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Consider the explicit midpoint method, i.e $$y_{n+1}-y_{n-1} = 2hf(y_n).$$

I'm asked to apply this method to the linear test equation, $f(y_n) = \lambda y_n,$ in order to find the method's stability region.

So I have $$y_{n+1}-y_{n-1} = 2h\lambda y_n.$$

In the problems I've done before I wouldn't have the $y_{n-1}$ term so I could just easily write $y_{n+1} = R(z)y_n.$ How would I solve this?

Lutz Lehmann
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Heuristics
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  • Since only $h \lambda$ appears here, clearly that's your $z$. Now if you had a one term recursion such as $y_{n+1} = R(z)y_n$, how would you determine the stability region? What is the next step? – Hans Engler Apr 22 '19 at 12:46
  • @HansEngler Just simple algebra? I would write it in that form, and the stability region is where $R(z) \leq 1$. What confuses me is the $y_{n-1}$ term. I want to expess the $y_n$ term in terms of $y_{n-1}$ but I just end up in a endless loop of expressions. – Heuristics Apr 22 '19 at 13:07
  • Note that the explicit midpoint method is a two-stage Runge-Kutta method $k_1=hf(x_n,y_n)$, $k_2=hf(x_n+h/2,y_n+k_1/2)$, $y_{n+1}=y_n+k_2$. – Lutz Lehmann Apr 22 '19 at 13:44
  • See also Deriving the central Euler method and intuition on the same method and its implementation. – Lutz Lehmann Apr 22 '19 at 13:48

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