Painlevé equations and orthogonal polynomials

Transcription

1 KU Leuven, Belgium Kapaev workshop, Ann Arbor MI, 28 August 2017

2 Contents Painlevé equations (discrete and continuous) appear at various places in the theory of orthogonal polynomials: Discrete Painlevé equations for the recurrence coefficients of orthogonal polynomials Painlevé differential equations for the recurrence coefficients (Toda flows) Rational solutions of Painlevé equations [Special function solutions of Painlevé equations] [Local asymptotics for orthogonal polynomials at critical points] In preparation: Orthogonal Polynomials and Painlevé Equations Australian Mathematical Society Lecture Series

3 Orthogonal polynomials p n (x)p m (x)w(x) dx = δ m,n, p n (x) = γ n x n + p n (x k )p m (x k )w k = δ m,n. k=0 Three term recurrence relation xp n (x) = a n+1 p n+1 (x) + b n p n (x) + a n p n 1 (x), orthonormal xp n (x) = P n+1 (x)+b n P n (x)+anp 2 n 1 (x), monic polynomials Orthogonal polynomials on the unit circle (OPUC) 1 2π 2π 0 ϕ n (z)ϕ m (z)v(θ) dθ = δ m,n, z = e iθ, ϕ n (z) = κ n z n + zφ n (z) = Φ n+1 (z) + α n Φ n(z), Φ n(z) = z n Φ n (1/z).

4 discrete Painlevé I Let us consider w(x) = e x4 +tx 2 on (, ). The symmetry w( x) = w(x) implies b n = 0, hence The structure relation is xp n (x) = a n+1 p n+1 (x) + a n p n 1 (x). p n(x) = A n p n 1 (x) + C n p n 3 (x). Compatibility between these two relations gives ( 4an 2 an an 2 + an 1 2 t ) = n 2 This is known as discrete Painlevé I (d-p I ).

5 Some history Already in old work of Laguerre (1885), Also in work of Shohat (1939), Independently found by Freud (1976), Positive solution analyzed by Nevai (1983). Asymptotic expansion by Máté-Nevai-Zaslavsky (1985) Recognized as d-p I by Fokas-Its-Kitaev (1991). Relation with continuous Painlevé IV (Magnus, 1995)

6 Asymptotic behavior of a 2 n Theorem (Freud) The recurrence coefficients for the weight w(x) = e x4 +tx 2 satisfy lim n a n n 1/4 =

7 Special solution Put x n = a 2 n, then (for t = 0) x n (x n+1 + x n + x n 1 ) = an, a = 1/4. (1) For orthogonal polynomials we want a solution with x 0 = 0 (because a 2 0 = 0) and x n > 0 for n 1. Theorem (Lew and Quarles, Nevai) There is a unique solution of (1) for which x 0 = 0 and x n > 0 for all n 1.

8 Langmuir lattice or Kac-van Moerbeke equations General orthogonal polynomials P n (x)p m (x) dµ(x) = 0, m n. R Theorem Let µ be a symmetric positive measure on R for which all the moments exist and let µ t be the measure for which dµ t (x) = e tx2 dµ(x), where t R is such that all the moments of µ t exist. Then the recurrence coefficients of the orthogonal polynomials for µ t satisfy the differential-difference equations d dt a2 n = a 2 n(a 2 n+1 a 2 n 1), n 1.

9 Painlevé IV Put a 2 n = x n, then n = 4x n (x n+1 + x n + x n 1 t/2), (2) x n = x n (x n+1 x n 1 ) (3) Eliminate x n+1 and x n 1 using (2) (3) x n = (x n) 2 + 3x 3 ( ) n n 2x n 2 tx n 2 + x n 2 + t2 n2 8 32x n This is Painlevé IV.

