Proof. Firstly, consider $s=2$. Substituting $s=2$ into $J_i(h)\,(i=1,\,2,\,3)$, we get from lemma 2.4 that

(2.8)\begin{equation} J_i(h)=\frac{3}{2}\displaystyle\oint_{\Gamma_{h}}f_i(x)y\ \mathrm{d}x,\quad i=1,2,3, \end{equation}

where $f_i(x)=\Psi '(x)\int _{\bar a}^{x}g_{i}(s)\mathrm {d}s$. Since $\,\{G_{1},\,G_{2},\,G_{3}\}$ is an ECT-system, using lemma 2.3 we deduce that $W[G_1],$ $W[G_1,\,G_2]$ and $W[G_1,\,G_2,\,G_3]$ are non-vanishing on $(\bar a,\,\nu ).$ Now let's verify the hypotheses $(\mathrm H_2)$, $(\mathrm H_3)$ and $(\mathrm H_4)$.

• • Verify hypotheses $(\mathrm H_2)$ and $(\mathrm H_3)$: Without loss of generality, suppose that $W[G_1]=G_1(x)>\,0$ for $x\in \,(\bar a,\,\nu )$ (otherwise, we use $-G_1(x)$ instead of $G_1(x)$). Since $f_i(x)=\Psi '(x)\int _{\bar a}^{x}g_{i}(s)\mathrm {d}s$, we get

  \[ F_i(x)=\frac{f_{i}(x)}{\Psi'(x)}-\frac{f_{i}(z)}{\Psi'(z)}=\int_{z}^{x}g_{i}(s)\mathrm{d}s. \]
  Notice that $F_i(\bar a)=0$ and
  \[ F'_i(x)=g_{i}(x)-g_{i}(z)\frac{\mathrm{d}z}{\mathrm{d}x}=\Psi'(x)\,G_i(x),\quad i=1,2,3. \]
  Then we have $F_1(x)>\,0$ and $F'_1(x)>\,0$ for all $x\in \,(\bar a,\,\nu ).$ Thus hypotheses $(\mathrm H_2)$ and $(\mathrm H_3)$ hold.

• • Verify hypothesis $(\mathrm H_4)$: Firstly, we prove that $\alpha \,F_{1}(x)+\beta \,F_{2}(x)+\gamma \,F_{3}(x)$ has at most two zeros on $(\bar a,\,\nu ).$ Since $\{G_{1},\,G_{2},\,G_{3}\}$ is an ECT-system on $(\bar a,\,\nu ),$ we deduce that $\alpha \,G_{1}+\beta \,G_{2}+\gamma \,G_{3}$ has at most two zeros on $(\bar a,\,\nu )$ counted with multiplicities. Combining

  \[ \alpha\,F_{1}(x)+\beta\,F_{2}(x)+\gamma\,F_{3}(x)=\int_{\bar a}^{x}\Psi'(s)\,\left(\alpha\,G_{1}(s)+\beta\,G_{2}(s)+\gamma\,G_{3}(s)\right)\mathrm{d}s \]
  and Rolle's theorem, we obtain that $\alpha \,F_{1}(x)+\beta \,F_{2}(x)+\gamma \,F_{3}(x)$ has at most three zeros on $[\bar a,\,\nu )$ counted with multiplicities. Since $\alpha \,F_{1}(\bar a)+\beta \,F_{2}(\bar a)+\gamma \,F_{3}(\bar a)=0$, it follows that $\alpha \,F_{1}(x)+\beta \,F_{2}(x)+\gamma \,F_{3}(x)$ has at most two zeros on $(\bar a,\,\nu )$ counted with multiplicities.

  Similarly, we have that $\alpha \,F_{1}(x)+\beta \,F_{2}(x)$ has at most one zero on $(\bar a,\,\nu )$ counted with multiplicities.

  Summarizing the above analysis, it follows from the definition of ECT-system and $F_1(x)>0$ that $\{F_1,\,F_2,\,F_3\}$ is an ECT-system on $(\bar a,\,\nu )$. By lemma 2.3 and remark 2.7, we get that $W[F_1,\,F_2]\neq \,0$ and $W[F_1,\,F_2,\,F_3]\neq \,0.$ So we can see that hypothesis $(\mathrm H_4)$ also holds.

When $s>2$, the same result can be obtained by using lemma 2.5 repeatedly.

Proof. Following the notations in theorem 2.6, let

\[ f_1(x)=x^{4},\quad f_2(x)=x^{2},\quad f_3(x)=1. \]

Notice that $\Psi (x)$ is an even function, thus the involution is $z=\sigma (x)=-x.$ Then by direct calculations, we have

\begin{align*} & (\mathrm H_1): \Psi'(x)x=x^{2}(1+bx^{2}-x^{4})>0,\quad x\in(-\nu(b),\nu(b))\setminus\{0\},\\ & (\mathrm H_2): F_1(x)=\frac{2\,x^{3}}{1+bx^{2}-x^{4}}>0,\quad x\in(0,\nu(b)),\\ & (\mathrm H_3): F'_1(x)=\frac{2x^{2}(x^{4}+bx^{2}+3)}{(1+bx^{2}-x^{4})^{2}}=\frac{4x^{6}+4x^{2}+2x^{2}\,(1+bx^{2}-x^{4})}{(1+bx^{2}-x^{4})^{2}}>0,\\ & \quad\qquad x\in(0,\nu(b)). \end{align*}

Furthermore, one has $(\mathrm H_4)$:

\[ \xi(x)=\frac{1}{x^{2}},\quad\xi'(x)={-}\frac{2}{x^{3}}\neq 0, \]

and

\[ \eta(x)=\frac{1}{x^{4}},\quad\eta'(x)={-}\frac{4}{x^{5}}. \]

Finally it follows that

\[ \left(\frac{\eta'(x)}{\xi'(x)}\right)'={-}\frac{4}{x^{3}}\neq 0. \]

Thus, according to theorem 2.6, we obtain that the integral $I(h)$ in (2.12) has at most two zeros in $(0,\,h_s)$ counted with multiplicities.

