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I am not sure if my question is asked with the correct terms.

I have a non-homogenous second order recurrence relation :

$$a_{n}+\beta \alpha a_{n-1} + \beta (1-\alpha) a_{n+1}+ C = 0 $$

where $\alpha,\beta \in (0,1) $, $C$ a finite constant, and $n \in Z=\{...,-1,0,1,...\}$.

Suppose the following two conditions are substituting two initial conditions :

$\lim_{n\rightarrow\infty}a_{n}=c_{1}$ and $\lim_{n-\rightarrow\infty}a_{n}=c_{2}$ where both $c_{1}$ and $c_{2}$ are finite constants.

In this case, does there exist a unique solution? can I characterize it as a closed-form solution?

Any formal references would be more than appreciated.

Anonymously lost student
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1 Answers1

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I think that under these circumstances, the sequence must be constant.

Consider the case where $n\geq 0$. The inhomogeneous equation has the solution $$a-N\equiv-\frac C{1+\beta}$$ so the general solution is of the form $$a_n=k_1r_1^n+k_2r_2^n-\frac C{1+\beta},\tag1$$ if the characteristic equation has two simple roots $r_1,r_2$, or of the form $$a_n=k_1r^n+k_2nr^n-\frac C{1+\beta},\tag2$$ if it has a double root $r$.

Suppose it is not the case that $k_1=k_2=0.$

I don't think that $\lim_{n\to\infty}a_n$ will exist unless the roots have modulus $<1$, but I haven't confirmed this. It should be simple to check, I think. (We have to consider the case where the roots are complex conjugates.)

The solution $(1)$ or $(2)$, as appropriate can be extended over the negative integers, and if $\lim_{n\to\infty}a_n$ then it appears to me that $\lim_{n\to-\infty}a_n$ will not exist.

If both limits exist, then I think that $k_1=k_2=0$ and the sequence is constant, and the limits must be equal. In this case, the solution is unique.

saulspatz
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