substituting γ into δ 7 = 0, we find β = α.the asymptotic expansions of W N for these four cases of (α, β, γ) are given by the following: (A) For (α, β, γ) =(α, α, N α), we have t N + α + b 0 + (N + α)b 0 +(N + α +)b 0 + (α )((N + α)+)(n + α +) 6(N + α) + O(t ), b 0 R. (A) For (α, β, γ) =(α, α, N α), we have t N + α + + b 0 + t (N + α +)b 0 (N + α)b 0 (α )((N + α)+)(n + α) 6(N ++α) (A) For (α, β, γ) =(α, α, N), we have t N + b 0 + Nb 0 +(+N)b 0 t + O(t ), b 0 R. +(α ) (N +)(N +) 6N t + O(t ), b 0 R

11 Exceptional solutions to the Painlevé VI equation Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. (A4) For (α, β, γ) =(α, α, N), we have (B) If δ 6 0,wehave t N + + b 0 + (α ) (N +)b 0 Nb 0 N(N +) 6(N +) t + O(t ), b 0 R. N ++α + β +γ N ++α + β + γ t δ 7 + O(t ). δ 6 (5) If B = 0, then the second term suggests B (9(N ++α + β)(n ++α + β +γ)+)+b B =0. (6a) In case B 0,wehave B t B 9(N ++α + β)(n ++α + β +γ)+t + O(). (6b) In case B = 0, the third term suggests B 0 (9α 9β +) B =0. (i) For B 0,wehave B t + (9α 9β +) ( (N ++α + β)(n ++α + β +γ) + (9(α + β) )(9(α β) ) 6B t (ii) For B = 0, the fourth term suggests (α + β)b 0 α(α β)b 0 α =0. (A) If β = ±α, α 0,thenwehave + O(t ). ) B (B) If β ±α, thenwehave or + O(t ). α α β + O(t ), α α + β + O(t )

12 S. Lyu & Y. Chen Finally, we shall adopt the similar procedure as in the preceding two cases to establish the expressions for W N as t +. Since setting t = in the non-derivative term of (.) yields finite solutions for W N (), we find that the expansion formula for W N (t) should start from c 0 (t ) 0. On inserting c j (t ) j, t +, j=0 into the left-hand side of (.), we find from the first term Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. (N ++α + β + γ)c 0 α(n ++α + β + γ)c 0 + α(c 0 ) 4 =0. α (7) If c 0 = N++α+β+γ, i.e. N ++β + γ 0, then the second term implies 4 α +N + β + γ +N + β + γ c =0. (7a) In case γ = 4 N β, wehave α α + c (t ) + c (α )(t ) + c (α ) α (9N(N + β)+6(n + β)+) (t ) 4(α ) ( + c 4 (α ) + κ ) c (t ) c 5 (α )4 +(α ) + O((t ) ), c R, ( 5κ ) c (t ) 5 where κ =(N + )((N + β) + )(α +). (7b) In case γ = N β, wehave α α + + c (t ) c (α +)(t ) + c (α α (9N(N + β)+(n + β) ) +) 4(α +) + c c (α +) + κ 6 + (t ) 4 + c (α +) c (α +) 5κ 4 (t ) O((t ) ), c R, (t ) where κ := (N + )((N + β) + )(α )

13 Exceptional solutions to the Painlevé VI equation (7c) In case γ 4 N β or N β, andα =0,wehave O((t ) ). Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. (7d) In case γ 4 N β or N β, andα 0,wehavec =0and the third term implies 5 +N + β + γ +N + β + γ c =0. (i) For γ = 5 N β, wehave α α + c (t ) α α 8 N(N + β)+(n + β)+ (t ) + c 5 (α )(t ) 4 c (6N + 5)(6N +6β + 5)(6α +) 5 0 (t ) 5 + O((t ) ), c R. (ii) For γ = N β, wehave α α + + c (t ) α α + ( ) 8 N(N + β)+(n + β) (t ) 5 c (α +)(t ) 4 c + (6N + )(6N +6β + )(6α ) 0 +5(t ) 5 + O((t ) ), c R. (iii) For γ 5 N β or N β, wehavec = 0 and the fourth term implies where δ 8 c δ 9 =0, (.5) δ 8 := (N ++β + γ)(n + β + γ), δ 9 := α(n(n +)+(N ++γ)(β + γ)). (A) If δ 8 =0,thenEq.(.5) givesδ 9 =0.Solvingforβ and γ from δ 8 =0andδ 9 =0yieldsβ = γ = N or β = γ = N. This is impossible for β> andn. (B) If δ 8 0,thenwehave α N ++α + β + γ + δ 9 δ 8 (t ) + O((t ) 4 )

14 S. Lyu & Y. Chen (8) If c 0 = α N++α+β+γ andc 0 0, i.e. N ++α + β + γ 0andα 0, then the second term implies 4 +N +α + β + γ +N +α + β + γ c =0. Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. (8a) In case γ = 4 N α β, wehave α α + + c (t ) c (α +)(t ) + c (α α +) ((N + α)(n +4+α +β) 4(α +) +(β +)) (t ) + c c (α +) + 9 α (N ++α + β) + α (N(N + β)+n + β) (N + )(N +β +) 6 + (t ) 4 + c c (α +) 4 45α ((N + α)(n +4+α 8 +β)+β +)+ α (5(N + β) + 6) (N + )(N +β +) (t ) O((t ) ), c R. (8b) In case γ = N α β, wehave α α + c (t ) + c (α ) (t ) + c (α ) α ((N + α)(n ++α +β)+β ) 4(α ) (t ) + a c (α ) 9 α (N ++α + β) α (N(N + β)+(n + β)+) 6 (N + )(N +β +)+ (t ) 4 + c c (α ) α ((N + α)(n ++α +β)+β) + α 5 (5(N + β) + 4) (N + )(N +β +) 6 (t ) 5 + O((t ) ), c R

15 Exceptional solutions to the Painlevé VI equation (8c) In case γ 4 N α β or N α β, wehavec =0and the third term implies 5 +N +α + β + γ +N +α + β + γ c =0. (i) For γ = 5 N α β, wehave Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. α α + + c (t ) α α + 6 (N + α)(n +5+α +β)+β + (t ) 5 c (α +)(t ) 4 + O((t ) 5 ), c R. (ii) For γ = N α β, wehave α α + c (t ) α 6(N + α)(n ++α +β)+β (t ) 5(α ) + c (α )(t ) 4 + O((t ) 5 ), c R. (iii) For γ 5 N α β or N α β, wehavec =0and the fourth term implies where δ 0 c + δ =0, (.6) δ 0 := (N +α + β + γ)(n ++α + β + γ), δ := α(n + α)(n ++α + β + γ)+(β + γ)(γ +). (N ++α + β + γ) (A) We find, due to (.6), that δ 0 = 0 implies δ =0.Solvingforγ from δ 0 = 0 and plugging it into δ =0,weobtain(N ++α)(n + +α + β) =0or(N + α)(n + α + β) = 0, which is impossible for α>, β>, and N. (B) If δ 0 0,thenwehave α N ++α + β + γ δ (t ) + O((t ) 4 ). δ

16 S. Lyu & Y. Chen (9) If c 0 =, then the second term implies ( ) 9 δ +5 c c c =0, where δ := (N + β + γ)(n ++α + β + γ)+α. Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. (9a) In case c 0,wehave ( ) 9 +c (t ) + δ +5 c (t ) + O(t ). (9b) In case c = 0, the third term implies c (9β 9γ +) =0. Hence we have (i) c 0and c +c (t ) + (9β 9γ +)(t ) + (9δ + ) c 8 + (9(β γ) )(9(β + γ) ) 6c (t ) (β γ )(δ +5)+ 6 + O((t ) ); or, (ii) c = 0 and the fourth term implies so that, we find (A) for γ = ±β, (β + γ)c β(β γ)c β =0, c (t ) 5 (B) for γ ±β, or + (t ) + O((t ) 4 ); + + β β γ (t ) + O((t ) 4 ), β β + γ (t ) + O((t ) 4 )

