I want to prove that the consistency order of the trapezoidal rule is actually second order, that means that the error of the actual solution $x(t_{k+1})$ where we can restrict ourselves to the equation $x'(t)=\lambda x(t)$ and the approximation in each step $x_{k+1}=x(t_k)+\frac{\lambda \tau}{2}(x(t_k)+x_{k+1})$ is $O(\tau^3)$. I guess one has to use taylor approximation somewhere, but this does not work.


This is a sketch which shows that solving an ODE with the trapezoidal rule has global error $O(h^2)$.

Consider the differential equation $\dfrac{dx}{dt}=\lambda x$. Integrating this from $t_n$ to $t_{n+1}=t_n+h$ gives $$ \int_{t_n}^{t_{n+1}}\dfrac{dx}{dt}dt=\int_{t_n}^{t_{n+1}}\lambda x\,dt \Longrightarrow x(t_{n+1})-x(t_n)=\lambda\int_{t_n}^{t_{n+1}} x\,dt.\tag{1} $$ Evaluating the integral on the RHS: $$ \int_{t_n}^{t_{n+1}} x\,dt=\frac{h}{2}(x(t_{n+1})+x(t_n))-\frac{h^3}{12} x''(\xi), $$ where $t_n\leq\xi\leq t_{n+1}$, provided that $x(t)\in \mathcal{C}^2$.

This gives the update formula $(1)$ as $$ x(t_{n+1})=x(t_n)+\dfrac{h\lambda}{2}\left(x(t_{n+1})+x(t_n)\right)-\dfrac{h^3}{12} x''(\xi). $$ Solving for $x(t_{n+1})$ gives $$ x(t_{n+1})=\dfrac{2+h\lambda}{2-h\lambda}x(t_n)-\dfrac{h^3}{6(2-h\lambda)}x''(\xi)=\dfrac{2+h\lambda}{2-h\lambda}x(t_n)+O(h^3). $$ Thus, the local error is $O(h^3)$.

Considering the global error, integrating the DE from $0$ to $t_n$: $$ \int_{t_0}^{t_{n}}\dfrac{dx}{dt}dt=\int_{t_0}^{t_{n}}\lambda x\,dt \Longrightarrow x(t_{n})-x(t_0)=\lambda\int_{t_0}^{t_{n}} x\,dt. $$ The integral can be computed using the trapezoidal rule $$ \int_{t_0}^{t_n}x\,dt=\frac{h}{2}\left[x(t_0)+2\sum\limits_{i=1}^{n-1}x(t_i)+x(t_n)\right]-\frac{(t_n-t_0)h^2}{12}x''(\xi), $$ where $t_0\leq\xi\leq t_n$. Following similarly as above, the global error can be shown to be $O(h^2)$.

This is one example of a Runge-Kutta method, this one from the second order family.

  • $\begingroup$ How do you get the third Order local error cause there is also h in the denominator? $\endgroup$
    – user66906
    Jun 20 '13 at 10:04
  • $\begingroup$ Expanding in a series about $h=0$ gives $\dfrac{h^3}{6(2-h\lambda)}=\dfrac{h^3}{12}+\dfrac{h^4\lambda}{24}+O(h^5)=O(h^3).$ $\endgroup$
    – Daryl
    Jun 20 '13 at 11:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy