AN OLD NEW CLASS OF MEROMORPHIC FUNCTIONS 6 NORBERT STEINMETZ 1 0 2 n Abstract a Based on the so-called re-scaling method, we will give a detailed de- J scription of the solutions to the Hamiltonian system (1) below, which 5 was discovered only recently by Kecker [4, 5], and is strongly related 1 to Painlev´e’s fourth differential equation. In particular, the problem to determinethosefourth Painlev´etranscendentswithpositiveNevanlinna ] V deficiency δ(0,w),is completely resolved. C Keywords. Hamiltonian system, Riccati differential equation, Painlev´e transcendent, . h asymptotic series, re-scaling, pole-free sector, Nevanlinnadeficiency t a 2010 MSC. 30D30, 30D35, 30D45 m 1. Introduction [ The Hamiltonian system 1 v (1) p = q2 zp α, q =p2+zq+β ′ ′ 2 − − − 3 with time-dependent Hamiltonian 0 (2) H(z,p,q)= 1(p3+q3)+zpq+βp+αq 4 3 0 hasbeendiscoveredbyKecker[4,5]whenquestingforsystemshavingtheso-called . 1 (pseudo) Painlev´e property. Roughly speaking, this means that the system has 0 “nomovablesingularitiesbut poles”,or,moreprecisely,thateverysolutionadmits 6 unrestrictedanalytic continuationin the plane exceptfor fixedsingularities(which 1 do not occur here); and “pseudo” means that, in addition, also moving algebraic : v singularities are admitted. It is not hard to verify, although more elaborate to i discover,that w =p+q z solves the differential equation X − r (3) 2ww′′ =w′2 w4 4zw3 (2α+2β+3z2)w2 (α β+1)2. a − − − − − Kecker [4] found some implicit second order equation for q. Obviously, equation (3) is closely related to Painlev´e’s fourth equation (4) 2yy =y2+3y4+8zy3+4(z2 αˆ)y2+2βˆ ′′ ′ − with parameters αˆ = i (α+β) and βˆ = 2(α β +1)2; one has just to consider √3 9 − y(z)=aw(bz) with b= 4 4 and a= 1b3. Since also w˜ =ωp+ω¯q z (ω3 =1) −3 −2 − is a solutionto (3) withq(α,β) replacedby (ωα,ω¯β), p and q may be re-discovered: (5) (ω ω¯)p=w˜ ω¯w (ω¯ 1)z and (ω¯ ω)q =w˜ ωw (ω 1)z. − − − − − − − − This couldbe the endof the story. The “oldnew” transcendentsp and q, however, have so many interesting properties that it seems justified to study them in their own right. 1 2 NORBERTSTEINMETZ 2. Notation and simple properties By Λ we denote the set of non-zero poles of p. It is easily seen that the poles λ aresimple,andp andq haveresidues̺ and ̺¯, respectively,where ̺ is somethird − rootofunity. ThusΛisdividedinanaturalwayintothreedisjointsetsΛ̺,̺3 =1. With Λ we associate the sets (6) (λ)= z : z λ <δ λ 1 and Λ = (λ). δ − δ δ △ { | − | | | } △ λ Λ [∈ If it is clear which solution (p,q) is under consideration, we will also write H(z)= H(z,p(z),q(z)). The following holds at poles in Λ1: p(z) = (z λ) 1+ 1λ+ 1+ 1(α 2β) 1λ2 (z λ) − − 2 3 − − 4 − + h 5 + 1(α β) λ (z λ)2+ (7) − 8 4 −(cid:0) − ··· (cid:1) q(z) = (z λ) 1+ 1λ+ 1+ 1(2α β)+ 1λ2 (z λ) − (cid:0)− −(cid:0) 2 (cid:1)3(cid:1) − 4 − + h+ 5 + 1(α β) λ (z λ)2+ 8 4 −(cid:0) − ··· (cid:1) (cid:0) (cid:0) (cid:1) (cid:1) (8) H(z)=(z λ) 1+[2h+ 1λ3+ 1(α+β)λ]+ 1(α+β)+ 3λ2 (z λ)+ − − 3 2 3 4 − ··· (cid:2) (cid:3) (9) p(z)+q(z) z =(1+α β)(z λ)+2h(z λ)2+ − − − − ··· Thecoefficienth=h(λ) remainsundetermined andfree. Actuallyλ andh maybe prescribedtodetermineauniquesolution. TodeterminetheLaurentseriesatpoles with residue ̺ (̺3 = 1), we replace p and q by x = ̺¯p and y = ̺q, respectively. Then x and y satisfy x = y2 zx ̺¯α, y = x2+zy+̺β, and we thus obtain ′ ′ − − − the Laurent developments for p=̺x and q =̺¯y by replacing (p,q,α,β) in (7) by (x,y,̺¯α,̺β): p(z) = ̺(z λ) 1+ 1̺λ+ ̺+ 1(α 2̺¯β) 1̺λ2 (z λ)+ (10) − − 2 3 − − 4 − ··· q(z) = ̺¯(z λ) 1+ 1̺¯λ+ ̺¯+ 1(2̺α β)+ 1̺¯λ2 (z λ)+ − − − 2 (cid:0) 3 − 4 (cid:1) − ··· In particular, this means that the deve(cid:0)lopment (8) remains va(cid:1)lid at poles with residue ̺, providedα andβ arereplacedby ̺¯α and̺β, respectively. Hence g(z)= exp( H(z)dz) is an entire function that has simple zeros at the poles of p, and no others. R 2.1. B¨acklund transformations. Trivial B¨acklund transformations are (ω3 = 1 arbitrary) p˜(z)=ω¯p(z), q˜(z)=ωq(z) M : ω (α,β) (ω¯α,ωβ) (cid:26) 7→ p˜(z)= iq(iz), q˜(z)= ip(iz) (11) R: − − (α,β) ( β, α) (cid:26) 7→ − − p˜(z)=p(z¯), q˜(z)=q(z¯) R: (α,β) (α¯,β¯) (cid:26) 7→ Nontrivial B¨acklund transformations were found by Kecker [4] ωα ω¯β+1 p˜(z)=p(z) ω¯ − − ωp(z)+ω¯q(z) z (12) B :  ωα ω¯β+1− ω  q˜(z)=q(z)+ωωp(z)+−ω¯q(z) z − (α,β) (ωβ ω¯,ω¯α+ω) 7→ −  AN OLD NEW CLASS OF MEROMORPHIC FUNCTIONS 3 (as long as the denominator does not vanish identically). By B¨acklund transfor- mation we mean any repeated application of the above transformations. B¨acklund transformationsactonpairs(p,q),componentspandq,andalsoparameters(α,β). Under the B¨acklund transformation B the residues ̺=res p change as follows: ω λ ̺=ω¯ : res B p=̺ λ ω ̺6=ω¯ : B p is regular at λ and ω ωp+ω¯q z =0 z=λ (13) ωp(λ˜)+ω¯q(λ˜) λ˜ =0 − | (λ˜ not a po−le of p) : res B p=ω¯  λ˜ ω ωp′(λ˜)+ω¯q′(λ˜)= ωα+ω¯β  − 3. Rescaling 3.1. Yosida functions. Let a and b > 1 be real parameters. By definition, the − class Y consists of all meromorphic functions f such that the family (f ) of a,b κ κ>1 | | functions (14) e fκ(z)=κ−af(κ+κ−bz) is normal on C in the sense of Montel, and all limit functions f = lim f are , at least one of them being non-constant. If, in addition, all liκmni→t∞funκcntions 6≡ ∞ are non-constant, then f is said to belong to the Yosida class Y . The functions a,b of class Y were introduced by Yosida [12], and for arbitrary real parameters 0,0 by the author [9]. The class Y is universal in the sense that it contains all 0,0 limit functions f = lim f for f Y . The functions f Y have striking κn→∞ κn ∈ a,b ∈ a,b properties, for example they satisfy f♯(z) = O(z a+b), T(r,f) = O(r2+2b), and | | | | e m(r,f)=O(logr), andevenT(r,f) r2+2b and m(r,1/f )=O(logr) if f Y . ′ a,b ≍ ∈ For notations and results in Nevanlinna Theory the readeris referredto Hayman’s monograph[3]. 