For the convenience of description, the original system proposed in [22] as follows:

$$\left\{\begin{array}{l}\dot{x}=a(y-x)\\ \dot{y}=-c+xz\\ \dot{z}=b-{y}^{2},\end{array}\right.$$(2.1)

when parameters *a*(*a* > 0), *b* and *c* are all real constants, and *x*, *y*, *z* are three state variables in this case. Apparently, when *c* = 0, system (2.1) is invariant while it is under the transformation T(*x*, *y*, *z*) → T(−*x*, −*y*, *z*). In the sense of this transformation, the system not itself invariant under the T transformation, and there is another orbit that corresponds to it. We calculate the next formulas:

$$a(y-x)=0,\phantom{\rule{2em}{0ex}}-c+xz=0,\phantom{\rule{2em}{0ex}}b-{y}^{2}=0.$$

*a*(*y* − *x*) = 0,−*c* + *xz* = 0, *b* − *y*_{2} = 0.

It is easily obtained that if *b*_{2} + *c*_{2} = 0, there are non-isolated equilibria *O*_{z}(0,0, *z*); if either *b* < 0 or *b* = 0 and c ≠ 0, there is not an equilibrium; if *b* > 0, there are two equilibria ${E}^{+}(\sqrt{b},\sqrt{b},\frac{c}{\sqrt{b}})\phantom{\rule{thickmathspace}{0ex}}\text{and}\phantom{\rule{thickmathspace}{0ex}}{E}^{-}(-\sqrt{b},-\sqrt{b},-\frac{c}{\sqrt{b}}).$.

In this part, based on the model (2.1), we design a controller to get the controlled system as below

$$\left\{\begin{array}{l}\dot{x}=a(y-x)\\ \dot{y}=-c+xz+m(x-y)\\ \dot{z}=b-{y}^{2}.\end{array}\right.$$(2.2)

As needed, we only have to control the second equation, and the other two equations remain unchanged. Obviously, the characteristic of the control law is to ensure that the structure of the equilibrium and the dimension of the original system (2.1) are unchanged. Especially, if *m* = 0, the controlled system will restore to the original system.

The linearized system (2.2) acts at *E*_{±}, and we can get the Jacobian matrix

$${J}^{\pm}=\left[\begin{array}{ccc}-a& a& 0\\ m+{z}^{\pm}& -m& {x}^{\pm}\\ 0& -2{y}^{\pm}& 0\end{array}\right],$$(2.3)

its characteristic equation being

$$\mathrm{\Delta}(\lambda )={\lambda}^{3}+(m+a){\lambda}^{2}+(2{x}^{\pm}{y}^{\pm}-a{z}^{\pm})\lambda +2a{x}^{\pm}{y}^{\pm}=0.$$(2.4)

Regarding the controlled system, we conclude that if $c>\frac{2bm\sqrt{b}}{a(a+m)}$ *E*^{+} is unstable while *E*^{−} is stable, and if $c<\frac{2bm\sqrt{b}}{a(a+m)}$, *E*^{+} is stable while *E*^{−} is unstable. The stability of the both is always opposite.

We just analyse the stability of equilibrium *E*^{+}. By using the Routh-Hurwitz criterion, we can easily prove the equilibrium *E*^{+} is stable while $c<\frac{2bm\sqrt{b}}{a(a+m)}$.

The formula (2.3) at equilibrium *E*^{+} is just as follows:

$${\mathrm{\Delta}}_{+}(\lambda )={\lambda}^{3}+(m+a){\lambda}^{2}+(2b-\frac{ac}{\sqrt{b}})\lambda +2ab=0.$$(2.5)

Let $A=m+a,B=2b-\frac{ac}{\sqrt{b}},C=2ab$. On the basis of the Routh-Hurwitz criterions, with the equation (2.4), while *A* > 0, *C* > 0, *AB* − *C* > 0, i.e. *a* > 0, *b* > 0, and $c<\frac{2bm\sqrt{b}}{a(a+m)}$, all of the real parts of the three roots are negative. We take into account the characteristic equation (2.3) at equilibrium *E*^{−} in the same way.

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