Algebraic General Solutions of Algebraic ODEs 1)

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1 MM Research Preprints, 1 16 MMRC, AMSS, Academia Sinica No. 23, December Algebraic General Solutions of Algebraic ODEs 1) J.M. Aroca, J. Cano, Ruyong Feng and Xiao-Shan Gao Department of Algebra, Geometry and Topology Fac. Ciencias. Univ. de Valladolid Valladolid 47011, Spain (aroca,jcano)@agt.uva.es Key Laboratory of Mathematics Mechanization Institute of Systems Science, AMSS, Academia Sinica Beijing , China (ryfeng,xgao)@mmrc.iss.ac.cn Abstract. In this paper, we give a necessary and sufficient condition for an algebraic ODE to have an algebraic general solution. For an autonomous first order ODE, we give an optimized bound for the degree of its algebraic general solutions and a polynomialtime algorithm to compute an algebraic general solution if it exists. 1. Introduction Finding the close form solution of an ODE can be traced back to the work of Liouvillian. For the algorithm consideration, the pioneer work is due to Risch. In [17], Risch gave an algorithm to find the elementary integration of udx where u is a rational function or a monomial over Q(x). In Trage s Ph.D thesis [24], he gave a method to compute the integration of algebraic function. In [1], Bronstein generalized Trager s results to elementary functions. For the higher order linear homogeneous ODEs, Kovacic presented an effective method to find the Liouvillian solutions for the second order ODEs [14]. In [21], Singer established a general framework for finding Liouvillian solutions for general linear homogeneous ODEs. Many other interesting results on finding Liouvillian solutions of linear ODEs were reported in [2, 3, 8, 25, 26, 27]. Most of these results are limited to the linear case or some special type nonlinear equations. Work on finding closed form solutions for nonlinear differential equations is not systematic as that for linear equations. With respect to the particular ODEs of the form y = R(x, y) where R(x, y) is a rational function, Darboux and Poincaré made important contributions [16]. More recently, Cerveau, Neto and Carnicer also made important progresses [6, 4, 5]. In particular, Carnicer gave the degree bound of algebraic solutions in the nondicritical case. In [22], Singer studied the Liouvillian first integrals of differential equations. In [12], Hubert gave a method to compute a basis of the general solutions of first order ODEs and applied it to study the local behavior of the solutions. In [19, 20], Feng and Gao gave a necessary and sufficient conditions for an algebraic ODE to have a rational type general solution and a polynomial-time algorithm to compute a rational general solution if it exists. 1) Partially supported by a National Key Basic Research Project of China and by a USA NSF grant CCR

2 2 J.M. Aroca, J. Cano, R. Feng and X.S. Gao In this paper, the idea proposed in [19] is generalized to compute the algebraic function solutions. In Section 2, we give a sufficient and necessary condition for an algebraic ODE to have an algebraic general solution, by constructing a differential equation whose solutions are all algebraic functions. In Section 3, by treating the variable and its derivative as independent variables, an first order autonomous ODE defines an algebraic plane curve. Using Riemann- Hurwitz formula, we give a degree bound of the algebraic function solutions of the equation. This degree bound is optimized in the sense that there is a class of first order autonomous ODEs, whose algebraic function solutions reach this bound. In Section 4, based on the above results and the theory of algebraic approximants we give a polynomial-time algorithm to find an algebraic function general solution for an autonomous first order ODE. A first order algebraic autonomous ODE F ( dy dx, y) = 0 can be reduced to the form, y) = 0, where G is also a polynomial. Then to find the solution of F = 0, we may first find x = φ(y) as a function in y by computing the integration of an algebraic function, and then compute the inversion y = φ 1 (x). Hence, our algorithm is equivalent to a polynomial-time algorithm for finding an algebraic integration for algebraic functions. The previously known algorithms for elementary function integration have exponential worst-case complexity. G( dx dy 2. Algebraic general solutions of algebraic ODEs 2.1. Definition of algebraic general solutions In the following, let K = Q(x) be the differential field of rational functions in x with differential operator d dx and y an indeterminate over K. Let Q be the algebraic closure of the rational number field Q. We denote by y i the i-th derivative of y. We use K{y} to denote the ring of differential polynomials over the differential field K, which consists of the polynomials in the y i with coefficients in K. All differential polynomials in this paper are in K{y}. Let Σ be a system of differential polynomials in K{y}. A zero of Σ is an element in a universal extension field of K, which vanishes every differential polynomial in Σ [18]. In this paper, we also assume that the universal extension field of K contains infinite number of arbitrary constants. We will use C to denote the constant field of the universal extension field of K. Let P K{y}/K. We denote by ord(p ) the highest derivative of y in P, called the order of P. Let o = ord(p ) > 0. We may write P as follows P = a d y d o + a d 1 y d 1 o a 0 where a i are polynomials in y, y 1,..., y o 1 and a d 0. a d is called the initial of P and S = P y o is called the separant of P. The k-th derivative of P is denoted by P (k). Let S be the separant of P, o = ord(p ) and an integer k > 0. Then we have P (k) = Sy o+k + R k (1) where R k is of lower order than o + k. Let P be a differential polynomial of order o. A differential polynomial Q is said to be reduced with respect to P if ord(q) < o or ord(q) = o and deg(q, y o ) < deg(p, y o ). For two

