Scientific Computing - Quiz 1 - Question 6

Problem:

Walk through one time-step of 4th-order Runge-Kutta with simple differential equation system $y'=y$, $y(0)=1$ and $\Delta t=2$.
Identify the correct values for f1; f2; f3; f4 and y(2).

Solution:

It is always good to know the solution to the differential equation first, in this case it is really easy.

$\begin{eqnarray*} \frac{dy}{dt} &=& y \\ \frac{1}{y}\frac{dy}{dt} &=& 1 \\ \int{\frac{1}{y}\frac{dy}{dt}dt} &=& \int{1dt} \\ \log y &=& t + C_1 \\ y &=& e^{t + C_1} \\ &=& C_2e^t \\ \end{eqnarray*}$

Now the initial condition decides $C_2 = 1$, making the solution simply $y = e^t$.

The analytic solution is not required, it is just that it is a perfect guide for us to evaluate whether our solution make sense. Now, looking up the Runge-Kutta equations, we have

$f_1 = f(t_n, y_n)$
$f_2 = f(t_n + \frac{\Delta t}{2}, y_n + \frac{\Delta t}{2} f_1)$
$f_3 = f(t_n + \frac{\Delta t}{2}, y_n + \frac{\Delta t}{2} f_2)$
$f_4 = f(t_n + \Delta t, y_n + \Delta t f_3)$

Next, we substitute $f(t, y) = y$ in the expressions above, then we get:

$f_1 = y_n$
$f_2 = y_n + \frac{\Delta t}{2} f_1$
$f_3 = y_n + \frac{\Delta t}{2} f_2$
$f_4 = y_n + \Delta t f_3$

Excel will just happy to crank those answers for us now.

The final answer is

$f_1 = 1$
$f_2 = 2$
$f_3 = 3$
$f_4 = 7$
$y(2) = 7$.

With excel, I fiddled around the parameter. In particular, I tried to tune $\Delta t$ to 0.1 and iterate 20 times, the result has $1.1 \times 10^{-5}$ error only. It is in fact quite accurate.

As an aside to the analytic solution - the integral of $\int{\frac{1}{y}dy} = \log{y}$ only make sense when $y > 0$. To make sense of cases when $y < 0$, we can take the sign out first before the integration and the expression become $\log|y| = sign(y)t + C_1 \implies e^{|y|} = e^{|sign(y)|}e^{t + C_1} \implies e^{y} = e^{t + C_1}$ and the rest follows as is.