To be complete, it is worth noting that there are many variants of Runge-Kutta (RK), including some new ones. These include, but are not limited to:
- Generalized RK, RK1, RK2, RK3, RK4, Simple RK, Euler-Cauchy, Optimal Method, RK5, New Variant RK, RK-Merson (RKM), RK-Fehlberg (RKF), Cash-Karp-RK (CKRK), Implicit RK (IRK) ...
For this problem, I will assume you want to typical fourth order RK (RK4).
Initialization
- $h = \dfrac{b-a}{N} = .1 = \dfrac{2-1}{N} \rightarrow N = 10$
- $x_0 = a = 1, y_0 = y(a) = y(1) = \alpha = 0 \rightarrow a = 1, \alpha = 0$
Iteration using RK4
- $y'(x) = f(x, y) = \dfrac{3 x -2 y}{x}, y(1) = 0$
- $k_1 = h f(x_i, y_i)$
- $k_2 = h f(x_i + h/2, y_i + k_1/2)$
- $k_3 = h f(x_i + h/2, y_i + k_2/2)$
- $k_4 = h f(x_i + h, y_i + k_3)$
- $y_{i + 1} = y_i + \dfrac{1}{6}\left( k_1+2k_2+2k_3+k_4 \right)$
- $x_{i + 1} = x_i + h = 1 + 0.1 i$
Results
- For $i= 1$, we have:
- $x_0 = 1, y_0 = 0$
- $k_1 = h f(x_i, y_i) = \dfrac{1}{10} \left(\dfrac{3 x_{i} - 2 y_{i}}{x_{i}}\right) = \dfrac{1}{10}\left(\dfrac{3 x(0) - 2y(0)}{x(0)}\right) = \dfrac{1}{10}\left(\dfrac{3 (1) - 2(0)}{1}\right) = 0.3$
- $k_2 = h f(x_i + h/2, y_i + k_1/2) = \dfrac{1}{10} \left(\dfrac{3 (x_{i}+(1/20)) - 2(y_{i} + (0.3/2))}{x_{i}+(1/20)}\right) = \dfrac{1}{10} \left(\dfrac{3 (1+(1/20)) - 2(0 + (0.3/2))}{1+(1/20)}\right) = 0.271428571428571428571428571428571428571428571428571428571428$
- $k_3 = h f(x_i + h/2, y_i + k_2/2) = \dfrac{1}{10} \left(\dfrac{3 (x_{i}+(1/20)) - 2(y_{i} + (0.27142857/2))}{x_{i}+(1/20)}\right) = \dfrac{1}{10} \left(\dfrac{3 (1+(1/20)) - 2(0 + (0.27142857/2))}{1+(1/20)}\right) = 0.274149659863945578231292517006802721088435374149659863945578$
- $k_4 = h f(x_i + h, y_i + k_3) = \dfrac{1}{10} \left(\dfrac{3 (x_{i}+(1/10)) - 2(y_{i} + 0.2741496598)}{x_{i}+(1/10)}\right) = \dfrac{1}{10} \left(\dfrac{3 (1+(1/10)) - 2(0 + 0.2741496598)}{1+(1/10)}\right)= 0.250154607297464440321583178726035868893011750154607297464440$
- $y_{i + 1} = y_i + \dfrac{1}{6}\left( k_1+2k_2+2k_3+k_4 \right) = 0.273551844980416408987837559266130694702123273551844980090476 \approx 0.273552$
- $x_{i + 1} = x_i + h = 1 + 0.1 i = 1.1$
Continuing with this itearation, we generate the following table.
$~~~~~\text{Step} ~~|~~~~ x ~~~|~~ y $
- $~~00 ~~|~~ 1.0 ~~|~~ 0. 00000$
- $~~01 ~~|~~ 1.1 ~~|~~ 0.273552$
- $~~02 ~~|~~ 1.2 ~~|~~ 0.505553$
- $~~03 ~~|~~ 1.3 ~~|~~ 0.708281$
- $~~04 ~~|~~ 1.4 ~~|~~ 0.889793$
- $~~05 ~~|~~ 1.5 ~~|~~ 1.05555$
- $~~06 ~~|~~ 1.6 ~~|~~ 1.20937$
- $~~07 ~~|~~ 1.7 ~~|~~ 1.35398$
- $~~08 ~~|~~ 1.8 ~~|~~ 1.49136$
- $~~09 ~~|~~ 1.9 ~~|~~ 1.62299$
- $~~10 ~~|~~ 2.0 ~~|~~ 1.75$
The exact solution is given by:
$$y(x) = \dfrac{x^3-1}{x^2}$$
At $x=2$, we have: $y(2) = \dfrac{7}{4} = 1.75$
Compare that to Runge-Kutta's-4 method, which has $1.75$.
