It should be noted that even in case of two points method we have to calculate values of the function in related to
, , this values could be evaluated by linear interpolation (because it is necessary to avoid oscillations), so estimation of integration error become very complicated process, but this error will be less or equal to trapezoidal rule.Mechanism of Gauss-Chebyshev method is almost the same like described above, and integration error will be almost the same, so there is no reason to use such methods for the current problem.
It is well known that the third order Runge-Kutta method is of the following form
,Suppose that you are asked to derived a new third order Runge-Kutta method in the following from
,Find determine the unknowns
, , and so that your scheme is a third order Runge-Kutta method.Generally Runge-Kutta method looks like:
where coefficients
could be calculated by the scheme:The error function:
Coefficients
, , must be found to satisfy conditions , consequently we can suppose that for each order of Runge-Kutta scheme those coefficients are determined uniquely, it means that there are no two different third order schemes with different coefficients. Now it is necessary to prove statement.For
, : ; ; ; ; ; ; ;Thus we have system of equations:
Some of coefficients are already predefined:
; ; ; ; ; ; ;Obtained results show that Runge-Kutta scheme for every order is unique.
Discuss the stability problem of solving the ordinary equation
, via the Euler explicit scheme , say . If , how to apply your stability restriction?The Euler method is 1st order accurate. Given scheme could be rewritten in form of:
If
has a magnitude greater than one then will tend to grow with increasing and may eventually dominate over the required solution. Hence the Euler method is stable only if or:For the case
mentioned above inequality looks like:Last result shows that integration step mast be less or equal to
.For the case
, for the iteration method coefficient looks like