Energy is conserved along solutions of the mechanical system with a configuration-dependent kinetic metric

Let $n\ge 1$, let $\mathrm{\Omega }\subseteq {\mathbb{R}}^n$ be open, let $M:\mathrm{\Omega }\to M_n(\mathbb{R})$ be a $C^2$ field of symmetric positive definite matrices, let $U:\mathrm{\Omega }\to \mathbb{R}$ be $C^2$, and let $E(q,v)=\frac{1}{2}{v}^{\mathsf{T}}M(q)v+U(q)$ be the energy of The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart. If $q:I\to \mathrm{\Omega }$ is a $C^2$ solution of the mechanical system $M(q)q^{\mathrm{\prime }\mathrm{\prime }}=\frac{1}{2}{\mathrm{\nabla }}_q(q^{\mathrm{\prime }\mathsf{T}}M(q)q^{\mathrm{\prime }})-(D_{q}M(q)[q^{\mathrm{\prime }}])q^{\mathrm{\prime }}-\mathrm{\nabla }U(q)$ on the interval $I$, then

$\frac{d}{dt}E(q(t),q^{\mathrm{\prime }}(t))=0(t\in I);$

that is, the energy is constant along every solution.

• The cancellation uses exactly the two terms produced by the configuration dependence of $M$: the derivative of the kinetic energy along the flow produces $v^{\mathsf{T}}(D_{q}M[v])v$, and substitution of the two configuration-dependent acceleration terms gives $\frac{1}{2}{v}^{\mathsf{T}}{\mathrm{\nabla }}_q(v^{\mathsf{T}}Mv)-v^{\mathsf{T}}(D_{q}M[v])v=-\frac{1}{2}{v}^{\mathsf{T}}(D_{q}M[v])v$, cancelling the kinetic derivative term $\frac{1}{2}{v}^{\mathsf{T}}(D_{q}M[v])v$. For constant $M$ the statement reduces to the familiar $\frac{d}{dt}(\frac{1}{2}{v}^{\mathsf{T}}Mv+U)=0$.

• The lemma is a statement about the model: it does not assert that any physical system conserves energy, nor that solutions exist globally, nor that the energy is a Liapunov function for any equilibrium; those uses are made in A strict local minimum of the potential gives a Lyapunov stable equilibrium of the mechanical system, by energy conservation under their own hypotheses.