Cm Math Balances — Examples

Case (i): the momentum relation reads $v_1^{+}+v_2^{+}=0$, while restitution gives $v_2^{+}-v_1^{+}=e$. Adding and subtracting these equations gives the unique pair $v_1^{+}=-e\mathrm{/}2$, $v_2^{+}=e\mathrm{/}2$. Thus $J=(-1\mathrm{/}2-1)e=-3e\mathrm{/}2$, and the other momentum jump is $(1\mathrm{/}2+1)e=3e\mathrm{/}2=-J$. By [L1], $\mathrm{\Delta }K=\frac{1}{2}(1\mathrm{/}4+1\mathrm{/}4)-\frac{1}{2}(1+1)=1\mathrm{/}4-1=-3\mathrm{/}4$. All claimed values follow from these two equations and this direct finite sum.

Case (ii): restitution gives $v_2^{+}-v_1^{+}=2e$, and momentum gives $v_1^{+}+3v_2^{+}=2e$. Substitute $v_1^{+}=v_2^{+}-2e$ to obtain $4v_2^{+}=4e$, so uniquely $v_2^{+}=e$, $v_1^{+}=-e$. Then $J=(-1-2)e=-3e$ and the other momentum jump is $3(e-0)=3e=-J$. By [L1], $\mathrm{\Delta }K=\frac{1}{2}(1)+\frac{3}{2}(1)-\frac{1}{2}(4)=0$. This preserves the elastic zero-loss conclusion without invoking a conditional distributional result.

Case (iii): the scalar map $t\mapsto 2\mathrm{/}(2-t)$ is positive and smooth on an open neighbourhood of $[0,1]$, so the displayed $v$ is differentiable on all of $[0,1]$, endpoints included. The quotient and chain rules of [L3] give $v^{\mathrm{\prime }}(t)=(2\mathrm{/}(2-t{)}^2\left)\mathrm{/}\right(2\mathrm{/}(2-t))e=e\mathrm{/}(2-t)$. Hence $m(t)=2-t\ge 1$, $m^{\mathrm{\prime }}=-1=r_{\mathrm{i}\mathrm{n}}-r_{\mathrm{o}\mathrm{u}\mathrm{t}}$, and $m{v}^{\mathrm{\prime }}=e$. The product rule gives $(mv{)}^{\mathrm{\prime }}=-v+e=-w_{\mathrm{o}\mathrm{u}\mathrm{t}}=F+r_{\mathrm{i}\mathrm{n}}{w}_{\mathrm{i}\mathrm{n}}-r_{\mathrm{o}\mathrm{u}\mathrm{t}}{w}_{\mathrm{o}\mathrm{u}\mathrm{t}}$, since the inflow term is zero. Thus all hypotheses of [L2] are met on $I=[0,1]$, and its derived law is $m{v}^{\mathrm{\prime }}=-r_{\mathrm{o}\mathrm{u}\mathrm{t}}(w_{\mathrm{o}\mathrm{u}\mathrm{t}}-v)=e$.

The rocket has $v(0)=0$, $m(1)=1$ and $v(1)=\log 2e$. Let $\bar{v}$ instead be a differentiable velocity with $\bar{v}(0)=0$ obeying the naive law $(m\bar{v}{)}^{\mathrm{\prime }}=0$ on $[0,1]$. For every $0<t\le 1$, applying [L4] to $m\bar{v}$ on $[0,t]$ gives $m\bar{v}=0$, hence $\bar{v}=0$ because $m\ge 1$. This candidate fails the prescribed material balance already at $t=0$: $(m\bar{v}{)}^{\mathrm{\prime }}(0)=0$, whereas $-w_{\mathrm{o}\mathrm{u}\mathrm{t}}(0)=e\mathrm{\ne }0$. Moreover, for every $0<t\le 1$, $2\mathrm{/}(2-t)>1$, so [L3] gives $\log (2\mathrm{/}(2-t))>\log 1=0$ and $v(t)\mathrm{\ne }\bar{v}(t)$. Thus the readings differ as soon as $t>0$, with the original endpoint value and material-balance conclusion retained.

The quotient rule [L1] gives $x_{+}^{\mathrm{\prime }}(t)=(1-t{)}^{-2}=x_{+}(t{)}^2=F_{+}(t,x_{+}(t))$ on $[0,1)$ and $x_{+}(0)=1$, and $x_{-}^{\mathrm{\prime }}(t)=-(1+t{)}^{-2}=-x_{-}(t{)}^2=F_{-}(t,x_{-}(t))$ on $[0,\mathrm{\infty })$ and $x_{-}(0)=1$.