10 Orthogonal polynomials on the imaginary line The weight e x4 +tx 2 behaves well on R, but it also tends to 0 along the imaginary axis ir. Orthogonal polynomials (Q n ) n on the imaginary axis +i i Q n (x)q m (x)e x4 +tx 2 dx = 0, n m, xq n (x) = Q n+1 (x) b n Q n 1 (x). One has Q n (x; t) = ( i) n P n (ix; t). Hence b n (t) = a 2 n( t). The (b n ) n is the unique negative solution of d-p I with b 0 = 0.

11 Orthogonal polynomials on the cross We can combine R and ir and look for orthogonal polynomials (R n ) n for which (n m) α R n (x)r m (x)e x4 +tx 2 dx+β +i i with α, β > 0. xr n (x) = R n+1 (x) c n R n 1 (x) R n (x)r m (x)e x4 +tx 2 dx = 0, The (c n ) n still satisfy d-p I, but with initial conditions where c 0 = 0, m 2k (t) = c 1 = αm 2(t) βm 2 ( t) αm 0 (t) + βm 0 ( t), x 2k e x4 +tx 2 dx are in terms of parabolic cylinder functions.

12 Orthogonal polynomials on the cross The solution of d-p I behaves very different. (t = 0, β/α = 1/2)

13 Orthogonal polynomials on the cross ( 4x n x n+1 + x n + x n 1 t ) = n 2 The case α = β and t = 0 is special: the initial values are x 0 = x 1 = 0. 0 is a singularity of d-p I and gives x 2 =, hence R 3 does not exist. The singularity is confined to a finite number of terms. Property For t = 0 and α = β one has that R 4n 1 does not exist for n 1. Furthermore R 4n (x) = r n (x 4 ), R 4n+1 (x) = xs n (x 4 ), R 4n+2 (x) = x 2 s n (x 4 ).

14 discrete Painlevé II Let v(θ) = e t cos θ on the unit circle Trigonometric moments: modified Bessel functions 1 2π 2π 0 e inθ v(θ) dθ = I n (t), v( θ) = v(θ) implies that α n (t) are real valued. v(θ) = ˆv(z), z = e iθ, ˆv(z) = exp ( t z + 1 z 2 ). Pearson equation ˆv (z) = t 2 ( 1 1 ) z 2 ˆv(z).

15 Structure relation Property The monic orthogonal polynomials for v(θ) = e t cos θ satisfy Φ n(z) = nφ n 1 (z) + B n Φ n 2 (z), for a sequence (B n ) n. In fact, one has B n = t 2 κ 2 n 2 κ 2. n

16 Compatibility zφ n (z) = Φ n+1 (z) + α n Φ n(x) Φ n(z) = nφ n 1 (z) + B n Φ n 2 (z) Theorem (Periwal and Shevitz) The Verblunsky coefficients for the weight v(θ) = e t cos θ satisfy with initial values t 2 (α n+1 + α n 1 ) = (n + 1)α n 1 αn 2, α 1 = 1, α 0 = I 1(t) I 0 (t). This is discrete Painlevé II (d-p II )

17 Discrete Painlevé II Let x n = α n 1, then x n+1 + x n 1 = αnx n 1 x 2 n (4) We need a solution with x 0 = 1 and x n < 1 for n 1. Such a solution is unique. Theorem Suppose α > 0. Then there is a unique solution of (4) for which x 0 = 1 and 1 < x n < 1. The solution corresponds to x 1 = I 1 (2α)/I 0 (2/α) and is positive for every n 0.

18 Asymptotic behavior This solution converges to zero (fast) x n 0. Property The solution of d-p II with x 0 = 1 and 0 < x n < 1 for n 1 satisfies 1 α n n! x n n 4n n! k=0 α k α n (2n)! 1 ( e ) n. k! 2 αn

19 The Ablowitz-Ladik lattice The lattice equations corresponding to orthogonal polynomials on the unit circle are Ablowitz-Ladik lattice equations (or Schur flow). Theorem Let ν be a positive measure on the unit circle which is symmetric (the Verblunsky coefficients are real). Let ν t be the modified measure dν t (θ) = e t cos θ dν(θ), with t R. The Verblunsky coefficients (α n (t)) n for the measure ν t then satisfy 2α n = (1 α 2 n)(α n+1 α n 1 ), n 0.

