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Yes. The Euler recurrence isWrite $y_0=mh+r$, where $m=\lfloor y_0/h\rfloor$ and $0\le r<h$. For $n\le m$, the numerical values agree with the exact linear descent. If $r=0$, the method reaches zero and stays there. If $0<r<h$, thenand the recurrence thereafter alternates between $r$ and $r-h$. The exact solution is then zero, while both numerical values have magnitude at most $h$. Thus, uniformly for $0\le n\le N$,This is the explicit Euler method for the sign-decay equation.