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 $\lambda_{1,2,3}$?


1 Answer 1


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.

  • $\begingroup$ Is there any way to isolate the real part of $\lambda_i$ from Cardano's formula (see updated question)? I would like to place some restriction on a constant (preferably $\beta$) by ensuring that $\Re(\lambda_j)\neq 0$ for all $j$. $\endgroup$
    – UNOwen
    Nov 19, 2021 at 5:08
  • 1
    $\begingroup$ If $\lambda = i t$ is an imaginary root, then $t^3 -\frac{ k-\beta+1}{\kappa} t = 0$ and $\frac{\kappa+1}{\kappa} t^2 - \frac{k}{\kappa} = 0$. Assuming $k \ne 0$ and $\kappa \ne -1$, we can eliminate $t$ and get $\beta = 1 + k/(\kappa+1)$ and $t = \pm \sqrt{k/(\kappa+1)}$ (which requires $k/(\kappa+1) > 0$). $\endgroup$ Nov 19, 2021 at 18:05
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    $\begingroup$ I suspect what you are really after is stability boundaries. For that, you do not need an explicit formula for the eigenvalues, you only need to know when there are purely imaginary eigenvalues. If you set $\lambda=i\omega$ and separately consider real and imaginary parts, you find the condition $(\kappa+1)/k=\kappa/(k-\beta+1)$. $\endgroup$ Nov 19, 2021 at 18:06

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