20 Painlevé V and III d-p II gives and Ablowitz-Ladik gives α n+1 + α n 1 = 2nα n t(1 α 2 n) α n+1 α n 1 = 2α n 1 αn 2. Eliminate α n+1 and α n 1 to find α α n n = 1 αn 2 (α n) 2 α n t α n (1 α 2 n) + (n + 1)2 t 2 α n 1 αn 2. If we put α n = 1+y 1 y, then y satisfies Painlevé V with γ = 0. The ratio w n = α n /α n 1 satisfies Painlevé III [Hisakado, Tracy-Widom]

21 Generalized Charlier polynomials Discrete orthogonal polynomials c k P n (k)p m (k) (β) k k! k=0 = 0, n m. b n + b n 1 n + β = cn an 2, (an+1 2 c)(an 2 c) = c(b n n)(b n n + β 1). This corresponds to a limiting case of discrete Painlevé with surface/symmetry D (1) 4 in Sakai s classification.

22 Toda Lattice Put c = c 0 e t, then c k (β) k k! = etk c k 0 (β) k k! Theorem Suppose µ is a positive measure on the real line and dµ t = e tx dµ(x), where t is such that all the moments of µ t exist. Then the recurrence coefficients b n (t) and a 2 n(t) of the orthogonal polynomials for the measure µ t satisfy d dt a2 n = an(b 2 n b n 1 ), n 1, d dt b n = an+1 2 an, 2 n 0.

23 generalized Charlier polynomials Put x n = a 2 n and y n = b n and x n = dx n /da, y n = dy n /da, then (x n a)(x n+1 a) = a(y n n)(y n n + β 1), y n + y n 1 n + β = an x n and the Toda lattice equations are ax n = x n (y n y n 1 ), ay n = x n+1 x n. Eliminate y n 1 and x n+1, and put x n = y = 1 2 ( 1 2y + 1 ) (y ) 2 y ( y)2 n 2 y +(1 y 1 a a 2 2 This is Painlevé V with δ = 0. This can be transformed to Painlevé III. a 1 y, then y(a) satisfies ) (β 1)2 2y 2y a

24 generalized Meixner polynomials Discrete orthogonal polynomials P n (k)p m (k) (γ) ka k = 0, n m. (β) k k! k=0 Put an 2 = na (γ 1)u n, and b n = n + γ β + a γ 1 a v n, then (u n + v n )(u n+1 + v n ) = γ 1 ( a 2 v n (v n a) v n a γ β ), γ 1 ( u n (u n + v n )(u n + v n 1 ) = u n an (u n + a) u n + a γ β ). γ 1 γ 1 Initial values a 0 = 0, b 0 = γa M(γ + 1, β + 1, a) β M(γ, β, a) This is asymmetric discrete Painlevé IV or d-p(e (1) 6 /A(1) 2 ).

25 generalized Meixner If we put v n (a) = ( ) a ay (1 + β 2γ)y 2 + (n + 1 a + β 2γ)y n, 2(γ 1)(y 1)y then y = with ( 1 2y + 1 ) (y ) 2 y y 1 a A = +(y 1)2 a 2 ( Ay+ B y ) Cy + 1) + +Dy(y a y 1 (β 1)2, B = n2 2 2, C = n β + 2γ, D = 1 2, which is Painlevé V.

26 Other examples Chen and Its (2010): w(x) = x α e x e t/x on [0, ) Put b n = 2n + α c n, an 2 = n(n + α) + y n + n 1 j=0 c j, and c n = 1/x n d-p((2a 1 ) (1) /D (1) 6 ) x n + x n 1 = nt (2n + α)y n y n (y n t) y n + y n+1 = t 2n + α + 1 x n 1 x 2 n c n = (c n) 2 c n c n t + (2n + α + 1)c2 n t 2 + c3 n t 2 + α t 1 c n which is Painlevé III.

