Analytically Integrable Centers of Perturbations of Cubic Homogeneous Systems

Abstract

We consider the analytically integrable perturbations of cubic homogeneous differential systems whose origin is an isolated singularity. We prove that are orbitally equivalent to the cubic vector field associated. We also characterize the analytically integrable centers. We apply the results to two families of degenerate vector fields.

Appendix

The following results will be used to compute a normal form of the vector fields considered.

The following technical lemma is a direct consequence of Hilbert’s Nussstellnsatz.

Lemma 5.10

Consider ${{\mathbf {F}}}_{n}\in {\mathcal {H}}_{n}$ irreducible (its components are coprime) and $f \in {\mathbb {C}}[x,y]$ an irreducible invariant curve at the origin of ${{\mathbf {F}}}_{n}$. If $F_n(p_k) \in \left<f\right>$ with $p_k\in {\mathscr {P}}_{k},$ then $p_k \in \left<f\right>$.

Lemma 5.11

Let $f\in {{\mathbf {C}}}[x,y]$ an irreducible polynomial invariant curve of ${{\mathbf {F}}}_{n},\ K\in {\mathscr {P}}_{n-1}$ its cofactor and $k,m\in {\mathbb {N}}$ with $n+k-1\ge m.$ Assume that the vector fields of ${\mathcal {H}}_{n}$, ${{\mathbf {F}}}_n+{\textstyle {\frac{jK}{k-j}}}{{\mathbf {D}}},\ 0\le j\le m-1,$ are irreducible. Then, if $F_n(p_k) \in \left<f^m\right>$ with $p_k\in {\mathscr {P}}_{k},$ it satisfies that $p_k \in \left<f^m\right>$.

Proof

Lemma 5.10 proves the statement for $m=1$.

We first consider the case $m=2$. If $F_{n}(p_k)\in \langle f^2\rangle$ then $F_{n}(p_k)\in \langle f\rangle$ and, by Lemma 5.10, we have that there exists $p_{k-1}\in {\mathscr {P}}_{k-1}$ such that $p_k=fp_{k-1}$. Therefore,

$$\begin{aligned} F_{n}(p_k)= & {} F_{n}(fp_{k-1})=p_{k-1}F_{n}(f)+fF_{n}(p_{k-1})=p_{k-1}Kf +fF_{n}(p_{k-1})\\= & {} f\left( {\textstyle {\frac{K}{k-1}}}D(p_{k-1})+F_{n}(p_{k-1}) \right) =f(F_{n} +{\textstyle {\frac{K}{k-1}}}D)(p_{k-1})\in \langle f^2\rangle . \end{aligned}$$

Hence $(F_{n}+{\textstyle {\frac{K}{k-1}}}D)(p_{k-1})\in \langle f\rangle .$ Applying Lemma 5.10 for the irreducible vector field ${{\mathbf {F}}}_{n}+{\textstyle {\frac{K}{k-1}}}{{\mathbf {D}}},$ we have that $p_{k-1}\in \langle f\rangle$ and consequently $p_k\in \langle f^2\rangle$.

Consider now the case $m=3$. If $F_{n}(p_{k})\in \langle f^3\rangle$ then $F_{n}(p_{k})\in \langle f^2\rangle$ and by the previous paragraph we have that there exists $p_{k-2}\in {\mathscr {P}}_{k-2}$ such that $p_k=f^2p_{k-2}$, therefore

$$\begin{aligned} F_{n}(p_k)= & {} F_{n}(f^2p_{k-2})=p_{k-2}F_{n}(f^2)+f^2F_{n}(p_{k-2}) =2p_{k-2}Kf^2+f^2F_{n}(p_{k-2})\\= & {} f^2\left( {\textstyle {\frac{2K}{k-2}}}D(p_{k-2})+F_{n}(p_{k-2}) \right) =f^2(F_{n} +{\textstyle {\frac{2K}{k-2}}}D)(p_{k-2})\in \langle f^3\rangle . \end{aligned}$$

Hence $(F_{n}+{\textstyle {\frac{2K}{k-2}}}D)(p_{k-2})\in \langle f\rangle$ and as ${{\mathbf {F}}}_{n}+{\textstyle {\frac{2K}{k-2}}}{{\mathbf {D}}}$ is irreducible, applying Lemma 5.10 we have that $p_{k-2}\in \langle f\rangle$ and consequently $p_{k}\in \langle f^3\rangle$. Reasoning by induction we get the result for all m natural number. $\square$

Lemma 5.12

The following statements are satisfied:

(i):

The vector field ${{\mathbf {F}}}_{3,ix}+\alpha K_1{{\mathbf {D}}}$ with $K_1:=-2q(d^2-1)xy$ (cofactor of $f_1=x^2+y^2$, invariant curve at the origin of ${{\mathbf {F}}}_{3,ix}$) and $\alpha \in {\mathbb {Q}}^+,$ is irreducible if and only if $2\alpha \ne \frac{p}{q}.$

(ii):

The vector field ${{\mathbf {F}}}_{3,ix}+\alpha K_2{{\mathbf {D}}}$ with $K_2:=2p(d^2-1)xy$ (cofactor of $f_2=x^2+d^2y^2$, invariant curve at the origin of ${{\mathbf {F}}}_{3,ix}$) and $\alpha \in {\mathbb {Q}}^+,$ is irreducible if and only if $2\alpha \ne \frac{q}{p}.$

Proof

We prove item (i), being the case (ii) analogous. The components of ${{\mathbf {F}}}_{3,ix}+\alpha K_1{{\mathbf {D}}}$ are of the form

$$\begin{aligned} \begin{array}{l} -y(a_{20}x^2+a_{02}):=-y\left( (2\alpha d^2q+d^2q-2\alpha q+p)x^2+d^2(p+q)y^2\right) ,\\ x(b_{20}x^2+b_{02}):=x\left( (p+q)x^2+(-2\alpha d^2q+d^2p+2\alpha q+q)y^2\right) . \end{array} \end{aligned}$$

Both polynomials are coprime if and only if $a_{20}b_{02}-a_{02}b_{20}\ne 0$, i.e.

$$\begin{aligned} q(d^2-1)^2(2\alpha +1)(2\alpha q-p)\ne 0, \end{aligned}$$

that is, $2\alpha \ne \frac{p}{q}.$ $\square$