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Let $(u_n)_{n\geq 1}$ be a sequence of real numbers satisfying a (non-constant) linear recurrence of order $2$ :

$$ u_{n+2}+a_nu_{n+1}+b_nu_n=0 \tag{1} $$

where $(a_n)$ and $(b_n)$ are both convergent sequences, with limit $a$ and $b$ respectively. Suppose further that any solution of the constant recurrence tends to zero (in other words, the two roots of $X^2+aX+b$ have modulus $<1$).

Question 1. Must $(u_n)$ converge to zero as in the constant case ?

Question 2. (independent of question 1) When $(u_n)$ converges to zero, must we have a geometric convergence $|u_n| \leq kc^n$ with $0 \lt c \lt 1$ as in the constant case ?

Update 12/09/2019 : The answer to both questions is yes when $|a|+|b| \lt 1$, because then there is a $r\in (0,1)$ such that $|a_n|+|b_n| \leq r$ for large enough $n$, say $n\geq n_0$ for some $n_0\in{\mathbb N}$. Then, (1) yields $|u_{n+2}| \leq r \times {\mathsf{max}}(|u_n|,|u_{n+1}|)$ for all $n\geq n_0$. Putting $k={\mathsf{max}}(|u_{n_0},u_{n_0+1}|)$, by induction we deduce that $|u_{n_0+p}| \leq kr^{\lfloor \frac{p}{2} \rfloor}$ for all $p\geq 0$.

Ewan Delanoy
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1 Answers1

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The answer is yes for both questions.

The scalar recursion of order 2 is equivalent to the following recursion of order 1 for vectors via $v_n=(u_n,u_{n+1})^T$. \begin{equation}\tag{1}\label{one} v_{n+1}=A_nv_n,\ \ A_n=\left(\begin{matrix}0&1\\-b_n&-a_n\end{matrix}\right) \end{equation} If $\lambda,\mu$ are the roots of $X^2+aX+b$ which satisfy $|\lambda|,|\mu|<1$ by assumption then there is a constant matrix $T$ such that $$\newcommand{\eps}{\varepsilon}T^{-1}A T= \left(\begin{matrix}\lambda&\eps\\0&\mu\end{matrix}\right)=:D \mbox{ for }A=\lim A_n=\left(\begin{matrix}0&1\\-b&-a\end{matrix}\right)$$ where $\eps=0$ unless $\lambda=\mu$ and the eigenvalue $\lambda$ of $A$ has algebraic multiplicity 1; in this case $\eps\neq0$ and it can be chosen such that $|\lambda|+|\eps|<1$. Putting $v_n=Tw_n$ transforms (\ref{one}) to $$w_{n+1}=(D+F_n)w_n,$$ where $F_n$ is a matrix of elements tending to 0.

Using the maximum norm for vectors and the corresponding matrix norm, we find $$||w_{n+1}||\leq\big(\max(|\lambda|+|\eps|,|\mu|)+||F_n||\big)||w_n||.$$

By assumption, the maximum is smaller than 1 and $||F_n||$ tends to 0. Therefore there exist some $0<M<1$ and some integer $N$ such that the big parenthesis is $\leq M$ for $n\geq N$. This means that $$||w_{n+1}||\leq M||w_{n}||\mbox{ for }n\geq N$$ and hence $||w_n||\leq ||w_N|| M^{n-N}$ for $n\geq N$. Therefore $w_n$ tends to 0 and therefore also $v_n$ and $u_n$ tend to 0 and there are constants $K,M$, $M<1$ such that $|u_n|\leq K M^n$ for all $n$.

Helmut
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  • Nice and simple. The argument will also work for orders greater than $2$, because one can modify the Jordan normal form to have $\varepsilon$'s rather than ones on the upper diagonal. I'll wait a little before awarding the bounty to this answer. – Ewan Delanoy Dec 10 '19 at 14:08
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    @EwanDelanoy The argument even works for nonlinear perturbations and gives a certain theorem of linearized stability: Given $$u_{n+k}+a_{1}(n,u_n,...,u_{n+k-1}),u_{n+k-1}+...a_k(n,u_n,...u_{n+k-1}),u_n=0,$$ where $a_j(n,v_k,...,v_1)\to A_j$, $j=1,...,k$, as $n\to\infty$ and $v_1,...,v_k\to 0$. If $X^k+A_1X^{k-1}+...+A_k=0$ has only roots $\lambda$ with $|\lambda|<1$ then, for sufficiently small initial values $u_1,...u_k$, the sequence $u_n$ tends to $0$ and there are $K,M, M<1$ such that $|u_n|\leq K ,M^n$ for all $n$. – Helmut Dec 14 '19 at 10:08
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    Actually, it is also needed that $(v_1,...v_k)\to a_j(n,v_k,...v_1)$ are bounded for fixed $j=1,...,k$, fixed $n$ and $(v_1,...,v_k)$ close to the origin. – Helmut Dec 15 '19 at 13:43