3.2. An application to the system (1). We quote fromShimomura’spaper [6], section 5.2, and also from section 6 in the author’s paper [8] the following facts aboutthesolutionsto(4)and,obviously,alsotothefunctionsw=p+q z solving − (3) and to w =̺¯p+̺q z (̺3 =1) which solve similar equations: − For δ >0 sufficiently small, the discs (λ) (λ Λ) are mutually disjoint; δ • △ ∈ w =O(z ) as z outside Λ (for the definition of and Λ cf. (6)). δ δ δ • | | →∞ △ It follows from (5) that the second condition is equivalent to (15) p + q =O(z ) (z ,z / Λ ), δ | | | | | | →∞ ∈ while (1) implies (16) p + q =O(z 2) (z ,z / Λ ). ′ ′ δ | | | | | | →∞ ∈ Ifλisanypolewithresidueres p=̺,thenitfollowsfrom(7)thatf =̺p+p2 ̺zp λ ′ − is regular on (λ), and f(z) = O(z 2) on ∂ (λ) continues to hold on (λ) δ δ δ △ | | △ △ by the Maximum Principle. This implies p = O(z 2 + p2) on (λ), and on ′ δ | | | | | | △ combination with (16) this gives p q ′ ′ (17) | | + | | =O(1) (z ) without restriction. z 2+ p2 z 2+ q 2 →∞ | | | | | | | | From (17) easily follows: 4 NORBERTSTEINMETZ Theorem 1. The solutions p and q to equation (1) belong to the class Y . The 1,1 limit functions p= lim p and q= lim q solve κn→∞ κn κn→∞ κn e (18) p = q2 p, q =p2+q, 1(p3+q3)+pq=c ′ − − ′ 3 for some constant c depending on (κ ). n Proof. Set f(z)=p(z)/z, z =κ+κ 1z and f (z)=f(z). Then (17) implies − κ p(z) z 1 p(z) z 2 p(z) f♯(z) | ′ || |− + | || |− = z | ′ | +O(1)=O(1+ z ) ≤ 1+ p(z)2 z 2 1+ p(z)2 z 2 | | z 2+ p(z)2 | | | | | |− | | | |− | | | | and fκ♯(z)=|κ|−1f♯(κ+κ−1z)=O(1+|κ|−2|z|)=O(|z|) (z→∞). ByMarty’swell-knowncriterion(seeAhlfors [1]), the family (f ) is normalon κ κ>1 C, and so is the family of functions p (z) = (1+κ 2z)f (z). The| |same is true for κ − κ the family (q ). The limit functions are finite since p and q have simple poles at κ z=0 with residues ̺ and ̺¯, respectively, if lim κ dist(κ ,Λ)=0, while p(0) n n − κn→∞| | and q(0) are finite if liminf κ dist(κ ,Λ)>0. Obviously, p and q solve (18) with n n κn→∞| | (19) c=nl→im∞κ−n3H(κn) if lκimn→in∞f|κn|dist(κn,Λ)>0 and (20) c= lim λ 3(2h(λ )+ 1λ3) if lim κ κ λ =0 n→∞ −n n 3 n κn→∞| n|| n− n| for some sequence of poles (λ ). (cid:4) n Remark. Constant limit functions are (0,0) and ( ω, ω¯) with ω3 = 1. They − − correspond to c=0 and c= 1, respectively. 3 3.3. The cluster set. By definition, the cluster set C(p,q) C of any non-trivial ⊂ solution (p,q) to (1) consists of all limits (19). Theorem2. TheclustersetC(p,q)isclosed,bounded, andconnected,andcontains all limits (20). Proof. For δ > 0 sufficiently small, the closed discs ¯ (λ) about the poles λ = 0 δ are mutually disjoint, hence any two points a,b D△= C ¯ (λ) may6 be ∈ δ \ λ Λ△δ joined by a curve that is contained in D z : z min a, b ∈. We denote the δ corresponding cluster set of z 3H(z) as z∩{ | o|n≥D b{y|C| |(Sp|,}q}), and note that − δ δ C(p,q) = C (p,q). The cluster sets C→(p∞,q) are closed, bounded in C since δ>0 δ δ z 3H(z) is uniformly bounded on D , and connected by the special property of − δ S D . Since C (p,q) C (p,q) for 0<η <δ, it remains to show that δ η δ ⊃ (21) C (p,q) C (p,q) (0<η <δ sufficiently small), η δ ⊂ and that the limits (20) belong to C (p,q). If δ η κ dist(κ ,Λ)= κ κ λ <δ n n n n n ≤| | | || − | holdsforsomeλ Λ,andifp= lim p ,q= lim q andc= lim κ 3H(κ ) exist,wereplacenκ∈ byλ withthκenf→ol∞lowκinngeffecκtn:→f∞romκnλ =κ +κκn→1∞z w−nith(zn) n n n n −n n n bounded, hence z z as we may assume, it follows that λ 1 = (1+o(1))κ 1, n → 0 −n −n p (z)=(1+o(1))p (z z +o(1))andq (z)=(1+o(1))q (z z +o(1)),hence κn λn − 0 κn λn − 0 ˆp(z)= lim p (z)=p(z+z ), ˆq(z)= lim q (z)=q(z+z ) λn→∞ λn 0 λn→∞ λn 0 AN OLD NEW CLASS OF MEROMORPHIC FUNCTIONS 5 and 1(ˆp3+ˆq3)+ˆpˆq=c= lim λ 3(2h(λ )+ 1λ3). 