27 Other examples Basor, Chen, Ehrhardt (2010): w(x) = (1 x) α (1 + x) β e tx tb n = 2n α + β t 2R n, t(t + R n )a 2 n = n(n + β) (2n + α + β)r n tr n(r n + α) R n 2t(r n+1 + r n ) = 4Rn 2 2R n (2n α + β t) 2αt, ( t 2 n(n + β) (2n + α + β)r n = r n (r n + α) + t + t ) R n R n 1 R n and for y = 1 + t/r n y = 3y 1 2y(y 1) (y ) 2 y t +2(2n+1+α+β)y t which is Painlevé V. R n 1 ( + 1) (y 1)2 α 2 ) y 2y(y + y 1 t 2 2 β2 2y

28 Other examples q-orthogonal polynomials: w(x) = x α ( x 2 ; q 2 ) ( q 2 /x 2 ; q 2 ), x [0, ) gives rise to q-discrete Painlevé III x n 1 x n+1 = (x n + q α ) 2 (q n+α x n + 1) 2. x α ( p/x 2 ; q 2 ) w(x) = ( x 2 ; q 2 ) ( q 2 /x 2 ; q 2, x [0, ) ) gives rise to q-discrete Painlevé V (z n z n 1 1)(z n z n+1 1) = (z n + q 2 α /p) 2 (z n pq α 2 ) 2 (q n+α/2 1 pz n + 1) 2. w(x) = x α (q 2 x 2 ; q 2 ), x {q k, k = 0, 1, 2, 3,..., } gives again q-discrete Painlevé V

29 Other examples Bi-orthogonal polynomials on the unit circle (Forrester and Witte, 2006) { w(z) = z µ ω (1+z) 2ω 1 (1+tz) 2µ 1, θ / (π φ, π), 1 ξ, θ (π φ, π)., t = eiφ, gives rise to g n+1 g n = t (f n + n)(f n + n + 2µ), f n (f n 2ω 1 ) f n + f n 1 = 2ω 1 + n 1 + µ + ω + g n 1 discrete Painlevé V (Sakai s surface D (1) 4 ). (n + µ + ω)t g n t

30 Other examples Biane (2014) worked out a q-version w(e iθ ) = (ae iθ ; q) (be iθ ; q) and found discrete Painlevé equations corresponding to A (1) 3 r n+2 (a bq n+2 )+qr n (a bq n ) = 2r ( ) n+1 1 rn+1 2 (1 q)(a+bq n+1 )β n+1 +(1 q n+1 )(ab+q), 2, with n β n = r r r k 1, r n = Φ n (0). k=1

31 Other examples Witte (2015) worked out the most general case (deformed Askey-Wilson polynomials): Pearson-type equation for the weight D x w(x) = 2V (x) W (x) M xw(x), where D x is a divided difference operator on a special nonuniform lattice obtained from a quadratic equation Ay 2 + 2Bxy + Cx 2 + 2Dy + 2Ex + F = 0, and M x is an average operator for the forward and backward moves on the lattice, and V, W are polynomials. W (z) ± yv (z) = z 3 6 (1 a j q 1/2 z ±1 ), gives the q-discrete Painlevé cases for E (1) 7 in Sakai s scheme. j=1

32 Rational solutions of Painlevé equations

33 Rational solutions of Painlevé II y = 2y 3 + xy + α has rational solutions if and only if α = n Z. The solutions are y n = d dz log Q n 1(z) Q n (z) where Q n are the Yablonskii-Vorobiev polynomials p 1 p 3 p 5 p 2n 1 p 1 p 3 p 5 p 2n 1 Q n (z) = c n det.... p (n 1) 1 p (n 1) 3 p (n 1) 5 p (n 1) 2n 1 with polynomial p k given by p k (z)λ k = exp(zλ 4 3 λ3 ). k=0

34 Rational solutions of Painlevé II The polynomials p k are multiple orthogonal polynomials on the 3-star 2 Γ = {z C : arg(z) = 2πk 3 }, k=0 p n (z)ai(2 2/3 z )z k dz = 0, 0 k n 2 1, Γ Γ p n (z) z z Ai (2 2/3 z )z k dz = 0, 0 k n 2 1.

