A stiff penalty equation defeats a fixed explicit step

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.