Sn+1Snâ1 = (t2 â t)(SnSâ²â² ... âH. âq where. H = H(b) = q(q â 1)(q â t) t(t â 1). {p2 â p( b1 + b2 q. + ... h(t) = t(t â 1)H(p(t)...

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A FOUR-PARAMETRIC RATIONAL ´ VI SOLUTION TO PAINLEVE Gert Almkvist November 8, 2018 Introduction The seminal paper by Okamoto [3] showed how to get a sequence of rational solutions to Painlev´e VI if you start with a rational seed solution. But Okamoto did not even write down the B¨acklund transformation. This is understandable since its denominator is of degree 6 in p and q. Today we have Maple to handle such things and the author computed hundreds of examples starting with rational solutions that come from a Riccati equation and can be expressed by hypergeometric functions (see [2] ). Soon a pattern emerged. The first τ -function, τ1 , had numerator 1 and τ2 was also rather simple. An explicit formula for τ2 was found and proved. As a consequence we have the following main result. Let m be a positive integer. Define m X s + m m−j j r+m+1−j (−1) t W (r, m, s) = m−j j j=0 Let further T1 = 1, T2 = W (r, m, s) S1 = 1, S2 = W (r − 1, m + 1, s − 1) and Tn+1 Tn−1 = (t2 − t)(Tn Tn′′ − Tn′ 2 ) + (2t − 1)Tn Tn′ + (n − 1)(n + r)Tn2 Sn+1 Sn−1 = (t2 − t)(Sn Sn′′ − S ′ 2 ) + (2t − 1)Sn Sn′ + (n − 1)(n + r − 1)Sn2 for n ≥ 2. Then the Tn and Sn are polynomials and T′ t(t − 1) Sn′ n+s−1 n+r−m−s qn = t + − n+1 − − n+r Sn Tn+1 t t−1 solves PV I 1 q = 2 ′′

1 1 1 + q q−1q−t

q

′2

1

−

1 1 1 + + t t−1 q−t

q′ +

q(q − 1)(q − t) t2 (t − 1)2

βt γ(t − 1) δt(t − 1) α+ 2 + + q (q − 1)2 (q − t)2

with α=

(n + r)2 2

β=−

(m + s)2 2

γ=

(r − s + 1)2 2

δ=

1 − (n + m)2 2

If also r, s are integers then Tn and Sn have integer coefficients, usually growing very fast with n. One can ask for the smallest integer coefficients since qn is independent of multiplicative constants in Tn+1 and Sn . As a result we get some intriguing conjectures. E.g. Let p ≥ 3 be a prime. Define T1 = 1, T2 =

W (p, p − 1, 1) p

and c(n) by p2 (p + n)(p + n − 2) if n ≡ 1 mod p (p + n)(p + n − 2) if n ≡ 0, 2 mod p p (p + n)(p + n − 2) otherwise Then c(n)Tn+1 Tn−1 = (t2 − t)(Tn Tn′′ − Tn′ 2 ) + (2t − 1)Tn Tn′ + (n − 1)(n + p)Tn2 for n ≥ 2 has solutions Tn in Z[t] where the coefficients have no common factor > 1. We have checked the conjecture for p = 3, 5, 7, 11 up to n = 20 or more. In [1] there are other rational solutions which hopefully can be used in the same way to produce sequences of rational solutions to PV I .

1.Deriving the results. We follow the notation in Okamoto’s paper [3]. Given a solution q = q(b) with parameters b = (b1 , b2 , b3 , b4 ) 2

i.e. q satisfies PV I (

(b1 + b2 )2 (b1 − b2 )2 (b3 + b4 )(b3 + b4 + 2) (b3 − b4 )2 ,− , ,− ) 2 2 2 2

This equation is equivalent to the system dq ∂H = dt ∂p dp ∂H =− dt ∂q where H

= H(b) = b1 − b2 b3 + b4 (b1 + b3 )(b1 + b4 ) b1 + b2 q(q − 1)(q − t) + + )+ p2 − p( t(t − 1) q q−1 q−t q(q − 1)

We want to find formulas for the solution q + = q(b+ ) where b+ = (b1 , b2 , b3 + 1, b4 ) Let h(t) =

t(t − 1)H(p(t), q(t), t, b) + (b1 b3 + b1 b4 + b3 b4 )t 1 − (b1 b2 + b1 b3 + b1 b4 + b2 b3 + b2 b4 + b3 b4 ) 2

and

1 h+ (t) = h(t) − pq(q − 1) + (b1 + b4 )q − (b1 + b2 + b4 ) 2 Following the notation of Okamoto [3], p.354 we have A+ = (

dh+ dh+ + (b3 + 1)2 )( + b24 ) dt dt

d2 h+ dh+ + (b + b + b + b ) − 1 2 3 4 dt2 dt (b1 b2 (b3 + 1) + b1 b2 b4 + b1 (b3 + 1)b4 + b2 (b3 + 1)b4 ) B + = t(t − 1)

C + = 2(t

dh+ −h+ )−{b1 b2 + b1 (b3 + 1) + b1 b4 + b2 (b3 + 1) + b2 b4 + (b3 + 1)b4 } dt

Then by (2.5) in Okamoto we have the B¨acklund transformation dh+ U 1 + + + (b3 + 1 + b4 )B + ( = q = − (b3 + 1)b4 )C 2A+ dt V 3

where U = t {p(q − 1)(q − t) − (b3 + 1)(t − 1) − (b1 + b4 )(q − 1)} · {pq(q − 1)(q − t) − (b1 + b4 )q(q − 1) − (b3 + 1)q(t − 1) + b1 t(q − 1) + b2 (q − t)} and 2 p q(q − 1)(q − t)2 − p(q − t)(2(b1 + b4 )q 2 − (b1 + b2 + 2b4 + 2b1 t)q+ V = (q−t) +(b1 + b2 )t) + (b21 − (b3 + 1)2 )t2 + (b1 b2 + b1 b4 + b2 b4 + (b3 + 1)2 )t +(b1 + b4 )2 q 2 − (b1 + b4 )(2b1 t + b2 + b4 )

The τ −function is defined by (up to a multiplicative constant) H(t) =

d log(τ (t)) dt

After doing the +-construction n times we obtain bn = (b1 , b2 , b3 + n, b4 ) qn Hn τn We have the Toda equation d τn+1 τn−1 d (t(t − 1) log(τn )) + (b1 + b3 + n)(b3 + b4 + n) = c(n) dt dt τn2 where c(n) is a constant which can be chosen to be 1. Let σn be the n-th τ -function obtained by replacing b = (b1 , b2 , b3 , b4 ) by e = (b1 , b2 , b3 , b4 + 1). Then by (4.16) in Okamoto we have b t(t − 1) d d qn = t + log(σn ) − log(τn+1 ) b3 + n − b4 dt dt Now we choose as seed solution the rational function (see [2]) q = t+

t(t − 1) z ′ r z

where z =2 F1 (r, −m, s; t) µ positive integer, which satisfies PV I (α, β, γ, δ) with α= β=−

r2 2

(m + s)2 2 4

γ=

(r − s + 1)2 2

δ= This corresponds to

1 − m2 2

m+r+1 2 m − r + 2s − 1 b2 = 2 r+m−1 b3 = 2 m−r−1 b4 = 2 b1 =

Observe that b1 = b3 + 1 We will use that q satisfies the Riccati equation (see [2] ) q′ =

1 (b4 − b3 )q 2 + (2b1 t + b2 − b4 − 1)q − (b1 + b2 )t t(t − 1)

Substituting b3 = b1 − 1 in the formidable expression for q + , it collapses to t(q − 1) pq(q − 1)(q − t) − (b1 + b4 )q 2 + (2b1 + b4 + b2 )q − (b1 + b2 )t + q = (q − t) {pq(q − 1)(q − t) − (b1 + b4 )q 2 + (2b1 t + b2 + b4 )q − (b1 + b2 )t} By the Hamiltonian equations we get b1 + b4 1 b1 + b2 b1 − b2 b3 + b4 t(t − 1)q ′ = + + + p= 2q(q − 1)(q − t) 2 q q−1 q−t q−t after using the Riccati equation for q . Substituting this into q + we obtain q1 = q + =

(b1 + b2 )t(q − 1) (2b1 t + b2 − b1 )q − (b1 + b2 )t

and H1 = H + = H(t) − −

pq(q − 1) − (b1 + b4 )(q − t) = t(t − 1)

(b1 + b2 )(b1 + b4 ) (b1 − b2 )(b1 + b4 ) − t t−1

which gives Z τ1 = exp( H1 dt) =

1 t(b1 +b2 )(b1 +b4 ) (t − 1)(b1 −b2 )(b1 +b4 )

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To start the induction we need also to find τ2 .Since we know σ1 (replace b4 by b4 + 1 in τ1 ), namely σ1 =

t(b1 +b2 )(b1 +b4 +1) (t

1 − 1)(b1 −b2 )(b1 +b4 +1)

we can use the formula for q1 q1 = t +

t(t − 1) b3 − b4 + 1

d d log(σ1 ) − log(τ2 ) dt dt

i.e. d log(τ2 ) = dt −

(b1 + b2 )(b1 + b4 + 1) (b1 − b2 )(b1 + b4 + 1) (b1 − b4 )(q1 − t) − − t t−1 t(t − 1)

Inspired by numerous experiments we put τn =

Tn t(b1 +b4 )(b1 +b2 +n−1) (t − 1)(b1 +b4 )(b1 −b2 +n−1)

so T1 = 1. We will later show that the Tn are polynomials. It follows that b4 − b2 b4 + b2 (b1 − b4 )(q1 − t) T2′ = + − T2 t t−1 t(t − 1) In order to compute T2 we have to use the explicit formula for q = t+ where

t(t − 1) z ′ r z

m X (−1)j z =2 F1 (r, −m, s, t) = j=0

Definition: W (r, m, s) =

r+j−1 m j j s+j−1 j

tj

m X r+m+1−j s + m m−j t (−1)j m−j j j=0

Lemma 1: We have the identity (s − r − 1)z ′ + (r + m + 1)(tz ′ + rz) = (−1)m

(s − 1)!m!r(r + 1) W (r, m, s) (s + m − 1)!

Proof: This is just an identity between binomial coefficients that is easily verified.

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Expressing everything in the parameters r, m, s and the function z we have T2′ b4 − b2 b4 + b2 (b1 − b4 )(q1 − t) = + − = T2 t t−1 t(t − 1) s m−r+s−1 r + 1 r(m + s)z − (r + m + 1)t((t − 1)z ′ + rz) − + − = t t−1 t(t − 1) (s − r − 1)z ′ + (r + m + 1)(tz ′ + rz) {(s − r − 1)z ′ + (r + m + 1)(z ′ + rz)}′ (s − r + 1)z ′ + (r + m + 1)(z ′ + rz) after some computations using the hypergeometric equation t(1 − t)z ′′ + (s − (r − m + 1)t)z ′ + mrz = 0 Hence we have shown Proposition 1.. We have (the constant is of no importance) T2 = W (r, m, s) Recall that σm is obtained from τm by the change b4 −→ b4 + 1 In the new parameters this corresponds to r −→ r − 1 m −→ m + 1 s −→ s − 1 We define Sn by σn =

Sn − 1)(b1 +b4 )(b1 −b2 +n−1)

t(b1 +b4 )(b1 +b2 +n−1) (t

Then we have S1 = 1 and S2 = W (r − 1, m + 1, s − 1) The Toda equation for τn and the corresponding one for σn imply that for n ≥ 2 we have Tn+1 Tn−1 = (t2 − t)(Tn Tn′′ − Tn′ 2 ) + (2t − 1)Tn Tn′ + (n − 1)(n + r)Tn2 Sn+1 Sn−1 = (t2 − t)(Sn Sn′′ − Sn′ 2 ) + (2t − 1)Sn Sn′ + (n − 1)(n + r − 1)Sn2 7

Thus we obtain our main result Theorem: For n,m positive integers we have that ′ Tn+1 t(t − 1) Sn′ n+s−1 n+r−m−s qn = t + − − − n+r Sn Tn+1 t t−1 satisfies

(n + r)2 (m + s)2 (r − s + 1)2 1 − (n + m)2 ,− , , ) 2 2 2 2 The Tn and Sn are polynomials. PV I (