17 Exceptional solutions to the Painlevé VI equation. The Hankel Determinant for Four Cases of (α, β, γ) Define ( (α 0,α,α,α )= µ, µ,µ, ) µ 4. (.) The following result is proved in. Proposition.. The complete list of exceptional solutions of P VI is the following: Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. () If α j =0, 0 j, they are Picard s solutions with real (µ, ν) R \Z. () If α 0 = α, α = α, then there is a solution y(t) = t. () If α 0 = α, α = α, then there is a solution y(t) =+ t. (4) If α 0 = α, α = α, then there is a solution y(t) =t + t t. (5) If α 0 =9α =9α =9α 0, then there is a unique solution defined by y 4 4ty 4y +6ty t =0. (.) (6) If 9α 0 = α =9α =9α 0, then there is a unique solution defined by y 4 6ty +4t(t +)y t =0. (7) If 9α 0 =9α = α =9α 0, then there is a unique solution defined by y 4 4y +6ty 4t y + t =0. (8) If 9α 0 =9α =9α = α 0, then there is a unique solution defined by y 4 4ty +6ty 4ty + t =0. Remark. We present a method of checking the solutions y(t) for the eight cases. We take the fifth case as an example, i.e. (α 0,α,α,α )=(9α,α,α,α ),α 0. On the one hand, we substitute {α 0 =9α, α = α, α = α } into the nonderivative term of (.) toget y(y )(y t) t (t ) ( 9 t y + t t( t) + (y ) (y t) ) α + t(t ) (y t). To get solution (.) which is independent of α from Eq. (.), the coefficient of α in the above term has to be 0, which gives (.). On the other hand, if y(t)satisfies(.), one of the four solutions of this algebraic equation has series expansion at t =as y(t) =+ ɛ + ɛ + ɛ 4 + ɛ 5 ɛ O(ɛ 7 ), ɛ = t 0 +. Inserting this into (.) and comparing the series expansions of both sides to yield α 0 =9α =9α =9α, which is the condition of the fifth case

18 S. Lyu & Y. Chen For our problem, we have ( ) (N ++α + β + γ) (α 0,α,α,α )=, α, β, γ. (.) Applying the above proposition, we find that there are four cases of (α, β, γ) where there are explicit exceptional solutions for W N (t). To do this, we need the next lemma. Lemma.. Let t (, 0) (, ), β> and β 0. Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. (i) If (t )β/(n ++α + β + γ) < 0, then (t )β W N (t) < + N ++α + β + γ <. (ii) If (t )β/(n ++α + β + γ) > 0, then Proof. Recalling that we have, for t D and β 0, so that RN (t) β W N (t) > + R N (t) β 0 R N (t) :=β P N = >. Noting that we complete the proof. W N (t) := = (t )β N ++α + β + γ >. (y)w(y; t) dy y 0 P, N (y)w(y; t)dy 0 P N (y; t)w(y; t) ydy y 0 P > 0, N (y; t)w(y; t)dy (t )R N (t) N ++α + β + γ + (t )β N ++α + β + γ RN (t) +, β In order to study the behavior of D N (t), we recall the following result of. Proposition.. Let H N (t) :=t(t ) d dt ln D N(t). Then the quantity H(t) defined by H(t) :=H N (t)+ d t + d,

