In the previous sections, we have discussed the conditions for which the periodic solutions of the system (2.2) bifurcate from coaxial equilibrium \(E_{2}\) at the critical values of \(\Delta _{k}\) via the Hopf bifurcation. In this section, we will analyze the direction, stability, and periods of the periodic solutions of the system (2.2) using normal theory and the center manifold theorem developed by Hassard et al. [35].
Let \(u_{1}(t) = x(t)  \bar{x}\), \(u_{2}(t) = y(t)\bar{y}\), \(u_{3}(t) = z(t)  \bar{z}\), \(x(t) = u_{1}(\Delta t)\), \(y(t) = u_{2}(\Delta t)\), \(z(t) = u_{3}(\Delta t)\), and \(\Delta = \Delta _{0} + \mu \), where \(\Delta _{0}\) is defined by (3.11), and \(\mu \in \mathbb{R}\). The system (2.2) can be written as a functional differential equation in \(\mathbb{C} = \mathbb{C}([1,0], \mathbb{R}^{3})\) as
$$ X^{\prime }= L_{\mu } (X_{t}) + f(\mu , X_{t}), $$
(6.1)
where \(X(t) = (x(t), y(t), z(t))^{T} \in {\mathbb{R}^{3}}\), and \(L_{\mu }: \mathbb{C} \rightarrow {\mathbb{R}^{3}}\), \(f: \mathbb{R} \times \mathbb{C} \rightarrow {\mathbb{R}^{3}}\) are given, respectively, as follows: for \(\Phi (t) = (\Phi _{1}(t), \Phi _{2}(t), \Phi _{3}(t))^{T} \in \mathbb{C}([\Delta , 0], \mathbb{R}^{3})\), we define
$$ L_{\mu }(\Phi ) = D_{1}\Phi (0) + D_{2}\Phi (1), $$
(6.2)
and
$$ f(\mu , \Phi ) = (\Delta _{0}+ \mu )M, $$
(6.3)
where
$$\begin{aligned}& D_{1} = (\Delta _{0} + \mu ) \begin{pmatrix} 1 & 0 & 0 \\ a_{1} & a_{2} & 0 \\ 0 & 0 & a_{4} \end{pmatrix} , \qquad D_{2} = (\Delta _{0} + \mu ) \begin{pmatrix} 0 & \frac{\bar{z}}{1+\bar{z}} & \frac{\bar{y}}{(1+\bar{z})^{2}} \\ 0 & 0 & 0 \\ 0 & \bar{z} & \bar{y} \end{pmatrix} , \end{aligned}$$
(6.4)
$$\begin{aligned}& M = \begin{pmatrix} \Phi _{2}(1)\Phi _{3}(1)  \Phi _{2}(1)\Phi _{3}^{2}(1) + \text{HOT} \\ 0 \\ \Phi _{2}(1)\Phi _{3}(1) \end{pmatrix} , \end{aligned}$$
(6.5)
HOT → higher order terms.
By the Riesz representation theorem, there exists a matrix function \(\eta (\theta ,\mu )\) of bounded variation for \(\theta \in [1, 0]\) such that
$$ L_{\mu }(\Phi ) = \int _{1}^{0} d\eta (\theta , \mu )\Phi (\theta ) \quad \text{for } \Phi \in \mathbb{C}. $$
(6.6)
For the Dirac delta function δ, choose
$$ \eta (\theta , \mu ) = D_{1}\delta (\theta ) + D_{2}\delta (\theta +1), $$
(6.7)
and for \(\Phi \in {\mathbb{C}}^{1}([1,0],\mathbb{R}^{3})\), define
$$ A(\mu )\Phi (\theta ) = \textstyle\begin{cases} \frac{d\Phi (\theta )}{d\theta }, & \theta \in [1,0), \\ \int _{1}^{0} d\eta (s,\mu )\Phi (s), & \theta = 0, \end{cases} $$
(6.8)
and
$$ R(\mu )\Phi (\theta ) = \textstyle\begin{cases} 0, & \theta \in [1,0), \\ f(\mu ,\theta ), & \theta = 0. \end{cases} $$
(6.9)
Hence, system (6.1) is equivalent to the operator equation
$$ X^{\prime }= A({\mu })X_{t} + R(\mu )X_{t}, $$
(6.10)
where \(X_{t}(\theta ) = X(t+\theta )\) for \(\theta \in [1,0)\).
