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Linearisation of the mechanical system at an equilibrium: M(q_0) xi'' + D²U(q_0) xi = 0, and the small-oscillation quadratic model

Statement

Let $n \geq 1$ , let $Ω \subseteq R^{n}$ be open, let $M : Ω \to M_{n} (R)$ be a $C^{2}$ field of symmetric positive definite matrices, let $U : Ω \to R$ be $C^{2}$ , and let $q_{0} \in Ω$ be a mechanical equilibrium, that is $\nabla U (q_{0}) = 0$ (The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart). Let $G$ be the mechanical vector field on $Ω \times R^{n}$ . Then:

$G$ is totally differentiable at $(q_{0}, 0)$ (The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder) and, writing $D^{2} U (q_{0})$ for the Hessian matrix (The Hessian matrix and critical points of a scalar field),

$D G (q_{0}, 0) (δ q, δ v) = (δ v, - M (q_{0})^{- 1} D^{2} U (q_{0}) δ q) (δ q, δ v \in R^{n});$

consequently the linearised equation at the equilibrium is

$ξ^{''} = - M (q_{0})^{- 1} D^{2} U (q_{0}) ξ, equivalently M (q_{0}) ξ^{''} + D^{2} U (q_{0}) ξ = 0,$

a constant-coefficient system whose mass matrix $M (q_{0})$ is symmetric positive definite and whose stiffness matrix $D^{2} U (q_{0})$ is symmetric (The Hessian of a $C^{2}$ scalar field is symmetric); 3. if in addition $q_{0}$ is a strict local minimum of $U$ , then $K : = D^{2} U (q_{0})$ is non-negative, so the hypotheses of Decoupling of the constant-coefficient quadratic system $M q^{''} + K q = 0$ into normal modes, with the zero-mode and energy identities are satisfied by $M (q_{0})$ and $K$ , and the linearised system is exactly the constant-coefficient quadratic system of that theorem with mass matrix $M (q_{0})$ and stiffness $K$ ; its modes are the normal modes of the linearisation.

Facts & Assumptions

Given: The $C^{2}$ kinetic metric field $M$ , the $C^{2}$ potential $U$ , the equilibrium $q_{0}$ , and the vector field $G (q, v) = (v, H (q, v))$ with $H (q, v) = M (q)^{- 1} (\frac{1}{2} \nabla_{q} f (q, v) - (D_{q} M (q) [v]) v - \nabla U (q))$ , $f (q, v) = v^{T} M (q) v$ .

[F1]

Regularity: under the stated hypotheses $G$ is of class $C^{1}$ on $Ω \times R^{n}$ (the inverse $M (q)^{- 1}$ has entries polynomial in the entries of $M$ divided by the nonvanishing continuous function $\det M (q)$ ), hence totally differentiable at every point (The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart, If $\det (A)$ is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ , If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative, $C^{k}$ Euclidean maps and diffeomorphisms).

[F2]

At $(q_{0}, 0)$ : $\nabla_{q} f (q_{0}, 0) = 0$ and its $q$ -derivative vanishes, because $\partial_{k} (v^{T} M (q) v) = v^{T} \partial_{k} M (q) v$ is a quadratic form in $v$ vanishing identically at $v = 0$ ; similarly $(D_{q} M (q) [v]) v = \sum_{k} v_{k} \partial_{k} M (q) v$ is quadratic in $v$ , so it and both of its first partial derivatives vanish at $v = 0$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes).

[F3]

At the equilibrium $\nabla U (q_{0}) = 0$ and the total derivative of $\nabla U$ is the Hessian matrix $D^{2} U (q_{0})$ , symmetric (The Hessian matrix and critical points of a scalar field, The Hessian of a $C^{2}$ scalar field is symmetric, The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder).

[F4]

At a strict local minimum, $\nabla U (q_{0}) = 0$ and $D^{2} U (q_{0})$ is non-negative (A C² local minimum has vanishing gradient and positive semidefinite Hessian); and the constant-coefficient system $M ξ^{''} + K ξ = 0$ with $M$ symmetric positive definite and $K$ symmetric non-negative is decoupled by Decoupling of the constant-coefficient quadratic system $M q^{''} + K q = 0$ into normal modes, with the zero-mode and energy identities.

