Equilibrium in system of non-linear difference equations

[math4everyone]

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
$$\begin{cases} x[n+1]=(x[n])^2y[n]+z[n]e^{-ax[n]} \\
y[n+1]= z[n]x[n]+x[n+1]y[n+1]\\
z[n+1]= \frac{x[n]}{1+x[n]}
\end{cases}$$
I would like to know how to calculate equilibrium points when $$n \rightarrow \infty$$

 

[S.G. Janssens]

Using the equation for $x_{n+1}$ first write $y_{n+1}$ as a function of $(x_n,y_n,z_n)$. (So, eliminate $x_{n+1}$). After this, the right-hand side of your system defines a function $f$ on $\mathbb{R}^3$, as
$$
\begin{bmatrix}
x_{n+1}\\
y_{n+1}\\
z_{n+1}
\end{bmatrix}
= f(x_n,y_n,z_n).
$$
Your equilibrium points are the fixed points of $f$.

 

[math4everyone]

Using the equation for $x_{n+1}$ first write $y_{n+1}$ as a function of $(x_n,y_n,z_n)$. (So, eliminate $x_{n+1}$). After this, the right-hand side of your system defines a function $f$ on $\mathbb{R}^3$, as
$$
\begin{bmatrix}
x_{n+1}\\
y_{n+1}\\
z_{n+1}
\end{bmatrix}
= f(x_n,y_n,z_n).
$$
Your equilibrium points are the fixed points of $f$.

Thanks for the reply, but this was just an example. I apologize for not stating the real problem.The real system of difference equations I have is the following
$$
\begin{cases}
X_j [n+1] = (1-m_j - \theta - \delta N[n])X_j[n]+\alpha X_a[n] e^{-\beta X_a [n]} \\
X_a[n+1]=(1-m_a-H[n]-\delta N[n])X_a[n] + \theta X_j[n]\\
N[n+1] = \frac{\gamma K X_a [n]}{K+\gamma X_a[n]} \\
H[n+1] = q \left( \frac{a(P_f q X_a [n]-C_f - \pi \omega_\tau)}{1+b(P_f q X_a [n]-C_f - \pi \omega_\tau)}+E_{min} \right)
\end{cases}
$$
I am really new to this topic, so I don't fully understand what I have to do.

 

[S.G. Janssens]

It's the same principle: This system corresponds to a mapping $f$ on (a domain in) $\mathbb{R}^4$ that takes as inputs the state $(X_j[n], X_a[n], N[n], H[n])$ at the (discrete) time $n \in \mathbb{Z}_+$ and produces the state
$$
\begin{bmatrix}
X_j[n+1]\\
X_a[n+1]\\
N[n+1]\\
H[n+1]
\end{bmatrix}
= f(X_j[n], X_a[n], N[n], H[n])
$$
at time $n + 1$. A stationary solution of the difference equations then corresponds to a state that does not change in time. Such a state is precisely a fixed point of $f$. I did not calculate the fixed points myself, that is up to you. Sometimes they can be found analytically, but sometimes you need to resort to numerical root finders such as Newton's (and Newton-like) method(s).

This and related topics (such as stability determination of the stationary solutions) are typically treated in courses and books on discrete-time dynamical systems ("iterated maps").