35 Rational solutions of Painlevé II Bertola and Bothner (2015) gave another determinant representation µ 0 (z) µ 1 (z) µ 2 (z) µ n 1 (z) Qn 1(z) 2 µ 1 (z) µ 2 (z) µ 3 (z) µ n (z) = d n det.... µ n 1 (z) µ n (z) µ n+1 (z)... µ 2n 2 (z) with µ n (z) = p n (2 2/3 z)2 2n/3. Advantage: this is a Hankel matrix, and hence allows to use the moment problem and orthogonal polynomials µ 0 (z) µ 1 (z) µ 2 (z) µ n (z) d µ 1 (z) µ 2 (z) µ 3 (z) µ n+1 (z) n P n (x) = Qn 1 2 det (z).... µ n 1 (z) µ n (z) µ n+1 (z)... µ 2n 1 (z) 1 x x 2 x n

36 Rational solutions of Painlevé II z is a zero of Q n 1 (and a pole of the rational solution) if and only if the monic orthogonal polynomial P n (x) does not exist. This allows to locate the zeros of Q n 1 by investigating the Riemann-Hilbert problem for the orthogonal polynomials (P k ) k N. Theorem (Bertola-Bothner, Buckingham-Miller) The zeros of Q n (n 2/3 z) accumulate on a triangular shaped region with corners x k = 3/ 3 2e 2πi 3 k (k = 0, 1, 2). The sides of the region are given by ( R 2 log a 3 ia + ) 1 + 2a 3 4a3 1 2a 3a 3 = 0, where a = a(x) is a solution of the cubic equation 1 + 2xa 2 4a 3 = 0.

37

38 Rational solutions of Painlevé III y = (y ) 2 y y z + αy 2 + β + y 3 1 z y Has rational solutions if and only if α + β = 4n or α β = 4n for n Z. For α = 2n + 2µ 1 and β = 2n 2µ 1 y n = 1 + d dz S n 1 (z; µ 1). S n (z; µ) and for α = 2n + 2µ 1 and β = 2n 2µ + 1 y n = 1 d dz S n 1 (z; µ) S n (z; µ 1).

39 Rational solutions of Painlevé III S n (z; µ) are Umemura polynomials S n (z; µ) = c n det L µ n n ( z) L µ n 1 n+1 ( z) L µ 2n+1 2n 1 n 2 ( z) L µ n+1 n 1 ( z) L µ 2n+3 L µ n+2. L µ+n 2 n+2 ( z) with L α n (x) the Laguerre polynomials. Lµ+n 3 n+3 2n 3 ( z) ( z).. ( z) Lµ 1 1 ( z)

40 Rational solutions of Painlevé III Wronskian formula: L µ 1 1 ( z) L µ 3 3 ( z) L µ 2n+1 2n 1 ( z) [L µ 1 1 ( z)] [L µ 3 3 ( z)] [L µ 2n+1 2n 1 ( z)] S n (z; µ) = c n det... [L µ 1 1 ( z)] (n 1) [L µ 3 3 ( z)] (n 1) [L µ 2n+1 2n 1 with L α n (x) the Laguerre polynomials. ( z)] (n 1)

41 Zeros of Umemura polynomial S 10 (µ), 9 µ 9

42 Rational solutions of Painlevé IV y = (y ) 2 2y Has rational solutions if and only if d dz log H m+1,n(z) H m,n, y 3 + 4zy 2 + 2(z 2 α)y + β y α = m, β = 2(2n m + 1) 2 α = m, β = 2(2n m )2. d dz log H m,n(z) H m,n+1 (z), with H m,n the generalized Hermite polynomials 2z + d d log H m,n+1(z) H m+1,n (z) 2 3 z+ d dz log Q m+1,n(z) Q m,n (z), 2 3 z+ d dz log Q m,n(z) Q m,n+1 (z), 2 3 z+ d dz log Q m,n+1(z) Q m+1,n (z) with Q m,n the generalized Okamoto polynomials

43 Rational solutions of Painlevé IV The generalized Hermite polynomials are H m (z) H m+1 (z) H m+n 1 (z) H m+1 (z) H m+2 (z) H m+n (z) H m,n (z) = det... H m+n 1 (z) H m+n (z) H m+2n 2 (z) and H m (z) are Hermite polynomials.

