#### Proof

Let us analyze now for (12) the critical points in both the finite and infinity planes. For this the Theorem 1 (see [9, §3.10-P 271]) is used, which is based on the study of the system, on the equator of the Poincaré sphere, that is when *x*^{2} + *y*^{2} = 1.

Remember that *P*(*x*, *y* ) = *y* y *Q* (*x*,*y* ) = 2(*a* + *b* )*y* − (*a*^{2} + *b*^{2} + *c*)*x* with *a*,*b*,*c* > 0, then the critical points are given by the system:

$$\begin{array}{r}\hfill y=0\\ \hfill 2(a+b)y-({a}^{2}+{b}^{2}+c)x=0\end{array}$$

then the only critical point in the finite plane is *(x*, *y*) = (0, 0). Now for the qualitative analysis of the critical point we find the Jacobian matrix evaluated in the point, given by:

$$D(P\mathrm{,}Q){|}_{(\mathrm{0,0})}=\left(\begin{array}{cc}0& 1\\ -({a}^{2}+{b}^{2}+c)& 2(a+b)\mathrm{.}\end{array}\right)$$

The characteristic polynomial is

$${\lambda}^{2}-2(a+b)\lambda +({a}^{2}+{b}^{2}+c)=\mathrm{0,}$$

then the eigenvalues are:

$${\lambda}_{\mathrm{1,2}}=\frac{2(a+b)\pm \sqrt{4(a+b{)}^{2}-4({a}^{2}+{b}^{2}+c)}}{2}=(a+b)\pm \sqrt{2ab-c}\mathrm{.}$$(13)

According to Theorem 1 (see [9, §3.10-P 271]) and Theorem 2 (see [9, §3.10-P 272, 273]), to find the critical points and their behavior we can use the system:

$$\begin{array}{l}\pm \dot{y}=y{z}^{r}P(1/z\mathrm{,}y/z)-{z}^{r}Q(1/z\mathrm{,}y/z)\hfill \\ \pm \dot{z}={z}^{r+1}P(1/z\mathrm{,}y/z)\mathrm{.}\hfill \end{array}$$

In our case *r* = 1 and when replacing we find:

$$\begin{array}{l}\pm \dot{y}={y}^{2}-2(a+b)y+({a}^{2}+{b}^{2}+c)p\hfill \\ \pm \dot{z}=zy\mathrm{.}\hfill \end{array}$$(14)

The critical points for this system result from solving the system:

$$\begin{array}{l}{y}^{2}-2(a+b)y+({a}^{2}+{b}^{2}+c)=0\hfill \\ z=\mathrm{0,}\hfill \end{array}$$

then

$${y}_{\mathrm{1,2}}=\frac{2(a+b)\pm \sqrt{4(a+b{)}^{2}-4({a}^{2}+{b}^{2}+c)}}{2}=(a+b)\pm \sqrt{2ab-c}\mathrm{.}$$(15)

Also, as we are in the equator of the Poincaré sphere *x*^{2} + *y*^{2} = 1 that is $x=\pm \sqrt{1-{y}^{2}}$.

We note that if we change the sings of *x* e *and* we obtain points called *Antipodes*, which have a contrary stability. Taking into account the above conditions, the critical points at infinity are:

$$(x\mathrm{,}y\mathrm{,0})=(\pm \sqrt{1-{y}^{2}}\mathrm{,}(a+b)\pm \sqrt{2ab-c}\mathrm{,0})\mathrm{.}$$

The Jacobian matrix (14) is:

$$D(P\mathrm{,}Q)=\left(\begin{array}{cc}2y-2(a+b)& 0\\ z& 0\end{array}\right)$$

in our case:

$$D(P\mathrm{,}Q){|}_{({y}_{\mathrm{1,2}}\mathrm{,0})}=\left(\begin{array}{cc}\pm 2\sqrt{2ab-c}& 0\\ 0& (a+b)\pm \sqrt{2ab-c}\mathrm{.}\end{array}\right)$$

We can notice then that the signs of the own values depend on the signs of the elements in the diagonal. The following indicates that several cases must be analyzed for both the finite and infinity planes, be *Δ* = 2*ab* − *c* .

Case 1. If *Δ* > 0 then in (13) λ_{1} > 0, but for λ_{2} we have three options. If λ_{2} < 0 and λ_{2} = 0, contradicts the fact that *a*, *b*, *c* > 0 then the only option is λ_{2} > 0. Since both eigenvalues are positive then the origin is a critical *repulsor*.

Regarding the critical points at infinity in this case, we can see the elements on the diagonal of (14) have two options *y*_{1},*y*_{2}. For *y*_{1}:

$$D(P\mathrm{,}Q){|}_{({y}_{1}\mathrm{,0})}=\left(\begin{array}{cc}2\sqrt{2ab-c}& 0\\ 0& (a+b)+\sqrt{2ab-c}\end{array}\right)$$

For this point, the values in the diagonal are positive and then in infinity this critical point is of the *repulsor* type and, therefore, its antipode is of an *attractor* type.

For *y*_{2}:

$$D(P\mathrm{,}Q){|}_{({y}_{2}\mathrm{,0})}=\left(\begin{array}{cc}-2\sqrt{2ab-c}& 0\\ 0& (a+b)-\sqrt{2ab-c}\mathrm{,}\end{array}\right)$$

where $-2\sqrt{2ab-c}<0$ and $(a+b)-\sqrt{2ab-c}={\lambda}_{2}>0$ Therefore, the critical point is of a *saddle* type. So the phase portrait for the system in this case is:

Global phase portrait for case 1 (12)

Case 2. If *Δ* = 0 then λ_{1} = λ_{2} > 0 that is, the origin is a *repulsor* critical point. For the points at infinity the system (14) is reduced to:

$$\begin{array}{l}\pm \dot{y}=(y-a-b{)}^{2}\hfill \\ \pm \dot{z}=zy\hfill \end{array}$$

with solution *y* = *a* + *b* > 0. Then the phase portrait for the system in this case is:

Global phase portrait for case 2 (12)

Case 3. If *Δ* < 0 then ${\lambda}_{1}=(a+b)+i\sqrt{c-2ab}$ and ${\lambda}_{2}=(a+b)-i\sqrt{c-2ab}$ with (*a* + *b* ) > 0 that is the origin is critical point of a *spiral* type. So the phase portrait for the system in this case is:

Global phase portrait for case 3 (12)

As regards infinity, the equation (15) has no real solutions, that is, it has no critical points at infinity.

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