On the history
See Butcher: A History of the Runge-Kutta method
In summary, people (Nystroem, Runge, Heun, Kutta,...) at the end of the 19th century experimented with success in generalizing the methods of numerical integration of functions in one variable $$\int_a^bf(x)dx,$$ like the Gauss, trapezoidal, midpoint and Simpson methods, to the solution of differential equations, which have an integral form $$y(x)=y_0+\int_{x_0}^x f(s,y(s))\,ds.$$
Carl Runge in 1895[1] came up with ("by some curious inductive process" - "auf einem eigentümlich induktiven Wege" wrote Heun 5 years later) the 4-stage 3rd order method
\begin{align}
k_1&=f(x,y)Δx,\\
k_2&=f(x+\tfrac12Δx,y+\tfrac12k_1)Δx\\
k_3&=f(x+Δx,y+k_1)Δx\\
k_4&=f(x+Δx,y+k_3)Δx\\
y_{+1}&=y+\tfrac16(k_1+4k_2+k_4)
\end{align}
[1] "Über die numerische Auflösung von Differentialgleichungen", Math. Ann. 46, p. 167-178
Inspired by this Karl Heun in 1900[2] explored methods of the type
$$
\left.\begin{aligned}k^i_m &= f(x+ε^i_mΔx,y+ε^i_mk^{i+1}_m)Δx,~~ i=1,..,s,\\ k^{s+1}_m&=f(x,y)Δx\end{aligned}\right\},~~ m=1,..,n\\
y_{+1}=y+\sum_{m=1}^n\alpha_mf(x+ε^0_mΔx,y+ε^0_mk^1_m)Δx
$$
He computed the order conditions by Taylor expansion and constructed methods of this type up to order 4, however the only today recognizable Runge-Kutta methods among them are the order-2 Heun-trapezoidal method and the order 3 Heun method.
[2] "Neue Methode zur approximativen Integration der Differentialgleichungen einer unabhängigen Veränderlichen", Z. f. Math. u. Phys. 45, p. 23-38
Wilhelm Kutta in his publication one year later in 1901[3] considered the scheme of Heun wasteful in the number of function evaluations and introduced what is today known as explicit Runge-Kutta methods, where each new function evaluation potentially contains all previous values in the $y$ update.
\begin{align}
k_1&=f(x,y)Δx,\\
k_m&=f(x+c_mΔx, y+a_{m,1}k_1+...+a_{m,m-1}k_{m-1})Δx,&& m=2,...,s\\[0.5em]
y_{+1}&=y+b_1k_1+...+b_sk_s
\end{align}
He computed order conditions and presented methods up to order $5$ in parametrization and examples. He especially noted the 3/8 method for its symmetry and small error term and the "classical" RK4 method for its simplicity in using always only the last function value in the $y$ updates.
[3] "Beitrag zur näherungsweisen Lösung totaler Differentialgleichungen", Z. f. Math. u. Phys. 46, p. 435-453
On the order dependence of the performance
The Euler method has global error order 1. Which means that to get an error level of $10^{-8}$ (on well-behaved example problems) you will need a step size of $h=10^{-8}$. Over the interval $[0,1]$ this requires $10^8$ steps with $10^8$ function evaluations.
The classical RK4 method has error order 4. To get an error level of $10^{-8}$ you will thus need a step size of $h=0.01$. Over the interval $[0,1]$ this requires $100$ steps with $400$ function evaluations.
If you decrease the step by a factor of $10$ to $h=0.001$, the RK4 method will need $1000$ steps with $4000$ function evaluations to get an error level of $10^{-12}$. This is still much less effort than used in the Euler example above with a much better result.
Using double precision floating point numbers you will not get a much better result with any method using a fixed step size, as smaller step sizes result in an accumulating floating point noise that dominates the error of the method.
Best Answer
Yes, one-step and multi-step methods look different.
The Adams-Bashford methods can also be derived via Taylor expansions. Comparing the Taylor expansions $$ y(x+h)=y(x)+y'(x)h+\tfrac12y''(x)h^2+O(h^3)\\ \text{ and }~y'(x-h)=y'(x)-y''(x)h+O(h^2), $$ the second derivative can be eliminated to give $$ y(x+h)=y(x)+y'(x)h+\tfrac12(y'(x)-y'(x-h))h+O(h^3)\\ =y(x)+\tfrac32y'(x)h-\tfrac12y'(x-h)h+O(h^3) $$ which is the A-B 2 formula.
You get equally degrees of freedom in multi-step methods if you include more previous $y$ values, in the explicit case $y_{k+1}+\sum_{j=0}^{s-1}\alpha_jy_{k-j}=\sum_{j=0}^{s-1}\beta_jf_{k-j}$. The difference is that with $s$ steps back you can get an order $s$ method, which is not true for $s$-stage explicit Runge-Kutta methods for $s>4$.
yes, see 1.
The relation of explicit and implicit methods is not as close in the Runge-Kutta case as in the multistep case. The Gauß 2-stage implicit method has order 4, there is no closely related explicit method, especially with order 4.
Numerical test
For a practical demonstration, take the standard example of the Lorenz system with its chaotic attractor, select the initial value close to the attractor
use
scipy.integrate.odeintto compute a reference solution with sufficiently tight tolerances and apply the Heun and A-B 2 methods for different step sizes, pairing A-B 2 with Heun of twice the step size to have the same "density" of ODE function evaluations. The first plot below shows the comparison of the $x$ components under relatively large step sizes, stopping shortly before all the graphs diverge completely.The second and third plot show the error profiles, that is, the difference to the reference solution divided by the square of the step size. What one can conclude is that overall the solutions appear similar in their divergence points from the reference solution for similar step sizes. However the error profiles show that the error coefficient of Adams-Bashford is almost 4 times larger than the error coefficient of the Heun method, so that the error is similar for solutions with the same number of ODE function evaluation.