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Computational Methods and Function Theory
, Volume 21 (1) – Feb 6, 2021

/lp/springer-journals/malmquist-type-theorems-for-cubic-hamiltonians-ZZ7ReBb5rJ

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- Springer Journals
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- Copyright © The Author(s) 2021
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- 1617-9447
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- 10.1007/s40315-020-00356-3
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The aim of this paper is to classify the cubic polynomials j k H (z, x , y) = a (z)x y jk j +k≤3 over the ﬁeld of algebraic functions such that the corresponding Hamiltonian system x = H , y =− H has at least one transcendental algebroid solution. Ignoring trivial y x subcases, the investigations essentially lead to several non-trivial Hamiltonians which are closely related to Painlevé’s equations P ,P ,P , and P . Up to normalisation I II 34 IV of the leading coefﬁcients, common Hamiltonians are 3 1 2 H :−2 y + x − zy 1 1 2 2 H : x y − y + zy + κ x II/34 2 2 2 2 x y + xy + 2zx y + 2κ x + 2λ y H : IV 1 3 3 (x + y ) + zx y + κ x + λ y, but the zoo of non-equivalent Hamiltonians turns out to be much larger. Keywords Hamiltonian system · Painlevé differential equation · Painlevé property · Malmquist property · Algebroid function Mathematics Subject Classiﬁcation 30D35 · 30D45 · 34M05 Dedicated to the memory of Stephan Ruscheweyh. Communicated by Vladimir V. Andrievskii. B Norbert Steinmetz stein@math.tu-dortmund.de http://www.mathematik.tu-dortmund.de/steinmetz/ Fakultät für Mathematik, Technische Universität Dortmund, Beethovenstrasse 17, 67360 Lingenfeld, Germany 123 44 N. Steinmetz 1 Introduction Malmquist’s so-called First Theorem [6] singles out (linear and) Riccati differential equations w = a (z) + a (z)w + a (z)w (1) 0 1 2 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 there- fore 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. 2 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 x = H (z, x , y), y =− H (z, x , y) (3) y x with cubic Hamiltonians j k H (z, x , y) = a (z)x y (4) jk j +k≤3 123 Malmquist-Type Theorems for Cubic Hamiltonians 45 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 solutionissaidtohavethe Malmquist property. We will start with polynomials 1 3 3 j k H (z, x , y) = (x + y ) + a (z)x y . jk j +k≤2 Replacing x and y with x − a and y − a , respectively, the corresponding Hamil- 20 02 tonian system may easily be transformed into x = y + cx + a (5) y =−x − cy − b with Hamiltonian 3 3 H (z, x , y) = (x + y ) + c(z)xy + b(z)x + a(z) y. (6) This system has the Painlevé property if and only if c = a = b ≡0(7) holds, see Kecker [4]. Of course, the trivial case a = b = c = 0 will be excluded. Theorem 1 Suppose the Hamiltonian (6) has the Malmquist property. Then the reso- nance condition c (z) + ωb (z) −¯ ωa (z) ≡0(8) holds for either one or two or all third roots of unity (ω = 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. Theorem 2 Suppose the necessary condition (8) holds ±2π i /3 (i) for ω = 1, say, but not for ω = e ; ±2π i /3 (ii) for ω = e , say, but not for ω = 1. 