The question I'm trying to solve is the following:
The implicit midpoint method is defined as:
$y_{n+1} = y_n + hf(t_{n+1/2},(y_n + y_{n+1}/2),$
where $t_{n+1/2} = t_n + {h/2}$
What is the stability region for the equation $y' = \lambda y$? In other words, for what values $\bar{h} = h \lambda \space \in \space \mathbb{C}$ is the method stable?
I really have no idea on where to even begin with this. What I tried to do was plug in $\lambda y_n$ into the trapezoid method so I can obtain $y_n$ as a function of it and $\bar{h}$, but I wasn't really getting any where. Honestly, I'm not even sure if there's supposed to be an explicit way of getting the answer here. Does anybody have any idea on how to do this? I'd really appreciate any help.
Best Answer
Let's apply the numerical method on the test equation \begin{align} y_{n+1} &= y_{n}+h\lambda\left(\frac{y_n+y_{n+1}}{2}\right)\\ (1-\frac{h\lambda}{2})y_{n+1}&=(1+\frac{h\lambda}{2})y_{n}\\ y_{n+1} & = \bigg(\frac{1+\frac{h\lambda}{2}}{1-\frac{h\lambda}{2}}\bigg)y_{n} \end{align} For absolute stability, we require $$ \bigg|\frac{1+\frac{z}{2}}{1-\frac{z}{2}}\bigg| \le 1 $$ where $z=h\lambda\in \mathbb{C}$. Check that this gives you the entire left half plane. If you plot a contour plot, this is confirmed.