Malmquist-Type Theorems for Cubic Hamiltonians

The aim of this paper is to classify the cubic polynomials H(z,x,y)=∑j+k≤3ajk(z)xjyk\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} H(z,x,y)=\sum _{j+k\le 3}a_{jk}(z)x^jy^k \end{aligned}$$\end{document}over the field of algebraic functions such that the corresponding Hamiltonian system x′=Hy,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$x'=H_y,$$\end{document}y′=-Hx\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$y'=-H_x$$\end{document} has at least one transcendental algebroid solution. Ignoring trivial subcases, the investigations essentially lead to several non-trivial Hamiltonians which are closely related to Painlevé’s equations PI\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathrm{P_{I}}$$\end{document}, PII\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathrm{P_{II}}$$\end{document}, P34\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathrm{P_{34}}$$\end{document}, and PIV\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathrm{P_{IV}}$$\end{document}. Up to normalisation of the leading coefficients, common Hamiltonians are HI:-2y3+12x2-zyHII/34:x2y-12y2+12zy+κxHIV:x2y+xy2+2zxy+2κx+2λy13(x3+y3)+zxy+κx+λy,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \begin{array}{rl} \mathrm{H_I}:&{}-2y^3+\frac{1}{2}x^2-zy\\ \mathrm{H_{II/34}}:&{} x^2y-\frac{1}{2}y^2+\frac{1}{2}zy+\kappa x\\ \mathrm{H_{IV}}:&{}\begin{array}{l} x^2y+xy^2+2zxy+2\kappa x+2\lambda y\\ \frac{1}{3}(x^3+y^3)+zxy+\kappa x+\lambda y,\end{array} \end{array} \end{aligned}$$\end{document}but the zoo of non-equivalent Hamiltonians turns out to be much larger.


Introduction
Malmquist's so-called First Theorem [6] singles out (linear and) Riccati differential equations w = a 0 (z) + a 1 (z)w + a 2 (z)w 2 (1) among the variety of differential equations w = R(z, w) (R rational) (2) by postulating the existence of some transcendental meromorphic solution (always in the whole plane). For a long time, Malmquist's theorem was viewed as a singular and isolated result in the field of complex differential equations. With Nevanlinna theory as a tool it became the template for various theorems of this kind. Instead of citing the legion of original papers the reader is referred to Laine's monograph [5] and the more recent book [8].
If R is merely rational in w with coefficients analytic on some planar domain, the very same result is obtained by postulating the absence of movable critical and essential singularities of the solutions. This is abbreviated by saying that among the Eqs. (2) only (1) has the Painlevé property. For the first-and second-order case (Fuchs and Painlevé, respectively) the reader is referred to the books of Ince [3] and Golubew [1]. It is quite reasonable to believe that (algebraic) differential equations having the Painlevé property may also be characterised by the aforementioned Malmquist property, although the situation is quite different: arbitrary analytic versus rational coefficients on one hand, and the totality of solutions versus one transcendental solution on the other.
The nature of the problem makes the appearance of 'many-valued' algebroid instead of 'single-valued' meromorphic functions inevitable. One of our main tools will therefore be the Selberg-Valiron theory of algebroid functions in place of Nevanlinna theory. The interested reader will find a rudimentary description in the appendix at the end of this paper.

Six Theorems of Malmquist-Type
The aim of this paper is to support the aforementioned duality principle by proving Malmquist-type theorems for two-dimensional Hamiltonian systems with cubic Hamiltonians over the field of algebraic functions. It is nothing more than an exercise to show that our results may be generalised insofar as the terms 'algebraic coefficients' and 'transcendental algebroid solutions' may be replaced with 'algebroid coefficients' and 'admissible algebroid solutions', that is, solutions that grow much faster than the coefficients measured in terms of the Selberg-Valiron characteristic. Any Hamiltonian (4) such that the corresponding Hamiltonian system possesses some transcendental algebroid solution is said to have the Malmquist property. We will start with polynomials Replacing x and y with x − a 20 and y − a 02 , respectively, the corresponding Hamiltonian system may easily be transformed into with Hamiltonian This system has the Painlevé property if and only if holds, see Kecker [4]. Of course, the trivial case a = b = c = 0 will be excluded. (6) has the Malmquist property. Then the resonance condition

