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Suppose I have the following system of differential equations: $$ \begin{cases} y'_{1} (t) = y_{2}(t) - y_{1}(t) \\ y'_{2}(t) = -y_{1}(t)y_{2}(t) \end{cases} $$ with the initial conditions: $$ y_{1}(0) = 1,\ y_{2}(0) = 2 $$

I would like to compute $y_{1}(1)$ and $y_2(1)$ using a single step of the Runge-Kutta method with the following Butcher tableau:

$$ \begin{array}{c|cc} 0 & 0 & \\ 1 & 1 & 0 \\ \hline & \frac{1}{2} & \frac{1}{2} \\ \end{array} $$

I've started by writing out my system of equations in vector form:

$\overrightarrow {y}(t) = \begin{bmatrix} y_{1}(t) \\ y_{2}(t) \end{bmatrix} $, $\overrightarrow {f}(t, \overrightarrow {y}(t)) = \begin{bmatrix} y_{2}(t) - y_{1}(t) \\ -y_{1}(t)y_{2}(t) \end{bmatrix} $, and $\overrightarrow {y}(0) = \begin{bmatrix} 1 \\ 2 \end{bmatrix} $

Now I am solving the following system:

$$ \overrightarrow {y}'(t) = \overrightarrow {f}(t, \overrightarrow y(t)) $$

IIUC, this is how you build the solution:

$$ \overrightarrow {y}_{1} = \overrightarrow {y}_{0} + \frac{1}{2} \overrightarrow {k_{1}} + \frac{1}{2} \overrightarrow {k_{2}} $$

where $k_{1}$:

$$ \overrightarrow {k_{1}} = h \cdot \overrightarrow {f}(0 + 0 \cdot h, \overrightarrow {y}(0)) = \overrightarrow {f}(0, \overrightarrow {y}(0)) = \begin{bmatrix} 2 - 1 \\ -1 \cdot 2 \end{bmatrix} = \begin{bmatrix} 1 \\ -2 \end{bmatrix} $$

and $k_{2}$:

$$ \overrightarrow{k_{2}} = h \cdot \overrightarrow{f}(0 + 1 \cdot h, \overrightarrow{y}(0) + 1 \cdot \overrightarrow{k_{1}}) = \overrightarrow{f}(1, \overrightarrow{y}(0) + \overrightarrow{k_{1}}) = \overrightarrow{f}(1, \begin{bmatrix} 2 \\ 0 \end{bmatrix}) $$

I think I've got it right so far (I'd appreciate the notation edits as I'm not really sure I got everything down correctly).

Now, I don't know how to compute this:

$$ \overrightarrow{f}(1, \begin{bmatrix} 2 \\ 0 \end{bmatrix}) $$

According to my understanding, this would lead to evaluating $y_{1}(1)$ and $y_{2}(1)$ as part of my computation. But these are the values I'm trying to compute with my method! Therefore, I must have done something wrong.

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1 Answer 1

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What you got in $\vec y(0)+\vec k_1$ is a first approximation of the value at $\vec y(1)$. But it is the Euler value, first order. With $\vec k_2$ and the composition formula you compute a better approximation of order 2.

So forget the context for a moment and just compute the function value of $\vec f$.

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  • $\begingroup$ So, $\overrightarrow k_{2} = (-2, 0)^{T}$? $\endgroup$ Commented Aug 21, 2024 at 6:21
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    $\begingroup$ Yes, simply that. Of course that means that $k_1+k_2=0$, so that $y(1)\approx y(0)$. For more accurate results you need a smaller step size. This ODE and method could be implemented in Excel or similar, if you have no BASIC or Python available. $\endgroup$ Commented Aug 21, 2024 at 16:33
  • $\begingroup$ $\overrightarrow {k_{1}} + \overrightarrow {k_{2}} = (-1, -2)^{T}$ according to my calculations. Precision of the result is not important to me at this time, only the correct procedure since this is an exam question. $\endgroup$ Commented Aug 21, 2024 at 16:57
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    $\begingroup$ Yes, that is correct, I was reading too fast and taking some close prominent vector. But still, with changes the same size as the components the step size is likely much too large. $\endgroup$ Commented Aug 21, 2024 at 21:50

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