Consider the vector field $V:\mathbb{R}^4\rightarrow\mathbb{R}^4$, defined by \begin{equation} V(x,v,M_0,M_1)=(v,\kappa^{-1}(\beta M_0-v-kx),-M_0+v M_1,-M_1+1-vM_0), \end{equation}

such that $\beta,\kappa,k$ are constants. The only equilibrium point occurs at $P^*=(0,0,0,1)$ and the Jacobian matrix of $V$ at $P^*$ is \begin{equation} JV_{P^*}= \begin{pmatrix} 0 & 1 & 0 & 0 \\ -\frac{k}{\kappa} & -\frac{1}{\kappa} & \frac{\beta}{\kappa} & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}. \end{equation}

So, we have $\lambda_0=-1$ and using Cardano's formula \begin{equation*} \begin{split} \Delta_0&=\left(\frac{\kappa+1}{\kappa}\right)^2-\frac{3(k+1-\beta)}{\kappa},\\ \Delta_1&=2\left(\frac{\kappa+1}{\kappa}\right)^3-\frac{9(k+1-\beta)(\kappa+1)}{\kappa^2}+\frac{27k}{\kappa},\\ \mathcal{C}&=\sqrt[3]{\frac{\Delta_1\pm \sqrt{\Delta_1^2-4\Delta_0^3}}{2}},\\ \lambda_i&=-\frac{1}{3}\left(\frac{\kappa+1}{\kappa}+\xi^{i-1} \mathcal{C}+\frac{\Delta_0}{\xi^{i-1}\mathcal{C}}\right). \end{split} \end{equation*}

The motivation is to find a solution to the linearised system $$\vec{\dot{x}}=JV_{P^*}\text{ }\vec{x},$$

whereby $\vec{x}=C_1 e^{\lambda_1 t}\vec{v}_1+...+C_4 e^{\lambda_4 t}\vec{v}_4$ iff $JV_{P^*}$ is diagonalisable. Due to the Hartman–Grobman Theorem, we rely on the property that $\Re(\lambda_j)\neq0$, $\forall j$. Is it possible then to isolate the real part of $\lambda_{1,2,3}$?