Nonsmooth contact time stepping and projections — Examples

Let $C=\{-1,1\}\subset \mathbb{R}$. It is nonempty. If $y\mathrm{\notin }C$, put $\delta =\min (\mathrm{\mid }y-1\mathrm{\mid },\mathrm{\mid }y+1\mathrm{\mid })>0$. For any $z$ with $\mathrm{\mid }z-y\mathrm{\mid }<\delta \mathrm{/}2$, [F3] gives $\mathrm{\mid }z-1\mathrm{\mid }\ge \mathrm{\mid }y-1\mathrm{\mid }-\mathrm{\mid }z-y\mathrm{\mid }>\delta \mathrm{/}2>0$ and $\mathrm{\mid }z+1\mathrm{\mid }>0$, so this ball misses $C$. Thus $\mathbb{R}\setminus C$ is open and $C$ is closed by [F2].

For input $y=0$, both points of $C$ have distance $1$ by [F1], and these are all the points in $C$. Hence the nearest-point set is exactly $\{-1,1\}$, which is not a singleton. The claimed unique projection fails for this nonconvex closed set.

The inverse reduced mass is $1\mathrm{/}{m}_1+1\mathrm{/}{m}_2=1\mathrm{/}2+1\mathrm{/}3=5\mathrm{/}6$, and $v_n=⟨-n-n,n⟩=-2$. Thus $J=-(3\mathrm{/}2)(-2)\mathrm{/}(5\mathrm{/}6)=18\mathrm{/}5$. The post-impact velocities are $v_1^{+}=-n+(9\mathrm{/}5)n=(4\mathrm{/}5)n$ and $v_2^{+}=n-(6\mathrm{/}5)n=(-1\mathrm{/}5)n$.

Total momentum is $2(-n)+3n=n$ before impact and $2(4n\mathrm{/}5)+3(-n\mathrm{/}5)=n$ after. The relative normal speed after impact is $4\mathrm{/}5-(-1\mathrm{/}5)=1=-e{v}_n$. The kinetic energies are $K^{-}=1+3\mathrm{/}2=5\mathrm{/}2$ and $K^{+}=16\mathrm{/}25+3\mathrm{/}50=7\mathrm{/}10$, so $K^{+}-K^{-}=-9\mathrm{/}5$, matching $-(3\mathrm{/}4)\cdot 4\mathrm{/}(2\cdot 5\mathrm{/}6)=-9\mathrm{/}5$ from [F1].

The free-flight gap is $g(s)=1-2s$, whose unique root on $(0,1]$ is $s=1\mathrm{/}2$ with $v^{-}=-2<0$. Hence the step returns an impact at contact for both restitution values.

If $e=0$, then $J=4$ and $v^{+}=0$; if $e=1$, then $J=8$ and $v^{+}=2$. In both cases $q_{\ast }J=0$ and $J\ge 0$. The pre-impact kinetic energy is $\frac{1}{2}(2)(-2{)}^2=4$; the post-impact energies are $0$ and $4$, respectively. The momentum identities are $2(-2)+4=0$ and $2(-2)+8=4=2(2)$. These equal the inelastic and elastic endpoints of [F2].

For $C=[0,\mathrm{\infty })$, $N_C(0)=(-\mathrm{\infty },0]$ because $⟨n,z⟩=nz\le 0$ for every $z\ge 0$. Thus $x(t)=0$ solves the projected inclusion: $f-\dot{x}=-1\in N_C(0)$. For $y_{\kappa }(t)=-(1-e^{-\kappa t})\mathrm{/}\kappa$, [F2] gives $y_{\kappa }^{\mathrm{\prime }}=-e^{-\kappa t}=-1-\kappa y_{\kappa }$; [F3]–[F4] give $0<e^{-\kappa t}<1$ for $t>0$, so $y_{\kappa }(t)<0$ and the penalty equation holds. Also $0\le 1-e^{-\kappa t}\le 1$, so ${\sup }_{0\le t\le 2h}\mathrm{\mid }{y}_{\kappa }(t)\mathrm{\mid }\le 1\mathrm{/}\kappa$ and the exact penalty paths converge uniformly to the constrained path as $\kappa \to \mathrm{\infty }$.

Explicit Euler gives $y_1=-h$, since the penalty is zero at $y_0=0$. At $y_1=-h$, the next value is $y_2=-h+h(-1+\kappa h)=\kappa h^2-2h$. For fixed $h>0$, $y_2\to +\mathrm{\infty }$ as $\kappa \to \mathrm{\infty }$, despite the continuous penalty solutions converging uniformly to zero. Therefore the timestep must be controlled as stiffness grows; this example makes no general penalty-to-rigid-impact limit claim.

For $0<x<1$, $N_C(x)=\{0\}$; at $x=0$, $N_C(0)=(-\mathrm{\infty },0]$, and at $x=1$, $N_C(1)=[0,\mathrm{\infty })$. Therefore $x(t)=\max (1-t,0)$ satisfies the projected inclusion for $t<1$ and $t>1$: before contact $\dot{x}=-1$ and $f-\dot{x}=0$; after contact $\dot{x}=0$ and $f-\dot{x}=-1\in N_C(0)$. At $t=0$, $f-\dot{x}=0\in N_C(1)$. The single nondifferentiability time $t=1$ does not affect the almost-everywhere condition. The data satisfy the hypotheses of [F3], using [A1] and [A2].

For $C=[0,1]$, $P_C(y)=\min (1,\max (0,y))$: if $y<0$, then $\mathrm{\mid }z-y\mathrm{\mid }\ge \mathrm{\mid }0-y\mathrm{\mid }$ for every $z\in C$; if $0\le y\le 1$, $y$ itself is the nearest point; if $y>1$, then $\mathrm{\mid }y-z\mathrm{\mid }\ge \mathrm{\mid }y-1\mathrm{\mid }$. By [F4], this exact projection supplies the zero-residual certificate, so the provider completes each step. Since all iterates are at most $1$, the recurrence is $x_{k+1}=\max (0,x_k-h)$. Starting with $x_0=1$, induction gives $x_k=\max (1-kh,0)$: if $1-kh\ge 0$, the next value is $\max (0,1-(k+1)h)$; if $1-kh<0$, both the current and next values are zero and $1-(k+1)h<0$. On cells entirely before $t=1$, interpolation equals $1-t$; on cells entirely after $1$, it is zero. If a cell straddles $1$, its left value is $1-t_k<h$ and its right value is zero; the difference from $\max (1-t,0)$ is linear on each side of $1$ and has maximum at $1$, at most $h$. Thus ${\sup }_t\mathrm{\mid }{X}_h(t)-x(t)\mathrm{\mid }\le h$, consistent with [F3].