Event location and hybrid trajectory error — Examples

Start immediately after a bounce with upward velocity $v_0>0$. Under $q^{\mathrm{\prime }}=v$, $v^{\mathrm{\prime }}=-g$, the next impact occurs after $2v_0\mathrm{/}g$, and the incoming impact velocity is $-v_0$. Its guard derivative is $q^{\mathrm{\prime }}=v^{-}=-v_0<0$, so it has the declared downward orientation; the reset produces upward velocity $e{v}_0$ and does not immediately re-trigger that oriented guard.

Repeating the calculation, the time between the $k$th and $(k+1)$st impacts is $2e^k{v}_0\mathrm{/}g$ for $k=0,1,2,\dots$. Every such interval is positive, so the impacts are distinct; at each one the incoming guard derivative is $q^{\mathrm{\prime }}=v^{-}=-e^k{v}_0\mathrm{\ne }0$, so each crossing is transverse and has the declared orientation.

The impact times are ${\tau }_n={\sum }_{k=0}^{n-1}2e^k{v}_0\mathrm{/}g$ for $n\ge 1$, and converge to $T_Z=2v_0\mathrm{/}(g(1-e))<\mathrm{\infty }$. Concatenating the exact mode solutions and resets defines a hybrid execution on the bounded half-open interval $[0,T_Z)$: every $t<T_Z$ lies before only finitely many impacts, while the event times ${\tau }_n$ accumulate at the excluded endpoint. It therefore has infinitely many transverse impacts in a bounded interval contained in its time domain, refuting the statement. The event separations tend to zero, so the finite separated-event theorem does not apply.

In mode $I$, the exact solution is $x(t)=-1+t$, so both guards first vanish at $t=1$ and have derivative $1>0$ there. Thus both transitions are enabled simultaneously.

Completing the specification with priority $g_1\prec g_2$ selects transition $g_1$ and the continuation in mode $A$ is $x(t)=0$ for $t\ge 1$. Completing it instead with priority $g_2\prec g_1$ selects transition $g_2$ and the continuation in mode $B$ is $x(t)=t-1$ for $t\ge 1$. Each completed system has a fixed priority and determines one continuation, but the incomplete specification alone does not determine which completion was intended.

The exact trajectory is $x(t)=0$, so the exact guard is $h(t)=(t-1{)}^2$. It has an interior root at $\tau =1$ and $h^{\mathrm{\prime }}(1)=0$. The guard does not cross either declared orientation, so this is a grazing guard contact, not an enabled hybrid event under [F1].

For any $\eta >0$, take the dense reconstruction $U_{\eta }(t)=-\eta$. Its uniform state error is $\eta$, while its guard $g(t,U_{\eta }(t))=(t-1{)}^2-\eta$ has zero residual at $\widehat{\tau }=1+\sqrt{\eta }$. Thus $\epsilon =0$ and $\mathrm{\mid }\widehat{\tau }-\tau \mathrm{\mid }=\sqrt{\eta }$.

If a fixed $C$ gave the refuted root-location bound, it would require $\sqrt{\eta }\le C\eta$ for every $\eta >0$, equivalently $1\mathrm{/}\sqrt{\eta }\le C$. The left side is unbounded as $\eta \downarrow 0$, so no such constant exists. Here $h^{\mathrm{\prime }}(1)=0$, so the positive derivative margin required by [F2] is absent; the calculation concerns sensitivity of a guard contact, not the time of an enabled event.

The exact incoming solution is $y(t)=(1-t^2,-2t)$, so its guard $h(t)=1-t^2$ has the unique root $\tau =1$ on the bracket. For $t\in [1,51\mathrm{/}50]$, $-h^{\mathrm{\prime }}(t)=2t\ge 2$, so the downward crossing is transverse with margin $\nu =2$.

The dense guard is $1-t^2+1\mathrm{/}25$, whose root in the bracket is $\widehat{\tau }=\sqrt{26\mathrm{/}25}$ and whose residual there is $\epsilon =0$. For $x\ge 0$, both sides are nonnegative and $(1+x\mathrm{/}2{)}^2-(1+x)=x^2\mathrm{/}4\ge 0$, so $\sqrt{1+x}\le 1+x\mathrm{/}2$. Taking $x=1\mathrm{/}25$ gives $0\le \widehat{\tau }-\tau \le 1\mathrm{/}50$.