3 Algebraic General Solution 3 differential polynomials P and Q, let R = prem(p, Q) be the differential pseudo-remainder of P with respect to Q. We have the following differential remainder formula for R [13, 18] JP = i B i Q (i) + R where J is a product of certain powers of the initial and separant of Q and B i, R are differential polynomials. Moreover, R is reduced with respect to Q. For a differential polynomial P with order o, we say that P is irreducible if P is irreducible when P is treated as a polynomial in K[y, y 1,..., y o ]. Let P K{y}/K be an irreducible differential polynomial and Σ P = {A K{y} SA 0 mod {P }}. (2) where{p } is the perfect differential ideal generated by P [13, 18]. Ritt proved that [18] Lemma 2.1 Σ P is a prime differential ideal and a differential polynomial Q belongs to Σ P iff prem(q, P ) = 0. Let Σ be a non-trivial prime ideal in K{y}. A zero η of Σ is called a generic zero of Σ if for any differential polynomial P, P (η) = 0 implies that P Σ. It is well known that an ideal Σ is prime iff it has a generic zero [18]. As a consequence of Lemma 2.1, we have Lemma 2.2 Let F K{y}/K be an irreducible differential polynomial with a generic solution η. Then for a differential polynomial P we have P (η) = 0 iff prem(p, F ) = 0. When we say a constant, we mean it is an element in C. general solution is due to Ritt. The following definition of Definition 2.3 Let F K{y}/K be an irreducible differential polynomial. A general solution of F = 0 is defined as a generic zero of Σ F. An algebraic general solution of F = 0 is defined as a general solution ŷ which satisfies the following equation n m j G(x, y) = a i,j x i y j = 0 (3) j=0 i=0 where a i,j are in C and n j=0 mj i=0 a i,jx i y j is irreducible in C[x, y]. When n = 1, ŷ is called a rational general solution of F = 0. For algebraic solutions of a differential equation F = 0, we have the following lemma. Lemma 2.4 Let G(y) C(x)[y] and irreducible in C(x)[y] where C is the algebraic closure of C. If one solution of G(y) = 0 is a solution of F = 0, then every solution of G(y) = 0 is the solution of F = 0. Proof: Since G(y) is irreducible in C(x)[y], every solution of G(y) = 0 is a generic zero of G(y) = 0. By Lemma 2.2, prem(f, G) = 0. That is, S k I l F = P G + QG

4 4 J.M. Aroca, J. Cano, R. Feng and X.S. Gao where S = G y, I is the initial of G and k, l Z. Since every solution of G(y) = 0 is a generic zero, it does not vanish S or I. Hence every solution of G(y) = 0 is a solution of F = 0 The proof is complete. A general solution of F = 0 is usually defined as a family of solutions with o independent parameters in a loose sense where o = ord(f ). The definition given by Ritt is more precise. Theorem 6 in section 12, chapter 2 in [13] tells us that Ritt s definition of general solutions is equivalent to the definition in the classical literature A Criterion for existence of algebraic general solutions For non-negative integers h, α, k, let A (h,α;k) (y) be the following (h + 1) (α + 1) matrix: ( k+1 ) ( k+1 ) yk+1 yk ( 0 1 k+2 ) ( k+2 ) yk+2 yk (. k+h+1 ) ( k+h+1 ) yk+h+1 yk+h 0 1 ( k+1 ) ( yk+1 α α k+2 ) yk+2 α α (. k+h+1 ) yk+h+1 α α Let α = (α 1,, α n ) Z n 0, α 0 Z 0 where Z 0 means the set of non-negative integers. Let A (α0 ;α)(y) be the (h + 1) (h + 1) matrix (A (h,α1 ;α 0 )(y) A (h,α2 ;α 0 )(y 2 ) A (h,αn;α 0 )(y n )) where n + α α n = h + 1. Let D (α0 ;α) be the determinant of A (α0 ;α)(y). Note that if n = 1, D (α0,α) is just equal to D n,m in [19]. Lemma 2.5 An element ȳ in the universal extension of K is a solution of D (α0 ;α) = 0 iff it satisfies equation (3) with m i α i where i = 0,, n. Proof: Assume that ȳ satisfies the equation (3) with m i α i where i = 0,, n. Then we have m n j a i,j (x i ȳ j ) (m0+1) = 0 j=1 i=0 where (x i ȳ j ) (m0+1) means the (m 0 + 1)-th derivative of x i ȳ j with respect to x. Since a i,j are constants, (x i ȳ j ) (m0+1) (i = 0,, m j, j = 1,, n) are linearly dependent over C. That is, the Wronskian determinant W ((x i ȳ j ) (m0+1) ) for (x i ȳ j ) (m0+1) vanishes where j = 0,, n, i = 0,, m j [18]. Then ȳ satisfies the equation (3) with m i α i iff W ((x i ȳ j ) (m0+1) ) = 0. By the computation process, W ((x i ȳ j ) (m 0+1) ) = D (α0 ;α)(ȳ) B1 B 2... B n, B i = 1 x x αi 0 1 α i x αi α i! for i = 0,, n. Hence W ((x i ȳ j ) (m 0+1) ) = 0 D (α0 ;α)(ȳ) = 0. The proof is complete. By the above Lemma, we can prove the following criteria theorem easily. Theorem 2.6 Let F be an irreducible differential polynomial. Then a differential equation F = 0 has an algebraic general solution ŷ iff there exist α = (α 1,, α n ) Z n 0, α 0 Z 0 such that prem(d (α0 ;α), F ) = 0.

