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

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

  • $\begingroup$ 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? $\endgroup$ – James Nov 20 '18 at 0:48
  • $\begingroup$ 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? $\endgroup$ – Moo Nov 20 '18 at 1:12
  • 1
    $\begingroup$ I got it. Thanks for answering $\endgroup$ – James Nov 20 '18 at 2:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.