Proof: To show that the Tn and Sn are polynomials we refer to the paper [1] where the more general difference equation Pn+1 Pn−1 = f (x)(Pn Pn′′ − Pn′ 2 ) + g(x)Pn Pn′ + hn (x)Pn2 is considered. If hn (x) = n2 + an + b + p(x) then −2hn−1 (x) + hn (x) = −hn−2 (x) + 2 and it follows that the condition f f ′′ − f ′ 2 + 3f ′ g − 2f g ′ − 2g 2 + 2f = 0 is sufficient for the Pn to be polynomials. One checks that f (x) = x2 − x and g(x) = 2x − 1 satisfy this relation. We have to show that two consecutive Tn are relatively prime. If not then Tn and Tn−1 have a common zero, say t0 . It follows from the difference equation that then also Tn′ (t0 ) = 0, i.e. t0 is a double root of Tn . Assume first that t0 6= 0, 1. We have 1 h(t) = t(t − 1)H(t) + σ ′ (b)t − σ(b) 2 where σ(b) is the second symmetric function of b1, b2, b3, b4 and σ ′ (b) is the same of b1, b3, b4. Assume that h(t) =

c + lower order terms (t − t0 )k

But h(t) satifies the differential equation 2

t2 (t − 1)2 h′ h′′ 2 + {(2h − (2t − 1)h′ )h′ + b1 b2 b3 b4 } =

4 Y

j=1

8

(h′ + b2j )

which gives after looking at the highest order terms k=1 c = t0 (t0 − 1) It follows that H(t) =

1 + lower order terms t − t0

and that after integration that t0 is a simple zero of τ (t). Contradiction. To treat the case t = 0 and t = 1 we consider the generic case, i.e.we consider r and s as indeterminates. One sees that Tn (0) and Tn (1) are nonzero polynomials of r and s.

We can also find a determinantal formula for the Tn . Define τen = (t(t − 1)

Then we have

n(n+r+1) 2 Tn+1

τe0 = 1

τe1 = W (r, m, s) · (t(t − 1))

and

δ 2 log(e τn ) = where

r+2 2

τen+1 τen−1 τen2

δ = t(t − 1)

d dt

Darboux’s formula gives τe1 δe τ1 2 δe τ δ τe1 1 τen = · · · · ·· n−1 δ τe1 δ n τe1

· · · δ n−1 τe1 ··· δ n τe1 ··· ··· · · · δ 2n−2 τe1

This formula should possibly be useful in proving the following

Conjecture 1.Given T1 = 1 T2 = W (r, m, s)

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and Tn+1 Tn−1 = Tn2 + Then qn = solves PV I

2 1 (t − t)(Tn Tn′′ − Tn′ 2 ) + (2t − 1)Tn Tn′ (n − 1)(n + r)

m + s Tn (r, m + 1, s) · Tn+1 (r − 1, m, s − 1) · n + r Tn+1 (r, m, s) · Tn (r − 1, m + 1, s − 1)

(n + r)2 (m + s)2 (r − s + 1)2 1 − (n + m)2 ,− , , 2 2 2 2

The discriminant of Tn defined above factors nicely as a polynomial in r and s .We make the Conjecture 2. Define h(k, j) = kj 2 −

j 3 + 2j 3

Then discrim(Tn (r, m, s)) = const· m−1 Y

{(r + n + m − j)(s + j)(s − r + m − 1 − j)}

h(n−1,j)

j=1

m−1 Y

{(r + 1 + j)(s + n + m − 1 − j)(s − r − n + j)}

n−1 Y

{(r + 1 + j)(s + m + m − 1 − j)(s − r − n + j)}h(j,m)

h(m,j)

j=1

j=m

The degree of the discriminant is 3

m(n − 1) 2

Example 1. We consider the special case when s = r + 2. Then one finds that Tn (t) = (t − 1)m(n−2) Tˇn (t) and where

Sn (t) = (t − 1)(m+1)(n−2) Sˇn (t) deg(Tˇn ) = m 10

and

deg(Sˇn ) = m + 1

One can find explicit formulas for Tˇn and Sˇn . Define V (a, m, b, n) =

b X

(−1)j+1

j=0

n+m+a a+j j t b−j j

Then we have the following result: Proposition 2. We have qn = solves PV I (

V (r + 1, m, m + 1, n)V (r, m, m, n) V (r + 1, m, m, n)V (r, m, m + 1, n)

(m + r + 2)2 1 1 − (n + m)2 (n + r)2 ,− ,− , ) 2 2 2 2

2.Some numbertheoretic conjectures. Experiments suggest that the Tn and Sn contain constant factors depending on r . Conjecture 3. Given T1 = 1, T2 = m!W (r, m, s) S1 = 1, S2 = (m + 1)!W (r − 1, m + 1, s − 1) Then define Tn and Sn for n ≥ 3 by Tn+1 Tn−1 −1 Tn2 d Sn+1 Sn−1 d − 1 (t(t − 1) log(Sn )) = (n − 1)(n + r − 1) dt dt Sn2 d d (t(t − 1) log(Tn )) = (n − 1)(n + r) dt dt

for n ≥ 2. Then Tn and Sn are in Z[r, s, t].

