Finite-slab geometric Cauchy stability — Examples

Lift the circle map to $F_{ϵ}(x)=x+ϵ\sin x$ on $\mathbb{R}$. By [F8], repeated differentiation makes $\sin$ smooth, and its first derivative gives $F_{ϵ}^{\mathrm{\prime }}(x)=1+ϵ\cos x\ge 1-\mathrm{\mid }ϵ\mathrm{\mid }>0$. The mean-value theorem [F11] makes $F_{ϵ}$ strictly increasing. Periodicity [F9] gives $F_{ϵ}(x+2\pi )=F_{ϵ}(x)+2\pi$, and boundedness [F10] gives $F_{ϵ}(x)-x$ bounded; hence its limits at the two ends of $\mathbb{R}$ are $\pm \mathrm{\infty }$. For any real target, choose endpoints with the target between their images and apply [F12]; strict increase gives uniqueness of the preimage. Thus $F_{ϵ}$ is bijective. Its derivative never vanishes, so [F4] makes the inverse smooth locally at every point, hence globally smooth. The translation identity makes the map and inverse descend to smooth inverse maps on the circle in [F14]; extending by the identity in the other coordinates gives a bijective smooth map with smooth inverse, hence the diffeomorphism ${\mathrm{\Phi }}_{ϵ}$ by [F13].

In these coordinates $d{\mathrm{\Phi }}_{ϵ}$ is diagonal with entries $1$, $1+ϵ\cos x^1$, $1$, $1$. Applying [F3] to the diagonal bilinear form gives $({\mathrm{\Phi }}_{ϵ}^{\ast }g)=-(d{x}^0{)}^2+L^2((1+ϵ\cos x^1{)}^2(d{x}^1{)}^2+(d{x}^2{)}^2+(d{x}^3{)}^2)\mathrm{.}$ At $x^1=0$, $(1+ϵ{)}^2\mathrm{\ne }1$ because $0<\mathrm{\mid }ϵ\mathrm{\mid }<1$, so the component arrays differ. Since $1+ϵ\cos x^1\ge 1-\mathrm{\mid }ϵ\mathrm{\mid }>0$, the displayed diagonal matrix has one negative and three positive entries and is Lorentzian by [F5].

The curve $\gamma (\lambda )=(\lambda ,x_{\ast })$ has tangent ${\mathrm{\partial }}_0$ and is future timelike for $g$ with $g(\dot{\gamma },\dot{\gamma })=-1$. Since ${\mathrm{\Phi }}_{ϵ}$ fixes $x^0$ and acts only on spatial coordinates, ${\mathrm{\Phi }}_{ϵ}\circ \gamma$ is also a unit-coordinate-time axis and $g(({\mathrm{\Phi }}_{ϵ}\circ \gamma {)}^{\mathrm{\prime }},({\mathrm{\Phi }}_{ϵ}\circ \gamma {)}^{\mathrm{\prime }})=-1$. By [F3], $({\mathrm{\Phi }}_{ϵ}^{\ast }g)(\dot{\gamma },\dot{\gamma })=g(({\mathrm{\Phi }}_{ϵ}\circ \gamma {)}^{\mathrm{\prime }},({\mathrm{\Phi }}_{ϵ}\circ \gamma {)}^{\mathrm{\prime }})=-1$. Formula [F6] therefore gives proper time $c^{-1}{\int }_0^{\mathrm{\ell }}1d\lambda =\mathrm{\ell }\mathrm{/}c$ in seconds on both curves.

The proper-time value is one of the finite-domain geometric observables treated in [F7]; here its equality is exact. The metric component change in step 2.1 is compatible with that equality because [F1] and [F2] define the analytic estimate in one fixed gauge, whereas [F3] compares the pulled-back geometric field. Thus the coordinate component difference alone is not a difference of the corresponding proper-time observable.

The coefficients in the Statement are smooth in $(t,x,z)$, each $A^{\mu }$ is symmetric, and $A^0=1$ is strictly positive. Thus [F1] makes this a symmetric-hyperbolic system. Its equation reduces to ${\mathrm{\partial }}_{t}z=2tz$ and contains no spatial derivatives.

By [F3] and the chain rule, $z_{\delta }(t,x)=\delta \exp (t^2)$ satisfies ${\mathrm{\partial }}_t{z}_{\delta }=2t\delta \exp (t^2)=2t{z}_{\delta }$ and $z_{\delta }(0,x)=\delta$. The zero solution corresponds to $\delta =0$. For each finite $T$, the function $\delta \exp (t^2)$ and its time derivative are bounded on $[0,T]$; its values lie between the endpoint values $\delta$ and $\delta \exp (T^2)$, hence in a compact interval. Since it is spatially constant, it has finite $H^s$ norm for every integer $s$. Its state and time-derivative bounds may depend on the selected finite $T$, as permitted by the finite-slab estimate [F5].

The difference from zero is exactly $\delta \exp (t^2)$. Homogeneity of the Sobolev norm on the spatially constant profile gives $\frac{\mathrm{\parallel }{z}_{\delta }(T)-z_0(T){\mathrm{\parallel }}_{H^{s-1}}}{\mathrm{\parallel }{z}_{\delta }(0)-z_0(0){\mathrm{\parallel }}_{H^{s-1}}}=\frac{\mathrm{\mid }\delta \mathrm{\mid }\exp (T^2)\mathrm{\parallel }\mathbf{1}{\mathrm{\parallel }}_{H^{s-1}}}{\mathrm{\mid }\delta \mathrm{\mid }\mathrm{\parallel }\mathbf{1}{\mathrm{\parallel }}_{H^{s-1}}}=\exp (T^2)\mathrm{.}$ Suppose one constant $C>0$ worked for all slab lengths in the displayed exponential estimate. Choose $T$ large enough that $y:=T^2-CT>C$. Then $y>0$, and [F2] gives $\exp (y)\ge 1+y>C$. By [F4], $\exp (T^2)=\exp (CT+y)=\exp (CT)\exp (y)>C\exp (CT),$ since [F2] also gives $\exp (CT)>0$. This contradicts that uniform estimate.

For each fixed $T$, the exact solution map is multiplication by $\exp (t^2)$ and is Lipschitz on $[0,T]$ with finite constant $\exp (T^2)$; this is consistent with [F5] and [F6], whose constants and conclusions are attached to a fixed finite slab. The calculation establishes only that finite-slab continuity alone does not supply a $T$-independent exponential constant for general symmetric-hyperbolic systems.