Is there a method to calculate the eigenvalues for this a n x n symmetrical matrix

Question

I'm working with a mechanics problem where I try to find the eigenmodes of the system. The system contains of $n$ masses all connected with springs to one another (same spring constant $k$), the outer most springs, meaning the first and last mass, have their springs connected to a wall. In principal, something like this:

Now, I want to calculate the eigenfrequencies of the system, and so I have managed to find the general equation for this system:

$$\begin{pmatrix} \ddot{x}_1 \\\vdots \\ \ddot{x}_n \end{pmatrix} + \frac{k}{m}\begin{pmatrix} 2 & -1 & 0 & 0 & 0 & 0\\ -1 & 2 & -1 & 0 & 0 & 0 \\ 0 & -1 & 2 & -1 & 0 & 0 \\ \vdots& \vdots& \vdots& \ddots & \vdots & \vdots \\ \vdots& \vdots& \vdots& \vdots & \ddots & \vdots\\ 0 & 0& 0& 0 & -1 & 2 \\ \end{pmatrix} \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix} = 0 $$

In order to find the solutions of this system, we make an ansatz of the form $$\begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix} = Me^{i\lambda \sqrt{k/m} \cdot t}$$ for some $n$ by $1$ matrix $M$, and for non zero solutions, we get the eigenvalue problem, meaning we have to find the determinant of the following matrix:

$$\begin{pmatrix} 2 - \lambda^2& -1 & 0 & 0 & 0 & 0\\ -1 & 2 - \lambda^2& -1 & 0 & 0 & 0 \\ 0 & -1 & 2 - \lambda^2& -1 & 0 & 0 \\ \vdots& \vdots& \vdots& \ddots & \vdots & \vdots \\ \vdots& \vdots& \vdots& \vdots & \ddots & \vdots\\ 0 & 0& 0& 0 & -1 & 2 - \lambda^2\\ \end{pmatrix}$$

Now, I managed to solve it numerically using a python code, for which I found the eigenvalues for some input for a general $n$ and thus the eigenfrequencies of the system. But is there any other way to find the exact solutions using analytical methods? I would think not since we get a polynomial of degree $2n$, but maybe the determinant simplifies itself in a neat way. It at least looked like that when I tried to calculate it for $n=3$.

Thanks for any feedback.

Answer 1

If I'm not mistaken, you have a tridiagonal Toeplitz matrix. The eigenvalues are known : https://en.wikipedia.org/wiki/Tridiagonal_matrix#Eigenvalues

Answer 2

Let $A_n$ be the determinant of an $n\times n$ matrix with $a$ on the diagonal and $b$ on the sub- and super-diagonal and zero everywhere else, so your matrix is the case $a=2-\lambda^2, b=-1$. Expanding using the first row and then the first column you get $A_n = aA_{n-1} -b^2A_{n-2}$. The general solution is $A_n= r \alpha_+^n + s \alpha_-^n$ where $\alpha_{\pm}$ are the roots of $x^2=ax-b^2$. With the initial conditions $A_0=1$ and $A_1=a$ you get $r=\frac{1}{2}\left( 1+ \frac{a}{\sqrt{a^2-4b^2}}\right)$ and $s=1-r$

Answer 3

Your system is $$ m\ddot x_n=k(x_{n-1}-2x_n+x_{n+1}), ~~~ n=1,...,N,~~~x_0=x_{N+1}=0. $$ Applying Fourier sums $X(f,t)=\sum_{n=1}^Nx_ne^{2\pi i\,fn}$ in space direction (at the moment in the sense of generating functions) gives $$ mX_{tt}(f,t)=2k(\cos(2\pi f)-1)\,X(f,t)=-4k\sin^2(\pi f)\,X(f,t). $$ This is now a scalar harmonic oscillator with solution $$ X(f,t)=c_+(f)e^{iω(f)t}+c_-(f)e^{-iω(f)t}, ~~~ ω(f)=2\sin(\pi f)\sqrt{k/m} $$

---

Now treat the DFT as such, restricting the frequencies to the discrete set $f_m=\frac{m}{N+1}$, $m=0,...,N$, that is required for a reconstruction of the $N$ points $x_n$. The inverse transform is then $$ x_n=\frac1{N+1}\sum_{m=0}^{N}X(f_m,t)e^{-2\pi i\,f_m n},~~~ $$ containing terms $$e^{\pm iω(f_m)t-2\pi i\,f_m n}.$$ This gives 4 real-valued basis functions for each $f_m$, and $x_n$ as a linear combination of these basis functions.

---

(The construction of the generating function and the resulting harmonic oscillator appears to be valid independent of the DFT mechanism for general frequency $f$, but that would give linear combinations of the solution functions that have arbitrary frequencies in time direction. In contradiction the DFT reconstruction has only a finite set of frequencies $ω(f_m)$ in time, this does not change under linear combinations. Somehow the generating function needs to be invalid unless $e^{2\pi i\,f(N+1)}=1$...)