In special cases one gets some remarkable difference equations. We give some examples.

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Example 2. Define c(n) by 9(n + 1)(n + 3) if n ≡ 1 mod 3 (n + 1)(n + 3) otherwise 3 Let further T1 = 1, T2 =

W (3, 2, 1) = 5t2 − 5t + 1 3

S1 = 1, S2 = W (2, 3, 0) = (2t − 1)(10t2 − 10t + 1) and c(n)Tn+1 Tn−1 = (t2 − t)(Tn Tn′′ − Tn′ 2 ) + (2t − 1)Tn Tn′ + (n − 1)(n + 3)Tn2 (n + 2)2 Sn+1 Sn−1 = (t2 − t)(Sn Sn′′ − Sn′ 2 ) + (2t − 1)Sn Sn′ + (n − 1)(n + 2)Sn2 for n ≥ 2. Then we conjecure that all Tn and Sn have integer coefficients and c(n) (and (n + 2)2 ) is best possible, i.e. the coefficients in the polynomials have no common factor other than one. Example 3.Define c(n) by (n + 2)(n + 4) if n is even 8 4(n + 2)(n + 4) if n ≡ 3 mod 4 16(n + 2)(n + 4) if n ≡ 1 mod 4 Let further T1 = 1 T2 =

W (4, 3, 1) = (2t − 1)(7t2 − 7t + 1) 4

and c(n)Tn+1 Tn−1 = (t2 − t)(Tn Tn′′ − Tn′ 2 ) + (2t − 1)Tn Tn′ + (n − 1)(n + 4)Tn2 for n ≥ 2. Then we conjecture that Tn has integer coefficients and c(n) is best possible. Example 4. Define c(n) by 25(n + 3)(n + 5) if n ≡ 1 mod 5 (n + 3)(n + 5) if n ≡ 0, 2 mod 5 5 12

(n + 3)(n + 5) otherwise Let further T1 = 1 T2 =

W (5, 4, 1) = 42t4 − 84t3 + 56t2 − 14t + 1 5

and c(n)Tn+1 Tn−1 = (t2 − t)(Tn Tn′′ − Tn′ 2 ) + (2t − 1)Tn Tn′ + (n − 1)(n + 5)Tn2 We conjecture that Tn has integer coefficients and c(n) is best possible.

Based on these examples we make the

Conjecture 4. Let p be a prime ≥ 3. Define c(n) by p2 (p + n)(p + n − 2) if n ≡ 1 mod p (p + n)(p + n − 2) if n ≡ 0, 2 mod p p (p + n)(p + n − 2) otherwise Then c(n)Tn+1 Tn−1 = (t2 − t)(Tn Tn′′ − Tn′ 2 ) + (2t − 1)Tn Tn′ + (n − 1)(n + p)Tn2 for n ≥ 2 has polynomial Tn with integer coefficients and c(n) is best possible.

We have checked the conjecture for p = 3, 5, 7, 11 and for n up to 20 (at least). Final remark.After this paper was finished the author found the polynomials Tm in [4] which up to a factor and some notation agree with our Tn . There is even a conjectured explicit formula for them ( Conjecture 3.5 ).

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References: 1. G.Almkvist, Polynomial solutions to difference equations connected to Painlev´e II-VI, CA/0208244. 2. G.Almkvist, Some rational solutions to Painlev´e VI. 3. K.Okamoto, Studies on the Painlev´e equations I, Sixth equation PV I ,Ann. Mat. Pura 146 (1987), 337-381. 4. M.Noumi, S.Okada, K.Okamoto, H.Umemura, Special polynomials associated with the Painlev´e equations II in: Integrable systems and algebraic geometry, Kobe-Kyoto 1997, World Science Publishing 1998.

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