123 46 N. Steinmetz 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 x + y − c = c + b − a ≡ 0 is true, but the Painlevé property fails. Remark In case (i), x satisﬁes some Riccati differential equation 2 2 x = x − cx + a − c . 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 second- order differential equation which is closely related to Painlevé’s fourth differential equation 2 4 3 2 2 2ww = w + 3w + 8zw + 4(z − α)w + 2β, (P ) IV ±2π i /3 see [7]. Moreover, in case (ii) the functions ω ¯ x + ω y − c (ω = e ) satisfy simple Riccati equations. Next we will consider the (already simpliﬁed) polynomials 1 2 3 H (z, x , y) = a(z) y + b(z)xy + x − 2 y (9) containing only one third power. Theorem 3 Suppose the Hamiltonian (9) has the Malmquist property. Then the coef- ﬁcients a and b are coupled, 2 2 4 a =−α + (−2b + 4bb + 3b + 2b b − b ), (10) 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 − b satisﬁes Painlevé’s ﬁrst equation w = z + 6w , (P ) 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. 123 Malmquist-Type Theorems for Cubic Hamiltonians 47 By 1 1 2 x = u + bβ, y = v − β(β = b − b ) 12 12 the Hamiltonian (9)istransformed into 1 2 1 2 1 2 1 2 3 K (z, u,v) = a − β + b β v + u + buv + βv − 2v . (11) 24 12 2 2 Our next Theorem shows that the Hamiltonian systems corresponding to (9) and (11) simultaneously can have the Malmquist (and Painlevé) property in very special cases only. In particular, this shows that in general neither the Malmquist nor the Painlevé property is invariant under a linear change of variables of the Hamiltonian. Theorem 4 Suppose the Hamiltonians (9) and (11) have the Malmquist property. Then b is either constant or has the form b(z) = ( R =−1or R = 6). z − z Under these circumstances both Hamiltonian systems have the Painlevé property. We note the different cases explicitly: b 1 2 3 − α(z) + y + x + bx y − 2 y ; 24 2 xy 1 2 3 −α(z) y + x − − 2 y ; 2 z−z 105 1 6xy 2 3 − α(z) + y + x + − 2 y ; 2 z−z 2(z−z ) 0 b and z are arbitrary constants, and α is constant or linear. 2 2 Finally we will consider cubic polynomials with dominating terms x y, xy ,again in simpliﬁed form: 2 1 2 2 H (z, x , y) = x y − y + a(z)x + b(z)xy + c(z)x + d(z) y (12) and 2 2 H (z, x , y) = x y + xy + c(z)xy + b(z) y + a(z)x (13) (a, b, c, d algebraic functions). 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 u = α + β u + 2u (14) 123 48 N. Steinmetz and 2 3 2 1 2 2vv − v = 4v − 2βv + (β − 2α )v − (β − 2α) , (15) respectively. The coefﬁcients 1 2 1 β = 2a + b − b + 2d and α =−ab + c − a + β (16) 2 2 satisfy either (i) α =±β /2 for one sign or else (ii) α = β ≡ 0. Remark In the second case of Theorem 5,Eq. (15) takes the form 2 3 2 2vv − v = 4v − 2βv − λ; λ = (2α − β ) /4 is constant. 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 2 3 2 1 2vv = v + 4v − 2zv + α − , (P ) which is closely related to equation XXXIV in Ince’s book [3, p. 340], and, of course, Painlevé’s second equation u = α + zu + 2u . (P ) II Theorem 6 Suppose the Hamiltonian (13) has the Malmquist property. Then either the Painlevé property c = a = b ≡ 0 or else x + y − c ≡ 0 holds. Remark In the most important case c(z) = z and a and b constant, x and y sepa- rately satisfy Painlevé equations P . In the exceptional case, x and y satisfy Riccati IV equations 2 2 x =−x + 3c(z)x + b(z) and y =− y − 3c(z) y − a(z). 