Proof. Taylor expansion of the associated vector field of system (1.2) is ${{\mathbf {F}}}={{\mathbf {F}}}_1+{{\mathbf {F}}}_3+\sum _{j\ge 2}{{\mathbf {F}}}_{2j+1},$ with ${{\mathbf {F}}}_1=(-y,\,x,\,0)^{T}$ and ${{\mathbf {F}}}_3=(-y+x(a_1(x^{2}+y^{2})+a_2z^{2}),$ $x+y(a_1(x^{2}+y^{2})+a_2z^{2}),\,z(b_1(x^{2}+y^{2})+b_2z^{2}))^{T}.$ If $I=I_2+\sum _{k\ge 2}I_{2k}$ with $I_{2k}\in \mathscr {P}_{2k},$ then $F(I)$ has only even-degree homogeneous terms. The equation $F(I)=0$ to degree two is satisfied. The $2k$-degree term of $F(I),\,\ k\ge 2$ is

\[ F(I)_{2k}=F_1(I_{2k})+F_3(I_{2k-2})+R_{2k} \]

where $R_{2k}=\sum _{j=1}^{k-2}F_{2k-2j+1}(I_{2j}).$

The procedure for obtaining the scalar function $I$ will have the following scheme: For each order $k$, we first compute $I_{2k}$ to get an expression more reduced of $R_{2k}$. Let us note that the term $I_{2k}$, in general, is not unique. Later on, in the following step, we will use the no-uniqueness of $I_{2k-2}$ for obtaining the simplest expression of $F(I)_{2k}.$

The above scheme suggests us to consider the following linear operators. The operator

(2.4)\begin{equation} \begin{array}{rcl}{{\ell}^{(3)}}_{2k} & : & \mathscr{P}_{2k} \longrightarrow \mathscr{P}_{2k}\\ & & \mu_{2k} \longrightarrow F_{1}(\mu_{2k}). \end{array} \end{equation}

It is easy to prove that $\mbox {Ker}({{\ell }^{(3)}}_{2k})=\mbox {Span}\{(x^{2}+y^{2})^{k},\,(x^{2}+y^{2})^{k-1}z^{2},\,\cdots,\, z^{2k}\}.$ Moreover, $\mathscr {P}_{2k}=\mbox {Range}({{\ell }^{(3)}}_{2k})\bigoplus \mbox {Ker}({{\ell }^{(3)}}_{2k}).$ Therefore, for all $k\ge 2,$ we can choose $\mbox {Cor}({{\ell }^{(3)}}_{2k})=\mbox {Ker}({{\ell }^{(3)}}_{2k}),$ a complementary subspace to $\mbox {Range}({{\ell }^{(3)}}_{2k}).$

We also consider the linear operator:

(2.5)\begin{equation} \begin{array}{rcl}\tilde{{\ell}}^{(3)}_{2k} & : & \mbox{Ker}({{\ell}^{(3)}}_{2k-2}) \longrightarrow \mbox{Cor}({{\ell}^{(3)}}_{2k})\\ & & \mu_{2k-2} \longrightarrow F_{3}(\mu_{2k-2}). \end{array} \end{equation}

The transformed by $\tilde {{\ell }}^{(3)}_{2k}$ of an element of the basis of $\mbox {Ker}({{\ell }^{(3)}}_{2k-2}),\,\ (x^{2}+y^{2})^{k_0}z^{2j_0}$ with $0\le k_0,\, j_0\le k-1,\,\ k_0+j_0= k-1,$ is

(2.6)\begin{equation} \tilde{{\ell}}^{(3)}_{2k}((x^{2}+y^{2})^{k_0}z^{2j_0})=2A_{j_0,k_0}(x^{2}+y^{2})^{k_0+1} z^{2j_0}+2B_{j_0,k_0}(x^{2}+y^{2})^{k_0}z^{2j_0+2}, \end{equation}

where $A_{j_0,k_0}=b_1j_0+a_1k_0$ and $B_{j_0,k_0}=b_2j_0+a_2k_0.$

Therefore, the operator $\tilde {{\ell }}^{(3)}_{2k}$ is well-defined.

We analyse each case:

Consider system (1.2) with $a_1=a_2=0$ and $b_1b_2\ne 0.$ The proof consists on the computation, degree to degree, of the homogeneous terms of $I=I_2+I_4+\cdots,$ with $I_2=x^{2}+y^{2},$ satisfying (2.1).

The transformed by $\tilde {{\ell }}^{(3)}_{2k}$ of an element of the basis of $\mbox {Ker}({{\ell }^{(3)}}_{2k-2})$ is

\[ \tilde{{\ell}}^{(3)}_{2k}((x^{2}+y^{2})^{k_0}z^{2j_0})=2b_1j_0(x^{2}+y^{2})^{k_0+1} z^{2j_0}+2b_2j_0(x^{2}+y^{2})^{k_0}z^{2j_0+2}. \]

Therefore, $\mbox {Ker}(\tilde {{\ell }}^{(3)}_{2k})=\mbox {Span}\{(x^{2}+y^{2})^{k-1}\}$ and we can choose $\mbox {Cor}(\tilde {{\ell }}^{(3)}_{2k})=\mbox {Span}\{(x^{2}+y^{2})^{k},\,z^{2k}\}.$

We write $I_{2k-2}=I_{2k-2}^{a}+I_{2k-2}^{b}$ with $I_{2k-2}^{b}$ fixed in the previous step and $I_{2k-2}^{a}\in \mbox {Ker}({{\ell }^{(3)}}_{2k-2})$. So, for $k\ge 2,$

\[ F(I)_{2k}={{\ell}^{(3)}}_{2k}(I_{2k})+\tilde{{\ell}}^{(3)}_{2k}(I_{2k-2}^{a})+F_3(I_{2k-2}^{b})+R_{2k}. \]