For \(\Psi \in {\mathbb{C}^{1}}([1,0), {(\mathbb{R}^{3})}^{*})\), define
$$ A^{*}\Psi (s) = \textstyle\begin{cases} \frac{d\Psi (s)}{ds}, & s \in (0,1], \\ \int _{1}^{0} d\eta ^{T}(t, 0)\Psi (t), & s = 0, \end{cases} $$
(6.11)
and the bilinear inner product
$$ \bigl\langle {\Psi (s), \Phi (\theta )} \bigr\rangle = \bar{ \Psi }(0)\Phi (0)  \int _{\theta = 1}^{0} \int _{\xi = 0}^{\theta }\bar{\Psi }(\xi  \theta )\,d\eta (\theta )\Phi (\xi )\,d\xi , $$
(6.12)
where \(\eta (\theta ) = \eta (\theta , 0)\). Then \(A(0)\) and \(A^{*}\) are adjoint operators. We already assume that \(\pm \iota \omega _{0}\Delta _{k}\) are the eigenvalues of \(A(0)\). Hence, the eigenvalues of \(A^{*}\) are \(\mp \iota \omega _{0}\Delta _{k}\). We need to compute the eigenvectors of \(A(0)\) and \(A^{*}\) corresponding to the eigenvalues \(\iota \omega _{0}\Delta _{k}\) and \(\iota \omega _{0}\Delta _{k}\) respectively.
Suppose that \(v(\theta ) = (1, v_{1}, v_{2})^{T}e^{\iota \omega _{0}\Delta _{k} \theta }\) is the eigenvector of \(A(0)\) corresponding to \(\iota \omega _{0}\Delta _{k}\). Then \(A(0)v(0) = \iota \omega _{0}\Delta _{k}v(0)\). It follows from the definition of \(A(0)\) and (6.3), (6.6) and (6.7) that
$$ \Delta _{k} \begin{pmatrix} 1 & 0 & 0 \\ a_{1} & a_{2} & 0 \\ 0 & 0 & a_{4} \end{pmatrix} v(0) + \Delta _{k} \begin{pmatrix} 0 & \frac{\bar{z}}{1+\bar{z}} & \frac{\bar{y}}{(1+\bar{z})^{2}} \\ 0 & 0 & 0 \\ 0 & \bar{z} & \bar{y} \end{pmatrix} v(1) = \iota \omega _{0}\Delta _{k}v(0). $$
Then, for \(v(1) = v(0)e^{\iota \omega _{0}\Delta _{k}}\), we obtain
$$ v_{1} = \frac{a_{1}}{a_{2}+\iota \omega _{0}} ; \qquad v_{2} =  \frac{a_{1}\bar{z}e^{\iota \omega _{0}\Delta _{k}}}{(a_{2}+\iota \omega _{0}) (\iota \omega _{0}+\bar{y}e^{\iota \omega _{0}\Delta _{k}})}. $$
In a similar manner, we can obtain the eigenvector \(v^{*}(s) = D(1, v_{1}^{*}, v_{2}^{*})^{T}e^{\iota \omega _{0} \Delta _{k}s}\) of \(A^{*}\) corresponding to \(\iota \omega _{0}\Delta _{k}\), where
$$ v_{1}^{*} = \frac{1\iota \omega _{0}}{a_{1}} ; \qquad v_{2}^{*} =  \frac{\bar{y}}{(1+\bar{y})^{2}(\iota \omega _{0} \bar{y}e^{\iota \omega _{0}\Delta _{k}}) e^{\iota \omega _{0}\Delta _{k}}}. $$
In order to guarantee \(\langle {v^{*}(s), v(\theta )}\rangle = 1\), we need to determine the expression of D
$$ \begin{aligned} &\bigl\langle {v^{*}(s), v(\theta )} \bigr\rangle \\ &\quad = \bar{D} \bigl(1, \bar{v_{1}}^{*}, \bar{v_{2}}^{*} \bigr) (1, v_{1}, v_{2})^{T} \\ &\qquad {}  \int _{\theta = 1}^{0} \int _{\xi = 0}^{\theta } \bar{D} \bigl(1, \bar{v_{1}}^{*}, \bar{v_{2}}^{*} \bigr)e^{\iota \omega _{0}\Delta _{k}( \xi  \theta )}\,d\eta (\theta ) (1, v_{1}, v_{2})^{T}e^{\iota \omega _{0} \Delta _{k}\xi } \,d\xi \\ &\quad = \bar{D} \biggl\{ \bigl(1 + \bar{v_{1}}^{*}v_{1} + \bar{v_{2}}^{*}v_{2} \bigr)  \int _{\theta = 1} ^{0} \bigl(1, \bar{v_{1}}^{*}, \bar{v_{2}}^{*} \bigr)\theta e^{ \iota \omega _{0}\Delta _{k}\theta }\,d\eta ( \theta ) (1, v_{1}, v_{2})^{T} \biggr\} \\ &\quad = \bar{D} \biggl\{ \bigl(1 + \bar{v_{1}}^{*}v_{1} + \bar{v_{2}}^{*}v_{2} \bigr) + \biggl\{ \frac{\bar{z}}{1+\bar{z}}v_{1}+\frac{\bar{y}}{(1+\bar{z})^{2}}v_{2} \bar{z}\bar{v_{2}}^{*}v_{1}\bar{y} \bar{v_{2}}^{*}v_{2} \biggr\} \Delta _{k}e^{ \iota \omega _{0}\Delta _{k}} \biggr\} . \end{aligned} $$
Therefore we can choose D as
$$ \bar{D} = \biggl[ \frac{1}{(1 + \bar{v_{1}}^{*}v_{1} + \bar{v_{2}}^{*}v_{2}) + \{\frac{\bar{z}}{1+\bar{z}}v_{1}+\frac{\bar{y}}{(1+\bar{z})^{2}}v_{2}\bar{z}\bar{v_{2}}^{*}v_{1}\bar{y}\bar{v_{2}}^{*}v_{2} \}\Delta _{k}e^{\iota \omega _{0}\Delta _{k}}} \biggr]. $$
On the other hand, due to adjoint property, we can write \(\langle {\Psi , A\Phi }\rangle = \langle {A^{*}\Psi , \Phi }\rangle \).
We have
$$ \begin{aligned} \iota \omega _{0}\Delta _{k} \bigl\langle {v^{*},\bar{v}} \bigr\rangle & = \bigl\langle {v^{*}, A\bar{v}} \bigr\rangle = \bigl\langle {A^{*}v^{*}, \bar{v}} \bigr\rangle \\ & = \bigl\langle {\iota \omega _{0}\Delta _{k}v^{*}, \bar{v}} \bigr\rangle \\ & = \iota \omega _{0}\Delta _{k} \bigl\langle {v^{*}, \bar{v}} \bigr\rangle . \end{aligned} $$
Therefore, \(\langle {v^{*}, \bar{v}}\rangle = 0\).
Now, we will compute the coordinates describing the center manifold \(C_{0}\) at \(\mu = 0\). Let \(X_{t}\) be the solution of (6.10) when \(\mu = 0\). We define
$$\begin{aligned}& Z(t) = \bigl\langle {v^{*}, X_{t}} \bigr\rangle , \\& W(t,\theta ) = X_{t}  z(t)v(\theta )\bar{z}(t) \bar{v}(\theta ) = X_{t}( \theta )2\operatorname{Re} \bigl\{ z(t)v( \theta ) \bigr\} . \end{aligned}$$
(6.13)
On the center manifold \(C_{0}\) we have \(W(t,\theta ) = W(z(t),\bar{z}(t), \theta )\), where
$$ W(z,\bar{z},\theta ) = W_{20}(\theta ) \frac{z^{2}}{2}+W_{11}(\theta )z \bar{z} + W_{02}( \theta )\frac{\bar{z}^{2}}{2}+W_{30}(\theta ) \frac{z^{3}}{6}+ \cdots, $$
(6.14)
where z and z̄ are the local coordinates for the center manifold \(C_{0}\) in the direction of \(\bar{v}^{*}\) and \(v^{*}\). Note that W is real if \(x_{t}\) is real. Here, we are only interested in real solutions. From (6.13), we have
