# How to find equilibria of a system of ODEs

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,

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

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, $$y_i* = \frac{a_i - \sum ^N_{j\neq i} (c_{ij} - b_{ij})y_j}{(c_{ii} - b_{ii})}$$

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

## 1 Answer

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

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

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