We now write $\tilde {R}_{2k}=F_3(I_{2k-2}^{b})+R_{2k}=\tilde {R}_{2k}^{(r)}+\tilde {R}_{2k}^{(c)}$ with $\tilde {R}_{2k}^{(r)}\in \mbox {Range}({{\ell }^{(3)}}_{2k})$ and $\tilde {R}_{2k}^{(c)}\in \mbox {Cor}({{\ell }^{(3)}}_{2k}).$

Reasoning as in the classical Normal Form Theory, we choose $I_{2k}$ such that ${{\ell }^{(3)}}_{2k}(I_{2k})=-\tilde {R}_{2k}^{(r)}$ and choose $I_{2k-2}^{a}$ in order to annihilate the part of $\tilde {R}_{2k}^{(c)}$ belonging to the range of the operator $\tilde {{\ell }}^{(3)}_{2k}.$ So, we have that $F(I)_{2k}=\eta _k(x^{2}+y^{2})^{k}+\nu _kz^{2k}\in \mbox {Cor}(\tilde {{\ell }}^{(3)}_{2k}),$ with $\eta _k$ and $\nu _k$ real numbers.

Last on, we note that the solutions of the equation (2.1) are $I_{2k}+\lambda _k(x^{2}+y^{2})^{k},$ where $I_{2k}$ is the unique solution chosen following the recursive procedure and $\lambda _k$ real, that is, $I_{2k}$ is unique module $(x^{2}+y^{2})^{k}.$

Consider system (1.2) with $b_1=b_2=0$ and $a_1a_2\ne 0.$ In this case, we derive, degree to degree, the homogeneous terms of $I=I_2+I_4+\cdots,$ with $I_2=z^{2},$ satisfying the equation (2.2).

It has that $\mbox {Ker}(\tilde {{\ell }}^{(3)}_{2k})=\mbox {Span}\{z^{2(k-1)}\}$ and we can choose $\mbox {Cor}(\tilde {{\ell }}^{(3)}_{2k})=\mbox {Span}\{(x^{2}+y^{2})^{k},\,z^{2k}\}.$ Reasoning as before, we obtain the result.

For systems (1.2) where $b_2(b_1-a_1),\,\ a_1(a_2-b_2)$ and $a_1b_2-a_2b_1$ are rational numbers different from zero with the same sign, we compute the homogeneous terms of $I=I_{2M}+\sum _{k>M}I_{2k},$ with $I_{2M}=(x^{2}+y^{2})^{p}z^{2q}((b_1-a_1)(x^{2}+y^{2})+(b_2-a_2)z^{2})^{s},$ satisfying the equation (2.3). It has that $\tilde {{\ell }}^{(3)}_{2k}((x^{2}+y^{2})^{k_0}z^{2j_0})$ is given by (2.6) where

\begin{align*} A_{j_0,k_0}& =b_1j_0+a_1k_0=\displaystyle\frac{1}{(b_2-a_2)m}\left((p+s)j_0-qk_0\right),\\ B_{j_0,k_0}& =b_2j_0+a_2k_0=\frac{1}{(a_1-b_1)m}\left({-}pj_0+(q+s)k_0\right), \end{align*}

with $0\le k_0,\, j_0\le k-1,\,\ k_0+j_0= k-1.$ Note that $(p+s)j_0-qk_0=Mj_0-q(k-1),$ i.e. $\{A_{j_0,k_0}\},\,\ j_0=0,\,\dots,\, k-1,$ is an arithmetic progression whose difference is $M\ne 0$, and $-pj_0+(q+s)k_0=Mk_0-p(k-1),\,$ i.e. $\{B_{j_0,k_0}\},\,\ k_0=0,\,\dots,\, k-1,$ is an arithmetic progression whose difference is $M$. Therefore, the numbers $A_{j_0,k_0}$ with $0\le k_0,\, j_0\le k-1,\,\ k_0+j_0= k-1,$ are different and the numbers $B_{j_0,k_0}$ also are different. Fixed $k$, this fact allows us to distinguish the following cases:

If $A_{j_0,k_0}\ne 0$ for all $0\le k_0,\, j_0\le k-1,$ with $k_0+j_0= k-1,$ (i.e. $q(k-1)\not \equiv 0(\mbox {mod}\, M)$), it has that $\mbox {Ker}(\tilde {{\ell }}^{(3)}_{2k})=\{ 0\}$ and we can choose $\mbox {Cor}(\tilde {{\ell }}^{(3)}_{2k})=\mbox {Span}\{z^{2k}\}.$

If there exists $j_1$ with $0< j_1< k-1$ such that $A_{j_1,k_1}=0,$ that is $Mj_1=q(k-1),$ and $B_{j_0,k_0}\ne 0$ for all $0\le k_0,\, j_0\le k-1,$ it has that $\mbox {Ker}(\tilde {{\ell }}^{(3)}_{2k})=\{ 0\}$ and we can choose $\mbox {Cor}(\tilde {{\ell }}^{(3)}_{2k})=\mbox {Span}\{(x^{2}+y^{2})^{k}\}.$

Otherwise, there exist $(j_1,\,k_1),\, (j_2,\,k_2)$ such that $A_{j_1,k_1}=0$ and $B_{j_2,k_2}=0.$ We are going to prove that $k-1$ is a multiple of $M$. Indeed, we have that $k-1=\frac {j_1}{q}M=\frac {k_1}{p+s}M=\frac {j_2}{q+s}M=\frac {k_2}{p}M,$ i.e. $\frac {p(k-1)}{M},\, = \frac {q(k-1)}{M}$ and $\frac {s(k-1)}{M}$ are natural numbers. Thus, $\frac {(n_1p+n_2q+n_3s)(k-1)}{M}$ is an integer number, for all $n_1,\,n_2,\,n_3\in \mathbb {Z}.$ On the other hand, from Bezout identity, there exist $m_1,\,m_2,\,m_3\in \mathbb {Z}$ such that $m_1p+m_2q+m_3s=1$ since $\mbox {gcd}(p,\,q,\,s)=1.$ Therefore, $\frac {k-1}{M}$ is a natural number. So, $k-1$ is a multiple of $M$.

