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 $X_{\max}$.

score 2 · Accepted Answer · answered Dec 01 '13 at 18:00

2

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 $x$.

answered Dec 01 '13 at 18:00

how about $X_{max}$? – valerie Dec 01 '13 at 18:02
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@valerie $X_{max}$ is a known quantity. You do not solve for $X_{\text{max}}$ – Dec 01 '13 at 18:03
do you mean like this $\beta \alpha x - \beta \alpha x^2- t=0 $ – valerie Dec 01 '13 at 18:10
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@valerie. Please look at the last equation given by user17762. It contains Xmax which is known and part of the quadratic. – Claude Leibovici Dec 01 '13 at 18:15
ahhh, I see...okay many thanks all. I just recognize the $X_{max}$ which is also constant known.. – valerie Dec 01 '13 at 18:19

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

1 Answers1