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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 $a_n =b_n=o(n)$

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    $\begingroup$ As $|\lambda|=1$ doesn't $\lambda^*=\lambda^{-1}$? $\endgroup$ – Antoine Labelle Feb 3 at 2:10
  • $\begingroup$ @AntoineLabelle thanks, is implemented now. $\endgroup$ – Kung Yao Feb 3 at 17:11
  • $\begingroup$ Do you have reasons to believe that such initial conditions exist? $\endgroup$ – Iosif Pinelis Feb 3 at 17:19
  • $\begingroup$ @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. $\endgroup$ – Kung Yao Feb 3 at 17:24
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    $\begingroup$ @KungYao : I think it could be helpful for some MO users to see the details of what you mentioned in your latter comment. $\endgroup$ – Iosif Pinelis Feb 3 at 19:55
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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.

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  • $\begingroup$ so what are these standard methods? $\endgroup$ – Pritam Bemis Feb 3 at 18:22
  • $\begingroup$ @PritamBemis: en.wikipedia.org/wiki/Matrix_differential_equation $\endgroup$ – Max Alekseyev Feb 3 at 18:25
  • $\begingroup$ @MaxAlekseyev well, I am convinced you will not be able to write down any explicit solution fo that ODE. In fact, at the moment, it seems you just reformulated the question in a more complicated way. $\endgroup$ – Kung Yao Feb 3 at 18:52
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    $\begingroup$ @KungYao: I did not try, and so I cannot comment on feasibility of getting explicit solution at the moment. However, we don't need the solution per se, but only asymptotic of its coefficients. Also, this formulation may be "more complicated" (although it's a matter of taste) but it captures all the coefficients at once and shows explicitly the dependence on the initial conditions, which was somewhat hidden in your original formulation. $\endgroup$ – Max Alekseyev Feb 3 at 19:05
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    $\begingroup$ @PritamBemis This system of ODE can be replaced by one ODE of second order on $A(x)$ or $B(x)$. $\endgroup$ – Alexey Ustinov Feb 5 at 4:14

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