For small oscillations about equilibrium, put $M_{ij}=T_{ij}(0)$ and $K_{ij}={\partial }_i{\partial }_{j}V(0)$. The potential has no linear term at equilibrium; to quadratic order the Lagrangian is $L_2={\dot{x}}^{\mathsf{T}}M\dot{x}/2-x^{\mathsf{T}}Kx/2$. Its equations are $M\ddot{x}+Kx=0$. A normal mode $x=v{e}^{i\omega t}$ satisfies the generalized eigenvalue problem $(K-{\omega }^{2}M)v=0$. For a positive-definite mass matrix, diagonalization $M^{-1/2}K{M}^{-1/2}$ gives the squared frequencies and independent normal coordinates. A strictly positive Hessian gives positive frequencies; zero eigenvalues require separate higher-order stability analysis.

Here the spring extensions are $x_1$, $x_2-x_1$, and $-x_2$. ThusThe stiffness matrix is $k\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$ and $M=mI$. Its eigenvectors $(1,1)$ and $(1,-1)$ have stiffness eigenvalues $k$ and $3k$, givingIn the first mode the masses move together with equal displacements and the middle spring keeps its length. In the second they move oppositely with equal amplitudes, stretching and compressing the middle spring twice as much as either end spring.

Independent endpoint-fixed variations of the two coordinates give the Euler-Lagrange equationsDifferentiating the sphere parametrization gives the Riemannian metric $d{s}^2=a^2(d{\theta }^2+{\sin }^2\theta d{\varphi }^2)$. Its energy functional is the stated $I$. Its stationary paths are geodesics with affine, constant-speed parametrization; conversely a geodesic can be so parametrized. This explains the use of energy rather than arc length.

For $F=a^2({\dot{\theta }}^2+{\sin }^2\theta {\dot{\varphi }}^2)/2$, the equations areThe second gives the conserved generalized momentum ${\sin }^2\theta \dot{\varphi }=h$. Since $F$ has no explicit time dependence, its Beltrami identity gives constant $\dot{\theta }{F}_{\dot{\theta }}+\dot{\varphi }{F}_{\dot{\varphi }}-F=F$. Dividing by $a^2/2$ gives the second first integral,Thus $E$ is the squared speed divided by $a^2$, while $h$ records the rotational symmetry about the polar axis. The coordinate equations apply away from the polar coordinate singularities.

The Burnside lemma states that the number of orbits of a finite group action isFor $m=2n-1\ge 3$, the dihedral group consists of $m$ rotations $r^j$ by $2\pi j/m$, and $m$ reflections $r^{j}s$, each across the line through one vertex and the midpoint of the opposite edge. Thus $|G|=2m=4n-2$.

For the pentagon, the identity fixes $v^5{e}^5$ colourings. Each of the four nonidentity rotations has a single cycle on vertices and a single cycle on edges, fixing $ve$ colourings. Each of the five reflections has one fixed vertex and two vertex pairs, and one fixed edge and two edge pairs, fixing $v^3{e}^3$ colourings. Applying the Burnside lemma to simultaneous vertex and edge colourings givesVertex colours and edge colours may be chosen independently; each symmetry must preserve both assignments.