Consider the initial-value problem: $$\left\{ \begin{array}{ll}y'(t) = f(t,y(t)), \ \ \ t \geq 0,\\ y(0) = y_0. \end{array} \right. $$
Now consider the numerical time-integration method: $$w_{n+1} = w_n + h \cdot f(t_n+ \alpha \cdot h, w_n + \alpha \cdot h \cdot f(t_n, w_n)), $$ where $h$ is the step size, $w_n$ is the numerical solution after $n$ steps and $\alpha \in [0,1]$.
What is the order of the local truncation error for the numerical method described above and for which $\alpha$ does it have maximal order? Also, how would we determine the amplification factor from the test equation $$y' = \lambda y?$$
For the local truncation error, I tried computing $$\frac{y(t_{n+1}) – z_{n+1}}{h}, $$ where $z_{n+1} = y(t_n) + h \cdot f(t_n + \alpha \cdot h, \alpha \cdot h \cdot f(t_n, y(t_n))),$ but the argument of the function seems difficult to work with. I tried Taylor expanding the function, but I got nowhere (maybe I did in the wrong point, I tried to do it in $t_n$). Now, I simply don't know how to proceed. The same can be said about the amplification factor.
Best Answer
The general two-stage explicit Runge-Kutta method \begin{array}{c|cc} 0\\ α&α\\ \hline &1-\beta&\beta \end{array} that is, $$ w_{n+1}=w_n+\Bigl[(1−β)f(t_n,w_n)+βf\bigl(t_n+αh, w_n+αf(t_n,w_n)h\bigr)\Bigr]h $$ satisfies automatically the order 1 order condition, and satisfies also the order 2 order condition if $βα=\frac12$. In your case, $β=1$.