Projected Euler for a constrained overdamped particle

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].