$$ \begin{aligned} \bigl\langle {v^{*}, W} \bigr\rangle & = \bigl\langle {v^{*}, X_{t}zv\bar{z}\bar{v}} \bigr\rangle \\ & = \bigl\langle {v^{*},X_{t}} \bigr\rangle z \bigl\langle {v^{*},v} \bigr\rangle z \bigl\langle {v^{*}, \bar{v}} \bigr\rangle \\ & = z  z \\ & = 0. \end{aligned} $$
For the solution \(x_{t} \in C_{0}\) in (6.10), since \(\mu = 0\), hence
$$ \begin{aligned} \dot{z}(t) &= \bigl\langle {v^{*}, \dot{X}_{t}} \bigr\rangle = \bigl\langle {v^{*},A(0)X_{t}+R(0)X_{t}} \bigr\rangle \\ & = \bigl\langle {A^{*}(0)v^{*},X_{t}} \bigr\rangle + \bar{v}^{*}(0)f(0,X_{t}) \\ & = \bigl\langle {\iota \omega _{0}\Delta _{k}v^{*}, X_{t}} \bigr\rangle + \bar{v}^{*}(0)f_{0}(z, \bar{z}) \\ & = \iota \omega _{0}\Delta _{k}z + \bar{v}^{*}(0)f_{0}(z,\bar{z}) \\ & = \iota \omega _{0}\Delta _{k}z(t)+g(z,\bar{z}), \end{aligned} $$
where
$$ \begin{aligned} g(z,\bar{z}) & = \bar{v}^{*}(0)f_{0}(z,\bar{z}) \\ & = g_{20}\frac{z^{2}}{2} + g_{11}z\bar{z}+ g_{02} \frac{\bar{z}^{2}}{2}+g_{21}\frac{z^{2}\bar{z}}{2} + \cdots. \end{aligned} $$
(6.15)
From (6.13) and (6.14), it follows that
$$ X_{t} = W(z,\bar{z},\theta ) + zv+ \bar{z}\bar{v}. $$
Thus,
$$ X_{t} = \begin{bmatrix} X_{1t}(\theta ) \\ X_{2t}(\theta ) \\ X_{3t}(\theta ) \end{bmatrix} = \begin{bmatrix} W^{(1)}(z,\bar{z},\theta ) \\ W^{(2)}(z,\bar{z},\theta ) \\ W^{(3)}(z,\bar{z},\theta ) \end{bmatrix} + z \begin{bmatrix} 1 \\ v_{1} \\ v_{2} \end{bmatrix} e^{\iota \omega _{0}\Delta _{k}\theta } + \bar{z} \begin{bmatrix} 1 \\ \bar{v}_{1} \\ \bar{v}_{2} \end{bmatrix} e^{\iota \omega _{0}\Delta _{k}\theta }, $$
where
$$ \left . \begin{aligned} &X_{1t}(\theta ) = ze^{\iota \omega _{0}\Delta _{k}\theta } + \bar{z}e^{ \iota \omega _{0}\Delta _{k}\theta } + W_{20}^{1}( \theta ) \frac{z^{2}}{2} + W_{11}^{1}(\theta )z \bar{z}+W_{02}^{1}(\theta ) \frac{\bar{z}^{2}}{2} + \cdots, \\ &X_{2t}(\theta ) = v_{1}ze^{\iota \omega _{0}\Delta _{k}\theta } + \bar{v_{1}}\bar{z}e^{\iota \omega _{0}\Delta _{k}\theta } + W_{20}^{2}( \theta )\frac{z^{2}}{2} + W_{11}^{2}(\theta )z \bar{z}+W_{02}^{2}( \theta )\frac{\bar{z}^{2}}{2} + \cdots, \\ &X_{3t}(\theta ) = v_{2}ze^{\iota \omega _{0}\Delta _{k}\theta } + \bar{v_{2}}\bar{z}e^{\iota \omega _{0}\Delta _{k}\theta } + W_{20}^{3}( \theta )\frac{z^{2}}{2} + W_{11}^{3}(\theta )z \bar{z}+W_{02}^{3}( \theta )\frac{\bar{z}^{2}}{2} + \cdots. \end{aligned} \right \} $$
Using these values and from (6.3) it follows that
$$ \left . \begin{aligned} g(z,\bar{z}) & = \bar{v}^{*}(0)f_{0}(z,\bar{z}) \\ & = \bar{v}^{*}(0)f_{0}(z,X_{t}) \\ & = \Delta _{k}\bar{D} \begin{bmatrix} 1, & \bar{v_{1}}^{*}, & \bar{v_{2}}^{*} \end{bmatrix} \begin{bmatrix} X_{2t}(1)X_{3t}(1)X_{2t}(1)X_{3t}^{2}(1) \\ 0 \\  X_{2t}(1)X_{3t}(1) \end{bmatrix} \\ & = \Delta _{k}\bar{D} \bigl[X_{2t}(1)X_{3t}(1)X_{2t}(1)X_{3t}^{2}(1)X_{2t}(1)X_{3t}(1) \bar{v_{2}}^{*} \bigr]. \end{aligned} \right \} $$