If we write $k=1+\hat {k}M,$ it has that $\mbox {Ker}(\tilde {{\ell }}^{(3)}_{2k})=\mbox {Span}\{I_{2M}^{\hat {k}}\}$. Moreover, $j_1=\frac {q}{q+s}j_2$. Thus $j_1< j_2$ and we can choose as a complementary subspace of the range of the operator $\tilde {{\ell }}^{(3)}_{2k}$ to $\mbox {Cor}(\tilde {{\ell }}^{(3)}_{2k})=\mbox {Span}\{(x^{2}+y^{2})^{k},\,z^{2k}\}.$ Reasoning as before, we obtain the result.

Proof. Consider the function $I$ introduced in proposition 2.1. The sufficient condition is trivial. If $\eta _k=\nu _k=0,$ for all $k,$ then the function $I$ is a formal first integral since $F(I)=0.$ From lemma 3.1, system (1.2) admits an analytic first integral. Moreover, from García [Reference García14] it is not completely analytically integrable because it is not linearizable. Therefore, system (1.2) has a unique functionally independent first integral.

Let us prove the necessary condition. If system (1.2) has an analytic first integral then, from theorem 1.1, according to each case, it admits an analytic first integral $\tilde {I}$ of the form $x^{2}+y^{2}+\cdots,\,\ z^{2}+\cdots$ or $(x^{2}+y^{2})^{p}z^{2q}((b_1-a_1)(x^{2}+y^{2})+(b_2-a_2)z^{2})^{s}+\cdots,$ having only even-degree homogeneous terms. Then, $I_p=\tilde {I}-\sum _{k\ge 2}\beta _k\tilde {I}^{k}$ is also a formal first integral, i.e. $F(I_p)=0.$ To complete the proof it is enough to choose $\beta _k$ such that $I_p$ is the unique scalar function given by proposition 2.1. Thus, by the uniqueness of $I_p$, $\eta _k=\nu _k=0,$ for all $k.$

Proof. Taylor expansion of the associated vector field of system (1.2) is ${{\mathbf {F}}}={{\mathbf {F}}}_1+{{\mathbf {F}}}_3+\sum _{k\ge 2}{{\mathbf {F}}}_{2k+1},$ with ${{\mathbf {F}}}_1=(-y,\,x,\,0)^{T}$ and ${{\mathbf {F}}}_3=(-y+x(a_1(x^{2}+y^{2})+a_2z^{2}),$ $x+y(a_1(x^{2}+y^{2})+a_2z^{2}),\,z(b_1(x^{2}+y^{2})+b_2z^{2}))^{T}.$ Thus $F(J)-J\mbox {div}({{\mathbf {F}}})$ has only odd-degree homogeneous terms. The term of degree five of $F(J)-J\mbox {div}({{\mathbf {F}}})$ is zero. The $2k+1$-degree term of $F(J)-J\mbox {div}({{\mathbf {F}}}),\,\ k\ge 3$ is

(2.10)\begin{equation} F_1(J_{2k+1})+F_3(J_{2k-1})-J_{2k-1}\mbox{div}({{\mathbf{F}}}_3)+R_{2k+1} \end{equation}

where $R_{2k+1}=\sum _{j=1}^{k-2}F_{2k-2j+1}(J_{2j+1})-J_{2j+1}\mbox {div}({{\mathbf {F}}}_{2k-2j+1}).$

Applying Euler Theorem for homogeneous function, it has that $F_3(J_{2k-1})-J_{2k-1}\mbox {div}({{\mathbf {F}}}_3)=(F_3-\frac {1}{2k-1}\mbox {div}({{\mathbf {F}}}_3)(x,\,y,\,z)^{T})(J_{2k-1})$

The above expression suggests us to consider the following linear operators. The operator

(2.11)\begin{equation} \begin{array}{rcl} {{\ell}^{(3)}}_{2k+1} & : & \mathscr{P}_{2k+1}\longrightarrow \mathscr{P}_{2k+1}\\ & & \mu_{2k+1} \longrightarrow F_{1}(\mu_{2k+1}). \end{array} \end{equation}

It is easy to prove that $\mbox {Ker}({{\ell }^{(3)}}_{2k+1})=\mbox {Span}\{(x^{2}+y^{2})^{k}z,\,(x^{2}+y^{2})^{k-1}z^{3},\,\cdots, z^{2k+1}\}.$ Moreover, we can choose $\mbox {Cor}({{\ell }^{(3)}}_{2k+1})=\mbox {Ker}({{\ell }^{(3)}}_{2k+1}).$

We also consider the linear operator:

(2.12)\begin{equation} \begin{array}{rcl}\tilde{{\ell}}^{(3)}_{2k+1} & : & \mbox{Ker}({{\ell}^{(3)}}_{2k-1}) \longrightarrow \mbox{Cor}({{\ell}^{(3)}}_{2k+1})\\ & & \mu_{2k-1} \longrightarrow (F_3-\dfrac{1}{2k-1}\mbox{div}({{\mathbf{F}}}_3)(x,y,z)^{T})(\mu_{2k-1}). \end{array} \end{equation}

The transformed by $\tilde {{\ell }}^{(3)}_{2k+1}$ of an element of the basis of $\mbox {Ker}({{\ell }^{(3)}}_{2k-1}),\, (x^{2}+y^{2})^{k_0}z^{2j_0+1}$ with $0\le k_0,\, j_0\le k-1,\,\ k_0+j_0= k-1,$ is

(2.13)\begin{align} \tilde{{\ell}}^{(3)}_{2k}((x^{2}+y^{2})^{k_0}z^{2j_0+1})=2A_{j_0,k_0}(x^{2}+y^{2})^{k_0+1} z^{2j_0+1}+2B_{j_0,k_0}(x^{2}+y^{2})^{k_0}z^{2j_0+3}, \end{align}

where $A_{j_0,k_0}=b_1j_0+a_1(k_0-2)$ and $B_{j_0,k_0}=b_2(j_0-1)+a_2(k_0-1).$ Therefore, the operator $\tilde {{\ell }}^{(3)}_{2k+1}$ is well-defined.

If we write $J_{2k-1}=J_{2k-1}^{a}+J_{2k-1}^{b}$ with $J_{2k-1}^{b}$ chosen in the previous step and $J_{2k-1}^{a}\in \mbox {Ker}({{\ell }^{(3)}}_{2k-1}),$ the expression (2.10), for $k\ge 3$, becomes

\[ {{\ell}^{(3)}}_{2k+1}(J_{2k+1})+\tilde{{\ell}}^{(3)}_{2k+1}(J_{2k-1}^{a})+\tilde{R}_{2k+1}, \]

where $\tilde {R}_{2k+1}=F_3(J_{2k-1}^{b})-J_{2k-1}^{b}\mbox {div}({{\mathbf {F}}}_3)+R_{2k+1}.$

We now write $\tilde {R}_{2k+1}=\tilde {R}_{2k+1}^{(r)}+\tilde {R}_{2k+1}^{(c)}$ with $\tilde {R}_{2k+1}^{(r)}\in \mbox {Range}({{\ell }^{(3)}}_{2k+1})$ and $\tilde {R}_{2k+1}^{(c)}\in \mbox {Cor}({{\ell }^{(3)}}_{2k+1}).$ We now choose $J_{2k+1}$ such that ${{\ell }^{(3)}}_{2k+1}(J_{2k+1})=-\tilde {R}_{2k+1}^{(r)}$ and choose $J_{2k-1}^{a}$ in order to annihilate the part of $\tilde {R}_{2k+1}^{(c)}$ belonging to the range of the operator $\tilde {{\ell }}^{(3)}_{2k+1}.$

Last on, we note that the solution of the equation (2.1) is $J_{2k+1}+J_{2k+1}^{b},$ for any $J_{2k+1}^{b}\in \mbox {Ker}(\tilde {{\ell }}^{(3)}_{2k+2})$ real and $J_{2k+1}$ is unique module $\mbox {Ker}(\tilde {{\ell }}^{(3)}_{2k+1})$.

We finish the proof, obtaining the expression of $\mbox {Ker}(\tilde {{\ell }}^{(3)}_{2k+1})$ and $\mbox {Cor}(\tilde {{\ell }}^{(3)}_{2k+1})$ for each case:

Consider system (1.2) with $a_1=a_2=0$ and $b_1b_2\ne 0.$ It has that $A_{j_0,k_0}=b_1j_0,\, B_{j_0,k_0}=b_2(j_0-1).$ So, $\mbox {Ker}(\tilde {{\ell }}^{(3)}_{2k+1})=\mbox {Span}\{(x^{2}+y^{2})^{k-2}z(b_1(x^{2}+y^{2})+b_2z^{2})\}$ and we can choose $\mbox {Cor}(\tilde {{\ell }}^{(3)}_{2k+1})=\mbox {Span}\{(x^{2}+y^{2})^{k}z,\,z^{2k+1}\}.$

Consider system (1.2) with $b_1=b_2=0$ and $a_1a_2\ne 0.$ It has that $A_{j_0,k_0}=a_1(k_0-2),\, B_{j_0,k_0}=a_2(k_0-1).$ So, $\mbox {Ker}(\tilde {{\ell }}^{(3)}_{2k+1})=\mbox {Span}\{(x^{2}+y^{2})z^{2(k-2)}(a_1(x^{2}+y^{2})+a_2z^{2})\}$ and we can choose $\mbox {Cor}(\tilde {{\ell }}^{(3)}_{2k+1})=\mbox {Span}\{(x^{2}+y^{2})^{k}z,\,z^{2k+1}\}.$

For the systems (1.2) with $b_2(b_1-a_1),\,\ a_1(a_2-b_2)$ and $a_1b_2-a_2b_1$ are rational numbers different from zero with the same sign, it has that $\tilde {{\ell }}^{(3)}_{2k}((x^{2}+y^{2})^{k_0}z^{2j_0+1})$is given by (2.13) where

\begin{align*} A_{j_0,k_0}=\displaystyle\frac{1}{(b_2-a_2)m}\left((p+s)j_0-q(k_0-2)\right),\\ B_{j_0,k_0}=\displaystyle\frac{1}{(a_1-b_1)m}\left({-}p(j_0-1)+(q+s)(k_0-1)\right). \end{align*}

Note that $(p+s)j_0-q(k_0-2)=Mj_0-q(k-3)$ i.e. $\{A_{j_0,k_0}\},\,\ j_0=0,\,\dots,\, k-1,$ is an arithmetic progression whose difference is $M\ne 0$, and $-p(j_0-1)+(q+s)(k_0-1)=M(k_0-1)-p(k-3),$ i.e. $\{B_{j_0,k_0}\},\,\ k_0=0,\,\dots,\, k-1,$ is an arithmetic progression whose difference is $M$.

Therefore, the numbers $A_{j_0,k_0}$ with $0\le k_0,\, j_0\le k-1,\,\ k_0+j_0= k-1,$ are different and the numbers $B_{j_0,k_0}$ also are different. Fixed $k$, this fact allows us to distinguish the following cases:

If $A_{j_0,k_0}\ne 0$ for all $0\le k_0,\, j_0\le k-1,$ with $k_0+j_0= k-1,$ then $\mbox {Ker}(\tilde {{\ell }}^{(3)}_{2k+1})=\{ 0\}$ and we can choose $\mbox {Cor}(\tilde {{\ell }}^{(3)}_{2k+1})=\mbox {Span}\{z^{2k+1}\}.$

If there exists $j_1$ with $0< j_1< k-1$ such that $A_{j_1,k_1}=0,$ that is $Mj_1=q(k-3),$ and $B_{j_0,k_0}\ne 0$ for all $0\le k_0,\, j_0\le k-1,$ it has that $\mbox {Ker}(\tilde {{\ell }}^{(3)}_{2k+1})=\{ 0\}$ and we can choose $\mbox {Cor}(\tilde {{\ell }}^{(3)}_{2k+1})=\mbox {Span}\{(x^{2}+y^{2})^{k}z\}.$

Otherwise, there exist $(j_1,\,k_1),\, (j_2,\,k_2)$ such that $A_{j_1,k_1}=0$ and $B_{j_2,k_2}=0.$ Reasoning as in the proof of proposition 2.1, we have that $k-3$ is a multiple of $M$.

