Forces, initial data and Newtonian flow — Examples

Model 1: the constant map $z\mapsto m\mathbf{g}$ is continuous by [L3]; and for any two states $\parallel F(t,z)-F(t,z^{\mathrm{\prime }}){\parallel }_2=\parallel m\mathbf{g}-m\mathbf{g}{\parallel }_2=0$, so $L=0$ is a state-Lipschitz constant on every cylinder, independent of time. Hence model 1 satisfies [L1].

Model 2: the map $(t,x,v)\mapsto m\mathbf{g}-cv$ is a constant plus a scalar multiple of the linear map $v\mapsto v$, hence continuous by [L3]; and for states $z=(x,v)$, $z^{\mathrm{\prime }}=(x^{\mathrm{\prime }},v^{\mathrm{\prime }})$, $\parallel F(t,z)-F(t,z^{\mathrm{\prime }}){\parallel }_2=\parallel c(v^{\mathrm{\prime }}-v){\parallel }_2=c\parallel v-v^{\mathrm{\prime }}{\parallel }_2\le c\parallel z-z^{\mathrm{\prime }}{\parallel }_2$ by [L2], so $L=c$ is a state-Lipschitz constant on every cylinder and does not depend on time; in particular model 2 satisfies [L1].

Model 3, Lipschitz bound on an annulus: let $0<r\le R$ and let $x,y\in {\mathbb{R}}^3$ with $r\le \mid x\mid ,\mid y\mid \le R$. Then $\parallel \frac{x}{\mid x{\mid }^3}-\frac{y}{\mid y{\mid }^3}{\parallel }_2\le \frac{\parallel x-y{\parallel }_2}{\mid x{\mid }^3}+\mid y\mid \mid \mid x{\mid }^{-3}-\mid y{\mid }^{-3}\mid \le \frac{\parallel x-y{\parallel }_2}{r^3}+R\cdot \frac{3R^2}{r^6}\parallel x-y{\parallel }_2,$ where the first inequality inserts $y\mathrm{/}\mid x{\mid }^3$ and uses the triangle inequality, and the second uses $\mid \mid x{\mid }^{-3}-\mid y{\mid }^{-3}\mid \le 3R^2{r}^{-6}\mid \mid x\mid -\mid y\mid \mid \le 3R^2{r}^{-6}\parallel x-y{\parallel }_2$, the factorization $\mid v^3-u^3\mid =\mid v-u\mid (v^2+vu+u^2)$ applied to $u=\mid x\mid$, $v=\mid y\mid$ and the reverse triangle inequality of [L2]; multiplying by $k$ gives the constant $L=k(r^{-3}+3R^3{r}^{-6})$ for $F=-kx\mathrm{/}\mid x{\mid }^3$ on the annulus (the state variable enters only through $x$).

Model 3, continuity and local-Lipschitz on the domain: every nonempty compact subset $K$ of $Q={\mathbb{R}}^3\setminus \{0\}$ lies in an annulus $r\le \mid x\mid \le R$ with $r={\min }_K\mid x\mid >0$ and $R={\max }_K\mathrm{\mid }x\mathrm{\mid }<\mathrm{\infty }$ by [L2] and [L4] (the norm is continuous by its reverse triangle inequality, and its attained minimum is positive because $0\mathrm{\notin }K$); by step 1.3 the map is Lipschitz with the finite constant above on each such annulus, the empty compact set imposes no inequality, and small closed balls around each point of $Q$ avoid the external source, hence the map is locally Lipschitz, and in particular continuous, on $Q$; the time variable does not occur, so the constant is locally uniform in time. Model 3 therefore satisfies [L1] on its domain.

Model 3, no extension to the excluded point: for $x\mathrm{\ne }0$ one has $\parallel F(x){\parallel }_2=k\mid x{\mid }^{-2}$, so for every bound $M>0$ there is $\delta =\sqrt{k\mathrm{/}M}>0$ with $\parallel F(x){\parallel }_2>M$ whenever $0<\mid x\mid <\delta$; hence no continuous extension of $F$ to $x=0$ exists (Continuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$: the $\epsilon$-$\delta$ condition, its agreement with ${\lim }_{x\to c}f(x)=f(c)$ at a limit point, and continuity at an isolated point: a continuous extension would carry a sequence $x_n\to 0$ to a convergent image sequence, but the values are unbounded). The external-source singularity exclusion $Q={\mathbb{R}}^3\setminus \{0\}$ is therefore not cosmetic: it is what places model 3 inside the force branch at all.

