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The role of algebraic solutions in planar polynomial differential systems

Published online by Cambridge University Press:  01 September 2007

HÉCTOR GIACOMINI ,
JAUME GINÉ  and
MAITE GRAU
HÉCTOR GIACOMINI
Affiliation:
Lab. de Mathématiques et Physique Théorique, CNRS UMR 6083, Faculté des Sciences et Techniques, Université de Tours, Parc de Grandmont, 37200 Tours, France. email Hector.Giacomini@lmpt.univ-tours.fr
JAUME GINÉ
Affiliation:
Departament deMatemàtica, Universitat de Lleida, Avda. Jaume II, 69, 25001 Lleida, Spain. email: gine@eps.udl.es, mtgrau@matematica.udl.es
MAITE GRAU
Affiliation:
Departament deMatemàtica, Universitat de Lleida, Avda. Jaume II, 69, 25001 Lleida, Spain. email: gine@eps.udl.es, mtgrau@matematica.udl.es
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Abstract

We study a planar polynomial differential system, given by . We consider a function , where gi(x) are algebraic functions of with ak(x) and algebraic functions, A0(x,y) and A1(x,y) do not share any common factor, h2(x) is a rational function, h(x) and h1(x) are functions of x with a rational logarithmic derivative and . We show that if I(x,y) is a first integral or an integrating factor, then I(x,y) is a Darboux function. A Darboux function is a function of the form , where fi and h are polynomials in and the λi's are complex numbers. In order to prove this result, we show that if g(x) is an algebraic particular solution, that is, if there exists an irreducible polynomial f(x,y) such that f(x,g(x)) ≡ 0, then f(x,y) = 0 is an invariant algebraic curve of the system. In relation with this fact, we give some characteristics related to particular solutions and functions of the form I(x,y) such as the structure of their cofactor.

Moreover, we consider A0(x,y), A1(x,y) and h2(x) as before and a function of the form . We show that if the derivative of Φ(x,y) with respect to the flow is well defined over {(x,y): A0(x,y) = 0} then Φ(x,y) gives rise to an exponential factor. This exponential factor has the form exp {R(x,y)} where and with B1/B0 a function of the same form as h2A1/A0. Hence, exp {R(x,y)} factorizes as the product Φ(x,y) Ψ(x,y), for Ψ(x,y): = exp {B1/B0.

Type
Research Article
Information
Mathematical Proceedings of the Cambridge Philosophical Society , Volume 143 , Issue 2 , September 2007 , pp. 487 - 508
Copyright
Copyright © Cambridge Philosophical Society 2007

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