Proof

technique · direct; differentiate the explicit components of $G$ at the equilibrium and substitute

1.1givenF1algebra

The linearisation of $G$ at $(q_{0}, 0)$ is $D G (q_{0}, 0) (δ q, δ v) = (δ v, D_{q} H (q_{0}, 0) δ q + D_{v} H (q_{0}, 0) δ v)$ . We compute the two partial derivatives of $H = M^{- 1} B$ with $B (q, v) : = \frac{1}{2} \nabla_{q} f (q, v) - (D_{q} M (q) [v]) v - \nabla U (q)$ .

2.1step 1.1F2F3algebra

At $(q_{0}, 0)$ , $B (q_{0}, 0) = \frac{1}{2} \nabla_{q} f (q_{0}, 0) - 0 - \nabla U (q_{0}) = 0$ by [F2] and the equilibrium condition. Hence $D_{q} H (q_{0}, 0) δ q = D_{q} (M^{- 1}) (q_{0}) [δ q] B (q_{0}, 0) + M (q_{0})^{- 1} D_{q} B (q_{0}, 0) δ q = M (q_{0})^{- 1} D_{q} B (q_{0}, 0) δ q,$ and by [F2] and [F3] the derivative of $B$ in $q$ at $(q_{0}, 0)$ is $- \frac{1}{2} \cdot 0 + 0 - D^{2} U (q_{0}) = - D^{2} U (q_{0})$ . Hence $D_{q} H (q_{0}, 0) = - M (q_{0})^{- 1} D^{2} U (q_{0})$ .

2.2step 1.12.1F2algebra

Similarly $D_{v} H (q_{0}, 0) = M (q_{0})^{- 1} D_{v} B (q_{0}, 0)$ because $B (q_{0}, 0) = 0$ again kills the derivative of $M^{- 1}$ along $v$ ; and $D_{v} B (q_{0}, 0) = 0$ because both $\frac{1}{2} \nabla_{q} f$ and $(D_{q} M (q) [v]) v$ are quadratic in $v$ with vanishing first $v$ -derivative at $v = 0$ [F2]. Therefore $D G (q_{0}, 0) (δ q, δ v) = (δ v, - M (q_{0})^{- 1} D^{2} U (q_{0}) δ q)$ , which is clause 1.

2.3step 1.12.2F3algebra

Clause 2: the linearised first-order system is $ξ^{'} = η$ , $η^{'} = - M (q_{0})^{- 1} D^{2} U (q_{0}) ξ$ ; eliminating $η$ and multiplying by the invertible matrix $M (q_{0})$ gives $M (q_{0}) ξ^{''} + D^{2} U (q_{0}) ξ = 0$ . The matrix $D^{2} U (q_{0})$ is symmetric by [F3], and $M (q_{0})$ is symmetric positive definite by hypothesis; the stiffness $K : = D^{2} U (q_{0})$ need not be non-negative, and no normal-mode conclusion is drawn at this point.

3.1step 2.3F4∎

Clause 3: if $q_{0}$ is a strict local minimum, [F4] shows that $K = D^{2} U (q_{0})$ is non-negative; with $M = M (q_{0})$ the hypotheses of the quadratic-mode theorem are exactly satisfied (symmetric positive definite mass and symmetric non-negative stiffness), and its clauses describe the normal modes of the linearisation.

Remarks

The linearisation is a mathematical object: it is the derivative of the model field at the equilibrium, and the replacement of the nonlinear system by its linearisation is an approximation whose error is not controlled here. In particular clause 3 does not assert that the nonlinear system's solutions remain close to the normal modes for all time; the exact quadratic calculation alone gives no long-time error bound.
If the equilibrium is not a minimum, $D^{2} U (q_{0})$ may have positive and negative eigenvalues; the linearised modes then include real exponential directions. Stability conclusions from such a linearisation are the subject of An equilibrium whose linearisation has all eigenvalues in the open left half-plane is locally exponentially stable and A hyperbolic equilibrium whose linearisation has an eigenvalue with positive real part is unstable, which require hyperbolicity and do not apply to the pure oscillation case of clause 3.

Linearisation of the mechanical system at an equilibrium: M(q_0) xi'' + D²U(q_0) xi = 0, and the small-oscillation quadratic model

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources