An external-source singularity reached in finite time: the inverse-square external model has a motion whose maximal interval is bounded above

Force from the potential: fix $x\mathrm{\ne }0$ and a coordinate index $i\in \{1,2,3\}$, and let $e_i$ be the corresponding standard coordinate vector. For $\mathrm{\mid }h\mathrm{\mid }<\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2$, [L3] gives $\mathrm{\mid }x+h{e}_i\mathrm{\mid }\ge \mathrm{\mid }x\mathrm{\mid }-\mathrm{\mid }h\mathrm{\mid }>0$, so the difference quotient is defined; also $\mathrm{\mid }x+h{e}_i\mathrm{\mid }\to \mathrm{\mid }x\mathrm{\mid }>0$. Rationalizing the difference of squares gives $\frac{\mathrm{\mid }x+h{e}_i\mathrm{\mid }-\mathrm{\mid }x\mathrm{\mid }}{h}=\frac{2x_i+h}{\mathrm{\mid }x+h{e}_i\mathrm{\mid }+\mathrm{\mid }x\mathrm{\mid }}\to \frac{x_i}{\mathrm{\mid }x\mathrm{\mid }}$. Therefore ${\mathrm{\partial }}_{i}U(x)={\lim }_{h\to 0}\frac{-k\mathrm{/}\mathrm{\mid }x+h{e}_i\mathrm{\mid }+k\mathrm{/}\mathrm{\mid }x\mathrm{\mid }}{h}={\lim }_{h\to 0}\frac{k(\mathrm{\mid }x+h{e}_i\mathrm{\mid }-\mathrm{\mid }x\mathrm{\mid })}{h\mathrm{\mid }x+h{e}_i\mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }}=\frac{k{x}_i}{\mathrm{\mid }x{\mathrm{\mid }}^3}\mathrm{.}$ By the definition of the Euclidean gradient in Restricted interaction models: pair decompositions, antisymmetry and centrality options, potential forces and the gravitational branch clause (iii), $-\mathrm{\nabla }U(x)=-kx\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^3$; this is the model's force map, continuous and locally Lipschitz on $Q$ by [L1] and Standing regularity of the page's force models checked: uniform field, linear drag and the inverse-square model on its singularity-excluded domain. Its dimension is $[k]\mathrm{/}[x{]}^2=\mu \mathrm{\ell }{\tau }^{-2}$, as required by [L5].

Set $\kappa :=k\mathrm{/}m>0$, let $r_0>0$ have dimension $\mathrm{\ell }$, and define $T:=\frac{\sqrt{2}{r}_0^{3\mathrm{/}2}}{3\sqrt{\kappa }},v_0:=-\sqrt{\frac{2\kappa }{r_0}}\mathrm{.}$ Then $[T]=\tau$ and $[v_0]=\mathrm{\ell }{\tau }^{-1}$ by [L5], and the definitions give $T^2=2r_0^3\mathrm{/}(9\kappa )$ and $2r_0\mathrm{/}(3T)=\sqrt{2\kappa \mathrm{/}{r}_0}=-v_0$. Take the initial datum $x(0)=r_0{e}_1$, $\dot{x}(0)=v_0{e}_1$.

For $t<T$ put $s(t):=1-t\mathrm{/}T>0$, let $q(t)$ be the unique positive cube root of $s(t{)}^2$, and set $r(t):=r_{0}q(t)$ and $x(t):=r(t)e_1$. This defines $q$ by [L6]. The cube-root map is continuous on $[0,\mathrm{\infty })$: for nonnegative $a,b$, if ${\alpha }^3=a$ and ${\beta }^3=b$, then $\mathrm{\mid }\alpha -\beta {\mathrm{\mid }}^3\le \mathrm{\mid }a-b\mathrm{\mid }$, since ${\alpha }^2+\alpha \beta +{\beta }^2\ge (\alpha -\beta {)}^2$; applying this to $a=s(t+h{)}^2$ and $b=s(t{)}^2$ gives $q(t+h)\to q(t)$. For $h\to 0$ with $t+h<T$, rationalizing and using $s(t+h)=s(t)-h\mathrm{/}T$ gives $\frac{q(t+h)-q(t)}{h}=\frac{(s(t+h{)}^2-s(t{)}^2)\mathrm{/}h}{q(t+h{)}^2+q(t+h)q(t)+q(t{)}^2}⟶-\frac{2s(t)}{3Tq(t{)}^2},$ so $q^{\mathrm{\prime }}(t)=-2s(t)\mathrm{/}(3Tq(t{)}^2)$. Differentiating this identity by [L2] and using $s^{\mathrm{\prime }}=-1\mathrm{/}T$ and $q^3=s^2$ gives $q^{\mathrm{\prime }\mathrm{\prime }}(t)=-2\mathrm{/}(9T^{2}q(t{)}^2)$. Thus $x$ is $C^2$ on $(-\mathrm{\infty },T)$ and $\dot{r}(t)=-\frac{2r_{0}s(t)}{3Tq(t{)}^2},\ddot{r}(t)=-\frac{2r_0}{9T^{2}q(t{)}^2}\mathrm{.}$ Moreover $q(t)\to 0$ as $t\uparrow T$: for each $\epsilon >0$, $q(t)\ge \epsilon$ would imply $s(t{)}^2=q(t{)}^3\ge {\epsilon }^3$, contradicting $s(t{)}^2\to 0$ once $t$ is sufficiently close to $T$.

The equation holds: from $T^2=2r_0^3\mathrm{/}(9\kappa )$ one has $2r_0\mathrm{/}(9T^2)=\kappa \mathrm{/}{r}_0^2$, so step 1.3 gives $\ddot{r}(t)=-\kappa \mathrm{/}(r_0^{2}q(t{)}^2)=-\kappa \mathrm{/}r(t{)}^2$. Hence $\ddot{x}(t)=-\kappa e_1\mathrm{/}r(t{)}^2=-k{e}_1\mathrm{/}(mr(t{)}^2)$ and $F(x(t))=-kr(t)e_1\mathrm{/}r(t{)}^3=-k{e}_1\mathrm{/}r(t{)}^2$, so $m\ddot{x}(t)=F(x(t))$ for every $t<T$. Also $q(0)=1$, $q^{\mathrm{\prime }}(0)=-2\mathrm{/}(3T)$, and step 1.2 gives $x(0)=r_0{e}_1$ and $\dot{x}(0)=-(2r_0\mathrm{/}(3T))e_1=v_0{e}_1$. Since $q(t)>0$ for $t<T$, $x(t)\in Q$ and $t\in I_0=\mathbb{R}$ throughout; by [L1], $x$ is a motion through the datum on $(-\mathrm{\infty },T)$.

No extension past $T$: suppose a motion $\tilde{x}$ with the same datum were defined on an interval containing some $t_{\ast }>T$. By the uniqueness clause of [L1] it agrees with $x$ on the intersection of their intervals; that intersection contains $[0,T)$ because the interval of $\tilde{x}$ contains both $0$ and $t_{\ast }$. It is continuous at $T$, and step 1.3 gives $q(t)\to 0$ as $t\uparrow T$, so $\tilde{x}(T)={\lim }_{t\uparrow T}x(t)=0$. But a motion's values must lie in $Q={\mathbb{R}}^3\setminus \{0\}$ at every time of its interval, a contradiction. Thus no motion through the datum extends past $T$, and by the maximality clause of [L1] the maximal motion is $x$ on $(-\mathrm{\infty },T)$, with finite right endpoint $b=T$.

Escape consistent with the breakdown criterion: as $t\uparrow T$, $q(t)\to 0$ while $\mathrm{\mid }\dot{r}(t)\mathrm{\mid }=2r_{0}s(t)\mathrm{/}(3Tq(t{)}^2)\to +\mathrm{\infty }$, since $(s(t)\mathrm{/}q(t{)}^2{)}^2=1\mathrm{/}q(t)\to +\mathrm{\infty }$ and the quotient is positive. Thus the state $(x(t),\dot{x}(t))$ leaves every compact subset of $U=Q\times {\mathbb{R}}^3$ near $T$ in both ways predicted by Maximal Newtonian motions and the compact-continuation criterion: finite-time breakdown requires escape from every compact subset of the time-state domain clause 4: its velocity is unbounded and its position approaches the excluded configuration. The refuted all-time-existence statement is false for this model.