Let k denote an effective field of characteristic zero. We consider planar vector fields and their associated differential equations that have the form
$ \frac{{{\text{d}}x}}{{{\text{d}}t}} = P\left( {x, y} \right), \frac{{{\text{d}}y}}{{{\text{d}}t}} = Q\left( {x, y} \right), $ | (1) |
where P and Q are polynomials with the coefficients in the real field. If the linearization of system (1) has purely imaginary eigenvalues, then, by a linear transformation, the system can be converted to the form
$ \begin{gathered} \frac{{{\text{d}}x}}{{{\text{d}}t}} = y + {P_2} + \cdots + {P_m}, \hfill \\ \frac{{{\text{d}}y}}{{{\text{d}}t}} =-x + {Q_2} + \cdots + {Q_m}. \hfill \\ \end{gathered} $ | (2) |
Distinguishing between a center and a focus is one challenge of nonlinear differential equations with long history.
In this paper we present a method to improve the methods of distinguishing between a center and a focus.
1 Study on the problem of center and focusThe theorem of Poincare-Liapunov says that the origin of polynomial system (2) is a center if and only if the system has a non-constant analytic first integral in a neighborhood. In searching for sufficient conditions for a center, both Poincare and Liapunov's works involve the idea of trying to find an analytic function F(x, y) in a neighborhood of O(0, 0), where
Darboux[3] showed how to construct the first integrals of planar polynomial vector fields possessing sufficient invariant algebraic curves. In particular, he proved that the planar vector field of degree m with at least
It is known that the origin is a center if system (2) has an analytic integrating factor of the form
Let
$ D\left( f \right) = P\frac{{\partial f}}{{\partial x}} + Q\frac{{\partial f}}{{\partial y}} = f{L_f}. $ | (3) |
The polynomial Lf is called the cofactor of f. Considering differential systems (2), we can obtain the special proposition of real invariant algebraic curve.
Theorem 2.1 Suppose real coefficient polynomial
Proof Let
$ y\frac{{\partial {f_s}}}{{\partial x}}-x\frac{{\partial {f_s}}}{{\partial y}} =-\lambda {f_s}, $ | (4) |
so that
$ \left( {\begin{array}{*{20}{l}} \lambda &1&0&0&0 \\ {-s}&\lambda &2&0&0 \\ 0&{-s + 1}&\lambda&\ddots &0 \\ 0&0& \ddots&\ddots &s \\ 0&0&0&{-1}&\lambda \end{array}} \right)\left( {\begin{array}{*{20}{l}} {{c_{0, s}}} \\ {{c_{1, s - 1}}} \\ \vdots \\ {{c_{s - 1, 1}}} \\ {{c_{s, 0}}} \end{array}} \right) = 0. $ | (5) |
No matter λ>0 or λ < 0, the coefficient matrix of equation (5) is through the primary transformation into an upper triangular matrix or a lower triangular matrix, whose diagonal elements are non zero. Then equation (5) has the unique zero solution, which leads to a contradiction. Hence Lf(0, 0)=λ=0. For equation (5), if i and j are odd numbers, ci, s-i=cs-j, j=0. From fs≠0, s is an even number. Let
An algebraic curve f=0 is said to be irreducible if it has only one component, and reduced if all components appear with multiplicity one. A reducible polynomial is split into its irreducible factors which also define invariant curves of system. Proposition 2.1 can be found in Ref.[10].
Proposition 2.1 f∈k[x, y] and n=deg(f). Let f=f1n1f2n2…frnr be its factorization in irreducible factors. Then, for a vector field D, f=0 is an invariant algebraic curve with the cofactor Lf if and only if fi=0 is an invariant algebraic curve for each i=1, 2, …, r with the cofactor Lfi and Lf=n1Lf1+n2Lf2+…+nrLfr.
In order to obtain more invariant algebraic curves, we can compute complex algebraic curves. Let
Theorem 2.2 Suppose
Proof Since the conjugate f(x, y) is also an invariant algebraic curve with cofactor
$ \left\{ \begin{gathered} D\left( {\operatorname{Re} f} \right) = \operatorname{Re} f\operatorname{Re} {L_f}-\operatorname{Im} f\operatorname{Im} {L_f} \hfill \\ D\left( {\operatorname{Im} f} \right) = \operatorname{Re} f\operatorname{Im} {L_f} + \operatorname{Im} f\operatorname{Re} {L_f}. \hfill \\ \end{gathered} \right. $ | (6) |
If Lf(0, 0)=0, then
Remark 2.1 Let
Garcia and Grau[11] stated that, if an invariant algebraic curve f(x, y), whose imaginary part is not null, appears in the expression of a first integral or integrating factors with exponent λ, then the conjugate f(x, y) appears in the same expression with exponent λ. Then,
$ \begin{gathered} {f^\lambda }{{\bar f}^{\bar \lambda }} = {\left( {{{\left( {\operatorname{Re} f} \right)}^2} + {{\left( {\operatorname{Im} f} \right)}^2}} \right)^{\operatorname{Re} \lambda }}\exp \\ \;\;\;\;\;\;\;\;\left\{ {-2\operatorname{Im} \lambda \arctan \left( {\frac{{\operatorname{Im} f}}{{\operatorname{Re} f}}} \right)} \right\}. \\ \end{gathered} $ | (7) |
If cofactor Lf(0, 0)=0 and the lowest order terms Refs(x, y)≠(x2+y2)p1, based on Theorem 2.2 and remark 2.1 the lowest order terms of Imf has the form of (x2+y2)p2, that is to say, Ref or Imf are not equal to zero in some deleted neighborhood U°(O) of the origin. Since
Given two coprime polynomials f, g∈k[x, y], the function e=exp(g/f) is called an exponential factor of the vector field
Proposition 3.1 If e=exp(g/f) is an exponential factor with cofactor Le for the vector field D, then f=0 is an invariant algebraic curve and g satisfies the equation
$ D\left( g \right) = g{L_f} + f{L_e}, $ | (8) |
where Lf is the cofactor of f.
