Implicit midpoint rule

The implicit midpoint rule advances an autonomous ordinary differential equation byIt is a second-order, A-stable, time-symmetric numerical method.

The implicit midpoint rule is the midpoint method with $y_{n+1}=y_n+hf(t_n+h/2,(y_n+y_{n+1})/2)$. It is a one-stage implicit Runge-Kutta method.

For a uniformly Lipschitz continuous state field $f$ with constant $L$, an implicit midpoint rule step is uniquely solvable when $hL/2<1$, by the contraction mapping theorem. Write its step map as ${\Psi }_h(u)=u+h{\Phi }_h(u)$, where ${\Phi }_h(u)=f(t+h/2,u+h{\Phi }_h(u)/2)$. Subtracting the increment equations givesTherefore $\mathrm{Lip}({\Phi }_h)\le L/(1-hL/2)$, and the triangle inequality gives the bound for ${\Psi }_h$. Combining it with the implicit midpoint consistency bound yieldsfor $L>0$. This follows by summing the geometric error recurrence with forcing $K{h}^3/(1-hL/2)$. For $hL\le 1$ and $nh\le T$, $R_h^n\le e^{2LT}$, proving second-order global error with exact starting data. If $L=0$, the direct bound is $\Vert e_n\Vert \le \Vert e_0\Vert +KT{h}^2$.

For an exact solution $y^{\prime }=f(t,y)$, define the normalized local truncation error of the implicit midpoint rule byIf $\Vert y^{\prime \prime }\Vert \le M_2$, $\Vert y^{\prime \prime \prime }\Vert \le M_3$ and the state Jacobian matrix of $f$ has operator norm at most $L$ on the intervening state segments, then the displayed bound holds uniformly on a fixed finite interval. To prove it, put $t=t_n+h/2$. Symmetric Taylor expansions giveFor the first bound, integrate the second-order remainder of $y^{\prime }(t+s)$ over the symmetric step; for the second, average the second-order remainders of $y$ at the endpoints. The odd terms cancel. The Lipschitz condition for $f$ and $y^{\prime }(t)=f(t,y(t))$ finish the proof by the triangle inequality. The unnormalized step residual is $h{\tau }_n=O(h^3)$.