I have a system of 4 ordinary differential equations with positive parameters, which has two stable fixed points. The Jacobian matrix evaluated at the first equilibrium leads easily to finding the conditions for linear stability. These conditions are necessary but are not sufficient for the second equilibrium to be stable. At the second equilibrium, the determinant of the Jacobian is:
\begin{equation*} \small \boldsymbol{J} = \begin{pmatrix} % pds of Capital 0 & 0 & 0 & 0 \\ % pds of Inventory 0 & - q - \frac{ b }{ 4 }& - \frac{ b }{ 4 } & 0 \\ % pds of Demand 0 & 0 & - c & - c \big( \frac{k b}{a + \frac{b}{2}} \big)^2 \\ % pds of Price 0 & - d \big( \frac{a + \frac{b}{2}}{\kappa b} \big)^2 & d \big( \frac{a + \frac{b}{2}}{\kappa b} \big)^2 & 0\\ \end{pmatrix} \end{equation*}
which has characteristic polynomial $\rho(\lambda)$:
\begin{equation} \rho(\lambda) = -\lambda \Big[ -\lambda^3 - (a + b + c) \lambda^2 - c(a+b+d) \lambda - cd(a+b) \Big] \end{equation}
Using the Routh-Hurwitz criterion, I know the the 3rd order polynomial inside the bracket has negative roots always (all parameters are positive), which means the remaining eigenvalue must control the second equilibrium's stability. However, the remaining eigenvalue is zero, so the linearisation fails.
I have been attempting to approximate the leading eigenvalue by taking the Taylor series approximation around the point $a \approx 0$ but I haven't been having much luck. That is, making $\lambda = \lambda_0 + \lambda_{1} a + \lambda_{2} a^2 + ...$ in $\rho(\lambda)$ above, and computing $\lambda_0$, $\lambda_1$, $\lambda_2$ etc. at $a = 0$, just returns each coefficient equal to zero, so it seems that the Taylor series approximation is also failing in higher powers.
Does anyone know how I can assess the stability of the second equilirium? Thank you
The fact that one row (or one column) of the matrix is $0$ implies that it is singular, and thus has an eigenvalue of $0$. That means that you can't decide the stability by just looking at this Jacobian. On the other hand, you do know in what direction to look for potential trouble: the first coordinate direction. Seeing what happens (in the nonlinear system) to points near the equilibrium in that direction may help decide.