Theorem 2.9 follows.

Proof Proof of lemma 1.1

By [Reference Iliev15], we just need compute the Abelian integrals associated to system (1.4). By Poincaré–Pontryagin theorem, one has

(3.1)\begin{equation} I(h)=\oint_{\Gamma_h^{k}}\frac{1}{x}\left(f(x,y)\ \mathrm{d}y-g(x,y)\ \mathrm{d}x\right), \quad k=1,2. \end{equation}

Since $f(x,\, y)$ and $g(x,\, y)$ are quadratic polynomials, we calculate each item directly. Obviously,

\[ \oint_{\Gamma_h^{k}}\frac{x^{i}}{x}\ \mathrm{d}x=0,\quad i=0,1,2,\qquad \oint_{\Gamma_h^{k}}\frac{xy^{j}}{x}\ \mathrm{d}y=0,\quad j=0,1. \]

Note that both the ovals $\Gamma _h^{1}$ and $\Gamma _h^{2}$ are symmetry with respect to the $x$-axis, which locate in the half plane $x>0$ and $x<0$, respectively. It follows that

\begin{align*} & \oint_{\Gamma_h^{k}}\frac{y^{2}}{x}\ \mathrm{d}x=({-}1)^{k}\iint_{H\leq h}-\frac{2y}{x}\ \mathrm{d}x\mathrm{d}y=0,\\ & \oint_{\Gamma_h^{k}}\frac{y}{x}\ \mathrm{d}y=({-}1)^{k}\iint_{H\leq h}-\frac{y}{x^{2}}\ \mathrm{d}x\mathrm{d}y=0. \end{align*}

Besides, we have

\begin{align*} \oint_{\Gamma_h^{k}}\frac{1}{x}\ \mathrm{d}y& ={-}\oint_{\Gamma_h^{k}}y\ \mathrm{d}\frac{1}{x} =\oint_{\Gamma_h^{k}}\frac{y}{x^{2}}\ \mathrm{d}x,\quad \oint_{\Gamma_h^{k}}\frac{x^{2}}{x}\ \mathrm{d}y={-}\oint_{\Gamma_h^{k}}y\ \mathrm{d}x,\\ \oint_{\Gamma_h^{k}}\frac{y^{2}}{x}\ \mathrm{d}y& ={-}\oint_{\Gamma_h^{k}}\left(bx-2(b-1)+\frac{b-2}{x}\right)\frac{y}{2x}\ \mathrm{d}x\\& ={-}\frac{b}{2}\oint_{\Gamma_h^{k}}y\ \mathrm{d}x+(b-1)\oint_{\Gamma_h^{k}}\frac{y}{x}\ \mathrm{d}x- \frac{b-2}{2}\oint_{\Gamma_h^{k}}\frac{y}{x^{2}}\ \mathrm{d}x. \end{align*}

Lemma 1.1 is completed by all the computations above and (3.1).

Let $\Psi (x)={bx^{2}}/{2}-2(b-1)x+(b-2)\ln |x|+{3b}/{2}-2,$ then $\Psi '(x)=b(x-1) (1+{r}/{x}),$ where $r={(2-b)}/{b}>0.$

Proof of theorem 1.2. Proof of theorem 1.2

Consider the centre $(1,\,0)$ first. There exists an involution $z=\sigma (x)$ satisfying that $\Psi (x)=\Psi (z),$ where $0<\,z<\,1<\,x< + \infty.$

We claim that $x+z>2,\,\ 0< xz<1$ hold for all $x\in (1,\, + \infty )$. Because $\Psi '(x)=b(x-1)(1+{r}/{x}),$ we have that $\Psi (x)-\Psi (1)>\,0$ for $x\in (0,\, + \infty )\setminus \{1\}.$ Write $\Psi (x)-\Psi (1)=\Psi (z)-\Psi (1)=\lambda ^{2}\,(\lambda >0).$ When $0<\,x-1\ll 1,$ a straightforward calculation yields to the inverse transformation

(3.2)\begin{equation} \begin{aligned} & x-1=\lambda+\frac{2-b}{6}\lambda^{2}+O(\lambda^{3}),\\ & z-1={-}\lambda+\frac{2-b}{6}\lambda^{2}+O(\lambda^{3}). \end{aligned} \end{equation}

According to (3.2), we have

(3.3)\begin{equation} \begin{aligned} & x+z=2+\frac{2-b}{3}\lambda^{2}+O(\lambda^{3}),\\ & xz=1-\frac{1+b}{3}\lambda^{2}+O(\lambda^{3}). \end{aligned} \end{equation}

So, when $0<\,x-1\ll \,1,$ we obtain that $x+z>\,2$ and $0<\,xz<\,1.$

Next, we prove that $x+z>\,2$ and $0<\,xz<\,1$ hold for all $x\in (1,\, + \infty ).$