3 n −n n 3 n →∞ Finally, if we start with sequences (p ) and (q ) with λ Λ, we may as well λn λn n ∈ consider (p ) and (q ) with κ κ λ =δ, hence κn κn | n|| n− n| c= lim λ 3(2h(λ )+ 1λ3)= lim κ 3H(κ ) C (p,q). λn→∞ −n n 3 n κn→∞ −n n ∈ δ Combining both arguments this also proves (21). (cid:4) 3.4. An algebraic curve. The algebraic curve (22) 1(u3+v3)+uv =c 3 is reducible if c= 1: u3+v3+3uv 1= (v+̺u ̺¯), • 3 − − ̺3=1 has genus zero if c=0, and Q • has genus one otherwise. • In the first case (c= 1) the corresponding Hamiltonian system 3 (23) u = v2 u, v =u2+v ′ ′ − − u+ω hassolutionsgivenby =ei√3t (ω = 1( 1+i√3))andv =1 u,andsimilar u+ω¯ 2 − − expressionsif v =̺¯ ̺u and v =̺ ̺¯u,respectively. It is just importantto know − − that the poles form a 2π/√3 periodic sequence with fixed residue. If the genus − 3ωe2t 3ω¯et is zero (c = 0), the system (23) is solved by u = − and v = − , with e3t+1 e3t+1 poles forming a 2πi/3 periodic sequence, this time with alternating residues. In − the remaining cases, the solutions to (23) are elliptic functions parametrising the curve (22); they have elliptic order three with corresponding lattice L depending c only on c. 4. Value distribution 4.1. The order of growth. We note as a corollary to Theorem 1 the following estimate: Theorem 3. The solutions to equation (1) have order of growth at most four: (24) T(r,p)+T(r,q)=O(r4). Proof. From f♯(z) = O(z ) and T(r,p) = T(r,f)+O(logr) for f(z) = p(z)/z we | | easily obtain (24) when using the Ahlfors-Shimizu form of the Nevanlinna charac- teristic, r dt 1 T(r,f)= A(t) with A(t)= f♯(z)2dxdy. (cid:4) t π Z0 Z|z|<t Remark. Thisresultcorrespondstothemeanwhilewell-establishedestimate[6,8] for the order of growth of the fourth Painlev´e transcendents. 6 NORBERTSTEINMETZ 4.2. Solutionsofmaximalgrowth. SupposethattheclustersetC(p,q)contains some parameter lim H(κ )κ 3 = 0,1. Since H(z)z 3 is varying slowly, this following from κn→∞ n −n 6 3 − d H(z)z 3 = 3H(z)z 4+p(z)q(z)z 3 =O(z 1) outside Λ , − − − − δ dz − | | given ǫ> 0 sufficiently small there exists η > 0 such that the re-scaling procedure for κ˜ κ <η κ leads to limit functions p and q with n n n | − | | | p3+q3+3pq=3c (min c, c 1 ǫ, c 1/ǫ). {| | | − 3|}≥ | |≤ Nowpandqareelliptic functionswithfundamentalparallelogramPc,whosediam- eter and area is uniformly bounded and bounded away from zero. It is thus easily deduced that the disc z κ <η κ contains at least constκ 4 poles of p (it is n n n | − | | | | | almost the same to say that the disc z <r contains πr2 lattice points m+in), | | ∼ hence n(2κ ,p) constκ 4 n n | | ≥ | | holds. On combination with n(r,Λ)=O(r4) we