The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart

Let $n\ge 1$, let $\mathrm{\Omega }\subseteq {\mathbb{R}}^n$ be open, let $M:\mathrm{\Omega }\to M_n(\mathbb{R})$ be of class $C^2$ ($C^k$ Euclidean maps and diffeomorphisms) with $M(q)$ symmetric and positive definite for every $q\in \mathrm{\Omega }$, that is $v^{\mathsf{T}}M(q)v>0$ for every nonzero $v\in {\mathbb{R}}^n$, and let $U:\mathrm{\Omega }\to \mathbb{R}$ be of class $C^2$. The matrix field $M$ is the kinetic metric field and $U$ the potential of the data in the chart $\mathrm{\Omega }$.

Coordinates and matrix entries are indexed by $0\le i,j,k<n$, as in The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$ and The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case. Vectors are read as columns: for $A\in M_n(\mathbb{R})$, $(Av{)}_i:={\sum }_{j=0}^{n-1}{A}_{ij}{v}_j$, and $v^{\mathsf{T}}Av:={\sum }_{i,j=0}^{n-1}{v}_i{A}_{ij}{v}_j$.

Define, for $(q,v)\in \mathrm{\Omega }\times {\mathbb{R}}^n$, the kinetic energy, Lagrangian and energy

$T(q,v):=\frac{1}{2}{v}^{\mathsf{T}}M(q)v,L(q,v):=T(q,v)-U(q),E(q,v):=T(q,v)+U(q),$

and, for $q\in \mathrm{\Omega }$ and $v\in {\mathbb{R}}^n$, the matrix

$D_{q}M(q)[v]:={\sum }_{k=0}^{n-1}{v}_k{\mathrm{\partial }}_{k}M(q),({\mathrm{\partial }}_{k}M(q){)}_{ij}={\mathrm{\partial }}_k{M}_{ij}(q)(0\le i,j,k<n)\mathrm{.}$

The mechanical vector field of the data is the map

$G:\mathrm{\Omega }\times {\mathbb{R}}^n⟶{\mathbb{R}}^n\times {\mathbb{R}}^n,G(q,v)=(v,M(q{)}^{-1}(\frac{1}{2}{\mathrm{\nabla }}_q(v^{\mathsf{T}}M(q)v)-(D_{q}M(q)[v])v-\mathrm{\nabla }U(q))),$

where ${\mathrm{\nabla }}_q$ and $\mathrm{\nabla }U$ are Euclidean gradients (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case) and $M(q{)}^{-1}$ is the inverse of the positive definite matrix $M(q)$ (Invertible matrices and the general linear group ${\mathrm{GL}}_n(F)$, Rectangular matrix multiplication and the identity matrix $I_n$, including zero-sized shapes). The mechanical system in the chart is the first-order system

$q^{\mathrm{\prime }}(t)=v(t),v^{\mathrm{\prime }}(t)=M(q(t){)}^{-1}(\frac{1}{2}{\mathrm{\nabla }}_{q}f(q(t),v(t))-(D_{q}M(q(t))[v(t)])v(t)-\mathrm{\nabla }U(q(t))),f(q,v):=v^{\mathsf{T}}M(q)v\mathrm{.}$

A point $q_{\ast }\in \mathrm{\Omega }$ is a mechanical equilibrium when $\mathrm{\nabla }U(q_{\ast })=0$.

• Equilibria. If $q_{\ast }\in \mathrm{\Omega }$ is a mechanical equilibrium, then $G(q_{\ast },0)=(0,0)$: the first component is $0$, and in the second component ${\mathrm{\nabla }}_{q}f(q_{\ast },0)=0$ (a quadratic form in $v$ evaluated at $v=0$), $(D_{q}M(q_{\ast })[0])0=0$ and $\mathrm{\nabla }U(q_{\ast })=0$. Thus $(q_{\ast },0)$ is an equilibrium of $G$ in the sense of Equilibria, Lyapunov stability and asymptotic stability of an autonomous first-order system.

• Equivalence with the Euler-Lagrange equation. A $C^2$ curve $q:I\to \mathrm{\Omega }$ with velocity $v:=q^{\mathrm{\prime }}$ satisfies the Euler-Lagrange equation $\frac{d}{dt}{\mathrm{\nabla }}_{v}L={\mathrm{\nabla }}_{q}L$ of the Lagrangian above if and only if $(q,v)$ solves the system $q^{\mathrm{\prime }}=v$, $v^{\mathrm{\prime }}=G_2(q,v)$. Indeed ${\mathrm{\nabla }}_{v}L=M(q)v$, so

$\frac{d}{dt}(M(q)q^{\mathrm{\prime }})=M(q)q^{\mathrm{\prime }\mathrm{\prime }}+(D_{q}M(q)[q^{\mathrm{\prime }}])q^{\mathrm{\prime }},{\mathrm{\nabla }}_{q}L=\frac{1}{2}{\mathrm{\nabla }}_q(q^{\mathrm{\prime }\mathsf{T}}M(q)q^{\mathrm{\prime }})-\mathrm{\nabla }U(q),$

and multiplying the Euler-Lagrange equation by $M(q{)}^{-1}$ gives the displayed second component. The computation is linear algebra plus the product rule for the matrix-valued curve $t\mapsto M(q(t))$.

• Regularity. Under the stated hypotheses the map $G$ is of class $C^1$ on $\mathrm{\Omega }\times {\mathbb{R}}^n$: $M$ is $C^2$, so the maps involving ${\mathrm{\partial }}_{k}M$, ${\mathrm{\nabla }}_{q}f$ and $D_{q}M[\cdot ]\cdot$ are $C^1$; $\mathrm{\nabla }U$ is $C^1$ because $U$ is $C^2$; and $q\mapsto M(q{)}^{-1}$ is $C^1$ wherever $\det M(q)\mathrm{\ne }0$, since $M^{-1}=\det (M{)}^{-1}\mathrm{a}\mathrm{d}\mathrm{j}(M)$ has entries that are polynomial in the entries of $M$ divided by the continuous nonvanishing function $\det M$. Hence the mechanical system satisfies the Picard-Lindelöf regularity hypotheses locally.

• Scope. This is a mathematical chart model: $M$ is an arbitrary $C^2$ positive definite symmetric matrix field and no physical law, identification with masses, units or empirical assertion is made by the definition. The constant case $M(q)\equiv M$ reduces to $M{q}^{\mathrm{\prime }\mathrm{\prime }}=-\mathrm{\nabla }U(q)$, and a mechanical equilibrium is a critical point of the potential.