Two declared linear models worked through: the harmonic flow has zero finite-time growth in every direction, while the repulsive linear model has exact rates $\pm 1$ and the positive Lyapunov exponent $1$ at its equilibrium

Declared model data and normalization. Take one label with $Q:={\mathbb{R}}^3$ and declare physical reference data $m_0:=1\mathrm{k}\mathrm{g}$, ${\mathrm{\ell }}_0:=1\mathrm{m}$, $t_0:=1\mathrm{s}$. For ${\sigma }_1=-1$, ${\sigma }_2=1$ define physical force maps $F_{i,\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}:\mathbb{R}\times Q\times {\mathbb{R}}^3\to {\mathbb{R}}^3$ by $F_{i,\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}(T,\mathbf{X},\mathbf{V})={\sigma }_i(m_0\mathrm{/}{t}_0^2)\mathbf{X}$, with $T$ in seconds, $\mathbf{X}$ in metres, $\mathbf{V}$ in metres per second, and outputs in newtons. These are smooth globally state-Lipschitz force maps of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with physical mass $m_0$. Put $t=T\mathrm{/}{t}_0$, $\mathbf{q}=\mathbf{X}\mathrm{/}{\mathrm{\ell }}_0$, $\mathbf{v}=t_0\mathbf{V}\mathrm{/}{\mathrm{\ell }}_0$ and $E_0=m_0{\mathrm{\ell }}_0^2\mathrm{/}{t}_0^2$. Then $m_0\ddot{\mathbf{X}}=F_{i,\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}$ becomes ${\mathbf{q}}^{\mathrm{\prime }\mathrm{\prime }}={\sigma }_i\mathbf{q}$, with ${\mathbf{q}}^{\mathrm{\prime }}=\mathbf{v}$, since $\ddot{\mathbf{X}}=({\mathrm{\ell }}_0\mathrm{/}{t}_0^2){\mathbf{q}}^{\mathrm{\prime }\mathrm{\prime }}$. The normalized force maps are $F_i(t,\mathbf{q},\mathbf{v})=F_{i,\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}(t_{0}t,{\mathrm{\ell }}_0\mathbf{q},({\mathrm{\ell }}_0\mathrm{/}{t}_0)\mathbf{v})\mathrm{/}(m_0{\mathrm{\ell }}_0\mathrm{/}{t}_0^2)={\sigma }_i\mathbf{q}$, with normalized mass $m=1$. The physical potentials are $U_{i,\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}(\mathbf{X})=-{\sigma }_i{m}_0\parallel \mathbf{X}{\parallel }_2^2\mathrm{/}(2t_0^2)$; normalizing their energies by $E_0$ gives, in the constant-metric chart $\mathrm{\Omega }:={\mathbb{R}}^3$ with $M:=I_3$, they have the potentials $U_1(\mathbf{q}):=\frac{1}{2}\parallel \mathbf{q}{\parallel }_2^2$ and $U_2(\mathbf{q}):=-\frac{1}{2}\parallel \mathbf{q}{\parallel }_2^2$ of The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart. The associated fields and complete motion maps are listed below; ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$ denote the coordinate basis, explicitly ${\mathbf{e}}_j(k)=1$ for $k=j-1$ and $0$ otherwise, for $k\in \{0,1,2\}$ and $j\in \{1,2,3\}$. All state and tangent norms below apply to the dimensionless numerical coordinates $(\mathbf{q},\mathbf{v})$.

1. The harmonic model has field $G_1(\mathbf{q},\mathbf{v})=(\mathbf{v},-\mathbf{q})$ and motion map ${\mathrm{\Phi }}_t^1({\mathbf{q}}_0,{\mathbf{v}}_0)=({\mathbf{q}}_0\cos t+{\mathbf{v}}_0\sin t,-{\mathbf{q}}_0\sin t+{\mathbf{v}}_0\cos t)\mathrm{.}$
2. The repulsive linear model has field $G_2(\mathbf{q},\mathbf{v})=(\mathbf{v},\mathbf{q})$ and motion map ${\mathrm{\Phi }}_t^2({\mathbf{q}}_0,{\mathbf{v}}_0)=(\frac{1}{2}({\mathbf{q}}_0+{\mathbf{v}}_0)e^t+\frac{1}{2}({\mathbf{q}}_0-{\mathbf{v}}_0)e^{-t},\frac{1}{2}({\mathbf{q}}_0+{\mathbf{v}}_0)e^t-\frac{1}{2}({\mathbf{q}}_0-{\mathbf{v}}_0)e^{-t})\mathrm{.}$

For a state $a$ and a nonzero tangent vector $u$ with $D_a{\mathrm{\Phi }}_t(a)u\mathrm{\ne }0$, the finite-time tangent growth rate is $t^{-1}\log (\parallel D_a{\mathrm{\Phi }}_t(a)u{\parallel }_2\mathrm{/}\parallel u{\parallel }_2)$; when the forward lifetime is unbounded and its finite limit as $t\to \mathrm{\infty }$ exists, that limit is the Lyapunov exponent (Finite-time tangent growth rates and Lyapunov exponents of the model's flow: the rate per unit time of a transported deviation, with its existence convention and its limits clauses (i)-(ii)). Both displayed flows are complete, so this lifetime condition holds for each state.

Assertions to check. (i) Both displayed maps are the local flows of their fields; (ii) for the harmonic model $D_a{\mathrm{\Phi }}_t^1(a)$ is the orthogonal block rotation $\left(\begin{array}{cc}\cos t{I}_3& \sin t{I}_3\\ -\sin t{I}_3& \cos t{I}_3\end{array}\right),$ so every finite-time tangent growth rate vanishes and every existing Lyapunov exponent is $0$; (iii) for the repulsive model the rates along the directions $({\mathbf{e}}_1,-{\mathbf{e}}_1)$ and $({\mathbf{e}}_1,{\mathbf{e}}_1)$ are $-1$ and $+1$, so the Lyapunov exponent of the equilibrium along the unstable eigenvector is $1$.

• What is genuinely checked. The example verifies that the variational equation holds with the displayed derivative matrices and that the growth rate is invariant under multiplication of the deviation by nonzero scalars (the logarithm of a positive factor cancels against the norm ratio), so the rate is a function of the direction, not of the length of $u$. Both flows are complete on the declared chart; no domain-escape phenomenon occurs.

• The unstable direction. For the repulsive model the directions $({\mathbf{e}}_j,{\mathbf{e}}_j)$ and $({\mathbf{e}}_j,-{\mathbf{e}}_j)$ are exactly the eigendirections of $A=D{G}_2(\mathbf{0},\mathbf{0})=\left(\begin{array}{cc}0& I_3\\ I_3& 0\end{array}\right)$, with eigenvalues $+1$ and $-1$, respectively, so the rates $\pm 1$ are the eigenvalues of the linearisation, in agreement with Hyperbolic linearisation of an equilibrium of the model: exponential dichotomy of the linearised flow, local exponential asymptotic stability in the stable case, and instability in the saddle case clause 1 with $c=1$.