thus have: Theorem 4. Let (p,q) be any solution to equation (1). Then C(p,q) 0,1 6⊂ { 3} implies n(r ,Λ) r4, at least on some sequence r . k ≍ k k →∞ 4.3. Thedistributionofresidues. Leavingtherationalsolutionsasidewehence- forth will consider only transcendental solutions. From q2 = p′ zp α= p(z+p′/p) α − − − − − itfollowsbytheusualrulesofNevanlinnaTheorythat2m(r,q) m(r,p)+O(logr), ≤ and in the same manner 2m(r,p) m(r,q)+O(logr) is obtained, hence m(r,p)+ ≤ m(r,q) = O(logr), T(r,p) = N(r,Λ)+O(logr), and T(r,q) = N(r,Λ)+O(logr) hold; here N(r,Λ) denotes the common Nevanlinna counting function of poles. Cauchy’s Residue Theorem yields 1 (25) p(z)dz =n(r,Λ1)+̺n(r,Λ̺)+̺¯n(r,Λ̺¯) (̺= 1( 1+i√3)), 2πi 2 − ZCr provided the circle C : z = r intersects no pole. If C intersects the disc (λ), r r δ | | △ we replace the arc C (λ) by the sub-arc of ∂ (λ) outside C if λ r, and r δ δ r ∩△ △ | |≤ inside C otherwise. This way we obtain a simple closed curve Γ such that r r p(z) + q(z) =O(z ) holds on Γ , r | | | | | | without changing the integral (25). The length of the part of Γ restricted to any r sector of central angle Θ is O(rΘ), hence the integral is O(r2). Taking real and imaginary parts, we obtain n(r,Λ1) 1(n(r,Λ̺)+n(r,Λ̺¯)) = O(r2) (26) − 2 (̺= 1( 1+i√3)). √3(n(r,Λ̺) n(r,Λ̺¯)) = O(r2) 2 − 2 − Theorem 5. For any transcendental solution (p,q) to equation (1), the counting functions n(r,Λ̺) are equal up to a term O(r2), that is, we have n(r,Λ̺)= 1n(r,Λ)+O(r2) (̺3 =1). 3 In particular, if one kind of poles is missing, p and q have order of growth at most two: T(r,p)+T(r,q)=O(r2). AN OLD NEW CLASS OF MEROMORPHIC FUNCTIONS 7 4.4. Strings of poles. Re-scalingalong any sequence of poles with corresponding limit c 0,1 leads to the following situation: given ǫ>0 and R>0 there exists ∈{ 3} r >0, such that for any pole λ in z >r , the disc (λ) contains the poles 0 0 0 R | | △ λk =λ0+k(̟+ǫk)λ−01 (|ǫk|<ǫ, −k1 ≤k ≤k2) with ̟ =2πi/3 and ̟ =2π/√3, respectively, and no others (k =k =[R/̟ ] if 1 2 | | R is large and not an integer multiple of ̟ ). In other words, for C(p,q) 0,1 , | | ⊂{ 3} eachpole λ with λ sufficiently largebelongs to some unique sequence (λ ), called n | | string of poles; it satisfies the approximative recursion (27) λn+1 =λn±(̟+o(1))λ−n1. Setting σ =λ2 we obtain σ =σ 2̟+o(1), σ =2n̟+o(n), n n n+1 n± n λ = 2n̟(1+o(1)) and argλ = 1arg̟+o(1) modπ; n | | n 2 2 the counting fupnction of λ is given by n(r, λ ) r2 . We note that the { n} { n} ∼ 2̟ estimate n(r,Λ)=O(r4) now can be completed: | | Theorem 6. Any transcendental solution (p,q) to (1) satisfies r2 =O(n(r,Λ)) and n(r,Λ)=O(r4). Proof. Thisis obviousifΛcontainsafull string. Inanycasethe re-scalingmethod shows that to any pole λ of sufficiently large modulus there exists some pole λ˜ satisfying λ < λ˜ < λ +O(λ 1), this implying r2 =O(n(r,Λ)). (cid:4) − | | | | | | | | 5. Asymptotic expansions and pole-free sectors 5.1. Pole-free sectors. Let (p,q) be any transcendental solution to (1). If p has no poles on some sector S : argz θˆ <θ, z >r , the re-scaling procedure with 0 | − | | | κ S : argz θˆ <θ δleadstosolutions(p,q)withoutpoles,hencetoconstant n δ ∈ | − | − solutions. The possible constantsare (0,0)and( τ, τ¯), againwith τ3 =1,hence − − we have either p(z) = o(z ) and q(z) = o(z ), or else p(z) = τz +o(z ) and | | | | − | | q(z) = τ¯z+o(z ) as z on S. In each case we will prove that this leads to − | | → ∞ asymptotic expansions on certain pole-free sectors. 