Decoupling of the constant-coefficient quadratic system $M{q}^{\mathrm{\prime }\mathrm{\prime }}+Kq=0$ into normal modes, with the zero-mode and energy identities

Let $n\ge 1$, let $M\in M_n(\mathbb{R})$ be symmetric positive definite and let $K\in M_n(\mathbb{R})$ be symmetric and non-negative, that is $v^{\mathsf{T}}Kv\ge 0$ for all $v\in {\mathbb{R}}^n$. Let $M^{1\mathrm{/}2}$, $M^{-1\mathrm{/}2}$ be the square roots of Positive definite symmetric matrices have positive definite square roots, and the congruence M^{-1/2} K M^{-1/2} preserves symmetry and positivity and let

$S:=M^{-1\mathrm{/}2}K{M}^{-1\mathrm{/}2}\mathrm{.}$

By the real spectral theorem choose an orthogonal matrix $O$ and a diagonal matrix $\mathrm{\Lambda }=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$ with $S=O\mathrm{\Lambda }{O}^{\mathsf{T}}$ and ${\lambda }_i\ge 0$ for all $i$.

Let $I\subseteq \mathbb{R}$ contain at least two points and satisfy $r,s\in I,r\le u\le s⟹u\in I$ for all $r,s,u\in \mathbb{R}$; suppose also $0\in I$. Let $q:I\to {\mathbb{R}}^n$ be twice differentiable with continuous first and second domain-relative derivatives, in the sense of The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral. Put $z:=O^{\mathsf{T}}{M}^{1\mathrm{/}2}q$. Then:

1. $q$ satisfies the quadratic system $M{q}^{\mathrm{\prime }\mathrm{\prime }}+Kq=0$ on $I$ if and only if $z$ satisfies the decoupled system

$z_i^{\mathrm{\prime }\mathrm{\prime }}+{\lambda }_i{z}_i=0(1\le i\le n)$

on $I$; 2. for every $i$ the solutions of the scalar equation in clause 1 on an interval containing $0$, with prescribed initial data, are exactly as follows:

$z_i(t)=a_i\cos ({\omega }_{i}t)+b_i\sin ({\omega }_{i}t)\text{with }{\omega }_i:=\sqrt{{\lambda }_i}>0\text{ when }{\lambda }_i>0,$

equivalently $z_i(t)=\mathrm{Re}(C_i{e}^{i{\omega }_{i}t})$ with $C_i=a_i-i{b}_i\in \mathbb{C}$, and

$z_i(t)=a_i+b_{i}t\text{when }{\lambda }_i=0;$

the coefficients are $a_i=z_i(0)$, $b_i=z_i^{\mathrm{\prime }}(0)\mathrm{/}{\omega }_i$ for ${\lambda }_i>0$ and $b_i=z_i^{\mathrm{\prime }}(0)$ for ${\lambda }_i=0$, so the displayed families cover every initial datum and are the complete solution sets for the first-order system $(z_i,z_i^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=(z_i^{\mathrm{\prime }},-{\lambda }_i{z}_i)$; 3. consequently every solution of $M{q}^{\mathrm{\prime }\mathrm{\prime }}+Kq=0$ is $q=M^{-1\mathrm{/}2}Oz$ for a unique choice of the modal solutions of clause 2, and the correspondence between initial data $(q(0),q^{\mathrm{\prime }}(0))\in {\mathbb{R}}^{2n}$ and the coefficient tuples $(a_i,b_i)$ is a bijection; 4. for every solution the modal energies $E_i:=\frac{1}{2}(z_i^{\mathrm{\prime }2}+{\lambda }_i{z}_i^2)$ are constant on $I$, and so are

$\frac{1}{2}{\dot{q}}^{\mathsf{T}}M\dot{q}+\frac{1}{2}{q}^{\mathsf{T}}Kq=\frac{1}{2}\mathrm{\mid }{z}^{\mathrm{\prime }}{\mathrm{\mid }}_2^2+\frac{1}{2}{\sum }_{i=1}^n{\lambda }_i{z}_i^2={\sum }_{i=1}^n{E}_i\mathrm{.}$

• The transformation is the one used for small oscillations: $O$ diagonalises the symmetric matrix $M^{-1\mathrm{/}2}K{M}^{-1\mathrm{/}2}$, whose eigenvalues ${\lambda }_i$ are non-negative by Positive definite symmetric matrices have positive definite square roots, and the congruence M^{-1/2} K M^{-1/2} preserves symmetry and positivity clause 4. A zero eigenvalue produces a free affine mode $a_i+b_{i}t$, not an oscillation, and negative eigenvalues cannot occur for non-negative $K$; for an indefinite $K$ the corresponding scalar equation has real exponential solutions.

• Clause 1 is an exact equivalence for the constant-coefficient system. It asserts nothing about a configuration-dependent kinetic metric or a nonlinear potential; the reduction of a small-oscillation linearisation to this theorem is the content of Linearisation of the mechanical system at an equilibrium: M(q_0) xi'' + D²U(q_0) xi = 0, and the small-oscillation quadratic model, and the theorem itself makes no long-time approximation claim about any nonlinear system.