If we write $k=3+\hat {k}M,$ it is easy to check that $\mbox {Ker}(\tilde {{\ell }}^{(3)}_{2k+1})=\mbox {Span}\{J_5I_{2M}^{\hat {k}}\}$ where $I_{2M}=(x^{2}+y^{2})^{p}z^{2q}((b_1-a_1)(x^{2}+y^{2})+(b_2-a_2)z^{2})^{s}$ and we can choose as a complementary subspace of the range of the operator $\tilde {{\ell }}^{(3)}_{2k+1}$ to $\mbox {Cor}(\tilde {{\ell }}^{(3)}_{2k+1})=\mbox {Span}\{(x^{2}+y^{2})^{k}z,\,z^{2k+1}\}.$

Proof. Its proof is similar to the one of theorem 2.2.

Proof of theorem 1.1. As, by hypothesis $(a_1,\,a_2,\,b_1,\,b_2)$ is non-zero, the system (1.1) is not orbitally equivalent to $(-y,\,x,\,0)^{T}.$ From García [Reference García14], the system (1.1) is not completely analytically integrable, it has at most one functionally independent analytic first integral.

We prove the sufficient condition. We see that system (1.4), satisfying one of the series of conditions (a) or (b), it has a polynomial first integral. In fact, we distinguish the cases separately:

1. (a) System (1.4) with $a_1=a_2=0,$ has the first integral $x^{2}+y^{2}.$ And system (1.4) with $b_1=b_2=0,$ has the first integral $z.$

2. (b) Assume that there is a rational number $m$ such that $p:=b_2(b_1-a_1)m,\,\ q:=a_1(a_2-b_2)m$ and $s:=(a_1b_2-a_2b_1)m$ are natural numbers. The polynomial $(x^{2}+y^{2})^{p}z^{2q}((b_1-a_1)(x^{2}+y^{2})+(b_2-a_2)z^{2})^{s}$ is a first integral of system (1.4).

Assume that system (1.1) is orbitally equivalent to system (1.4), i.e. there exists a change of variables $\phi$ and a reparameterization of the time-variable such that the first one becomes the second one. Let $\tilde {I}$ a polynomial first integral of system (1.4). Undoing the change of variables, system (1.1) is formally integrable and by applying lemma 3.1, there exists a scalar function $\hat {l}$ such that $I=\hat {l}\circ \tilde {I}\circ \phi ^{-1}$ with $I({{\mathbf 0}})=0$ is an analytic first integral of the system (1.1).

We see the necessary condition. We assume that the analytic system (1.1) has one functionally independent analytic first integral and $(a_1,\,a_2,\,b_1,\,b_2)$ is non-zero, performing a formal change of variables and a re-parameterization of the time-variable, system (1.1) can be transformed into

(3.1)\begin{align} \dot{x}& ={-}y+xf(x^{2}+y^{2},z^{2}), \nonumber\\ \dot{y}& =x+yf(x^{2}+y^{2},z^{2}),\\ \dot{z}& =zg(x^{2}+y^{2},z^{2}), \nonumber \end{align}

where $f,\,g$ are formal functions with $f(x^{2}+y^{2},\,z^{2})=a_1(x^{2}+y^{2})+a_2z^{2}+\mbox {h.o.t.}$ and $g(x^{2}+y^{2},\,z^{2})=b_1(x^{2}+y^{2})+b_2z^{2}+\mbox {h.o.t.},$ see [Reference Algaba, Freire and Gamero1, Reference Guckenheimer and Holmes17].

This system is a Poincaré–Dulac normal form, by [Reference Yamanaka22] a formal first integral of system (3.5) is a first integral of its linear part, i.e. $I(x,\,y,\,z)=I(x^{2}+y^{2},\,z^{2}).$

By using cylindrical coordinates, the system is

(3.2)\begin{equation} \dot{r}=rf(r^{2},z^{2}),\quad \dot{z}=zg(r^{2},z^{2}),\ dot{\theta}=1. \end{equation}

Doing the change $R=r^{2},\, Z=z^{2},\, \Theta =2\theta$ and $\tau =2t,$ system (3.2) is transformed into

(3.3)\begin{equation} (R',Z',\Theta')^{T}:=\left(\frac{\mbox{d}R}{\mbox{d}\tau}, \frac{\mbox{d}Z}{\mbox{d}\tau},\frac{\mbox{d}\Theta}{\mbox{d}\tau}\right)^{T}=(Rf(R,Z),Zg(R,Z),1)^{T} \end{equation}

and if the system (3.3) admits some first integral, then it is of the form $I=I(R,\,Z),$ i.e. it is invariant under rotations.

Removing the azimuthal component, we obtain the planar system

(3.4)\begin{equation} \left(\begin{array}{@{}l@{}} R'\\ Z' \end{array}\right)=\left(\begin{array}{@{}c@{}}R(a_1R+a_2Z)\\ Z(b_1R+b_2Z)\end{array}\right)+\left(\begin{array}{@{}l@{}}R\hat{f}(R,Z)\\ Z\hat{g}(R,Z)\end{array}\right), \end{equation}

with $\hat {f},\,\hat {g}$ having terms of degree greater than or equals to two in $R,\,Z,$ that is $R=0$ and $Z=0$ invariant curves of the system.

Hence, the analysis of the integrability problem for system (3.3) (or for system (1.2)) is equivalent to the corresponding one for planar system (3.4).

In summary, the system (1.2) with $a_1^{2}+b_1^{2}\ne 0$ and $a_2^{2}+b_2^{2}\ne 0,$ admits some formal first integral if, and only if, system (3.4) is formally integrable. The integrability problem of these systems is studied in § 4.