Conclusion: models 1-3 satisfy the standing hypotheses of clause (iii) of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance, with the constants computed above; hence the associated first-order fields of The associated first-order system, motions, maximal motions and the Newtonian flow of a force model satisfy the Picard–Lindelöf hypotheses and the local existence, uniqueness and maximal-motion conclusions of Local existence and uniqueness for the Newtonian initial-value problem, with the quantitative Picard-Lindelöf cylinder, as well as the continuation criterion of Maximal Newtonian motions and the compact-continuation criterion: finite-time breakdown requires escape from every compact subset of the time-state domain, apply to each of them on its stated domain.

The datum's coordinate triples are $x_0=0\in {\mathbb{R}}^3$ and $v_0^{\mathrm{♭}}=v_0\cos \alpha e_1+v_0\sin \alpha e_3\in {\mathbb{R}}^3$; the constant-field model with $N=1$, $m_1=m$ and $\mathbf{g}=-g{e}_3\in {\mathbb{R}}^3$ is a force model of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance by The uniform-force model: explicit global motion of a label under a constant declared field, with uniqueness, dimensions and frame covariance claim 1, and by its claim 2 the unique maximal motion through $(x_0,v_0^{\mathrm{♭}})$ is $x(t)=v_0^{\mathrm{♭}}t-\frac{1}{2}g{t}^2{e}_3$, which is the displayed formula; its maximal interval is all of $\mathbb{R}$.

Direct verification of the formula, independently of [L1]: differentiating componentwise by [L2] gives $\dot{x}(t)=v_0\cos \alpha e_1+(v_0\sin \alpha -gt)e_3$ and $\ddot{x}(t)=-g{e}_3=F(t,x(t),\dot{x}(t))\mathrm{/}m$; the initial values are $x(0)=0$ and $\dot{x}(0)=v_0(\cos \alpha e_1+\sin \alpha e_3)$, so the curve is a motion of the model by The associated first-order system, motions, maximal motions and the Newtonian flow of a force model clause (c), and by Local existence and uniqueness for the Newtonian initial-value problem, with the quantitative Picard-Lindelöf cylinder clause 2 it is the only motion through the datum on any interval containing $0$.

Planarity: the coefficient of $e_2$ in $x(t)$ is $0$ for every $t$, so the coordinate trajectory lies in the plane spanned by $e_1$ and $e_3$; equivalently $⟨x(t),e_2⟩=0$ for all $t$ for the standard Euclidean inner product on ${\mathbb{R}}^3$. Its spatial displacement therefore lies in $\mathrm{span}(b_1,b_3)\subseteq K$ under the declared coordinate map. Since $v_0\cos \alpha >0$, eliminating $t=x_1\mathrm{/}(v_0\cos \alpha )$ from the $e_3$-component gives $x_3=\tan \alpha x_1-\frac{g}{2v_0^2{\cos }^2\alpha }{x}_1^2$; its quadratic coefficient is nonzero because $g>0$, so this is a parabola in the stated plane.

Return time and range: the height above the launch level is $h(t):=v_0\sin \alpha t-\frac{1}{2}g{t}^2$; for $g>0$ and $v_0\sin \alpha >0$ one has $h(t)=0$ exactly when $t=0$ or $t=\frac{2v_0\sin \alpha }{g}=:T_{\mathrm{f}}$, so $T_{\mathrm{f}}$ is the positive return-to-launch-height time; the first coordinate at that time is $v_0\cos \alpha T_{\mathrm{f}}=\frac{2v_0^2\sin \alpha \cos \alpha }{g}=\frac{v_0^2\sin (2\alpha )}{g}$ by [L3], the displayed range $R_{\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}}$.

