Runge-Kutta method for higher-order differential equations

Question

I am studying Numerical Analysis with the book of Richard L.Burden. A question which I'm struggling with right now is following.

Transform the second-order initial-value problem

$y'' - 2y' + 2y = e^{2t}\sin t$ for $0 \leq t \leq 1, $ with $y(0) = -0.4, y'(0) = -0.6, h=0.1$

into a system of first order initial-value problems, and use the Runge-Kutta method ith h=0.1 to approximate the solution.

Then, $$u_1(t) = y(t), u_2(t) = y'(t)$$ $$u_1'(t) = u_2(t)$$ $$u_2'(t) = e^{2t}\sin t - 2u_1(t) + u_2(t)$$ $$u_1(0) = -0.4, u_2(0) = -0.6$$

This initial conditions give $w_{1,0} = -0.4, w_{2,0}=-0.6$

I can understand that $k_{1,1} = hf_1(t_0, w_{1,0}, w_{2,0}) = hw_{2,0}$

$f_1 = u_1'= u_2(t)$, So $f_1(t_0, w_{1,0}, w_{2,0}) = u_2(t_0, w_{1,0}, w_{2,0}) = w_{2,0}$ (By definition of $w_{i,j}$)

However, I can't understand the following. $$k_{2,1} = hf_1(t_0 + \frac{h}{2}, w_{1,0} + \frac{1}{2}k_{1,1}, w_{2,0} + \frac{1}{2}k_{1,2}) = h\left[w_{2,0} + \frac{1}{2}k_{1,2}\right]$$

Why does $f_1(t_0 + \frac{h}{2}, w_{1,0} + \frac{1}{2}k_{1,1}, w_{2,0} + \frac{1}{2}k_{1,2})$ equal to $w_{2,0} + \frac{1}{2}k_{1,2}$? It seems that third argument in the function comes out, but there is no detailed explanation in this book.

Answer 1

In this problem, I think what you might be confusing is that we have

$$\begin{align} u_1'(t) &= u_2(t) = f_1(t, u_1, u_2) \\ u_2'(t) &= e^{2t} \sin t - 2 u_1(t) + 2 u_2(t) = f_2(t, u_1, u_2) \end{align}$$

From this, we can see that for all the iterations of $j$ on $f_1$, we have

$$f_1(t, u_1, u_2) = f_1(u_2) = w_{2,j} \tag{1}$$

That is, $f_1$ will only ever have $u_2$ terms, which do not depend on $t$ explicitly or $u_1$, thus the iteration formula reduces to

$$k_{2,1} = hf_1\left(t_0 + \frac{h}{2}, w_{1,0} + \frac{1}{2}k_{1,1}, w_{2,0} + \frac{1}{2}k_{1,2}\right) = hf_1\left(w_{2,0} + \frac{1}{2}k_{1,2}\right)$$