# Two-term recurrence relation

We consider the following system of recurrence relations for $$n \in \mathbb Z$$ and $$\vert \lambda \vert=1$$ with $$\lambda \in \mathbb{C}$$

$$a_{n+1} = \lambda a_{n-1}+ \lambda^* a_n + \lambda^* n b_n$$ $$b_{n+1} = \lambda^* b_{n-1}+ \lambda b_n - \lambda n a_n.$$

Here, $$\lambda^*$$ is the complex conjugate of $$\lambda.$$

I am interested in non-zero initial conditions under which $$a_n,b_n$$ tend to zero for $$n \rightarrow \pm \infty.$$

Observation: If there is a limit $$a_n \rightarrow a$$ and $$b_n \rightarrow b$$, then indeed by the last term in each row, it has to be zero and $$a_n =b_n=o(n)$$

• As $|\lambda|=1$ doesn't $\lambda^*=\lambda^{-1}$? – Antoine Labelle Feb 3 at 2:10
• @AntoineLabelle thanks, is implemented now. – Kung Yao Feb 3 at 17:11
• Do you have reasons to believe that such initial conditions exist? – Iosif Pinelis Feb 3 at 17:19
• @IosifPinelis I do, this comes from the matching conditions of coefficients in an ODE and numerically that ODE has a solution (for certain choices of $\lambda$ at least.) In particular, when $\lambda$ is the third root of unity. – Kung Yao Feb 3 at 17:24
• @KungYao : I think it could be helpful for some MO users to see the details of what you mentioned in your latter comment. – Iosif Pinelis Feb 3 at 19:55

I assume that the initial conditions $$a_0,a_1,b_0,b_1$$ and that $$n\to +\infty$$.
Let $$A(x):=\sum_{n\geq 0} a_n x^n$$ and $$B(x):=\sum_{n\geq 0} a_n x^n$$. Then the recurrence relations become: $$\begin{cases} A(x) - a_1x - a_0 = \lambda x^2 A(x) + \lambda^* x (A(x)-a_0) + \lambda^* x^2 B'(x), \\ B(x) - b_1x - b_0 = \lambda^* x^2 B(x) + \lambda x (B(x)-b_0) - \lambda x^2 A'(x). \end{cases}$$ That is, we have a system of 2 linear first-order ODE: $$\begin{bmatrix} A'(x)\\ B'(x)\end{bmatrix} = \begin{bmatrix} 0 & -\lambda^* x^{-2}+x^{-1}+\lambda^{*2}\\ \lambda x^{-2} - x^{-1} - \lambda^2 & 0\end{bmatrix} \cdot \begin{bmatrix} A(x)\\ B(x)\end{bmatrix} + \begin{bmatrix} \lambda^*(b_0x^{-2} + (b_1-\lambda b_0)x^{-1})\\ -\lambda(a_0x^{-2} + (a_1-\lambda^*a_0)x^{-1})\end{bmatrix},$$ which may be analyzed with the standard methods.
• @PritamBemis This system of ODE can be replaced by one ODE of second order on $A(x)$ or $B(x)$. – Alexey Ustinov Feb 5 at 4:14