5.2. Asymptotics on pole-free sectors with C(p,q)= 0 . In the first case { } the following holds in more generality: Theorem 7. Suppose that p(z)=o(z ) and q(z)=o(z ) hold as z on some | | | | →∞ ray σˆ :argz =θˆ π modπ. Then p, q, and H have asymptotic expansions 6≡ 4 2 α α+β2 3α+β2+2α2β p(z) + ∼ −z − z3 − z5 ··· β β α2 3β α2 2αβ2 (28) q(z) + − + − − + ∼ −z z3 z5 ··· αβ α3+β3 H(z) + (z ) ∼ − z − 3z3 ··· →∞ on the sector Σ : argz νπ < π that contains the ray σˆ. ν | − 2| 4 Proof. From our hypothesis it also follows that H(z)= o(z 3) holds along σˆ. Re- | | scaling along any sequence (κ ) on σˆ yields limit functions p and q satisfying n p′ = q2 p, q′ =p2+q, p(0)=q(0)=0, and p3+q3+3pq=0, − − 8 NORBERTSTEINMETZ hence p = q 0. Thus there are pole-free discs z reiθˆ < r 1κ(r) such that − ≡ | − | κ(r) as r . To identify the maximal pole-free sector that contains σˆ (if → ∞ → ∞ any),westartwithr >0sufficientlylargeanddefinethesequence(r )inductively 0 n by r =r +4r 1. By θ we denote the largest number such that n+1 n n− n A = z :r z r , θˆ argz <θ n n n+1 n { ≤| |≤ ≤ } contains no pole of p, noting that r θ κ(r ) . We may assume that there n n n ∼ →∞ exists some pole λ˜ on ∂A , hence argλ˜ =θ . The same is true at least for some 0 0 0 0 sub-sequence∂A . Re-scalingalongthesequence(λ )thenyieldslimitfunctions nk nk ˜pand˜q thathavesimple polesatz=0 andsatisfy˜p3+˜q3+3˜p˜q=3c. Since A is nk large with respect to the metric ds = z dz and contains no pole, ˜p and ˜q cannot | || | beelliptic functions,andthisand lim H(z)z 3 =0on A implies c=0,hence − n n z the algebraic curve (22) has genus→ze∞ro. It follows from Hurwitz’ Theorem on the S poles of limit functions that to each pole λ˜ there exist five poles nk z =λ˜ +ν(̟+o(1))λ˜ 1 ( 2 ν 2, ̟ = 2πi/3) k,ν nk −nk − ≤ ≤ ± on z λ˜ < 5λ˜ 1, and no others. Since z does not belong to the annulus n n − k, 2 | − | | | ± r z r , it follows that A and A , hence each A contains sonmk e≤po|le| λ˜≤ onnk+it1s boundary, and (λ˜ n)k−is1a sub-senqku+e1nce of some sequnence (λ ) n n n of poles satisfying the approximate recursion (27). Thus argλ and θ approach n n (2ν+1)π >θˆforsomeν. Thesameargumentappliestotheregions z :r z 4 { n ≤| |≤ r , θ < argz θˆ , this showing that the sectors Σ are the natural pole-free n+1 n ν ≤ } sectors in the sense that for every δ >0, S = z : argz νπ < π δ contains δ { | − 2| 4 − } only finitely many poles. For z S sufficiently large we set max p(z), q(z) =ǫz . Since δ ∈ {| | | |} | | (29) p(z) + q(z) =o(z ) (z on S ), δ | | | | | | →∞ 2 we may assume ǫ = ǫ(z) < 1. From (1) and p(z) + q (z) 0 as z on S 2 | ′ | | ′ | → → ∞ δ (this