Past exam of the numerical analysis course of the University of Oxford / 2021 / 3 / a

Here $y^{\prime }=x/(1+y^2)$ and $y(0)=1$. Since $x\ge 0$, the solution is nondecreasing, so $y(x)\ge 1$. In fact $0\le y^{\prime }\le 1/2$ on $[0,1]$, which also bounds the solution and ensures it extends across this entire interval. The smooth right-hand side makes $y$ twice continuously differentiable, with second derivativeFor $y\ge 1$ and $0\le x\le 1$,so $y^{\prime \prime }\ge 0$. The subtracted term in the first expression is nonnegative, giving the useful boundThe integral form of the Taylor remainder now givesTherefore the normalized local truncation error satisfies the stronger estimateThis proves the requested bound for $n=1,\dots ,N-1$, and also for $n=0$, which will be needed to start the global error estimate. For $N=1$ the stated range is empty, but the estimate at $n=0$ still applies.

For $0\le x\le 1$ and every real $z$,where $2|z|\le 1+z^2$ was used. The mean value theorem therefore proves Lipschitz continuity in $y$ with constant $1$.

Let $e_n=y(x_n)-y_n$. The explicit Euler method and the exact-solution identity from part (i) giveUsing the requested local truncation error bound $|{\tau }_n|\le h/2$, including at $n=0$, yieldsIteration and the finite geometric series giveSince $1+h\le e^h$ and $Nh=1$,Keeping the stronger bound $|{\tau }_n|\le h/4$ from part (i) improves the same proof to $(e-1)h/4$. The requested inequality follows already from the weaker bound.