5 Algebraic General Solution 5 Proof: ( ) Let ŷ be an algebraic general solution of F = 0 which satisfies the equation (3). Assume that each m i is minimal. Let α = (m 1, m 2,, m n ) and α 0 = m 0. Then from Lemmas 2.2 and 2.5 D (α0,α)(ŷ) = 0 D (α0,α) Σ F prem(d (α0,α), F ) = 0. ( ) By Lemma 2.1, prem(d (α0,α), F ) = 0 which implies that D (α0,α) Σ F. Then all the zeros of Σ F must satisfy the equation (3). In particularly, the generic zero of Σ F satisfies the equation (3). The proof is complete. Given an algebraic differential equation F = 0, if we know the degree bound of the equation (3) with respect to x and y which defines an algebraic general solution of F = 0, then we can decide whether it has an algebraic general solution by computing prem(d (α0,α), F ) step by step. However for ODEs with order greater than one or with variate coefficients, we do not know this bound. Even for the case y = P (x,y) Q(x,y) where P (x, y), Q(x, y) Q[x, y], we have no effective method to get the bound [4, 5, 16]. In the following, for the first order autonomous ODEs, we give a degree bound for its algebraic function solutions. 3. Degree bound for first order autonomous ODEs In the following, we will always assume that F = 0 is a first order autonomous ODE in Q{y} and irreducible in Q{y} and G(x, y) Q[x, y] which is irreducible. We say G(x, y) is nontrivial if deg(g, x) > 0 and deg(g, y) > 0. From now on, we always assume that G(x, y) is nontrivial. When we say that G(x, y) = 0 is an algebraic solution of F = 0, we mean that one of the algebraic functions ŷ(x) defined by G(x, ŷ(x)) = 0 is a solution of F = Structure for the algebraic general solutions It is a trivial fact that for an autonomous ODE, the solution set is invariant by a translation of the independent variable x. Moreover, we have the following fact. Lemma 3.1 Let G(x, y) = 0 be an algebraic solution of F = 0. Then G(x + c, y) = 0 is an algebraic general solution of F = 0, where c is an arbitrary constant. Proof: Assume that ȳ(x) is a formal power series solution of G(x, y) = 0. Then ȳ(x + c) will be a solution of G(x + c, y) = 0. Because ȳ(x) is a solution of F = 0, ȳ(x + c) is still a solution of F = 0. Hence G(x+c, y) = 0 is an algebraic solution of F = 0. For any T K{y} satisfying T (ȳ(x + c)) = 0, let R = prem(t, F ). Then R(ȳ(x + c)) = 0. Suppose that R 0. Since F is irreducible and deg(r, y 1 ) < deg(f, y 1 ), there are two differential polynomials P, Q K{y} such that P F +QR K[y] and P F +QR 0. Thus (P F +QR)(ȳ(x+c)) = 0. Because ȳ(x + c) / Q and c is an arbitrary constant which is transcendental over K, we have P F + QR = 0, a contradiction. Hence R = 0 which means that T Σ F. So ȳ(x + c) is a generic zero of Σ F. Hence G(x + c, y) = 0 is an algebraic general solution. The proof is complete. Lemma 3.1 reduces the problem of finding an algebraic general solution to the problem of finding a nontrivial algebraic solution. In what below, we will show how to find a nontrivial algebraic solution in Q[x, y]. First of all, we decide the degree of an algebraic solution.

6 6 J.M. Aroca, J. Cano, R. Feng and X.S. Gao 3.2. Degree bound of algebraic function solution Assume that G(x, y) = 0 is an algebraic solution of differential equation F = 0. In this subsection, we will give a bound for deg(g, x) and deg(g, y). First of all, we introduce some concepts on algebraic function field in one variable. Definition 3.2 Q(x, α) is called an algebraic function field in one variable, if x is transcendental over Q and α is algebraic over Q(x) [10]. For an irreducible algebraic curve G(x, y) = 0 where G(x, y) Q[x, y], it corresponds to a unique algebraic function field Q(α, β) under isomorphism where α, β satisfies G(α, β) = 0 and α or β is transcendental over Q. It is well known that the genus of an algebraic curve is invariant under birational transformation. About birational transformation, we have the following lemma [10]. Lemma 3.3 Two varieties are birationally equivalent iff their function fields are isomorphic Assume that n = deg(g, y). For fixed c Q { }, let t = x c or 1 x if c =. It is known that 1 k=1 Q((t k )) is algebraically closed where Q((t 1 k )) is the field of the Puiseux series [29]. Then G(x, y) = 0 has n roots y 1,, y n in 1 k=1 Q((t k )). For each y i, there exists a smallest positive integer e such that y i Q((t 1 e )). e is called ramification index of y i and if e > 1, c is call ramified points. Assume that y i (t 1 e ) = a 0 + a 1 t 1 e + a 2 t 2 e +. It is easy to know that y i (εt 1 e ) is still a roots of G(x, y) = 0 in Q((t 1 e )) where ε e = 1. We say that y i (t 1 e ) and y i (εt 1 e ) are equivalent. Hence if y i has ramification index e, then it will be equivalent to e roots of G(x, y) = 0. Each equivalent class is call a place [7, 29]. e is also called the ramification index of the place which includes y i. Therefore, for each c Q { }, we can associate the list of the ramification indices of the places e 1,c,, e h,c. Note that h i=1 e i,c = n. It is known that ramified points is finite. We have the following result, known as Riemann-Hurwitz formula, or simply the Hurwitz formula [7, 15]. Theorem 3.4 Let g be the genus of G(x, y) = 0 and n = deg(g, y). Then we have g = 1 n + 1 (e i,q 1) 2 Q i where e i,q is ramification index at Q and the summation spreads all of the ramified points. Now we are ready to give the degree bound of the algebraic solution of F = 0. First, we could determine the degree deg(g, x) exactly from the degree of F. Theorem 3.5 Let G(x, y) Q[x, y] be irreducible and G(x, y) = 0 is an algebraic solution of F = 0. Then we have deg(g, x) = deg(f, y 1 ) Proof: Assume that deg(g, x) = s and deg(f, y 1 ) = d. Let us write G(x, y) = A 0 (y) + A 1 (y)x + + A s (y)x s F = F 0 (y) + F 1 (y)y F d (y)y d 1 where A i (y), F j (y) Q[y]. and B with respect to z. We use Res(A, B, z) to denote the Sylvester-resultant of A Let S = Zero(A s (y)) Zero(F d (y)) Zero(Res(G, G x, x))

