Runge Kutta Method using quadrature formula

Question

I am trying to solve this exercise

the following Runge-Kutta method as a butcher tableau is given:

$$\begin{array} {c|cccc} 0\\ \frac{1}{2} & \frac{1}{4} &\frac{1}{4}\\ 1& 0& 1& 0\\ \hline & \frac{1}{6} &\frac{2}{3} &\frac{1}{6} \end{array} $$

(i) Rewrite the Butcher tableau in step form and outline the derivation of the Runge Kutta method using known square formulas

(iii) Apply the step form of the Runga Kutta method to the differential equation.

Lutz Lehmann · Accepted Answer · 2020-05-18 14:04:07Z

I do not see what you are computing. Obviously the quadrature method underlying the method is the Simpson method, this fixes both the $b_k$ as also the $c_k$. The computation of $k_2$ is by the implicit trapezoidal method, $k_2=y'(x+\frac12h)+O(h^2)$. $k_3$ is computed by the midpoint method, so that $k_3=y'(x+h)+O(h^2)$. Up to this point this gives second order. The remaining condition for order $3$ is (see for instance the first and second set of slides from J. Butcher's tutorials) $$ \frac16=\sum b_ia_{ij}c_j=b_2a_{22}c_2+b_3a_{32}c_2 \\ \implies \frac23\frac14\frac12+\frac16 a_{32}\frac12=\frac16 \\ \implies a_{32}=1\implies a_{31}=0 $$ Now one could also check the 4th order conditions $$ \frac18=\sum b_ic_ia_{ij}c_j=b_2c_2a_{22}c_2+b_3c_3a_{32}c_2=\frac1{24}+\frac1{12} \\ \frac1{12}=\sum b_ia_{ij}c_j^2=(b_2a_{22}+b_3a_{32})c_2^2=\frac13\frac14 \\ \frac1{24}=\sum b_ia_{ij}a_{jm}c_m=(b_2a_{22}+b_3a_{32})a_{22}c_2=\frac13\frac18 $$ Which confirms that this method is a 4th order method.

For a linear system $\dot z=f(t,z)=A(t)z+b(t)$ one can implement the method as \begin{align} k_1&=A(t)z+b(t)\\ k_2&=(I-\tfrac14hA(t+\tfrac12h))^{-1}\Bigl(A(t+\tfrac12h)(z+\tfrac14hk_1)+b(t+\tfrac12h)\Bigr)\\ k_3&=A(t+h)(z+hk_2)+b(t+h)\\ \hline z_{+1}&=z+\frac{h}6(k_1+4k_2+k_3) \end{align} In the current case, $b(t)=0$, as the right side of the equation is another term that is linear in $y$.

Thank you! Do you have a source/literature for part 3. ? I don't quite understand how you figure it out, and I don't find it in my script, I'd be interested in how you figure it out. — Mufasa
– Mufasa, Commented May 18, 2020 at 11:01
Just write out the equation as you had it and replace $f$ with the linear expression. In the second equation you get $k_2$ on both sides, so solve for it. That's all. — Lutz Lehmann
– Lutz Lehmann, Commented May 18, 2020 at 11:41
It seems I did not read the equation correctly, I did not register the $y$ on the right side. I'm going to change that. — Lutz Lehmann
– Lutz Lehmann, Commented May 18, 2020 at 14:02

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