### Solutions of the Equations of Motion for the Two Oscillator System  
We use the Morse theory to determine the critical points of the potential \(U\) and to discuss the solutions of Equation (27), with an aim to fully characterize the dynamical stability of the coupled oscillator system. Then  
\[z\bigg{(}\lambda^{2}-\frac{1}{r^{3}}\bigg{)}=0\, \tag{28}\]  
which leads to a number of two possible cases:  
**Case 1.**\(z=0\). The first equation of the system (27) can be rewritten as  
\[\rho-\frac{\
u^{2}}{\rho^{3}}-\frac{\rho}{r^{3}}=0\,\]  
which gives us \(\rho=r\) for \(z=0\). In such case, a function results  
\[f(\rho)=\rho^{4}-\rho-\
u^{2}\,\ f^{\prime}(\rho)=4\rho^{3}-1. \tag{29}\]  
The second relationship in Equation (29) shows that \(\rho=\sqrt[3]{\frac{1}{4}}\) is a point of minimum for \(f(\rho)\). In case when \(\rho_{0}>0\), for \(\
u\
eq 0\) and \(z_{0}=0\):  
\[f(\rho)=\rho_{0}^{4}-\rho_{0}-\
u^{2}=0\.\]  
In case when \(\
u=0\) we obtain \(f(\rho)=\rho\big{(}\rho^{3}-1\big{)}=0\). Then, the solutions are \(\rho_{1}=0\) and \(\rho_{2}=1\), where only \(\rho_{2}=1\) is a valid solution. Moreover, for \(\
u=0\) and \(z_{0}=0\), \(\rho_{0}=1\) is a solution.  
**Case 2.**\(r^{3}=1/\lambda^{2}\Rightarrow\)  
\[r=\lambda^{-2/3}. \tag{30}\]  
We return to the system of Equations (27) and infer  
\[\rho=\frac{\sqrt{|\
u|}}{\sqrt[4]{1-\lambda^{2}}}\, \tag{31}\]  
for \(\lambda<1\). In case when \(\lambda\leq 1\) and \(\
u\
eq 0\), the system admits no solutions. In the scenario when \(\lambda<1,\
u=0\) we find \(\rho=0\), while in case when \(\lambda=1,\ \
u=0\) it results that any \(\rho\geq 0\) represents a solution. As \(r=\sqrt{z^{2}+\rho^{2}}\), then  
\[z=\pm\sqrt{r^{2}-\rho^{2}}=\pm\sqrt{\lambda^{-4/3}-\rho^{2}}. \tag{32}\]  
We differentiate among three possible sub-cases  
_Subcase (i):_\(\lambda<1\,\
u\
eq 0\) and  
\[z_{12}=\pm\sqrt{\lambda^{-4/3}-\frac{|\
u|}{\sqrt{1-\lambda^{2}}}}\,\]  
provided that \(1-\lambda^{2}>\
u^{2}\lambda^{8/3}\) or  
\[z=0\ \text{for}\ 1-\lambda_{c}^{2}=\
u^{2}\lambda_{c}^{8/3}\,\]  
where the \(c\) index of \(\lambda\) refers to critical.  
_Subcase (ii):_\(\lambda<1,\
u=0\) which leads to \(\rho=0\) and \(z_{12}=\pm\lambda^{-2/3}\).  
_Subcase (iii):_\(\lambda=1,\
u=0\) which results in \(z_{12}=\pm\sqrt{\lambda^{-4/3}-\rho^{2}}\), with \(\rho\geq 0\).  
These are the solutions we find for the equations of motion corresponding to the two coupled oscilators system. After doing the math, the Hessian matrix of the potential \(U\) appears as  
\[\mathsf{H}=\left|\begin{array}{cc}1+\frac{3\
u^{2}}{\rho^{4}}-\frac{1}{r^{3}}+ \frac{3\rho^{2}}{r^{5}}&\frac{3\rho\varepsilon}{r^{5}}\\ \frac{3\rho\varepsilon}{r^{5}}&\lambda^{2}-\frac{1}{r^{2}}+\frac{3\kappa^{2} }{r^{5}}\end{array}\right|\,. \tag{33}\]  
The determinant and the trace of the Hessian matrix result as  
\[\begin{split}\det\mathsf{H}&=\frac{3\
u^{2}}{\rho^{4}}\bigg{(} \lambda^{2}-\frac{1}{r^{3}}+\frac{3z^{2}}{r^{5}}\bigg{)}+\lambda^{2}\bigg{(}1- \frac{1}{r^{3}}+\frac{3\rho^{2}}{r^{5}}\bigg{)}-\frac{1}{r^{3}}\bigg{(}1+ \frac{2}{r^{3}}-\frac{3z^{2}}{r^{2}}\bigg{)}\,\\ \text{Tr}\,\mathsf{H}&=1+\lambda^{2}+\frac{3\
u^{2}}{\rho^{4}}+ \frac{1}{r^{3}}\.\end{split} \tag{34}\]  
From Equation (34) we infer that \(\text{Tr}\,\mathsf{H}=0\). Thus, the Hessian matrix \(\mathsf{H}\) has at least a strictly positive eigenvalue.