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

Write the increment function of the one-step method as $\Phi (x,v,h)=f(x+\beta h,v+\beta hf(x,v))$. Its consistency error isAs $h\to 0$, the first term tends to $y^{\prime }(x_n)$ and the second tends to $f(x_n,y(x_n))=y^{\prime }(x_n)$ by continuity. Both limits are uniform along the exact-solution curve, a compact set: its small stage perturbations lie in a fixed compact neighborhood, so uniform continuity applies there. Thus ${\max }_{n<N}|T_n|\to 0$, proving consistency for every $\beta \in [0,1]$.

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.

For $f(x,y)=y$ and $\beta =1/2$, the explicit midpoint method multiplies the current value by $R=1+h+h^2/2$. Starting from one, induction givesLet $A=e^h$ and $B=R$. The Taylor theorem with Lagrange remainder for the exponential function givesThe difference of powers identity gives $A^n-B^n=(A-B){\sum }_{k=0}^{n-1}{A}^{n-1-k}{B}^k$. Every summand lies between one and $A^{n-1}$. Since $nh=x_n$, the global error therefore satisfiesThe same inequalities hold at $n=0$, where the error vanishes. At the fixed endpoint $x_N=1$, the lower bound $h^2/6$ prevents any higher order of a numerical method, while the upper bound is $e{h}^2/6$. Thus the convergence order is exactly two.