1
$\begingroup$

I was asked to work out a differential equation using the Euler method and then followed by the Runge-Kutta method. Based on the theory I have come across it says that the Euler method agrees with the Taylor series solution up to the term $h$ and therefore the Euler method is the Runge-Kutta method of first order. Now based on the formula for both the Euler method as well as the first order Runge-Kutta method:

$$y_{n+1} = y_n + hy'_n$$

This means both methods are the same and if I have to use the Runge-Kutta method to work out the question it would make sense to use some higher order for the Runge-Kutta method to arrive at my answer. I am hoping that I haven't not confused both the R-K method of first order and the Euler's method of being the same.

$\endgroup$
2
  • $\begingroup$ what was my confusion was the question stated to use the runge kutta method but previous to that it said to work the question using the eulers method so i wanted to make sure that the first order runge kutta method and the eulers method is the same and i was not mixing up anyhting @Lutz Lehmann. Therefore i decided to work the question using a higher oder R-K method. $\endgroup$
    – John
    Apr 17, 2020 at 7:00
  • $\begingroup$ I transferred my comments to an answer. $\endgroup$ Apr 17, 2020 at 8:36

1 Answer 1

1
$\begingroup$

What you say about the Euler method is correct in the context of explicit methods.

Runge-Kutta methods are a class of methods by Martin Wilhelm Kutta. The Euler method is one of them. Then there is THE Runge-Kutta method of 4th order, or classical RK4, that Kutta constructed to simultaneously fit the type of methods of Karl Heun.

There are lots of 1-stage first order RK methods $$ k=f(x_n+αh,y_n+αhk), \\ y_{n+1}=y_n+hk, $$ but the only explicit one is the explicit Euler method. Additionally, for $α=\frac12$ (and only that) you get the implicit midpoint method which has order $2$.

$\newcommand{\D}{\mathit{\Delta}}$ If told to use the Runge-Kutta method without further specification, what is meant is the method \begin{align} y_{n+1}&=y_n+\frac{\D y+2\D'y+2\D''y+\D'''y}6\\ \D y&=f(x_n+\D x,y_n+\D'y)\D x\\ \D'y&=f(x_n+\tfrac12\D x,y_n+\tfrac12\D''y)\D x\\ \D''y&=f(x_n+\tfrac12\D x_n,y_n+\tfrac12\D'''y)\D x\\ \D'''y&=f(x_n,y_n)\D x \end{align}

$\endgroup$

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.