Edit: figure it out, made an error

We are given that $$y' = f(y,t), y(t_0) = y_0$$ We want to use the following method $$ \begin{align} u^1_{n+1/2} &= u_n + \frac{h}{2}f(u_n, t_n)\\ u^1_{n+1} &= u_{n+1/2} + \frac{h}{2}f(u^1_{n+1/2}, t_n+h/2)\\ u^2_{n+1} &= u_n + hf(u_n, t_n)\\ u_{n+1} &= \alpha u^1_{n+1} + \beta u^2_{n+1} \end{align} $$ The goal is to choose $\alpha$ and $\beta$ such that the above method is second order. I used Taylor expansions for everything and assumed that $f$ is Lipschitz. I can add more details, but assuming I did it correctly I found that the local truncation error $\tau$ is given by $$\tau = (1-\alpha-\beta)u_n + (1-\alpha-\beta)hu_n' +(1-\alpha)\frac{h^2}{2}u_n'' + O(h^3)$$

Edit: expression for $\tau$ should have $$(1-\alpha/2)\frac{h^2}{2}u_n''$$ Expression above was incorrect.

So if we let $\alpha=1, \beta=0$, then we get $\tau= O(h^3)$, thus giving second order convergence. I then implemented this method for two different functions $f$ and only got 1st order convergence.

Any ideas? Is those the correct values of $\alpha$ and $\beta$? Thanks.


1 Answer 1


Once I found my mistake (see edit in post), that led to different values of the constants, namely $$\alpha = 2, \beta = -1$$ The posted method then reduces down to the midpoint method, which is second order.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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