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Given the Runge-Kutta Method given by:

$y_{i+1} = y_{i} + \frac{h}{9}(2 K_1 + 3 K_2 + 4K_3)$

with:

$K_1 = f(x_i, y_i) \qquad K_2 = f(x_i + \frac{h}{2}, y_i + \frac{h}{2} K1) \qquad K_3 = f(x_i + \frac{3}{4}h, y_i + \frac{3}{4}h K_2)$

Prove that it's order 3.

If it's order 3 we'll have:

$T_3 - \Phi(xi,yi,h) = O(h^3)$

Where $T_3 = f(x,y(x)) + \frac{h}{2} [f_t + f_x f] + \frac{h^2}{2} [f_{tt} + 2f_{tx} +f_x f_t + f_{xx}f^2 + f_x f_x f]$ is the order 3 taylor expansion.

So my idea was to taylor expand each of the $K_i$:

$K_2 = f + f_t \frac{h}{2} + f_x K_1 \frac{h}{2} \qquad K_3 = f + f_t \frac{3}{4} h + f_x K_2 $

And in the expansion of $K_3$ plug in the expansion of $K_2$, doing that I can match between the two methods the first terms but I'm missing the terms that have $f_{tt}$. So what is missing?. Should I expand $K_3 = f + f_t \frac{3}{4} h + f_x K_2 + (\frac{3}{4}h)^2 [f_{tt} + f_{tx}f + K_2(f_{xt} + f_{xx}f)]$??

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  • $\begingroup$ You can always reduce to the check of an autonomous system, since by adding a variable with derivative $1$ every non-autonomous system can be transformed into an autonomous one. $\endgroup$ – LutzL Nov 8 '16 at 22:33
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You should expand more as follows:

$$K_2=f+\frac{h}{2}f_x+\frac{h}{2}ff_y+\frac{1}{2}\left[\frac{h^2}{4}f_{xx}+2\cdot\frac{h}{2}\frac{h}{2}ff_{xy}+\frac{h^2}{4}f^2f_{yy}\right]+O(h^3)\\ K_3=f+\frac{3}{4}hf_x+\frac{3}{4}hf_yK_2+\frac{1}{2}\left[\frac{9}{16}h^2f_{xx}+2\cdot\frac{3}{4}h\frac{3}{4}hK_2f_{xy}+\frac{9}{16}h^2K_2f_{yy}\right]+O(h^3).$$

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