# Numerical Analysis/Order of RK methods/Derivation of a third order RK method

Let the recurrence equation of a method be given by the following of Runge Kutta type with three slope evaluations at each step

$y_{n+1}=y_{n}+h(a_{1}k_{1}+a_{2}k_{2}+a_{3}k_{3})\,$ (2.1)

with

$k_{1}=f(t_{n},y_{n}),\,$ $k_{2}=f(t_{n}+p_{1}h,y_{n}+q_{11}hk_{1}),\,$ $k_{3}=f(t_{n}+p_{2}h,y_{n}+q_{21}hk_{1}+q_{22}hk_{2}),\,$ Taylor series expansion of $y(t_{n}+h)\,$ about $t_{n}\,$ is the same as in Example 1. Therefore, we will just use the final expression (1.7), since the procedure of the derivation is the same. For convenience, the final expression is repeated, which is going to be a reference equation for the comparison with the method's recurrence equation. Since the formulas for the given form of recurrence equation will get complicated, we will use the compact symbolic notation for the derivatives, which is shown in Example 1.

${\overline {y_{n+1}}}=y_{n}+hf(t_{n},y_{n})+{\frac {h^{2}}{2}}(f_{t}+f_{y}f)_{(t_{n},y_{n})}+{\frac {h^{3}}{6}}\left(f_{tt}+2f_{ty}f+f_{t}f_{y}+f_{yy}f^{2}+f_{y}f_{y}f\right)_{(t_{n},y_{n})}+O(h^{4})\,$ (2.2)

The Taylor expansion of the terms in (2.2) is shown up to $O(h^{4})\,$ , rather then up to $O(h^{5})\,$ , as we should in order to check that eventually next higher order terms cancel out, but we will assume that the method cannot achieve better local accuracy then fourth order, or equivalently, the global error of the third order. This will save us getting into the third level expansion of the two variable function f, which has 18 terms and would not be appropriate due to its length (even if the compact symbolic notation is used).

After we prepared the Taylor series expansion, we need to adjust the method's recurrence equation such that it can be compared with the Taylor series (2.2).

Now, we need to group the terms in the similar way they are grouped in the Taylor series (2.2), such that we can establish the conditions on the parameters that will yield the same terms as in the Taylor expansion up to the terms containing $h^{4}\,$ .

$y_{n+1}=y_{n}+h(a_{1}+a_{2}+a_{3})f(t_{n},y_{n})+h^{2}(a_{2}(p_{1}f_{t}+q_{11}f_{y}f)+a_{3}(p_{2}f_{t}+q_{21}ff_{y}+q_{22}ff_{y}))_{(t_{n},y_{n})}+h^{3}({\frac {a_{2}}{2}}(f_{tt}p_{1}^{2}+f_{yy}q_{11}^{2}f^{2}+2f_{ty}p_{1}q_{11}f)+\,$ ()

$+a_{3}(q_{22}(p_{1}f_{t}f_{y}+q_{11}f_{y}^{2}f)+{\frac {1}{2}}(f_{tt}p_{2}^{2}+f_{yy}(q_{21}+q_{22})^{2}f^{2}+2f_{ty}p_{2}f(q_{21}+q_{22}))))_{(t_{n},y_{n})}+\,$ ()

$+\underbrace {[hO_{2}(h^{3})+h^{2}O_{3}(h^{2})(f_{y}q_{22}a_{3})+f_{yy}q_{22}a_{3}fh^{3}O_{5}(h)+a_{3}f_{ty}p_{2}q_{22}h^{3}O_{5}(h)]_{(t_{n},y_{n})}} _{O_{6}(h^{4})}\,$ (2,4)

By comparing the two expressions (2,4) and (2,2), the following system of equations is obtained.

$a_{1}+a_{2}+a_{3}=1\,$ (2.5)

$p_{1}a_{2}+p_{2}a_{3}=1/2\,$ (2.6)

$q_{11}a_{2}+a_{3}(q_{21}+q_{22})=1/2\,$ (2.7)

$a_{2}p_{1}^{2}+a_{3}p_{2}^{2}=1/3\,$ (2.8)

$a_{2}p_{1}q_{11}+a_{3}p_{2}(q_{21}+q_{22})=1/3\,$ (2.9)

$a_{3}q_{22}p_{1}=1/6\,$ (2.10)

$a_{2}q_{11}^{2}+a_{3}(q_{21}+q_{22})^{2}=1/3\,$ (2.11)

$a_{3}q_{22}q_{11}=1/6\,$ (2.12)

At the first glance, the system is closed, the number of equations is (2.5 through 2.12) which matches the number of undetermined parameters. However, only 6 equations are independent, the rest of them can be obtained from those 6 equations. By dividing (2.10) with (2.12), we can obtain that $q_{11}=p_{1}\,$ . Similarly, by substracting (2.11) from (2.9) equation, we see that $p_{2}=q_{21}+q_{22}\,$ . When we replace these two results into the rest of the equations, it is evident that the (2.6) and the (2.7) are the same, and (2.8) and the (2.9) equations are the same. Therefore, two equations can be obtained from other six, and we have to choose two variables in order to obtain a solution for the parameters.

For example, we can choose that $p_{2}=1,q_{11}=1/2\,$ , then we obtain the following recurrence equation.

$y_{n+1}=y_{n}+{\frac {h}{6}}(k_{1}+4k_{2}+k_{3})\,$ (2.13)

where

$k_{1}=f(t_{n},y_{n})\,$ $k_{2}=f(t_{n}+1/2h,y_{n}+1/2hk_{1})\,$ $k_{3}=f(t_{n}+h,y_{n}-hk_{1}+2hk_{2})\,$ The recurrence equation (2.13) is known Runge Kutta third order method [1,3] (List of Runge–Kutta methods), which indicates that our approach was correct.