Dimensions and scope: $[g]=\mathrm{\ell }{\tau }^{-2}$ so $[T_{\mathrm{f}}]=\mathrm{\ell }{\tau }^{-1}\mathrm{/}(\mathrm{\ell }{\tau }^{-2})=\tau$ and $[R_{\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}}]={\mathrm{\ell }}^2{\tau }^{-2}\mathrm{/}(\mathrm{\ell }{\tau }^{-2})=\mathrm{\ell }$, consistent with Dimensions and SI unit conventions of the non-relativistic classical mechanics framework; the computation uses no drag term, so it does not describe motion through a medium (compare the drag model The linear drag model: a phenomenological velocity-proportional resistance with a declared coefficient), and a realistic gravitational field is constant only as an approximation, so treating a measured local field strength as the parameter $g$ is an empirical premise with its own conditions and uncertainty.

The vector equation $m\ddot{x}=m\mathbf{g}-c\dot{x}$ is, in the frame's coordinates, the three scalar equations $m{\ddot{x}}_j=m{g}_j-c{\dot{x}}_j$ for $j=1,2,3$, each the scalar model of [L2] with $k=0$, $c>0$ and constant forcing $f_0=m{g}_j$; in particular each scalar initial value problem has exactly one solution on $\mathbb{R}$ by [L2].

Candidate and its derivatives: set $v_{\mathrm{\infty }}=(m\mathrm{/}c)\mathbf{g}$ and ${\xi }_j(t):=x_{0,j}+v_{\mathrm{\infty },j}t+\frac{m}{c}(v_{0,j}-v_{\mathrm{\infty },j})(1-e^{-ct\mathrm{/}m});$ then by [L3] ${\xi }_j^{\mathrm{\prime }}(t)=v_{\mathrm{\infty },j}+(v_{0,j}-v_{\mathrm{\infty },j})e^{-ct\mathrm{/}m}$ and ${\xi }_j^{\mathrm{\prime }\mathrm{\prime }}(t)=-\frac{c}{m}(v_{0,j}-v_{\mathrm{\infty },j})e^{-ct\mathrm{/}m}$, so $m{\xi }_j^{\mathrm{\prime }\mathrm{\prime }}+c{\xi }_j^{\mathrm{\prime }}=-c(v_{0,j}-v_{\mathrm{\infty },j})e^{-ct\mathrm{/}m}+c{v}_{\mathrm{\infty },j}+c(v_{0,j}-v_{\mathrm{\infty },j})e^{-ct\mathrm{/}m}=c{v}_{\mathrm{\infty },j}=m{g}_j,$ the scalar equation of step 1.1; moreover ${\xi }_j(0)=x_{0,j}$ and ${\xi }_j^{\mathrm{\prime }}(0)=v_{\mathrm{\infty },j}+(v_{0,j}-v_{\mathrm{\infty },j})=v_{0,j}$.

By uniqueness in [L2] each ${\xi }_j$ is the unique scalar solution on $\mathbb{R}$ with the datum, so the vector curve $x(t):=({\xi }_1(t),{\xi }_2(t),{\xi }_3(t))$ displayed in the Example satisfies the vector equation for all $t\in \mathbb{R}$ with values in $Q={\mathbb{R}}^3$ and the stated datum; by The associated first-order system, motions, maximal motions and the Newtonian flow of a force model clause (c) it is a motion of the model, and it is defined on all of the time window. By Local existence and uniqueness for the Newtonian initial-value problem, with the quantitative Picard-Lindelöf cylinder clauses 2 and 4 every motion with the same datum is the restriction of this one, so it is the unique maximal motion, with maximal interval $\mathbb{R}$; this proves claim 1.

