I have a little question. I want to know if there is a process in which I can find equilibrium solutions to some system of difference equations. For example, if I have something crazy like \begin{cases} x[n+1]=(x[n])^2y[n]+z[n]e^{-ax[n]} \\[3mm] y[n+1]= z[n]x[n]+x[n+1]y[n+1] \\[2mm] z[n+1]= \dfrac{x[n]}{1+x[n]} \end{cases} I would like to know how to calculate equilibrium points when $n \rightarrow \infty$
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To find the equilibrium for the system
\begin{align*}
x[n+1] &=(x[n])^2y[n]+z[n]e^{-ax[n]},
\\[3mm]
y[n+1] &= z[n]x[n]+x[n+1]y[n+1], \mbox{ and }
\\[2mm]
z[n+1] &= \dfrac{x[n]}{1+x[n]} \\
\end{align*}
of equations, set
\begin{align*}
x[n+1] &= x[n]=x_*, \\
y[n+1] &= y[n]=y_*, \\
z[n+1] &= z[n]=z_*. \\
\end{align*}
Then the system of equations becomes
\begin{align*}
x_* &=x_*^2 y_*+ z_* e^{-a x_*},
\\[3mm]
y_* &= z_* x_* + x_* y_*,
\\[2mm]
z_* &= \dfrac{x_*}{1+x_*}. \\
\end{align*}
For simplicity, you can remove the lower stars and write $x = x_*$, $y = y_*$, and $z = z_*$ to obtain:
\begin{align*}
x &=x^2 y+ z e^{-a x}, \hspace{6mm} \clubsuit
\\[3mm]
y &= z x + x y, \hspace{6mm} \diamondsuit
\\[2mm]
z &= \dfrac{x}{1+x}. \hspace{6mm} \spadesuit \\
\end{align*}
The final step to find the equilibrium points is to solve for $x$, $y$, and $z$ such that the three equations $\clubsuit$, $\diamondsuit$, and $\spadesuit$ are satisfied.
Mee Seong Im
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