This is a preprint of: “Weierstrass integrability of complex differential equations”, Jaume Llibre, Clàudia Valls, Acta Math. Sinica (N.S.), vol. 37(10), 1497–1506, 2021. DOI: [10.1007/s10114-021-8374-4] WEIERSTRASS INTEGRABILITY OF COMPLEX DIFFERENTIAL EQUATIONS JAUME LLIBRE1 AND CLÀUDIA VALLS2 Abstract. We characterize the complex differential equations of the form dy = an (x)y n + an−1 (x)y n−1 + · · · + a1 (x)y + a0 (x), dx where aj (x) are meromorphic functions in the variable x for j = 0, . . . , n that admit either a Weierstrass first integral or a Weierstrass inverse integrating factor. 1. Introduction and statement of the main results Let x and y be complex variables. In this paper we study the differential equations of the form dy (1) = an (x)y n + an−1 (x)y n−1 + · · · + a1 (x)y + a0 (x) with an (0) ̸≡ 0, dx where aj (x) are meromorphic functions of x for j = 0, . . . , n. In particular, the differential equation (1) contains the well-known Abel differential equations when n = 3, the Riccati differential equations when n = 2, and the linear differential equations when n = 1. In what follows instead of working with the differential equation (1) we shall work with the equivalent differential system (2) ẋ = 1, ẏ = an (x)y n + an−1 (x)y n−1 + · · · + a1 (x)y + a0 (x) with an (0) ̸≡ 0, in C2 , where the dot denotes derivative with respect to the time t, real or complex. The objective of this paper is to study the integrability of the differential equations (1) restricted to a special kind of first integrals. For such systems the notion of integrability is based on the existence of a first integral, and we want to characterize when the differential equations (1) have either a Weierstrass first integral or a Weierstrass inverse integrating factor. When one studies the integrability of a differential system, the first class of functions to look for first integral is the polynomial functions. Then, one can go a step further and try to look for analytic first integrals. Usually, this is a very hard task and instead of this, one studies the first integrals that can be described by formal series. The use of formal series in the study of differential equations is a classical tool (see for instance [4], where the author used formal series to prove the Dulac’s conjecture). Here, guided by the fact that the equations in (1) are polynomial in the variable y, we study the first integrals that are polynomials in the variable y and formal series in the variable x, called Weierstrass first integrals. 2010 Mathematics Subject Classification. Primary 34C05, 34A34, 34C14. Key words and phrases. Weierstrass first integrals, Weierstrass inverse integrating factor, complex differ- ential equations. 