# Eigenvalue equation for kinetic and potential energy

In Boas' Mathematical Methods there is a section on linear algebra in which it is stated that we can write the eigenvalue equation for a set of springs using the kinetic energy and the potential energy where $$V = \frac{1}{2}kr^{T}Vr$$ and $$T = \frac{1}{2}m\dot r^{T}V\dot r$$

Then it's stated that we can write the equations of motion as $$\lambda T r = Vr$$ where $$\lambda = \frac{mw^{2}}{k}$$

My question is, what is the logic of setting up an eigenvalue equation in which we set the potential energy equal to the kinetic energy times an eigenvalue factor?

On the top of Figure we have $\:n+1\:$ ideal springs and $\:n\:$ particles in equilibrium. The constants of the springs are $\:k_{\rho}\: (\rho=1,2,\cdots, n+1) \:$ with equilibrium lengths $\:\ell_{\rho}\:(\rho=1,2,\cdots, n+1 )\:$ and the particle masses $\:m_{\rho}\:(\rho=1,2,\cdots, n)$. Disturbing the system from this equilibrium, the equation of motion of the particle $\:m_{\rho}\:$ is

$$m_{\rho}\ddot{x}_{\rho} =-k_{\rho}\left(x_{\rho}-x_{\rho-1} \right)+k_{\rho+1}\left(x_{\rho+1}-x_{\rho} \right) \tag{01}$$

where $\:x_{\rho}(t)\:$ the displacement of this particle from its equilibrium position, see in the middle of above Figure. We set $\:x_{0}(t)=0\:$ and $\:x_{n+1}(t)=0\:$ for the extreme fixed points A and B respectively.

Equation (01) may be written as $$m_{\rho}\ddot{x}_{\rho}-k_{\rho}x_{\rho-1} +\left(k_{\rho}+k_{\rho+1} \right)x_{\rho}-k_{\rho+1}x_{\rho+1}=0 \tag{02}$$

or

$$\mathrm{M}\ddot{\mathbf{x}}+\mathrm{K}\mathbf{x}=0 \tag{03}$$

where

$$\mathbf{x}= \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ \vdots\\ x_{n-1}\\ x_{n} \end{bmatrix} \in \mathbb{R}^{n} \tag{04}$$

$\:\mathrm{M}\:$ the $\:n \times n\:$ diagonal matrix

$$\mathrm{M}= \begin{bmatrix} m_{1} & 0 & 0 & \cdots & 0 & 0 \\ 0 & m_{2} & 0 & \cdots & 0 & 0 \\ 0 & 0 & m_{3} & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m_{n-1} & 0 \\ 0 & 0 & 0 & \cdots & 0 & m_{n} \end{bmatrix} \tag{05}$$ and $\:\mathrm{K}\:$ the $\:n \times n\:$ tridiagonal symmetric matrix

$$\mathrm{K}= \begin{bmatrix} (k_{1}+k_{2}) & -k_{2} & 0 & \cdots & 0 & 0 \\ -k_{2} & (k_{2}+k_{3}) & -k_{3} & \cdots & 0 & 0 \\ 0 & -k_{3} & (k_{3}+k_{4}) & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & (k_{n-1}+k_{n}) & -k_{n} \\ 0 & 0 & 0 & \cdots & -k_{n} & (k_{n}+k_{n+1}) \end{bmatrix} \tag{06}$$

Equation (03) yields

$$\ddot{\mathbf{x}}+\left(\mathrm{M}^{-1}\mathrm{K}\right)\mathbf{x}=0 \tag{08}$$

or

$$\ddot{\mathbf{x}}+\mathrm{S}\mathbf{x}=0, \qquad \mathrm{S}\equiv \mathrm{M}^{-1}\mathrm{K} \tag{09}$$

Now if $\:\mathrm{S}=\mathrm{M}^{-1}\mathrm{K}\:$ is diagonalizable with eigenvalues $\:\lambda_{\rho}\:(\rho=1,2,\cdots, n)$ and $\:\mathrm{P}\:$ a invertible matrix which diagonalizes it then

$$\mathrm{P}^{-1}\mathrm{S} \mathrm{P} = \rm{diag}\left(\lambda_{1},\lambda_{2},\cdots,\lambda_{n}\right) \tag{10}$$ Defining $$\mathbf{y}\equiv\mathrm{P}^{-1}\mathbf{x} \tag{11}$$ and multiplying (09) by $\:\mathrm{P}^{-1}\:$, we have $$\ddot{\mathbf{y}}+\left(\mathrm{P}^{-1}\mathrm{S} \mathrm{P}\right)\mathbf{y}=0 \tag{12}$$

that is $\:n \:$ independent differential equations

$$\ddot{y}_{\rho}+\lambda_{\rho}y_{\rho}=0, \quad \rho=1,2,\cdots, n \tag{13}$$ Note that taking the inner product of (03) with the "velocity" $\:n-$vector $\:\dot{\mathbf{x}} \:$ $$\dot{\mathbf{x}}= \begin{bmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{3}\\ \vdots\\ \dot{x}_{n-1}\\ \dot{x}_{n} \end{bmatrix} \in \mathbb{R}^{n} \tag{14}$$

we have

$$\langle\mathrm{M}\ddot{\mathbf{x}},\dot{\mathbf{x}}\rangle+\langle\mathrm{K}\mathbf{x},\dot{\mathbf{x}}\rangle=0 \tag{15}$$

that is the equation of conservation of energy

$$\dfrac{ \mathrm{d} }{\mathrm{d}t }\left[\frac12\langle\mathrm{M}\dot{\mathbf{x}},\dot{\mathbf{x}}\rangle+\frac12\langle\mathrm{K}\mathbf{x},\mathbf{x}\rangle\right]=0 \tag{16}$$

related to the question.

For the special case of a common spring constant $\:k_{\rho}=k \: (\rho=1,2,\cdots, n+1) \:$ and common particle mass $\:m_{\rho}=m \: (\rho=1,2,\cdots, n) \:$, equation (08) gives

$$\ddot{\mathbf{x}}+\omega_{o}^{2}\,\mathrm{\Xi}\,\mathbf{x}=0 \tag{17}$$

where $$\omega_{o}\equiv \sqrt{\dfrac{k}{m}}= \textrm{fundamental frequency} \tag{18}$$

and $\:\mathrm{\Xi}\:$ the following $\:n \times n\:$ tridiagonal symmetric matrix (a special case of the so-called Toeplitz matrices)

$$\mathrm{\Xi}= \begin{bmatrix} 2& -1 & 0 & \cdots & 0 & 0 \\ -1 & 2 & -1 & \cdots & 0 & 0 \\ 0 & -1 & 2& \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 2 & -1 \\ 0 & 0 & 0 & \cdots & -1& 2 \end{bmatrix} \tag{19}$$ with real positive eigenvalues

$$\xi_{\rho}= 4\sin^{2}\left[ \rho\dfrac{\pi}{2(n+1)} \right]=2\Bigg(1-\cos\left[ \rho\dfrac{\pi}{(n+1)} \right]\Bigg), \quad \rho=1,2,\cdots, n \tag{20}$$

and eigenvectors(1) $\:\mathbf{e}_{\rho}\:$ with $\:\sigma-$component

$$\left(\mathbf{e}_{\rho} \right)_{\sigma}=\sqrt{\dfrac{2}{n+1}}\sin\left( \rho \sigma \dfrac{\pi}{n+1} \right) , \quad \rho,\sigma=1,2,\cdots, n \tag{21}$$ In this special case the system of independent equations (13) is

$$\ddot{y}_{\rho}+\left(\xi_{\rho}\omega_{o}^{2}\right)y_{\rho}=0, \quad \rho=1,2,\cdots, n \tag{22}$$

that is :

The motion of a system of $\:n\:$ particles of the same mass $\:m\:$ connected by $\:n+1\:$ ideal springs of the same constant $\:k\:$, see Figure above, is the superposition of $\:n\:$ independent harmonic oscillations with frequences $$\omega_\rho=\sqrt{\xi_{\rho}}\omega_o=2\omega_o \sin\left[\rho\dfrac{\pi}{2(n+1)} \right],\quad \omega_{o}\equiv \sqrt{\dfrac{k}{m}} , \quad \rho=1,2,\cdots,n-1,n \tag{23}$$ as shown in Figure below.

(1) Any $\:n \times n\:$ tridiagonal symmetric Toeplitz matrix has the same eigenvectors !!!

EDIT

For other more general cases a useful theorem from "Matrix Theory" by Joel N.Franklin, is given below unchanged :

Theorem Let $\:\mathrm{M}\:$ and $\:\mathrm{K}\:$ be $\:n \times n\:$ Hermitian matrices. If $\:\mathrm{M}\:$ positive definite, then there is a $\:n \times n\:$ matrix $\:\mathrm{C}\:$ for which $$\mathrm{C}^{*}\mathrm{M}\mathrm{C}=\mathrm{I} \quad \textrm{and} \quad \mathrm{C}^{*}\mathrm{K}\mathrm{C}=\Lambda= \rm{diag}\left(\lambda_{1},\lambda_{2},\cdots,\lambda_{n}\right) \tag{t-17}$$

The numbers $\:\lambda_{j}\:$ are real. If $\:\mathrm{K}\:$ is positive definite, the $\:\lambda_{j}\:$ are positive. The $\:\lambda_{j}\:$ are generalized eigenvalues satisfying

$$\mathrm{K}\,c^{j}=\lambda_{j}\,\mathrm{M}\,c^{j}, \quad c^{j}\ne 0 \quad (j=1,\cdots, n) \tag{t-18}$$

If $\:\mathrm{K}\:$ and $\:\mathrm{M}\:$ are real, then a real matrix $\:\mathrm{C}\:$, with columns $\:c^{j}\:$, may be found satisfying (t-17) and (t-18).

• I must say I really am impressed with your figures and formatting; one of the best here. +1.
– user36790
Nov 22 '16 at 8:04
• @MAFIA36790 : I have always in my mind your kindness and your friendly to me comments in the past : I joined Physics SE as diracpaul in June'15 and I quit the site in Sep'15 for personal reasons. I came back as Frobenius in Mar'16. Nov 22 '16 at 8:21
• That's great! I was about to mention that yours were similar to someone I knew who unfortunately left Phys.SE a year ago; but that would make the above comment chatty. Nevertheless, welcome again @Frobenius.
– user36790
Nov 22 '16 at 8:35