(6.16)
Putting the values of \(X_{2t}(1)\), \(X_{3t}(1)\), and \(X_{3t}^{2}(1)\), computing the above expressions (6.16), and comparing the coefficients of \(z^{2}\), zz̄, \(\bar{z}^{2}\), and \(z^{2}\bar{z}\) with (6.15), we have
$$ \left . \begin{aligned} &g_{20} = 2 \Delta _{k}\bar{D} \bigl(v_{1}v_{2}e^{2\iota \omega _{0} \Delta _{k}} v_{1}v_{2}\bar{v}_{2}^{*}e^{2\iota \omega _{0}\Delta _{k}} \bigr), \\ &g_{11} = \Delta _{k}\bar{D} \bigl(v_{1} \bar{v}_{2}+\bar{v}_{1}v_{2}v_{1} \bar{v}_{2}\bar{v}_{2}^{*} \bar{v}_{1}v_{2}\bar{v}_{2}^{*} \bigr), \\ &g_{02} = 2\Delta _{k}\bar{D} \bigl( \bar{v}_{1}\bar{v}_{2}e^{2\iota \omega _{0}\Delta _{k}}  \bar{v}_{1}\bar{v}_{2}\bar{v}_{2}^{*}e^{2 \iota \omega _{0}\Delta _{k}} \bigr), \\ &g_{21} = 2\Delta _{k}\bar{D} \biggl[v_{1}e^{\iota \omega _{0}\Delta _{k}}W_{11}^{3}(1) + \frac{\bar{v}_{2}}{2}e^{\iota \omega _{0}\Delta _{k}}W_{20}^{2}(1) + v_{2}e^{\iota \omega _{0}\Delta _{k}}W_{11}^{2}(1) \\ &\hphantom{g_{21} ={}}{} +\frac{\bar{v}_{1}}{2}e^{\iota \omega _{0}\Delta _{k}}W_{20}^{3}(1) 2v_{1} \vert v_{2} \vert ^{2}e^{\iota \omega _{0}\Delta _{k}} \bar{v}_{1}v_{2}^{2}e^{ \iota \omega _{0}\Delta _{k}} \\ &\hphantom{g_{21} ={}}{}  \bar{v}_{2}^{*} \biggl\{ v_{1}e^{\iota \omega _{0}\Delta _{k}}W_{11}^{3}(1)+ \frac{\bar{v}_{2}}{2}e^{\iota \omega _{0}\Delta _{k}}W_{20}^{2}(1) + v_{2}e^{\iota \omega _{0}\Delta _{k}}W_{11}^{2}(1) \\ &\hphantom{g_{21} ={}}{}+ \frac{\bar{v}_{1}}{2}e^{\iota \omega _{0}\Delta _{k}}W_{20}^{3}(1) \biggr\} \biggr], \end{aligned} \right \} $$
(6.17)
and
$$ \left . \begin{aligned}& W_{20}(\theta ) = \frac{\iota g_{20}}{\omega _{0}\Delta _{k}}q(0)e^{ \iota \omega _{0}\Delta _{k}\theta }+ \frac{\iota \bar{g}_{02}}{3\omega _{0}\Delta _{k}} \bar{q}(0)e^{ \iota \omega _{0}\Delta _{k}\theta }+ E_{1}e^{2\iota \omega _{0} \Delta _{k}\theta }, \\ &W_{11}(\theta ) = \frac{\iota g_{11}}{\omega _{0}\Delta _{k}}q(0)e^{ \iota \omega _{0}\Delta _{k}\theta }+ \frac{\iota \bar{g}_{11}}{\omega _{0}\Delta _{k}}\bar{q}(0)e^{\iota \omega _{0}\Delta _{k}\theta }+E_{2}, \end{aligned} \right \} $$
(6.18)
where \(E_{1} = (E_{1}^{(1)},E_{1}^{(2)},E_{1}^{(3)})\) and \(E_{2} = (E_{2}^{(1)}, E_{2}^{(2)}, E_{2}^{(3)})\) are a constant vector in \(\mathbb{R}^{3}\) satisfying the following equations:
$$\begin{aligned}& \begin{pmatrix} 1\iota \omega _{0} & \frac{\bar{z}}{1+\bar{z}}e^{\iota \omega _{0} \Delta _{k}} & \frac{\bar{y}}{(1+\bar{z})^{2}}e^{\iota \omega _{0} \Delta _{k}} \\ a_{1} & a_{2}\iota \omega _{0} & 0 \\ 0 & \bar{z}e^{\iota \omega _{0}\Delta _{k}} & a_{4}\iota \omega _{0} \bar{y}e^{\iota \omega _{0}\Delta _{k}} \end{pmatrix} \begin{pmatrix} E_{1}^{(1)} \\ E_{1}^{(2)} \\ E_{1}^{(3)} \end{pmatrix} = 2 \begin{pmatrix} \Gamma _{11} \\ \Gamma _{21} \\ \Gamma _{31} \end{pmatrix} , \end{aligned}$$
(6.19)
$$\begin{aligned}& \begin{pmatrix} 1& \frac{\bar{z}}{1+\bar{z}}& \frac{\bar{y}}{(1+\bar{z})^{2}} \\ a_{1} & a_{2} & 0 \\ 0 & \bar{z}& a_{4}\bar{y} \end{pmatrix} \begin{pmatrix} E_{2}^{(1)} \\ E_{2}^{(2)} \\ E_{2}^{(3)} \end{pmatrix} = 2 \begin{pmatrix} \Gamma _{12} \\ \Gamma _{22} \\ \Gamma _{32} \end{pmatrix} , \end{aligned}$$
(6.20)
$$\begin{aligned}& \left . \begin{aligned} &\Gamma _{11} = v_{1}v_{2}e^{2\iota \omega _{0}\Delta _{k}}, \\ &\Gamma _{12} = v_{1}\bar{v}_{2}+ \bar{v}_{1}v_{2}, \\ &\Gamma _{21} = 0, \\ &\Gamma _{22} = 0, \\ &\Gamma _{31} = v_{1}v_{2}e^{2\iota \omega _{0}\Delta _{k}}, \\ &\Gamma _{32} = v_{1}\bar{v}_{2} \bar{v}_{1}v_{2}. \end{aligned} \right \} \end{aligned}$$
(6.21)
Furthermore, we can compute \(g_{21}\) with respect to parameters and delay. Hence, from the above analysis we can conclude that in order to find each \(g_{ij}\) we have to use the parameters and delay in system (2.2). Thus we can compute the following values:
$$ \left . \begin{aligned} &c_{1}(0) = \frac{\iota }{2\Delta _{k}\omega _{0}} \biggl(g_{11}g_{20}2 \vert g_{11} \vert ^{2} \frac{ \vert g_{02} \vert ^{2}}{3} \biggr)+ \frac{g_{21}}{2}, \\ &\mu _{2} =  \frac{\operatorname{Re}\{c_{1}(0)\}}{\operatorname{Re}\{\lambda ^{\prime }(\Delta _{k})\}}, \\ &\beta _{2} = 2\operatorname{Re} \bigl\{ c_{1}(0) \bigr\} , \\ &T_{2} =  \frac{\operatorname{Im}\{c_{1}(0)\}+\mu _{2}\operatorname{Im}\lambda ^{\prime }(\Delta _{k})}{\Delta _{k}\omega _{0}}. \end{aligned} \right \} $$
(6.22)
Based on our analysis, by the result of Hassard et al. [35], we have the following result.
Theorem 6.1
The sign of \(\mu _{2}\), \(\beta _{2}\), and \(T_{2}\) determines the directions of Hopf bifurcations, stability of the bifurcating periodic solutions, and the period of bifurcating periodic solutions respectively for \(\Delta = \Delta _{k}\). In view of (6.22), the following results hold for system (2.2):

(a)
If \(\mu _{2} < 0\) (\(\mu _{2} > 0\)), the Hopf bifurcation is subcritical (supercritical).

(b)
If \(\beta _{2} > 0\) (\(\beta _{2} < 0\)), the bifurcation periodic solutions are unstable (stable).

(c)
If \(T_{2} < 0\) (\(T_{2}> 0\)), the period of the bifurcated periodic solution decreases (increases).