1 http://www.gsd.uab.cat 2 J. LLIBRE AND C. VALLS The integrability of the Abel differential equations (i.e. of equation (1) with n = 3) has been studied by several authors, see [2, 3, 6, 7] to cite just a few. For instance in [2], [3] and [6], the authors provide a list of the known integrable Abel differential equations with aj (x) for j = 0, 1, 2, 3 rational functions, and in [8] the Weierstrass integrability of the Abel differential equations has been characterized. As usual C[[x]] is the ring of formal power series in the variable x with coefficients in C, and C[y] is the ring of polynomials in the variable y with coefficients in C. A function of the form ∑n (3) wi (x)y i ∈ C[[x]][y], i=0 is called a formal Weierstrass polynomial in y of degree n if and only if wn (x) = 1 and wi (0) = 0 for i < n. A formal Weierstrass polynomial whose coefficients are convergent is called Weierstrass polynomial, see [1]. Let V : W → C be a function satisfying ∂V ∂V + (an (x)y n + an−1 (x)y n−1 + · · · + a1 (x)y + a0 (x)) ∂x ∂y = (nan (x)y n−1 + (n − 1)an−1 (x)y n−2 + · · · + a1 (x))V. Then V is an inverse integrating factor of system (2), and it is known that there exists a first integral H such that 1 ∂H an (x)y n + an−1 (x)y n−1 + · · · + a1 (x)y + a0 (x) ∂H = , =− . V ∂y V ∂x We say that a differential system (2) is Weierstrass integrable if it admits a first integral or an inverse integrating factor which is a Weierstrass polynomial. We recall that an analytic first integral H : U → C of system (2) where U is an open subset of C2 is a non-locally constant analytic function such that it is constant on the solution of system (2) contained in U . In [5] the definition of Weierstrass integrability is given in a more general context. The main objective of this paper is to provide the differential equations (2) that have Weierstrass first integrals, or Weierstrass integrating factors. More precisely: How to recog- nize functions aj (x), for j = 0, 1, . . . , n for which the differential equation (1) is Weierstrass integrable? Our first result is the following. Proposition 1. System (2) has no Weierstrass first integrals. The proof of Proposition 1 is given in section 2. We look for inverse integrating factors given by the Weierstrass polynomial s ∑ s s−1 (4) V = y + Vs−1 (x)y + · · · + V1 (x)y + V0 (x) = Vi (x)y i i=0 with Vs (x) = 1. Then our main result is the following. Theorem 2. System (2) admits a Weierstrass inverse integrating factor of the form (4) if and only if an−k (0) = 0 for k = 1, . . . , n − 1, WEIERSTRASS INTEGRABILITY 3 1 a0 (0) = (an−1 (0)a′n (0) − a′n−1 (0)an (0)), nan (0)2 ∏k l=2 (n − l + 1) an−1 (x)k an−k (x) = , for k = 2, . . . , n − 1, nk−1 k! an (x)k−1 and ∏n−1 1 d an−1 (x) (n − l + 1) an−1 (x)n a0 (x) = − + l=2 n−1 . n dx an (x) n n! an (x)n−1 In this case the inverse Weierstrass integrating factor is n−1 ∑ n V =y + Vi (x)y i i=0 with ai (x) Vi (x) = for i = 1, . . . , n − 1 an (x) and a0 (x) 1 d an−1 (x) V0 (x) = + . an (x) nan (x) dx an (x) The proof of Theorem 2 is given in section 2. 