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 $\lambda_{1,2,3}$?
Best Answer
The characteristic polynomial is $$P(\lambda) = (\lambda+1)\left(\lambda^3 + \frac{\kappa+1}{\kappa} \lambda^2 + \frac{k-\beta+1}{\kappa} \lambda + \frac{k}{\kappa}\right) $$ Since $\kappa \lambda^3 + (\kappa+1) \lambda^2 + (k-\beta+1) \lambda + k$ is irreducible over the rationals, there's no further factorization possible: if you want explicit expressions for the roots, you will indeed have to use the formulas for roots of a cubic. Of course, for numerical values of the constants you can use numerical methods to get approximate eigenvalues, or you might try series expansions.