Past exam of the numerical analysis course of the University of Oxford / 2023 / 3 / c / Solution

Let $f$ be continuous in its first variable and uniformly Lipschitz continuous in its second variable, with constant $L\ge 0$. Retain the finite-interval smoothness assumptions and the bound $\Vert {\tau }_n\Vert \le K{h}^2$ from the preceding subpart. We first establish solvability of each implicit step. For a proposed starting state $u$, the maphas Lipschitz constant $hL/2$. If $hL/2<1$, the contraction mapping theorem on the complete state space gives a unique next state $v={\Psi }_{n,h}(u)$.

To compute a useful Lipschitz constant for the method, introduce its increment function byFor two starting states $u,w$,Rearranging gives the implicit midpoint Lipschitz estimateThus, for example, $hL\le 1$ gives an increment Lipschitz constant at most $2L$, uniformly as $h\to 0$.

For an explicit global error estimate, set $e_n=Y_n-y_n$. Subtract the numerical formula from the exact-value identity containing $h{\tau }_n$:The Lipschitz condition and the triangle inequality implyand thereforeIterating this inequality is the elementary discrete Gronwall inequality argument. For $L>0$, summing the geometric series and using $R_h-1=hL/(1-hL/2)$ givesFor $hL\le 1$, $R_h=1+hL/(1-hL/2)\le 1+2hL\le e^{2hL}$. If $nh\le T$, this provesWith the exact initial value, $e_0=0$, the global error is $O(h^2)$ and tends uniformly to zero on every such finite interval. If $L=0$, the recurrence instead gives $\Vert e_n\Vert \le \Vert e_0\Vert +KT{h}^2$, with the same conclusion. Starting values merely converging to the exact initial value also suffice for numerical convergence; second-order accuracy requires their errors to be $O(h^2)$. The result concerns exact implicit solves and convergence on fixed finite intervals, not uniform accuracy over an unbounded time interval.