Let $a\geq 0,b>0$, $N\in \mathbb N$ and $$ A_N = \begin{pmatrix}a(-N) & b && 0\\b & \ddots & \ddots & \\ &\ddots & \ddots & b\\0 && b & a N \end{pmatrix}\in \mathbb R^{(2N+1)\times(2N+1)}.$$ The diagonal elements are $a(-N), a(-N+1), \dots, a(N-1), aN$. Let vector $y_N$ be the unique solution of the ODE initial value problem:

$$\frac{\mathrm{d}}{\mathrm{d}t}y_N = A_{N} y_N, \quad y_N(0)=\begin{pmatrix}1\\\vdots\\1 \end{pmatrix}.$$

Explicitly, $$y_N(t)= e^{tA_N}y_N(0)\mbox{, for }t\in[0,1].$$

It can be shown that all its components are positive. Let us denote $y_N^n(t)$ the $n$-th component of vector $y_N(t)$.

Numerically, I get that for each fixed $t\in [0,1]$ the successive quotients of components of $y_N(t)$ are bounded independently of $N$, i.e., for each fixed $t\in [0,1]$ there exist $C_1,C_2>0$ (independent of $N$), such that

$$ y_N^{n+1}(t)/y_N^{n}(t)\leq C_1\mbox{ and }y_N^{n}(t)/y_N^{n+1}(t)\leq C_2$$ for all $n=-N,\dots N-1$.

Any idea on how to prove this?

In case it is useful, other properties I am getting numerically are:

1) $\left(y_N^{n+1}(t)/y_N^{n}(t)\right)_{n=-N}^{N-1}$ is decreasing in $n$ (and therefore, $\left(y_N^{n}(t)/y_N^{n+1}(t)\right)_{n=-N}^{N-1}$ is increasing). In particular, the maximums for each finite sequence of quotients (the sequence that takes the first component as denominator and the one that takes the first one as a numerator) are attained at the first and last term respectively.

2) $A_{N}$ has eigenvalues of the form $-\lambda_N^N< -\lambda_N^{N-1}<\dots<\lambda_N^0=0<\dots\lambda_N^{N-1}<\lambda_N^N$ (upper indices are not powers, just indices), where $\lambda_N^N\to \infty$ when $N\to \infty$ and there exists $K>0$ (independent of $N$) such that $\lambda_N^{n+1}-\lambda_N^n\geq K$ for all $n=0,\dots,N-1$.