Since ${\mathrm {d}z}/{\mathrm {d}x}={\Psi '(x)}/{\Psi '(z)}$, some calculation shows that

(3.4)\begin{equation} \frac{\mathrm{d}(x+z)}{\mathrm{d}x}=\frac{2(r-1)-(x+z){(r-xz)}/({xz})}{(z-1)\left(1+r/z\right)} \end{equation}

and

(3.5)\begin{equation} \frac{\mathrm{d}(xz)}{\mathrm{d}x}=\frac{(x+z)(x+z+r-1)-2(xz+r)}{(z-1)\left(1+r/z\right)}. \end{equation}

Let $\mathbb {A}=\{x:\,x+z\leq \,2,\,x>1\}$ and $\mathbb {B}=\{x:\,xz\geq \,1,\,x>1\}.$

1. (I) If $\mathbb {A}=\mathbb {B}=\varnothing,$ then our claim is established.

2. (II) If $\mathbb {A}\neq \,\varnothing,\,\mathbb {B}=\varnothing,$ denote $x_{1}=\inf \{x:\,x\in \mathbb {A}\}.$ It is easy to check that

   (3.6)\begin{equation} \frac{\mathrm{d}(x+z)}{\mathrm{d}x}\Big|_{x=x_{1}}\leq 0,\quad x_{1}z_{1}<1,\ x_{1}+z_{1}=2, \end{equation}
   where $z_{1}=\sigma (x_{1}).$ Substituting (3.6) into (3.4), we have $2(r-1)-2{(r-x_{1}z_{1}})/({x_{1}z_{1})}\geq 0.$ A direct calculation shows that $x_{1}z_{1}\geq 1,$ which contradicts with $x_{1}z_{1}<1.$

3. (III) If $\mathbb {A}=\varnothing,\,\mathbb {B}\neq \,\varnothing,$ denote $x_{2}=\inf \{x:\,x\in \mathbb {B}\}.$ Using this and (3.3) we have

   (3.7)\begin{equation} \frac{\mathrm{d}(xz)}{\mathrm{d}x}\Big|_{x=x_{2}}\geq 0,\quad x_{2}z_{2}=1,\ x_{2}+z_{2}>2, \end{equation}
   where $z_{2}=\sigma (x_{2}).$ Substituting (3.7) into (3.5), we have $(x_{2}+z_{2})(x_{2}+z_{2}+r-1)-2(1+r)\leq 0,$ further we obtain $x_{2}+z_{2}\leq 2.$ This leads to a contradiction.

4. (IV) If $\mathbb {A}\neq \,\varnothing,\,\mathbb {B}\neq \,\varnothing,$ denote $x_{1}=\inf \{x:\,x\in \mathbb {A}\},\,x_{2}=\inf \{x:\,x\in \mathbb {B}\}.$

   If $x_{1}< x_{2}$, similar to the proof in $(\mathrm {II}),$ we will arrive at a contradiction.

   If $x_{2}<\,x_{1}$, similar to the proof in $(\mathrm {III}),$ we will obtain the contradiction.

   If $x_1=x_2,$ then we have $x_1+z_1=x_2+z_2=2$ and $x_1z_1=x_2z_2=1.$ According to mean value inequality, we obtain $x_1=z_1=x_2=z_2=1$, which contradicts with $x>1>z>0.$

From all the discussions above, we get $\mathbb {A}=\mathbb {B}=\varnothing.$ Hence $x+z>\,2,\,0< xz<\,1$ hold for all $x\in (1,\, + \infty ).$

After a linear combination, the integral $I(h)$ given in (1.5) can be rewritten as

\[ I(h)\!=\!{-}\frac{\gamma}{b} \oint_{\Gamma^{1}_{h}}b \left(\frac{1}{x}-\!1\right)y\ \mathrm{d}x-\frac{\gamma\!+\!\beta}{b}\oint_{\Gamma^{1}_{h}}\frac{b}{x} \left(\frac{1}{x}-\!1\right)y\ \mathrm{d}x+\frac{\alpha\!+\!\beta\!+\!\gamma}{b}\oint_{\Gamma^{1}_{h}}\frac{b}{x^{2}}y\ \mathrm{d}x. \]

For the sake of shortness, we denote

\[ f_1(x)=b \left(\frac{1}{x}-1\right),\quad f_2(x)=\frac{b}{x} \left(\frac{1}{x}-1\right),\quad f_3(x)=\frac{b}{x^{2}}. \]

Now, we verify the hypotheses ($\mathrm H_1$)–($\mathrm H_4$).

• • Verify the hypothesis $(\mathrm H_1)$

  \[\Psi'(x)(x-1)=b(x-1)^{2}\left(1+\frac{r}{x}\right)>0, \ \ x\in(0,\,+\infty)\setminus\{1\}\]
  .

