2

\begin{align} w_0 &= \alpha \\ w_{i+1}&=w_i +a_1 f(t_i,w_i) + a_2 f(t_i+\alpha_2,w_i+\delta_2 f(t_i,w_i)) \end{align} for each $i =0,1,2,...,N-1$, cannot have local truncation error $O(h^3)$ for any choice of constants $\ a_1,a_2,\alpha_2 $ and $\delta_2 $

my text book do not explain why

and this form had not been seen ever before

usually Runge-Kutta method has the form

$$ \ w_{i+1}=w_i +h(a_1 f(t_i,w_i) + a_2 f(t_i+\alpha_2,w_i+\delta_2 f(t_i,w_i)))$$

it is replaced by Taylor 2 order form, which is defined to 2 order Runge-Kutta method

but first form seemed not correct form and do not understand why this form has local truncation error with not having $O(h^3)$ regardless of choice of $ a_1,a_2,\alpha_2,\delta_2 $

Lutz Lehmann
  • 126,666
seyunkim
  • 51
  • How do you define the truncation error? The Heun second order method has $O(h^3)$ in the local difference of exact and numerical solution over one step. – Lutz Lehmann Jul 24 '19 at 09:53

1 Answers1

2

Yes, if the task is correctly reproduced, then it is strange, the iteration should be written as \begin{align} w_0 &= \alpha \\ w_{i+1}&=w_i +a_1h f(t_i,w_i) + a_2h f(t_i+\alpha_2h,w_i+\delta_2h f(t_i,w_i)) \end{align} if $a_1$, $a_2$, $α_2$ and $δ_2$ are really constants independent of $h$. Then $δ_2=α_2$, and if $w(t)$ is the exact solution with $w(t_i)=w_i$ $$ f(t_i+α_2h,w_i+α_2h f(t_i,w_i))=f(t_i+α_2h,w(t_i+α_2h)+O(h^2))=w'(t_i+α_2h)+O(h^2) $$ so that \begin{align} w_{i+1}&=w_i+a_1h w'(t_i) + a_2h w'(t_i+α_2h)+O(h^3) \\&=w_i+a_1h w'(t_i)+a_2h [w'(t_i)+α_2hw''(t_i)+O(h^2)]+O(h^3) \\&=w_i+(a_1+a_2)w'(t_i)+a_2α_2h^2w''(t_i)+O(h^3) \end{align} Now compare with the quadratic Taylor polynomial for $w(t+h)$ to find the conditions $a_1+a_2=1$ and $a_2α_2=\frac12$ so that $w_{i+1}-w(t_{i+1})=O(h^3)$.

Lutz Lehmann
  • 126,666