Finite-time linearization bound for the planar pendulum with explicit constants

Quadratic remainder: for $z=(\theta ,\omega )$ with $\mathrm{\parallel }z{\mathrm{\parallel }}_2\le 1\mathrm{/}2$, $F(z)-Az=(0,-\sin \theta )-(0,-\theta )=(0,\theta -\sin \theta ),$ so $\mathrm{\parallel }F(z)-Az{\mathrm{\parallel }}_2=\mathrm{\mid }\sin \theta -\theta \mathrm{\mid }\le \mathrm{\mid }\theta {\mathrm{\mid }}^3\mathrm{/}6\le \frac{1}{2}\mathrm{\mid }\theta {\mathrm{\mid }}^2\mathrm{/}6\le \frac{1}{12}\mathrm{\parallel }z{\mathrm{\parallel }}_2^2$ by [L2] and $\mathrm{\mid }\theta \mathrm{\mid }\le \mathrm{\parallel }z{\mathrm{\parallel }}_2\le 1\mathrm{/}2$; hence $C=1\mathrm{/}12$ is admissible.

Regularity and growth: $U={\mathbb{R}}^2$ is open and contains $\bar{B}(0,1\mathrm{/}2)$. The differentiable sine and coordinate functions make $F$ continuous. For $z=(\theta ,\omega )$ and $w=(\eta ,\nu )$, [L6] and [L4] give $\mathrm{\mid }\sin \theta -\sin \eta \mathrm{\mid }\le \mathrm{\mid }\theta -\eta \mathrm{\mid }$, hence $\mathrm{\parallel }F(z)-F(w){\mathrm{\parallel }}_2^2=(\omega -\nu {)}^2+(\sin \theta -\sin \eta {)}^2\le \mathrm{\parallel }z-w{\mathrm{\parallel }}_2^2$. Thus $F$ is globally $1$-Lipschitz and in particular locally Lipschitz in the state variable. Growth bound: if $\theta \mathrm{\ne }0$, the mean value theorem (The mean value theorem, as the case $g(x)=x$ of Cauchy's: for $f$ continuous on $[a,b]$ with $a<b$ and differentiable on $(a,b)$ there is $c\in (a,b)$ with $f(b)-f(a)=f^{\mathrm{\prime }}(c)(b-a)$) applied to $\sin$ between $0$ and $\theta$, with $\mathrm{\mid }\cos \mathrm{\mid }\le 1$ from [L4], gives $\mathrm{\mid }\sin \theta \mathrm{\mid }=\mathrm{\mid }\sin \theta -\sin 0\mathrm{\mid }\le \mathrm{\mid }\theta \mathrm{\mid }$; this inequality is immediate at $\theta =0$. Thus $\mathrm{\parallel }F(z){\mathrm{\parallel }}_2^2={\omega }^2+{\sin }^2\theta \le {\omega }^2+{\theta }^2=\mathrm{\parallel }z{\mathrm{\parallel }}_2^2$, so $L=1$ is admissible. With $\delta =1\mathrm{/}100$ and $T=1$, [L4] gives $\delta e^{LT}=e\mathrm{/}100<3\mathrm{/}100<1\mathrm{/}20<1\mathrm{/}2=r$, so the hypotheses of [L1] hold.

Substitute: by [L1] and [L3], for $0\le t\le 1$ the error is at most $\frac{1}{12}{\delta }^2{\int }_0^t{e}^{(1-s)}{e}^{2s}ds\le \frac{1}{12}{\delta }^2{\int }_0^1{e}^{(1-s)}{e}^{2s}ds$, since $e^{t-s}\le e^{1-s}$ for $s\le t\le 1$ and the resulting integrand is positive by [L4]. The addition formula gives $e^{(1-s)}{e}^{2s}=e^{1+s}$; since $0\le s\le 1$, monotonicity gives $e^{1+s}\le e^2<9$ by [L4]. Monotonicity of the integral [L5] therefore yields ${\int }_0^1{e}^{1+s}ds\le 9$, and the error is at most $\frac{1}{12}\cdot {10}^{-4}\cdot 9=\frac{3}{4}\cdot {10}^{-4}<{10}^{-4}$.