Constructing Modified Euler

Using the same strategy as done in the construction of Modified Euler. Starting from Trapezoidal Method $$y_1 = y_0 + \dfrac{h}{2}\left(f(x_0,y_0) + f(x_1,y_1)\right)$$ then approximating $f(x_1,y_1)$ using the explicit euler $y_1 = y_0 + hf(x_0,y_0)$ , thereby constructing the modified euler method $$y_1 = y_0 + \dfrac{h}{2}\left(f(x_0,y_0) + f(x_1,y_0 + hf(x_0,y_0))\right)$$

This is a valid method that is often seen in most textbooks.

New Method: Combining both backward and forward Euler methods

However, I wanted to ask why haven't I seen a similar approach done for the backward euler?

starting with the implicit euler:
$$y_1 = y_0 + hf(x_1,y_1)$$

then using the same approach as before, we can replace $y_1$ $$y_1 = y_0 + hf(x_1,y_0 + hf(x_0,y_0))$$

Come to think of it, maybe it is not used because it defeats the purpose of the implicit euler method. But I am not sure if this really the case. And, Is this even consistent?

I was thinking that it may be equivalent to some other method, and I should look for its equivalent RK-Method/Butcher Tableau?

  • 1
    $\begingroup$ See the last paragraph: en.wikipedia.org/wiki/Heun%27s_method#Derivation So apparently this is a thing yes. $\endgroup$
    – lightxbulb
    Mar 24 at 23:48
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    $\begingroup$ It's also like you used one step in a fixed-point iteration for implicit Euler: en.wikipedia.org/wiki/Backward_Euler_method#Description $\endgroup$
    – lightxbulb
    Mar 25 at 0:02
  • $\begingroup$ Usually the combination of backwards and forwards Euler as alternating method steps gives the midpoint method. The implicit method can be made explicit in the same way as above and remains second order (but with limited stability). $\endgroup$ Mar 25 at 8:31
  • $\begingroup$ I don't recognize the Butcher Tableau of your new method, but it is 1st order accurate, fully explicit, and has has a region of absolute stability which includes some region of the pure imaginary axis. $\endgroup$ Mar 25 at 9:10
  • $\begingroup$ One downside I can see is that the method is first order and requires 2 function evaluations. That's twice the work of explicit Euler! The local truncation error would have to be half that of explicit Euler for this to be worth using, ignoring the stability properties. $\endgroup$
    – whpowell96
    Mar 29 at 15:02

1 Answer 1


You can construct an ODE solver out of basically any set of function evaluations you want. The better question to ask is what makes a good combination? This is a topic that has had a lot of research, but the short answer is that the proposed method doesn't seem to have any of the properties one would look for (high order, low leading coefficient etc).


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