# Some Dynamic Aspects of a Sextic Galactic Potential in a Rotating Reference Frame
## 4 Stability Analysis
### Application of the Lagrange Theorem  
To apply the Lagrange Theorem 4.1, it is essential to find the effective potential of the problem. This requires a re-expression of the Hamiltonian in Equation (1) in terms of thegeneralized velocities \(\hat{x},\hat{y}\) rather than the generalized momentum. Using the equations in Equation (3), we obtain  
\[H(x,y)=\frac{1}{2}(\hat{x}^{2}+\hat{y}^{2})+V(x,y), \tag{18}\]  
where \(V(x,y)\) is the effective potential given by  
\[V(x,y)=\frac{1}{2}(1-\Omega^{2})(x^{2}+y^{2})+\frac{1}{6}(ax^{6}+by^{6})+\frac{ 1}{2}x^{2}y^{2}(cx^{2}+dy^{2}). \tag{19}\]  
Notice that when \(\Omega=0\), the effective potential and potential in Equation (2) are identical. Since the system of equations \(\frac{\partial V}{\partial x}=\frac{\partial V}{\partial y}=0\) is the same as in Equation (17), the equilibrium positions of the Hamiltonian system in Equation (1) are the critical points of the effective potential (Equation (19)). The nature of these positions is determined in Theorem 6 below by examining the Hessian matrix for the effective potential and using the Lagrange theorem. The following remark is needed:  
**Remark 2**. : _If \(A\) is a non-singular \(2\times 2\) matrix, then the matrix \(A\) is positive definite if and only if \(\det(A)>0,tr(A)>0\). The matrix \(A\) is negative definite if and only if \(\det(A)>0,tr(A)<0\)._  
**Theorem 6**. : _Under the existence conditions for the equilibrium points in Theorem 4, the following are satisfied:_  
1. _The equilibrium_ \(\mathcal{E}_{0}\) _is stable if_ \(0\leq\Omega<1\)_. If_ \(\Omega>1\)_, then_ \(\mathcal{E}_{0}\) _is an unstable maximum critical point;_
2. _The equilibrium_ \(\mathcal{E}_{1,2}\) _is unstable when_ \(\frac{d}{b}<1\)_. If_ \(\frac{d}{b},\Omega>1\)_, then_ \(\mathcal{E}_{1,2}\) _is a stable position. If_ \(\frac{d}{b}>1\) _and_ \(0\leq\Omega<1\)_, then the point_ \(\mathcal{E}_{1,2}\) _is unstable and maximally critical;_
3. _The equilibrium_ \(\mathcal{E}_{3,4}\) _is unstable when_ \(\frac{d}{a}<1\)_. If_ \(\frac{d}{a},\Omega>1\)_, then_ \(\mathcal{E}_{3,4}\) _is a stable position. The positions_ \(\mathcal{E}_{3,4}\) _are maximally unstable if_ \(\frac{d}{a}>1\) _and_ \(0\leq\Omega<1\)_;_
4. _The equilibrium_ \(\mathcal{E}_{5,6}\) _is unstable when_ \((a-c)(\Omega^{2}-1)<0\)_. The case where_ \((a-c)(\Omega^{2}-1)>0\) _implies that the point_ \(\mathcal{E}_{5,6}\) _is stable if_ \(a+c>0\)_. The positions_ \(\mathcal{E}_{5,6}\) _are unstable and maximally critical if_ \((a-c)(\Omega^{2}-1)>0\) _and_ \(a+c<0\)_._  
Proof. : The Hessian matrix for the effective potential in Equation (19) takes the form  
\[Hess_{V}(x,y)=\left( \begin{array}{cc}1-\Omega^{2}+5 ax^{4}+6 cx^{2}y^{2}+dy^{ 4}&4 xy(cx^{2}+dy^{2})\\ \\ 4 xy(cx^{2}+dy^{2})&1-\Omega^{2}+5 by^{4}+6 dx^{2}y^{2}+cx^{4}\end{array} \right). \tag{20}\]  
We examine the Hessian matrix at the equilibrium positions calculated in Theorem 4. For \(\mathcal{E}_{0}=(0,0)\), the Hessian matrix in Equation (20) takes the form  
\[Hess_{V}(\mathcal{E}_{0})=\left( \begin{array}{cc}1-\Omega^{2}&0\\ 0&1-\Omega^{2}\end{array} \right), \tag{21}\]  
with \(\det(Hess_{V}(\mathcal{E}_{0}))=(1-\Omega^{2})^{2}>0\). Using Remark 2, the point \(\mathcal{E}_{0}\) is either a local minimum if \(0\leq\Omega<1\) or local maximum if \(\Omega>1\). Based on the Lagrange theorem, \(\mathcal{E}_{0}\) is stable if \(0\leq\Omega<1\) and unstable if \(\Omega>1\). The Hessian matrix in Equation (20) at \(\mathcal{E}_{1,2}\) is  
\[Hess_{V}(\mathcal{E}_{1,2})=\left( \begin{array}{cc}(\Omega^{2}-1)(\frac{d} {b}-1)&0\\ 0&4(\Omega^{2}-1)\end{array} \right), \tag{22}\]  
whose determinant is \(4(\Omega^{2}-1)^{2}(\frac{d}{b}-1)\) and whose trace is \((\Omega^{2}-1)(\frac{d}{b}+3)\). This implies the following:  
* If \(\frac{d}{b}<1\), then \(Hess_{V}(\mathcal{E}_{1,2})\) is an indefinite matrix, and the point \(\mathcal{E}_{1,2}\) is a saddle and hence unstable. * If \(\frac{d}{b}>1\), the point \(\mathcal{E}_{1,2}\) is a local minimum and hence stable if \(\Omega>1\).