3 Proof of Theorem 1 From x = y + cx + a (17) y =−x − cy − b 123 Malmquist-Type Theorems for Cubic Hamiltonians 49 it follows that our algebroid solutions satisfy 2m(r , y) ≤ m(r , x ) + O(log(rT (r , x ))) and 2m(r , x ) ≤ m(r , y) + O(log(rT (r , y))), hence m(r , x ) + m(r , y) = O(log(rT (r , x )) + log(rT (r , y))) (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 = 1. Assuming x =−ω¯ + ξ + ξ t + ξ t + ··· 0 1 2 (t = z − p) y = ω + η + η t + η t + ··· 0 1 2 it turns out that ξ ,η ,ξ , and η , but not ξ and η may be computed (and one of 0 0 1 1 2 2 these numbers may be prescribed), but the resonance condition c ( p) +¯ ωb ( p) − ωa ( p) = 0[ ] 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: ∞ ∞ −¯ ω ω j /q j /q x = + ξ t and y = + η t (t = z − p). j j t t j =−q+1 j =−q+1 Let n and m denote the ﬁrst index, if any, such that n ≡ 0mod q, ξ = 0 and m ≡ 0 mod q, η = 0. Then the ﬁrst branched terms n m −1+n/q −1+m/q ξ t and η t n m q q on the left hand sides of the Hamiltonian system (17) are equal to the ﬁrst branched terms −1+m/q −1+n/q 2ωη t and 2ωξ ¯ t m n 2 2 on the right hand sides corresponding to y and −x , respectively. This implies n = m, nξ = 2qωη , and nη = 2qωξ ¯ , n n n n 2 2 hence also n = 4q , which contradicts n ≡ 0mod q and proves Theorem 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. 123 50 N. Steinmetz 4 Proof of Theorem 2 In the ﬁrst case, (x , y) has simple poles with residues (−1, 1) and almost no others, hence −1 1 1 2 1 2 x =−(z − p) + c( p) + c( p) + c ( p) − a( p) + b( p) (z − p) + ··· 2 4 3 3 −1 1 1 2 1 y = (z − p) + c( p) − c( p) − c ( p) + a( p) − b( p) (z − p) + ··· 2 4 3 3 holds, and we obtain x + y − c(z) = (c ( p) + b( p) − a( p))(z − p) + ··· 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 2 2 2 2 x = a + c − cx + x and y =−b − c + cy − y , (18) respectively. Conversely, starting with any solution x to the ﬁrst equation (18), the pair (x , c − x ) solves the Hamiltonian system provided c + b − a ≡ 0 holds. This yields the Malmquist property. In the second case the substitution u =¯ ωx + ω y − c,v = ωx +¯ ω y − c, ζ = iz/ 3 transforms the given Hamiltonian system (17)into u =−c + A − 3cu − 2uv − u (19) v =−c + B + 3cv + 2uv + v with A = (ω − 1)a + (ω ¯ − 1)b, B = (1 −¯ ω)a + (1 − ω)b, and Hamiltonian 2 2 (c − B )u + ( A − c )v − 3cuv − u v − uv . The functions u and v can have only ﬁnitely many poles in common. The second equation (19) shows that v even vanishes at almost every pole of u (since uv has to be regular), and from res u = 1 it follows that −v ( p) = B( p) − c ( p) and so uv ≡ c − B since m(r , uv) = O(log r +log T (r , x )+log T (r , y)). Similarly u (q) = A(q)−c (q) is obtained at poles of v (res v =−1), hence uv ≡ A − c , 123 Malmquist-Type Theorems for Cubic Hamiltonians 51 and √ √ 2c = A + B = i 3(a − b) and 2c = i 3(a − b ) ±2π i /3 holds. On the other hand the hypothesis c + ωb −¯ ωa ≡ 0for ω = e implies c = b =−a , hence c = b = a ≡ 0, this proving the Malmquist and Painlevé property. We note that (19) together with uv = A − c = c − B imply the Riccati differential equations 2 2 u = c − A − 3cu − u and v = c − B + 3cv + v . 