2. Proof of the results Proof of Proposition 1. Imposing that system (2) has a first integral given by the Weier- strass polynomial s ∑ s s−1 H = y + Hs−1 (x)y + · · · + H1 (x)y + H0 (x) = Hi (x)y i i=0 (with Hs (x) = 1) we obtain a polynomial in y whose coefficients must be zero. Hence we get that s−1 ∑ s ∑ (5) Hi′ (x)y i + iHi (x)y i−1 (an (x)y n + an−1 (x)y n−1 + · · · + a1 (x)y + a0 (x)) = 0. i=0 i=0 The highest power in (5) of the variable y is s+n−1 and its coefficient is san (x) = 0. Since an (x) ̸= 0 we get s = 0. So, from the definition of Weierstrass polynomial, H = H0 (x) = 1 in contradiction with the fact that a first integral H cannot be constant. This completes the proof.  Now we shall give some preliminary results that we shall need for proving Theorem 2. Imposing that system (2) has an inverse integrating factor of the form (4) we obtain a polynomial in y whose coefficients must be zero. Hence we get that ∑s s ∑ Vi′ (x)y i + iVi (x)y i−1 (an (x)y n + an−1 (x)y n−1 + · · · + a1 (x)y + a0 (x)) i=0 i=0 (6) (∑ s ) = (nan (x)y n−1 + (n − 1)an−1 (x)y n−2 + · · · + 2a2 (x)y + a1 (x)) Vi (x)y i . i=0 Now, computing the coefficients in (6) of y s+n−1 we get sVs (x)an (x) = nVs (x)an (x). So we have that san (0) = nan (0). Since an (0) ̸= 0, it follows that s = n. 4 J. LLIBRE AND C. VALLS Lemma 3. Equation (6) can be written as n−1 n−1 2n−2 min{l,n} ∑ ∑ ∑ ∑ (7) Vi′ (x)y i +a0 (x) (i+1)Vi+1 (x)y i + yl (2i−1−l)Vi (x)al−i+1 (x). i=0 i=0 l=0 i=max{0,l+1−n} Proof. Since Vn (x) = 1 and s = n, equation (6) can be written as n−1 ∑ n ∑ 0= Vi′ (x)y i + a0 (x) iVi (x)y i−1 i=0 i=1 n ∑ + iVi (x)y i−1 (an (x)y n + an−1 (x)y n−1 + · · · + a1 (x)y) i=0 (∑ n ) − (nan (x)y n−1 + (n − 1)an−1 (x)y n−2 + · · · + 2a2 (x)y + a1 (x)) Vi (x)y i i=0 n−1 ∑ n−1 ∑ = Vi′ (x)y i + a0 (x) (i + 1)Vi+1 (x)y i i=0 i=0 (8) n ∑ + iVi (x)y i−1 (an (x)y n + an−1 (x)y n−1 + · · · + a1 (x)y) i=0 n ∑ − Vi (x)y i−1 (nan (x)y n + (n − 1)an−1 (x)y n−1 + · · · + a1 (x)y) i=0 n−1 ∑ n−1 ∑ = Vi′ (x)y i + a0 (x) (i + 1)Vi+1 (x)y i i=0 i=0 ∑n + Vi (x)y i−1 ((i − n)an (x)y n + (i − n + 1)an−1 (x)y n−1 + · · · + (i − 1)a1 (x)y). i=0 We can write the last sum in (8) as ∑n Vi (x)y i−1 ((i − n)an (x)y n + (i − n + 1)an−1 (x)y n−1 + · · · + (i − 1)a1 (x)y) i=0 n−1 ∑∑ n (9) = (i − n + j)Vi (x)an−j (x)y n+i−1−j j=0 i=0 2n−2 min{l,n} ∑ ∑ l = y (2i − 1 − l)Vi (x)al−i+1 (x). l=0 i=max{0,l+1−n} Now, the proof follows immediately from (8) and (9).  Let l−1 ∑ (10) Sn,l (x) = (l − 2j)an−j (x)an−l+j (x). j=1 Note that l ≤ n + 1. Lemma 4. We have Sn,l (x) = 0. WEIERSTRASS INTEGRABILITY 5 Proof. We consider two cases. Case 1: l even. In this case l = 2m and Sn,l (x) = Sn,2m (x) becomes 2m−1 ∑ 2 (m − j)an−j (x)an−2m+j (x) j=1 m−1 ∑ 2m−1 ∑ =2 (m − j)an−j (x)an−2m+j (x) + 2 (m − j)an−j (x)an−2m+j (x) j=1 j=m+1 2m−1 ∑ 2m−1 ∑ =2 (l − m)an−2m+l (x)an−l (x) + 2 (m − j)an−j (x)an−2m+j (x) l=m+1 j=m+1 = 0. Case 2: l odd. In this case l = 2m + 1 and Sn,l (x) = Sn,2m+1 (x) becomes 2m ∑ (2m + 1 − 2j)an−j (x)an−2m−1+j (x) j=1 m ∑ = (2m + 1 − 2j)an−j (x)an−2m−1+j (x) j=1 2m ∑ + (2m + 1 − 2j)an−j (x)an−2m−1+j (x) j=m+1 2m ∑ =− (2m + 1 − 2l)an−2m−1+l (x)an−l (x) l=m+1 2m ∑ + (2m + 1 − 2j)an−j (x)an−2m−1+j (x) = 0. j=m+1 This concludes the proof of the lemma.  Lemma 5. For k = 1, . . . , n − 1 we have that an−k (x) (11) Vn−k (x) = . an (x) Proof. We compute in (6) with s = n the coefficients of y l for l = n, . . . , 2n − 2. Note that in this case, by Lemma 3 (see (7)) for l = n, . . . , 2n − 2 we have n ∑ (12) (2i − 1 − l)Vi (x)al−i+1 (x) = 0. i=l+1−n For k = 1, . . . , n − 1 we write l = 2n − 1 − k and (12) becomes n ∑ k ∑ (13) 0= (2i − 2n + k)Vi (x)a2n−k−i (x) = (k − 2j)Vn−j (x)an−k+j (x), i=n−k j=0 6 J. LLIBRE AND C. VALLS where in the last sum we have taken j = n − i. We rewrite (13) as (recall that Vn (x) ≡ 1) k−1 ∑ 0 = kVn (x)an−k (x) − kVn−k (x)an (x) + (k − 2j)Vn−j (x)an−k+j (x) j=1 k−1 ( ) ∑ (14) = k Vn (x)an−k (x) − Vn−k (x)an (x) + (k − 2j)Vn−j (x)an−k+j (x) j=1 k−1 ( ) ∑ = k an−k (x) − Vn−k (x)an (x) + (k − 2j)Vn−j (x)an−k+j (x). j=1 Now we proceed by backwards induction. For k = 1 in (14) we have an−1 (x) an−1 (x) − Vn−1 (x)an (x) = 0 that is Vn−1 (x) = , an (x) which proves (11) for k = 1. Now assume that (11) is true for k = 1, . . . , l − 1 and we shall prove it for k = l. Thus, using the induction hypothesis in (14) we get l−1 ( ) ∑ 0 = l an−l (x) − Vn−l (x)an (x) + (l − 2j)Vn−j (x)an−l+j (x) j=1 l−1 ( ) ∑ an−j (x) = l an−l (x) − Vn−l (x)an (x) + (l − 2j) an−l+j (x) an (x) (15) j=1 l−1 ( ) 1 ∑ = l an−l (x) − Vn−l (x)an (x) + (l − 2j)an−j (x)an−l+j (x) an (x) j=1 ( ) 1 = l an−l (x) − Vn−l (x)an (x) + Sn,l (x), an (x) where Sn,l (x) we introduced in (10). By Lemma 4, Sn,l (x) = 0, and thus (15) becomes ( ) an−l (x) l an−l (x) − Vn−l (x)an (x) = 0 that is Vn−l (x) = . an (x) This concludes the proof of the lemma.  Lemma 6. We have that a0 (x) 1 d an−1 (x) V0 (x) = + . an (x) nan (x) dx an (x) Proof. We compute in (6) with s = n the coefficients of y n−1 . By Lemma 3 with l = n−1 < n we get n−1 ∑ ′ Vn−1 (x) + na0 (x)Vn (x) + (2i − n)Vi (x)an−i (x) = 0. i=0 WEIERSTRASS INTEGRABILITY 7 Now, using Lemma 5 we get n−1 ∑ ′ 0= Vn−1 (x) + na0 (x)Vn (x) + (2i − n)Vi (x)an−i (x) i=0 n−1 ′ 1 ∑ = Vn−1 (x) + na0 (x)Vn (x) − nV0 (x)an (x) + (2i − n)ai (x)an−i (x) an (x) i=1 ′ Sn,n (x) = Vn−1 (x) + na0 (x) − nan (x)V0 (x) − an (x) ′ = Vn−1 (x) + na0 (x) − nan (x)V0 (x), where in the last equality we have used (10) and Lemma 4. Therefore 1 ( ′ ) a0 (x) 1 d an−1 (x) V0 (x) = na0 (x) + Vn−1 (x) = + . nan (x) an (x) nan (x) dx an (x) This concludes the proof of the lemma.  