7 Algebraic General Solution 7 Zero(Res(G, G F y, x)) Zero(Res(F, y 1, y 1 )). Then S is a finite set. Hence we can choose a c Q such that c / S. Then we have 1. The set {z Q F (c, z) = 0} = {z 1, z 2,, z d } has exactly d elements. 2. The set {x Q G(x, c) = 0} = {x 1, x 2,, x s } has exactly s elements. 3. Since G y (x i, c) 0, there exists a unique formal power series y i (x) = c + g i,1 (x x i ) + g i,2 (x x i ) 2 + such that G(x, y i (x)) = 0 for each i = 1,, s. From Lemma 2.4, y i (x) is a solution of F = 0. That is, F (y i (x), y i (x)) = 0 which implies that F (c, g i,1 ) = 0. Suppose that s > d. Then there exist two of g i,1 which are equal to each other, without lost of generalization, assume that they are g 1,1 and g 2,1. It is not difficult to certify that g i,j is uniquely decided by c and g i,1 for j 2, because we have g i,j = y i(x) (j) and G(x, y i (x)) (j) x=xi = G y (x i, c)y i (x) (j) x=xi + = 0. Hence y 1 (x) = y 2 (x + x 2 x 1 ). Then we have G(x, y 1 (x)) = G(x, y 2 (x + x 2 x 1 )) = 0 j! x=xi which implies that G(x + x 1 x 2, y 2 (x)) = 0. However, since G(x, y 2 (x)) = 0 and G(x, y) is irreducible, G(x, y) = λg(x + x 1 x 2, y) where λ is a nonzero element in Q. So A i (x) = λa i (x + x 1 x 2 ). Since x 1 x 2 0, A i (x) Q which implies that s = 0, a contradiction. So s d. Let G G = y 1. We have that y + G x H(y, y 1 ) = Res(G, G, x) = y s 1Res(G, G y, x) + terms of lower order in y 1. Since Res(G, G y, x) 0, we have deg(h, y 1) = s. Assume that ȳ(x) is a solution of G(x, y) = 0. Then we have H(ȳ(x), ȳ (x)) = F (ȳ(x), ȳ (x)) = 0. Because F is irreducible, we have that deg(h, y 1 ) deg(f, y 1 ). In the other word, s d. The proof is complete. Since F is first order and autonomous, we can regard F = 0 as an algebraic curve and we will use F (y, y 1 ) to denote F. Lemma 3.6 Assume that G(x, y) = 0 is an algebraic solution of F = 0. Then the genus of G(x, y) = 0 equals to that of F (y, y 1 ) = 0. Proof: Let α satisfy G(x, α) = 0. It is clear that α is transcendental over Q. Then Q(x, α) and Q(α, α ) are the algebraic function fields of G(x, y) = 0 and F (y, y 1 ) = 0 respectively. By Lemma 3.3, we need only to prove Q(x, α) = Q(α, α ). From Theorem 3.5, we have [ Q(x, α) : Q(α)] = [ Q(α, α ) : Q(α)]. Since G(x, α) = 0, α = G x (x,α) α Q(x, α). Hence Q(x, α) = Q(α, α ). The proof is complete. For convenience, we consider a new differential equation (G) y (x,α). which implies that F (y, x 1 ) = x deg(f,y 1) 1 F (y, 1 x 1 ) = 0 (4)