From proposition 4.1, if system (3.4) is formally integrable, then one of the following conditions is satisfied:

1. (a) $a_1=a_2=0$ (or $b_1=b_2=0).$

   We study the first case $a_1=a_2=0$ and $b_1b_2\ne 0$ (the other case is analogous changing $R$ by $Z$).

   From theorem 4.2, if system (3.4) is formally integrable, then there exist a change of variables and a reparameterization of the time such that system (3.4) is transformed into

   \[ R'=0,\qquad Z'=Z(b_1R+b_2Z), \]
   i.e., the three-dimensional system (3.3) is transformed into a system of the form
   (3.5)\begin{equation} (R',Z',\Theta')^{T}:=(0,Z(b_1R+b_2Z),1+\Psi(R,Z))^{T} \end{equation}
   where $\Psi$ is a formal function and $\Psi (0,\,0)=0.$

   Undoing the change, $(x,\,y,\,z,\,t)=(\sqrt {R}cos(\Theta /2),\,\sqrt {R}sin(\Theta /2),\,\sqrt {Z},\,\tau /2),$ we obtain system (1.4). These changes transform the first integral $R$ of (3.3) into a first integral of system (1.1) which has the expression $I=I(x^{2}+y^{2},\,z^{2})=x^{2}+y^{2}+\cdots$

2. (b) Assume that $p:=b_2(b_1-a_1)m,\,\ q:=a_1(a_2-b_2)m$ and $s:=(a_1b_2-a_2b_1)m$ are natural numbers and $\mbox {gcd}(p,\,q,\,s)=1.$ By theorem 4.3, if system (3.4) is formally integrable at the origin then it is orbitally equivalent to

   \[ R'=R(a_1R+a_2Z),\quad Z'=Z(b_1R+b_2Z), \]
   or equivalently, system (1.1) is is orbitally equivalent to system (1.4). Moreover, in such a case, it has an analytic first integral of the form $(x^{2}+y^{2})^{p}z^{2q}((b_1-a_1)(x^{2}+y^{2})+(b_2-a_2)z^{2})^{s}+\cdots$

Proof of theorem 1.3. We assume that $a_1^{2}+b_1^{2}\ne 0$ and $a_2^{2}+b_2^{2}\ne 0.$ From García [Reference García14], the system (1.1) is not completely analytically integrable, it has at most one analytic first integral.

We see the necessary condition. We assume that system (1.1) has one, and only one, functionally independent analytic first integral. From theorem 1.1, it is orbitally equivalent to system (1.4) satisfying one of the series of conditions (a) or (b). It is easy to check that system (1.4) has the polynomial inverse Jacobi multiplier $J_5=(x^{2}+y^{2})z((b_1-a_1)(x^{2}+y^{2})+(b_2-a_2)z^{2}).$ From lemma 3.2, system (1.1) has an inverse Jacobi multiplier of the form $J=J_5+\cdots$

We prove the sufficient condition. We assume that the analytic system (1.1) has an inverse Jacobi multiplier of the form $J=J_5+\cdots$ and $(a_1,\,a_2,\,b_1,\,b_2)$ satisfies one of the series of conditions (a) or (b).

Reasoning as in proof of theorem 1.1 and applying lemma 3.2, system (1.1) has an inverse Jacobi multiplier of the form $J=J_5+\cdots$ if system (3.4) has an inverse integrating factor of the form $\hat {J}=\hat {J}_3+\cdots$ with $\hat {J}_3=RZ((b_1-a_1)R+(b_2-a_2)Z).$

We notice that $\hat {J}_3=RZ(b_1R+b_2Z)-RZ(a_1R+a_2Z).$ Thus, applying theorem 4.4, we have the result.

Proof. Let $I=I_M+\mbox {h.o.t.}$ a formal first integral of system (3.4) and let ${{\mathbf {F}}}={{\mathbf {F}}}_2+\cdots$ be the associated vector field to system (3.4). Equation $F(I)=0$ for degree $M+1$ is $F_2(I_M)=0,$ i.e. ${{\mathbf {F}}}_2(R,\,Z)=(R(a_1R+a_2Z),\,Z(b_1R+b_2Z))^{T}$ is polynomially integrable and $I_M$ is a first integral of ${{\mathbf {F}}}_2.$ So, if ${{\mathbf {F}}}$ is formally integrable then ${{\mathbf {F}}}_2$ is polynomially integrable.

We study the polynomial integrability of ${{\mathbf {F}}}_2$ (necessary condition of formal integrability). We distinguish the following cases separately:

• • Assume $a_1=a_2=0$ (or $b_1=b_2=0).$ In such a case, $I=R$ (or $I=Z$) is a polynomial first integral of ${{\mathbf {F}}}_2$.

• • Assume $a_1= b_1\ne 0.$ If $a_2\ne b_2,$ the origin of ${{\mathbf {F}}}_2$ is an isolated singular point and $(R,\,Z)^{T}\wedge {{\mathbf {F}}}_2$ has multiple factors, thus ${{\mathbf {F}}}_2$ is not polynomially integrable, by [Reference Algaba, García and Reyes6, theorem 3.1].

  Otherwise, $a_1=b_1\ne 0$ and $a_2=b_2\ne 0,$ the vector field ${{\mathbf {F}}}_2$ is of the form $(a_1R+a_2Z)(R,\,Z)^{T}$ which is not polynomially integrable.

• • Assume $a_2= b_2\ne 0.$ In this case, the result follows by changing $R$ by $Z$.