Lemma 7. For k = 1, . . . , n − 1 we have that an−k (0) = 0, and 1 a0 (0) = (an−1 (0)a′n (0) − a′n−1 (0)an (0)). nan (0)2 Proof. From the definition of Weierstrass polynomial, Vk (0) = 0 for k = 0, . . . , n − 1. Then, and since by assumption an (0) ̸= 0, it follows from Lemma 5 that an−k (0) = 0 for k = 1, . . . , n − 1. Now using that V0 (0) = 0 it follows from Lemma 6 that a0 (0) 1 0= + (a′ (0)an (0) − an−1 (0)a′n (0)), an (0) nan (0)3 n−1 which clearly concludes the proof of the lemma.  Lemma 8. We have ∏k l=2 (n − l + 1) an−1 (x)k (16) an−k (x) = , for k = 2, . . . , n − 1. nk−1 k! an (x)k−1 Proof. We compute in (6) with s = n the coefficients of y n−k with k = 2, . . . , n − 1. By Lemma 3 with l = n − k < n we get n−k ∑ ′ (17) Vn−k (x) + (n − k + 1)a0 (x)Vn−k+1 (x) + (2i − 1 − n + k)Vi (x)an−k−i+1 (x) = 0. i=0 8 J. LLIBRE AND C. VALLS Using Lemma 5 we get that (17) becomes n−k d an−k (x) an−k+1 (x) ∑ 0= + (n − k + 1)a0 (x) + (2i − 1 − n + k)Vi (x)an−k−i+1 (x) dx an (x) an (x) i=0 d an−k (x) an−k+1 (x) 0= + (n − k + 1)a0 (x) − (n − k + 1)V0 (x)an−k+1 (x) dx an (x) an (x) n−k 1 ∑ + (2i − 1 − n + k)ai (x)an−k−i+1 (x) an (x) i=1 d an−k (x) an−k+1 (x) = + (n − k + 1)a0 (x) − (n − k + 1)V0 (x)an−k+1 (x) dx an (x) an (x) Sn−k+1,n−k+1 (x) − . an (x) Now using Lemma 4 we get d an−k (x) an−k+1 (x) + (n − k + 1)a0 (x) − (n − k + 1)V0 (x)an−k+1 (x) = 0, dx an (x) an (x) and by Lemma 6, d an−k (x) an−k+1 (x) a0 (x) 0= + (n − k + 1)a0 (x) − (n − k + 1) an−k+1 (x) dx an (x) an (x) an (x) n−k+1 d an−1 (x) − an−k+1 (x) nan (x) dx an (x) d an−k (x) n − k + 1 d an−1 (x) = − an−k+1 (x) . dx an (x) nan (x) dx an (x) That is d an−k (x) n−k+1 d an−1 (x) (18) = an−k+1 (x) . dx an (x) nan (x) dx an (x) Now we proceed by induction on k. For k = 2 we have d an−2 (x) n − 1 an−1 (x) d an−1 (x) = dx an (x) n an (x) dx an (x) n − 1 d ( an−1 (x) )2 = . 2n dx an (x) Solving this equation we get an−2 (x) n − 1 ( an−1 (x) )2 = K2 + . an (x) 2n an (x) Using Lemma 7 we get that K2 = 0 and thus n − 1 an−1 (x)2 an−2 (x) = , 2n an (x) which proves (16) for k = 2. WEIERSTRASS INTEGRABILITY 9 Now we assume that (16) holds for k = 2, . . . , j − 1 and we shall prove it for k = j. Thus, using the induction hypothesis, it follows from (18) with k = j that d an−j (x) n−j+1 d an−1 (x) = a (x) dx an (x) nan (x) n−(j−1 dx an (x) ∏j−1 n − j + 1 l=2 (n − l + 1) an−1 (x)j−1 d an−1 (x) = nan (x) nj−2 (j − 1)! an (x)j−2 dx an (x) ∏j (n − l + 1) an−1 (x)j−1 d an−1 (x) = l=2 nj−1 (j − 1)! an (x)j−1 dx an (x) ∏j (n − l + 1) d ( an−1 (x) )j = l=2 j−1 . n j! dx an (x) Solving this equation we get ∏j an−j (x) l=2 (n − l + 1) ( an−1 (x) )j = Kj + . an (x) nj−1 j! an (x) Using Lemma 7 we get that Kj = 0 and thus ∏j (n − l + 1) an−1 (x)j an−j (x) = l=2 j−1 , n j! an (x)j−1 which proves (16) for k = j. This concludes the proof of the lemma.  