8 8 J.M. Aroca, J. Cano, R. Feng and X.S. Gao where x 1 = dx dy = 1 y 1. F is irreducible in Q[y, x1 ] and deg( F, y) = deg(f, y), deg( F, x 1 ) = deg(f, y 1 ). If G(x, y) = 0 is an algebraic solution of F = 0, then we will prove that it is also an algebraic solution of F = 0. Lemma 3.7 Let F be defined as in (4) and G(x, y) = 0 an algebraic solution of F = 0. Then G(x, y) = 0 also defines an algebraic function (in y) solution of F (y, x 1 ) = 0. Proof: From the proof of Theorem 3.5, we know that Res(G, G, x) = A(y)F (y, y 1 ) where G G = y 1 y + G x. In the other word, there exist two polynomials P, Q Q[x, y, y 1 ] such that P G + QG = A(y)F (y, y 1 ). Substituting y 1 by 1 x 1 and multiplying some power of x 1, we have P G + G Q( y + x G 1 x ) = xk 1A(y) F (y, x 1 ) (5) where P, Q Q[x, y, x 1 ] and k Z 0. Suppose that β satisfies G(β, y) = 0. Substituting x by β and x 1 by β in (5), we have that F (y, β ) = 0. Hence G(x, y) = 0 is an algebraic solution of F = 0. Let y(x) = y 0 + a 1 (x x 0 ) p 1 q + a 2 (x x 0 ) p 2 q + satisfy G(x, y(x)) = 0 where x 0, y 0, a i Q and p i, q Z, q > 0. By the method introduced in Section 2, Chapter 3 in [29], we can get another Puiseux series x(y) = x 0 + b 1 (y y 0 ) r i p 1 + b 2 (y y 0 ) r 2 p 1 + which satisfies G(x(y), y) = 0 where b i Q and r i Z. For more details, let t q = x x 0. Then we have that y(t) = y 0 + i=1 a i t p i. There exists a series d 1 t + d 2 t 2 + such that substituting t by d 1 t + d 2 t 2 +, we have y( t) = y 0 + t p 1. This series can be computed by the method given in [29]. Now let x x 0 = (d 1 t + d 2 t 2 + ) q. Since t = (y y 0 ) 1 p 1, x x 0 = b 1 (y y 0 ) q p 1 + b 2 (y y 0 ) r 2 p 1 +. It is not difficult to certify that G(x 0 + b 1 (y y 0 ) q p 1 +, y) = 0. If the Puiseus series y(x) has other form, we can do the same thing by the above method. Theorem 3.8 Assume that G(x, y) = 0 is an algebraic solution of F = 0. Then we have that deg(g, y) deg(f, y) + deg(f, y 1 ) Proof: Let F be as in (4). Since deg( F, y) = deg(f, y), deg( F, x 1 ) = deg(f, y 1 ), we need only to prove that deg(g, y) deg( F, y) + deg( F, x 1 ). In this proof, when we regard F = 0 as an algebraic curve, we will use F (y, x 1 ) = 0 to denote it. Let g G and g F be the genus of G(x, y) = 0 and F (y, x 1 ) = 0 respectively. By Riemann- Hurwitz formula, to decide deg(g, y), we need only to know the genus of G(x, y) = 0 and all

9 Algebraic General Solution 9 of the ramification indices. The ramification index can be gotten by computing the Puiseux expansion of the branch of an algebraic curve passing through the ramified point. If y(x) is a Puiseux series expansion of G(x, y) = 0 at some ramified point, then we can get a new series x(y). Furthermore, we have dx(y) dy is a Puiseux series solution of F (y, x 1 ) = 0. Now by Riemann-Hurwitz formula again, we can bound the ramification index in this series. In the following, we give more details. By Riemann-Hurwitz formula, we have that g G = 1 deg(g, y) q i (q i 1) (6) where q i are the ramification indices and the summation extends all of the ramified points. All these q i come from one of the following four types of Puiseux series. 1. Puiseux series expansion at x 0 Q. (a) y y 0 = a 1 (x x 0 ) p q i (b) y = a p (x x 0 ) p q i + i 1 a i(x x 0 ) i+p q i + i 1 a i(x x 0 ) i p q i 2. Puiseux series expansion at x 0 =. (a) y y 0 = a 1 ( 1 x ) p q i (b) y = a p ( 1 x ) p q i + i 1 a i( 1 x ) i+p q i + i 1 a i( 1 x ) i p q i where a i Q and p, q i Z >0, q i > 1. Then we change all of these series into Puiseux series solution of F (y, x 1 ) = From (1-a) 2. From (1-b) x x 0 = b 1 (y y 0 ) q i p + b 2 (y y 0 ) q i +1 p + x 1 = c 1 (y y 0 ) q i p p for q i > p, + c 2 (y y 0 ) q i p+1 p + (7) p p+1 q i p q y y 0 = d 1 x i p 1 + d 2 x1 + (8) x x 0 = b 1 ( 1 y ) q i p + b 2 ( 1 y ) q i +1 p + x 1 = c 1 ( 1 y ) p+q i p + c 2 ( 1 y ) q i +p+1 p + p p+1 q i +p y = d 1 x q i +p 1 + d 2 x1 + (9) 3. From (2-a) x = b 1 (y y 0 ) q i p + b 2 (y y 0 ) q i +1 p + x 1 = c 1 (y y 0 ) p+q i p + c 2 (y y 0 ) q i p+1 p + y y 0 = d 1 ( 1 x 1 ) p q i +p + d 2 ( 1 x 1 ) p+1 q i +p + (10)

