# Show this Runge-Kutta method is order 3

We have been asked to prove the following Runge-Kutta method is of order 3

$y_{i+1} = y_i + \frac{K_1}{4} + \frac{3K_3}{4}$

with

$K_1 = h \times f(x_i,y_i)$

$K_2 = h \times f(x_i + \tfrac{h}{3}, y_i + \tfrac{K1}{3})$

$K_3 = h \times f(x_i + \tfrac{2h}{3}, y_i + \tfrac{2K_2}{3})$

We have ideas from this answer -Determining Runge-Kutta Order - but are struggling to apply our system to this method.

Thanks

According to the man the Butcher tables are named after (see these slides), the conditions for order 3 are \begin{align} b_1+b_2+b_3&=1\\ b_2c_2+b_3c_3&=\frac12\\ b_2c_2^2+b_3c_3^2&=\frac13\\ \text{and}\quad b_3a_{32}c_2=\frac16 \end{align} which gives $$b_2c_2(c_3-c_2)=\frac12c_3-\frac13$$ which for $$c_2=\frac13$$ and $$c_3=\frac23$$ results in $$b_2=0$$ and $$b_3=\frac34$$ which finally has $$b_1=\frac14$$, $$a_{32}=\frac23$$, $$a_{31}=0$$.
This is a (the) third order Heun method, as the type of method Karl Heun (1900) considered were based on combining slope iterations of the form $$Δ^m_\nu y = f(x+ε^m_\nuΔx,y+ε^m_\nuΔ^{m+1}_\nu y)Δx$$, $$m=0,...,s_\nu$$ with $$ε^{s_\nu}_\nu=0$$ into a final update $$Δy=\sum \alpha_\nuΔ^0_\nu y$$. This third order method Heun, p. 30 gave as
und hieraus resultiert die für die Anwendungen sehr bequeme Formel IV)\quad\left\{\begin{aligned} Δy &= \frac14\left\{f(x,y)+3f\left(x+\frac23Δx,y+Δ'y\right)\right\}\cdotΔx\\ Δ'y &= \frac23f\left(x+\frac13Δx,y+\frac13f\cdotΔx\right)\cdotΔx \end{aligned}\right.