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Solve the following cauchy problem: $y'=(x-1)y$ and also given $y(0)=2$

using the runge kutta 2 method $(RK2)$ for $\alpha=0.2$ on the interval $[0,0.6]$ with step $h=0.2$

The answer in the book: for $\alpha=0.2$ and $h=0.2$ we have $k_1=f(x_m,y_m)$,$k_2=f(x_m+0.04,y_m+0.04k_1)$ and $y_{m+1}=y_m+0.2(-1.5k_1+2.5k_2)$

then there is a table which goes on from $m=0$ to $m=2$ I will just put the first line $m=0$ , $x_m=0$ , $y_m=2$ , $k_1=-2$ , $x_m+0.004=0.04$ , $y_m+0.04 \cdot f(x_m,y_m)=1.92$ , $k_2=-1.8432$ and lastly $y_{m+1}=1.6784$.

Solving this question is not really important as the answer is in the book as shown above.. but I am trying to understand some things:

I found that $k_1=f(x_m,y_m)$ and $k_2=f(x_n+c_2 \cdot h,y_n+(a_{21} \cdot k_1) \cdot h)$ and I got a bit confused here, I assume $c_2= \alpha$ here but why? what does it actually represent and same goes for $a_{21}$ I can't understand how they got the values here $(k_2=f(x_n+c_2 \cdot h,y_n+(a_{21} \cdot k_1) \cdot h))$

also a similar question for $y_{m+1}=y_m+0.2(-1.5k_1+2.5k_2)$, I know that $y_{m+1}=y_m+h \cdot \sum_{i=1}^s b_i \cdot k_i$ but again what does $b_i$ represent here? or in my question how did they get to those numbers in $y_{m+1}$

last question is why $x_m+0.04$ and $y_m+0.04f(x_m,y_m)$ was taken in the table? does it mean something or just randomly taken?

Sorry for my English hopefully it was understandable and hopefully the question is acceptable but I just can't seem to understand how they solved no matter how much I search.

Adamrk
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1 Answers1

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This is an older tradition, here going back to the original papers of W. Martin Kutta in 1901 and Karl Heun in 1900, to name the parameters with the greek or latin alphabet letters. See https://archive.org/details/zeitschriftfrma12runggoog/page/438/mode/2up

It is the consistency condition that $c_i=\sum_ja_{ij}$. In an explicit method in the second row there is only one non-zero $a_{21}$, so $c_2=a_{21}=\alpha$.

In both occurrences you have the combination $\alpha h=0.04$.

For the final update step the conditions are $b_1+b_2=1$ and $b_2c_2=\frac12$, so $b_2=\frac52$ and $b_1=1-b_2=-\frac32$.

The structure of the order conditions becomes clearer if you consider those for order 3 or 4 methods, see How to derive 4th Runge-Kutta?

Lutz Lehmann
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