10 10 J.M. Aroca, J. Cano, R. Feng and X.S. Gao 4. From (2-b) x = b 1 ( 1 y ) q i p + b 2 ( 1 y ) q i +1 p + x 1 = c 1 ( 1 y ) p q i p + c 2 ( 1 y ) p q i +1 p + (11) for p < q i, y = d 1 x p p q i p+1 p q i 1 + d 2 x1 + (12) where b i, c i, d i Q and x 1 = dx dy. Each Puiseux series which has one of the above four types can be changed into one of the series (7),(9),(10),(11). Hence if we have bound q i (q i 1) in (7),(9),(10),(11), we can bound q i (q i 1) in (6). In the case (8) and (9), they are the Puiseux series expanded at x 1 = 0. Hence we have (q i + p) deg( F, y) (13) (8) (q i p) + (9) In the case (10) and (12), they are the Puiseux series expanded at x 1 =. Hence we have i + p) + (10)(q (q i p) deg( F, y) (14) (12) In the case (7) and (11), by Riemann-Hurwitz formula, we have (p 1) 2(g F + deg( F, x 1 ) 1) (15) (7) (p 1) + (11) From (15), we have that (p 1) + (7), (11) q i = p (7), (11) q i > p (p 1) + (7), (11) q i < p (p 1) 2(g F + deg( F, x 1 ) 1) (16) Note that the terms in (8) (q i p) and the terms in (7),q i >p (p 1) correspond one to one. (12) (q i p) and (11),q i >p (p 1) are the same. Combine (13), (14), (16), we have that (q i 1) 2(g F + deg( F, y) + deg( F, x 1 ) 1) (7),(9),(10),(11) Since (6) (q i 1) = (7),(9),(10),(11) (q i 1), (q i 1) 2(g F + deg( F, y) + deg( F, x 1 ) 1) (6) Because F (y, y 1 ) = 0 and F (y, x 1 ) = 0 are birational equivalent by (4), g F = g F. From Lemma 3.6, g G = g F. Then by (6), our proof is complete. The following example shows that the degree bound given in Theorem 3.8 is good. Example 3.9 Assume that n > m > 0 and (n, m) = 1. Let G(x, y) = y n x m which is irreducible. We have that G(x, y) = 0 is an algebraic solution of F = y n m y m 1 ( m n )m = 0. In this case, we have that deg(g, y) = deg(f, y) + deg(f, y 1 ).

11 Algebraic General Solution A polynomial-time algorithm The simple degree bounds given the the preceding section allows us to give a polynomialtime algorithm to compute the algebraic function solutions of a first order autonomous ODE Algebraic approximant Algebraic approximant is a special type of Hermite-Padè approximant. It uses algebraic function to approximate a function [11]. Definition 4.1 Let G(x, y) be an irreducible polynomial in Q[x, y]. An algebraic function ȳ(x) satisfying G(x, ȳ(x)) = 0 is called an algebraic approximant to a function f(x) if where m = deg(g, x) and n = deg(g, y). More generally, we will find G(x, y) such that G(x, f(x)) = O(x (m+1)(n+1) 1 ) G(x, f(x)) = O(x N+1 ) (17) where N is a positive integer. We can get the coefficients of G(x, y) with respect to x and y by solving linear equations. Let n m G(x, y) = b i,j x i y j j=0 i=0 and f(x) = a 0 + a 1 x + + a N x N + O(x N+1 ). Let M 0 = I (m+1) (m+1) 0 (N m) (m+1) (18) where I (m+1) (m+1) is (m + 1) (m + 1) unit matrix, 0 (N m) (m+1) is (N m) (m + 1) zero matrix. Let M i = T M i M 0 for i = 1,, n where a a 1 a T M = a 2 a 1 a 0 0 (19)..... a N a N 1 a N 2 a 0 and a i are the coefficients of f(x). Then by the computation process, we can write (17) as the matrix form B 0 b 0,i B 1 b 1,i (M 0 M 2 M n ). B n = 0, B i = for i = 1,, n. Let ȳ(x) = a 0 + a 1 x + be a formal power series. When we say ϕ(x) is the first N + 1 terms of ȳ(x), we mean that ϕ(x) = a 0 + a 1 x + + a N x N. The following lemma will be used in our algorithm.. b m,i (20)