19 Exceptional solutions to the Painlevé VI equation with d := N(N + α + β + γ) (α + β), 4 d := N(N + α + β + γ)+β(α + β) γ(α β), 4 satisfies the Jimbo Miwa Okamoto σ-form of Painlevé VI t (t ) H ( H ) +( ( H ) +(t H H) H ν ν ν ν 4 ) =( H + ν )( H + ν )( H + ν )( H + ν 4) Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. with ν = α + β, ν = β α, ν = N + α + β, ν 4 = N + γ + α + β. From it follows the equation for H N (t), which is mentioned but not presented in. Corollary.4. This second-order differential equation for H N (t) holds (t ) t (H N ) +(th N H N ) 4H N (N + α + β) 4Nγ +(th N H N ){ 4(H N) +(N +(N + α)(α + β + γ) βγ)h N +Nγ(β α)(n + α + β + γ)} (α + γ)h N Nγ(N + α + β + γ) =0. To achieve the relationship between H N (t) andw N (t), we make use of certain results of. Lemma.5. The quantity H N (t) is expressed in terms of W N (t) and W N (t) as H N (t) = 4(t W N )(W N )W N α t t (t ) (W N ) +t(t )(W N )W N W N +(N + α + β + γ + )(N + α + β + γ )W 4 N t(4n +(α + β + γ)(4n + α + β)) +N +(N + α + γ)(α + β + γ) WN +t ((N + α + β) +4Nγ) + t(8n +4(N + α)(α + β + γ)+βγ)+(α + γ) WN tt(n(n + α + β + γ)+α(α + β)) + α(α + γ)w N. Proof. First, as is pointed out by (5.) of, we can express r N and r N by R N and R N. Then, we combine Eqs. (.5) and (.45) of to deduce the expression of t d dt ln D N in terms of R N and R N. Finally, on account of the definition of H N (t) and W N (t), we come to the desired result

20 S. Lyu & Y. Chen.. Case : (α, β, γ) =(α, α, N α), α 0 In this case, we have ( (N + (α 0,α,α,α )= + α), α, α, (N + + α) which corresponds to the fourth case of Proposition., so that there are two possible exceptional solutions for W N (t), i.e. W N (t) =t ± t t. ), Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. Since β = α and N ++α + β + γ = N + + α>0, we conclude from Lemma. that W N (t) has the same sign with (t )α. We use this fact to determine the exact form of W N (t). Lemma.6. Define the two choices of W N (t) as Then W () I (t) :=t + t t = t W () I (t), W () I (t) :=t t t = t W () I (t). (i) for <α<0 and t<0, we have (t )α >0, W () I < 0, and W () I < 0; (ii) for <α<0 and t>, we have (t )α <0, W () I > 0, and W () I < 0; (iii) for α>0 and t<0, we have (t )α <0, W () I < 0, and W () I < 0; (iv) for α>0 and t>, we have (t )α >0, W () I > 0, and W () I < 0. By plugging the chosen W N (t) into the expression of H N (t) in Lemma.5 and using the fact that H N (t) =t(t ) d dt ln D N (t), we obtain H N (t) andd N (t). We summarize this as a theorem. Theorem.7. For (α, β, γ) =(α, α, α N ),α 0, the quantities W N (t), H N (t), and D N (t) are given as follows: (i) For <α<0 and t<0, there are no exceptional solutions for W N (t). (ii) For <α<0 and t>, we have W N (t) =t t t, H N (t) = N 4 (N +) N (N +)t + N (N +α) t t, D N (t) =d ( t + t ) N(N+α) (t t) N 4 (N+)