following from (29) and Cauchy’s Theorem), hence p(z) + q (z) < 1, say, ′ ′ | | | | we obtain zp(z) <ǫ2 z 2+ α +1 and zq(z) <ǫ2 z 2+ β +1. | | | | | | | | | | | | This yields (ǫ ǫ2)z 2 <K and ǫ<2K z 2, hence − − | | | | p(z) + q(z) =O(z 1) and p(z) + q (z) =O(z 2) (z on S ), − ′ ′ − δ | | | | | | | | | | | | →∞ again by Cauchy’s Theorem. From (1) it then follows that zp(z)+α=O(z −2) and zq(z)+β =O(z −2) | | | | as z on S . Now assume that δ →∞ n p(z) = αz 1+ a z 2ν 1+O(z 2n 3)=φ (z)+O(z 2n 3) − ν − − − − n − − − | | | | (30) Xν=n1 q(z) = βz−1+ bνz−2ν−1+O(z −2n−3)=ψn(z)+O(z −2n−3) − | | | | ν=1 X has already been proved. From (1) it then follows that zp(z)+α = q(z)2 p(z)= ψ (z)2 φ (z)+O(z 2n 4) − − ′ − n − ′n | |− − zq(z)+β = p(z)2+q (z)= φ (z)2+ψ (z)+O(z 2n 4) − ′ − n n′ | |− − AN OLD NEW CLASS OF MEROMORPHIC FUNCTIONS 9 holds. Since φ , ψ , φ2, and ψ2 are even functions, (30) holds with n replaced by ′n n′ n n n+1. A more detailed computation(1) then gives (28). (cid:4) 1 5.3. Asymptotics on pole-free sectors with C(p,q)= . The other cases 3 { } are yet easier to deal with, since the principal terms τz and τ¯z (τ3 = 1) are − − already known. Details are left to the reader. There is, however, one remarkable difference: the natural pole-free sectors now are Σ : νπ < argz < (ν +1)π, and ν 2 2 these sectors are bordered by sequences of poles λ satisfying (27) with ̟ = 2π n √3 and having counting function n(r, λ ) √3r2. We also note that in the present { n} ∼ 4π case the algebraic curve (22) is reducible (c= 1). 3 Theorem 8. Suppose that p(z) = τz+o(z ) and q(z) = τ¯z+o(z ) (τ3 = 1) − | | − | | hold as z on some ray σˆ : argz = θˆ 0 modπ. Then p, q, and H have → ∞ 6≡ 2 asymptotic expansions p(z) ∼ −τz+ α+23τ¯zβ+τ + 3α+2τ¯β+2τ¯α29+z33β2+4ταβ−2τ +··· (31) q(z) ∼ −τ¯z+ β+23τzα−τ¯ − 3β+2τα−2τβ29−z33α2−4τ¯αβ+2τ¯ +··· H(z) ∼ z33 −(τα+τ¯β)z+ τα2+τ¯β32z+αβ−1 +··· as z on the sector Σ :νπ <argz <(ν+1)π that contains σˆ. →∞ ν 2 2 5.4. Asymptotics on adjacent pole-free sectors. Boththeoremsmaybe com- pleted as follows: Theorem 9. Suppose that the same asymptotics (28) and (31) hold on adjacent sectors Σν and Σν+1, respectively. Then this is true on Σν Σν+1 ◦. ∪ Proof. Inbothcases,g(z)=exp H(z)dz isanentirefu(cid:0)nctionoffin(cid:1)iteorder( 4) ≤ having simple zeros at the poles of p and q. Taking into account the asymptotics R (28) and (31) of H, we consider f(z)=zαβg(z) and f(z)=g(z)z (τα2+τ¯β2+αβ 1)/3 exp( z4 + 1(τα+τ¯β)z2) − − −12 2 in the respective cases. Then on argz θ¯ < π, say (argz = θ¯ denotes the ray | − ν| 8 ν that separates Σ and Σ ), f has order of growth at most four, ν ν+1 loglog f(z) limsup | | 4 ( argz θ¯ < π), log z ≤ | − ν| 8 z →∞ | | and satisfies rlim f(rei(θ¯ν±δ)) = C± for δ = 1π0, say. The Phragm´en-Lindel¨of Principle(2) th→en∞shows that C =C = C, and f(z)=C+o(1) holds as z + on argz θ¯ <δ. Hencef andalso−g hasonlyfinitelymanyzeros,andphas→on∞ly ν | − | finitely many poles on that sector, that is, p and q have an asymptotic expansion on argz θˆ <δ which coincides with those on the sectors Σ and Σ . (cid:4) ν ν+1 | − | 1Manyofthecomputations atvariousplaces wereperformedwiththehelpofmaple. 