I don't know Extrapolation Method for initial value ODE

Question

my text book burden

$ y'(t)=f(t,y), a \leq t \leq b ,y(a)= \alpha$

apply for the extrapolation technique

we let $ h_0 = \frac{h}{2}$ and use Euler method with $w_0=\alpha$

$ w_1 = w_0+h_0f(a,w_0)$

and then midpoint method

$ t_{i-1}=a ,t_i=a+h_0=a+\frac{h}{2}$ to produce

approximation to $ y(a+2h_0)$

$ w_2=w_0+2h_0f(a+h_0,w_1)$

$y_{1,1} = \frac{1}{2}[w_2+w_1+h_0f(a+2h_0,w_2)] $

then why this form results $ O(h_0^2)$ approximation to $y(t_1)$

Answer 1

The solution to the initial value problem has the series

$$ y(a+h) = \alpha + f(a,\alpha) h + \frac{f_t(a,\alpha) + f(a,\alpha) f_y(a,\alpha)}{2} h^2 + O(h^3)$$

where $f_t = \partial f/\partial t$ and $f_y = \partial f/\partial y$. On the other hand, compute the series for $y_{11}$ and you should also find

$$y_{1,1} = \alpha + f(a,\alpha) h + \frac{f_t(a,\alpha) + f(a,\alpha) f_y(a,\alpha)}{2} h^2 + O(h^3)$$

Answer 2

This looks more like an order-reduced RK3 method. Using the standard Runge-Kutta notation, you get \begin{align} k_1&=hf(a,w_0),& w_1&=w_0+\tfrac12k_1\\ k_2&=hf(a+\tfrac12h,w_0+\tfrac12k_1),& w_2&=w_0+k_2\\ k_3&=hf(a+h,w_0+k_2) \\\hline y_{1,1}&=\tfrac12(w_1+w_2+\tfrac12k_3)\\ &=w_0+\tfrac14k_1+\tfrac12k_2+\tfrac14k_3 \end{align} This then gives the Butcher tableau $$ \begin{array}{c|ccc} c_1 & \\ c_2 & a_{21}\\ c_3 & a_{31} & a_{32}\\\hline &b_1&b_2&b_3 \end{array} = \begin{array}{c|ccc} 0 & \\ \tfrac12 & \tfrac12\\ 1 & 0 & 1\\\hline &\tfrac14&\tfrac12&\tfrac14 \end{array} $$ where the first and second order conditions $$ b_1+b_2+b_3=1\\ c_1b_1+c_2b_2+c_3b_3=\frac12 $$ are both satisfied.

See the Butcher's tutorial on low-order methods