What is the steady state for this difference equation: $X_{n+1}-X_{n}+\beta \alpha X_{n-1}(1-\frac {X_{n-1}}{X_{max}})=t$

Question

This is my self study, as I know the steady state from an difference equation should satisfy $x=X_{n+1}=X_{n}$

What is the steady state for this difference equation?

$$X_{n+1}-X_{n}+\beta \alpha X_{n-1}\left(1-\frac {X_{n-1}}{X_{\max}}\right)=t$$ $\beta$, $\alpha>0$ $~t$ is constant, so this equation look like non-homogenous.

I confused while I was trying to derive $X_{\max}$.

Answer 1

Steady state, if it exists, should satisfy: $$\lim_{n \to \infty} X_n = x \in \mathbb{R}$$ Hence, we have $$\lim_{n \to \infty} \left(X_{n+1}-X_{n}+\beta \alpha X_{n-1}\left(1-\frac {X_{n-1}}{X_{max}}\right) \right)=t$$ $$\lim_{n \to \infty} X_{n+1}- \lim_{n \to \infty} X_{n}+\beta \alpha \lim_{n \to \infty} X_{n-1}\left(1-\frac {\lim_{n \to \infty} X_{n-1}}{X_{max}}\right)=t$$ $$x -x + \beta \alpha x(1-x/x_{\max}) = t \implies \beta \alpha x(1-x/x_{\max}) = t$$ Solve the quadratic to obtain $x$.