An old new class of meromorphic functions. Norbert Steinmetz. June 13, 2015

Transcription

1 Norbert Steinmetz June 13, 2015

2 NEW & OLD NEW: Hamiltonian system p = H q = q 2 zp α q = H p = p 2 + zq + β with Hamiltonian H(z, p, q) = 1 3 (p3 + q 3 ) + zpq + αq + βp OLD: Almost Painlevé IV 2ww = w 2 w 4 4zw 3 (2ωα+2 ωβ+3z 2 )w 2 (ωα ωβ+1) 2. OLD & NEW: w = ωp + ωq z (ω 3 = 1).

3 Three reasons to consider p = q 2 zp α q = p 2 + zq + β The solutions are meromorphic in the plane, while any known and essentially different system with Hamiltonian H(z, p, q) = 1 m pm + 1 n qn + nµ+mν<mn a µν (z)p µ q ν (m, n 3) has movable algebraic poles or more complicated singularities (Kecker). The solutions have extremely nice properties New insight in Painlevé IV There exists a method

4 Re-scaling (also re-normalisation) Set p h (z) = h 1 p(h + h 1 z), q h (z) = q(h + h 1 z). p = q 2 zp α q = p 2 + zq + β p = lim h p h q = lim h q h (formal) p = q 2 p q = p 2 + q 1 3 (p3 + q 3 ) + pq = c?! properties of (p, q) (???) properties of (p, q) W 2 The problem (???) disappears if one can show that the families (p h ) h >1 and (q h ) h >1 are normal in the plane.

5 Re-scaling constant solutions to ( ) p = q 2 p q = p 2 + q give rise to asymptotic expansions on Stokes sectors for the solutions to p = q 2 zp α q = p 2 + zq + β non-constant (periodic or elliptic) solutions to ( ) yield information on the distribution of poles of (p, q).

6 Poles Trivial and non-trivial properties p and q have simple poles λ Λ with residues ω and ω, resp. (ω 3 = 1). H has simple poles with residue 1: H(z) = (z λ) 1 + [2h λ3 + 1 (α + β)λ] + 2 (h = h(λ) is unknown and may be prescribed once!) The discs δ (λ) = {z : z λ < δ λ 1 } (λ 0, δ > 0 sufficiently small) are mutually disjoint. p + q = O( z ) and p + q = O( z 2 ) outside the exceptional set Λ δ = λ 0 δ(λ).

7 Poles H(z) = O( z 3 ) outside Λ δ. The order of growth is at most 4: n(r, Λ) = 1 H(z) dz = O(r 4 ) 2πi Γ r (Γ r = approximately the circle z = r). The residues are equi-distributed: n(r, Λ ω ) = 1n(r, Λ) + O(r 2 ) (ω 3 = 1) follows from 3 ωn(r, Λ ω ) = 1 p(z) dz = O(r 2 ). ω 3 =1 2πi Γ r n(r, Λ) = O(r 2 ) if at least one residue ω is missing.

8 Generalised spherical derivative Lemma. p z 2 + p + q 2 z 2 + q = O(1). 2 Proof. Res z=λ p = ω: f = ωp + p 2 ωzp is regular on δ (λ) and satisfies f (z) = O( z 2 ) on δ (λ): Maximum principle. More examples: (R) w = z n w 2 : (P I ) w = z + 6w 2 : (P II ) w = α + zw + 2w 3 : (P IV ) 2ww = w 2 + 3w 4 + 8zw 3 + 4(z 2 α)w 2 + 2β : w z n + w = O(1) 2 w z + w = 2 O( z w z + w = O(1) 2 w z 2 + w 2 = O(1) 1 4 )

9 Generalised spherical derivative Lemma. Suppose f is meromorphic in the z-plane and satisfies f (z) z 2a + f (z) 2 = O( z b a ) (z ) for some a R, b > 1. Then the re-scaled family (f h ) h >1 f h (z) = h a f (h + h b z) is normal in the z-plane. (non-constant f = lim f h k h k function). Proof. Part of the afternoon exam. Yosida

10 Re-scaling The families (p h ) h >1 with p h (z) = h 1 p(h + h 1 z) and (q h ) h >1 are normal on the plane. The limit functions p = lim p h k and q = lim q h k satisfy h k h k p = q 2 p, q = p 2 + q with constant Hamiltonian H(p, q) = 1 3 (p3 + q 3 ) + pq c = lim h k h 3 k H(h k) (inf h k dist(h k, Λ) > 0) lim λ k 2λ 3 k h(λ k) + 1 ( lim h 3 k h k λ k = 0) k the constants c form the cluster set C(p, q); it is closed, bounded and connected.

11 The algebraic curve H(x, y) = 1 3 (x 3 + y 3 ) + xy c = 0 either is reducible (c = 1 3 ), H(x, y) = (y + x 1)(y + ωx ω)(y + ωx ω) or else has genus 0 (c = 0), and is parametrised by x = 3ωt2 3 ωt and y = t t 3 + 1, or else has genus 1 (c 0, 1 ) and is parametrised by a 3 pair of elliptic functions: (x ) 3 + 9x(x ) x 2 x + (x 6 + (16 6c)x 3 + 9c 2 ) = 0 (y ) 3 9y(y ) y 2 y (y 6 (16 6c)y 3 + 9c 2 ) = 0

12 The hamiltonian system p = q 2 p q = p 2 + q has non-constant solutions q = 1 p, p + ω p + ω = 3z ei (ω = e 2πi/3 ), and similar expressions if q = ω ωp in the reducible case p = 3ωe2z 3 ωez and q = in case of genus 0, e 3z + 1 e 3z + 1 p(z) = ωζ(z z ω (c)), q(z) = ωζ(z z ω (c)) in ω 3 =1 ω 3 =1 case of genus 1 with lattice depending on c.

13 Distribution of poles: reducible case p = q 2 p p = q 2 zp α q = p 2 + q q = p 2 + zq + β 1 3 (p3 + q 3 ) + pq = z-plane ( z < 8) λ z-plane ( z λ < 8 λ 1 ) ( λ large) The discs have radii o( λ 1 ).

14 Distribution of poles: genus zero p = q 2 p p = q 2 zp α q = p 2 + q q = p 2 + zq + β 1 3 (p3 + q 3 ) + pq = 0 0 z-plane ( z < 7) z-plane ( z λ < 7 λ 1 ) The residues alternate λ

15 Distribution of poles: genus one p = q 2 p p = q 2 zp α q = p 2 + q q = p 2 + zq + β 1 3 (p3 + q 3 ) + pq = c 0, 1 3 z-plane z-plane

16 The cluster set and sub-normal solutions Transcendental solutions satisfy C 1 r 2 n(r, Λ) C 2 r 4. Solutions satisfying n(r, Λ) r 2 are called sub-normal. C(p, q) {0, 1/3} n(r k, Λ) r 4 k Guess: n(r, w) r 4. Hinkkanen and Laine: C(p, q) {0, 1/3} n(r, Λ) r 2

17 Strings of poles Any sequence (λ k ) satisfying some parabolic-like recursion λ k+1 = λ k + (c + o(1))λ s k (c 0, s > 1 rational) is called a (c, s)-string. λ k ((1 + s)c)k) 1/(1+s) (consider µ k = λ 1+s k ) n(r, {λ k }) r 1+s (1 + s) c arg λ k 1 2π arg c modulo 1+s 1+s Sub-normal solutions have finitely many (± 2π 3, 1)-strings of poles if C(p, q) = {1/3} (± 2πi, 1)-strings of poles if C(p, q) = {0} 3

18 Much ado about nothing: Sub-normal solutions with C(p, q) = {0} do not exist. Proof. There are finitely many strings of poles, each asymptotic to some ray arg z = (2ν + 1)π/4 p, q and H have asymptotic expansions on each sector (2ν 1)π/4 < arg z < (2ν + 1)π/4, in particular H(z) = αβ z α3 + β 3 + O( z 5 ) 3z 3 g(z) = exp( H(z) dz) is entire of order 4, with zeros at the poles of (p, q) Phragmén-Lindelöf argument on arg z 2ν+1 4 π < δ g is a polynomial (of degree αβ), hence p and q are rational functions.

19 C(p, q) = {1/3} Type-1 sub-normal solutions Transcendental solutions(p, q) having only poles with fixed pair of residues (ω, ω) are sub-normal with cluster set C(p, q) = {1/3}. In case of ω = 1: p + q z = 0 α β + 1 = 0 p = α z 2 + zp p 2, q = β + z 2 zq + q 2 One string of poles in four [two adjacent] Stokes directions p ϱz p ϱz p ϱz p ϱz p ϱz p ϱz Generic and degenerate distribution of poles and asymptotics; Res p = 1; ϱ = 1 2 ( 1 + i 3).

20 C(p, q) = {1/3} Type-2 sub-normal solutions Transcendental solutions having only poles with pairs of residues (ω 1, ω 1 ) and (ω 2, ω 2 ) are sub-normal with cluster set C(p, q) = {1/3}. In case of ω 1,2 = 1 2 ( 1 ± i 3): α = β p 2 + q 2 + z 2 pq + z(p + q) + 3β p 2 + A 2 (z, p)p + A 4 (z, p) = 0 (Fuchsian) Two strings of poles in 4 [2 adjacent] directions p ωz p ωz p ωz p ωz p ωz p ωz Res p = ω 1, Res p = ω 2 ; in the narrow domains between two strings, p(z) = z + o( z ) holds.

21 C(p, q) = {1/3} Type-3 sub-normal solutions

22 C(p, q) = {1/3} Type-n sub-normal solutions Theorem. Every sub-normal solutions has some type. Type-n solutions have 4 n [in exceptional cases only 2 n] strings of poles. They exist if n 0 modulo 3 (for known parameters), but never if n 0 modulo 3. p z p ωz p ωz p z p ωz p ωz p ωz p ωz n = 3k + 1 (ω = 1 2 ( 1 + i 3)) n = 3k + 2 4[2] ((k + 1) + k + k) strings 4[2] ((k + 1) + (k + 1) + k) strings

23 Asymptotics and distribution of residues p(z) ωz + p(z) = z + o( z ) p(z) = ωz + o( z ) p(z) = ωz + o( z ) p z + p(z) ωz + (1 st quadrant) p(z) = z + o( z ) p(z) = ωz + o( z ) p(z) = ωz + o( z ) p(z) = z + o( z ) p ωz + (4 th quadrant) Res = 1, Res = ω = 1 2 ( 1 + i 3), Res = ω

24 Bäcklund transformations (Kecker) The Bäcklund transformations (ω 3 = 1) M ω : (p, q) ( ωp, ωq) and ( ) ωα ωβ + 1 p(z) ω p B ω : ωp(z) + ωq(z) z q ωα ωβ + 1 q(z) + ω ωp(z) + ωq(z) z transform p = q 2 zp α, q = p 2 + zq + β into the same Hamiltonian system with resp. parameters ( ωα, ωβ) (ωβ ω, ωα + ω).

25 Main Result Theorem. To any Type-n (n > 1) sub-normal solution (p, q) there exists some Bäcklund transformation B 1 M ω which maps (p, q) onto some Type-(n 1) sub-normal solution if n 2 modulo 3; Type-(n 2) sub-normal solution if n 1 modulo 3. Corollary. Every sub-normal solution has its source in a Type-1 subnormal solution; p and q satisfy first order differential equations P(z, p, p ) = Q(z, q, q ) = 0 of Fuchsian type and are algebraically dependent, K(z, p, q) = 0 (genus zero).

26 Proof of the Main Result p and q have asymptotic expansions p(z) τ ν z +, q(z) τ ν z + (τ 3 ν = 1) on (ν 1)π/2 < arg z < νπ/2; shortly written τ 2 τ 1 τ 3 τ 4. Reason: p = q 2 p, q = p 2 + q, 1 3 (p3 + q 3 ) + pq = 1 3 has the constant solutions (p, q) = (τ ν, τ ν ) under B 1 M ω, old strings of poles may disappear or may change their residue, and new strings may be created n(r, Λ) = n(r, M ω Λ). 81 configurations τ 2 τ 1 τ 3 τ 4 may be reduced to 5.

27 Proof of the Main Result τ 1 τ 1 = τ 1 τ 1 is ruled out (Phragmén-Lindelöf). τ 3 τ 3 τ 3 τ 3 Set p ω = M ω p, Λ ω = M ω Λ and H ω = M ω H. n(r, Λ) n(r, B 1 Λ ω ) = n(r, Λ ω ) n(r, B 1 Λ ω ) = 1 (H ω (z) B 1 H ω (z)) dz 2πi Γ r (H ω B 1 H ω = p ω B 1 p ω ) = 1 (p ω (z) B 1 p ω (z)) dz 2πi Γ r 3r 2 (ω, {τ ν }) 4π (ω, {τ ν }) assumes integer values 1, = 0, and 1 if ω varies. Choose ω with (ω, {τ ν }) 1 to decrease the number of strings.

28 Symbolic illustration ( display poles with different residues, not necessarily the same in different pictures)

29 Application to Painlevé IV 2ww = w 2 w 4 4zw 3 (2α + 2β + 3z 2 )w 2 (α β + 1) 2 is equivalent to Painlevé IV. Known: α β m(r, 1/w) = O(log r) α β + 1 = 0 m(r, 1/w) 1 T (r, w) + O(log r), 2 except when α = 1( 1 ± i 3), β = α + 1, and 2 w = i 3 (3zw + w 2 ) : N(r, 1/w) = 0. Theorem. δ(0, w) = 0, except when w = p + q z and (p, q) is a Type-(3k + 1) sub-normal solution with parameters α k = 1 2 ( 1 ± (2k + 1) 3i), β k = α k + 1, and δ(0, w) = 1 2k + 1. For k = 0, zero is a Picard value.

30 Application to Painlevé IV n = 3k + 1: w = p + q z has 4(k + 1) and 4k strings of simple poles and, respectively, and 4k strings of 2k double zeros : δ(0, w) = 1 k k = 1 2k + 1. n = 3k + 2: α β + 1 = 0 is impossible.

31 Thank you for listening