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Consider the ODE $\dot y=f(t,y)$.

  • Explicit Euler says: $$y_{n+1}=y_n+hf(t_n,y_n)\tag{1}$$
  • Implicit Euler says: $$y_{n+1}=y_n+hf(t_{n+1},y_{n+1})\tag{2}$$

The above schemes are well-known. In a way similar to what is done in Heun's method, we could make Explicit the Implicit Euler scheme as follows:

  • Plug (1) in the right handside of (2) to get: $$y_{n+1}=y_n+hf(t_{n+1},y_n+hf(t_n,y_n))=0\tag{3}$$
  • We could also plug (2) (possibly multiple times) into the right handside of (2) to get $$y_{n+1}=y_n+hf(t_{n+1},y_n+hf(t_{n+1},y_{n+1}))=0\tag{4}$$ or $$y_{n+1}=y_n+hf(t_{n+1},y_n+hf(t_{n+1},y_n+hf(t_{n+1},y_{n+1})))=0\tag{5}$$

We could finally make (5) explicit by plugging (1) into its right handside, as follows: $$y_{n+1}=y_n+hf(t_{n+1},y_n+hf(t_{n+1},y_n+hf(t_{n+1},y_n+hf(t_n,y_n))))=0\tag{6}$$ I have never heard of the schemes above. Maybe they bear no specific good numerical features. Are there known results on this?

pluton
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2 Answers2

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Repeated iterations do not offer improved convergence. Note that any solution to $(2)$ is also a solution to $(4)$ and $(5)$. Provided that the iterations converge, this is just a fixed-point estimate of $(2)$ and is thus no more accurate than $(2)$. This approach is mentioned on Wikipedia.

Simply Beautiful Art
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With 3 evaluations of the ODE function you can implement an explicit 3-stage order 3 method. Your combinations of Euler steps however only give order 1. This should be the main reason that you do not find them widely discussed.

This pattern may seem negative for implicit methods in general, but for higher order methods the explicit methods need more stages than the order number, and implicit methods can be constructed to need only close to half of the order as stages. With step size control there is also a tendency that implicit methods allow larger steps, so that over the same time interval the number of overall function evaluations can become competitive.

Lutz Lehmann
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