5 Proof of Theorem 3 From the Hamiltonian system x = a + bx − 6 y (20) y =−by − x the second-order equation 2 2 y =−a + (b − b ) y + 6 y is easily obtained: differentiate the second equation in (20) and then eliminate x and x . The substitution w = y + β with β = b − b leads to w = α(z) + 6w (21) with 1 1 α =−a + β − β . (22) 12 24 To proceed we will derive the resonance condition α = 0, which holds for alge- braic 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 |w | |w | 6|w| ≤|α(z)|+|w| |w | |w| it follows that m(r,w) = O(log(rT (r , w))) as r →∞ outside possibly some excep- tional set, hence w has inﬁnitely many poles. Assuming w(z) = + c (z − p) , (z − p) j =−1 123 52 N. Steinmetz an elementary computation gives c = c = c = 0, c =−α( p)/10, and c = −1 0 1 2 3 −α ( p)/6, while c remains undetermined (and free), but 2 1 2 3 w − α(z) − 6w =− α ( p)(z − p) + O((z − p) )(z → p) 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 j /q w(z) = + c (z − p) (z − p) j =−2q+1 and let n denote the smallest index, if any, such that c = 0 and n ≡ 0mod q. Then the ﬁrst branched terms on the left and right hand side of the differential equation (21) are n n −2+n/q −2+n/q − 1 c (z − p) and 12c (z − p) . n n q q Thus ξ = n/q satisﬁes ξ −ξ = 12, which is absurd since the roots ξ = 4 and ξ =−3 are integers. This ﬁnishes the proof of Theorem 3 since 2 2 2 4 2β − β =−2b + 4bb + 3b + 2b b − b . 6 Proof of Theorem 4 From 1 2 1 2 2 u = a − β + b β + bu + βv − 6v 24 12 v =−u − bv the second-order differential equation 1 2 1 2 2 v = −a + β − b β + 6v 24 12 easily follows, hence 1 1 2 2 α =−a + β − b β (23) 24 12 is also linear. In combination with (22) we obtain 2 2 β − b β − β = 12(α − α), (24) hence b satisﬁes the differential equation 2 2 = b − 2bb − b − b b = κ z + λ. (25) 123 Malmquist-Type Theorems for Cubic Hamiltonians 53 The proof of Theorem 4 then follows from the subsequent Proposition Algebraic solutions to (25) do not exist if κ z + λ ≡ 0.If κ = λ = 0,the non-constant algebraic solutions have the form b(z) = ( R =−1or R = 6, z arbitrary). (26) z − z Proof Let b be any algebraic solution with p-fold pole at z = z . 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(z) ∼ R/(z − z ) as z → 0 with 2 2 3 − 6 R − 4 R − R + R = 0, (27) hence R =−1or R = 6. Now suppose q q b(z) = + c(z − z ) + o(|z − z | ) as z → z (c = 0) 0 0 0 z − z is any local solution, where q > −1 is some rational number. Then (25) yields 3 2 −4 q−3 q−3 (z) = ( R − 5 R − 6 R)z + J ( R, q) c(z − z ) + o(|z − z | ) 0 0 as z → z with 2 2 2 3 J ( R, q) = 4 R − 2 R − (2 + 4 R − R )q + (3 + 2 R)q − q . This requires (27) and 2 3 J (6, q) =− 48 + 10q + 15q − q =0resp. 2 3 J (−1, q) =− 6 + 3q + q − q = 0 if κ = λ = 0. Apart from q =−2 < −1 in both cases, the roots are irrational √ √ (q = (17 ± 193)/2) and non-real (q = (3 ± i 3)/2), respectively, which is absurd. Non-constant algebraic solutions have either ﬁnite poles or a pole at inﬁnity of order 2 3 p−1 p > 0, say. Here the term −b b ∼ cz dominates the other terms which are 2 p−2 3 p−1 O(|z| ) = o(|z| ) as z →∞ and inhibits ≡ 0. This proves (26) for non- constant solutions in case of κ = λ = 0. We have to exclude the case |κ|+|λ| > 0. Here q−3 q−3 (z) = J ( R, q) c(z − z ) + o(|z − z | ) = κ z + λ + κ(z − z ) 0 0 0 0 Like many other computations also these were performed by my favourite tool maple. 