Lemma 9. We have ∏n−1 1 d an−1 (x) l=2 (n − l + 1) an−1 (x)n a0 (x) = − + . n dx an (x) nn−1 n! an (x)n−1 Proof. We compute in (6) with s = n the coefficients of y 0 . By Lemma 3 with l = 0 we get (19) V0′ (x) + a0 (x)V1 (x) − a1 (x)V0 (x) = 0. Using Lemmas 5 and 6 we get that (19) can be written as ( ) d a0 (x)a1 (x) a0 (x) 1 d an−1 (x) 0= V0 (x) + − a1 (x) + dx an (x) an (x) nan (x) dx an (x) d a1 (x) d an−1 (x) = V0 (x) − . dx nan (x) dx an (x) Now using Lemma 8 with k = n − 1 we get d a1 (x) d an−1 (x) V0 (x) = dx nan (x) dx an (x) ∏n−1 (n − l + 1) an−1 (x)n−1 d an−1 (x) = l=2 n−1 n (n − 1)! an (x)n−1 dx an (x) ∏n−1 (n − l + 1) d ( an−1 (x) )n = l=2 n−1 . n n! dx an (x) Therefore, integrating it we get ∏n−1 l=2 (n − l + 1) ( an−1 (x) )n V0 (x) = K0 + . nn−1 n! an (x) 10 J. LLIBRE AND C. VALLS Using Lemma 7 (i.e. that V0 (0) = 0) we get that K0 = 0 and thus ∏n−1 (n − l + 1) ( an−1 (x) )n V0 (x) = l=2 n−1 . n n! an (x) Now, using Lemma 6 we have ∏n−1 a0 (x) 1 d an−1 (x) (n − l + 1) ( an−1 (x) )n + = l=2 n−1 , an (x) nan (x) dx an (x) n n! an (x) which yields ∏n−1 1 d an−1 (x) (n − l + 1) an−1 (x)n a0 (x) = − + l=2 n−1 , n dx an (x) n n! an (x)n−1 as we wanted to prove.  Proof of Theorem 2. It follows readily from Lemmas 5, 6, 7, 8 and 9.  Acknowledgements The first author is partially supported by a MCYT/FEDER grant MTM2008-03437, a CIRIT grant number 2009SGR-410, an ICREA Academia, and two grants FP7-PEOPLE- 2012-IRSES 316338 and 318999. The second author has been supported by FCT (grant PTDC/MAT/117106/2010 and through CAMGSD). References [1] E Casas-Alvero, Singularities of Plane Curves, London Math. Soc. Lecture Note Ser., vol. 276, Cambridge University Press, Cambridge, 2000. [2] E.S. Cheb-Terrab and A.D. Roche, Abel ODE’s: equivalence and integrable classes, Comput. Phys. Comm. 130 (2000), 204–231. [3] E.S. Cheb-Terrab and A.D. Roche, An Abel ordinary differential equation class generalizing known integrable classes, Eur. J. Appl. Math. 14 (2003), 217–229. [4] J. Écalle, Introduction deux fonctions analysables et preuve constructive de la conjecture de Dulac, Hermann, 1992. [5] J. Giné and M. Grau, Weierstrass integrability of differential equations, Appl. Math. Lett. 23 (2010), 523–526. [6] J. Giné and J. Llibre, On the integrable rational Abel differential equations, Z. Angew. Math. Phys. 61 (2010), 33–39. [7] E. Kamke, Differentialgleichungen ”losungsmethoden und losungen”, Col. Mathematik und ihre an- wendungen vol. 18, Akademische Verlagsgesellschaft Becker und Erler Kom-Ges., Leipzig (1943). [8] J. Llibre and C. Valls, Generalized Weierstrass integrability of the Abel differential equations, to appear in Mediterr. J. Math. 1 Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bellaterra, Barcelona, Catalonia, Spain E-mail address: jllibre@mat.uab.cat 3 Departamento de Matemática, Instituto Superior Técnico, Universidade Técnica de Lis- boa, Av. Rovisco Pais 1049–001, Lisboa, Portugal E-mail address: cvalls@math.ist.utl.pt