• • Verify the hypothesis $(\mathrm H_2)$

  Since $f_{1}(x)=b ({1}/{x}-1),$ it is easy to get that $F_{1}(x)={1}/{(z+r)}-{1}/{(x+r)}>0.$

• • Verify the hypothesis $(\mathrm H_3)$

  After some calculation, we get

  \[ F'_{1}(x)={\frac{ \left( x-z \right) \left( 3\,{r}^{2}xz+3\,r{x}^{2}z+3\,rx {z}^{2}+{x}^{3}z+{x}^{2}{z}^{2}+x{z}^{3}+{r}^{3}-3\,rxz-{x}^{2}z-x{z}^ {2} \right) }{ \left( z+r \right) ^{3} \left( x+r \right) ^{2}x \left( 1-z \right)}}. \]
  Let $x+z=u,\,xz=v,$ then we have $u>2$, $0< v<1$, and
  \[ F'_{1}(x)={\frac{\left( x-z \right) \left( {r}^{3}+3\,{r}^{2}v+ \left( 3\,uv-3\,v \right) r+{u}^{2}v-uv-{v}^{2} \right) }{ \left( z+r \right) ^{3} \left( x+r \right) ^{2}x \left( 1-z \right) }}. \]
  It is easy to check that $3\,uv-3\,v >0$ and ${u}^{2}v-uv-{v}^{2}>0$ for all $u>2,\,0< v<1.$ Thus, we get that $F'_1(x)>0$ for all $x\in (0,\,+\infty ).$

• • Verify the hypothesis $(\mathrm H_4)$

  Since $F_{2}(x)={{1}/{(z ( z+r )) }}-{{1}/{(x ( x+r )) }},$ after some calculation, we obtain

  \[ \xi'(x)\!=\!\left(\frac{F_2(x)}{F_1(x)}\right)^{'}\!=\!{\frac{{r}^{2}x\!+\!{r}^{2}z\!+\!2r{x}^{2}\!+\!2r{z}^{2}\!+\!{x}^{3}\!+\!{z}^{3}\!-\!2 {r}^{2}\!-\!2xr\!-\!2zr\!-\!{x}^{2}\!-\!{z}^{2}}{z\,{x}^{2} \left( 1-z \right) \left( z+r \right) }}. \]
  Taking $x+z=u,\, \,xz=v$ into $\xi '(x)$, we get
  \[ \xi'(x)={\frac{\left( u-2 \right) {r}^{2}+ \left( 2\,{u}^{2}-2\,u-4\,v \right) r+{u} ^{3}-{u}^{2}-3\,uv+2\,v}{z\,{x}^{2} \left( 1-z \right) \left( z+r \right) }}. \]
  It is easy to check that all the coefficients of $r^{i}$ are positive for $u>2$ and $0< v<1,$ thus $\xi '(x)>0.$

  Furthermore, by (2.7)

  \[ \left(\frac{\eta'(x)}{\xi'(x)}\right)'={\frac{A_{5}(u,v)r^{5}+A_{4}(u,v)r^{4}+A_{3}(u,v)r^{3}+A_{2}(u,v)r^{2}+v A_{1}(u,v)r+v A_{0}(u,v)}{\xi'^{2} (x)(x-1)^{3} x^{4} (z-1)^{6} z (r+z)^{3}}}, \]
  where
  \begin{align*} A_{5}(u,v)=& 2\,{u}^{4}v+{u}^{4}-15\,{u}^{3}v-4\,{u}^{2}{v}^{2}-3\,{u}^{3}+ 34\,{u}^{2}v+30\,u{v}^{2}-4\,{v}^{3}\\ & +6\,{u}^{2}-48\,uv-32\,{v}^{2}-6\, u+36\,v,\\ A_{4}(u,v)& = 10\,{u}^{5}v+3\,{u}^{5}-65\,{u}^{4}v-38\,{u}^{3}{v}^{2}-6\,{u} ^{4}+135\,{u}^{3}v\\ & \quad +208\,{u}^{2}{v}^{2}+22\,u{v}^{3}+9\,{u}^{3}-174\,{u }^{2}v-288\,u{v}^{2} -92\,{v}^{3}-6\,{u}^{2}\\ & \quad +114\,uv+192\,{v}^{2}-36\,v,\\ A_{3}(u,v)& = 20\,{u}^{6}v+3\,{u}^{6}-115\,{u}^{5}v-108\,{u}^{4}{v} ^{2}-6\,{u}^{5}+242\,{u}^{4}v+491\,{u}^{3}{v}^{2}\\ & \quad +162\,{u}^{2}{v}^{3}+ 6\,{u}^{4}-276\,{u}^{3}v-808\,{u}^{2}{v}^{2}-424\,u{v}^{3}-80\,{v}^{4} -3\,{u}^{3}\\ & \quad +174\,{u}^{2}v +600\,u{v}^{2}+368\,{v}^{3}-48\,uv-192\,{v}^{ 2},\\ A_{2}(u,v)& =20\,{u}^{7}v+{u}^{7}-105\,{u}^{6}v-136\,{u}^{5}{v}^{2}-3\,{u}^{6}+226 \,{u}^{5}v+567\,{u}^{4}{v}^{2}\\ & \quad +293\,{u}^{3}{v}^{3}+3\,{u}^{5}-242\,{u} ^{4}v-992\,{u}^{3}{v}^{2} -798\,{u}^{2}{v}^{3}\\ & \quad -216\,u{v}^{4}-{u}^{4}+ 135\,{u}^{3}v+808\,{u}^{2}{v}^{2}\\ & \quad+888\,u{v}^{3}+240\,{v}^{4}-34\,{u}^{ 2}v-288\,u{v}^{2}-368\,{v}^{3}+32\,{v}^{2},\\ A_{1}(u,v)& = 10\,{u}^{8}-50\,{u}^{7}-80\,{u}^{6}v+105\,{u}^{6}+334\,{u}^{5 }v+211\,{u}^{4}{v}^{2}-115\,{u}^{5}\\ & \quad-567\,{u}^{4}v-686\,{u }^{3}{v}^{2}-186\,{u}^{2}{v}^{3}+65\,{u}^{4}+491\,{u}^{3}v+798 \,{u}^{2}{v}^{2}\\ & \quad +450\,u{v}^{3}-12\,{v}^{4}-15\,{u}^{3}-208\,{u}^{2}v-424\,u{v}^{2}-240\,{v}^{3}\\ & \quad +30\,uv+92\,{v}^{2},\\ A_{0}(u,v)& = 2\,{u}^{9}-10\,{u}^{8}-18\,{u}^{7}v+20\,{u}^{7}+80\,{u}^{6}v+54\,{u}^{5}{v}^{2}-20\,{u}^{6}\\ & \quad -136\,{u}^{5}v-211\,{u}^{4} {v}^{2}-53\,{u}^{3}{v}^{3}+10\,{u}^{5}+108\,{u}^{4}v+293\,{u}^{3}{v}^{2}\\ & \quad +186\,{u}^{2}{v}^{3}-6\,u{v}^{4}-2\,{u}^{4}-38\,{u}^{3}v\\ & \quad-162\,{u}^{2}{v}^{2}-216\,u{v}^{3}+12\,{v}^{4}+4\,{u}^{2}v+22 \,u{v}^{2}+80\,{v}^{3}+4\,{v}^{2}. \end{align*}
  In order to verify that $A_i(u,\,v)>0\,(i=0,\,1,\,2,\,3,\,4,\,5)$ for all $(u,\,v)\in \,\{(u,\,v)\mid u>2,\,0< v<1\}$, we will consider $A_i(u,\,v)$ in two regions $\{(u,\,v)\mid u\geq 200,\,0< v<1\}$ and $\{(u,\,v)\mid 2< u<200,\,0< v<1\}$. Firstly, choose some functions $B_i(u)$ such that
  \[ A_i(u,v)>B_i(u)\quad (i=0,1,2,3,4,5), \]
  for $u>2$ and $0< v<1$, where
  \begin{align*} B_0(u)& = 2\,{u}^{9}\!-10\,{u}^{8}\!-18\,{u}^{7}\!-20\,{u}^{6}\!-136\,{u}^{5}-213\,{u}^{ 4}-91\,{u}^{3}-162\,{u}^{2}-222\,u,\\ B_{1}(u)& = 10\,{u}^{8}\!-50\,{u}^{7}\!-80\,{u}^{6}\!-115\,{u}^{5}\!-567\,{u}^{4}\!-701\,{u} ^{3}\!-394\,{u}^{2}-424\,u-252,\\ B_2(u)& ={u}^{7}-108\,{u}^{6}-136\,{u}^{5}-243\,{u}^{4}-992\,{u}^{3}-832\,{u}^{ 2}-504\,u-368,\\ B_3(u)& =3\,{u}^{6}-121\,{u}^{5}-108\,{u}^{4}-279\,{u}^{3}-808\,{u}^{2}-472\,u- 272,\\ B_4(u)& =3\,{u}^{5}-71\,{u}^{4}-38\,{u}^{3}-180\,{u}^{2}-288\,u-128,\\ B_5(u)& ={u}^{4}-18\,{u}^{3}-4\,{u}^{2}-54\,u-36. \end{align*}
  Then, it is easy to verify that $B_i(u)>0\,\,(i=0,\,1,\,2,\,3,\,4,\,5)$ for all $u\geq 200$ by the estimate step by step. For example,
  \begin{align*} B_5(u)& \geq 200\,{u}^{3}-18\,{u}^{3}-4\,{u}^{2}-54\,u-36\\ & \geq 182\times 200\,{u}^{2}-4\,{u}^{2}-54\,u-36\\ & \geq 36\,396\times 200\,u-54u-36>0, \quad u\geq 200. \end{align*}
  Hence, $A_i(u,\,v)>0\,(i=0,\,1,\,2,\,3,\,4,\,5)$ for $u\geq 200,\,\ 0< v<1$.

  On the other hand, according to the polynomial-ring method in algebraic geometry in Maple, it is concluded that $A_i(u,\,v)\,(0\leq i\leq 5)$ has no extreme points in $\{(u,\,v)\mid 2< u<200,\,0< v<1\}.$ And it is easy to check that $A_i(2,\,v)\geq 0,$ $A_i(u,\,1)\geq 0,$ $A_i(u,\,0)\geq 0$ and $A_i(200,\,v)>0.$ This implies that all the coefficients of $r^{i}$ are positive for $2< u<200$ and $0< v<1$. Therefore, we deduce that $A_i(u,\,v)>0\,(i=0,\,1,\,2,\,3,\,4,\,5)$ for all

  \[ (u,\,v)\in \,\{(u,\,v)\mid u>2,\,0< v<1\}. \]
  Summarizing the above analysis, we get $({\eta '(x)}/{\xi '(x)})'\neq 0.$

To sum up, by theorem 2.6, the integral $I(h)$ in (1.5) has at most two zeros in $(0,\, + \infty )$ counted with multiplicities.

Finally, we consider the number of limit cycles bifurcated from the centre $({(b-2)}/{b},\,0)$. Use the transformation $\tilde {x}={b}/{(b-2)}x,\,\tilde {y}=\sqrt {{b}/{(2-b)}} y,\, \tilde {t}=\sqrt {{(2-b)}/{b}}t,\, \,c=2-b$, and rewrite $(\tilde {x},\,\tilde {y},\,\tilde {t})$ as $(x,\,y,\,t)$, then system (1.2) has the following form

(3.8)\begin{equation} \begin{aligned} & \frac{\mathrm{d}x}{\mathrm{d}t}=2xy,\\ & \frac{\mathrm{d}y}{\mathrm{d}t}={-}c x^{2}+2(c-1) x+2-c. \end{aligned} \end{equation}

A first integral of system (3.8) is

(3.9)\begin{equation} \tilde H(x,y)=y^{2}+\frac{cx^{2}}{2}-2(c-1)x+(c-2)\ln|x|+\frac{3c}{2}-2. \end{equation}

It is easy to verify that the centre $({(b-2)}/{b},\,0)$ of system (1.2) becomes the centre $(1,\,0)$ of system (3.8) and the continuous family of ovals $\Gamma _h^{2}$ surrounding the centre $({(b-2)}/{b},\,0)$ defined in (1.3) becomes $\Gamma _h=\{(x,\,y)\mid \tilde H(x,\,y)=h,\,\ h\in (0,\,+\infty )\}$ if we take $\tilde h=({(2-c)}/{c})h+(2-c)\ln ({c}/{(2-c)})+2-{2}/{c}$ in (1.3) and rewrite $\tilde h$ as $h$. Similarly, the cyclicity of period annulus around the centre $({(b-2)}/{b},\,0)$ is equal to the number of zeros of the following Abelian integrals in $(0,\,+\infty )$ by lemma 1.1

\[ I(h)={-}\alpha\sqrt{\dfrac{2-c}{c}}\displaystyle\oint_{\Gamma_{h}}\frac{y}{x^{2}}\ \mathrm{d}x +\beta\sqrt{\frac{c}{2-c}}\displaystyle\oint_{\Gamma_{h}}\frac{y}{x}\ \mathrm{d}x -\gamma\left({\frac{c}{2-c}}\right)^{3/2}\displaystyle\oint_{\Gamma_{h}}y\ \mathrm{d}x, \]

where $\Gamma _h$ is the periodic orbit surrounding the centre $(1,\,0)$ of system (3.8) defined by $\tilde H(x,\,y)=h$ with $\tilde H(x,\,y)$ given in (3.9). Using the conclusion on the centre $(1,\,0)$, thus we can see that at most two limit cycles generate from the period annulus around the centre $({(b-2)}/{b},\,0)$ counted with multiplicities.

The proof is finished.

Proof. It is easy to check that $P(h)>0$, $Q(h)>0$ and we only prove that $P'(h)>0$ and $Q'(h)<0.$

Since $l_1(x)=1,$ $l_2(x)={1}/{x}$ and $l_3(x)={1}/{x^{2}}$, it is easy to obtain that

\begin{align*} & L_1(x)={\frac{(1+r) \left( z-x \right) \left( xz+r \right) }{ 2\left( x-1\right) \left( x+r \right) \left( z-1 \right) \left( z+r \right) }},\\ & L_2(x)={\frac{ (1+r)\left( z-x \right) \left( x+r-1+z \right) }{ 2\left( x-1\right) \left( x+r \right) \left( z-1 \right) \left( z+r \right) }} \end{align*}

and

\[ L_3(x)={\frac{ (1+r)\left( z-x \right) \left( xr+zr+{x}^{2}+xz+{z}^{2}-r-x-z \right) }{2x \left( x-1 \right) \left( x+r \right) z \left( z-1\right) \left( z+r \right) }}. \]

After some calculation, we get

\begin{align*} \left(\frac{L_{1}(x)}{L_{2}(x)}\right)'& ={\frac{1}{ \left( x\!+\!r-1\!+\!z \right) ^{2}x \left( z-1 \right) \left( z\!+\!r \right) }}\left({r}^{2}{x}^{2}z\!+\!{r}^{2}x{z}^{2}\!+\!2\,r{x}^{3}z+2\,rx{z}^{3}+{x}^ {4}z\right.\\ & \quad\left.+x{z}^{4}-4\,{r}^{2}xz-4\,r{x}^{2}z-4\,rx{z}^{2}-2\,{x}^{3}z-2\,x{ z}^{3}+{r}^{2}x+{r}^{2}z+4\,rxz\right.\\ & \quad\left.+{x}^{2}z+x{z}^{2}\right). \end{align*}

Let $x+z=u,\,xz=v,$ then

\[ \left(\frac{L_{1}(x)}{L_{2}(x)}\right)'={\frac{ \left( uv\!+\!u-4v \right) {r}^{2}\!+\! 2\left( {u}^{2}-2u-2{v} \!+\!2 \right)v r\!+\!\left({u}^{3}-2{u}^{2}-3u{v}\!+\!u\!+\!4{v}\right)v}{ \left( x\!+\!r-1\!+\!z \right) ^{2}x \left( z\!-\!1 \right) \left( z\!+\!r \right) }}. \]

Similar to the proof of $({\eta '(x)}/{\xi '(x)})'\neq 0$ in theorem 1.2, we obtain $({L_{1}(x)}/{L_{2}(x)})'<0.$ Using theorem 1 in [Reference Li and Zhang19], we know $Q'(h)=({I_{1}(h)}/{I_{2}(h)})'<0.$

