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When I have this butcher table:

\begin{array} {c|cccc} 0\\ c & c\\ \hline & (1-d) &d \end{array}

I got this equation:

$\phi(x,y,h) = (1-d)*f(x,y) + d*f(x+ch, y+ch*f(x,y))$

Now I need to know the $\phi(x,y,h)$ function for this table:

\begin{array} {c|cccc} 0\\ \frac{1}{3} & \frac{1}{3}\\ \frac{2}{3} &0 &\frac{2}{3} \\ \hline & \frac{1}{4} &0 &\frac{3}{4} \end{array}

1 Answers1

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Just read the tableau from bottom to top, \begin{align} y_{+1}&=y+h\Phi(x,y,h)\\ &=y+\frac h4 f(x,y)+\frac{3h}4 f\left(x+\frac{2h}3,y+\frac{2h}3 f\left(x+\frac{h}3,y+\frac{h}3f\left(x,y\right)\right)\right) \end{align} In the methods that are in the intersection of Heun's and Kutta's approaches, that is, $a_{i,i-1}=c_i$, $a_{i,j}=0$ for $j\ne i-1$, you get a simple nested expression like this. There just are not many methods of that class, classical RK4 and some lower order methods.

In the 3-stage case \begin{array}{l|lll} 0\\ c_2&c_2\\ c_3&0&c_3\\ \hline &b_1&b_2&b_3 \end{array} the order conditions reduce to $$ I)~b_1+b_2+b_3=1,~~ II)~b_2c_2+b_3c_3=\frac12,~~ III)~b_2c_2^2+b_3c_3^2=\frac13,~~ \text{and}~~ IV)~b_3c_3c_2=\frac16. $$ so that (II&IV) $b_2c_2^2=\frac12c_2-\frac16$, (..&III) $b_3c_3^2+\frac12c_2=\frac12$ and then (..&IV) $\frac16c_3=\frac12c_2(1-c_2)$, (&IV) $b_3c_2^2(1-c_2)=\frac1{18}$. Thus one has one degree of freedom to choose $c_2\in (0,1)$, the other coefficients then follow, $$ c_3=3c_2(1-c_2),~b_3=\frac1{6c_2c_3},~b_2=\frac1{2c_2}-\frac1{6c_2^2},~b_1=1-b_2-b_3. $$

Lutz Lehmann
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  • How come there is a y at the beginning unlike the other table? And why in the first table does it have (1-d)f(x,y) whereas this has (1/4)f(x,y)*h (i.e. why is there a h here?) –  Feb 28 '19 at 16:11
  • What if I was doing this in the same way as the order 2 table, i.e. numbers were c and d, not 3/4, etc. @LutzL –  Feb 28 '19 at 16:18
  • Yes, you can do that, replace $\frac13$ with $c_1$, $\frac23$ with $c_2$, $\frac34$ with $d$ and $\frac14$ with $1-d$. The order conditions will greatly reduce the freedom in the choice of parameters. – Lutz Lehmann Feb 28 '19 at 16:26