2The version we will use can easily be derived from the following standard version: Let f be holomorphic on the domain D : |z| > r0, Imz > 0, with continuous boundary values log|f(z)| on (−∞,−r0)∪(r0,+∞). If f satisfies limsup = 0 on D, lim f(x) = a and z→∞ |z| x→+∞ lim f(x)=b, then a=b and lim f(z)=a holds on D. x→−∞ z→∞ 10 NORBERTSTEINMETZ 5.5. B¨acklund transformations and asymptotics. Under the B¨acklundtrans- formation B the asymptotics (31) changes as follows: ω ω =τ¯ τz (32) p(z)∼−τz and ω =τ¯ implies Bωp(z)∼ −τ¯z. (cid:26) 6 (cid:26) − 5.6. Existence and uniquenessofasymptoticexpansions. InTheorem7and 8itwasshownthatspecificsolutionshaveasymptoticexpansionsoncertainsectors ofcentralangle π. Thisshouldnotbeconfusedwithournextresultaboutexistence 2 and uniqueness of solutions having asymptotic expansions. Theorem 10. To any half-plane argz (2ν +1)π < π and argz νπ < π | − 4| 2 | − 2| 2 there exists a unique solution to (1) with prescribed asymptotics (28) and (31), respectively. Proof. To prove existence we set, in the first case (28), t = z2, p(z) = t−21u(t)2, and q(z)=t−12v(t)2 to obtain α+u2 u2 v4 β+v2 u4+v2 u˙ = + − , v˙ = + . − 4u 4tu 4v 4tv Then given any half plane H, Theorem 14.1 in Wasow’s monograph [11] applies to the corresponding system for x = u √ α and y = v √ β in H; it yields − − − − existence of some solution having asymptotic expansion u √ α + ∞ a t k, k − ∼ − k=1 v √ β+ ∞ b t k ast onH. Thus(1)has somesolutionwithasPymptotic k − ∼ − →∞ k=1 expansion (nPecessarily given by (28)) on S = √H. If, however, S contains some ray σˆ :argz = νπ, then by Theorem 7, the asymptotic expansion extends to the ν 2 half-plane argz νπ < π. Wenote,however,thattheproofonlyworksifαβ =0. | − 2| 2 6 In the trivial case α= β = 0 we have p q 0. In case of α= 0 and β =0, say, ≡ ≡ 6 we set t=z2, p(z)=t 3/2u(t)2, and q(z)=t 1/2v(t)2 to obtain the system − − u2+v4 3u v2+β v u u˙ = + , v˙ = + + . − 4u 4t 4v 4t 4t3v Again, Theorem 14.1 in [11] applies to the system for x = u iβ, y = v √ β − − − (u2+v4 =v2+β =0 is solved by v =√ β, u= iβ). − ± In the same way one may deal with the second case (31): just set t = z2, p(z) = t12u(t)2, and q(z)=t12v(t)2 to obtain u2+v4 α+u2 u4+v2 β v2 u˙ = , v˙ = + − , − 4u − 4tu 4v 4tv and apply Theorem 14.1 in [11] to the system for x=u √ τ, y =v √ τ¯ (the − − − − non-trivial solutions to ξ+η2 =ξ2+η =0 are (ξ,η)=( τ, τ¯) with τ3 =1). − − Toproveuniquenessweconsidertwosolutions(p ,q )and(p ,q )havingthesame 1 1 2 2 asymptotic expansion on some sector S. Then u = p p and v = q q 1 2 1 2 − − u tend to zero faster than every power z n, and u = solves z 1u = A(z)u − v − ′ (cid:18) (cid:19) 1 0 1 2τ¯ with A( ) = − (eigenvalues 1) and A( ) = − − (eigenvalues ∞ 0 1 ± ∞ 2τ 1 (cid:18) (cid:19) (cid:18)− (cid:19) √5), respectively. In every sector argz νπ < π and argz (2ν+1)π < π, ± | − 2| 4 | − 4| 4 respectively, systems of this kind have some solution u that tends to infinity, 1