• • Assume $a_1\ne b_1$ and $a_2\ne b_2.$ From [Reference Algaba, García and Reyes3, proposition 1.7], as the factors of $(R,\,Z)^{T}\wedge {{\mathbf {F}}}_2$ are $R,\, Z$ and $(b_1-a_1)R+(b_2-a_2)Z,$ if it exists a polynomial first integral of ${{\mathbf {F}}}_2$, then it has the expression $I_M=R^{p}Z^{q}((b_1-a_1)R+(b_2-a_2)Z)^{s}$ with $p,\,q,\,s$ natural numbers. By imposing $F_2(I_M)=0,$ we have that

  \[ a_1(p+s)+b_1q=0,\quad a_2p+b_2(q+s)=0. \]
  Thus, $a_1,\,a_2,\,b_1$ and $b_2$ are different from zero. The natural exponents of the invariant curves in the expression of the first integral $I_M$ are of the form
  \[ p=b_2(b_1-a_1)m,\quad q=a_1(a_2-b_2)m,\quad s=(a_1b_2-a_2b_1)m \]
  with $m$ a rational number. So, $I_M$ is polynomial if $b_2(b_1-a_1),\,\ a_1(a_2-b_2)$ and $a_1b_2-a_2b_1$ are rational numbers with same sign.

Proof. We consider the system 3.4 with $a_1=a_2=0$ and $b_1b_2\ne 0$. The other case is analogous, it is enough to change $R$ by $Z$.

The sufficient condition is trivial because $R$ is a polynomial first integral of system $(\dot {R},\,\dot {Z})^{T}=(0,\,Z(b_1R+b_2Z))^{T}$. Undoing the change of variables system (3.4) is formally integrable.

We see now the necessary condition. Assume that system (3.4) with $a_1=a_2=0$ and $b_1,\,b_2\ne 0$, is formally integrable. By theorem A.2, system (3.4) is orbitally equivalent to $\tilde {{{\mathbf {G}}}}={{\mathbf {G}}}_2+\cdots$ given in (A.3) which is also formally integrable. We assume that there exists $k_0:=\mathrm {min}\{k\in \mathbb {N},\, k\ge 2 : \alpha _{k}^{2}+\beta _k^{2}\neq 0\}.$ We study two cases separately:

• • If $\frac {\alpha _k}{b_1}=\frac {\beta _k}{b_2}=\lambda \ne 0,$ for all $k\geq k_0$ then $\tilde {{{\mathbf {G}}}}=(b_1R+b_2Z)\bar {{{\mathbf {G}}}}$ with $\bar {{{\mathbf {G}}}}=(\lambda R^{k_0}+\cdots ),\, Z(1+\lambda R^{k_0-1}+\cdots )^{T}.$ The origin of $\bar {{{\mathbf {G}}}}$ is an isolated saddle node point (the linear part evaluated at origin only has one eigenvalue zero). From [Reference Li, Llibre and Zhang18, Reference Li, Llibre and Zhang19], $\bar {{{\mathbf {G}}}}$ is not analytically integrable. Therefore, $\tilde {{{\mathbf {G}}}}$ is not analytically integrable.

• • Otherwise, If $\frac {\alpha _k}{b_1}\ne \frac {\beta _k}{b_2}$ for some $k\geq k_0$ then $b_1R+b_2Z=0$ is an invariant curve of $\tilde {{{\mathbf {G}}}}$ which pass by the origin and whose cofactor is $b_2Z+\sum _{k\geq k_0}(\alpha _k R^{k}+\beta _kR^{k-1}Z)$. Therefore $b_1R+b_2Z$ is a factor of any analytic first integral at the origin $I$ of $\tilde {{{\mathbf {G}}}}$. On the other hand, if $I=I_M+\cdots$ is a first integral of $\tilde {{{\mathbf {G}}}}$ then $I_M$ is a polynomial first integral of ${{\mathbf {G}}}_2$ first homogeneous component of $\tilde {{{\mathbf {G}}}}$. This fact is a contradiction since $I_M=R.$

Proof. Let $m$ be an integer number such that $p=b_2(b_1-a_1)m,\,\quad q=a_1(a_2-b_2)m$ and $s=(a_1b_2-a_2b_1)m$ are natural numbers with $\mbox {gcd}(p,\,q,\,s)=1.$

Performing the scaled-change $(R,\,Z,\,\tau )\rightarrow (-pa_1R,\,-qb_2Z,\,pq\tau ),$ system (3.4) turns into

(4.1)\begin{equation} (R',Z')^{T}=(R({-}qR+(q+s)Z+\mbox{h.o.t.}),Z((p+s)R-pZ+\mbox{h.o.t.}))^{T}. \end{equation}

The formal integrability of system (4.1) has been studied in [Reference Algaba, García and Reyes3]. By [Reference Algaba, García and Reyes3, theorem 2.11], if system (4.1) is formally integrable at the origin then it is linearizable (orbitally equivalent to its leading homogeneous term).

Proof. System (3.4) is of the form $(R',\,Z')^{T}=(P_2+\cdots,\,Q_2+\cdots )^{T}$, a perturbation of a quadratic system, where $P_2=R(a_1R+a_2Z),\,\ Q_2=Z(b_1R+b_2Z).$

The polynomial $xQ_2-yP_2=RZ[(b_1-a_1)R+(b_2-a_2)Z]$ under the series of condition (a) or (b) has only simple factors on $\mathbb {C}[x,\,y]$. From Algaba et al. [Reference Algaba, García and Reyes4, theorem 6], system (3.4) has a formal inverse integrating factor $(xQ_2-yP_2)+\mbox {h.o.t.}$ if, and only if, it is orbitally equivalent to $(P_2,\,Q_2)^{T}$. From theorems 4.2 and 4.3, the result follows.

Proof. System (5.1) is a perturbation of system (1.2) with $a_1=-1,\,a_2=3,\, b_1=3,\, b_2=-1.$ Therefore, it has at most one functionally independent first integral. We prove the necessary condition. For that, we apply Theorem 2.4. System (5.1) satisfies condition $(b)$ of theorem 2.4 since $b_2(b_1-a_1)=-4,\,\ a_1(a_2-b_2)=-4$ and $a_1b_2-a_2b_1=-8$ are rational numbers different from zero with the same sign.

