I am having difficulties in showing the order and Local truncation of the following implicit method:
The implicit method, to solve a 1st order initial value problem $\frac{dy}{dt}=f(t,y)$ with t $\in$ $[t_0,T]$ and $y(0)=t_0$ where:
$Y_0=y(t_0)$
$Y_{n+2}=\frac{8}{3}Y_{n+1}-\frac{2}{3}Y_n+Y_{n+2}-\frac{h}{3}(f(t_{n+2},Y_{n+2}))$
To show that this method is consistent of order 2 and have Local truncation of $O(h^3$)
For the order 2, we use the operator:
$L_hY_n=\frac{8}{3}Y_{n+1}-\frac{2}{3}Y_n+Y_{n+2}-\frac{h}{3}(f(t_{n+2},Y_{n+2}))$
I did the following:
$L_hY_n=\frac{8}{3}y(t+h)-\frac{2}{3}y(t)+y(t+2h)-\frac{h}{3}y'(2t+2h)$
Using taylor series to expand:
Expanding $(t+h)$:
$\frac{8}{3}((y(t)+hy'(t)+\frac{h^2}{2}y''(t)+\frac{h^3}{6}y'''(t)+...)$
Expanding: $y(t+2h): $
$y(t)+2hy'(t)+2h^2y''(t)+\frac{4}{3}y'''(t)+...)$
Now I am having dificulties in expanding this:$ -\frac{h}{3}y'(2t+2h)$
After that some terms should cancel and I probably can show that the roder is 2 and the Local Truncation is $O(h^3)$
Can anyone help me on this?
Thank you
Hint:
Instead of Taylor expansion, use the pretty-fast-derived formula $$C_p = \sum_{j=0}^{m} j^p\alpha_j + pj^{p-1}\beta_j $$
In case it holds that $C_p = 0$ and $C_{p+1} \neq 0$ the linear $m$-step method is of consistency order $p$. Here $\alpha_i$ and $\beta_i$ are the coefficients of the multistep method.