12 12 J.M. Aroca, J. Cano, R. Feng and X.S. Gao Lemma 4.2 Let ȳ(x) be a formal power series such that G(x, ȳ(x)) = 0. Assume that m = deg(g, x) and n = deg(g, y). Let ϕ(x) be the first 2mn + 1 terms of ȳ(x). If Q 0 (x), Q 1 (x),, Q n (x) Q[x] such that Q 0 (x) + Q 1 (x)ϕ(x) + + Q n (x)ϕ(x) n = O(x 2mn+1 ) where deg(q i (x), x) m and not all of them are zero. Then where λ Q is nonzero. G(x, y) = λ(q 0 (x) + Q 1 (x)y + + Q n (x)y n ) (21) Proof: Let Q(x, y) = Q 0 (x) + Q 1 (x)y + + Q n (x)y n. There exist S, T Q[x, y] such that SG(x, y) + T Q(x, y) = Res(G(x, y), Q(x, y), y) (22) where deg(s, y) < n and deg(t, y) < n. If Q(x, ȳ(x)) = 0, then (21) is true. Assume that Q(x, ȳ(x)) 0 and Res(G, Q, y) 0. Then it is easy to know that deg(res(g, Q, y), x) 2mn. However, substituting ȳ(x) to the left side of (22), the left side will become a series and its order will greater than 2mn, a contradiction. Hence Res(G(x, y), Q(x, y), y) = 0 which implies (21) is true, because G(x, y) is irreducible An algorithm to compute algebraic solutions Firstly, we give an algorithm to compute the first N + 1 terms of a formal power series solution of F = 0 for a given positive integer N. Regarding F = 0 as an algebraic curve, find a point (z 0, z 1 ) on it such that it does not vanish the separant S(y, y 1 ) of F (y, y 1 ). Then we can compute y i = z i step by step from (1). Then ȳ(x) = z 0 + z 1 x + z 2 2! x2 + is a formal power series solution of F = 0. Moreover, if z 1 0, then ȳ(x) / Q. Algorithm 4.3 Input: F = 0 and a positive integer N. Output: the first N + 1 terms of a formal power series solution of F = 0 which is not in Q. 1. Find a point (z 0, z 1 ) Q 2 on F (y, y 1 ) = 0 such that S(y, y 1 ) 0 and y i := 2 and ϕ(x) := z 0 + z 1 x. 3. while i N do (a) Substitute y by ϕ(x) and y 1 by ϕ (x) in F (y, y 1 ). (b) c := the coefficient of x i 1 in F (ϕ(x), ϕ (x)). (c) z i := (i 1)!c S(z 0,z 1 ) and ϕ(x) := ϕ(x) + z ix i i!. (d) i := i Return(ϕ(x)).

13 Algebraic General Solution 13 The correctness of the algorithm comes from the following fact. Let ȳ(x) be a formal power series solution of F = 0. Then by (1), (F (ȳ(x), ȳ 1 (x))) (i 1) = Sȳ i (x) + R(ȳ(x),, ȳ i 1 (x)) = 0. Since ȳ k (x) x=0 = z k for k = 1, 2,, we have that S(z 0, z 1 )z i + R(z 0,, z i 1 ) = 0. Now assume that ϕ(x) = z 0 + z 1 x + + z i 1 (i 1)! xi 1. Then (F (ϕ(x), ϕ (x))) (i 1) = R(ϕ(x),, ϕ (i 1) (x)) Since ϕ (k) (x) x=0 = z k for k = 1,, i 1, we have that R(z 0,, z i 1 ) = (F (ϕ(x), ϕ (x))) (i 1) x=0 which equals to (i 1)! times the coefficient of x i 1 in F (ϕ(x), ϕ (x)). Let T = tdeg(f ), the total degree of F. Theorem 9 given in ([19]) shows that there is at most T 2 points on F (y, y 1 ) = 0 which satisfies S(y, y 1 ) = 0 or y 1 = 0. In most cases, it is very difficult to find a rational point on F (y, y 1 ) = 0. Hence, the basic field of Algorithm 4.3 is an algebraic extension field of Q. Assume that Q(α) is the basic field and L(z) is the minimal polynomial of α. Then deg(l) T. Let β, γ Q(α). Then there exist P (z), Q(z) Q[z] such that β = P (α), γ = Q(α) where deg(p ) T 1, deg(q) T 1. We have β γ = R(α) where R = prem(p Q, L). R can be computed in O(T 2 ). Hence the complexity of the algorithm in Q(α) equals to at most T 2 times that in Q. The complexity of Algorithm 4.3 is polynomial. In Step 1, we can find a point (z 0, z 1 ) as the following: set z 0 = 0, ±1, ±2,... and z 1 is an algebraic number determined by F (z 0, z 1 ) = 0. We can compute z 1 by using the factorization of a polynomial in Q[y 1 ]. Since there are at most T 2 points which do not satisfy the assumption in Step 1, the complexity of Step 1 is polynomial [28]. In Step 3, the complexity of computing (a 0 + a 1 x + + a N x N ) T is at most O(N 2 T 4 ). The multiplication of two univariate polynomials with degree NT needs O(N 2 T 4 ) at most. The number of monomials in F is at most T 2. Hence the complexity of Step 3 is O(N 3 T 6 ). Here we only consider the number of multiplications over Q in the algorithm. Now we can give the algorithm to compute an algebraic solution of F = 0. Algorithm 4.4 Input: F = 0. Output: an algebraic solution of F = 0 if it exists. 1. d := deg(f, y 1 ) and e := deg(f, y). 2. k := 1 while k d + e do (a) Compute the first 2dk + 1 terms ϕ(x) of a formal power series solution of F = 0 by Algoritm 4.3. (b) a i := the coefficient of x i in ϕ(x) for i = 0,, 2dk.