Remark 3.1 Let f(x, y),
Example 3.1 (see Ref.[7]) Consider the system
$ \begin{gathered} \frac{{{\text{d}}x}}{{{\text{d}}t}} = y + a{x^3} + b{x^2}y + c{y^3}, \hfill \\ \frac{{{\text{d}}y}}{{{\text{d}}t}} =-x-c{x^3} + a{x^2}y + \left( {b-2c} \right)x{y^2}. \hfill \\ \end{gathered} $ | (9) |
System (9) has an invariant line given by f(x, y)=x+iy=0 with cofactors Lf(x, y)=-i+ax2+bxy-c(ix2+xy+iy2). Since D(1+(b-c)x2-axy)=2(1+(b-c)x2-axy)(ax2+bxy-cxy)+(x2+y2)(-a+acx2+2c(b-c)xy-acy2), the function
If we have
$ {\lambda _1}\operatorname{Re} \left( {{L_f}} \right) + {\lambda _2}\operatorname{Im} \left( {{L_f}} \right) + \rho {L_e} = 0, $ |
that is to say,
$ \left\{ \begin{gathered} a{\lambda _1} + 2c{\lambda _2} = 0 \hfill \\ \left( {b-c} \right){\lambda _1} + 2c\left( {b-c} \right)\rho = 0 \hfill \\ {\lambda _2} = a\rho . \hfill \\ \end{gathered} \right. $ |
Then the function
In order to construct the explicit first integral or integrating factors which can be used to distinguish the types of singularitie, we need to obtain more invariant curves. Christopher[8] used invariant curves, which did not vanish at the origin, in seeking local integrals or integrating factors. In this work, more invariant algebraic curves which may vanish at equilibrium point are considered. Sometime, the constructed function may be undefined at the singular point. Considering system (2), we know that the invariant algebraic curves of system (2) have the proposition in Theorems 2.1 and 2.2. We improve the methods of distinguishing between a center and a focus by using more invariant algebraic curves. Hence, we have the theorem given below.
Theorem 4.1 Suppose system (2) admits p distinct invariant algebraic curves fi=0 with cofactor Lfi which vanishes at equilibrium point for each i=1, …, p, and q independent exponential factors ej for j=1, …, q. If there exist λi, ρj∈k not all zero such that
Proof Since
Remark 4.1 Since
If
Theorem 4.2 Suppose system (2) admits p distinct invariant algebraic curves fi=0 with cofactor Lfi which vanishes at equilibrium point for each i=1, …, p, and q independent exponential factor ej with cofactor Lej for j=1, …, q. If there exist λi, ρj∈k not all zero such that
Example 4.1 Consider the systems
$ \frac{{{\text{d}}x}}{{{\text{d}}t}} = y + axy, \frac{{{\text{d}}y}}{{{\text{d}}t}} =-x + \frac{{a-3}}{2}{x^2} + \frac{{3a}}{2}{y^2}. $ | (10) |
The invariant algebraic curve x2+y2+x3 vanishes at equilibrium point and the function
$ \begin{gathered} \oint\limits_{{x^2} + {y^2} = {r^2}} {{{\left( {B\left( {x, y} \right)\overline {B\left( {x, y} \right)} } \right)}^{\frac{1}{2}}}\left( {xP + yQ} \right){\text{d}}s} = \hfill \\ \;\;\;\;\;\;\;\oint\limits_{{x^2} + {y^2} = {r^2}} {\frac{3}{2}\frac{{\left( {a-1} \right){x^2}y + a{y^3}}}{{{{\left( {{x^2} + {y^2} + {x^3}} \right)}^{\frac{4}{3}}}}}{\text{d}}s = 0.} \hfill \\ \end{gathered} $ |
Example 4.2 Consider the systems
$ \begin{gathered} \frac{{{\text{d}}x}}{{{\text{d}}t}} = y + {x^2}y-x{y^2} + {y^3}, \hfill \\ \frac{{{\text{d}}y}}{{{\text{d}}t}} =-x-{x^3} - x{y^2} - {y^3}. \hfill \\ \end{gathered} $ | (11) |
Since X(F)=-(x2+y2)2+o(r4) where
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