Past exam of the numerical linear algebra and numerical solution of differential equations course of the University of Oxford / 2020 / 1 / b

By the chain rule, $y^{\prime \prime }=f_x+f_{y}f$ along the exact solution. Applying Taylor's theorem to the exact increment and the two arguments of $f$ givesThe assumed differentiability makes the remainders uniform on $0\le x\le 1-h$. Subtracting gives the consistency errorHere $O$ denotes Big O notation. The general order of a numerical method is therefore one for $\beta \ne 1/2$ and at least two for $\beta =1/2$. For example, on $y^{\prime }=y$ with unit initial value, $y_N=(1+h+\beta h^2{)}^{1/h}$. Expanding its logarithm gives $y_N=e[1+(\beta -1/2)h+O(h^2)]$, so the first-order term is actually nonzero when $\beta \ne 1/2$.

To control accumulation, the mean value theorem and the bound on $f_y$ givePut $K_h=L(1+\beta hL)$, $q_h=1+h{K}_h$ and $e_n=y(x_n)-y_n$. The exact and numerical updates yield $|e_{n+1}|\le q_h|e_n|+h|T_n|$, with $e_0=0$. Iterating this recurrence, as in the Lipschitz one-step error bound, givesBecause $K_h\le L(1+L)$ for $h\le 1$, the prefactor is bounded independently of the mesh. When $\beta =1/2$, the consistency error is $O(h^2)$, hence the maximum global error is $O(h^2)$. This proves at least second-order numerical convergence of the explicit midpoint method.