Local Lipschitz continuity of $F_{\pm }$: by [L1], $G_{\pm }$ are differentiable and continuous, with $G_{\pm }^{\mathrm{\prime }}(z)=\pm 2z$; consequently $F_{\pm }$ are continuous, since $(t,z)\to (t_0,z_0)$ implies $z\to z_0$ and $F_{\pm }(t,z)=G_{\pm }(z)$. Here a compact time-state cylinder means $C=T\times [a,b]$, where $T\subseteq \mathbb{R}$ is a compact interval and $a\le b$; this includes the one-dimensional closed-ball cylinders used in [L2] of Compact continuation for locally Lipschitz ODEs: a solution whose graph stays in a compact subset of the domain near a finite endpoint extends past it. Put $S:=[a,b]$. Set $R:=\max \{\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }b\mathrm{\mid }\}$, so $\mathrm{\mid }{G}_{\pm }^{\mathrm{\prime }}(z)\mathrm{\mid }\le 2R$ on $S$. For $(t,z),(t,w)\in C$ with the same time and $z\mathrm{\ne }w$, the mean value inequality [L3] on the interval between $z$ and $w$ gives $\mathrm{\mid }{F}_{\pm }(t,z)-F_{\pm }(t,w)\mathrm{\mid }=\mathrm{\mid }{G}_{\pm }(z)-G_{\pm }(w)\mathrm{\mid }\le 2R\mathrm{\mid }z-w\mathrm{\mid }$, since that interval lies in $S$; for $z=w$ the inequality is immediate. If $C$ is empty it holds with $L=0$. This verifies the local Lipschitz condition of Local Lipschitz continuity in the state variable, locally uniform in time and parameters on every such cylinder.

Non-extendability in (i): suppose a solution $\tilde{x}$ of the IVP $x^{\mathrm{\prime }}(t)=F_{+}(t,x(t))$, $x(0)=1$ existed on $[0,1+\epsilon )$ for some $\epsilon >0$. Both $\tilde{x}$ and $x_{+}$ solve the same IVP on the common interval $[0,1)$ containing $0$, and $F_{+}$ is continuous and locally Lipschitz in the state variable by step 1.2. The uniqueness clause of Picard-Lindelöf (Picard-Lindelöf local existence and uniqueness for first-order systems) gives $\tilde{x}=x_{+}$ on $[0,1)$, so $\tilde{x}(t)=1\mathrm{/}(1-t)\to +\mathrm{\infty }$ as $t\uparrow 1$. But $\tilde{x}$ is continuous [L1] on the compact interval $[0,1]$, so [L2] bounds $\mathrm{\mid }\tilde{x}\mathrm{\mid }$ there, a contradiction. Hence no extension past $b=1$ exists.

Escape in (i): let $K\subseteq D$ be compact and $\eta >0$. Put ${\eta }_0:=\min \{\eta ,1\mathrm{/}2\}>0$, so $(1-{\eta }_0,1)\subseteq (1-\eta ,1)$ and all times in the smaller interval lie in $[0,1)$. By [L2], $K$ is bounded, so there is $R>0$ with $\mathrm{\mid }z\mathrm{\mid }\le R$ for every $(t,z)\in K$. Choose $t\in (1-{\eta }_0,1)$ with $\frac{1}{1-t}>R$, which is possible since $\frac{1}{1-t}\to +\mathrm{\infty }$; then $x_{+}(t)>R$, so $(t,x_{+}(t))\mathrm{\notin }K$. This is the escape asserted by A non-extendable solution leaves every compact subset of the domain near its finite endpoint; an autonomous solution trapped in a compact subset of its state domain cannot break down there for this solution.

Trapping in (ii): $K_0=[0,1]\subseteq \mathbb{R}={\mathbb{R}}^1$ is closed and bounded, hence compact by [L2], and $0<x_{-}(t)\le 1$ for every $t\ge 0$, so $x_{-}(t)\in K_0$ throughout its domain. For any finite $b>0$, its restriction to $[0,b)$ is a solution trapped in $K_0$ near $b$; clause 2 of A non-extendable solution leaves every compact subset of the domain near its finite endpoint; an autonomous solution trapped in a compact subset of its state domain cannot break down there therefore extends that restriction past $b$. Thus no finite right endpoint can be a nonextendable endpoint for this solution, and the displayed formula gives its global extension.