21 Exceptional solutions to the Painlevé VI equation (iii) For α>0 and t<0, we have W N (t) =t + t t, H N (t) = N 4 (N +) N (N +)t N (N +α) t t, or D N (t) =d ( t + t) N(N+α) (t t) N 4 (N+), W N (t) =t t t, Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. H N (t) = N 4 (N +) N (N +)t + N (N +α) t t, d D N (t) = ( t + t) N(N+α) (t t). N 4 (N+) (iv) For α>0 and t>, we have W N (t) =t + t t, H N (t) = N 4 (N +) N (N +)t N (N +α) t t, d 4 D N (t) = ( t + t ) N(N+α) (t t). N 4 (N+) (v) It turns out with these choices (ii) (iv) of α and t H N (t) = N ( 4 (N +)+N α ) t N (N +α)w N (t). Here {d i, i =,,, 4} are t independent constants. Remark. The case α =, i.e. (α, β, γ) =(,, N), corresponds to the Hilbert series of N = SQCD with SO(N) gauge group with N f = N flavors. The case α =, i.e. (α, β, γ) =(,, N ), corresponds to the Hilbert series of N = SQCD with Sp(N) gauge group with N f = N+ flavors, N odd. Paper provides a detailed account of the moduli space of suppersymmetric QCD... Case : (α, β, γ) =(α, α, N ), <α<,α 0 We have ( (N + (α 0,α,α,α )= ), α, α, (N + ) ), which again belongs to the fourth case of Proposition.. (t )β As N++α+β+γ = (t )α, it follows from Lemma. that W N+ N (t) has opposite sign with (t )α. Owing to this fact, we can apply Lemma.6 to find the

22 S. Lyu & Y. Chen explicit form of W N (t). Then, in view of Theorem.7, weobtainh N (t) andd N (t) instantly. Theorem.8. In case (α, β, γ) =(α, α, N ), <α<,α 0, we find (i) for <α<0 and t<0, W N (t) =t + t t, Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. or D N (t) =d ( t + t) N(N+α) (t t) N 4 (N+), W N (t) =t t t, d D N (t) = ( t + t) N(N+α) (t t) ; N 4 (N+) (ii) for <α<0 and t>, W N (t) =t + t t, d 4 D N (t) = ( t + t ) N(N+α) (t t) ; N 4 (N+) (iii) for α>0 and t<0, there are no exceptional solutions for W N (t); (iv) for α>0 and t>, W N (t) =t t t, ( t + t ) N(N+α) D N (t) =d ; (t t) N 4 (N+) (v) as what we have expected, H N (t) = N ( 4 (N +)+N α ) t N (N +α)w N (t)... Case : (α, β, γ) =(N +,N +, N ) In view of (.), we have, for α = N +, ( ) 9 (α 0,α,α,α )= α, α, α, α, which corresponds to the fifth case of Proposition., sothatw N (t) satisfies(.) with t D, D := (, 0) (, ), i.e. W 4 N 4(t +)W N +6tW N t =0. (.4) Thus, on account of Lemma., weget W N ( ) = +, W N () =

23 Exceptional solutions to the Painlevé VI equation And, for t D\{, }, W N (t) isgivenby W N (t) = t + ± a A (t) ± b A (t) ± a A (t), where the index a or b indicates the dependence of the sign, the meaning of which will become clear below. Here A (t) := 9 (t t +) 6 t (t ) > 0, Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. A (t) := 9 (t t +)+ t (t ) 6, A (t) := (t +)(t )(t ). 7A (t) For ease of notations, we shall not display the t dependence of A i unless we have to. To find real solutions for W N (t), we choose the sign in ± a to make A ± a A non-negative. We then apply Lemma. to determine the sign ± b in the chosen solution. And finally, we are led to the following conclusion. Theorem.9. We have (i) for t<, W N (t) = t + (ii) for <t<0, (iii) for <t<, (iv) for t>, W N (t) = t + W N (t) = t + A A A := W () III (t); + A A + A := W () III (t); A + A A := W () III (t); W N (t) = t + + A + A + A := W (4) III (t). Proof. Since A ( ) = A () = 0, we know that A (t) makes no sense when t = ort =. However, we can study the limit of A (t) ast ort. To continue, we rewrite A (t) as A (t) = (t )(t ). 7 A(t) t