We impose the existence of a formal function $J=J_5+\cdots$ with $J_5=(x^{2}+y^{2})z(x^{2}+y^{2}-z^{2})$ introduced in proposition 2.3 satisfying (2.9) with $p=q=1,\,s=2$ that is $M=4.$ Equation (2.9) to degree 19 is

\begin{align*} F(J)-\mbox{div}({{\mathbf{F}}})J& =\nu_3z^{7}+\nu_4z^{9}+\nu_5z^{11}\\& \quad +\nu_6z^{13}+\nu_7z^{15}+\eta_7z(x^{2}+y^{2})^{7}+\nu_8z^{17}+\nu_{9}z^{19}. \end{align*}

Following the procedure given in proof of proposition 2.3, we compute the coefficients of $F(J)-\mbox {div}({{\mathbf {F}}})J$ and the proof consists on the vanishing of its terms:

The coefficient $\nu _3$ is null and $\nu _4=A_{02}+10b_{02}+A_{11}+a_{02}+a_{11}+2b_{11}.$ Solving the equation $\nu _4=0$, we have $A_{02}=-10b_{02}-A_{11}-a_{02}-a_{11}-2b_{11}.$ For order 11, we have $\nu _5= (b_{11}+3b_{02})(2b_{11}+A_{11}+a_{11}).$ Imposing $\nu _5=0$, we distinguish two cases: Case 1. $b_{11}=-3b_{02}.$ In this case, $\nu _6=\eta _7=0$ and $\nu _7= (A_{11}+a_{02}+2b_{02})(A_{11}+2a_{02}+a_{11}+4b_{02}).$

Case 1.1. $A_{11}=-a_{02}-2b_{02}.$ To order 17, we have $\nu _8= (a_{02}+a_{11}+2b_{02})^{2}(a_{02}-a_{11}+4b_{02}).$ If $a_{02}=-a_{11}-2b_{02},$ we arrive to case (i). Otherwise, $a_{02}=a_{11}-4b_{02},$ we have $\nu _9= (a_{11}-b_{02})^{2}.$ Therefore $a_{11}=b_{02},$ i.e. case (i).

Case 1.2. $A_{11}=-2a_{02}-a_{11}-4b_{02},$ we have $\nu _8= (a_{02}+a_{11}+2b_{02})^{2}(a_{02}+3b_{02}).$ If $a_{02}=-a_{11}-2b_{02},$ we arrive to case (i). Otherwise, $a_{02}=-3b_{02},$ we have $\nu _9=(a_{11}-b_{02})^{2}$. Imposing $\nu _9=0$, we have $a_{11}=b_{02},$ i.e. case (i).

Case 2. $b_{11}=-\frac {1}{2}(A_{11}+a_{11}).$ In this case, $\nu _6=\eta _{7}=0$ and $\nu _7= (a_{02}+5b_{02})(2a_{02}+A_{11}-a_{11}+10b_{02}).$ Imposing the vanishing of $\nu _7$, we have:

Case 2.1. $a_{02}=-5b_{02}.$ To order 17, $\nu _8= (A_{11}-a_{11})^{2}(A_{11}+a_{11}+6b_{02}).$ If $A_{11}=a_{11},$ we are in the case (ii). Otherwise, $A_{11}=-a_{11}-6b_{02},$ we have $\nu _9=(5b_{02}^{2}+96)(3b_{02}+a_{11})^{2}.$ Thus, $\nu _{9}=0$ if $a_{11}=-3b_{02},$ i.e. case (ii).

Case 2.2. $a_{02}=\frac {1}{2}(-A_{11}+a_{11}-10b_{02}).$ To order 17, we have $\nu _8= (A_{11}-a_{11})^{2}(3A_{11}+3a_{11}-32b_{02}).$ For $A_{11}=a_{11},$ we have the case (ii). Otherwise, $A_{11}=-a_{11}+\frac {32}{3}b_{02},$ we have $\nu _9=(1152+115b_{02}^{2})(16b_{02}-3a_{11})^{2}$. So, $a_{11}=\frac {16}{3}b_{02},$ i.e. case (ii).

We prove the sufficient condition. It is easy to check that $(x^{2}+y^{2})z(x^{2}+y^{2}-z^{2})(1-b_{02}z^{2})$ is an inverse Jacobi multiplier of system (5.1) case (i) and $(x^{2}+y^{2})z(3x^{2}+3y^{2}-3z^{2}-a_{11}(x^{2}+y^{2})z^{2}+3b_{02}z^{4})$ is an inverse Jacobi multiplier of system (5.1) case (ii). Thus, both are of the form $J_5+\cdots$. The result follows applying theorem 1.3.

Proof. System (5.2) belongs to the family of systems (1.2) with $a_1=a_2=0$ and $b_1=1,\, b_2=-1.$ Therefore, it has at most one functionally independent first integral.

We prove the necessary condition. For that, we apply theorem 2.2. We impose the existence of a formal function $I=(x^{2}+y^{2})+\cdots$ introduced in proposition 2.1 satisfying (2.1). Following the procedure given in proof of proposition 2.1, we compute the coefficients of $F(I)$ and we obtain $\eta =0,\,\ \nu _2=0$ and to degree 6, it has that $\eta _3=A_{20}$ and $\nu _3=a_{11}.$ Thus, we impose that $A_{20}=a_{11}=0.$ In this case, the equation $F(I)=0$ to order 8 is zero and to order 10 and 12, we have $\eta _5=0,\,\ \nu _5= 5a_{02}b_{11}$ and $\eta _6=0,\, \nu _6=-10a_{02}b_{02}b_{11}-10a_{02}^{2}-\frac {1}{4}A_{11}^{2}.$

On the one hand, if $a_{02}=0,$ it has that $\nu _6$ is zero if $A_{11}$ is zero. On the other hand, if $b_{11}=0$, then $\nu _6=-10a_{02}^{2}-\frac {1}{4}A_{11}^{2}$. So, if $\nu _6=0$, we have that $A_{20}=A_{11}=a_{11}=a_{02}=0.$

The sufficiency is trivial since $x^{2}+y^{2}$ is a first integral of system (5.2) with $A_{20}=A_{11}=a_{11}=a_{02}=0.$