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We have the following Runge Kutta Butcher tableau:

$$ \begin{array}{c|ccccc} \tau_1 =0 & a_{11}=0 & a_{12} = 0\\ \tau_2 =\frac{3}{2} & a_{21} = \frac{3}{2} & a_{22} = 0\\ \hline & b_1 = \frac{2}{3} & b_2 = \frac{1}{3} & \ \end{array} $$

$$y'=-10y \\ y(0)=1$$

For which $h$ do we have the smallest possible number of steps?

  1. $h=0.5$
  2. $h=0.01$
  3. $h=0.02$
  4. $h=0.1$

I found that the approximation is $$y^{n+1}=y^n+\frac{2}{3}hf(t^n, y^n)+\frac{1}{3}hf(t^n+\frac{h}{3}, y^n+\frac{3h}{2}f(t^n, y^n)) \\ \Rightarrow y^{n+1}=y^n-\frac{2}{3}h10y^n-\frac{h}{3}10(y^n+\frac{3h}{2}(-10y^n)) \\ \Rightarrow y^{n+1}=y^n(1-10h+50h^2)$$

Is this correct?

So the method is stable when $$|1-10h+50h^2| \leq 1 \\ \Rightarrow h<\frac{1}{5} $$ The options 2,3,4 satisfy this inequality.

Since we want the smallest number of steps we take the largest h, so $h=0.1$

Is this correct?

Or do we not use the absolute stability to find the $h$ ?

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  • $\begingroup$ Is it a typo that you have $3/2$ in your tableau, but $1/3$ in your equation? $\endgroup$ – KittyL Jun 19 '15 at 10:39
  • $\begingroup$ I edited my post... @KittyL $\endgroup$ – evinda Jun 19 '15 at 10:50
  • $\begingroup$ Should be $1-10h+50h^2$. $\endgroup$ – KittyL Jun 19 '15 at 10:54
  • $\begingroup$ @KittyL I reedited my post. $\endgroup$ – evinda Jun 19 '15 at 11:09
  • $\begingroup$ I think it is correct now. $\endgroup$ – KittyL Jun 19 '15 at 13:10

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