0
$\begingroup$

Given the two-step Adams-Bashforth method $$ u_{n+1} = u_n + \tfrac{h}{2}(3f_n - f_{n-1}) $$ find its order.

Some notation: $t_n = t_0 + nh$ is the $n$-th node and $y_n = y(t_n)$; $f_n$ stands for $f(t_n,y_n)$ and $u_n$ is an approximation of $y_n$. Being $u^*_{n+1} = y_n + \tfrac{h}{2}(3f_n - f_{n-1})$, what one can say about the local truncament error is that

$$ \tau_{n+1} = \frac{y_{n+1} - u^*_{n+1}}{h} $$

A first step in getting the LTE is expanding $y_{n+1}$ in $y_n + hy'_n + \tfrac{h^2}{2}y''_n + \tfrac{h^3}{6}y'''_n(\xi)$, however I can't properly explain to myself why

$$ u^*_{n+1} = y_n + \tfrac{3}{2}hy'_n - \tfrac{h}{2}\left(y'_n - hy''_n + \tfrac{h^2}{2}y'''_n(\chi)\right) $$

How does one get there? What I'm asking is basically an explanation about how should I do the Taylor expansion of $3f_n - f_{n-1}$.

The order of the method is 2. The solution I'm trying to understand is the one proposed by my teacher.

$\endgroup$
0
$\begingroup$

I assume you are exploring the standard first order equation $\frac{dy}{dt}=f(t,y)$. Then

$$ \frac{dy}{dt}\Big|_{t=t_k}=f(t_k,y_k) $$

for each $k$. Hence

$$ f_n=y'_n $$

and

$$ f_{n-1}=y'_{n-1}=y'_n-hy''_n+\frac{h^2}{2}y'''(\chi) $$

for some $\chi$ between $t_{n-1}$ and $t_n$. The expression for $f_{n-1}$ is determined by finding the Taylor series for $y'(t_n-h)$ centered at $t_n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy