Find Stability-Function and -Interval of Runge Kutta method

Question

We're given this Butcher Tableau

$$ \renewcommand\arraystretch{1.2} \begin{array} {c|cccc} 0\\ 1 & 1\\ \hline & \frac{1}{2} &\frac{1}{2} \end{array}$$

We want to:

1. Write it as a one-step process
2. Find its stability-function and -interval

Answer 1

1. The one step process is: $$k_1=f(t_k, y_k)$$ $$k_2=f(t_k + h, y_k + hk_1)$$ $$y_{k+1}=y_k + \frac{h}{2}(k_1 + k_2)$$
2. For the stability function, we let $f(t_k, y_k) = \lambda y_k$ and $f(t_k + h, y_k + hk_1) = \lambda (y_k + hk_1)$.
   Thus $$y_{k+1} = y_k + \frac{h}{2}((\lambda y_k) + (\lambda (y_k + hk_1))) = y_k(1 + \lambda h + \frac{\lambda^2 h^2}{2})$$. Now, if $z := h \lambda$ then our stability function can be written as $$g(z) = 1 + z + \frac{z^2}{2}$$
   For the stability interval, we need $|g(z)| < 1$. In our case, we can rearrange that as $z + \frac{z^2}{2} < 0$ or $z(2+z) < 0$. Our stability interval is thus $-2 < z < 0$.