Is it possible that a Runge-Kutta method has a global error of 0?

In a quiz of a course I am taking, there was the following multiple choice question:

Decide which of the following statements is false, and justify your election showing a counterexample:

(a) There are numerical methods for solving an Initial Value Problem of fifth order

(b) There are integrals that cannot be computed exactly using the midpoint rule.

(c) Every numerical solution of an Initial Value Problem generated by a Runge-Kutta method will have a global error greater than zero

(d) Newton's method allows finding solutions for a non linear system of equations.

(b) and (d) are obviously true (I can easily find examples of these). (a) Every numerical method we covered in the class are for first order differential equations, but for higher order you can do variable substitution to get a system of first order equations that can be solved using the methods. So I can say (a) is true.

Discarding every other choice, I came to the conclusion that (c) is false, but I cannot find a counter example to this. It is supposed that the global error for a $RK_{pq}$ is proportional to the step $h$, $global\_error = Ch^p$, this will always be greater than zero unless C is 0, but I cannot come up with an example to this. Is this statement true or false and could someone give me an example?

• Consider a trivial equation like $x'=0$. What error does RK have with it? Aug 29, 2016 at 17:33

I am pretty sure irregardless of which method you use, numerically solving $\dot{y} = 0$ will not give you any numerical errors.