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Energy stability and instability of the mechanical model at an equilibrium: the strict minimum is Lyapunov stable, a negative Hessian eigenvalue gives linear instability, a negative definite potential Hessian gives nonlinear instability, and exact energy conservation rules out attraction

Physical scope

The stability consequences of the mechanical chart data at a mechanical equilibrium q0: a strict local minimum of the potential gives a Lyapunov stable equilibrium; a negative Hessian eigenvalue gives an unbounded solution of the associated linearised vector equation (linear instability, with no nonlinear conclusion); a negative definite Hessian (strict local maximum) makes the equilibrium unstable in the nonlinear sense through the hyperbolic instability theorem; and exact energy conservation along the model's motions prevents asymptotic stability whenever states of different energy accumulate at the equilibrium, so a strict minimum is stable but not attracting. No global stability, no asymptotic stability and no conclusion is inferred from a zero Hessian eigenvalue alone.

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, and let $U : Ω \to R$ be of class $C^{2}$ ; let $G$ be the mechanical vector field and $E (q, v) = \frac{1}{2} v^{T} M (q) v + U (q)$ the energy of The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart. Let $q_{0} \in Ω$ be a mechanical equilibrium, $\nabla U (q_{0}) = 0$ , and put $K : = D^{2} U (q_{0})$ (The Hessian matrix and critical points of a scalar field, The Hessian of a $C^{2}$ scalar field is symmetric). All chart data, state norms and times are numerical in the fixed reference scales of Stability vocabulary of the model: Lyapunov, asymptotic and orbital stability, linear stability of equilibria and map fixed points, and hyperbolicity of equilibria and fixed points' numerical-state convention. Fix also an energy reference $E_{r e f} > 0$ , so that physical energy is $E_{p h y s} = E_{r e f} E$ and physical time is $t_{r e f} t$ . Then:

(Strict minimum.) If $q_{0}$ is a strict local minimum of $U$ , then $(q_{0}, 0)$ is a Lyapunov stable equilibrium of $G$ (Equilibria, Lyapunov stability and asymptotic stability of an autonomous first-order system, Stability vocabulary of the model: Lyapunov, asymptotic and orbital stability, linear stability of equilibria and map fixed points, and hyperbolicity of equilibria and fixed points clause (i)).
(Linear instability from a negative eigenvalue.) If $K$ has a negative eigenvalue (Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism), then there are $κ > 0$ and $w \neq 0$ in $R^{n}$ such that $Y : R \to R^{2 n}$ , $Y (t) : = e^{κ t} (w, κ w)$ , solves the associated vector equation $Y^{'} = D G (q_{0}, 0) Y$ along the constant solution $z \equiv (q_{0}, 0)$ , with $Y (0) = (w, κ w)$ . This vector equation is the equation for the action on an initial deviation of the normalized matrix equation $A^{'} = D G (q_{0}, 0) A$ , $A (0) = I_{2 n}$ defined in The variational equation along an ODE solution; in particular $(q_{0}, 0)$ is linearly unstable in the sense of Stability vocabulary of the model: Lyapunov, asymptotic and orbital stability, linear stability of equilibria and map fixed points, and hyperbolicity of equilibria and fixed points clause (iii).
(Nonlinear instability from a negative definite Hessian.) If $K$ is negative definite, then every complex eigenvalue of $D G (q_{0}, 0)$ is real and nonzero, at least one is positive, and $(q_{0}, 0)$ is unstable: it is not Lyapunov stable.
(No attraction under exact energy conservation.) The energy $E$ is constant along every maximal solution of $G$ . If there are states $z = (q, v)$ with $∥ z - (q_{0}, 0) ∥_{2}$ arbitrarily small and $E (z) \neq E (q_{0}, 0)$ , then $(q_{0}, 0)$ is not asymptotically stable. In particular, if $q_{0}$ is a strict local minimum of $U$ , then $(q_{0}, 0)$ is Lyapunov stable but not asymptotically stable.
(Scope.) A zero eigenvalue of $K$ alone gives no stability conclusion; clause 1 still applies to strict minima with degenerate Hessians. Clause 2 gives no nonlinear conclusion when $K$ has both a negative and a positive eigenvalue: the linearisation then also has purely imaginary eigenvalues and is not hyperbolic, so the instability theorem used in clause 3 does not apply. No global stability, no basin of attraction and no statement about the size of any neighbourhood is claimed.

Facts & Assumptions

Given: the chart data $M, U, G, E$ , the equilibrium $q_{0}$ with $\nabla U (q_{0}) = 0$ , and $K = D^{2} U (q_{0})$ ; the energy $E$ is continuous on $Ω \times R^{n}$ , being a composition and product of the continuous maps $M$ and $U$ ( $C^{k}$ Euclidean maps and diffeomorphisms).

[L1]

Energy stability and conservation (A strict local minimum of the potential gives a Lyapunov stable equilibrium of the mechanical system, by energy conservation, Energy is conserved along solutions of the mechanical system with a configuration-dependent kinetic metric): under the chart hypotheses, a strict local minimum $q_{0}$ of $U$ makes $(q_{0}, 0)$ a Lyapunov stable equilibrium of the mechanical field, and the energy $E$ is constant along every maximal solution of that field.

[L2]

Linearisation and the instability theorem (Linearisation of the mechanical system at an equilibrium: M(q_0) xi'' + D²U(q_0) xi = 0, and the small-oscillation quadratic model, A hyperbolic equilibrium whose linearisation has an eigenvalue with positive real part is unstable): the mechanical field is differentiable at $(q_{0}, 0)$ with $D G (q_{0}, 0) (δ q, δ v) = (δ v, - M (q_{0})^{- 1} K δ q)$ , and the linearised equation is $M (q_{0}) ξ^{''} + K ξ = 0$ . If a $C^{1}$ field has an equilibrium at $0$ whose derivative has no eigenvalue with real part $0$ and at least one eigenvalue with positive real part, then $0$ is not Lyapunov stable.

[L3]

Spectral and square-root tools (Positive definite symmetric matrices have positive definite square roots, and the congruence M^{-1/2} K M^{-1/2} preserves symmetry and positivity, Real spectral theorem: a self-adjoint endomorphism of a finite-dimensional real inner product space has an orthonormal eigenbasis): $M_{0} : = M (q_{0})$ has a symmetric positive definite square root $M_{0}^{1 / 2}$ and a symmetric positive definite inverse square root $M_{0}^{- 1 / 2}$ with $M_{0}^{- 1 / 2} M_{0} M_{0}^{- 1 / 2} = I_{n}$ ; every self-adjoint endomorphism of $R^{n}$ has an orthonormal basis of eigenvectors with real eigenvalues.

[L4]

Calculus of the exponential (The exponential function is smooth and $(\exp)^{'} = \exp$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ): the real exponential function is its own derivative, $(\exp)^{'} = \exp$ , and for a differentiable scalar function $α$ the composite $t \mapsto \exp (α (t))$ has derivative $α^{'} (t) \exp (α (t))$ .

[L5]

Matrix and vector conventions (The variational equation along an ODE solution): along the equilibrium the supplier's variational equation is $A^{'} = B A$ , $A (0) = I_{2 n}$ , where $B : = D G (q_{0}, 0)$ and $A$ takes values in $M_{2 n} (R)$ . For a constant initial deviation $u \in R^{2 n}$ , $Y (t) : = A (t) u$ obeys $Y^{'} = B Y$ , $Y (0) = u$ , by differentiating the finite sums defining matrix multiplication. We call $Y^{'} = B Y$ with specified initial vector its associated vector equation; a vector solution is not itself a normalized matrix solution.

Proof

technique · direct

1.1givenL1

Clause 1: the chart data satisfy the hypotheses of [L1] first clause (M a $C^{2}$ field of symmetric positive definite matrices, U of class $C^{2}$ , $q_{0}$ a strict local minimum, $\nabla U (q_{0}) = 0$ ); [L1] therefore gives Lyapunov stability of the equilibrium $(q_{0}, 0)$ of the mechanical field.

1.2givenL3algebra

Clause 2, spectral reduction: put $M_{0} : = M (q_{0})$ and $S : = M_{0}^{- 1 / 2} K M_{0}^{- 1 / 2}$ . By [L3] $M_{0}^{- 1 / 2}$ is symmetric, and $K$ is symmetric; hence $S^{T} = (M_{0}^{- 1 / 2})^{T} K^{T} (M_{0}^{- 1 / 2})^{T} = S$ , so $S$ is self-adjoint and [L3] second clause gives an orthonormal basis $u_{1}, \dots, u_{n}$ of $R^{n}$ with $S u_{i} = λ_{i} u_{i}$ , $λ_{i} \in R$ . If $K$ has a negative eigenvalue, then some $λ_{j} < 0$ : otherwise $λ_{i} \geq 0$ for all $i$ and, for any nonzero $e$ with $K e = η e$ , $η < 0$ , the vector $c : = M_{0}^{1 / 2} e$ is nonzero and satisfies $c^{T} S c = e^{T} M_{0}^{1 / 2} M_{0}^{- 1 / 2} K M_{0}^{- 1 / 2} M_{0}^{1 / 2} e = η ∥ e ∥_{2}^{2} < 0$ , while expansion in the orthonormal eigenbasis gives $c^{T} S c = \sum_{i} λ_{i} c_{i}^{2} \geq 0$ , a contradiction.

1.3givenL2L3algebra

Clause 3, hyperbolicity: assume $K$ negative definite. For $v \neq 0$ , $v^{T} S v = (M_{0}^{- 1 / 2} v)^{T} K (M_{0}^{- 1 / 2} v) < 0$ , so $S$ is negative definite; hence every eigenvalue of $S$ is negative. Let $ρ \in C$ be a complex eigenvalue of $A : = D G (q_{0}, 0)$ with eigenvector $ζ = (a, b) \neq 0$ . By [L2] clause 1, $b = ρ a$ and $- M_{0}^{- 1} K a = ρ b = ρ^{2} a$ ; if $a = 0$ then $b = 0$ , so $a \neq 0$ , and with $c : = M_{0}^{1 / 2} a \neq 0$ the identity $S c = - ρ^{2} c$ holds because $S c = M_{0}^{- 1 / 2} K M_{0}^{- 1 / 2} M_{0}^{1 / 2} a = M_{0}^{- 1 / 2} K a = - ρ^{2} M_{0}^{1 / 2} a = - ρ^{2} c$ . Thus $- ρ^{2}$ is an eigenvalue of $S$ ; as $S$ is negative definite, $- ρ^{2} < 0$ , that is $ρ^{2} > 0$ , so $ρ$ is real and nonzero. Every complex eigenvalue of $A$ is therefore real and nonzero.

2.1step 1.2L2L4L5algebra

Clause 2, the growing solution: fix $j$ with $λ_{j} < 0$ , put $κ : = \sqrt{- λ_{j}} > 0$ and $w : = M_{0}^{- 1 / 2} u_{j} \neq 0$ . Then $K w = λ_{j} M_{0} w$ : indeed $K w = K M_{0}^{- 1 / 2} u_{j} = M_{0}^{1 / 2} M_{0}^{- 1 / 2} K M_{0}^{- 1 / 2} u_{j} = M_{0}^{1 / 2} S u_{j} = λ_{j} M_{0}^{1 / 2} u_{j} = λ_{j} M_{0} w$ . Define $Y (t) : = e^{κ t} (w, κ w) \in R^{2 n}$ . By [L4], $Y_{1}^{'} (t) = κ e^{κ t} w = Y_{2} (t)$ and $Y_{2}^{'} (t) = κ^{2} e^{κ t} w = - λ_{j} e^{κ t} w = - e^{κ t} M_{0}^{- 1} K w = - M_{0}^{- 1} K Y_{1} (t)$ ; by clause 1 of [L2] and the convention [L5], the map $Y$ therefore solves the associated vector equation $Y^{'} = D G (q_{0}, 0) Y$ with $Y (0) = (w, κ w)$ . Since $∥ Y (t) ∥_{2} = e^{κ t} ∥ (w, κ w) ∥_{2}$ is unbounded as $t \to \infty$ , the equilibrium $(q_{0}, 0)$ is linearly unstable.

2.2step 1.3givenL2L3algebra

Clause 3, instability: by [L3] second clause and the negativity of all eigenvalues of $S$ , choose any eigenvalue $λ_{j} < 0$ of $S$ with eigenvector $u_{j}$ and put $κ : = \sqrt{- λ_{j}} > 0$ , $w : = M_{0}^{- 1 / 2} u_{j}$ . Then $A (w, κ w) = (κ w, - M_{0}^{- 1} K w) = (κ w, - λ_{j} w) = (κ w, κ^{2} w) = κ (w, κ w)$ , so $A$ has the positive real eigenvalue $κ$ . By step 1.3 no eigenvalue of $A$ has real part $0$ . The associated field $G$ is $C^{1}$ in the state near the equilibrium because $M$ is $C^{2}$ and $U$ is $C^{2}$ (The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart), and $G (q_{0}, 0) = 0$ . On the translated open state domain define $F (y) : = G ((q_{0}, 0) + y)$ , so $F (0) = 0$ and $D F (0) = A$ ; translation preserves Euclidean distances and maximal-solution intervals. Thus [L2] second clause applies to $F$ and shows that $(q_{0}, 0)$ is not Lyapunov stable.

3.1step 1.1givenL1algebra∎

Clause 4: [L1] second clause gives $E (z (t)) = E (z (0))$ along every maximal solution. Suppose $(q_{0}, 0)$ were asymptotically stable with attracting radius $δ_{0} > 0$ and let $z_{0}$ with $∥ z_{0} - (q_{0}, 0) ∥_{2} < δ_{0}$ and $E (z_{0}) \neq E (q_{0}, 0)$ , which exists by the clause's hypothesis. Its maximal solution is defined for all $t \geq 0$ and converges to $(q_{0}, 0)$ ; by the continuity of $E$ recorded in the Given, $E (z (t)) \to E (q_{0}, 0)$ , while $E (z (t)) = E (z_{0})$ constantly, so $E (z_{0}) = E (q_{0}, 0)$ , a contradiction. Hence $(q_{0}, 0)$ is not asymptotically stable. If $q_{0}$ is a strict local minimum with radius $r_{0} > 0$ , then in every radius $r > 0$ there is $q \in Ω$ with $0 < ∥ q - q_{0} ∥_{2} < \min (r, r_{0})$ (use $n \geq 1$ and openness); the state $z_{0} : = (q, 0)$ satisfies $E (z_{0}) = U (q) > U (q_{0}) = E (q_{0}, 0)$ , so such states accumulate at the equilibrium; combining clause 1 with the first part of this clause gives a Lyapunov stable but not asymptotically stable equilibrium.

Remarks

Why clause 3 needs negative definiteness. The instability theorem of [L2] requires the linearisation to be hyperbolic. Strict concavity alone need not give a negative definite Hessian: $U (q) = - q^{4}$ is strictly concave but has $U^{''} (0) = 0$ . In the rate formulas, $λ_{i}$ denotes an eigenvalue of $S = M_{0}^{- 1 / 2} K M_{0}^{- 1 / 2}$ from step 1.2, not of $K$ . Negative and positive eigenvalues of $S$ give respectively real rates $\pm \sqrt{- λ_{j}}$ and purely imaginary rates $\pm i \sqrt{λ_{k}}$ ; a mixed-sign Hessian therefore leaves centre directions, which the cited hyperbolic theorem does not cover. Clause 2 is the linear statement in that case, and clause 5 records the scope.
Comparison with the mathematical suppliers. Clause 1 is exactly A strict local minimum of the potential gives a Lyapunov stable equilibrium of the mechanical system, by energy conservation; clause 3 is A hyperbolic equilibrium whose linearisation has an eigenvalue with positive real part is unstable applied after the eigenvalue computations of steps 1.3 and 2.2; clause 4 is the physical reading of Energy is conserved along solutions of the mechanical system with a configuration-dependent kinetic metric and is the reason A strict local minimum of the potential gives a Lyapunov stable equilibrium of the mechanical system, by energy conservation asserts no asymptotic stability.
Units and coordinates. $[E_{p h y s}] = μ ℓ^{2} τ^{- 2}$ (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework) and the physical rate corresponding to $κ$ is $κ / t_{r e f}$ , of dimension $τ^{- 1}$ . State norms apply to dimensionless numerical coordinates; fixed positive reference scales preserve the topology and stability predicates. At an equilibrium of a $C^{1}$ field, a $C^{2}$ chart transition gives similar derivative matrices, as Stability vocabulary of the model: Lyapunov, asymptotic and orbital stability, linear stability of equilibria and map fixed points, and hyperbolicity of equilibria and fixed points clause (vi)(b) records.

Energy stability and instability of the mechanical model at an equilibrium: the strict minimum is Lyapunov stable, a negative Hessian eigenvalue gives linear instability, a negative definite potential Hessian gives nonlinear instability, and exact energy conservation rules out attraction

Statement

Facts & Assumptions

Proof

Remarks

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