On the bracket, $\mathrm{\parallel }U(t)-y(t){\mathrm{\parallel }}_2=1\mathrm{/}25$. Since $g(q,v)=q$, $\mathrm{\mid }g(q_1,v_1)-g(q_2,v_2)\mathrm{\mid }=\mathrm{\mid }{q}_1-q_2\mathrm{\mid }\le \mathrm{\parallel }(q_1,v_1)-(q_2,v_2){\mathrm{\parallel }}_2$, so $L_g=1$; the reported guard residual is zero. Apply [F2] with $\nu =2$ to obtain $\mathrm{\mid }\widehat{\tau }-\tau \mathrm{\mid }\le (1\mathrm{/}25)\mathrm{/}2=1\mathrm{/}50$, agreeing with the explicit bound in step 2.1.

The reset is defined on all of ${\mathbb{R}}^2$, is $L_R=1\mathrm{/}2$-Lipschitz in state, and has $L_{R,t}=0$. On the incoming bracket, $\mathrm{\parallel }U(t)-y(t){\mathrm{\parallel }}_2=1\mathrm{/}25$ and $\mathrm{\parallel }{f}^{-}(t,y(t)){\mathrm{\parallel }}_2=\sqrt{4t^2+4}<3$. Step 3.1 gives $\mathrm{\mid }\widehat{\tau }-\tau \mathrm{\mid }\le 1\mathrm{/}50$. The outgoing field $(q,v)\mapsto (v,-2)$ is globally $1$-Lipschitz in state. The exact outgoing path through $\widehat{\tau }$ and the reported reset state lie in $[-1\mathrm{/}10,1\mathrm{/}10]\times [4\mathrm{/}5,6\mathrm{/}5]$, where the field norm is less than $3$; by local existence and continuity both outgoing solutions remain in the bounded convex rectangle $[-1\mathrm{/}5,1\mathrm{/}5]\times [3\mathrm{/}4,5\mathrm{/}4]$ for a short common forward interval. Thus the reset and outgoing tube hypotheses of [F3] hold with $M^{-}=M^{+}=3$ and $L^{+}=1$. By [F3], the aligned post-reset error is at most $\frac{1}{2}\cdot \frac{1}{25}+(\frac{1}{2}\cdot 3+3)\frac{1}{50}=11\mathrm{/}100$.

Directly, the exact outgoing state at $\widehat{\tau }=1+\mathrm{\Delta }$ is $(\mathrm{\Delta }-{\mathrm{\Delta }}^2,1-2\mathrm{\Delta })$ and the reported reset state is $(0,1+\mathrm{\Delta })$. Their difference has components bounded by $1\mathrm{/}50$ and $3\mathrm{/}50$, so its Euclidean norm is at most $\sqrt{(1\mathrm{/}50{)}^2+(3\mathrm{/}50{)}^2}<11\mathrm{/}100$. This verifies the reset estimate for this instance.

The incoming-mode continuation is $x(t)=t$, but the guard is independent of $x$, so $h(t)=(t-1\mathrm{/}3)(t-2\mathrm{/}3)$. Its endpoint values are $h(0)=h(1)=2\mathrm{/}9>0$. A rule that checks only endpoint signs therefore reports no crossing. The roots $1\mathrm{/}3$ and $2\mathrm{/}3$ are not dyadic, so neither lies on a midpoint boundary of the isolator's dyadic bisection of $[0,1]$.

The two roots are $1\mathrm{/}3$ and $2\mathrm{/}3$, and $h^{\mathrm{\prime }}(t)=2t-1$. Thus $h^{\mathrm{\prime }}(1\mathrm{/}3)=-1\mathrm{/}3$ is a downward crossing while $h^{\mathrm{\prime }}(2\mathrm{/}3)=1\mathrm{/}3$ is an upward crossing. Under the declared positive orientation, the first root is nontriggering and the second is enabled.

With exact polynomial range and derivative enclosures and a finite box budget sufficient for a complete result, the interval isolator in [F2] finds both roots and classifies their derivative signs. It cannot certify no event, because the root at $2\mathrm{/}3$ is enabled and is the earliest enabled event. Thus equal endpoint signs do not justify an event-absence conclusion; a certified absence report requires complete root isolation and orientation classification with no enabled root.