24 S. Lyu & Y. Chen By taking the series expansion of A (t) att =, we find Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. A (t) csgn(t +) (t +), 6 t, so that A (t) t + csgn(t +), t. 6 Here the csgn function is defined by {, z > 0, csgn(z) =, z < 0, for z R. Hence we have, as t, and, due to A ( ) =, Similarly, we can show that, for t, A (t) csgn(t +), A (t) ± A (t) ± csgn(t +). A (t) ± A (t) ± csgn(t ). From Fig., weseethata (t) +A (t) anda (t) A (t) are non-negative according as t (, 0) (, ) andt (, ) (, ). Therefore, to find real solutions for W N (t), we choose W () () III or W III,fort (, ) (, ), and W () (4) III or W III,fort (, 0) (, ). (a) (b) Fig.. A (t) ± A (t)

25 Exceptional solutions to the Painlevé VI equation Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. (a) Fig.. W N t+. β Since N++α+β+γ = > 0, it follows from Lemma. that W N (t) t+ < 0 for t<0andw N (t) t+ > 0fort>. With the aid of Fig., we apply this fact to arrive at our conclusion. Corollary.0. The solution presented in Theorem.9 satisfies this relation: W N (t)+w N ( t) =. Moreover, W N (t) has the same series expansion at t =+ and t =. Proof. On replacing t by t in (.4), we see that W N ( t) is a solution to (.4) with the domain t D unchanged. This fact suggests that W (i) III ( t), i = {,,, 4}, are closely related. Noting that A ( t) =A (t), A ( t) =A (t), A ( t) = A (t), we find W (4) () III ( t) =W III W () () III ( t) =W III (t), W () () III ( t) =W III (t), W () (4) III ( t) =W III which lead to W N ( t) =W N (t). (t), t <, <t<0, <t<, (t), t >, (b)

26 S. Lyu & Y. Chen We observe that tw () III ( (4) t ),t (, 0), and tw III ( t ),t (0, ), can both be simplified to the same expression as W (4) () III (t). This indicates that tw III ( t )and tw (4) III ( t ) have the same series representation at t = 0. Hence the expansion formula of W () (4) III (t) att = and W III (t) att =+ are identical. This completes the proof. Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. Corollary.. The solution given in Theorem.9 has the following series expansions near ±, 0, and : W N (t) = 4t 6 64 t 8 t t t 4 W N (t) = t 0945 s t 5 + O(t 6 ), t ±, + t 8 t t 5 + t 64 6 t t 8 + t t t ( ) + 79t O t, t 0, and, due to the previous corollary, we can readily obtain the expression for t + from the one for t 0. These formulas combined with Lemma.5 give rise to the initial conditions for H N (t). According to Corollary.4, we establish the following results. Corollary.. H N (t) satisfies this second-order differential equation: 6(t ) t (H N ) + 6(tH N H N )((t )H N H N )(4H N 6N(N +) ) N (N +) (4N +) =0, with initial conditions H N () = 4 = H N(0), N H N (± ) + N 4 5 ( t). 8 Here the symbol refers to the limiting procedure t ±..4. Case 4: (α, β, γ) =(N +,N +, (N + )) In light of (.), we have, for α = N +, ( α (α 0,α,α,α )=, α, α, 9 ) α,

27 Exceptional solutions to the Painlevé VI equation which fits into the eighth case of Proposition.. Hence W N (t) satisfies this algebraic equation W 4 N 4tW N +6tW N 4tW N + t =0, (.5) with t D, D =(, 0) (, ). Its four solutions are given by Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. Here W N (t) =t B (t) B (t) B (t) :=W () IV (t), W N (t) =t + B (t) B (t)+b (t) :=W () IV (t), W N (t) =t B (t)+ B (t) B (t) :=W () IV (t), W N (t) =t + B (t)+ B (t)+b (t) :=W (4) IV (t). B (t) := t(t ) + t (t ) > 0, B (t) :=t(t ) t (t ), B (t) := t(t )(t ). B (t) For the sake of brevity, we shall only display the dependence of B i necessary for latter discussion. on t when Fig.. B ± B (t D)

28 S. Lyu & Y. Chen Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. Fig. 4. W N t. On account of Figs. and 4, we apply Lemma. to establish the following statement: Theorem.. We find (i) for t<0, (ii) for t>, W N (t) =W () IV (t); W N (t) =W (4) IV (t). Proof. To find real solutions for W N (t), from Fig., weseethatw N (t) =W () IV or W () IV for t<0, and W N (t) =W () (4) IV or W IV for t>. β Since N++α+β+γ =, it follows from Lemma. that W N t<0ift<0, and W N t>0ift>. Based on this fact, we obtain the desired unique solution for W N (t) fromfig.4. Corollary.4. This relation holds for the solution given in Theorem.: W N (t)+w N ( t) =. In addition, W N (t) has the same series expansion at t =+ and t =. Proof. Replacing t by t in (.5), we find that W N ( t) isasolution to (.5) with the domain t D retained. This observation motivates us to find out W () (4) N ( t) and W N ( t)

29 Exceptional solutions to the Painlevé VI equation Noticing that B ( t) =B (t), B ( t) =B (t), Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. B ( t) = B (t), we get W (4) () IV ( t) =W IV (t), t < 0, W () (4) IV ( t) =W IV (t), t >. Hence we conclude that W N ( t) =W N (t). We observe that tw () IV ( (4) t ),t (, 0), and tw IV ( t ),t (0, ), can both be converted to W (4) () IV (t). This implies that tw IV ( (4) t )andtw IV ( t ) have the same series representation at t = 0. Thus the expansion formula of W () IV (t) att = and W (4) IV (t) att =+ are identical. This completes the proof. Corollary.5. The solution given in Theorem. has the following series expansions near ±, 0, and : W N (t) =4t 64 t 8 t t t t 5 + O(t 6 ), t ±, W N (t) =+ (t ) + (t ) (t ) + 4 (t ) (t ) (t ) (t ) (t ) 04 (t ) + O((t ) 0 ), t +, and, according to the preceding corollary, we can immediately obtain the expression for t 0 from the one for t +. Combining these formulas with Lemma.5, we obtain the initial conditions for H N (t). Then from Corollary.4, we deduce the following results. Corollary.6. The second-order differential equation for H N reads 6(t ) t (H N ) +4(4H N N(N +) )(( t)h N +H N ) (H N + N)(H N +N(N + ))(4H N +(N +))=0, with initial conditions H N (0) = N + N = H N(), H N (± ) N(N + )(t ). 4 Here the symbol refers to the limiting procedure t ±

30 S. Lyu & Y. Chen Table. List of (α, β, γ) corresponding to seven cases of Proposition.. Case Case Case 4 (α, β, β),n = ; (α, β, α),n = ; ( β N 0,β,β),N < β; (α, N 0, N 0), (α, α N 0,α),N < α; ( N 0,β, N 0), (α, α, γ),n = ; ( N 0, N 0,γ),N < ; Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. ( N 0,β, β),n < ; (α, α, α),n = ; (α, N 0, α),n < ; Case 5 Case 6 Case 8 (N 0, N 0,N 0), ( N 0,N 0,N 0), <N< ; N0, N0, N0 N0, N0, N0 5 <N < ; «, «, N0, N0, N0 N0, N0, N0 «,N < ; «,N < 5 ; ( β,β, β),n = ; N 0, N0, N0 N0 N0, N0, N0, N0, N0 N0, N0, N0 5 <N < 6 ; «,N < ; «, «,N < 6 ; «, ( N 0, N 0,N 0),N < 6 ; ( N 0,N 0, N 0), ( N 0,N 0,N 0), <N < 6 ; (α, α, α),n = ; «N0, N0, N0,N < 5 ; «N0, N0, N0,N < ; N0, N0, N0 N0, N0, N0 5 <N< ; (N 0, N 0, N 0), ( N 0,N 0, N 0), <N< ; Note: Here N 0 := N +. Note that exchanging α and β in the list of (α, β, γ) for Case 6 yields the one for Case 7. Remark. Approximations of D N (t) near0, ±, for Cases and 4 can be easily obtained from Corollaries. and.6, respectively. The normalization constants in there and also d i in Cases and will be computed in a subsequent paper. Remark 4. Since the exceptional solutions of the P VI equation, which are found by Eremenko etc. and given by Proposition., do not rely on the random matrix model, we formally extend the integer N toanyrealnumber.inthisprocess,we discover a number of interesting cases of (α, β, γ) wherethep VI equation (.) with the parameters (µ,µ,µ,µ 4 ) = ( (N++α+β+γ), α, β, γ ), α >,β >,γ R, has exceptional solutions. The results are listed in Table. «, «, Remark 5. The methods of this paper do not deal with the case 0 <t<. Acknowledgments The financial support of the Macau Science and Technology Development Fund Under Grant Number FDCT 077/0/A, FDCT 0/04/A is gratefully

31 Exceptional solutions to the Painlevé VI equation acknowledged. We also like to thank the University of Macau for generous support: MYRG FST, MYRG FST. Random Matrices: Theory Appl Downloaded from by on 0/05/8. For personal use only. References E. Basor and Y. Chen, Perturbed Laguerre unitary ensembles, Hankel determinants, and information theory, Math. Methods Appl. Sci. 8 (05) E. Basor, Y. Chen and N. Mekareeya, The Hilbert series of N =SO(N c)and Sp(N c)sqcd, Painlevé VI and integrable systems, Nucl. Phys. B 860 (0) M. Chen and Y. Chen, Singular linear statistics of the Laguerre unitary ensemble and Painlevé III. Double scaling analysis, J. Math. Phys. 56 (05) Y. Chen and M. Ismail, Thermodynamic relations of the Hermitian matrix ensembles, J. Phys. A: Math. Gen. 0 (997) Y. Chen and M. Ismail, Ladder operators and differential equations for orthogonal polynomials, J. Phys. A: Math. Gen. 0 (997) Y. Chen and M. Ismail, Jacobi polynomials from compatibility conditions, Proc. Amer. Math. Soc. () (004) Y. Chen and N. Lawrence, On the linear statistics of Hermitian random matrices, J. Phys. A: Math. Gen. (998) Y. Chen and S. M. Manning, Distribution of linear statistics in random matrix models (metallic conductance fluctuations), J. Phys.: Condens. Matter 6 (994) Y. Chen and M. Mckay, Perturbed Hankel determinants: Applications to the information theory of MIMO wireless communications, preprint (00), arxiv: v. 0 Y. Chen and M. R. McKay, Coulomb fluid, Painlevé transcendents, and the information theory of MIMO systems, IEEE Trans. Inform. Theory 58 (0) D. Dai and L. Zhang, Painlevé VI and Hankel determinants for the generalized Jacobi weight, J. Phys. A: Math. Theor. 4 (00) Article ID: 05507, 4pp. F. J. Dyson, Statistical theory of the energy levels of complex systems. I, II, III, J. Math. Phys. (96) A. Eremenko, A. Gabrielov and A. Hinkkanen, Exceptional solutions to the Painlevé VI equation, preprint (06), arxiv: N. J. Higham, Functions of Matrices: Theory and Computation (SIAM, 008). 5 L. K. Hua, Harmonic Analysis of Functions of Several Complex Variables in the Classical Domains (American Mathematical Society, 96). 6 S. Lyu and Y. Chen, The largest eigenvalue distribution of the Laguerre unitary ensemble, preprint (05), arxiv: A. P. Magnus, Painlevé-type differential equations for the recurrence coefficients of semi-classical orthogonal polynomials, J. Comput. Appl. Math. 57 (995) M. L. Mehta, Random Matrices, rd edn., Pure and Applied Mathematics, Vol. 4 (Elsevier/Academic Press, 004). 9 G. Szegö, Orthogonal Polynomials, American Mathematical Society Colloquium Publications, Vol. (American Mathematical Society, 99)