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I'm asked to determine the butcher tableau of the implicit midpoint method, i.e $$y_{n+1} = y_n + hf\left(t_n + \frac{h}{2}, \frac{y_n}{2} + \frac{y_{n+1}}{2}\right).$$

A general runge-kutta method can be written as $$y_{n+1} = y_n + h\sum_{i=1}^sb_ik_i$$ where $$k_i = f\left(t_n + c_ih, y_n + h\sum_{j=1}^sa_{ij}k_j\right), i=1,...,s.$$

The butcher tableau is supposed to look like this: Butcher tableau of implicit midpoint method.

I get why the $b^T$ array consist of only $1$ and why our only $c$ value is $\frac{1}{2}$. But I can't figure out why $a_{11} = \frac{1}{2} $ gives $$h\sum_{j=1}^sa_{ij}k_j = \frac{y_n}{2} + \frac{y_{n+1}}{2}$$

homocomputeris
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IdiotWithNoShame
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1 Answers1

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So since $s=1$ for your case, You have two equations namely

$$ k_1=f(t_n+c_1h,y_n+ha_{11}k_1)\\ y_{n+1}=y_n+hb_1k_1 $$ To get a "nicer" formula, the $k_1$ is substituted from using the second equation $$ k_1=\frac{y_{n+1}-y_n}{hb_1}\\ \Rightarrow \frac{y_{n+1}-y_n}{hb_1}=f(t_n+c_1h,y_n+ha_{11}\frac{y_{n+1}-y_n}{hb_1})\\ \Leftrightarrow y_{n+1}=y_n+hb_1\cdot f(t_n+c_1h,y_n+\frac{a_{11}}{b_1}(y_{n+1}-y_n)) $$

From here, can you see how the coefficients have to be what they are?

maxmilgram
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