As for ${L_{3}(x)}/{L_{2}(x)},$ some computations show that

\[ \left(\frac{L_{3}(x)}{L_{2}(x)}\right)'={\frac{M(x,z,r)}{z{x}^{2} \left( x+r-1+z \right) ^{2} \left( 1-z \right) \left( z+r \right) }}, \]

where

\begin{align*} M(x,z,r)=& \left( {u}^{2}-2\,u-2\,v+2 \right) {r}^{3}+ \left( 3\,{u}^{3}-7\,{u}^ {2}-7\,uv+7\,u+6\,v-2 \right) {r}^{2}\\ & + \left( 3\,{u}^{4}-8\,{u}^{3}-8 \,{u}^{2}v+7\,{u}^{2}+16\,uv-2\,{v}^{2}-2\,u-6\,v \right) r\\ & +{u}^{5}-3 \,{u}^{4}-3\,{u}^{3}v+3\,{u}^{3}+8\,{u}^{2}v-u{v}^{2}-{u}^{2}-7\,uv+2 \,{v}^{2}+2\,v. \end{align*}

Similarly, we have $M(x,\,z,\,r)>0.$ This implies that $({L_{3}(x)}/{L_{2}(x)})'>0.$ Therefore, by theorem 1 in [Reference Li and Zhang19], we get $P'(h)=({I_{3}(h)}/{I_{2}(h)})'>0.$ The proof is finished.

Proof. By Green formula and mean value theorem, there exist $\rho _{1},\,\,\rho _{2}$ such that

\[ Q(h)=\frac{\iint_{\{(x,y)\mid H(x,y)\leq h, x>0\}}1\ \mathrm{d}x\mathrm{d}y}{\iint_{\{(x,y)\mid H(x,y)\leq h, x>0\}}({1}/{x})\ \mathrm{d}x\mathrm{d}y}=\rho_{1} \]

and

\[ P(h)=\frac{\iint_{\{(x,y)\mid H(x,y)\leq h, x>0\}}({1}/{x^{2}})\ \mathrm{d}x\mathrm{d}y}{\iint_{\{(x,y)\mid H(x,y)\leq h, x>0\}}({1}/{x})\ \mathrm{d}x\mathrm{d}y}=\rho_{2}. \]

When $h\rightarrow 0^{+},$ $\{(x,\,y)\mid H(x,\,y)\leq h,\, x>0\}$ tends to the point $(1,\,0),$ then

\[ \rho_{1},\,\rho_{2}\rightarrow 1,\,P(h),\,Q(h)\rightarrow 1. \]

When $h\gg 1$, from $\Psi (\nu (h))=h$, we obtain

\[ \sqrt{\frac{b}{2}}\nu(h)\left[1-\frac{2(b-1)}{\nu(h)}\frac{2}{b}+O\left(\frac{\ln \nu(h)}{\nu^{2}(h)}\right)\right]^{{1}/{2}}=\sqrt{h}, \]

which implies that

(4.1)\begin{equation} \nu(h)=\sqrt{\frac{2}{b}}h^{{1}/{2}}+\frac{2(b-1)}{b}+o(1),\quad h\gg 1. \end{equation}

On the other hand, by $\Psi (\mu (h))=h$, we get $0<\mu (h)\ll 1$ and

\[ \exp\left\{\frac{b\mu^{2}(h)}{2}-2(b-1)\mu(h)+(b-2)\ln \mu(h)+\frac{3b}{2}-2\right\}=\exp(h). \]

After some calculation

(4.2)\begin{equation} \mu(h)\exp\left\{\frac{b\mu^{2}(h)}{2b-4}-\frac{2(b-1)\mu(h)}{b-2}\right\}=\exp\left\{\frac{3b-4}{4-2b}\right\}\exp\left\{\frac{h}{b-2}\right\}. \end{equation}

Combining (4.2) and $\exp (z)=1+z+o(z)$ for $|z|\ll 1$, we have

\[ \mu(h)\left(1-\frac{2(b-1)\mu(h)}{b-2}+o(\mu(h))\right)=\exp\left\{\frac{3b-4}{4-2b}\right\}\exp\left\{\frac{h}{b-2}\right\}. \]

Further, we get that

(4.3)\begin{equation} \mu(h)=\exp\left\{\frac{3b-4}{4-2b}\right\}\exp\left\{\frac{h}{b-2}\right\} +o\left(\exp\left\{\frac{h}{b-2}\right\}\right),\quad h\gg 1. \end{equation}

In the following we estimate the order of $I_1(h),\,I_2(h)$ and $I_3(h)$ for $h\rightarrow +\infty.$

Let us consider the integral $I_1(h)$ firstly. Since $y^{2}+\Psi (x)=h$, we get

\[ I_1(h)=\oint_{\Gamma^{1}_{h}}y\ \mathrm{d}x=2 \int_{\mu(h)}^{\nu(h)}\sqrt{h-\Psi(x)}\ \mathrm{d}x. \]

Then

(4.4)\begin{equation} I_1(h)\leq 2 \int_{\mu(h)}^{\nu(h)}\sqrt{h}\ \mathrm{d}x=2\sqrt{\frac{2}{b}}h+o(h). \end{equation}

