Solving state-space function with using of Runge-Kutta method

Question

I need to implement my own integration routine that will take state space function $f$, free variable $t$, and initial state $x(0)$ as input and produce the solution $x(t)$ as output. I thought that using Runge-Kutta method will be great. But I cannot understand how to apply it to the state-space function matrix.
$$\dot{x}=f(t,x)$$
$$\dot{x}=A\cdot{x}$$
I have only $A$ matrix. How can I apply Runge-Kutta method?
For example, could you provide step-by-step solution for:
$$
\begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
-10 & -5 & -2 \\
\end{bmatrix}
\quad
$$
Thanks in advance!

Answer 1

You have a system $\dot{x} = f(x)$ with

$$ f(x) = A x = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -10 & -5 & -2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} x_2 \\ x_3 \\ -10 x_1 - 5 x_2 - 2 x_3 \end{pmatrix} $$

The formula for Runge Kutta with step size $h$ is:

$$ \begin{align} k_1 &= f(x(t)) \\ k_2 &= f(x(t) + \frac{h}{2}k_1) \\ k_3 &= f(x(t) + \frac{h}{2}k_2) \\ k_4 &= f(x(t) + h k_3) \\ x(t + h) &= x(t) + \frac{h}{6}(k_1 + 2 k_2 + 2 k_3 + k_4) \end{align} $$

So all you need to do is choose a step size $h$ and initial condition $x(0) = x_0$.

---

For example use $h = 0.01$ and

$$ x_0 = \begin{pmatrix} 5 \\ -2 \\ 3 \end{pmatrix} $$

Insert this:

$$ \begin{align} k_1 &= f(x_0) = \begin{pmatrix} -2 \\ 3 \\ -46 \end{pmatrix} \\ k_2 &= f(x_0 + \frac{h}{2}k_1) = \begin{pmatrix} -1.985 \\ 2.77 \\ -45.515 \end{pmatrix} \\ k_3 &= f(x_0 + \frac{h}{2}k_2) = \begin{pmatrix} -1.98615 \\ 2.772425 \\ -45.51485 \end{pmatrix} \\ k_4 &= f(x_0 + h k_3) = \begin{pmatrix} -1.97227575 \\ 2.5448515 \\ -45.02970925 \end{pmatrix} \\ \end{align} $$

And so:

$$ x(0.01) = x(0) + \frac{0.01}{6}(k_1 + 2 k_2 + 2 k_3 + k_4) = \begin{pmatrix} 4.98014237375 \\ -1.972283830833333 \\ 2.544850984583333 \end{pmatrix} $$

If you repeat this 500 times you get:

If you simulate even longer you can see this is an undamped oscillation which makes sense because the three eigenvalues of $A$ are $-2, \sqrt{5}i, -\sqrt{5}i$.