123 54 N. Steinmetz requires q = 3 and J ( R, 3) c = κ z + λ if κ z + λ = 0, and q = 4 and J ( R, 4) c = κ 0 0 if κ z + λ = 0. This way we get a unique formal solution + c (z − z ) q 0 z − z q=3 by successively solving rather elaborate equations J ( R, q) c = (c ,..., c )(q = 4, 5,...) q q 3 q−1 for c ; this is possible since J ( R, q) = 0for 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 value problem = κ z + λ, b(z ) = b , b (z ) = b , b (z ) = b has a unique local 0 0 0 1 0 2 solution. Assume ν/q b(z) = c (z − z ) ν 0 ν=0 and let n denote the smallest integer, if any, such that c = 0 and n ≡ 0mod q; call 2 2 this n index of b. Then b has index n − 3q, while the indices of bb , b , and b b are ≥ n − 2q, which is not possible. Also at z =∞, 2 2 −b b + o(|b b |) = κ z +λ(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 =− H resp. u = K ,v =−K , y x v u 1 2 7 −2 w = y + b ,w = y,w = y + z resp.v 12 2 2 2 solve Painlevé’s ﬁrst equation w = α(z) + 6w resp. v = α(z) + 6v (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. 7 Proof of Theorem 5 Starting with the Hamiltonian system x = x − y + bx + d y =−2xy − 2ax − by − c 123 Malmquist-Type Theorems for Cubic Hamiltonians 55 we obtain 1 1 2 2 u = u − v + a + d + b − b 2 4 (28) v =−2uv + ab + a − c by the transformation u = x + b,v = y + a. Then (14) and (15) with coefﬁcients (16) are obtained in the usual way from (28): differentiate the ﬁrst and second equation and replace the variable v and u with the help of the second and ﬁrst equation, respectively. To proceed we note that any algebroid solution to (14) has inﬁnitely many poles, almost all of them simple with residues ±1. The latter follows by inspection, while the former stems from the dominating term 2u , which immediately implies m(r , u) = O(log(rT (r , u))). Proposition Suppose the differential equation (14) has some transcendental algebroid solution having inﬁnitely many poles with residue ∈{−1, 1}. Then α = β (29) holds, and α = β ≡0(30) if u has inﬁnitely many poles also with residue −. Otherwise u satisﬁes 1 1 u =− β − u and α = β . (31) 2 2 Proof Let p be a pole of u with residue and assume that α and β are regular at z = p. Then −1 1 1 2 3 u(z) = (z − p) − β( p)(z − p) − (α( p) + β ( p))(z − p) + c (z − p) + ··· 6 2 and 3 1 u − α(z) − β(z)u − 2u =− (2α ( p) − β ( p))(z − p) + O(|z − p| ) hold. Thus 3 2 1 u − α(z) − β(z)u − 2u , 2α − β and u + u + β 123 56 N. Steinmetz 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, = u + u + β vanishes identically since has at most ﬁnitely many poles, vanishes at almost each pole of u with residue , and has characteristic T (r,) ≤ 2m(r , u) + O(log r ) = O(log(rT (r , u))) as r →∞, possibly outside some set of ﬁnite measure. 8 Proof of Theorem 6 Hamiltonians H (z, x , y) and K (z, u,v) = kH (z, au + bv, cu + dv) with a, b, c, d, k ∈ C and k(ad − bc) = 0 simultaneously have or fail to have the 2π i /3 Malmquist resp. Painlevé property. Choosing a = d = 1, b = c = λ = e , and k =−1/3 we obtain 1 1 1 2 2 1 3 1 3 K (z, u,v) =− (λa + b)u − (a + λb)v − λc(u − uv + v ) + u + v 3 3 3 3 3 and 1 2 1 u =− (a + λb) − λcv + cu + v 3 3 3 (32) 1 2 1 2 v = (λa + b) + λcu − cv − u . 3 3 3 By u = U + λc/3 and v = V + λc/3, system (32) is transformed into U = A + CU + V (33) V =− B − CV − U with 1 1 1 A =− (λa + b + λc ), B =− (a + λb − λc ), and C = λc. 3 3 3 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 C = B = A ≡ 0, 123 Malmquist-Type Theorems for Cubic Hamiltonians 57 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 U + V − C = C + B − A ≡ 0. It is easily seen that this is equivalent with x + y − c = c + b − a ≡ 0. Funding Open Access funding enabled and organized by Projekt DEAL. Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. 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 P(z,w) = A (z)w ( A (z) ≡ 0) κ k κ=0 be any irreducible polynomial in w over the ring of entire functions. Then the solutions w = f (z) (1 ≤ κ ≤ k) to the equation P(z,w) = 0 admit unrestricted analytic continuation into C \ S , where S denotes the set of singularities; it consists of the P P zeros of A 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 form the algebroid function f ={ f ,..., f }. 1 k For algebroid functions, Selberg and Valiron independently developed an analogous Nevanlinna theory as follows (see, for example [8]): 2π 1 1 + i θ m(r , f) = log | f (re )| dθ, N (r , f) = N (r , 1/ A ) κ k 2kπ k κ=1 and T (r , f) = m(r , f) + N (r , f) 123 58 N. Steinmetz denote the proximity function,the counting function of poles, and the Selberg charac- teristic of f, respectively. Up to a bounded term the latter coincides with the Valiron characteristic 2π i θ T (r , f) = U (re ) dθ, 2kπ where 2π i φ U (z) = log | P(z, e )| dφ 2π = log | A (z)|+ log | f (z)| k κ κ=1 = max log | A (z)|+ O(1); 1≤κ≤k the deﬁnition of T is similar to the Ahlfors–Shimizu formula in ordinary Nevanlinna theory (see, for example, Hayman’s monograph [2]). We will not distinguish between both characteristics and just write T (r , f).The First Main Theorem T r , = T (r , f) + O(1) f − c k κ follows from T (r , 1/f) = T (r , f) (based on w P(z, 1/w) = A (z)w ) V V k−κ κ=0 combined with T (r , f − c) = T (r , f) + O(1) (based on || f |−| f − c|| ≤ |c|). S S κ κ Algebraic functions have characteristic T (r , f) = O(log r ) as r →∞, while for transcendental (non-algebraic) algebroid functions log r = o(T (r , f)) holds. The fun- damental result in Nevanlinna theory, the Lemma on the proximity function of the logarithmic derivative remains valid (f /f has branches f / f ): m(r , f /f) = O(log(rT (r , f))) (r →∞) holds (possibly) outside some set E ⊂ (0, ∞) of ﬁnite measure. References 1. Golubew, W.W.: Differentialgleichungen im Komplexen. Dt. Verlag d. Wiss, Berlin (1958) 2. Hayman, W.K.: Meromorphic Functions. Oxford Clarendon Press, Oxford (1964) 3. Ince, E.L.: Ordinary Differential Equations. Dover, New York (1956) 4. Kecker, T.: Polynomial Hamiltonian systems with movable algebraic singularities. J. d’Analyse Math. 129, 196–218 (2016) 5. Laine, I.: Nevanlinna Theory and Complex Differential Equations. de Gruyter, Berlin (1993) 6. Malmquist, J.: Sur les fonctions à un nombre ﬁni de branches satisfaisant à une équation différentielle du premier ordre. Acta Math. 36, 59–79 (1913) 7. Steinmetz, N.: An old new class of meromorphic functions. J. d’Analyse Math. 134, 616–641 (2018) 8. Steinmetz, N.: Nevanlinna Theory, Normal Families, and Algebraic Differential Equations, Springer UTX. Springer, Berlin (2017) 123 Malmquist-Type Theorems for Cubic Hamiltonians 59 9. Wittich, H.: Eindeutige Lösungen der Differentialgleichung w = P(z,w). Math. Ann. 125, 355–365 (1953) Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional afﬁliations.

Computational Methods and Function Theory – Springer Journals

**Published: ** Feb 6, 2021

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