44 Zeros of generalized Hermite polynomials

45 Zeros of generalized Hermite polynomials Theorem (Buckingham) The zeros of the generalized Hermite polynomial H m,n (m 1/2 z) for the rational solution of Painlevé IV accumulate in a square shaped region with corners given by four solutions of the equation r 4 x 8 24r 2 (r 3 +r+1)x 4 +32r(2r 3 +3r 2 3r 2)x 2 48(r 2 +r+1) 2 = 0 (not real or imaginary), and sides given by the curves R ( (1 + r)r 1/2 ( ) x 4 R (1 + r) log 2R 2 (1 + r)q S + (r 1) log ( (1 + r)q 3 + (1 + r)q 2 R + s ) + log(s 2 4Q 2 ) ) = 0 where r = m/n 1.

46 Zeros of generalized Hermite polynomials Theorem (continued) Q = Q(x, r) is the unique solution of 3(1 + r) 2 Q 4 + 8(1 + r)r 1/2 xq 3 + 4(r 1 + rx 2 )Q 2 4 = 0 for which Q(x, r) = x + O(x 2 ) as x, S = (1 + r)q 3 + 2r 1/2 Q 2 and ( (1 + r) 2 Q 4 + 2(1 + r)qs + 4 ) 1/2 R =. (1 + r)q

47 Zeros of generalized Hermite polynomials

48 Rational solutions of Painlevé IV The Okamoto polynomials are Q n (z) = Q n+1,0 (z) = Wr(H 3n 1, H 3n 4,..., H 2 ), R n (z) = Q n,1 (z) = Wr(H 3n 2, H 3n 5,..., H 1 ) The generalized Okamoto polynomials are Q m, n (z) = H λ (z), λ = (m + 2n, m + 2n 2,..., m + 2, m, m, m 1, m 1,..., 1, 1), Q n,m+1 (z) = H λ (z), λ = (m + 2n 1, m + 2n 3,..., m + 1, m, m, m 1, m 1,..., 1, 1). H λ1 H λ1+1 H λ1+2 H λ1+k 1 H λ2 1 H λ2 H λ2+1 H λ2+k 2 H λ (z) = det H λ3 2 H λ3 1 H λ3 2 H λ3+k H λk k+1 H λk k+2 H λk k+3 H λk

49 Zeros of generalized Okamoto polynomials

50 Rational solutions of Painlevé V y = ( 1 2y + 1 ) (y ) 2 y ( 1)2 +(y y 1 z z 2 αy + β ) + γy + 1) y(y y z 2(y 1) Has rational solutions if 1 α = 1 2 (m ± 1)2 and β = n2 2, with n > 0, m + n odd, α 0 when m < n; 2 α = n2 2, β = 1 2 (m ± 1)2, with n > 0, m + n odd, β 0 when m < n; 3 α = a 2, β = 1 2 (a + n)2, and γ = m, with m + n even; 4 α = (b + n) 2, β = b2, and γ = m, with m + n even; 5 α = 1 8 (2m + 1)2, β = 1 8 (2n + 1)2. 2

51 Rational solutions of Painlevé V In terms of Umemura polynomials L r n(x) L r n+1 (x) Lr 2n 1 (x) L r n 2 U 0,n (x; µ) = C n det (x) Lr n 1 (x) Lr 2n 3 (x)..., L r 2 n (x) Lr 3 n (x) Lr 1 (x) L r m( x) L r m+1 ( x) Lr 2m 1 ( x) L r m 2 U m,0 (x; µ) = D m det ( x) Lr m 1 ( x) Lr 2m 3 ( x)..., L r 2 m ( x) Lr 3 m ( x) Lr 1 ( x) with L α n (x) the Laguerre polynomials. r = µ+n 1 r = µ m 1

