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.

$\endgroup$
1
$\begingroup$

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.