Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 1 / 17B (Numerical Analysis)

The function is not Lipschitz on $\mathbb{R}$. For $\epsilon >0$,which is unbounded as $\epsilon \downarrow 0$.

The functionis continuous, satisfies the equation away from $\tilde{t}=y_0$, and has unequal one-sided derivatives there.

It is unique in the stated class. While a solution is positive its derivative is $-1$, so continuity forces it to follow $y_0-t$ until it first reaches zero at $t=y_0$. It cannot subsequently make a positive excursion: on any connected interval $(a,b)$ where $y>0$ and $y(a)=0$, integration of $y^{\prime }=-1$ gives $y(t)=-(t-a)<0$. A negative excursion similarly contradicts $y^{\prime }=1$. Hence the solution remains zero. This is the sign-decay differential equation.

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.