Theorem 1 Suppose the Hamiltonian
holds for either one or two or all third roots of unity (ω 3 = 1). In the third case, (6) certainly has the Malmquist and Painlevé property (7).
The question whether or not just one or two of the necessary conditions (8) are also sufficient for the polynomial (6) to have the Malmquist resp. Painlevé property will be answered in the following theorem.
Then (6) has the Malmquist property in both cases. In case (ii), the necessary condition b = −a holds in addition to the Painlevé property, while in case (i) even is true, but the Painlevé property fails.
Remark In case (i), x satisfies some Riccati differential equation Then (x, c − x) solves (5), but is 'too weak' to enforce the Painlevé property. In any other case with c(z) = z, say, and a and b constant, x + y − c satisfies some secondorder differential equation which is closely related to Painlevé's fourth differential equation see [7]. Moreover, in case (ii) the functionsωx + ωy − c (ω = e ±2πi/3 ) satisfy simple Riccati equations.
Next we will consider the (already simplified) polynomials containing only one third power.
where α is linear; in that case the corresponding Hamiltonian system has the Painlevé property without any restriction on b.
Remark If α is non-constant we may assume α(z) = z by a linear change of the independent variable. Then w = y + β/12 with β = b 2 − b satisfies Painlevé's first equation hence y is transcendental algebroid and so is x = −y − by, and the Malmquist and Painlevé property hold. Nevertheless the occurrence of the Hamiltonians with b ≡ 0 is really surprising.
the Hamiltonian (9) is transformed into Our next Theorem shows that the Hamiltonian systems corresponding to (9) and (

Under these circumstances both Hamiltonian systems have the Painlevé property.
We note the different cases explicitly: b and z 0 are arbitrary constants, and α is constant or linear. Finally we will consider cubic polynomials with dominating terms x 2 y, x y 2 , again in simplified form: and . The reader will not have any difficulty to adapt the proofs to more sophisticated cases.
Theorem 5 Suppose the Hamiltonian (12) has the Malmquist property. Then u = x + b/2 and v = y + a separately solve second-order differential equations and respectively. The coefficients Remark In the second case of Theorem 5, Eq. (15) takes the form In particular, each Hamiltonian with α = β ≡ 0 has the Malmquist and Painlevé property. In the most important case β(z) = z (remember β = 0) we obtain Painlevé's equation which is closely related to equation XXXIV in Ince's book [3, p. 340], and, of course, Painlevé's second equation Theorem 6 Suppose the Hamiltonian (13) has the Malmquist property. Then either the Painlevé property

Remark
In the most important case c(z) = z and a and b constant, x and y separately satisfy Painlevé equations P IV . In the exceptional case, x and y satisfy Riccati equations

Proof of Theorem 1
From it follows that our algebroid solutions satisfy 2m(r , (for notations and results in Nevanlinna-Selberg-Valiron theory see the appendix). In particular, x and y have infinitely many poles. It is easily seen that the poles of (x, y) are simple with residues (−ω, ω) restricted to ω 3 = 1. Assuming it turns out that ξ 0 , η 0 , ξ 1 , and η 1 , but not ξ 2 and η 2 may be computed (and one of these numbers may be prescribed), but the resonance condition 1 is obtained instead. Thus (8) holds if infinitely many poles with residues (−ω, ω) exist, and a = b = c ≡ 0 if this is true for each third root of unity. It is, however, not at all clear that the poles are regular (not branched)! To exclude this case let p be any pole of (x, y) with residues (−ω, ω) and assume that a, b, and c are regular at z = p, but x and/or y have a branched pole there: Let n and m denote the first index, if any, such that n ≡ 0 mod q, ξ n = 0 and m ≡ 0 mod q, η m = 0. Then the first branched terms n q ξ n t −1+n/q and m q η m t −1+m/q on the left hand sides of the Hamiltonian system (17) are equal to the first branched terms 2ωη m t −1+m/q and 2ωξ n t −1+n/q on the right hand sides corresponding to y 2 and −x 2 , respectively. This implies n = m, nξ n = 2qωη n , and nη n = 2qωξ n , hence also n 2 = 4q 2 , which contradicts n ≡ 0 mod q and proves Theorem 1. 1 Assuming the Painlevé property this holds for every p in the domain of the coefficients and every third root of unity, see [4]. This illuminates the difference between both concepts.

