TITLE: Using matrix to analyze an ODE system $\begin{cases}\dot a_k(t)=2(b_k^2-b_{k-1}^2)\\\dot b_k(t)=b_k(a_{k+1}-a_{k})\end{cases}$ with $b_0(t)=b_n(t)=0$.
QUESTION [5 upvotes]: Let $a_k(t), b_k (t) \in \mathbb{R} \ (k = 1, 2, \dots, n)$ satisfy the following differential equations:
$$\begin{aligned} \frac{d}{dt}a_k(t) &= 2 \left( b_k^2 - b_{k-1}^2 \right) \\ \frac{d}{dt} b_k(t) &= b_k \left( a_{k+1} - a_{k} \right)\end{aligned} \qquad k = 1, 2, \dots , n $$
where $b_0(t) = b_n(t) = 0$. Consider the $n \times n$ tri-diagonal matrix
$$L(a,b)=\begin{bmatrix}
    a_1 & b_1 & 0 & \dots  & 0 &0\\
    b_1 & a_2 & b_2 & \dots  & 0 &0\\
    \vdots & \vdots & \vdots & \ddots & \vdots& \vdots \\
0 & 0 & 0 & \dots  & a_{n-1} & b_{n-1}\\
    0 & 0 & 0 & \dots  & b_{n-1} & a_n
\end{bmatrix}$$
show that:

The eigenvalues of $L(t) = L(a(t), b(t))$ are independent of $t$.
$\lim_{t \to \infty}b_k(t) = 0$ for $k=1,2,\dots,n-1$.



This problem comes from the Applied Math track in S.-T. Yau College Student Mathematics Contests 2021 as a final problem in the final round.
My idea is that you have to use the condition by taking derivatives, so for the first problem I tried to prove that the derivative of $\vert \lambda I-L(a(t),b(t))\vert$ is zero, but it seems not working, and for the second problem I have no ideas whether to use the properties of $L(a(t), b(t))$ or not, and I'm stuck here.

Edit1. The problem didn’t consider $k=n$ case, I think $a_{n+1}$ should be $1$, since after that if we naturally extend this matrix to $(n+1)\times(n+1)$ the corresponding problem still works and matrix $\lambda I-L(a(t),b(t))$ still satisfies the differential equations. I’m not sure if this is helpful️

Edit2. Edit1 is wrong and now hidden, $b_n(t)=0$ so $a_{n+1}$ won't appear.
Any help, hint or solution would be appreciated!

REPLY [2 votes]: These are the equations of motion of the Toda lattice . It is an integrable classical mechanical system, with a Lax matrix given by $L(a,b)$ as you define. The eigenvalues of a Lax matrix can be shown to be time-independent, see here.