# Fluid Net Models: From Behavioral Properties to Structural Objects
## 3 Bifurcations: Operation Mode Properties
### Bifurcations in Timed Join Free and Choice Free Nets  
In a Join Free TCPN (JF TCPN), every transition has only one input place (\(\forall t\in T\), \(|^{\star}t|=1\)); thus, JF TCPNs have only one configuration. In a Choice Free TCPN (CF TCPN), every place has only one output transition (\(\forall p\in P,|p^{\star}|=1\)); thus, CF TCPNs have no decision places. If the net is Join-Free (JF) or Choice-Free (CF), bifurcations of the equilibrium marking cannot occur in the TCPN system when the nets are consistent and conservative [14]. Bifurcations of the equilibrium marking in strongly connected JF TCPN systems may appear due to the variation of some firing rates. As already said, that will be reflected in the eigenvalues of its _unique_ dynamic matrix \(\mathbf{C}\mathbf{\Lambda}\mathbf{\Pi}\). Let us illustrate this through the following example, in which the equilibrium marking of the system changes from being unstable to stable. **Example 1**. : _Consider the JF TCPN in Figure 7a with \(\boldsymbol{\lambda}= \begin{bmatrix}\lambda_{1}&1&1&1\end{bmatrix} ^{T}\) and \(\boldsymbol{m}_{0}= \begin{bmatrix}10&0&0\end{bmatrix} ^{T}\). It is strongly connected and consistent, but non-conservative (in fact, it has no P-semiflows). Its unique dynamic matrix is:_  
\[\boldsymbol{A}=\mathbf{C}\mathbf{\Lambda}\mathbf{\Pi}= \begin{bmatrix}-\lambda _{1}-1&0.5&2\\ \lambda_{1}&-1&0\\ 1&0&-1\end{bmatrix} .\]  
* _If_ \(0<\lambda_{1}<2\)_, then_ \(\boldsymbol{A}\) _is a full rank matrix, so none of its eigenvalues rely on the imaginary axis (i.e., the only equilibrium marking,_ \(\boldsymbol{m}_{e}=\boldsymbol{0}\)_, is hyperbolic). In this case, one of them has a positive real part; thus,_ \(\boldsymbol{m}_{e}\) _is unstable._
* _If_ \(\lambda_{1}=2\)_, then_ \(\boldsymbol{A}\) _is singular, and one of its eigenvalues is zero. That is, the equilibrium marking (_\(\boldsymbol{m}_{e}=\beta \begin{bmatrix}1&2&1\end{bmatrix} ^{T}\)_, with_ \(\beta\geq 0\)_) is non-hyperbolic._
* _If_ \(\lambda_{1}>2\)_, then_ \(\boldsymbol{A}\) _is a full rank matrix, so_ \(\boldsymbol{m}_{e}=\boldsymbol{0}\) _is hyperbolic. Specifically, all the eigenvalues have a negative real part;_ \(\boldsymbol{m}_{e}\) _is stable._ _Consequently, a bifurcation occurs for_ \(\lambda_{1}=2\)_._  
Nevertheless, if the JF TCPN is conservative, a bifurcation of the equilibrium markings is not possible. It follows from the fact that it cannot have closed paths gaining/losing tokens. **Proposition 2** ([14]). : _Let \((N,\boldsymbol{\lambda},\boldsymbol{m}_{0})\) be a strongly connected JF TCPN system. If \(N\) has a P-semiflow, then the equilibrium markings of the system do not bifurcate._  
The equilibrium markings of the TCPN system \((N,\boldsymbol{\lambda},\boldsymbol{m}_{0})\) for the net in Figure 7b do not bifurcate, because \(N\) is consistent and conservative (therefore, structurally live). Figure 7: Three strongly connected JF Petri nets: (**a**) its TCPN system exhibits bifurcations; (**b,c**): their TCPN systems do not exhibit bifurcations. The converse of Proposition 2 is not true; consider the net in Figure 6(c). For every \(\mathbf{\lambda}>\mathbf{0}\) the dynamic matrix of the TCPN system, \(\mathbf{A}=\mathbf{C}\mathbf{\Lambda}\mathbf{\Pi}\), has only eigenvalues with a negative real part. Therefore, \(\mathbf{m}_{\varepsilon}=\mathbf{0}\) is always hyperbolic; thus, there is no bifurcation value for any \(\mathbf{\lambda}>\mathbf{0}\), and the net has no P-semiflow. When the assumption of strongly connectedness is removed in Proposition 2, \(N\) must be conservative to guarantee that the equilibrium markings of the reduced subsystem do not bifurcate. This is because if \(N\) is not strongly connected and conservative, then it should be not consistent (this is a classical result: Ct & Cv imply strongly connectedness of the net [2]). Hence, the net is not structurally live, which guarantees that the eigenvalues of the reduced system are in the left half of the complex plane.