# Runge-Kutta method for higher-order differential equations

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.

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

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

• Why does f_1 depend only on u_2? Does the formula u_2(t) = f_1(t, u_1, u_2) mean that f_1 depends only on u_2? – James Nov 20 '18 at 0:48
• Yes, because $f_1 = u_2$, there is nothing like $f_1 = e^t(u_1 + u_2)$, for example, in which case $f_1$ would depend on all three, namely $t, u_1 ~ \text{and}~ u_2$. In this problem, we have $f_1 = u_2$ only - it is how it was setup in the very beginning. Clear? – Moo Nov 20 '18 at 1:12
• I got it. Thanks for answering – James Nov 20 '18 at 2:15