Proof of Theorem 2
In the first case, (x, y) has simple poles with residues (−1, 1) and almost no others, hence holds, and we obtain at almost every pole. Then x + y − c has at most finitely many poles, and from m(r , x + y−c) = O(log r +log T (r , x)+log T (r , y)) and m(r , c +b−a) = O(log r ) it follows that x + y − c and c + b − a vanish identically. Also x and y satisfy Riccati differential equations To proceed we will derive the resonance condition α = 0, which holds for algebraic α and algebroid solutions to (21) as well as for rational α and transcendental meromorphic solutions; here the argument is due to Wittich [9]. From  T (r , w))) as r → ∞ outside possibly some exceptional set, hence w has infinitely many poles. Assuming an elementary computation gives c −1 = c 0 = c 1 = 0, c 2 = −α( p)/10, and c 3 = −α ( p)/6, while c 4 remains undetermined (and free), but holds instead. This requires α ( p) = 0, and since w has infinitely many poles and α is algebraic, the assertion α ≡ 0 follows. Again we have to assure that the poles p of w are not branched, at least when α is regular at z = p. To this end write and let n denote the smallest index, if any, such that c n = 0 and n ≡ 0 mod q. Then the first branched terms on the left and right hand side of the differential equation (21) are n q n q − 1 c n (z − p) −2+n/q and 12c n (z − p) −2+n/q .

Proof of Theorem 4
From u = a − 1 24 β 2 + 1 12 b 2 β + bu + βv − 6v 2 v = −u − bv the second-order differential equation v = −a + 1 24 β 2 − 1 12 b 2 β + 6v 2 easily follows, hence is also linear. In combination with (22) we obtain hence b satisfies the differential equation The proof of Theorem 4 then follows from the subsequent Proposition Algebraic solutions to (25) do not exist if κz + λ ≡ 0. If κ = λ = 0, the non-constant algebraic solutions have the form Proof Let b be any algebraic solution with p-fold pole at z = z 0 . Then the single terms in (25) have poles of order p + 3, 2 p + 2, 2 p + 2, and 3 p + 1, respectively. For p > 1, 3 p + 1 dominates the other orders, while for p < 1 this role is taken by p + 3. This means p = 1 and b is any local solution, where q > −1 is some rational number. Then (25) yields This requires (27) and  for c q ; this is possible since J (R, q) = 0 for q ∈ N and R = −1, 6. Thus our algebraic function b has no finite algebraic poles. To exclude other algebraic singularities, that is, to prove that b is a rational function, it does not suffice to indicate that every initial and let n denote the smallest integer, if any, such that c n = 0 and n ≡ 0 mod q; call this n index of b. Then b has index n − 3q, while the indices of bb , b 2 , and b 2 b are ≥ n − 2q, which is not possible. Also at z = ∞, holds, and again we obtain a contradiction since |κ| + |λ| > 0 was assumed. This finishes the proof of the proposition.
To finish the proof of Theorem 4 we just note that for any (local) solution to the Hamiltonian systems x = H y , y = −H x resp. u = K v , v = −K u , solve Painlevé's first equation w = α(z) + 6w 2 resp. v = α(z) + 6v 2 (with the very same α). Thus y and v are meromorphic in C, and so are u = −v − bv and x = −y − by, hence H has the Malmquist (and Painlevé) property.

Proof of Theorem 5
Starting with the Hamiltonian system by the transformation Then (14) and (15) Proof Let p be a pole of u with residue and assume that α and β are regular at z = p. Then hold. Thus u − α(z) − β(z)u − 2u 3 , 2α − β and u + u 2 + 1 2 β vanish at z = p. If u has infinitely many poles with residue , (29) follows at once, and also (30) if u in addition has infinitely many poles with residue − . If, however, only finitely many poles with residue − exist, vanishes identically since has at most finitely many poles, vanishes at almost each pole of u with residue , and has characteristic as r → ∞, possibly outside some set of finite measure.
Again (32) and (33) simultaneously have or fail to have the Malmquist resp. Painlevé property. By Theorem 1, the latter holds for (33) if and only if that is, if and only if c = b = a ≡ 0. And so the circle is complete, since by Theorems 1 and 2 this is true except in one particular case, namely when It is easily seen that this is equivalent with x + y − c = c + b − a ≡ 0.
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Appendix: Algebroid Functions and the Selberg-Valiron Theory
For the convenience of the reader we will give a short overview of the Selberg-Valiron theory. Let be any irreducible polynomial in w over the ring of entire functions. Then the solutions w = f κ (z) (1 ≤ κ ≤ k) to the equation P(z, w) = 0 admit unrestricted analytic continuation into C \ S P , where S P denotes the set of singularities; it consists of the zeros of A k and the discriminant of P w.r.t. w. The singularities (including poles) are algebraic; ordinary poles will not be viewed as singularities. The branches f κ form the algebroid function For algebroid functions, Selberg and Valiron independently developed an analogous Nevanlinna theory as follows (see, for example [8]): m(r , f) = 1 2kπ