14 14 J.M. Aroca, J. Cano, R. Feng and X.S. Gao (c) In (18) and (19), let m = d, n = k and N = 2dk. We construct the linear equations (20) (d) If (20) has no nonzero solution or the dimension of the solution space of (20) is great than one, then go to Step (h). (e) Otherwise, choose one of nonzero solutions b i,j where i = 0,, d and j = 0,, d + e. (f) G(x, y) := d+e j=0 di=0 bi,j x i y j. (g) Let R = prem(f, G). If R = 0, then return(g(x, y) = 0). (h) k := k If the algorithm does not return G(x, y) = 0 in Step 2, F = 0 has no algebraic solution and the algorithm terminates. By the proof of Lemma 3.1, from a nontrivial formal power series solution of F = 0, we can get a general solution. From Theorem 2.6 and Lemma 2.5, we know that if F = 0 has a nontrivial algebraic solution, then every formal power series solution is algebraic. From Lemma 4.2, we need only to compute the first 2dk + 1 terms of a nontrivial formal power series solution. From Theorems 3.5, 3.8, if F = 0 has an algebraic solution G(x, y) = 0, then there is a k which satisfies that k 1 and k d + e such that deg(g, x) = d and deg(g, y) = k. In (18) and (19), let m = d, n = k and N = 2dk. We construct the linear equations (20). From Lemma 4.2 again, the dimension of the solution space of (20) equals to one. At last, by Lemma 2.2, G(x, y) = 0 is an algebraic solution if prem(f, G) = 0. Algorithm 4.4, the complexity is polynomial in T = tdeg(f ). In Step 2(a), the complexity is polynomial. In Step 2(c), we need only to compute T M 2T M 0 which needs O(T 8 ), because T M is a l l matrix with l 2T and M 0 is a p q matrix with p 2T 2 + 1, q T + 1. In Step 2(d), we need only to solve at most 4T linear equations with at most 2T 2 + 3T + 1 variables. Hence its complexity is polynomial. In Step 2(g), for deciding whether prem(f, G) = 0, we compute R 1 = prem(f, G ) firstly. Since R 1 = ( G y )k F (y, G x / G y ) where k T, we can compute it in O(T 12 ) and have that deg(r 1, x) 2T 2 and deg(r 1, y) 4T 2 + T. Then we compute the GCD(R 1, G) which can be computed in O(T 10 ) ([28],p152). If GCD(R 1, G) = G, then prem(f, G) = 0, otherwise prem(f, G) 0. The number of the circulation in Step 2 is at most 2T. Hence the complexity of Step 2 is also polynomial. Example 4.5 Consider F = (y 6 + 2y + 1)y 3 1 (12y 5 + 9y 4 1)y y y y 6 + 4y 3 1. Let d = 3 and e = (1, 2) is a point on F (y, y 1 ) = 0 which satisfies the assumption in Step 1 in Algorithm For the case k = 1, we get a G(x, y) = 0 which is not the solution of F = 0. For saving the space, here we only give the process in the case k = 2.

15 Algebraic General Solution The first 13 terms of the formal power series solution of F = 0 is ϕ(x) = 1 2x x2 9 4 x x x x x x x x x x12 5. let m = 3, n = 2 and N = 12. We construct the linear equations (20). Solving it, we get a nonzero solution. ( 1, 1, 0, 0, 0, 3, 3, 1, 1, 0, 0, 0) 6. Let G(x, y) = 1 + x + 3xy 3x 2 y + x 3 y + y prem(f, G) = 0. Hence G(x, y) = 1 + x + 3xy 3x 2 y + x 3 y + y 2 = 0 is an algebraic solution of F = 0. References [1] Bronstein, M., Integration of Elementary Functions, J. Symobolc Computation, 9, , [2] Bronstein, M. and Lafaille, S., Solutions of linear ordinary differential equations in terms of special functions, Proc. ISSAC2002, ACM Press, [3] Barkatou, M. A. and Pflügel, E., An algorithm computing the regular formal solutions of a system of linear differential equations, J. Symbolic Computation, 28(4-5), , [4] Campillo, A. and Carnicer, M.M., Proximity inequalities and bounds for the degree of invariant curves by foliations of PC 2, Trans. Amer. Math. Soc. 349, , [5] Carnicer, M.M., The Poincaré problem in the nondicritical case, Ann. Math., 140, , [6] Cerveau, D. and Lins Neto, A., Holomorphic foliations in CP (2) having an invariant algebraic curve, Ann. Inst. Fourier, 41(4), , [7] Cormier, O. Singer, M.F., Trager, B.M. and Ulmer, F., Linear Differential Operators for Polynomial Equations, J. Symbolic Computation, 34, , [8] Cormier, O., On Liouvillian solutions of linear differential equations of order 4 and 5, Proc. ISSAC2001, , ACM Press, [9] Davenport, J.H., On the Integration of Algebraic Functions, Lecture Notes in Computer Science, 102, Springer-Verge, New York, [10] Fulton, W., Algebraic Curves, Benjamin/Cummings Publishing Company, Inc, [11] George A. Baker, Jr and Graves-Morris, P. R. Existence of certain sequences of Hermite-Padè approximants J. Computational and Applied Mathematics,V.54, Issue 1, 20 Sep. 1994, Pages [12] Hubert, E., The general solution of an ordinary differential equation, Proc. ISSAC1996, , ACM Press, [13] Kolchin, E.R., Differential algebra and algebraic groups, ACM Press, New York, [14] Kovacic, J.J., An algorithm for solving second order linear homogeneous differential equations, J. Symbolic Computation, 2(1), 3-43, [15] Lang, S., Introduction to Algebraic and Abelian functions, second edition, Springer-Verlag, New York, [16] Poincaré, H., Sur l integration algebrique des equations diff. du premier ordre, Rendiconti del circolo matematico di Palermo, t.11, , 1897.

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