2022-07-20

Approximating ${y}^{\prime }\left({t}^{n}\right)$ at the relation ${y}^{\prime }\left({t}^{n}\right)=f\left({t}^{n},y\left({t}^{n}\right)\right)$ with the difference quotient $\left[\frac{y\left({t}^{n+1}\right)-y\left({t}^{n}\right)}{h}\right]$ we get to the Euler method.
Approximating the same derivative with the quotient $\left[\frac{y\left({t}^{n}\right)-y\left({t}^{n-1}\right)}{h}\right]$ we get to the backward Euler method
${y}^{n+1}={y}^{n}+hf\left({t}^{n+1},{y}^{n+1}\right),n=0,\dots ,N-1$
where ${y}^{0}:={y}_{0}$.
In order to find the formula for the forward Euler method, we use the limit $\underset{h\to 0}{lim}\frac{y\left({x}_{0}+h\right)-y\left({x}_{0}\right)}{h}$ for ${x}_{0}={t}^{n},h={t}^{n+1}-{t}^{n}$.
In order to find the formula for the backward Euler method, could we pick $h={t}^{n-1}-{t}^{n}$ although it is negative?
Or how do we get otherwise to the approximation:
${y}^{\prime }\left({t}^{n}\right)\approx \frac{y\left({t}^{n}\right)-y\left({t}^{n-1}\right)}{h}$

iljovskint

Expert

obviously, you still choose ${t}_{n+1}={t}_{n}+{h}_{n}$, thus $h={h}_{n}={t}_{n+1}-{t}_{n}$ for the step from ${y}_{n}$ to ${y}_{n+1}$.

The mean value theorem only tells us that
$\frac{y\left({t}_{n+1}\right)-y\left({t}_{n}\right)}{{t}_{n+1}-{t}_{n}}={y}^{\prime }\left({t}_{n}+\theta \left({t}_{n+1}-{t}_{n}\right)\right)$
where for most of the usual functions $\theta \in \left(0,1\right)$ can be found close to 1/2. Setting, for approximation purposes, $\theta =0$ or $\theta =1$ has thus about the same degree of inaccuracy.

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