# One Degree-of-Freedom (DoF) Hamiltonian Bifurcation of Equilibria

(ADD INTRODUCTORY LANGUAGE ABOUT PROBLEM DEVELOPMENT)
We will now consider two examples of bifurcation of equilibria in two dimensional Hamiltonian system; in particular, the Hamiltonian saddle-node and Hamiltonian pitchfork bifurcations.  

## Hamiltonian saddle-node bifurcation

We consider the Hamiltonian:

\begin{equation}
H (q, p) = \frac{p^2}{2} - \lambda q + \frac{q^3}{3}, \quad (q, p) \in \mathbb{R}^2.
\label{eq:hamApp13}
\end{equation}


where $\lambda$ is considered to be a parameter that can be varied. From this Hamiltonian, we derive Hamilton's equations:

\begin{eqnarray}
\dot{q} & = & \frac{\partial H}{\partial p} = p, \nonumber \\
\dot{p} & = & -\frac{\partial H}{\partial q} =\lambda - q^2.
\label{eq:hamApp14}
\end{eqnarray}

### Revealing the Phase Space Structures and their implications for Reaction Dynamics

The fixed points for \eqref{eq:hamApp14} are:

\begin{equation}
(q, p) = (\pm\sqrt{\lambda}, 0),
\end{equation}


from which it follows that there are no fixed points for $\lambda <0$, one fixed point for $\lambda =0$, and  two fixed points for $\lambda >0$. This is the scenario for a saddle-node bifurcation. 

Next we examine the stability of the fixed points. The Jacobian of \eqref{eq:hamApp14} is given by:

\begin{equation}
J =\left(
\begin{array}{cc} 
0 & 1\\
-2 q & 0
\end{array}
\right).
\label{eq:hamApp15}
\end{equation}


The eigenvalues of this matrix are:

\begin{equation}
\Lambda_{1, 2} = \pm \sqrt{-2q}.
\end{equation}


Hence $(q, p) = (-\sqrt{\lambda}, 0)$ is a saddle, $(q, p) = (\sqrt{\lambda}, 0)$ is a center, and $(q, p) = (0, 0)$  has two zero eigenvalues. The phase portraits are shown in Fig. [fig:1](#fig:appC_fig3).



<a id="fig:appC_fig3"></a>
<figcaption style="text-align:center;font-size:14px"><b>fig:1 </b><em> The phase portraits for the Hamiltonian saddle-node bifurcation.</em></figcaption><hr>

## Hamiltonian pitchfork bifurcation

We consider the Hamiltonian:

\begin{equation}
H (q, p) = \frac{p^2}{2} - \lambda \frac{q^2}{2} + \frac{q^4}{4},
\label{eq:hamApp16}
\end{equation}


where $\lambda$ is considered to be a parameter that can be varied. From this Hamiltonian, we derive Hamilton's equations:

\begin{eqnarray}
\dot{q} & = & \frac{\partial H}{\partial p} = p, \nonumber \\
\dot{p} & = & -\frac{\partial H}{\partial q} =\lambda q  -  q^3.
\label{eq:hamApp17}
\end{eqnarray}

### Revealing the Phase Space Structures and their implications for Reaction Dynamics

The fixed points for \eqref{eq:hamApp17} are:

\begin{equation}
(q, p) = (0, 0), \, (\pm\sqrt{\lambda}, 0),
\end{equation}


from which it follows that there is one fixed point  for $\lambda \leq 0$, and  three fixed points for $\lambda >0$. This is the scenario for a pitchfork  bifurcation.

Next we examine the stability of the fixed points. The Jacobian of \eqref{eq:hamApp17} is given by:

\begin{equation}
J = \left(
\begin{array}{cc} 
0 & 1\\
\lambda-3q^2 & 0
\end{array}
\right).
\label{eq:hamApp18}
\end{equation}


The eigenvalues of this matrix are:

\begin{equation}
\Lambda_{1, 2} = \pm \sqrt{\lambda - 3q^2 }.
\end{equation}


Hence $(q, p) = (0, 0)$ is a center for $\lambda <0$, a saddle for $\lambda >0$ and  has two zero eigenvalues for $\lambda =0$. The fixed points $(q, p) = (\sqrt{\lambda}, 0)$ are centers for $\lambda >0$. The phase portraits are shown in Fig. [fig:2](#fig:appC_fig4).



<a id="fig:appC_fig4"></a>
<figcaption style="text-align:center;font-size:14px"><b>fig:2 </b><em> The phase portraits for the Hamiltonian pitchfork bifurcation.</em></figcaption><hr>

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