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Table 1 Summary of eigenvalue analysis on the simplest bistable system [41]

From: Using sensitivity analyses to understand bistable system behavior

 

Solution 1

Solution 2

Solution 3

Steady States (x, y)

(0,0)

(2, 0.5)

(6, 4.5)

Jacobian \(\textbf{J}\)

\(\left( \begin{matrix} -1.5 &{} 16\\ 0 &{} -8 \end{matrix}\right) \)

\(\left( \begin{matrix} -6 &{} 14\\ 4 &{} -8 \end{matrix}\right) \)

\(\left( \begin{matrix} -18 &{} 10\\ 12 &{} -8 \end{matrix}\right) \)

Eigenvalues \((\lambda _1, \lambda _2)\)

\((-1.5,-8)\)

\((-14.5,0.54)\)

\((-25.04,-0.95)\)

Stability Property

Stable

Saddle node

Stable

  1. For each steady state solution, the corresponding Jacobian, eigenvalues, and stability are listed. There are two stable steady states, which make this system bistable