Firstly, I am not asking anyone to solve my equations for me. I am just looking for guidance on how to find the equilibria of my system and point me in the correct direction with analysing stability.

My system is as follows,

\begin{equation} \frac{dy_i}{dx} = a_i y_i + \sum ^N_{j=1}(b_{ij}-c_{ij})y_i\cdot y_j\qquad i=1,\dots ,N \end{equation}

Here $N$ can be quite large. $a_i$, $b_{ij}$ and $c_{ij}$ are just positive parameters. I understand that there are in general four types of equilibrium solution.

  • $y_i = 0\qquad$ $\forall i\in N$
  • $y_i=y_i*\qquad$ $\forall i\in N$
  • $(y_1 = 0, \dots , y_{i-1} =0, y_i=y_i*, y_{i+1}=0,\dots,y_N =0)$
  • $(y_1 = 0, \dots , y_{i-1} =0, y_i=y_i*, y_{i+1}=0,\dots, y_{j-1} =0, y_j=y_j*, y_{j+1}=0,\dots,y_N =0)$

Where $y_i*$ is an equilibrium value. To find these we set each $dy_i/dx =0$ and where the nullclines intersect we have an equilibrium point.

So in the case of simply re-arranging the above equation for $y_i$ we find that, \begin{equation} y_i* = \frac{a_i - \sum ^N_{j\neq i} (c_{ij} - b_{ij})y_j}{(c_{ii} - b_{ii})} \end{equation}

However I am a little lost as how to proceed from here! i.e how do I find the other equilibria?


1 Answer 1


Since the sum is taken over $j$, you can factor out $y_i$, and rewrite the equilibrium conditions as

$$\begin{equation} y_i(a_i + \sum ^N_{j=1}(b_{ij}-c_{ij})y_j) = 0\qquad i=1,\dots ,N \end{equation}$$

Now for a non-trivial solutions you only have a system of linear equations.


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