52 Rational solutions of Painlevé V and generalized Umemura polynomials U m,n (x; µ) L r 1 ( x) Lr 0 ( x) Lr m+2 ( x) Lr m+1 ( x) Lr m n+2 ( x) L r 3 ( x) Lr 2 ( x) Lr m+4 ( x) Lr m+3 ( x) Lr m n+4 ( x)..... L r det 2m 1 ( x) Lr 2m 2 ( x) Lr m( x) L r m 1 ( x) Lr m n ( x) L r n m (x) Lr n m+1 (x) Lr n 1 (x) Lr n(x) L r 2n 1 (x) L r n m 2 (x) Lr n m 1 (x) Lr n 3 (x) Lr n 2 (x) Lr 2n 3 (x)..... L r n m+2 (x) Lr n m+3 (x) Lr n+1 (x) Lr n+2 (x) Lr 1 (x) where r = µ m + n 1.

53 Zeros of U 8,10 (µ = 2k + 1), 11 k 9

54 Rational solutions of Painlevé VI Special function solution in terms of terminating hypergeometric functions These are determinants with Jacobi polynomials Little research has been done on these rational solutions.

55 Special function solutions of Painlevé equations

56 Special functions Painlevé I Painlevé II Airy functions Painlevé III Bessel functions Painlevé IV Parabolic cylinder functions Painlevé V confluent hypergeometric functions Painlevé VI hypergeometric functions seed function: comes from a Riccati equation Bäcklund transformations give all the special function solutions in terms of the seed function.

57 Connection with orthogonal polynomials Hankel determinant with moments of a measure µ m 0 m 1 m 2 m n 1 m 1 m 2 m 3 m n n = det m 2 m 3 m 4 m n+1, m k =.... m n 1 m n m n+1 m 2n 2 m 0 = and n is a Wronskian w(x)e xt dx, m k = d k m 0 dt k x k dµ(x). n = det d i+j m 0 dt i+j.

58 Connection with orthogonal polynomials Recurrence coefficients an 2 = n+1 n 1 2, b n = n+1 n, n n+1 n where m 0 m 1 m 2 m n 2 m n m 1 m 2 m 3 m n 1 m n+1 n = det m 2 m 3 m 4 m n m n+2 = d dt n m n 1 m n m n+1 m 2n 3 m 2n 1

59 Examples w(x) = e x4 +tx 2 m 0 = 2 1/4 πe t2 /8 U(0, t/ 2) P IV w(x) = e x3 /3+tx m 0 = 2πiAi(t) P II w k = ak (β) k k! on N m 0 = a (β 1)/2 Γ(β)I β 1 (2 a) P III w(x) = x α e x e s/x m k K k 1 (2 t) P III w(x) = x a 1 (1 x) b a 1 e xt m 0 = Γ(a)Γ(b a)m(a, b, t) P V w k = (γ) ka k (β) k k! on N m 0 = M(γ, β, a) P V w(x) = x b 1 (1 x) c b 1 (1 xt) a m 0 = Γ(b)Γ(b c) 2 F 1 (a, b; c; t) P VI

60 References P.A. Clarkson, Painlevé equations nonlinear special functions, Lecture Notes in Mathematics 1883, Springer, Berlin, 2006, pp K. Kajiwara, M. Noumi, Y. Yamada, Geometric aspects of Painlevé equations, J. Phys. A: Math. Theor. 50 (2017), no. 7, (164. pp.) W. Van Assche, Discrete Painlevé equations for recurrence coefficients of orthogonal polynomials, in Difference Equations, Special Functions and Orthogonal Polynomials, World Scientific, 2007, pp W. Van Assche, Orthogonal Polynomials and Painlevé Equations, Australian Mathematical Society Lecture Series, Cambridge University Press (to appear).