The role of algebraic solutions in planar polynomial differential systems
Open Access
- 1 September 2007
- journal article
- Published by Cambridge University Press (CUP) in Mathematical Proceedings of the Cambridge Philosophical Society
- Vol. 143 (2), 487-508
- https://doi.org/10.1017/s0305004107000497
Abstract
International audienceWe study a planar polynomial differential system, given by $\dot{x}=P(x,y)$, $\dot{y}=Q(x,y)$. We consider a function $I(x,y)=\exp\!\{h_2(x) A_1(x,y) \diagup A_0(x,y) \}$ $ h_1(x)\prod_{i=1}^{\ell} (y-g_i(x))^{\alpha_i}$, where gi(x) are algebraic functions of $x$, $A_1(x,y)=\prod_{k=1}^r (y-a_k(x))$, $A_0(x,y)=\prod_{j=1}^s (y-\tilde{g}_j(x))$ with ak(x) and $\tilde{g}_j(x)$ 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 $\alpha_i \in \mathbb{C}$. 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 $f_1^{\lambda_1} \cdots f_p^{\lambda_p} \exp\{h/f_0\}$, where fi and h are polynomials in $\mathbb{C}[x,y]$ 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 $\Phi(x,y):= \exp \{h_2(x)\, A_1(x,y)/A_0 (x,y) \}$. 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 $R=h_2 A_1/A_0 + B_1/B_0$ and with B1/B0 a function of the same form as h2 A1/A0. Hence, exp {R(x,y)} factorizes as the product Φ(x,y) Ψ(x,y), for Ψ(x,y): = exp {B1/B0
Keywords
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