Secondly, consider the integral $I_2(h)$. Because

\[ I_2(h)=2 \int_{\mu(h)}^{\nu(h)}\frac{\sqrt{h-\Psi(x)}}{x}\mathrm{d}x=2\int_{\mu(h)}^{1}\frac{\sqrt{h-\Psi(x)}}{x}\mathrm{d}x +2\int_{1}^{\nu(h)}\frac{\sqrt{h-\Psi(x)}}{x}\mathrm{d}x, \]

we have

\[ I_2(h)>2\int_{\mu(h)}^{1}\frac{\sqrt{h-\Psi(x)}}{x}\mathrm{d}x=2\int_{0}^{\bar{\mu}(h)}\sqrt{h-\Psi(\exp\{{-}w\})}\ \mathrm{d}w, \]

where

\[ \bar{\mu}(h)={-}\ln(\mu(h))=\frac{h}{2-b}+o(h) \]

follows from (4.3). Let $w=-\ln x,$ one has

\[ \Phi(w):=\Psi(\exp\{{-}w\})=\frac{b}{2}\exp\{{-}2w\}-2(b-1)\exp\{{-}w\}+(2-b)w+\frac{3b}{2}-2. \]

Then for $0< w<{\bar {\mu }(h)}/{2}$,

\[ \Phi(w)<\frac{b}{2}+2+(2-b)w+\frac{3b}{2}-2=(2-b)w+2b<(2-b)\frac{\bar{\mu}(h)}{2}\!+\!2b=\frac{h}{2}\!+\!o(h). \]

Consequently

(4.5)\begin{equation} \begin{aligned} I_2(h) & >2\int_{0}^{\bar{\mu}(h)}\sqrt{h-\Phi(w)}\ \mathrm{d}w\\ & >2\int_{0}^{{\bar{\mu}(h)}/{2}}\sqrt{h-\Phi(w)}\ \mathrm{d}w\\ & >2\int_{0}^{{\bar{\mu}(h)}/{2}}\left(\sqrt{\frac{h}{2}}+o\left(\sqrt{h}\right)\right)\ \mathrm{d}w\\ & =\frac{\sqrt{2}h^{{3}/{2}}}{2(2-b)}+o\left(h^{{3}/{2}}\right). \end{aligned} \end{equation}

On the other hand, we have

\begin{align*} I_2(h)& =2\int_{0}^{\bar{\mu}(h)}\sqrt{h-\Phi(w)}\ \mathrm{d}w+2\int_{1}^{\nu(h)}\frac{\sqrt{h-\Psi(x)}}{x}\ \mathrm{d}x\\ & \leq 2\int_{0}^{\bar{\mu}(h)}\sqrt{h}\ \mathrm{d}w+2\int_{1}^{\nu(h)}\sqrt{h}\ \mathrm{d}x\\ & = \frac{2h^{{3}/{2}}}{2-b}+o\left(h^{{3}/{2}}\right). \end{align*}

To sum up, we obtain

(4.6)\begin{equation} I_2(h)\sim h^{{3}/{2}},\quad h\rightarrow +\infty. \end{equation}

Thereby, by (4.4) and (4.6),

\[ Q(h)=\frac{I_1(h)}{I_2(h)}\rightarrow 0,\text{ when } h\rightarrow +\infty. \]

Similar as the estimate in (4.5), we have

\begin{align*} I_3(h)& >2\int_{0}^{\bar{\mu}(h)}\exp\{w\}\sqrt{h-\Phi(w)}\ \mathrm{d}w\\ & > 2\int_{0}^{{\bar{\mu}(h)}/{2}}\exp\{w\}\left(\sqrt{\frac{h}{2}}+o\left(\sqrt{h}\right)\right)\ \mathrm{d}w\\ & = \sqrt{2h}\exp\left\{\frac{h}{2}\right\}+o\left(\sqrt{h}\exp\left\{\frac{h}{2}\right\}\right). \end{align*}

Thus,

\[ \lim_{h\rightarrow +\infty}P(h)=\lim_{h\rightarrow +\infty}\frac{I_3(h)}{I_2(h)}={+}\infty. \]

The proof is finished.

Proof of theorem 1.3. Proof of theorem 1.3

Put two centroid curves $\Sigma$ and $\bar {\Sigma }$ in the same plane, thus the lines $L_{\alpha \beta \gamma }$ and $\bar L_{\alpha \beta \gamma }$ are the same. The graphs of $\Sigma$ and $\bar {\Sigma }$ are shown in figure 3 by propositions 4.1 and 4.2 and (4.7). Obviously, the configurations of limit cycles $(0,\,0),\,(1,\,0),\,(0,\,1)$ and $(1,\,1)$ are trivial (see figure 3), where $(u,\,v)$ represents that exactly $u$ and $v$ limit cycles simultaneously bifurcate from the periodic orbits surrounding the centres $(1,\,0)$ and $({(b-2)}/{b},\,0)$ respectively. Without loss of generality, suppose that $L_{\alpha \beta \gamma }$ has two intersections with the centroid curve $\Sigma$. As is shown in figure 3, the slope of $L_{\alpha \beta \gamma }$ must be negative, thus $\bar L_{\alpha \beta \gamma }$ has no intersection with $\bar \Sigma$, and the configuration of limit cycles must be $(2,\,0)$. Similarly, the configuration of limit cycles $(0,\,2)$ can be achieved.

Figure 3.

Number of intersection points of the curve $\Sigma \ (\bar \Sigma )$ with the line $L_{\alpha \beta \gamma }\ (\bar L_{\alpha \beta \gamma })$.