Velocity and approach to the terminal velocity: from step 2.1, $\dot{x}(t)-v_{\mathrm{\infty }}=(v_0-v_{\mathrm{\infty }})e^{-ct\mathrm{/}m}$, so by [L4] $\mid \dot{x}(t)-v_{\mathrm{\infty }}{\mid }_2=\mid v_0-v_{\mathrm{\infty }}{\mid }_2{e}^{-ct\mathrm{/}m}$ for every $t$. The factor decays to $0$: given $\epsilon >0$, for $t>\frac{m}{c}\cdot \frac{1}{\epsilon }$ one has $\frac{ct}{m}>\frac{1}{\epsilon }$ and hence, using $\exp (s)\ge 1+s$ at $s=\frac{ct}{m}>0$, the reciprocal comparison of [L3] and $\frac{m}{ct}<\epsilon$, $0<e^{-ct\mathrm{/}m}=\frac{1}{e^{ct\mathrm{/}m}}\le \frac{1}{1+ct\mathrm{/}m}<\frac{m}{ct}<\epsilon \mathrm{.}$ Therefore $\mid \dot{x}(t)-v_{\mathrm{\infty }}{\mid }_2\le \mid v_0-v_{\mathrm{\infty }}{\mid }_2{e}^{-ct\mathrm{/}m}\to 0$ as $t\to +\mathrm{\infty }$, i.e. $\dot{x}(t)\to v_{\mathrm{\infty }}$ (Limits at $+\mathrm{\infty }$ and $-\mathrm{\infty }$, and infinite limits at a point); this proves claim 2.

The map $F(t,x,v)=(\lambda \sqrt{\mathrm{\mid }{x}_1\mathrm{\mid }\mathrm{/}{\mathrm{\ell }}_{\ast }},0,0)$ is continuous on $\mathbb{R}\times {\mathbb{R}}^3\times {\mathbb{R}}^3$: its first component is the composition of the continuous coordinate map $x\mapsto x_1$, absolute value, division by the positive constant ${\mathrm{\ell }}_{\ast }$, and the square root on $[0,\mathrm{\infty })$ ([L2]); its other components are constant. By [L4] its force components have dimension $\mu \mathrm{\ell }{\tau }^{-2}$, so it is a dimensionally consistent continuous force map for the weakened model.

The zero curve $x_0(t):=(0,0,0)$ is a motion through the datum on $\mathbb{R}$: it is $C^2$ with ${\ddot{x}}_0=0$ and $F(t,x_0(t),{\dot{x}}_0(t))=(0,0,0)$.

Let $t_1:=1\mathrm{s}$ and $A:=(\lambda \mathrm{/}(12m\sqrt{{\mathrm{\ell }}_{\ast }}){)}^2>0$. Define $x_1:\mathbb{R}\to {\mathbb{R}}^3$ by $x_1(t):=0$ for $t\le t_1$ and $x_1(t):=(A(t-t_1{)}^4,0,0)$ for $t\ge t_1$. The coefficient $A$ has dimension $\mathrm{\ell }{\tau }^{-4}$, so $A(t-t_1{)}^4$ has dimension $\mathrm{\ell }$. For $t>t_1$ the first component has derivatives $4A(t-t_1{)}^3$ and $12A(t-t_1{)}^2$ by [L3]. At $t=t_1$ its first difference quotient is $A{h}^3\to 0$ from the right and $0$ from the left; the difference quotient of its first derivative is $4A{h}^2\to 0$ from the right and $0$ from the left. Thus $x_1$ is $C^2$ on $\mathbb{R}$, with $x_1(0\mathrm{s})=0$ and ${\dot{x}}_1(0\mathrm{s})=0$.

$x_1$ satisfies the equation: for $t<t_1$ both sides vanish; for $t>t_1$ its first acceleration component is $12A(t-t_1{)}^2$, while the first component of $F(t,x_1(t),{\dot{x}}_1(t))$ is $\lambda \sqrt{A\mathrm{/}{\mathrm{\ell }}_{\ast }}(t-t_1{)}^2$. The definition of $A$ gives $12mA=\lambda \sqrt{A\mathrm{/}{\mathrm{\ell }}_{\ast }}$, so $m{\ddot{x}}_1=F$; at $t=t_1$ both sides vanish. Hence $x_1$ is a motion on $\mathbb{R}$ through the datum by [L1].

The two motions differ at $t_2:=2\mathrm{s}$: since $t_2-t_1=1\mathrm{s}$, $x_1(t_2)=A(1\mathrm{s}{)}^4{e}_1\mathrm{\ne }0=x_0(t_2)$. Thus the initial value problem for the continuous force map $F$ has two distinct solutions on every interval containing $[0\mathrm{s},2\mathrm{s}]$, and the refuted statement is false.